INTRODUCTION TO THlE NATIONAL ARITHMETIC, ON THE INDUCTIVE SYSTEM; COMBINING THE ANALYTIC AND SYNTHETIC METHODS. IN WHICH THE PRINCIPLES OF THE SCIENCE ARE FULLY EXPLAINED AND ILLUSTRATED, DESIGNED FOR COMMVON SCHOOLS AND ACADEMIES. By BENJAMIN GREENLEAF, A. M., PIRINCIPAL OF BRADFORD TEACHERS' SEMINARY. NEW STEREOTYPE EDITION, WITH ADDITION.S AND IMPROVEMENTS. B O S T ON: PUBLISHED BY ROBERT S. DAVIS & CO. NEW YORK: D. APPLETON & CO., AND DANIEL BURGESS & Co. PHIIADELPHIA: LIPPIiNCOTT, GRAMSBO, & CO. And sold by the trade generally. 1856. Entered according to Act of Congress, in the year 1848, by B. GREgENLBAP, in the Clerk's Office of the District Court of the District of Massachusetts. - GR EEI LEAt F'2S Ii[RES O}F ARI'Dlgi ETI[CS. 1. MIENTAL ARITHMETIC, upon the Inductive Plan, designed for Beginners. By Benjamin Greenleaf, A. M., Principal of Bradford (Mass.) Teachers' Seminary. 2. INTRODUCTION TO THE NATIONAL ARITHMETIC, designed for Common Schools. Fifteenth improved stereotype edition, revised and enlarged. 3, THE NATIONAL ARITHMETIC, for advanced Scholars in Common Schools and Academies. Twenty-fifth improved stereotype edition. 360 pages. full bound. COMPLETE KEYS TO THE INTRODUCTION AND NATIONAL ARITHMETICS, containing Solutions and Explafiations, for Teachers only. (In separate volumes.) *** The attention of Teachers and Superintendents of Schools generally is respectfully invited to this popular system of Arithmetic, which is well adapted to all classes of students. The whole or a part of this series has been recommended and adopted by the superintending school committees of the principal towns in New England, and is also used in the best public and private schools in various sections of the United States. GREENLSEAF'S NATIONAL ARITHiMETIC is now extensively used as a text-book in many distinguished seminaries of learning, including the following:- The several STATE NORMAIL SCHOOLS in Massachusetts, under the direction of the State Board of Education; the NORMAL SCHOOLS in New York city; Rutgers Female Institute, New York; Brooklyn (N. Y.) Female Academy; Abbott Female Academy, and Phillips Academy, Andover; Chauncey Hall School, Boston; Bradford Female Seminary; Phillips Academy, Exeter; Young Ladies' Institute, Pittsfield; Worcester County High School, Worcester; Williston Seminary, East Hampton, Mass.; together with the best schools in Boston, New York, Philadelphia, Richmond, Charleston, Savannah, Mobile, New Orleans, and other cities; and wherever the work has been introduced, it ts still used iwith great success, - which is deemed a sufficient recommendation. Parker's Progressive Exercises in Enlgllsh Conipositolla. New stereotype edition, revised, enlarged, and improved. 144 pages. Price, 3S cents. Claes-Book of Prose and Poetry: consisting of Selections from the best English and American Authors; designed as Exercises in Parsing, for the use of Common Schools and Academies. By T. Rickard, A. M., and H. Orcutt, A. MN (Teachers). Price 20 cents single,$ 1 per dozen. *i* A cheap work like the above (comprised in a small volume) has long been needed. The Classical Reader: A Selection of Lessons in Prose and Verse, from the most esteemed English and American Writers. Intended for the Use of the Higher Classes in Public and Private Seminaries. By F. W. P. Greenwood, D. D., and George B. Emerson, A. M., of Boston. Tenth edition, stereotyped. With an engraved frontispiece. Smit'lhs ClasE-]Book of Asatomsy: Explanatory of the First Principles of Human Organization as the Basis of Physical Education; with numerous Illustrations, a full Glossary, or Explanation of Technical Terms, and Practical Questions at the Bottom of the Page. Designed for Schools and Families. Tenth stereotype edition, revised and enlarged. A OGramanar of the CGc reek Lasnglage. By Benjamin Franklin Fisk, A. M. Twenty-ninth improved stereotype edition. *** Fisk's Greek Grammar is used in Harvard University, and in many other dis tinguished collegiate and academic institutions in various parts of the United States. Fisk's lGreek Exerceses. ['ew EDi11tiao.] Greek Exercises: containing the Substance of tihe Greek Syntax, illustrated by Passages froin the best Greek Authors, to be,writt{,n'out from the Words given in their simplest Form. By Benjamin Franklin F:slr., A. M.' Consuetudo et exercitatio facilitatem maxime parit." -Quintil. Adapted to the Author's "Greek Grammar." Leverett's Cvesar's Commnentaries. Caii Julii Caesaris Commentarii do Bello Gallico ad Codices Parisinos recensiti, a N. L. Achaintre and N. E. Lemaire. Accesserunt Notulra Anglicee, atque Index' Historicus et Geographicus. Curavit F. P Leverett, A. M. ]Folsom's Cicero's Orations. M. T. Ciceronis Orationes Quaedam Selectme, Notis Illustrataw. [By Charles Folsom, A. M.] In Usum Academim Exoniensis. Editio stereotypa, Tabulis Analyticis instructa. [These editions of Caesar and Cicero are highly recommended by Prof. John J. Owen.] Published by ROBERT S. DAVIS, 118 Washington Street, BOSTON, and sold by all the principal Booksellers throughout the country. - Also constantly for sctle (in addition to his owen publications), a complete assortment of School-books and Stationery, which n are o,fTred to Boolksellers, School Coslsmittees., and'7eacher.e on c?' limi/,eral terrz.s PREFACE. THE present edition of this work has been thoroughly revised and rewritten, and also enlarged by the addition of nearly one hundred and fifty pages of new matter; rendering it sufficiently extensive for the majority of pupils in common schools. The arrangement is strictly progressive; no question for solution requiring the use of a rule, or the knowledge of a principle, which has not been previously explained. In this respect it is believed the work will be found to differ from most other Arithmetics. The author has endeavored to make the language simple, precise, and accurate, and such as in all cases to render the rules, definitions, and illustrations intelligible to the pupil. The examples are of a practical character, and adapted not only to fix in the mind the principles which they involve, but also to interest the pupil, exercise his ingenuity, and inspire a love for mathematical science. The reasons for the operations are explained, and an attempt is made to secure to the learner a knowledge of the philosophy of the subject, and prevent the too prevalent practice of merely performing, mechanically, operations which he does not understand. Analysis has been made a prominent subject, and employed in the solution of questions under most of the rules in which it could be used with any practical advantage; and it cannot be too strongly recommended to the pupil to make use of this mode of operation, where it is recommended by the author. Thd subject of cancellation, also, is more extensively treated, it is believed, than in any other work of the kind. The principles on which it depends, and their application, are fully developed, and the whole subject, it is thought, is made perT fectly clear to the comprehension of the learner. rv P'IPEFACE, Questions have been inserted at the bottom of each page, designed to direct the attention of teachers and pupils to the most important principles of the science, and fix themi in the mind, it is not intended, however, nor is it desirable, that the teacher should servilely confine himself to these questions, but vary their form, and extend tnem at pleasure, and invariably require the pupil thoroughly to understand the subject, and give the reasons for the various steps in tile operation by which he arrives at any result in the solution of a question. The object of studying mathematics is not only to acquire v knowledge of the subject, but also to secure mental discipline, to induce a habit of close and patient thought, and of persevering and thorough investigation. For the attainment of this object, the examples for the exercise of the pupil are numerl ous, and variously diversified, and so constructed as necessarily to require careful thought and reflection for the right application of principles. The author would respectfully suggest to teachers, who may use this book, to require their pupils to become, familiar with each rule before they proceed to a new one; and, for this purpose, a frequent review of rules and principles will be of service, and will greatly facilitate their progress. If the pupil has not a clear idea of the principles involved in the solution of questions, he will find but little pleasure in the study of the science; for no scholar can be pleased with what he does not understand. The article on weights, measures, and money, will be found, it is believed, to contain valuable information, and such as no similar work places within the reach of pupils. This addition, it is hoped, will be found interesting to teachers and scholars BENJAMIN GREENLEAF Bradford Teachers' Seminary, Nov. 15th, 1848. NOTICE. Two editions of this work, and also of the NATIONAL ARITHMETIC, one containing the answers to the examples, and the other with — out them, are now published. CONTENTS. Page tPagu SEC ION I. SEC rION X. NOTATION AND NUMERATION,. 7 REDUCTION,.......... S4' Notation.. 7 English Money, Table,...... 84 Table of Roman Letters,...... 8 Troy Weight Table,.. 86 Exercises in Roman Notation,. 8 Apothecaries' Weight, Table,. Numneration,..... 9 Avoirdupois Weight, Table,.... 89 French Numeration Table,..... 11 Cloth Measure, Table, 91 Exercises in French Numeration,. 12 Long MIeasure, Table, 92 Exercises in French Notation and Nu- Surveyors' Measure, Table,.. 94 meration,..... 13 Sluare Measure, Table..,... 95 English Numeration Table,. 14 Cubic or Solid Measure, Table,. 97 Exercises in English Numeration,. 15 Wine Measure, Table,.99 Exercises in English Notation and Ale and Beer Measure, Table,.. 101 Numeration,...... 15 Dry Measure, Table,....... 102 SECTION II. Measure of Time, Table,..... 103 Circular Measure or Motion, Table, 105 rIDDTION - Mental Exercises,. 16 Miscellaneous Table,....... 106 Addition Table,...... 16 Miscellaneous Exercises in ReducExercises for the Slate,....... 20 tion,.......... 107 Examples for Practice,... 21 SECTION XI. SECTION XI. SECTION III. ADDITION OF COMPOUND NUMBERS, - SUBTRACTION- Mental Exercises,. 26 English Money,......... 110 Subtraction Talble....... 27 Examples for Practice in the different Weights and Measures,. 111 SECTION IV. MULTIPLICATION - Mental Exercises, 34 SECTION XII. Multiplication Table,....... 35 SUBTRACTION OF COMPOUND NUMBERS, English MBoney......... 114 SECTION V. Examples for Practice..... 115 DIVISION - Mental Exercises,.... 46 SECTION XIII. Division Table,....... 46 MISCELANEOUS EXE}CISES IN ADDISECTION VII TION AND SUBTRACTION OF COMsPOUND NUMBERS....... 119 CONTRACTIONS IN MULTIPLICATION AND DivisoN,............ 59 SEC`TION XIV. Contract.ions in Multiplication,... 59 MULTIPLICATION OF COIFPOUND NUMContractions in Division,...... 61 BERS............. 121 SECTION VII. Examples for Practice,...... 122 MiIScE LLANEOUS EXAMPLES INVOLVING SECTION XV. THE FOREGOING RULES,.... 63 DIVISION OF COMPOUND NUmnsEIS,. 125 SECTION VIII. Examples for Practice,..... 126 ~SECTION VIII. SECTIlON XVI. UNITED STATES MONEY,....... 67 Reduction of United States Money, 68 MicLANous EAMPLE IN MULTIAddition of United States Money,. 69 PLICATION AND DIVISION OF Cos,Subtraction of United States Money, 71 POUND NUMBERs.1.. 129 Mlultiplication of U. States Money, 72 SECTION XVII. I)ivision of United States Money,. 73 CANCELLATION......130 Practical Questions by Analysis,.. 74 Bills, Exercises in,...... 77 SECTION XVIIL TIO I PROFPEtRTIES AND RELATIONS Or NUmSECTION IX. IeS.... n134 lios INV0s LvNof, FR-Ci'-IOT, 7,A. Prime FEar....... 134 VI CONTENTS. Page Page A Common Divisor....... 135 SECTION XXXI. The Greatest Commlon Divisor,.. 136 ASSESS&MENT OF TAXES..... 229 A Common Multiple,....... 13 SE ION II. SECTIION XIX. EQUATION OF PAYMENTS,..... 232 FRACTIONS -VULGAR FRACTIONS. 140 Reduction of Vulgar Fractions,. 141 SECTION XXXIII. A Common Denominator.. 146 RATIO,....... 236 Addition of Vulgar Fractions,. 148 SECTION XXXIV. Subtraction of Vulgar Fractions, 150 Multiplication of Vulgar Fractions, 155 PROPORTION........... 238 Division of Vulgar Fractions,... 160 Simple Proportion,. 239 Complex Fractions,....... 165 Compound Proportion,...... 245 Miscellaneous Exercises in Vulgar SECTION XXXV. Fractions.1.......... 168 Reduction of Fractions of Compound PARTNERSHIP, OR COMPANY BUSINESS, 248 Numbers,.... 170 SECTION XXVI Addition of Fractions of Compound Numbers,........... 174 PROFIT AND Loss.......... 252 Subtraction of Fractions of Com- Miscellaneous Examples in Profit and pouid Numbers,,....... 175 Loss,............. 253 Questions to be performed by Analy- SECTION XXVII. sis,... 176 SiS,.............. 176 SECTION XXXV76. Miscellaneous Questions by Analy- DUODrECIMALS,. 258 sis.............. 179 Addition and Subtraction of Duodeci.SECTION XX. mals258 SDECMAL FRACTIONS.. 181 Multiplication of Duodecimals, 259 DECIMAL FRACTIONS,........ 181 Numeration of Decimal Fractions,. 182 SECTION XXXVIII. Notation of Decimal Fractions.. 183 INVOLUTION,............261 Addition of Decimals,..... 184 Subtraction of Decimals,.... 185 SECTION XXXIX. Mlultiplication of Decimals,. 186 EVOLUTION,....... 263 Division of Decimals,....... 188 Extraction of the Square Root,. 264 Reduction of Decimals,.. 190 Application of the Sqtuare Root,. 268 Miscellaneous Exercises in Decimals, 192 Extraction of the Cube Root,.. 273 SE~CTION XXI. IApplication of the Cube Root,.. 277 REDUCTION OF CURRENCIES..... 193 SECTION XL. SECIO XXIARITHMETICAL PROoRESSlON,... 279 SECTION XXII, Annuities at Sim-lple Interest by ArithPERCENTAGE,.......... 196 metical Progression,...... 284 SECTION XXIII. SECTION XLI. SIMPLE INTEREST,......... 198 GEOMETRICA, PROGRESSION,. 286 Miscellaneous Exercises in Interest, 206 Annuities at Comnpound Interest by Partial Payments,........ 207 Geometrical Progression, Table,. 290 Problems in Interest,....... 212 SECTION XLI. SECTION XXIV. ALLIGATION,. 292 COMPOUND INTEREST........ 214 Alligationl Medial,.. 292 Table,.............. 216 Alligation Alternate,....... 293 SECTION XXV. SECTION XLIII. DISCOUNT,............. 218 PERMUTATION........... 297 SECTION XXVI. SECTION XLIV. BANK DISCOUNT,.......... 220 MENSURATION OF SURFACES,... 298 SECTION XXVII. SECTION XLV, COMMISSION AND BROKERAGE,... 222 MENSURATION OF SOLIDS...... 304 SECTION XXVIII. SECTION XLVI. STOCKS,............ 224 MENSUTR.ATION OF ILUMBER AND TIMSECTION XXIX. BSE,. 31C INSURANCE,..226 SECTION XLVII. SECTION XXX. MISCELLANEOUS QUESTIONS,... 311 Du)rI,... 227 WgTSITs, MbAZTJUtS A-VD MMONE,. 318 ARITHM E TIC ARTICLE.o ARITHIMETIC is the science of numbers and the art of computing by them. A number is a unit or an assemblage of units. A unit or unity is the number one, and signifies an individuai thing or quantity. The introductory and principal rules of arithmetic are Now tation, Numeration, Addition, Subtraction, Multiplication, and Division. The last four are called the principal or jiundamental rules because all arithmetical operations depend upon them. ~ I NOTATION AND NUMERATION. NOTATION. ART. 2. NOTATION is the art of expressing numbers by figures or other symbols. There are two methods of notation in common use; the Roman, and the Arabic or Indian.@ ART.. The Roman notation employs seven capital letters, viz.: 1, for one; V, for five; X, for ten; L, for fifty; C, for one hutnred; D, forfive hundred; M, for one thousand. The inte(lrmeditate numbers and the numbers greater than one thouS."mud are expressed by the use of these letters in various comb miattions; thus, II expresses two; IV, four; VI, six; IX, aitne; XV, fifteen; &c. Foor the origin of our present numeral characters, see the History of Arithmetic in the larger work of the author. OmQs'rIows. — Art. 1. What is arithmetic? What is number? What is a unit or unity? Which are the principal or fundamental rules of arithmetic? Why are they called the principal rules? - Art. 2, What is notation? How many and what methods of notation are in common use' - Art. 3. What are used to express numnbers in the Roman notation? What are their names? 8 NOTATIONL [LSECT. A. When two or more equal numbers are united, or a less number follows a greater, the sum of the two represents their value; as, XX, twenty; VI, six. But when a less number is placed before a greater, the difference of the two represents their value; as, IV, four; IX, nine. TABLE OF ROMAN LETTERS. I one. LX sixty. II two. LXX seventy. III three. LXXX eighty. IV four. XC ninety. V five. C one hundred. VI six. CC two hundred.'VII seven. CCC three hundred. VIII eight. CCCC four hundred. IX nine. D, or I1 five hundred. X ten. DC six hundred. XX twenty. DCC seven hundred XXX thirty. DCCC eight hundred. XL forty. DCCCC nine hundred. L fifty. M, or CID one thousand. Any number between unity and two thousand may be ex pressed by the letters in the preceding table,By first writing down the largest part of the required number, found in the table, and then annexing to this the next less, that will not make a number greater than the one required and thus proceeding until the number is complete. EXERCISES IN ROMAN NOTATION. The learner may write the following numbers in letters: —1. Ninety-six. Ans. XCVJ. 2. Eighty-seven. 3. One hundred and ten. 4. One hundred and sixty-nine. 5. Two hundred and seventy-five. 6. Five hundred and forty-two. 7. One thousand three hundred and nineteen. 8. One thousand eight hundred and forty-eight. qUESTIONS. - When is the sum of two letters taken for their value? When the difference? Repeat the Table of Roman Letters. What direc~ tion is given for writing numbers in the Roman notation-? SECT. 1.j NUMEE AT ION. 9 ART. -o The Arabic or Indian notation employs ten distinct characters or figures, sometimes called digits, viz.: — 1, 2, 3, 49 5, 6, 7, 8, 9, 0. one, two, three, four, five, six, seven, eight, nine, cipher. The first nine are called significant figures, because each one has a value of itself when standing alone. The cipher is also sometimes called naught or zero; and, when standing alone, it has no value and signifies nothing. NU MERATION. ART. 5o NUvIMERATION is the art of reading numbers, or naming the value of figures in the order of their places. ART. G. The Arabic figures have two values, a simple and a local, and, from their convenience, are now universally used in arithmetical calculations. AnT. 7o The simpZe value of a figure is the value it has when, standing alone, thus, 6; or when standing in the right;-hand place of whole numbers, thus, 26. In either case the 6 denotes six units or ones. ART. 8. The local value of a figure is the value it has when it is removed from the right-hand place toward the left, and depends on the place the figure occupies. For example, 6 standing at the left hand of 5, thus, 65, expresses ten times the value it does when standing alone, or in the right-hand place, and denotes six tens or sixty; the five at the right hand of it denotes five units, and the two figures together express sixtylfive. When placed at the left of two figures, thus, 678, it expresses one hundred times its simple value, or ten times its value when standing in the second or tens' place; its value being always increased tenfold, when it is removed one place to the left. Therefore, while the 8 denotes eight'units, and the 7, seven tens, the 6 denotes six hundreds, and the whole together, 678, six hundred and seventy-eight. QuEsrioNs. - Art. 4. How many characters are employed in the Arabic or Indian notation? What are the first nine called? Why? What is the tenth called? What does it represent or signify when standing alone? - Art. 5. What is numeration? - Art. 6. What two values have the Arabic figures? - Art. 7. What is the simple valle of a figure? - Art. 8. What is the local value? Why is this value called its local value? Wrhat effect has the removal of a figure oane plac. to the left upon, its rvalue i? Two places &ce. 1-0 ANUMERATION. [SECT. I. ART. 9. The cipher becomes significant when connected with cther figures; as in 10 (ten), where it gives a tenfold value to the 1; and 120 (one hundred and twenty), where it gives a tenfold value to the 12; and 304 (three hundred and four), where it has the same influence on the 3, causing it to represent three hundreds instead of three tens. The local value of figures will be made plain by the following table and its explanation. aS he igures in this table are read thus. 5, o g. The figures in this table are red thus. 9 Nine. 9 8 Ninety-eight. 9 8 7 Nine hundred eighty-seven. 9 8 7 6 Nine thousand eight hundred seventy-six. 9 8 7 6 5 Ninety-eight thousand seven hundred sixty-five. 39 8: 7 6 5 94t Nine hundred eighty-seven thousand six hundred ififty-four. 9 7 6 f5 L 3 i Nine millions eight hundred seventy-six thousand 2 five hundred forty-three. It will be noticed in the above table, that each figure in the right-hand or units' place expresses only its simple value, or so many units; but, when standing in the second place, it denotes so many tens, or ten times its simple value; and when in the third place, so many hundreds, or one hundred times its simple value; when in the fourth place, so many thousands, or a thousand times its simple value, and so on; the value of any figure being always increased tenfold by each removal of it one place to the left hand. QUESTIONS. - Art. 9. When does the cipher become significant? What is its effect, when placed at the right hand of a figure 1 What is the design of this table? What value has a figure standing in the right-hand or units' place? What, in the second place? What, in the third? How do figures increase from the right toward the left? SECT. I.] NUMERATION. 11 ART..l0. There are two methods of numeration in common use; the French and the English. The former is more generally used on the continent of Europe and in the United States. in the French method, a new name is given to every third figure above millions, and in the English, to every sixth figure above millions. FRENCH NUMERATION TABLE. Un -no - fW Q~ -'W.'- A. C) QW'H ~" k; C);:j pi W- 0.Q s i2 7, 894, 2 37 867, 143, 678, 478, 63S. Period of Period of Period of Period of Period of Period of Pexiod of Period of Sextil- QCuintil-,ua 0dril- Trillions. Billions. Millions. Thousands. Units. lions. lions. lions. The value ofthe numbers in this table, expressed in words is, Onehundred twenty-seven sextillions, eight hundred ninety4 tour quintillions, two hundred thirty-seven quadrillions, eight hundred sixty-seven trillions, one hundred twenty-three billions, six hundred seventy-eight millions, four hundred seventy-eight thousand, six hundred thirty-eight. The preceding table may be extended to any number of figures by supplying the names of the periods above sextillions, in their order; viz. Septillions, Octillions, Nonillions, Decillions, Undecillions, Duodecillions, Tredecillions, Quatuordecillions, Quindecillions, Sexdecillions, Septendecillions, Octodecillions, Novemdecillions, Vigintillions, &c. QUESTIONS.- Art. 10. What are the two methods of numeration in common use? Where is the Fre nch method more generally used? How does the FTench method differ from the English? Repeat the French Numeration Table, giving the names of all the places or orders, beginning at the right. What are the names of the different periods in the table? What is the value of the nmin mbers in the table expressed in words Repeat the names of the periods above sextillions. 12 NUMERATION. (SECT. x. ART. 1 fl The successive places occupied by figures are often called orders. Hence, a figure in the right-hand or units' place is called a figure of the first order, or of the order of units; a figure in the second place is a figure of the second order, or of the order of tens; in the third place, of the order of hundreds, and so on; thaus, in the number 1847, the 7 is of the — order of units, 4 of the order of tens, 8 of the order of hundreds, and 1 of the order of thousands, each figure expressing so many units of that order to which it belongs; so that we read the whole number one thousand eight hundred and fortyseven. ART. 19o From the preceding table and explanation, we deduce the following rule for numerating and reading numbers according to the French method. RULE. - Begin at the right hand, and divide the number into periods of THREE figures each, remembering the name of each period. Then, commencing at the left hand, read the figures of each period in the same manner as the period of units, giving the name of each period excepting the last. ExERCISES IN FRENcHI NTUMERATION. The learner may read orally, or write in words, the following numbers: - 1. 152 11. 592614 21. 20463162486135 2. 276 12. 400619 22. 63821024711802 3. 998 13. 610711 23. 44770630147671 4. 1057 14. 3031671 24. 3761700137706M71 5. 2254 15. 4869021 25. 242173562357421 6. 4384 16. 637313789 26. 870037637471078635 7. 7932 17. 39461928 27. 8216243812706381 8. 42198 18. 427143271 28. 2403172914376931 9. 84093 19. 6301706716 29. 3761706137706167138 1.0. 98612 20. 143776700333 30. 610167637896430607761607 QUESTIONS. - Art. 11. What are the successive places of the figures in the table called? Of what order is the first or right-hand figure? The second? The third? &c - Art. 12. What is the rule for numerating and reading numbersaccording to the French umethod 7 SECT. s.] NUIJ UEMI TION. 13 ART. V, To write numbers -according to the French method, we have the following RuLEn. -Begin at the ieft, and write the figure of the highest order to be written, and place in each successive order the figures belonging to it, observing to fill the place by a cipher, when no number is mentioned to be written. EXERCISES Ixi FR ENCH. NOTATION AND NUIERATION. The learner may write in figures, and read, the following rnumbers: -- 1. Forty-seven. 2. Three hundred fifty-nine. 3. Six thousand five hundred seventy-five, 4. Nine hundred and eight. 5. Nineteen thousand. 6. Fifteen hundred and four 7. Tweunty-seven millions five hundred. 8. Ninety-nine thousand ninety-nine. 9. Forty-two millions two thousand and five. 10. Four hundred eight thousand ninety-six. 11. Five thousand four hundred and two. 12. Nine hundred seven millions eight hundred five thou sand and seventy-four. 13. Three hundred forty-seven thousand nine hundred and fifteen. 14. Eighty-nine thousand fbrty-seven. 15. Fifty-one thousand eighty-one. 16. Seven thousand three hundred ninety.five. 17. Fifty-seven billions fifty-nine millions ninety-nine thousand and forty-seven. AnRT. 1zo The following table exhibits the English method of numeration, in which it will be observed that the figures are separated by commas into divisions or periods of six figures each. The first or right-hand period is regarded as units and thousands of units; the second as millions and thousands ot millions; and so on, six places being assigned to each division designated by a distinct name. QUESTIONs. —Art. 13. What is the rule for writing numbers according to the French mnethod At which hand do you begin to numerate figures? Where do you begin to read them? At which hand do you begin to write numbers? Why? - Art. 14. How many figures in each period in the English method of numeration?'What orders are found in the Elnglish method that are not in the French? 2_ :14 PaNUMERA'IOI. (SECT. i ENGLISH NUM ERATION TABLE. Ir.,QsQo.a~a5 *ai3 7 8 9 0 7 1 1 7 16, 3 7 1 7 12, 4 5 6 7 1 1. Period of Tril- Period of Bil- Period of Mil- Period of lions. lions. lions. Units. The value of the figures in the above table, expressed in words according to the English method, is, One hundred thirty. seven thousand eight hundred ninety trillions; seven hundred eleven thousand seven hundred sixteen billions; three hundred seventy-one thousand seven hundred twelve millions; four hundred fifty-six thousand seven hundred eleven. NOTE. - Although there is the same number of figures sn the English and in the French table, yet it will be observed that in the French table we have the names of three higher divisions than in the English. It will also be observed that the variation commences after the ninth place, or the place of hundreds of millions. If, therefore, we would know the value of numbers higher than hundreds of millions, when we see them written in words, or hear them read, we need to know whether they are expressed according to the French or the English method of numeration. ART. MQ To numerate and read numbers according to the English method we have the following RULE. -Begin at the right hand and divide the number into periods QUESTIONS. - Give the names of the periods in the English Numeration Table, beginning with the period of units. Repeat the table, giving the names of all the orders or places. What is the value of the numbers in the table expressed in words? How do the figures in the English and French table compare as to numbers? How as to periods? Why is this difference? ias a million the same value reckoned by the French table as when reckoned by the English? Has a billion the same value? Why not? By which table has it the greater value?- Art. 15. What is the rule for numeratii,. and reading numbers according to the English method 7 SECT.!.] NUMERATION. 15 of six figures each, remembering the name of each period. Then, commencing at the left hand, read the figures of each period in the same manner as the period of units, giving the name of each period excepting the last. ExERCIsEs IN ENGLISH NUMERATION. The learner may read orally, or write in words, the following numbers:. — 1. 125 5. 27306387903 2. 1063 6. 531470983712 3. 25842 7. 4230578032765038 4. 904357 8. 716756378807370767086389706473 ART. Gee, To write numbers according to the English method, we have the follpwing RULE. -.Begin at the left, and write the figure of the highest order to be written, and place in each successive order of the periods the figure belonging to it, observing to fill the place by a cipher, when no number is mentioned to be written. EXERCISES IN ENGLISH NOTATION AND NUMERATIONo The learner may write in figures, and read, the following numbers - 1. Three hundred- twenty-five thousand four hundred and twelve. 2. Two hundred fourteen thousand, one hundred sixty-five millions seventy-eight thousand and fifty-six. 3. Forty-two billions, six hundred seventeen thousand thirtyone millions, forty-one thousand three hundred forty-two. 4. Two thousand eight billions, nine thousand eighty-two millions, seven hundred one thousand, nine hundred and eight. 5. One hundred sixty-eight thousand two hundred fortyseven trillions, three hundred twenty-four thousand three hundred forty-one billions, four hundred seventy-two thousand three hundred nineteen millions, eight hundred sixteen thousand four hundred and twenty-one. QUESTION.- Art 16. What is the rule for writing numbers according to the English method? 16 ADDITION1. ISECT. 1t. ~ II. ADDITION. MENTAL ExERCISES. ART. ]7. WHEN it is required to find a single number to express the sum of the units contained in several smaller numbers, the process is called Addition. Ex. 1. James has 3 pears and his younger brother has 4, how many have both? ILLUSTRATION. - To solve this question, the 3 pears and 4 pears must be added together; thus, 3 added to 4 makes 7. Therefore James and his brother have 7 pears. ADDITION TABLE. 2 and 0are 2 3 and 0are 3 4and 0are 4 5 and 0are 5 2 and lare 3 3and l are 4 4 and lare 5 5 and lare 6 2 and 2 are 4 3 and 2 are 5 4 and 2 are 6 5 and 2 are 7 2and 3are 5 3and 3 are 6 4and 3are 7 5and 3are 8 2and 4-are 6 3and 4are 7 4and 4are 8.5and 4are 9 2 and 5 are 7 3 and 5 are 8 4 and 5 are 9 5 and 5 are 10 2 and 6 are 8 3 and 6 are 9 4 and 6 are 10 5 and 6 are 11 2and 7are 9 3and 7are 10 4and 7 are l 5 and 7are 12 2and 8are10 3 and 8 are 11 4 and 8 are 12 5 and 8 are 13 2 and 9 are 11 3 and 9 are 12 4 and 9 are 13 S and 9 are 14 2 and 10 are 12 3 and 10 are 13 4 and 10 are 14 5 and 10 are 15 2 and 11 are 13 3 and 11 are 14 4 and 11 are 15 5 and 11 are 16 2 and 12 are 14 3 and 12 are 15 4 and 12 are 16 5 and 12 are 17 6and 0are 6 7and 0are 7 8and 0are 8 9 and 0are 9 6 and I are 7 7 and 1 are 8 8 and 1 are 9 9 and 1 are 10 6and 2 are 8'7 and 2are 9 8and 2are 10 9and 2are 11 6and 3are 9 7 and 3 are 10 8and 3are l 9and 3are 12 6and 4are 10 7and 4 are 11 8and 4are 12 and 4are 13 6and 5 are 11 7and 5are 12 8and 5 are 13 9and 5are 14 6 and 6 are 12 7 and 6 are 13 8 and 6 are 14 9 and 6 are 15 6 and 7 are 13 7 and 7 are 14 8 and 7 are 15 9 and 7 are 16 6 and 8 are 14 7 and S are 15 8 and 8 are IS Sand 8 are!7 6and 9are 15 7 and 9 are 16 8 and 9 are 17 9 and 9 are 18 6 and 10 are 16 7 and 10 are 17 5 and 10 are 1S 9 and 10 are 19 6 and 11 are 17 7 and 11 are 1S 8 and II are 19 9 and 11 are 20 6 and 12 are 18 7 and 12 are 19 8 and 12 are 20 9 and 12 are 21 10 and 0 are 10 11 and 0 are 11 12 and 0 are 12 13 and 0 are 13 10 and 1 are 15 11 and 1 are 12 12 and 1 are 13 13 and S are 14 10 and 2 are 12 11 and 2 are 13 12 and 2 are 14 13 and 2 are 15 10 and 3 are 13 11 and 3 are 14 12 and 3 are 15 13 and 3 are 16 10 and 4 are 14 11 and 4 are 15 12 and 4 are 16 13 and 4 are 17 10 and 5are 15 11 and 5are 16 12 and are 17 13 and 5are 18 10 and 6 are 16 11 and 6 arer 17 12 and 6 are 18 13 and 6 are 19 10 and 7 are 17 11 and 7 are 18 12 and 7 are 19 13 and 7 are 20 10 and 8 are 18 II and 8 are 19 12 and 8 are 20 13 and 8 are 21 10 and 9 are 19 11 and 9 are 20 12 and 9 are 21 13 and 9 are 22 10 and 10 are 20 11 and 10 are 21 12 and 10 are 22 13 and 10 are 23 10 and 11 are 21 11 and 11 are 22 12 and 11 are 23 13 and 15 are 24 10 and 12 are 22 11 and 12 are 23 12 and 12 are 24 13 and 12 are 25 qUESTI.oO. - Art, 17. What does addition teach 7 SECT. II.. ADDITION. 17 2. Thomas had three nuts, and James gave him two more; how many had he then? 3. A boy had four apples, and he found two more; how mk any in all? 4. 1 have six dollars, and a man has paid me three more how many have I? 5, Enoch had seven marbles, and John gave him two more; how many had he then? 6. Benjamin has four dollars, and his sister, has three; how Umany have bothh? 7. Paid five dollars for a barrel of flour, and seven dollars mor sugar; how much for both S. James had two cents, and Samuel gave him six more; how many had he in all? 9. How many are five apples and six apples? 10. How many are four dollars and eight dollars? 11. HIow many are 2 and 3? 2 and 5? 2 and 7? 2 and 9? 2 and 4? 2 and 2? 2 and S? 2 and 6? 12. How many are 3 and 3? 3 and 5? 3 and 7? 3 and 9? 3 and 4? 3 and 6? 3 and 8? 3 and 3? 13. How many are 4 and 3? 4 and 5? 4 and 8? 4 and 9? 4and l? 4and2? 4and4? 4and7? 14. How many are 5 and 3? 5 and 4? 5 and 7? 5 and 8? 5and9? 5and2? 5and5? 5 and6? 5and I? 15. H3ow many are 6 and2? 6 and4? 6and3? 6 and 5? 6 and 7? 6 and 9? 6 and 1? 6 and 6? 6 and 8? 16. How many are 7 and 3? 7 and 5? 7 and 7? 7 and 6? 7 and 8? 7 a a 9 and2? 7 and 4? 7 and 10? 17. How many are 8 and 2? 8 and 4? 8 and 5? 8 and 7? 8 and 9? 8 and 8? 8 and 1? 8 and 3? 8 and 6? 18. How many are 9 and 1? 9 and 3? 9 and 5? 9 and 4? 9 and 6? 9 and 8? 9 and9? 9 and 2? 19. How many are 1 and 3? 11 and 2 11 and 4 11 and 6? 11 and 7? 11 and 9?. 11 and I I? 11 and 13? lI and 12? 11 and 2 and 3? 11 and 4 and 4? I Iand 15? 12 and 7 and 3? 12 and 6 and 3? 8 and 8 and 4? 9 and 5 and 6? 9 and 8 and 8? 20. Gave nine cents for a pound of cheese, and seven cents for a quart of molasses; what did I give for both? 21. if you buy a picture-book for eleven cents, and a knife for nine cents, what is the cost of both? 22. John paid Luke seven cents for marbles, and twelvld, cents for gingerbread; how much money was received P in i ADDITION. [saCT. lI. 23. Thomas paid twelve cents for a top, and eight cents for cherriio; what did both cost? 24. A merchant sold three barrels of flour to one man, and thirteen to another; what was the quantity sold? 25. I have two apple-trees; one bears twelve bushels of apples, and the other eleven; how many bushels do both trees produce? 26. How many are 4 and 2 and 3? 5 and 7 and 1? 3 and 4 and 3? 6 and 6 and 5? 2 and 2 and 8? 2 and 3 and 9? 3 and 5 and6? 2 andS and S? 27. How many are 2 and 6 and 7? 2 and 7 and 7? 2 and 8and9? 2and7and4? 2and 5and9? 2and9and6? 2 and 3 and 10 10 and 10 and 9? 28. How many are 3 and 2 and 2? 3 and 3 and 2? 3 and 5 and 5? 3 and 4 and 7? 3 and 6 and7? 3 and7and 10? 3 and 8 and 9? 3 and 9 and? 9 9 and 12? 29, How many are4 and 2 and 2? 4 and 3 and 3? 4 and 4 and 5? 4 and 6 and 7? 4 and 7 and 7? 4 and 8 and 3? 4and 9 and3? 4andSand? 12 and 12? 30. How many are 5 and 3 and 3? 5 and 4 and 4? 5 and 5 andil? 5 and 6 and 7? 5 and 7 and 8? 5 andS and 7? 5and9and9? 5and 10and3? 3and 112 31. HIow many are 6 and 2 and 7? 6 and 3 and 6? 6 and 5 and i? 6 and 7 and-5? 6 and 8 and 7? 6 and 9 and 8? 6 and 1O and 10? and 6 and 6 and 6? 32. How many are 7 and 2 and 3? 7 and3 and 3? 7 and 5and9? 7and6and6? 7and8andS? 7and9and 8? 7 and 10 and 1 1? 8 and 1I and 7? 33. How many are 8 and 2 and 9? 8 and 4 and 3? 8 and 7and7? 8and 9and 10? Sand 7 and9? 8and 10and 10? 8 and 9 and 12? 7 and 7 and 11? 34. How many are 9 and 5 and 2? 9 and 4 and 3? 9 and 9and6? 9and 10 and 3? 9 and 8 and 8? 9 and 4 and 9 9 and 9 and 9? 8 and 8 and 12? 35. How many are 2 and 2 and 4 and 5? 3 and 4 and 5 and 6? 4and5and 6 and 7? 5 and 5 and 4 and 4? 9 and I and 2 and 3 and 5? 7 and 7 and 7 and 7? 36. James had 4 apples, Samuel gave him 5 more, and John gave him 6; how many had he in all? 37. Gave 7 dollars for a barrel of flour, 5 dollars for a hundred weight of sugar, and 5 dollars for a tub of butter; what did I give for the whole? 38. Paid 5 dollars for a pair of boots, 12 dollars for a coat, and 6 dollars for a vest; what was trhe twhole qost? EaCT. IHX: ADDI'T1OI. 19 39. I have 7 apple-trees, 9 cherry-trees, 6 pear-trees, and 8 olum-trees; how many in all? 40. Paid 50 dollars for a horse and 70 dollars for a chaise; what was the cost of both? ILLUSTRATION. - We may first divide the dollars into tens, and.hen add them together. Thus, 50 equals 5 tens, and 70 equals 7 tens; 5 tens and 7 tens are 12 tens or 120 units. Therefore the cost of the horse and chaise was 120 dollars. 41. A man performed a journey in 4 days; the first day he travelled 10 miles, the second day 20, the third day 30, and the fourth day 40 miles; what was the whole distance? 42. Gave 10 cents for an almanac, 30 cents for paper, 50 cents for quills, and 80 cents for a penknife; what did I give for the whole? 43. Gave 75 cents for an arithmetic and 67 for a geography; what was the cost of both? ILLUSTRATION.- We may divide the cents into tens and units. Thus, 75 equals 7 tens and 5 units; 67 equals 6 tens and 7 units; 7 tens and 6 tens are 13 tens; and 5 units and 7 units are 12 units or 1 ten and 2 units; 1 ten and 2 units added to 13 tens - make 14 tens and 2 units, or 142. Therefore the arithmetic and geography cost i42 cents, or 1 dollar and 42 cents. 44. Bought eight yards of broadcloth for 32 dollars, and forty yards of carpeting for 46 dollars; what did they both cost? 45. In a certain school 9 scholars study grammar, 12 arithmetic, 7 logic, 2 rhetoric, and 17 punctuation; how many are there in the school? 46. Paid 2 dollars for a cap, 3 dollars for shoes, 7 dollars for pantaloons, 6 dollars for a vest, and 22 dollars for a coat; what was the cost of the whole? 47. On the fourth of July 20 cents were given to Emily, 15 cents to Betsey, 10 cents to Benjamin, and none to Lydia; what did they all receive? 48. Bought four loads of hay; the first cost 15 dollars, the second 12 dollars, the third 20 dollars, and the fourth 17 dollars; what was the price of the whole? 49. Gave 55 dollars for a horse, 42 dollars for a wagon, and 17 dollars for a harness; what did they all cost? 50. Sold 3 loads of.wood for 17 dollars, 6 tons of timber for 19 dollars, and a pair of oxen for 67 dollars; what sum did I receive? 20 ADDITION. [sECT. It. ART. I S. From the solution of the preceding questions the learner will perceive, that ADDITION is the process of collecting several numbers into one sum, which is called their amount. ART.:!9. SIGNS. - Addition is commonly represented by this character, +, which signifies plus, or added to. The expression 7 + -5 is read,7 plus 5, or 7 added to 5. This character, -, is called the sign of equality, and signifies equal to. The expression 7 + 5 = 12 is read,7 plus 5, or 7 added to 5, is equal to 12. EXERCISES FOR THE SLATE. ARLT. 20. The method of operation, when the numbers are large, and the sum of each column is less than 10. Ex. 1. A man bought a watch for 42 dollars, a coat for 16 dollars, and a set of maps for 21 dollars; what did he pay for the whole? Ans. 79 dollars. OPERATION. In this example, having arranged the Dollars. numbers, units under units, and tens un4 2 der tens, in regular columns, we first add 1 6 the column of units; thus, 1 and 6 are 21 and 2 are 9 (units), and set down the amount under the column of units. We Amount 7 9 then add the column of tens; thus, 2 and I are 3 and 4 are 7 (tens), which we set under the column of tens, and thus find the amount of the whole to be 79 dollars. ART. 21. First Method of Proof. - Begin at the top and add the columns downward in the same manner as they were before added upward; and if the two sums agree, the work is presumed to be right. The reason of this proof is, that, by adding downward, the order of the figures is inverted; and, therefore, any error made in the first addition would probably be detected in the second. NoTa. - This method of proof is generally used in business. QUESTIONS. - Art. 18. What is addition I - Art. 19. What is the sign of addition, and what does it signify? What is the sign of equality, and what does it signify? - Art. 20. Hlow are numbers arranged for addition? Which column must first be added? Why? Where do you place its sum? Where must the sum of each column be placed? What is the whole sum called 7 -Art. 21. flow is addition proved? What is the reason for this method of proof? s, this method in common use? SEC'. a1.) ADDITIOI. 21 EXAMPLES FOR PRACTICE. 2. 3. 4. 5. -Miles. Furlongs. Days. Weeks. 151 234 472 121 212 423 315 516 321 321 102 361 Ans. 684 6. What is the sum of 231, 114, and 324? Ans. 669. 7. Required the sum of 235, 321, and 142. Ans. 698. 8. What is the sum of 11, 22, 505, and 461? Ans. 999. 9. Sold twelve ploughs for 104 dollars, two wagons for 214 dollars, and one chaise for 121 dollars; what was the amount of the whole? Ans. 439 dollars. 10. A drover bought 125 sheep of one man, 432 of another, and of a third 311; how many did he buy? Ans. 868 sheep. ART.!W. Method of operation, when the sum of any column is equal to, or exceeds, 10. Ex. 1. I have three lots of wild land; the first contains 246 acres, the second 764 acres, and the third 918 acres; I wish to know how many acres are in the three lots Ans. 1928 acres. Having arranged the numbers as in cOPE s.ATIO. the preceding examples, we first add Acres. the units; thus, 8 and 4 are 12, and 2 4 6 6 are 18. In 18 units there are 1 ten 7 6 4 and 8 units; we write the 8 units un9 1 8 der the column of units, and, reserving the 1 ten in the mind, we carry or add Amount 1 9 2 8 it to the column of tens; thus, 1 carried to 1 makes 2, and 6 are 8, and 4 are 12 (tens), equal to 1 hundred and 2 tens. We write the 2 tens under the column of tens, and, reserving the 1 hundred in the mind, carry it to the column of hundreds; thus, 1 carried to 9 makes 10, and 7 are 17, and 2 are 19 (hundreds), equal to I thousand and 9 hundreds. We write the 9 under the column of hundreds; and there being no other column to be added, we set down the 1 thousand in thousands' place, and find the amount of the several numbers to be 1928. QUESTIONS. -Art. 22. When the sum of any column exceeds ten, where are the units written? What is done with the tens? What is meant by carrying the tens? Why do you carry one for every 10? Why not for every 112 7 How is the sum of the last column written? 22 ADDITION. [SECT. 11. ART. s3. From the preceding examples and illustrations in addition, we deduce the following general RULE. - 1. Write down the numbers to be added, units under units, tens under tens, 4dc., and draw a line underneath. 2. Then, beginning at the right hand, add, upward, all the figures,in the column of units, and, if the amount be less than ten, set it down under the column added. But, if the amount be ten or more, write down the unit figure only, and carry the figure denoting the ten or tens in the mind, and add it to the next column. 3. Proceed in this way with each column, until they are all added, ob. serving to write down the whole amount of the last column. ART. Ql. Second Method of Proof. —Separate the numbers to be added into two parts by drawing a horizontal line between them. Add the numbers below the line and set down their sum. Then add this sum and the number or numbers above the line together; and, if their sum is equal to the first amount, the work is presumed to be right. The reason of this proof depends on the obvious principle, That the sum of all the parts into which any number is divided is equal to the whole. EXAMPLES FOR PRACTICE. 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROOF. 526 526 241 241 317 37 532 532 132 3 9 13 91 3 Ans. 1504 First at1504 Ans. 1893 Firstamt 9 93 978 1652_ Ans. 1504 Ans. 1893 4: 5. 6. 7. 8. 9. Dollars. lMiles. Pounds. Rods. Inches. Feet. 11 47 127 678 789 1769 23 87 396 971 478 7895 97 58 787 147 719 7563 86 83 456 716 937 8765 2 17 275 1766 2512 2923 25992 QUESTIONS. —Art. 23. What is the general rule for addition?-Art. 24. What is the second method of proving addition? What is the reason of this method of proof? SECT. f.1l ADDITION. G10I L 12. 12. 14. Ounces. Drachms, Cents. Eagles. Degrees. 876. 789 123 471 1234 376 567 478 617 3456 715 743 716 871 6544 678 435 478 317 7891 910 678 127 899 8766 15. 16. 17, 18. Feet. inches. Hours. Minutes. 78956 71678 71 123 98 7 6 5 37667 12345 45678 1234 12345 67890 34680 67i1 1 67890 34567 56777 33333 78999 89012 67812 7-1 345 13579 789 17 71444 99999 19. 20. 21. 22. Days. Years. IVionths. Hogsheads 17875897 789567 37 30176 71 67512 7613 1378956 31 876567 123123 700714 8601 98765 70071 367 11 7896 475 76117 991 1 789 1069 4'611779 89120 78 374176 9 1 7 1 710 7 761 13176.5 4325 23. Add 1001, 76, 10078, 15, 8761, 7, and 1678. Ans. 21616. 24. Add 49, 761, 3756, 8, 150, 761761, and 18. Ans. 766503. 25. Required the sum of 3717, 8, 7, 10001, 58, 18, and 5. Ans. 13814. 26. Add 19, 181, 5, 897156, 81, 800, and 71512. Ans. 969754. 27. What is the sum of 999, 8081, 9, 1567, 88, 91, 7, and 878? Ans. 11720. 28. Add 717 18765, 9111, 1471 678, 9, 1446, and 71. Ans. 31622. 29. Add 51, 1, 7671, 89, 871787, 61, and 70001. Ans. 949661 24 A&DDITION. [SECT. It. 30. What is the sum of 71, 8956, 1, 785, 587, and 76178? Ans. 86578. 31. Add 9999, 8008, 8, 81, 4777, and 516785. Ans. 539658. 32. Add 5, 7, 8911, 467, 47895, and 87. Ans. 57372. 33. Add 123456, 71, 8005, 21, and 716787. Ans. 848340. 34. Add 47, 911111, 717, 81, 88767, and 56. Ans. 1000779e 35. What is the sum of 71, 8899, 4, 7111, and 678679? Ans. 694764. 36. Add 81, 879, 41, 76789, 42, 1, and 78967. Ans. 156800. 37. Add 917658, 75, 876789, 46, and 8222. Ans. 1802790. 38. Add 91, 76756895, 76, 14, 3, and 76378. Ans.'76833457. 39. Add 10, 100, 1000, 10000 100000, and 1000000. Ans. 1111110. 40. What is the sum of 9, 99, 99, 1111, 8000, and 5? Ans. 9323. 41. Add 41, 76517 7678956, 43, 15, and 6780. Ans. 7693486. 42. Add 1234, 7891, 3146751, 27, 9, and 5. Ans. 3155917. 43. What is the sum of 19, 91, 1, 1, 1478, 1007, and 46? Ans. 2643. 44. Add four hundred seventy-six, seventy-one, one hundred five, and three hundred eighty-seven. Ans. 1039. 45. Add fifty-six thousand seven hundred eighty-five, seven hundred five, thirty-six, one hundred seventy thousand and one, and four hundred seven. Ans. 227934. 46. Add fifty-six thousand seven hundred eleven, three thou. sand seventy-one, four hundred seventy-one, sixty-one, and three thousand and one. Ans. 63315. 47. What is the sum of the following numbers: seven hundred thousand seven hundred one, seventeen thousand nine, one million six hundred thousand seven hundred six, forty-seven thousand six hundred seventy-one, seven thousand forty-seven, four hundred one, and nine? Ans. 2373544. 48. Gave 73 dollars for a watch, 15 dollars for a cane, 119 dollars for a horse, 376 dollars for a carriage, and 7689 dollars for a house; how much did they all cost? Ans. $S72 dollars. SECT. I.] AD9DITIOIN. 25 49. In an orchard, 15 trees bear plums, 73 trees bear apples, 29 trees bear pears, and 14 trees bear cherries; how many trees are there in the orchard? Ans. 131 trees. 50. rhe hind quarters of an ox weighed 375 pounds each, the fore quarters 315 pounds each; the hide weighed 96 pounds, and the tallow 87 pounds. What was the whole weight of the-ox? Ans. 1563 pounds. 51. A man bought a farm for 1728 dollars, and sold it so as to gain 375 dollars; how much did he sell it for? Ans. 2103 dollars. 52. A merchant bought five pieces of cloth. For the first he gave 376 dollars, for the second 198 dollars, for the third 896 dollars, for the fourth 691 dollars, and for the fifth 96 dollars. How much did he give for the whole? Ans. 2257 dollars. 53. A merchant bought five hogsheads of molasses for 375 dollars, and sold it so as to gain 25 dollars on each hogshead.; for how much did he sell it? Ans. 500 dollars. 54. John Smith's farm is worth 7896 dollars; he has bank stock valued at 369 dollars, and he has in cash 850 dollars. How much is he worth? Ans. 9115 dollars. 55. Required the number of inhabitants in the New England States by the census of 1840, there being in Maine 501,793, in New Hampshire 284,574, in Massachusetts 737,699, in Rhode Island 108,830, in Connecticut 309,978, and in Vermont 291,948. Ans. 2,234,822. 56. Required the number of inhabitants in the Middle States in 1840, there being in New York 2,428,921, in New Jersey 373,306, in Pennsylvania 1,724,033, in Delaware 78,085, and in Maryland 469,232. Ans. 5,073,577. 57. Required the number of persons in the Southern States in 1840, there being in Virginia 1,239,797, in North Carolina 753,419, in South Carolina 594,398, in Georgia 691,392, in Alabama 590,756, in Mississippi 375,651, and in Louisiana 352,411. Ans. 4,597,824. 58. How many inhabitants in the Western States in 1840, there being in Tennessee 829,210, in Kentucky 779,828, in Ohio 1,519,467, in Indiana 685,866, in Illinois 476,183, in Missouri 383,702, in Arkansas 97,574, and in Michigan 212,267? Ans. 4,984,097. 59. How many inhabitants in 1840 in the following Territories and the District of Columbia, there being in Florida 54,477, in Wisconsin 30,945, in Iowa 43,112, and in the District of Columbia 43,712? Ans. 172,246. UP 26 SUBTRACTION. (SECT. lIR 60. 61. 62. 63. 64. Ounces. Yards. Feet. Inches. Chaldrons. 1234 2345 3456 789 1 5678 5678 6789 7891 1356 3215 9012 1023 3456 7891 6789 3456 4456 7891 2345 3214 7890 7890 3456 6789 1234 1345 1234 7890 1234 3789 6 7 89 5678 1378 5678 1379 3216 9012 8123 9123 9008 7(590 3456 4567 4567 1071 1030 7890 8912 8912 7163 7055 1345 3456 3456 6781 5678 6789 7891 7812 1780 1234 3456 3456 3456 3007 5678 7890 7891 7812 5617 9001 5678 3783 3713 4456 2345 9012 1237 7891 3456 6789 3456 7891 1357 789 1 1030 7890 1007 9009 3070 781'6 1234 5670 8765 4567 1781 5678 1234 4321 3456 ~ III. SUBTRACTION. MENTAL EXERCISES. ART. W5. \WHEN it iS required to find the difference be. tween two numbers, the process is called Subtraction. The operation is the reverse of addition. Ex. 1. John has 7 oranges, and his sister but 4; how many more has John than his sister? ILLUSTRATION. - To solve this question, we first inquire what number added to 4 will make 7. From addition we learn that 4 and 3 are 7; consequently, if 4 oranges be taken from 7 oranges, 3 will remain. Hence John has 3 oranges more than his sister. QUESTIONS. Art. 25. What does subtraction teach? Of what is it the reverse? SECT, 11.) ] SUBTRACTION. 27 The following table will be of service to facilitate the progress of the learner in the solution of questions in subtraction. SUBTRACTION TABLE. 1 from 1 leaves 0 2 from 2 leaves 0 3 from 3 leaves 0 4 from 4 leaves 0 1 from 2 leaves 1 2 from 3 leaves 1 3 from 4 leaves 1 4 from 5 leaves 1 1 from 3 leaves 2 2 from 4 leaves 2 3 from 5 leaves 2 4 from 6 leaves 2 1 from 4 leaves 3 2 from 5 leaves 3 3 from 6 leaves 3 4 from 7 leaves 3 1 from 5 leaves 4 2 from 6 leaves 4 3 from 7 leaves 4 4 from 8 leaves 4 1 from 6 leaves 5 2 from 7 leaves 5 3 from 8 leaves 5 4 from 9 leaves 5 1 from 7 leaves 6 2 from 8 leaves 6 3 from 9 leaves 6 4 from 10 leaves 6 1 from 8 leaves 7 2 from 9 leaves 7 3 from 10 leaves 7 4 from 11 leaves 7 1 from 9 leaves 8 2 from 10 leaves 8 3 from 11 leaves 8 4 from 12 leaves 8 1 from 10 leaves 9 2 from 11 leaves 9 3 from 12 leaves 9 4 from 13 leaves 9 1 from 11 leaves 10 2 from 12 leaves 10 3 from 13 leaves 10 4 from 14 leaves 10 l from 12 leaves 11 2 from 13 leaves 11 3 from 14 leaves 11 4 from 15 leaves 11 1 from 13 leaves 12 2 from 14 leaves 12 3 from 15 leaves 12 4 from 16 leaves 12 5 from 5 leaves 0 6 from 6 leaves 0 7 from 7 leaves 0 8 from 8 leaves 0 5 from 6 leaves 1 6 from 7 leaves 1 7 from 8 leaves 1 8 from 9 leaves 1 5 from 7 leaves 2 6 from 8 leaves 2 7 from 9 leaves 2 8 from 10 leaves 2 5 from 8 leaves 3 6 from 9 leaves 3 7 from T0 leaves 3 8 from 11 leaves 3 5 from 9 leaves 4 6 from 10 leaves 4 7 from 11 leaves 4 8 from 12 leaves 4 5 from 10 leaves 6 6 from 11 leaves 5 7 from 12 leaves 5 8 from 13 leaves 5 5 from 11 leaves 6 6 from 12 leaves 6 7 from 13 leaves 6 8 from 14 leaves 6 5 from 12 leaves 7 6 from 13 leaves.7 7 from 14 leaves 7 8 from 15 leaves 7 5 from 13 leaves 8 6 from 14 leaves 8 7 from 15 leaves 8 8 from 16 leaves 8 5 from 14 leaves 9 6 from 15 leaves 9 7 from 16 leaves 9 8 from 17 leaves 9 5 from 15 leaves 10 6 frorm 16 leaves 10 7 from 17 leaves 10 8 from 18 leaves 10 5 from 16 leaves 11 6 from 17 leaves 11 7 from 18 leaves 11 8 from 19 leaves 11 5 from 17 leaves 12 6 from 18 leaves 12 7 from 19 leaves 12 8 from 20 leaves 12 9 from 9 leaves 0 10 from 10 leaves 0 1 from 11 leaves 0 12 from 12 leaves 0 9 from 10 leaves 1 10 from 11 leaves 1 11 from 12 leaves 1 12 from 13 leaves 1 9 from 11 leaves 2 10 from 12 leaves 2 11 from 13 leaves 2 12 from 14 leaves 2 9 from 12 leaves 3 10 from 13 leaves 3 11 from 14 leaves 3 12 from 15 leaves 3 9 from 13 leaves 4 10 from 14 leaves 4 11 from 15 leaves 4 12 from 16 leaves 4 9 from 14 leaves 5 10 from 15 leaves 5 11 from 16 leaves 5 12 from 17 leaves 5 9 from 15 leaves 6 10 from 16 leaves 6 11 from 17 leaves 6 12 from 18 leaves 6 9 from 16 leaves 7 10 from 17 leaves 7 11 from 18 leaves 7 12 from 19 leaves 7 9 from 17 leaves 8 10 from 18 leaves 8 11 from 19 leaves 8 12 from 20 leaves 8 9 from 18 leaves 9 10 from 19 leaves 9 11 from 20 leaves 9 12 from 21 leaves 9 9 from 19 leaves 10 10 from 20 leaves 10 11 from 21 leaves 10 12 from 22 leaves 10 9 from 20 leaves 11 10 from 21 leaves 11 11 from 22 leaves 11 12 from 23 leaves 11 9 from 21 leaves 12 10 f aor, 22 leaves 12 11 from 23 leaves 12 12 from 24 leaves 12 2. Thomas had five oranges, and gave two of them to John, how many had he left? 3. Peter had six marbles, and gave two of them to Samuel; how many had he left? 4. Lydia had four cakes; having lost one, how many had she left? 5. Daniel, having eight cents, gives three to Mary; how many has he left? 6. Benjamin had ten nuts; he gave four to Jane, and three to Emily; how many had he left? 7. Moses gives eleven oranges to John, and eight to Enoch; how many more has John than Enoch? ~2 ~ SUBTRACT1ION.. [SECT. a. 8. Agreed to labor for a man twelve days; how many remain, after I have been with him five days? 9. 1 owed Thomas nine dollars, and, having paid him seven, how many remain due? 10. From ten dollars, I paid four dollars to one man, and three dollars to another; how much have I left? i1. Timothy had eleven marbles, and lost seven; how many hlad he left? 12. John is thirteen years old, and his brother Thomas is seven; how much older is John than Thomas? 13. From fifteen dollars, I paid five; how many had I left? 14. Sold a barrel of flour for eight dollars, and a bushel ofi wheat for two dollars; what was the difference in the prices? 15. Paid seven dollars for a pair of boots, and two dollars for shoes; how much did the boots cost more than the shoes? 16. How many are 4 less 2? 4 less 1? 4 less 4? 17. How many are 4 less 3? 5 less 1? 5 less 5? 18. How many are 5 less 2? 5 less 3? 5 less 4? 19. How many are 6 less 1? 6 less 2 6 less4? 6 less 5? 20. Hlow many are 7 less 2? 7 less 3? 7less4? 7 less 6? 21. Hlow many are 8 less 6? 8 less 5? 8 less 2? 8 less 4? 8 less 1? 22. Hlow many are 9 less 2? 9 less 4? 9 less 5? 9 less 7? 9 less 3? 23. How many are 10 less 8? 10 less 7? 10 less 5? 10 less 3? 10 less 1? 24. HIow many are 11 less 9 11 less 7? 111 less 5? 11 less 3?:I less 4? 25. How many are 12 less 10? 12 less 8? 12 less 6? 12 less 4? l2 less 7? 26. How many are 13 less 11? 13 less 10? 13 less? 83 less 9? 13 less 5? 27. Fow many are 14 less 11? 14 less 9? 14 less 8? 14 less 6? 14 less 7? 14 less 3? 28. How many are 15 less 2? I5 less 4? 15 less 5? 15 less 7? 15 less 9? 15 less 13? 29. How many are 16 less 3? 16 less 4? 16 less 7? 16 less 9? 16 less 11? 16 less 15? 30. How many are 17 less 1? 17 less 3? 17 less 5? 17 less 7? 17 less 8? 17 less 12? 31. How many are 18 less 2? 18 less 4? 18 less 7 18 less 8? 18 less 10? 18 less 12? 32. How many are 19 less I? 19 less 3? 19 less 5? 19 less 7? 19 less9? 19 less 16 SECT. Ill. SUBTRACTION. 29 33. How many are 20 less 5? 20 less 8? 20 less 9? 20 less 12? 20 less 15? 20 less 19? 34. Bought a horse for 60 dollars, and sold him for 90 doltars; how much did I gain? ILLUSTRATION. - We may divide the two prices of the horse into tens, and subtract the greater from the less. Thus 60 equals 6 tens, and 90 equals 9 tens; 6 tens from 9 tens leave 3 tens or 30. Therefore I gained 30 dollars. 35. Sold a wagon for 70 dollars, which cost me 100 dollars; how much did I lose? 36. John travels 30 miles a day, and Samuel 90 miles; what is the difference? 37. I have 100 dollars, and after I shall have given 20 to Benjamin, and paid a debt of 30 dollars to J. Smith, how many dollars have I left? 38. John Smith, Jr. had 170 dollars; he gave his oldesi daughter, Angeline, 40 dollars, his youngest daughter, Mary, 50 dollars, his oldest son, James, 30, and his youngest son, William, 20 dollars; he also paid 20 dollars for his taxes; how many dollars had he remaining? ART..0. The pupil, having solved the preceding questions, will perceive, that SUBTRACTION is the taking of a less number from a greater to find the difference. The greater number is called the Minuend, and the less number, the Subtrahend.@ The answer, or number found by the operation, is called the Diference or Remainder. ART. 7. SIGNS. - Subtraction is denoted by a short horizontal line, thus -, signifying minus or less. It indicates that the number following is to be taken from the one that precedes it. The expression 6 — 2 4 is read, 6 minus, or less, 2 is equal to 4. The words minuend and szubtrahend are derived from two Latin words, the former from minuendum, which signifies to be diminished or made less, and the latter from subtrahendunm, which means to be subtracted or taken away. QUESTIONS. - Art. 26. What is subtraction? What is the greater number called? What is the less number called? What the answer? - Art. 27. What is the sign of subtraction? What does it signiffy and indicate? 8 * 30 SUBTRACTIONI [sc'r. xx XERCISES FOR THE SLATE. A rT. S i:Method of operation, when the numbers are lal'ge and each figure in the subtrahend is less than the figure above it in the minuend. Ex. 1. Let it be required to take 245 from 468, and to find their difference. Ans. 2923. OPERATION. eIn this operation, we place the less numrt inuend 4 68 tber under the greater, units under units, tens under tens, &c., and draw a line beSubtrahend 2 4 5 low them. We thenr begin at the right Remainder 2 2 3 hand, and say, 5 from 8 leaves 3, anld write the 3 directly below. We then say, 4 from 6 leaves 2, and write the 2 below the line, as before, and proceed with the next figure and say, 2 fromi 4 leaves 2, which we also write below. W e thus find -the difference to be 223. ART. QN;. First Il.ethod of Proof. - Add the remainder and the subtrahlend together, and their sum will be equal to the minuend, if the work is right. This method, of proof depends on the obvious principle, That the greater of any two numbers is equal to the less added to the difference between them. E~XAMPLES FOR PRACTICE. -24. 32 2 3. OPERATION. OPERATION AND PROOF. OPERATION. OPERATION AND PROO00 Minuend 5 4 7 5 4 7 98 6 986 Subtrahend 235 35 763 763 Remainder 31 2 312 223 223 Min. 5 4 7 Min. 9 86 4. 5. 6. 7. From 6 8 4 7 3 5 8 6 4 9 48 Take 462 523 65-1 746 8. A farmer paid 539 dollars for a span of fine horses, and sold them for 425 dollars; how much did he lose? Ans. 114 dollars. 9. A farmer raised 896 bushels of wheat, and sold 675 bushels of it; how much did he reserve for his own use? Ans. 22!1 bushels. QUESTIONS. - Art. 28. How are numnlbers arranged for subtraction? Where do you begin to subtract? Why? Where do you write the difference 7? Art. 29. What is the first method of proving subtraction? What is the reason of this proof, or on what principle does it depend? 8r cT. 111.] $UBTRACTION. 31 0.%O A gentleman gave his son 3692 dollars, tand hiis daughter 1.212 dollars less than his son; how mnuch did his daughter receive? Ans. 2480 dollars. ART., 0S Method of operation when any figure in. the subtrahend is greater than the figure above it in the minuend~ Ex. 1. If I have 624 dollars, and lose 342 of them, how many remain? Ans. 282. OPERATION. In performing this examlple, we first talke IMinuend 6 2 4 the 2 units from the 4, and find the difference Subtrahend 3 42 to be 2, which we write directly under the - figure subtracted. VWe then proceed to take Remainder 2 8 2 the 4 tens from the 2 tens above it; but we here find a difficulty, since the 4 is greater than 2, and cannot be subbtracted from it. We therefore add 10 to the 2, which Imakes 12, and then subtract 4 froi. 12 and 8 remains, which we write directly below Then, to compensate for the 10 thus added to the 2 in the minuend, we add one to the 3 in the next higher place in the subtrahend, which makes 4, and subtract 4 from 6, anu 2 remains. The remainder therefore, is 282. The reason of this operation depends upon the self-e-vident truth'Thal, if any to nza.mbers are equally increased, their difference.re nmae-ns the sname. Tn this example 10 tens, equal to I hundred, Awere added to the 2 tens in the upper number, and 1 was added to the 32 hundreds in the lower nanber. 3Now, since the 3'stands in the hundreds' place, the 1 added was in fact 1 hundred. 7-:ence, the two aurmbers being equally increased, the difference is the same. NorE. - This addition of 10 to the minuend is sometimes called bor rowing 1(0, and the addition of 1 to the subtrahend is called carrying 1. ARnT. 3g1. From the preceding examples and illustrations in subtraction, we deduce the following general RUaE. -= 1. Place the less number undler the greater,.units nder units, tens un.der tens, 4c. anl d draw a line untCer thens. 2. Then, commencing with the'units, subtract each figure of the subtrahcLd from the figure above it in the wzinuend, and write the icfference below. QUEsrsfioNs. - Art. 30. How do you. proceed when a figure of the subtrahend is larger than the one above it in the mninuend 1? How do you comnpensate for the 10 which is added to the minuend 1 What is the reason for this addition to the minueld and subtrahend? How' does it appear that the I added to the subtrahend equals the 10 added to the minuend? What is the addition of 10 to the -minuend sonmetimes called'! The additionl of 1 to the3 subtrahend'? - Art. 31. What is the general rule for subtraction'i SUBTRACTIOJ, [Sue. IaI 3. If any figure of the subtrahend as larger than the figure above it in the minuend, add 10 to that figure of the minuend, and jfrom their sum subtract the lower figure; then carry 1 to the next fig-ure of the subtrahend, and subtract as before, till all the figures of the subtrahend are subtracted, and the result will be the difference or remainder. ART. 30 Second 1llethod of Proof. - Subtract the remainder or difference from the minuend, and the result will be like the subtrahend if the work is right. This method of proof depends on the principle, That the smaller of any two numbers is equal- to the remainder obtained by subtracting their difference from the greater, EXAMPLES FOR PRACTICE. 2. 2. 3. 3. OPERATION. OPERATION AND PROOF. OPERATION, OPERATION AND PROOFI. Minuend 376 3 7 6 5 3 1 5 3 1 Subtrahend 167 67 389 389 Remainder 2 0 09 2 0 9 142 1 42 Sub. 1 6 7 Sub. 389 4. 5. 6. 7, Tons. Gallons. Pecks. Feet. From 978 67158 14711 1 00000 Take 199 14339 9197 90909 Ans. 7 7 9 528 19 55 14 9091 S. 9. 10. I1. Miles. Dollars. Minutes. Seconds. From 67895 456798 765321 555555 Take 19999 190899 177777 177777 12. 13. Rods. Acres. From 100200300400500 1000000000000 Take 90807060504030 999999999999 14. From 671111 take 199999. Ans. 471 112. 15. From 1789100 take 808088. Ans. 981012. 16. From 1000000 take 999999. Ans. 1. 17. From 9999999 take 1607. Ans. 9998392. 18. From 6101507601061 take 3806'790989. Ans. 6097700810072. QUESTIONS. -Art. 32. What is the second method of proving subtraction? What is the reasona for this method of proof, or on what principle does it dopend? 'iUcT. Mx~.] SUBTRACTIOIN, 33 19. From 8054010657811 take 76909748598. Ans. 7977100909213. 20. From 7100071641115 take 10071178. Ans. 7100061569937. 21, From 501505010678 take 794090589, Ans, 500710920089, 22. Take 9999999 froim 100000000. Ans. 1. 23. Take 444444444 from 500000000. Ans. 455555556. 2M4 Take 1234 567890 from 9987654321. Ans. 875308643. 25. From 800700567 take 1010101. Ans. 79969046G. 26. Take twenty-five thousand twentyRfive fiom twenty-five iillions. Ans. 24974975. 27. Take nine thousand ninetymnine from ninety-nine thousand. Ans. 89901. 28. From one hundred one millions ten thousand one hundred one tale ten millions one hundred one thousand and ten. Ans. 90909091. 29. From one million take nine. Ans. 999991. 30. From three thousand take thirty-three. Ans. 2967. 31. From one hundred millions take five thousand. Ans. 99995000. 32. From 1,728 dollars, I paid 961 dollars; how many re mnain? Ans. 767 dollars. 33. Our national independence was declared in 1776; how many years from that period to the close of' the last war with Great Britain in 1815? Ans. 39 years. 34. The last transit of Venus was in 1769, a.nd the next will be in 1874; how many years will intervene? Ans. 105 years. 35. In 1830, the number of inhabitants in Bradford was 1,856, and in 1840 it was 22222; what was the increase? Ans. 36Go 36. How many more inhabitants were there in New York city than in Boston, in 1840, there being, by the census of that year, 312,710 inhabitants in the former, and 93,383 in the latter city? Ans. 219,; 27 inhabitants. 37. In 8l21 there were imported into the United States 1,2 a73,659 pounds of coffee, and in 1839, I06,696,992 pounds; what was the increase? Ans. 85,423,333 pounds. 38. By the census of 1840, 1[1,853,50f bushels off wheat were raised in New York, and 13,029,756 bushels in Pennsylvania; how many bushels in the latter State more than in the for-'mer _ nsr. 1,i76,2149 bushels, 34 MULTIPLICAT10N. [SECT. IV. 390 The real estate of James Dow is valued at 3,769 dollars, and his personal estate at 2,648 dollars; he owes John Smith 1,728 dollars, and Job Tyler 1,161 dollars; how much is Dow worth? Ans. 3,528 dollars. 40. If a man receive 5 dollars per day for labor, and it cost him 2 dollars per day to support his family, what will he have accumulated at the close of one week? Ans. 16 dollars. 41. The city of New York owes 9,663,269 dollars, and Boston owes 1,698,232 dollars; how much more does New York owe than Boston? Ans. 7,965,037 dollars. 42. From five hundred eighty-one thousand take three thousand and ninety-six. Ans. 577,904. 43. E. Webster owns 6,765 acres of land, and he gave to his oldest brother 2,196 acres, and his uncle Rollins 1,981 acres; how much has he left? Ans. 2,588 acres. 44. It was ascertained by a transit of Venus, June 3, 1769, that the mean distance of the earth from the sun was ninety, five millions one hundred seventy-three thousand one hundred twenty-seven miles, and that the mean distance of Mars from the sun was one hundred forty-five millions fourteen thousand one hundred forty-eight miles. Required the difference of their distances from the sun, Ans. 49,841,021 miles, ~ Iv. MULTIPLICATION. MENTAL EXERCISES& ART. S E. W NHr\ any number is to be added to itself several times, the operation may be shortened by a process called M1fulti lication. Ex. 1. If a ian can earn 8 dollars in 1 week, what will he earn in 4 weeks? ILLUSTRATION. - It is evident, if a man can earn 8 dollars in 1 week, in 4 weeks he will earn 4 times as much, and the result may be obtained by addition; thus, 8 + 8 + 8 -- 8 - 32; or, by a more convenient process, by setting down the 8 but once, and multiplying it by 4, the number of times it is to be repeated; thus, 4 times 8 are 32 1 Hence in 4 weeks he will earn 32 dollars. SECT. IV.] MULTIPLICATION. 35 The following table must be thoroughly committed to memory before any considerable progress can be made in this rule: - MULTIPLICATION TABLE. 2 times 1 are 2 3 times 1 are 3 4 times 1 are 4 5 times 1 are 5 2 times 2 are 4 3 times 2 are 6 4 times 2 are 8 5 times 2 are 10 2 times 3 are 6 3 times 3 are 9 4 times 3 are 12 5 times 3 are 15 2 times 4 are 8 3 times 4 are 12 4 times 4 are 16 5 times 4 are 20 2 times 5 are 10 3 times 5 are 15 4 times 5 ale 20 5 times 5 are 25 2 times 6 are 12 3 times 6 are 18 4 times 6 are 24 5 times 6 are 30 2 times 7 are 14 3 times 7 are 21 4 times 7 are 28 5 times 7 are 35 2 times 8 are 16 3 times 8 are 24 4 times 8 are 32 5 times 8 are 40 2 times 9 are 18 3 times 9 are 27 4 times 9 are 36 5 times 9 are 45 2 times 10 are 20 3 times 10 are 30 4 times 10 are 40 5 times 10 are 50 2 times 11 are 22 3 times 11 are 33 4 times 11 are 44 5 times 11 are 55 2 times 12 are 24 3 times 12 are 36 4 times 12 are 48 5 times 12 are 60 6 times 1 are 6 7 times 1 are 7 8 times 1 are 8 9 times are 9 6 times 2 are 12 7 times 2 are 14 8 times 2 are 16 9 times 2 are 18 6 times 3 are 18 7 times 3 are 21 8 times 3 are 24 9 times 3 are 27 6 times 4 are 24 7 times 4 are 28 8 times 4 are 32 9 times 4 are 36 6 times 5 are 30 7 times 5 are 35 8times 5 are 40 9 times 5 are 45 6 times 6 are 36 7 times 6 are 42 8 times 6 are 48 9 times 6 are 54 6 times 7 are 42 7 times 7 are 49 8 times 7 are 56 9 times 7 are 63 6 times 8 are 48 7 times 8 are 56 8 times 8 are 64 9 times 8 are 72 6 times 9 are 54 7 times 9 are 63 8 times 9 are 72 9 times 9 are 81 6 times 10 are 60 7 times 10 are 70 8 times 10 are 80 9 times 10 are 90 6 times 11 are 66 7 times 11 are 77 8 times 11 are 83 9 times 11 are 99 6 times 12 are 72 7 times 12 are 84 8 times 12 are 96 9 times 12 are 108 10 times I are 10 10 times 11 are 110 11 times 8 are 88 12 times 4 are 48 10 times 2 are 20 10 times 12 are 120 11 times 9 are 99 12 times 5 are 60 10 times 3 are 30 11 times 10 are 110 12 times 6 are 72 10'times 4 are 40 ii times 1 are 11 11 times 11 are 121 12 times 7 are 84 10 times 5 are 50 11 times 2 are 22 11 times 12 are 132 12 times 8 are 96 10 times 6 are 60 11 times 3 are 33 - 12 times 9 are 108 10 times 7 are 70 11 times 4 are 44 12 times l'are 12 12 times 10 are 120 10 times 8 are 80 11 times 5 are 55 12 times 2 are 24 12 times 11 are 132 10 times 9 are 90 11 times 6 are 66 12 times 3 are 36 12 times 12 are 144 10 times 10 are 100 11 times 7 are 77 2. What cost 5 barrels of flour at 6 dollars per barrel? ILLUSTRATION. - If 1 barrel of floul cost 6 dollars, 5 barrels will cost 5 times as much; 5 times G are 30. Hence 5 barrels of flour at 6 dollars per barrel will cost 30 dollars. 3. Vhat cost 6 bushels of beans at 2 dollars per bushel? 4. What cost 5 quarts of cherries at 7 cents per quart? 5. What will 7 quarts of vinegar cost at 12 cents per quart? 6. What cost 9 acres of land at 10 dollars per acre? 7. If a pint of currants cost 4 cents, what cost 9 pints? 8. If in 1 penny there are 4 farthings, how many in 9 pence? In 7 pence? In 8 pence? In 4 pence? In 3 pence? 9. If 12 pence make a shilling, how many pence in 3 shi!tings? In 5 sillins? n ln 9 shilli hi 9 hill 38 MULTIPLICATION. [sEcr. xv. 10. If 4 pecks make a bushel, how many pecks in 2 bushels? In 3 bushels? In 4 bushels? In 6 bushels? In 7 bushels? In 9 bushels? 11. If 12 inches make 1 foot, how vmanv inches in 3 feet? In 4 feet? In 5 feet? In 7 feet? In 8 feet? In 9 feet? In lO feet? In 12 feet? 12. If there be 9 feet in a square yard, how many feet in 4 yards? In 5 yards? In 6 yards? In 8 yards? In 9 yards? In 12 yards? 13. What cost 3 yards of cloth at 5 dollars per yard? 4 yards? 5 yards? 6 yards? 7 yards? 8 yards? 9 yards? 10 yards? II yards? 12 yards? 14. If 1 pound of iron cost 7 cents, what cost 2 pounds? 3 pounds? 5 pounds? 6 pounds? 7 pounds? 8 pounds? 9 pounds? 12 pounds? 1,. If l pound of raisins cost 6 cents, what cost 4 pounds? 5 pounds? 6 pounds? 7 pounds? 8 pounds? 9 pounds? 10 pounds?.2 pounds? 16. In I acre there are four roods how many roods in 2 acres? In 3acres? In 4 acres? In 5 acres? In 6 acres? In 9 acres? 17. A good pair of boots is worth 5 dollars; what must I give for 5 pairs? For 6 pairs? For 7 pairs? For 8 pairs? 18. A cord of good walnut wood may be obtained for 8 dollars; what must I give for 4 cords? For 6 cords? For 9 cords? 19. What cost 4 quarts of milk at 5 cents a quart, and 8 gallons of vinegar at 10 cents a gallon? 20. If a man earn 7 dollars a week, how much will he earn in 3 weeks? In 4 weeks? In 5 weeks In 6 weeks? In 7 weeks? In 9 weeks 21. If I thousand feet of boards cost 12 dollars, what cost 4 thousand? 5 thousand? 6 thousand? 7 thousand? 9 thousand? 12 thousand? 22. If 3 pairs of shoes buy 1 pair of boots, how many pairs of shoes will it take to buy 7 pairs of boots? 23. If 5 bushels of apples buy I barrel of flour, how many bushels of apples are equal in value to 12 barrels of flour? 24. If 1 yard of canvas cost 25 cents, what will 12 yards cost? ILLUSTRATION. - The number 25 is composed of 2 tens and 5 units; 12 times 2 tens are 24 tens; and 12 times 5 units are s~:oT..~v.I ~ UT I'TAT"ON'. Sq S "r, - t V.3 ",TIPLIC 17MiN. 60 nmits, or ~ tens; 24 tens added to 6 tens mrake 30 tens, or 300 q1 Therefore, 12 yards 2 iit cost 330 cents, or 3 dollars, 25. In _I poulned *lY m iore e- shhillinigs'Iow r-aDy shilings in 3 1ounds In 4 pounds? In G pounds? )2. A gallon of moolasses is worth':' 4ents; -w'vhat is the value of L2 gallons? fO 3 glls Of 4 gallons? Of 5 gallo7s? Of G gallons Of 9 gallorns 2)7'. Ii'f 1m -men can do a piece of w or1e in 1 0 days, how long w-fill it t:ake 2 mhian to do it? 3. If a steam-eongine run s -8 m3 iles itt i hour, how far will r nt n ira 4 hou rs) in o hours in 9 hours? 29. If the earth turns on ints axis!5 degrees in I hour, how Car w ill it turn in' hours? in! bO 1 h? In 12 hours? 30. In a certain regiment there ace 6 companies, in each company 6 platoons, and in each platoon 13 soldiers; how many soldiers are the're in the regilment > ArT. QII.' The learner, having performed the foregoing questions, will perceive that iu:rTIPrt icArToN is the repetition ofl a nunmber any proposed lm)ber of tines, and is therefore a compendious method of addition. in multiplication, three termls are employed, called the M:iutdtiptticand, the ]lultiper, and the Product. T'he au ltit liccand i the number to be moultiplied or repeated. The mnzzitZplier is tlle nunber by which we mnultiply, and denotes'the number of repetitions to be made. The product is the answer, or number produced by the multiplication. Thle ma ultiplicand and multiplier vare ofen called factors. AET. 2, GeNs. -The sign of multiplication is formed by two short lines erossing each other obliquelyr thus, X. it shosws that the numbers between which it is placed are to be multiplied together; thus, the expression 7 X 5 5 5 is read, 7 multipliecd by 5 is equal to 35. QCUSrSTroNS. —Airt 3d1. What is rmultiplication? "What three terms are em!loyed'? What is the mrultipicrand? What is the mullltiplier? WVhat is the product?'That are the multiplicarnd and multiplier often called-'!-.rt. 35. What is tile siig of multiplication? What does it show'' 38 MULTIPLICATION, [SECT. IV. EXERCISES FOR THIE SLATE. ART. 36o Method of operation, when the multiplier does not exceed 12. Ex. 1. Let it be required to multiply 175 by 7. Ans. 1225. OPERATION. After having written the multiplier unMultiplicand 1 7 5 der the unit figure of the multiplicand, and Multiplier 7 drawn a line below it, we rultiply the 5 Product 1 2 2 5 in the multiplicand by 7, saying, 7 times 5 are 35, and set down the 5 units directly under the 7, and reserve the 3 tens in the mind. We then multiply the 7 in the multiplicand, saying 7 times 7 are 49, and, adding the 3 tens which were reserved, we have 52 tens, or 5 hundreds and 2 tens. Setting down the 2 tens, and reserving the 5 hundreds, we multiply 1 by 7, and, adding the reserved 5 hundreds, we have 12 hundreds, which, as it completes the multiplication, we set down in full, and the product is 12s25. EXAIMPLES FOR PRACTICE. 2. 3. 4. Multiply 8 7 5 6 4567 7896 By 4 3 5 Ans. 35024 1 370 1 39480 5. 6. 7. 8. 56807 47893 61657' 89765 5 6 7 9 284035 287358 43 1 599 807885 9. Multiply 767853 by 9. Ans. 6910677. 10. Multiply 876538765 by 8. Ans. 7012310120. 11. Multiply 7654328 by 7. Ans. 53580296. 12. Multiply 4976387 by 5. Ans. 24881935. 13. Multiply 8765448 by 12. Ans. 105185376. 14. Multiply 4567839 by 11. Ans. 50246229. 15. What cost 8675 barrels of flour at 7 dollars per barrel? Ans. 60725 dollars. QUESTIONS. - Art. 36. 1-low must numbers be written for multiplication? At which hand do you begin to multiply 7 Why? Where do you write the first or right-hand figure of the product of each figure in the multiplicand? Why? What is done with the tens or left-hand figure of each product? How, then, do you proceed when the multiplier does not exceed 12? SECT. XV.J MULTIPLICATION. 39 16. What cost 25384 tons of hay at 9 dollars per ton? Ans. 228456 dollars. 17. If on 1 page in this book there are 2538 letters, how many are there on 11 pages? Ans. 27918 letters. ART. 37~ Method of operation, when the multiplier exceeds 12. Ex. 1. Let it be required to multiply 763 by 24. Ans. 18312. OPERATION. Here we write the multiplicand and Mu Oltiplicand 7A6 3 mO. multiplier as before, and proceed to multiply the multiplicand by 4, the unit figMultiplier 2 4 ure of the multiplier, precisely as in Art. o 0 5 2 36. We then, in like manner, multiply 1 5 2 6 the multiplicand by the 2 tens in thle multiplier, taking care to set the first figure Product 1 8 3 1 2 obtained by this multiplication directly under the 2 of the multiplier, and, adding together the products obtained by the two multiplications and placed as in the operation, we have the full product of 763 multiplied by 24, which is 18312. ARnT. 3 o The preceding examples sufficiently illustrate the principle and method of multiplication; and the learner is now prepared to understand and apply the following general RULE. - 1. Place the larger number uppermost for the multiplicand, and the smaller number under it for a multiplier, arranging units under units, tens under tens, f-c. 2. Then multiply each figure of the multiplicand by each figure oJ the multiplier, begmnning with the right-hand figure, and carrying for every ten as in addition. 3. If the multiplier consists of more than one figure, the right-hand figure of each product must be placed directly under the figure of the multiplier that produces it. The sum of the several products will be the whole product required. NOTE. - When there are ciphers between the significant figures of the multiplier, pass over them in the operation, and multiply by the significant figures only, remembering to set the first figure of the product directly under the figure of the multiplier that produces it. QU.STIONS. -Art. 37. How do you proceed when the multiplier exceeds 12? Where do you set the first figure of each partial product? Why? -How is the true product found? -Art. 38. What is the general rule for multiplication? When there are ciphers between the significant figures of the multiplier, how do you proceed? 40 itd ULTI' LICA TIOI, [ScT. iv. ART., 39 ir.st M[etlhod of 7rooj. - Miultiply the multiplier by the nuiltiplicaod, a.otl if' thie result is like the first product, thie work is supposeld to hbe right. The reason of this proof' depends on the principle, /izfit, when two or more nulmbers a(re vmul.tiplied together, the -roduct is the sanme, whatever the ord der multilying them. Ex. 2. Multiply 7895 by 56. Ans. 442120. OPiEATION, PROe,. Mlultiplicand 7 S 9 5 5 6 Multiplier 5 J 7 8 95 4'7370 So8 394l75 50z4 4 4486 Product 4 4 2 0 Prodiouct 4 4 2 1 2 0 No'r. - The common rmrode of proof in busless is to divide the product by the multiplier, and, if the work is right, the quotient will be like the multiplicand. This mode of proof anticipates the principles oi division, and therefore cannot be employed without a previous knowledge of that rule. ART. Oo Second ie,'aod of oo —-j Begin at the left hand of the multiplicand, and add together its successive figures toward the right, till the sum obtained equals or exceeds the number nineo If it eqtualt i, drop the nine, and begin to add again at this point, and proeeed till you obtain a sum equal to, or greater than, nine. If it exceedcls inet drop the nine as before, and carry the excess to the next figure, and then con-' tinue the addition as before. Proceed in this wvay till you have added all the figures in the multiplicand and rejected all the nines contained in it, and write the final excess at the right hand of the multiplicand. Proceed in the same ma -nner with the multiplier, and write the final etcess under that of the multiplicand. Multiply these excesses together, and place the excess of nines in tlheit product at the riglht. Then proceed. to find the excess of nines in the product obtained by the, origin.al operationq and, if the work is right, QUESTIONr — Art. 39. How is multiplication proved by the first method? What is the reason for this method of proof? What is the common mode of proof in business? —Art. 40. What is the second me4thod of proving mu! tipiication 7 SECT. IV.] MULTIPLICATION. 41 the excess thus found will be equal to the excess contained in the product of the above excesses of the multiplicand and multiplier. Ex. 3. OPERATION. Multiplicand 12 345 = 6 excess. Multiplier 2 2 3 1 8 excess. 12345 4 8 3 37035 24690 Product 27541695- 3 NoTE. - This method of proof, though perhaps sufficiently sure for comnmon puiposes, is not always a test of the correctness of an operation. If two or more figures in tile work should be transposed, or the value of one figure be just as mnuch too great as another is too small, or if a nine be set down in the place of a cipher, or the contrary, tile excess of nines vill be the same, and still the work may not be correct. Such a balance of errors will not, however, be likely to occur. EXAMIPLES FOR PRACTICE. 4. 5. Multiply 67 8 95 78956 3y 36 47 407370 552692 203685 315824 Arns. 2444220 3710932 6. 7. Multiply 89 3 2 5 47 8 96 By 901 2008 89325 383168 803925 95792 Ans. 80481825 96175168 8. What cost 47 hogsheads of molasses at 13 dollars pet hogshead? Ans. 611 dollars. 9. What cost 97 oxen at 29 dollars each? Ans. 2813 dollars. QUESTIONS. —Is this method of proof always a true test of the correctness ot an operation? What is the reason for this method of proof? tBil e k[2 ryMl ~ULTiPLItCATION [SECT'. V1 10. Sold a farm containing 367 acres, at 97 dollars per acre; what was the amoun-t? Ans. 35599 dollars. 11. An army of 17006 men receive each 109 dollars as their annual pay; wlhat is the amount paid the whole army e Ans. 18536564 dollars. 12. If a mechanic deposit annually in the Savings Bank 407 dollars, what t will be the sum deposited in 27 years? Ans. 10989 dollars. I3. if a man travel 37 miles in I day, how far will he travel na 365 days? Ans. 1'3505 miles. 14. If there'e 24 hours in I day, how rmany hours in 365 days? Ans. 8'760 hours. 15. IHow many gallons in 87 hogsheadsu there being 63 gallons in each?n 4 Agallons. L16. If the expenses -f the 1Iassachusetts Legislature be 1839 dollars P er day, ar will be the amoult in a session of 109 days? Ans. 0045 1 dollars. 17. If a hogsiead' of sucar cornta.sinr; 368 pounds, how many pounds in 187 hogsheads 688Gl pounds. 18. M11ultiply 675 by 47, Ans. 321300. 19. MTultiply 679 by 7 A ns. 5 018 77 20 IMultiply 899 by o 93! Ai s. 803719 9. 2~. Multiply 7854 by "n 99:. A n 83 2.i Multiply 300i 93by 607i. Al!S. 1 829071o 233. Mkluiltiply 71i 7 by 9i 7,- Ans. e,:2.8749",c; ci h V0203409. 24. Multiply:r t6u546 by 4 i07t091 hu. Ais. a f388i87 6b8y 2' Mul.tiply tD'7'0i1009 by 7007867. Ans. 49062 139s'803. 56. 1Multiply five hundred atnd d eighty-sx by nine hundred. arnd eighta A.ans. $ 3208,8 1;b,/7. Multiply tiree thousand eight hunldred and five by one e thou toisand and seven. Ans. g83 635,S 28. Multiply t-iyo t1housand and seventy-one by' sevean i'ro-1drked and six. P Anr. S L4-G,,,12,.. 5J9. Multjly, eiplqy-ight t;ho'usa-nd a.ld eigrht by three thikil sand and seven. Ans. 290S33037, ifO, Mlfultiply niyraty tih ousand eight hundred and seven bi w, oe thousand and a nd niinetyoneJ. Ans. 99017043'. SQl. Mulftiply lninety thousand eight hundred and seven by tonie thnousand one hundlred and sit, Ans. 8268' 852-~ 2. IMultiply fifty thousaid and one by five thousand eight hundred and sevenR Ans. 290355807. I3, Miultiply:,'*"~ tchou rsand and nine by nline thousand and sL'teem' A/-Xa. 721i36 i44o SErCT. I7.1 My/IULTIPhLICTiOIN0 4. 34. MJultiply forty-seven thousand and thirteen by eighty thousand eight hundred and seven. Ans. 37989 7949ao ART. 4 o 4. A COM'fPOSITE number is the product of two or more factors greater thanl unity or one; thus 12 is a composite number, since it is tlih product of 3 X 4; and also'24 is a ompnosite number, since it is the product off 2 X 3 )<( 4A FACTOR of any number is a namrne grciven to one of two or mnore numbe:rs which, being multipliecd together, produce that'number; thus 3 and 4 ar e fitetors of' 12, since 3 >\ 4 =- 12 _AT. a;, o mrultiply y a composite number. Ex. Io )~ merchant bought 15 pieces of broadcloth at 96 dollars per piece; how mnuch did he pay for the whole? Ans. 1440 dollars.:1The factors of 15 are 3 and Or>Lr dii s. 5.. \Now if we mnultiply the 9 6 dolls., price of 1 piece puce of oit pie y ite factor 3 3, we get the costc of 3 pieces; 2 8 9 do'ls., p.ice of 3 pieces. and theri, by multiplying the 8 8 notfs, fti ce of pise cost of 3 pieces by the factor 5_ it is evidenlt we obtain the cost of 15 the numibes of pieces I 4 4 0 dolls., price o-' -1 icca.$ """t Of F5? %he nul-l9 Ff pieces bought, since 15 is equal to 5 times 3. Hence we, adopt the following ULE. - -h1flti //thei muliticapcnd bTy onze of thee Jactor-s f te n. aulplier,, and this prodluct by anoothicer, and so on untll all tie ticlt(rs have been used as nmulltplieis, anzd the last prodluct wZll be the oas7wer. NOTE. -- PThe product of any number of flctors is the samne in whaaever order they are, multiplied. Thus, 3 < 4 = 12, and 4 \< 3 =1 12. EXAMnIPLES ItOR PRACTICrE. Multiliy p 3013 by 25 5 ns. 765`326-, l Multiply j14G9 by o. >.S. P2339.9 4. iulcticply 75,45) by 81, using its flfctors. Ans. 61 1226. i, 5fultirply 7r901 by 5o, using its factors. Ans. 9876,5. 6. In I mile thlere are 6330O inc'htes, how irmany incics' 4s5 miles.Azns.,~ 00, incl es.. If in year there are 86,7 hlours, s. how manaiy hlours in Y2, years? As. 63!5 hu,. QUESTIONF. - Art. 41. Wlhat is a cormposite number "! ihat is a fctoiu oany number? -Art. 12. What are the factors of 15 I How do we multiply by a composite unumber? Repeat the rule. 1n what order Tiust tle, facturs of a composit; - nuabem t ei muipililed? 44 MULTIPLICATION. [SECT. Iv. 8. If sound moves 1142 feet in a second, how far will it move in one minute? Ans. 68520 feet. ART. CK, To multiply by 10, 100, 1000, &c. Ex. 1. In I day there are 24 hours; how many hours in 10 days? In 100 days? Answers 240, 2400 hours. OPERATION. By a principle of notation, Multiplicand 2 4 2 4 the removal of a figure one Mlultiplier 1 0 1 0 0 place to the left increases its Product 2 4 0 2 4 0 0 value ten times (Art. 9). If then we annex a cipher to the Or thus, 240, 2400. right of 24, the multiplicand each figure is removed one place to the left, and its value is increased ten timles, or multiplied by 10. If two ciphers are annexed, eac(h figure is removed two places to the left, and its value is increased 1(10 times, or multi)lied by 100; every additional cipher increasing the value ten times. Hence the propriety of the following RULE. - Annex to the multiplicand all the ciphers of the multiplier, and the result will be the product required. EXAMIPLES FOR PRACTICE. 2. Multiply 2356 by 10. Ans. 23560. 3. Multiply 5873 by 100. Ans. 587300. 4. Multiply 7964 by 1000. Ans. 7964000. 5. Multiply 98725 by 100000. Ans. 9872500000. AnRT. 4 o. Method of operation when there are ciphers on the right hand of the multiplier or multiplicand, or both. Ex. 1. What will 600 acres of land cost at 20 dollars per acre? Ans. 12,000 dollars. OPERATION. In this example the multiplicand may be Multiplicand 6 0 0 resolved into the factors 6 and 100, and the Multiplier 2 0 multiplier into the factors 2 and 10. Now Producet 12 0 0 0 it is evident (Art. 42), if these several factors be multiplied together, they will produce the same product as the original factors, 600 and 20. Thus 6 X 2 = 12, and 12 X 100 1 1200, and 1200 X 10 - 1S000, the same result as in the operation. Hence the followinog QESTIONS. -- Art. 43. What is the effect of removing a figure one place to the left? What is the effect of annexing a cipher to any figure or nurnber? Two ciphers? &c. What is the rule for multiplying by 10, 100, &c.?Art. 44. How do you arrange the figures for multiplication, when there are ciphers on the right hand of either the multiplier or multiplicand, or both? VWhy does multiplying the significant figures and annexing the ciphers produce the true product?: suCT. Xv,] c/tIUL3TJPLIYCATIOi,..tULrE. e&t down the significant figures of th'e l?.vliphier ns dt.r those oj' the mit:ictzand, anld inoultldy them tolethitr. flthen annexe as,ainuy cip/he.rs to their product as there are o7 the ri'hit of zhthe nmullip]. 6,/rd antd mllilJltier. EXAMIPLES FOR P RACTICE, Multiply 38 7 8 5 3 1 4'7 1 3 3'78 9 3 0 BYy 3200 V 00 80 1757004 57 003 12. 26355972 49936523 &ns,, 28113036800 499935933 12000 4. Multiply 8010700'by 9000909. AnSo 721083581'220300 5. Multiply 700110000 by 700110000. Ans. 490154012100000000. (6 Multiply 4020VG0 by 7007000. Ans. 23527443249000. 7~ Multiply 4110000 by 101701.0. Anso 417991 100000,, 83. Multiply twenty-nine rafillions two thousand nine hundred and nine by four hundred and four thousand. ins. 1 71211725236000. 9. Multilply eighty-seven millions by eight hundred thousand senven hundred. Ans. C9660900000000. 10. IMiultiply one million one thousand one hundred by nine hundred nine thousand and ninety. Ans. 910089999000. 1. iTiultiply forty-nine millions and forty-nine by four'lundred and ninety thousand. Ans. 240100240100008 l,~. Multiply two hundred millions two hunidred by' two nilions two thousand and two. Ans. 400400800400400. M3. Multiply four millions forty thousand four hundred by hree hundred thlree thousand.- A.ns. 1224241200000~. 14. Multiply three hundred thousand thirty by forty-seveA thousand seventy. Ans. 14122412t106 QersZslicer —What is the rule? 46 DIVISION. [SECT. V. ~ V. DIVISION. IMENTAL EXERCISES. Aar. 45. WVIEN it is required to find how many times one number contains another, the process is called Division. Ex. 1. A boy has 32 cents, which he wishes to give to 8 of his companions, to each an equal number; how many must each receive? ILLUSTRATION. - It is evident that each boy must receive as many cents as the number 8 is contained times in 32. We therefore inquire what number 8 must be multiplied by to make 32. By trial, we find that 4 is the number; because 4 tines 8 make 32. Hence 8 is contained in 32, 4 times, and the boys receive 4 cents apiece. The following table should be studied by the learner to aid him in solving questions in Division: — DIVIS1ION TABLE. 2 in 2 1 timhe 3 in 3 1 time 4 in 4 1 time 5 in 5 1 time 2 in 4 2 times 3 in 6 2 times 4 in 8 2 tines 5 in 10 2 timles 2 in 6 3 times 3 in 9 3 timles 4 in 12 3 timnes 5 in 1i 3 timles 2 in 8 4 times 3 in 12 4 times 4 in 16 4 tines 5 in 20 4 times 2 in I0 5 timtes 3 in 31 5 5 times 4 in 20 5 times 5 in 25 5 tim}es 2 in 12 6 tilnes 3 in 18 6 tirsles 4 in 24 6 times 5 in 30 6 tilnes 2 in 14 7 times 3 in 21 7 times 4 in 28 7.times 5 in 35 7 times 2 in 16 8 timles 3 in 24 8 tinmles 4 in 32 8 timles 5 in 40 8 tiles 2 in 18 9 times 3 in 27 9 timlnes 4 in 36 9 timnes 5 in 45 9 timles 2 in 20 10 titnes 3 in 30 10 timles 4 in 40 10 times 5 in 50 10 times 2 in 22 11 timles 3 in 33 11 times 4 in 44 11 times 5 in 55 11 tilmes 2 in 24 12 tinmles 3 in 36 12 times 4 in 43 12 times 5 in 60 12 times 6 in 6 1 timne 7 in 7 1 ti me 8 in 8 1 time 9 in 9.1 tinse 6 in 12 2 titmles 7 in 14 2 tim1es 8 in 16 2 times 9 in 18 2 times 6 in 18 3 tinles 7 in 21 3 times 8 in 24 3 tinmes 9 in 27 3 times 6 in 24 4 times 7 in 23 4 times 8 in 32 4 times 9 in 36 4 times 6 in 30 5 times 7 in 35 5 times 8 in 40 5 times 9 in 45 5 tines 6 in 36 6 titnes 7 in 42 6 times 8 in 48 6 tiules 9 in 54 6 timhes 6 in- 42 7 times 7 in 49 t7 7imes 8 in 56 7 timles 9 in 63 7 titles 6 i;1 48 8 timles 7 in 56 8 timiles 8 in 64 8 tinles 9 in 72 8 timiles 6 in 54 9 titmles 7 in 63 9 times 8 in 72 9 tiles 9 in 81 9 timtes 6 in 6(0 10 times 7 in 70 10 times 8 in 80 10 times 9 in 90 10 tinmes 6 in 66 11 tinmes 7 in 77 11 times 8 in 83 11 timtes 9 in 99 11 tines 6 in 72 12 ti111es 7 in 8-1 12 times 8 in 96 12 times 9 in 103 12 t2ines 10 in 10 I time 10 in 110 11 times 11 in 83 8 times 12 in 43 4 times 10 in 20 2 times 10 in 120 12 tines 11 in 99 9 times 12 in 60 5 tines 10 in 30 3 timies 11 in 110 10 times 12 in 72 6 times 110 in 40 4 timles I1 in 11 1 time 11 in 121 11 timies 12 in 84 7 timies i 0 in 10 5 timlles 11 in 22 2 timnes 11 in 132 12 tinmes 12 in 96 8 Limes 10 in 60 6 times 11 in 33 3 times 12 in 10 9 timnes 10 in 70 7 tirmues I in 44 4 timies 12 in 12 1 tinme 12 in 120 10 tiines 10 in 80( 8 times 11 ian 55 5 tirles 12 in 24 2 times 12 in 132 11 timles 10 in 90 9 timles 11 in 66 6 timles 12 in 36 3 times 12 in 144 12 times 10 in 100 10 tinles 11 ir 77 7 timnes SECT. v.] DIVISION. 4'7 2. A farmer received 8 dollars for 2 sheep; what was the price of each? tLLUSTRATION. -It is evident, if he. received 8 dollars for 2 sheep, for 1 sheep he must receive as many dollars as 2 is contained times in 8; 2 is contained in 8, 4 times, because 4 times 2 are S. Hence 4 dollars was the price of each sheep. 3. A man gave 15 dollars for 3 barrels of flour; what was the cost of each barrel? 4. A lady divided 20 oranges among her 5 daughters; how many did each receive? 5. If 4 casks of lime cost 12 dollars, what costs I cask? 6. A laborer earned 48 shillings in 6 days; what did he receive per day? 7. A man can perform a certain piece of labor in 30 days, how long will it take 5 men to do the same?,. When 72 dollars are paid for 8 acres of land, what costs I acre? What cost 3 acres? 9. If 21 pounds of flour can be obtained for 3 dollars, how much can be obtained for 1 dollar? EHow much for 8 dollars? Hlow much for 9 dollars? 10. Gave 56 cents for 8 pounds of raisins; what costs I pound? What cost 7 pounds? 11. If a man walk 24 miles in 6 hours, how far will he walk in 1 hour? How far in 10 hours? 12. Paid 56 dollars for 7 hundred weight of sugar; what costs 1 hundred weight? What cost 10 hundred weight? 13. If 5 horses will eat a load of hay in 1 week, how long would it last I horse? 14. In 20, how many times 2? How many times 4? How many times 5? I-low many times 10? 15. In 24, how many times 3? How many times 4? I-ow many times 6? -low imany times 8? 16. I-How many times 7 in 21? In 28? In 56? In 35? In 14? n 63? In 77? In 70? In 84? 17. fHow many times 6 in 12? In 36? In 18? In 54? In 60? In 42? In 48? In 72? In 66? 18. Elow many times 9 in 27? In 45? In 63? In81? In 99? In 108? 19. I-low many times 11 in 22? In 55? In 77? In 88? In 110? In 132? 20. How many times 12 in 36? In 60? In 72? In 84? In 120? In 144? i RI p4j',;.8i'!S %TViSO seCT. V ArTr. P he pupiln will now perceive, tihat DivesIot:. is tie ptrocess of ftndiing how imany ti-mes one numb.lner is ~- oiaihled in anotieud in division r ereL are t ree principacl tenrnms teis e Divitdend,." Dle Div sor, and the Quoziezt, or ascziEe'.''e dZ'-renld is the nunlber to be divideali nLile fvir;ror is the number by which ye divide. T'lte uo"ioebZ en is the nnumber f times tile divisor is con kiined in the dividend,'Wihejan the divicend does noL contain dtie divisor an exaei ni:ber of times, the excess is called a reclnina er, and ay lbe icag1crde'd as afourrt term in the division. The remainder, being part of the dividend, will always be of thle same denomnination or kind as the dividend, and must a1ways be less than tihe divisor. PRT. o SIGNS. - The siCa of division is a short horizontal line, witoh's dot above it and another below; thus, -.o t shows tdhat the nulmber bef7ore it is to be divided by the number cier ii. Th e expression C -e- n d, 3 is readG 6 divi vdeid by 2 is equal to 2,% Divisionl is also indicated bby writing the dividend above a short horizontal line and the divisor below, thus c(. T-he,x pression 3 is read, 6 divided by 2 is equal to 3. There is Da third Imetiod of indicating division by a curved line placed betiween the divisor and dividend. Thlus the e'x, pression 6) 2)1 shiows tlat'SJ is to be divided by C. jIEXIrCISES 1OI0 THE SLATE. A~RT. _4!0 Tlhe imethod of operation by Shor Division, or when the divisor does not exceed 12.:EX. 1. Divide 7554 dollars equally among 6 men. Ans. 1259 dollars.'h o.oti.ent is derived fron the Latin word quoties, which signifies;5ow-,'ien, or hoe mo)an7y times. QuESTTONS. S Art. 46. What is division? WhIat are the three principal terms in division? What is the dividend? NWV1hat is the divisor? W7hlJat is the quotient? What the remainder?'What will be the detnonrination of' the remnaiinder? How doles it comnare with the divisor? - Art. it7. WJhat is the first sign of division, aIld what does it show? Wthat is the secondl, and what does it show 7?What is tie thihd, arid wthat does it show'A...A 4. V P rhat irs short divisre.'.-n i SECT. v.] I)VSI S ON. OTPRgATIO:4. In performing this question, we Divisor 6) 7 5 5 4 Dividend. first inquire how many times 6, the 1 2 5 9 Quotienit. divisor, is contained in 7, the first figure of thle dividend, and find it to be 1 time, and I remaining. WeT write the qulotient figure, 1, directly under the 7, and then imagine the I remainin to be placed before the next figure of the dividend, which is 5, thus forming the nlimbe-r 15. We then inquire how many times 6, the divisor, is contained in 15, and find it to be 2 times, with a'emainder of 3. The 02 beingc wtritten under the 5, we next imagine thle 3 (remainder) to be pleaced before the next figure of the dividend, which is 5, and we haive 35, which, being divided by 6, gives 5 for the quotient, with 5 for a remainder. Writing down the quotient figure as before, and imatrining the remainder to be plaed before the 4 in the dividend, we have 55, which, divided by 6, gives 9 as a quotient, which we write under the figure, and which completes the operation, giving 1259 as the quotient, or the number of times which the dividend 7554 contains the divisor 6. ART. 4%i From the foregoing illustration we deduce the osllowing RULE. - 1. WATrite the divisor at the left hand of the dividend, vwith a curved line betwzeen thlem, and draw a horizontal line under the dividend. 2. Then inquire hoiz many times the divwisor is contained in the lefthand fi guere of the dividend, or fi gures if.zore than one is necessary to contain it, and place the result below the line, directly under the list fiure qf the dividend takien, as the first fi glure of the quotient. If there be no remainder, proceed in the same mannzer wit/h each of the subseguent fifures of the dividend. 3..But if there be a ren.alrinder after cdividin- the first or any subsequent fig ure of the dividecnd, re "ard that remainler as prefixecd to the. nzext figure of the dividcend, ard tlhen. inqutire how many times the divisor is contained in the number th.ls forimed, and place the quotient figu'e zndcrirncatlh as biore..Procecd i7, J1 cis t;'iay, until all thie figures of' the divcidezzd are divided~ 4. i ar, y is azn/ ce, the divsor is greater thane the figure to be rdivided in the dividend, and therefore cannaot be contained in it, a ci her v:n?,ss be written in the quotient, and the undivided figure must be re-.-ordaed as prejxed to te next fiegure of the dividend. 5..f there is a remainer ar ajn dividilg thv last figutre of the d7ivi2ite;rd, it may be placed at tie'ight hand of the quotient az-d marked Remainder, or be written over the divisor, with a horizontal line betwoeen thIen, anld ariexed to th7e qtotient. QUEsrIONs. — How are the numibers arranged for short division 7 At which hand do you begin to divide? Why not begin at the right, where you begin to multiply? Where do you write the quotient? If there is a remainder iter dlivi!ing a flre, what isn do-o with it? - Art. 49. VWhat is the r.le for short dividiong 7 n? 50 DIVISION. LsECT.'V. ARTy. @. First $Method of Proof. —Multiply the divisor and quotient together, and to the product add tihe remainder, iI there is any, an.d, if the work is right, the sum thus obtained will be equal to the dividend. NOTE. - it will be seen, from this method of proof, that division is the reverse of multiplication. The dividend answers to the product, the divisor to one of the factors, and the quotient to the other. EXAMPLES FOR PRACTICE. 2. Divide 6375 by 5. OPERATION. PROOF. Divisor 5) 6 3 7 5 Dividend. 12 7 5 Quotient. 1 27 5 Quotient. 5 Divisor. 6 3 7 5 Dividend. 3, 4. 5. 3) 7893762 4)4763256 5)3789565 2631254 1190814 6. 7. 8. 6) 8765389 7)987635 8)378532 9. 10. 11. 9)8953-7 8 4 1 1)7678903 12)634532 Quotients. 12. Divide 479956 by 6. 79992& 13. Divide 3 p5678 by 7. 55096ti 14. Divide 458789 by 8. 54848 - 15. Divide 1678767 by 9. 1865'296 16. Divide 11497583 by 12. 9581311 Quotients. Rem. 17. Divide 5678956 by 5. 1 18. Divide 1135791 by 7. 6 19. Divide 1622550 by 8. 6 20. Divide 2028180 by 9. 3 2l. Divide 2253530 by 12. 2 2. Divide 1877940 by 11. 9 Sum of the quotients, 2084732 27 QUESTIONS. -Art. 50. How is short division proved? Of what is division the reverse? To what do the three terms in division answer in multiplication? VWhat, then, is the reason for this proof of division? SET. v. DIVISION. 51 23. Divide 944,580 dollars equally among 12 men, and what will be the share of each? Ans. 78,715 dollars. 24. Divide 154,503 acres of land equally among 9 persons. Ans. 17,167 acres. 25. A plantation in Cuba was sold for S7,011,608 dollars, and the amount was divided among 8 persons. 3What was paid to each person? Ans. 876,451 dollars. 26. A prize, valued at 178,656 dollars, is to be equally divided among 12 men; what is the share of each? Ans. 14,888 dollars. 27. Among 7 men, 67,123 bushels of wheat are to be distributed; how many bushels does each man receive? Ans 9,589 bushels. 28. If 9 square feet alake I square yard, how many yards in 895,347 square feet? Ans. 99,483 yards. 29. A township of 876,136 acres is to be divided among 8 persons; how many acres will be the portion of each? - Ans. 109,51 acres. 30. Bought a farm for 5670 dollars, and sold it for 7896 dollars, and I divide the net gain among 6 persons; what does each receive? Ans. 371 dollars. 31. If 6 shillings make a dollar, how many dollars in 7890 shillings? A. 1315. ART. 5t. The method of operation by Long Division, or when the divisor exceeds 12. Ex. 1. A gentleman divided 896 dollars equally among his 7 children; how much did each receive? Ans. 128 dollars. OPERATION. Having set down the divisor Divide-nd. and dividend as in short divisDivisor 7) 8 9 6 (1 2 8 Quotient. ion, Awe draw a curlved line at ~7 the right of the dividend, to rmark the place for tihe quo1 9 tient. We tllen inquire how 1 4 many times 7, the divisor, is - 6 contained in 8, tile left-hand firgure of tile dividend; and 5 6 finding it to be 1 time, we write the I in the quotient, and QUE.STIONS.- Art. 51. What is long division? What is the ditference between long division and short division? How do you arrange the numbers for long division 7 What do you first do after arranging the numbers for long division'! Where do you place the figures of the quotient? 17),IViStOfN. [,s3:c r. s. amultiply the divisor, 7, by it, placing the product, 7, under the 8 from which we subtiact it, and to the right of the remainder, 1, bring down 9,' the next firure of the dividend, makingr 19. W-Ve now inquire how many tirTs 7 is contained in 19, alld place the number, 2, at the righlt of the quotient figure befaore obtained. ~We then multiply the divisor by it, and place the product under the 09, and subtract as before and to the remainder, 5, we bring down 6, the next rand last figure of tile dividend, makling 56. VWe proceed, as before, to find the next quotient figure, and, after subtractinc the product o:f the divisor nultiplied by it, from 56, find therea is no remainder left, H-ence we learn that each one of thle 7 children must receive 12 dollars. NoTrE. - The preceding example, and the ihur that follow, are osually performed by short division, but are hiere introduced to illustrate rnoro clearly the method of operation by long division. EXAsIr'LES FOR PRaAcTICE. 2. Divide 172i8 by 8. Ans. 216. 3. Divide 987656 by 110 Ans. 89786-;14. DiVide 123456789 by 9. Ans. 137171421. 5. Divide 390413609 by 12. Ans. 32534467- fa. Ex. 6. A gentleman divided 4712 dollars equally among Ins 19 sons; what was the share of each? Ans. 248 dollars. Having arranged the di-. DisideAri. visor and divideind as beDivisid end 9 4 7 2, 2 4 Q n. fore, we first inquire how Divisor 1 9) 4 7 1 2 (2 4 8 Quotient. many times 19, the diviso'r, 3 8 is contained in 47, the two 9 1 left-hand figures of the div7 idend; aind finding it to be 7 6 2 times, we write the 2 in 1 52 lthe quotient, multiply the 1 5 2 divisor by it, and subtract the product from the 47; and to the right of the remiainder, 9, bringr down 1, the next figure of the dividend, making 91. NWe next inquire how many times 19 is contained in 91, place thle:number, 4, in the quotient, then multiply and subtract as before., aid to the remainder, 15, bring down 2, the last figure of the dividend, and, proceeding as before, after finding the quotient figure, Lno rermainder is left. HeLnce the share of each of the 1.9 sons is 248 dollIars. QUESTIONS.- After the quotient figure is found, what is the next thing you do? Where do you place the product? What do you next do? What is the next step? How do you then proceed? Is long division the same in principle as short division 7 vscr. v.] DIVISION..3 ART. P0. From the preceding illustrations, the pupil will perceive the propriety of the following general RULE. - I. TWrite down the divisor and dividend as en short diviszon, artd draw a curved line at the right hand of the dividend. 2. Then inquire how many times the divisor is contazned in an equal number of fiurues on the left hand of the dividend, or in one more, ip' an equal nulmber uill not contain the divisor, and place the result en the quotient at the right hand -of the dividend. 3. Aultiply the divisor by the-quotient figure, writing the product under the figu.res of the dividend that twere taken. Subtract this product from the jigures of the dividend above it, and to the right of the renaindcer bring down the next fig ure of the dividend. 4. IFind how many tzmes the divisor is contained in the number thus formzed; place the figure denoting it at the riz lAt hand of the former qluolent fiure; multiply the divisor by it, and subtract the product from the nuurdler divided, and to the remainder lbring down the ne.xt figure of the dividend, as before. Thius proceed until all the figures of the dividend are divided; and if there is a remainder, write it as dzreced in the preceding rule. I-olTE 1. - Thee proper remainder is in all cases less than the divisor. [, in the course of' the operation, it is at any time found to be larger than tile divisor, it will show that there is an error in the worlk, and that the quotient figure should be increased. ]NOTE 2.- If, at any time, the divisor, multiplied by the quotient figure, produces a product larger than the part of the dividend used, it shows that the quotient figure is too laroe, and must be diminished. NOTE 3. -It will oft-en happen, that, when a figure is brought down, the nuimber will not contain the divisor, and in that case a cipher must be placed in the quotient, and another figure of the dividend brought down, and so on until the number is large enough to contain the divisor. ART. 6,o Second Method of Proof. - Add together the remainder, if any, and all the products that have been produced by multiplying the divisor by the several quotient figures, and the result will be like the dividend, if the work is right. ART. 54. Third SMethod. - Subtract thile remainder, if any, from the dividend, and divide the difference by the quotient. The result will be like the original divisor, if the work is right. NOTE. -The first method of proof (Art. 50) is iusually most convenlent, and is most cornmmonly employed. QUESTloNS. - Art. 52. What is the general rule for long division? H-low mnay you know when the quotient figure is too srnall 7 Itow may you know iwhen it is too large? What do you do when the part of the dividend used will not contain the divisor — Art. 53. What is the second method of proof for division?- Art. 54. What is the third method? Can long division be proved by the first method of proof (Art. 50)? 54 L'tSECTr. V. xEXAMPLES FOR PRACTICE. Ex. 7. It is required to find how many times 48 is contained in 28618. Ans. 596. OP-IRATION. Dividend. PROOF BY MiULTIPLiCATI'i',_I, Divisor 4 8) 2 8 6 1 8 (5 9 8 Quotient. 5 9 6 Quotient. 24 0 48 Divisor. I,6. =7A 298 328608 2 8 C'li ca {enmainder. 1 0 leinainder. 2t' 6 I 8 Dividiend Divtidendo Divisor 2 6) 5 6 9 8 (2 1 9 Quotient AI 2- 6 2> Produacss, -- 6 4 ltemainder. 2 3 8 5 6 9 8Di-',idend. i _.,4..t- 4 I't-rma-inder - i, OPETRATION. PI'e00'F B'Y Dl SIO.i. Di v iidend. 13ividend. Divisor i 4 4) 1 q 8 2 4 (9 6 Quotient. 9 L) I 8 8 2 ) ( 4 4 4 Divisor 296 9 6 864 422 864 388 3 8 A 384 Quotients. Rem 10. Divide 38276 by I4. 234 i, Divide 6205 by 17. 3I65 This sign of addition denotes the several products to be added. se,'rcT. vi.1 DIVSlOiO. Quotients, tem. i2,. Divide. Si5 by A1. I45'!3, Divide 190850 by 25 76:4 0 14. Divide 2 18579 by 42, 5-204- 11 t5. Divide 90't!2345 by 31- 9590'720 25 t. Divide 6'717890 by 83 68549 883 i5' Divirde 456,7890 by 19, 4041 5 2,i. Iwivide u157DO91f by 87.I F5 08 5 1.9. Divide 9988891 by 77. 2 9725 68 2 i0. Divide 9999999 by 69. 4497 36 21. Divide 867%532 by 59 4703 55 Divide!67008 by 8v 189 9 55 I2. Divide,%5S78 by 3i9. 9 12 30 2'4. Divide 3456789567 by 987. 3S50231.9 7 14 ia- v Divide S 97'74.-4441 by 3, 145. O 3 Divide 4500700701 by 407, 9 iO0582'''77H:2 D. Divide 67895i63 by -934. 9u5 298!)ividl 78'$ ~ I 435 y B3... it,-O'9. Divid-e SL(! 533G69 by 9999. 71 253 30. Divide:,,99,9c99 by 3333, 0:31. Divide 4785-'71.2 by 1789. 982 D3. Divide 3455789f3017363 by 4007I, 236 875i5 4-80 83. Divide 478656785178 by 56789. 8428688 f82346 34. Divide 678957000107 by 107895G61. 2(32 92 2950'30 35. Divide 990070171009 by 900700601. 1099 2002O105 10 3fL Divide three hundred twventy-one thousand three hundree dollaurs equally amiong six hundred seventy-five men. Ans. 476 dollars. 37. Four hundred seventy-one men purchase a township containincg one hundred eighty-six thousand fortye acres; what is the share of each? Ans. 03''acres. 38. A railroad, which cost five hundred eighteen tehousand seventy-seven dollars, is divided into six hunldrcd seventy-nine shares; what is the value of each share A ns. 763 dollars. 39. Divide forty-two thousand four hundred thirty-five bushels of wheat equally among one hlundred twent y-thre-e men. Ai-s:. b45 busrels each. 401 A. p'rize, valued at one hundrse d eightty-fthur tlhousandt seven hiundred seventy-five dollars, is to be divided equaly, amlnOtO four Ilunred seventy-five nmen;ill-t is the sha-re, o h? ns. S39 dollars. / 1. A certain comrpany purchased a tvaluable towntship for r'in io i hundred nine v-one thouiinsd eigsi hundred 56 DIVISION. [SECT. V thirty-six dollars; each share was valued at seven thousand eight hundred fifty-four dollars;.of how many men did the company consist? Ans. 1234 men. 42. A tax of thirty millions fifty-six thousand four hundred sixty-five dollars is assessed equally on four thousand five hundred ninety-seven towns; what sum must each town pay? Ans. 6538k 5 79 dollars. ART. do. Method of operation, when the divisor is a cornposite numnber. Ex. 1. A merchant bought 15 pieces of broadcloth for 1440 dollars, what was the value of each piece? Ans. 96 dollars. OPECATION. The factors of 15 are 3 3) 1 4 4 0 dolls., cost of 15 pieces. and 5. Now, if we divide the 1440 dollars, the cost 5) 4 8 0 dolls., cost of 5 pieces. of 15 pieces, by 3, we oh9 6 dolls., cost of 1 piece. tain 480 dollars, which is evidently the cost of 5 pieces, because there are 5 times 3 in 15. Then, dividing 480 doltars, the cost of 5 pieces, by 5, we get the cost of 1 piece. 1Hence we deduce the following RULE. - Divide the dividend by one of the factors, and the quotient thus found by another, and thus proceed till every factor has been made a divisor, and the last quotient will be the true quotient required. EXAIPLES FOR PRACTICES Quotients. 2. Divide 765325-by 25 5 X 5. 306'13 3. Divide 123396 by 84 7 X 12. 1469 4. Dividc' 611226 by 81, using its factors. 7546 5. Divide 987625 by 125, using its factors. - 7901 6. Divide 17472 by 96, using its factors. 182 7. Divide 34848 by 132, using its factors. 264 ART. &_5Go Method of finding the true remainder, when here are several in the operation. Ex. 1. How many months of 4 weeks each are there in 298 days, and how many days remaining? Ans. 10 months and 18 days. QUESTIONS.- Art. 55. What are the factors of 15? What do you get the cost of, in this example when you divide by the factor 3? What, when you divide by 5 1 Why? What is the rule for dividing by a composite number? s:EcC. T DIVSION. 5 OPERAtTION, Since there are 7 days in t /); 9 8 week, we first (livile the 298 days by 7, and have 42 weeiks 4 4o- ___ 4 dais ys 0 18ee d3ays?land a remlaiider of 4 days. -i- - days' lTheni, since 4 veekl-s mniake 1 morn1th, w e divide the 42 Swees by 4, and 1hae 10 months and a renlnainder o 2 wveeks. Ntowg to find the true eremainder in days, it is evident that we moust multiSply the 2 wedekis by 7, because 7 days make a weeli and to the product add the 4 days; thus 2 X 7 —= 14, and 14 —4- - 18 days for.he rernainder~ Sienee the propriety of the followinog Pura? -- ir -Le.ltny acich remarinder by all the divisors )r ecedincr tih on~e ulilics piroduced it; and the first 9'emainder being added to the sum oJf the prolilcts, the amount wuill be the trie remainzder. NI'OT'r. - There will be but one product to add to the first remuainder, when there arc only two divisors anrid two remlainders. i. -Diivide d89 by 26, using the factors 2, 3, and 6, and find the true remnainder. Ans. "33 OPERATIOr. FINDING THE'TRUE REiVAINDER. 2) 7 8 9 5 X 3 X 2 — 30P, Ist Product. 3) >( 2 2, 2, d Product. 3) 8 9 4 --!, 1st!:tom. lest Rellainder. 6) 1 I t, 2d Ie m. 23g true.Reln. 2. I -$ 5Sd Rem. EXAliPLEs Foir PrAC'TICE. 3. Divide 924 by 55, using the lactoers 5 and 11, -ancd find the true remainder. Ans. 54. 4. Divide 5348 by 48, using the factors 6 and 8, and find the true remnainder..Ans. 20. 5. -Divide 5873 by 84, using the factors 3, 4, and 7, and find the- true remainder. Ans. 77. 6. Diide 249237 by 1728, using the factors 12, 6, 6, and 4, and find the true remainder. Ans. 405. A.rnr, 4O''o, divide by 10, 100, &c., or I with ciphers at the ril;it Ex. 1. Dividee'35 dollars equally among 10 men, -what w-ill each man have? Ans. 35 6 dollars. -QUijsIws. - Art. 56. IWhen there are several remnainders, what is the rule icr fbldlinig the trie rerrmainder? Will you give the reason for this rule? B8 DIVISION. [SECT. V. OPERATION. It will be remembered, that, to mulJ0O) 3 516 tiply by 10, we annex one ciphler, which removes the figures one place to the Quotient 3 5 - 6 Rem. left, and thus increases their value ten r t 3 5ties. Now it is obvious, that, if we vOr thus, 3 %. reverse the process and cut off the righthand figure by a line, we remove the remain:ilng figures one place to the right, and consequently dclinish the value of each ten times, and thus divide the whole number by 10. The figures on the left of the line are the quotient, and the one on the right is the remainder, which may be written over the divisor and annexed to the quotient. Hence the share of each man is 35 6 dollars. RULE. - Cut off as many figures from the right hand of thie diti dend as there are ciphers in the divisor, and the figures on the left hand of the separatrix will be the quotient, and those on the right hand the remainder. EXAMPLES FOR0 PRACTICE. Quotient. Rem. 2. Divide 6892 by 10. 689 2 3. Divide 4375 by 100. 75 4. Divide 24815 by 1000. 815 5. Divide 987654321123 by 100000000. 54321123 ART. 5S. Method of operation when the divisor has ciphers on the right. Ex. 1. If I divide 5832 pounds of bread equally among 600 soldiers, what is each one's share? Ans. 94 2 pounds. OPMATION. The divisor, 600, may 110 0) 5 8f13 2 be resolved into the factors 6 and 100. We first 6) 5 8 - 3 2, Ist Remrl. divide by the factor 100, 9 m. by cutting off two fig-, 2d Rem. ures at the right, and Or thus, 6f1(0 0) 5 813 2 get 58 for the quotient and 32 for a remainder. 9 — 4 3 2 We then divide the quotient, 58, by the other factor, 6, and obtain 9 for the quotient and 4 for a remainder. The last remainder, 4, being multiplied by the divisor, 100, and 32, the first remainder added, we obtain 432 for the true remainder (Art. 56). Hence each soldier receives 94g2 pounds. QUESTIONS. - Art. 57. How do you divide by 10? How does it appear that this divides the number by 10? What is the rule for dividing by 10, 100, &c.? Art. 58. How do you divide by 600 in the example? How does it appear that this divides the number? SrECT. V.] CONTRACTIONS IN PMULTIPLICATION. 59 RULE. — Cut off the ciphers from the divisor, and the same number of figures figu rom the right hand of the dividend.''hen divide the re maining figures f the dide the dividend by the rermaining rfigures of the divisor, and the result will be the quotient. To comprlete the work, annex to the last remainder found by the operation the figures cut off from the div idend, and the whole will form the true remainder. EXABMPLES FOR PRACTICE. Quotients. Rem. 2. Divide 3594 by 80. 44 74 3. Divide 79872 by 240. 332 192 4. Divide 467153 by 700. 667 253 5. Divide 13112297 by 8900. 2597 6. Divide 71897654325 by 700000000. 497654325 7. Divide 3456789123456787 by 990000. 306787 8. Divide 4766666000000 by 55550000000. 44916000000 VI. CONTRACTIONS IN MULTIPLICATION AND DIVISION.r CONTRACTIONS IN MULTIPLICATION. ART. 59. To multiply by 25. Ex. 1. Multiply 876581 by 25. Ans. 21914525. OPERATION. We first multiply by 100, by 4) 8 7 6 5 8 1 0 0 annexing two ciphers to the multiplicand, and since 25, the mul2 1 9 1 4 5 2 5 Product. tiplier, is only one fourth of 100, we divide by 4 to obtain the true product. RULE. - Annex two ciphers to the multiplicand, and divide it by 4, and the quotient is the product required. If the principles on which these contractions depend are considered too difficult for the young pupil to understand at this stage of his progress, they may be omitted for the present, and attended to whenf he is further advanced. QUESTIONS. - What is the rule for dividing when there are ciphers on the right of the divisor?- Art. 59. What is the rule for multiplying by 25? What is the reason for the rule? 60 CO NTRACTIO.N'S ITNi MULTIPLICATIO2P, [Secr. vx E'XAMIPLES FOR PRACTICE. 2. Multiply 76589 58 by 25. Ans. 1914741450, 3. Multiply 567898717 by 25. Ans. ^ 14l1974670926 4. _Multiply 1.23456789 by 250 Acns. 3086419725.ART. G To multiply by 33. Ex.:. M ultiply 87678963 by 33- Ac ns.'922 632lO. OPERATION.' We first multiply by 100 3) 8 7 6 7 8 9 6 3 00 as before, and since 33~, tlhe multiplier, is only one lhird 2 M P9 2 2 6 8 1 00 Plodluct, of 100, we divide by 3 to obtain the true product. RULE. -- Annee two ciphlers to the mulfiplicand, and dlivide it by 3, and the quotient is the product required. ExlXAMPLES FOR PRACTICEo I. Multiply 356789541 by 33~. Ans. 11892984700. 3. Multiply 871132182 by 33~. Ans. 2903t739400. 4. Multiply 583647912 by 33-.. Ans. 19454930400, ART. C io To multiply by 125. 7Ex..'ilMultiply 7896538 by 125. Ans. 987067250. OPERATION. We multiply by 1000, by 8) 7 8 9 6 5 3 8 0 0 0 annexing three ciphers to-tlhe multiplicand, and since 125, 9 8 7 0 6 7 2 5 0 Product. the multiplier, is onlUy on0 eigohth of 1000, we divide by 8 to obtain the true producto Rur,E. - Annex three ciphers to the multiplicand, and divide by 8, and the quolient is thie product. EXAMPLES FOR PRACTICE.. Iultiply'7965325 by 125. Ans. 995665 25. $. Multiply 1234567 by 1 25. 15ce. 14320875. 4. Miultiply 3049862 by 25. Ac ns,. 381232750, QUESTIONS. - Art. 60. What is the rulel for nmultiplying by 33;}? i hbat is the reason for this rule? - Art. 61. What is t'he rule for mnultiplying, by 125 Givo the reason bor th.e ru11le? E.CT,'V.] CGONTRACTIONS IEN DIVIOSBl Ib,T. 2, To multiply by ay by number bof 9's, 2.1J J.. M:Tfultiply 4789653 by 9999999. Ans. 4 7,390605 034'. eO0dIGnArihOT. EBy adding i to any number 4 7 8 9 6 5 3 8 0 0 00 coposed of nines, we obtain 4' 7 8 9 6 5 3 a number expressed by 1 with as many ciphers annexed ai 4'7 83 9 0 5 Pr 4 7 9roduct, there are nines in the number to which e is added. Thus, 999 -1- 1 L000 Therefore annexing to the multiplicand as many ciphers as there are nines in the mnultiplier is the same ting as multiplying the number by a multiplier too large by 1, and subtracting the number to be multiplied from thjls ehlarged product will give the true product. RIJLE - Annex as many ciphers to the muiliplicand as there are 9's'n the multiplier, and from thzs number subtract the number to be multiplied, and the remainder is the product required. XAXIIPLES J'OR PRACTICE. 2. Miultiply 1234567 by 999. Ans. 12338'324 3. 3. Multiply 876543 by 999999. Ans. 876542123457 4. TM'ultiply 999999 by 99999. Ans. 999998000 G01. CONiTRACTIONS IN DIVISION. ART. G;3, To divide by 25. Ex. 1. Divide 1234567 by 25. Ans. 19382 6 8% OPERATI ON..Multiplying the dividend by 4 makes 123456 7 it four times too great; therefore, to obtain the true quotient, we must divide by 100, a divisor four times greater than 9 3 1 ~.6 8 uotienta the true one. This we do by cutting off two figures on the rig''. Ru LE. l- Mizltiply the dividend by 4, and the product, except the last tto figures at the rig ht, is the quotient. T/he last two are hundredths.,XAI AMPLES FOR PRACTICE. 2. Divide 9876525 by 25. Ans. o5`06,,s 3. Divide 1378925 by 25. Ans. 55157. 4. Divide 699999 by 25, Ans. 3a599 9s qUoESTIONS. Art. 62. What is the rule for multiplying by any number of 9's? lhat is the reason for the rule? - Art. 63. What is the rule tbr dividinr: by 92? Give the reason for the rule. -62 6CONTRACTIONS IN DIVISION. i[ECT. vI. ART. 64o To divide by 33]-. Ex. 1. Divide 6789543 by 33 -. Ans. 2C3686-29. opunATION]. Multiplying the dividend by 3 makes 6 89543 it three times too great; tilerefore, to 39 ~obtain the true quotient, we must divide by 100, a divisor three times greater than 2 0 3 6 8 6.2 9 Quotient. the true one. This is done by cutting off two figures on the rirght. RULE.- Multiply the dividend by 3, and the product, except the last two figures at the right, is the quotient. Thle last two are hundredths. ExAMPLES FOR PRACTICE. 2. Divide 987654321 by 33~. Ans. 2962962963%. 3. Divide 8712378 by 33~. Ans. 261371-4,y%. 4. Divide 4789536 by 33k. Ans. 143686r8-. 5. Divide 89676 by 333. ~ Ans. 2690%. 28 6. Divide 17854 by 334. Ans. 535.T0 o ART. 65. To divide by 125. Ex. 1. Divide 9874725 by 125. Ans. 78997- 18. Multiplying the dividend by 8 maLkes OPERATION.. 9 ON 7 <. it eight times too great; therefore, to 987 4 7 2 5 8 7 4 7 25 oobtain the true quotient, we must divide by 1000, a divisor eight times greater 7 8 9 9 7.8 0 0Quotient. than the true one. We do this by cutting off three figures on the right. RULE. - PM ultiply the dividend by 8, and the product, c'^ept the last three figures, is the quotient. The last three figures are thousandths. EXAMPLES FOR PRACTICE. 2. Divide 1728125 by 125. Ans. 13825. 3. Divide 478763250 by 125. Ans. 3830106. 4. Divide 591234875 by 125. Ans. 4729879. 5. Divide 489648 by 125. Ans. 391 7 rTllo4f. 6. Divide 836184 by 125. Ans. 6689 4701. QuFSTroNs. —Art. 64. What is the rule for dividing by 33:. Give the reason for the rule. - Art. 65. What is the rule for dividing by 125? What is the reason for the rule? SECT. VIx.] MISCELLANEOUS EXAMPLES. 63 Q VII. MISCELLANEOUS EXAMPLES, INVOLVING THE FOREGOING RULES. 1. A. bought 73 hogsheads of molasses at 29 dollars per hogshead, and sold it at 37 dollars per hogshead; what did he gain? Ans. 584 dollars. 2. B. bought 896 acres of wild land at 15 dollars per acre, and sold it at 43 dollars per acre; what did he gain? Ans. 25088 dollars. 3. N. Gage sold 47 bushels of corn at 57 cents per bushel, which cost him only 37 cents per bushel; how many cents did he gain? Ans. 940 cents. 4. A butcher bought a lot of beef weighing 765 pounds at 11 cents per pound, and sold it at 9 cents per pound; how many cents did he lose? Ans. 1530 cents. 5. A taverner bought 29 loads of hay at 17 dollars per load, and 76 cords of wood at 5 dollars a cord; what was the amount of the hay and the wood? Ans. 873 dollars. 6. Bought 17 yards of cotton at 15S cents per yard, 46 gallons of molasses at 28 cents per gallon, 16 pounds of tea at 76 cents a pound, and 107 pounds of coffee at 14 cents a pound; what was the amount of my bill? Ans. 4257 cents. 7. A man travelled 78 days, and each day he walked 27 miles; what was the length of his journey? Ans. 2106 miles. 8. A man sets out fiom Boston to travel to New York, the distance being 223 miles, and walks 27 miles a day for 6 days in succession; what distance remains to be travelled? Ans. 61 miles. 9. What cost a farm- of 365 acres at 97 dollars per acre? Ans. 35405 dollars. 10. Bought 376 oxen at 36 dollars per ox, 169 cows at 27 dollars each, 765 sheep at 4 dollars per head, and 79 elegant horses at 235 dollars each; what was paid for all? Ans. 42884 dollars. 11. J. Barker has a fine orchard, consisting of 365 trees, and each tree produces 7 barrels of apples, and these apples will bring him in market 3 dollars per barrel; what is the income of the orchard? Ans. 7665 dollars. 12. J. Peabody bought of E. Ames 7 yards of his best broadcloth at 9 dollars per yard, and in payment he gave Ames a tc:SCELLA' EOUS EXAMPLES.'S2:CT, v one T-altnlred-dollar bill; how many dollars must Anmes return to Peiabody? Ans. 37 dollars, 13, Bought of P. P1arker a cooking-stove for 3'2 dollatrs 7 quintals of his best fish at 6 dollars per quintal, 14 bushels of rye at I dollar per bushel, and 5 mill-saws at 1 6 dollars each; in part payment for the above articlesi I sold him eiAlht thousand feet of boards at 15 dollars per tholusand; how much must I pay him to balance the account? Ans. 47 dollars14. In 1 day there are 24 hours; how many in 57 days? s, 13533 hous, 15. in one pound avoirdupois weightt there are 16 ounces, how mlany ounces are there in 3869 pounds? Ans. 0a041 ounces. 116. in a square mile there are 540 acres; how many acres are there in a town which contains 89 sqlluare mniles Amns. 56990 acres. 17.r Wihat cost'18 ba.-rels of applesl aet 3 dollars per bar. rel P Ans. 234 dollars. 18. Bought.500 barrels of flour at 5 dollars per barrel, 47' hundred weighlt of cheese at 9 dollars per hundred weight, and 15 barrels of salmon at 17 dollars per barrlel; wvhat was the amount of my purchase? Ans. 178 dollars. 19. Bought'760 acres of land at 47 dollars per acre, and sold J'. Emery 171 acres at 56 dollars per acre, J, Smith 275 acres at 37 dollars per acre, and the remainder I sold to J. Kimball at 75 dollars per acre; how miuch did I gaein by my sales? Ans..581 dollars. 920 13oughit a hogshead of oil containing 184 gallons at'5 cents per gallon; but,23 gallons having leaked outi, I sold the remainder at 98 cents per gallon; did I gain or lose by my bargain? Ans. 1180(3 cennts gain. 21o Bought a quantity of flour, for which I gave u.728 diie Lars, there being 288 barrels; I sold the saie e t 8 do'larS p)e barrel; how much did I gain?. ns, 57rG dtollarsi 22. Purchased a cargo of molasses o:r 9'.i2 doai-rs thera, being 190 hogsheads; I sold the same at 7 doollars pel hle head; how much did i gain on each hogshead? Ans. 20 dollars. 23. A flaruer bought 5 yoke of oxen at 8U dollars a yoke'7 cows at 3'7 dollars each; 89 sheep at 3 dollhrs apiece, HTe sold the oxen at 98S dollars a yoke; for thei cows hoe received 4-0 dollars each; and for the sheep he lad 4 dollars apiiece, t-ow mr uch did he- gain by his trade? Ans, 2'55 dollars5 sECT. vwL.] iMISCELLANEOUS EXAMPLES. 65 24. The sum of two numbers is 5482, and the smaller number is 1962; what is the greater? Ans. 3520. 25. The difference between two numbers is 125, and the smaller number is 1482; what is the greater? Ans. 1607. 26. The difference between two numbers is 1282, and the greater number is 6948; what is the smaller? Ans. 5666. 27. If the dividend is 21775, and the divisor 871, what is the quotient? Ans. 25 28. If the quotient is 482, and the divisor 281, what is the dividend? Ans. 135442. 29. If 144 inches make 1 square foot, how many square feet in 20736 inches? Ans. 144 feet. 30. An acre contains 160 square rods; how many rods in a farm containing 769 acres? Ans. 123040 rods. 31. A gentleman bought a house for three thousand fortyseven dollars, and a carriage and span of horses for five hundred seven dollars. He paid at one time two thousand seventeen dollars, and at another time nine hundred seven dollars. }Low much remains due? Ans. 630 dollars. 32. The erection of a factory cost 68,255 dollars; supposing thlis sum to be divided into 365 shares, what is the value of each? Ans. 187 dollars. 33. IBought two lots of wild land; the first contained 144 acres, for which I paid 12 dollars per acre; the second contained 108 acres, which cost 15 dollars per acre. I sold both lots at 18 dollars per acre; what was the amount of gain? Ans. 1188 dollars. 34. Sold 17 cords of oak wood at 6 dollars per cord, 36 cords of maple at 3 dollars per cord, and 29 cords of' walnut at 7 dollars per cord, -What was the amount received? Ans. 413 dollars. 35. Daniel Bailey has a fine farm of 300 acres, which cost him 73 dollars per acre. IHe sold 83 acres of this farm to Minot Thayer, for 97 dollars per acre; 42 acres to J. Russel, fior 87 dollars per acre; 75 acres to J. Dana, at 75 dollars per acre; and the remainder to J. Webster, at 100 dollars per acre. What was his net gain? Ans. 5430 dollars. 36. J. Gale purchased 17 sheep for 3 dollars each, 19 cows at 27 dollars each, and 47 oxen at 57 dollars each. HIe sold his purchase for 3700 dollars. What did he gain? Ans. 457 dollars. 37. Purchased 17 tons of copperas at 32 dollars per ton. I sold 7 tons at 29 dollalrq per ton, 8 tons at 36 dollars per ton, {C)UA MISC"ELLAN=EOUS EXAM1 PLES. [ sesCr. vIE. and the remainder at 25 dollars per ton. Did I gain or lose,, and how much? Ans. 3 dollars, loss. 38. John Smith bought 28 yards of broadcloth at 5 dollars per yard; and having lost 10 yards, he sold thLe remrainder at 9 dollars per yard. Did he gain or lose5 and how much? Ans. 22 dollars, gain. 39. Which is of the greater value, SSJ acres of land at 7' dollars per acre, or 968 hogsheads of moIa.sses at 25 dollars per hogshead? Als. The land, by 5!3G dollars. 40. Bought of J. Low 387; tons of hay at D.O dollars per ton. I piaid him 75 dollars, and 12 yards of broadcdcloth,;t 4o - dole Iars per yard. I-eow much remains due to Lo-\w? An-,s. 5,12S doilar::. 41. A purchased of B 41 cords of wood at 5 dollars per cord, 9 tons of hay at 17 dollars per ton, i9 grmindstones at 2 dollars apiece, 37 yards of' broadcloth at 4 dollars per ytarl,l and 16 barrels of flour at 6 dollars per blarrel; what is t'e amount of A's bill? AnSo`s5 dollars. 42. John Smith, SiJr., bouglht of LZ S. a:vis IS dozen of Na, tional Aiitlimetics at G dollars per dozen, 23 dozen of Menta. Arithmetics at 1 dollar per dozen, 17 dozen Farnly Hibes at 3 dollars per copy; what is the amount of the bill? AjsI.'743 dollars. 43. it.:asseltine sold to John James 1659 tons of timber at'7 dollas per ton, 116 cords of of oak woodl at 6 dollars per cord, atrd 37 cords of maple wood at 5 dollars per cord; James has paid Hasseltine 144 dollars in cash, andl 23 yards of cloth at - dollars per yard; what remains due to Hasseltine? Acins. 18-28 dollars. 44.. F'rost owes me on account 375 dollars, and he has paid me 6 cords of wood at 5 dollars per cord, 15.tons of hay at 12 dollars per ton, and 32 bushels of rye at i dollar per bushel. How much remains due to me? Ans. 1'lt dollars. 45. Cave 1.9 dollars for a chaise, 8'7 dCllars for a haraesse aUld 176 dollars for a horse. I sold the chaise for 187 dollars, the harress for 107 dollars, and the horse for 165 dollars,.'W~hat sum have.l gained? Ans. 27 dollars. 48. Boughlt a farm of'J. C Bradbury for 1728 dollars, for which I paid him 75 barrels of flour at 6 dollars per barrel, 9 cords of wood at 5 dollars a cord, 17 tons of hay at 25 dollars a ton, 40 bushels of wheat at 2 dollars a bushel, and 65 bushels of beans at 3 dollars a bushel; how 3 an.3.y dollars remain dlue EC~'r. vS.] U.ITED STATES MON&EY. ~ VIII. UlNIT ED STATES MONEY/. A-RT. 6')9 UNITED STATES ] iONSEY iS the legal currency of dee United Sates,, i0) iils'. a.e e Cent, marakied c. 10 Cents m; 1 Dimne,..0 Dimes. Dol1g i 0 01ls4-s'6 - }w Dollars I L. lO Dollars " I E-ao-lec 9' E. C:c.nts~ Iruiill, Dollars, a... -- 100N i:,_..'. 0 100 == 1000 tI.. Gtb iO.. I 0 0 --- 0000 7 ihm d enominationsl of United States mnoney increase from riCtlht;o lef, and decrease from left to right, in the same ratio a snn!ple numbers Tihely may thlerefore be added, subtracted, i.:7:lltilliedl anrd divided according to tle saime rules. I:n thli wc rk dollars are separated from cenlts by a period or dot, an cents fi'om mills by a comma thiusm, 16.25,3 is read, siiLteen dolltzas, trwenty five cents, three mills. Since cents occupy two places, the place of dimes aind of tnits, johen thle number of cents is less than 10, a cipher must be written before thelm in thei place of dimes; thus.03,.07, &c. This wass frnerly, and is now friequently, called Fedrera l Money, because, on the adoption of the Federal Constitution, it was made the cur reney of t'he Federal Union. i Thle word iMI.L is finom the Latin word snilie (one thoulsand); the word CENT frOlm the Latin centum (one hundred); the word D.IMEX filom a F'rench word signifying a tlithe or tenth; and the reason of tlese names, as a)pplied to our coins, is Iound in the proportion whichI they respectively bear to the dollar. The terin )OLLAR is said to be derived from the DPanish worTd Doler, ao dI this fiorn Dale, the nanme of a town wlhere it was first coined. No coin of the denomination of Mills has ever been struckr at the mint; while, in addition to the pieces rnamed in the foregoing table, half-dines,!le quarter and half dollars, and the quarter and half" eagles, are in conmnro!ea use. QuE:sTIoNs.Art. 66. YVhat is United States money?'What is it frequently called? Ilepeat the Table of United States loney. What are tihe denorninnations of United States money? 7 -ow do they increase fiom right to lift? How are they added, subtracted, multiplied, and divided 7 How are dollars, cents, and nills separated 7 Why must a cipher be. placed before cents, when the nrmber is less than l' I "V- s- are two,ra, 1el foi cants!-e 3 1.1...'g......i.] 68 UNITED STATES MONEY. [SECT. VIII REDUCTION OF UNITED STATES MONEY. ART. 6W0. REDUCTION of United States Money is changing the units of one of its denominations to the units of another, either of a higher or lower denomination, without altering their value. ART. GS, To reduce dollars to cents and mills. Ex. 1. Reduce 25 dollars to cents and mills. Ans. 2500 cents, 25000 mills. OPERATION, 2 5 dollars. We first multiply the 25 dollars by 1 0 0 100 to reduce.them to cents, because 100 cents make 1 dollar; and then 2 5 0 0 cents. multiply the cents by 10 to reduce I 0 them to mills, because 10 mills make 2 5 0 0 0 mills. Or thus, 2 5 0 0 0 mills. RULE. - To reduce dol7ars to cents, annex TWO ciphers; and to reduce dollars to mills, annex THREE ciphers. ART. (09o To reduce dollars and cents to cents, or dollars, cents, and mills to nills. RULE, - Remove the separating point, or points, between the dollars, cents, and mills, and give the name qf the lowest denomination to the whole sum. ART. @0o To reduce mills to cents, cents to dollars, and mills to dollars. Ex. 1. Reduce 25000 mills to cents and dollars. Ans. 2500 cents,, 25. OPRATI'ON., 10) 2 5 0 0 0 mills. tWe first divide the mills by 10 to reduce them to cents, because 10 1 0 O ) 2 5 0 0 cents. mills make I cent; and then the cents 2.) dollars. by 100, to reduce them to dollars, because 100 cents make 1 dollar. Or thus, 2 510 0', millso QUESTIONS. - Art. 67. What is reduction of United States money? —. Art. 68, What is the rule for reducing dollars to cents and mills? Give the reason for the rule. - Art. 69. Ilow do you reduce dollars and cents to cents, or dollars, cents, and raills to mills 7 What is the reason for this rule? s C',r v xi..] UJiIT~rT-ED STATES V[1iONEY. R.ULE I. -- To reduce mills to cents, point off ONE figure on the ri/ght; the figures on the left of the point will be cents, the figure on the right mizlls, -tLE: 1I. --'0o reduce cents to dollars, point o'' TWO fig'urC on the rig/i-t; the figu;res on the left of the point will be dollars, those on the rig-ht cents. RULE III. - 7o reduce mills to dollars, point off TIREE figures on the rig ht; the figures on the left of the point will be dollars; the first two on the right cents,.an d the third mills. EXnAMPLEs FOR PRACTICE. 1, Reduce 8 125 to cents. Ans. 12500 cents. 2. Reduce $ 345 to -mills. Anns. 345000 nills. 30 Reduce 297 mills to cents..Ans. 29,7 cents. 4. Reduce 2682 olills to dollars. Ans. $ 2.68,o. 5. Reduce 4123 cents to dollars, Ans. $ 41.23. educe 156.29 to cents Ans 1629 cents, 7, lReduce (16.42,8 to mills. Ans. 16428 mills. 8 eRduce 8 9.7 to mills, A.ns, 9870 mills. APRT,.,.ADDITTiOni! OF UNITED STATES MONEY,..ULE. - Writle dtolTars under dollars, cents under cents, and mils under mills, and then proceed as tn simple addition. T7he result will be the sum, in the lowest denomination. added, which must be pointed off fs in reduction of United States money. (Art. 70.) Proojf — The proof is the same as in simple addition.,EXA.M&PLES FO:R IPACTICE. ~1,~ 2. 3. 4.. rcts..n. cts. Mi, e0. cts. 11n, i. c.s.,- 2 7. 5.6 4,3. 6.7 0,5 1 4 7.8 6 I 3. 8 9,1 I.8 97V 1 4.0 03) 7 8 9.5 8 9 3,5 1 4 3.8!iG 1 8.7 1,9 4 9 6.3 7 5 2.3,3 5 8.31 1,3 9 7.0 0,9 9 1 1.3 4 Ans. 2 0 4.9 9,8 9 4.6 t9 4 6.4 3y6 2; 3 4 5.1 5 Qur.srTIONs. - Art. 70. dWhat is the rule for reducing mills to cents' For reducina cents to dollars 1 For reducing mills to dollars? Give the reason for each of these rules. — Art. 7l. Iowv jmust the nulmbers be written down in addition of United States money? -Iow added? C' what denomination is tie stc 1M't flow nointed off'i Repeat thse rule. 70 UNITED STATES MONEY. [SECT. vln 5. 6(. 7 8. cts. ni. S. cts. m, In. ta,. In.' cts. m. 7 8 6.7 1,3 8 7.0 5,9 9 1.7 6,3 7 8 6.7 1.3 1 7 6.0 7,1 3 7.8 1,0 8 4.1 6,1 3 4 5.6 7,8 5 6 7.8 1,9 8 1.4 7,5 1 0.0 7,0 9 0 7.0 1,7 1 2 3.4 5,6 4 0.0 7,8 5 3.6 1,5 8 6 1.0 9,0 7 8 9.0 1,2 2 1.1 5,6 8 1.1 7,6 1 2 3.4 7,6 3 4 5.6 7,8 8 1.1 7,7 3 2.8 1,7 9 8 7.0 1,6 9 0 1.23, 3 3.6 2,1 5 3.1 9,6 3 4 5.7 0,5 7 1 8.9 0,5 28.0 9,3 4 1.5 7,0 3 5 7.0 9,1 9. Bought a coat for $ 17.81, a vest for $ 3.75, a pair of pantaloons for ( 2.87, and a pair of boots for 9 7.18; what was the amount? Ans. $ 31.61. 10. Sold a load of wood for seven dollars six cents, five bushels of corn for four dollars seventy-five cents, and seven bushels of potatoes for two dollars six cents; what was received for the whole? Ans. 8 13.87. 11. Bought a barrel of flour for ( 6.50, a box of sugar for $ 9.87, a ton of coal for 8 12.77, and a box of raisins for 8 2.50; what was paid for the various articles? Ans. $ 31.64. 12. Paid 4.62 for a hat, $ 9.75 for a coat, $ 5.75 for a pair of bobots, and $ 1.50 for an umbrella; what was: paid for the whole? Ans. 8 21.62. 13. A grocer sold a pound of tea for $ 0.62,5; 4 pounds of butter for 80.75; 4 dozen of lemons for $ 0.87,5; 9 pounds of sugar for $ 0.80; and 3 pounds of dates for 9 0.37,5. What was the amount of the bill? Ans. $ 3.42,5. 14. A student purchased a Latin grammar for 80.75, a Virgil for 3.75, a Greelk lexicon for 4.75, a Homer for 1.25, an English dictionary for ( 3.75, and a Greek Testamerit for $ 0.75; what was the amount of the bill? Ans. $15.00. 15. Bought of J. H. Carleton a China tea-set for ten dollars eighty-two cents, a dining set for nine dollars sixty-two cents five mills, a solar lamp for ten dollars fifty cents, a pair of vases for four dollars sixty-two cents five imills, and a set of silver spoons for twelve dollars seventy-five cents; what did the whole cost? Ans. 48.32. SECT. vYI.] UNITEDI STATES MfSONEY. 71 ART. 72o SUBTRACTION OF UNITED STATES MONEY. RULE. - Wrrite the several denominations of the subtrahend under the corresponeing ones of the minuend, and then proceed as in simple subtraction; the result will be the difference, in the lowest denomination in the questibn, which must be pointed off as in reduction of United States money. (Art. 70.) Proof. -The proof is the same as in simple subtraction. EXAMSPLES FOR PRACTICE. 1. S. 3. 4. $. cts. m. ct$. ta. 5. ct. t. Cts. Min. 6 1.58,5 47 1.81 156.00,3 14 1.7 0 Sub. 1 9.1 9,7 1 5 8.1 9 1 9.00,9 9 0.91 Rem. 4 2.3 8,8 3 13.6 2 1 3 6.99,4 5 0.7 9 5. a. 7. 8. e Ct. ct. Cts. In, etc., i. cts. in. From 71.86,1 9 1.0 7,1 8 1 5.7 0,1 1 0 78 1.3 0,3 Take 1 9.1 9,7 1 9.0 9,5 9 0.8 0,3 9 9 9 9.0 b97 9. From $ 71.07 take $ 5.09. Ans. $ 65.98. 10. From $ 100 take @ 17.17. Ans. $ 82.83. 11. From one hundred dollars, there were paid to one man seventeen dollars nine cents, to another twenty-three dollars eight cents, and to another thirty-three dollars twenty-five cents; how much cash remained? Ans. $ 26.58. 12. From ten dollars take nine mills. Ans. $ 9.99,1. 13. A lady went'" a shopping," her mother having given her fifty dollars. She purchased a dress for fifteen dollars seven cents; a shawl for eleven dollars ten cents; a bonnet for seven dollars nine cents; and a pair of shoes for two dollars. Hlow much money had she remaining'? Ans. $ 14.74. 14. From one hundred dollars, there were taken at one time thirty-one dollars fifteen cents seven mills; at another time, seven dollars nine cents five mills; at another time, five dollars five cents; and at another time, twenty-two dollars two cents seven mills. How much cash remained of the hundred dol. lars Ans. $ 34.67,1. QUESTIONS.- Art. 72. How do you write down the numbers in subtraction of United States money? H-low subtract 7 Of what denoxmination is the remi!t or difference? lTow pointed o.ff Repeat the rule. ~7 8JAUNIT'ED STATES MONE1S'F [tesr. wn. ART.-r FJ ULTPL!CAT[ON PUNTITED.0A..E'... B.... IOiE Y XIULE. hVitlh dollars, centse 4-c., for the multiplicand, proceed as in simple multiplicatlion; the result will be tlhe product, in the terrns of the lowest denomination contained in the multiplicand, which must be pointed off as in reduction of United States money. (Art. 70.) Proof. The proof is the same as in simple multiplication. EXAnPLES FOR IPRlAC'T.ICE, 1. WAVha.t will 143 barrels 2. VWhat will 144 gallons of' flour cost at 0 7.25 per of oil cost at, 1.62,5 a gal. barrel? Ans.o 1036.75, Ion? Ans. 0 234. OPlIRATION. OPERATION. Multiplicand 0 7.2 5 Multiplicand 0 1.6 2,5 Multiplier 1 4 3 Multiplier 1 4 4 1'7 5 6 5 00 2900 65 00 725 1625 Product 0 3 6.7 5 Product 0 2 3 4.0 0,0 3. What will 165 gallons of molasses cost at 0 0.27 a gal. ]on? Ans. $ 44.55. 4. Sold 73 tons of timber at 0 5.68 a ton; what was the amount? Ans. $ 414.64. 5. What will 43 rakes cost at 0 0. 17 apiece? Ans. 0 7.31. 6. What will 19 bushels of salt cost at $ 1.62,5 per bushel? Ans. $ 30.87,5. 7. What will 47 acres of land cost at 0 37.75 per acre? Ans. 0 1774.25. 8. What will 19 dozen penknives cost at $ 0.37,5 apiece? Ans. 0 85, 50. 90 What is the value of 17 c ts of cests of souchong tea, each weighing 59 pounds, at, 0.67 per pound? Ans. $ 672.01. 10. When 19 cords of wood are sold at i 5.63 per cord. what is the armount? Ans. 8 10.9D7. 11. A. merchant sold. 18 barrels of pork, each weighing 200 pounds, at 192 cents 5 mills a pound; what did he receive? Ans. 0 450. QUESTIONS. - At. 73. How do you arrange the rmultiplicand and multiplier in multiplication of United States Money? How multiply? Of wvhat denomnination is the product? How uslt it be pointed off? Repeat thle r,.I.in GRCT. vInI.] UNITED S'.PAES MO NEY. 12. W7hat cost 132 tons of hay at $ 12.I12,5 per ton? Ans. $ 1600.50. 13. A farmer sold one lot of land, containing 187 acres, at S 37.50 per acre; anotlher lot, containing 89 acres, at $ 137.37 pert acre; and another lot, containing 57 acres, at $ 89.29 per acre; what was the amount received obr the whole? Ans. $ 24,327.98. ART. 72oL DIvISION OF UNITEED STATES MIONEY. RULE. - WTith the sum given.for the dividend, proceed as in simple divtiswin; the result wrill be the q;ltient, in the lowest denomination contained in the dividend, which must be pointed off as in reduction of United Sates tonieey. (Art. 7O.) If the dividend consist of dollars only, and be either smaller than the divisor, or not divisible by it without a remnaiuder, reduce it to a lower denomiination by annexing two or three ciphers, as the case may require, and the quotient will be cents or mills accordingly. Proof. - The proof is the same as in simple division. EXAMPLES FOR PRACTICE, 1. If 59 yards of cloth cost 2. Purchased 68 ounces of $ 90.27, what will I yard indigo for $ 17. What did I cost Ans. 8 1.53. give per ounce Ans. $ 0.25. OPEtATION. OPERATION. Dividend. Divi('dend. $. Divisor 5 9) 9 0.2 7 (1.5 3 Quotient. Divisor 6 8) 1 7.0 0 (0.2 5 Quotieon. 59 136 340 I acre? Ans. $ 137.37. 4. When 19 yards of cloth are sold for $ 106(.97, what should be paid for 1 yard? Ans. $ 5.63. QusTIONS. - Art. 74. How do you arrange the dividend and divisor in division of United States money? How divide? Of what denolinnation is the quotient I How pointed oil? How do you proceed when the dividend is dollars only, and is either smaller than the divisor, or not divisible by it without a remainde, 7 Repeat the rule. 74 UNITED STATES MONEY. [SECT. VII1. 5. Gave $ 22.50 for 18 barrels of apples; what was paid for 1 barrel? For 5 barrels? For 10 barrels? Ans. $ 20 for all. 6. Bought 153 pounds of tea for, 90.27; what was it per pound? Ans. $ 0.59. 7. A merchant purchased a bale of cloth, containing 73 yards, for $ 414.64; what was the cost of 1 yard? Ans. $ 5.68. 8. If 126 pounds of butter cost $ 16.38, what will 1 pound cost? Ans. $0.13. 9. If 63 pounds of tea cost $ 58.59, what will 1 pound cost? Ans. $ 0.93. 10. If 76cwt. of beef cost $ 249.28, what will lcwt. cost? Ans. $ 3.28. 11. If 96,000 feet of boards cost $ 1120.32, what will a thousand feet cost? Ans. $ 11.67. 12. Sold 169 tons of timber for $ 790.92; what was received for 1 ton? Ans. $ 4.68. 13. When 369 tons of potash are sold for $ 48910.95, what is received for 1 ton? Ans. $ 132.55. 14. For 19 cords of wood I paid $ 109.25; what was paid for 1 cord? Ans. $ 5.75. PRACTICAL QUESTIONS BY ANALYSIS. ART. 75. ANALYSIS is an examination of a question by resolving it into its parts, in order to consider them separately, and thus render each step in the solution plain and intelligible. ART. 76. The price of one pound, yard, bushel, &c., being given, to find the price of any quantity. RULE. - Multiply the price by the quantity. Ex. 1. If 1 ton of hay cost $ 12, what will 29 tons cost? Ans. $ 348. ILLUSTRATION. - Since 1 ton costs $12, 29 tons will cost 29 times as much; $ 12 X 29 - $ 348. 2. If 1 bushel of salt cost 93 cents, what will 40 bushels cost? What will 97 bushels cost? Ans. $ 90.21. QUESTIONS. —Art. 76. The price of 1 pound, &c., being given, how do you fiad the price of any quantity? Give the reason for this rule. SECT. vISI.] UlNITED STATES MONEY. 75 3. If 1 bushel of apples cost $ 1.65, what will 5 bushels cost? What will IS bushels cost? Ans. $ 29.70. 4. If I ton of clay cost 8 0.67, what will 7 tons cost? What will 63 tons cost? Ans. 8 42.21. 5. When $ 7.83 are paid for lecwt. of sugar, what will 12cwt. cost? What will 93cwt. cost? Ans. $ 728.19. 6. When $ 0.09 are paid for lib. of beef, what will 121b. cost? What will 7601b. cost? Ans. 8 68.40. 7. A gentleman paid $ 38.37 for 1 acre of land; what was the cost of 20 acres. What would 144 acres cost? Ans. $ 5525.28. 8. Paid $ 6.83 for 1 barrel of flour; what was the value of 9 barrels? What must be paid for 108 barrels? Ans. $ 737.64. ART. 77. The price of any quantity and the quantity being given, to find the price of a unit of that quantity. RULE.- Divide the price by the quantity. 9. If 15 bushels of corn cost $ 10.35, what will 1 bushel cost? Ans. $ 0.69. ILLUSTRATION. - If 15 bushels cost $ 10.35, 1 bushel will cost as many cents as 15 is contained times in $ 10.35; $ 10.35 _ 15 =$ 0.69. 10. Bought 65 barrels of flour for $ 422.50, what cost one barrel? What cost 15 barrels? Ans. $ 97.50. 11. For 45 acres of land, a farmer paid $ 2025; what cost one acre? What 180 acres? Ans. $ 8100. 12. For 5 pairs of gloves, a lady paid $ 3.45; what cost 1 pair? What cost 11 pairs? Ans. $ 7.59. 13. If 11 tons of hay cost $ 214.50, what will 1 ton cost? What will 87 tons cost? Ans. $ 1696.50. 14. When $ 60 are paid for 8 dozen of arithmetics, what will 1 dozen cost? What will 87 dozen cost? Ans. $ 652.50. 15. Gave $ 5.58 for 9 bushels of potatoes; what will 1 bushel cost? What will 43 bushels cost? Ans. $ 26.66. 16. Bought 5 tons of hay for $ 85; what would 1 ton cost? What would 97 tons cost? Ans. $ 1649. QUESTIONS, —Art. 77. How do you find the price of I pound, &c., the price of any quantity and the quantity being given? What is the reason for this rule? U'NITED STATES MONEY, [SECT. V. 17. If J. Ladd will sell 20lb. of butter for 83.80, what sbhould Ihe chlarge for 591lb.? Ans. 43 Il.21. 18. Sold 27 acres of land for $ 472.50; what was the iprice of 1 acre? NVlatI shIould be given fo)r 12 acres? Ans. 2 10. 19. Paid $ 39.69 lbr 7 cords of wood? What will I cord cost? AXVhat wIill 57 cords cost? Ans. 8 023. 9 20. Paid $ 10.08 for 141-1). of pepper; vwh at was tIle pIc: of 1 pound? vWhat cost 3591b.? Ans. 25 1 3. 21. Paid,- 77.13 for 8$571.b. of rie9 what cost llb.? ~What cost 3591b.? A ns. 6 s2.3. 22. J. Johnson paid 187.53 for 9'87gtl. of mojiasses what cost Igal,? What cost 329a1..? Al. Ans 62.51. 23. For 47 bushels of salt, J. Ingersoll paid 6$,4 2860.32; what cost 1 bushel? What cost 39 bushels Ans. 8 24.84. AnT. 78. The price of any (quantity and the price of a unit of'that quantity being given, to find the quantity. RULE. Divide the w zole przce lj t/.e price of a Znit of te quantity required. 24. If I expend 3 150 for coal at D 6 per ton, how nmany tons can I purchalse? Ains. 25 tons. ILLUSTRATION.- Since I pay 66 for 1 ton, I can purchase as many tons with $ 150 as 6 6 is contained tirnes in $ I150; 8 t50 e — $ 6 = 25 therefore I can purchase 25 tons. 25. At $5 per ream, how many reams of paper can be bouglht for $ 175? Ans. 35 ream-s. 26. At 8 7.50 per barrel., how many barrels of follor can be obtained for $ 2 1.50 P A ns. 29 barrels. 27. At $W5s per ton, how many tonrs of iron can be purchased for' 48`,5? Ans. 65 tons. 28. At $ 4 per yard, how nmany yards of cloth can be bolught for i 1728? Ans. 432 yard's. 29. I-low many hundred weight of hay can be bought for $ 9.66, if 6 0.69 are paid for 1 hundred weight? Ans. 14 hluindred weil.olt. 20. If 6.6.51 are paid for flour at 6 7.39 per barrel, how many barrels can be bought? Ans. 9 barrels. 31. Paid t; 136.50 fbr wood, at, 3.25 per cord; how many cords did I buy A1ns. 42 cords. qUESTm'INS. - Art. 78. How do you find the quantity, the price of one pound, &c., being given? Give the reason for die vtle Sr;CT. VIII ] UNITED STATES MIIONEY. BILLS. ART. 79. A BILL is a paper, given by merchants, containing a statement of goods sold, and their prices. What is the cost of each article in, and the amount of, each of the following bills? (1.) New York, May 20, 1842. Dr. John Smith, Bought of Somes & Gridley, 82 gals. Temperance Wine, at $ 0.75 89 " Port do. ".92 24 pairs Silk Gloves, C".50 $ 155.38. Received payment, Sonmes & Gridley. (2.) Newburyport, March 7, 1842. Mr. Levi Webster, Bought of James Frankland, 6 lbs. Chocolate, at $ 0.18 12 " Flour, ".20 6 pairs Shoes, 6 1.80 30 lbs. Candles,.26 $ 22.08. Received payment, James Franklando (3.) Baltimore, July 19, 1842. Mr. John Kimball, Bought of Simon Grey, 14 oz. Gum Camphor, at $0.63 12 " Laudanum, 4.88 23 " Gum Elastic,.62 16 " Emetic Tartar, " 1.27 17 " Cantharides, 2.25 $ 92.21. Received payment, Simon Grey, by Enoch Osgood. QUESTIoN. - Art. 79. What it a bill in mercantile transactions? 7* 78 UNTITED STATES MOiNEY. [SECT. Vill. (4.) Haverhill, March 19, 1842. Mr. William Greenleaf, Bought of Moses Atwood, 86 Shovels at' 0.50 90 Spades, 66.86 IS Ploughs, 11.00 23 [andsaws, " 3.50 14 H[ammers,';.62 12 Aill-saws, 6 12.12 46 cwt. Iron,'" 2.00 _'3 1105.02 (5.) Lowell, June 5, 1842 Mr. Aim-os Dow, Bought of Lord & Greenleaf,; 37 Chests Green Tea., at 23.75 42' 66 Black do. t" 7.50 43 Casks Wine, 66 99.00 12 Crates Liverpool Ware," 175.00 19bbl. Genesee Flour, " 7.00 23bu. Rye 6 o1.52 _ 8138.71 (G.) Salem, May 13, 1842 MIr. Noah vWebster, Bought of Ayer, Fitts, & Co., 80 pairs HTose, at $.20 67 "" Boots, 3.00 19 Shoes, 66 1.03 23 Gloves,.75 $ 184.77. (7.) Baltimore, June 30, 18f2 Mr. Samuel Osgood, Bought of Stephen Barnwell, 27 Young Readers, at $ 0.20 l0 Greek Lexicons, 6 3.90 7 Ainsworth's Dictionaries eS 4. 75 19 Folio Bibles, 2.93 20 Testaments,.37 4 7, sECT. lx.] QUESTIONS INVOLVING FRACTIONS. 1 I~. QUESTIONS INVOLVING FRACTIOCNS. ART. -o& If a unit or individual thing is divided into parts, one of the parts is called a Fraction of the number or thzing divided. JHence FRACTloNS are parts of whole nuanmbers. IrLUSTRATAONS. - 1. If any number or thing is divided into two equal parts, one of the parts is called one half, and is written thus~ -!. 2. If any number or thing is divided into thAree equal parts, on^e of the parts is called one third (4); two of the parts are called two thirds (d-)o 3. When any numrber or thing is divided into four equal parts, o0ne of the parts is called one fourth (I) three of the parts, three fourths (s-). 4. If any number or thing is divided into five equal parts, one of the parts is called one Jifth ( 3); two parts, two fifihs (~); three parts, three fifths ( -); and four parts, four fifths ( ). 5. When any number or thing is divided into six equal parts, vwhat is one of the parts called? Two parts? Five parts? 6. If a number or thing is divided into 7 equal parts~, what is 1 part called? 2 parts? parts? 4 parts? 5parts? 6 parts? 7. If a nu:mnber or thing is divided into 9 equal parts, what is 1 part called? 2 parts? 4 parts? 5 parts? 7 parts? 8 parts? 8. What is hal of4? Of f? Of 0? Of 28? Of 32? 9. What is 1 tird of? Of 12 Of 15? Of 27 Of S0? Of 36?Of 6O? 10. What is I fourth of 8 Of 13? Of 20? Of 24? Of 40? Of 48? Of 100 11. What is lfifh of 107? Of 25? Of 30? Of 35.' Of 45 f O 5f550? Of 5 f8 65? 12. Whatt is I sixth of 12 Of 1? Of 3? Of 42? Of 60? O f 72 Of 90 O,U r.s'riowas. - Art. 80. What is a fraction? What is meant by one half of any number or thing?. How is it writtein? What is neaiit by one thlird, and now is it written? Wh tt by one fourth, arid how written? \hat by one fifth, and how written 1 What by fuiir fifths, and how written? What by five sixths, and how written I How do you find one half of any number? [tow one third 1 1How one fourth? &c. How many halves matke a whole onle I o-.wo many thirds? Hlow many fburtis 1? low lany fifthIs 1 Hlow many evihtths'?-How m.a ny fifteenthe? Ie,. 80 QrUESTIONS IINVOLVING FRACTIONS. (SECT. IX. 13. How many fourths in 1 apple? 14. How many fourths in 2 apples? In 3 apples? In 8 apples? In 16 apples? 15. How many fifths in 1 barrel of flour? In 3 barrels? — In 5 barrels? In 7 barrels? In 9 barrels? 16. flow many sixths in 1 bushel of wheat? In 4 bushels? In 7 bushels? In 9 bushels? In 12 bushels? 17. James owns 3 fifths of a kite, and his brother Thomas the remainder. HIow many fifths does Thomas own? ILLUSTRATION.- Since there are 5 fifths in the kite, if James owns 3 fifths, there will remain for Thomas 5 fifths (5) less 3 fifths (a) = 2 fifths. Ans. 2 fifths. 18. From a load of hay I sold 4 sevenths; how many sevenths remain? 19. Bought a hogshead of molasses, and 4 fifths of it leaked out; how much remained? 20. John Jones found a large sum of money; he gave 5 eighths of it to the poor of the parish; how much did he reserve for himself? 21. If the captain and crew have 3 elevenths of the income of a ship, what part remains for the owners? 22. John Smith gave 2 ninths of his farm to his son, 3 ninths to his daughter, and the remainder to his wife; how many ninths did his wife receive? ILLUSTRATION. - If he gave 2 ninths (i) to his son, and 3 ninths (]) to his daughter, he gave them both - + -= -; and since there are 9 ninths (9) in the farm, he must have given his wife 9 — - W-. Ans. a. 23. In a certain school -r of the pupils study grammar, -1 study arithmetic, A geography, and the remainder philosophy. What part of the school study philosophy? 24. J. Dow spends + of his time in reading, 4 in labor, and + in visiting. How large a portion of his time remains for eating and sleeping? 25. Ira Thomas gave I of his money for apples, and 4 for pears; what part has he left? 26. If a yard of cloth. cost $ 8, what costs J of a yard? What cost i of a yard? ILLUSTRATION. - If 1 yard cost $ 8, 4 of a yard will cost v of $ 8 = 2; and if a of a yard cost $3 2, J will cost three times as much; 3 times $ 2 $ 6. Ans. $ 6. scrT. Ix.] QUESTIONS INVOLVING FRACTIONS. 27. fI an acre of land cost 824, what will -- of an acre cost? Wl.aUt wVill cost?:23. Wtlhen 96 cents are paid for a bushel of rye, what cost * of a bustlel? 29P. Paid 0 630 for a piano; what arle a of it wxorth? 0. S.. nkins has a firne farm, val1uld at 0 8767. AF of this farm he wills to hlis son, A to lis dauplghter, and the renainud;ier eo'his wife; what i)art docs. she receive, and -what is tihe val-ue of her share? 311. if of a barrel of flour cost 2 dollars, what cost -, of a LI,IJSTRATIOX, -— f 5 cost 2 dollars, 5 will cost 4 times 2 dollars -- *3 8, Ans. 0 8. $321. If of an acre ofrland cost 0t24, what will Z of an acr-e cs't? 3:. f i- of a ho-ishead or molasses cost 0 I i what will a ogshtead cost? 3$; If ~- oaf a y-ard of broadcloth cost 03 2.70, what will -~ cost? 33. Bouht I o a el or poMrk or 0 1.43, what must be paid for o-t f a b' re, - -36. If - of acre of land cost 0 21, what cost 1 of an acre? V hat cost an acre? vWhat cost 10 acres? ILLUSTRATIOT. -- -i- cost 8 21, } will cost 1- of 0 21, alnd - olf $2G is 3, and - will cost 8 times 3o- to' 24,and 10 acres will cost O0 times 0424 -- 0 240. Ans. 0 240. 327, If -T of a hogshead of sugcar cost $ 18, what costs 1 Ibogshead, WVhat cost 4 hogshetads? 38. if 5 of a barrel of apples cost S 1.50, what costs a barrel? N\1ha cost 10 barrels? 29. Whuen 49 are paidl for xiT of a ton of potash, what must be paid for 2 tons? 40. WheLn a lady pays 18 cents for 3 of a yard of ribbon, what shoutld she pay for 17 yards? Ans. 0 13.26. 41 P ail G for of' a barrel of flour; what was tihe value of 6 b-arrels? t2. 6Snamuel Jones 1boughl t of L. Clark ~ c of a ton of coal for 04.26; v t at would(l be the priceof 1 1 tons? Ans. $ 109.34. 43.,Boutrght a cwt. of sulgar, and sold 1A-L of it for x 2.4-; waiat is tte v;llue of the remainder at the same price? 4-. Mly second-best chlaise is valued at $ 64.47, whiclh is T the value of mry best chaise. What is the vaIlue of botll A ns 150AI347 82 QUESTIONS INVOLVING FRACTIONS. ISECT. I1 45. How many half-barrels of flour are there in 2 and a half (21) barrels? ILLUSTRATION. — Since 1 barrel contains 2 halves, 2 barrels will contain 2 times 2 - 4 halves, and the I half added makles 5 halves. Ans. 2. 46. How many half-bushels in 4~ bushels of oats? In 56 bushels? In 7~ bushels? In 9~ bushels? 47. How many thirds of a pound in 2~ pounds? In 4} pounds? In 72 pounds? In 12% pounds? 48. FHow many fifths of a gallon in 3- gallons? In 72 gallons? In S8 gallons? In 113 gallons? 49. IHow many eighths of a dollar in 2- dollars? In 43 dollars? In 7% dollars? In 97 dollars? In 12% dollars? 50. HIow many tenths of an ounce in 4r3, ounces? In 5-7ounces? In 8rd ounces? In 10-9 ounces? 51. Flow many barrels of wine in 6 half (%) barrels? ILLUSTRATION. -Since it takes 2 halves to make one whole one, there will be as many whole barrels in 6 halves (6) as 2 is contained times in 6. 2 is contained in 6, 3 times. Ans. 3 barrels. 52. Nfow many firkins of butter in 4 firkins? In 2o firkins? 53. I-low many whole numbers in 1O? In JA5? In 2-5-? 54. flow many whole numbers in 6? In —? In _ In 48? 55. If a skein of silk is worth 31 cents, what are 6 skeins worth? ILLUsTRATION. - If 1 skein is worth 3g cents, 6 skeins are worth 6 times as much; 6 times 3% are equal to 6 times 3 and 6 times; 6 times 3 18; 6 times ~ — _ 3; 18 + 3 = 21. Ans. 21 cents. 56. Bought one pair of boots for $ 62; what must I pay for 4 pairs? For 8 pairs? For 10 pairs? For 12 pairs? 57. Paid 12% cents for one pound of cloves; what will 6 pounds cost? 10 pounds? 12 pounds? 58. Gave 2% cents for one pound of raisins; what cost 2 pounds? 6pounds 86pounds? 10 pounds? pounds? 10 pounds? 59. If one pound of butter is worth 12 cents, what are 4% pounds worth? ILLUrSTRATION. -— f 1 pound is worth 12 cents, 4. pounds are worth 4} times as much; 4~ times 12 cents are equal to 4 SECT. ix.] QUESTIONS INVOLVING FRACTIONS. 83 times 12 and 4 of 12; 4 times 12 are 48, and ~ of 12 is 6; 48 cents and 6 cents are 54 cents. Ans. 54 cents. 60. If 1 ton of hay is worth $ 10, what are 8I tons worth? 61. When lard is sold for 9 cents per pound, what must be paid for 73 pounds? For 8I pounds? For 94 pounds? 62. Bought 1 pound of coffee at 16 cents; what will 51 pounds cost? 3 pounds? 51 pounds? 6- pounds? 63. If 1 yard of cloth is worth 20 cents, what is the value of 164 yards? 124 yards? 81 yards? 11 yards? 64. If 1 bushels of corn cost $ 1.20, what will 1 bushel cost? ILLUSTRATION. -1: bushels -4 bushels. Now, if 4 cost $1.20, A will cost i of $ 1.20 - $ 0.40; and -2 or a whole bushel will cost 2 times $ 0.40 - $ 0.80. Ans. $ 0.80. 65. If 2- pounds of coffee cost 60 cents, what will 1 pound cost? ILLUSTRATION. 22- pounds — t- pounds. If -,2 cost 60 cents, 1 will cost -t of 60 cents = 5 cents; and 5, or a pound, will cost 5 times 5 cen.ts 25 cents. Ans. $ 0.25. 66. How many times will 60 contain 2i? 67. Paid $ 54 for 75 barrels of oil; what cost 1 barrel? Ans. $ 7. 68. How many times is 74- contained in 54? 69. How many cords of wood, at $ 54 per cord, can be bought for $ 66? 70. How many times will 66 contain 5? 71. Gave $ 40 for 642 yards of broadcloth; what cost 1 yard 72. How many times is 6- contained in 40? 73. The distance between two places is 110 rods. I wish tG divide this distance into spaces of 5~ rods each. Required the number of spaces. 74. flow many times will 110 contain 54? 75. If 164 hundred weight of'hay cost $ 33, what costs I hun dred weight? What cost 9 hundred weight? Ans. $ 18. 76. flow many times is 164 contained in 33? 77. If 44 pounds of sugar cost 46 cents, what will 2 pounds cost? What will 7 pounds cost? 78. How many times will 46 contain 44? 79. Paid $ 90 for 74 tons of coal; what cost 20 tons? Ans. 8 240. 80. How many times is 74 contained in 90? tfijL~Z4 aREDUCTIONJ. Zs~cT. X. X. REDUCTION. ART.. g RE:DUCTIo N is changing numbers, either simp,e olr coryll1poundl froml one denomination to another, without alteriltnr tltcir valIues. It is of' two kinds, Reduction Descending, and Reduction Ascendi tng. B:letvtdn rtion Descending is cilancgin numbers of a higler de. nomnittion to a lower denomination; as pounds to shillings, de. It is performed by multiplicatio1n. henduction Ascending is chalngring numbers of a lower denomlination to a higher denoilnation n; as farthings to pence, &c. It is the -reverse of RJeduction Descending, and is performed by division. ART. 8S. A SIMIPLE numlber is a number which expresses things of the sanze kind or denomination; thus, 126. 90 apples, are simple numbers,.ART. c i$ ACOMP'OUNt D number4] a number which ex. presses things of di/ferent kinds or denominations, taken collectively; thus, ~12.. 1Ss. 9d. is a compound number. ENGLISH BIONIEY. ART. SI. English Money is the currency of England. 4 Farthings (qr.) make I Penny, raarked d. 12 Pence 1 Shillinig, " s. 20 Silillings I" Pound, ". 21 Shillings sterling 1 Gulinea, 6 gin. 20 Shillings " " 1 Sovereign,' soy. 28 Shillings tD. Eg. currency I Guinea, " guin. d. qr. ~. I; - 19 - 48 1 20 2 240 - ~60 QueTsrloIs. -. Art. 81. What is reduction? HIow many kinds of reduction? UV/-:ihat are they? What is reduction descending 7 WVhat is reduction ascend inA?- Art. 82. What is a simple nulber? - Art. 83. What is a compound number? - Art. 84. What is English money? Repeat the tRble, SECT. X.] REDUCTION. 65 IMENTAL EXERCISES. 1. HTow many farthings in 3 pence? In 4 pence? In 6 pence? In pence? In9pence? In 1 pence? 2. How Nmany pence in 2 shillings? In 3 shillings? In 4 shillings? In 5 shillings? In 6 shillings? In 10 shillings? 3. IHow many shillings in 2 pounds? In 7 pounds? In 10 pounds? In 12 pounds?. How many pence in 8 farthings? in 16 farthings? In 24 farthings? in 80 farthings? In 144 farthings? 5. How many shillings in 36 pence? In 60 pence? In 9b pence? G. H-tow many pounds in 40 shillings? In 80 shillings? In 100 shillings? EXERCISES FOR THIE SLATE. ART. Sg. To reduce higher denominations to lower. 1. How many farthings in 17X~. Ss. 9d. 3qr. P OP A TIOI.In this question, we multiply the i 7I. 8s. 9d. 3ir. q 17V. by 20, becaus 20 shillirngs 20 make I pound, and to this product 34 8 shilliwnes vswe add the 8 shillings in the question. We then multiply by 12, because 12 pence make 1 shilling, and 4 1 8 5 pence. to the product we add the 9d. in the question. Again, we multiply by 4, because 4 farthings make i Ans. 1 6 7 4 3 farthings. penny, and to this product we add the 3qr. in. the question, and we fi:d the answer to be 16743 farthings. RuLE.- - lultiply the highest denomination given by the number required of the ne xt lower denomination to make one in the denomination multiplied, aed add to the product thus obtained the corresponding denomination of the multiplicand. Proceed in this way, till the reduction as brought to the denomination required by the question. A.&RT. G To reduce lower denominations to higher. 2. I-ow many pounds in 16743 farthings? QUF.STIOrS.-Art. 85. How do you reduce pounds to shillings? Why multiply by 20? How do you reduce shillings to pence? Why 1 Pence to farthings? Why.? Guineas to shillings? What is the general rule for reduction deseending 7 36 REDUCTION. [SECT. x. OPERATION. We first divide by 4, because 4 far4 ) 1 6 7 4 3 qr. things make 1 penny, and the result is 12 ) 4 1 8 5 d. Oqr. 4185 pence, and tile remainder, 3, is farthings. We then divide by 12, be2 0 ) 34 8 s. 9d..cause 12 pence make 1 shillinrr, and the result is 348 shillings, and the 9 re. I 7~. 8s. maining is pence. Lastly, we divide by Ans. 1'7.1. Ss. 9d. 3qr. 20, because 20 shillings make 1 pound, Anse. 17~~. 8s. 9 d. 3E % * and the result is 17~. 8s. Therefore, by annexing all the remainders to the last quotient, we find the answer to be 17.L. 8s. 9d. 3qr. RULE. - Divide the lowest denomination given by the number which zt takes of that denomination to make one of the next higher; and so proceed, until it is b)rougrht to the denomination required. Any remainders occurring in the successive divisions will be of the same denominations with the dividends to which they respectively belong. NOTE. - In the following exercises in reduction, for the slate, many of the questions in reduction ascending are the anszwers for the questions in reduction descending, and conversely. 3. In 9~. 18s. 7d. how many pence? 4. In 2383d. how many pounds, &c.? 5. How many farthings in 14~. Ils. 5d. 2qr.? 6. How many pounds in 13990qr.? TROY WEIGHT. ART. S7. Troy Weight is the weight used in weighing gold, silver, and jewels. TABLE. 24 Grains (gr.) make 1 Pennyweight, marked dwt 90 Pennyweights " 1 Ounce, o. 12 Ounces " I Pound, " lb. dwt. gr. on. I 24 lb. 1 = 0 - 480 l1 12 -'40 - 5760 QUESTIONS. —Art. 86. How do you reduce farthings to pence? Why divide by 4? How.do you reduce pence to shillings?'Why Shillings to pounds? Why? What is the general rule for reduction ascending - Art. 87. For what is troy weight used? Repeat the table. SECT. X.1 REDUCTION. 87 No'rE. -The original of all weights used in England was a grain or corn of wheat, gathlered out of the middle of the ear; and being well dried, 32 of them were to make one pennyweight, 20 pennyweiglits one ounce, and 12 ounces one pound. But in later times, it was thought sufficient to divide the same pennyweighit into 24 equlal parts, still called grains, being-the least weight now in common use, from which the rest are computed. MENTAL EXERCISES. 1. How many grains in 2dwt.? In 4dwt.? In 10dwt.? 2. How many pennyweights in 4oz.? In 6oz.? In 200oz.? 3. How many ounces in 21b.? In 51b.? In 10lb.? In 50ib.? 4. How many pennyweights in 48gr.? In 96gr.? In 144gr.? 5. I-ow many ounces in 40dwt.? In 120dwt.? 3:n 720dwt.? 6. How many pounds in 24oz.? In 60oz.? In 12Ooz.? In 480oz.? In 96oz.? In l80oz.? In 132oz.? In 144oz.? 7. How many grains in 2oz.? In 4oz.? In 5oz.? In 10oz.? In 20oz.? In 30oz.? In 40oz.? In 50oz.? 8. How many pennyweights in llb.? In 21b.? In4lb. In 51b.? In'7lb.? In 101b.? In 12lb.? In 201b.? EXERCISES FOR THE SLATE. 1. How many grains in 721b. 2. In 419887 grains, how 0)oz. 15dwt. 7gr.? many pounds? OPERATION. OPERATION. 721lb. 10oz. 15dwt. 7gr. 24)4 1 9 887 gr. 1 2 220)17 495 dwt. 7gr. 8 7 4 ounces. 1 2) 8 7 4 oz. 15dwt. 721 b. lOoz. 1 7 49 5 pennyweights. 2 4 Ans. 721b. 10oz. 15dwt. 7gr. 69987 34990 Ans. 4 1 98 7 grains. Qu.iSTIONS.- What was the original of all weights in England? How many of these grains did it take to make a pennyweight? How Imxany grains in a pennyweight now? ow do you reduce poutnds to grains? Give the reason of the operation. Hiow do you reduce grains to pounds? Give the reawon of the operation. 8J 3 RIEDUCTiOIO. (sc 3. HTow many grains in 76dwt. l2gr.? 4. HIow many pennyweightts in 1836gr.? 5. Inl 761). 5oz. how many grains? G In 440160 grains how many pounds? 7. H-Iow many pennyweiglits in 1441b. 9oz.? 3. I-low many pounds in 34740dwt.? 9. IHow many pounds in 17895gr.? 10. In 31b. loz. 5dwt. 15gr. how many grains? 1l. A valuable gem weighing 2oz. 1Sdwt. 192gr. was sold for $ 1.37 per grain; what was the sum paid? Ans. $ 1923.48. APOTH-ECARIES' WVEIGHT. ART. SS. Apothecaries' Weight is used in mixing medicines. But medicines are usually bought and sold by avoirdupois weight, TABLE. 20 Crains (gT.) make I Scruple, marked se. or "D 3 Scruples " 1 I)ram, " dr. or 3 S Drarns " 1 Ounce, ~ " oz. or 12 Ounces I' L Pound, " lb. or b dr, SC20 a.~t ~ 3. 60 lb. I 1 8 24 480 1 12 96 288 5760 IMENTAL EXERCISES. 1. in 40 grains how many scruples? In 6Ogr.? In 120gr.? In 140gr.? In 200gr.? In 240gr.? In 260gr.? In 300(r.? 2. in 5 scrulples how many grains? In 10sc.? In 40sc.? In 100se.? In Ssc.? In l2sc.? In 20sc.? In 30se.? 3. In 3 dramns how many scruples? In lOdr.? In 17dr.? in 20(r.? In 30dr.? In 40dr.? In 50dr.? In 60dro? 4. Ilow many pounds in 48oz.? In 96oz.? In 144oz.? In 17'2Soz.? I 32oz.? In G64oz.? In S4oz.? 5. low many ounces in 24dr.? In 6-4dr.? In 96dr.? In 144dr.? In l2Odr.? QUES TIOs. — Art. 813. For what is apothecaries' weight used? By what weight are medicines usually bought and sold R Repeat the table. SECT. X.] REDUCTION. 04 EXERCISES FOR} TIHE SLATE. 1. In 401b. 8oz. 5dr. Isc. 2. How many pounds in 7gr., how many grains? 234567 grains? OPERATION. OPERATION. 4 0 lb. Soz. 5dr. Isc. 7gr. 20)234 5 67 gr. 1 2 3) 1 1 7 2 8sc. 7gr. 4 8 8 ounces. 8) 3 9 0 9 dr. Isc. 8 -390O 9 drams. 1 2)4 8 oz. 5dr. 3 9 0 9 drams. 3 40 lb. 8oz. 1 1 7 2 8 scrulples. Ans. 401b. 8oz. 5dr. lsc. 7gr. Ans. 2 3 4 5 6 7 grains. 3. How many scruples in 761b.? 4. hIow many pounds in 21888D? 5. How many grains in 1441b.? 6. IHow many pounds in 829440gr.? 7. In 12lb 85 33 1D 18gr. how many grains? 8. In 73178 grains how many pounds? 9. I-ow many doses are there in 7S 65 2D of tartar emetic, admitting 20 grains for each dose? Ans. 188. AVOIRDUPOIS WEIGHT. ART. S9. Avoirdupois Weight is used in weighing almost every kind of goods, and all metals except gold and silver. TABLE. 16 Drams (dr.) make 1 Ounce, marked oz 16 Ounces " 1 Pound, " lb. 28 Poilnds 1 Quarter,' " qr. 4 Quarters " 1 IHtundred Weight, "' cwt. 20 Hundred Weight " 1 Ton, " T. oz. dr. lb. 1 = 16 qr. 1 - 16 256 cwt. I A 28 448 7168 T. 1 4 112 - 1792 - 28672 1 20 80 = 2240 35840 ~ 573440 QUESTIONS. - How do you reduce pounde to grains? What is the reason for the operation? How do you reduce grains to pounds? Give the reason of the operation. - Art. 89. For what is avoirdupois weight used? Recite the tahb. 8* (90 REDUCTION. [SECT. X No'Tr. - Bi a late law of Massachnsetts, the cwt. contains 100lb., instead of I'12b., and 251b. are considered a (liiarler of a cwt. in moost of thle UIiited Set;tes, except at the custoll-houses, where gross weight is used in collecting duties. MENTAL EXERCISES, l. 1-ow many drams in 3oz.? In 7oz.? In lOoz.? In 12oz.. [low many ounces in 1Olb.? In 151b.? In 121b. > In l,01[b. 3. -low mrany pounds in 2 quarters? In 3r.? In 20(qr.? 4. low many quarters in 10cwt.? In 16cwt.? In 17cwt.. 5. How many tons in 0cwt.? In 1OOcwt.? In 600cwt.? 5.,low many hundred weight in 16qro? in 4Sqr.? In 96(jr? 7. HI-ow many quarters in 56 pounds? In 1401b.? In il61b.? EXERCISES FOR TIlE SLATE. 1. Iuow many pounds in 176T. 2o In 3962431b., how!7cwt. 3qr. 15lb.? many tons? OPERATION. OPERATIO,. 1 7 6 T. 17cwt. 3qr. 15I1b. 2 8) 3 9 6 2 4 3 lb. 2 0 4 4~) %14151 qr. 15lb. 3 5 3 7 hundred weight. 2 ) 3 4 3 7 cwt. 3q.'20) 3 7 cwt. 3qr. i 4 1 5 1 quarters 7 6 T. 17ct 2 8 Ans. 176T. 17cwt. 3qr. 151b 132 3 28303 Ans. 396243 pouais. 3. In G6T. 19cwt. Oqr. Bl b. 11oz. 5dr. how many drams? 4. In 9722549 drams how many tons? 5. In 679cwt. how many pounds? 6. In 760481b. how many cwt.? 7. WVhat cost l7cwt. 3qr. 181b. of beef, at 7 cents per pound? Ans. 0 140.42. 8. What cost 48T. 17cwt. of lead, at 8 cents per I)pound? A:ns. $ 8753.9P, QFTsTrioNs.-How many pounds are now allowed for a cwt., and how many tfor a quarter of a cwt., in most of the United States, in buying and sellin!s articles by weight H [ow riany at the customi-houses? How do you reduce ton= to dramls? Give the reason for the operation. How do you reduce iramu to tob 7? What is the reason i r the opera.ion? OACT. X.] REDUCTION. 92 CLOTH MEASURE. ART. D 0. Cloth AMeasure is used in measuring cloth, rib. bons) lace, and other articles sold by the yard or ell. TABLE. 22 inches (in.). make I NaiI9 marked na, 4 Nails 6 1 Quarter of a yard, 6" qr. 4 Qlarters 66 1 Yalrd, yld 3 Qua;rters 1 I Eli Flemish, " E. F. 5 Quarters 1 Ell English, E. E. NOTE. — The Ell French is not in use. ns. in. E.F. i = 4 9 yd 1 3n 12 27 E.. 1i, 1 — I 4 16 - 36 1 n g 1i5 2 0 45 MIiENTAL EXERCISES. 1. In 2 quarters how many nails In I qr.? In 8qr.? In O0qr.? In'25qr.? In 30qr.? In 40qr.? 2. In 3 yards how many quarters? In 7yd.? In Syd.? In 14yd.? In 19yd.? In 100yd.? In 200yd.? 3. I-low many quarters in 8 nails? In 20na.? In 4Sna.? 4. How many yards in 20 quarters? In q I40 qr.? 00qr.? XE ICISES FOR THIE SLATEo T. ITow many nails in 47yd. 2. In 765 nails how many 3qr. 1 na.? yards? OPUIATTON. OPMRATIO-N. 4 7 yd. 3qr. 1 na. 4) 7 6 5 na. 4) 1 9 1 qr. na. 41 qAartet Ans. 4 7 yd. 3qr. Ina. Ans. 7 65 nails. QurTlsro s. -Art. 90. For what is cloth measure used? Repeat th abie. Is tile eli "renich now in use? How do you reduce yards to natlus? (Give the reason for the operation. How do you reduce nails to yards? VWhat is ULei reason for the operation? 92 REDUCTION. [SECT. x. 3. In 144yd. 3qr. how many quarters? 4. In 579 quarters how many yards? 5. In 17 E. E. 4qr. 3na. how many nails? 6. In 359 nails how many ells English'? 7. In 126yd. Oqr. 3na. how many nails? 8. In 2019 nails how many yards? 9. What cost 49yd. 3qr. of cloth, at $ 2.17 per quarter of a yard? Ans. $ 431.83. 10. What cost 144yd. lqr. 3na. of cloth, at 25 cents per nail? Ans. $ 577.75. LONG MEASURE. ART. 91. Long Measure is used in measuring distances, or where length is required without regard to breadth or depth. TABLE. 12 Tnches (in.) make 1 Foot, marked ft. 3 Feet " 1 Yard, " yd. 5A Yards, or 16A Feet, " 1 Rod, or Pole, " rd. 40 Rods " 1 Furlong, I fur 8 Furlongs, or 320 Rods, " 1 Mile, m. 3 Miles " 1 League, " lea. 69) Miles (nearly) " 1 Degree, " deg. or~ 360 Degrees " 1 Circle of the Earth. ft. in. yd. 1 12 rd. i 3 = 36 fur. 1 5. -~ 16 198 m. 1 40 - 220 - 660 79'20 1 8 == 320 - 1760 5 5280 63360 MENTAL EXERCISES. 1. How many inches in 4 feet? In lOft.? In 12ft.? In 2Oft.? 2. flow many feet in 2 yards? In 5yd.? In 20yd.? In ISyd.? 3. Elow many rods in 2 furlongs? In Sfur.? In lfur.? In 30fur.? In 100fur.? In 200fur.? In 400fur.? 4. -low many leagues in 9 miles? In 21m.? In 81m.? In 144m.? In 40m.? In 50m.? In 80m.? 5. I-low many furlongs in 120 rods? In 360rd.? In 1440rd.? 6. I-ow many yards in 99 feet? In 66ft.? In 144ft.? 7. Ilow many feet in 108 inches? In 144in.? In 172Sin.? QUESTIONS. -- Art, 91. For what is long measure used I Repeat the table. SECT. X.] REDUCTION. 93 EXERCISES FOR THE SLATE. 1. In 87dec. 56m. 7fur. 37rd. 12ft. 9in. how many inches? OPERATION. 8 7 deg. 5Gm. 7fur. 37rd. i2ft. 9in. ~69 2. In 386717319 inches how many 7 8 9 degrees? 5 2 7 OPERATION. 43_ _ 12)3867 17319inches. 6 102' miles. I6 ) 32226443ft. 3in. 8 2 2 48827 fur. 33 )64452886 [12ft.6in. A40 40) 1953117rd. 25. 2 1 9 5S l3 1 7 rods. 8) 48827 fur. 37rd. l64 I 1-7 1 8 7 04 9 L) 6 1 0 3m. 3ur. 2 2 1953 118 2 976558 139) 12206 32226442, f 8 7 deg. 113 — 2= * 2 [56m. 4fur. 38 6 7 17 3 1 9 in. Ans. 8/deg. 56m. 4fur. 37rd 12ft. 6in. 3 3 87 56 7 37 12 9 Ans. NoT' - To multiply by w, e take! of the mfultiplicand.-To divide by 16., we first reduce both the divisor and dividend to i1halves, and tlien divide, and tie remainder beintg 25 halw:feet, we take h11a' of it for thei trte remainder = 1211. 6in. We adopt the same princiilo in dividing by 6!}~; the reimainder being 113 hlalf-miles, we dividle them by 2, and tlhe quotient is 56 miles 4 urtlongs. fly adding. the 3 incles to the 6 inches, and the 3 furlongs to the 4 furlongs, we obtain the true answer. 3. In 47 miles how many feet? 4. In 248160 feet how rnmany miles? 5. In 78deg. 50m. 7fur. 308rd. 5yd. 2ft. 10in. how many inches? 6. Hlow many degrees in 346704154 inches? Qtv:sTIOs.s- -flow do you reduce degrees to inches? -Give the reason of the operatior. I-low do you reduce inclies to degrees? What is the reason for the oper.ation? How (0o you miultiply by 2? [How do you divitle by 1bi a;el find the true renlainder? 7 flow do you obtain the true answer in exampleas eo Ahis kind 1 94 REDUCTION. [SECT. X. SURVEYORS' MEASURE. ART. 92. This measure is used by surveyors in measuring land, roads, &c. TABLE. 7?09 Inches (in.) make 1 Link, markced Ii. 25 Links 1I Pole, " p 100 Links, 4 Poles, or 66 Feet, i 1 Chain, " cha. 10 Chains " 1 Furlong, " furo 8 Furlongs, or 80 Chains, " 1 Mile, m. i, in. p. 1 ~ 792 hlla. 1 25 - 198 fur. I 4 100 - 792 m. 1 10 -40 -- 1000 792(0 1 = 8 80 -- 320 8000 8 63360 NOTE. - An engineer's chain is usually 100 feet in length, containing 120 links, each 10 inches long. MENTAL EXERCISES. 1. In 2 poles how many links? In 4 poles? In 7 poles? In 20 poles? In 30 poles? In 50 poles? In 100 poles? 2. In 5 chains how many links? In 8cha.? In 1Ocha.? In 15cha.? In 20cha.? In 24cha.? In 36cha.? 3. -In 4 furlongs how many chains? In 6fur.? In 1 fur.? 4. I-ow many poles in 50 links? In 751i.? In 1251i.? In 2001i.? In 2251i.? In 3001i.? In 4001i.? EXERCISES FOR THE SLATE. 1. I-low many links in 7m. 2. In 61630 links how many 5fur. 6cha. 301i.? miles. OPERATION. OPERATION. 7m. 5fur. 6cha. 301i. 1 0 0) 6 1 6 3 0 li. 8 0) 6 16 cha. 301i. 6 1 furlongs. C.1 furlongs.8) 6 1 fur. 6cha. 10 eX m. 5fur. 6 1 6 chains. 1 0 0 Ans. 7m. 5fur. 6cha. 301i. 6 1 6 3 0 links, Ans. QUESTIONS.- Art. 92. For what is surveyors' measure used 1 Recite the table. How do you reduce miles to links? What is the reason for the oper ation I How do you reduce inches to chains I To miles? Give the reason of the operationm SRCT. X.) REDUCTION. 95 3. How many miles in 4386 chains? 4. In 54m. 66cha. how many chains? 5. In 75m. 49cha. how many poles? 6. I-How many miles in 24196 poles? 7. How many links in 7rn. 4fur. 30rd.? 8. H-low many miles in 60750 links? SQUARE MEASURE. ART. 93. Square Measure is used in measuring surfaces of all kinds, such as land, flooring, plastering, &c., where length and breadth only are considered. TABLE. 144 Square inches make 1 Square foot, marked ft. 9 Square feet " 1 Square yard,. " yd. 304 Square yards " 1 Square rod or pole, " p. 2724 Square feet " I Square rod or pole, " p 40 Square rods or poles " 1 Rood, " R. 4 Roods, or 160 Poles, " 1 Acre,'" A. 640 Acres " 1 Square mile, "S. M At. in. yd. 1 144 p. 1~ 9- 1296 R. i= 304 - 272i - 39204 A. 1 40 1210 -- 10890- 1568160 S.. 1 = 4- 160 = 4840- 43560 = 6272640 1 640 - 2560 - 102400 = — 3097600 = 27878400 = 4014489600 3ft. -. lyd. A square is a figure having four equal Square sides, and four right angles. | 0t. In this diagram, the large square repre- sents a square yard, and each of the smaller squares within it represents one squarefoot. X Now, since there are three rows of small squares, and three square feet in each row, -there will be 3 times 3 — =9 sq. ft. in the large square. But the large square is 3ft. in length, and 3ft. in breadth; hence, To find the contents of a square, multiply its length by its breadth. QUESTIroNS.-Art. 93. For what is square measure used? Repeat the table. What is a square? How may the contents of a square be found? Explain by the diagram the reason of the operation. I-low do you find the contents of all figures having four right angles? Describe a square foot. 9:6 REDUCTION. [fsCT. 3, hENTAL EXERCISES. 1. In 2 square feet how many square inches? In 4 square feet? In 5 square feet? In 20 square feet? 2. In 3 square yards how many square feet? in 10 sq. yd.? in 20 sq. yd.? In 50 sq. yd.? In 100 sq. yd.? 3. In 5 roods how mnany poles? In 20 roods? In 30 roods? 4. In 7 acres how many roods? In 24 acres? In 40 acres 5. Hlow many roods in 80 sq. rods? In 160 sq. rods? ]EXERCISES FOR TIIE SLATE. l. How many square inches in 12A. 3R. 24p. 144ft. 72in.? OPERATION. 1 2 A. 3R. 24p. 144ft. 72in. 4 5 1 roods..40 2 0 6 4 poles. ) 7 2 1 NOTE.- To multiply by w, we 2 take 4 of the multiplicand. 41 32 14452 41 29 516 562068 feet. 144 2248274 2 248279 562068 Ans. 809 37 8 6 4 inches. 2. In 80937864 square inches how many acres? OPERATIOM. 1 44)8 0 9 3 78 6 4 inches. 272{1) 562 0 68 ft. 72in, 4 4 1 0 8 9) 2 2 48 272 fourths of a foot. 40)2 0 64poles. 576 - 4 — 144ft. 4)5 1 R. 24p. Ans. 12 A. 3R. 24p. 144ft. 72in. QUESTiONS. - How do you reduce acres to square inches? Give the reason for the operation. How do you reduce square inches to acres? What is the ea.on for th3 operati on? 7 ow do you multiply by {? E5CT. xo. R1EDUCTIO2. 97 NOTE. - To divide by 272:-, we first reduce both the divisor and dividend to fourths, and tflenr divide. Tlhe quotient is wllr0e nuIlbihers, a1nd lthe remlainder fourths, wllich we reduce to whole numbesbrs by dividiig by 4. 3. Tn 49A 3R.. Gp. how mrany square feet? 4. In 2171466 square feet howv mlany acres? 5. What is the -value of' 365A. 3R. 17p. at' 1.75 pet square rod or pole Ans. $ 102,439.75. 6. Sold a valuable piece of land, containing 3A. 1R. 30p., at 0 1.25 per square foot; what was received for the land? Ans, $ 1S7,171.87,5. 7. In a tract of land 12 miles square, hrow many square miles? flow many acres? Ans. 92160 acres. 8. In 18A. 02.. 16p. how many square feet? IAns. 78843S square feet. 9. Purchased 48A. 3R. 14p. of land for $ 2.25 per square rod, and sold the same for 3.1l5 per square rod; what did I gain by mry bargain? Ans. 0$ 7032.60. CUBIC OR SOLID MEASUitE. AnR. 9&4o Cubic or Solid Measure is used in measuring such bodies or things as have length breadth, anlld thickness; as timber, stone, &ct, TABLE. 1728 Cubic inches (cu. in.) make 1 Cubic Toot, marled cu. ft. 27 6 feet 6S 66 yard, c6 c. yd. 40 " feet 6 1 Ton,' T. 16 "6 feet 6" 1 Cord foot " C. ft. 8 Cord feet, or I Cord of wood, C 128 Cubic feet R,. in. yd. _ - 1728 1T 27 ut:46,56 uC. 1 ]1a3 40 69120 -- 33 == _2. 28 4 221184 NOTE. - A pile of wood 8R. in length, 4d. in breadth, and 4f1. in height, contains a cord. - Also, one ton of timber, as usually surveyed, contains 50 90 coubic or solid feet, QUFSTIONS. - How do you divide by 272-? Of what denomination is the remainder?'low is the true remainder found? - Art. 94-. 1'For what is cubic measure used? Recite the table. What are the dimensions of a pile of wood containing 1 cord? How umany solid feet does a ton of timlber contain, - usually surveyed'? 98 REDUCTION. LSECT. X A cube is a solid body with six square and equal sides. If the sides of a cube are 3ft. long, 3ft. wide, and 3ft. thick, as reipre. ii'illini sented in this diagcram, it is called a I -~~ cubic or solid yard. Now, since each side of a cubic yard contains 9 sq. ft. H:. of surface (Art' 93), it is plain, if a X|ll!I 5~ l block be cut off from-one side, one foot! 11 lill l 1thick, it can be divided into 9 solid 3 ft. =1 yd. blocks with sides 1 foot in length, breadth, and thickness, and therefore will contain 9 solid feet; and since the whole block or cube is THrREe feet thick, it must contain 3 times 9 = 27 solid feet; for 3ft. X 3ft. X 3ft. = 27 solid feet. Hence, To find the contents of a cubic or solid body, multiply its length, breadth, and thickness together. MENTAL EXERCISES. 1. In 2 cubic feet how many cubic inches? In 4 cu. ft.? 2. In 3 cubic yards how many cubic feet? In 10 cu. yd.? 3. In 5 cord feet how many cubic feet? In 8 cord feet? 4. In 2 cords of wood how many cubic feet? In 6 cords? 5. How many cords of wood in 64 cord feet? In 96 c. ft.? 6. How many tons in 80 cu. ft. of timber? In 160 cu. ft.? EXERCISES FOR THE SLATE. 1. In 48 cu. yd. and 15 cu. 2. In 2265408 cubic inches ft. how many cubic inches? how many cubic yards? OPERATION. OPERATION. 4 8 yd. 15ft. 1 7 2 8) 2 2 6 5 4 0 8 cu. in. 27 27) 1311 cu. ft. 34 Ans. 4 8 yd. 15ft. 97 1 3 11 feet. 1728 10488 2622 9177 1311 Ans. 2 2 6 5 4 0 8 inches. QUESTIONs.- What is a cube 7 How do you find the contents of a cube? Give the reason for the operation. Describe a cubic foot. How do you reduce a ton to cubic inches? Give the reason for. the operation. How do you reduce cubic inches to cubic yards? Give the reason for the operation. GECT. X.] REDUCTION. 99 3. In 45 cords of wood how many cubic inches? 4. In 9953280 cubic inches how many cords of wood? 5. H-low many cubic feet in a pile of wood 15ft. long, 4ft. wide, and 6 ft. high? How many cords? Ans. 3C. 6 cu. ft. 6. How many cubic inches in a block of marble 4ft. long, 31ft. wide, and 2ft. thick? Ans. 44928. 7. In a room 14ft. long, 12ft. wide, and Sft. high, how many cubic feet-? Ans. 1344. WINE MEASURE. ART. 945. Wine Measure is used in measuring all kinds of liquids, except milk, ale, and beer. This measure is also used by government in collecting duties. TABLE. 4 Gills (gi.) make 1 Pint, marked pt 2 Pints " 1 Quart, 6 qt. 4 Quarts 4" 1 Gallon, " gal. 32 Gallons " 1 Barrel, "' bar. 42 Gallons " 1 Tierce, " tier. 63 Gallons 4 I Hogshead, "' hhd. 2 Tierces " 1 Puncheon, " pun 2 Hogsheads "' I Pipe or Butt, i" pi.'2 Pipes, or 4 Hogsheads, "' 1 Tun, " tun pt. gi,qt. 1 4 gal. 1 2 8 tier. I 4 = 8 32 hhd. I 42 168 = 336 -- 1344 pun. I 1a - 63 - 252 - 504 2(016 pi. I1 I - 2 - 84 - 336 = 672 = 2688 tun. I - 1 2 - 3 126 = 504 = 1008 4032 2 = 3 4 - 6 252 1008 - 2016 -- 8064 NOTE. —We have no statute specifying how many gallons a hogshead, tierce, or pipe shall contain. The standard unit of Liquid JMeasure adopted by the government of the 3United States is the Winchester XVWine Gallon, which contains 231 cubic inches. The Imperial Gallon now adopted in Great Britain contains 277.274 cubic inclhes. * Called "Winchester gallon " because the standard measures were kept at Winchester, England. QuESTIONS.- Art. 95. For what is wine measure used 1 Repeat the table. Does the law specify the number of gallons in a hogshead, tierce, or pipe? How many cubic inches in the standard wine gallon in the United States? [n G-eat Britain 1 1800 REDUCMiTON. [SECT. 1. MENTAL EXERCISES. I. in 3 pints how many gills? In 5 pints? In 9 pints? in 11 pints? 2. In 4 quarts how many pints? In 6 quarts? In 8 quarts? In 12 qularts? 3. In 5 gallons how many quarts? In 7 gallons? In 10 gallons? 4. In 2 hogsheads how many gallons? In 4 hogsheads? in 6 hogsheads? 5. ifow nmany quarts in 8 pints? n 10 pintss? In 16 pints? In 20 pints? 6. flow many gallons in 12 quarts? In 18 quarts? In 24 quarts In 32 quarts? EXERCISES FOE TIE SLATEo 1. In 47 tuns of wine how 2. in 3790083 gills how many gills? many tuns? OPERATTON. OPR. ATTO; 4 7 tuns. 4) 3 7 9 0 0 8 gi. e)94752 pt. 8 8 hogsheads. ) 4 7 3 7 6 qt. 6') I 8 4 4 gal. 128 4) 8 8 hhd. 1 1 8 4 4 gallons. Ans. 4 7 tuns. 4 7 3 7 6 quarts. 9 4 5i 5 pints. 4 Ans. 3 79008 gilts. 3. Reduce 197 tuns 3hnd. 60 al. 3qt. Ipt. to gills. 4. In 1596864 gills how many tuns? 5. -Vh t will 7 hogsheads of wine cost, at 5 cents a pint? Ains. l' 176.40. g. What cost s 8 tuns lhhd. 47gaI. of oil, at 8 1.25 per gallon? PAns. $ 5SO'7.50. QU!tSTioNs. -- How do you rediuce tuns to gills? Give the reason of the operation. HEow do you reduce gills to gallons? To hogsheads T'o tuna? Give the reason for the operationm SECT. X.] REDUCTION. 101 ALE AND BEER MEASURE. ART. 96. Ale and Beer Measure is used in measuring milk, ale, and beer. TABLE. 2 Pints (pt.) make 1 Quart, marked qt. 4 Quarts " 1 Gallon, " gal. 32 Gallons, c 1 Barrel, " bar. 54 Gallons " 1 Hogshead, " hhd 2 Hogsheads " 1 Butt, " butt. 2 Butts, or 4 Tlogsheads, " 1 Tun, " tun. qt. pt. gal. 1 2 bar. 1 4 = 8 hhd. 1 7 32 = 128 = 256 butt. i = lig = 54 -- 16 - 432 tun. 1 2 = 3 --- 108 432 = 864 1 - = 4 = 6 = 216 = 864 = 1728 NOTE. - The Barrel in Massachusetts contains 32 gallons. The Ale Gallon contains 282 cubic or solid inches. MENTAL EXERCISES. 1. In 6 quarts how many pints? In 11 quarts? In 13 quarts? In 15 quarts? 2. In 3 barrels how many gallons? In 4 bar.? In 5 bar.? 3. In 2 hogsheads how many- gallons? In 10 hogsheads? In 20 hogsheads? 4. How many quarts in 10 pints? In 12 pints? In 16 pints? In 18 pints? EXERCISES FOR TIIE SLATE. 1. How many quarts in 76 2. In 16416 quarts how hogsheads? many hogsheads? OPERATION. OPERATION. 7 6 hhd. 4) 1 64 1 6 qt. 304 54) 4 104 gal. 30 o4 0Ans. 76 hhd. 380 4 1 0 4 gallons. 4 Ans. 1 6 4 1 6 quarts. QvUESTIONS. - Art. 96. For what is ale and beer measure used? Repeat the table. How many cubic inches in a gallon of ale? How do you reduce hogsheads to pints? Give the reason of the operation. How do you reduce pints to barrels?` To tuns? What is the reason for the operation 7 REDUCTION. LSECT. A3. In 4 tuns Ihhd. l7gal. Ipt. how many pints? 4. H-Iow many tuns in 7481 pints? 5. Whlat cost 7hhd. 18gal. of beer, at 4 cents a quart Ans. $ 63.36. 6. At 15 cents per gallon, what will 1Shhd. of ale cost? Ans. 8 145.80. DRY BMIEASURE. ART.'97o This measure is used in measuring grain, fruit, salt, coal, &c. TABLE. 2 Pints (pt.) make I Quart, marked qt 8 Quiarts 66 1 Peek, 6 pk. 4 Peeks I6 J Bushel9 66 bu. 8 BJiushei s 6 1 Quarter, 66 qr 36 Bushels 6 1 Clhaldron9 c6 oh. qt.a pt. bu. I8 16 ch. 1 4 32 64 1 36 = 144 1152 2304 NOTE. - A Winchester Bushel is 181 inches in diameter, and 8 inches deep. The Standaud Gallon in dry measure conLains 2684 cubic inhies. M NTAL EXERC ISES. I. In 2 quarts how many pints? in 5 quarts? In 7 quarts? In 13 quarts? 2. In 3 pecks how many quarts? In 6 pecks? In 9 pecks? In 12 peclks? 3. In 5 bushels how many pecks? In 10 bushels? In 15 bushels? In 20 bushels? 4. H-low many pecks in 16 quarts? In 25 quarts? In 36 quarts? In 64 quarts? 5. flow many chaldrons in 72 bushels? In 144 bushels? In 180 bushels? QUEnssTIos. Art. 97. For what is dry measure used? Recite the table. kVttl is the size of the Winchester bushel'i SHo s many cubic inches in a gallon, dry measure Which is the larger, the gallon, dry measure, or the wine gallon of the United 6tatoe? ~Ecv. r. R 11EDUCTION. 103 EXERCISES FOR THE SLATE. 1. THow many quarts in 2. In 56731 quarts how 49ch. Sbu. 31k. and 3qt.? many chaldrons? P.ERPATIO1. OPERATIOI. 4 9 ch. 8bu. 3pk. Sqt. ) 5 6 7 3 1 qt.,) 7 0 9 1 pk. 3qt, 302 4 L47 386) 1 7 72bu. 3pk, 4 9 ch. S bu. 1 7 7 2 bushels. Ans,, 49 ch. 8bu. 3pk. 8qt. 0 9 1 pecks. Ans. 5 6 7 3 1 quarts. 3. Reduce 97ch. 80bu. 2pk. to quarts. 4. In 112720 quarts how many chaldrons? 5. flow nmany pints in 35bu. Ipt.? G. Reduce 2241 pints to bushels. 7. Reduce 18qr. 3pk. qtk. to quarts. 8. flow nmany quarters in 4637 quarts? 9. In'19})u. 3pk. 7qt. Ipt. how many pints? 10. In 1279 pints how mlany bushels? MEASURE OF TIMiE. ART. OSo This measure is applied to the various divisions and subdivisions into which time is divided. TABLEo 60 Seeonds (sec.) make I Minute, markedke m 60 Mlinutes 6 Hon'lur,, h 24 -ours 1 I ay, da 7 I)avs " Week, 5 w. 4 Weeks 6I Month i " mo. 13 lIt.hs, I day, 6 hours, or I Julian ear, 36i5 days, 6 hours, Julian Year, 12 Calendar Months " 1 Year y, QUESTIONS.- How do you reduce chaldrons to pints? Give the reason. How do you reduce pints to bushels 7 Give the reason, - Art, 98. To what ts the measlar orf time applied? Repeat tho table. 104 REDUCTION. ISECT. X. ALSO, w. d. h. mo., da. h. 52 1 6 = 13 1 6 = I Julian Year. da h. M. sec. 365 5 48 67 Ia 1 Solar Year. da. h, mn. see. 365 6 9 14 1= I Sidereal Year. m. sec. h. I 60 da, I 60 - 3600 w. I 24 1440 - 86400 mo. 1 = 7 - 168 = 10080 = 604800 y. 1,== 4 - 28:= 672 = 40320 = 2419200 I ~ 365 - 8766 -- 525960 - 31557600 NOTE. -The true solar year is the time measured from the sun's leaving either equinox or solstice to its return to the same again. A periodical year is the time in which the earth revolves round the sun, and is 365d. 6h. 9m. 141sec., and this is often called the sidereal year. The civil year is that which is in common use among the different nations of the world, and contains 365 days for three years in succession, but every fourth year it contains 366 days. When any year can be divided by four, without any remainder, it is leap year, and has 366 days, except the last year of those centuries which cannot be divided by 4; as the fifteenth, seventeenth, and nineteenth centuries. The days in each month are stated in the following lines: - "Thirty days hath September, April, June, and November; And all the rest have thirty-one, Save February, which alono Hiath twenty-eight; and this, in fine, One year in four hath twenty-nine." MIENTAL EXERCISES. 1. In 3 minutes how many seconds? In 5 minutes? In 10 minutes? 2. In 2 hours how many minutes? In 4 hours? In 8 hours? 3. In 4 weeks how many days? In 6 weeks? In 9 weeks? In 12 weeks? 4. In 2 days how many hours? In 3 days? In 7 days? In 11 days? 5. How many weeks in 21 days? In 30 days? In 50 days? In 84 days? Qis ESTONS. - What is a true solar year? What is a sidereal year? What is a civil year - How many days in a solar year? In a sidereal year? In a civil vcar? What is leap year? -lHow often do we have leap year? Repeat the lines on the days of the month. aECT.. R IDUCTIO. 105 ~XERCISES FOR THE 1 SLATE~ 1. }Iow many seconds in 2.. In 3156937 seconds 3S5da. 511. 4,3n. S7sec., or h-ow many days? one solar year? OPERATION. O'PERAT'fON. 8 6 5da. 5h. 48sr. 59sec. 6O) 3 I 5 5 6 9 3 7 2 41 60) 5 2 5 9 4 8 m. 57ec.,3 24)876 5h. 48m. 3 6 5 d3s. 5h. 8 7 6 5 hours. 6 da. 5h. Ans. 365da. 5h. 48m. 57see. 5 2 5: 4 8 minutes. 6' 60 Ans 3 5 5 6 9 3 7 seconds, 3.,Reduce 2.da.'Sh. S32mn to minutes. 4. In 427352 minutes how many days? 5. How mrany seconds in 30 solar years 262da. 17h. 28m. 42sec.? 6. In 96940'7832 seconds hows m'any solar years? 7. H-ow mntany weeks in 684592 minutes? 8. Tn 67w. 6da. 9h. 52nm. how manly minutes? CTRCULART) I.~EASU!E 0 OTION. ART. 03o. Circular Measure is used in reckoning latitude and longitude, in computing the revolutions of the planets round the sun, and in the division of circles, 60 Seconds ("' make lIinute maiked 60} Minutes 66 " De)erc, o 0 I)evgrees I Sig 6n, " g 12 Signs, or 360 Degrees, he Circle of ithe Zodiac,' C.,': 1...-. 60 a, G;. O & 60 3600,. 1 *= 3a a= lo I800 10= N800!,== t2 2 360 -21600 1296000 QuesTIon s. - lo do you reduce years to ecconds? Give the reason of the operation. How do you 1reduce seconds to days 7 To years? Give the reason of the operation0 Art. 99. F1or what is circular measure used? Repeat the tahble i06 REDUCTI1A. [SECT.. -MENTAL EXERCISES. 1. In 3 minutes how many seconds? In 5? In 8? In 10? 2. In 2 signs how many degrees? In 4 signs? In 6 signs? In 9 signs? 3. How many degrees in 120'? In 180'? In 360'? 4. How many signs in 90~? In 120~? In 600~? EXERCISES FOR TIHE SLATE. 1. How many minutes in 2. In 20937 minutes how 11S. 18~ 57'? many signs? OPERATION. OPERATION. l 1 S. 180 57' 60)20937' 3 0 830) 3480 57 348 degrees. 1S. 18 60 Ans. 2 0 9 3 7 minutes. Ans. 11 S. 18~ 57'. 3. In 27S. 190 51'28!" how many seconds? 4. How many signs in 2987488 seconds? MISCELLANEOUS TABLE. ART. i 00 This table embraces a variety of things of practical importance to the pupil, all of which are more or less used in business. 12 Units, or things, make I Dozen. 12 Dozen, or 144, 1 Gross. 12 Gross, or 144 Dozen,' 1 (Great Gross. 20 Units, or things, " 1 Score. 24 Sheets of paper " 1 Quire. 20 Quires 4 1 Realm. 56 Pounds " 1 Firkin of butter. 112 Pounds' 1 Barrel of raisins. 106 Pounds " I Barrel of flour. 200 Pounds 6' 1 Barrel of pork. 200 Pounds 4 1 Barrel of beef. 200 Pounds of shad or salmon " 1 IBarrel in N. Y. and Conn. 30 Gallons I5 1;alrrel of fish in Massachusetts. QUESTIoNs. —How do you reduce signs to seconds? Give the reason of the operation. Hlow do you reduce seconds to degrees? To signs? Give the reason of the operation. How many degrees in a circle? -Art. 100. What is embraced in the Miscellaneous Table? Recite the table. SECT..] REDUCTION. 1 14 Pounds of lead or iron maake 1 Stone. 21 Stone "6 1 Pig. 8 Pigs, or 19,cwt., "' 1 Fother. 70 Pounds "' 1 Bushel of salt. 60 Pounds " 1 Bushel of wheat. 56 Pounds I" 1 Bushel of Indian corn or rye. 46 Pounds 1 Bushel of barley or buckwheat 30 Pounds " 1 Bushel of oats. OF BOOKS. A sheet folded in two leaves is called a folio. A. sheet folded in four leaves is called a quarto, or 4to. A sheet folded in eight leaves is called an octavo, or 8vo. A sheet folded in twelve leaves is called a duodecimo, or 12mo. A sheet folded in eighteen leaves is called an 18Imo. A sheet folded in twenty-four leaves is called a 24tmo. MISCELLANEOUS EXERCISES IN REDUCTION. 1. In $ 345.18 how many mills? 2. How many dollars in 345180 mills? 3. In 46L. 18s. 5d. how many farthings? 4. How many pounds in 45044 farthings? 5. Reduce 611b. Ooz. 17dwt. 17gr. troy to grains. 6. In 351785 grains troy how many pounds? 7. How many scruples in 271b 35 15 ID? 8. In 7852 scruples how many pounds? 9. In 83T. Iecwt. 3qr. 1Slb. how many ounces? 10. How many tons in 2996064 ounces? 11. How many nails in 97yd. 3qr. 3na.? 12. In 1567 nails how many yards? 13. In 57 ells English how many yards? 14. flow many ells English in 71yd. lqr.? 15. I-low many inches in 15m. 7fur. 18rd. 10ft. 6in.? 16. In 1009530 inches how many miles? 17. In 95,000,000 of miles how many inches? 18. HI-ow many miles in 6,019,200,000,000 inches? 19. In 48dego 1Sm. 7fur. 1Srd. how many feet? 20. In 17714037 feet how many degrees? 21. HIow many square feet in 7A. 3R. 16p. 218ft.? 22. In 432164 square feet how many acres? 23. flow many square inches in 25 square miles? QUEsTION. - What gives name to the size or form of books ]0o RtEDUCTION. [SECT. x. 24. In 10036224~000 square inches how many qluare miles? 25. I-How many cubic inches in'5 tons of timber? 26. In i03(6S00 cubic inch1es' how rnmay ton3s? 27. How imany gills of wine in S5lihhd. i7gal. 3qt.? 28. In 10648 gills how mnaniy hogslheadas of wine? 29. low many quarits of beer in 29hh11d. 30gal. 3qt.? 30. In 6387 quarts of beer how nmany hogsheads? 31. flow many pints in 15ch. IGbu. glSpk. of wheat? 32. In 35632 pints of wheat how many chaldrons? 33. IHow many seconds of time in 365 days 6 hours? 34. In 31557600 seconds how miany days? 35. How many hours in 1842 years (of 3655da. 6h. each)? 36. In 16146972 hours how many years? 37. I-low many seconds in 8S. 140 18' I7/? 88O. In 915497" ho-w>.n many silgns? 9. What will be the cost of 13 gross of steel pens, at 24 cents per pen? Ans. 8 46.80. 40. Bought 12 reams of paper, at 20 cents per quire; hovwv much did it cost? Ans. $ 48. 41. 1 wish to put 2 hogsheads of wine into bottles that will contain 3 quarts each, how mnany bottles are required? Ans. 168 bottles. 42. When $ 1480 are paid for 25 acres of iand, what costs i acre? What costs I rood? Whlat cost 37A, 21. Sp. Ans, 2 2'22~.66. 43. John Webster bought 5cwt., l1b. of sugar at 9 cents per lb., for which he paid 25 barrels of apples at $ 1.75 per barrel; how much remains due?. Ans. 0 15V.83. 440 Boughlt a silver tankaird weiehing 21b 7oz. for 46.50; what did it cost per oz.? 1ow mri elh per lb.?.Ans. $ 18 45. Bought ST. Ic,-zt. 81b. of leather at J0 cents per lb. and sold it at 9 cernt:>e lb.; 9 hat did I lose? Ans, $ 205.50. 46. Phineas Bailey has agreed to grade a certain railroad at $ 5.75 per rod; what will he receive for grading a road between two cities, whose distance from each other is 37m. 7fur. 29rd.? Ans,. 69856.75. 47. ITf it cost 8 17.29 per rod to grade a certain piece of railroad, what will be the expense of grading 5Irn. fur. 37rd.? Ans. 8 87,781.33. 48. What is the value of a house-lot, containing 40 square rods and 200 sauare feet, at 6 1.50 per square foot?.AnS. 8 16635, t'.r. x.] REDUCTION. 109 49. How many yards of carpeting, that is one yard in width, will be required to carpet a room 18ft. long and i5ft. wide? Ans. 30 yards. 50. A certain machine will cut 120 shingle-nails in a minute; how many will it cut in 47 days 7,hours, admitting the machine to be in operation 10 hours per day? Ans. 3434400 nails. 51. In a field 80 rods long, and 50 rods wide, how many square rods? How many acres? Ans. 25 acres. 52. How long will it take to count 18 millions, counting at the rate of 90 a minute? Ans. 138da. 21h. 20m. 53. A merchant purchased 9 bales of cloth, each containing 15 pieces, each piece 23 yards, at 8 cents per yard; what was the amount paid? Ans. ~ 248.40. 54. Suppose a certain township is 6 miles long, and 4~ miles wide; how many lots of land of 90 acres each does it contain? Ans. 192 lots. 55. The pendulum of a certain clock vibrates 47 times in one minute; how many times will it vibrate in 196 days 49m.? Ans. 13267583 times. 56. I-low many shingles will it take to cover a building that is 36 feet long and 24 feet wide, with rafters 16 feet long, supposing one shingle to cover 27 square inches? Ans. 6144 shingles. 57. I-low many times will the large wheels of an engine turn round in going from Boston to Portland, a distance of 10 miles, supposing the wheels to be 12 feet and 6 inches in circumference? Ans. 46464 times. 58. In a certain house there are 25 rooms, in each room 7 bureaus, in each bureau 5 drawers, in each drawer 12 boxes, in each box 15 purses, in each purse 178 sovereigns, each sovereirgn valued at.8 4.84; whlat is the amount of the money? Ans. $.135689400. 59. In 18rd. 5yd. 2ft. l1in. how many inches? Ans. 3779 inches. 60. In 3779 inches how many rods? Ans. 1Srd. 5yd. 2ft. 11in. 61. Sold 5T. 17cwt. 3qr. 181b. of potash for 3 cents per pound; what Was the amount? Ans. $ 396. 18. 62. A gentleman purchased a house-lot that was 25 rods long and 16 rods wide for $100,000, and sold the samne for e 1.25 per square foot; what did he gain by his purchase? Ans. $ 36,125. 110 ADDITION OF COMPOUNID NUMBERS. (SECT. MH XI. ADDITION OF COMPOUND NUMBERS. ART. Z101. Addition of Compound Numbers is the adding together of two or more numbers of diffbrent denominations. ENGLISH MONEY. Ex. 1. Paid a London tailor 7~. 13s. 6d. 2qr. for a coat, 2C. 17s. 9d. lqr. for a vest, 3~. Ss. 3cd. 3qr. for pantaloons, and 9X. 11s. Sd. 3qr. for a surtout; what was the amount of the bill? Ans. 23Z. 1 s. 4d. lqr. OPERATION. Having written farthings under farthings, ~. d. dqr. pence under pence, &c., we find tie sum 7 1 3 6 2 of the farthings in the right-hand column 2 1 I7 9 1 to be 9 farthings, equal to 2d. and iqr. 3 8 3 3 over; we write the 1 farthing under the 9 1 JL 1, <o column of farthings, and carry the 2d. to 19 1 8 the column of pence, the sum of whici is Ans. 2 3'11 4 1 28d., equal to 2s. 4d.; we write the L d. under the column of pence, and carry',e 2s. to the column of shillings, the sum of which is 51is., equal to'2/. 1 Is.; having written the 1 Is. under the column of shillings, we carry the 2~X. to the column of pounds, and find the whole amount to be 23/~. Ils. 4d. lqr. From the above process we deduce the following RULE. - 1. Write all the given numbers of the same denomznatwn under each other; as pounds under pounds, shillings under shillings, 4~c 2. Then add together the numbers of the lowest denomination, and divide their sum by that number which it requires of the column added to make oNE of the next higher denomination; set the remainder under the column added, and carry the quotient to the next left-hand column, the sum of which divide by the appropriate number, as bejfore; and thus proceed to the highest denomination, under which place its whole sum. Proof. - The proof is the same as in simple addition. QUESToNS.- -Art. 101. What is addition of compound numbers7 How do you arrange compound numbers for addition? 7 hy? Will you explain the operation? What is the rule for the addition of compound numbers? The proof? What is the di.fference between simple addition and addition of compound numbers? BSECT. SI.] ADDITION OF COMPOUND NUMBERS. 13l EXADIPLES FOR PRACTICE. 3.. S. d. qr... r. 6 19 11 3 11 9 7 2 9 6 3 34 10 1 1 13 18 3 1 31 4 6 3 67 0 8 I 73 0 8 1 97 5 3 0 TROY WEIGHT. 4. 5, lb. oz. dwt. gr. lb. oz. dt. gr 15 11 19 22 10 lO lO 10 71 10 1 17 8 3 I 1 19 23 65 9.7 14 47 7 8 19 73 11 13 13 16 9 10 14 14 8 9 9 33 10 9 21 242 4 14 3 APOTHECARIES' WEIGHIT. 7 t 3 5 D gr. S 5 D gr. 81 11 6 1 19 35 9 6 2 19 75 10 7 2 13 71 1 1,1 I 11 14 9 7 1 12 37 3 3 2 12 37 8 1 11 14 4 7 1 13 6 1 ii 3 2 3 75 5 6 1 17 272 4 3 0 18 AVOIRDUPOIS WEIGHT. 8. 9, T. cwt. qr. lb. oz. dr. T. cwt. qr, lb o. o. r. 71 19 271413 27 14 13 2 15 15.15 14 13 1 11 13 12 13 17 3 13 11 13 39 9 3 13 9 9 46 16 3 11 13 10 15 17 3 16 10 14 4 14 5 2 7 6 9 61 16 3 13 7 8 11 17 3 16 15 11 203 17 3 27 8 8 112 ADDITION OF COMPOUND NUMBEIRS [sECT. X. CLOTH MEASURE. 10. 11. yd. qr. na. in. E. E. qr. na. in. 5 3 3 2 16 3 2 1 7 1 1 2 71 1 1 2 8 3 3 1 1 3 3 2 1 9 1 2 2 47 3 2 2 4 3 3 2 39 2 3 2 36 3 0 0 LONG MEASURE. 12. 13. deg. rn. fur. rd. ft. in. m. fur. rd. yd. ft. In. 18 19 7 15 11 1 12 7 35 5 2 11 61 47 6 39 10 11 13 6 15 3 1 10 78 32 5 14 9 9 16 1 17 1 2 5 17 59 7 36 16 10 13 4 13 2 1 9 98 56 1 30 16 1 17 7 36 5 2 7 205 71 5 17 141 8 ~=4 1=6 205 8 1 17 15 2 SURVEYOR'S MEASURE. 14. 15. m. fur. ch. p. I. m. fur. ch. p. 1. 17 5 8 3 24 14 7 9 3 21 16 3 7 1 21 37 1 0 3 16 47 7 9 3 19 17 7 8 3 17 19 6 6 1 16 61 6 5 3 16 31 7 1 0 20 47 I 1 0 23 133 7 4 0 0 LAND OR SQUARE MEASURE. 16. 17. A. R. p. ft. in. A.. R. p. yd. ft. -in 67 3 39 272 143 43 1 15 30 8 1t 78 3 14 260 116 16 3 39 19 7 141 14 2 31 167 135 47 1 16 27 5 79 67 1 17 176 131 38 3 17 18 8 17 49 3 31 69 117 15 1 32 11 1 117 278 3 1 15 131 66 278 35 10 278 3. 15 131 102 SECT. XIe. ADDITION OF COMPOUND NUMBERS. 113 SOLID MEASURE. 18. 19. Tun. ft. in. Cord. ft. in. 17 39 1371 14 116 1169 61 17 1711 67 113 1711 47 16 1666 96 127 969 71 38 1711 19 98 1376 47 17 1617 14 37 1414 246 11 1164 WINE MEASURE 20. 21. Tutn. hhd. gal. qt. pt. Tun. hhd.- gal. It. pt. 61 1 62 3 1 14 3 18 3 0 71 3 14 1 1 81 1 60 3 1 60 0 17 3 0 17 3 61 3 0 14 5 1 51 1 1 61 3 57- 3 1 57 3 14 3 17 1 17 17 1 265 2 35.1 0 ALE AND BEER MEASURE. 22. 23. Tun. hhd. gal. qt. pt. - Tun. hhd. gal. qt. pt. 15 3 50 3 1 67 I 51 1 1 0 67 3 17 3 1 15 3 16 3 1 17 1 44 1 0 44 1 45 1 I 71 3 12 3 1 15 2 12 2 1 81 1 18 1 0 67 -3 35 1 0 254 1 36 0 1 DRY MIEASURE. 24. 25. clh. bu. pk. qt. pt. ch. bu. pk. qt. pt 15 35 -3 7 1 71 17 1 1 1 61 16 3 6 1 16 31 3 3 0 51 30 1 5 0 41 14 3 1 1 42 17 2 2 1 71 17 1 0 1 14 14 1 4 1 10 Il:10 2 3 0 186 7 1 2 0 1Oa 114 SUBTRACTION OF COMPOUND NUMBERS. [SECT. XII. TIME. 26. 27. y. da. h. m. S. w. da. h. m. s. 57 300 23 59 17 15 6 23 15 17 47 169 15 17 38 61 5 15 27 18 29 364 23 42 17 71 6 21 57 58 18 178 16 38 47 18 5 19 39 49 49 317 20 52 57 87 6 19 18 57 203 237 4 30 56 CIRCULARt MOTION. 28. 29. I 1 2 8 56 58 S. 17' 17 1 8 10 21 51 37 7 0 9 19 51 8 13 39 57 8 18 57 45 8 19 38 49 4 17 16 39 7 17 47 4 8 7 27 38 48 1L1 11 55 09 NOTE.- The sum of the signs, in circular motion, must always be divided by 12, and the remainder only be written down, as in Ex. 28. XII. SUBTRACTION OF COMPOUND NUMBERS. ART. O,2. SUBTRACTION of Compound Numbers is finding the difference between two numbers of different denominations. ENGLISH MONEY. Ex. 1. From 87~. 9s. 6d. 3qr., take 52 9. Ils. 7d. lqr. Ans. 34. 17s. 11id. 2qr. OPERATION. H-aving placed the less number un ~. B. d. qr. der the greater, farthings under far Min. 8 7 9 6 3 things, pence under pence, &c., wto Sub. 5 2 1 ]1: 7 1 begin with the qrs. or farthings thus: Rem 3 4 1 7 1 1 2 lqr. from 3qr. leaves 2qr., which we set under the column of qrs. Now, QUESTIONS. - Art. 102. What is subtraction of compound numbers? How do you arrange the numbers for subtraction? SECT. xII.] SU1BT- CTION OF. COMPOUND kN.F ~BRS. 115 as we cannot take 7d. from 6d., we add 12d. = is. to the 6d., making 18d., and then subtract the 7d. from it, and set the remainder, lld., under the column of pence. We then add ls- =12d. to the lls. in the subtrahend, making 12s., to compensate for the 12d. we added to the 6d. in the minuend. (Art. 30.) Again, since we cannot take 12s. from 9s., we add 20s. = 1~'. to the 9s., making 29s., from which we take the 12s. and set the remainder, 17s., under the column of shillings. Having added 1~.=-20s. to the 52~., to compensate for the 20s. added to the 9s. in the minuend, we subtract- the pounds as in subtraction of simple numbers, and obtain 34~. for the remainder. RULE. — 1. Write those numbers under each other which are of the same denomination, the less compound number under the greater. 2. Then begin with the lowest denomination, and subtract each lower number from the one above it, and write the difference underneath. 3. If any lower number is larger than the upper, suppose as many to be added to the upper number as would make one of the next higher denomination, then subtract the lower figure, remembering to carry one to the next lower number before subtracting it; and proceed thus till all the numbers are subtracted. PROOF. -The proof is the same as in simple subtraction. EXAMPLES FOR PRACTICE. 2. 3. S~. d. qr. d. qr 78 11 5 2 765 16 10 1 41 13 3 3 3 713 17 11 3 36 1 8 1 3 TROY WEIGHT. 4. 5. lb. on. dwt. gr. lb. oz. dwt. gr. 15 3 12 14 711 1 3 17 9 11 17 21 19 3 18 19 5 3 14 17 APOTHECARIES' WEIGHT. 6. 7. l ~ 5 9 g. lb 5 3 D gr. 15 7 1 2 15 161 6 3 1 17 11 9 7 1 19 97 7 1 2 18 3 9 2 0 16 QuESTIONS. — hat do you do when the upper number is smaller than the lower? How many do you carry to the next denomination? What is the rule for subtraction 7 The proof? 116 SUBTRACTION OF COMPOUND NUMBERS. SEcT, flI. AVOIRDUPOIS WEIGHT. 8. 9. T. cwt. qr. lb. oz. dr. T. cwt. qr. lb. oz. dr117 16 1 13 O 14 11 1 0 1 1 13 19 17 3 27 1 15 9 18 3 1 13 15 97 18 1 13 14 1 5 CLOTH IMEASURE. 10. 11. yd. qr. na. in. E.. qr. na. in. 15 1 1 2 171 2 2 1 9 3 3 1 i1 9 3 0 2 5 1 2 1 LONG MEASURE. 12. 13. aeg, m. fulr. rd, yd. ft. in. deg. m. fur. rd. ft. in 97 3 7 31 1 1 3 18 19 1 1 3 7 19 17 1 39 1 2 7 9 28 7 1 16 9 77 55k 5 3 1 4~ 1 8 -=4 -=z1 6 77 56 1 31 5 0 2 SURVEYORS' MEASURE. 14. 15. n. filr. cha. p. ii. m. fur. cha. ). li. 21 3 52 17 31 7 1 1 19 9 5 8 1 20 18 1 7 3 23 11 5 7 0 22 LAND OR SQUARE MEASURE. 16. 17. A. R. p. ft. in. A. IR. p. yd. t. in 116 1 13 100 113 139 1 17 18 1 30 87 3 17 200 117 97 3 18 30 1 31 28 1 35 171k 140 = 36 28 1 35 172 32 SECT. xI'1. SUBTRACTION OF COMPOUND NUMBERS. 117 SOLID MEASURE. 1is. 19. T. ft. in. Cords. ft. in. 171 30 1000 571 18 1234 98 37 1234 199 19 1279 72 32 1494 WINE MEASURE. 20. 21. T. hhd. gal. qt. pt. gi. T. hhd. gal. qt. pt. gh. 171 3 8 1 1 1 71 1 1 1 1 1 99 1 19 3 1 3 9 3 3 3 1 3 72 1 51 1 1 2 ALE AND BEER MEASURE. 22. 23. T. hhd. gal. qt. pt. T. hhd. gal. qt. pt. 15 1 17 1 0 79 2 2 2 0 9 3 19 3 1 19 3 13 3 1 5 1 51 1 1 DRY MEASURE. 24. 25. ch. bu. pk. qt. pt. ch. bu. pk. qt. pt. 716 1 2 1 0 73 13 3 0 1 19 9 3 1 1 19 18 1 3 1 696 27 2 7 1 TIME. 26. 27. y. da. h. m e. w:. W. da. h.. s. ec, 375 15 -13 17 5 14 1 3 4 15 199 137 15 1 39 9 6 17 37 48 175 242 22 15 26 CIRCULAR MOTION. 28. 29. S. o o -, S. o, 1 7 1 3 15 1 23 37 39 9 29 17 36 9 15 38 47 1 7 55 39 4 - 7 58 52 NOT.. - In Circular Motion, the minuend is sometimes less than the subtrahend, as in Ex. 29,- in which case it must be increased by 12 signsi 115 SUBTRACTION OF COMPOUND NUMBERS. [SECT. Xil. AnT. O@,3 To find the time between two different dates. Ex. 1.'What is the difference of time between October 16th, 1842, and August 9th, 1844? Ans. ly. 9mo. 23da. FIRST OPERATION. Commencing with January, the first y. mo. da. month in tile year, and counting the lin. 1 8 4 4 7 9 months and days in the later date up to Sub. 1 8 4 2 9 1 6 August 9th, we find that 7mo. and 9da. have elapsed; and counting the -months Pem. 1 9 2 3 and days in the earlier date up to OctoSECOND OPERATION. ber 16th, we find that 9mo. and 16(la. M.in 1 44 4 8 9 have elapsed. Then, setting down the earlier date under the later, and placing Sub. 1 8 4 2 1 0 1 6 the months and days at the right of the Rem. 1 9 2 3 year of each date respectively, as in the. example, we subtract the. lower number from the upper, and the remainder is the time between the dates. RULE. - Set down the earlier date under the later, writing the year of each on the left, next after this the number of months that have elapsed since the beginning of the year, and on the right the day of the month. Then subtract as in the preceding rule. NOTE. - 1. A month in legal transactions is reckoned fiom any day in one month to the same day of the following month; but in computing interest for less time than a month, and in finding the difference between two dates, 30 days are considered a month, and 12 months a year. 2. Some prefer reckoning the number of the given month, instead of the number of months that have elapsed since the beginning of the year; tile result is the same in both cases. See 2d operation. EXAMPLES FOR PRACTICE. 2. What is the time from March 21st, 1843, to Jan. 6th, 1847? Ans. 3y. 9mo. 15da. 3. A note was given ]Nov. 15th, 1832, and paid April 25th, I837; how long was it on interest? Ans. 4y. 5mo. 10da. 4. John Quincy Adams was born at Braintree, iM5ass., July 11th, 1767, and died at Washington, D. C., Feb. 23, 1848; to what age did he live? Ans. 0Sy. 7mo. 12da. 5. Andrew Jackson was born at Waxaw, S. C., March 15th, 1767, and died at Nashville, Tenn., June 8th, 1845; at what age did he die? Ans. 78y. 2mo. 23da. QUEsTIONS. —Art. 103. From what period do you count the monthls and days in preparing dates for subtraction I How do you arrange the dates for subtraction? How subtract? How many days are considered a month in business transactions? What is the second method of preparing dates for subtraction? SECT. X1XrI. MIISCELLANEOUS EXERCISES. 119 q XIII. MISCELLANEOUS EXERCISES IN ADDITION AND SUBTRACTION OF COMPOUND NUMBERS. 1. What is the amount of the following quantities of gold 41b. Soz. 13dwt. Sgr., 51b. lloz. 19dwt. 23gr., 81b. Ooz. 17dwt, 15gr., and 18lb. 9oz. 14dwt. 10gr.? Ans. 371b. 7oz. 5dwt. Sgr. 2. An apothecary would mix 7%b 3 25 2D 1 gr. of rhubarb, 21th 106 03 1D 13gr. of cantharides, and 21h 35 75 2D 17gr. of opium; what is the weight of the compound? Ans. 12tb 53 33 03 1agr. 3. Add together 17T. 11cwt. 3qr. 111b. 12oz., liT. 17cwt. Iqr. 191b. l1oz., 53T. 19cwt. lqr. 171b. Soz., 27T. 19cwt. 3qr. 181b. 9oz., and 16T. 3cwt. 3qr. Olb. 13oz. Ans. 127T. 12cwt. lqr. 121b. 50oz. 4. A merchant owes a debt in London amounting to 767 1S., what remains due after he has paid 172Sf. 17s. 9d.? Ans. 5942~. 2s. 3d. 5. From 731b. of silver there were made 261b. 11oz. 13dwt. 14gr. of plate; what quantity remained? Ans. 461b. Ooz. 6dwt. 10gr. 6. From 711b 83 15 I D 14gr. take 71b 93 15 1D 17gr Ans. 631b 103 75 2D 17gr. 7. From 28T. l3cwt. take lOT. 17cwt. 191b. 14oz. Ains. 17T. 15cwt. 3qr. 81b. 2oz. 8. A merchant has 3 pieces of cloth; the first contains 37yd. 3qr. 3na., the second iSyd. lqr. 3na., and the third 31yd. lqr. 2na.; what is the whole quantity? Ans. 87yd. 3qr. Ona. 9. Sold 3 loads of hay; the first weigned 2T. 13cwt. iqr. 171b., the second 3T. 271b., and the third IT. 3qr. Il1b.; what did they all weigh? Ans. 6T. 14cwt. lqr. 271b. 10. WVhat is the sum of the following distances: 16m. 7fur. 1Srd. 142ft. liin., 19m. lfur. 13rd. 16ft. 9in., 97m. 3fur. 27rd..3ft. 3in., and 47m. 5fur. 37rd. 13ft. lOin.? Ans. 181mn. 2fur. ISrd. 9ft. 3in. I 1. From 76yd. take 18Syd. 3qr. 2na. Ans. 57yd. 0qr. 2na. 12. From O20m. take 3m. 4fur. lSrd. 13ft. Sin. Ans. 16m. 3fur. 21rd. 2ft., -Oin. 13, From 144A4 3R. take. 8iSA. 1R. 17p. 200ft. lOOin. An s. 123 6 A. 1T, P.2p. 7 ft. SG0i, 120 MISCELLANEOUS EXERCISES, [scsr. sxtz 14. From 18 cords take 3 cords 100ft. 100lOin. Ans. 14 cords 27ft. 728in, 15. A gentleman has three farms; the first contains 169A. 3R. 15p. 227ft., the second 187A. IR. 15p. 165ft., and the third 217A. 2R. 28p. 165ft.; what is the whole quantity? Ans. 574A. 3R. 20p. 12ift. 16. There are 3 piles of wood; the first contains 18 cords 116ft. 1000in., the second 17 cords llft. 1600in., and the third 21 cords 109ft. 1716in.; how much in all? Ans. 58 cords 82ft. 860in. 17. From 17T. take 5'T. 18ft. 765in. Ans. 11T. 21ft. 963in. 18. From 169gal. take 76gal. 3qt. Ipt. Ans. 92gal. Oqt. ipt. 19. From 17ch. 18bu. take 5ch. 20bu. lpk. 7qt. Ans. llch. 33bu. 2pk. lqt. 20. From 83y. take 47y. 10mo. 27d. 18h. 50m. 14s. Ans. 35y. Imo. 2d. 5h. 9m. 46s. 21. From 11S. 15~ 36' 15" take 5S. 180 50' 18". Ans. 5S. 26~ 45' 57". 22. John Thomson has 4 casks of molasses; the first con tains 167gal. 3qt. ipt., the second 186gal. lqt. Ipt., the third 108gal. 2qt. ipt., and the fourth 123gal. 3qt. Opt.; how much is the whole'quantity? Ans. 586gal. 2qt. Ipt. 23. Add together 17bu. lpk. 7qt. Ipt., 18bu. 3pk. 2qt., 19bu. lpk. 3qt. Ipt., and 51bu. 3pk. Oqt. Ipt. Ans. 107bu. lpk. 5qt. Ipt. 24. James is 13y.4mo. 13d. old, Samuel is 12y. lImo. 23d., and Daniel is 18y. 9mo. 29d.; what is the sum of their united ages? Ans. 45y. 2mo, 5d. 25. Add together 18y. 345d. 13h. 37m. 15s., 87y. 169d. 12h. 16m. 28s., 316y. 144d. 20h. 53m. 18s., and 13y. 360d. 21h. 57m. 15s. Ans. 436y. 290d. 20h. 44m. 16s. 26. A carpenter sent two of his apprentices to ascertain the length of a certain fence. The first stated it was 17rd. 16ft. 11in., the second said it was 18rd. 5in. The carpenter finding a discrepancy in their statements, and fearing they might both be wrong, ascertained the true length himself, which was 17rd. 5yd. Ift. 1 lin.; how much did each differ from the other? 27. From a mass of silver weighing 106lb., a goldsmith made 36 spoons, weighing 51b. I loz. 12dwt. 15gr.; a tankard, 31b. Ooz. 13dwt. 14gr.; a vase, 71b. lloz. 14dwt. 23gr.; how much unwrought silver remains? Ans, Slb. lb. loz. 8dwt. 20Og. s:ECr. xxv.] MULTIPLICATION OF COMPOUND 1NUMBERS. 121 28. From a piece of cloth, containing 17yd. 3qr., there were taken two garments, the first measuring 3yd. 3qr. 2na., the second 4yd. lqr. 3na.; how much remained? Ans. 9yd. lqr. 3na. 29. Venus is 38. 18~ 45t 15"/ east of the Sun, Mars is 7S. 150 36' 18" east of Venus, and Jupiter is 5S. 21~ 38' 27" east of Mars; how far is Jupiter east of the Sun? Ans. 4S. 26~. 30. The longitude of a certain star is 3S. 1S~ 14' 35", and the longitude of Jupiter is IIS. 25~ 30' 50"; how far will Jupiter have to move in his orbit to be in the same longitude with the star? Ans. 3S. 22 43/ 43' 45. ~ XIV. MULTIPLICATION OF COMPOUND NUMBERS. ART. l04. MULTIPLICATION of Compound Numbers is re peating numbers of different denominations any proposed number of times. ART. 105. To multiply when the multiplier is not more than 12. Ex. 1. If an acre of land cost 14~,. 5s. 8d. 2qr., what will 9 acres cost? Ans. 128. 1 Ils. 4d. 2qr. OPERAT1o0X. In performing this question, ~. 8. d. qr. we first write the multiplier un. Multiplicand 14 5 8 2 der the lowest denomination of Multiplie`r 9 the multiplicand, and then say 9 times 2qr. are 18qr., equal to Product 1 2 8 1 1 4 2 4d. and 2qr. We set down the 2qr, under the number multiplied, reserving the 4d. to be added to the next product. We then say 9 times 8d. are 72d., and the 4d. make 76d., equal to 6s. and 4d., and set the 4d. under the column of pence, reserving the' 6s. to be added to the next product. Then 9 times 5s. are 45s., and 6s. make 51s., equal to 2~. and uls. We place the l1s. under the column of shillings, reserving the 2 ~. to be added to the next product. Again, 9 times 14~. are 126~., and 2$. make 128~. This,placed under the coluimn of pounds, gives us 128~. I s. 4d. 2qr. for the answer. QuvsTroNs. -Art. 104. What is multiplication of comround numbers? -Art. 105. Explain the operation. By what do you divide the product of each denomination? What do you do with the quotient and remainders thus obtained 7 l. [22 MULTIPLICATION OF COMPOUND NUMBEtRS. [SECT. XIV, RuLE. - Multiply each denominatorn of the compound number by the multiplier separately, beginning with the lowest, and ca rry as in adldi-ton. of compound numbers. XATM'PLES FOt PPRACTICE 2. 2. 4, 5, ~. s. d... d s. d. ~. d. 5 6 8 19 11 7 25 1711 18 15 8* 10 13 4 58 149 1299 9 7 12 14 46o 7. cwt. qr. lb. oz. Ton. cuwt. qr. lb. cwt. qr. lb. oz. 1 83 17 1 1 5 3 142 1 9 1 8 5 6 8$ 1 3 1 21 12 103 11 0 0!54 2 1 5 8 9.. 10. I1. lb. oz. dr. in. fur. rd. ft. deg. m. fur. rd 15 14:13 97 7 14 13 18 12 6 18 9 6. 8 143 5 5 587 74 8 1 2 145 3 2 7 24 12. 13. rd. yd. ft. in. fur. rd. ft. in. 23 3 2 9 9 31 16 11 9 10 213 2 0 9 98 0 4 2 NOTE. -The answers to the following questions are found in the cor. responding questions in Division of Compound Numbers, p. 128. 14. What cost 7 yards of cloth at 18s. 9d. per yard? 15. If a man travel 12m. 3fur. 29rd. in one day, how far will he travel in 9 days? 16. If 1 acre produce 2 tons 13cwt. 191b. of hay, what will 8 acres produce? rThe expression 4d. is equivalent to lqr.; 4d. to 2qr.; id. to 3qr. QUEsTIONS - What is the rule? How are farthings sometimes written 7 szcr. Xiv.] MULTIPLICATION OF' COMPOUND NUMBERS. 123 17. If a family consume 49ga1 Sqt. lpL. of' molasses in one month, what quantity will be sufficient for one year? 18. John Smith has 12 silver spoons, each weighing 3oz. 17dwt. 1gr.; what is the weight of all? *19. Samuel Johnson bought 7 loads of timber, each mneasuring 7 tons 87ft.; what was the whole quantity? 20. If the moon umove in her orbit 13~ 11' 35" in 1 day, how far will she move in 10 days? 21. If 1 dollar will purchase 21b S8'75 ID 10 gr. of ipecacuanha, what quantity would 9 dollars buy? 22. If 1 dollar will buy 2A. 3R. 15p. 30yd. SfL. 10in. of wild land, what quantity may be purchased for 12 dollars? 23. Joseph Doe will cut 2 cords 97ft. of wood in 1 day; how much will he cut in 9 days? 24. If I acre of land produce 8ch. 6bu. 2pk. 7qt. 1 pt. of corn, what will 8 acres produce? ART. -@. When the multiplier is a composite number, and all its factors are within the table. Ex. 1. What cost 24 yards of broadcloth at 2.~. 7s. 1 ld. per yard? Ans. 57X. 1Qs. Od. OPERATION. In this question, we find the. s. d. o y numbher 24 equal to the prod2 7 11 — price of 1;yard..uct of 4 arnd 6; we therefore 4 miultiply the price first by 4, 9 1 1 8 price of 4 yards. and then that Iroduct by 6, 6 and the last product is the n alswer. 5 7 10 O 0 —price of 24 yards. sx. 2. What cost 360 tons of iron at 17E.~ 1s. ld. per ton? Ans. 6409,. 10s. 0d. OPERATION. ~. d. 1 7 1 3 1-price of 1 ton. In this question, we fild the factors of 360 to be 6, 1 0 6 1 6 6= price of 6 tons. 6, andt 10. We first multiG ply by 6, and then that ~,priduict by 6, and then 6 4 0 1 9 0 = price of 36 tons. again the I ast product by I 0 10. 6 4 0 9 10 0 = price of 360 tons. 124 MUffLTIPLICATION OF COMPOUND NUMBERS. [sxcT. xIv. RULE. -- ultiply the compound number first by one of the factors of the multiplier, and this product by another, and so on, until all the factors have been used as multipliers. The last product will be the answer. EXAMIPLES FOR PRACTICE. 3. If a man travel 3m. 7fur. 18rd. in one day, how far would he travel in 30 days? 4. If a load of hay weigh 2 tons 7cwt. 3qr. 181b., what would be the weight of 84 similar loads? 5. When it requires 7yd. 3qr. 2na. of silk to make a lady's dress, what quantity would be sufficient to make 72 similar dresses? 6. A tailor has an order from the navy agent to make 132 garments for seamen; how much cloth will it take, supposing each garment to require 3yd. 2qr. lna.? ART. A@e. When the multiplier is not a composite number, and is greater than 12, or, if a composite number, when all its factors are not within the table. Ex. 1. What cost 379cwt. of iron at 3~. 16s. 8d. per cwt.? Ans. 1452 ~. 16s. Sd. OPMATION. O~E 2RtSTION. Since 379 is not a com83 16 8 x 9 unitr posite number, we cannot re3 1 06 8 NX 9 units. solve it into factors; but we may separate it into parts, and 3 8 6 8 X 7 tens. find the value of each part 1 0 separately; thus, 379 = 300 — + 70 + 9. In the opera3 8 3 6 8 tion, we first multiply by 10, 3 hundreds. and then this product by 10, to get the cost of lOOcwt. 1 1 5 0 0 0 cost of 300cwt. To find the cost of 300cwt. we 2 6 8 6 8 cost of 70cwt. multiply the last product by 3 4 1 0 0 cost of 9cwt. 3; and to find the cobt of 70cwt., we multiply the cost.t 4 5 2 1 6 8 cost of 379cwt. of l0cwt. by 7; and then, to Aind the cost of 9cwt., we multiply the cost of 1cwt. by 9. Adding ihe several products, we obtain 1452L.C.16s. 8d. for the answer. RULE. - Multiply first by 10, and, if the multiplier contains hundreds, multiply this PRoOucT by 10 to get the productfor 100; then multiply the product of 100 by the number of HUNDREDS, the product of 10 Xnj the number of TENS, and the multiplicand by the number of UNITS; the sum of the several products will be the answer required. QuEsTIorNs.- Art. 106. xWhat is the rule for multiplying by a composite nunber? Give the reason for the rule. - Art. 107. flow do you find the cost of 300cwt. in the example? Of70cwt.? Of 9cwt.? What is the rule when the multiplier is large and is not a composite number 7 SECT. XV.] DIVISION OF COMPOUND IUMBERS. 125 EXAMPLES FOR PRACTICE. 2. If 1 dollar will buy 171b. 10oz. 13dr. of beef, how much may be bought for 62 dollars? 3. Wha.t cost 97 tons of lead at 2X. 17s. 9~d. per ton? 4. If a man travel 17m. 2Sur. 19rd. 3yd. 2ft. 7in. in one day, how far would he travel in 38 days? 5. If 1 acre will produce 27bu. 3pk. 6qt. Ipt. of corn, what will 98 acres produce? 6. If it require 7yd. 3qro 2na. to make 1 cloak, what quantity would it require to make 347 eloalks? 7. One ton of iron will buy 13A. 3R. 14p. 18yd. 7ft. 76in. of land; how many acres will 19 tons buy? 8. If I ton of copper ore will purchase 17TT. 14cwt. 3qr. 18ib. 14oz. of iron ore, how much can be purchased for 451 tots? Ans. 8003T. 8cwt. lqr. Olb. 10oz. ~ XV. DIVISION OF COMPOUND NUMBERS. ART. 1@89 DiVISXON of Compound Numbers is the process of dividing numbers of different denominations into any proposed number of parts. ART. 9,. To divide when the divisor does not ex. ceed 12. Ex. 1. If 9 acres of land cost 128~. 11s. 4d. 2qr., what is the value of 1 acre? Ans. 14&. 5s. 8d. 2qr. OPERATION. Having divided the l28~. by 9, we of 2. S. d. qr. o9 1 2 8 1I 4 1r find the quotient to be 14&. and 2.. remaining. We place the quotient 14iC. 1 4 5 8 2 under the 128~'., and to the remainder 2X'., equal to 40s., we add the Iu s. in the question, and divide the amount, 51s., by 9. We write the quotient 5s. under the lls., and to the remainder 6s., equal to 7'2d., we add the Id. in the question, making 76d., which we divide by 9, and write the quotient 8d. under the 4d. To the remainder 4d., equal to 1Gqr., we QUESTIONS. - Art. 108. What is division of compound numbers?- Art. 109. Where do you begin to divide? Why? When there is a rem-ainder after dividing any one denomination, what must be done with it? I1 * 126 DIVISION OF COMPOUiND NUMBERS. [SECT. XV. add the 2qr. in the question, and divide the amount, 18qr., by 9, and obtain 2qr. for a quotient, which we place under the 2qr. in the dividend, Thus we find the answer to be 14~C. 5s. 8d. 2qr. RULE. — 1. Divide the highest denomination of the dividend by the divisor, and, if there be a remainder, reduce it to the next lower denomination, adding to the number thus found the number in the dividend of the same denomination. 2. Divide the result thus obtained by the divisor; and, if there be a remainder, proceed as before, till all the denominations of the dividend are taken, or till the work is finished. The successive quotients will be of the same denominations with the successive nunmbers divided, or wil correspond with the several denominations of the dividend. EXAiIPLES FOR PRACTICF. ~. 8~3. 4. d. S5 a. 5. S. 5. 2) 10__3 4 3)58 14 9 5) 129 9 7 5 6 8 19 1 17 225 17 1 5. 6. 7.. a. d. qr. cMwtqr.. lb. o. Ton. cwt. qr.. b. )ll 1 2 14 4 2 6)1 3 1 21 12 7)1 3 11 0 0 18 15 8 3 18 3 17 10 14 15 312 8. 9. 10. cwt. qr. lb es. lb. eo. dr. m. fur. rd. ft. 8)154 2 15 8 9)143 5 5 6)587 4 8 12 19 1 8 15 151413 1I. 12. 13. deg. m. fur. rd. rd. yd. A. ia. fur. rd. ft. in. 8)1 45 3 2 7 24 9)5213 2 0 9 1 0)98 O 4 2 NOTE. -The answers to the following questions are found in the corresponding numbers in Multiplication of Compound Numbers. 14. What costs 1 yard of cloth, when 7yd. can be bought for 6;. 1Is. 3d.? 15. If a man, in 9 days, travel 112m. Ifur. 21rd., how fat will he travel in 1 day? 16. If 8 acres produce 21T. 5cwt. 1 qr. 121b. of hay, what will I acre produce? 17. If a family consume in I year 598gal. 2qt. of molasses, how much will be necessary for 1 month? qzsUTos. -- What is the rule i'or division of compound number?I SECT. xv.] DIVISION OF COMPOUND NUMBERS. 127 18. John Smith has 12 silver spoons, weighing 31b. 10oz. 11dwt.; what is the weight of each spoon? 19. Samuel Johnson bought 7 loads of timber, measuring 55T. 19ft.; what was the quantity in each load? 20. If the moon, in 10 days, move in her orbit 4S. 11 55' 50', how far does she move in 1 day? 21. If $9 will buy 24b 85 35 1 D 10gr. of ipecacuanha, how large a quantity will $1 purchase? 22. When $12 will buy 34A. OR. 32p. Syd. 5ft. 48in. of wild land, how much will $ 1 buy? 23. Joseph Doe will cut 24 cords 105 feet of wood in 9 days; how much will he cut in I day? 24. When 8 acres of land produce 25ch. 17bu. 3pk. 4qt. of grain, what will 1 acre produce? ART. I l @o When the divisor is a composite number, and all its factors are within the table. Ex. 1. When 24 yards of broadcloth are sold for 57~. 10s. Od., what is the price of 1 yard? Ans. 2X. 7s. lId. PrnarTION. In this question we find 6) 5 7 1 0 ice of 24 yards. the component parts, or factors, of 24 are 6 and 4) 9 1 1 8 - price of 4 yards. 4. We therefore first di-2 7 11 price of 1 yard. vide the price by one of these numbers, and then the quotient by the other. RULE. - Divide the dividend by one of the component parts, and the quotient thence arising by another, and so on until all the factors have been used as divisors; the last quotient twill be the answer. EXAMPLES FOPR PRACTICE. 2. If 360 tons of iron cost 6409i6. 10s. Od., what is the cost of 1 ton? 3. If a man travel 117m. 7fur. 20rd. in 30 days, how far will he travel in 1 day? 4. If 84 loads of hay weight 201 tons 4cwt. 2qr. Olb., what wnll 1 load weigh? 5. When 72 ladies require 567yd. Oqr. Ona. for their dresses, how many yards will be necessary for 1 lady? QUEsTION. -- Art. 110. How does it appear that dividing by 6 in Ex. 1 gies the price of 4 yards? What is the rule for dividing by a composite aumDCr'? 128 DIVISION OF COMPOUND NUMBERS. [SECT. xv. 6. When 132 sailors require 470yd. lqr. of cloth to make their garments, how many yards will be necessary for 1 sailor? ART. I I E. When the divisor is not a composite number, and is greater than 12, or, if a composite number, when all its factors are not within the table. Ex, 1. If 23cwt. of iron cost 171~. Is. 3d., what cost lcwt.? Ans. 7~. 8s. 9d. OPERATION... s.d. 2 3) 1 7 1 1 3 (7X. In this question we first divide the 1 6 1 pounds by 23, and obtain 7 for the quotient, 10 and 10)'. remaining, which we reduce to 2 0 shillings, and add the Is., and again divide by 23, and obtain 8s. for the quotient. 2 3) 2 0 1 ( 8s. The remainder, 17s., we reduce to pence, 1 8 4 and add the 3d., and again divide by 23, 1 J7 ~and obtain 9d. for the quotient. Thus, by 1 2 uniting the several quotients, we find the answer to be 7~'. 8s. 9d. 2 3)207 (9d. 207 RULE.- Divide in the same manner as when the divisor does not exceed 12 (Art. 109), and write down the whole operation as in the precedin, example. 2. If $ 62 will buy 10951b. 14oz. 6dr. of beef, how much may be obtained for $ 1? 3. Paid ~ 280. 5s. 9~d. for 97 tons of lead; what did it cost per ton? 4. If a man travel 662m. 4fur. 28rd. 3yd. 2ft. 2M. in 38 days, how far will he travel in 1 day? 5. When 98 acres produce 2739bu. lpk. 5qt. of grain, what will I acre produce? 6. A tailor made 347 garments from 2732yd. 2qr. 2na. of cloth; what quantity did it take to make 1 garment? 7. Wheni 19 tons of iron will purchase 262A. 3R. 237p. 25yd. ] t. 40in. of land, how much may be obtained for 1 ton? 8. if 45 I tons of copper ore will purchase 8003T. Scwt. lqr. 01b. 10oz. of iron ore, how much will I ton purchase? Ans. 17T. 14cwt. 3qr. 18b. 14oz. QvUEs'r-:Ois. -Art. 111. What ics the rule hen the divisor is large, an& not Z ~:orspomte eamaber?, SECT. XVi.] MISCELLANEOUS EXAMPLES. 129 ~ X~T. MISCELLANiEOIUS EXAMPLES IN MIILTTIPLICATION AND DIVISION OF COMPOUND NUMBEIRS. 1. Bought 30 boxes of sugar, each containing 8cwt. 3qr. 201b., but having lost 67cwt. 3qr. 121b., I sold the remainder for l~. 17s. 6d. per cwt.; what sum did I receive? Ans. 375.~.,. A company of 4I4 persons purchased a tract of land containing 110O7A 1R. 8p. John Smith, who was one of the comnpany and owned an equal share with the others, sold his part of the land for Is. 9I-d. per square rod; what sum did he receive?.Ans. 1101S. 12s. lId. 3. The exact distance from Boston to the mouth' of the Columbia River is't644m. 3fur. 12rd. A man, starting from Boston, travelled 100 days, going 18m. 7fur. 32rd. each day; required his distance from the mouth of the Columbia at the end of that time. Ans. 746m. 7fur. 12rd. 4. James Bent watts born J-uly 4, 1798, at 3h. 17m. A. M.; how long had he lived Sept. 9, 1807, at l1h. 19m. P. M., reckoning 365 days for each year, excepting the leap year 1L04, which has 366 days? Ans. 3353da. 20h. 2m. 5. The distance from Viera Cruz, in a straight line, to the city of Mexico, is 121m. Sfur. if a man set out from Vera Cruz to travel this distance, on the first day of January, 1848, which was Saturday, and travelled 3124rd. per day until the eleventh day of Ja nuary, omitting, however, as in duty bound, to travel on the Lord's Day, how far would he be from the city of Mexico on the morning of that day? Ans. 43m. 4fur. 8rd. 6. Bought 16 casks of potash, each containing 7cwt. 3qr. 181b., at 5 cents per pound. I disposed of 9 casks at 6 cents per pound, and. sold the remainder at' cents per pound; what did I gain? Ans. $ 203.78. 7. A merchant purchased in'London 17 bales of clothl for 17~. 1Ss. 10d. per bale. 1He disposed of the cloth at Havana for sugar at IX. %s. 6d. per cwt. Now, if he purchased 44cvwt. of sugar, what balance did he receive? Ans. 35~. Os. 2d, S. A and B commenced travelling, the same way, round an island 50 miles in circumfrn rerlce. A travels 17m. 4fur. 30rd. a day, antd B travels 12mn. 3ur. 20rd. a day; required how far they are apart at the end of 10 days. Ans. 1m. 4fUr. 2Ord. 130C CANCELLATION. [SECT. XVIi 9. Bought 760 barrels of flour at $ 5.75 per barrel, which i paid for in iron at 2 cents per pound. The purchaser afterwards sold one half of the iron to an axe-manufacturer; what quantity did he sell? Ans. 48 tons 15cwt. lqr. 221b. 10. Bought 17 house-lots, each containihng 44 perches, 200 square feet. From thlis purcchase I sold 2A. 2R. 240ft., and the remaining quantity I disposed of at Is. 24do. per square foot; what amount did I receive for the last sale? Ans. 59 14'. 19s. 5~d. 1l. J. Spoflord's farm is 100 rods square. From this he sold to H. Spaulding a fine house-lot and garden containing 5KV 3R. 17p.; and to D. Fitts a farm 50rd. square; and to R. Thornton a farm containing 3000 square rods; what is the value of the rernainder, at 1.75 per square rod? Ans. $ 6235.25. ~ XVII. CANCELLATION. ART. I f, CANCELLATION, as commonly used in arithlretic, signifies erasing or striking out any factor or factors common to the divisor and dividend. It can be employed in most rules involving multiplication and division of whole numbers, but is more especially important in abridging operations in mnultiplication and division of vulgar fractions, and in simple and compound proportions. ART. r Ii, If the diridend and divisor are botah dir7ilvae by the same number, the quotient is?not alltered. Thus, if the -dividend is 20 and the divisor 4, the quotient will be 5. Now, if we divide the dividend and divisor by some number, as 2, their proportion is not altered, and we obtain 10 and 2 respectively; and 10 -- 2 -_ 5, the same as the original quotient. ART. 4 Il. Iff a jhctor is cancelled in any number, the number is divided by that factor. Thus, if 15 is the dividend and 5 the divisor, the quotient will be 3. Noow, since the diviQe:STlONS.- Art. 112. What dow cancellaition sivnify In what rules is it o(1st advanitagteously employed 1 - Art. 113. What is the effect on the quotiient when the dividend and disvisor are both divtided by the same nusriber7 -Art. 114. What is the eiffect of cancelling a factor of tany number 7 SECT. xvI1. CANCELLATION. 131 sor and quotient are the two factors, which, being multiplied together, produce the dividend (Art. 50), it is plain, if we cancel the factor 5, thus 5 X 3 -- 15, the remaining factor 3 is the quotient, and the dividend 15 has been divided by 5. ART. Z. Method of cancelling, when there is one factor, or more, common to the dividend and divisor. Ex. 1. A man sold 25 hundred weight of iron at 5 dollars per hundred weight, and expended the money for flour at 5 dollars per barrel; how many barrels did he purchase? Ans. 25 barrels. OPER ATION. The price of the iron per hundred Dividend Y X - 25 25 weight must be multiplied by the Divisor b number of hundred weight sold, to obtain the value of the whole; and this product, being divided by the cost of the flour per barrel, will give the number of barrels bought. But we may indicate this multiplication and division by their signs, without actually performing the operations. In this example, 5 is a common factor of the divisor and dividend; therefore, we divide the divisor and dividend by this factor, or, which is the same thing, cancel it in both, and obtain 25 for the quotient. 2. Divide the product of 15, 2, 4, and 6, by the product of 4, 3, 2, and 5, and find the quotient. Ans. 6. OPERATION. Dividend i5 X 0 X 4 X 6 90 r 3 =6 Quotient. Divisor: X 3 >< /0 X 5 _5 In this example we cancel the common factors in the divisor and dividend, and divide the product of the remaining factors in the divideind by the product of those in the divisor, and obtain the quotient 6. RULE. - 1. Write the dividend above the divisor, with a horizontal line between them, as in division. (Art. 47.). 2. Cancel the actor or factors common to the dividend and divisor, and, if there is only one factor remaining in the dividend and one QUF.sTIoNs.-Art. 115. How do you arrange the dividend and divisor for cancellation? H-Tow do you then proceed? Is the factor 5, in E.x. -, reduced to 0, or 1, by being cancelled? If all tile factors, both in the dividend and divisor, are cancelled, what will the quotient be? If more than one factor remain in the dividend and divisor after cancelling, howdo you obtain the quotient? What is the rule for cancellation? 132 CANCELLATION. LsECT. XVIT. in the divisor, divide the factor of the dividend by the factor of the divisor. 3. If there are two or morefactors remaining in the dividend, and two or more in the divisor, divide the product of the factors of the dividend by the product of thefactors of the divisor. NOTE. - When a factor is cancelled, it is not reduced to 0, but to unity or 1. Therefore, when all the factors are cancelled, either in the divi dend or divisor, the factor I remains, and must be used as a factor of the divisor or dividend, as the case may be. EXAMPLES FOR PRACTICE. 3. Divide 27 X 16 by 27. Ans. 16. 4. Divide 42 x 19 by 19. Ans. 42. 5. Divide the product of 8, 6, and 3, by the product of 6, 3, and 4. Ans. 2. 6. Divide the product of 17, 6, and 2, by the product of 6, 2, and 17. Ans. 1. 7. Sold 15 pieces of shirting, and in each piece there were 30 yards, for which I received 10 cents per yard; expended the money for 10 pieces of calico, each containing 15 yards; what was the calico per yard? Ans. 30 cents. ART. 1 1L6. When a number in the dividend and another in the divisor have a common factor. 8. Divide the product of 12, 7, and 5, by the product of 2, 4, and 3. Ans. 17g. OPERATION. Dividend SO X 7 X 5 35 Divisor 2 X A X - It will be seen, in this example, that 4 in the divisor is a factor of 12 in the dividend. Therefore we divide 12 by 4, cancelling these numbers, and use the quotient:3 instead of 12. The operation may be carried still further by cancelling this factor 3, and a similar one in the divisor. 9. Divide the product of 20, 13, and 9, by the product of 13, 16, and 1. Ans. 11}. QuESTIONss.- Art. 116. How do you proceed when a number in the dividend and another in the divisor have a common factor? Is the common fac. tor always one of the two numbers? SECT. XVII.] CANCELLATION. 133 OPER.ATION. Dividend M0 X X - X 9 45 Divisor X X 1 4 Quotient. 4 In this example, 20 in the dividend and 16 in the divisor may be divided by 4. We therefore cancel these numbers and use their quotients in the operation. RULE. - When there is a factor in the dividend and another in the divisor which may be divided by the smaller factor, or by some other number, without a remainder, cancel these factors, and use the quotients arising from the division instead of them. EXAMPLES FOR PRACTICE. 10. Divide the product of 9, 8, 2, and 14, by the product of 3, 4, 6, and 7. Ans. 4. 11. Divide the product of 16, 5, 10, and 18, by the product of 8, 6, 2, and 12. Ans. 12a. 12. Divide the product of 22, 9, 12, and 5, by the product of 3, 11, 6, and 4. Ans. 15. 13. Divide the product of 25, 7, 14, and 36, by the product of 4, 10, 21, and 54. Ans. ljg. 14. Divide the product of 26, 72, 81, and 12, by the product of 36, 13, 24, and 54. Ans. 3. ART. 117. When the product of two or more factors of the dividend is equal to the product of two or more factors of the divisor, or conversely. 15. Divide the product of 8, 5, 3, 16, and 28, by the product of 10, 4, 12, 4, and 7. Ans. 4. OPERATION. 4 Dividend ~ X' X $ X r X ~ - 4 Quotient. Divisor 0~ X 4 X.X' X 4 X f The product of the factors 8 and 5 in the dividend is equal to the product of 10 and 4 in the divisor; therefore we cancel these factors. Again, the product of the factors 3 and 16 is equal to the product of QUESTIONS. - What is the rule for cancelling when a number in the dividend and another in the divisor have a common factor 7 —Art. 117. How do you proceed when the products of two or more factors in the dividend and divisor are alike? 12! 134 PROPlERTIES AND RELATIONS OF NUMBERS. [SECT. XVIIl. the factors 12 and 4, which we also cancel, and there remains 28 di vided by 7, which is equal to 4. RULE. - When the product of two or more factors in the dividend is equal to the product of two or more factors in the divisor, cancel these factors in both. NOTE.-If the product of two or more factors in the dividend is equak to any one factor in the divisor, or conversely, these factors may be cancelled in both. EXAMPLES FOR PRACTICE. 16. Divide the product of 8, 4, 9, 2, 12, 16, and 5, by the product of 4, 6, 6, 3, 8, 4, and 20. Ans. 2. 17. Divide the product of 6, 15, 16, 24, 12, 21, and 27, by the product of 2, 10, 9, 8, 36, 7, and 81. Ans. 8. ~ XVIII. PROPERTIES AND RELATIONS OF NUMBERS. ART. L X S. ALL numbers are either odd or even. An odd number is a number that cannot be divided by 2 without a remainder; thus, 3, 7, 11. An even number is a number that can be divided by 2 without a remainder; thus, 4, 8, 12. Numbers are also either prime or composite. A prime number is a number which can be divided only by itself or a unit; as 1, 3, 5, 7. Numbers are said to be prime to each other when no number greater than a unit will divide them without a remainder, thus, 7 and 11 are prime to each other. NOTE. -For definition of composite numbers, see Art. 41. ART. 1 ~9. A prime factor of a number is a factor which can be divided only by itself or a unit; thus, the prime factors of 21 are 3 and 7. QUESTIONS.-What is the rule for cases of this kind? If the product ol two or more factors is equal to any one factor, how do you proceed? - Art. 113. What are all numbers? What is an odd number? What an even number? What other distinctions of numbers are mentioned? What is a prime number? When are numbers prime to each other? What is a corn polite numbrer? -l Art. 119. What is a prime factor of a number SECT. xv1l.]) PROPERTIES AND RELATIONS OF NUMBERS. 135 NOTE. - Unity or 1 is sometimes regarded as a prime factor: but since multiplying any number bv 1 does not alter its value, we shall omit it when speaking of the prime factors of numbers. ART. l2@. To find the prime factors of a number. Ex. 1. It is required to find the prime factors of 24. Ans. 2, 2, 2, 3. OPEATIO.N. Since 24 is not a prime number, we divide 212 4 it by 2, the least prime number greater than 21 1 2 1, and obtain the quotient 12. And since 1:2 21 6 is a composite number, we divide this also 1]-'6 by2, and obtain a quotient 6. We next 3 divide 6 by 2, and obtain 3 for a quotient, which is a prime number, and therefore the 2 X 2 X 2 x 3 24 division ends. The several divisors and the last quotient constitute all the prime factors of 24. RULE. - Divide the given number by the least prime number, greater than 1, that will divide it without a remainder, and then this quotient, if a composite number, in the same manner, and thus continue the division until a prime number is obtained for a quotient. The several divisors and the last quotient are its prime factors. NOTE. - The composite factors of any number may be found by multiplying together two or more of its prime factors. EXAMPLES FOR PRACTICE. 2. What are the prime factors of 36? Ans. 2, 2, 3, 3. 3. What are the prime factors of 48? Ans. 2, 2, 2, 2, 3. 4. What are the prime factors of 56? Ans. 2, 2, 2, 7. 5. What are the prime factors of 144? Ans. 2, 2, 2, 2, 3, 3. 6. What are the prime factors of 3420? Ans. 2, 2, 3 3, 5, 19. 7. What are the prime factors of 18500? Ans. 2, 2, 5, 5, 5, 37 A COMIEMON DIVISOR. ART. 121. A common divisor of two or more numbers QUESTIONS.- Is 1 usually considered a factor when speaking of the prime factors of a number? — Art. 120. Vhat is the rule for finding the prime factors of a number? -low can the composite factors of a number be found 1 - Art. 121. What is a conimon divisor of two or more numbers? 136 PROPERTIES AND RELATIONS OF NUMBERS. [SECT. XVUnX. is any number that will divide them without a remainder; thus, 2 is a common divisor of 2, 4, 6, and 8. ART. 12t. To find a common divisor of two or more numbers. Ex. 1. What is the common divisor of 10, 15, and 25? Ans. 5. OPERATION. We resolve each of the given numbers into two 1 0- 5 X 2 factors, one of which is common to all of them. 1 5 5 X 3 In the operation 5 is the common factor, and there2 5 = 5 X 5 fore must be, a common divisor of the numbers. RULE. - Resolve each of the given numbers into two factors, one of which is common to all of them, and this common factor is a common divisor. NOTE.- A number is divisible by 2, when the last figure is even; by 4, when the last two figures are divisible by 4; by 8, when the last three figures are divisible by 8; by 5, when the last figure is 5; and by 5 ox 10, when the last figure is 0. EXAMPLES FOR PRACTICE. 2. What is the common divisor of 3, 9, 18, 24? Ans. 3. 3. What is the common divisor of 4, 12, 16, 28? Ans. 2 or 4. 4. What is the common divisor of 12, 16, 3', 48? Ans. 4. 5. W:Vhat is the common divisor of 14, 21, 35, 49? Ans. 7. 6. What is the common divisor of 8, 16, 24, 48, 64, 96, 128, and 144? Ans. 4. THE GREATEST COMMON DIVISOR. ART. 13o. The greatest common divisor of two or more numbers is the greatest number that will divide each of them without a remainder. ART. 1124. To find the greatest common divisor of two or more numbers. Ex. 1. What is the greatest common divisor or measure of 84 and 132? Ans. 12. QUESTIONs. — Art. 122. What is the rule for finding a common divisor? Wbhen is a number divisible by two? By 4? By 8? By 5? By 101Art. 123. What is the greatest csmamon divisor of two or more numbers 1 SECT. XVIIl.] PROPERTIES AND RELATIONS OF NUMBERS. 137 OPERATION. As 12 will divide 36, it is evident 8 4) 1 3 2 (1 it will also divide 48, which is equal 8 4 to 12 + 36. It will also divide 84; because 84 is equal to 36 +- 48; for, 4 8) 8 4 (i as 12 will divide each of these numn4 8 bers, it is evident it will divide their sum.'For, the same reason, it will 3 6) 4 8 (1 also divide 132, which is equal to 84 3 6 + 48. We therefore find, that 12 1 2) 3 6 (3 is a common divisor of 84 and 132. 3 6 To prove that 12 is the greatest common divisor, we resolve 84 and 132 into their prime factors; thus, 84 2= X 2 X 3 X 7, and 132 - 2 X 2 X 3 X 11. Now it is evident that 84 cannot be divided by any number except by one of its -prime factors or the product of two or more of them. The same is true of 132. Both these numbers, therefore, can be divided by 12, since it is the product of the first three prime factors of. each of them; thus, 2 X 2 X 3 -= 12. Again, if any number greater than 12 will divide both of these numbers, it must be a number common to the factors 7 and 11; but 7 and 11 are prime to each other, and therefore can have no common factor greater than 1. Hence 12 is the greatest common divisor of 84 and 132. RUJLE. - 1. Divide the greater number by the less, and if there is a remainder, divide the last divisor by it, and so continue dividing the last divisor by the last remainder until nothing remains, and the last divisor is the greatest common divisor. 2. If there are more than two numbers, first find the greatest common divisor of two of them, and then of that common divisor and one of the other numbers, and thus proceed until all the numbers are used. The last common divisor, thus obtained, is the greatest common divisor required. NOTE. -From the preceding demonstration it may be seen, that the greatest common divisor can also be found by resolving the given numbers into their prime fhetors, and multiplying together those which are common to all the numbers; thus 2, 2, 3, are factors, common to 84 and 132, and their product, 12, is their greatest common divisor. EXA1MXPLES FOR PRACTICE. 2. What is the greatest common divisor of 85 and 95? Ans. 5. 3. What is the greatest common divisor of 72 and 168? Ans. 24. 4 What is the greatest common divisor of 119 and 121? Ans. 1. QUE.STIONS. - Art. 124. How does it appear that 12 in the example is the,greatest common divisor of 84 and 132? WVhat is the rule for finding the greatest common divisor? What other mode of finding the greatest common divisor 7 138 PROPERTIES AND RELATIONS OF NUMBERS. [SECT. XVIII. 5. What is the largest number that will divide 324 and 586? Ans. 2. 6. What is the largest number that will divide 582 and 684? Ans. 6. 7. What is the greatest common divisor of 32 and 172? Ans. 4. 8. What is the largest number that will divide 84 and 1728? Ans. 12. 9. What is the greatest common divisor of 16, 20, and 26? Ans. 2. 10. What is the greatest common divisor of 12, 18, 24, and 30? Ans. 6. A' COMMON MULTIPLE. ART. 125o A multiple of a number is a number that can be divided by it without a remainder; thus 6 is a multiple of 3. ART. 126. A common multiple of two or more numbers is a number that can be divided by each of them without a remainder; thus 12 is a common multiple of 3 and 4. ART. 12P7. The least common multiple of two or more numbers, is the least number that can be divided by each of hem without a remainder. ART. 12fSe To find the least common multiple. Ex. 1. What is the least common multiple of 6, 9,; 12? Ans. 36. OPERATION. In this operation we first divide the 3 6 9 12 numbers by 3, a number that will divide x 2 3 4 most of them without a remainder, and 1 3 2 write the quotients in a line below. We next divide by 2, writing down the quotients and undivided number, as before. 3 X 2 X 3 X 2 = 36 Then, since these numbers are prime to each other, we multiply together the divisors, last quotients, and undivided number, which are all the prime factors of 6, 9, and 12, and thus obtain 36 for the least common multiple. To prove that 36 is the least common multiple, we resolve 6, QUESTIONS. - Art. 125. What is a multiple of a number? - Art. 126. What is a common multiple of two or more numbers 7 - Art. 127. What is the least common multiple of two or more numbers?- Art. 123. How does it appear that 36 is the least common multiple of 6, 9, and 12? SECT. xv1I1.] PROPERTIES AND RELATIONS OF NUMBERS. - 9, and 12 into their prime factors; thus, 6=2 X 3; 9=3 X 3; 12 = 2 X 2 X 3. Now, since the least common multiple of two or more numbers must contain all the factors of those numbers, it is evident that, if any different prime factor occurs more than once in any one of the numbers, it must be taken the same number of times as a factor of the multiple, or it will not contain all the factors of the numbers. In the example, 2 and 3 are the different prime factors, and as 3 occurs twice in 9, and 2 twice in 12, each must be taken twice as factors of the least common multiple; thus, 2 X 2 X 3 X 3 = 36, the same as in the operation. RULE. - Divide by such a number as will d&vide MOST of the given nuttmbers without a remainder, and set the several quotients with the several undivided numbers in a line beneath; and so continue to divide, until no number greater than unity will divide two or- more of them. mi7hen nultiply all the divisors, the last quotients, and undivided numbers together, and the product is the least common multiple. NoTE 1.- Care must be taken always to divide by a number that will divide most of the given numbers, or a multiple may be obtained which is not the least common multiple. NOTF 2.- When one or more of the given numbers are factors of any one of the other numbers, the factor or factors may be cancelled, and a common multiple of the remaining numbers found as in other examples. Thus, if the common multiple of 5, 15, 30, 7, 14, and 28 were required, we might cancel the 5, 15, 7, and 14, because 5 and 15 are factors of 30, and 7 and 14 are factors of 28. EXAMPLES FOR PRACTICE. 2. What is the least common multiple of 8, 4, 3, and 6. Ans. 24. 3. What is the least common multiple of 7, 14, 21, and 15? Ans. 210. 4. What is the least common multiple of 3, 4, 5, 6, 7, and 8? Ans. 840. 5. What is the least number that 10, 12, 16, 20, and 24 will divide without a remainder? Ans. 240. 6. What is the least common multiple of 9, 8, 12, 18, 24, 36, and 72? Ans. 72. 7. Five men start from the same place to go round a certain island. The first can go round it in 10 days; the second in 12 days; the third in 16 days; the fourth in 18 days; the fifth in 20 days. In what time will they all meet at the place from which they started? Ans. 720 days. QUESTIONS -- What is the rule for finding the least common multiple What caution is given in the note 7 VULGAR FRACTiONS. [SECT XIX ~ XIX. FRACTIONS. AnT. -9. A FRACTION is an expression denoting a part (f any number or thing. The term fraction is derived from the Latin word frango, which signifies to break; from the idea that a number or thing is broken or separated into parts. Fractions are of two kinds, Vulgar and Decimal. TVULGAR FRACTIONS. ART. 12,. A VULGAR FRACTION is any part of an integer or whole number, expressed by two numbers one above the other with a line between them. The number below the line is called the denominator, and the number above, the numerator; Numerator 3 Thrce Thus, Denominator 5Fiftl. The denominator shows into how many parts the whole number is divided, and gives a name to the fraction. The numerator shows how many of these parts are taken, or expressed by the fraction. The numerator and denominator together are called the terms of the fraction. ARTn. E31o There are six kinds of vulgar fractions, viz. proper, improper, mixed, simple or single, compound, and complex. A proper fraction is one whose numerator is less than the denominator; as, W. An improper fraction is one whose numerator is equal to, or greater than, the denominator; as, —, g. T* he word vulgar here means common, and is employed in this connection to denote the kind of fractions in most common use. QUEoSTIO Ns. - Art. 129. What is a fraction 7 From what is the term derived, and what does it signify? How many kinds of fractions, and what are they called? - Art. 130. What is a vulgar fraction, and how expressed? What is the number ahove the line called?'The number below the line? What does the denomninator show? What the numerator? What are the numerator arid denominator together called - Art. 131. How many kinds of vulgar fractiou, ansud what are their inmei 7 Give the definition of each. EcT. xMx.] REDUCTION OF VULGAR FRACT1ONSS 141 NOTE. - A fraction, strictly speaking, is less than a unit; hence, if the numerator is equal to, or greater than, the denominator, it expresses a unit or more than a unit, and is therefore-called an improper fraction. A mixed number is a whole number with a fraction, as, 7 T,, 5f A simple or single fraction has but one numerator and one denominator, and may be either proper or improper; as, a, s. A compound fraction is a fraction of a fraction, connected by the word of; as, i of -I of -. A complex fraction is a fraction having a fraction or a mixed number for its numerator or denominator, or both; as 7 8sk7 i' 9i' al' 9'ET ART. 132. When we divided 479956 by 6 (Art. 49, Ex. 12), we had a remainder of 4, which could not be divided by 6, and therefore we wrote it over the divisor with a line between them. This is an expression for division without performing the operation, and is called a fraction; the number above the line being the numerator, and the one below the denominator. Hence, Fractions originate from division; the numerator answvers to the dividend, and the denominator to the divisor. ART. 133. From what has preceded, we perceive that the value of afraction is the quotient arising from the division of the numerator by the denominator, or from the expression of this division. Thus the quotient of 6 or 6 -+- 2 is 3; and the quotient of 3 or 3 * — 4 is ~. REDUCTION OF VULGAR FRACTIONS. ART. 134. REDUCTION of Fractions is changing theirform or terms without altering their value. ART. 135. To reduce a fraction to its lowest terms. Ex. 1. Reduce -a to its lowest terms. Ans. i. QUESTIONS. - When the numerator is equal to, or greater than, the denominator, is the expression, strictly speaking, a fraction? - Art. 132. From what do fractions originate? To what does the numerator answer? To what the denominator 1 - Art. 133. What is the value of a fraction? - Art. 134. What is reduction of fractions? 142 REDUCTION OF VULGAR FRACTIONS. [SECT. XIX. OPERATION. We first divide both terms of the fraction by 2, 3) a number that will divide them both without a 2) - --- - = - remainder, and obtain W. We next divide this result by 3, and obtain i, which cannot be divided by any number greater than 1, and therefore the fraction is in its lowest terms. The result would have been the same, if we had first divided by 6, the greatest common divisor. By dividing the numerator and denominator of a fraction by the same number, it is evident we cancel equal factors in both (Art. 113),. and diminish them in the same proportion; consequently, their relation to each other is not changed, and the value of the fraction remains the same. Therefore, Dividing the numerator and dencmnnator of a fraction by the same number does not alter the value of the fraction. RULE I. - Divide the numerator and denominator by any number greater than 1, that will divide them both without a remainder, and thus proceed with the successive results until the operation can be carried no farther. Or, RULE II. -Divide both the numerator and denominator by their greatest common divisor, and the result will be the fraction in its lowest terms. NOTE. -A fraction is in its lowest terms, when its numerator and denominator are prime to each other. (Art. 118.) EXAMPLES FOR PRACTICE. 2. Reduce - to its lowest terms. Ans. -. 3. Reduce 3 to its lowest terms. Ans. l. 4. Reduce J2 to its lowest terms. Ans. -. 5. Reduce T96T to its lowest terms. Ans. 2. 6. Reduce 1Of to its lowest terms. Ans. -. 7. Reduce -2 to its lowest terms. Ans. a 8. Reduce "Ir to its lowest terms. Ans. A. 9. Reduce s191 to its lowest terms. Ans. is89c. 10. What is the lowest expression of —? Ans. II. Art. 1"0. To reduce a mixed number to an improper fraction. Ex. 1. In 73 how many fifths? Ans. 3_o. QU ESTIONS. - Art. 135. How do you reduce a fraction to its lowest terms 7 WVhy does dividing both terms of a fraction by the same number not alter the value of the fraction? Has the same value as i? Why? What is the rule for reducing a fraction to its lowest terms? How may you know when a fraction is in its lowest terms? srcrT. xIx.] REDUCTION OF VULGAR FRACTIONS.; 143 OPERATION. 5 Since there are 5 fifths in 1 whole one, there will be -35 ffths. 5 times as many fifths as whole ones; therefore, in 7 35 fifths. there are 35 fifths, and the 3 fifths being added makie 38 3 fifths, which are expressed thus, Wi. -- RULE. - Multiply the whole number by the denominator of the fraction, and to the product add the numerator, and place their sum over the denominator of the fraction. NOTE. - 1. Any whole number may be expressed in the form of a fraction, by taking the number itself for a numerator, and a unit for the denominator; thus, 5 may be written I. 2. To reduce a whole number to a fraction of the same value, having a given denominator, we multiply the whole number by the given denominator, and make the product the numerator; thlus 5, reduced to a fraction, having 3 for a denominator, becomes 1. EXABIrLES FOR PRACTICE. 2. In 8. dollars how many sevenths Ans. Ah. 3. In 31 oranges how many fourths? Ans. At. 4. In 9T-r gallons how many elevenths? Ans. -O-. 5. Reduce 8 r to an improper fraction. Ans. a6. Reduce 15 7 to an improper fraction. Ans. -its. 7. In 187 how many ninths? Ans. _19. 8. In 161T1-M how many one hundred and seventeenths? Ans.t-a$s 9. Change 43an,1 to an improper fraction. Ans. a. 10. What improper fraction will express 27-T? Ans. 3a6 0 11. Change 11 T{-1 to an improper fraction? Ans. 12l 312T 12. Change 125 to an improper fraction. Ans. 1-_ 13. Change 25 to an improper fraction, having 6 for a de. nominator. Ans. _5Q. 14. Reduce 75 to ninths. Ans. 675.15. Change 343 to the form of a fraction. Ans. I-41. 16. Reduce 84 to fifteenths. Ans. 1_62a. ART. 137. To reduce improper fractions to whole or mixed numbers. QuESTIONS. -Art. 136. What is the rule for reducing a mixed number to an improper fraction i Give the reason. How can a whole number be expressed in the form of a fraction? How do you reduce a whole number to a fraction of the same value, having a given denominator? 144 RKRE1UCTIION OF VULGAR FRACTIONS. tSECT XIX, Ex. 1. How many dollars in 37 dollars Ans. $ 2y%. OPERATION. This question may be analyzed by saying, As 16 I ) 3 7 (25I sixteenths make one dollar, there,.will be as many 3 2 dollars in 37 sixteenths as 37 contains 16, which is ~ 5 2j9 times, = $ 2 6. RULE. -- Divide the numerator by the denominator, and, if there be a remainder, place it over the denominator at the right hand of the whole number. EXAMPLES FOR PRACTICE. 2. Reduce,s6- to a whole number. Ans. 12. 3. Change 11_R! to a mixed number. Ans. 10O1. 4. Change -1-il-l to a mixed number. Ans. 10T+-. 5. Change J-1T3?- to a mixed number. Amrs 1-5. 6. Reduce _?onvQ to a mixed number, Ans. 142". 7. Reduce T 7 to a whole number. Ans. 1. 8. Change 46L to a whole number. Ans. 567. 9. Reduce.774gw to a mixed number. Ans. 932 10. Reduce a0.4s to a mixed number. Ans. 4 —t-. ART. 13S. To reduce a compound fraction to a simple fraction. Ex. 1. Reduce t of 17T to a simple fraction. Ans. ~~. OPERATION. 8 To show the reason of the operation, this t X T1 6-7T question may be analyzed by saying, that, if rT of an apple be divided into 5 equal parts, one of these parts is As of an apple; and, if - of 1-1T be 5s, it.is evident that i of dT will be 7 times as much. 7 times -~ is -72; and if - of iT be {7~, { of 7T will be 4 times as much. 4 times -{ are -5. Or, by multiplying the denominator of AT by 5, the denominator of A, it is evident we obtain 5 ofr T - -s, since the parts into which the number or thing is divided are 5 times as many, and consequently only i as large, as before. Again, since - of -T 7 {, of T7 will be 4 times as much; and 4 times If1 =1 a. This process will be seen to be precisely like the operation. QUESTIONS. —Art. 137. What is the rule for reducing improper fractions to whole or mixed numbers? Give a reason for the rule. - Art. 138, How do you reduce a compound fraction to a simple one? Give the reason fior the operation. SECT. XIX.] tREDUCTION OF VULcAR FRACTIONS. 145 Ex 2. R, educe T of A of o of 6off to ta simple fraction. AAns. -r OPERATION BY CANCELLATION. Since some of the nunmerators 2 X x4 >< 2 X X and denominators to be multiplied 2 together are alike, we may cancel m X X / X - X 1 f. these common factors, according ~ 5to the rules of cancellation. FUtLE. - 1. Multiply all the numerators together for a new numerator, and all the denominators for a nezv denominator, and then reduce the fraction to its lolwest terms. 2. If there are factors in the numerator similar to those in the de-,nozinator, cancel them in the operaton. NOTE. - All whole and mixed numbers in the compound fraction must be reduced to improper fractions, before multiplying the nusmerators and denominators together.!,~XABIPLES FORt PRACTICE. 3. What is I of 4of? An — ~%. 44. WVNhat is o- f 9T of'7? Ars, 5a~. 5. What is 7- of -9T of of of -? Arns T2 6 Change 1 of - of 3 of U% of 7 to a simple fi-cticr. Ans. 93O. 7. Required the value of 3 of' -T of 1I of 17 of' &5. Ans. k 8. Reduce I of 8- of of o f 7- to a simple fraction. -5 _9 T T?'1; -A 3 9. Reduce ~ of T of 7 of -% of 4-4 to a simple firaction..Ans. 6" 55n 10. Reduce 1-{ of - of -7r to a simple fraction. Ans.'-. 11. Reduce AT of 22 of 15 of 95 to a whole number. AlsS. I. 12. Rteduce - of A 4 of of S3 of It to a simple fraction. Ans. 1 QrEsrTIoNs. -When there are common factors in the nu-ilerator aned denominlator, how may the operation be shortened'! What is the rule? What must be done with all whole and mlixed numbers in the compound fraction i 146 REDUCTION OF VULGAR FRACTIONS. Ls cT. XIz, A CoMBIoN D NOML1_NATOR AIRT. B9. A coimmion denominator of two or more fractions is a common multiple of their denominators. The least common denominator is the least.common multiple. NOTE. -Fractions have a comnmon denominator, when all their de nominators are alike. ART. L1@0. To reduce fractions to a common denominator. Ex. 1. Reduce a, t, and 7- to a common denominator. Ans 1 4:4160 169 OPERATION. 3 X X 8 = 1 4 4 new numerator for -4 5X4X8-160 5 1 6 0 7X4X6=168 6 " 6=8Tl 4 X 6 X 8 - 1 9 2 common denominator. We first multiply the numerator of - by the denominators 6 and 8, and obtain 144 for its numerator. WVe next multiply the numerator of } by the denominators 4 and 8, and obtain 160 for its numerator; and then we multiply the numerator of I by the denominators 4 and 6, and obtain 168 for its numerator. Finally, we rnultiply all the denominators together for a common denominator, and write it under the several numerators, as in the operation. By this process the numerator and denominator of each fraction are multiplied by the same numbers, and consequently, both being irrereased an equal nurrmber of times, their relation to each other is not changed, and the value of the fiaction remains the same. (Art. 133.) Therefore, tJultipiyin. the numerator and adenoiZinator of a fraction by the same number does not alzer the value of the j'raction. RULE-. MuiteIplyp eachr numerator Ainto all the denolinators except its own for a new numeralor; and tall the denominators into each other for a common denominator, NoTE.- Fractions of this form may often be reduced to lower terms, without destroying their commlon denominator, by dividing all their numerators and denominators by a common divisor. QUESTIONS.- Art. 139. What is a conllmon denominator of two or more fractions? What is the least comrmon denomninator? When have fractions a common denominator? - Art. 1410. How do you find a common denominator of two or more tfractions? Give the reason of the operation. W;'hat inference- is drawn from it? 5V-hat is the rule for finding a common denominator'T Ilow may fractions having a common denominator be redu.lced to lower terms t EZC'T. xsx.] REDUCTION OF VULGAR FRACTIONS. 147 EXAMPLES FOR PRACTICE. 2. Reduce 3- and - to common denominators. Ans. I- -f2, or9 2, 9 3. Reduce 4, iS and -j to a common denominator. 4. Reduce -5-, and -I to a common denominator. Ans. 352 231 2B0 5 Reduce s-, -- and to a commnon denominator. Ans. 44Q, 445, 456, Q 44, 3, 3 6 5. Reduce -, 9, -, and } to a common denominator. dA iso60 384 840 240 4 20 20 o S0 nSaT.,OG -, -g- ~, or _-%, 0 orFI ART.,, l. To reduce f-actions to their least common denominator. Ex.i, leduce 4, -', and * to the least common denominator. OPERATI[ON. 3 6 3 t",! co mmons deno-ninator. 1 2 4 4 X 2 8 nume rator for =:3 0I 63 ~2 X -- 10 numerator fbr -. 1 21 1 X 7- 7 numerator for -i _ +, X 2 X 2 = 12, tle leaist common denonminator. Having first obtained a coomnmon multiple, or denominator of the given fractions, we take the part of it expressed by each of these fractions separately for their new numerators. Thus, to gret a new numerator for j, we take? of 1s, the common denominator, by dividingr it by 3, aud multiplying the. quotient 4, by 2. We proceed in this manner with each of the fractions, and write, the numerators thus o'btained' over -the common denominator, NOTE. - The change in tilhe terms of the fractions, in reducing themn to the least common denominator by this process, depends upon the same principle as explained in the preceding article. RurLE. Find the least common multiple of the denominators or the several fractions, and it will be their least common denominator. 2. Divide the least common dc.rnominator by the denominator of each of the given fractions, and mnultiply the quotients by their respecti.ve ustmerators, -nd t izir products.witsl be the numerators of the fr3actions required. ITorT.- Compound fiaetions in4'-mt bla reduced to simple ones, whole Qus'rioNss. Art. 141. I-ow do you find the least common denominator of two or more fractions? Upon what principie does this process depend? What is the rule for reducing fractions to their least comnron denominator? What must be done with compound fractious, whole numhbers. and mixed nurrbers [148 ADDITION OF VULGAR F'RACTIONS. [ SET. XI. and mixed numbers to improper fractions, and all to their lowest terms, before finding the least common denominator. EXAMIPLES FOR PRACTICE. 2. Reduce,3 4-, 4, and 7 to the least common denominator. Ans. 90 96 100 105 3. Reduce 2 4, and r2 to the least common denominator. Ans. 1485, 92 880 6e0 4. Reduce, -, and 7X to the least cornmon enominator. Aos. 3, 310 5. [Reduce 9 - 1 and 53 to the least common denomninator. Ans. 8-1!~', 25s 1 5- 2 6. Reduc3 2- 3- 5 5 and - 7 to the least common denominaltor. Ans. 4 48 2 1, 5 2I 1, o Ans. 47 24 0 2 IC 2 21T 9.:Reduce'7, s5i 79, and 5 to the least common denominator. Ans. 4 44 1 30 2 32. 6. Reduj e 5 4 7 east common neator. nians. 12, 1 6 2 1 1. Reduce 7, 4 55, 7, and 9 to the least common denomi inator. Ans. 3 I4 20v t 3l:. _10. Reduce 3, 4, 5, 74 and 9 to the least common denomino,ADDns.I jTIr' G LA ow l a 6 20 28 36 A-DDITIPION OF VULGAR~ FRAkOT-,iO'S. ART. 4 ~. ADDITION of Vulgar Fractions is the process of finding the value of two or more fractions in one sum. ART. 2 11. To add factions that have a commilon (t.nanl nator. Ex.. Add, 7:-, s5 and _- together. Ans 2. eOPER ATION. Thoese fractions all being 1 — 2 -+4 -t+ 5 —- 6 s euenlhs, that is, havinlg 7 fbs 7, 7 7 7 7 p a common denominatcor, we sinrply add their numerators together, and write the sum over the common denomina.tor. RujLE. Add together the nnmerator.s o the fractions, and place their sumn ove"r the commaon denZonator, ani redurce the fr.ction if necessary. (-; l:s lor4's. -Art. 14". What is addition of vulgaLr fractions? -- Art. 143. mia.t. is the rnle for adding fracti.ions having a comr;.ixonl denomin.ator' Give th:e reason f;)r the. tle. SECT. XIx.] ADD'TIOI F0' VULGAR' FRACTIONS. 149 EXAMPLES FOR PRACTICE. 2. Add 1,'T, -T, T7', 9T, and -11 t-ogether. Ans. 3j-. 3. Add Pt, -Tr 7-1 9, and I-I together. Ans. 2T1r. 4. Add 3, 2 19, and 2j together. Ans. 2. 5. Add 7, is, N, and ~4 together. Ans. 2r. 6. Add I11 9, j}~,, and 1v9 together. Ans. 1112 7. Add 1 68, 1, and TT9 together. Ans. 11-06. TYTYD TTI 1TT o 6 T ART. 44.], To add fractions that have not a common de. nominator. Ex. 1. What is the sum of, a, and J-Z? Ans. 1~. OPaERATION. 216 8 12 i24 common denominator. 3 3 4 6 6 4x5=20 2| 1 4 2 S 3X3-= 9 new numerators. ___-"8 12 2x7= —14 2 1 Sum of numerators 43 2 X 3 X 2 X 224. Comn denominator 2-4 = Having found the common denominator and new numerators, as in Art. 141, we add the numerators together and place their sum over the common denominator, and reduce the fraction. RULE. - Reduce mixed numbers to improper frcttons, and compound fractions to simple fractions; then reduce all the fractions to a common denominator; and the sum of their numerators, written over the common denominator, will be the answer required. NOTE. - In adding mixed numbers, it is sometimes more convenient to add the fractional parts separately, and then to add their sum to the amount of the whole numbers. EXAMPLES FOR PRACTICE. 2. What is the sum of ~-, I, and? Ans. 21-. 3. What is the sum of 9,, 1, and 5? Ans. 1XO. 4. What is the sum of:.- and -? Arns. I-n. 5. What is the sum of,, 3 and t? Ans. 2g~. 6. Add 4- 8, T, and ~ together. Ans. I —39~5. 7. Add +I, 5 1 and 7 1 together. Ans. t23. S. Add 3S, 9 7, 4 and 8 — together. Ans. 22,&. QUESTIONS. - Art. 144. What is the rule for adding fractions not having a common denominator? How may mixed numbers be conveniently added 1 150 SUBTRACTION OF VULGAR TRACTIONS. [SECT. XIx 9. Add ~, 5 -, 3 5, 6, and ltogether. Ans. 5S7S. 10. Add s T9 1 7 1 9 12, and }~ together. if) T9 lT~ TI29 I-IT 1 Ans. 614401. 11. Add 2 of 3 to r of }. Ans. 1-I. 12. Add - of 8 to 1fi of'. Ans. l' 13. Add -of 2 to -' of. Ans. TX8S9' 14. Add 2 of 3 of - to 5 of 6 of i. Ans., 15. Add, of 3 o'f I to f of. Ans. I7f 16. Add t3i- to 41' Ans. 83T 17. Add 4L-a to 56- Ans. 10. 18. Add 1i7 -o I 8-)i Ans. 36n. ART. AB;1, To add any two fractions whose numerators are a unit. Ex. 1. Add -d to -}. Ains. 9' OI-RATIO'; WVe first find the Sum of the denominators 4+ 5 9 product of the deProduct of the denominators 4 X 5 O nominators, 4 X 5ich is 20, and then their sum, which is 9, and write the former foir t-he denomlinator of the required fraction, and the latter for the numerator. TIhe reason of this operation will be seen, when we consider, that the process reduces the fractions to a common denominator, and then adds their numerators. RuLE. - Add together the gwven denominators, and place the sum over thetr product. LEXA1.rPLES FOR PRACTTCG. 2. Ad d to 1-j- r to ", 7to 7, - o to 4. Add d to } t to T -I Yto -, tt to -l,&toU tt 6. Add I to,r - to'I, I to -, - to -, - to -,' to to 2, 1 -, I T2 14 5 TY TY -G- 5a a4 7 5. Add I to x7,' to J-, _-I to 9o-, - to' % t 7 1 49 Y L 51 Y -, to I to 7. Adtd t, to 9 Ito'1 t I t:-o r. 1 t o I t - 7Y _9 - to _9~, SUII;TRACTION OF VJULGAI A1 FR ACTLOf S. ArL.' I4A1 SUBTRACTION of Vuilgar Fractions is the protess of finding the difierence between two fractions of unequal values. QUESTIOgS.- Art. 145. Xtihlat is the rule for adding two fractions whenl the un-umeraturs are a nlitl -Whtat is the reason ibr this rule 1 - Art. 146. What is subhu-actijn of vulg'ar fractions'? S:CT.XIX.] SUBTRACTION OF VULGAR FRACTIONS. 151 ART.,ost7 To subtract fractions that have a common denominator. Ex. 1. From 7 take s. Ans. -. OPERATION!ff. In this operation we take the less numerator from 7 ~ —2 the greater, and write the difference over the common 9 9 - denominator. Hence the following RULE.- Subtract the less numerator from the greater, and write the difference over the common denominator, and reduce the fraction if aecessary. EXAMPLES FOR PRACTICE. 2. From j1 take f2w Ans.,T. 3. From l, take g7. Ans. 4-. 4. Prom 2 4 take 4 Ans. 23 5. From 267 take 19 Ans, 148 6. From T6 2g take T-f-z1A% Ans. 2. 7. From ~ take i5%. Ans. I. 8. From 97, take'17 Ans. 5. ART. l4. To subtract fractions that have not a common denominator. Ex. 1. From 1 take 7-. Ans. {-. OPERATION. 4116 12 48 common denominator. 4 3 16 3x13 -- 3 9 } new nurnerators. 121 4x 7 =283 4 x 4 x 3- 48. 4 4 311 difierence of numerators, 48 common denominator. laving found the common denominator and new numlerators as in Art. 141, we subtract the less numerator firom the greater, and place the difference over the conlmmon denominator. RULE. - Reduce the fractions to a common denominotor, then wrilte the dlifference of the numerators over the cormmon7 denomizinlaor. NoTE. - If the minuend or subtrahend, or both, are comporlund fractions, they must he reduced to siniple ones. QVUESTION. Art. 147. XWhat is tfhe rule for subtractigia fractions bhaving a corrniion denominator?- Art. 148. What is the rule fotr subtractinog fractionts not having a commnon denominator ( If the mninlenld or subtrahelnd is a compound fraction, what must be done? 152 SUBTRACTION OF VULGAR FRACTIONS. LSECT. XIx. EXAMPLES FOR PRACTICE. 2. From -Tw take 2AT. Ans. ~-. 3. From ~ take d-L. Ans. {-. 4. From Ad take 27.. Ans. 45% 5. From ~I take ad. Ans..9 6. From A take T,. Ans. TT. 7. From I" take xa. Ans. WF'7T S. From I take 1. Ans 90 9. From Al take of. Ans. 9. 10. From; of 9 take I of. Ants. 7 11. From I of T9 take 1l of jj. Ans. 3e.. 12. From,f of 12X- take - of 94.. Ans. 4jr ART. 11zg9. To subtract a proper fraction or a mixed number from a whole number. Ex. 1. From 16 take 2~. Ans, 13M. OPERATION. Since we have no fraction from which to subFrom 1 6 tract the A, we must add 1, equal to 4, to the Take 21 T~Take 2: minuend, and say 4: from t leaves:. We Rem. 1 3T write the: below the line, and carry 1 to the 2 in the subtrahend, and subtract as in subtraction of simple numbers. The same result will be obtained, if we adopt the following RULE. - Subtract the numerator from the denominator of the fraction, and under the remainder write the denominator, and carry one to the subtrahend to be subtracted from the minuend. NOTE. - If the subtrahend is a mixed number, we may, if we choose, reduce it to an improper fraction, and change the whole number in the minuend to a fraction having the same denominator, and then proceed as in Art. 148. EXAMPLES FOR PRACTICE. 2. 3. 4. 5. 6. From 12 1 9 13 14 1 7 Take 43- 3r 9ST 8 6_ j Ans. 7} 1 5+ 311 5} 1 0~, 7. From 23 take 13-i. Ans. 9. 8. From 47 take -3-. Ans. 46,47,. 9. From 139 take 754 1- Ans. 63-4. QU s SrT IONus. - Art. 149. What is the rule for subtracting a proper fraction or mixed number from a whole number? Give the reason for this rule. :cjT. xsx.] SUBTRACTION OF VULGAR FRACTIONS. 153 AnRT. I@. To subtract a mixed number from a mixed number. Ex. 1. From 92 take 3-. Ans. 52. FIRST OPERATION. In this example, we first reduce the From 92 -- 9~IO TFom T - %5 fractions to a common denominator by Take 5- 3 multiplying the terms of the upper fracRiem. 3-'x tion 2, by 5, the denominator of the lower, thus,2 5 1; and then the terms of the lower fraction -, 3 X 7 21 by 7, the denominator of the upper, thus, 3x 7 -,5. Now, since we cannot take 23 from ~o,Q we add 1, equal to x5, to the 1 5 in the minuend, and obtain 45. We next subtract 21 from 5, and write the remaindIer, 4, below the line, and carry 1 to the 3 in the subtrahend, and subtract as in simple numbers. SECOND OPeRATION. From 9 S- 6S 5 O- 32R 5 In this operation, we first rom T7 - 178 = 1W2reduce the mixed numbers to Take 3 = J?5- improper fractions, and these Rem. -9 5a24 fractions to a common denom. inator, as in the first operation. We then subtract the less fraction fiom the greater, and, reducing the remainder to a mixed number, obtain the same answer as before. RULE I. - Reduce the fractions, if necessa;y, to a common denomt nator, and if the lower fraction is greater than the upper, subtract the numerator of the lowerfractwon from the common denominator, and to the remainder add the numerator of the upper fraction. VIlrite this sum over the common denominator, anzd carny I to the subtrahend, and subtract as in simple numbers. But if the upper fraction is greater than the lowVer, subtract the less from the greater, and the whole numbers as beifore. Or, Ru LE Ii. - Reduce the mixed numbers to improper fractions, then to a common denominator, and subtract the less fJaction from the greater. WiVrite the remainder over the common denominator, and if the fractions is improper, reduce it to a whole or mixed number. EX AIMZPLES FOR PRACTICE. 2. 3. 4. 5. 6. From 97 7- 87 9'I 03 Take 5-1 3A 44- 3:- l ) As' 7 -7 5: 3 Ans. 32 37 5.. QUESTIONS. — Art. 150. How do you reduce the fractions of mixed mumbers to. comnmon denolninator? How does it appear that this process reduces theml to a common denominator? How do you then proceed? What other mtethod of subtractillg nixed numbers't What are thtl; two rules? 154 SUBTRACTION OF VULGAR FRACTIONS. [SECT. XIX. 7. 8. 9. 10. 1. L From 1 2 1 613 19 9 9 7- 83 7.1 Take 9~ 5 5`. 1 s1 1 9 Ans. 21 1 0 3~6 7 84 6 7 12. From 19*- take 7 T. Alns. 119. 13. From 151 take 8. Ans. 62, 14. From 911 take 3'-. Ans. 5z21. 15. From 71 -l take 13-k7. Ans. 57 s87. 16. From 61 T take 331i. Ans. 27z7. 17. From a hogshead of wine there leaked out 12j- gallons; how much remained? Ans. 505 gallons. 18. From $ 10, $ 2- were given to Benjamin, $ 3I to Lydia, $ 1 to Emily, and the remainder to Betsey; what did she receive? Ans.. 3-. ARnT. ~ To subtract one fraction from another, when both fractions have a unit for a numerator. Ex. L What is the difference between - and? Ans. 2To OpIiRATION. Difibrence of the denomrinators 7 - 3 - 4 Product of' the denominators 7 X 3 -= 1 We first find the product of the denominators, which is 21, and then their difference, which is 4, and write the former for the denominator of the required fraction, and the latter for the numerator. RULE. -- WTrite the divterence of the denominators of the fractions over their p'oduct. EXALrPLE S FOR PRACTICE. 2T 0'ke fio nm 1-S $f' lrol j, - from -}I 7 prom -X - from, fi4om7 fom fom 3.'ake f'or o' nT from.1 from in from 4-,.ra.('- t01 f0mrom. ~'r~ake3 -~ fro ~,-z-, fr om ~, I- from,-, 6 To 1. Take I frot5 4-, from j7 from -, 7 from 4v-a'f from Y-. 7'5. Tarle - -0mno, - fom 1,f -' g from 1 fro QUESTIONS. -Art. 151. What is the rule for subtracting one fraction from another-whea both fractions have a ulnit for a numerator? What is tihe reaason for the rule? SECT. XMX.] MULTIPLICATION OF VULGAR FRACTIONS. 155 MULTIPICATION OF VULGAPL, FIRACTIONS. ARTY. 1O. MULTIPLICATION of Fractions is the process of multiplying fractions together, or whole numbers and fractions into each other. ART. REA5 To multiply a fraction by a whole numbe~. Ex. 1. Multiply 7 by 4. Ans. 3y. FIRST OPERATION. In the first operation, we multiply the 7X4 23 8 ----- 3~ numerator of the fraction by the whole number, and obtain 3, for the answer. It is evident, that the fraction 7 is multiplied by multiplying its numerator by 4, since the parts taken are 4 times as many as before, while the parts into which the number or thing is divided remain the same. Therefore, Multiplying the numerator of a fraction by any number multiplies the fraction by that number. SECOND OPERATION. In the second operation, we divide the ~4 3 -31 denominator of the fiaction by the whole number, and obtain 3- for the answer, as before. It is evident, also, that the fraction - is multiplied by dividing its denominator by 4, since the parts into which the number or thing is divided are only } as many, and consequently 4 times as large, as before, while the parts taken remain the same. Therefore, Dividing the denominator od a fraction by any number multiplies the fraction by that number. RULE I.- Mfultiply the numerator of the fraction by the whole number, and under the product write the denominator. Or, RULE ILo - Divide the denominator of the fraction by the whole number, when it can be done without- a r'emainder, and write the quotient under the numerator. EXAIIPLES FOR PRACTICE. 2. Multiply - by 9. Ans. 6.3 3. Multiply 1e- by 5. Ans. 2-. 4. Multiply 16 by A. Ans. 1d. 5. Multiply 4-9 by 85. Ans. 49. QuEsTIONS. — Art. 152. What is multiplication of fractions? -- Art. 1.53. fHow is a fraction multiplied by the first operation? Give the reason of the operation. What inference is drawn friom it? flow is a friaction multiplied by bhe second operation? What is the reason of the operation? What inference is drawn from it? What is the first rali for m ultiplyinga fraction by a whole number? The second 7 156 MULTIPLICATION OF VULGAR FRACTIONS. [SECT. XiX. 6. Multiply -{1 by 83. Ans. 76 1 2 7. Multiply) { by 189. Ans. 166r1. 8. Multiply I-l - by 365. _ns. 352l6l. 9. Multiply s7 by 48. Ans. 43. 10. If a man receive 3 of a dollar for one day's labor, what will he receive for 21 days' labor? Ans. $ 7-. 11. What cost 561b. of chalk at Y of a cent per lb.? Ans. $0.42. 12. What cost 3961b. of copperas at 9T of a cent per lb.? Ans. $ 3.24. 13. What cost 79 bushels of salt at 7 of a dollar per bushel? Ans. $ 69-. ART. 1l. To multiply a whole number by a fraction. Ex. 1. Multiply 15 by A. Ans. 9. FIRST OPERATION. In the first operation we divide the whole 5) 1 5 nnumber by the denominator of the fraction, 3 X 3- 9 and obtain - of it. We then multiply this quotient by 3, the numerator of the fraction, and thus obtain - of it, which is 9. SECOND OPERATION. In the second operation we multiply the 1 5 - whole number by the numerator of the fraction, and divide the product by the denomina 4 5_ 5 5 9 tor, and obtain 9 for the answer, as before Therefore, Multiplying by a fraction is taking the part of the multiplicand denoted by the multiplier. RULE I. -- Divide the whole number b/y the denominator of the frac lion, when it can be done without a remainder, and multiply the quotient by the numerator. Or, RULE 11. - Multiply the whole number by the numerator of the fraction, and divide the product by the denominator. EXAM.PLES FOR PRACTICE. 2. Multiply 36 by -. Ans. 28. 3. Multtiply 144 by 4I. Ans. 88. 4. Multiply 375 by {1. Ans. 325. 5. Multiply 2277 by s. Ans. 1610. 6. Multiply 376 by 11. Ans. 24357. QES'I'~IoNS. A —art. 154. I-low do you multiply a whole number by a fraection accordlin( to the first operation? HIow by the second? What inference is dr iawIn troml the operations? What is the first rule for multiplying a whole number by ia friaction 1 The secoild? tECT. XIX.] MULTIPLICATION OF VULGAR FRACTIONS. T17 7. Multiply 471 by -2-T Ans. 83 8. Muiltily 871 by IJ-. Ans. 2:-~-. 9. Multiply 867 by. Ans. 6-i-balS. ART. gJ. mTo multiply a whole and mixed number together, Ex. 1. Multiply 17 by 6a. A.ns. 114-. OPERATION. 1 63 MWe first multiply 17 by 6, the whole 6 number of the multiplier, and then by tlhe 1 0 2 fractional part, %, which is simply taklinr % of 17 - 1 2- of it, and add the two products together 143Ex. 2. Multiply 75 by 4. Ans, 30-. OPERATION. We first multiply 3 in the multiplicanpd by 5 4, the multiplier; thus, 4 times s are 2 equal to 2%, which is in effect taking x- of o of 4 = 2%- the multiplier, 4. We then multiply the 2 8 whole number by 4, and add the two prod3 0% ucts together. R.ULE. - Wi/rite the less number under the greater, and if the FRAcTION is in the multiplier, take a part of the multiplicand denoted by thle fraction; but if it is in the multiplicand, talce a part of the multiplicr denoted by thefraction, and in each case add the product thus obtained to the product of the whole numbers. XAMIPLES FOR PRACTICE. 3. Multiply 9% by 5. Ans. 46r1 4. Multiply 1~2-a by 7. Ans. 881k. 5. Multiply 9 by 81. Ains. 80}. 6. Mtultiply 10 by 7%. Ans. "7 I. 7. Multiply 116 by 8. Ans. c-a e S. What cost 7.rlb. of beef at 5 cents per pound? An. 8 0. t37 9. What cost 237 bbli. of flour at 6g p er barrel r? Ans. ~ 141,r 10. WJhat cost 83-yd. of cloth at $ 5 per yard? rns. 8 41Z. 11. What cost 9 barrels of vinegar at $ G, per bavrel? A.n'. 87 3. QUESTIONS. -Art. 155. What is the rule for multiplyirg a whole eind mixed number together i Does it make any difference whichi is taken for the multihlie.r? ci~ 158 MULTIPLICATION OF VULGAR FRACTIONS. [SECT. XX 12. What cost 12 cords of wood at $ 6.371 per cord? Ans. $ 76.50. 13. What cost I lcwt. of sugar at $ 9~ per cwt.? Ans. $ 103114. What cost 4` bushels of rye at $ 1.75 per bushel? Ans. $ 7.65kz 15. What cost 7 tons of hay at $ 11- per ton? Ans. $ 83-, 16. What cost 9 doz. of adzes at $10- per doz.? Ans. $ 95~17. What cost 5 tons of timber at $ 3 per ton? Ans. $ 15-. 18. What cost 15cwt. of rice at $ 7.621 per cwt.? Ans. $ 114.372. 19. What cost 40 tons of coal at $ 8.37- per tonl? Ans. $ 335 ART. 156. To multiply a fraction by a fraction. Ex. 1, Multiply; by i. Ans. T7r, OPERATION B3Y CANCELTATION. OPERATION. 7 X - - X X - <7 = 9f 4 12 To multiply 7- by; is to take X of the multiplicand; (Art. 154). Now, to obtain 3 of 7, we simply multiply the numerators together for a new numerator, and the denominators together for a new denominator (Art. 138). Therefore, liutl -tiplying one fraction by another is the same as reducing compound fractions to simple ones. RULE I. -- elfZtiply the numerators together for a new numeralor, and the denominators t gether for a new denominator; then reduce the fraction.to its lowuest terms. Or, RULE II. - Cancel all the common factors in the numerators and denominators, and then multiply the remaining factors together as before. EXAMPLES FOR PRACTICE. 2. Multiply -E by b ry Ans. -7T. 3. Multiply a by'U. Anss J QUErSTIONS. -Art. 156. What is the first rule for multiplying one f-action by another? Hiio doea it appear that this operation multiplies the fraciion of the multiplicad? What is the inference drawn fromn it? What isi tie.t secnrd rtnl y? sCT. Rxx.] MULTIPLICATION OF VULGAR FRACTIONS. 159 4. Multiply 8 by 1`. Ans. ~. 5. Multiply t by any. Ans. -. 6. Multiply 15 by ma. Ans. I. 7. Multiply 5 by Anr. Ans. 8. 8. Multiply 62- by 23. Ans.. 9. What cost; of a bushel of corn at a- of a dollar per bushel? Ans. - of a dollar. 10. If a man travels 1T of a mile in an hour, how far would ae travel in 1 of an hour? Ans. v of a mile. 11. If a bushel of corn will buy 7 of a bushel of salt, how much salt might be bought for ~ of a bushel of corn? Ans. 2 of a bushel. 12. If ~ of 3 of a dollar buy one bushel of corn, what will -@ of 9 of a bushel cost? Ans. OF of a dollar. 13. If 3 of 4 of 9fT of an acre of land cost one dollar, how much may be bought with 2 of $ 18? Ans. 1 acres. A.T. 157'. To multiply a mixed number by a mixed number, it is only necessary to reduce them to improper fractions, and then proceed as in the foregoing rule. Ex. 1. Multiply 43 by 6}. Ans. 302. OPERATION. 4 S- 3; 6 3-=S 3 ~. 4 23 X20 92 EXAMPLES FOR PRACTICE. 2. Multiply 7A by 8s. Ans. 60j%. 3. Multiply 47 by 91- Ans. 45 32. 4. Multiply 117 by 8-. A1ps. 99~g. 5. Multiply 12 by 11-. Ans. 147-. So What cost cost 73 cords of wood at.$5 per cord? Ans. 8 412.. What cost 7yd. of clotli at $ 3- per yard? Ans. 25-1, 3 c What cost 63 gallo0s of molasses at 283 cents per galion Ans. 15i2s" 9.:? a man travel 34 miles in one hour, how far will he travel 1;: 9- hours? Ans. 34r-'-v. UQE sTIO 0hs- Art. 157. How do you multiply a maixed number by a mi.ed. wimb)er? O160 DIVISION OF VULGAR i-./,1'-',"';, [SBECT, X s 10. What cost 351-1 acres of lanei -; r acre?,::!,.I~;. 9167~ 13 11. How many square rods of land in a garden, whlich is 97-1 rods long, and 493 rods wide? Ans. 4810 51 rods. DIVISION OF VULGAR FRACTIONS. ALT. ]1 SI. DiVISION of Vulgar Fractions is the process of dividing fractions by fractions, or whole numbers and fractions by each other. ART. V 9D: To divide a fraction by a whole number. Ex. 1. Divide 9 by 4. Ans. ~. FIRST OPRRnTION. In this operation we divide the nunerator of the 8 4 2 fraction by 4, and write the quotient'2 over the de-'-9 9 —- nomninator. It is evident this process divides the fraction by 4, since the numbern and size of the parts illto which the whole nulnber is djiviied remnain the samle, while only ~ of the number. of parts is expressed bv the fraction. Therefore, Dividing the nun'erator of a frtactzn by any numnber divides the fraction by that num,,nber. Ex. 2. Divide by 9. Ans. a-. FECON PA TIECS'.ON. WVe multiply the denominator of the fraction by ~ +,he divisor, 9, and write the product under the nu7 X 9 e- Wi3 i S' It is evident this process divides the fraction, since multiplying the denominator by 9 makes the number of parts inIo v litch thle whole numbler is divided 9 times as many as before, iand (cotr;steqietly each part can have but t of its fornmer value. Notv, if each part has but } of its former:value, -while only the saime nulni,(:. of iart. is eexpressed ( by the fraction, it is piain thl'e fhiacuon hait 1fecn dividedt by 9. Therlf'ire, i/lttlply/ing the'denoineatr of a jcr;io by acly'numTber diridc:.; the fraction by that numiber. Rt U L EI D.ivide tlhe numerator of the Jfiactiont by the who.,-':;, oe,, ivhe itI Calt b-e dZone wilhout a remntainder, and write the't,,i:7::.over the citnor/nmit tr. Or, R IJut If. -- JMltuliply tie denominator J' the fraction by tihe c S,::,;:[ ll.mIt'e, aid 1,rite the product under th.e numcr7ator. t c;s'itS. r. ti.. t.158. W.Vlhat is division of vulgar fractions?-': i59.,lo)w is the friation divided bv the first operation? How loes it a!,l;5a.r tilat this rtc). Ps s dilides the fraction? iWhat inftrence niav be drawn eir)ln thiis vcni>ri,tion, -ow is, a frictioni divided by the sec-nd operatioln? Wil vout xiCil aiu hw t!`; pnr,,cess divides the fiactionr? \Vhat inference iV, drawn rooin this o ter 3ai,,? hlati is the lirst rule for dividing a fractle' by a whole nue:.'.seOr??sie secund' sE:cT xXx. DIVISION OF VULGAR FRACTIONS. 161 EXAMPLES FOR PRACTICE. 3. Divide 6- by 3. Ans. 2 4. Divide s by 6. Ans. 5. 7 7 5. Divide T by 12. Ans. -rT6. Divide -1- by 8. Ans. 4, 7. Divide I7 by 9. Ans. 3. 8. Divide 7 by 15. Ans. -5. 9. Divide — 0 by 75. Ans. 5. 10. Divide i by 12. Ans. TST. 11. John Jones owns $ of a share in a railroad valued at $ 117; this he bequeathes to his five children. What part of a sliare will each receive? Ans. -. 12. JDivide 2 by 15. Ans. 3-%. 13. Divide rT by 28. Ans. xinu 14. James Padge's estate is valued at $10,000, and he whas given; of it to the Seamen's Society; ~ of the remainder he gave to his good minister; and the remainder he divided eqally amonl his 4 sons and 3 daughters. What sum -will eiach of his children receive? Ans. I$ 6$-? 4-. ART, a1 O. To divide a whole number by a fraction, Ex. 1. How many times will 13 contain -. Ans. S0&. For convenience, we invert the OPERATION,. terms of the divisor, and thle 3 13 X 7 91 multiply the whole nu-mber by 13 - 30- 3 the original denominator, and divide the product by the numerator. T1he ieason of this operation is evident, since 13 will coontain, as ma.ay times as there are sevenths in 13, equal 91 sevenths. Now if.[3 icontain 1 seventh 91 times, it will contain 3 as many tirmes as 91 1oir contain 3, equal to 301. ARui,. - )Aultiply the whole.;nmber by the dConoininator C/ e j raec t:.u, cFud divide the product by the numerator-. ExxAMPLES FOR PRACTICE. o Divide I8 by -. A.s. 24. 3. Divide 27 by -. Ans. ~91%. -. Divide 23 by.- Ans. 92. QUESTONS. - Art. 160. VWhat is the rule for d.i-dirg a whole nuramber by a fra'ction? Give the reason for the rule. 162 DIVISION OF VULGAR FRACTIONS. ISECT. X1x. 5. Divide 5 by A. Ans. 25. 6. Divide 12 by }. Ans. 16. 7. Divide 16 by 3. Ans. 32. 8. Divide 100 by 1{. Ans. 1113'. 9. I have 50 square yards of' cloth; how many yards, 3 of a yard wide, will be sufficient to line it? Ans. 83- yards. 10. A. Poor can walk 3T7 miles in 60 minutes; Benjamin cal walk r9 as fast as Poor. I-low long will it talie Benjamin to walk the same distance? Ans. 73~ minutes. ArZT. -311. To divide a mixed number by a whole number. Ex. 1. Divide 173 by 6. Ans. 0'3 OPERATION. Having divided the whtole number 6)14743 as in simple division, we have a re4 o3 43 ~ mainder of 5], which we reduce to 8~a- -1; 8 X6==4; an improper fraction, and divide it by the divisor as in Art. 159. Annexing 2 ~l 4- 24. a 2+ this fraction to the quotient 2, we obtain 243 for the answer. tRULE. — Divide as in whole numbers, as far as the division can be carried, and, if the remainder is a mixed nudmber, reduce it to an i7n proper fraction, and then divide it by the divisor; bt if the remrninder is a simple fj action only, merely divide it by the divisor. (Art. 159. ) E:XAVIPLES FOR PRACTIC,. 2. Divide 14 3 by 7. Ans, 2s. 3. Divide 18 by 8. Ans. 2)r7 4. Divide 27B- by 9. Ans. 83 -l-1 5. Divide 31i,0 by I1. Ans. 2 1.1 6, Divide r78 by 12. cins. 6 j 7. Divide 1891- 1 by 4 Ans. 47 I 3 d. Divide 10T1L' y 3..Ans.'35. 9. Divide *, 14{} amo1g 7 men. Ans. $ 29-,-3 i.0, iivide 18I4-0 anmong 8 boys. Ans. 13-a I. -W}at is tl-e vrale of 2 of a dollar. Ans..34 -. t2, Thivide e if1-_11 -among 4 boys and 3 girls, and give t;e girls twice as rrmucl as the boys..Ars Boy'r share-, 1 04; Girl's share $ 21 -. i3. If t$14 will purchase L- of a ton of copperas, what quantity will, 1;, Ipurchase A Ans. Icwt. Oqr. 241b. Qenr:; r.-.:.t. 1l, What is the rule for dividing a mixed number by a w hole nt::ma.f?e ECTo. xix.] DIV ISlON OF VULGAR F(ACTIONo. 163 ART. il2. To divide a whole number by a mixed number Ex. 1. Divide 25 by 4`. Ans. 5.( OPEIRATION. WVe first reduce the divisor and dividend to fifths |2 and then divide as in whole numbers. 5 I 5 TThe reason why the answer is in whole numbers, and not in fifths, is because thie divisor and dividend 23)'3( -2' srwere both multiplied by the same number, 5, and 115 therefore their rela.tion to each other is the same as 10 before, and the quotient will not be altered. R uVLE. - Reduce the divisor and dividend to the same parts as are denoted by tihe den-lominator of the frJaction in the divisor, and then divide as in whole numbers. ~XA3IAPLES FOR PRgACTICEo 2. Divide 36 by 97. Ans. "2:3. Divide 97 l)y 1 - Ans. 6'r. 4. Divide 113 by 21o. Ans. 5Y5'. 5. Divide 34"2 by P1454ko Ans. 23-~90 6. There is a board 19 feet in length, which I wish to saw into pieces 29 feet long; what will be the number of pieces, and how many feet will remain? Ans. 7 pieces, 2 feet. ART. 360. To divide a fraction by a fraction.'Ex. 1 Divide' by-. Ans. 1 IOPERATION. In this operation, we invert the - 7- X 9 - 6, -- 32, terms of the divisor and then proceed as in Art. 156. The reason of this process will be seen, when we consider that the divisor 4 iS an expression denoting that 4 is to be divided by 9. Now, regarding 4 as a whole number, we divide the fraction 1 by it, by 7 7 multiplyin;l the denomiinator; thus, 8X4: —';* But the divisor 4 is 9 times too great, since it was to be divided by 9, as seen in the origi-. nal fraction therefore the quotient,, is 9 times too small, an.Id must 7 x 9 63] be multiplied by 9; thus, -p 1' - II. B3y this operation, we have multiplied the denominator of the dividend by the numrera,-tor of the divisor, and the numerator of the dividend by the deniurimixator of the divisor. Hlence thle irTIO)NS.. -- Art. 162. 2. What is the rule for dividing a;xhole by a mFix.ed nubeher i -ow does it appear that this process does rot aiter the quotient? - Art. 163. How do you divide a fiaction by a fraction? Give the reason why this process divides the fraction of tthe dividend. 164 DIVISION OF VULGAR FRACTIONS. [SErC. XIX. RU -E. - Invert the divisor, and cancel all the factors common to the aurnerators and denominators, and then proceed as in multiplication of fractions. NOTE. - When the divisor and dividend have a common denominator, t3heir denominators cancel each other, and the division may be performed by simply dividing the numerator of the dividend by the numerator of the divisor. EXAMPLES FOR PRACTICE. 2. Divide -F by -. Ans, 1W-53. Divide I by T. Ans. 3.3. Divide -T by 1-. Ans. s5 5. Divide 1 by II Ans. 2-.2 5. Divide 2 by if. Ans. 2-2-. 6. Divide -9 by 7..Ans. 6-rj. 7. Divide - by 2T Ai-s. 4-. 8. Divide -9 by 23. Ans. 6. 9. Divide ~- by 7- Ains. 2~. 10. Divide 2 of 7 by } of A. Ans. 18-.o!ll Divide - of 6of by of by of Af Ans.. 12. D)ivide of of by of of-. Ans. 3a ART. 4: C, To divide a mirined number by a mixed nuntmbox it is only'necessary to reduce them to improper fractions and proceed as in the foregoing rule. (Art. 163.) Ex. 1. Dividle 7` by 3M. Ans. 2-I OPERATION. A X.T=1G72=q 6 EXAMPLEPS FOR PRACTIrCE,. Divide-7r tby 4. A, Is.!). 4. i)ivide ~i by 7~. Ans. 7,A. 4, U)ivide 411 by 5i. Ans. 2T- -z. 6 Di videi ItS by 14.b. Ans. 8 —. 7 D)ivide 81f by 9I Arns, n 3' 8 ). D vi d e 3 of E A of 7 by ~ of 3t. Ans. 11-...;si'S-x s. —- \'at is the rule for dividing one fraction by another? How ayl fi.-t';(c!s be divided when they have a common denominator'1 Does this pr- cess.lf1i'r in principie fromn dte other?- Art. 164. How do you divide a cmixed number ty a Tixed ilumbuer r EECT. IXX.] COMPLEX FRACTIONS. 165 COPLEX - FRACTIONS. ART. 6 @. To reduce complex to simple fractions. Ex. 1. Reduce -~ to a simple fraction. Ans. nz. OPERATION. Since the numerator of a fraction is - = - X s the dividend, and the denominator the di5- X visor (Art. 132), it will be seen by this operation, that we simply divide the numerator - by the de. nominator 3, as in division of fractions. (Art. 163.) Ex. 2. Reduce X8 to a simple fraction. Ans. 7!. OPERATION. We reduce the numerator, -8 -_ 1 _ >< - _:Z 8, and the denominator, 4(, to 4 2 ~ ~ - - improper fractions, and then proceed as in Ex. 1. Ex. 3. Reduce t to a simple fraction. Ans. 1H. OPERATION-. We here reduce the f -- Y denominator, of R, to Y of'IS -4 a simple fraction, and then proceed as before. From the preceding illustrations we deduce the followinga RIuL:E. - Reduce whole and rizxed numbers to improper 3fractions, a:nd comni-poun(dfractio'ns to simple ones, and then divide the ntuirzertor o/' the conplea fraction, by the denominator, according to the rule / 3r the diCisioz ofj'actions. EXAMPLES FOR PRACTICE. 4, Redauce- to a whole number. Ans, 28. 5. Reduce -, to a simple fraction. Ans. s-. 6. Reduce to a simple fraction. Ans. a QuvEsrToI. — As. 165. What is the rale for redIlcif5g comple:- to simPlo tianctious 1 How does this process dihtr fionsa division of'tracltiotns? 166 COMPLEX FRACTIONS. [slECT. XI,]. 7. Reduce 9 to a simple fraction. Ans. s-9-S 8. Redlce s-4 to a simple fraction. AnS 7 9. Reduce 6 — to a whole number. Ans. 9. 10. Reduce 5} to a simple fraction. Ans..'3 11, Reduce 4. to a simple fraction. Ans. A 15 i.2. Reduce to a whole number,.ns. 20.,:J3. Chlange to a simple fraction. A-i. An'. 53. 14. [Reduce, 7 -524 to a simple fraction. Ans. -:> i 0 15, Change &I to a simple fraction. Ans. ~-_,.. (Changre - to a simple fraction. 2hs S. 3 i8 Jea~lne'-.; to a simple fractiona. A.ns's6 1.9. If 7' is the denorminator of the following fraction, *i, what iS its value when -reduced to a simple fractioun 20. ]if i is the numerator of the following fraction 1what i5s ASit';alue whern'-educed to a simple.fraction? Ains, 7s -T, - GM Complex fractions, after beinu reduced to seinvii', or s, e ma:ry be added, subtracted. nmulliplied,,,ld divided,.n....di,- t;- i"-i respective i'ules foi simnle fiiactiions. Qiu'.si:,,s. —rt 16, -G. How d.o you Y d ud r, utrA.ct,:ltiprl, ad divide Io ulex fr,-.ti::? iBCT. xIX ] COMPLEX FRACTIONS.!.~ EXIAMPLES FOR PRACTICE. I. Add - and, together. ns. I-i-.3 A dd 2. Add and - together. Anso 25-. T T' 4 15 Add and together. Ans. 245 4. Add - to -- Ans. 209-t 5. From t'ake n. A -n. 6. Forn IBI! take A5'3 An,, 3-7 L ~tj~ke ~Ans. h,~s. S. From'~ take s Ans. S~. 1 3'2.' ultiply bY Arns, ~. i'. Mutilpl y ]f of -? by X f' 10' I. Mivtidepl T by -r Ans. I-, 5i A. DivMine 75 by JnJS. i2. Duipl a of 62- by of 8Ans. i3. Multiply by1 An: 0-7, A,1 12 14. D -ivide r.. _ 5 s T Y 4 I 51 q 2I ~~:Zt2 16~8 MJSCELLAJNEOUS EXERCISES.. [se:T. x' s BIPSCELLANEOUS E XER>CISES IN VUIGA FGACTIONS. l. What: are the contents of a field 76 rods in length an( M8i rods in breadth.? Ans. 8A. >: ~ 3 2. W'hat are the contents of 10 boxes which are 73 f}fei:. long, 1- feet wide, and 1- feet in height Ans. 169 17 ciubic feet. 3. From -~r of an -acre of land there were sold 00 poles an< o200 square feet. What quantity remained? A.ns. o20T75ft. 4. What cost -1 of an acre at $ 1.75 per square rod?:Ans. $ 0g;3.("' 4 5 1iiA.t Cost -o3 of a tonr at 15-, per ciwt. c Ans. 9 49.7-'31-. 6, VWhat is the continued product of the following numbers: I. 4 9II, and 0? Ans. 9i14. i. i-ro 7- c of a cwt. of sugar there was sold - of it; wWh, is the value of the remainder at $ 6 0.12,3 per noud? Ans. O 3.57. 8. Vt-lat cost 9'3 hbarrels of flour at $ 73- per barrel? Ails.''-...4. 9. WhVl'at cost 1l28 3- qutintals of fish at 3$ -3 per quint? Ans. - 1 2. 3 — 10 1 have two paircls of'!and, one containig'" —7 aroei~ and the otiher 19 a- acres.'What is their value at 78t- p:: acre: A n. $1 P00. 705 J~;'rc~)-ni a, ua, t' e1 wnna'tIe tifblir r73. T gave'~ oJl. ItV'es qSt 7lF xx -'s ()t 14)>, I jmng I, o5]b. I4a Snowx {',i it iof' tile riemai~_,der sd t'1 o o! Cohn. VV I l~ie value o 1'.i- r emaiandr ait -} cets' CO,c.Ans. al. x n i... olexande rren bi't f' goha For"iune a be'::,! sUQ;Irs- coni[iin: a..... f' 30'- - ft at, I..' _im:i i i; o.... S it l d I 0 i.pes pot ld? ran.d — o'f 11e rl.i ae I cents per i's tl~,e:-vae of vhat st ill 5ea':tins at -'> 12 cents per po D iC!, a t,i-,:":~. Gre,* e-rn ma!;~ <i hJs b?:S a~?l x eni t1 * 0 o what Yroa c,. -.., i si~~"!P C.-c t' bl argai i,. i.' of,~'~ a s f' 1 acre at p........ "-'' i,T per ac %;'e:.:,:nin'e' s o('a i:,;,r I i incq!te 1..;'-: SECT. xix.] MISCELLANEOUS EXERCISES. 169 16. How many square rods in a garden which is 183 rods in lengthl and 9 —7 rods wide? Ans. 78 rods 17.c Wlhat cost 19`r acres of land at $17U per acre? Ans. 3509,-!8. %that cost 14.71 tons of coal at 0 7a per ton: Ans. II 1 7u 19. What cost 138~ tons of hay at $ 81 per ton? Ans. $120 1-r. 20. What cost 17 bushels of corn at $ 1 per bushel? -n9. $ 3`1,'3 21. Wthat is the vallue of 9 of a. dollar? Ans. $ 050 22. LWhat is the value of 9-' of a dollar? Ans. 8 0.2]1 23. What is the value of A4-7 of a dollar? Arns. 8'0.25-. 24i-. What is the value of -- of a dollar? Ans. $ 0.5 i J-. 25,. Bought a cask of molasses, containing 871- gallons' - of it haviing leaked out, the remainder was sold at 27,1 cents per gallon; what was the sum received? Ans. * 15.03-2. 26. Bought of L. Johnson 7-yd. of broadcloth at, 3- per yard, and sold it at $ 4:- per yard; what was gained? A.ns. $,3.68'. 27. Bought a piece of land that was 47T1- rods in length and 29 -* in breadth; and from this land there were sold to Abijah Atwood 5 square rods, and to IHazen Webster a piece that was 5 rods square; how much remains unsold? Ans. 1366J-i square rods. 28, Bought a tract of land that was 97 rods long and 48-1rods wide, and from this I sold to John Ayer a house-loct, 18;:! rods long, and 143 rods wide; the remainder oi ry purchase was sold to John Morse, at $. 3.75 per square rod; what sum shall I receive? Ans. 816717. 90-3-. 29. What are the contents of 10 boxes, each of which is 74 fe, long, 4 feet wide, and 43 fcet high P Ans. 1l 1: c' 7 te*.'- n.;7 i:' s 30. Mary Brow rad $1, 7.87; half of this su Was iveni t,1t th}e missionary societoy and' orf the remainder she gaSve to tihe Bible society; what surtn has she left' Ans. 3 57. 31. What number shall be taken from 1a nd.ih reai der muitipued bny'o.0, that', the product shahll ]-.''i n XiVI. VThat num.b' ewr lmus ie o nfltipied by 7T, Ihat the?ro d it miav S,'c2? Ar.s. 2 -. -, 3 o. Bought of John Dow 9 - yards ef' ciodl at p, 4.62: per yar i.; what was the whole cost? Ans..- 45.''-c -. 3 170 Ft.CTIOF5 OF' ( JOMIPOUPI'D NUMiBERS. [s CT. XNX. 34. Bought of John Appleton 4T7- gallons of molasses for,1l2.371; what cost one gallon? W hat cost 127 gallons? Ans. $ 3.3 3 35. When 8!5.87}1 are paid for 12: bushels of wheat, what costs one bushel? What cost 11 bushels? Ans. $ 14.11 -. 36. When $ 9.18- are paid for 3- cords of wood, what cost one cord? What cost - of a cord? Ans. $ 2.138~. 37. What are the contents of a box 8Tz feet long, 3H1 feet wide, and 2 1. feet high?.ns. 6j 7 1 feet. 38. On 2 of my field, I plant corn; on 2 of the remainder, I sow wheat; potatoes are planted on i of what still remains, and I have left two small pieces, one of which is 3 rods square, and the other contains 3 square rods. Hlow large is my field? Ans. IA. OR. 29p. R.EDEUCTION OF FRACTION'S OF COMPOUND NUIMBERS. Axr, 16To To reduce a fraction of a higher denomination to a fraction of a lower. Ex. 1. Reduce rG of a pound to the fraction of a farthing. Ans. tqr. OPERATION. 1 X20_ 20.20 X12 240 240 240X4 960 4... -s2.; s. d.; _s. -- _. q. = r. 2160 2160 2160 2160 2160 2160 9 OPERA.TION BY CANCELLATION. i D.. x 0 x P' x 4 4 Since 20s. make a pound, - __;. -, -qr. there mnust be 20 times as many parts of a shilling as parts of a 9 pound; we therefore multiply fi~. byh 203 and obtain ~s; and since 12d. make a shilling, there will be 12 tinrles as many parts of a penny as parts of a shilling; helnce we multiply 26os. by 12, and obtain 2",d. Again; since 4qr. make a penny, there will be 4 times as many parts of a farthing as parts of penny; we threfore iultiply ",Od. by 4, and obtain "60.qr, Ans. TIutLl. i-v.bizltp/y!e gEien fraction by the numbers that would be em t1ios Ied zn reduction of whole nzumbers to reduce t/1e denomrinalion, of the,eaclion to ihe lower denominatioen required. eU:sirios. -- Asrt. A.67. What is the rute for reducing a fraction of a hither dlenom;v.!ation to the fraction ofi' lower 1 Will you explain the operations i Does thii prorcs. d..t er in p;rinciple fi-rom reduction of whole compound unm. bers-? NECT. xIx] FRACTIONS OF COMPOUND NUMBERS. 171 EXAMPLES FOR PRACTICE. 2. Reduce -r, of a.pound to the fiaction of a farthing. Ans. 2-4 3. What part of a penny is,s of a shilling P Ans.. 6 4. What part of a grain is t of a pound Troy Ans. -. 5. W7hat part of an ounce is lTX of a cwt.? Ans. I2x. 6. Reduce T~-~x y of a furlong to the fraction of a foot. Ans. -. 7. What part of a square foot is - of an acre? Ans. ~. 8. What part of a second is —,~-} of a day? Ans. -3. 9. lWhat part of a peck is 3 of a bushel? A-ns.. 10. Wihat part of a pound is IFr of a cwt. r Ains..e A.rT. 1 G~. To reduce a fra.ction of a lower denomination to a fraction of a higher. Ex. 1. Reduce 4 of a farthing to the fiactionr of a pound. Ans. YFfr. OPER.ATION. 4 4 4 4 4 4 1 9qr'x 4 = 36d; 6d.X 12 -- 432s 432s' 20~ S OPERATION BY CANCELLATION. 4A~ ~ ~~~~ 1 ~Since 4qr. malke a poelnny 9 X X 1>2 X 20 2160 there will be - as manny pence as farthings; tllenrfore we divide the 4qr. by 4, and obtain Ad. And, siie!2d. make a shilling, there will be ~ as many shillings as pene'e hence we divide A~d. by 12, and obtain 4 s Again, since 20s. make a pound, there will be "2- as many pounds as shilli s, therefore we divide:-s. by 20, and obtain XB -- o - ~. ro the answer. RULE. - Divide the given Jraction by the numbers th.at would be erat — piloyed in reduction of whole numbers to reduce the denomination of the fraction to the higher denomination requiired EXABIPLES FOR PRACTICE~. 2. Reduce $ of a grain Troy to the fraction of a pound. Ans. g QUIrSTIONS. - Art. 168. Do youl multiply or divide to red-ouce a fractionr of a lower denom infationii to tie fraction of a highler? VVWh~lat is the ru!a? 172 FRACTIONS OF COMPOUND NUMBERS. [SECT. XIX 3. What part of an ounce is - of a scruple? Ans. Of. 4. What part of a ton is -5 of an ounce? Ans. cFjut.. 5. What part of a mile is 1- of a rod? Ans. ap. 6. What part of an acre is - of a square foot? Ans. 1 7. What part of a day is 24 of a second? Ans. r64'6U. S. What part of 3 acres is t of a square foot? Ans. 9:1U.O 9. What part of 3hhd. is 4 of a quart? Ans. TIo. 10. What part of 6 of a solid foot is * of a yard solid? Ans. 8. ART. 169. To find the value of a fraction in whole numbers of a lower denomination. Ex. 1. What is the value of J-f of 1~.? Ans. 7s. 9d. 1jqr. OPERATION. 7 20 1 8) 1 4 0 (7s. 1 2 6 The reason of this operation will be 1 4 seen, if we analyze the question according 1 2 to Art. 167. Thus, 7X.' 7 X20 _ A140s. 1 8) 1 6 8 (9d. - s.; and 14s. 4 12 _ d 1 62 -- TS' - 6X4 9Ard.; and -d. = -8 X Tqr. 1qr. 4 Ans. 7s. 9d. 1qr. 8) 2 4 (lqr. 6TsI RuLE. - Multiply the numerator of the given fraction by the number required of the next lower denomination to make ONE of the denomination of the given fraction, and divide the product by the denominator. Then, if there is a remainder, proceed to multiply and divide in this manner, until it is reduced to the denomination required; and the several quotients will be the answer. QUE.snTIo. - Art. 169. %What is the rule for finding the value of a fraction in whole numbers of a lower denomination? SECT. xix.] FRACTIONS OF COMPOUND NUMBERS. 173 EXAMPLES FOR PRACTICE. 2. What is the value of ~ of a cwt.? Ans. 3qr. 31b. loz. 124dr. 3. What is the value of - of a yard? Ans. 3qr. 04na. 4. What is the value of of an acre'? Ans. 1R. 28p. 155ft. 82;in. 5. What is the value of 2- of a mile? Ans. Ifur. 31rd. Ift. 10in. 6. What is the value of Yjr of an ell English? Ans. lqr. l-5Tna. 7. What is the value of 2 of a hogshead of wine? Ans. ISgal. Oqt. Opt. S. What is the value of 7T of a year? Ans. 232da. 10h. 21m. 49 1rsec. ART. 170. To reduce a simple or compound number to the fractional part of any other simple or compound number of the same kind. Ex. 1. What part of li. is 3s. 6d. 22qr.? Ans. Xs. OPERATION. In performing this operation, we 3s. 6d. 22qr. = 512 reduce the 3s. 6d. 2iqr. to thirds iL,- 25 $80>) of qr., the lowest denomination in the question, for the numerator of the required fraction, and 1IE. to the same denomination for the denominator. We then reduce this fraction to its lowest terms, and obtain 85.~. for the answer. RULE. - Reduce the given numbers to the lowest denomination mentioned in either of them; then write the number which is to become the fractional part for the numerator, and the other number for the denominator of the required fi'action. EXAIMPLES rOR PRACTICE. 2. Reduce 4s. Sd. to the fraction of.? A. 3. What part of a ton is 4cwt. 3qr. 12ib.? Ans. 75. 4. What part of 2m. 3fur. 20rd. is 2fur. O3rd.? Ans. 4t. 5. What part of 2A. 2R. 32p. is 3R. 24p.? Ans. 3. 6. What part of a hogshead of wine is 18gal. 2qt.? Ans. QUESTION. -Art. 170. What is the rule for reducing a simple or compound number to the fractional part of any other simple or comlpound number of the vamme kind?'5I 174 FRACTIONS OF COMPOUND NUMBERS. [sECT. XXX. 7. What part of 30 days are 8 days 17h. 20nm.? Ans. 5. S. From a piece of cloth containing 13yd. Oqr. 2na. there were taken 5yd. 2qr. 2na. What part of the whole piece was taken? Ans. -. 9. What part of 3 yards square are 3 square yards? Ans.. ADDITION OF FRACTIONS OF COMPOUND NUMBERS. ART. 17. To add fractions of compound numbers. Ex. 1. Add 6 of a pound to t of a shilling. Ans. 17s. 1ld. 0?'4~qr. FIRST OPERATION. dWe find the value of each Value of "6. d. 7 1 qr fraction separately, and add the!e Of 7 -7- 9 g1 tuwo values together according to aue of 9s the rule for adding compound 1 7 1 1,0TI numbers. (Art.101.) SECOND OPERATION. lWe first reduce g- of a shilling =7 20- -zX~. the fraction of a ~' - T-,s. l d. - _. shilling to the fracI *T 1S.2lq r O -6 9tion of a pounId, then add the two fractions together, and find the value of their sum. (Art. 169.) EXAMIPLES FOR PRACTICE. 2. Add -' of a pound to - of a shilling. Ans. 7s. ld, 3-9qr. 3. Add together - l of a ton, - of a ton, and 4 of a cwt. Ans. 11T. 14cwt. lqr. 60ilb. 4. Add together 2 of a yard, ~ of a yard, x4x of a quater. Ans.. l yd t. q. 2a. 0 in 5. Add together x4r of a mile, 4 of a mile, f3 of a furlorng antd A of a yard. Atns. fifur. 29rd. 3yd. Ift. 0-i 6. Add together -1- of an acre, } Of a rood, and 5 of a square rod. Als. 1A. OR. 2p. 1691it. 0'26'in. 7. Sold t house-lots, the first ~- of an acre, the,secornd - of an acre, the thirtd -:2 of an acre, and the fourth 3 of an acre, what was the quantity of land in the four lots? Ans. 3R. SRp. 34p,t5-2- ft..vii S'Tios -- Art. 1'li. What is the first mevthod of adding fractiwois of con. Iounld iuah wers'? What is the second? SEcr. xSx.1 FRACTIONS OF COMPOUND JUIBttERS. i75 SUBTRACTION OF FRACTIONS OF COMPOUND NUMBERS. ART. Echo To subtract fractional parts of compound numbers. Ex. 1. From 6 of a pound take 41 of a pound. Ans. 9s. 10d. lg9qr. FIRST OPERATION. We find the value of eacl a. d. qr. fraction separately, and subValue of 1-. =17 21 tract one from the other, acValue of X. = 7 3 Ia cording to the rule for sub9 1 - 1- tracting compound numbers. Ta1Yg (Art. o102.) SECON'D OPERATION, 30 4 a,_ ae _ We first subtract the less fraction " *X'- -r~ I Y' -- from the greater, and then find the 9s. 1Od. %1-qqr. value of their difference. (Art. 169.) EXAMPLES FOR PRACTICE. 2. From ~ of a ton take 6- of a cwt. Ans. 11cwt. Oqr. GTA8lb. 3. From. of a mile take 17 of a furlong. Ans. 5fur. 33rd. 5ft. 6in. 4. From,- of an acre take -2 of a rood. Ans. 3R. 16p. 154ft. 5. From a hogshead of molasses containing 100 gallons, -T of it leaked out; 2 of the remainder I kept for my family; what quantity remained for sale? Ans. 24fgal. Oqt. 1 3 Apt. 6. The distance from Boston to Worcester is about 41 miles. A sets out from VWorcester and travels ~- Of this distance towards Boston; B then starts from Boston to meet A, and having travelled'- of the remaining distan ce it is required tce find the distance between A and B. PAns. 12m. Cfuir 9rd. t"Ift. 94in. 7. A agrees to labor for B 365 days, but he was absent on account of sickness { part of the time; lie was also obliged to be employed in his own business -A of the remaining time; required the timte lost. Ans. 137dai lh. 13Im. 14V-2'sec Q usrI o Ns.- Art. 172. What is the first meethod of subtracting fractions of comnpound nlunbers The second? i76 QUESTIONS TO BE PERFORMED BY ANALYSIS. [Lsc.x. QJUESTIONS TO BE PERFORMED BY ANALYSIS... If one yard of cloth cost $ 4.40, what will A of a yard ~ost? ILLUSTRATION. - If 1 yard cost 8 4.49, 1 of a yard will cost 5 of 4.40, equal to $ 0.88; and -4 will cost 4 times $0 0.88, equal to 8 3.527 Ans 2. If a barrel of flour cost 8 7.80, what will -T of a barrel cost? Ans., 2.34. 3. If a load of hay cost $ 17.84, what will - of a load cost? Ans. $15.61. 4. If $ 786.63 are paid for a cargo of wheat, what is the cost of t- of'the cargo? Ans. $ 665.61. 5. What is {2 of 0 87.50? Ans. $ 80.20.6. Wthat is 3- of 17X. 18s. 9d.? Ans. 13'. 9s. 0,d. 7. What is ~ of 3T. 16cwt. 3qr. 231b.? Ans. 2T. 3cwt. 3qr. 25}lb. 8. AWhat is 4} of 27A. 3R. 33p.? Ans. 12A. IR. 28p. 9. If v 3.52 are paid for 4 of a yard of cloth, what is the price of I yard? Ans. $4.40. ILLUSTRATION. - If 4 of a yard cost. 3.52, -1 will cost 4 of 8 3.52, equal to $ 0.88; and s5 or a whole yard, will cost 5 times 8 0.88, equal to $ 4.40, Ans. 10. If J of a barrel of flour cost $ 2.34, what will be the cost of a Ywhole barrel? Ans. $ 7.80. HI. *Wben $ 15.572 are paid for -7 of a ton of hay, what will I ton cost? Ans. $ i7.80. 12. When -11 of a cargo of flour cost l 665.50, what sum will pay for the whole cargo? Ans. $ 786.50. 13. If A$ 73.60-k are paid for - of a ton of potash, what sum must be paid for a ton? Ans. 80. 30. 14. Bought ~ of a bale of broadcloth for 13. 9s. 0-d.; what would have been the cost of the whole bale? Ans. 17. _18s. 9d. 15.::If> - of an acre produce 18cwt. Oqr. lib, of hay, what quantity will a whole' acre produce Ans. 76cwt. 3qr. 23lb. -15 Bought 4 of a lot of land containing 12A. 1R. 30'p.; w-hat wer e the contents of the whole lot? Ans. 27A. 3. 239 p. 7. If -A of a iton of potaslh cost $ 80.205,- what is the value of a lton? Ans. MSi50. SEc T. xx.] QUESTIONS TO BE PERFORtMED BY ANALYSIS. 177 18. If 3 of a cwt. of sugar cost $ 5.40, what is the value of - of a cwt.? ILLLUSTRATION. - If - of a cwt. cost 1 5.40 -1 will cost 3 of $ 5.40, equal to $ 1.80; and -, or a cwt., will cost 4 times $ 1.80, equal to $7.20. Now, if lcwt. cost $7.20, - of a cwt. will cost I of $ 7.20, equal to 2 0.80; and - will cost 7 times $ 0.80, equal to ~ 5.60, Ans. 19. If -T Of a pound of ipecacuanha cost V 2.52, what is the value of 4- of a pound? Ans. 1.76. 20. When 580 are paid for 3 of an acre of land, what cost, of an acre? Ans. $ 93.33 l. 21. If 9 of a carding-mill are worth 0 63 1.89, what are - of it worth? Ans. $ 401.20. 22. Ifr - of a ship and cargo are valued at $ 141.52, what are ~ir of them worth? Ans.,V 30.50. 23. If the value of 3-of a farm containing 1785- acres is $ 1728, what is the price of - of the remainder? Ans. V 2304. 24. E. Carter's garden is i7~4 rods long, and 119T rods wide. I-le disposes of e of it for 5 82.S 0; what is the value- of - of the reim.ainder? Ans. $ 41.40. 25. When 26~. 12s. 6d. a-e paid for -- of a bale of cloth, what sum should be paid for -7 of the remainder? Ans. 1S~. 12s. 9d. 26. If 7cwt. of sugar cost 0 28.14, what will 9-cwt. cost? ILLUSTtATION. -— If 7cwt. cost $2o.14, lcwt. will cost~ of $ 28.14, equal to * 4.02. In 9-5-cwt. there are 9cwt.;and if lcwt. cost $ 4.021, -cwt. will cost I of $ 4.02, equal to $ 0.67, and 5-9 will cost 59 times,p 0.67, equal to V 39.53, Ans. 2'7. If three tonls of hay cost V, 49, what will 7-T tons cost? Ans. o 120.27-3T. 23. Gave $ 78.80 for 11 tons of cocal; what should I give for 34 tons?.Ans. 24.67- -. 29. IPaid 37. 1!8s. 10d. for 3 bales of velvet; what iwas the cost of 5 balces? Ans. "o..9s. 6-i l30. Gave. 40 fr 5 yards of broadcloth; what was tlhe price of 191- yards Arns. $ 1 -.57,L. 31. Paid - 360 for 20 barrels of beer; what m 7ust be given for 43h barrels?.Ans. $3 79. 32. If 7 bushels of rye cost $ 8.75, what cost i8-1 bushels? Ano. i 23.29-._6 178 QUESTOINS 10 BE PERFORATED BfY ANALYSIS. [sEC. XMI, 33. -Paid 19.380 for 3 yards of broadcloth; what sum must be given for 11- yards? Ans. 8 76.37k, 34. if 9 ewt. of sugar cost $ 39.53, what must be paid fobr "cwt.?. ILLUSTRATION.- In 95-cwt. there are -~9cwt. if a59-wt. cost 8 39.523 cwt. will cost -, of. 8 39.53, equal to 80.67; and 6-, or iewt., will cost G times $ 0.67, equal to $ 4.02; and 7cwt. will cost 7 times p 4.02, equal to $ 28.14, Ans. 35. When $3 187 are paid for cwt. of -sugar, how much may be purchased for 1? How much for 78? Ans. 12T.-cwt. 36. If 3 f tons of peltah cost8 27 6.13,,h at 1,, ill be the value of 1 ton? Of 75 tons? _ns. i$ 6041.43-. 37. If 7 a cres of land cost $ 875, what will one acre cost? What will 75 acres cost? Ans. S919 2.03. 9 28. If 4` tons of coal cost $ 70, what will 1 ton cost.? What will 86 tons cost? Ans. $ 1376. 39. For 271 acres of land there were paid 375;, what cost 1 acre? WVhat cost 69 acres? Ans.' 932'.4.3-9. 40. If 4, tons of hay cost $ 80.50, what costs 1 ctn C What cost 15 tons? Ans. 8 2 2. C 41. If 7-A-cwt. of sugar cost $ 62.37, what will Icwt. cost? What cost C9wt,? ns?., 160.93. 42. If 7- yards of cloth cost O w3.95, what will be the vatue of 11- yards? ILLTrsTATorr -. - In 7-, yards there are -31- of a yard. If 3' of a yard cost $ 13.95, ~ will cost I-.J of J 123.95, equal to $ 0.45; mod a, or 1. yard, will cost 4 timies r 0.45, equtal to 1.$0i In I]-i- yards there are 3_ of a yard yard cost 1.30, ~ of a yard will cost 4 of 2!.c0, equal to;$ 0.0, and 4a1 will cost.!03 times 8, 0.20, equal to $ 20CVA1) An2. 43. Whln 8 G68650 are paid for 7I.-7 acres, Twhat would be the value of 8991 acres? Ans. 8$ 4 57. n44 ITf _ 17983 are given fbor ai9 tons of iron, wiat will be1 []Le cost o0f g7~ t,_ms? A ans. O 3288. 4,5. Paid Ih l- for 1128 feet of boards; how many could X have purchased f r,$ IItD T-? Ans. 14;0S feet. 4 6. For s tons of potash I received n11cwt. Of sugar; required the un.tin:y ofn suair that may be received for:I t1 os of potash li- Ans 376'ewt. 4:7, VFPorl-P tuon of'ocs in h I- received. 36c'wt of se tvgar sITca'. dIX.] MISCELLANEOUS QUES:aIONS BY ANALYSI1S. 17 9 required tl'e quantity of,sugar that should be received for 3tons.o A-ns. 1.6c wrt. 48. When 08 aire paid r'j-'yard of broadcloth, how much must be given for 3 ryards? Ans, 8 49. 49. Gave t' 4 Ibfor 20- -7 acres of laud; what shall be givwe for 1143 5acres ~ Anus. t 236, MISCELLANEOUS QUESTIONS BY ANALYSIS. 1. Sold a small farm for 8 896.50; what was received for 1 of it? For -7 of it For I~ of it? Ans. f 815. 2 -,. G-ae') l,7~ for 3 bar'els of flour; what cost 1 barrel? What 37 barrels? Ans. $ 21 3.03k1ff. 3. Sold a house for 3687; what sum was received for - of it? Ans. $ 3226. 12. 4. BouLight 1' 7If tons of hay for $0 187; whVat is the cost of 5 of a ton? A~ns. $ 7.6!~7. 5. Bought a hogshead of mLolasses for $ oa-; what cost -? of it? WhXat cost - What cost 1-5? Ans. *p 30.52~. 6. When $ L T- are paid for 100 gallons of molasses, what cost - of a gallon? Ans. $ 0.2132. 7. When 12 cents are paid for -1 of a gallon of molasses, what will 4 -17i gallons cost? Ans. $ 16.0 11. S. If, of a barrel of flour cost 83-&, what will 6A barrels cost? Ans. $ 48 1' 9. When 0 236 are paid for 114 acres, what will be paid for 20-7 acres? Ans. 8 414. 10. Paid in Liverpool 97g~. for 3 bales of cloth; how many bales should be received for 1073z,.? Ans. 33 bales. 11, If 6- barrels of flour cost 8 4a5"-', what will:- of a barrel cost? A ns. $, 3.28,. 1i2. I.f 2- pounds of coffee cost 34 cents, what sum must be pLaid fbr 74- pounds? Ans. 8 6.909-.. 13. If 2-}- tons of hay cost $ 63, what will be the cost of 1.6 tons? Ans. $ 3!1-. 14. i a piece of land'' rods square cost what will be the cost of 4 square rods Ans. $ 7- -. 15. Paid 3 3 b1 for 2 5cwt. of iron; required the sum to be paid for 669? cwt~? Ans. $ 76860.!1. For 682 cords of wood, J Jolt paid $ 63 what sum mnust be paid for 18 cords? Ansr $ 170.. 1i7. Gave 0 243r1 fbr 96 barrels of tar; what quantity b-ould be purclhased for S 1.000? Ans94. 43';22 barrels......~~~~~~~~~~~~~~TP 9 o.Z,~T+~-~V 180 MISCELLANEOUS QUESTIONS BY AINALYSIS. [scla. 1xi. 18. Paid $ 7888.30 for 83-9 acres of wild land; what sum did I pay- for each acre, and what would be the cost of 7 acres? Ains. Tiy 660.80. 19. -axe s2,=p. 12s. -for T7g tons of statrch; vhat. cost 127tons? Alns. 2:42. Sc, 20. For 17' days' work I paid $ 25.44; what should be paid for 89- days' labor? Ans. $ 128.64. 21. Sold 7 - bushels of apples for $ 7.28; what should I receive for 191-L bushels? Ans. 8 19.12. 22. Paid 84355.52 for 496 pieces of carpeting; what did 375 pieces cost? Ans. $ 3294.72. 23. If X of -a of the cost of the Capitol at Washington was 300,000, what was the whole cost? Ans. [ 2,000,000. 24. Purchased 7-6i thousand of boards for $ 135.80; what must be paid for 1943 thousand? Ans. 359.45. 25. My wood-pile contains 6 cords and 76 cubic feet. If I dispose of - of it; what is the value of the remainder at 4} cents per cubic foot? Ans. $ 23.1434-. 26. I have a field 30 rods square, and having sold 18 square rods to S. Brown, and 82 square rods to J. Smith, what part of the field remained unsold? Ans. 8. 27. Bought 7T. 12cwt. 3qr. 181b. of iron, and having sold 3T. 1Scwt. lqr. 20lb., what is the value of -a of the remainder at 5- cents per lb.? Ans. $ 271.71 3. 28. Boughl-t 37 tons of iron at $ 68.50 per ton, for 3 of which I paid in coffee at $ 8.50 per cwt., and for the remainder I paid cash. Required the amount of cash paid, and also the value of the coffee. Ans. Cash, $ 633.62-}; Value of the coffee, $ 1900.87~. 29. A man having received a legacy of $ 7896, spent 3 of it in speculations, and the remainder he put in the savings' bank, where it continued 15 years. It was then found that the sum deposited had doubled. R.equired the sum in the bank. Ans. i 3948. 30. Bought a piece of broadcloth for $' 88, and sold. A of it to J. Smith, and 5- of the remainder to 0. Lake; what is the value of the part unsold? Ans. 0 3'7.49 19. 31. A gentleman gave s of his estate to his wife,,- of the remainder to his oldest son, and;: of what then remained to his daughter, who received $ 750; required the whole estate. Ans. $12,000. SEcT. xx.l DECIMAL_,'RACTIONS. S]. ~ XX. DECIMAL FRACTIONS. ART~. -17o A DECIMnAL'RACTION is a fraction whose delonminator is 10, 1i00, 1000, &c. ART.: 74/ Decimal fiactions are commonly expressed by writing the numerator only with a point before it, called the decmcal point or separatrix; thus, -? is written and read.9 tenths. r9o- 6S 66.99 hundredths. T9 99 C9 4 C.999 thousandths. ARtT., 7. By examining thPe foregoinr g fractions, it will be seen that -T -.9 can occupy only one place while it remains a proper fraction; 9 -.99, only two places; and ~99 9.999, only three places; for, if their numerators are increased by I.1 =. 1 - 0,1.001, respectively, each fraction becomes a unit or whole number. Hence, The fyrst figure or place of any decimal on the right of the point is tenths, the second hundredths, the third thousandths, 4.c. NoTE. -When a decimal place has no significant figure, it must be filled with a cipher. ART. -6o. The denominator of,.9 is 1 with one cipher annexed; the denominator of 99,-.99 is I with two ciphers annexed; the denominator of -9- =.999 is I with three ciphers annexed. Hence, The denominator of a decimal fraction is 1 with as many ciphers annexed as the numerator has places. ART. 1:~7 Decimal fiactions originate from dividing the unit, first, into 10 equal parts, and then each of these parts into 10 other equal parts, and so on indefinitely. Thus, 1 -- 10'.001. Hene; 1,f 1 L= 9 -.01; I 1-0 1.0Q1. Hlence, The unit in decimal ffractions is divided into 10, 100, 1000, 1c., equal parts. qUESTIONS. - Art. 173. What is a decimal fraction? - Art. 174. How are decimal fractions commonly expressed? - Art. 175. What is the first figure o0 place of any decimal? The second? The third? &c. Why? What must be done when a decimal place has no signiicant figure to fill it? - Art. 176. What is the denominator of a decimal fraction? - Art. 177. -ow do decimal fractionr oricrnlate? Ui.9 182 INUMiERATION OF DECIMAL FRACTIO]NS. [SECT. XN. ART. _ @. If ciphers are placed on the left hand of deci. mal figures, they change their places, each cipher removing them one place to the right; thus,.3 ==-, but.03 - 1, and.003 T -r. Hence, Ciphers placed on the left hand of decizals decrease their vaZue in a ternfold proportion. ART. 1719 If ciphers are placed on the right hand of decimal figures, their places are not changed; thus, 0.3 - 3, and.30 — 30 - 3 3. Hence, Ciphers placed on the right hand of decimals do nzot aIter their value. NUMERATION1 F DECIMAL FRACTIONS. ART'. ]~') The relation of decimals to whole numbers and to each other, and also the names of their different orders and places, may be learned from the following TABLE. em 4 o o ~ o.,~ 7 6 5 4 3 2 o 3 4 5 7 8 9 3 H 0 0 0 - a PIS~ a0 a V — ~ P ~ g4 O) O P P4 P,, O a 4 A 4;, 0 q --; 4 a Q HSTOHNS.-At? 78 What efet ha cihr pae at thec Hef h-'7654321 52 r4 ~7,93 of decimals? Why?-Art. 179. What effect if placed at the right ha W 7hy? - Art. 180. What may be learned from the table 7 Why?~r - Ar4-. 189. What,2 may bs learned frPom th~ - able I SECT. XX.] NOTATION OF DECIMAL FRACTIONS. 1-83 The preceding table consists of a whole number and decimal, which, taken together, are called a mixed number. Tlhe part on the left of the separatrix is the whole number, and that on the right the decimal. The value of the decimal is expressed in words thus: — Two hundred thirty-four millions five hundred sixty-seven thousand eight hundred ninety-three billionths. And the mixed number thus:- Seven millions six hundred fifty-four thousand three hundred twenty-one, and two hundred thirty-four millions five hundred sixty-seven thousand eight hundred ninety-three billionths. From the table and explanations we have the following rule for numerating and reading decimals. RUL E. -Be ginning at the left hand, name the order of each figure of the given decimal, and then read it as in whole numbers, giving the name of the last order to the whole. The pupil may write in words, or read orally, the following numbers: 1..s i..3001 9..72E59 2..42 6..0984 10. 12.02003 3..01 7,.00013 11. 121.000386 1i,.908 So.82007 12, 2.3058217 NOTATION OF DECIMAL FRACTIONS. ART. IS. By examining the decimal table, we perceive that tenths occupy the first place, hundredths the second, &c., and that each figure takes its value by its distance from the place of units; therefore, to write decimals, we h'ave the following RULE. - Wdrrite the deciaalfigures in the place of the order denoted by their names, supplying,$ with ciphers such places as have no significant fi'gu7res. The pupil may write in figures the following numbers 1. Three hundred seven and twenty-five hundredths. 2. Forty-seven and seven tenths. QursT'rioNs. - Of what does it consist? What is the number called, when taken together? iWhat is the part on the left hand of the separatrix? The part on the right? What is the value of the decimal 7 What is the ialue of the mixed number' What is the rule for nhumeratinlg and reading.decimals? Art. 181. Upon what does the value of a decimal figure depend? What is the rule for writing decimals? lf84 ADDITION OF DECIMALS. [sECT. XX. 3. Eighteen and five hundredths. 4. Twentyv-nine and three thousandths. 5. Forty-nine ten thousandths. 6. Eight and eight millionths. 7. Seventy-five and nine tenths. 8. Two thousand and two thousandths. 9. Eighteen and eighteen thousandths. 10. Five hundred five and one thousand six millionths. 11. Three hundred and forty-two ten millionths. 12. Twenty-five hundred and thirty-seven billionths. AnRT. 1 ~. It will be seen that decimals increase Ir'om right to left, and decrease from left to right, in the same ratio as simple numbers; hence they may be added, subtracted, multiplied, and divided in the same manner. ADDITION OF DECIMALS. AnT. R SN Ex. 1. Add together 5.018, 171.16, 88.133, 1113.6,.00456, and 14.178. Ans. 1392.09356. OPERATION. 5.0 1 8 7 1.1 6 We write the numbers, units under units, tenths 8 8.1 3 3 under tenths, hundredths under hundredths, &c., t 1 1 3.6 and then, beginning at the right hand, add them.0 0 4 5 6 as whole- numbers, and place the decimal point in 1 4.1 7 8 the result directly under those above. 1 392.09356 RuLE. — /Vrite the numbers under each other according to their value, add as in whole numbers, and point of from the right hand as many places for decimals as there are in that number which contains the greatest number of decimals. Proof. -'he proof is the same as in simple addition. EXAMPLES FOR PRACTICE. 2. Add together 171.61111, 16.7101,.00007, 71.0006, and 1.167895. Ans. 260.489775. 3. Add together.16711, 1.766), 6111.1, 167.1,.000007, and 1476.1. Ans. 77756.233117. QUESTIONS. - Art. 182. How do decimals increase and decrease? How may they be added, subtracted, multiplied, and divided? - Art. 183. How are decimals arranged for addition? What is the rule for addition of' decimals? What is thee proof? SECT. xx.] SUBTRACTION OF DECIMALS. 185 4. Add together 151.01, 611111.01, 16.5, 6.7, 46.1, and.67896. Ans. 611331.99896. 5. Add fifty-six thousand and fourteen thousandths, nineteen and nineteen hundredths, fifty-seven and forty-eight ten thousandths, twenty-three thousand five and four tenths, and fourteen millionths. Ans. 79081.608814. 6. What is the sum of forty-nine and one hundred and five ten thousandths, eighty-nine and one hundred seven thousandths, one hundred twenty-seven millionths, forty-eight ten thousandths? Ans. 138.122427. 7. WVhat is the sum of three and eighteen ten thousandths, one thousand five and twenty-three thousand forty-three millionths eighty-seven and one hundred seven thousandths, fortynine ten thousandths, forty-seven thousand and three hundred nine hundred thousandths? Ans. 48095.139833. SUBTRACTION OF DECIMALS. ART. Ex. 1. From 74.806 take 49.054. Ans. 25.752. OPERATION. Having written the less number under the greater, units 7 4.8 0 6 under units and tenths under tenths, &c., we subtract as 4 9.0 5 4 in whole numbers, and place the decimal point in the re2 5.7 5 2 sult, as in addition. RULE. - Write the less number under the greater, units under units, tenths under tenths, 4-c.; then subtract as in whole numbers, and point off as many places jor decimals as there are in that number which contains the greatest number of decimals. Proof. -The proof is the same as in simple subtraction. E.XABUrLES FOn PRACTICE. 1. 2. 3. 4. 1 1.078 47.117 46.1 3 87.107 9.8 1 8.781.9 5 7.891. 5 1.1 1986 1.268 3 8.33505 38.2385 85.987 1 4 5. From 81.35 take 11.678956. Ans. 69.671044. 6. From 1 take.876543. Ans..123457. 7. From 100 take 99.111176. Ans..888824. 8. From 87.1 take 5.6789. Ans. 81.4211. QUESTIONs. — Arte 184. What is the rule for subtraction of decirials? What is the proof 7 16' 186 MULTIPLICATION OF DECIMALS. [SECT. XX 9. From 100 take.001. Ans. 99.999. 10. From seventy-three take seventy-three thousandths. Ans. 72.927. 11. From three hundred sixty-five take forty-seven ten thousandths. Ans. 364.9953. 12. From three hundred fifty-seven thousand take twentyeight and four thousand nine ten millionths. Ans. 356971.9995991. 13. From.875 take.4. Ans..475. 14. From.3125 take.125. Ans..1875. 15. From.95 take.44. Ans..51 16. From 3.7 take 1.S. Ans. 1.9. 17. From 8.125 take 2.6875. Ans. 5.4375. 18. From 9.375 take 1.5. Ans. 7.875. 19. From.666 take.041. Ans..625. ]MULTIPLICATION 0F' DECIMALS. ART. 1185 Ex. 1. Multiply 18.72 by 7.1. Ans. 132.912. OPERATION. l9We multiply as in whole numbers, and point off 18.72 on the right of the product as many figures fbr 7.1 decimals as there are decimal figures in the multi1 8 7 2 plicand and multiplier counted together. 1 3 l 0 4 The reason for pointing orf decimals in the prod1 3 2.9 1 2 uct as above will be seen, if we convert the multiplicand and multiplier into vulgar fractions, and multiply them together. Thus, 18.72 = 1872S 1 872,- and r~lL - 7_ T hen 1872 Y T 7 -13 29l2. 12 5 7 _-. ~ lo TI U -- I UT9 -_ 132.912, Ans., the same as in the operation. Ex. 2. Multiply 5.12 by.012. sOPERATIOR. Since the number of figures in the product 5.1 2 is not equal to the number of decimals in the.0 1 2 multiplicand and multiplier, we supply the 1 0 2 4 deficiency by placing a cipher on the left 5 1 2 hand..0 6 1 4 4 Ans. The reason of this process will appear, if we perform the question thus: 5.12- 5 2 QUESTIoNs.-Art. 185. In multiplication of decimals how do you point off the product? Will you give the reason for it? When the number of figures in the product is not equal to the number of decimals in the multiplicand and multiplier, what must be done? BECT. XX.] MULTIPLICATION OF DECIMALS. 187 _= 5 29 and.012 In Then 12X 1-2 a= i44.06144, Ans., the same as before. Hence we deduce the following RULE. - /fltiply as in whole numbers, and point off as many igures for decimals, in the product, as there are decimals in the multiplicand and multiplier; but if there are not so many figures in the product as in the multpl'icand and multiplier, supply the defect by prefixing ciphers. Proof. -- The proof is the same as in simple multiplication. ExAMPLEs FOR PRACTICE. 3. Multiply 18.07 by.007. Ans..12649. 4. Multiply 18.46 by 1.007. Ans. 18.58922. 5.' Multiply.00076 by.0015. Ans..00000114. 6. Multiply 11.37 by 100. Ans. ]1137, 7. Multiply 47.01 by.047. Ans. 2.20947. S. Multiply.0701 by.0067. Ans..00046967. 9. Multiply 47 by.47. Ans. 22.09. 10. Multiply eighty-seven thousandths by fifteen millionths Ans..000001305. 11. Multiply one hundred seven thousand, and fifteen ten thousandths by one hundred seven ten thousandths. Ans. 1144.90001605. 12. Multiply ninety-seven ten thousandths by four hundred, and sixty-seven hundredths. Ans. 3.886499. 13. Multiply ninety-six thousandths by ninety-six hundred thousandths, Ans..00009216. 14. Multiply one million by one millionth. Ans. 1. 15. Multiply one hundred by fourteen ten thousandths. Ans..14, 16. Multiply one hundred one, thousandths by ten thousand one hundred one, hundred thousandths. Ans..01020201. 17. Multiply one thousand fifty, and seven ten thousandths by three hundred five, hundred thousandths. Ans. 3.202502135. 18. Multiply two million by seven tenths. Ans. 1400000. 19. Multiply four hundred, and four thousandths by thirty, and three hundredths. Ans. 12012.12012. QUESTIONS. - What is the rule for multiplication of decimals'! What is the proof? l88 DIVISION OF DECIMALS. [SECT'. x. 20. What cost 46lb. of tea at $ 1.125 per pound? Ans. $ 51.75. 21. What cost 17.125 tons of hay at $ 18.875 per ton? Ans.,$ 323.234375. 22. What cost 181b. of sugar at $ 0.125 per pound? Ans. $ 2.25. 23. What cost 375.25bu. of salt at $ 0.62 per bushel? Ans. $ 232.655. DIVISION OF DECIMALS. ART.. 3,e Ex. 1. Divide 45.625 by 12.5. Ans. 3.65. OPERATION. bWe divide as in whole numbers, 1 2.5) 4 5.6 2 5 (3.6 5 and since the divisor and quotient are 3 7 5 the two factors, which, being rnultiN 8 1 2 plied together, produce the dividend, 7 5 0 we point off two decimal figures in 6 2 - the quotient, to make the number in 6 2 5 the two factors equal to the product or dividend. The reason for pointing off will also hDe seen by performing the question with the decimals in the form of vulgar fractions. Thus, 45.625 - % -4 5 and 12.5 12Y U- f2-. Then 5I%6 ~ 15 — 4 5 l - -45625- 45 31 —5' 3.65, Ans., as before. Ex. 2. Divide.175 by 2.5. Ans..07. OPERATION. We divide as in whole numbers, and 2.5).1 7 5 (.07 since we have but one figure in the 1 7 5 quotient, we place a cipher before it which removes it to the place of hun. dredths, and thus makes the decimal places in the divisor and' quotient equal to those of the dividend. The reason for prefixing the cipher will appear more obvious by solving the question with the decimals in the form of vulgar fractions. Thus,.175- 175 and 2.5 I2- T2yThen 7sJ5U. TU -= TUT9F X -- T7U00.07, Ans., as before. Hence the following RULE.- 1. Divide as in whole numbers, and point off as many deciQUESTIONS.-. Art. 186. In division of decimals how do you point off the quotient? What is the reason for it? If the decimal places of the divisor and quotient are not equal to the dividend, what must be done? What is the rule for division of decimals? SECT. XX.I DIVISION OF DECIMALS. 189 reals on the right of the quotient as the number of decimals in the dividend exceeds that of the divisor,; but if the number of decimals in the quotient and divisor together is not equal to the number in the dividend, supply the defect by prefixing ciphers to the quotient. 2. When the number of decimals in the divzsor exceeds that of the dividend, reduce the dividend to the same denomination as the divisor by annexing ciphers, and the quotient will be a whole number. If there is a remainder after the division of the given dividend, ciphers may be annexed to it, and the division continued at pleasure; the ciphers thus annexed being regarded as decimals of the dividend. NOTE. - It is not usually necessary that decimals should be carried to more than six places. Proof. - The proof is the same as in simple division. EXAMIPLES FOR oPRACTICE. 3. Divide 183.375 by 489. Ans..375~ 4. Divide 67.8632 by 32.8. Ans. 2.069. 5. Divide 67.56785 by.035. Ans. 1930.51. 6. Divide.567891 by 8.2. Ans..069255. 7. Divide.1728 by 12. Ans..01448. Divide 13.50192 by 1.38, Ans. 9.784. 9. Divide 783.5 by 6.25. Ans. 125.36. 10. Divide 983 by 6.6. Ans. 148.939-+-, 11. Divide 172.8 by 1.2. Ans. 12. Divide 1728 by.1. Ans. g13 Divide.1728 by.12. Ans. 14. Divide 1.728 by 12. Ans. 15. Divide 17.28 by 1.2. Ans. 16. Divide 1728 by.0012. Ans. 17. Divide.001728 by 12, Ans. 18. Divide one hundred forty-seven, and eight hundred twenty-eight thousandths by nine and seven tenths. Ans. 15.24. 19. Divide six hundred seventy-eight thousand seven hundred sixty-seven millionths by three hundred twenty-eight thousandths. Ans. 2.069+L 20. Divide seventy-five, and sixteen hundredths by five, andc forty-two thousand eight hundred one, hundred thousandths. Ans. 13.846+-. 21. Divide four, and one million twenty thousand three hundred four, hundred millionths by thirty one, and seventy-six thousandths. Ans.. 129045+ 0 O REDUCTION OF DECIMALS. [SECT. xR. REDUCTION OF DECIAMALS. AR T. 11j7. To reduce a vulgar fraction to a decimal. Ex. 1. Reduce - to a decimal. Ans..625. OPERATION. Since we cannot divide the 6) 5.0 (6 tenths. numerator 5 by 8, we reduce 4 8 it to tenths by annexing a cipher, and then dividing, we ob8) 2 0 (2 hundredths. tain 6 tenths and a remainder I 6 of 2 tenths. Reducing this re. - mainder to hundredths by an8) 4 0 (5 thousandths. nexing a cipher, and dividing, 4 0 Ans..625. we obtain 2 hundredths and a remainder of 4 hnndredths, Or thus' 8) 5.0 0 0 which being reduced to thou___ __ _ sandlths by annexing a cipher, and then dividing again, gives a quotient of 5 thousandths. The sum of the several quotients,.625, is the answer. To prove that.625 is equal to D-, we write it in the form of a vu!gar fraction and reduce it to its lowest terms. Thus, 625 --- g, An,. Hence the following RULE. - Divide the numerator by the denominator, anne.in' one er mzore ciphers to the numerator, and the quotient wvill be the decimal; e quired. EXAIMIPLES FOR PRACTICE. 2. Reduce s to a decimal. Ans..75. 3. Reduce - to a decimal. Ans..875. 4. Wahat decimal fraction is equal to 6? Ans..4375. 5. Reduce -~ to a decimal. Ans..363636-+. 6. Reduce -z to a decimal. Ans..416666+. 7. Reduce ~r to a decimal. Ans..235294+-. ART. t g8, To reduce a compound number to a decimr o0f a higher denomination.' Ex. 1. Reduce 8s. 6d. 8qr. to the decimal of a pound. Ans..428125. QUESTIONS. -Art. 187. How do you reduce a vulgar fraction to a decimal? How can you prove the answer correct? What is the rule for reducing a vulgar fraction to a decimal? SECT. xx.] REDUCTION OF DECIMALS. i9][ OPELRATION. Wle commence with the 3qr., and first reduce 4 3.0 0 them to hundredths by annexing two ciphers; and then, to reduce these to the decimal of a penny, we divide by 4, since there will be.] as 2 0 8 5 2m6 2 o many hundredths of a penny as of a fhrthi-ng, and obtain.75d. Annexing this decimal to the.4 2 -1 2 5 6d., we divide by 12, since there will be, -1 as many shillings as pence; and then the 8s. and this quotient by 20, since there will be I as many pounds as shillings, and obtain.428125~. for the answer. Hence the following RULE. - 1. TV rile the given numbers perpendicularly under each other for dividends, proceeding orderly from the least to the orealest; opposite to each dividend on the LEFT hand, place such a nu.mber j;fr a divisor as wrill brin-g it to the next superior denomznation, ande draw a line between them. 2. Begin to divide (at the lowest denomination, annexing ciphers,f necessary, and write the quotient of each division, as decimal parts, on the RIGHT of the dividend next below it, and so on, until they are all divided; and the last quotient will be the decimal required. NOTE. — A compound number may also be reduced to a decimal by first reducing it to a vulgar fraction (Art. 170), and then this fraction to a decirnal (Art. 187). Tbus, 2s. 6d. s -= I -.125~o -EXA3MPLES FOPR PRACTICE. 2. Reduce 15s. 6d. to the fraction of a pound. Ans..775. 3. Reduce 5cwt. 2qr. 141b. to the decimal of a ton. Ans..28125. 4. Reduce 3qr. 2llb. to the decimal of a cwt. Ans..9375, 5. Reduce ffur. 8rd. to the decimal of a mile. Ans..775. 6. Reduce 3R. 19p. 167ft. 72in. to the decimal of an acre. Ans..8725 95 —+ ART. I 9o. To find the value of a decimal in whole numbers of a lower denomination. Ex. 1. 5What is the value of.9875 of a pound. Ans. 19s. 9d. OPERATION. There will be 20 times as many ten thousandths of a.9 8 7 5 shilling as of a pound; therefore, we multiply the deci2 0 mal.9875 by 20, and reduce the improper fraction to a 1 9.7 5 0 mixed number by pointing off four figures on the right, 2 which is dividing by its denominator 10000. The figures on the left of the point are shillings, and those 9.0 0 0 0 on the right decimals of a shilling. This decimal of a QusjlisroNs. — Art. 188. VWill you explain the operation for reducing a compound number to a decimal of a higher denomination? Repeat the rule. By what other method can this be done? - Art. 189. Explain the operation for finding the vatue of a decimal in whole numbers of lower denominations. 192 MISCELLANEOUS EXERCISES. LSECT. XX. shilling we multiply by 12, and, pointing off as before, obtain 9d., which, added to the 19s., gives 19s. 9d. for the answer. RULE. -.lfultiply the g.iven decimal by the number required of the next lower denomznation to make oNE of the given denomination, and point qff on the RIGEIT, Jbr a REMAINDER, as many places as there are places in the given decimal. M.lultiply this remainder by the number that will reduce it to the next lower denominration, pointing off for a remainder as before, and thus proceed, until the reduction is carried to the denlomination required. The several numbers standing at the left hanzd of the point will be the answer, in whole numnbers, of the different lower denominations. EXABIPLES FOR PRACTICE. 1. What is the value of 628125 of a plound? Ans. 12s. 6-d. 2. What is the value of.778125 of a ton? Ans. 15cwt. 2qr. 71b. 3. What is the value of.75 of an ell English? Ans. 3qr. 3na. 4. What is the value of.965625 of a mile? Ans. 7fur. 29rd. 5. What is the value of.94375 of an acre? Ans. 3R. 31p. 6. What is the value of.815625 of a pound Troy? Ans. 9oz. 15dwt. M8gr. 7. What is the value of.5555 of a pound apothecaries' weight? Ans. 6S 55 0Q 19zgr. IISCELLANEOUS EXERCISES IN DECIMALS. 1. What is the value of l5cwt. 3qr. 141b. of coffee at $ 9.50 per cwt.? Ans. $ 150.81-]-. 2. What cost 17T. 18cwt. lqr. 71b. of potash at $ 53.80 per ton? Ans. $ 963.86+. 3. What cost 37A. 3R. 16p. of land at $ 75. 16 per acre? Ans. $2844.80-]-. 4. What cost 15yd. 3qr. 2na. of cloth at $ 3.75 per yard? Ans. $ 59.53Jr. 5. What cost 15- cords of wood at $ 4.621 per cord? Ans. $ 71.10 —. 6. What cost the construction of 17m. 6fur. 36rd. of railroad at $ 3765.60 per mile? Ans. $ 67263.03+. QUsTrON. - What is the rule. SECT. XI.)] REDUCTION OF CURRENCIES. a 93 7. What cost 27hhd. 21 gal. of temperance wine at $ 15.372 per hogshead? Ans. [$ 420.24-Jr. 8. What are the contents of a pile of wood, 18ft. 9in. long, 4fit. Gin. wide, and 7ft. 3in. high? Ans. 61lft. 1242in. 9. What are the contents of a board 12ft. Gin. long, and 2ft. 9in. wide? Ans. 34ft. 54ino 10. Bought a cask of vinegar containing 25gal. 3qt. lpt. at; 0.37~ per gallon; what was the amount? Ans. 9.70+-. 11. Bought a farm containing 144A. 3R. 30p. at $97.62~ per acre; what was the cost of the farm? Ans. 14149.52+. 12. Sold Joseph Pearson 3T. 1Gcwt. 211b. of salt hay, at $ 9.37- per ton. He having paid me $20.25, what remains due? Ans. 5$16.40+-{. 13. If - of a cord of wood cost $ 5.50, what cost one cord? VWhat cost 7- cords? Ans. $ 48.71+. 14. If 4` yards of cloth cost $ 125, what cost 17-3 yards? Ans. 8 46.18+. ~ XXI. REDUCTION OF CURRENCIES. ART. Ii9~o REDUCTION OF CURRENCIES is finding the value of the denominations of one currency in the denominations of another. The nominal value of the dollar, expressed in shillings and pence, differs in the different States of the Union and in different countries, as may be seen by the following TABLE. In New England, Indiana, Illinois, Missouri, Virginia, Kentucky, Tennessee, Mississippi, Texas, Alabama, and Florida, the dollar is valued at 6 shillings; $ = -.. = -3. In New York, Ohio, and Michigan, the dollar is valued at 8 shillings; $ l = Is3. - gX. In New Jersey, Pennsylvania, Delaware, and Maryland, the dollar is considered 7 shillings and 6 pence; $1- x 9.e QUESTIONS. - Art. 190. What is reduction of currencies? What is the value of a dollar, in the different States, expressed in shillings and pence? ]~94 REDUCTION OF CURRENCIES. [SECT. XXX, In North Carolina the dollar is reckoned at 10 shillings, O 1- - Io 1 — In South Carolina and Georgia 4 shillings 8 pence is the value of a dollar; 1 =- ~56. 7= e In Canada and Nova Scotia the dollar is valued at 5 shillings 8 1 — =t s.X-~ x In English or sterling money, the dollar is valued at 4s; G.6d. nearly; = - 2,5lo NOTE. — The value of a pound English or sterling money is very nearly fi 4.84. AnT. 5 o To reduce pounds, shillings, pence, and far things, of the different currencies, to United States money. Ex....Reduce 18S. 15s. 6d., Newr England currency, to United States money. Ans., 62.5s8W. OPERAsTION.. t tWe first reduce the shillings 18,X WS f * - ga 11 18.775s. 6d. - 18.758. and pence to the decimal of a!8.756. -: ~. 258 pound (Art. 188), and then annexing it to the pounds, we divide the sum by ~3, because 6s. or a dollar in this currency is -~ of a pound, and thus obtain the answer in dollars and the decimal of a dollar. Hence the RULE. -Reduce the shillings, pence, and farthings, if anty, to the decimal of a pound, and annex it to the pounds; then divide this number by the value of the dollar in the given currency, expressed as a firaction of a pound. The quotient is the answoer in dollars, and the decimal of a dollar.'- EXATPLES FOn PRACTICE. 2. Change 1414,. 7s. 6d. of the old New England currency to United States money. Ans. $ 481.25. 3. Change 74X. Is. 6d. of the old currency of'Tew York to United States money. Ans. $ 185. lS. 4. Change 129CC- of the old currency of Pennsylvania to United States money. Ans. $ 344. a. Change 144g,. 6s. 3d. 2qr. of the old North Carolina currency to United States money. Ans. $ 288.62,9. QUFSTIONS. - What is the value of a dollar in Canada?' In English or sterling money 1 What is the value of a pound, English or sterling money, expressed in United States money? -Art. 191. How do you reduce pounds, shillings, pence, and farthings, New England currency, to United States money? aWhy divide by 7 ~.I How reduce them if in New York currency? How, if in Georgia currency? What is the general rule? SECT. XXI.l REDUCTION OF CURRENCIES. 195 6. Change 84X. of the old currency of South Carolina to United States money. Ans. $ 360. 7.- Change 1441f. 4s. of Canada and Nova Scotia currency to United States money. Ans. $ 576.80. 8. Change 257. Ss. 6d. English or sterling money to United States money. Ans. Q$i245.93,7. A RT. 92o To reduce United States money to pounds, shillings, pence, and farthings of the difierent currencies. Ex. 1. Reduce $152.625 to old New England currency. Ans. 45.' 5s. 9d. OPERATION. Since 6s. or a dollar in this cur 1;752.625 3 _ - 4C57,;,;,S7& rencyis 3 of a pound, we multiply 52.2 >X ~ -- 45.7 *. the given sum by the fraction T-aq, and reduce the decimal to shillings 45.7875X. - 45w. 15s. 9d. and pence. (Art. 189.) Htence the following RULE. — Multiply thie dollars, cents, q'c., of the iven sum by the value of the dollar in the required currency E1XPRESSED AS A FRACTION OF A POUND. The p7roluct is the anszwer in pounds and the decimal of a pound, which must be reduced to shillings, pence, 4-c. (Art. 189.) EXAMPLES F OP PRACTICE. 2. Change $ 481.25 to the old currency of New England. Ans. 144X. 7s. 6d. 3. Change $ 1S5.18E to the old currency of New York. Ans. 74~. Is. 6d. 4. Change 0 344 to the old currency of Pennsylvania. Ans. 129X~. 5. Change $ 288 to the old currency of 3North Carolina. Ans. 144. 6. Change $ 360 to the old currency of South Carolina. Ans. 84. 7. Change 8 576,50 to Canada and Nova Scotia currency. Ans. 144~. 2s. 6d. 8. Change $1245.93,7 to English or sterling money. Ans. 257X. 8s. 6d. QUESTIONS. -Art 192. H-ow do you reduce United States money to pounds, shillings, pence, andl farthings, New England currency? Why multiply by -3i~-? How would you reduce United States money to pounds, &c., Ohio currency? How, to Pennsylvania currency? What is the general rule? PERCENTAGE. ISecT. XS)l. ~ XXII. PERCENTAGE. ART. 1 93lb. PERCENTAGE andper cent. are terms derived fiom the same Latin words, per and centurn, which signify by the hundred. Percentage, therefore, is any rate or sum on a hundred, or it is any number of hundredths. Thus, if an article is bought for $ 100 and sold for $105, the gain is 5 per cent., because 8 5 are TK-5 of $ 100, or of the original cost. Again, if an article is bought for $ 25 and sold for $ 30, the gain is 20 per cent., because $ 5 are 5 __ 0 of $ 25, or of the original cost. Since per cent. is any number of hundredths, it is a decimal written in the same manner as hundredths in decimal. fractions. Thus, 5 per cent., 25 per cent., &c., are written.05,.25, respectively. (Art. 175.) WVhen the per cent. is more than 100, it is an improper fraction, and if expressed declmally, it becomes a mixed number, thus, 103 per cent., equal to 10o 3 is written 1.03. If the per cent. is a vulgar fraction, or contains a vulgar fraction, the fraction is a part of one hundredth, and if expressed decimally, must be written at the right of hundredths in the place of thousandths, &c. Thus, a- per cent., 3 per cent., 12jper cent., are written.005,.0075,.122, respectively. EXAMPLES iN WRITING PERCENTAGE~ Write decimally 2 per cent.; 3 per cent.; 5 per cent.; 6 per cent.; 7 per cent., 8 per cent.; 10 per cent.; 15 per cent.; 25 per cent.; 50 per cent.; 100 per cent.; 105 per cent.; 115 per cent.; 6.- per cent.; 83 per cent.; 20- per cent.; i- p er cent.; per ce n t.; pe cent.; per cent. ART. il941, To find the percentage on any sum or quantity. Ex. 1. Bought a house for $ 625, and sold it at 6 per cent. advance; what did I gain by the sale? Ans.,$ 37.50. QUESTIONS.- Art. 193. From what are the terms percentage and per cent derived, and what do they signify? How then will you define percentage? How will you illustrate it? How is per cent. written, when less than 100? How, when more than 100? If the per cent. is a fraction, or contains a frac tion, what is the fraction, and if expressed decimally, what place must it oo cupy? SECT. XXI.] PERCENiTAGE. 197 OPERATION. Since 6 per cent. is x D- =.06 Sum, $ 6 2 5 of the original cost, we multiply IRate per cent.,.0 6, 625 by the decimal expression.06, and point off as in multipliPer cent. or gain, $ 3 7.5 0 cation of decimal fractions. RuILE. Miultiply the given quantity or number by the rate per cent. considered as a decimal, and point off the product according to the rule for multiplication of decimal fractions. (Art. 185.) NOTE. -If the per cent. contains a vulgar fraction that cannot be ex pressed decimally, or, if thus expressed, would require several figures, it is more convenient to multiply by it as a mixed number. (Art. 155.) EXABUPLES FOR PRACTICE. 2. WThat is 2 per cent. of $ 325? Ans. $.5I0. 3. What is 5 per cent. of $ 789? Ans. $ 39.45. 4. What is 6 per cent. of O. 856.49? Ans.;8 51.38,9. 5. What is 7g per cent. of 765 tons? Ans. 57.375 tons. 6. That is 9- per cent. of $ 5000? Ans. $ 490. 7. What is - per cent. of 81728? Ans. 815.12. 8. What is 4} per cent. of 5S7 yards of cloth? Ans. 26.415 yards. 9. I lost 10 per cent. of $ 975; how much have I remaining? Ans. $ 877.50. 10. A piece of cloth containing 32 yards, after being spung. e-d, shrunk 8 per cent. in length; what was the length of the piece after shlrinking? Ans. 29.44 vyards. 11. A man received a legacy of $ 10000, but he lost 15 per cent. of it in speculation; what sum had he remaining? Ans. $ 8500. 12. Bought 25 shares in the Boston and Maine Railroad, at $100 each; but soon after I sold them at 12 per cent. advance; what did I gain? Ans. $ 300. 13. Bought 1728 acres of land at, 25 per acre, and sold my bargain at 15 per cent. advance; what did I gain by my purchase? Ans. $ 6480. 14. Sent to Liverpool 5000 bushels of wheat, which cost me $ 1.25 per bushel; but 25 per cent. of the wheat was thrown QUESTIONS.- Art. 194. Will you explain the operation for finding the percentage on any sum or quantity? Give the reason for the process. What is the rule? 198 SIMPLE INTEREST. [SECT. XXIIX. overboard in a storm, and the remainder was sold at S 2.00 per bushel; what was gained on the wheat? Ans. $1250. 15. T. Page received a legacy of 8 8000; he gave 19 per cent. of it to his wife, 37 per cent. of the remainder to his sons, and $ 2000 to his daughters; what sum had he remaining? Ans. $ 2082.40. 16. The nominal value of a share in a certain railroad is 8 100; if I purchase 17 shares at 15 per cent. below their nominal value, and sell them for 15 per cent. above, what do I gain? Ans. $ 510. 17. My tailor informs me, it will take 10 square yards of cloth to make me a full suit of clothes. The cloth I am about to purchlase is 13 yards wide, and on spunging, it will shrink 5 per cent. in width and 5 per cent. in length. H-low many yards of the above cloth must I purchase for my " new suit"? Ans. 6yd. lqr. 1Z3-Jna. ~ XXIII. SIMPLE INTEREST. ARTr. Gil.NTEREST iS the compensation which the borrower of money makes to the lender. The rate per cent. is the sum paid for the use of $100, 100 cents, or 100S., for one year. The principal is the sum lent, on which interest is computed. The amount is the interest and principal added together. Legal interest is the rate per cent. established by law, Usury is a higher rate per cent. than is allowed by law. ART. 19. TIhe legal rate per cent. varies in the different States and in.different countries. In the New England States, NewJersey, Pennsylvania, DeIaware, Maryland, Virginia, North Carolina, Tennessee, Kentucky, Ohio, Indiana, Illinois, Missouri, Arkansas, District of Columbia, and on debts or judgments in favor of the United States, it is 6 per cent. QuEsTroNs. -Art. 195. What is interest? What is rate percent.'-'What is the principal? What is the amount? What is legal interest? What is usury?.:Art. 196. What is the legal rate per cent. in the different States? SECT. xxII.] SIMPLE IINTEREST. 199 In New York, Michigan, Wisconsin, Iowa, and South Carolina, it is 7 per cent. In Georgia, Alabama, Mississippi, Texas, and Florida, it is 8 per cent. In Louisiana, it is 5 per cent. In Canada, Nova Scotia, and Ireland, it is 6 per cent. In England and France, it is 5 per cent. ART. 97. To find the interest of $1 at 6 per cent. for any given time. if the interest of $ 1 is 6 cents f6r one year, or 12 months, for 1 month it will be I2 of 6 cents, or A a cent, equal to 5 mills; and for 2 months, twice 5 mills, or 1 cent. Now since the interest for 1 month, or 30 days, is 5 mills, the interest for 6 days, or - of 30 days, will be 1 mill. And as 1 day, 2 days, &e., are a, -, &c., of 6 days, the interest for any number of days less than 6 will be as many sixths of a mill as there are days. Hence the INTEREST OF $ 1 AT 6 PER CENT., For 1 year, is 6 cents or $ 0.06, " 2 months, " 1 cent or.01; " 1 month, "a ~ cent or.005; " 6 days, " 1 mill or.001; " 1 day, a;- of a mill or.000g. Ex. 1. What is the interest of $ 1 for 2yr. 7mo. I3da. Ans. $0.157i.t OP ERATrLO. The interest for 2 years will be [nterest for 2y. — 1 2 twice as much as for 1 year, equal 12 7mo. --.0 3 5 cents; and since the interest for 2 " " 13da..03 0 2- months is I cent, for 7 months it will be 3A cents. And as the interest for Ans. $ 0.1 5 7- 6 days is 1 mill, for 13 days it will be 2- mills. Adding the several sums together, we have $0.157 for the answer. RULE I. - Reckon 6 cents for every YEARu, cent for every TWO MONTHS, 5 mills for the odd month, I mill for every 6 days; and for QUESTIONS. —In Canada, Nova Scotia, and Ireland? In England and France? - Art. 197. Will you explain, by analysis, the reason of the rule for finding the interest of $1 at 6 per cent. for any given time! Explain the operation. What is the first rule? 200 SSISMPLE INTEREST. LSeCT. XXXlr any number of days less than six, as many sixths of a mill as there are days. The sum of the cents and mills, expressed decimally, is the interest required. Or, RULE II. - Reckon 6.cents for every YEAR, and call HALF the number of months so many cents additional, and one SIXTH of the days so many mills. Th'e sum of the cents and mills, expressed decimally, is the interest required. NOTE. — If half the number of months contain the fraction A, it must be reckoned 5 mills. EXAMIPLES FrPo PRACTICE. 2. What is the interest of $1 for ly. 4mo. 6da.? Ans. 0.08 1,. 3. What is the interest of 0 1 for ly. 9mo. 12da.? Ans.,.0. o107. 4. What is the interest of $ I for 3y. Smoo. 19da.? Ans. $ 0.223 —. 5. What is the interest of $ 1 for 2y. Imo. 2Oda? Ans. $ 0.!228' 6. What is the interest of- 1 for 7y. 15da.? Ans. 0.422'2. 7. What is the interest of 1 for 3mo. 2daL.? Ans. f 0.0192. S. VWhat is the interest of 1 for 4y. 2mo. 5da.? Ans. $ 0.2506, 9. What is the interest of $1 fo r 4ml. 3da.? Ans. 0..020", 10. What is the interest of $ 1 for 17y. 2da.? Ans. 8 1.020, 11. What is the interest of, 1 for 21y. 1mo. 29da.? Ans. $ 1. 39-. 12, What is the interest of $ 1 for 20y. 4da.? Ans. 8 1.2002. 13. What is the interest of 81 f or 35y. 3mo. 2da.? Ans. $ 2. 115-1 ART. 9. To find the interest on any sum of money al 6 per cent. for any given time. Ex. 1. What is the interest of $ 926 for 3y. inmo. 15da.? What -s the amount? Ans. Interest, $ 219.92,5; Amount, $ 1145.92,5. QUEsTION. - What is the second rule? SECT. XXIIX.] SIMPLE INTEREST. 201 OPERATION. Principal, $ 9 2 6 Interest of $ 1,.2 3 7~ We find the interest of $ 1 for the giv6 4 8 2 en time to be $ 0.237i (Art. 197). Now, 2 7 7 8 since the interest of $ 1 is $ 0.237,, the in — 1 8 5 2 terest of $ 926 will be 926 times as much, 4 6 e therefore we multiply them together. T'o find the amoulnt, we add the principal to Int. 2 1 9.9 2 5 the interest. Hence the following Prin. 9 2 6 Amt. $ 1 4 5.9 2,5 RuLE. - 1. Find the interest of $ 1 for the given time; then multiply the principal by this interest, and point off as in multiplication of decimal fractions. (Art. 185.) 2. To find the amount, add the principal to the interest. NOTE. - If the interest of $ 1 contains a vulgar fraction; the fraction may be reduced to a decimal, if preferred. The interest may also be multiplied by the principal, when it is more convenient. EXADIPLES FOR PRACTICE. 2. What is the interest of $ 197 for 1 year? Ans. $ 11.82. 3. What is the interest of $ 1728 for 3 years? Ans. $ 311.04. 4. WA\hat is the interest of $ 69 for 2 years? Ans. S8.28. 5. What is the interest of $ 1728 for 1 year, 6 months? Ans. $ 155.52. 6. What is the interest of t 16.87 for I year, 8 months? Ans. $ 1.6 8,7. 7. Required the interest of $ 118.15 for 2 years, 6 months Ans. $ 17.72,2. 8. Required the interest of $ 97.16 for 1 year, 5 months. Ans. $ 8.25,8. 9. Required the interest of $ 789.87 for 1 year, 11 months. Ans. $ 90.83,5. 10. Required the amount of $ 978.18 for 2 years, 3 months. Anis. $ 1110.23,,!. 11. Required the amount of $ 87.96 for I month. Ailns. $ 38.39,9. 12. Required the amount of $ 81.81 for 8 years, 4 months. Ans. 5 122.71,5. QuEs'TIors. —Art. 198. Explain the operation for finding the interest on any sum of money at 6 per cent. for any given time. What is the rule? How do you find the amount? 202 SIMPLE IINTEREST. [tsax'. xii 13. Required the amount of $ 0.87 for 7 years, 3 montlls. Ans. 0 1.24,8. 14. What is the interest of $ 1.71 for 2 years, 2 days? Ans. 0 0.20j5. 15. Required the interest of $ 100 for 8 years, 4 ronths, 1 day. Ans. $ 50.01.,6 16. Required the interest of $3.05 for 2 months and 2 days? Ans. 0 0.0351. 17. What is the interest of 76 1.75 for i year, 2 months, 16 days? Ans. 55.60,7. IS, What is the interest of 0 17286.19 for 1 year, 5 months, 10 days? Ans. 0 149.77,6. 19. What is the interest or $ 88.96 for 1 year, 4 months, 6 days? Ans. 0 7.20,5. 20. What is the interest of 8 107.50 -for I month, 29 days? Ans. 0 1.05,7, 21. What is the interest of 6 87.25 for 1 year, 8 months, 5 days Ans. $ 8.7 97. 22. What is the interest of 8 73.16 for 1 yearx 7 months, 23 days?Ans. q0 7.23. 23. What is the anmount of 371.15 for 3 years, G onths, 10 days? Ans. 0 16617.37,. ART. R9jo To find the interest of any sum of money at atny rate per cente for any given time. Ex.!. 5What is the interest of $ 26.25 for 2 years, 4 months, at 7 per cent.? Ans. 0 4.28,75. oCP2e.TION V~We first find the interest on the given siun. at 6 per Pirinzzcipa $ 6'WI 2 5 cent., and then add to this Interest of 1 at 6 per cent.,.1 4 interest the fractional part of itsel` denoted.by the I 0 5 0 0 excess of tihe rate above 6 2 2 25 per cent. The excess is per cent.; thelerefore we Interest at 6 per cent., 0. 3.6 7 5 0 add -of the inierest at 6 1 r vlper cent. to itself, for the of interest at 6 per cent.,.6 1 2 5 f answer.!T the:ate per cent. had been less than 6, Interest at 7 per cent., 04.2 S,7 5 we should have subtraeted the fractional palrt. liI-ence the following QUESTIONs.- Art. 1I99. Explain the operation for finding the interest on any sum of money at any rate per cent, sEeCr. xx. SIMPLLE INT'EREST. 203 RULE.- -Find the interest of the given sum at 6 per cent., and then add to this interest, or subtract from it, such a fractional part of itself as the given rate is greater or less than 6 per cent., and poinlt oqg as in nultiplication cf decimal fractions. (Art. 185.) NOTE 1. -- I the rate per cent. is 12 per cent., the interest at 6 per ~ent. Tmust be doubled. NoTE< 2. - If the interest is for years only, it may be found by multiplying the principal by the interest of 1 for the given time and rate. EXAMIPL:ES FOR PRACTICE. I. WFhat is the interest of 0 1A44 for one year at 7 per cent.? Ans. $ 10.08. 2. What is the interest of 69 850 for I year, 7 months, 18 days, at V per cent.? Ans.. 97. IS. 3. What is the interest of 8065.75 for 3 years, 9 months, 24 days, at 7 per cent.? Ans. $ 231.29,9. 4. eWhat is the interest of $ 960.108 for year, ~2 months, at 7 per cent.? Ans. 6 78.414. 5. What is the interest of- 1728.19 for 3 years, 8 mornths, 10 days, at 7 per cent.? Ans. 6 446.92,9. 6. tWhat is the interest of 17.90 for 0 months, 4 days, at 7 per cent.? Ans. 6 0.84,9. 7. What is the interest of $ 1165.50 for 5 years, 0 months, 9 days, at 7 per cent.? Ans. 43 0.39. 8. lWhiat is the interest of 8 1237.90 for I year, 7 months, $3 days, at'7 per cent.? Aris. 1t37.92,2. 9.'What is the interest of $ 156.80 for 3 years and 3 days, at 3 per cent.?.Ans. 1 4.!5,1. 10. W5hat is the interest of 8 579.75 for I year, 2 months, 2 days, at 5 per cent.? Afns. 6 83.97,9. Il, Whlat is the interest of 7671.09 for 2 years, 8 months, 5 days, at 8 per cent.? Ans. 8 1645.02. 12. W,(hat is the interest of $ 943. I11 for I month, 29 days, at 9 per cent.? Ans. 6 13. 9 1. 13. VWhat is the interest of $ 975.06 for 2 years, 7 months, 9 days, at 80 per cent.? Ans. 0 209.82. 14. What is the amount of $ 1000 for 3 years, 3 months, 29 days, at 54 per cent.? Ans. $ 11 83.18. 15. What is the interest of $ 765 for 2 years, 9 months, at I per cent.? Ans. $ 21.03,7. 16. What is the interest of 8 979.15 for 3 years, 2 months, 4 days, at 12~ per cent.? Ans. $ 388.94. QUSTION. — What is the rule? What is note first? ote second? 204 SIMPLE INTEREST, ISECT. XXI1I ART. S@O. Second method of finding the interest of any sum of money, at any rate per cent., for any time. Ex. 1. What is the interest of E 26.25 for 3 years, 5 months, and 15 days, at 8 per cent.? Ans. $ 7.26,2. Having found the inteoOPERATION. est for 1 year and then foi Principal,,9 2 6.2 5 3 years, the interest for 5 Rate per cent.,.0 8 months is obtained by first taking ~ of i year's interInterest for 1 year, 2.1 0 0 e0tfor 4monthsandlhen 3 ~ of this last interest, for 1 Interest for 3 years, 6.3 0,00 month. Interest for 4mo., 1 of y.,.7 0 0 0 And since 15 days are ofs 4mo..175 o of 1 month, we take Interest for lmo.,; of 4no.,.1 7 5 0 of 1 month's interest for Interest for 15da., - of lmo.,.0 8 7 5 the interest of 15 days; and add the several sums Int. for 3y. 5mo. 15da., i 7.2 6,2 5 together for the answer. Hence the following RULE. - 1. First find the interest for one year by multiplying the prin cipal by the rate per cent., and pointing off as in multiplication of decimalifractions; and for two or more years multiply this product by the number of years, and point off as before. 2. Find the interest for the months by taking the most convenient fractional part or parts of ONE year's interest, and then of any PART of one year's interest, if necessary, denoted by the months, or any PART of the months. 3. Find the interest for the days by taking the most convenient fractional part or parts of ONE month's interest, and then of any PART of one month's interest, if necessary, denoted by the days or any part ofthe days. OTE. -- Many practical men prefer this method of casting interest to any other, but in most questions it is not.so expeditious as the preceding. lThe pupil may be required to solve the questions in interest by both methods. EXABMPLES FOR PRACTICE. 2. What is the interest of $ 1775 for 7 years? Ans. 8 745.50. 3. What is the interest of $ 987 for 3 years, 6 months? Ans. $ 207.27. QUESTIONSs.-Art. 200. Explain the operationforfinding the interest of any sum of money, at any rate per cent., for any time. What is the rule? BEsT. XXI11.] SIMPLE 1T1EREST. 205 4. Required the interest of $ 69.17 fbr 4 years, 9 months. Ans. $ 19.71,3. 5. Required the interest of 8 96.87 for 10 years, 7 months 15 days. Ans. $ 61.75,4. 6. Required the interest of $1.95 for 15 years, 11 months,'20 days. Ans. 8 1.86,8. 7. Required the interest of $1789 for 20 years, I month, 25i days. Ans. 8 2163.19,9. S. Required the interest of $ 666.66 for 6 years, 10 months, i 3 days. Ans. $ 274.77,5. 9. What is the amount of $ 98.50 for 5 years, 8 months? Ans. 8131.99. 10. What is the amount of $ 168.13 for 8 years, 5 months, 3 days? Ans. $ 253.11,9. 11. What is the amount of $ 75.75 for 4 years, 2 months, 27 days? Ans. 8 95.02,8. 12. Required the amount of $ 675.50 for 30 years, 3 months, 23 days? Ans. 81904.12,1. ART. 2@0 To find the interest on pounds, shillings, pence, and farthings, at any rate per cent., for any time. Ex. 1. What is the interest of 25~. 2s. 6d. for 2 years, 6 months, at 6 per cent.? Ans. 3~. 15s. 4d. 2qr. OPERATION. \We reduce the 2s. 6d. to the 25~. 2s. 6d. =2 5.1 2 5 ~. decimal of a pound (Art. 188), Interest of 1~..1 5 and, annexing it to the pounds, multiply this principal by the in1 2 5 6 2 5 terest of 1~. for the given time. 2 5 1 2 5 The product is the answer in 3.7 6 8 7 5s. pounds and the decimal of a pound, which must be reduced 3~. 15s. 4d. 2qr. to shillings, pence, and farthings. (Art. 189.) RULE. - Reduce the shillings, pence, and farthings to the decimal of a pound, and annex it to the pounds; then proceed as in United Stlates money, and reduce the deci mal in the result to a compound number. EXAMPLES FoR PRACTICE. 2. WVhat is the interest of 264. 10s. for 2 years, 4 months, at 5 per cent., Ans. 3~. Is. 10d. QUESTIONS. — Art. 201. How do you find the interest o01 pounds, shillings, pence, and farthings I Repeat the rule.!1a 6' SITMPLE INTEREST. [ sRc r. EX K. 3. hlxha.t is the interest of 42X. 18s. for 1 year, 9 mionths, 25 days, at 6 per cent.? Ans. 4X. 13s, I7'd. 4. VWhat is the interest of 94~. 12s. d. r 4 years, G months, 7 days, at 8 per cent. A ns. 344, 4s, 2- 1. 5. What is the amount of 110'. 7s, 6d. for 6 years~, months, at 5 per cent.? Ans. 1 48l -. s. t d i S CEALLANEOUS EXERCISES IN INTEREST. 1. What is the interest of $ 172.50 from Sept. P5, 1840, to July 9, 184:2? Ans. 18.51,5. 2. Wfhat is the interest of 9 169.75 from Dec. 190,:8sI, to May 5, 1841? Ans. 2,4.47 a2. 3. What is the interest of 1 i7.18 from July 29, 1837, to Sept. 1, 1841? Ans. $ 4.2.1. 4. What is the interest of 6 67.07 from April 7', 183, to Dec. 11, 1841? Ans. $!0.77,5. 5. Required the interest of 117.75 forom Jan. 7, 189, to Dec. 19, 1841. Ans. 20.84, t. 6. Required the interest of 8847.15 from Oct. 9, I839, to Jan. 1.1, 1843. Ans. 0$ 165.47,6 7. Required the interest of:'7.18 from March 1, 1841, to Feb. 1, 1842. Ans. $O 0.4N Le S. What is the interest of~ N 976.18 from May 29, 842, to io . %, 184L5? Ans. 5) a0~ o,~1, 9. I have John Smith's note for 8 144, dated July i5,!839; what is due March 9, 1842? ns. t 16.64,)0, i0. George Cogswell has two notes against j. Doe; the eirst is for 375.83, and is dated Jan. 19, 8340; the other is ~or 0 76.19, dated April 23, 1841; what is the amount of both notes Jan. 1, 842? Ans. $ 499.14, i. I1. What is the interest of $ 68.19, at 7 per cent., from June' 5, 1840, to June 11, 841? Ans. 4.2. 12. Required the amount of 8 79.15 from Feb. t7, 1839, to'Dec. 20, 1842, at 7- per cent. Ans. $ 102.1,9. 13. -What is the amount of $89.96 from June 19,.840, to Dec. 9, 1841, at 8S per cent. Ans. $ 100.88,6. 14. A. Atwood has J. Smith's note for $ 325, dated June 5 1839; what is due, at P4 per cent., July 4, 1841? Ans. 6 3-4.02,2. 15 J. Ayer has D. How's note for $ 1728, dated Dec. 29 1839; what is the amount Oct. 9, 1842, at 9 per cent.? Ans. $ 2160. BOCT, X1In.] PARTIAL PARTIAL YMENTS. 207 16. What is the interest of 976.18 from Jan. 29, S841, to 3uly 4, 1842, at 12 per cent.? Ans. 167.57,7. 11i What is the interest of 8 176.1.7 fron June 19, 1S39, to Sept. 7, 1843, at 9'- per cent? Ans. $ 72.42,7. 18. \hat is the arnount of $ 379.78 from Dec. 3, $089, to August *23, 1547, at 71 per cent.? Ans. 8 1519.48,9. W9. What is the amount of $ 175.08 from May 7, IS41, to Sepl. 25, 1843, at 7 per cent? A ns. 8 204.28,9. 20. Whbat is the amount of 8 160 from Dec. 11, 1843, to Sept. 9, 1S44, at 7 per cent.? Ans. $ 16S.33,7. PARTIAL PAYM ENTS. AaT. ~2,Q PARTIAL, PAYmIENTS are indorsements, or paymen ts made at different times, of a part of a note or other obligation. AR'T. 282~'SWhen notes are paid within one year from the time they become due, it hars been the usual custom to compute the interest by the following RuLe. — Find te amount of the pirincipa! from the Zimre it became due until the time of payment, and theLe find the amount of each indorsement fromi the time it was paid until settlement, and subtract their sum jr'om lhe amount of the principal..Ex.. t 1 23 4. Boston, Jan. 1, 1843. For value received, I promise to pay John Smith, or order, on demand, one thousand two hundred thirty-four dollars, with interest. John Y. Jones. Attest, Samuel Emerson. On this note are the following indorsements: - March 1, 1843, received ninety-eight dollars. June 7, 1843, received five hundred dollars. Sept. 25, 1843, received two hundred ninety dollars. Dec. 8, 1843, received one hundred dollars.'What remains due at the time of payment, Jan. 1, 1844. A.s,. 293.12. QUTESTIONS. - Art. 202. What a are ptial payments? - Art. 203. What is the rule for computing the interest wlien there are partial payments, and all are made within one year? 208 PARTIAL PAYMENTS. ECT. XX. II OPERATION, Principal, - - - $ 1234.00 Int. from Jan. I, 1843, to Jan 1, 1844, (ly.) 74- 4.04 Amount, - -. 1308.04 First payment, March 1, 1843, $ 98.00 Int. from March 1, 1843, to Jan. 1, 1844, (lOmo.) 4.90 Second payment, June 7, 1843, - - - 500.00 Int. from June 7, 1843, to Jan. 1, 1844, (6mo. 24da.).-.- 17.00 Third payment, Sept. 25, 1843, 290.00 Int. from Sept. 25, 1843, to Jan. 1, 1844, (3mo. 6da.) - - - 4.64 Fourth payment, Dec. 8,9 1843, - - 100.00 lnt. from Dee. 8, 1843, to Jan. 1, 1844, (23da.).38 Amount of payments to be deducted, - $ 1014.9il Balance, remains due Jan. 1. 1844, $ 293.12 2. $ 987.75. Danvers, Jan. 11. 1842. For value received, we jointly and severally promise to pay Fitch Pool, or order, on demand two months from date, nine hundred eighty-seven dollars seventy-five cents, with interest after two months. John T. Johnson. Attest, Isaiah Webster. Samuel Jones. On this note are the following indorsements: - Mlay 1, 1842, received three hundred dollars. June 5, 1842, received four hundred dollars. Sept. 25,'1842, received one hundred and fifty dollars. What is due Dec. 13, 1842? Ans. 6 156.94. 3. $ 800. Bradford, July 4, 1842. For value received, I promise to pay Leonard Johnson, or order, on demand, eight hundred dollars, with interest. Attest, Enoch True. Samuel Neverpay. On this note are the following indorsements: - Aug. 10, 1842, received one hundred fobrty-four dollars. Nov. 1, 1842, received ninety dollars. Jan. 1, 1843, received four hundred dollars. MIarch 4, 1843, received one hundred dollars. What remains due June 1, 1843? Ans. $ 88.02. ART.!T0i~. In the United States court, and in most of the courts of the several States, the following rule is adopted for computing the interest on notes and bonds, when partial payments have been made. QUESTION.- Exptain the operation. ECT. XXIII.j PARTIAL PAYMENTS. 209 RULE. -- Compute the interest on the principal sum, from the time.when the interest commenced to the time when the first payment was made which exceeds, either alone or in conjunction with the preceding payments, if any, the interest at that time due; add that interest to the principal, and from the sum subtract the payment made at that time, tgoether with the preceding payments, if any, and the remainder forms a new principal; on which compute the interest, as upon the first principal, and proceed in the same manner to the time of judgnment. This rule is illustrated in the following question. Ex. 1. 365.50. Lynn, Jan. 1, 1842. For value received, I promise to pay John Dow, or order, on demand, three hundred, sixty-five dollars fifty cents, with interest. John Smith. Attest, Samuel Webster. On this note are the following indorsements:June 10, 1842, received fifty dollars. Dec. 8, 1842, received thirty dollars. Sept. 25, 1843, received sixty dollars. July 4, 1844, received ninety dollars. Aug. 1, 1845, received ten dollars. Dec. 2, 1845, received one hundred dollars. What remains due Jan. 7, 1847? Ans. $ 92.53. OPERATION. Principal carrying interest from Jan. 1, 1842, to June 10, 1842, - ~ $ 365.50 Interest from Jan. 1, 1842, to June 10, 1842, (5 months, 9 days,). - D 9.68 Amount, 375.18 First payment, June 10, 1842, 5 50.00 Balance for new principal, - - 325.18 nterest from June 10, 1842, to Dec. 8, 1842, (5 months 28 days,) - - 9.64 Amount, 334.82 Second payment, Dec. 8, 1842, - 30.00 Balance for new principal, - - - 304.82 Interest from Dec. 8, 1842, to Sept. 25, 1843, (9 months, 17 days,).- - 14.58 Amount, 319.40 Third payment, Sept. 25, 1843, - 60.00 Balance for new principal, - - 259.40 Interest from Sept. 25, 1843, to July 4, 1844, (9 months, 9 days,) - - - 12.06 Amount, 271.46 QUEsrIONS. - Art. 204. What is the rule generally adopted by the several States for computing the interest on notes and bondsi when partial payments hrave been made? 18*94 210 PARTIAL PAYBiENiTS. sErCT. XXIII. Amount brought up, 971.46 Fourth payment, July 4, 1844, - 90.00 Balance for new principal,. 181.46 Interest from July 4, 1844, to Dec. 2, 1845,'(16 months, 28 days,) -. 15.36 Amount, 1996.82 Fifth payment, Aug. 1, 1845, (a sum less than the interest,) - - - $10.00 Sixth payment, Dec. 2, 1845, (a sum greater -than the interest,). - 100.00 110.00 Balance for new principal, - - 86.82 Interest from Dec. 2, 1845, to Jan. 7, 1847, (13 months, 5 days,) - - 5.71 Remains due Jan. 7, 1847, -1847.53 2. 5 1666. Newburyport, June 5, 1838. For value received, I promise to pay John Boardman, or order, on demand, one thousanrl six hundred sixty-six dollars, with interest. John J. Fortune. Attest T. WVebstelro On this note are the following indorsements: — July 4, 1839, received one hundred dollars. Jan. 1, 1810, received ten dollars. July 4, 1840, received fifteen dollars. Jan. 1, 1841, received five hundred dollars. Feb. 7, 1842, received six hundred filty-six dollars. What is due Jan, 1, 1843? Ans. $ 767.08. 3,. $ 960. Newark, N. J., Oct. 23, 1840. On demand, I promise to pay S. S. St. John, or order, nine hundred sixty dollars, for value received, with interest at seven per cent. John Q. Smith. Attest, 1. F. WVilcox. On this note are the following indorsements:Sept. 25, 1841, received one hundred forty dollars. July 7, 1842, received eighty dollars. Dec. 9, 1842, received seventy dollars. Nov 8, 1843, received one hundred dollars. What is due Oct. 23, 1844? Ans. $ 807.76. 4. s 1000. New York, January 1, 1839. Two months after date, I promise to pay S. Durand, or QuE sTIos. - Explain the operation. .ELCT. xxIS.] PA'RTIAL PAYMIENTS. 24 1 order, one thousand dollars, for value received, with interest after, at seven per cent. Paul Samps'on, Jr. Attest, William S. Hall. On this note are the following indorsements:tlarch I, 1840, received one hundred dollars. Sept. 25, 1841, received two hundtred dollars. Oct. 9, 1842, received one hun(lred fifty dollars July 4, 1843, received twenty dollars. Oct. 9, 1843, received three hundred dollars. 5What is due Dec. 1, 1844? Ans. 8 567.49. ART.;-j4 A. The following is the rule established by the Supreme Court of the State of Connecticut. 1. " Compute the interest to the time of the first payment; if that be one year or more frorn the time the interest commenced, acidd it to the principal, and deduct the payment from the sum total. if there be after payments made, conipute the interest on the b-allce due to the next payment, and then de(duct the payment as above; and in like manner from one payment to another, till all the payments are absorbed; provided the time between one payment and another be one year or more." 2. "But if any payments be made before one year's interest hath accrued, then complute the interest on the principal sumn due on the obligation for one year, * add it to the principal, and compute the interest on the sum paid fromn the time it was paid up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added together.'' 3. " If any payments be made of a less sum than the interest arisen at the time of such paymnent, no interest is to be computed, but only on the principal sumn for any period." Ex. 1. S 500. Hartford, July 1, 1844. For value received, I promise to pay J. Dow, or order, on demand, five hundred dollars, with interest. D. P. Pa ge. On this note are the following inldorsements: Sept. 1, 1845, received one hundred dollars. April 1, 1846, received one hundred tfrty-tour dollars. Jan. 1, 1847, receited ninety dollars fifty cents. Dec. 1, 1848, received one hundred sixty-eight dollars five cents. What is due Oct. 1, 1849? Ans. $ 92.40. * If a year extends beyond the time when the note becomes clue, find the amount of the relllaintiag principal to the tine of settlestcnt; fill, alI-o tile aItutnIt of the inllorseennlt or ilndorseelln ts, if any, frlnl tie tine tley v were paid to the time of'setlenlent, and subtract their suln from the amlount of the principal. kUUEST1ON. - Art. 204 A. What is the Connecticut rule for computiing iuerest on notes and bouds, when partial payxment; have been made'? ,S12 PROBLEMS IN INTEREST. [SECT. EXXU, PROBLEMS IN INTEREST. ART. 2-,. A PROBLEM in arithmetic is a question, or prop)Sition, which requires some unknown truth to be investigated. ART. 2@G. In the preceding questions in interest, five terms or things have been mentioned;- viz. the Interest, Amount, Rate per cent., Time, and Principal. The investigation of these involves five problems: I. to find the interest; II. to find the amount; III. to find the rate per cent.; IV. to find the time; V. to find the principal. With one exception, any three of the preceding terms being given, a fourth may be found by the rules deduced from the solution of the problems. But if the rate per cent., time, and amount are given, an additional rule is necessary to find the principal, which will form a sixth problem; but from its connection with Discount, its solution will be deferred until that subject is considered. The Problems I. and II. have already been examined, and we now proceed to an examination of those remaining. AnRT. 207. Problem III. To find the rate per cent., the principal, interest, and time being given. Ex. 1. The interest of $ 300 for 2 years is 48; what is the rate per cent.? Ans. 8 per cent. OPERATION. We find the interest on $3 00 the principal for 2 years.0 2 at 1 per cent., and divide the given interest by it. $ 6.0 0) 4 8.0 0 (8 per cent. Since the interest of 4 8.00 $1 at 1 per cent. for 2 years is 2 cents, the interest of $ 300 will be 300 times as much, equal to $ 6. Now if $ 6 is I per cent., $ 48 will be as many per cent. as $ 6 is contained times in $ 48, which gives 8 per cent. for the answer. RULE. - Divide the given interest by the interest of the given sum at I per cent. for the given time, and the quotient will be the rate per cent, required. QUES'rIOs.- Art. 205. What is a problem in arithmetic? - Art. 206. How many terms or things have been given in the preceding questions in interest 1 Name them. What does an investigation of these terms involve.? Name them. How many terms are given in each problem in order to find a fourth? What two problems have been examined?-Art. 207. What is Problem 1II.? Explain the operation. What is the rule for findinge the rate per cent., the principal. interest, and time being given? SEc'r. xXIx.] PROBLEMS 1N INTEREST. EXA-MPLEIS FOR PRACTICE. 2. The interest of $ 250 for 1 year, 3 months, is 8 28.125; what is the rate per cent.? Ans. 9 per cent. 3. If I pay 8 8.82 for the use of. 1p 72 for 1 year, 9 months, what is the rate per cent.? Ans. 7 per cent. 4. A note of 8 500, being on interest 2 years, 6 months, amounted to $ 550; what was the rate per cent.? Ans. 4 per cent. ART. 2@0. Problem IV. To find the time, the princlpal, interest, and rate per cent. being given. Ex. 1. For how long a time must 8 300 be on interest at 6 per cent. to gain ~ 36? Ans. 2 years. OPERATION. 3 0 0 We find the interest on the.0 6 given principal for 1 year, by which we divide the given in-', 1 8.0 0) 3 6.0 0 (2 years. terest. 3 6.0 0 Since the interest of $ 1 for 1 year is 6 cents, the interest of $ 300 will be 300 times as much, equal to $ 18. Now, if it require 1 year for the given principal to gain $ 18, it will require as many years to gain $ 36 as $ 18 is contained times in $ 36. Thus, $ 36.' $ 18 = 2 years for the answer. RULE. - Divide the given interest by the interest of the given principal for 1 year, ahd the quotient is the time. EXAMP'LES FOR PRACTICE. 2. If the interest of $ 140 at 6 per cent. is $ 42, for how long a time was it on interest? Ans. 5 years. 3. How long a time must. 165 be on interest at 6 per cent. to gain $ 14.85? Ans. 1 year, 6 months. 4. How long must 8 98 be on interest at 8 per cent. to gain. 25.48? Ans. 3 years, 3 months. 5. A note of $ 680 being on interest at 4 per cent. amounted to $ 727.60; how long was it on interest? Ans. 1 year, 9 months. QUESTIONS. Art. 208. What is Problem IV.? Explain the operation. What is the rule for finding the time, the principal, interest, and rate per cent. being given? 214 COMPOUiND INTEREST. [SECT. XXIv. A.n. OtK9. Problem V. To find the principal, the interest, time, and rate per cent. being given. Ex. 1.'What principal at 6 per cent. will gain $ 36 in 2 years? Ans. 8 300. OPIIrAT10I O. ifWe find thle interest of $ 1 for.0 intt. of $1 for Iy. 2 years, by which -we divide the given interest..1 2)3 6.0 0( 38 0 0 principal. Since it requires 2 years for a principal of $ 1 to gain 12 cents, it will require a principal of as many dollars to gain $ 36 as 12 cents are contained times in $ 36. Thus, $ 36.00.12 - 300 for the answer. Ruiu. -.Divide the given interest or amount ky the irterest or amount of $ 1 for the given rate and time, and the quotient is the principal. EXAMPLES FOE PRACTICaE 2. What principal will gain $ 24.225 in 4 years, 3 months, at 6 per cent.? An s. P 95. 3, What principal will gain 5.B11;m 3 years, 6 months, at 8 per cent.? - Ans, 1.25. 4. The interest on a certain note at 9 per cent. in I year and 8 months amounted to $ 42; what was the full amount of the note? Ans. $280. X~ XIV. COMPPOUND INTEREST. ART1. ~[~. COMPOUND INTEREST is interest on the principal and interest, when the interest is not paid at the end of the year, or when it becomes due. The law specifies, that the borrower of money shall pay the lender a certain sum for the use of $ 100 for a year. Now, if he does not pay this sum at the end of the year, it is no more than just that he should pay interest for the use of it as long as he shall keep it in his possession. The computation of compound interest is based upon this principle. QUESTIONS. - Art. 209. What is Problem V.? Explain the operation. What is the rule for findingl the principal, the interest, time, and rate per cent. being given 7 - Art. 210. What is compound interest? On what principe is it based t SICT. xasv.] CO21POUND LN'EREST.,5 ~AT. p g, To find the compound interest of any sum of money atany rate per cent. for any given timeo Ex. 6. Wh'at is the compound interest of $ 500 for 3 years, 7 montrhs, and 1s2 days, at 6 per cent.s Ans. $ 117.54,1. OPERATIONI. P rincipal, 5 0 0 interest of $9 for i year,.0 6 tuterest for'at year, 6 0.0 0 Amount for Ist year,.5 3 0.0 0.0 6 ~interest for 2d year, 3.80 0 0 5 3 030 0 Amount for 2d year, 56 S.80.06 Interest for 3d year 3 3.7 0 8 0 56 1.8O Amount for 3d year, 59 5.5 0 8 Interest of $ I for 7mo. I da., 4 1 6 85 6 1786524 Interest for 7mo. 12da., 2 2.0 3 379 6 5 9 5.5 08 A mount for 3y. 7mo. 12da., 6 7.5 4 1 7 9 6 Principal subtracted, 5 0 0 Compound interest, $ 1 I 7.5 4,1 7 9 6 We first multiply tLe principal by the interest of $ 1 f;o:r I year, and add the interest thus found to the principal for the amount, or new principal. We then find the interest on this amount for I year, and proceed as before; and so also -with the third year. For the months and days we find the interest on the amount for the last year, anld, adding it as before, we subtract the original principzal firom the last amount for the answer. RULE. - Find the interest of the given sum for one year, and add it to the principal; then find the amount of this amzount for the next year; and so continue, until the'time of settlement. If there are months and days in the given time, find the amount for them on the amount for the last year. Subtract the principal from the last amount, and the remainder is the compound interest. QUESTIONS. - Art. 211. Explain the operation in comoputing compound inu terest. What is the rule 7 216 COMPOUND INTEREST. [SECT. XXSv. NOTE. - 1. If the interest is to be paid semiannually, quarterly, monthly, or daily, it must be computed for the half-year, quarter-year, nmonth, or day, and added to the prixicipal, and then the interest computed on this, and each succeeding amount thus obtained, up to the time of settlement. 2. When partial payments have been made on notes at compound in:terest, the same rule is adopted as given in Art. 204. EXAMPLES FOR PRACTICE. 2. What is the compound interest of $ 761.75 for 4 years? Ans. $ 199.94,1. 3. What is the amount of $ 67.25 for 3 years, at compound interest? Ans. $ 80.09,5. 4. What is the amount of 0 78.69 for 5 years at 7 per cent.? Ans. 1 110.369,4 5. What is the amount of $ 128 for 3 years, 5 months, and 18 days, at compound interest? Ans. $~ 156.71,7. 6.. What is the compound interest of $ 76.18 for 2 years, 8 months, 9 days? Ans. $ 12.96,7. ART. -[. There is a more expeditious method of computing compound interest than the preceding, by means of the following TABLE, Showing the amount of$1, or ~1, for any number of years not exceeding 20, at 3, 4, 5, 6, and 7 per cent., compound interest. Years. 3 per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent, Years. 1 1.030000 1.040000 1.050000 1.060000 1.07000- 1 2 1.060900 1.181600 1.102500 1.12361)0 1.144900 2 3. 1 009727 1.124864 1.157625 1.191016 1.225043 3 4 1.125508 1.169858 1.215506 1.262476 1.310795 4 1.159374.2 16652 1.276281.3385 1.402552.5 6 1.1940.52 1_.26 53-1 9 1.3-40095 1.4:]851.9 1.500730 6 7 1.229873 1.315931 1.407100 1.503630 1.605781 7 8 1.266770 1.3685639 1.477455 1.5o93848 1.7181-86 8 9 1.3047731.42331. 1.551328 1.689478 1.838459 9 910 1.343916'1.4809284 1.6 28894 1,790847 1.967151 10 11 1.384233 1.539454 1.710339 1.898298 2.104852 11 12 1.425760 1,601.032 1.795856 2.012196 2.252191 12 13 1.468533 1.665i173 1.885649 2.1132928 2.409845 1o 1.4 1.512589 1.731676 1.979931 2.260903 2.578534 14 15 1.557967 1.800943.2.078928 2.3936558 2.750032 15 I 16 1.604706 1.872981 2.182874 2.540351 2.9521.64 16 17 1.652847 1.947900 2.292018 2.61J2772 3.158815 17 18 1.702433,2.0258 16 2.406619 2.854339 3.379932 18 1.9 1.753506 2.1006849 2.5269050 3.02,5599 3.616528 19 ] 20 1.806111 2.191123 2.653297 3.207135 3.869685 20 QuTEsTIONS. -If the interest is to be paid semianually, quarterly, &c,., how is it computed? How, when partial payments have been made? - Art. 212. ErXplain the method ocf cQ.omp.tin compound inter e;t by mnans of the table. ECT. XxIV.] COMPOUND INTEREST. 217 Ex. 1. What is the interest of $ 240 for six years, 4 months, and six days, at 6 per cent.? Ans. 0$ 107.59,93 OPERIATION. Amount of $ 1 for 6 years, 1.4 1 8 5 1 9 Principal, 2 4 0 56740760 2837038 Amount of principal for 6 years, 3 4 0.4445 60 Interest of $ 1 for 4 mo. 6da.,.0 2 1 34044456 68088912 interest of amount for 4mo. 6da., 7.1 4 933576 Amount added, 340.444560 Amount for 6y. 4mo. 6da., 3 47.5 9 3 8 95 7 6 Principal subtracted, 2 4 0 Interest for given time, $ 10 7.5 9,3 8 9 57 6 NVe find the amount of $ 1 for 6 years in the table, and multiply it by the- principal, and obtain the amount for 6 years. WVe then find the interest on this amount for the 4 months and 6 days, and add it to the first amount, and from this sum subtract the principal for the answer. RULE. - Multiply the amount of $ 1 for the given rate and time, as found in the table, by the principal, and the product is the amount. Subtract the principal from the amount, and the remainder is the compound interest. If there are months and days, proceed as in the foregoing rule. EXABMPLES FOR PRACTICE. 2. What is the interest of 9 884 for 7 years, at 4 per cent.? Ans. $ 279.28,3. 3. What is the interest of 8 721 for 9 years, at 5 per cent.? Ans.. 397.50,7. 4. pWhat is the amount of 9C0 for 12 years, 6 months, a: 3 per cent.? Ans. $1389.26. 5. Wrhat is the amount of $ 25.50 for 20 years, 2 months, and 12 days, at 7 per cent.? Ans.; 100.05,8. 6. What is the amount of $ 12 for 6 monlths, the interest eo be added each month? Ans. n 12.36,4-[. 7. What is the amount of $100 for 6 days, the interest to be added daily? Ans. $ 100.10,004. QUESreio. - What is the ruile? 19 218 DISCOUNT. [SECT. XXV ~ XXV. DISCOUNT. ART. ]:~. DISCOUNT is an allowance or deduction, according to the rate per cent., made for the payment of money before it becomes due. The present worth of any sum is the principal which, being put at interest, will amount to the given sum in the time for which the discount is made. Thus, 8 100 is the present worth of $ 106 due one year hence at 6 per cent.; for $ 100 at 6 per cent. will amount to $ 106 in this time; and $- 6 is the discount. ART. 214. The interest or percentage of any sum cannot properly be taken for the discount; for we see from the preceding illustration, that the interest for one year is the fractional part of the sum at interest, denoted by the rate per cent. for the numerator, and $ 100 for the denominator; and the discount for one year is the fractional part of the sum on which discount is to be made, denoted by the rate per cent. for the numerator, and the amount of $ 100 for the denominator. Thus, if the rate per cent. of interest is 6, the interest for one year is T:-A6 of the sum at interest; but if the rate per cent. of discount is 6, the discount for one year is T-OW of the sum on which discount is made. ART. A1d. In discount, the rate per cent., time, and the sum on which the discount is made, are given to find the present worth. These terms correspond precisely to Problem VI. in interest, in which the time, rate per cent., and amount are given to find the principal. (Art. 206.) ART. G. To find the present worth and the discount on any sum, at any rate per cent., for any given time. Ex. 1. What is the present worth of 8 25.44 due one year hence, discounting at 6 per cent.? What is the discount? Ans. 824 present worth; $ 1.44 discount. QUESTIONS. - Art. 213. What is discount? What is the present worth of any sum of money? How illustrated? - Art. 214. Are interest and discount the same 1 Explain the difflerence. Which is the greater, the interest or discount on any sum, for a given time? - Art. 215. What terms are given in discount, and what is required? To what do these correspond in interest? SECT. XXV.] DISCOUrNT. 219 OPERATION. Amount of $ 1, 1.0 6) 2 5.4 4 ($ 2 4 present wortn. 212 4 2 4 $2 5.4 4 sum or amount. 4 2 4 2 4.0 0 present worth.! 1.4 4 discount. We find the amount of $1 for the given time, by which we divide the given sum and obtain the present worth. Then, subtracting the present worth from the given.sum, we obtain the discount. Since the present worth of $ 1.06 due one year hence, at 6 per cent,, is $ 1, it is evident the present worth of $- 25.44 is as many dollars as $ 1.06 is contained times ill $ 25.44. $ 25.44 — $1.06 24. Htence the following RULE. — Find the amount of $ 1 for the given time and rate per cent., by which divide the given sum, and the quotient is the present worth. Subtract the ~present worth from the given sum, and the remainder is the discount. No'V. — The discount may be found directly by making the interest of $ 1 for the given rate and time the numerator of a fraction, and the amount of $ 1 for the given rate and time the denominator, and then multiplying the given sum by this fraction. EXAMPLES FOR PRACTICE. 2. WThat is the present worth of $ 152.64, due one year hence? Ans. $ 144. 3. What is the present worth of $477.71, due four years hence? Ans. $ 385.25. 4. What is the discount of $ 172.86, due 3 years, 4 months, hence? Ans. $ 28.81. 5. What is the discount of $ 800, due 3 years, 7 months, and 18 days hence? Ans. $ 143.18,6. 6. Samuel Heath has given his note for $ 375.75, dated Oct. 4, 1842, payable to John Smith, or order, Jan. 1, 1844; what is the real value of the note at the time given? Ans. $ 349.69,7. 7. Bought a chaise and harness of Isaac Morse, for $ 125.75, for which I gave him my note, dated Oct. 5, 1842, to be paid in six months; what is the present value of the note, Jan. 1, 1843? Ans. $ 123.81. QUESTrIns. - Art. 216. Explain the operation for finding the present worth and discount. Give the reason of the operation. What is the rule? What other method is given? 220 BANK DISCOUNT. [SECT.:XBX3. ~ XXVI. BANK DISCOUNT. ANT. p2'ne. BANK DISCOUNT is the simple interest of a note, draft, or bill of exchange, deducted from it in advance, or before it becomes due. The interest is computed, not only for the specified time, but also for three days additional, called days of grace. Thus, if a note is given at the bank for 60 days, the interest, which is called the discount, is computed for 63 days; and if the note is paid within this t mne, the debtor complies with the requirements of the law. The legal rate of discount is usually the same as the legal rate of interest; and the difference between bank discount and true discount is the same as the difference between interest and trite discount. A note is said to be discounted at a bank, when it is received as security for the money that is paid for it, after deducting the interest for the time it was given. The sum paid is called the avails or present worth of the note. ART. ~ go To find the present worth and the bank discount of any note or sum of money for any rate per cent. and time. Ex, 1. What is the bank discount on $ 842 for 90 days, at 6 per cent.? What is the present worth? Ans. $ 13.05,1 discount; 0 828.94,9 present worth. OPERATION. Sum discounted, 8 4 2 4 $ 42.000 interest of *l,.0 1 5 5 1 3.0 5 1 42 1 0 $ 8 28.9 4,9 present worth. 4210 842 Bank discount, t$ 1 3.0 5,10 We find the interest of $1 for 93 days, by which we multiply the: sum discounted, and the product is the discount. We then subtract the discount from the given sum, and obtain the present worth. QUESTIONS. - Art. 217. What is bank discount? When is it paid? Is interest computed for more than the specified time? What are these three additional days called? l-Towl will you illustrate this? Wlen must a note be paid, that is givens for sixty day's, and still comply with the requirements of the law? What is the legal rate of discount? lWhat is the difference between bank discount and true discount? When is a note said to be discounted at a bank? What is the sum paid for it called? - Art. 218. Explain the operation for finding the bank discount on any sum. sBCT. xxvIl.3 BANK DISCOUNT. "22X RULE.-. 1. Find the interest on the note, or sum discounted, for the given rate and time, including THREE days of grace, and this interest is the discount. 2. Subtract the discount from the note or sum discounted, and the remainder is the present wzorth. EXAMPLES OR PRACTICE. 2. What is the bank discount on $ 478 for 60 days? Ans. $ 5.01,9. 3. What is the bank discount on 7 r80 for 30 days? Ans. 8 4.29. 4. What is the bank discount on $ 1728 for 90 days? Ans. $ 26.78,4. 5. How much money should be received on a note of 1,000, payable in 4 months, discounted at a bank where the interest is 6 per cent.? Ans. $ 979.50. 6. What sum must a bank pay for a note of $ 875.35, payable in 7 months and 15 days, discounting at 7 per cent.? Ans. $ 836.54,2. 7. What are the avails of a note of $ 596.24, payable in 8 months and 9 days, discounted at a bank at 8 per cent.? Ans. $ 562.85. 8. What is the bank discount of a draft of $ 1350.50, payable in 1 year, 4 months, at 5 per cent.'? Ans. $ 90.59,6. ART. 919o To find the amount for which a note must be given at a bank, to obtain a specified sum for any given time. Ex. 1. For what amount must a note be given, payable in 90 days, to obtain 8 500 from a bank, discounting at 6 per cent.? Ans. $ 507.87,2. OPERATION. We subtract the interest of $ 1.0 0 00 $1 for 93 days, at 6 per cei)t., Int. of $ 1 for 93da.,.0 1 5 5 from $ 1, and divide the given Present worth of 1,.9 8 4 5 sum by the remainder, for the answer. $ 500 -~..9845 507.87,2. Since $ 0.9845 requires i 1 principal for the given time $ 500 will require as many dollars principal as $ 0.9845 is contained times in $ 500; and $ 500 +- $ 0.9845 = $ 507.87,2. Hence the uESTrIONS. - What is the rule 7 - Art. 219. Explain the operation for finding the amount for which a note must be given at a bank to obtain a specified sum tor a riven time. 222 COMMISSION AND BROKERAGE. [sECT. XXV1s RULE. - Divide the given sum by the present worth of $ 1 for the given time and rate per cent. of bank discount, including THRiEE days of grace, and the quotient will be the answer required. EXAMPLES FOR PRACTICE. 2. For what sum must I give my note at a bank, payable in 4 months, at 6 per cent. discount, to obtain $ 300? Ans. $ 306.27,8. 3. A merchant sold a quantity of lumber, and received a note payable in 6 months; he had his note discounted at a bank, at 6 per cent., and received $4572.40. What was the amount of his note? Ans. 8 4716.24,5. 4. A gentleman wishes to take $ 1000 from the bank; for what sum must he give his note, payable in 5 months, at 6 per cent. discount? Ans. $ 1026.16.7 5. The avails of a note, discounted at the bank for 8 months at 7} per cent., were $ 483.56; what was the face of the note? Ans. $ 509.34,5. XXVII. COMMISSION AND BROKERAGE. ART. 26. COIrMlISSION is the percentage paid to commisslon merchants and. agents for buying and selling goods and transacting other business. Brokerage is the percentage paid to brokers for making exchanges of money, negotiating different kinds of bills of credit, and transacting other business. The rate per cent. of commission or brokerage is not regulated by law, but varies in different places, and with the nature of the business transacted. Commission and brokerage are computed in the same manner ART. 21.O To find the commission or brokerage on any sum of money. Ex. 1. A commission merchant sells goods to the amount of 8 879; what is his commission, at 3 per cent.? Ans. $ 26.37. QUESTIONS. - What is the rule? - Art. 220. What is commission? What is brokerage? How is the rate per cent. regulated? H ow are commlliseiou and brokerage computed? S.ac'. xxvs.J Co)51MISSIOIN AND BROKERAG1I. 223 Since commission is a percentage on the given sum, the commission on $ 879 at 3 per cent. will be 8 879 X.03 = $26.37. RULE. - Find the percentage on the given sum at the given rate per cent., and the result is the commission or brokerage. (Art. 194.) EXAMPLES FOR PRACTICE. 2. What is the commission on the sale of a quantity of cotton goods valued at $ 5678, at 3 per cent.? Ans. $170.34. 3. A broker sells goods to the amount of $ 7896, at 2 pei cent.; what is his commission? Ans. $157.92. 4. My agent in Lowell has purchased goods for me to the amount of $1728; what is his commission, at 1 per cent.? Ans. 825.92. 5. My factor advises me, that he has purchased, on my account, 97 bales of cloth at $15.50 per bale; what is his commission, at 2~ per cent.? Ans. $ 37.58,7. 6. My agent at New Orleans informs me, that he has disposed of 500 barrels of flour at $ 6.50 per barrel, 88 barrels of apples at $2.75 per barrel, and 56cwt. of cheese at $ 10.60 per cwt.; what is his commission, at 3a per cent.? Ans. $153.21. 7. A broker negotiates a bill of exchange of $ 2500 at I pec cent. commission; what is his commission? Ans. $12.50. 8. A broker in New York exchanged $ 46256 on the Canal Bank, Portland, at - per cent.; what did he receive for his trouble? Ans. $ 57.82. AnT. 222. To find the commission or brokerage on any sum of money, when it is to be deducted from the given sum. Ex. 1. A merchant in Cincinnati sends $1500 to a commission merchant in Boston, with instructions to lay it out in goods, after deducting his commission of 2A per cent.; what is his commission? Ans..36.58,6. Since the agent cannot justly receive a commission on the money he reserves to himself, but only on the amount actually expended for the goods, his commission will be the same as the discount on the given sum at the given rate per cent. Hence the following qUESTIONS. - Art. 221. What is the rule? - Art. 222. How do you find the commission or brokerage on any sum when it'is to be deducted from the given sum? Give the reason for the operation. 224 STOCKS. [SECT. xxvsA1 RULE. -Find the discount on the given sum, at the given rate per cent., and the result is the commission. (Art. 216.) EXAMIPLES FOR PRACTICE. 2. A town agent has $ 2000 to invest in bank stock, after deducting his commission of 1~ per cent.; what will be his commission, and what the sum invested? Ans. $ 29.55,7 commission; $1970.44,3 sum invested. 3. A shoe-dealer sends $ 5256 to his agent in Boston, which he wishes him to lay out for shoes, reserving his commission of 3 per cent.; what is his commission? Anlls. $153.08,8. 4. A broker expends $ 3865.94 for merchandise, after deducting his commission of 4 per cent.; what was his commission, and what sum did he expend? Ans. $148.69 commission; $ 3717.25 sum expended. 5. I have sent to my agent at Buffalo, N. Y., $10000, which i wish him to expend for flour, after deducting bis commission of 3; per cent.; what will be his commission, an'd also the value of the flour purchased? Ans. $ 314.76+commission; $ 9685.23+ value of flour. ~ XXVIII. STOCKS. ART. 23. SrocK is a general name given to governmen* funds, and to the capital of incorporated institutions, such as banks, railroad companies, &c. Stocks are usually divided into equal shares, the market value of which is often variable. When stocks sell for their original value, they are said to be at par; when for more than their original value, above par, or in advance; when for less than their original value, below par, or at a discount. The premium or advance, and the discount on stocks, are generally computed at a certain per cent. on the original value of the shares. Qu srToNs. - What is the rule? —Art. 223. What is stock? IHow are stocks divided? Is their value uniform? When is stock said to be at par? When at an advance 7 When at a discount? Hoow are premiums ant the discount on stocks computed? What is the rule? What is the premium or discount? dECT. XXVIII.j] ST OCK 225 ART. 22.4 To find the value of stocks, when at an advance or at a discount. Ex. 1. What is the value of $ 2150 railroad stock, at 7 per cent. advance? Ans. $ 2300.50. OPERATION. 2150 X.07 - $150.50; $ 2150 -+ 150. 50 2300.50. 2. What is the value of $ 975 bank stock, at 5 per cent. discount? Ans. 926.25. OPERATION. 975 x.05 - 48..7; 975 - 48.75 92625. RULE. — Find the percentage on the given sum, and add or subtract, according as the stock is at an advance or at a discount. (Art. 194.) NOTE. - The percentage is the premium or discount. EXAMPLES FOR PRACTICE. 3. W;hat must be given for 10 shares in the Boston and Maine Railroad, at 15 per cent. advance, the shares being $ 100 each? Ans. $ 1150. 4. What must be given for 75 shares in the Lowell IRailroad, at 25 per cent. advance, the original shares being $ 100 each? Ans. ) 9375. 5. What is the purchase of,$ 8979 bank stock, at 12 per cent. advance? Ans. 10056.48. 6. What is the purchase of $1789 bank stock, at 9 per cent. below par? Ans. $1627.99. 7. A stockholder in the Boston and Maine Railroad sells his right of purchase on 5 shares of $100 each at 12 per cent. advance; what is the premium? Ans. $ 60. 8. What is the value of 20 shares canal stock, at 12~ per cent. discount, the original shares being $100 each? Ans. $1750. 9. What is the value of 15 shares in the Livingston County Bank, at 86 per cent. advance, the original shares being $ 100 each? Ans. 1623.75. 10. Bought 87 shares in a certain corporation, at 12 per cen-. below par, and sold the samne at 191 per cent. above par what sumn did I gain, the original shares being $ 175 each? Ans. $ 4795.8&7V 226 I1NSURANCE [SECT. XXI3 ~ XXIX. INSURANCE. ART. 2,5. INSURANCE is a security obtained by paying a certain sum for protection against such losses of property or of life as are specified in the policy. Premium is the amount of percentage paid on the property insured for one year, or any specified time. As a security against fraud, property is not usually insured for its whole value, nor is the insurer or underwriter bound to indemnify the insured for a loss more than is specified in the policy. ART. 2~6o To find the premium on any amount of property insured. Ex. 1. What is the premium on O 485 at 2 per cent.? Ans. $ 9.70. OPERATION. $ 485 X.02 -- $9.70. RULE. - Find the percentage on the given sum, and the result is the premium. (Art. 194.) EXAMPLES FOR PRACTICE. 2. What is the premium on $ 868 at 12 per cent.? Ans. $104.16. 3. What is the premium on $1728 at 15 per cent.? Ans. $ 259.20. 4. A house, valued at $3500, is insured at 1 per cent.; what is the premium? Ans. 9 61.25. 5. A vessel and cargo, valued at $ 35000, are insured at 33 per cent.; now, if this vessel should be destroyed, what will be the actual loss to the insurance company? Ans. $ 33687.50. 6. A cotton factory and its machinery, valued at $ 75000, are insured at 21 per cent.; what is the yearly premium? and if it should be destroyed, what loss would the insurance company sustain? Ans. $1875 premium; $ 73125 loss. QUESTIONS. - Art. 225. What is insurance? What is a policy? What ii the premium? Is property usually insured for its whole value? Is the insure, bound to make up for a loss more than is specified in the policy? - Art. 22.6 What is the rule for finding the premium on any amount of property insured? SE:cT. XX. DUTIES. 227 ~ XXx. DUTIES. ART. 27., DUTIES are taxes imposed by government on imported goods. Duties are either specific, or ad valorem. A specific duty is a certain sum paid on a ton, hundred weight, yard, gallon, &c. An ad valoremn ditty is a certain per cent. paid on the actual cost of the goods in the country from which they are imported. Draft is an allowance for waste, made in the weight of goods. Tare is an allowance made for the weight of the cask, box, &c., containing the commodity. Leakage is an allowance of 2 per cent., for waste, made on liquors. Gross weight is the weight of the commodity together with the cask, box, bag, &c., containing it. INTet weight is what remains after all allowances have been made. ALLONWANCE FORP DRAFT. lb. lb On 112 1 Above 112 and not exceeding 224 2 224 " " 336 3 " 336 " " 1120 4,i 1120 - t, 2016 7 " 2016 9 NOTE. — It is.not customary to mention in the question, either tne draft or the leakage, since it is always the same, and must be deducted from each box, bag, or cask, before the other allowances, named in the question, are made. ART. 92S. To find the specific duty on goods or merchandise. Ex. 1. What is the duty on 1 hogshead of sugar, weighing 12761b. gross, at 21 cents per pound; tare 12 per cent.? Ans. $ 27.92,5. QUESTIONS. - Art. 227. What are duties? What is a specific duty? What is an ad valorem duty? What is draft? What is tare? lWhat is leakage? What is gross weight? What is net weigh1'? 228 DUTIES. [SEac. Xcx. OPERATION. Gross weight, 12 7 6 lb. Draft deducted, 7 lb. 12691b. 12 per ct. of 12691b. tare, 15 21 lb. Net weight, 1 1 1 7 lb. X.021 =$ 27.92,5, duty. RIULE. - Deduct all allowances to be made from the given quantity, and multiply the remainder by the duty on a unit of the given quantity, and the product will be the duty required. NOoT E.- In reckoning allowance for tare or leakage, a fraction equal to, or greater than, one halJ; is reckoned 1; when less, it is omitted. EXAMIIPLES FOR PRACTICE. 2. WVhat is the duty on 144 casks of nails, each weighing 56 01b., at 3 cents per pound; tare 8 per cent.? Ans. $ 221 1.84. 3. Required the duty on 760 bags of coffee, each weighing 36Slb., at 2 cents per pound; tare 12 per cent. Ans. $ 4864. 4. What is the duty on 4 pipes of Lisbon wine, gross gauge as follows: No. 1, 187 gallons; No. 2, 196 gallons; No. 3,216 gallons; No. 4, 150 gallons; the actual wants or quantity necessary to fill each pipe 5 gallons, and the duty at 25 cents per gallon? Ans. $ 178.50. ART. J9. To find the ad valorem duty on goods or maerchandise. Ex. 1. At 25 per cent., what is the ad valorem duty on 165 yards of broadcloth, at $ 5 per yard?. Ans. $ 206.25. OPERATION. 165X $5 $- 825; $825 X.255 - 206.25, duty. RULE.- Find the per centage on the cost of the goods, and the result is the ad valorem duty, (Art. 194.) EXAMIPLES FOR PRACTICE. 2. W1hat is the duty on 17281b. of copper sheathing, invoiced at $ 3200, at 20 per cent. ad valorem? Ans. $ 640. 3. What is the net weioght of one ton of Russia iron, and also the duty at 30 per cent., ad valorem; the cost of the iron being 4 cents per lb.? Ans. 22311b. net; $ 26.77,2. duty. UtrSTIONS. -Art. 228. What is the rule for finding the specific duty on goods 7 -.. Art. 229. What is the rule for finding the ad valorem (ldut?' SECT. XXXI.j ASSESSMENT OF TAXES. 229 4. What is the net weight of 16981b. of lead, and also the duty at 20 per cent. ad valorem; the value of the lead being 5 cents per pound? Ans. 16911b. net weight; $ 16.91 duty. 5. What is the duty on 10 hogsheads of molasses, each hogshead gauging 150 gallons gross, the actual wants being 5 gallons to each hogshead, and the cost of the molasses 25 cents per gallon; duty 20 per cent. ad valorem? Ans. $ 71. duty. 6. What are the net weight and duty, at 30 per cent. ad valorem, on 13 boxes of sugar, weighing gross 450 pounds each; tare 15 per cent., and the cost of the sugar being 8 cents per pound? Ans. 49271b., net weight; 118.24 duty. 7. What is the duty on an invoice of woollen goods, which cost in Liverpool 1376 ~ sterling, at 33 per cent. ad valorem, the pound sterling being S 4.84? Ans. $ 2197.74,7. ~ XXXI. ASSESSMENT OF TAXES. ART. 23o A TAX is a duty laid by government, for public purposes, on the property of the inhabitants of a town, county, or state, and also on the polls of the male citizens, liable by law to assessment. NOTE.- Poll is said to be a Saxon word, meaning head. In the constitution of Massachusetts, it means a male person who is liable to taxation..Taxes may be either direct or indirect. A direct tax is one imposed on the income or property of an individual; an indirect tax is one imposed on the articles for which the income or property is expended. Immovable property, such as lands, houses, &c., is called real estate. All other property, such as money, notes, cattle, furniture, &c., is called pe?rsonal property. The method of assessing town taxes is not precisely the same in all the states, yet the principle is virtually the same. The following is the law regulating taxation in MIassachusetits, (Revised Statutes, p. 79,): — QUESTIONs. - Art. 230. What is a tax? What does poll mean, as derived friom the Saxon? What does it mean in the constitution of Massachusetts? What is a direct tax? What, an indirect tax? What is real estate? What is personal property? Is the method of assessing taxes the same in all the tlates? 230 ASSESSMENT OF TAXES. [SECT. XXXI. "The assessors shall assess upon the polls, as nearly as the same can be conveniently done, one sixth part of the whole sum to be raised; provided the whole poll tax assessed in any one year upon any individual for town and county purposes, except highway taxes, shall not exceed one dollar and fifty cents; and the residue of said whole sum to be raised shall be apportioned upon property;" that is, on the real and personal estate of individuals which is taxable. ART. I23. To assess a town or other tax. Ex. 1. The tax to be assessed on a certain town is $ 2200. The real estate of the town is valued at $ 60000, and the personal property at $ 30000. There are 400 polls, each of which is taxed $ 1.00. WVhat is the tax on $ 1.00'What is A's tax, whose real estate is valued at $ 2000, and his personal property at $ 1200, and who pays for 2 polls? OPERATION. $ 1.00 X 400 = $ 400, amount assessed on the polls. $ 2200- $ 400 —$ 1800, am't to be assessed on the property $ 60000 + 8 30000 $ 90000, amount of taxable property. $ 1800 -. $ 90000 -- $ 0.02, tax on $ 1.00. $ 2000 X.02 - $ 40, A's tax on real estate. $ 1200 X.02 = $ 24, A's tax on personal property. $1.00 X 2- S 2, A's tax on 2 polls. $ 40 + $ 24 + $ 2= $ 66, amount of A's tax. RULE. - 1. Take an inventory of all the taxable property, real and personal, in the town or county, as the case requires, and also the number of polls liable to taxation. ~Multiply the sumn assessed on each poll by the number of taxable polls in the town, and subtract this amount from the sum to be raised by the town. 2. Divide this remainder by the whole valuation of the town, and the quotient will be the sum to be paid on $ 1 of each individual's real or personal estate. zMultiply each man's property by this sum, and the product, with his poll tax added, is the amount of his tax. EXAMPLES FOR PRACTICE. 2. The town of L. is taxed $ 3600. The real estate of ihe town is valued at $560,000, and the personal property at $ 152,500. There are 600 polls, each of which is taxed $ 1.25. What is the per cent. or tax on $ 1.00? and what is B's l.ax, QUESTIONS. — What is the law regulating taxation in Massachuset-s'? Art. 231. What is the ulle for assessing taxes 7 BEoT. xxxi.l ASSESSMENT OF TAXES. 231 whose real estate is valued at $ 4100, and his personal property at $ 1800, he paying for 4 polls? Ans. $.004, tax on I 1; $ 28.60, B's tax. 3. What is C's tax, who, living in the same town, pays for 1 poll, and is worth $ 15800? Ans. $ 64.45. 4. What is D's tax, who pays for 3 polls, and whose real estate is valued at 8 40000, and his personal property at $ 23600? Ans. $ 258.15. ART. 23e. The operation of assessing taxes may be facilhtated by the use of a table, which can be easily made after having found the tax on $1. Ex. 1. A tax of $ 3900 is to be assessed on the town of P. The real estate is valued at $ 840000, and the personal property at $ 210000; and there are 500 polls, each of which is taxed $ 1.50. What is the assessment on $ 1.00? Ans. $.003. Having found the tax on $1 to be 8.003, before proceeding to make the assessment on the inhabitants of the town, we find the tax on $ 2, $ 3, &c., and arrange the numbers as in thb following TABLE. $1 gives $.003 $20 gives $.06 300 gives $.90 2.006 30 ".09 400 " 1.20 3 ".009 40 ".12 500 I" 1.50 4 ".012 50 4.15 600 " 1.80 5 ".015 60 (.18 1700 " 2.10 6 4.018 70 "d.21 800 it 2.40 7.4.021 80 "4.24 900 id 2.70 I A.024 90 ".27 1000 " 3.00 9.027 100 ".30 2000 " 6.00 10 4.030 200 ".60 3000 " 9.00 2. What is E's tax, by the above table, whose property, real and personal, is valued at $ 1860, and who pays 3 polls? Ans. $ 10.08. OPERATION. Tax on $10 0 0 O is $ 3.0 0 We find the tax on $1000 c " " 800 "' 2.4 0 in the table, and then on'i't 60 ".18 $ 800, and then on $ 60, i "s 3 ~polls6 " 0 ~ and to these sums add the ic 3 poll~s 1 4.5 0 tax on the 3 polls for the Valuation, S 1 8 6 0 $1 0.08, Tax. answer. QUESTIONS. - Art. 232. How may the operation of assessing taxes be facilitated? How is the above table formed? 232 EQUATION OF PAYMENTS. S.ECT. XXXII. 3. What is F's tax, whose real estate is valued at 8 6535, and his personal property at $ 3175; and who pays for 6 polls? Ans. $ 38.13.'4. What is G's tax, who pays for 1 poll, and has property to the amount of $ 7480? Ans. 8 23.94. 5. If IH pays for 2 polls, and has property to the amount of $ 4790, what is his tax? Ans.,$ 17.37. 6. Al's real estate is valued at 8 9280, and his personal property at 8 3600, what is his tax, if he pays for 4 polls? Ans. 8 44.64. ~ XXXII. EQUATION OF PAYMENTS, ART. S SS EQUATION OF PAYMENTS is finding the averag-e or 7mean time when the payment of several sums of money, due at different times, may all be made at once, without gain or loss either to the debtor or creditor. ART;.14L, To find the average or mean time of payment, when the several sums have the same date. Ex. 1. John Jones owes Samuel Gray 8 100; $ 20 of which is to be paid in 2 months; $ 40 in 6 months; $ 30 in 8 months, and $ 10 in 12 months; what is the average time for the payment of the whole sum? Ans. 6 mo. 12 d.l OPERATION. $ 2 0 X 2 _ 4 0 It is evident that th te interest of s$40 X( 6~ = o40, 20 for 2 months is the same as $3 X 8 = 24 0 the interest of $ 1 for 40 months; and of $ 40 for 6 mo. the same as I 10 X 12 -— 12 0 of $ 1 for 2410 mo.; and of $ 30 for $ 100 10 0)64 0(6 mo. 8 mo the same as of $1 for 240 6 00 imo.; and of $ 10 for 12 mo. the same as of $1 for 120 mo. Hence 4 0 the interest of all the sums to the 3 0 times of their payment is the same as the interest of $ 1 for 40 -+- 2140 1 0 0 ) 1 2 0 0 ( 1 2 da - +240+ 12so = =640 mo. Now if 1 2 00 0$ 1 require 640 mo. to gain a certain sum, $ 20 -- $40 - $ 30 - $ 10 — 100 will require only T-6- of this time; and 640 mo. --- 100 QUESTION S. - Art. 233. What is equation of payments? - Art. 234. Why in the example, do we multiyily the $ 20 by 2?7 SECT. XXXII.] EQUATION OF PAYMENTS. 233 = 6 mo. 12 da., the average or mean time for the payment of the whole. Hence the following RULE. - Multiply each payment by the time before it is due, then divide the sum of the products by the sum of the payments, and the quotient will be the true time required. NOTE 1. —This is the rule usually adopted by merchants, but it is not perfectly correct; for if I owe a man $ 200, $ 100 of which 1 was to pay downL, and the other $ 100 in two years, the equated time for the payment of both slmns would be one year. It is evident, that, for deferring the payrnent of the first $100 for 1 year, I ought to pay the amount of, 100 for that time, which is $ 106; but for the other $ 100, which I pay a yeai bef ore it is due, I ought to pay the present worth of$ 100, which is $ 94.33%4, whereas, by equation of payments, I only pay $ 200. NOT E 2. -- When a payment is to be made dlown it has no product, but it must be added with the other payments in finding the average time. EXAMPLES FOR PRACTICE. 2. John Smith owes a merchant, in Boston, $ 1000, $ 250 of which is to be paid in 4 months, $ 350 in 8 months, and the remainder in 12 months; what is the average time for the payment of the whole sum? Ans. 8 mo. 18 da. 3. A gentleman purchased a house and lot for $ 1560, X of which it to be paid in 3 months, - in 6 months,, in 8 months, and the remainder in 10 months; what is the average time of payment? Ans. 7 9,: months. 4, Samuel Church sold a farm for $ 4000; 8 1000 of which is to be paid down, $ 1000 in one year, and the remainder in 2 years; but he afterwards agreed to take a note for the whole amount; for what time must the note be given? Ans. 15 months. 5. A wholesale merchant in Boston sold a bill of merchandise, to the amount of $ 5000, to a retail merchant of Exeter, N. FH.; he is to pay J of the money down, 1 of the remainder in 6 months, - of what then remains in 9 nmonths, and the rest at the end' of the year. If he wishes to pay the whole at once, what will be the average time of payment? Ans. 6 mo. 27 da. ART. 235. To find the average or mean time of payment, when the several sums have different dates. Ex. 1. Purchased of James Brown, at sundry times, and on QUESTIOMNS. - What is the rule for equation of paymlnents I Is the rule per-.bctty correct? Explain why it is not. When a payment is to he made down, vwhat is to be done with it? QA0 @ 234 EQUATION OF PAYMENTS. [SECT. XXXII. various terms of credit, as by the statement annexed. When is the medium time of payment? Jan. 1, a bill amounting to $360, on 3 months' credit. Jan. 15, do. do. 186, on 4 months' credit. March 1, do. do. 450, on 4 months' credit. May 15, do. do. 300, on 3 months' credit. June 20, do. do. 500, on 5 months' credit. Ans. July 24th, or in 115 da OPERATION. Due April 1, 360 AMay 15, $186X 44- 8184 July 1,$450X 91- 40950 Aug. 15, $300 X 136 — 40800 Nov. 20, $500 X 233 — 16500 1 796 )206434( 1 14aa days 1796 2684 1 796 8874 7184 1 690 We first find the time when each of the bills will become due. Then, since it will shorten the operation and bring the same result, we take the first time when any bill becomes due, instead of its date, for the period from which to compute the average time. Now, since April 1 is the period from which the average time is computed, no time will be reckoned on the first bill, but the time for the payment of the second bill extends 44 days beyond April 1, and we multiply it by 44, Art. 234. Proceeding in the same manner with the remaining bills, we find the average time of payment to be 114 days and a fraction, from April 1, or on the 24th of July. RULE.- Find the time when each of the sums becomes due, and multiply each sum by the number of daysJfrom the time of the earliest payment to the payment of each sum respectively. Then proceed as in the last rule, and the quotient will be the average time required, in days,Jfrom the earliest payment. NOTE. - Nearly the same result may be obtained by reckoning the time hi months. QUESTIONS. - Art. 235. What is the rule for finding the average time, when there are different dates? By what other method can you obtain nearly the same result? SECT. XXXII.] EQUATION OF PAYMENTS. 235 EXAMPLES FOR PRACTICE. 2. I have purchased several parcels of goods at sundry times, and on various terms of credit, as by the following statement. What is the average time for the payment of the whole? Jan, 1, 1848, a bill amounting to $ 175.80, on 4 months' cr. " 16, " do. do. 96.46, on 90 days' " Feb. 11, " do. do. 78.39, on 3 months'" 6" 23, " do. do. 49.63, on 60 days' " Mar. 19, " do. do. 114.92, on 6 months' " Ans. XMay 31st, or in 45 da. 3. Sold S. Dana several parcels of goods, at sundry times, and on various terms of credit, as by the following statement. Jan. 7, 1841, a bill amounting to $375.60, on 4 months' cr. April 1S, " do. do. 687.25, on 4 months'" June 7, "6 do. do. 56S.50, on 6 months''" Sept. 25, " do. do. 300.00, on 6 months' Nov. 5, " do do, do. 675.75, on 9 months' Dec. 1, " do. do. 100.00, on 3 months' " What is the average time for the payment of all the bills? Ans. Dec. 25, or in 232 da. 4. The following is my account against G. M. Holbrook, and I wish to ascertain the average time of payment. Jan. 1, 1847, 97 yards of broadcloth, at $ 4.50, on 3 mos.' cr. Feb. 10, " 7 bales of cotton cloth," 18.50, on60 days' " May 1, " 9 tons of iron, " 45.00, on 4 mos.''6 June 15, " 11 hhds. of molasses, " 12.00, on30days' " July 5, " 8 doz. shovels, " 9.00, on 2 mos.'" Sept. 25, " 14cwt. of sugar, " 6.50, on 1 mo.'s" Dec. 1, " 8 chests of tea,' 15.00, on 90 days'" Ans. July 17, or in 107 da. 5. The following is an account of my bills against J. Crowell: jan. 1, 1844, a bill amounting to $ 300, on 6 months' credit. June 1, " do. do. 500, on 5 months' " Sept. 1, " do. do. 200, on 6 months' " Feb. 1, 1845, do. do. 800, on 8 months''" July 1, 1846, do. do. 400, on 9 months' Dec. 1, " do. do. 900, on 7 months' " May 1, 1847, do. do. 100, on 3 months' " What is the average time of payment on the above bills? Ans. March 10, 1846, or in 20 mo. 9 da. '236 UATIO. [SECT. XXXIII. ~ XXXIII. RATIO. ART. 3e'6. RATIO is the relation, in respect to magnitude or value, which one quantity or number has to another of the same lknd, or the quotient arising from the division of one number by another. Thus, the ratio of 6 to 3 is 2. Of the two numbers necessary to form a ratio, the first is called the antecedent, the last the consequent. Thus, in the example given, 6 is the antecedent and 3 the consequent. When there is but one antecedent and one consequent, the ratio is called a simple ratio. The antecedent and consequent are also called the terms of the ratio. ART. 237. A ratio may be expressed in two ways. The ratio of 6 to 3 may be expressed by two dots between the terms, thus, 6: 3; or in the form of a fraction, by making the antecedent the numerator and the consequent the denominator, thus, 6. The terms of a ratio must be of the same kind, or such as may be reduced to the same denomination, in order that they may have a ratio to each other. Thus, shillings have a ratio to shillings, and shillings to pounds, &c.; but shillings have not a ratio to gallons, nor pounds to days, because they are not commensurable. ART. 23.s A ratio may be either direct or inverse. A direct ratio is when the antecedent is divided by the consequent; an inverse ratio is when the consequent is divided by the antecedent. Thus, the direct ratio of 6 to 3 is 6, and the inverse ratio of 6 to 3 is A, or -2. The direct ratio of one quantity or number to another zs Jbund by dividing the number whose ratio is required, which is the antecedent, by the number with which it is comlpared, which/ is the consequent. The inverse ratio is found by reversing this process. QUESTIONS. - Art. 236. What is ratio? How many numbers are necessary to form a ratio? What is the first called? Wllat the second? What is a simple ratio? What are the antecedent and consequent called? - Art. 237. What two ways are there of expressing a ratio? How must the terms of a ratio compare' Have pounds any ratio to days? Why? — Art. 238. What is a direct ratio? -What an inverse ratio? How is the direct ratio of one number to another found? How the inverse ratio? SECT. XxxxII.] RATIO. 237 EXAlIPLES FOPr PRACTICE. 1. What is the direct ratio of 9 to 3? Ans. 3. Of 18 to 6? Of 16 to4? Of 24 to 12? Of 20 to5? Of 15 to3? Of 100 to 25? Of 144 to 12? 2. What is the direct ratio of 7 to 21? Ans. 1. Of 4 to 28? Of 6 to 30? Of 9to 11 Of 9to 99? Of 0to90? Of 12 to 10S? 3. What is the direct ratio of 60 to 12? Of 132 to 11? Of 40 to 120? Of 32 to 96? Of 200 to 50? Of 144 to 17283 Of 360 to 60? 4. What is the inverse ratio of 10 to 5? Ans. 1 Of 27 to 81? Of 16 to48? Of 72 to9? Of I1 to SS.? OfT7to35? Of 150 to 75? 5. What is the direct ratio of 2~. 5s. to 9s.? Ans. 5. Of 9 in. to 1 ft. 6 in.? ART. 239 A compoun ratio consists of two or more simple ratios, whose corresponding terms are to be multiplied together. Thus, The simple ratio of S: 4 is 2;: And " " of 12:3'is 4 The compound ratio of 8 X 12: 4 X 3 is 2 X 4 Or " " of 96: 12 is 8 When a compound ratio is composed of two equal ratios, it -s called a duplicate ratio; when of three, it is called a triplicate ratio, &c. The simple ratio of 4: 2 is 2 "' " it" of 6: 3 is 2 " it "6 of 8: 4 is 2 The triplicate ratio of 4 X 6 X 8: 2 X 3 X 4 is X 2 X 2 X 2 Or " " of 192: 24 is 8 ART. 240@ If the terms of a ratio are both multiplied or di-vided by the same number the ratio is not altered. Thus, the ratio of 8: 2 is 4; the ratio 8 X 2: 2 X 2 is 4; and the ratio of 8+ 2. 2+2 is 4. C1UESTIONS. -Art. 239. What is a compound ratio? What a duplicate ratio? What a triplicate ratio? - Art. 240. What is the effect of multiplying or dividing the terms of a ratio? 238 PROPORTION. ISECT. XXXIV ~ XXXIV. PROPORTION. ART. SEEo PROPORTION is the equality of ratios. Thus the ratios 9 3 and 12 ~ 4 are equal, and when united form a proportion. Proportion is usually expressed by four dots between the two ratios; thus, the proportion in the preceding example is written 9 3: 3::12: 4, and is read, 9 is to 3 as 12 to 4. The numbers, which form a proportion, are called proportionals. The first and third are called antecedents, the second and fourth are called consequer~!s; also, the first and last are called extremes, and the remaining two the means. ART. 2S. Any four numbers are said to be proportional to each other when the first contains the second as many times as the third contains the fourth; or when the second contains the first as many times as the fourth contains the third. Thus, 9 has the same proportion or ratio to 3 that 12 has to 4, because 9= contains 3 as many times as 12 contains 4. iART.:4l3. If the antecedents or consequents of a proportion, or both, a:re divided by thle same number, they are still proportionals. Thus, dividing the antecedents of the proportion 4: 8: 10:20 by 2, we have 2:8:: 5: 20; dividing the consequents by 2, we have 4: 4: 10 10; and dividing both the consequents and antecedents by 2, we have 2: 4:: 5: 10; each of which is a proportion, since if we divide the second term of each by the first, and the fourth by the third, the two quotients will be equal. The effect is the same when the terms are multiplied by the same number. ART. 924: The product of the extremes of a proportion is equal to the product of the means. Thus, the proportion 14: 7:18: 9 may be expressed fractionally, = -. Now, if wve reduce these fractions to a common denominator we have 2 6 1- _ 2-6_; but in this operation we multiplied togetherthe QUESTIONS. - Art. 241. Whatis proportion? How is proportion expressed? What are the numbers called that form a proportion? Which are called the antecedents? Which the consequents? Which the xtremes? Whic:h the means' - Art. 242. When are numblers said to be in pt,portion to each other? - Art. 243. What is the effect of dividing the antecedents or conse(quents of a proportion? Of multiplying them?- Art. 244. How does the product of the extremes compare with that of the means? How is it shown that the product of the extremes is equal to that of the means? SECT. XXXIV. SIMPLE PROPORTION. 239 two extremes of the proportion, 14 and 9, and the two means, IS and 7, thus 14 X 9- i8 X7. ART. 24. If the extremes and one of the means are given the other mean may be found by dividing the product of the extremes by the given mean. Thus, if the extremes are 3 and 24,. and the given mean 6, the other mean is 12; because 24 X 3 -72; and 72 - 6 = — 12. AnT. 246. If the means and one of the extremes are gmven, the other extreme may be found by dividing the 9product of the means by the given extreme. Thus, if the means are 8 and 16, and the given extreme 4, the other extreme is 32; because 16 X 8 -128; and 128 - 4=32. SIMPLE PROPORTION. ART. 247. SIMPLE PROPORTION is an expression of the equality between two simple ratios. NOTE. - Simple Proportion is sometimes called the Rule of Three. ART. 24.. Method of stating and solving questions in Simple Proportion. Ex. 1. If 71b. of sugar cost 56 cents, what will 361b. Yost? Ans. $ 2.88. OPERATION. Since 71b. have the same ratio Extreme. Mean. Mean. to 361b. as 56 cents, the cost of 7 lbb. 3 6 lb. 5 6 cts. the former, have to the cost of the 3 6 latter, we have the first three terms of a proportion given, viz., one of 3 3 6 the extremes and the two means. 1 6 8 Now, to ascertain which of these terms are the means, and which the 7) 2 0.1 6 extreme, we arrange them in the $ 2.8 8 Extreme. order of a proportion, or state the question, by making 56 cents the third term, because it is of the same kind, and has the same proportion to the required answer or fourth term as the first has to the second. QUESTION-s. -Art. 245. If the extremes and one of the means are given, how can the other mean be fouud? - Art. 246. When the means and one of the extremes are given, how can the other extreme be found? —Art. 247. What is simple proportion? By what other name is it sometimes called'! - Art. 248. How many terms are given in questions in simple proportion? Wvhat are they? 240 SIMPLE PROPORTION. [SECT. XXXIT. And from the nature of the question, since the answer will be more than 56 cents, or the third term, the second term must be greater than the first; we therefore make 361b. the second term, and 7lb. the first, and then proceed as in Art. 246. BY ANALYSIS. - If 7lb. cost 56 cents, lib. will cost,- of 56 cents, which is 8 cents. Then, if 1lb. cost 8 cents, 361b. will cost 36 times as much; that is, 36 times 8 cents, which are $ 2.88, Ans. as before. Ex. 2. If 76 barrels of flour cost $ 456, what will 12 barrels cost? Ans. $ 72. OPERATION. bar. bar. $ We state this question by maling 16: 122 456 $456 the third term, because it is of 1 2 the same kind of the required answer. - Then, since the answer must be less 7 6 ) 5 4 7 2 ($ 7 2 than $ 456, because 12 barrels will cost 5 3 2 less than 76 barrels, we make 12 barrels, the smaller of the other two terms, 1 5 2 the second term, and 76 barrels the first 1 5 2 term, and proceed as before. BY ANALYSIS.- If 76 barrels cost $ 456, 1 barrel will cost y~ of $ 456, which is $ 6. Then, if 1 barrel cost $ 6, 12 barrels will cost 12 times as much, that is, $ 72, Ans. as before. Ex. 3. If 3 men can dig a well in 20 days, how long will it take 12 men? Ans. 5 days. OPERATION. imen. men. days. Since the required answer is days, 1 2 * 3:: 2 0 we make 20 days the third term. And 3 as 12 men will dig the well in less time than 3 men, the answer must be less 12)6 0 than 20 days. Therefore we make 3 men the second term and 12 men the first, 5 dcays. and proceed as in the other examples. BY ANALYSIS.- If 3 men dig the well in 20 days, it will take one man 3 times as long, that is, 60 days. Again, we say, If one man dig the well in 60 days, 12 men would dig it in -1- of 60 days, that is, 5 days, Ans. as before. From the preceding examples we deduce the following QUESTIONS. - What is meant by stating the question? Which of the terms given in the exalnple do you make the third? Why? Which the second? Why? Which the first? Why? After the question is stated, how do you obtain the arswer? SECT. XXXIV. SItAPLE PROPORTION. 241 RuLE. - 1. State the question by making that number, which is of the samie name or quality as the answer required, the third term; then, if the ansiwer required is to be greater than the third term, make the second term' greater than the first; but, if the answer is to be less than the third term, make the second less than. thefirst. 2. Reduce the first and second terms to the lowest denomination mentioned in either, and the third term to the lowest denomination mentioned in it. 3. Mzldtiply the second and third terms together, and divide their product by the first, and the quotient is the answver in the same denomination to which the third is reduced. 4. If anything remains after division, reduce it to the next lower denomination, and divide as before. 5. If any or all of the terms are fractions, state the question as in whote numbers, and then multiply and divide according to the rules for multiplication and division offractions. (Arts. 152, 158.) NOTE. - The pupil should perform these questions by analysis as well as by proportion, and introduce cancellation, when it will abbreviate the operatlon. Ex. 4. If 16 bushels of wheat are worth S24, what are 96 bushels worth? Ans. $ 144. OPERATION BY CANCELLATION, We first state the question as dibu. bu. $ rected in the rule, and then write the 1 6: 9 6' 24 second and third terms above a hori6 zontal line, with the sign of multiplicat 3X 2 4 tion between them, for a dividend, and 144 the first term below the line, for a divisor, and cancel the common factors.,Y ANALYSIS AND CANCELLATION. By this method of analysis we first 6 place the $24, which is of the same $24 X kind of the required answer, above 2 4 X 144 a line for a dividend; and then say, I[ ~ since $ 24 is the price of 16 bushels, 1 bushel will cost wa of $ 24, and express the division by placing the 16 below the line for a divisor. Now, since we have an expression for the price of 1 bushel, we next express the multiplication of it by 96 bushels, the price of which is required, and then cancel as before. EXAMIPLES FOX PRACTICE. 5. What cost 9 gallons of molasses, if 63 gallons cost $ 14.49? Ans. $ 2.O07. QuESTIONs. -What is the rule for sinple proportion? How should the pupil perform the questions? How do you state the question and ara-nge the terms for cancellation? What do you cancel' How do you arrange the yto.rums for cancellation by analysis? 21~ 242 SIMPLE PLROPORTION. i ECT. XXXIV. 6. What cost 97 acres of land, if 19 acres can be obtained for $ 33T.25? Ans. $ 1721.75. 7. If a man travel 319 miles in 11 days, how far will he travel in 47 days. Ans. 1363 miles. S. If 71b. of beef will buy 41b. of pork, how much beef will be sufficient to buy 481b. of pork? Ans. 841b. 9. Paid for 87 tons of iron $ 5437.50, how many tons will $ 7687.50 buy? Ans. 123 tons. 10. When $ 120 are paid for 15 barrels of mackerel, what will be the cost of 79 barrels? Ans. $ 632. 11. If 9 horses eat a load of hay in 12 days, how many horses would it require to eat the hay in 3 days?. Ans. 36 horses. 12. When $ 5.88 are paid for 7 gallons of oil, what cost 27 gallons? Ans. $ 22.68. 13. When $ 10.80 are paid for 91b. of tea, what cost 1471b? Ans. $ 176.40. 14. What cost 27 tons of coal, when 9 tons can be purchased for $ 85.95? Ans. $ 257.85. 15. If 15 tons of lead cost $ 105, what cost 765 tons? Ans. $ 5355.00. 16. If 16hhd. of molasses cost $ 320, what cost 176hhd.? Ans. $ 3520.00. 17. If 15cwt. 3qr. 171b. of sugar cost $ 124.67, what cost 76cwt. 2qr. 191b.? Ans. $ 601.09. 18. If 7s. 6d. of the old Pennsylvania currency are equal to $ 1, what is the value of ~76 19s. 11d.? Ans. $ 205.322-. 19. If 8s. of the old currency of New York are equal to 1, what is the value of ~ 19 19s. Sd.? Ans. $ 49.95 -. 20. If 4s. 8d. of the old currency of South Carolina and Georgia are equal to $ 1, what is the value of ~ 176 18s. 4d.? Ans. $ 758.21 +. 21. As 4s. 6d. sterling of the English currency are equal to one dollar in the United States, how many dollars are there in ~ 769 18s. 9d.? Ans. $ 3421.94 +. 22. If the cars on the Boston and Portland Railroad go one mile in 2 minutes and 8 seconds, how long will they be in passing firom Haverhill to Boston, the distance being 32 miles? Ans. lbh. Sin. 16sec. 23. If one acre of land cost $ 37.86, what cost 144A. 3R. 17p.? Ans. $ 5484.25 4-. SECT. XXXIV.1 SIMPLE PROPORTION. 243 24. If a man travels 3m. 7fur. 18rd. in one hour, how far will he travel in 9h. 45min. 19sec.? Ans. 3Sm. 2fiur. 32rd+. 25. A fox is 96 rods before a greyhound, and while the fox is running 15 rods the greyhound will run 21 rods; how far will the dog run before he can catch the fox? Ans. 336 rods. 26. If 5 men can reap a field in 12 hours, how long would is take them if 4 men were added to their number? Ans. 62 hours. 27. Ten men engage to build a house in 63 days, but 3 of their number being taken sick, how long will it take the rest to complete the house? Ans. 90 days. 28. If a 4 cent loaf weighs 5oz., when flour is $ 5 per barrel, what should it weigh when flour is $ 7.50 per barrel? Ans. 31oz. 29. If 7 men can mow a field in ten days, when the days are 14 hours long, how long would it take the same men to mow the field, when the days are 13 hours long? Ans. 10 ~ days. 30. If 291b. of butter will purchase 401b. of cheese, how many pounds of butter will buy 791b. of cheese? Ans. 57111b. 31. If 4 of a yard cost ~ of a dollar, what will 1- of a yard cost? Ans. $ 0.76 7-8. yd. yd. $. 3. T2..aj 4 21 6; V1_ X eXg 2o-~= $ 0.76 7, Ans. 32. If 7T of a gallon of oil cost -9% of a dollar, what cost t of a gallon? Ans. 8 1.12A. gal. gal. I. X t 9 T7 iT: j 9 X X =;1.12, Ans. 33. If 4~ yards of cloth cost $ 2, what will 19l yards cost? Ans. $ 11.50. yd. yd.' 23 47:9 2; 191 2- X- - $ 11.50, Ans. 34. If for 4rT- yards of velvet, there be received 114 yards of calico, how many yards of velvet will be sufficient to pur chase 100 yards of calico? Ans. 39 4 9 yards. 35. If 14- ells English of broadcloth will pay for 5 —ficwt. of sugar, how many yards will 25j7Tcwt. buy? Ans. 85yd. 3qr. 3:- na. 36. A certain piece of labor was to have been performed by 244 SIIMPl, PROPORTION. SF CT. XXXIV 144 men in 36 days, but a number of them having been sent away, the work was performed in 48 days; required the num ber of men discharged. Ans. 36 men. 37. James can mow a certain field in 6 days, John can mow it in 8 days; how long will it take John and James both to mow it? Ans. 3~ days. 38. Samuel can reap a field of barley in 9 hours; but with the assistance of Alfred he can reap it in 4 hours; how long would it take Alfred to reap it alone? Ans. ~71 hours. 39. A. Atwood can hoe a certain field in 10 days, but with the assistance of his son Jerry, he can hoe it in 7 days; and he and his son Jacob can hoe it in 6 days; how long would it take Jerry and Jacob to hoe it together? Ans. 9-3- days. 40. Bought a horse for $ 75; for what must I sell him to gain 10 per cent.? $ 100: 8 110:: 75' $ 82.50, Ans. 41. Bought 40 yards of cloth at $ 5.00 per yard; for what must I sell the whole amount to gain 15 per cent.? Ans. $ 230.00. 42. Mly chaise cost $ 175.00, but, having been injured, I am willing to sell it at a loss of 30 per cent.; what should I receive? Ans. $ 122.50. 43. Bought a cargo of flour on speculation at $ 5.00 per barrel, and sold it at $ 6.00 per barrel; what did I gain per cent.? Ans. 20 per cent. 44. Bought a hogshead of molasses for $ 15.00, but, it not proving so good as I expected, I sell it for $ 12; what do I lose per cent. Ans. 20 per cent. 45. Sold a pair of oxen for 20 per cent. less than their value, whereas, I might have sold them so as to have gained 20 per cent., and, by so doing, I have lost $ 60.00; what was the price for which they were sold? Ans. $ 120.00. 46. Bought a hogshead of molasses for 8 27.50, at 25 cents per gallon; how much did it contain? Ans. 110 gallons. 47. A certain farm was sold for $ 1728, it being 8 15.75 per acre; what was the quantity of land? Ans. 109A. 2R. 342p. 48. A certain cistern has 3 cocks the first will empty it in 2 hours, the second in 3 hours, and the third in 4 hours; in what time would they all empty the cistern together? Ans. 55T5 minutes. SECT. XXXiv.] COMPOUND PROPORTION. 245 COMPOUND PROPORTION. ART. tB,9. COMPOUND PROPORTION is an expression of the equality between a coznptond and simple ratio. It is employed in performing such questions as require two or more operations in Simple Proportion. ART. 250. Method of stating and solving questions in Compound Proportion, sometimes called the Double Rule of' Three. Ex. 1. If 8 100 will gain $ 8 in 12 months, what will $ 600 gain in 10 months? Ans. $ 40. OPERATION. Extreme. Mean. ean. In stating this $ 1 0 0 $ 6 0 0 -ca n.:question, we make 1 2 mo.: 1 0 mo. $8, the gain, which is the same name 1 ~0 0~XI0X8 X 1 420 1 0 0 = $ 4 0, Extreme. of the required an100X12 - 1200 swer, the third term. Then, taking $ 100 and $ 600, two of the remaining terms of the same kind, we inquire if the answer, depending on these alone, must be greater or less than the third term; and since it must be greater, because $ 600 will gain more than $ 100 in the same time, we make $ 600 the second term, and $ 100 the first. Again, we take the two remaining terms, and make 10 mo. the second term and 12 mo. the first, since the same sum would gain less in 10 mo. than in 12 mo. WVe then find the continued products of the second and third terms, and divide it by the continued product of the first terms, for the answer. Hence the following RULE. - 1. Make that number, which is of the same kind as the answer required, the third term; and, of the remaining numbers, take any two, that are of the same kind, and consider whether an answer, depending upon these alone, would be greater or less than the third term, and place them -is directed in Simple Proportion. 2. Then take any other two, and consider whether an answer, deicnding, only upon them, would be greater or less than the third term, wnd arrange them accordingly; and so on until all are used. QUESTIONs. - Art. 249. What is compound proportion? For what is it employed? - Art. 250. By what other name is it sometimes called? In stating the question, which of the numbers do you make the third term? Wrhy? W;hat do you do with the remlaining terms? I-How do you know which of the two to take for the second term? Which for the first? After all the terms have been arranged, how do you find the answer? What is the rule for compound proportim? * ~~~21 246 COMPOUND PDROPORTIO~N. [SECT. XXXiv 3. Multiply the continued product of the second terms by tile third, and divide by the continued product of the first, and the quotient is the answer. NOTE. - The following questions should be performed not only by the roue, but by analysis and cancellation. Ex. 2. If $ 100 will gain $ 6 in 12 months, what will S80O gain in 8 months? OPERATION SBY CANCELLATION. 1 0 0 sos 0 6 12 mo.: 8 mo. 6 We state the question according to the rule, and then write the second and third terms for a dividend and the first 0 0 X 0 XX_ S 3 2 terms for a divisor, and cancel the comn 00 X0 X, mon factors. BY ANALYSIS ANLD CANCELLATION. 4 8 By this method of analysis $ 0 X 0 mo.y $ 0 0 we say, if $ 6 are the gain of Mo. x 00 3 2$32 $100 in 12 mo.,in I mo. the gain of $ 100 will be 4- as much, or TI2, and in 8 mo. 8 times as much, or $ -60t-. Again, if $ 100 gain $ 6Ixs- in 8 mo., 1 will gain i8 $ 1 will gain T- of it, or $ 12 X 100' and $ 800 will gain 800 times as much, or $ 6 X 8 X 800 as much, or 100, the same as in the operation. Cancelling the common factors, we obtain $ 32 for the answer. EXAIPLES FOR PRACTICE. 3. If $ 100 gain $6 in 12 months, in how many months will $ 800 gain $ 32? Ans. 8 months. 4. If $ 100 gain $6 in 12 months, how large a sum will it require to gain $ 32 in 8 months? Ans. $ 800. 5. If $ 800 gain $ 32 in 8 months, what is the per cent.? Ans. 6 per cent. 6. If 15 carpenters can build a bridge in 60 days, when the days are 15 hours long, how long will it take 20 men to build the bridge when the days are 10 hours long? Ans. 6r7 days. QuEsTtIons. -How does the author say the questions under this rule should be performed? How are questions stated for cancellation? Whichllterms are taken for the dividend'? Which for the dlivisor? What are cancelled? SECT. XXXXIl COMPOUND PROPORTION'. 47/ 7. If a regiment of soldiers, consisting of 939 men, can eat 351 bushels of wheat in 3 weeks, how many soldiers will it require to eat 1404 bushels in 2 weeks? Ans. 5634 soldiers. 8. If 8 men spend $64 in 13 weeks, what will 12 men spend in 52 weeks? Ans. $ 384. 9. If S horses consume 42 bushels of grain in 24 days, how many bushels will suffice 32 horses 48 days? Ans. 336 bushels. 10. If 6 men in 16 days of 9 hours each build a wall 20 feet long, 6 feet high, and 4 feet thick, in how many days of 16 hours each will 24 men build a wall 200 feet long, 16 feet high, and 6 feet thick? Ans. 90 days. 11. If a man travel 117 miles in 15 days, employing only 9 hours a day, how far would he go in 20 days, travelling 12 hours a day? Ans. 208 miles. 12. If 12 men in 15 days can build a wall 30 feet long, 6 feet high, and 3 feet thick, when the days are 12 hours long, in what time will 30 men build a wall 300 feet long, 8 feet high, and 6 feet thick, when they work 8 hours a day? Ans. 240 days.'13. If the carriage of 5cwt. 3qr., 150 miles cost $24.58, what must be paid for the carriage of 7cwt. 2qr. 251b. 32 miles at the same rate? Ans. $ 7.04 —. 14. A received of B $ 9 for the use of $ 600 for 6 months; now B wishes to hire of A $ 1800 until the interest shall amount to the same sum. How long may he keep it? Ans. 2 months. 15. If 15 oxen or 20 cows will eat 3 tons of hay ins8 weeks, how much hay will be sufficient for 15 oxen and 8 cows 12 weeks? Ans. 6~3~ tons. 16. If 5 men, by laboring 10 hours a day, can mow a field of 30 acres in 10 days, how long will it require 8 men and 7 boys, provided each boy can do jT as much as a man, to mow a field containing 54 acres? Ans. 7- T1 days. 17. If 2 men can build 124 rods of wall in 6- days, ho-w long will it take 18 men to build 247-~2 rods? Ans. 14 days. 18. If 248 men, in 5-1 days of 11 hours each, dig a trench of 7 degrees of hardness, and 232- feet long, 32 feet wide, and 21 feet deep, in how many days of 9 hours each will 24 men dig a trench of 4 degrees of hardness, and 337A feet long, 5feet wide, and 3- feet deep? Ans. 132 days. 248 PARTNERSHIP. [SECT. XXXV. ~ XXXV. PARTNERSHIP, OR COMPANY BUSINESS. ART. 5]1. PARTNERSHIP is the association of two or more persons in business, with an agreement to share the profits and losses in proportion to the amount of capital stock, or the value of the labor and experience of each. The association is called a Firm or Company, the money oi property invested is called the Joint Stock or Capital, each of the owners is called a _Partner, and the profit or gain the Diridend. Partnership is of two kinds. First, when the stock is employed for the same time. Secondly, when the stock is employed for unzeqal times. The former is sometimes called Single Fellowship, and the latter Double Fellowship. ART. 252o To find each partner's share of the profit or loss when the stock is employed for the same time. Ex. 1. John Smith and Henry Gray enter into partnership for three years; Smith puts in $4000, and Grav $2000. They gain $ 570. What is each man's share of the gain? Ans. Smith's gain, 5 380; Gray's gain, $ 190. OPERATION. $ 0 0 0 00, Smith's stock, o0o0 - Smith's part of the stock. 1$20007, Gray's " = 2, -1 Gray's part of the stock. $ 6 0 O 0, Whole stock. Then 2 of $ 5 7 0, the whole gain, $ 3 8 0, is Smith's share of the gain. And 3 of $ 5 7 0, " " $19 0, is Gray's share of - the gain. Proof, $570 Since the sum of $ 4000 and $ 2000, equal to $ 6000, is the whole stock, it is evident that Smith's part of the stock is 4 - 23; aond that Gray's part is o = -. Then, since each man's gain must be m proportion to his stock, - of $ 570, = $ 380, is Smith's share of the gain; and i of $ 570, = $ 190, is Gray's share of the gain. OuTESTIONS. -Art. 251. What is partnership? What is the association called'? What the property inrested? What are the owners called? What tie profit or loss? What two kinds of partnership are there? What is the,iis-tlrwtlon btelweel theni?7 What are they sometimnes called 7 SECT. XxxV.l PA aTNERSHIP. 249 RULE. - Find each partner's fractional part of the whole stock, by making each one's stock the numerator of afraction, and the whole stock the denominator. Then multiply the whole gain or loss by each man's fractional part of the stock, and the product will be the gain or loss of each. EXAMPLES FOR PRACTICE. 2. Three merchants, A, B, and C, engaged in trade. A put in $ 6000, B put in $ 9000, and C put in $ 5000. They gain 8 840. What is each man's share of the gain'? Ans. A's gain $ 252, B's gain $ 378, C's gain $ 210. 3. A bankrupt owes Peter Parker 8750, James Dole $ 3610, and James Gage $ 7000. His effects, sold at auction, amount to $ 6875; of this sum $ 375 are to be deducted foi expenses, &c. What will each receive of the dividend? Ans. Parker, $ 2937.75a1; Dole, $ 1212.03-%6; Gage, 8 2350.202~ox. 4. A merchant, failing in trade, owes A $ 500, B $ 386, C 988, and D $ 126. His effects are sold for $ 100. What will each man receive? Ans. A receives $ 25.00, B $ 19.30, C $ 49.40, D $ 6.30. 5. A, B and C engaged in trade. A put in $ 700, B put im 300, and C put in 100 barrels of flour. They gained $ 90; of which sum C took $ 30 for his part; what will A and B receive, and what was C's flour valued per barrel. Ans. A receives $ 42, B $ 18, C's flour $ 5 per barrel. ART. SLO. To find each partner's share of the profit or loss, when the stock is employed for unequal times. Ex. 1. Josiah Brown and George Dole trade in company. Brown put in $ 600 for 8 months, and Dole put in $ 400 for 6 months. They gain $ 60. What is each man's share of the gain? OPERATION. $ 6 0 0 X 8 _ $ 48 0 0, Brown's money for 1 month. 4-o O 0- 2, Brown's part of stock. 8 4 0 0 X 6 = $ 2 4 0 0, Dole's money for 1 month. o= o a, Dole's part of stock. $7 2 0 0,'Whole stock for 1 month. Then - of $60, the whole gain, = $ 40, is Brown's share of gain. And ~ of $60,"'" " -$ 8 20, is Dole's QUESTION. - Art. 252. What is the rule for finding the shares of profit or loss when the stock is employed for the same time? 250 PARTNERSHIP. [SeCT. XXXI It is evident that $ 600 for 8 months is the same as S 600 X 8 $4800 for 1 month, because $4800 would gain as much in 1 mlolth as 600) in 8 months. And for the same reason $ 400 for 6 months is the same as $ 400 X 6 - $'2400 for 1 month. The question is, therefore, the same, as if Brown had put in $ 4800 and Dole $ 2400 for 1 month each. The whole stock would then be $ 4800 +-$ 2400 _ $ 7200, and Brown's share of the gain would be $so = 2 of $ 60 840. Dole's share will be 24-0 = of $60 20. Hence the propriety of the following RULE. - Multiply each man's stock by the time it continued in trade, and consider each product a numerator, to be written over their sum, as a common denominator; then multiply the whole gain or loss by each fraction, and the several products will be the gain or loss of each man. EXA tMPLE S FOR PRACTICE, 2. A, B and C trade in company. A put in $ 700 for 5 months; B put in $ 800 for 6 months; and C put in $ 500 for 10 months. They gain $ 399. What is each man's share of the gain z Ans. A's gain $ 105, B's gain $144, C's gain 8 150. 3. Leverett Johnson, William Hyde, and William Tyler, formed a connection in business under the firm of Johnson, Hyde & Co.; Johnson at first put in $ 1000, and, at the end of 6 months, he put in $ 500 more. Hyde at first put in $ 800, and, at the end of 4 months, he put in $ 400 more, but, at the end of 10 months, he withdrew $ 500 from the firm. Tyler at first put in $ 1200, and, at the end of 7 months, he put in $300 more, and, at the end of 10 months, he put in $200. At the end of the year they found their net gain to be $ 1000. What is each man's share? Ans. Johnson's gain S 348.0 3 Hyde's $ 273.789r1, Ty ier's $ 378.1941T. 4. George Morse hired of William Hale, of Haverhill, his best horse and chaise for a ride to Newburyport, for $ 3.00, with the privilege of one person's having a seat with him. Having rode 4 miles, he took in John Jones, and carried him to Newburyport, and brought him back to the place from which he took him. What share of the expense should each pay, the distance from Haverhill to Nexvburyport being 15 miles? Ans. Morse pays $ 1.90, Jones pays $ 1.10. 5. J. Jones and L. Cotton enter into partnership for one. QUEsTIONs.- Art. 253. WVhat is the rule for finding the shares of profit or loss, when the stock is employed for unequal times? Wiry do you multiply each man's stock by the time it was in trade? SECT. xxxv.1 PARTNERSHIP. 251 year. January 1, Jones put in $1000, but Cotton did not put in any until the 1st of April. What did he then put in to have an equal share with Jones at the end of the year? Ans. $ 1333.333. 6. S, C and D engage in partnership, with a capital of $ 4700. S's stock was in trade 8 months, and his share of the profits was $ 96; C's stock was in the firm 6 months, and his share of the gain was $ 90; D's stock was in the firm 4 months, and his gain was $ 80. Required the amount of stocli which each had in the firm. ( S's stock $ 1200. Ans. C's stock $ 1500. D's stock 8 2000. 7. P and H engage in trade, and it was mutually agreed that each should receive of the profits in proportion to his stock, and the time it was continued in trade. P put in $ 4000 for 5 months, and H put in $ 6000 for 8 months, and they gained $ 680; what was each man's share of the gain? Ans. P's share $ 200, H's share $ 480. 8. A, B and C engage in trade. A put in $300 for 7 months, B put in $ 500 for 8 months, and C put in $ 200 for 12 months; they gain $ 85; what share of the gain does each receive? Ans. A 8 21, B $ 40, and C $ 24. 9. A and B engage in trade, with $ 500. A put in his stock for 5 months, and B put in his for 4 months. A gained $ 10, and B gained $ 12; what sum did each put in? Ans. A $ 200, B $300. 10. A and B trade in company; A put in $ 3000, and at the end of 6 months put in $ 2000 more; B put in $ 6000, and at the end of 8 months took out 8.3000; they trade one year, and gain $ 1080; what is each man's share of the gain e Ans. A's share is $ 480, B's $ 600. 11. Four men hired a pasture for $ 50. A put in 5 horses for 4 weeks; B put in 6 horses for 8 weeks; C put in 12 oxen for 5 weeks, calling 3 oxen equal to 2 horses; and D put in 3 horses for 14 weeks. How much ought each man to pay? Ans. A $6.662, B $16.00, C $13.331, and D 8 14.00. 12. A, B and C contract to build a piece of rail-road for S 7500 dollars. A employs 30 men 50 days; B employs 50 men 36 days; and C employs 48 men and 10 horses 45 days, each horse to be reckoned equal to one man and he is also to have 8 112.50 for overseeing the work. How much is each man to receive? Ans. A receives 81875: P $ 2250; C 8 337.5. 252 PROFIT AND LOSS. [srcT. XXXVr. ~ XXXVI. PROFIT AND LOSS. ART. 254. PROFIT and Loss is a rule by which merchants and other traders estimate their gain or loss in buying and sell ing goods. The following questions may be performed either by analysis or by proportion. ART. 255. To find the profit or loss per cent. in buying and selling goods, having the cost and selling prices given. Ex. 1. If I buy flour at $ 4 per barrel, and sell it at $5 per barrel, what is the gain per cent.? Ans. 25 per cent. OPERILATION. $5-$4= - 1; == 1.00 - 4.25, or 25 per cent. By subtracting the cost from the selling price, we find the gain per barrel to be $ 1, or l of the cost, which fraction being reduced to a decimal, (Art. 187,) we obtain.25, or 25 per cent., for the gain. OPERATION BY PROPORTION. 5 —$ $ 1; $ 4 $ 1 00 $ 1 2 5, that is, 25 per ct. 2. If I buy flour at $ 5 per barrel, and sell it at $4 per barrel, what is the loss per cent.? Ans. 20 per cent. OPERATION, $ 5-$4 =; - 5- 1.00 5 =.20, or 20 per cent. By subtracting the selling price from the cost, we find the loss per barrel to be $ 1, or 1 of the cost, which fraction being reduced to a decimal, (Art. 187,) we obtain.20, or 20 per cent. for the loss. OPERATION BY PROPORTION. $5;- $ a4 —= $ 1; 5: $ 1 0 0 1:: $: X 20, that is, 20 per cent. RULE I. - First find what fractional part the gain or loss is of thel cost, by making the gain or loss the numerator of the fraction, and thi; cost the denominator; and then reduce this fraction to a decimal for t/zzy answer. Or, RULE II. -As the cost of the goods is to 5 100, so is tlhe gai.n or loss to the gain or loss per cent. NOTE. - Since per cent. is a certain number of hundredths, the figures cenoting it can properly occupy only two places; hence those at the right Qu STIONS. -Art. 254. What is Profit and Loss? Art. 2.55. What is the first rule for finding the profit or loss in buying or selling goods? What is tlhe second rl, T1., SECT. xxxv1.] PROFIT AND LOSS.'253 of hundredths are a fractional part of 1 per cent., and may be expressed either as a decimal or vulgar fraction. EXAMPLES FOR. PlRACTICE. 3. Bought 40 yards of broadcloth, at $ 5.40 per yard, and I sell I of it at $ 6 per yard, and the remainder at $ 7 per yard; what do I gain per cent.? Ans. 152 per cent. 4. A merchant purchas6d for cash 50 barrels of flour, at $ 5 per barrel, and immediately sold the same on 8 months' credit, at $ 5.98 per barrel; what does he gain per cent.? Ans. 15 per cent. 5. A grocer bought a hogshead of molasses containing 100 gallons, at 30 cents per gallon; but 30 gallons having leaked out, he disposed of the remainder at 40 cents per gallon. Did he gain or lose, and how much per cent.? Ans. Lost 62 per cent. 6. A gentleman in Rochester, N. Y., purchased 3000 bushels of wheat, at $ 1.121 per bushel. He paid 5 cents per bushel for its transportation to N. Y. city, and then sold it at $ 1.37k per bushel; what did he gain per cent.? Ans. 17w- per cent. 7. J. Morse bought, in Lawrence, a lot of land 7-Tp' rods square, for $5 per square rod. He sold the land at 5 cents per square foot; what did he gain per cent.? Ans. 1721 per cent. ART. 256. To find at what price goods must be sold to,gain or lose a given per cent. Ex. 1. If I buy flour at $ 4 per barrel, for how much must I sell it per barrel to gain 25 per cent.? Ans. $ 5. OPERATION. $4 X.25 $1.00; then, $4 + $1 = $5, Ans. It is evident, if I sell the flour for 25 per cent. gain, I sell it for.25 more than it cost. Therefore, if I add.25 of the cost to itself, it will give the price per barrel for which the flour must be sold; as seen in the operation. OPERATION BY PROPORTION. $100 + $25 = $125; $100: $125:: 4: $5, Ans. 2. If I buy flour at $ 5 per barrel, for what must I sell it per barrel to lose 20 per cent.? QUESTIONS. -HOW many places can the figures denoting per cent. occupy? What is the next place below hundredths? How are the figures below hun h,-rodths reg-rdel? How may they be expressed 7 92 254 PROFIT AND LOSS. [SECT. XXXVY. OPERATION. $5 X.20 $1.00; $5 —$1=- 8 4, Ans. It is evident if I sell the flour for 20 per cent. loss, I sell it for.20 less than it cost. Therefore, if I subtract.20 of the cost from itself, it will give the price per barrel for which the flour must be sold; as seen in the operation. OPERATION BY PROPORTION. $100- $20 - $80; $100 $80 $5: $5 4, Ans. RULE I. — Find the percentage on the cost of the goods at the given rate per cent., and add it to the cost, or subtract it from it, according as the goods are sold at a profit or loss. Or, RULE II. - As $ 100 are to $ 100 with the profit added or loss subtracted, so is the given price to the price required. EXAMPLES FOR PRACTICE. 3. Bought a hogshead of molasses, containing 120 gallons, for 30 cents per gallon, but it not proving so good as was expected, I am willing to lose 10 per cent. on the cost; what shall I receive for it? Ans. $ 32.40. 4. A grocer bought a hogshead of sugar, weighing net 7cwt. 3qr. 121b., for $88; for what must he sell it per pound to gain 20 per cent.? I Ans. 12 cents per pound. 5. J. Simpson bought a farm for $ 1728; for what must it be sold to gain 12 per cent., provided lie is to wait 8 months, without interest, for his pay? Ans. $ 2012.77+. 6. J. Fox purchased a barrel of vinegar, containing 32 gallons, for $ 4; but 8 gallons having leaked out, for how much must he sell the remainder per gallon to gain 10 per cent. on the cost? Ans. $ 0.181 per gallon. 7. Bought a horse for $ 90, and gave my note to be paid in 6 months without interest; what must be my cash price to gain 20 per cent. on my bargain? Ans. $ 104.84+. 8. H. Tilton bought 7cwt. of coffee, at $ 11.50 per cwt., but finding it injured, he is willing to lose 15. per cent.; for how much must he sell the 7cwt.? Ans. $ 68.42+. ART. 257. To find the cost when the selling price and the gain or loss per cent. are given. Ex. 1. If I sell flour at $ 5 per barrel, and by so doing malie 25 per cent., what was the cost of the flour? Ans. $4 per barrel. QUESTION. — Art. 256. What is the first rule for finding at what price goods must he sold to gain or lose a given per cent.? What is the second rule? .SECT. xxxvx. ] PROFIT AND LOSS. 255 OPERATION. Let 11 ( ) represent the cost of the flour per barrel; then slace$ 5 is 25 per cent., or 2 more than the cost, it is equal to {- + P f - of the actual cost. Again, if 5 is l of the cost, a:o will be $5 125 -- $.04; and tok or the cost of the flour per barrel, will be $.04 X 100 $ 4.00. OPERATION Bi' PROPORTION. $100 + $25 $ 125; $1i25: $100: $5 $4, Ans. 2. If I sell flour at $ 4 per barrel, and by so doing lose 20 per cent., what was the cost of the flour? Ans. $ 5 per barrel. OP:3RAT11OR. Let }Ad (= 1) re-present the cost of the flour per barrel; then, since $ 4 is 20 per cent., or -2 less than the cost, it is equal to 1 d - 2 d 8 0 of the actual cost. Again, if $ 4 is -20' of the cost, i &r will be $ 4 -, 80 =.05, and t8-, or the cost per barrel, will be $.05 X 100 - $ 5.00. OPERATlON BY PROPORTION. $ 100- 2 0 u 8 0; S0 o'$100: o " $4'5, Ans. RULE I. - Find what fr<actional part the selling price is of the cost, by making 100, with the gain per cent. added, or the loss per cent. subIracted, the numerator of a fraction, and 100 the denominator; then divide the selling price by this fraction, and the quotient will be the cost. ar, RuLE.-. - As $ 100 with the gain per cent. added or loss per cent. ntbtracted is to $ 100, so is the selling price to the cost. Exi~PILE s FoR PeRAeCTmCE..3. Having used my chaise 16 years, I am willing to sell it orl 80 SO; but by so doing I lose 62- per cent.; what was the cost of the chaise? Ans. $ 213.33k. 4. If I sell wood at $ 7.20 per cord, and gain 20 per cent., whlat did the wood cost me per cord? Ans. $ 6 per cord. 5. J. Adams sold 40 cases of shoes for $ 1600, and gained iS per cent.; what was the first cost of the shoes? Ans. $ 1355.93 —. -tQusTIoN. - Art. 257. What is the first rule for finding the cost, when the selling price and the gain or loss per cent. are given? What is the second rule? 256 PROFIT AND LOSS. [SECT. XXXVI. 6. Sold 17 barrels of flour at $8 per barrel, for which I received a note payable in 3 months. This note I had discounted at the Granite Bank, but on examining my account, I find I have lost 10-per cent. on the flour; what was the cost of it? Ans. $ 148.76+. ART. 258. The selling price of goods and the rate per cent. being given, to find what the gain or loss per cent. would be, if sold at another price. Ex. 1. If I sell flour at $ 5 per barrel, and gain 25 per cent., what should I gain, if I were to sell it for $ 7 per barrel? OPERATION. The solution of this question involves two principles: First, to find the cost of the flour per barrel, (Art. 257.) Thus, $5. 125 = $.04; $.04 X 100 - $ 4.00, the cost per barrel. Second, to find the gain per cent. on the cost when sold at $ 7 per barrel, (Art. 255.) Thus, $ 7 - $4 $3; 3 _- 3.00 4 4=.75, or 75 per cent. OPERATION BY PROPORTIOIM $ 100+$25=$ 125; $5: $7S: 125 - $ 175, $ 1 75- $ 1 0 0 $= 75, that is, 75 per cent. RULE I. -Find the cost of the goods, (Art. 257,) and then the gain of loss per cent. on this cost at the last selling price. (Art. 255.) Or, RULE 11. -As the first price is to the proposed price, so is $ 100 with the profit per cent. added, or the loss per cent. subtracted, to the gain or loss per cent. at the proposed price. NOTE. - If the answer exceeds $ 100, the excess is the gain per cent. but, if it is less than $ 100, the deficiency is the loss per cent EXAMPLES FOR PRACTICE. 2. Sold a quantity of oats at 28 cents per bushel, and gained 12 per cent.; what per cent. should I gain or lose, if I were to sell them at 24 cents per bushel? Ans. Lose 4 per cent. 3. S. Rice sold a horse for $ 37.50 and lost 25 per cent.; what would have been his gain per cent. if he had sold him for $ 75? Ans. 50 per cent. 4. S. Phelps sold a quantity of wheat for $ 1728, and took QUESTIONS. - Art. 258. What is the first rule for finding what gain or loss is made by selling goods at another price when the selling price and rate per cent. are given? 1What is the second rule? If the answerexceeds 0 100, what is the excess? If it is less than $ 100, what is the deficiency? B;ac'r. xxizl.1J PROFIWT AND LOSS. 2t7 a note payable in 9 months without interest, and made 10 per cent. on his purchase; what would have been his gain per cent. if he had sold it to James Wilson for $ 2000 cash? Ans. 33+ per cent. MISCELLANEOUS EXERCISES IN PROFIT AND LOSS. 1. A horse that cost $ 84, having been injured, was sold for $ 75.60; what was the loss per cent.? Ans. 10 per cent. 2. Sold a horse for $ 75.60, and lost 10 per cent. on the cost. but I ought to have sold him for $ 97.44 to have made a reasonable profit; what per cent. did I lose on the price for which I ought to have sold the horse? Ans. 16 per cent. 3. M. Star sold a horse for $ 97.44, and gained 16 per cent.; whlat would have been his loss per cent. if he had sold the horse for $ 75.60, and what his actual loss? Ans. Loss 10 per cent. $ 8.40 loss. 4. If 1 buy cloth at'$ 5 per yard, on 9 months credit, for what must I sell it per yard for cash to gain 12 per cent.? Ans. $ 5.35+. 5. A. Pemberton bought a hogshead of molasses, containing 120 gallons, for $ 40; but 20 gallons having leaked out, for what must he sell the remainder per gallon to gain 10 per cent. on his purchase? Ans. $ 0.44. 6. H. Jones sells flour, which cost him $ 5 per barrel, fbr $ 7.50 per barrel; and J. B. Crosby sells coffee for 14 cents per pound, which cost him 10 cents per pound; which makes the greater per cent.? Ans. A. Jones makes 10 per cent. most. 7. J. Gordon bought 160 gallons of molasses, but having sold 40 gallons, at 30 cents per gallon, to a man who proved a bankrupt, and could pay only 30 cents on the dollar, he disposed of the remainder at 35 cents per gallon and gained 10 per cent. on his purchase; what was the cost of the molasses? Ans. $ 41.45+. 8. D. Bugbee bought a horse for 6 75.60, which was 10 per cent. less than his real value, and sold him for 16 per cent. more than his real value; what did he receive for the horse, and what per cent. did he make on his purchase? Ans. Received $ 97.44, and made 28E per cent. 9. A merchant bought 70 yards of broadcloth, that was 1~ yards wide, for $ 4.50 per yard, but the cloth having been wet, it shrunk 5 per cent. in length and 5 in width; for what must the cloth be sold per square yard to gain 12 per cent.? Ans. $ 3.19+. $"n L5 258 DUODECIMALS. [sECT. XXXVII. ~ XXXVII. DUODECIMALS. ART. 259. Duodecimals are a kind of mixed numbers in which the unit, or foot, is divided into; 12 equal parts, and each of these parts into 12 other equal parts, and so on indefinitely; thus, T'y, TT, &c. Duodecimals decrease from left to right in a twelvefold ratio, and the different orders, or denominations, are distinguished from each other by accents, called indices, placed at the right of the numerators. Hence the denominators are not expressed. Thus, 1 inch or prime, equal to -fa of a foot, is written 1 in. or 1 I second ":-i " " 1". 1 third " I- " " 1'". 1 fourth " xfx~ " " 1""'. Hence the following TABLE. 12 fourths make 1'". 12 thirds " 1". 12 seconds " 1'. 12 inches or primes " ift. ADDITION AND SUBTRACTION OF DUODECIIALS. ART. 6@o. Duodecimals are added and subtracted in the same manner as compound numbers. EXAMPLES FOR PRACTICE. 1. Add together 12ft. 6' 9", 14ft. 7' 8", 165ft. 11' 10". Ans. 193ft. 2' 3". 2. Add together 1S2ft. 1.1' 2" 4"', 127ft. 7' 8" 11"', 291ft. 5' 11" 10'". Ans. 602ft. 0' 11" 1'". 3. From 204ft. 7' 9" take 114ft. 10' 6". Ans. 89ft. 9' 3". 4. From 397ft. 9' 6' 11"' 7"" take 201ft. 11' 7" 8"' 10"". Ans. 195ft. 9' 11" 2"' 9"". QUESTIONS.- Art. 259. What are duodecinals? 7Ilto how many parts is the unit or foot divided? In what ratio do dnu(decimals decrease from left to right I How are the different denominations distiaguished from each other? Are the denominartors of duodecimals expressed? Repecat the table?- Art. 260. How are duodecimals added and subtracted? BECT. XXXVII.] DUODECIMALS. 259 MULTIPLICATION OF DUODECIDIALS. ART. 261. To find the denomination of the product of any two numbers in duodecimals, when multiplied together. Ex. 1. What is the product of 9ft. multiplied by 3ft.? Ans. 27ft. OPERATION. 9ft. X 3ft. - 2'7t. 2. What is the product of 7ft. multiplied by 6'? Ans. 3ft. 6'. OPERATION. 6'= -z of a foot; then 7ft. X - ft. = 42 42' 42'; - 12 -- 3ft. 6'. 3. What is the product of 5' multiplied by 4'? Ans. 1' S". OPERATION. 5'-, and 4' -=4; then x - X: 2 —- 20"; 20' 12 1 1' 8". 4. What is the product of 9' multiplied by 11"'? Ans. S"' 3"'". OPERATION. 9'= 1z, and 11' Tri; then 9 X T I79T - - 99'"'; 99'"'. 12 8S"' 3'"". It will be observed, in the examples above, that feet multiplied by feet produce feet; feet multiplied by primes produce primes; primes multiplied by primes produce seconds, &c.; and that the several products are of the same denomination as denoted by the sum of the indices of the numbers multiplied together. Hence, When two numbers are multiplied together, the sum of their indices annexed to their product denotes its denomination. ART. F62. To multiply duodecimals together. Ex. 1. Multiply 8ft. 6in. by 3ft. 7in. Ans. 30ft. 5' 6". OPERATION. We first multiply each of the terms in the 8ft. 6' multiplicand by the 3ft. in the multiplier; thus,'ift. 7' 3ft. into 6' = 18'= lft. and 6'. We write the f —-- - 6' under the primes, and add the Ift. to the pro25ft. 6' duct of the 3ft. into 8ft., making 25ft. We 4ft. 1 1' 6" then multiply by the 7'; thus, 7' into 6' = 42" = 3' and 6". Placing the 6" at the right of the 3 Oft. 5' 6" primes, we add the 3' to the product of 7' into QUESTIONS. - Art. 261. How is the denomination of the product denoted when duodecimals are multiplied together? - If feet are multiplied by teet, what is the product? What, if feet are multiplied by primes? If primes are multiplied by primes? Is it absolutely necessary to commence the multiplication with feet? 260 DUODECIMALS. [sc'r. xxxvI. Sft. = 59' = 4ft. and 11', which we write under the feet and inches, and the two products being added together we obtain 30ift. 5' 6" for the answer. ULE. - 1. Under the multiplicand write the same names or denominatiOns of the multiplier; that is, feet under feet, inches under inc/hes, 4'c. iulltiply each term in the mnultiplicand, beginning at the lowest, by the fcet of the multiplier, and write each result undler its respective term observing to carry a unit for every 12 from each denomination to its?lext superior. 2. In the same manner nmultiply the multiplicand by the inches oJ ihe 9multiplier, and write the result of each term one place fuitrther towards the right than the corresponding terms in the precedinlg product. 3. Proceed in the same mrLanner wvilh ite seconds and all the rest o' the denomi^nations, and the sum of the seve'ral products will be the product retluired. ExAMPLES roR PRACrICE. 2. Mlultiply 8ft. 3in. by 7ft. 9in. Ans. 63ft. 11' 3". 3. Multiply 12tft. 9' by 9ft. 11'. Ans. 126ft. 5' 3". 4. ilultiply 14ft. 9' 11" by 6ft. 11' 8". Ans. 103ft. 4' 5" 8"' 4"'. 5. Multiply 161ft. 8' 6" by 7ft. 10'. Ans. 1266ft. 8' 7". 6. Multiply 87ft. 1' 11" by 5ft. 7' 5". Ans. 4S9ft. 8' 0" 2"'...7"". 7. WVhat are the contents of a board 1Sft. long and Ift. 10in wide? Ans. 33ft. 8. What are the contents of a board 19ft. 8in. long and 2ft. Ilin wide? Ans. 57ft. 4' 4". 9. W hat are the contents of a floor 18ft. 9in. long and 10ft. 6in. wide Ans. 196ft. 10' 6". 10. How many square feet of surface are there in a room 14ft. 9in. long, 12ft. 6in. wide, and 7ft. 9in. high? Ans. 791ft. 1' 6'. 11. John Carpenter has agreed to make 12 shoe-boxes of boards that are one inch thick. The boxes are to be, on the outside, 3ft. 8in. long, ift. 9in. wide, and Ift. 2in. high. How rmany square feet of boardls will it require to make the boxes, and how many cubic feet wi-ll they hold? Ans. 280 square feet; 66 cubic feet, 864 inches. 12. My garden is 18 rods long and 10 rods wide; a ditcl is dug round it 2 feet wide and 3 feet deep; but the ditch not being of a sufficient breadth and depth, I have caused it to be dug 1 foot deeper and lft. 6in, wider. How many solid feet will it be necessary to remove? Ans. 7540 feet. QuEerIoN. -- Art... What is the rule for ul] tiplication of Duodecinals? SECT. XXXVIII.J INVOLUTION. 261 13. How many cords of wood in a pile 56 feet long, 4 feet wide, and 5 feet 6 inches high? Ans. 9j cords. 14. How many cords of wood in a pile 23 feet 8 inches long, 4 feet wide, and 3 feet 9 inches high? Ans. 2 9 9 cords. 15. How much wood in a pile 97 feet long, 3 feet 8 inches wide, and 7 feet high? Ans. 19 cords 3Z9 feet. 16. If a pile of wood be 8 feet long, 3 feet 9 inches wide, how high must it be to contain one cord? Ans. 4-f4 feet. 17. I have a room 12 feet long, 11 feet wide, and'7 feet high; in it are 2 doors, 6 feet 6 inches high, and 30 inches wide, and the mop-boards are 8 inches high; there are 3 windows, 3 feet 6 inches wide, and 5 feet 6 inches high; how many square yards of paper will it require to cover the walls? Ans. 251e9s square yards. ~ XXXVIII. INVOLUTION. ART. 263. INVOLUTION is the method of finding any required power of any given number or quantity. A power is a quantity produced by multiplying any given number, called a root, a certain number of times continually by itself. The number denoting the power is called the index or exponent of the power, and is a small figure placed at the right of the root. Thus, the second power of 6 is written 62; the third power of 4 is written 43, and the fourth power of I is written ART. 2641. To raise a number to any required power. 3 — 3, the first power of 3, is written 31 or 3. 3 X 3 = 9, the second power of 3, is written 32. 3 X 3 X 3- 27, the third power of 3, " " 33. 3 X 3 X 3 X 3 81, the fourth power of 3, "1 " 34. 3 X 3 X 3 X 3 X 3 243, the fifth power of 3, " ( 35. QUESTIONs.-Art. 263. What is Involution? What is a power? What is the number called that denotes the power? Where is it placed? - Art. 264. To what is the index in each power equal? 262 INVOLUTION. [-srCT. xXXVII1. By examining the several powers of 3 in the examples given, we see that the index of each power is equal to the number of times 3 is used as a factor in the multiplications producing the power, and that the number of times the number is multiplied into itself is one less than the power denoted by the index. Hence the RULE. - Multiply the given number continually by itself, till the numObe of multiplications is one less than the index of the power to be found. and the last product will be the power required. NOTEv 1. -A fraction may be raised to any power by this rule, by mul tiplying its terms continually together. Thus, the second power of A is 2 NIOTE 2. - A mixed number may be either reduced to an improper fiaction, or the fractional part reduced to a decimal, and then raised to the required power. EXAMIPLES FOR PRACTICE. 1. What is the 2d power of 6? Ans. 36. 2. W hat is the 3d power of 5? Ans. 125. 3. A-What is the 6th power of 4 Ans. 4096. 4. What is the 4th power of 3? Ans. 81. 5. What is the 3d power of 4- Ans.,t. 6. Wrhat is the 4th power of 1. Ans. A.L 7. What is the 5th power'of 3? e Ans. 662) 1 -. S. What is the 3d power of.25? Ans..015625. 9. What is the 1st power of 17? Ans. 17. ART. s 5.o To raise a number to any required power without producing all the intermediate powers.. Ex. 1. What is the 8th power of 4? Ans. 65536. OPERA.TION. 1 2 3 3 3 + + 2 13. 4, 16, 64; 64 X 64X 16 65536. We first raise 4 to the 3d power, and write the exponents denoting each power directly above it. We then add the explent 3 to itself, and increasing the sum by the exponent 2, obtain 8, a number equal to the power required. WVe next multiply 64, the power belonging to the exponent 3, into itself, and this product by 16, the power belonging to the exponent 2, and obtain 65536, for the 8th power. }Hence the. following RULE.. -Raise the given t nunLer to any convenient power, and write the ex onents denoting the respective powers directly above ther. Then Quesriox-s. - What is the rule for raising a number to any required power? How m;tv a vulgar fraction be raised to a required power? How a mixed number?Z- Art. 265. What are the numbers placed over the several powers of 4 called, and whiat do thev denote? SECT. XXXIX]. E'VOLUTION. 263 add together sach exponents as will make a number equal to the required power, repeating any one when it is more conveieent; and the product of the powers belonging to these exponents will be the required answer. NOTE.- When a power has been found, double that power may be ob tained by multiplying it into itself. EXAMPLES FOR PRACTICE. 2. What is the 7th power of 5? Ans. 78125. 3. What is the 9th power of 6? Ans. 10077696, 4. What is the 12th power of 7? Ans. 13841287201. 5. What is the 8th power of 8? Ans. 16777216. 6. What is the 20th power of 4? Ans. 1099511627776. 7. WVhat is the 30th power of 3? Ans. 205891132094649. 8. What is the 50th power of 2? Ans, 1125899906842624. ~ XXXIX. EVOLUTION. ART. 6G6. EVOLUTION is the method of finding the root of a given power or number, and is therefore the reverse of Involution. A root of any power is a number which, being multiplied into itself a certain number of times, produces the given power. Thus, 4 is the second or square root of 16, because 4 X 4 16; and 3 is the third or cube root of 27, because 3 X 3 X 3 27. The root takes the name of the power of which it is the root. Thus, if the number is a second power, the root is called the second or square root; if it is a third power, the root is called the third or cube root; and, if it is a fourth power, its root is called the fourth or biquadrate root. Those roots which can be exactly found are called rational roots; those which cannot be exactly found, but approximate towards the true root, are called surd roots. QUESTIONS. - What is the rule for involving a nunlter without producing all the intermediate powers? -Art. 266. What is Evolution? What is a root? From what does the root take its name? What are rational roots? What surd roots'? 264 EVOLUTION. |SECT. XXXIX Roots are denoted by writing the character,/ before the power, with the index of the root over it, or by a fractional index or exponent. The third or cube root of 27 is expressed thus,,/27 or 271; and the second or square root of 25 is expressed thus, V25 or 251. NOTE. - The index 2 over,/ is usually omitted, when the square root is required. Thus,,/ 64 denotes the square root of 64. EXTRACTION OF THE SQUARE ROOT. ART. 267. The Square Root is the root of any second power, and is so called because the square or second power of any number represents the contents of a square surface, of which the root is the length of one side. ART. 26S. To extract the square root of any number is to find a number which, being multiplied by itself, will produce the given number. The following numbers in the upper line represent roots, and those in the lower line their second powers or squares. Roots, 1 2 3 4 5 6 7 8 9 10 Squares, 1 4 9 16 25 36 49 64 81 100 It will be observed that the second power or square of each of the numbers above contains twice as many figures as the root, or twice as many wanting one. Hence, To ascertain the number of figures in the square root of any given number, it must be divided into periods, beginning at the right, each of which, excepting the last, must always contain two figures; and the number of periods will denote the number of figures of which the root will consist. Ex. 1. I wish to arrange 625 tiles, each of which is 1 foot square, into a square pavement; what will be the length of one of the sides? Ans. 25 feet. OPERATION. It is evident, if we extract the square root 6i 25 (25, Ans. of 625, we shall obtain one side of the pavement, in feet. (Art. 267.) 4 Beginning at the right hand, we divide the 45)22 5 number into periods, by placing a point over 22 5 the right hand figure of each period; and then find the greatest square number in the CQUESTIONS. -How are roots denoted? How is the third or cube root denoted? How the second or square root? What is said of the index 2? - Art. 267. What is meant by the square root, and why is it so called? - Art. 268. What is meant by extracting the square root? How many more figures in the square of any number than in the root? How do you ascertain the number of figures in the square root of any number? Why do you point off the numbers into periods of two figures each? What is found by extracting the square root of the number in the question 7 SECT. XXXIX.] EXTRACTION OF THE SQUARE ROOT. 265 left hand period, 6 (hundreds) to be 4 (hundreds), and that its root is 2, which we write in the quotient. As this 2 is in the place of tens, its value is 20, and represents the side of a square, the area or superficial contents of which are 400 square feet, as seen in Fig. 1. W~e now subtract 400 feet from 625 feet, Fig. 1. and have 225 feet remaining, which must 20 feet. lbe added on two sides of Fig. 1, in order that it may remain a square. We therefore double the root 2 (tens) or 20, one side of D the square, to obtain the length of the two 2 0 sides to be enlarged, making 40 feet; and fa - 2 0 C then inquire how many times 40, as a divi- CD sor, is contained in the dividend 225, and find 400 it to be 5 times. This 5 we write in the quotient or root, and also on the right of the 20 -feet. divisor, and it represents the width of the additions to the square, as seen in Fig. 2. The width of the additions being multiplied by 40, the length of the two additions, makes 200 square feet, the contents of the two additions E and F, which are 100 feet for each. The space G now remains to be filled, to 25 feet. complete the square, each side of which is E20 It - 5 feet, or equal to the width of E and F. 100 25 If, therefore, we square 5, we have the contents of the last addition, G, equal to 25 D F t square feet. It is on account of this last 2 0 2 0 addition that the last figure of the root is c, 2 0 5 t placed in the divisor, for we thus obtain 45 - - o0 o feet for the length of all the additions made, which, being multiplied by the width, (5ft.,) the last figure in the root, the product, 225 25 feet. square feet, will be the contents of the three additions E, F and G, and equal to the feet remaining after we had found the first square. Hence we obtain 25 feet for the length of one side of the pavement, since 25 X 25 - 625, the number of tiles to be arranged, and equal to the sum of the several parts of Fig. 2; thus, 400 -+- 100 +- 100 + 25 - 625. From this solution and explanation we deduce the following QUESTIONS. - What is first done after dividing the number into periods? What part of Fig. 1 does this greatest square number represent? What place does the figure of the root occupy, and what part of the figure does it represent? Why does it have the place of tens? Why do you double the root for a divisor? What part of Fig. 2 does the divisor represent? What part does the last figure of the root represent? Why do you multiply the divisor by the last figure of the root? What parts of the figure does the product represent? Why do you square the last figure of the root'! What part of the figure does this square represent? What other way of finding the contents of the additions without multiplying the parts separately by the width? 23 266 EXTRACTION OF THE SQUARE ROOT. LsECT. XXXI9. RULE.- 1. Separate the given number into periods of two figures each, by putting a point over the place qf units, another ofver the piace of hundreds, and so on, and these points will show the number of Jigurce of which the root will consist. 2. Find the greatest square number in the first or left hand perinois, placing the root of it at the right hand of the given num-ber, a2ter the manner of a quotient in division, for the first figure of the root, and the square number under the period, subtracting it therefrom; and to the remainder bring down the next periodfor a dividend. 3. Place the double of the root already found on the left hand of the dividend for a divisor. 4. Find how often the divisor is contained in the dividend, omitting the right hand figure, and place the answer in the root for the secolnd figure of it, and likewise on the right hand of the divisor.* Multiply the divisor with the figure last annexed by the figure last placed in the root, and subtract the product from the dividend. To the remainder join the next period for a new dividend. 5. Double the figures alreadyfound in the rootfor a new divisor, or bring down the last divisor for a new one, doubling the right hand figure of it, and from these find the next figure in the root as last directed, and continue the operation in the same manner till all the periods have been brought down. NOTE. - 1. If the dividend does not contain the divisor, a cipher must be placed in the root, and also at the right of the divisor; then, after bringing down the next period, this last divisor must be used as the divisor of the new dividend. 2. When there is a remainder after extracting the root of a number, periods of ciphers may be annexed, and the figures of the root thus obtained will be decimals. 3. If the given number is a decimal, or a whole number and a decimal, the root is' extracted in the same manner as in whole numbers, except in pointing off the decimals either alone or in connection with the whole number, we begin at the separatrix and place a point over every second figure toward the right, filling the last period, if incomplete, with a cipher. 4. The square root of any number ending with 2, 3, 7, or 8, cannot be exactly found. EXAMPLES FOR PRACTICE. 2. What is the square root of 14S996?' The figure of the root must generally be diminished by one or two units, on account of the deficiency in enlarging the square. QUESTIONS. - What is the rule for extracting the square root? What is te be done if the dividend does not conitain the divisor? What must be done if there is a remailnder after extracting the root? What do you do if the given number is a decimal? Of what numbers can the square root not be found? .OVtT. xxxix.] EXTRACTION OF THE SQUARE ROOT. 267 OPERATION. 148996(386 9 68)589 544 766)4596 4596 3. What is the square root of 516961? Ans. 719. 4. What is the square root of 182329? Ans. 427. 5. What is the square root of 23S04641? Ans. 4879. 6. What is the square root of 106'732S9? Ans. 3967. 7. "What is the square root of 20894041? Ans. 4571. 8. What is the square root of 42025? Ans. 205. 9. What is the square root of 1014049? Ans. 1007. 10. What is the square root of 535? Ans. 23.194+. 11. What is the square root of 71 Ans. 8.426+. 12. What is the square root of 7? Ans. 2.645+. 13. What is the square root of.1024? Ans..32. 14. What is the square root of.3364? Ans..A8. 15. What is the square root of.895? Ans..946+-. 16. What is the square root of.120409? Ans..347. 17. What is the square root of 61723020.96? Ans. 7856.4. 18. What is the square root of 9754.60423716? Ans. 98.7654. ART. ~t o If it is required to extract the square root of a v-ulgar fraction, or of a mixed number, the mixed number must be reduced to an improper fraction; and in both cases the froactions must be reduced to their lowest terms, and the root of the numlerator and denominator extracted. NOTE. -- hen thte exact root of the terms of a fraction cannot be found, it must be reduced to a decimal, and the root of the decimal extracted. EXAMPLES FOR, PRACTICE. 1. What is the square root of -%-? Ans. WYx. 2. What is the square root of 9 6? Ans. s. 3. What is the square root of 37 t% Ans. 6G 4. What is the square root of,s8A49u? Ans.4 3 5. WVat is the square root of 60-1 Ans. 7i. 6. What is the square root of 2835 7 Ans. 5-. IUEsTIONS. — Art. 269. What tdo you do when it is required to extract the square root of a vulgar fraction or of a nlixed1 nuatber? 268 APPLICATION OF THE SQUARE ROOT. [SECT. XXXIX. 7. What is the square root of 47~-? Ans. 6-. 8. What is the square root of 4-? Ans..58-+-. 9. What is the square root of 832? Ans. 9.14 —. 10. What is the square root of 121l-? Ans. 11.042+-. 11. What is the square root of 362 Ans. - 462' 12. What is the square root of 76? Ans. APPLICATION OF THE SQUARE ROOT. ART. 27@. The square root may be applied to finding the dimensions and areas of squares, triangles, circles, and other surfaces. 1. A certain general has an army of 226576 men; how many must he place rank and file to form them into a square? Ans. 476. 2. A gentleman purchased a lot of land in the form of a square, containing 640 acres; how many rods square is his lot? Ans. 320 rods. 3. I have three pieces of land; the first is 125 rods long, and 53 wide; the second is 62A rods long, and 34 wide; and the third contains 37 acres; what will be the length of the side of a square field whose area will be equal to the three pieces? Ans. 121.11+ rods. 4. W. Scott has 2 house lots; the first is 242 feet square, and the second contains 9 times the area of the first; how many feet square is the second? Ans. 126 feet. 5. There are two pastures, one of which contains 124 acres, and the area of the other is to the former as 5 to 4; how many rods square is the latter? Ans. 157.48+ rods. 6. I wish to set out an orchard containing 216 fruit trees, so that the length shall be to the breadth as 3 to 2, and the distance of the trees from each other 25 feet; how many trees will there be in a row each way, and how many square feet of ground will the orchard cover? Ans. 18 in length; 12 in breadth; 116875 sq. ft. ART. ~71. A TRIANGLE is a figure having three sides and three angles. A right angled triangle is a figure having three sides and three angles, one of which is a right angle. OQUESTIONS. - Art. 270. To what may the square root be applied? —Art. 271. What is a triangle? What is a right angled triangle? What is the longest side called? What the other two? SECT. xxxix.] APPLICATION OF THE SQUARE ROOT. 269 C d, co The side A B is called the hase of the triCD angle, the side B C the perpendicular, the AwLa side A C the hypot1henuse, and the angle at A B/ j B is a right an, le. Base. ART. ~72. In every right angled triangle the square of the h ypothenuse is equal to the sum of the squares of the base and perpendicular, as shown by the following diagram. C It will be seen by examining this diagram that the large square, formed on the hypothenuse A C, contains the same nunmber of small squares as the other two countA _ _ XB ed together. Hience, the propriety of the following rules. ART. 7e3. To find the hypothenuse, the base and perpendicular being given. RULE. - Add the square of the base to the square of theperpendicular, and extract the square root of their sum. ART. 271. To find the perpendicular, the base and hypothenuse being given. tRULE. - Subtract the square of the baseJ fom the square of the hypothenuse, and extract the square root of the remainder. ART. 27g. To find the base, the hypothenuse and perpen(licular being given. RuiL. - Subtract the square of the perpendicular fro m the square of the hypothenuse, and extract the square root of the remainder. EXAMPLES FOR PRACTICE. 1. What must be the length of a ladder to reach to the top of a house 40 feet in height, the bottom of the ladder being placed 9 feet from the sill Ans. 41 feet. QUESTION-S. - Art. 272. How does the square of' the hyplothenuse compare with the base and perpenlticular? How doues this fact appear from Fig. 2? - Art. 273. What is the rule for finding the hypothenuse? —Art. 274. WVIlhat for finding the perpendicular' - Art. 275. WVh-at for finding the base? 27r0 APPLICATION OF THE SQUARE ROOT. LaECT. XXXIX. 2. Two vessels sail from the same port; one sails due north 360 miles, and the other due east 450 miles; what is their distance from each other? Ans. 576.2+ miles. 3. The hypothenuse of a certain right angled triangle is 60 feet, and the perpendicular is 36 feet; what is the length of the base.? Ans. 48 feet. 4. A line drawn from. the top of the steeple of a certain meeting-house to a point at the distance of 50 feet on a level from the base of the steeple, is 120 feet in length; what is the height of the steeple? Ans. 109.08+ feet. 5. The height of a tree on an island in a certain river, is 160 feet. The base of the tree is 100 feet on a horizontal line from the river, and is elevated 20 feet above its surface. A line extending from the top of the tree to the further shore of the river is 500 feet. Required the width of the river. Ans. 366.47+ feet. 6. On the summit of a hill there is a tower 160 feet high whose base is 90 feet, on a level, from a certain road that is 110 feet wide; the length of a line extending from the top of the tower to a point on the opposite side of the road is 300 feet. What is the elevation of the base of the tower above the road? Ans. 63.64+ feet. 7. John Snow's dwelling is 60 rods north of the meetinghouse, James Briggs's is 80 rods east of the meeting-house, Samuel Jenkins's is 70 rods south, and James Emerson's 90 rods west of the meeting-house; how far will Snow have to travel to visit his three neighbors, and then return home? Ans. 428.4+ rods. 8. A certain room is 24 feet long, 18 feet wide, and 12 feet high; required the distance from one of the lower corners to an opposite upper corner. Ans. 32.3+ feet. ART. 276 A CIRCLE is a plane figure bounded by a curved line, every part of which is equally distant from a point called the centre. The circumference or periphery of a circle is the line which bounds it. A - - B The diameter of a circle is a line drawn through the centre, and terminated by the circumference: as A B. (QUESTIONS. - Art. 276. What is a circle? What is the ci;cumfernce, of a circle?I What the diameter? SECT. xxxix.] APPLICATION OF THE SQUARE ROOT. 271 ART. 277. All circles are to each other as the squares of mhei-r diameters, semidiameters, or circumferences. All similar triangles and other rectilineal figures are to each other as the squares of their homologous or corresponding sides ART. 7S. To find the side, diameter, or circumference of any surface, which is similar to a given surface. RULE. - State the question as in Proportion, and square the given sides, diameters, or circumferences, and the sqIuare root of the fourth term of (he proportion will be the required answer. ART. 279. To find the area of any surface which is similar to a given surface. RULE. - State the question as in Proportion, and square the given sidles, diameters, or circumferences, and thefourth term of the proportion is the reyuired answer. EXAMPLES FOR PRACTICE. Ex. 1. I have a triangular piece of land containing 65 acres, one side of which is 100 rods in length; what is the length of the corresponding side of a similar triangle containing 32acres? Ans. 70.71+ rods. OPERATION. 65: 3 2g' 1002: 5000; V/5000= 70.71 +rods. 2. 1 have a board in the form of a triangle, the length of one of Its sides is 16 feet. My neighbor wishes to purchase one half the board; at what distance from the smaller end must it be divided parallel to the base or larger end? Ans. 11.31+ feet. 3. There is a triangular piece of land, the length of one side of which is 11 rods; required the length of the corresponding side of a similar triangle containing three times as much. Ans. 19.05+ rods. 4. The diameter of a circle is 6 feet, and its area is 28.3 feet; what is the diameter of a circle whose area is 42.5 feet? Ans. 7.35+ feet. 5. If an anchor, which weighs 20001b., requires a cable 3 inches in diameter, what should be the diameter of a cable, when the anchor weighs 40001b.? Ans. 4.24+ inches. 6. A rope 4 inches in circumference will sustain a weight ~f 1000l b.; what must be the circumference of a rope that will sustain 50001b? Ans. 8.94+ inches. 7. There is a triangle containing 72 square rods, and one of QUESTIONS. - Art. 277. What proportion do circles have to each other?-.irt. 278. What is the rule for finding the side, diameter, &c., of a surface similar to a given surface? - Art. 279. What is the rule for finding the area.f a surface similar to a given surface? 272 APPLICATION OF THE SQUARE ROOT. [SECT. xxxix. its sides measures 12 rods; what is the area of a similar triangle whose corresponding side measures 8 rods? Ans. 32 rods. 8. A gentleman has a park, in the form of a right angled triangle, containing 950 square rods, the longest side or hypothenuse of which is 45 rods. HIe wishes to lay out another in the same form, with an hypothenuse I the length of the first; required the area. Ans. 105.55+ square rods. 9. If a cylinder 6 inches in diameter contain 1.178+ cubic feet, how many cubic feet will a cylinder of the same length contain, that is 9 inches in diameter? Ans. 2.65+ feet. 10. If a pipe, 2 inches in diameter, will fill a cistern in 204 minutes, how long would it take a pipe, that is 3 inches in diameter? Ans. 9 minutes. 11. A tube 4 of an inch in diameter will empty a cistern in 50 minutes; required the time it will empty the cistern, when there is another pipe running into it ~ of an inch in dianmeter? Ans. 62' minutes. ART. 2S@, To find the side of a square that can be inscribed in a circle of a given diameter. A square is said to be inscribed in a circle when each of its angles or corners touches the circuinference. It may be conceived to be composed of two right angled triangles, the base and perpendicular of each being equal, and their hypothenuse the diameter of the circle, as seen in the diagram. Hence the RULE. -Extract the square root of haof the square of the diameter and it is the side of the inscribed square. ExAMIPLES FOR PRACTICE. 1. What is the length of one side of a square that can be inscribed in a circle, whose diameter is 12 feet? Ans. 8.4-8+ feet. 2. How large a square stick may be hewn from a round one, which is 30 inches in diameter? Ans. 21.2-+- inches square. 3. A has a cylinder of lignum-vitoe, 194- inches long and 1j inches in diameter; how large a square, ruler may be made from it? Ans. 1.06 inches square. QUESTIONS. - -Art. 280. When is a square said to be inscriled in a circle'! Of what mlay the inscribed square be conceived to be comipoused? What part of the circle is the hypothenuse of the two triangles? NWhat is the rule foi tfiling the side of the inscribed square? BECT. xxxIx. EXTRACTION OF THE CUBE ROOT. 273 EXTRACTION OF THE CUBE ROOT. ART..IS]. The CUBE ROOT is the root of any third power, and is so called because the cube or third power of any number represents the contents of a cubic body, of which the cube root is one of its sides. ART. QS$. To extract the cube root is to find a number, which, being multiplied into its square, will produce the given number. The following numbers in the upper line represent roots, and those in the lower line their third powers or cubes. Roots, 1 2 3 4 5 6 7 8 9 10 Cubes, 1 8 27 64 125 216 343 512 729 1000 It will be observed that the cube or third power of each of the numrnbers aboxe contains three times as many figures as the root, or three times as many wanting one, or two at most. Hence, To ascertain the number of fioures in the cube root of any given number, it must be divided into periods, beginning at the right, each of which, excepting the last, must always contain three figures, and the number of periods will denote the number of figures, which the root will contain. Ex. 1. I have 17576 cubical blocks of marble, which measure one foot on each side; what will be the length of one of the sides of a cubical pile, which may be formed of them? Ans. 26 feet. OPERATION. 1 75 7 (26 ( 2 Root. It is evident that 8 the number of blocks or feet on a 2' X 3 0 0 = 1 2 0 ) 95 7 6 1st dividend. side will be equal to the cube root of 7 2 0 0 1st addition. 17576. (Art. 281.) 6 2 X 2 X 3 2 1 6 0 2d addition. Beginning at the 6 -= 2 1 6 3d addition. right hand, we divide the number 9 5 7 6 Subtrahend. into periods, by placing a point over the right hand figure of each period. We then find the greatest cube number in the left hand period 17 (thousands) to be 8 (thousands), and QUESTroNS.-Art. 281. What is the cube root, and why so called?-Art. 2-i2. What is meant by extracting the cube root? How rmany nmoore figures in the cube of any nutmber than in the root? 7lHow do you ascertain the anouber of' figures in the cube root of any nulnber? What is fotuld by extracting the cute root of the number in the example? What is first done after separating the number into periods? 274 EXTRACTION OF THE CUBE ROOT. [SEvr. Exxl, its root 2, which we place in the quotient or root. As 2 is in the place of tens, because there is to be another figure in the root, its value is 20, and it represents the side of a cube (Fig. 1), tile contents of which are 8000 cubic feet; thus 2() X 290 X 20 = 8000. We now subtract the cube of 2 (tens' F 8 (thousands) from the first period, 17 20 (thousands), and have 9 (thousand) feet 20 -- i remaining, which, being increased by the 4-c o,next period, makes 9576 cubic feet. This 20 mu st be added to three sides of the cube, Fig. 1, in order that it may remain a cube.?:0o.i~To do this, we must find the superficial contents of the three sides of the cube, to which the additions are to be made. Now, 2 since one side is'2 (tens) or 220 feet square, its superficial contents will be 20 X 20 400 square feet, and this multiplied by 3 will be the superficial contents of three sides; thus, 20 X 20 X 3 = 1200, or, which is the same thing, we multiply the square of the quotient figure, or root, by 300; thus, 22 X 300 = 1200 square feet. Making this number a divisor, we divide the dividend 9576 by it, and Fig. 2. obtain 6 for the quotient, which we place in the root. This 6 represents, the ~20 ~ -- 6 thickness of each of the three additions to be made to the cube, and their.,uperficial contents beingf multiplied by it we 20 have 1200 X 6 = 7200 cubic feet for 2the contents of the three additions, A, B3 ~20llll(llll ~and C, as seen in Fig. 2. HIavinig made these additions to the 1(1 I I! I v 6cube, we find that there are three other 20u deficiencies, n n, o o, and r r, the length of which is equal to one side of the additions, 2 (tens) or 20 feet; and their breadth and thickness, 6 feet, equal to the thickness of the additions. Therefore, to find the solid contents of the additions, necessary to supply these deficiencies, we multiply the product of their length, breadth and thickness by the number of additions; thus, 6 X 6 X 20 X 3 = 2160, or, which is the same thing, we multiply the square of the last quotient figuree by the former figure of the root, and that product by 30; thus, 62 X 2 CUESTIONs. - What is done with this greatest cube number, and what part of Fig. I does it represent? What is done with the root? What is its value and what part of the figure does it represeiit? How are the culical contents of the figure found? What constitutes the remainder after subtracting the cuble nnitilrer from the left hand periodl? To how many sides of the cule miust this retnaii(ler be added? Why? How do you find the divisor? What parts of the figure ldoes it represent? How do you obtain the last figure of tle root? What part of Fig. 2 does it represent? Why do you multiply the divisor by the last quotient figure? What parts of the figure does the product represent? What three other Reficiencies in the fiaure? SECT. XXXIX.] EXTRACTION OF THE CUBE ROOT. 275'K 30 = 2160 cubic feet for the contents of the additions s s, u u, and v v, as seen in Fig. 3. These additions being made to the 20 20 G @cube, we still observe another deficien-, 0 6 cy of the cubical space x x x, the 6/ X A'.. t. 1~ij length, breadth, and thickness of which,q~ 11~ —~~'':I: lI; 1!20 are each equal to the thickness of the ~ l ll l l tl llll other additions, which is 6 feet. ThereIii fore, we find the contents of the addition necessaryto supply this deficiency by multiplying its length, breadth, and thickness together, or cubing the last 20 6 figure of the root; thus, 6 X 6 X 6 216 cubic feet for the contents of the,addition z z z, as seen in Fig. 4. The cube is now complete, and if we'ig. 4. add together the several additions that 26 have been made to it, thus, 7200 +- 2160..... ~ + -i- 216 = 9576, we obtain the number of 26 L-==: 0 cubic feet remaining after subtracting the first cube, which, being subtracted from 26 the dividend in the operation, leaves no remainder. Hence, the cubical pile formed 9z~~~s ~ is 26 feet on each side; since 26 X 26 X 26 =2 17576, the given number of 2G blocks, and the sum of the several parts of Fig. 4. Thus, 8000 + — 7200 -- 26 2160 -+-216 = 17576. IHence the following RULE.- 1. Separate the given number into periods of three figures each, by placing a point over the unit figure, and every third figure beyond the place of units. 2. Find by the table the greatest cube in the left hand period, and put its root in the quotient. 3. Subtract the cube, thus found, from this period, and to the remnainder bring down the next period; call this the dividend. 4. Mizultiply the square of the quotient by 300 for a divisor, by which divide the dividend and place the quotient, usuall?/y diminished by one or tfwo units, for the next figure of the root.5. Multiply the divisor by this last quotient figure, and write the prohduct under the dividend; then multiply the square of the last quotient figure by the former quotient figure or figures, and this product by 30, and place the product under the last; under all, set the cube of the last quotient figure, and call their sum the subtrahend. QUESTIONS. - How do you find their contents? What parts of Fig. 3 does the product represent 7? What other deficiency do you observe? To what are tis length, breadth and thickness equal? How do you find its contents? What part of Fig. 4 does it represent? What is the rulle or extracting the cube root? 276 EXTRACTION OF THE CUBE ROOT. [SECT. XXXIX 6. Subtract the subtrahend from the dividend, and to the remainder rraing down the next period for a new dividend. with which proceed as before, and so on, till the whole is completed. NOTE. - The observations made in Notes 1, 2, and 3, under square root are equally applicable to the cube root, except in pointing off decimals each period must contain three figures, and two ciphers must be placed at the, right of the divisor when it is not contained in the dividend. EXAMPLES FOR PRACTICE. 1. What is the cube root of 78402752? Ans. 428 OPERATION. 78402752(428 Root. 64 480 0 ) 1 4 40 2 1st dividend. 9600 480 10 0 8 8 - 1st subtrahend. 529200 )43 14752 -- 2d dividend. 4233600 80640 512 4 3 1 4 7 5 2 -- 2d subtrahend. 2. What is the cube root of 74088? Ans. 42. 3. What is the cube root of 185193? Ans. 57Y. 4. What is the cube root of 80621568? Ans. 432. 5. What is the cube root of 176558481? Ans. 561. 6. WVhat is the cube root of 257259456? Ans. 636. 7. What is the cube root of 1860867? Ans. 123. 8. What is the cube root of 1879080904? Ans. 1234. 9. What is the cube root of 41673648.563? Ans. 346.7. 10. What is the cube root of 483921.516051? Ans. 78.51. 11. What is the cube root of S.144865728? Ans. 2.012. 12. What is the cube root of.075686967? Ans..423. 13. What is the cube root of 25? Ans. 2.92+-. QUESTION. - How many ciphers must be placed at the right of the divisor when it is not contained in the dividend? SECT. xxxlx.J EXTRACTION OF THE CUBE ROOT. 277 ART. 282. When it is required to extract the cube root of a vulgar fraction, or a mixed number, it is prepared in the same manner as directed in square root. (Art. 269.) EXAMPLES FOR PRACTICE. 1. What is the cube root of 81Tr? Ans. 4.334+-. 2. What is the cube root of r42792 Ans. - 3. What is the cube root of 49.: Ans. 32 4. What is the cube root of 166 2? Ans. 5-. 5. What is the cube root of 85-12Z? Ans. 42. APPLICATION OF THE CUBE ROOT. ART. 283. THE cube root may be applied in finding the dimensions and contents of cubes and other solids. 1. A carpenter wishes to make a cubical cistern that shall contain 2744 cubic feet of water; what must be the length of one of its sides? Ans. 14 feet. 2. A farmer has a cubical box that will hold 400 bushels of grain; what is the height of the box? Ans. 7.92+ feet. 3. There is a cellar, the length of which is 18 feet, the width 15 feet, and the depth 10 feet; what would be the depth of another cellar of the same size, having the length, width, and depth equal? Ans. 13.92+ feet. ART. 284. A SPiHERE is a solid bounded by one continued tonvex surface, every part of which is equally distant from a point within, called the centre. A The diameter of a sphere is a straight line passing through the centre, and terminated by the surface; as A B. B ART. 285I. A CONE is a solid having a circle for its base, and its top terminated in a point, called the vertex. QUESTIoNS.- Art. 282. How is a vulgar fraction or a mixed number prepared for extracting the square root? — Art. 283. To what may the cube root he applied 7 - Art. 284. W8hat is a sphere? What is the diameter of a sphere? - Art. 285. What is a cone? 24 '278 APPLICATION OF THE CUBE ROOT. [I:CT. XXXIx The altitzude of a cone is its perpendicular height,'4'/i \ or a line drawn from the vertex perpendicular to the A' \\1 plane of the base; as B C. ART. 2S6. Spheres are to each other as the cubes of their diameters, or of their circumferences. Similar cones are to each other as the cubes of their altitudes, or the diameters of their bases. All similar solids are to each other as the cubes of their homologous or corresponding sides, or of their diameters. ART. eS7. To find the contents of any solid which is sim. ilar to a given solid., RULE. - State the question as in Proportion, and cube the given sides diameters, altitudes, or circumferenlces, and the fourth term of the pro vortion is the required answer. ART. 28S. To find the side, diameter, circumference, oi altitude of any solid, which is similar to a given solid. RULE. - State the question as in Proportion, and cube the given sides, diameters, circumferences, or altitudes, and the cube root of the fourth term of the proportion is the required answer. EXAMPLES FOR PRACTICE. 1. If a cone 2 feet in height contains 456 cubic feet, what are the contents of a similar cone, the altitude of which is 3 feet? Ans. 1539 cubic feet. OPERATION. 2 33: 456: 1539. 2. If a cubic piece of metal, the side of which is 2 feet, is worth $ 6.25; what is another cubical piece of the same kind worth, one side of which is 12 feet? Ans. $ 1350. 3. If a" ball, 4 inches in diameter, weighs 01b., what is the weight of a ball 6 inches in diameter? Ans. 168.7+ lb. QUESTIONS. - What is the altitude of a cone? Art. 286. What proportion do spheres have to each other? WThat proportion do cones have to each other? What proportion do all similar solids have to each other? - Art. 287. Witat is the rule for finding the contents of a solid similar to a given solid? - Art. 288. What is the rule for finding the side, diameter, &c., of a solid similar to a given solil? SECT. XL.] ARITHMETICAL PROGRESSION. 279 4. If a sugar loaf, which is 12 inches in height, weighs 161b., how many inches may be broken from the base, that the residue nmay weigh 81b.? Ans. 2.5+ in. 5. If an ox, that weighs S001b., girts 6 feet, what is the weight of an ox that girts 7 feet? Ans. 1270.31b. 6. If a tree, that is one foot in diameter, make one cord, how many cords are there in a similar tree, whose diameter is two feet? Ans. 8 cords. 7. If a bell, 30 inches high, weighs 1000l b., what is the wei(ht of a bell 40 inches high? Ans. 2370.31b. 8. If an apple, 6 inches in circumference, weighs 16 ounces, what is the weight of an apple 12 inches in circumference? Ans. 12S ounces. 9. A and B own a stack of hay in a conical form. It is 15 feet high, and A owns 2 of the stack; it is required to know how many feet he must take from the top of it for his share. Ans. 13.1+ feet. ~ XL. ARITHMETICAL PROGRESSION. AnT. 8 l9. WrEN a series of numbers increases or decreases by a constant difference, it is called Arithmetical Progression, or Progression by Difference. Thus, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 29, 26, 23, 20, 17, 14, 11,, 2. The first is called an ascending series or progression. The second is called a descendlin, series or progression. The numbers which form the series are called the terms of the progression. Thefirst and last terms are called the extremes, and the other terms the means. The constant difference is called the commnzon difference of the progression. Any three of the five following things being given, the other two may be found: — rUESTIONS. - Art. 289. What is arithmetical progression? VCbat is an ascetldinlg series? What a descending series? What are the terms of a progression? What the extremnes? What the means? 280 ARITHMETICAL PROGRESSION. [SECT. XL. 1st. The first term, or first extreme, 2d. The last term, or last extreme; 3d. The number of terms; 4th. The common difference; 5th. The sum-of the terms. ART. 29@. To find the common difference, the first term, last term, and number of terms being given. ILLUSTRATION. - In the following series, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 2 and 29 are the extremes, 3 the common difference, 10 the number of terms, and the sum of the series 155. It is evident that the number of common differences in any series must be 1 less than the number of terms. Therefore, since the number of terms in this series is 10, the number of common differences will be 9, and their sum will be equal to the difference of the extremes; hence, if the difference of the extremes (29 - 2 - 27) be divided by the number of common differences, the quotient will be the common difference. Thus, 27. 9 - 3, the common difference. Hence the following' RULE. - Divide the difference of the extremes by the number of terms less one, and the quotient is the common difference. EXAMPLES FOR PRACTICE. 1. The extremes of a series are 3 and 35, and the number of terms is 9; what is the common difference? Ans. 4. OPERATION. 35-3 =- 4 common difference. 9 1 2. If the first term is 7, the last term 55, and the number of terms 17, required the common difference. Anis. 3. 3. If the first term is 4, the last term 14, and the number of terms 15, what is the common difference? Ans. a4. 4. If a man travels 10 days, and the first day goes 9 miles, QUESTIONS. - Art. 290. What is the common difference? What five things are named, any three of which being given the other two can be found? What is the rule for finding the common difference, the first term, last term, and number of terms being given'! RECT. XL. ARITHMETICAL PROGRESSION. 281 and the last 17 miles, and increases each day's travel by an equal difference, what is the daily increase? Ans. ~ miles. ART. 29 1. To find the sum of all the terms, the first term, last term, and number of terms being given. ILLUSTRATION. - Let the two following series be arranged as follows: - 2, 5, 8, 11, 14, 17, 20,. 77, sum of first series. 20, 17, 14, 11, 8, 5, 2, —- 77, sumof inverted series. 22, 22, 22, 22, 22, 22, 22,= 154, sum of both series. From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series. It is evident that, if 22 in the above series be multiplied by 7, the number of terms, the product will be the sum of both series; thus 22 X 7 -- 154; and, therefore, the sum of either series will be 154- 2 —-'77. But 22 is the sum of the extremes in each series, thus, 20 +- 2 = 22. Therefore, if the sum of the extremes be multiplied by the number of terms, the product will be double the sum of either series. Hence, RULE I. -Multiply the sum of the extremes by the nunmer of terms, and half the product will be the sum of the series. Or, RULE II. - Multiply the sum of the extremes by half the number of terms, and the product is the sum required. EXAMIPLES FOR PRACTICE. 1. If the extremes of a series are 5 and 45, and the number of terms 9, what is the sum of the series? Ans. 225 OPERATION. (45+ 5)X9 ( 4 5 5 ) X 9 _ 225, sum of the series. 2 2. John Oaks engaged to labor for me 12 months. For the first month I was to pay him $ 7, and for the last month $ 51. In each successive month he was to have an equal addition to his wages; what sum did he receive for his year's labor? A ns. 8' 348. QUESTION. -Art. 291. What is the rule for finding the suin of all the terms, the first term, last term and number of ternms being given? 24 282 ARITHMTETICAL PROGRESSION. ISECT. XL. 3. I have purchased from W. Hall's nursery 100 fruit trees, of various kinds to be set around a circular lot of land, at the distance of one rod from each other. Having deposited them on one side of the lot, how far shall I have travelled when I have set out my last tree, provided I take only one tree at a time, and travel on the same line each way? Ans. 9801 rods. ART. 292. To find the number of terms, the extremes and common difference being given. ILLUSTRATION. — Let the extremes of a series be 2 and 29, and the common difference 3. The difference of the extremes will be 29 - 2 =_ 27. Now, it is evident, that, if the difference of the extremes be divided by the common difference. the quotient will be the number of common differences; thus, 27 _ 3 _ 9. It has been shown, (Art. 2S9,) that the number of terms is I more than the number of differences; therefore, 9 + 1 = 10, is the number of terms in this series. Hence the following RULE. - Divide the difference of the extremes by the common dijference, and the quotient, incresecd by 1, wi:.- be the number oJ terms required. EXAMPLES FORP PRACTICE 1. If the extremes of a series are 4 and 44, 2nd the common difference 5, what is the number of termn? Ans. 9. OPERATION. 44 4 -4+ 1 9, number of terms. 5 2. A man going a journey travelled the first day 8 miles, and the last day 47 miles, and each day increased his journey by 3 miles. How many days did he travel? Ans. 14 days. ARLT. 293. To find the sum of the series, the extremes and common difference being given. ILLUSTRATION.-Let the extremes be 2 and 29, and the common difference 3. The difference of the extremes will be 29 - 2 = 27; and it has been shown, (Art. 292,) that if the difference of the extremes be divided by the common difference, the QUESTION. -Art. 292. What is the rule for finding the number of terms, the extremes and common difference being given? SECT. XL.] ARITHMETICAL PROGRESSION. 2S3 quotient will be the number of terms less one. Therefbre, the number of terms less one will be 27 - 3 = 9, and the number of terms 9 + 1 _ 10. It was also shown, (Art. 291,) that, if the number of terms be multiplied by the sum of the extremes, and the product divided by 2, the quotient will be the sum of the series. Hence the RULE.- Divide the difference of the extremes by the common differ ence, and add 1 to the quotient; multiply this sum by the sum of th extremes, and half the product is the sum of the series. EXAOMPLES F01R PRACTICE. 1. If the two extremes are 11 and 74, and the common dif ference 7, what is the sum of the series? Ans. 425. OPERATION. 74 —11 1 (74-{- 1 I) >< l0 + 1 10;( + 1) < 4 2 5, sum of series. 7 2 2. A pupil commenced Virgil by reading 12 lines the first day, 17 lines the second day, and thus increased every day by 5 lines, until he read 137 lines in a day. How many lines did he read in all? Ans. 1937 lines. ART. 94. To find the last term, the first term, the numn ber of termus, and the common difference being given. ILLUSTRATION. -Let the first term of a series be 2, the nunmber of terms 10, and the common difference 3. It has been shown, (Art. 290,) that the number of common differences is always 1 less than the number of terms; and that the sumn of the common differences is equal to the difference of the extremes; therefore, since the number of terms is 10, and the common difference 3, the difference of the extremes will be (10 - 1) X 3 27; and this difference, added to the first term, must give the last term; thus, 2 + 27 = 29. H1lence the foltowing RULE. - Multiply the number of terms less 1, by the common differ ence, and add this product to the first term for the last term. NOT. — If the series is descending, the product must be subtracted froin the first term. QUESTIONS. - Art. 293. What is the rule for finding the sum of the series, the extremes and commnlon difference being given?- Art 294. What is the rule for finding the last term, the first term, the nuimber of terms and common diflerence being given? 284 ANNUITIES AT SIMPLE INTEREST. [SECT. X,l EXAMPLES FOR PRACTICE. 1. If the first term is 1, the number of terms 7, and the common difference 6, what is the last term? Ans. 37. OPERATION. 1 + (7- 1 ) X 6- 3 7, last term. 2. If a man travel 7 miles the first day of his journey, and 9 miles the second, and shall each day travel 2 miles further than the preceding, how far will he travel the twelfth day? Ans. 29 miles. 3. If A set out frlom Portland for Boston, and travel 20} miles the first day, and on each succeeding day 11 miles less than on the preceding, how far will he travel the tenth day? Ans. 61 miles. ANNUITIES AT SIMIPLE INTEREST BY ARITHMETICAL PROGRESSION. ART. ~95I. AN ANNUITY is a sum of money to be paid annually, or at any other regular period, either for a limited time or forever. The present woqrth of an annuity is that sum which being put at interest will be sufficient to pay the annuity. The amount of an annuity is the interest of all the payments added to their sum. Annuities are said to be in arrears when they remain unpaid, after they have become due. ART. 29D. To find the amount of an annuity at simjple interest. Ex. 1. A man purchased a farm for $ 2000, and agreed to pay for it in 5 years, paying $ 400 annually; but finding him self unable to make the annual payments, he agreed to pay the whole amount at the end of the 5 years, with the simple interest, at 6 per cent., on each payment, from the time it became due till the time of settlement; what did the farm cost him? Ans. $ 2240. ILLUSTRATION. — It is evident the fifth payment will be $ 400, without interest; the fourth will be on interest 1 year, and will amount to $424; the third will be on interest 2 years, and will amount to $448; the second will be on interest 3 QUESTIONS. - Art. 295. What is an annuity? What is meant by the present worth of an annuity? By the amount? When are annuities said to be in arrears? SECT. XL.J,.NNT1ITIES AT SIMPLE INTEREST. 285 years, and will amount to $472; and the first will be on interest 4 years, and will amount to $ 496. Therefore, these several sums form an arithmetical series; thus, 400, 424, 448, 472, 496, of which the fifth payment, or the annuity, is the first term, the interest on the annuity for one year the common difference, the time in years, the number of terms, and the amcount of the annuity the sum of the series. The sum of this series is found by Art. 291; thus, (400 + 496) X 5 - $ 2240. Hence the 2 RULE. - First find the last term of the series, (Art. 294,) and then the sum of the series, (Art. 291.) NOTE. - If the payments are to be made semi-annually, quarterly, &c., these periods will be the number of terms, and the interest of the annuity for each period the common difference. EXAMPLES FOR PRACTICE. 2. What will an annuity of $ 250 amount to in 6 years, at 6 per cent., simple interest? Ans. $ 1725. 3. What will an annuity of $ 380 amount to in 10 years, at 5 per cent., simple interest? Ans. $4655. 4. An annuity of $825 was settled on a gentleman, Jan. 1, 1840, to be paid annually. It was not paid until Jan. 1, 1S48; how much did he receive, allowing 6 per cent. simple interest? Ans. $ 7986. 5. A gentleman let a house for 3 years, at $ 200 a year, the rent to be paid semi-annually, at 8 per cent., per annum, simple interest. The rent, however, remained unpaid until the end of the three years; what did he then receive? Ans. $ 660. 6. A certain clergyman was to receive a salary of $ 700, to be paid annually, but for certain reasons, which we fear were not very good, his parishioners neglected to pay him for 8 years, but he agreed to settle with them and allow them $ 100, if they would pay him his just due with interest; required the sum received? Ans. $ 6676. 7. A certain gentleman in Boston has a very fine house, which he rents at $ 50 per month. Now, if his tenant shall omit payment until the end of the year, what sum should the owner receive, reckoning interest at 12 per cent.? Ans. $ 633. QUESTIONS. -ART. 296. What forms the first term of a progression in an annuity? What the common difference? What the number of terms? What the sum of the series? What is the rule for finding the amount of an annuity at simple interest? If the payments are made semi-annually, quarterly, &c., what constitute the terns? What the common difference? 286 GEOMETRICAL PROGRESSION. L[sacT. xa. ~ XL1. GEOMETRICAL PROGRESSION. ART. 297. WHEN there are three or more numbers, and the same quotient is obtained by dividing the second by the first, the third by the second, and the fourth by the third, &c., these numbers are in Geometrical Progression, and may be called a Geometrical Series. Thus, 2, 4, 8, 16, 32, 64, 64, 32, 16, 8, 4, 2. The former is called an ascending series, and the latter a descending series. In the first series the quotient is 2, and is called the ratio; in the second, it is 1. Hence, if the series is ascending, the quotient is more than unity; if it is descending, it is less than unity. The first and last terms of a series are called extremes and the other terms, means. Any three of the five following things being given, the other two may be found: — Ist. The first term, or first extreme; 2d. The last term, or last extreme; 3d. The number of terms; 4th. The ratio; 5th. The sum of the terms, or series. ART. So. One of the extremes, the ratio, and the number of terms being given, to find the other extreme. ILLUSTRATION.- Let the first term be 2, the ratio 3, and the number of terms 7. It is evident, that, if we multiply the first term by the ratio, the product will be the second term in the series; and, if we multiply the second term by the ratio, the product will be the third term; and, in this manner, we may carry the series to any desirable extent. By examining the following series, we find that 2, carried to the 7th term, is 145S; thlUS, tUESTIONS. - Art. 297~ When are numbers in geometrical progression? What is an ascending series? What a descending series'? What is the ratio of a progression? Is the ratio greater or less than unity in an ascenlding series? In a descending series? What are the extremes oi a series? What the means? What five things are mentioned, any three of which being given, the other two mray ie found!? rIT. XLI.J GEOMETRICAL PROGRESSION. 287 1 2 3 4, 5 6 7 2 6 1s 54 162 4S6 1458 The factors of 1458, are 3, 3, 3, 3, 3, 3, and 2, the last of which is the first term of the series, and the others the ratio repeated a number of times 1 less than the number of terms. But multiplying these factors together is the same as raising the ratio to the sixth power, and then multiplying that power by the first term. Hence the following RULE. - Raise the ratio to a power whose index is equal to the number of terms less one; then multiply this power by the first term, and th, product is the last term, or other extreme. NOTE..-This rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year, the ratio, the time, in years, one less than the number of terms, and the amount the last term. EXAMIPLES FOR PRACTICE. 1. The first term of a series is 1458, the number of terms 7, and the ratio I; what is the last term? Ans. 2 OPERATION, Ratio (~)6 y= Th; ad~ X 1 4 5 8- 14 -_ 2, the last term. 2. If the first term of a series is 4, the ratio 5, and the number of terms 7, what is the last term? Ans. 625C0. 3. If the first term of a series is 28672, the ratio I, and the number of terms 7, what is the last term? Ans. 7. 4. The first term of a series is 5, the ratio 4, and the number of terms is 8; required the last time. Ans. 81920. 5. If the first term of a series is 10, the ratio 20, and the number of terms 5, what is the last term? Ans. 1600000. 6. If the first term of a series is 30, the ratio 1.06, and the number of terms 6, what is the last term? Ans. 40.146767328. 7. What is the amount of $ 1728 for 5 years, at 6 per cent., compound interest? Ans. $ 2312.453798+. 8. What is the amount of $328.90, for 4 years, at 5 per cent., compound interest? Ans. $ 399.78+. 9. A gentleman purchased a lot of land containing 15 acres, agreeing to pay for the whole what the last acre would come OQUESTIONS. - Art. 298. What is the rule for finding the other extreme, one (of the extremes, the ratio, and number of terms being given? To what may this rule he applied? 288 GEOMETRICAL PROGRESSION. [SECT. XLI. to, reckoning 5 cents for the first acre, 15 cents for the second, and so on, in a three-fold ratio. What did the lot cost him? Ans. $ 23914S.45. ART. 299. To find the sum of all the terms, the first term the ratio, and the number of terms being given. ILLUSTRATION. — Let it be required to find the sum of the following series:2, 6, 18, 54. If we multiply each term of this series by the ratio 3, the products will be 6, 18S, 54, 162, forming a second series, whose sum is three times the sum of the first series; and the difference between these two series is twice the sum of the first series. Thus, 6, 18, 54, 162, the second series. 2, 6, 18, 54, the first series. 2, 0, 0, 0, 162-2 — 160, difference of the two series. Now, since this difference is twice the sum of the first series, one half this difference will be the sum of the first series; thus, 160 -.- 2 80. It will be observed, by examining the operation above, that, if we had simply multiplied 54, the last term of the first series, by the ratio 3, and subtracted 2, the first term, from it, we should have obtained 160; and this being divided by the ratio, 3 less 1, would have given 80, the same number as before, for the sum of the first series. Hence the RULE.- Find the last term as in the preceding article, multiply it by the ratio, and from the product subtract the first term. Then divide this remainder by the ratio less 1, and the quotient will be the sum of the series. Or, RULE II. — Raise the ratio to a power whose index is equal to the number of terms,from which subtract 1; divide the remainder by the ratio less 1, and the quotient, multiplied by the given extreme, will be the sum of the series required. NOTE 1. — If the ratio is less than a unit, the product of the last term multiplied by the ratio must be subtracted from the first term; and to obtain the divisor, the ratio must be subtracted from unity, or 1. QUESTIOrs. —Art. 299. What is the rule for finding the sum of all the terms, the first term, ratio, and number of terms being given? If the ratio is less than a unit, what must be done with the product of the last term multiplied by the ratio? How is the divisor obtained when the ratio is less than 1? SECT. XLI.1 GEOMETRICAL PROGRESSION. 289 NOTE 2. -If the second rule is employed, when the ratio is less than one, its power, denoted by the number of terms, must be subtracted from 1, and the remainder divided by the difference between 1 and the ratio. EXAMPLES FOR PRACTICE. 1. If the first term of a series is 12, the ratio 3, and the number of terms 8, what is the sum of the series? Ans. 39360. OPERATION. Ratio 37 X 12 — 26244, the last term; 26244 X 3 —- 78732; 78732 - 12 = 78720; 78720 + (3 - 1) = 39360, the sum of the series. 2. The first term of a series is 5, the ratio J, and the number of terms 6; required the sum of the series. Ans. 134w. OPERATION. Ratio (2)5 X 5-=-~, the last term; 1 X i = 0~; 5-2 -9 2%5; a o(1 2_ 5 _ = 136~, the sum of the series. 3. If the first term of a series is 8, the ratio 4, and the number of terms 7, required the sum of the series. Ans. 43688. 4. If the first term is 10, the ratio j, and the number of terms 5, what is the sum of the series? Ans. 30z6. 5. If the first term is 18, the ratio 1.06, and the number of terms 4, what is the sum of the series? Ans. 78.743+. 6. When the first-term is $ 144, the ratio $ 1.05, and the number of terms 5, what is the sum of the series? Ans. $ 795.6909. 7. D. Baldwin agreed to labor for E. Thayer for 6 months. For the first month he was to receive $3, and each succeeding month's wages were to be increased by l of his wages for the month next preceding; required the sum he received for his 6 months' labor. Ans. $ 9177. 8. A lady, wishing to purchase 10 yards of silk for a new dress, thought $ 1.00 per yard too high a price; she, however, agreed to give I cent for the first yard, 4 for the second, 16 for the third, and so on, in a four-fold ratio; what was the cost of the dress? Ans. $ 3495.25. 25 290 ANNUITIES AT COMPOUND INTEREST. [sECT. XL1. ANNUITIES AT COMPOUND INTEREST BY GEOMETRICAL PROGRESSION. ART. 00.0 WHEN compound interest is reckoned on an annuity in arrears, the annuity is said to be at compound interest; and the amounts of the several payments form a geometrical series, of which the annuity is the first term, the amount of $ 1.00 for one year the ratio, the years the number of terms, and the amount of the annuity, the sum of the series. Hence, ART. 301. To find the amount of an annuity at compound interest, we have the following RULE I. - Find the sum of the series by either of the preceding rules, (Art. 299.) Or, RULE II. - Multiply the amount of $ 1.00, for the given time, found in the table, by the annuity, and the product will be the required amount. TABLE, Showing the amount of $ 1 annuity fiom 1 year to 40. Yearn.. 5 per cent. 6 per cent. Years. 5 per cent. 6 per cent. 1 1.000000 1.000000 21 35.719252 39.992727 2 2.050000 2.060000 22 38.505214 43.392290 3 3.152500 3.183600 23 41.430475 46.995828 4 4.310125 4.374616 24 44.501999 50.815577 5 5.525631 6.637093 25 47.727099 54.864512 6 6.801913 6.975319 26 51.113454 59.156383 7 8.142008 8.393838 27 54.669126 63.705766 8 9.549109 9.897468 28 58.402583 68.528112 9 11.026564 11.491316 29 62.322712 73.639798 10 12.577893 13.180795 30 66.438847 79.058186 11 14.206787 14.971643 31 70.760790 84.801677 12 15.917127 16.869941 32 75.298829 90.889778 13 17.712983 18.882138 33 80.063771 97.343165 14 19.598632 21.015066 34 85.066959 104.183755 15 21.578564 23.275970 35 90.220307 111.434780 16 23.657492 25.672528 36 95.836323 119.120867 17 25.840366 28.212880 37 101.628139 127.268119 18 28.132385 30.905653 38 107.709546 135.904206 19 30.539004 33.759992 39 114.095023 145.055458 20 33.065954 36.785591 40 120.799774 154.761966. QUESTIONS. - Art. 300. When is an annuity said to be at compound interest? What do the amounts of the several payments form? What is the first term of the series? What the ratio? What the number of terms? What the sum of the series?- Art. 301. What is the first rule for finding the amnountof an lnnuity? What the second? Whllat does the table show? SECT. XTI.] ANNUITIES AT COMPOUND INTEREST. 291 EXAMPLES FOn PRACTICE. 1. What will an annuity of $ 378 amount to in 5 years, at 6 per cent. compound interest? Ans. $ 2130.821+. OPERATION BY RULE FIRST. 1.0 6 1 1.06 1 X378=$2130.821+. OPERATION BY RULE SECOND. 5.637093 X 3 78=$2 13 0.82 1+-. 2. What will an annuity of 6 1728 amount to in 4 years, at 5 per cent. compound interest? Ans. $7447.89,6-+. 3. What will an annuity of $ 87 amount to in 7 years, at 6 per cent. compound interest? Ans. $ 730.26,3+-. 4. What will an annuity of $ 500 amount to in 6 years, at 6 per cent. compound interest? Ans. $ 3487.65,9+. 5. What will an annuity of $ 96 amount to in 10 years at 6 per cent. compound interest? Ans. $ 1265.35,6+. 6. What will an annuity of $ 1000 amount to in 3 years at 6 per cent. compound interest? Ans. 8 3183.60. 7. July 4, 1842, H. Piper deposited in an annuity office, for his daughter, the sum of $ 56, and continued his deposites each year, until July 4, 1848. Required the sum in the office July 4, 1848, allowing 6.per cent. compound interest. Ans. $ 470.05,4+-t 8. C. Greenleaf has two sons, Samuel and William. On Samuel's birth-day, when he was 15 years old, he deposited for him, in an annuity office, which paid 5 per cent. compound interest, the sum of 825, and this he continued yearly, until he was 21 years of age. On William's birth-day, when he was 12 years old, he deposited for him, in an office which paid 6 per cent. compound interest, the sum of $20, and continued this until he was 21 years of age. Which will receive the larger sum, when 21 years of age? Ans. $ 60.06,5+ William receives more than Samuel. 9. I gave my daughter Lydia, $ 10, when she was 8 years old, and the same sum on her birth-day each year, until she was 21 years old. This sum was deposited in the savings bank, which pays 5 per cent. annually. Now, supposing each deposite to remain on interest until she is 21 years of age, required the amount in the bank. Ans. $ 195.98,6+. 292 ALLIGATION. LIeCT. XLI. ~ XLII. ALLIGATION. ART. 30o. ALLIGATION signifies the act of tying together, and is a rule employed in the solution of questions relating to the mixture of several ingredients of different prices- or qualities. It is of two kinds, Alligation Medial and Alligation Alternate ALLIGATION MEDIAL. ART. 3e3. Alligatzon Miedial is the method of finding the mean price of a mixture composed of several articles or ingredients, the quantity and price of each being given. ART. 3O4. To find the mean price of several articles or ingredients, at different prices, or of different qualities. RULE. — Find the value of each of the ingredients, and divide the amount of their values by the sum of the ingredients; the quotient will be the price of the mixture. EXAMPLES FOR PRACTICE. 1. A grocer mixed 201b. of tea worth $ 0.50 a pound, with 301b. worth $ 0.75 a pound, and 501b. worth $ 0.45 a pound; what is 1 pound of the mixture worth? Ans. $ 0.55. OPERATION. $ 0.5 0 X 2 0 — $1 0.0 0 $0.75 X 30==$22.50 $0.45 X 50= $22.50 Sum of ingredients, 1 0 0 $ 5 5.0 0, value. Then, $ 5 5.0 0- 1 0 0 $ 0.5 5 per pound. Proof, $ 0.5 5 X 2 0 lb. $ 1 1.00 ) $0.55 X 301b.=$16.50 =$ 5 5.00. $0.55 5 01b. —$27.50 ) 2. I have four kinds of molasses, and a different quantity of each, as follows: 30 gal. at 20 cents; 40 gal. at 25 cents; 70 gal. at 30 cents, and S0 gal. at 40 cents; what is a gallon of the mixture worth? Ans. $ 0.31 T. 3. A farmer mixed 4 bush, of oats at 40 cents; 8 bush. of QUESTIONS. - Art. 302. What is alligation? What two kinds are there? Art. 303. What is alligation medial? - Art. 304. What is the rule for finding the mean price of several articles at different prices? How does it appear that this process will give the mean price of the mixture? SECT. XLIi.J ALLIGATION ALTERNATE. 293 corn at 85 cents, 12 bush. rye at $ 1.00; and 10 bush. of wheat at $ 1.50 per bushel. What will one bushel of the mixture be worth? Ans. $ 1.04-1?r. 4. I wish to mix 301b. of sugar at 10 cents a pound; 251b. at 12 cents; 41b. at 15 cents; and 501b. at 20 cents; what is 1 pound of the mixture worth? Ans. $ 0.15-2 r. ALLIGATION ALTERNATE. ART. 3@5. Alligation Alternate is the method of finding what quantity of two or more ingredients or articles, whose prices or qualities are given, must be taken, to coimpose a mixture of any given price or quality. ART. 3 4G. To find what quantity of each ingredient must be taken to form a mixture of a given price. Ex. 1. I wish to mix two kinds of spice, one at 21 cents, and the other 26 cents per pound, so that the mixture may be worth 24 cents per pound. How many pounds of each must I take? Ans. 21b. at 21 cents; 31b. at 26 cents. OPEIRATION. VrWe connect the price that is O21ER 2 TO less than the mean price, with Mean price 24 ]26 Ans. the price which is greater, and 26- 3 ) set the difference between each price and the mean price opposite the price with which it is connected; these numbers denote the quantity of each ingredient to be taken. It will be seen that the value of 1lb. of the spice at 2lets. is 3 cents less than of Ilb. of the mixture at the mean price, 24 cents, and that the value of lib. at 26 cents is 2 cents more than the mean price. Now if one of these prices were as much greater than the mean price as the other is less, the differences would balance each other, and the mixture of the two in equal quantities would be 24 cents per lb., the given mean price. But since the deficiency is more than the excess, we must take more pounds at 26 cents than at 21 cents per lb., to balance the deficiency. If we multiply the 2 cents by some number, as 3, and the 3 cents by some number, as 2, the product of the excess will be just equal to the product of the deficiency; and 31b. at 26 cents and 21b. at 21 cents per pound, will form a mixture of 51b., worth QUESTIONS. - Art. 3G5. What is alligation alternate?- Art. 306. How do you (connect the prices? Where do you set the differences between each price and the mean price? What do these differences denote? How does it ap)pear, front the explanation, that the differences denote the quantity of each kind to be taken? 25 294 ALLIGATION ALTERNATE. [SECT. XLI1. $ 1.20, of which lib. will be worth 24 cents, the price of the required mixture. Therefore, we must take 31b. at 26 cents, and 2lb. at 21 cents, to make a mixture worth 24 cents per pound, which is the same result as was obtained in the operation. Hence the RULE. - 1. Place the prices of the ingredients under each other, in the order of their value, and connect the price of each ingredient which is less in value than the price of the mixture, with one that is greater. 2. Then place the difference between the price of the mixture, and that of each of the ingredients, opposite to the price with which it is connected, and the number set opposite to each price is the quantity of the ingredient to be taken at that price. NOTE. -There will be as many different answers as there are different ways of connecting the prices, and by multiplying and dividing these answers they may be varied indefinitely. EXAMPLES rFR PmRACTICE. 2. A farmer wishes to mix corn at 75 cents a bushel, with rye at 60 cents a bushel, and oats at 40 cents a bushel, and wheat at 95 cents a bushel; what quantity of each must be take to make a mixture worth 70 cents a bushel? FIRST OPERATION. SECOND OPERATION.. THIRD OPERATION. Ans. Ans. f40 - 5 402-l 25t 5 30 1 60-1 60 25 K 60 5 5=5 30 0 751 10 75J.530 752, 10 30 40 95 30 95- 10 9)521 30 + 10 40 3. I have 4 kinds of salt worth 25, 30, 40, and 50 cents per bushel; how much of each kind must be taken, that a mixture might be sold at 42 cents per bushel? Ans. 8 bushels at 25, 30, and 40 cents, and 31 bushels at 50 cents. 4. My swamp hay is worth 8 12 per ton, my salt hay $ 15, and my English hay $ 20; how much of each kind must be taken, that a ton may be sold at $ 18? Ans. 2 tons of swamp hay, 2 tons of salt hay, and 9 tons of English hay. ART. g@7. When the quantity of one ingredient is given to find the quantity of each of the others. QUESTIONs. - What is the rule for alligation alternate? How can you obtain different answers? Are they all true 7 SECT. XLII.] ALLIGATION ALTERNATE. 295 Ex. 1. How much sugar, that is worth 6, 10, and 13 cents a pound, must be mixed with 201b. worth 15 cents a pound, st, that the mixture will be worth 11 cents a pound? OPERATION. 6-f 4 101 2 Then, 5:1: 20: 4 l13 1 G 5:5:2:: 20: 8 Ans. 15- 5 5:4:: 20: 16 By the conditions of the question we are to take 201b. at 15 cents a pound; but by the operation we find the difference at 15 cents a pound to be only 51b., which is but ~ of the given quantity. Therefore, ii we increase the 51b. to 20, the other differences must be increased in the same proportion. Hence the propriety of the following RULE.. Take the difference between each price and the mean price, as before; then say, as the difference of that ingredient whose quantity is given is to each of the differences separately, so is the quantity given to the several quantities required. EXAMPLES FOR PRACTICE. 2. A farmer has oats at 50 cents per bushel, peas at 60 cents, and beans at $1.50. These he wishes to mix with 30 bushels of corn at $ 1.70 per bushel, that he may sell the whole at $ 1.25 per bushel; how much of each kind must he take? Ans. 18 bushels of oats, 10 bushels of peas, and 26 bushels of beans. 3. A merchant has two kinds of sugar, one of which cost him 10 cents per lb., and the other 12 cents per lb.; he has also 100lb. of an excellent quality which cost him 15 cents per lb. Now, as he ought to make 25 per cent. on his cost, how much of each quantity must be taken that he may sell the mixture at 14 cents per lb. Ans. 3S311lb. at 10 cents, and 10Olb. at 12 cents. ART. 3@S. When the sum of the ingredients and their mean price are given, to find what quantity of each must be taken. Ex. I. I have teas at 25 cents, 35 cents, 50 cents, and 70 QUESTION. - Art. 307. What is the rule for finding the quantity of each of the other ingredients when one is given? 296 ALLIGATION ALTERNATE. [SECT. XLIIo cents a pound, with which I wish to make a mixture of IS01b., that will be worth 45 cents a pound. How much of each kind must I take? OPERATION. 25 25 Then, 60:25::180: 75 l 35 5 60: 5::180:15 Ans. 45 50 10 60 10:18030 70 20 60: 20: 180 60 Sum of differences, 60 Proof, 180 By the conditions of the question, the weight of the mixture is 1801b.. out by the operation we find the sum of the differences to be only 601b., which is but ~ of the quantity required. Therefore, if we increase 0Glb. to 180, each of the differences must be increased in the same proportion, in order to make a mixture of 1801b., the quantity required. Hence the RULE. - Find the differences as before; then say, as the sum of the differences is to each of the differences separately, so is the given quantity to the required quantity of each ingredient. EXAMPLE S FOR PRACTICE. 2. John Smith's "great box" will hold 100 bushels. He has wheat worth $ 2.50 per bushel, and rye worth $ 2.00 per bushel. How much chaff of no value must he mix with the wheat and rye, that if he fill the box, a bushel of the mixture may be sold at $ 1.80? Ans. 40 bushels of wheat and rye, and 20 bushels of chaff. 3. I have two kinds of molasses, which cost me 20 and 30 cents per gallon; I wish to fill a hogshead, that will hold SO gallons, with these two kinds. How much of each kind must be taken, that I may sell a gallon of the mixture at 25 cents per gallon and make 10 per cent. on my purchase? Ans. 58-zT of 20 cents, and 21T7- of 30 cents. 4. A lumber merchant has several qualities of boards; and it is required to ascertain how many, at $ 10 and $ 15 per thousand feet, each, shall be sold on an order for 60 thousand feet, that the price for both qualities shall be $ 12 per thousand feet. Ans. 36 thousand at $ 10, and 24 thousand at $ 15. QUESTION. - Art. 308. How do you find what quantity of each ingredient must be taken when the sum and mean price are given 7 SECT. XLIII. PERMUTATION. 297 ~ XLIII. PERMUTATION. ART. 309. PERMUTATTON is the, method of finding how many different changes or arrangements may be made of any given number of things. ART. 31o. To find the number of different arrangements that can be made of any given number of things. Ex. 1. How many different numbers may be formed from the figures of the following number, 432, making use of three figures in each number. Ans. 6. FIRST OPERATION. In the 1st operation, we have made all the X432, 423, 342, 324-, 243, 234. different arrangements SECOND OPERATION. that can be made of the 1 X 2 X 3_=6. given figures, and find the number to be 6. In the second operation, the same result is obtained by simply multiplying together the first three of the digits, a number equal to the number of figures to be arranged. Hence the RULE. — Multiply all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product will be the answer required. EXAMPLES FOR PRACTICE. 2. My family consists of nine persons, and each person has his particular seat around my table. Now, if their situations were to be changed once each day, for how many days could they be seated in a different position? Ans. 362880 days, or 994 years 70 days. 3. On a certain shelf in my library there are 12 books. If a person should remove them without noticing their order, what would be the probability of his replacing them in the same position they were at first? Ans. 1 to 479001600. 4. How many words can be made from the letters in the word "Embargo," provided that any arrangement of them may be used, and that all the letters shall be taken each time? Ans. 5040 words. QUESTIONS. -Art. 309. What is permutation? — Art. 310. What is the rule for finding the number of arrangements that can be made of any given number of things? 298 MENSURATION OF SURFACES. [SECT. XLIV. ~ XLIV. MENSURATION OF SURFACES. ART. SI 1. A SURFACE is a magnitude, which has lengtb and breadth without thickness. The surface or superficial contents of a figure, are called its area. ART. 3 It An ANGLE is the inclination or opening of two lines, which meet in a point. I A right angle is an angle formed by one line falling perpendicularly on another, and it contains 90 degrees. An acute angle is an angle less than a right angle, or less than 90 degrees. An obtuse angle is an angle greater than a right angle, or more than 90 degrees. THE TRIANGLE. ART. 313. A TRIANGLE is a figure having three sides and three angles. It receives the particular names of an equilateral triangle, isosceles triangle, and scalene triangle. It is also called a right angled triangle when it has one right angle; an acute angled triangle, when it has all its angles acute; and an obtuse angled triangle, when it has one obtuse angle. The base of a triangle, or other plane figure, is the lowest side, or that which is parallel to the horizon; as, C D. The altitude of a triangle is a line drawn from one of its angles perpendicular to its opposite side or base; as, A B. A An equilateral triangle is a, figure which has its three sides equal. c: D B 1QUESTIONS. - Art. 311. What is a surface? What are the superficial contents of a figure called? - Art. 312. What is an angle? What is a right angle? An acute angle? An obtuse angle? - Art. 313. What is a triangle? What particular names does it receive 1 When is it called a right angled triangle? When an acute angled triangle? When an obtuse angled triangle? What is the base of a triangle? What the altitude? What is an equilateral triangle? SECT. XLIV.] MENSURATION OF SURFACES. 299 n o is eA An isosceles triangle is a figure which has two of its sides equal. A scalene triangle is a figure which has its three sides unequal. A right angled triangle is a figure having three sides and three angles, one of which is a right angle. ART. 3141. To find the area of a triangle. RULE I. —Multiply the base by hay the altitude, and the product will be the area. Or, RULE II. —Add the three sides together, take half that sum, and from this subtract each side separately; then multiply the half of the sum and these remainders together, and the square root of this product will be the area. 1. What are the contents of a triangle, whose base is 24 feet, and whose perpendicular height is 18 feet? Ans. 216 feet. 2. What are the contents of a triangular piece of land, whose sides are 50 rods, 60 rods, and 70 rods? Ans. 1469.69 +-rods. TiHE QUADRILATERAL. ART. 315. A QUADRILATERAL is a figure having four sides, and consequently four angles. It comprehends the rectangle, square, rhombus, rhomboid, trapezium, and trapezoid. ART. 310. A PARALLELOGRAM is any quadrilateral whose opposite sides are parallel. It takes the particular names of rectangle, square, rhombus, and rhomboid. The altitude of a parallelogram is a perpendicular line drawn between its opposite sides; as C D in the rhomboid. QUESTIONS. - What is an isosceles triangle? A scalene triangle? A right angled triangle? —Art. 314. What is the first rule for finding the area of a triangle? What the second?-Art. 315. What is a quadrilateral? What figures does it comprehend? -Art. 316. What is a parallelogram? What particular names does it take? What is the altitude of a parallelogram? 300 MEZNSURATION OF SURFACES. [sBCT. XLIV. A rectangle is a right angled parallelogram, whose opposite sides are equal. A square is a parallelogram, having foun equal sides and four right angles. C A rhomboid is an oblique angled parallelo gram, whose opposite sides are equal. D A rhombus is an oblique angled parallelogram, having all its sides equal. ART. 317. To find the area of a parallelogram. RULE. - Multiply the base by the altitude, and the product will be the area. 1. What are the contents of a board 25 feet long and 3 feet wide? Ans. 75 square feet. 2. What is the difference between the contents of two floors; one is 37 feet long and 27 feet wide, and the other is 40 feet long and 20 feet wide? Ans. 199 square feet. 3. The base of a rhombus is 15 feet, and its perpendicular height is 12 feet; what are its contents? Ans. 180 square feet. ART. 31 S. A TRAPEZOID is a quadrilateral, ~/ —-~ which has only one pair of its opposite sides ~/ I parallel. ART. 319. To find the area of a trapezoid. RULE.- Multiply half the sum of the parallel sides by the altitude, and the product is the area. 1. What is the area of a trapezoid, the longer parallel side QuEsTrIoss. - What is a rectangle 7 A square? A rhomboid? A rhombus? -Art. 317. What is the rule for finding the area of a parallelogram? — Art. 318. What is a trapezoid 7- Art. 319, What is the rule for finding the ares of a trapezo;d? SECT. XLIV.] MENSURATION OF SURFACES. 301 being 482 feet, the shorter 324 feet, and the altitude 216 feet? Ans. 87048 square feet. 2. What is the area of a plank, whose length is 22 feet, the width of the wider end being 28 inches, and of the narrower 20 inches? Ans. 44 square feet. E>s -1 ART. 32. A TRAPEZIUM is a quadrilateral / which has neither two of its opposite sides F parallel. A diagonal is a line joining any two opposite angles of a quadrilateral; as, E F. ART. 321E. To find the area of a trapezium. RULE. - Divide the trapezium into two triangles by a diagonal, and then find the areas of these triangles; their sum will be the area of the trapezium. 1. What is the area of a trapezium, whose diagonal is 65 feet, and the length of the perpendiculars let fall upon it are 14 and 18 feet? Ans. 1040 square feet. 2. What is the area of a trapezium, whose diagonal is 125 rods, and the length of the perpendiculars let fall upon it are 70 and 85 rods? Ans. 9687.5 square rods. THE POLYGON. ART. 32.O A POLYGON is any rectilineal figure having more than four sides and four angles. It takes the particular names of pentagon, which is a polygon of five sides; hexagon, one of six sides; heptagon, one of seven sides; octagon, one of eight sides; nonagon, one of nine sides; decagon, one of ten sides; undecagon, one of eleven sides; and dodecagon, one of twelve sides. ART. $23. A REGULAR POLYGON is a plane rectilineal figure, which has all its sides and all its angles equal. \,/ The perimeter of a polygon is the sum of all its sides. ART. 324. To find the area of a regular polygon. QUESTIONS. - Art. 320. What is a trapezium? What is a diagonal?Art. 321. What is the rule for finding the area of-a trapezium?-Art. 322. What is a polygon? What particular names does it take? - Art. 323. What is a regular polygon? 26 302 MENSURATION OF SURFACES. [SECT. XLIV. RULE. —Multiply the perimeter by half the perpendicular letfall from the centre on one of its sides, and the product will be the area. 1. What is the area of a regular pentagon, whose sides are each 35 feet, and the perpendicular 24.08 feet? Ans. 2107 square feet. 2. What is the area of a regular hexagon, whose sides are each 20 feet, and the perpendicular 17.32 feet? Ans. 1039.20 square feet. THE CIRCLE. ART. 32t.o A CIRCLE is a plane figure bound ed by a curved line, every part of which is CG l H equally distant from a point, called its centre. The circumference or periphery of a circle is the line which bounds it. The diameter of a circle is a line drawn through the centre, and terminated by the circumference; as, G H. ART. 326. To find the circumference of a circle, the diameter being given. RULE. -Multiply the diameter by 3.141592, and the product is the circumference. 1. What is the circumference of a circle, whose diameter is 50 feet? Ans. 157.0796+ feet. 2. A gentleman has a circular garden whose diameter is 100 rods; what is the length of the fence necessary to enclose it? Ans. 314.15+ rods. ART. 327. To find the diameter of a circle, the circumference being given. RULE. - Multiply the circumference by.318309, and the product will be the diameter. 1. What is the diameter of a circle, whose circumference Is 80 miles? Ans. 25.46- miles. 2. If the circumference of a wheel is 62.84 feet, what is the diameter.? Ans. 20+ feet. QUESTIONS. - What is the perimeter of a polygon? - Art. 324. What is the rule for finding the area of a regular polygon? - Art. 325. What is a circle? What is the circumference of a circle? The diameter of a circle?- Art. 326. What is the rule for finding the circumference of a circle, the diameter being given? - Art. 327. What is the rule for finding the diameter of a circle, the circumference being given? SECT. XLIV.1 MENSURATION OF SURFACES. 303 ART..2S. To find the area of a circle, the diameter, the circumference, or both, being given. RULE I. - Multiply the square of the diameter by.785398, and the product is the area. RULE II.- Multiply the square of the circumference by.079577, and the product is the area. RULE III. -Multiply half the diameter by half the circumference, and the product is the area. 1. If the diameter of a circle be 200 feet, what is the area? Ans. 31415.92 square feet. 2. There is a certain farm, in the form of a circle, whose circumference is 400 rods; how many acres does it contain? Ans. 79A. 2R. 12p. - ART. 329o To find the side of a square, equal in area to a given circle. The square in the figure is supposed to have the same area as the circle. RULE I. - M1fultiply the diameter by.886227, and the product is the side of an equal square. Or, RULE II.- Multiply the circumference by.282094, and the product is the side of an equal square. 1. We have a round field 40 rods in diameter; what is the side of a square field, that will contain the same quantity? Ans. 35.44+ rods. 2. I have a circular field 100 rods in circumference; what must be the side of a square field, that shall contain the same area? Ans. 2S.2+- rods. ART. 330. To find the side of a square, inscribed in a given circle. A square is said to be inscribed in a circle ) when each of its angles touches the circumference or periphery of the circle. QUESTIONS. -Art. 328. What is the rule for finding the area of a circle wVhen the diameter is given? When the circumference is given? When the diameter and circumference are both given?- Art. 329. What is the first rule for finding the side of a square, equal in area to a given circle? WlWhat the qecond? - Art. 330. When is a square said to be inscribed in a circle. What is the first rule for finding the side of a square inscribed in a circle The second? 304 MENSURATION OF' SOLIDS. [SECT. XLV. RULE i. - MIultiply the diameter by.707106, and the product is the side of the square inscribed. Or, RULE II. - Multiply the circumference by.225079, and the product is the side of the square inscribed. 1. What is the thickness of a square stick of timber that may be hewn from a log 30 inches in diameter? Ans. 21.21+ inches. 2. How large a square field may be inscribed in a circle, whose circumference is 100 rods? Ans. 22.5+ rods square..THrE ELLIPSE. ART. 331. An ELLIPSE is an oval figure having two diameters or axes, the longer of which is called the transverse, and the shorter the conjugate diameter. ART. 33. To- find the area of an ellipse. RULE. —Multiply the two diameters together and their product by.785398; the last product is the area. 1. What is the area of an ellipse, whose transverse diameter is 14 inches, and its conjugate diameter 10 inches? -Ans. 109.95+ square inches. 2. What is the area of an elliptical table, 8 feet long and 5 feet wide? Ans. 31 square feet 59+ square inches. ~ XLV. MENSURATION OF SOLIDS. ART. 3. A SOLID is a magnitude which has length, breadth and thickness. Mensuration of solids includes two operations; first, to find their superficial contents, and, second, their solidities. THE PRISm. ART..4 A PRISM is a solid whose ends are any plane figures which are equal and similar, and whose sides are parallelograms. It takes particular names, according to the figure QUESTIONs. -Art. 331. What is an ellipse? What is the longer diameter called'? The shorter?- Art. 332. What is the rule for finding the area of an ellipse?-Art. 333. What is a solid? What two operations does mensuration of solids include? - Art. 334. What is a prism? What particular names does it take? 4SCT. XLV.j MENSURATION OF SOLIDS. 305 of its base or ends, viz., triangular prism, square prism, pentagonal nPrism, &c. The base of a prism is either end; and of solids in general, the part upon which they are supposed to stand. All prisms whose bases are parallelograms, are comprehended under the general name parallelopipedons or parallelopi-.eds. I A triangular prism is a solid whose base is a triangle. A sqzare prism is a solid whose base is a square, and when all the sides are squares it is called a cube. j A pentagonal prism is a solid whose base is a pentagon. ART. 33r. To find the surface of a prism. RULE: - Multiply the perimeter of its base by its height, and to this product add the area of the two ends; the sum is the area of the prism. 1. What are the superficial contents of a triangular prism, the width of whose side is 3 -feet, and its lengrth 15 feet? Ans. 142.79+- square feet. 2. What is the surface of a square prism, whose side is 9 net wide, and its length 25 feet? Ans. 1062 square feet. ART. 336. To find the solidity of a prism. RULE. - Multiply the area of the base by the height, and the prolzduct the solidity. QIuESTIONs. - What is the base of a prism and of solids in general? QWhat is a parallelopiped or parallelopipedlon? What is a trialgular prism? s sillt;are prisn? A pentagonal prism? - Art. 335. What is the rule for tind4g the surface of a prism? -Art. 336. What is the rule for finding the solid-,ty of a prism? 3f - 306 MENSURATION OF SOLIDS. [SECT.'.'V 1. What are the contents of a triangular prism, whose length is 20 feet, and the three sides of its triangular end or base 5, 4, and 3 feet? Ans. 120 cubic feet. 2. How many cubic feet are there in a cube, whose sides are 8 feet? Ans. 512 cubic feet. 3. What is the number of cubic feet in a room 30 feet long, 20 feet wide, and 10 feet high? Ans. 6000 cubic feet THE CYLINDER. ~=:-; ART..37. A CYLINDER is a long, circular solid of uniform diameter, and its extremities form equal parallel circles. Thie axis of a cylinder is' a straight line drawn I lthrough it from the centre of one end to the centre E ihlL of the other. ART. 33S. To find the surface of a cylinder. RULE. - Multiply the circumference of the base by the altitude, and to the product add the areas of the two ends; the sum will be the whole sulface. 1. What is the surface of a cylinder, whose length is 4 feet, and the circumference 3 feet? Ans. 13.43+ square feet. 2. John Snow has a roller 12 feet long and 2 feet in diameter; what is its convex surface? Ans. 75.39- square feet. ART. 339. To find the solidity of a cylinder. RULE. - Multiply the area of the base by the altitude, and the product will be the solidity. 1. What is the solidity of a cylinder, 8 feet in length and 2 feet in diameter? Ans. 25.13t- cubic feet. 2. What is the solidity of a cylinder, whose diameter is 5 feet and its altitude 20 feet? Ans. 392.69+ cubic feet. TEE PYRAMID AND CONE. ART. $4 A PYRA1MIID is a solid, standing on a tri\ angular,square,or polygonal base, and its sides are triangles, meeting in a point at the top, called the vertex The slant height of -a pyramid is a line drawn from iI! the vertex to the middle of one of the sides of the base. QUESTIOS. - Art. 337. What is a cylinder? What is the axis of a cylinder? - Art. 338. What is the rule for finding the surface of a cylinder? - Art. 339. What is the rule for finding the solidity of a cylinder? - Art. 340. What is a pyrarrnil? 7 hat is the slant height of a pyramid? BECT. XLV.] MENSURATION OF SOLIDS. 307 A ART. 3$1. A CONE is a solid, having a circle for its base, and its top terminated in a point, called the 7viI vertex. The altitude of a pyramid and of a cone, is a line drawn from the vertex perpendicular to the plane of A the base; as, B C. The slant height of a cone is a line drawn from the vertex to the circumference of the base; as, A C. ART. 34,o To find the surface of a pyramid and of a cone. RULE. - Multiply the perimeter or the circumference of the base by half its slant height, and the product is the surface. 1. How many yards of cloth that is 27 inches wide, will it require to cover the sides of a pyramid whose slant height is 100 feet, and whose perimeter at the base is 54 feet? Ans. 400 yards. 2. Required the convex surface of a cone, whose slant height is 50 feet, and the circumference at its base 12 feet? Ans. 300 square feet. ART. 343e To find the solidity of a pyramid and of a cone. RULE. - Multiply the area of the base by one third of its altitude, and the product will be its solidity. 1. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. Its height is 500 feet. Now, supposing it to come to a point at its vertex, what are its solid contents, and how many miles in length of wall would it makle, 4 feet in height and 2 feet thick? Ans. 80,041,500 cubic feet; 1S94.9 miles in length. 2. What are the solid contents of a cone, whose height is 30 feet and the diameter of its base 5 feet? Ans. 196.3+ feet. ~\\k\ ART. 44*. A FRUSTUM OF A PYRAMID is the part next to the base, that remains after cutting off the top, by a plane parallel to the base. "UE:STIoSS. — Art. 341. What is a cone? What is the altitude of a pyrani-l and of a cone? What is the slant height of a cone? -Art. 342. WlWhat is the rule for findlig the surface of a pyramid and of a cone? -Art. 343. What is the rule for finding the solidity of a pyramid and of a cone 7 - Art. 344. Whlat is the frustum of a pyramidl 308 MENSURATION OF SOLIDS. [SECT. XLY. A ART. 345. A FRUSTUlM OF A CONE is the part next g X to the base, that remains after cutting off the top, by a plane parallel to the base. ART. 34Lg To find the surface of a frustum of a pyramid or of a cone. RULE. -Add the perimeters or the circumferences of the two ends together, and multiply this sum by half the slant height. Then add the areas of the two ends to this product, and their sum will be the surface. 1. There is a square pyramid, whose top is broken off 20 feet slant height from the base. The length of each side at the base is 8 feet, and at the top 4 feet; what is its whole surface? Ans. 560 square feet. 2. There is a frustum of a cone whose slant height is 12 feet, the circumference of the base 18 feet, and that of the upper end 9 feet; what is its whole surface? Ans. 194.22+ squarr feet. ART. 347. To find the solidity of a. frustumn of a pyramid or of a cone. RULE. - Find the area of the two bases of the frustum; multiply these tlwo areas together, and extract the square root of the product. To this root add the two areas, and multiply their sum by one third of the altitude of the frustum; the product will be the solidity. 1. What is the solidity of the frustum of a square pyramid, whose height is 30 feet, and whose side at the bottom is 20 feet, and at the top 10 feet? Ans. 7000 cubic feet. 2. What are the contents of a stick of timber 20 feet long, and the diameter at the larger end 12 inches, and at the smaller end 6 inches? Ans. 9.162+ feet. THE SPHERE. ART. 348. A SPHERE is a solid, bounded by one continued convex surface, every part of which is equally distant from a point within, called the centre. The axis or diameter of a sphere is a line passing through the centre, and terminated by the surface. QUESTIONs. - Art. 345. What is a frustum of a cone? -Art. 346. What is the rule for finding the surface of a frustum of a pyramid or of a cone? — Art. 347. What is the rule for finding the solidity of a frustum of a pyramid or of a cone? - Art. 348. What is a sphere? What is the diameter or axis f'a sphro't SECT. XLV.] MENSURATION OF SOLIDS. 309 ART. 349. To find the surface of a sphere. RULE..- Multiply the diameter by the circumference, and the product will be the surface. 1. What is the convex surface of a globe, whose diameter is 20 inches? Ans. 1256.6+ square inches. 2. If the diameter of the earth is s000 miles, what is its convex surface? Ans. 20106188 square miles. ART. 50. To find the solidity of a sphere. RULE. - Multiply the cube of the diameter by.523598, and the pro duct is the solidity. 1. What is the solidity of a sphere, whose diameter is 20 inches? Ans. 418.7+ inches. 2. If the diameter of a globe or sphere is 5 feet, how many cubic feet does it contain? Ans. 65.44+ cubic feet. ART. 51,. To find how large a cube may be cut from any given sphere, or be inscribed in it. RULE. - Square the diameter of the sphere, divide -the product by 3, and extract the square root of the quotientfor the answer. 1. How large a cube may be inscribed in a sphere 10 inches in diameter? Ans. 5.773+ inches. 2. What is the side of a cube that may be cut from a sphere 30 inches in diameter? Ans. 17.32+ feet. THE SPHEROID. ART. 93*t. A SPHEROID is a solid, generated by the revolution of an ellipse about one of its diameters. If the ellipse revolves about its longer or translverse diameter, the spheroid is prolate or oblong; if about its storter or conjugate diameter, the spheroid is oblate or flattened. ART. 353. To find the solidity of a spheroid. RULE. - 1. Multiply the square of the shorter axis by the longer axis, eznd this product by.523598, if the spheroid is prolate, and the product wJill he its solidity. QU ESTIONS.- Art. 349. What is the rule for finding the surface of a sphere? - Art. 350. What is the rule for finding the solidity of a sphere? -- Art. 351. What is the rule for finding how large a cube canl be cut from a given sphere?-. Art. 352. What is a spheroid? What is a prolate spheroid? What an oblate spheroid?7-Art. 353. What is the rule for finding the solidity of a spheroid? 310 MENSURATION OF LUMBER, ETC. [sECT. XLVT. 2. If it is oblate, multiply the square of the longer axis by the shorter axis, and this product by.523598; the last product will be the solidity. 1. What is the solidity of a prolate spheroid, whose transverse axis is 30 feet, and the conjugate axis 20 feet? Ans. 6283.17+ cubic feet. 2. What is the solidity of an oblate spheroid, whose axes are 30 and 10 feet? Ans. 4712.38+ cubic feet. ~ XLVI. MENSURATION OF LUMBER AND TIMBER. ART. 21.s All rectangular and square lumber and timber, as planks, joists, beams, &c., are usually surveyed by board measure, the board being considered to be one inch in thickness. Round timber is sometimes measured by the ton, and sometimes by board measure. ART. 5ca. To find the contents of a board. RULE. - Multiply the length of the board, taken in feet, by its breadth taken in inches, and divide this product by 12; the quotient is the corntents in squarefeet. 1. What are the contents of a board 18 inches wide and 16 feet long? Ans. 24 feet. 2. What are the contents of a board 24 feet long and 30 inches wide? Ans. 60 feet. ART. 356. To find the contents of joists, beams, &c. RULE. - Multiply the depth, taken in inches, by the thickness, and this product by the length, infeet; divide the last product by 12, and the quotient is the contents in feet. 1i. What are the solid contents of a joist 4 inches wide, three inches thick, and 12 feet long? Ans. 12 feet. 2. What are the contents of a square stick of timber 25 feet long and ten inches thick? Ans. 2081 feet. ART. 57t. To find the contents of round timber. RULE. - Multiply the length of the stick, taken in feet, by the square 9f one fourth the girt, taken in inches; divide this product by 144, and the quotient is the contents in cubic feet. QUESTT.ON. - Art. 354. By what measure are planks, joists, &c., usually surveyed? What is the usual thickness of a board? How is round timber measured? - Art. 355. What is the rule for finding the contents of a board? - Art. 356. What is the rule for finding the contents of joists, &c.? SECTrr. XLVII.] MISCELLANEOUS QUESTIONS. 311 NOT'E 1. -The girt is usually taken about one third the distance from the larger to the smaller end. NOTE 2. - A ton of timber, estimated by this method, contains 501-9 cubic feet. 1. How many cubic feet of timber in a stick, whose length is 50 feet and whose girt is 60 inches? Ans. 781 cubic feet. 2. What are the contents of a stick whose length is 30 feet and girt 30 inches? Ans. 11.7+ solid feet. ~ XLVII. MISCELLANEOUS QUESTIONS. 1. WHAT is the difference between 7 pence and 10 cents? Ans..d. 2. What number is that, to which fif A be added, the sum will be'7.? Ans. 7 3 3. What number is that, from which if 32 be taken, the remainder will be 4 2. Ans. F7A. 4. What number is that, to which if 32 be added, and the,,,sum divided by 53, the quotient will be 5? Ans. 236. 5. From v of a mile take 7 of a furlong. Ans. 4fur. 12rd. Sft. Sin. 6. From 7 acres take 9:T of a rood. Ans. 6A. 3R. 7p. 74ft. 36in. 7. John Swift can travel 7 miles in - of an hour, but Thomas Slow can travel only 5 miles in -7: of an hour. Both started from Danvers at the same time for Boston, the distance being 12 miles. How much sooner will Swift arrive in Boston than Slow? Ans. 123 seconds. 8. If Ad of a ton cost $ 49, what will lcwt. cost? Ans. $ 3.92. 9. How many bricks, 8 inches long, 4 inches wide, and 2 inches thick, will it take to build a wall 40 feet long, 20 feet high, and 2 feet thick? Ans. 43200 bricks. -10. How many bricks will it take to build the walls of a house, which is 80 feet long, 40 feet wide, and 25 feet high the wall to be 12 inches thick; the brick being of the same dimensions as in the last question? Ans. 159300 bricks. 11. How many tiles, 8 inches square, will cover a floor 18 feet long, and 12 feet wide? Ans. 486 tiles. 12. If it cost $ 18.25 to carry 11cwt. 3qr. 191b. 46 miles, 312 MISCELLANEOUS QUESTIONS. [sECT. XLVI how much must be paid for carrying 83cwt. 2qr. l llb. 96 miles? Ans. $ 267. 12 r.5Ty. 13. A merchant sold a piece of cloth for $ 24, and thereby lost 25 per cent.; what would he have gained, had he sold it for $ 34? Ans. 61 per cent. 14. Bought a hogshead of molasses, containing 120 gallons, for $ 30; but 20 gallons having leaked out, for what must I sell the remainder per gallon to gain $ 10? Ans. $ 0.40. 15. How many acres are there in a piece of land 117a rods long, and 1122 rods wide? Ans. 82A. 1R. 18p. 2yd. 7ft. 1335in. 16. Bought a quantity of goods for $128.25, and, having kept them on hand 6 months, for what must I sell them to gain 6 per cent.? Ans. $ 140.02. 17. If 27 bushels of potatoes cost $ 8.75, what must be paid for 36 bushels? Ans. $ 11.66+. 18. How many bushels of oats, at 50 cents per bushel, must I give Moses Webster for 93 bushels of corn, at $ 1.25 per bushel? Ans. 232- bushels. 19. How many bushels of salt, at $ 1.30 per bushel, must be given in exchange for 75 bushels of wheat, at $ 1.25 per bushel? Ans. 72.A3 bushels. 20. If a sportsman spends I of his time in smoking,. in "' gunning," 2 hours per day in loafing, and 6 hours in eating, drinking, and sleeping, how much remains for useful purposes? Ans. 2 hours. 21. If a lady spend X of her time in sleep, - in making calls, r at her toilet, + in reading novels, and 2 hours each day in receiving visits, how large a portion of her time will remain for improving her mind, and for domestic employments? Ans. 32 7 hours per day. 22. What will a piece of land, 7- rods long, and 5- rods wide, come to at $ 25.75 per acre? Ans. $ 6.651. 23. If 53 ells English cost $ 15.16, what will 711 yards cost? Ans. $ 155.39. 24. If a staff 4 feet long cast a shadow 53 feet, what is the neight of a steeple whose shadow is 150 feet? Ans. 1071 feet. 25. Borrowed of James Day $ 150 for six months; afterwards I lent him $ 100; how long shall he keep it to compensate him for the sum he lent me? Ans. 9 months. 26. A certain town is taxed $ 6045.50; the valuation of SECT. XLVI. MISCELLANEOUS QUESTIONS. 313 the town is $ 293275.00; there are 150 polls in the town, which are taxed $ 1.20 each. What is the tax on a dollar, and what does A pay, who has 4 polls, and whose property is valued at $ 3675? Ans. $ 0.02. A's tax $ 78.30. 27. What is the value of 97 pigs of lead, each weighing 2cwt. 3qr. llb., at 3~. 17s. 9d. per cwt.? Ans. 1074C. Os. 6T1272 d 28. What is the interest of $ 17.86, from Feb. 9, 1S40, to Oct. 29, 1842, at 71 per cent.? Ans. $ 3.52 —. 29. What is the interest of $ 97.87, from Jan. 7, 1840, to Sept. 25, 1842, at 9 per cent.? Ans. $ 23.92+. 30. Required the superficial surface of the largest cube that can be inscribed in a sphere 30 inches in diameter? Ans. 1800 inches. 31. $ 1000. Salem, N. H., Oct. 29, 1836. For value received, I promise to pay Luther Emerson, Jr., or order, on demand, one thousand dollars with interest. Emerson Luther. Attest, Adams Ayer. On this note are the following endorsements: Jan. 1, 1837, was received - 125.00, June 5, 1837, do. $ 316.00, Sept. 25, 1837, do. $ 417.00, April 1, 1838, do. $ 100.00, July 7, 1838, do. $ 50.00. What is due, at compound interest, Oct. 29, 1842? Ans. $53.79. 32. D. Sanborn's garden is 234 rods long, and 13'. rods wide, and is surrounded by a good fence 74 feet high. Now if he shall malke a walk around his garden within the fence 7-1 feet wide, how much will remain for cultivation? Ans. 1A. 3R. 7p..85ts-ft. 33. J. Ladd's garden is 100 feet long and 80 feet wide; he wishes to enclose it with a ditch 4 feet wide; how deep must it be dug, that the soil taken from it may raise the surface one foot. Ans. 5}} feet. 34. How many yards of paper that is 30 inches wide, will it require to cover the walls of a room that is 15- feet long, 11I feet wide, and 7- feet high? Ans. 551 yards. 35. Charles Carleton has agreed to plaster the above room at 10 cents per square yard; what will be his bill? Ans. $ 6.54-. 27 314 MISCELLANEOUS QUESTIONS. [SECT. XLVI. 36. How many cubic inches are contained in a cube that may be inscribed in a sphere 40 inches in diameter? Ans. 12316.8+ - inches. 37. The dimensions of a bushel measure are 18 inches wide, and 8 inches deep; what should be the dimensions of a similar measure that would contain 4 quarts? Ans. 9- inches wide, 4 inches deep. 38. A gentleman willed ~ of his estate to his wife, and i of the remainder to his oldest son, and - of the residue, which was $ 151.331, to his oldest daughter; how much of his estate is left to be divided among his other heirs? Ans. $ 756.662. 39. A man bequeathed ~ of his estate to his son, and 5 of the remainder to his daughter, and the residue to his wife; the difference between his son and daughter's portion was $ 100; what did he give his wife? Ans. $ 600.00. 40. A young man lost I of his capital in speculation; he afterwards gained $ 500; his capital then was $ 1250; what was the sum lost? Ans. $ 250.00. 41. From + of a yard, there was sold f of it; how much remained z Ans. s3 yard. 42. Sold a lot of shingles for $ 50, and by so doing I gained 1.2! per cent.; what was their value? Ans. $ 44.449. 43. If tallow be sold at 71d. per lb., what is the value of 17cwt. 3qr. ISIb.? Ans. $ 208.955. 44. If -- of a yard cost $ 5.00, what quantity will $ 17.50 purchase? Ans. 21 yard. 45. If a man travel 17rd. lOft. in 7 of an hour, how far will he travel in 8 hours? Ans. 1 mile 928 feet. 46. When $ 11.75 are paid for 22 acres, what quantity will $ 100.00 purchase 2 Ans. 19A. 1R. 32152p. 47. John Savory and Thomas Hardy traded in company; Savory put in for capital $ 1000; they gained $ 128.00; Hardy received for his share of the gains $ 70; what was his capital? Ans. $ 1206.89~-1-. 48. E. Fuller lent a certain sum of money to C. Lamson, and, at the end of 3 years, 7 months, and 20 days, he received interest and principal $ 1000; what was the sum lent? Ans. $ 820.7921. 490 Lent $ 88 for 18 months, and received for interest and principal $ 97.57; what was the per cent.? Ans. A1 per cent. 50. When - of a gallon cost $ 87, what cost 71 gallons? Ais. $ 1051.225. BECT. XLVI.] MISCELLANEOUS QUESTIONS. 315 51. When $ 71 are paid for 183 yards of broadcloth, what cost 5 yards? Ans. $19.2646-. 52. How many yards of cloth, at $ 4.00 per yard, must be given for 18 tons l7cwt. 3qr. of sugar at $ 9.50 per cwt.? Ans. 897 2 yards. 53. How much grain, at $ 1.25 per bushel, must be given for 98 bushels of salt, at $ 0.45 per bushel? Ans. 3527, bushels. 54. How many acres of land, at $ 37.50 per acre, must be given for 86 tons 18cwt. 3qr. 201b. of coal, at $ 8.50 per ton? Ans. 19A. 2R. 33 9 p. 55. A person, being asked the time of day, replied, that 7 of the time passed from noon was equal to — 1L of the time to midnight. Required the time. Ans. 40 minutes past 4. 56. How many cubic feet of water in a pond, that contains 200 acres, and is 20 feet deep? Ans. 174,240,000 feet. 57. On a certain night, in the year 1842, rain fell to the depth of 3 inches in the town of Haverhill; the town contains about 20,000 square acres. Required the number of hogsheads of water fallen, supposing each hogshead to contain 100 gallons, and each gallon 2S2 cubic inches. Ans. 13346042hhd. 55gal. lqt. Opt. 2 ~gi. 58. If the sun pass over one degree in 4 minutes, and the longitude of Boston is 710 4' west, what will be the time at Boston, when it is llh. 16m. A. M. at London? Ans. 6h. 31m. 44sec. A. M. 59. When it is 2h. 36m. A. M. at the Cape of Good Hope, in longitude 18~ 24' east, what is the time at Cape Horn, in longitude 670 21' west? Ans. Sh. 53m. P. M. 60. Yesterday my longitude, at noon, was 160 18' west; to-day I perceive by my watch, which has kept correct time, that the sun is on the meridian at 1 lh. 36m.; what is my longitude? Ans. 100 18' west. 61. Sound, uninterrupted, will pass 1142 feet in 1 second; how long will it be in passing from Boston to London, the distance being about 3000 miles? Ans. 3h. 51m. 10+- sec. 62. The time which elapsed between seeing the flash of a gun, and hearing its report, was 10 seconds; what was the distance? Ans. 2 miles 860 feet. 63. If a globe of silver, 2 inches in diameter, be worth $ 125, what would be the value of a globe 3 inches in diameter? Ans. $421.87-. 64. J. Pearson has tea, which he barters with 3I. Swift, at 316 MIISCELLANEOUS QUESTIONS. [SECT.- XLv1. 10 cents per lb. more than it costs him, against sugar, which costs Swift 15 cents per pound, but which he puts at 20 cents per pound; what was the first cost of the tea? Ans. $ 0.30 per lb. 65. Q and Y barter; Q makes of 10 cents 122 cents; Y makes of 15 cents 19 cents; which makes the most per cent. and how much? Ans. Y makes 1. per cent. more than Q. 66. A certain individual was born in 1786, September 25 at 27 minutes past 3 o'clock, A. M.; how many minutes old will he be July 4, 1844, at 30 minutes past 5 o'clock, P. MI., reclkoning 365 days for a year, excepting leap years, which have 366 days each? Ans. 30,386,287 minutes. 67. The longitude of a certain star is 3s. 14~ 26' 14", and the longitude of the moon at the same time is Ss. 190 43' 28"; how far will the moon have to move in her orbit to be in conjunction with the star? Ans. 6s. 240 42' 46". 68. From a small field containing 3A. 1R. 23p. 200ft., there were sold 1A. 2R. 37p. 30yd. Sft.; what quantity remained? Ans. 1A. 2R. 25p. 21yd. 5ft. 36in. 69. What part of` of an acre is a of an acre? Ans. Z-. 70. A thief was brought before a certain judge, and it was proved that he had stolen property to the value of 1~. 19s. 11:3d. He was sentenced either to one year's imprisonment in the county jail, or to pay 1~. 19s. 113d. for the value of every pound he had stolen; required the amount of the fine? Ans. 3~ 19s. 1ld. 0,,qr. 71. My chaise having been injured by a very bad boy, I am obliged to sell it for $ 68.75, which is 40 per cent. less than its original value; what was the cost? Ans. $ 114.5SI. 72. Charles Webster's horse is valued at $120, but he will not sell him for less than $ 134.40; what per cent. does he intend to make? Ans. 12 per cent. 73. Three merchants, L. Emerson, E. Bailey, and S. Curtiss, engaged in a cotton speculation. Emerson advanced $ 3600, Bailey $ 4200, and Curtiss $2200. They invested their whole capital in cotton, for which they received $ 15000 in bills on a bank in New Orleans. These bills were sold to a Boston broker at 15 per cent. below par; what is each man's net gain? Ans. Emerson $990.00, Bailey $ 1155.00, Curtiss $4605.00. 74. Bought a box made of plank, 31 inches thick. Its length on the outside is 4ft. 9in., its breadth 3ft. 7in., and its SECT. XLVI.] MISCELLANEOUS QUESTIONS. 317 height 2ft. 11in. How many square feet did it require to make the box, and how many cubic feet will it hold? Ans. 70 6 square feet, 29, cubic feet. 75. How many bricks will it require to construct the walls of a house, 64 feet long, and 32 feet wide, and 28 feet high 2 The walls are to be ift. 4in. thick, and there are also three doors 7ft. 4in. high, and 3ft. 8in. wide; also 14 windows 3 feet wide and 6 feet high, and 16 windows 2ft. Sin. wide and 5ft. Sin. high. Each brick is to be 8 inches long, and 4 inches wide, and 2 inches thick. Ans. 167,480 bricks. 76. John Brown gave to his three sons, Benjamin, Samuel, and William, $ 1000, to be divided in the proportion of 3, a, and 1, respectively; but William, having received a fortune by his wife, resigns his share to his brothers. It is required to divide the whole sum between Benjamin and Samuel. Ans. Benjamin $ 571.426; Samuel $ 42S.57. 77. Peter Webster rented a house for one year to Thomas Bailey, for $ 100; at the end of four months, Bailey rented one half of the house to John Bricket, and at the end of eight months, it was agreed by Bricket and Bailey to rent one third of the house to John Dana. What share of the rent must each pay? Ans. Bailey $ 61, Bricket $ 27 -, and Dana $ 11}-. 78. I have a plank 42~ feet in length, 24 inches wide, and 3 inches thick; required the side of a cubical box that can be made from it? Ans. 48 inches.'9. D. Small purchased a horse for 10 per cent. less than his value, and sold him for 16 per cent. more than his value, by which he gained $21.84; what did he pay for the horse? Ans. $75.60. SO. Minot Thayer sold broadcloth at $4.40 per yard, and by so doing he lost 12 per cent.; whereas he ought to have gained 10 per cent.; for what should the cloth have been sold per yard? Ans. $5.50. S1. A gentleman has five daughters, Emily, Jane, Betsey Abigail, and Nancy, whose fortunes are as follows. The first two and the last two have $19,000; the first four $19,200; the last four $20,000; the first and the last three $20,500; the first three and the last $21,300. What "was the fortune of each? Ans. Emily has $5,000; Jane $4,500; Betsey $6,000; Abigail $3,700; and Nancy $5,$00. WEIGHTS, MEASURES AND MONEY. TnrE tables in this work are intended to aff'ord the learner a knowl edge of the various weights, measures and moneys used ill different countries, sufficient for the ordinary purposes of business and of practical arithmetic. It is here proposed to supply some items of information, such as are not found in popular works of this kind, nor, it is believed, in any compact or easily accessible form. WEIGHTS AND I TEASURES. The use of weights and measures can be traced back to a very early period of the world. Josephus, the Hebrew historian, asserts, that they were invented by Cain, the tiller of the ground and the first builder of a city. Whatever authority is to be attached to this statement, we learn from the Boo00 of Genesis that the cubit was employed in designating the dimensions of Noah's ark; and it is reasonable to suppose that several other measures, and a few simple weights, such as were demanded by the common intercourse and employnments of mankind, were in use among the antediluvians. In the time of Abraham, we find mention made of measures of capacity, (measures of meal,) and also of money. With the latter, the patriarch bought a field of Ephron, the Hittite, for which he paid him four hundred shekels of silver. Tliis sum was weighed out to Ephron, a circumstance plainly indicating that the value of money was then reckoned by its weight, as has been that of coins in all ages. But though the use of weights and measures can be referred to an origin thus remote in time, we are not to suppose that they were at first employed with the accuracy and uniformity of modern times. On the contrary, as men's ideas of distance, quantity and value were, in the early stages of society, vague and indefinite, so also were their standards of comparison. When it was first proposed to establish some measure by which smnall distances should be estimated, it was natural to have recourse to some parts of the human body, as the arm, the foot, the hand; and hence the origin of the cubit, the lelngth of the arm from the elbow to the end of the longest finger; of the foot, the length of a man's foot; and of the palm or handbreadth, the width of a man's hand. The span was the distance fiom the end of the thumb to that of the little finger, when extended; and the fathom the space between the extremities of the outstretched arms. WEIGHTS AND MEASURES. I19 When a longer distance was to be measured, the mind would easily aix upon some familiar object of greater length, as a rod or pole, cut from the forest; and when a shorter distance was to be expressed, it could be done either by dividing the foot or palm into any number of small equal parts, or by employing, as a unit, some minute natural object, as a grain of wheat or barley. In this way the pole, perch or rod, probably came into use, as -it is certain did the barley-corn and inch, the latter being the twelfth part of a foot.* It may also be mentioned in this place, that among the measures of the Hindoos, a people with whom there has been little change' for many centuries, we find the bambu-pole and the staff, which doubtless originated in a manner similar to the use of the modern rod or pole. The former of these is reckoned at twenty cubits and the latter at four. The names of several other modern measures clearly indicate their origin, as a mile, from the Latin ille - one thousand - that is, one tlollusand paces; furlong, from the Saxon, far or fur, and long, or, as sonme etymologists say, from furrow and long, that is, the length of a furrow. In some instances, distances have been reckoned by the space through which an arrow could be shot, or a stone thrown, and hence the -terms bow-shot and stone's-east or stone's-throw, with which we occasionally meet. One of the most indefinite standards ever in use among any people, would seem to be the Chinese unit of linear measure, which is said to be the lih, and to denote the distance which a man's voice will reach in a level country, when thrust forth with all hiis might. The instances thus adduced are sufficient to establish two points, first, that the measures of antiquity originated from the use of some familiar natural objects, as standards of length or distance; and, secondly, that men's ideas and estimate of space were, at an early period, vague, inaccurate, and destitute of uniformity. But it is evident that such vagueness and diversity could not con tinue. They could neither satisfy the desire of the human mind for accuracy, nor meet the demand for it created by advancing civilization. As men came to have more intercourse and business with each other; to exchange commodities, fix upon the boundaries of lands, and erect numerous and contiguous buildings; in a word, to live in society, and satisfy their necessities by the barter, sale and purchase of different articles in common use, there would be needed more definite and exact standards by which they might compare one commodity with another, and express its relation and value. Such standards would be necessary in order that equality and justice might be had in common ousiness transactions. And the history of weights and measures is little else than the history of the human mind, in its efforts to devise means and instruments by which commercial intercourse might be conducted on principles of reciprocity and just equivalents. The importance of uniformity in weights and measures has been felt by all civilized nations; and to prevent fraud by any alteration of the * The word inch is said to be ftron the Latin, uwcia, which signifies a twelfth part of anything. 320 WEIGHTS AND MEASURES. proper standards, or any departure from them in practice, these has usually been kept in the custody of the government. Among the Jews they were committed to the care of the, priests, and in Rome they were deposited in the temple of Jupiter. In England they are kept in the exchequer, and in the United States they are in the charge of the national treasury. So careful are the people in the different states of the Union that due weight and measure shall be given in all trade, that in most, if not in all of them, laws have been enacted requiring those who sell to have their weights and measures sealed, that is, tried or adjusted by some standards kept by public authority for the purpose. In some states this is done as often as once a year. With the desire of introducing a uniform standard of measure among the people of his kingdom, Henry I., in the year 1101, ordered that the ulna, or ancient ell, should be of the exact length of his own arm. On this all other measures were to be founded; and it is worthy of remark, that this very measure, the length of king Henry's arm, has remained, without sensible variation, to this day, and is the present English and American standard yard. For the last one hundred years, science has been at work to devise some system of weights and measures which should be accurate and intelligible in all its parts, and not liable to change or variation. For this purpose the Royal Society and the parliament of England have combined their efforts, and the result of their labors is a system of AMetrology in which the " IMPERIAL YARD" is made the standard of all measures, linear, superficial and solid; and ultimately, as we shall see, of all weights. The law by which this was made the legal standard was passed in 1824, and is entitled, the " Act of Uniformity." This yard is represented by a solid brass rod, kept in the exchequer, about an inch square, in which, about an inch and a half from each end, is inserted a gold pin or stud, the space between these pins being 36 inches. This distance is the Imperial yard. That this standard mal not be lost or mutilated, it was enacted that it should bear a fixed and definite proportion to the length of a pendulum, vibrating seconds, in a vacuum, in the latitude of London, at the level of the sea, and at the temperature of 62~ Fahrenheit. This proportion was to be that of 36 to 39.1393. A third part of this yard was to be the legal foot; a thirty-sixth part the legal inch; five and a half such yards a rod, &c. The manner, in which the legal measures of capacity are founded upon the legal yard is as follows. The " imperial gallon," which is the proximate standard of capacity, contains 277.'274 cubic inches, each solid inch being raised upon a thirty-sixth part of the Imperial yard. The law, however, allows that the gallon may be determined by weight, and then the measure containing a gallon must hold lOlbs. Avoirdupois weight, of distilled water, weighed in air, at 62~ Fahren heit, with the barometer at 30 inches. By the present law of England, the pound Troy contains 5760 grains: and this pound is the standard of all legal weights. But this pound itself is determined by a reference to the Imperial yard. The comnpar WEIGHTS AND MEASURES. 321 ison is effected by talking a cubic inch of water, weighed as above, and reckonina it at 252.458 of these grains. The pound Avoirdupois contains 70)00 such grains. We thus see that, ny the English system, all weights and measures being based upon the standard yard, and this yard being made to bear a fixed proportion to a pendulum vibrating seconds under the abovenamed conditions, the ultimate standard of reference is TIME, which is measured by the revolutions of the earth Strictly speaking, therefore, the whole system is as accurate and unvarying as the motion of the planet we inhabit. It must be admitted, however, that there are mechanical difficulties in the way of obtaining entire accuracy ill the length of the pendulum. Of all the systems of Metrology which have ever been constructed, that of the French is the most complete and scientific. In theory it seems to approach as nearly as possible to perfection, and the practical difficulties in the way of applying the theory to common use, are perhaps not greater than would be found in any complete and comprehensive system which could be devised. According to this system, the mretre is the standard measure of length. This metre is one ten-millionth part of a quadrant of a meridian, or of the distance of the equator from the poles, and does not differ materially from the common yard, being 3.281 English feet. On this standard are founded all other measures, whether linear, superficial, or solid. This also, like the imperial yard of the English system, is made the basis of all weights. A brief account of this system, together with some tables of French we;ights and measures, is given in the Appendix to the National Arithmetic. In the United States, the power of fixing the standard of weights and measures, like that of coining money, is conferred by the constitution on Congress. Nor has the subject been wholly neglected by the national legislature. At the request of the two houses, John Quincy Adams, when Secretary of State, prepared a report on weights arnd measures, which he presented, Feb. 22, 1821, nearly three years fromnt the time when his attention was called to the subject by the resolution of the senate. This report, when it appeared, was found to be oine of the most learned and elaborate treatises on the subject that had ever been written, and though the effect of it was, for the time, rather to restrain than to encourage legislation in this important particular, future legislators may yet avail themselves of the principles and facts which it contains, in giving to the country a more complete and uniform syst:em of Metrology. From the appearance of this report to the present time, but little has been done towards perfecting the weights and measures of the nation. By an act of Congress, in June, 1836, a set of standard weights and measures was prepared for the use of the different customhouses, and for eacth state; and these constitute the legal standard of the country, so far as any such standard exists. Several of the states have standards of their own, but nluch remains to be done before there can be anytlhin.l wl rthy tihe name of a uniform system. The stand 322 MIONE. ards now generally used do not differ essentially from those in use in England before the passage of the act of uniformity. In the state of New York, the standard of linear measure is the yard, which bears a definite proportion to a pendulum vibrating seconds, in a vacuum, at Columbia college, in the city of New York, at the temperature of 320 Fahrenheit. The proportion is that of 1000 to 1086. In the same state, the standard bushel contains 2218.192 cubic inches, and is equal to 80 pounds, Avoirdupois weight, of distilled water. In Massachusetts, the standard weights and measures kept in the treasury are such as were early sent over from England, with a certificate from the exchequer that they were approved Winchester measures. I) Massachusetts thie old weight of 1121b. for 100 is abolished, and the " hundred weight" is the net weight of an hundred pounds, Avoirdu pois. MONTEY. THE term Money* is used to denote any commodity which the inhabitants of a country employ as a universal medium of exchange, or which they accept as an equivalent for whatever is bought and sold. In civilized nations it has generally been composed of the precious metals, but this is not essential; other commodities might be employed for the same purpose, though with less convenience. In the early ages of the world, the material used for money varied with the characters and employments of different tribes and nations. In some countries cattle and sheepj were the common medium of exchange. Thus, according to Homer, the armor of Diomedes cost nine oxen, and that of Glaucus, one hundred. In nations where the people have been given to the chase, as the ancient Russians and the aboriginal inhabitants of America, the skins of wild beasts have been used for the purpose. The common money of Abyssinia was, at one time, salt; that of Newfoundland and Iceland, dried fish; and that of some of the West India Islands, sugar. Among the North American Indians the circulating medium was what they called wampum, which consisted of small beads, made of different colored shells. Nor is it necessary to look to distant times and countries to find other articles than gold, silver, and copper, employed as the common standard of value. In the state of Virginia, for more than a century, tobacco was not only merchandise, but money, - that is, the circulating medium of exchange. * The word money is of somewhat doubtful derivation. Some etymologisti regard it as coming from the Latin word monere, which signifies to admonish or inform. These contend that it is so called because the stamp upon it admonishes or informs us of its value; while others, and perhaps the majority. attribute the origin of the word to the name of the temple in which silver was first coined at iome, the temple of Juno l~ionetac, admonishing Juno. t The Latin word for money (pecunia) is derived from pecus, cattle, sheep, &c r, and from this comes the adjective pecuniary. MONEY. 323 But of all the substances which have ever been used as money, the precious metals are the best fitted for the purpose. To gold, silver, and copper, now in so general use, some statesmen and scientific men have proposed that platinum should be added, which would form a coin intermediate in value between. gold and silver. The value of all coins is dependent upon their purity and weight; and the stamp impressed upon them is designed to indicate these two qualities. For instance, the English sovereign, the American eagle, and the Mexican dollar, have each their peculiar impression; and this impression is the pledge of the government by which they are coined, that the metal of which they are composed is of a given degree of fineness, and of a certain weight, both of which are fixed by law, and essential to the particular value of the gold or silver piece. Were not the metal stamped or coined, its value could be known only by assaying and weighing it. This, especially the former, would be a difficult process, and would greatly diminish the convenience, and impede the circulation of coin as a medium of exchange or a representative of value. Hence the origin. of coinage; and as this is a work which needs to be conducted with the greatest care and the most scrupulous fidelity, governments have generally taken it into their own hands. The gold and silver wrought into coins are not usually pure. They contain more or less of some foreign substance, which is called alloy. In England, twelve ounces, or a Troy pound, of the metal of which silver coins are made, contain lloz. 2dwt. of fine silver, and 18dwt. alloy, which makes the standard silver coin to consist of 37 parts of pure silver and 3 parts of alloy. This alloy is copper. In like man ner, the gold coins of Great Britain contain but 11 parts out of 12 of pure gold, the remaining part being alloy. Gold is not estimated by the weights in common use, but by a weight called a carat.* When pure, or of the highest degree of fineness, it is said to be 24 carats fine. The present gold coin of the realm is therefore 22 carats fine and 2 carats alloy. The alloy, in this case, as in that of silver, is copper. The present standard for both gold and silver coins in the United States, as fixed by an act of Congress, in 1837, is 900 parts, by weight, of pure metal, to 100 parts of alloy. The alloy of the gold is composed of silver and copper, of which the former is not to exceed the latter in weight. The alloy of the silver is pure copper. By the mint regulations of the United States, the eagle, the equivdlent of 10 dollars, contained, previously to July 31, 1834, 270 grains of standard gold; but by an act of Congress, which took effect at that date, it is reduced to 258 grains. In consequence of this change, the sovereign, which was formerly valued at 4 dollars and 57 cents, is now reckoned at 4 dollars and 87 cents. Originally, the coins of all countries appear to have had the same * The word carat is of Abyssinian origin, and was the name of a certain kind of bean, which, from the time of its being gathered, varied little in its weight. It seems to have been used in the earliest ag'es as a weight for gold in Afriea. Tn India it is sai(d to be ulsed for weighing diamnonds. 324 MiONEY. denominations as the weights in common use in them, and to have contained the quantity of metal denoted by their names. Thus, in Greece, the talent was a weight, and the coin of that name a quantity of silver or gold which would balance it in an equal scale. In like manner, the pondo of Rome, the licre of France, and the pound of England, were weights, and the coins originally in use in these countries, which bore the same names, answered exactly to them in weight The pound sterling of 20 shillings was at first a veritable pound ot silver, though a pound of standard silver is now worth 66 shillings, or more than three times the present lawful pound. In France, in the time of the revolution, the livre (pound) contained less than a seventy-eighth part of the silver contained in the coin of the same name in the reign of Charlemagne, which was a pound, by weight, of silver. A similar reduction of the standard coin has frequently been made, both in ancient and modem times. Among modern nations, the means and facilities of business have been much enlarged by the establishment of banks and the introduction of a paper currency. The institution of banks, it is said, originated with certain Jews, in Lombardy, who, in the twelfth century, kept benches in the market-places, for the exchange of money and of bills. The Italian word for bench is banco, and hence the name bank. The modern public banks were originally deposite banks. Of these, the first was the celebrated Bank of Venice, which was instituted in 1171. Banks, as they are now constituted, serve the double purpose of a place of deposite for money, and of a company for the issue of notes. Bank bills are simply promissory notes, which the bank is bound to redeem on demand, by paying the sum specified in specie. These notes constitute what is called paper money, and are of value simply because they can, at any time, be converted into coin or cash. In a country whose business transactions are extensive, the convenience of such notes, as a portion of the circulating medium, is very great. They are more easily kept than coin; more easily handled, especially in large sums; more easily sent or carried from one place to another, for the payment of debts or other purposes; and if they are lost, as large sums frequently are, by the destruction of a ship at sea, the community sustains no real loss of property. In such a case, what the owner of the bills loses is gained by the bank or issuer,- whereas, in the loss of coin or bullion, under such circumstances, there is so much real wealth taken, not only from the owner, but from the community.