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 Agricultural arithmetic. 
 
 3 1924 002 954 992 
 
Cornell University 
 Library 
 
 The original of this book is in 
 the Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924002954992 
 
Agricultural Arithmetic 
 
 GEORGE C. SHUTTS 
 
 Department of Mathematics, Normal School 
 Whitewater, Wisconsin 
 
 WILBERT WALTER WEIR, B.S.(A.) 
 
 Department of Soils, College of Agriculture 
 
 University of Wisconsin 
 
 WEBB PUBLISHING COMPANY 
 ST. PAUL, MINNESOTA 
 
 1917 
 
Copyright, 1916 
 
 webb publishing co. 
 
 all rights reserved'. 
 
 W-2 
 
FOREWORD 
 
 Arithmetic, as a subject in the school curriculum, is 
 of no particular value in itself. It is not one of the satis- 
 factions of life. It was so regarded, however, in the past, 
 and much pleasure to the keen witted, as well as sorrow of 
 soul to the dull, was wrought by its puzzles which certainly 
 did not grow out of any of the occupations of man. But 
 tradition is so strong that it has taken a long time to elimin- 
 ate this theory concerning arithmetic with its attendant 
 conundrums. 
 
 Arithmetic is best regarded as an instrument, a tool, 
 with which to accomplish various desired ends and hence 
 should be studied in adaptation to those ends. 
 
 Facility in the use of arithmetic depends upon two 
 things: first a knowledge of the business relations to which 
 it is to be applied, and, second, a knowledge of the pure 
 arithmetical processes together with skill in performing 
 them. Hence this book has been organized into two parts. 
 
 In Part I has been gathered into a brief space the es- 
 sential things to be considered and drilled upon to enable 
 one to develop accuracy and rapidity in computation by 
 arithmetical processes. As it presupposes a considerable 
 knowledge of the processes of arithmetic, Part I does not 
 attempt to develop fully the general subject, but is simply 
 a rational review to strengthen some of the weak points. 
 
 Part II is an application of arithmetic to farm exper- 
 iences. The problems are not inventions, but are drawn 
 from life in its various phases upon the farm. Necessary 
 conditions in problems are sometimes omitted in order 
 that the pupil may collect his own data from observation, 
 experience, etc., at home; hence results will differ or only 
 
6 AGRICULTURAL ARITHMETIC 
 
 approximate one another. These varying results will 
 naturally lead to a study of the causes that produce them, 
 and thus may supply an incentive to improve unfavorable 
 conditions. 
 
 Lack of space has limited the number and a greater 
 variety of problems. It is hoped that the teacher will 
 formulate additional problems to give a further drill on 
 points not fully understood by the respective classes or 
 individuals. 
 
 Farm operations have been determined too largely by 
 tradition; and the prejudice against scientific methods has 
 been so strong that progress has been slow. It is believed 
 that the study of topics and selection of problems herein 
 will prove an incentive for improvement. 
 
 Indebtedness is acknowledged to the Wisconsin Agricul- 
 tural Experiment Station for the use of illustrations repre- 
 sented in figures 8, 10, 12, 13, 14, 17, 18, 20, 21, 22, 24, 25, 
 26, 27, 36, 37, 40, 41; and to the Wisconsin Live Stock As- 
 sociation for those in figures 4, 5, 6, 7, 9, 39. 
 
 The Authors 
 
CONTENTS 
 
 PART ONE 
 
 Page 
 
 Definitions 9 
 
 Addition, Subtraction, Multiplication, Division, Checking 12 
 
 Factoring 16 
 
 Fractions 18 
 
 Least Common Multiple 23 
 
 Addition, Subtraction, Multiplication and Division of Fractions . . 24 
 
 Area of Rectangle and Rectangular Survey 31 
 
 Volumes of Rectangular Solids 35 
 
 Decimals 38 
 
 Methods of Abbreviation 42 
 
 Measurement of Lumber 51 
 
 Square Root and Right Triangle 53 
 
 Area of Triangle and Trapezoid 67 
 
 The Circle 59 
 
 Volumes of Cylinders 62 
 
 The Cone 64 
 
 Percentage 65 
 
 Interest 69 
 
 Ratio and Proportion 72 
 
 PART TWO 
 
 Farm Crops ._ 77 
 
 Farm Animals 86 
 
 Feeds and Feeding Practice 91 
 
 Theory of Feeding 100 
 
 The Dairy Ill 
 
 The Soil 120 
 
 What Crops Require 124 
 
 The Soil a Great Storehouse 132 
 
 Balance Sheet of Soil Fertility 137 
 
 Manure and Commercial Fertilizers 147 
 
 Farm Management 160 
 
 Farm Measurements 189 
 
 Orchard and Garden 204 
 
 Household Economy and Human Feeding 213 
 
 Miscellaneous Problems 221 
 
 Appendix 233 
 
 7 
 
PART ONE 
 
 INTRODUCTION 
 
 To satisfy his wants in his conquest over matter, civilized 
 man, or even the savage, needs to determine the size of 
 the elements with which he deals. To do this he must 
 select a part of the whole, an amount with which he is 
 familiar, and by this estimate or measure the size of the 
 whole. 
 
 Mass or magnitude is the simple notion of size. It is 
 indefinite and always suggests the question, How much? 
 
 Measurement is the process of determining how much 
 there is of the magnitude. The process of measurement 
 consists in determining the number of units it contains. 
 
 A unit is a single thing or a portion of a magnitude into 
 which it is divided in the process of measurement. 
 
 Number tells how many units there are in a given mass 
 or magnitude. 
 
 Quantity is the complete answer to the question how 
 much there is of the mass or magnitude that is measured. 
 
 To illustrate: A person desires to know how large is 
 a given piece of land. He takes a plat of land of a given 
 size and calls it, for instance, a square rod. In the funda- 
 mental way of measuring, he separates the land into square 
 rods and counts them. His result, say, is 15 square rods. 
 The magnitude is the amount of land that prompts the 
 question, How much? The unit is the square rod, the 
 portion into which the whole is divided in the process of 
 measurement. The number is 15, and tells how many rods 
 there are. The quantity is 15 rods and tells the size of the 
 
10 AGRICULTURAL ARITHMETIC 
 
 magnitude. The measurement is the whole process of 
 division and counting by which the quantity is determined. 
 
 Any process, however complicated, by which any magni- 
 tude can be expressed as quantity is called measurement. 
 Usually short methods are used; as, I wish to know the 
 length of a board or a fence. I use a tool that, has been 
 invented for the purpose, namely a rule or tape line, and 
 lay it along the board or fence and read the result of some- 
 body else's counting marked upon it. In the case of the 
 land, we have learned by study and experience that the 
 process of division and counting can be shortened, if the 
 land is rectangular in form, by measuring two adjacent 
 sides and multiplying together the results. (Page 32.) 
 
 Measurement, in its simplest and probably its earliest 
 form, consists simply in counting. One desires to know the 
 size of a flock of sheep. The natural unit is the individual 
 animal; hence a division of the flock is not made, and the 
 process of measurement consists simply in counting. But 
 even here experience suggests that the counting can be 
 done more rapidly if the flock is divided into groups of two, 
 four, or some other number for a unit, and these groups 
 counted. It is said that, when Xerxes wanted to know the 
 size of the army with which he tried to conquer Greece, he 
 drove his men into pens holding ten thousand each, and 
 counted the pens. 
 
 The subject of arithmetic can be more easily under- 
 stood, if it can be appreciated that in all its essential pro- 
 cesses it grows out of and in reality is, in all its details, 
 but a continuation or an application of the results of the 
 process of measurement. The unit in its relation to number 
 must not be lost sight of, or, as too often happens, arithmetic 
 becomes a maze, a labyrinth in which the pupil is hopelessly 
 lost. 
 
 Arithmetic has been called the science of numbers and 
 
INTRODUCTION 11 
 
 the art of computation. It is rather the science of measure- 
 ment. It deals with quantity and number; and consists 
 of the processes of reduction, combination, and division of 
 quantity and number so as to express the results of measure- 
 ment in the best forms for man's use. 
 
 This book deals with arithmetic in so far as it relates 
 to man's needs in the occupation of agriculture. 
 
 Part I is intended as a review of the essentials to prepare 
 for the treatment of agricultural arithmetic in Part II. 
 Some topics of arithmetic are omitted, and others are 
 treated with more or less assumption of knowledge of the 
 subject on the part of the student. 
 
REVIEW OF ESSENTIALS 
 
 ADDITION 
 
 Facility in addition depends upon a thorough memoriz- 
 ing of the addition table. 
 
 Step 1. The addition table consists of forty-five com- 
 binations; viz., the sums of all possible pairs of digits. 
 Leaving out the combinations with 1, there are but thirty- 
 six to memorize; as, 
 222222223333333444 
 234567893456789456 
 
 4 
 
 4 
 
 4 
 
 5 
 
 5 
 
 5 
 
 5 
 
 5 
 
 6 
 
 6 
 
 6 
 
 6 
 
 7 
 
 7 
 
 7 
 
 8 
 
 8 
 
 9 
 
 7 
 
 8 
 
 9 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 6 
 
 7 
 
 8 
 
 9 
 
 7 
 
 8 
 
 9 
 
 8 
 
 9 
 
 9 
 
 One should not be satisfied with results when commit- 
 ting these combinations to memory until any given com- 
 
 c 
 bination is a symbol for the sum; as, g should suggest 
 
 to the mind seventeen rather than 8 and 9. These combin- 
 ations are similar to words in reading; as, for example, when 
 one sees the symbol "lion" he thinks of the meaning of the 
 word lion rather than the separate symbols "1-i-o-n." 
 
 Step 2. In learning the addition table, drill much on 
 combinations of numbers of two orders and a digit; as, 
 17 54 37 26 24 36 48 29 
 98657374 
 
 and the addition of numbers from 11 to 18 inclusive to 
 
 larger numbers of two orders; as, 
 
 28 34 46 79 63 57 85 91 
 12 17 15 18 13 16 14 19 
 
 12 
 
SUBTRACTION 13 
 
 The former group is necessarily involved when a column 
 of figures is being added; the latter, in more rapid addition. 
 
 3 Thus, if the combinations of the addition table 
 
 7 have been learned and facility in Step 2 acquired, 
 
 5 the column can be added doubly fast. 
 
 9 In adding up the column at the left, one sees 
 
 4 eleven in 6 and 5, then sixteen in 9 and 7, and 
 
 6 thinks 27; then seeing seven in 4 and 3, thinks 
 
 8 34; then 49, 59, 73, 83 in order. Adding 
 
 7 downward, he sees the sums in order: 10, 
 
 3 24, 34, 49, 56, 72, 83. 
 
 4 Thus addition depends (1) upon a quick 
 7 recognition of the thirty-six elements of the addi- 
 
 9 tion table, and (2) facility in applying them. 
 
 5 Additional drill material should be supplied 
 
 6 and utilized until facility is acquired. See p^ge 
 39, note. 
 
 SUBTRACTION 
 
 Subtraction is the process of finding how much greater 
 one of two numbers is than the other; as, 8 — 3 = 5, means 
 that eight is five greater than three. 
 
 The answer in subtraction is the difference, or remainder. 
 
 The greater number is the minuend. 
 
 The lesser number is the subtrahend. 
 
 Facility in subtraction depends upon skill in recognizing 
 what digit added to another digit will produce a given 
 number; as, 8 and what are 12? 6 and what axe 13? 
 
 Find the difference between 3,426 and 9,878. 
 
 9878 Solution: 6 and 2 are 8; 2 and 5 are 7; 
 3426 4 and 4 are 8; 3 and 6 are 9. 
 
 6452 
 Subtract 3,623 and 1,747 
 
 3623 Solution: 7 and 6 are 13; 5 (one to carry) and 7 are 12; 8 
 
 1747 (one to carry) and 8 are 16; 2 and 1 are 3. 
 
 1876 
 
14 AGRICULTURAL ARITHMETIC 
 
 Subtract 30,000 and 93 
 30000- Solution: 3 and 7 are 10; 10 and = 10; 1 and 9 = 10 
 93 1 and 9 = 10; 1 and 2 = 3. 
 
 29907 
 
 The process is the same as in adding the subtrahend to the re- 
 mainder. The merchant in making change subtracts in the same 
 way. 
 
 The advantage of this method is that one utilizes one's knowledge 
 of the addition table instead of learning a new table, and hence econo- 
 mizes energy. There are many methods of "borrowing" in subtrac- 
 tion, but experience has demonstrated beyond the possibility of a 
 doubt that the addition method is the easiest to learn and the most 
 rapid in practice. Pupils should drill until they are skillful in this 
 work. 
 
 MULTIPLICATION 
 
 Facility in multiplication depends upon a quick applica- 
 tion of the multiplication table. Hence it is necessary 
 that the combinations of the multiplication table be known 
 at sight. (The table on page 12 constitutes the multiplica- 
 tion table, if the numbers are used as factors.) The addition 
 and multiplication tables should never be taught together; 
 for confusion may arise. The subtraction form of the addi- 
 tion table and the division and factoring forms of the mul- 
 tiplication table make excellent drills for the respective 
 tables, besides being valuable in themselves. (Page 17.) 
 
 Sufficient material for drill should be provided to pro- 
 duce efficiency. 
 
 DIVISION 
 
 Step 1. Provide many exercises upon the division form 
 of the multiplication table, followed by such examples as 
 18 -5-4, 63 4- 8, 42 -5- 5. 
 
 Step 2. Application of Step 1 in short division with 
 the usual form: 6) 2573 
 
 428-5 
 
203 
 186 
 
 CHECKING 15 
 
 Step 3. Long division in the following form: 865 
 Two essentials should here be observed: the 31 ) 26834 
 
 proper placing of the first figure of the quotient, 24s 
 and facility in the use of the trial divisor. 
 
 The first figure of the quotient should be 
 placed directly over the last figure of 
 
 the first partial dividend and the other figures 174 
 
 should follow in order. The value of this method 1_5J> 
 
 will be seen in division of decimals. (Page 41.) 19 
 
 Much drill will be needed in acquiring facility in esti- 
 mating the quotient figures by means of a trial divisor. 
 The larger the second figure from the left of the divisor, the 
 more allowance must be made for carrying in multiplying 
 by the quotient figure. When the second figure is 6 or 
 more, it is usually better to use the first figure plus one for 
 the trial divisor. As, in dividing 9,376 by 387, try 4 into 9, 
 in estimating the quotient figure. 
 
 CHECKING OR PROVING 
 
 Mistakes due to careless figuring are costly in any bus- 
 iness. He is a poor business man indeed who does not 
 check his figures carefully before acting, upon them. Con- 
 fidence and reliability arise from the habit of proving the 
 correctness of one's own work, not in depending upon 
 others. A person can be instructed wherein he is ignorant, 
 but the careless habit is fatal. 
 
 Addition is usually checked by adding the column in 
 the opposite direction. The two sums should tally. 
 
 Subtraction is best checked by adding the subtrahend 
 and remainder and noting whether the sum tallies with the 
 minuend. 
 
16 AGRICULTURAL ARITHMETIC 
 
 Checks for Multiplication. 
 
 Multiplication may be checked by (1) reviewing the 
 work; (2) dividing the product by one of the factors and 
 noting whether the quotient agrees with the other factor; 
 (3) "ca,sting out nines." The latter is the shortest and 
 best method. In teaching, select one of the checks and 
 omit the others. 
 
 In the process of casting out the nines from a number, 
 drop from the sum of the digits a nine whenever nine is 8863 
 
 reached in the process of addition. 237 
 
 Cast out the nines in 4,379,326. 62041 
 
 Solution: 4 + 3 + 7 - 9 = 5. 5+3+2-9 = 1. 26589 
 
 1+6 = 7. The remainder is 7. Or, beginning at the 17726 
 right, drop 6 and 3, the 9 and 2 and 7 which leaves the 2100531 
 remainder of 3 + 4 or 7. As a check in multiplication, 
 cast out the nines in each factor and out of the product 
 of the remainders thus obtained. Note whether the result agrees with 
 the remainder after casting out the nines of the product of the multi- 
 plier and multiplicand. 
 
 As, in the above example, the remainder for the multiplicand is 
 7, for the multiplier is 3, for 21, the product of 3 and 7, is 3, as the 
 remainder for the product, 2,100,531, is 3, the work is correct. 
 
 Checks for Division. 
 
 The most common check for division is tp 
 multiply the divisor by the quotient, add the 
 remainder, and compare the result with the divi- 21)79637 
 dend. _63_ 
 
 The shortest check for division is the process of cast- 166 
 
 ing out the nines. 147 
 
 Cast out the nines in divisor and quotient separately 193 
 
 and find the product of the remainders thus obtained and 189 
 
 add the division remainder. Cast out the nines from this 47 
 
 sum. This remainder should agree with the remainder 42 
 
 obtained by casting out the nines from the dividend. "" 5 
 
 As, 3X3 + 5—9=5. The remainder obtained 
 by casting out the nines from the dividend is 5. 5 = 5. 
 .". the division is correctly performed. 
 
 FACTORING 
 
 Skill in factoring is helpful in acquiring facility in the 
 solution of many problems. This facility is valuable in 
 
 3792 
 
FACTORING 17 
 
 abbreviating the processes in problems involving con- 
 tinued operations of multiplication and division of simple 
 numbers and fractions (called cancellation), least common 
 multiple, computation of areas and contents, and many 
 others. Long processes can sometimes be so abbreviated 
 that the results can be obtained mentally or with but little 
 written work. 
 
 A composite number is one that can be divided exactly 
 by some other number than one or itself; as, 18, 36, 152, etc. 
 
 A prime number is one that is not exactly divisible by 
 any number except one and itself; as, 2, 7, 29, 37, 101, etc. 
 
 A composite factor is a composite number used as a factor; 
 as, 8 is a composite factor of 48; 16 of 96; 6 of 42, etc. 
 
 A prime factor is a prime number used as a factor; as, 7 
 is a prime factor of 42; 11 of 44; 13 of 91, etc. 
 
 The essentials for facility in factoring are a memory of 
 the two factors that produce the composite numbers from 1 
 to 100 and a few numbers frequently used above 100, the 
 squares of numbers from 1 to 20, the cubes of the digits, 
 and the prime numbers from 1 to 100. These tables of 
 combinations should be thoroughly memorized. A part of 
 them can be learned with the multiplication table, and 
 they will serve as an excellent drill in learning that table. 
 (Page 14.) 
 
 When the factoring table is learned, any of the above 
 numbers can readily be reduced to prime factors; as, 
 
 72 = 8 X 9 (table) = 2 X 2 X 2 X 3 X 3, or 2 s X 3 2 . 
 
 96 = 8 X 12 (table) = 2X2X2X2X2X3, or 2 s X 3 
 
 Such facility should be acquired that one can tell at 
 sight the number of times a given factor is found in any 
 number of two orders; as, 24 contains three 2's and one 3; 
 56 contains three 2's and one 7; 84 = one 7, one 3 and two 
 2's, etc. Pupils should learn to use the exponential form 
 
 2— 
 
18 AGRICULTURAL ARITHMETIC 
 
 for expressing the number of times any given factor is found 
 in a number; as, 8 = 2 3 ; 27 = 3 3 ; 16 = 2 4 ; 64 = 2 6 ; 72 = 
 2 3 X 3 2 ; 28 = 2 2 X 7. 
 
 If taught to read such an expression as 2 3 X 3 2 , as 3 factor two's 
 times 2 factor three's, the meaning would be more apparent and the 
 later work in algebra would not prove to be such a puzzle. The sub- 
 ject of factoring is too often ignored. Careful attention to its details 
 will add much to facility in computation in later work. 
 
 FRACTIONS 
 
 A fraction is an expression of number in which the unit 
 is any given part of the unit one. 
 
 f of a foot means 3 units, each of which is a fourth of a 
 foot; | of an apple can be expressed as 2 thirds of an apple, 
 in which 2 is the number and a third of an apple is the 
 unit. (Page 9.) 
 
 The numerator is the number above the line. It is 
 always in all its relations a number (page 9), and tells how 
 many times the unit, one over the denominator, is taken. 
 In -f , the 3 tells how many times the unit \ is taken. It is 
 to be dealt with as any other simple number. 
 
 The denominator is the number below the line, and 
 indicates the unit by telling into how many parts the original 
 unit or one is divided. In £ , the 4 tells that one was divided 
 into 4 equal parts, and, therefore, the unit of the numerator 
 3 is \. The number is 3 and the unit is \, and the expres- 
 sion f is a quantity (page 9) which should be thought of and 
 treated like any other quantity. It is in its nature like 3 
 pints. The denominator 5 indicates that the unit is \. The 
 denominator 7 indicates that the unit is \, etc. The fact 
 that its unit is indicated by a denominator, however, makes 
 it appear to be different from other quantities. But by 
 carefully considering its number and unit relations the dim- 
 
FRACTIONS 19 
 
 culties in fractions can largely be eliminated. The following 
 comparisons illustrate this idea: 
 
 | + f = \ . 3 pints + 2 pints = 5 pints. 
 
 f — J = \. 3 pints — 2 pints = 1 pint. 
 
 7X1=^-. 7X3 pints = 21 pints. 
 
 One third of \ = f , one third of 6 pints = 2 pints. 
 
 3 quarts and 2 gallons can be added only by reducing 
 both quantities to the same unit; as, 3 quarts + 8 quarts = 
 11 quarts; or 6 pints + 16 pints = 22 pints. The answer 
 in either case can be reduced to the desired unit. In adding 
 4 and f the same is true; they can be added only as they are 
 expressed in the same unit, f of anything equals £ of the 
 same thing. Then the addition becomes 1 + 1 = ^- 
 
 The same principle holds in subtraction. 
 
 To understand the processes of fractions, certain prin- 
 ciples need to be established: 
 
 Principle I. If the numerator of a fraction is multiplied 
 or divided by any number, the fraction is multiplied or divided, 
 respectively, by that number. 
 
 This is apparent when one thinks of a fraction as a 
 quantity consisting of a number and a unit. 6 fourths is 
 twice as large as 3 fourths, just as 6 bushels is twice as large 
 as 3 bushels, or 6 units of any kind is twice as large as 3 
 units of the same kind. In general: If the numerator is 
 increased, the fraction is increased, if the numerator is dimin- 
 ished, the fraction is diminished. 
 
 . Principle II. If the denominator of a fraction is mul- 
 tiplied or divided by a number, the fraction is divided or multi- 
 plied, respectively, by that number* 
 
 2§j = f. To show why f- is less than f : £ means 
 that one was divided into 4 equal parts and one of them 
 
 ♦This principle must be thoroughly worked out and understood, for herein 
 lies the apparent difference between' fractions and other numbers. 
 
20 AGRICULTURAL ARITHMETIC 
 
 expressed. % means that one was divided into 8 equal 
 parts and one expressed." As the one, or the whole, was 
 divided in the latter case into twice as many parts, each 
 part must be one half as large; therefore, £ is half as large 
 as \. Since £ is half as large as -J, 3 of the eighths must be 
 half as large as 3 of the fourths. 
 
 In a similar manner, it can be shown that the fraction is 
 multiplied by a number when the denominator is divided 
 by the number. 
 
 In general: If the denominator of a fraction is made 
 smaller, the fraction is made larger; and, if the denominator 
 is made larger, the fraction is made smaller. 
 
 From these principles the following rules are derived: 
 
 Rules : To multiply a fraction by a number, multiply the 
 numerator or divide the denominator by that number. 
 
 To divide a fraction by a number, divide the numerator or 
 multiply the denominator by that number. 
 
 It is always better to divide one of the terms, when the 
 division is exact, than to multiply, as the terms of the re- 
 sulting fraction will be smaller. 
 
 . 4X|=f, or 4 XI = ¥= H- f-3=f,orf4-3 
 = ^, which is equal to f- . It will be observed that the 
 division process in each case is the shorter. 
 
 Sometimes one term can be divided by one factor of a 
 number and the other term multiplied by the other factor. 
 
 12 X | = ¥• The 8 is divided by 4 and the 5 is mul- 
 tiplied by the other factor, 3. See footnote on next page. 
 
 f -f- 15 = ^. 6 is divided by the factor 3 and 7 is mul- 
 tiplied by the other factor. See footnote on next page. 
 
 2X3 pints = 6 pints, or 3 quarts. In the former 
 result the number was doubled, in the latter the unit was 
 doubled. 
 
KINDS OF FRACTIONS 21 
 
 2 X | = 4 or {. In a similar manner in £ the num- 
 ber, 3, was doubled; in f the unit, 1 fourth, was doubled. 
 The result is the same, £ = f . The process of operating 
 upon the unit in fractions is different from that in other 
 quantities, but the meaning is the same.* 
 
 Principle HI. If both terms of a fraction are multiplied or 
 divided by the same number, the value of the fraction is not 
 changed. 
 
 This is apparent by application of Principles I and II, 
 for to multiply both terms of a fraction by the same number 
 is to both multiply and divide the fraction by that number; 
 hence its value remains unchanged. 
 
 The division of both terms can be shown in a similar 
 manner to produce no effect upon the value of the fraction. 
 
 8 _ * _ 2 _ 1 . 5 -_1.3_8_12 o+f, 
 
 4 — 8 — 16 of/. 
 Z — TO" ~ TO") euu 
 
 KINDS OF FRACTIONS. 
 
 When fractions are classified in relation to the unit one, 
 they are divided into two classes, Proper and Improper 
 Fractions. 
 
 Proper fractions are those that are less than one; their 
 numerators are less than their denominators. 
 
 § is less than one. f means 2 of the third parts of one. 
 f of an apple means 2 parts of an apple, each a third of one 
 apple, f, T 9 T , \, f , are proper fractions. Why should their 
 numerators be less than their denominators? 
 
 Improper fractions are those that are equal to one or are 
 greater than one; their numerators are equal to their denom- 
 inators or are greater than their denominators. 
 
 f = 1, £ = 1, |- of a quart = 1 quart. Show why? 
 
 *The fact that there are two ways of multiplying or dividing a fraction is 
 not a new idea. By again considering the unit and number signification of the 
 fraction, and comparing with other quantities, the meaning of the processes will be 
 seen to be the same. 
 
22 AGRICULTURAL ARITHMETIC 
 
 f of a dollar equals 2 dollars. How much is \ of a dollar? 
 f of a dollar? What is the value of all fractions in which 
 the numerator equals the denominator? Compare such a 
 fraction with one in which the numerator is greater than the 
 denominator. Why, in an improper fraction, should the nu- 
 merator be equal to or greater than the denominator? 
 
 When fractions are classified with relation to their form, 
 they are usually divided into three classes: simple fractions, 
 mixed numbers, and complex fractions. 
 
 Simple fractions are those that have one numerator and 
 one denominator; as f , ■£, -^f , ^ft, etc. 
 
 Mixed numbers are fractions that are composed of an 
 integer and a fraction; as, 3f, 5f,"l£, 18f, etc. 
 
 Complex fractions are those that have a fraction in the 
 numerator or a fraction in the denominator or in both; as, 
 
 §f, i 4 _, 5, ?A 2f,etc. 
 4 6 3| f 4f 3^- 
 
 Which fractions are proper fractions in the illustrations 
 given under the classifications according to their forms? 
 Which improper? Classify the following according to form: 
 
 18 325 c2 15 0±1 8f 16, 3£ 125 36 
 | 17 r ' 112 *' 4 7 f; 28 142. 
 
 Classify the above list of fractions in relation to the unit 
 one. 
 
 REDUCTION OF FRACTIONS 
 
 A fraction is in its simplest form when it is a proper 
 fraction with its numerator and denominator prime to each 
 other (i. e., have no common factor), or when it consists of 
 an integer and a fraction in its simplest form. 
 
 An answer to a problem should always be expressed in 
 its simplest form, unless the problem specifically states 
 otherwise. 
 
LEAST COMMON MULTIPLE 23 
 
 Reduce ty to its simplest form. 
 
 Solution: Since 7 sevenths = 1, 24 sevenths will equal as many 
 ones as 7 sevenths is contained times m 24 sevenths, which is 3 times, 
 with 3 sevenths remaining. Therefore V = 3f . 
 
 Reduce 5f to an improper fraction. 
 
 Solution: Since 1 = |, 5 will equal 5 X I = ¥■ Y + I = 
 
 Reduce to improper fractions the following: 1\, 5f , 4f, 
 17*, 128|, 91f 
 
 Reduce to their simplest form the following: -^-, ±£-, *£-, 
 
 128 24, 21 162 356 512 
 ~~S~l ~5~1 ~T~> ^~~ 1 "16 I ~6T~- 
 
 Reduce f-| to its simplest form. 
 
 Solution: As 24 and 28 have a common factor, If can be re- 
 duced to lower terms by Principle III. 1$ = ?. 
 
 Reduce the following fractions to lowest terms: %%, ^, 
 
 5 7 i 3 18 12 19 65 34 
 
 TSi TS> ~5i T2~> TSi ~S~S> TT> FT) TS" - 
 
 LEAST COMMON MULTIPLE 
 
 In order to add or subtract fractions they must be ex- 
 pressed in terms of a common unit; hence a common unit 
 for these fractions must be found. As they cannot be ex- 
 pressed in a common unit of lower terms, they must be 
 reduced to higher terms. Hence both terms of each frac- 
 tion must be multiplied by such numbers respectively as 
 will make the denominators alike. (Principle III.) 
 
 A multiple of a number is one that exactly contains the 
 given number. 
 
 16 is a multiple of 4 because 16 exactly contains 4. It is 
 also a multiple of 2 and 8. 
 
 A common multiple of two or more numbers is one that is 
 a multiple of each of them. 
 
 24 is a common multiple of 4, 8, and 12. 48, 72 and 96 
 are common multiples of 4, 8 and 12. Name other com- 
 mon multiples of these numbers. 
 
24 AGRICULTURAL ARITHMETIC 
 
 A least common multiple of two or more numbers is the 
 smallest number that is a common multiple of them. 
 
 36 is the least common multiple of 4, 6, 9, and 12. 
 Why? Apply the definition. 
 
 The process of finding the least common multiple of 
 numbers depends upon the following principle: 
 
 Principle: The multiple of any number must contain 
 each prime factor of that number. 
 
 Any multiple of 24 must contain three factor two's and 
 one three. Any multiple of -36 must contain two factor 
 three's and two factor two's, etc. 
 
 To find the 1. cm. of 8, 12, and 15, it is necessary to have the pro- 
 duct of 2 3 , 3, and 5, which equals 120, for there are 3 two's (factors) 
 in 8, a 3 in 12, and a 5 in 15. /. the 1. c. m. of 8, 12, and 15 is the 
 product of 2 3 , 3 and 5, or 120. 
 
 Rule: To find the I. c. m. of any set of numbers select 
 each prime factor found in them the greatest number of times it 
 occurs in any of them and multiply the factors together. 
 
 If factoring is understood and this rule followed, the 
 1. c. m. of all numbers of one or two orders can be worked 
 mentally. 
 
 *Find the 1. c. m. mentally of the following: 
 
 1. 8, 15, 20. 3. 42, 18, 14. 5. 24, 15, 20, 40. 
 
 2. 16, 24, 12. 4. 36, 18, 12. 6. 14, 28, 56, 42. 
 
 7. 4, 12, 15. 8. 56, 63, 36, 42. 
 
 ADDITION AND SUBTRACTION OF FRACTIONS 
 
 Add f, 1, and ■&. 
 
 Solution: I = if, | = if, A = \\ ■ They can now be added, 
 as they have a common unit; viz. jfe. if + if + if =■= li = If}. 
 
 Subtract f from £. 
 
 Solution: J = ft. f = tf. H - tt - II- 
 
 * This work can be taught and drilled upon while teaching factoring. 
 
SPECIAL DEVICES 25 
 
 Required to add £ , f , -&. Reduce to twentieths. 
 
 2 _1_ 3 _L 7 _ 8 J. IS J. 14 _ 87 _ 117 
 
 Required to subtract £ from f. Reduce to fifteenths. 
 
 8 5 . . 1 
 
 Such problems should be added or subtracted mentally. 
 
 MENTAL PROBLEMS 
 
 AAA- 2 3 1.1 3 11.15 02 05.03 01.5 3 5 
 Add. 3-, -j, ^, -$, T , l-£, 1 T , *%, £> 7 , 4f, 4f, -g, -j, ^-. 
 
 B_ 7 . 7 _ 8 . 42 13.K3_97._S| 3 . 05. 020 
 
 In the following add and express only the answers: 
 
 (1) (2) (3) (4) (5) (6) 
 
 128| 196£ 172^ 52£ 27£ 821| 
 
 16| 216| 18£ 632^ 387^ 426f 
 
 132J 81^. 316f 412^ 24| 127| 
 
 284i 36f 508^ 386£ 498f 84| 
 
 961£ 183| 113| 496A 346£ 119£ 
 
 342| 14f 249i 82£ 27£ 16| 
 
 Subtract: 
 
 
 
 
 
 
 (1) 
 
 (2) 
 
 (3) 
 
 (4) 
 
 (5) 
 
 (6) 
 
 24f 
 
 136* 
 
 316^ 
 
 365| 
 
 l,324f 
 
 242| 
 
 16| 
 
 18f 
 
 (8) 
 
 UH 
 
 113| 
 (10) 
 
 876^ 
 
 156f 
 
 (7) 
 
 (9) 
 
 (11) 
 
 (12) 
 
 356| 
 
 42J 
 
 318| 
 
 412# 
 
 856^ 
 
 219| 
 
 138f 
 
 28f 
 
 227^ 
 
 319f 
 
 416f 
 
 218| 
 
 Special Devices 
 
 i + Ti r + h 
 In similar examples, in which the numerators are 1, by 
 observing the operation it can be seen that one adds the 
 denominators for the numerator and multiplies the denom- 
 inators for the denominators. 
 
 i _ i i _ i 
 
 T 41 T p 
 
 Make a similar rule for subtraction. 
 
26 AGRICULTURAL ARITHMETIC 
 
 f +f ; f +1; i+r- To shorten the process, when the 
 numerators are alike, multiply the sum of the denominators 
 by the numerator for the numerator and take the product 
 of the denominators for the denominator; as, £f, ff, £f. 
 Reduce these results to simplest form, f — f , f — f, £— -fc. 
 
 Make a rule for abbreviating the process of subtraction. 
 
 Practice for speed in addition and subtraction using the 
 following list; as, % + i, y + i. i~h t ~ h etc - Repeat 
 the process several times. 
 
 (1) (2) (3) (4) 
 
 1 
 
 1 
 
 1 
 
 1 
 
 2 
 
 i 
 
 3 
 
 3 
 
 ■5- 
 
 T 
 
 4 
 
 T 
 
 ¥ 
 
 T 
 
 ¥ 
 
 1 
 
 l 
 
 1 
 
 l 
 
 3 
 
 3 
 
 2 
 
 2 
 
 7 
 
 "S" 
 
 s 
 
 T 
 
 7 
 
 ¥ 
 
 T 
 
 T 
 
 1 
 
 l 
 
 1 
 
 1 
 
 5 
 
 ♦ 
 
 5 
 
 5 
 
 4 
 
 S 
 
 ¥ 
 
 7 
 
 ¥ 
 
 ¥ 
 
 7 
 
 1 
 
 1 
 
 1 
 
 I 
 
 4 
 
 4 
 
 1 
 
 1 
 
 ¥ 
 
 ¥ 
 
 ■3" 
 
 4 
 
 5 
 
 9 
 
 4 
 
 T2~ 
 
 1 
 
 l 
 
 1 
 
 1 
 
 3 
 
 3 
 
 1 
 
 1 
 
 ■g- 
 
 ¥ 
 
 7 
 
 s 
 
 T 
 
 7 
 
 T 
 
 TT 
 
 1 
 
 1 
 
 1 
 
 - 1 
 
 5 
 
 5 
 
 1 
 
 l 
 
 4 
 
 7 
 
 T 
 
 r 
 
 7 
 
 ¥ 
 
 "ff 
 
 ■9" 
 
 1 
 
 1 
 
 1 
 
 i 
 
 2 
 
 2 
 
 i 
 
 1 
 
 T 
 
 TT 
 
 T 
 
 ttf 
 
 S 
 
 T 
 
 TIF 
 
 WRITTEN PROBLEMS 
 
 If it becomes necessary to add several large fractions, 
 with numerators and denominators of two orders, the fol- 
 lowing form for the computation will assist in abbreviating 
 the work: Add 2f, 15^, 5^, and 21^. 
 
 The new denominator is set over the new 2 3 X 3 X 5 = 120 
 numerators to be out of the way for the addi- 
 tion of the numerators; the factors of the new 
 denominator, 2 3 , 3, and 5, are recorded to assist 
 in the division of the new denominator by the 
 denominators of the factions. To divide 120 by 
 8; by looking at the factors itcan be seen men- 
 tally that 3X5 is the quotient; to divide 120 by 
 24, by looking at the factors, it is clear at sight 
 that 5 is the quotient. 
 
 The reduction of f^gg to simplest terms is ac- 
 complished without unnecessary written work. 
 To test whether j% is in its lowest terms, test 
 
 whether any factor 2, 3, or 5 is contained in 86; finding 2, reduce the 
 fraction, and express it in the answer. 
 
 The above is sufficient recorded work for the solution of the problem. 
 
 2| 
 
 45 
 
 15ft 
 
 50 
 
 5ft 
 
 56 
 
 21H 
 
 55 
 
 44|§ 
 
 206 
 
 
 120 
 
 
 86 
 
3 2 X 5 X' 2 
 
 =."90 
 
 327/ g 
 223[| 
 
 42 
 132 
 
 85 
 
 103J5 
 
 
 47 
 
 MULTIPLICATION OF FRACTIONS 27 
 
 Solve the following in the given form using no more 
 written work than is necessary. 
 
 1. 14H , 21«, 43||. 4. 6^,4^,22^ 
 
 2- 8-^-, 26^-, 126ff . 5. f, yig-, f, x%. 
 
 3- Ai Tifi M; If- 6- 3f, 4^, 7f^. 
 
 Much greater facility can be acquired by a large use of 
 factors which this form encourages. 
 
 A similar form for subtraction of fractions is given. 
 Subtract 327^ and 223H- 
 
 The numerator of |B, borrowed, is added to 
 the numerator of }g and the result expressed just 
 above the numerator of |{J convenient for the sub- 
 traction. 
 
 Solve in the above form the following: 
 
 1. 8A - 5,V 3. 26^ - 15^. 5. 17« - 4fV 
 
 2. 16A - m- 4. 31H - 4A- 6. 25^ - llff. 
 
 MULTIPLICATION OF FRACTIONS 
 
 3X# should be read 3 times f . How much is 3 X f ? 
 
 2 X 1? 4 X f . 
 
 Many such examples should be worked by applying Examples I and 
 II (page 17). This should be continued until the pupil is conscious 
 of the meaning of multiplication of a fraction by an integer. 
 
 6 X * = ■¥ = 4f- 5 X A = i = H. 
 
 Solve: 
 
 7 X f , 3 X f, 2 X I, 8 X XT) 16 X A, 5 X -fs, 
 
 3 X A, 14 X I, 7 X 1, 11 X A- 
 
 Note : In problems of this kind it is better to divide the denomi- 
 nator when it can be done without a remainder. 
 
 In almost all cases of multiplication of fractions the 
 simplest rule to be employed is the following: 
 
 Rule : Multiply the numerators together for the numerator 
 of the product and the denominators together for the denomi- 
 nator of the product, abbreviating by cancellation, and reduce 
 the answer to its simplest form. 
 
28 AGRICULTURAL ARITHMETIC 
 
 If there are mixed numbers or integers among the factors, 
 reduce the former to improper fractions and regard the 
 latter as numerators, or as having one for the denominator. 
 
 11 
 7| X 9 X 1- Solution: ZJ x 9XJ5 = g5 
 
 13 
 24 X 9f X A- Solution: ?£ X^-f- X ~ = ^ = 32$ 
 
 2 
 Solve the following: 
 
 1. 3 X f X 51 X f. 3. 28f X 6. 5. 8 X 14f. 
 
 2. 51XtX3fXl. 4. 16fX5. 6. 1 X f ; If X 5. 
 
 7. 'jXnX 3^-. 
 8. 3|X8X A- I 2 - 1 X f X A X 2f 
 
 "• t X ^ X ■%• 13. 8 X f X j-g- X 5^. 
 
 10. 8i X A X 7£. 14. 9f X | X |f . 
 
 11. 241 X 18 X A X f 15. fi X 21 X f 
 
 3121 
 If two mixed numbers are to be multiplied, it is some- 15| 
 
 times better to multiply them as mixed numbers. 156J 
 
 In this problem each step in the multiplication is ex- 10 
 
 1560 
 312 
 
 16. 216 X 31. 18. 1246 X 1231. 20. 824 X 141. 
 
 17. 2331 X 16f. 19. 1421 X 3161. 21. 481 X 16f. 
 
 DIVISION OF FRACTIONS 
 
 Divide 18 by f. 
 
 Solution: 18+ $ = 18 X I = 27. 
 
 51 -f- 21. 
 
 Solution: «?Xj = V = 2|. 
 
PROBLEMS OF FRACTIONS 29 
 
 The best general rule for division of fractions is the fol- 
 lowing: 
 
 Rule: Express mixed numbers as improper fractions, 
 think whole numbers as having one for a denominator, invert 
 the divisor and proceed as in multiplication of fractions. 
 
 The pupil should be grounded in the thought, however, 
 
 that to divide a fraction by an integer he must divide the 
 
 numerator or multiply the denominator by the integer. 
 
 (Principles I and II, p. 19.) 
 
 It is always wise in problems of this kind to divide the numerator 
 rather than to multiply the denominator when it can be done without 
 a remainder. 
 
 Compare this method with the rule and note whether 
 
 the two methods are the same. 
 
 Solve : 
 
 1. 18f -f- |. 4. 13# H- 25£. 7. 24 -f- f. 
 
 2. f -h f . 5. 4vf. 8. 18* hh 4f . 
 
 3. 23f + 14*. 6. 23| - § ■ 9. % -=- 5. 
 Division of a mixed number by an integer can sometimes 
 
 be abbreviated, if the divisor is not large, by proceeding as 
 in short division; as, 
 
 Divide837|by7. 
 
 Solution: 7) 837| 
 
 H9jf 
 The 4 remainder = ^ which added to f = ¥■ V ' + 7 = li- 
 Divide 147f by 6; 352| by 7; 1,422 by 13£. 
 
 THE THREE PROBLEMS OF FRACTIONS 
 
 A problem that occurs and recurs in various phases is 
 one involving two factors and their product. In its sim- 
 plest form all three numbers are integers. From it arise 
 three problems; viz: If three numbers are in the relation of 
 two factors and a product, given any two of them to find 
 
30 
 
 AGRICULTURAL ARITHMETIC 
 
 the other; as, (1) Given 5 and 6, to find the product. (2) 
 Given the factor 5 and the product 30, to find the other fac- 
 tor. (3) Given 6 and 30, to find the other factor. 
 
 When one of the factors is a fraction, the following forms 
 of the respective problems arise: 
 
 I. To find a fractional part of a number; as, 
 
 To find 1 of 24. Solution | of 24 = 6, f of 24 = 3 X 6 
 = 18. 
 
 The pupil must realize that the numbers given are 
 factors and the product is to be found. (See multiplication 
 of fractions.) 
 
 II. To find what fractional part one number is of an- 
 other; as, 9 is what part of 12? This involves the mean- 
 ing of a fraction; as, 1 = -fa of 12; then 9 = &, or f of 12. 
 Verify by finding f of 12. 
 
 2 is what part of 5? 2 = f of 5. 
 
 8 is what part of 17? 9 is what part of 28? 372 is what 
 part of 976? 
 
 III. To find the number of which a given number is a 
 certain fractional part; as, 12 is y of what number? (-f- X ? 
 
 = 12.) 
 
 Since 12 is 4 parts or sevenths of a number, one of the 
 sevenths is -J of 12 or 3, and the whole number, or 7 sevenths 
 of it, is 7 X 3 or 21. Therefore, 12 is a f 21. Verify by 
 finding T of 21, or by finding what part 12 is of 21. 
 
 II and III are but phases of the problem, given a 
 product and one factor to find the other factor, and hence 
 are problems in division. 
 
 It is important to understand these three problems thoroughly 
 and practically to reduce their solution to the automatic; for, in addi- 
 tion to their value in the realm of fractions, the solution of all prob- 
 lems of percentage rests upon one or the other of them. 
 
AREA OF RECTANGLE 
 
 31 
 
 PROBLEMS 
 
 1. I sold | of a farm for $1,260. What is the whole 
 farm worth? 
 
 2. How much should I receive for the f of my farm at 
 the same price per acre, if the whole is worth $1,650? 
 
 3. How many pounds of nitrogen in 5 tons of a com- 
 mercial fertilizer testing -fa nitrogen? 
 
 4. Two men planted a field of corn and agreed to divide 
 the corn in proportion to the number of days' work done by 
 each. One worked 27 days and the other 36 days. What 
 part of the product should each have? If the yield is 1,264 
 bushels, how many bushels should each have? 
 
 5. In example 4, if one worked f- as many days as the 
 other, how many bushels should he get? 
 
 THE AREA OF A RECTANGLE 
 
 Area is a quantity (page 9) which names the unit of 
 measurement and tells how many times the unit is con- 
 tained in a given surface; as 25 square inches, 87 square 
 rods, 36 acres. 
 
 The unit of area is a square with a given linear unit for 
 a side; as, a square inch, square rod, etc. 
 
 Note: The one exception to this definition is the acre, which 
 means 160 square rods, without regard to shape. 
 
 In measuring a rectangular surface, the problem is to 
 find the number of given units in the rectangle. The fun- 
 damental method is to divide the 
 rectangle into units, or to lay off the 
 unit repeatedly upon the rectangle and 
 count the number thus obtained. This 
 counting can be shortened by count- 
 ing by rows rather than by squares; 
 in short, multiplying the number in a 
 Figure l. row by the number of rows. 
 
32 AGRICULTURAL ARITHMETIC 
 
 It can be seen readily that there are as many squares 
 in a row as the number of linear units on one side, and as 
 many rows as there are linear units on an adjacent side. 
 Hence the number of square units in the rectangle equals 
 the product of the number of linear unitsin the length 
 multiplied by the number of linear units in its breadth; 
 or, for short. 
 
 Rule: The area of a rectangle equals the product of its 
 length by its breadth. (Page 10) 
 
 The length and breadth of a rectangle are its dimensions. 
 
 The sign X between dimensions should be read by. It 
 has no relation to multiplication. A rectangle 3' X 8' 
 should be read, "a rectangle 3' by 8'," and means that the 
 rectangle is 3' wide and 8' long. 
 
 PROBLEMS 
 
 In each of the following problems indicate the operation and 
 abbreviate by cancellation. (Page 49, problem XII.) 
 
 1. The dimensions of a rectangular field are 80 rods 
 and 160 rods. How many square rods in the field? 
 
 2. A field is 40 X 80 rods, Determine the number of 
 acres. 40 X 80 = 2Q . = 2Q 
 
 160 
 
 3. A room is 12' X 16'. How much will it cost to floor 
 it at $60 per thousand square feet, allowing one quarter 
 extra for matching? Regard a square foot as the unit. 
 
 nioo^ x * X 60 =14 t ••• = $ 14 - 40 - 
 
 4. If the walls are 8' high in example 3, what would it 
 cost to plaster the walls and ceiling at 25 cents a square 
 yard, no allowance for doors and windows? 
 
RECTANGULAR SURVEY 
 
 33 
 
 5. A farm is \ mile X 1 mile. How many acres? What 
 is the meaning of a "forty"*? Usually what is its shape 
 and what are its dimensions? 
 
 6. A farm is 40 X 160 rods. How many acres? If 
 60 X 160 rods? If 160 X 160 rods? 
 
 7. A ball park is 24 X 24 rods. What will the boards 
 cost for a tight fence 8' high around it at $30 per 1,000 feet? 
 
 8. A fair ground is 80 X 120 rods. Find cost of boards 
 for fencing with conditions of problem 7. 
 
 9. A street to be paved is one mile long and 40' wide. 
 What does it cost to grade and pave it at $3.20 a square 
 yard? 
 
 RECTANGULAR SURVEY 
 
 6 
 
 5 
 
 4 
 
 3 
 
 2 
 
 1 
 
 7 
 
 8 
 
 9 
 
 10 
 
 11 
 
 12 
 
 18 
 
 17 
 
 16 
 
 15 
 
 14 
 
 13 
 
 19 
 
 20 
 
 21 
 
 22 
 
 23 
 
 24 
 
 30 
 
 29 
 
 28 
 
 27 
 
 26 
 
 25 
 
 31 
 
 32 
 
 33 
 
 34 
 
 35 
 
 36 
 
 Figure 2. 
 
 
 
 a. 
 Z 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 A 
 
 
 
 
 
 
 
 
 
 
 BASE 
 
 
 
 
 
 
 UHE 
 
 
 
 
 
 
 
 
 
 B 
 
 
 
 
 
 
 
 
 
 
 
 
 
 fl=ras.n.aw 
 
 
 i 
 
 i 
 a. 
 
 
 
 
 
 Figure 3. 
 
 In the older eastern states the farms are usually irreg- 
 ular in shape, and the lines are indicated by metes and 
 bounds. In the western states a new and simpler system 
 was adopted in 1785. The country is laid out into districts, 
 usually a state, and this is divided into strips running north 
 
 *A square containing 40 acres. 
 
34 
 
 AGRICULTURAL ARITHMETIC 
 
 and south, six miles wide, called ranges (R), which are num- 
 bered east and west from some designated line called the 
 principal meridian. These ranges are divided into squares, 
 called townships, six miles on a side, by east and west lines. 
 These townships (T) are numbered north and south from a 
 designated line called the base line. A township is described, 
 then, for example, T 108 N., R 17 W., state of Minn. The 
 townships are divided into sections one mile square, which 
 are numbered consecutively beginning in the northeast cor- 
 ner and counting as indicated in the 
 diagram. A section is known by its 
 number. A section with a given 
 number always occupies the same 
 relative position in a township. 
 
 The sections are divided, down 
 to ten acres, by bisecting the divi- 
 sion lines and drawing lines parallel 
 to the adjacent sides. 
 
 Division D is described as the 
 N E i of Sec 17, T 5 N., R 7 W. 
 B is the S W i of N W i of Sec. 17, etc.; 4,N| 
 of S W i of Sec. 17, etc. Describe C and E. 
 
 
 D 
 
 B 
 
 C 
 
 A 
 
 
 
 E 
 
 Figure 4. A Section. 
 
 PROBLEMS 
 
 1. How many acres in El In D1 In A1 
 
 2. The NW^ofSE^ofa section is worth how much 
 at $70 per acre? 
 
 3. The N i of S E i of N E | is worth how much at $90 
 per acre? Make a drawing to illustrate. 
 
 4. What would it cost to fence in the last named piece 
 of land if posts at 20 cents are set a rod apart, the wire three 
 strands high costs $3.50 a hundred rods single strand, and 
 the labor costs $30. What is the cost of the fence per rod? 
 
RECTANGULAR SOLIDS 35 
 
 5. How many acres in a piece of land consisting of the 
 S | of the N W i and the N \ of S W \ of a section? How 
 many rods of fencing to surround it? How many miles? 
 
 6. How many rods of fencing are required to surround 
 the N i of the N E \ and the N i of the N W i of a section? 
 How many acres are enclosed? How could you arrange the 
 same amount of land so as to require the least amount of 
 fencing? How many rods would be saved? 
 
 VOLUMES OF RECTANGULAR SOLIDS 
 
 Volume is a quantity (page 9) that names the unit of 
 measurement and tells how many times the unit is con- 
 tained in a given solid; as, 16 cu. in., 37 cu. yds. 
 
 The unit of a solid is a cube having a given linear unit 
 on an edge; as, a cubic inch, a cubic yard, etc. 
 
 In measuring a given solid, the problem is to find the 
 number of times it contains a given cubic unit. (Page 10) 
 The best way to develop a rule is to imagine the solid filled with 
 the cubic units and count them. This requires the laying 
 of a row along an edge, then the number of rows in a layer 
 on the bottom, then layer upon layer until full. Hence it 
 must be determined how many cubes there are in a row 
 along an edge, then how many rows in the bottom or base, 
 then how many layers. As the edge of the unit and the 
 linear unit of the edge of the solid are the same, the number 
 of linear units in the edge of the base is the same as the 
 number of cubic units in a row, the number of linear units in 
 another edge of the base is the same as the number of rows in 
 the base, and the number of linear units in the height of 
 the solid is the same as the number of layers in the solid. 
 Therefore the number of cubes in a layer is the product of 
 the number in a row by the number of rows; and the number 
 
36 AGRICULTURAL ARITHMETIC 
 
 of cubes in the solid is the number of cubes in a layer mul- 
 tiplied by the number of layers. 
 
 Given a rectangular solid 5X7X8. Find its volume. 
 
 Solution: Along the edge AB a row of 8 cubic units can be laid, 
 for the edge of the unit is 1 unit long and AB is 8 units long. The 
 layer AC will contain 7 rows, for the row is 1 unit wide and BC the 
 edge of the layer is 7 units long. The solid will contain 5 layers, for 
 the layer is 1 unit high and the edge BE is 5 units long. Hence the 
 layer contains 7 X 8 or 56 cubic units and the solid contains 5 X 56, 
 or 280 cubic units, i. e., 5 X 7 X 8 cubic units. Hence the rule: 
 
 Let the pupil make a drawing from this description, or build a 
 rectangular solid with inch cubes and verify. 
 
 Rule: Multiply together the number of linear units in 
 each of the three dimensions and give the product the name of 
 the required cubic unit. Or, as it is commonly expressed, 
 Multiply together the three dimensions. 
 
 The length, breadth, and thickness of a rectangular solid 
 are its dimensions. Sometimes the product of two of them 
 is called the base, and the dimensions are then termed the 
 base and altitude. 
 
 The dimensions of a solid 7' long, 4' wide, and 3' high 
 are usually expressed as 7' X 4' X 3' and read 7' by 4' by 3'. 
 
 PROBLEMS 
 
 1. The dimensions of a grain bin are 6', 8' and 10'. 
 Find its volume in cubic feet. 
 
 2. How many bushels in the bin in example 1?* 
 
 3. I. H. Sears built a vat If X 2f -X 6' in which to 
 make a spraying solution; allowing 2" of the depth (If) to 
 prevent the solution from boiling over, how many gallons 
 does the vat hold?* How many barrels? 
 
 *A foot = ^j or |bu. 6X8X10X4 = 384. 
 
 For approximate number of bushels in bins when dimensions are 
 given in feet use \\' = 1 bu.; for exact number use 2,150.4. The fol- 
 lowing, also are convenient approximations: 7| gals. = 1 cu. ft. and 
 4J ft. = 1 bbl. (Indicate the operation and abbreviate by cancellation.) 
 
RECTANGULAR SOLIDS 37 
 
 4. A cellar is 18' X 12' X 7'. The owner desires to 
 lay the bottom in cement or grout 4" deep and the side walls 
 6" thick.- How many cubic feet of grout are required? If 
 the grout is mixed in the relation 1 part cement, 3 parts 
 sand, and 5 parts gravel, how many bags of cement and cubic 
 yards of gravel are required?* 
 
 5. I wish to make a cement watering trough 2%' X 6' 
 X 8', sides 8" thick and bottom 4". How much cement, 
 sand, and gravel are required, approximately, if mixed in 
 the proportions 1, 3, and 5? Use outside measurements. 
 
 6. How many barrels will the trough hold with the above 
 as inside measurements? 
 
 7. If, in 5, the measurements are from the outside, 
 except the depth, how many gallons will the trough hold? 
 
 8. How many cords of wood in a pile 4 ft. high, 8 ft. 
 wide, and 80 ft. long? Work mentally, regarding a cord as 
 a pile 4' X 4' X 8'. 
 
 9. Find the value of a pile _ of wood 56' X 12' X 6' 
 at $6 per cord. Work mentally, using cancellation. 
 
 10. A pile of wood is 4 ft. high, 4 ft. wide, 18 ft. long at 
 the top and 22 ft. long at the bottom. What is its real length 
 and how many cords in the pile? 
 
 11. A flat roof is 26' X 36'; how many inches of rain 
 would have to fall upon it in order to fill a cistern 13' X 9' 
 X 12'? 
 
 12. If one and one half inches of rain should fall on the 
 roof in 11, what part of the cistern could be filled from it? 
 
 13. Regarding 12 ft. as the depth of the cistern, how 
 deep would f of an inch of rain fill it? 
 
 14. How many cubic feet of ice 16 in. thick can be cut 
 from an acre ice field, allowing ^ for waste? 
 
 *Approxnnately, 1 bag of cement = 1 cu. ft. 
 
38 AGRICULTURAL ARITHMETIC 
 
 15. How deep would the ice in example 14 fill an ice 
 house 80' X 66', inside dimensions? 
 
 16. Allowing that the waste of ice in harvesting would 
 be offset by waste of space in packing, how large an ice 
 field expressed in acres would be required to fill an ice house 
 33' X 66' X 80' if the ice is 16 in. thick? 
 
 17. My coal bin is 4' X l'O' X 10'. How many tons 
 will it hold, assuming that a cubic foot of anthracite coal 
 weighs 55 lbs. and bituminous coal 50 lbs.? Solve for each 
 kind of coal. 
 
 18. Measure the dimensions of your bin of coal after 
 leveling the top. Weigh a bushel and determine whether 
 your dealer gave you what he charged for. 
 
 19. My wagon box is 3 ft. wide and 10 ft. long. How 
 deep shall I fill it to contain a cubic yard of sand? * 
 
 20. My team can readily draw 2 tons over a given road. 
 How deep shall I fill my box with sand to contain a load?* 
 
 21. How much does a cubic yard of sand weigh? 
 
 DECIMALS 
 
 Decimals are like integers in form and meaning, the 
 only difference being, that the units expressed are tenths, 
 hundredths, etc., instead of ones, tens, hundreds, etc. 
 In the number .8273 the unit of 8 is a tenth, of 2 a hundredth, 
 of 7 a thousandth, and of 3 a ten-thousandth; while in 8,273 
 the unit of 8 is a thousand, of 2 a hundred, of 7 a ten, and 
 of 3 a one. 
 
 The plan of reading is the same; viz, read the number 
 as if it were an integer and give it the name of the unit of 
 right-hand figure. .8273 is read eight thousand two hundred 
 seventy-three ten-thousandths; while 8,273 is read eight 
 thousand two hundred seventy-three ones. 
 
 *Sand weighs nearly 1^£ times as much as water. 62}4 lbs. is the weight of a 
 cubic foot of water. 
 
MULTIPLICATION OF DECIMALS 39 
 
 In the latter case the "ones" is usually omitted, as it is 
 understood, for the unit is always the same, while in decimals 
 the unit of the right-hand figure is variable. Hence facility 
 in reading decimals requires that the names of the orders 
 be memorized, by noting their number from the decimal 
 point; -viz., the first, tenths; the second, hundredths; the 
 third, thousandths; the fourth, ten-thousandths, etc. It is rare 
 that any order need be considered above the third or fourth, 
 and in an answer above the second. 
 
 The units at the right of the decimal point are dealt with 
 in the processes just the same as those at the left, for they 
 are organized on the same decimal system. As the names 
 of the various units or orders are known by their distance 
 from the decimal point, great care should be exercised to 
 keep the decimal point in its proper place, and the rules 
 governing the location of the point should be carefully learned. 
 
 In addition, as the name of the unit of the sum is. the 
 same as the name of the unit added, the point of the sum 
 should be expressed under the point of the addends. 
 
 In subtraction, a corresponding rule applies. 
 
 Note : As addition and subtraction of decimals are exactly the same 
 as addition and subtraction of simple numbers, and the rule for the 
 point is so simple, this subject can be taken up with the work of ad- 
 dition and subtraction of simple numbers and completed in less time 
 than if treated separately. 
 
 MULTIPLICATION OF DECIMALS 
 
 In multiplication, to understand the rule for the point, 
 express the multiplier and multiplicand in the form of com- 
 mon fractions, and multiply. It can readily be seen that the 
 point should be located so as to give the product as many 
 places as the sum of those in the multiplier and multiplicand. 
 
 2.37 X 4.3 -fjgx I 3 = S =10.191 
 
40 AGRICULTURAL ARITHMETIC 
 
 Rule: Multiply as in simple numbers and point off as 
 many places in the product as there are decimal places in both 
 multiplier and multiplicand. 
 
 As the numbers in the respective orders mean the same 
 as in integers; i. e., they tell how many units there are, the 
 one new thing to learn is how to determine the unit of the 
 answer, or, in other words, how to locate the point. This 
 process should be thoroughly memorized and carefully ap- 
 plied. 
 
 The check for the answer is the same as in multiplication 
 of simple numbers. (Page 16.) 
 
 Solve the following and check the results: 
 
 1. 18.73 X .043 4. 353.86 X .24 7. 36.8 X 34.7 
 
 2. 5.43 X 3.24 5. .0236 X 48.96 8. 368 X .347 
 
 3. 1.893 X .18 6. 36.8 X 3.47 9. .07 X .007 
 
 10. Barbed wire costs $.03 a pound and runs a pound 
 to a rod. How much does it cost to supply the wire for a 
 fence 5 strands high and 75.23 rods long? Retain only two 
 decimal places in your answer. 
 
 11. If four-inch drain tile costs $.04 a foot, how much 
 would the tile cost for a ditch 23.6 rods long? 
 
 12. A field is 23.5 rods long by 18.3 rods wide. How 
 many square rods in the field? 
 
 DIVISION OF DECIMALS 
 
 To understand the rule for the point in division of deci- 
 mals, consider the problem in its relation to multiplication. 
 For instance, divide 10.191 by 2.37. 
 
 Solution: 10.191 is the product of 2.37 by the required quo- 
 tient. Hence, by the rule of multiplication, there must be as many 
 decimal places in the divisor and quotient as in the dividend. So after 
 performing the division as in simple numbers, one place should be 
 marked off in the quotient, as, 4.3. 
 
DIVISION OF DECIMALS 41 
 
 Rule: Point off as many places in the quotient as are 
 
 necessary to make the number of decimal places in the divisor 
 
 and quotient equal to the number in the dividend. 
 
 To apply the rule, ciphers should be annexed to the dividend to 
 give it as many places as the divisor, if it has fewer. 
 
 Probably the method for division of decimals least likely 
 to result in error is the following: 
 
 Divide 8912.4 by .024. 371350 
 
 Solution: Move the decimal point to the right in 024)8912400 
 
 the divisor so as to make the divisor a whole number; 72 
 
 move the decimal point in the dividend the same num- ~yj\ 
 
 ber of places. to the right, annexing ciphers if necessary; jgg 
 
 place the decimal point in the guotient directly over the — 5b 
 
 point in the dividend in long division and directly under — 
 
 it in short division. Show that this method accords' 84 
 
 with the other rule for division of decimals. 72 
 
 ■I Ml 
 
 The principle involved in moving the point in divisor iZi 
 
 and dividend is: multiplying both divisor and dividend by 
 
 the same number does not change the value of the quotient. 
 
 The check for division of decimals is the same as for 
 division of simple numbers. (Page 16.) 
 Solve the following and check the results: 
 
 1. 89,036 -^ 4.7 3. 3.964 -h 8.44 5. 3.6942 -f- 8.6 
 
 2. 324.6 -f- 36 4. 369.42 -=- .86 6. 36.942 + 86 
 
 7. Find the number of acres in a field in example 12, 
 page 40. 
 
 8. If a grain bin is 8.4 ft. by 6.5 ft. by 4.2 ft., how 
 many bushels does it hold? Work example 2, page 36, using 
 .8 instead of £ . 
 
 9. If an apple barrel holds 2.5 bu., how many barrels 
 are required to hold 356 bu. of apples? 
 
 10. A bunch of choice steers weigh, respectively, 1,847, 
 1,472, 1,563, 1,606, 1,656, 1,476, 1,396, 1,410, and 1,673 lbs.; 
 what are they worth at $8.37? Look up the present quo- 
 tations and find their value. 
 
42 AGRICULTURAL ARITHMETIC 
 
 11. A wagon box is 10.5 ft. long, 3 ft. wide, and holds 
 63 bu. by measure. How deep is it? Suppose it holds 
 95 bu.? Express the answer correct to one decimal place.* 
 
 SOLUTION OF INDICATED EXPRESSIONS 
 
 A term in arithmetic, as in algebra, is part of an expres- 
 sion set off from the rest of the expression by the plus or 
 minus sign. 3X4 — 2 + 6-f-4isan expression of three 
 terms: 3 X 4, 2, and 6 -5- 4. 
 
 In the solution of such indicated expressions, proceed 
 frtom left to right, solving the indicated operation in the terms 
 before the operation between the terms. 
 
 1. 3X4 + 2 + 6-5-4 = 16J 
 
 2. 10-2Xi-3X2-i-4 = 7i 
 
 3. 4 X (8 - 3) + 7 = 27 
 
 A. 2 _|_ 3 y 2 3 _^ 3 _ 2 _]_ 1 4 _ 1 
 
 Solve the following indicated expressions: 
 1. 6 + 8 X 3 - 15 2. 5j_2.3 + * + 8-i-s-t 
 
 8X1-2 
 
 3 ' 9-3X2 
 
 METHODS OF ABBREVIATION OF VARIOUS 
 OPERATIONS 
 
 In the following classes of abbreviated methods many 
 additional examples should be given and frequently be 
 reviewed until the pupil naturally uses the method sug- 
 gested. 
 
 I. To multiply by 10, 100, 1,000, etc. 
 
 Multiply 866 by 100, and 472 by 10,000. 
 
 Solution: 100 X 866 = 86,600. 10,000 X 472 = 4,720,000. 
 
 *If the hundreths order is larger than five, increase the tenths by one, if not, 
 drop it. 
 
METHODS OF ABBREVIATION 43 
 
 Rule: To multiply any number by 1 with ciphers an- 
 nexed, annex as many ciphers to the number as are annexed to 
 the 1. 
 
 Note: 3X4 may be read 3 multiplied by 4 or 3 times 4. The 
 former is the original usage, but the latter to-day generally is regarded 
 the better form. 
 
 II. To multiply a number by any number with ciphers 
 annexed. 
 
 Multiply 342 by 600, and 73 by 12,000. 
 
 Solution: 600X342 = 205,200. 12,000 X 73= 876,000. 
 
 Rule : To multiply a number by any number with ciphers 
 annexed, multiply the given number by the significant figures 
 of the multiplier and annex to the result the number of ciphers 
 annexed to the multiplier. 
 
 If in I or II the multiplicand has decimal orders, move 
 the decimal point to the right instead of annexing ciphers. 
 
 Multiply 3.24 by 10, and 2.34 by 300. 
 Solution: 10 X 3.24 = 32.4 300 X 2.34 = 702. 
 
 III. To divide by any number with ciphers annexed. 
 
 Divide 837 by 100, and 837 by 30. 
 Solution: 837 -r 100 = 8.37. 837 -i- 30 = 27.9. 
 
 Rule: To divide by any number with ciphers annexed, 
 place a decimal point as many places to the left as there are 
 ciphers annexed to the divisor and divide by the significant 
 figures of the divisor. 
 
 1. 83.24 H- 3,200. 3. 692.7 -5- 5,600. 
 
 2. 9,637 -J- 500. 4. 89.6 -f- 24,000. 
 
 IV. To multiply any number by aliquot parts of 100. 
 Multiply 8,361 by 33$. 
 
 Solution: Since 33J = X P, annex two ciphers and divide by 3. 
 3 )836100 
 278700 
 Rule: Annex two ciphers and divide by the number which 
 represents the aliquot part of 100. 
 
44 AGRICULTURAL ARITHMETIC 
 
 1. Multiply the following numbers by 33£, also by 
 .33£ and by 3.3|: 7,932, 6,329, 83.46, 756.1, 437.61, 325, 
 46.5, 82, 743. 
 
 2. Multiply each of the above numbers, correct to two 
 decimal places, by 14f, 16f , 8£, 12|, and 25. 
 
 If the multiplicand has decimal orders, move the decimal point 
 the required number of places to the right instead of annexing ciphers; 
 as, 83.426 X 14f. 
 
 Solution: 7) 8342.6 142 _ u^ 
 
 1191.8 7 7 
 
 3. Multiply 834 by 3|. 
 
 4. Multiply 729.6 by 20 in two ways. Which is the 
 better? 
 
 V. To multiply any number by the multiple of an aliquot 
 part of 100. 
 
 Multiply 8,352 by 66|. 
 
 Solution: Since ,66} = *$», annex two ciphers, 835200 
 
 3)1670400 
 
 multiply by 2, and divide by 3. 
 
 556800 
 
 Rule: Annex two ciphers, multiply the given number by 
 the number which represents the multiple and divide by the 
 number which represents the aliquot part. 
 
 1. Multiply the following numbers by 66f, 6.6f, and 
 .661: 487,38.6,5.923,843,76.29. 
 
 Make a rule for multiplying by 75, 37£, 83|, 62£, 87£, 
 284-, 25, 42f, and 11|. 
 
 Multiply the numbers in V (1) by each of them. 
 
 VI. To divide any number by aliquot parts of 100. 
 
 Divide 837 by 33|. 
 
 Solution: Since 33i = ig", 837 '+ 33J = 8.37 X 3 = 25.11. 
 
 Rule: Divide by 100 and multiply by the number which 
 represents the aliquot part of 100. 
 
METHODS OF ABBREVIATION 45 
 
 1. Divide the following numbers by 331, 3.31, and by 
 .331: 829, 634.2, 79.54, 6,943, 625.69, 837, 9,625, 47.81. 
 
 Make a rule for dividing by the following aliquot parts 
 of 100 and divide each of the above numbers by them: 
 12J, 8i, 16|, 14f 9A, 11*. 
 
 2. Divide 836.4 by 66|. 8.364 
 Explain the solution at the right. § 
 
 2) 25.092 
 
 12.546 
 
 Divide the numbers in VI (1) by the following numbers 
 and express the answers correct to two decimal places: 
 
 371, 831, 28f, m , 57+, 831, 87*. 
 
 VII. To multiply mixed numbers having identical frac- 
 tions. 
 
 Multiply 8} by 12*. 
 
 In the solution the process of finding * of 8 and 1 of 12 
 can be abbreviated by taking \ of 8 + 12, or 20, when the 
 answer can be written with other recorded work. 
 
 Multiply: 
 
 241 
 91 
 
 Solution: 
 1 X 33 = 11 
 
 
 2271 
 
 9 X 24 + 11 = 227 
 
 
 
 161 28f 181 631 
 
 821 
 
 
 41 Of 121 7i 
 68J|r 2771 
 
 _§± 
 
 Rule : Multiply the fractions and set down the product. 
 Multiply the sum of the whole numbers by the identical frac- 
 tion and add the product to the product of the whole numbers. 
 
 VIII. To multiply numbers of two orders having identical 
 units or tens. 
 
 83 Solution: 3X3 = 9. 3 X 14 (3 X 6 + 8) = 42. 
 
 63 Express the 2. 6X8 = 48. 
 
 48 + 4 = 52. 
 
 5229 It can be seen that 3X3 added to 3 X 6 is the same as 3 X 9. 
 
46 AGRICULTURAL ARITHMETIC 
 
 Rule: Multiply the units by the units, express the units 
 and carry the tens. Multiply the sum of the unlike numbers 
 in any order by one of the identical numbers in the other order, 
 set down the right-hand figure and carry the left; multiply the 
 tens by the tens and set down the product. 
 
 Record only the answers of the following: 
 
 27 X 57 42 X 62 18 X 58 82 X 42 
 
 24 Solution: 3X4 = 12. 
 
 23 Express the 2. 2X7 (2X3+ 4) = 14. 
 
 14 + 1 = 15. 
 
 552 Express the 5. 2X2 + 1=5. 
 
 23 X 25 46 X 44 92 X 93 36 X 34 
 
 In the following, in the usual way, perform the process 
 
 mentally and record only the answer; as, 
 62 82 X 26 63 X 27 18 X 24 
 
 25 16 X 23 43 X 55 28 X 53 
 
 1,550 28 X 43 56 X 42 84 X 33 
 
 IX. To square a mixed number whose fraction is one half. 
 
 Multiply 3* X 3£, or square 3*. 
 
 Solution: 3| 1X1=1- 1 of 3 + 1 of 3 or 1 X 3 = 3. 
 
 _3J_ 1X3 + 3X3=4X3 = 12. 
 
 12J 12 + 1 = 121; or 1 X i = J. 
 
 3 X 4 = 12 /. 31 X 3| = 121. 
 
 Rule : To square a mixed number in which the fraction is 
 one half, square the fraction, add one to one of the integers and 
 multiply the result by the other integer. 
 
 51 Solution: 1X1 = 1 5 X 6 = 30. 
 
 51 In multiplying, 1x5 + 1X5=1X5. (1+5)X5 
 
 =6X5. 
 
 30} 
 
 Solve the following: 
 
 8* X 8* 
 
 6* X 6* 
 
 20* X 20* 
 
 4* X4* 
 
 9* X 9* 
 
 7*X 7* 
 
 12* X 12* 
 
METHODS OF ABBREVIATION 47 
 
 X. To multiply sets of factors. 
 
 From the fact that 2 X 5 = 10, where 2's and 5's are 
 found in sets of factors the multiplication can be abbreviated 
 by setting aside the pairs of 2's and 5's, multiplying the re- 
 maining factors together, and annexing as many ciphers as 
 there are pairs of 2's and 5's; as, 25 X 16 X 75 = (2 X 5) 
 X (2 X 5) X (2 X 5) X (2 X 5) X 3 = 30,000. 
 
 With a knowledge of factoring it is easily seen that 
 there are four pairs of 2 and 5 and one 3; hence the answer 
 is 3 with four ciphers annexed. 
 
 Multiply 24 by 15 by 5. Ans. 1,800. 
 
 By this method solve the following mentally: 
 
 1. 8 X 15 X 75. 3. 8 X 75 X 15 X 4. 5. 42 X 15. 
 
 2. 24 X 25 X 15. 4. 56 X 75 X 15. 6. 828 X 75. 
 To multiply such a group as 8, 9 and 24. It is shorter 
 
 to multiply in the order 24, 8, and 9 than 8, 9, and 24. Mul- 
 tiply 24 by 8 or 9 simply setting down the answer, then the 
 result by the other factor. But to multiply 24 by 72 takes 
 longer. Try it. 
 
 Multiply: 1. 8, 3, 24 and 5 4. 20, 6 and 18 
 
 2. 734, 6 and 10 5. 62, 4, 8 and 12 
 
 3. 46, 9, 12 and 3 6. 7, 12 and 13 
 
 XI. Abbreviation by Cancellation. 
 
 It has been shown that, if the numerator and denominator 
 of a fraction are divided by the same number, the value of 
 the fraction is not changed. (Prin. Ill, page 21.) It can be 
 shown in the same way that, if the divisor and dividend 
 are both divided by the same number, the quotient remains 
 the same. 
 
 Show by experiment that this is true. 
 
 If the product of several factors is to be divided by the 
 product of other factors, the division can be abbreviated by 
 striking out common factors from both dividend and divisor. 
 
48 AGRICULTURAL ARITHMETIC 
 
 Divide 27X 8 X 16 X 42 by 24 X 28 X 12 
 
 2 
 
 9 I 
 
 Solution: 27X8X10X42 
 
 24X28XZ2 
 
 $ 1 % 
 
 = 18 
 
 Divide 26 X 36 X 72 by 56 X 7 X 18 
 
 13 9 
 
 Solution; 20X36X72! 13X36 468 
 
 00X7XZ8 7X7 49 WTir 
 
 7 9 
 
 The problem, after cancellation, simply requires that 36 
 be multiplied by 13 and divided by 7 X 7, which requires 
 much less computation than to perform the operations orig- 
 inally indicated. 
 
 Such problems as the following can be solved mentally. 
 36 X 42 -r- 63 = 24 
 
 Write the answers at sight of the following: 
 
 1. 84 X 63 -5- 108 5. 57 X 85 4- 51 
 
 2. 36 X 34 -f- 72 6. 
 
 3. 24 X 18 -7- 72 7. 
 
 4. 57 X 85 -J- 95 8. 
 
 9. 35 X 75 X 6 h- 5 
 10. 45 X 75 4- 125 
 
 In performing the operation of cancellation it is usually 
 more convenient to place the dividend over the divisor. 
 (Page 15.) When one has a quotient factor either above or 
 below the line, it does not pay to set it down if it can be im- 
 mediately canceled. There is sufficient written work in the 
 following examples. A different order of cancellation, of 
 course, would give a different solution, but the same answer. 
 
 49 X 24 ■*■ 
 
 56 
 
 52 X 18 4- 
 
 9 X 13 
 
 72 X 15 h- 
 
 40 
 
 X 15 X 30 
 
 
METHODS OF ABBRE VIA TION 49 
 
 2 2 
 Care must be exercised to Z8 X 20 X 30 X %& 
 set down a factor that cannot 72 X 00 X 03 " = 
 
 be canceled, to avoid error in 3 
 
 result. 2 3 2 7 4 
 
 Z3X70X21XJM _ Z3X20X30X24 _ 
 
 0ZX 30X00 7 02X00X03 ¥ 
 
 7 7 3 
 
 Solve the following: 
 
 1. 57 X 91 X 85 X 42 -^ 34 X 95 X 52 
 
 2. 56 X 24 X 85 X 32 -f- 84 X 16 X 15 
 
 3. 27 X 72 X 63 X 20 -f- 54 X 56 X 76 
 
 4. 91 X 87 X 76 X 24 -^ 58 X 29 X 56 
 
 XII. Straight Line Analysis. 
 
 In the solution of many concrete problems in more or 
 less numerous operations of multiplication and division, 
 much time can be saved by indicating the operations with 
 the multipliers above the line, the divisors below the line 
 and abbreviating the computations by cancellation. If 
 fractions occur as multipliers, put numerators above the line 
 and denominators below. If a fraction occurs as a divisor, 
 invert the fraction and proceed as above. Page 30. 
 
 1. Find the capacity in barrels of a cement watering 
 trough 1^' X 7' X 11' inside measurement. 
 
 Solution: 
 
 3 X 7 X 11 X 12 s X 2 = 192 = 
 2X231X63 7 
 
 Suggestions: 12 X 12 X 12 fl2fj = number of gallons in a cubic foot. 
 
 Z\\ = — = number of gallons in a barrel. To divide by this fraction, 
 
 12 3 
 invert and multiply. — = 4i nearly. 
 
50 AGRICULTURAL ARITHMETIC- 
 
 A close enough approximation can be obtained by using 
 4| cu. ft. = 1 bbl. 
 
 Solution: X 7 X 11 X 5 _ 55 _ _ 
 
 2 X U ~ 2 ~ Z1 * 
 
 2. Find the number of tons in a rectangular block 
 of granite 6' X 6' X 4', specific gravity 2.6. (Page 241.) 
 
 3 3 13 
 
 Solution: X X I XZZ0 X 20 _ u _ 
 
 2000 X 2 X10 " U -' 
 
 3. The state of Massachusetts requires by law that 
 school buildings shall furnish each pupil 30 cu. ft. of fresh 
 air per minute. A school building seating 40 pupils has 
 an exit register 2£' X 2J'. With air moving through the 
 register 5 lineal feet per second, is the ventilation up to the 
 legal requirements, assuming that obstructions in the reg- 
 ister reduce the opening £? If air moves 5 ft. per sec, 
 the volume is (2£) 2 X 5 cu. ft. per sec. 
 
 Solution: 5X0X0X4X00 = 5 = . 
 
 2 x 2 x $ x m X 30 4 l¥ 
 
 4. If corn is planted 3' 8" each way, how many hills 
 are there per acre? 
 
 40 3 3 
 Solution: Jf00 %$ $$ 3 3 
 
 — x f x f x n x zz = 324a 
 
 In reductions involving square and cubic units, it is enough to 
 remember the linear table and use the number involved twice for area 
 and three times for volume. For instance: 
 
 12 3 cu. in. = 1 cu. ft. 
 
 16| X 161 sq- ft. = 1 sq. rd. 
 
 4' X 4' X 8' = a cord. 
 
 5. How many bushels of corn in a 20-acre field, if planted 
 3' 8" apart each way, averaging 3 stalks to the hill, 4 good 
 ears to 5 stalks, and 120 ears to the bushel? 
 
 6. What is the value of 15 acres of such corn at 70 
 cents per bushel? 
 
MEASUREMENT OF LUMBER 51 
 
 7. Work mentally: 
 
 (a) A pile of cordwood is 80' X 6' X 4'. How many 
 cords? 
 
 (b) What is the value of a pile of wood 40' X 8' X 6' 
 at $7 per cord? 
 
 (c) Find the value of a pile 96' X 4' X 4' at $6 per cord. 
 8*. If trees are planted 20 ft. apart by the rectangular 
 
 system, how large a square is devoted to each tree? Make 
 a drawing to illustrate. 
 
 9. By the rectangular system, if cherry trees are plant- 
 ed 16£ ft. apart, how many trees in 12 acres? 
 
 10. If apple trees are planted by the rectangular sys- 
 tem, 22 feet apart, how many trees can be planted to the 
 acre? If planted 18' X 20'? 
 
 11. At what distance apart would trees be planted by 
 the rectangular system, if 196 plum trees were planted to 
 the acre? 
 
 12. How many tons of water in an inch of rainfall on an 
 acre of ground? On 12 acres? 
 
 13. What is the value of a piece of land 40 X 60 rods at 
 $150 an acre? 
 
 MEASUREMENT OF LUMBER 
 
 The unit of lumber, primarily, is the board foot, though 
 lumber is bought and sold by the thousand feet. 
 
 The board foot is a magnitude 1' X 1' X 1". Twelve 
 board feet make a cubic foot. 
 
 Lumber is always sawed at the mills an even number of 
 
 feet in length; in the main, 12', 14', 16', and 18' long, though 
 
 longer and shorter lengths can be obtained, and it is always 
 
 *The rectangular system, 20' apart, means parallel rows each way 
 
 at right angles to each other 20' apart. 
 
52 AGRICULTURAL ARITHMETIC 
 
 an integral number of inches wide, usually an even number. 
 In computing amounts of lumber, anything less than 
 an inch thick is regarded as an inch, though the price of 
 the lumber usually varies according to the thickness. 
 
 To find the number of board feet in lumber. 
 
 As a board foot is 1' X 1' X 1", the number of board feet 
 in a given piece of timber is the product of the length and 
 width in feet by the number of inches in thickness. 
 
 How many board feet in a board 14' X 12" X 1"? Think 
 of it as sawed up into boards 1' X 1' X 1" and note how 
 many there are. 
 
 How many board feet in a stick 2" X 10" X 18'? 
 In solving this it becomes necessary by above statement to think 
 the product of 2 X {§ X 18; but as the product of 2 X \% X 18 is the 
 
 same as r~ it is simpler to cancel 12 wherever possible; 
 
 hence the rule: 
 
 Rule: To find the number of board feet in a- piece of 
 lumber, multiply the dimensions as given and divide by twelve, 
 abbreviating by cancellation when possible. 
 
 How many board feet in 20 pieces (pes.) of lumber 
 6" X 8" X 16'? 
 
 Solution: Cancel mentally the factors of 12 in 6 and 8, leaving 
 to be multiplied 20, 4, and 16, which equals 1,280 board feet. 
 
 Work mentally the following: 
 
 Find amount of lumber in: 
 
 1. 12 pes. 2" X 4" X 16'. 4. 24 pes. 2" X 8" X 10'. 
 
 2. 8 pes. 2" X 6" X 20'. 5. 13 pes. 3" X 4" X 12'. 
 
 3. 18 pes. 2" X 4" X 16'. 6. 16 pes. 2" X 6" X 12'. 
 
 Solve mentally or with as little pencil work as is neces- 
 sary for facility: (Page 48.) 
 
 7. 24 pes. 2" X 8" X 18'. 9. 43 pes. 2" X 6" X 20'. 
 
 8. 16 pes. 3" X 8" X 16'. 10. 18 pes. 4" X 4" X 16'. 
 
SQUARE ROOT 53 
 
 To find the cost of lumber, multiply the number of board 
 feet by the price per thousand and point off three places. 
 
 In practice, as one does not wish to have more than two decimal 
 
 E laces to express dollars and cents, it is customary to drop the num- 
 er 10 from among the factors when it occurs and point off two places; as, 
 
 20 pes. 3" X 8" X 16' @ $25 = 16 X 100 = $16. (Page 45., X.) 
 Compute mentally, applying appropriate abbreviations. 
 
 1. 12 pes. 1" X 12" X 16' @ $30. 4. 30 pes. 2" X 12" X 16' @ $20. 
 
 2. 8pcs.2"X10"X18'@$25. 5. 18 pes. 2" X 4" X 12' @ $25. 
 
 3. 24 pes. 1" X 10" X 16' @ $50. 6. 25 pes. 2" X 12" X 18' @ $23. 
 
 Solve with as little pencil work as possible. Find value of: 
 
 7. 16 pes. 2"X4"X 16'@$24. 9. 40 pes. 4" X 8" X 18' @ $21. 
 
 8. 36 pes. 2"X8"X 16'@$27. 10. 33 pcs.2"X6".X16'@$30. 
 
 SQUARE ROOT 
 
 In many problems it is necessary to extract the square 
 root of numbers. By having a table of the square roots of 
 the prime numbers to 19 or higher inclusive, most of the 
 problems can be solved without resorting to the "long rule." 
 (See Appendix, page 249.) 
 
 To extract square root is to find one of two equal factors 
 of a number. If a number can be divided into pairs of 
 equal factors, the process is very simple; as 256 = 2 8 ; hence 
 2 4 or the product of half of the 2's is the square root. 
 V576 = V9~ X V2« = 24. 
 
 But in case there is an odd number of factors, the root 
 of the odd factors must be found, or the table be used. 
 
 Solution: 72 = 6v"2. V2 = 1.414.. 6 XV 2 = 8.484. 
 96=4V6.=4XV2XV3. V2 = 1.414. V 3 = 1.73. 4X1.414 
 X 1.73 = V96. 
 
 But without a table of square roots at hand it is neces- 
 sary to know how to extract the square root of numbers.* 
 
 *For pupils under the second year of the high school probably the best 
 method is to commit the rule to memory and work enough examples by it to get 
 a fair degree of facility. 
 
54 AGRICULTURAL ARITHMETIC 
 
 Rule for Square Root 
 
 Divide the number into periods of two 621.1 
 
 figures each counting both ways front the 385824.00 
 
 decimal point. Subtract the largest second 36 
 
 power from the left hand period, and bring 
 
 down the next period. Double the root fig- \aoa 
 
 urefor a trial divisor and find how many . . 
 
 times it is contained in the new partial 18300 
 
 power, omitting the right hand figure, or 12421)12421 
 
 regarding the trial divisor as tens. Annex 5879 
 
 the new root figure to the trial divisor; 
 
 multiply by the root figure, subtract from the partial power, and 
 
 bring down the next period. Double the root already found and 
 
 proceed as before. 
 
 For those who would like to look into the reasons for the rule, 
 an explanation is appended on page 249. 
 
 The matter of its use is left to the judgment of the teacher. 
 
 1. Extract the square root of 27 in two ways (sugges- 
 tion: 27 = 3 V 3); of 892.534; of 6374.5. 
 
 2. Extract the square root of 2, 3, 5, 7, 11, 13, 17 and 
 19 correct to two decimal places, and compare with the 
 table (page '249). It is well to commit to memory V 2, V 3, 
 and V5. 
 
 3. Extract the square root of 89.7324; of .639; of 
 2973.642; of 18. 
 
 THE RIGHT TRIANGLE PROBLEM 
 
 I. Given two sides of a right triangle to find the hypote- 
 nuse. 
 
 Let ABC represent a right triangle, with AC the hy- 
 potenuse. AB and BC are called sides or legs of the right 
 triangle. 
 
RIGHT TRIANGLE 
 
 55 
 
 B 
 
 Figure 6. 
 
 The hypotenuse is the side opposite the right angle. 
 
 If AB = 4 in., BC = 3 in. and 
 the triangle be drawn to a scale 
 and squares be drawn on the three 
 sides, it will be found that the sum 
 of the areas of square AB and 
 square BC will equal the area of 
 square AC; viz, 25. Therefore, 
 side A C is 5. 
 
 Draw a right triangle with 
 legs 5 and 12. What is the hypo- 
 tenuse? Test it by measurement. 
 
 This principle is established in general in geometry and 
 may be stated : The square on the hypotenuse equals the sum 
 of the squares on the other two sides of a right triangle. 
 
 I. Given the two sides of a right triangle to find the 
 hypotenuse. 
 
 Rule: Extract the square root of the sum of the squares 
 of the two sides. 
 
 II. Given the hypotenuse and one side to find the third side. 
 Rule: Extract the square root of difference between the 
 
 square of the hypotenuse and the square of the remaining side. 
 
 PROBLEMS 
 
 1. A carpenter wishes to know that the frame of a 
 building he is laying has right angles as it lies upon the 
 foundation. He measures 6 ft. from a corner along one 
 timber, 8 ft. from the corner along the other, and then 
 measures across between the two points located. He then 
 adjusts the frame until the distance across is 10 ft. Is he 
 justified in so doing? 
 
56 
 
 AGRICULTURAL ARITHMETIC 
 
 Figure 6. 
 
 2. I wish to build a port- 
 able A-shaped pig cot 6 ft. wide, 
 8 ft. long, and 6 ft. high; what 
 length of boards shall I need for 
 the slope? How much waste 
 lumber, if I use 14-foot boards 
 in making the roof? 
 
 3. If I make the length of slope just 7 ft., how high 
 will the cot be, the width remaining the same? 
 
 4. In planning a building it is decided to make the 
 distance between the eaves 30 ft. and that the roof be one 
 third pitch. What is the length of the rafter? 
 
 One third pitch means that 
 the perpendicular distance from 
 ridge to the line between the eaves 
 is one third the latter line. If 
 CD is the perpendicular to AB, 
 the roof would be one third pitch, 
 if CD is one third of AB; one 
 half pitch, if CD is one half AB. 
 
 5. I want to use a 16-ft. rafter, and the line between 
 the eaves is 24 ft. What is the pitch of the roof? 
 
 6. What is the length of the diagonal of a square 5 
 ft. on a side? Knowing V2 = 1.41, solve without extraction 
 of a root. 
 
 7. Solve by extracting the root, as a check. 
 
 8. Solve examples 4 and 5 by use of the table of roots. 
 
 9. How many square feet in the gable end of a house 
 18 ft. wide, 16 ft. high to the plate, and 25 ft. high to the 
 ridge? (The area of a triangle = \ base X altitude.) 
 
 10. If the house in example 9 is 24 ft. long, and rec- 
 tangular, how many square feet must be clapboarded? 
 Take no account of doors and windows. 
 
AREA OF TRIANGLE 57 
 
 11. I wish to plant apple trees on A . . . . B 
 
 the hexagonal system 24 ft. apart in C D 
 
 the row; how far is row AB from row E F 
 
 CD? How many trees can be planted 
 
 to the acre? (See page 206.) 
 
 The distance between rows AB and CD is the perpendicular from 
 one tree to the mid-point of the line between two opposite trees. 
 
 AREA OF A TRIANGLE 
 
 The dimensions of a triangle are its base and altitude. 
 
 Any side of the triangle may be taken for the base. 
 
 The vertex of a triangle is the vertex opposite the base. 
 
 The altitude is the perpendicular distance from the ver- 
 tex of the triangle to the base. 
 
 I. To find the area of a triangle when the altitude and 
 base are given. 
 
 Solution: Find the area of the triangle 
 ABC. Study the figure and compare parts, I and 
 II, III and IV of th« rectangle in figure 8. 
 Compare triangle ABC with rectangle BCEF. 
 
 The area of the rectangle is BC X AD. 
 
 -m. * a * • i BC X AD 
 The area of the triangle = = . 
 
 & Figure 8. 
 
 Rule : The area of a triangle equals one half of the product 
 of the base by the altitude. 
 
 PROBLEMS 
 
 1. The altitude of a triangle is 24 ft., the base 40 ft. 
 What is its area? If the dimensions are 17 and 24, 13 and 
 18, 17 and 23, find the respective areas. 
 
 2. A piece of land in triangular form has a base of 20 
 rds. and an altitude of 12 rds. How many acres in the area? 
 
 3. Drive three stakes for the vertices of a triangle, make 
 the required measurements and find the area inclosed. 
 
 II. To find the area of a triangle, if the three sides are 
 , given. 
 
58 
 
 AGRICULTURAL ARITHMETIC 
 
 Rule.: Multiply one half the sum of the sides by the three 
 factors formed by subtracting the three sides, respectively, from 
 one half of the sum of the sides and extract the square root. 
 
 The sides of a triangle are 12, 13, and 15. Find its area. 
 
 Solution: (13 + 15 + 12) -i- 2 = 20; 20 - 13 = 7; 20 - 15 = 
 5; 20 - 12 = 8; 20 X 7 X 5 X 8 = 5,600. The V5600 = 74.83+. 
 
 PROBLEMS 
 
 1. Find the number of acres in a tract of land in triangu- 
 lar form whose sides are respectively 20, 24, and 30 rds. 
 
 2. A swamp triangular in shape has sides 7, 10 and 
 11 miles respectively. How many acres in the swamp? 
 
 THE AREA OF A TRAPEZOID 
 
 A trapezoid is a figure of four sides having two and only 
 two sides parallel. The parallel sides are the bases of the 
 trapezoid and the perpendicular distance between the bases 
 is the altitude of the trapezoid. 
 
 To find the area of a trapezoid. 
 
 Soi/cttion: Draw a di- 
 agonal of the trapezoid, as 
 AC. ABC and ADC are 
 two triangles having the alti- 
 tude MN and the bases AB 
 and DC respectively. The 
 area of the trapezoid is: 
 AB XMN CD X MN 
 2 2 
 (AB + CD) X MN. 
 or g 
 
 Rule: To find the area of a trapezoid multiply the altitude 
 by the sum of the bases and divide by two. 
 
 The bases of a trapezoid are 9 ft. and 12 ft. and the 
 altitude is 8 ft. Find its area. 
 
 Solution: (9 + 12) X 8 = 21 X 8 = g4 
 
THE CIRCLE 
 
 59 
 
 PROBLEMS 
 
 1. A tract of land in the form of a trapezoid has an alti- 
 tude of 20 rds. and the parallel sides are 40 and 60 rds. re- 
 spectively. How many acres in the tract? 
 
 2. A section of land has a fence running obliquely 
 across it. The parallel sides of one part are, respectively, 
 247 rods and 133 rods. How many acres in each piece? 
 
 3. A man owned a quarter section of land divided by a 
 highway cutting two opposite sides 40 rds. and 100 rds., 
 respectively, from two adjacent corners. He sold the 
 smaller piece. How many acres did he sell? 
 
 320 rods n 4. ABCD represents a section of 
 
 land. AE is 44 rds. and FC is 96 rds. 
 How many acres in each piece? How 
 long is line EF? 
 
 B r c 
 
 Figure 10. 
 
 5. How many acres 
 in the trapezoid indi- 
 cated by the figure at 
 the right? 
 
 &fcO rods 
 
 Figure 11. 
 
 THE CIRCLE 
 
 A circle* is a portion of a plane bounded by a closed 
 curved line every point of which is equally distant from a 
 point within called the center. 
 
 A circumference (C) is the line which bounds a circle. 
 
 The radius (R) of a circle is a straight line drawn from 
 the center to the circumference. 
 
 The diameter (D) of a circle is a straight line through 
 the center and limited by the circumference. 
 
 * Thia definition is the one generally used in arithmetics, but in other mathe- 
 matics the circle is usually described as a curved line synonymous with circumference. 
 
60 AGRICULTURAL ARITHMETIC 
 
 The area of a circle is the area of the surface bounded by 
 the circumference. 
 
 I. To find the ratio of the circumference to the diameter. 
 
 This ratio is called Pi (ir). 
 
 Measure carefully the distance around a barrel,, tin pail, 
 any cylindrical object, and also the diameter of the same. 
 Divide the former measurement by the latter. The quo- 
 tient will approximate 3\. 
 
 Let each member of class make from two to six such measure- 
 ments and bring the averages to class. Throw out of the average 
 any result that manifestly is in error by being much different from 
 the others. Then average the remaining averages. In this way a 
 fairly close approximation to the true ratio can be obtained. 
 
 The purpose of this work is not so much to obtain the true ratio 
 as, by his experience in obtaining it, to have the pupil appreciate its 
 meaning. 
 
 The value of ir to four decimal places is 3.1416; to five, 
 3.14159. For all practical purposes for the farmer or 
 mechanic 3+ is sufficiently correct. Expressed in symbols, 
 C = irD, or C = 2irR. 
 
 1. The diameter of a circle is 14 ft. Find the circum- 
 ference. 
 
 Solution: 14' X 22 _ AA , 
 7 ~ 44 - 
 
 2. The diameter of a circle is 6 in. What is the length 
 of the circumference? 
 
 3. What is the distance around a 1^-in. strawberry? 
 A 3|-in. apple? A 15-in. pumpkin? A 3-ft. tree? For 
 such approximations let Pi = 3. Obtain the answer with- 
 out a pencil. 
 
 4. If the distance around a tree is 37 ft., what is its 
 diameter? Express the answer in the nearest integer. 
 
 5. The circumference of an apple is 9 in. What is its 
 diameter? 
 
THE CIRCLE 
 
 61 
 
 6. Find the length of a circle 21 ft. in diameter by 
 using 3| for ir. Then find its length by using 3.1416, and 
 subtract to find the difference. Is the first answer too 
 large or too small? 
 
 II. To find the area of a circle. 
 
 Cut a slice half an inch thick from a 
 potato, turnip, apple, or other round object, 
 and cut it into halves through a diame- 
 ter. Then cut each piece from the center 
 along the radii, not severing the skin; as, 
 
 Fit the two pieces together with 
 the skin outward; as, 
 
 This form approximates a par- 
 allelogram or rectangle, the area of 
 which is the base times the altitude 
 (B X A) (Page 31). Now express the base and altitude 
 in terms of the circle. 
 
 Another way to think out the area is to compute the 
 area from the triangles formed. If cut into 10 triangles, 
 each triangle would equal $R X TgC. All the ten trian- 
 gles would equal %R X C = — = — . 
 
 III. To express area in terms of radius: 
 
 Solution: „,, . . , R X C 
 
 The area of a circle = — = — . 
 
 Since C = 2wR, the area = R * 2lnR = ^2. 
 
 These formalae, C = n-D, C = 27rR, area equals- 
 
 and area = ?iR 2 , should be thoroughly memorized, for it is often 
 convenient to use one or more of them. 
 
 1 . What is the area of a circle whose radius is 7? Whose 
 radius is 12? Whose diameter is 8? 21? 24? 
 
 R X C 
 
62 AGRICULTURAL ARITHMETIC 
 
 2. What is the area of the cross-section of a tree 36 in. 
 around? 91 in. around? Measure several trees and find 
 the areas of cross-sections. 
 
 3. A sequoia tree is 21 ft. in diameter. What is the 
 area of a cross-section? 
 
 4. A horse is tied to a stake in an open field with a 
 tether 4 rds. long. What part of an acre can he graze? 
 
 5. One circle has a diameter of 3 in., another a diam- 
 eter of 6 in. How many times as much area in the second 
 as in the first? 
 
 6. If the second circle in example 5 has 3 times the diam- 
 eter of the first, how does its area compare with the first? 
 If 4 times? If 5 times? 
 
 Formulate a statement as to how areas of circles compare 
 in terms of their diameters. 
 
 7. Two drains with tiles 3 in. and 4 in. respectively 
 unite. What diameter of tile should be used to carry a full 
 head of water from the two drains. 
 
 The principle is the same as in the Right Triangle problem. 
 (Page 54.) 
 
 8. If three lines of tile 4, 6, and 8 in. unite, what size of 
 tile should be used in the continuation drain? 
 
 VOLUMES OF CYLINDERS 
 
 A cylinder is a solid bounded by two circular faces and a 
 curved surface. A section of a stove pipe" a baking powder 
 can, etc., illustrate a cylinder. The circular faces are the 
 bases of the cylinder. 
 
 The altitude of a cylinder is the perpendicular distance 
 between the bases. 
 
 Cylinder is a much broader term than above defined, but the 
 definition given covers general usage. 
 
VOLUME OF CYLINDERS 63 
 
 The base and altitude of a cylinder are'its dimensions. 
 
 To find the volume of a cylinder, take a root, as a carrot, 
 potato, turnip, from which a fairly good cylinder can be 
 cut, having the skin for the curved surface. Cut it as you 
 did in finding the area of a circle. Place it, as before, in the 
 form of a rectangular solid. Use the cut surfaces for bases. 
 What is its volume? Express the volume in formulae; i. e., 
 in terms of altitude, R, and it. 
 
 Rule : To find the volume of a cylinder, multiply the area 
 of its base by the altitude. Expressed in a formula, the 
 volume = it X R 2 X h, h being the symbol for altitude, or 
 height. 
 
 PROBLEMS 
 
 1. A circular watering tank is 7 ft. in diameter and 2£ 
 ft. deep. How many cubic feet of water does it hold? 
 How many gallons? How many barrels (31£ gal.)? 
 
 For approximations of this kind 4J cu. ft. or even 4 cu. ft. per 
 barrel is near enough. 
 
 Solution: 
 
 4 £ 4 
 
 22 V 7 V 7 V 5 Y n x n x n v 2 _ ieo _ 99 „ 
 
 T x % x ■% x 2 x — m — u ~ r— 22t 
 
 7 
 
 11 
 
 - f x H x I x l = 7? = 23 - 
 
 3 
 
 What part of a barrel is the difference between the two 
 results? Which is the approximation? 
 
 2. A circular cistern is 6 ft. in diameter and 8 ft. deep 
 inside measurements; how many barrels does it hold? 
 
 3. A circular silo is 14 ft. in diameter and 30 ft. deep; 
 how many cubic feet of silage does it hold? 
 
64 AGRICULTURAL ARITHMETIC 
 
 4. How many cubic feet in a section of a tree trunk 
 45 in. around and 28 ft. high? Allowing 4 in. of diameter 
 for slabs and \ for sawdust, how many board feet approx- 
 imately? If ^ of the slabs can be utilized for shingles, 
 laths, etc., how many feet of waste? What part of the log 
 is waste? Use 3 for ir. 
 
 5. A steam boiler is 4 ft. in diameter and 18 ft. in length, 
 and has 24 fire tubes 4 in. in diameter and 18 ft. long run- 
 ning through it. How many gallons does the boiler hold 
 when £ full? 
 
 6. Find the specific gravity of a piece of rock. 
 Suggestion: Immerse the piece of rock in a cylinder partly full 
 
 of water and carefully measure the amount the water rises in the 
 cylinder. How can its volume be found from this result? See defi- 
 nition of Spec. Gr. (page 241) and complete the problem. 
 
 THE CONE* 
 
 A cone is a solid bounded by a circle and a curved surface 
 from all points of which straight lines can be drawn to a 
 point called the vertex. 
 
 The circle is the base of the cone. 
 
 The perpendicular distance from the vertex to the base 
 is the altitude of the cone. 
 
 If grain is poured from the same point upon a level 
 surface the shape of the pile will approximate a cone. A 
 cornucopia is cone-shaped. 
 
 The dimensions of a cone are its base and altitude. 
 
 The volume of a cone is one third that of a cylinder having 
 the same base and altitude. 
 
 To find the volume of a cone. 
 
 Solution: Construct a cylinder out of cardboard. It is not 
 necessary to make the base. Set it upon a level surface and fill with 
 sand, corn, or other convenient material. Construct a cone with the 
 
 *This subject may be omitted in the discretion of the teacher as its practical 
 value is small. 
 
PERCENTAGE 65 
 
 same base and altitude as the cylinder and see how many times the 
 material in the cylinder will fill the cone. If made accurately, the 
 cylinder will fill the cone three times. 
 
 Rule: To find the volume of a cone multiply the base by 
 
 one third the altitude. 
 
 1. A cone-shaped pile of wheat has an altitude of 5 ft. 
 and a diameter of 9 ft.; how many bushels, approximately, 
 in the pile? Use 3 for ir. 
 
 2. What measurements would you make to approx- 
 imate the number of bushels in a heap of grain piled as high 
 as possible and with a circular base? 
 
 3. A circular crib of chicken wire 8 ft. in diameter and 
 4 ft. high is filled with ear corn to a point 7 ft. from the 
 ground. How many bushels, approximately, of shelled 
 corn does it hold? How many heaped baskets of ear corn?* 
 
 PERCENTAGE 
 
 A considerable portion of the arithmetical operations 
 demanded in the business world is included in a system 
 termed Percentage. 
 
 This system is but an application of the three problems 
 of fractions (page 29) in which the fraction is one or more 
 hundredths. The fraction is usually expressed decimally. 
 Hence Percentage is based upon the subject of fractions and 
 decimals. For the term hundredths, per cent is substituted, 
 i-fj is termed 7 per cent and may be expressed 7%, .07, or 
 t^. 120 per cent is expressed 120%, 1.20 or \$$. 
 
 Rate is a term used to designate this fraction whether 
 expressed as per cent or hundredths; as, 8%, yfj or .08. 
 
 The base is the number upon which the per cent is reck- 
 oned. In the example to find 7% of 160, 160 is the base. 
 
 *1.6 cu. ft. make a heaped basket of ear corn. l.S cu. ft. of ear corn make 
 1 bu. of shelled corn. 
 
66 AGRICULTURAL ARITHMETIC 
 
 Percentage is the product of the rate by the base. In 
 the example to find 5% of 125, the answer, 6.25, is the per- 
 centage. 
 
 PROBLEM I 
 
 The simplest and most commonly used problem of per- 
 centage is: 
 
 To find a given per cent of a number. 
 
 Find 5% of 125. 
 
 Solution: 1% of 125 means T J of 125 which equals 1.25; 
 5% of 125 equals 6.25; or, 5% of 125 = .05 X 125 = 6.25. 
 
 Rule : To find a given per cent of a number, multiply 
 the number by the given rate. 
 
 MENTAL PROBLEMS 
 
 Find 3% of 12; 6% of 24; 10% of 17; 25% of 40; 11% of 
 36; 8% of 360; 331% of 96; 14f% of 21. 
 
 In the last two examples the work can be abbreviated by reduc- 
 ing the rate to a common fraction; as, 33J% of 96 = 4 of 96 = 32.-- 
 14?% of 21 = i of 21 = 3. 
 
 The equivalents of the following rates should be mem- 
 orized: 
 
 ■331 — 1, 
 
 .66f = f, 
 
 •25 = h 
 
 •75 = f , 
 
 •16f = h 
 
 801 — 5 
 ■OO-g- — -g-, 
 
 142 = A 
 
 ■"7 7) 
 
 • 12^ 7= 1, 
 
 071 — 3 
 
 iii_i 
 • L1 ? — ?> 
 
 •621 = | j 
 
 ■871 = h 
 
 C)-L- — _1_ 
 ■"11 11) 
 
 S 1 — 1 
 
 71 _ 1 
 ■>r — TT> 
 
 ft 2 — 1 
 
 • D ^— IT) 
 
 ■6i = A- 
 
 
 
 
 In solving problem I when these rates are involved, the work 
 should be shortened by using the fractional equivalent. (Compare 
 with page 44.) 
 
 Find 6|% of 300; of 75; of 45. 
 
 Find 14f% of 168; of 315; of 18; of 76. 
 
 Find 831% of 36; of 42; of 143; of 564. 
 
 Find 871%, 621%, 66|%, 331%, 83%, of 1728. 
 
PERCENTAGE 67 
 
 WRITTEN PROBLEMS 
 
 1. A Guernsey cow averaged 45 lbs. of milk a day during 
 June, which tested 4%. How many pounds of butter-fat 
 did she produce during the month? 
 
 2. A farmer delivered at a factory during a month 7,516 
 lbs. of milk which tested 4-|%. For what amount of butter- 
 fat should he be paid? What was it worth @ 28|-c. a pound? 
 
 3. Find 23% of 1,684, 133, 8,432, 146. 
 
 •4. Find 1.32 of the numbers in the preceding example. 
 
 5. Ear corn, while drying, shrinks in weight in the crib 
 on an average 3% a month. A man cribbed 27.5 tons of 
 corn, Oct. 1st. If sold January 1, what could it be ex- 
 pected to weigh? If offered $20 a ton October 1st, for 
 how much must it be sold per ton January 1st to make a 
 profit of $1 per ton for holding it? 
 
 6. Nitrate of soda, a commercial fertilizer, has 15% of 
 nitrogen. How many pounds of nitrogen are in 12.5 tons? 
 
 PROBLEM n 
 
 II. To find what per cent one number is of another. 
 
 What per cent is 3 - of 4? 
 
 Solution: 3 = J of 4. (Page 30 Problem II.) \ = 75% .-. 3 
 = 75% of 4. Verify by finding 75% of 4. 
 
 5 is what per cent of 6? 
 
 Solution: 5 = f of 6. f = 831%. •'■ 5 = 831% of 6. 
 
 13 is what per cent of 17? 
 
 Solution: 13 = Jf of 17, Jf reduced to hundredths =76 I 'V%. 
 
 Rule: To find what per cent one number is of another, 
 find what part the first is of the second and reduce to hun- 
 dredths. 
 
68 AGRICULTURAL ARITHMETIC 
 
 PROBLEMS 
 
 1. Find what per cent 8 is of 15; 9 is of 20; 4 is of 9; 3 
 is of 10; 16 is of 15; 12 is of 7; 18 is of 11; 1,234 is of 6,325. 
 
 Find the per cent 2f is of 3$. 
 
 Solution: The part 2| is of 3| = ff = SX&=I? = 70i?% •'• 2| is 
 70J?%of3J. 3 i 
 
 2. Find the per cent 3$ is of 4f ; f is of f; 5£ is of 3f ; 
 12 is of 41, 
 
 3. One cow yields 24 lbs. of 3£% milk daily, and an- 
 other cow 45 lbs. of 4%-milk. From the standard of fat pro- 
 duction what per cent is the value of the first cow compared 
 with the value of the second? The second is worth how 
 many times the first, not considering beef value? 
 
 4. If a farm is valued at $10,000 and the net income is 
 $800, what per cent does the farm yield on the investment? 
 
 5. If it costs $75 to raise a colt to the age of 4 yrs., 
 what is the per cent of profit if it is sold for $140? 
 
 PROBLEM III 
 
 III. To find a number when a certain per cent of it is 
 given. 
 
 4 is 8% of what number? (Page 30, problem III.) 
 Solution: Since 8% of the number is 4, 1% of the number = | 
 of 4 or J. Since I is 1% of the number, the number, or }gg of the num- 
 ber, = 100 X I = 50, /. 4 is 8% of 50. Verify by finding 8% of 50, 
 or by finding what per cent 4 is of 50. 
 
 15 is 75% of what number? 
 
 Solutjon: 15 X 10 ° = 20. 
 75 
 
 Rule : To find a number when a certain per cent of it is 
 given, multiply the number by 100 and divide by the per cent. 
 
INTEREST 69 
 
 PROBLEMS 
 
 1. 30 is 10% of what number ? 45 is 9% of ? 18 is 
 3% of ? 16 is 8% of ? 24 is 12% of ? 750 is 15% of ? 
 
 2. Solve, having the answer correct to two decimal 
 places: 36 is 7% of? 18 is 11% of? f is 13% of? 143 
 is 36% of? 216 is 13% of? 
 
 3. What sum must be invested to produce $1,000 at 
 4%? 
 
 4. As an income producer at 6%, what is a cow worth 
 
 that produces on an average 11,325 lbs. of 4£% milk wcrth 
 
 30c. per pound of butter-fat for 6 yrs., at a total expense 
 
 of $80 per year? 
 
 Suggestion: She must produce her value in 6 years, together 
 with 6% on her value as an investment. 
 
 5. A man rented his farm for $720, which was 8% of 
 its valuation. At what did he value the farm? 
 
 6. Received $640 interest at 5%. Find the principal. 
 
 INTEREST 
 
 When a person loans money, he receives for its use or 
 rent a certain sum of money. 
 
 The money loaned is the principal and money paid for 
 its use is the interest. 
 
 The interest is usually reckoned at a certain per cent for 
 a year. This per cent is the rate. 
 
 Where many problems in computing interest have to be solved, 
 interest tables are used. But the following general method is probably 
 the best for all kinds of examples. It is better to teach one general 
 rule than to direct the pupil's attention to several in order that he may 
 select the best one when he needs to apply it. 
 
 MENTAL PROBLEMS 
 
 Find the interest on $600 at 5% for 1 yr.; $350 at 4% 
 for 2 yrs.; $150 at 8% for 1 mo.; $375 at 6% for 6 mo. 
 
70 AGRICULTURAL ARITHMETIC 
 
 WRITTEN PROBLEMS 
 
 Find the interest on $875 for 2 yrs., '6 mo. at 7%. 
 Solution: 875 
 
 M 
 61.25 1 yr. 
 2 
 
 122.50 2 yrs. 
 30.62 6 mo. 
 
 153.12 
 2. Find the interest on $311.10 for 2 yrs., 3 mo. at 4%. 
 
 3. $327 at 5%, 1 yr. and 9 mo. 
 
 4. $1,436 at 7% for 2 yrs., 7 mo. 
 
 5. $875 for 3 yrs. 6 mo. at 5% 
 
 6. Find the interest on $500 for 2 yrs., 3 mo. at 4%. 
 
 7. $2,630.80 at 5%, 1 yr. and 9 mo. 
 
 8. Find the interest on$l,436at7%for 2yrs. ; 7 mo., 11 da. 
 
 Solution: 1436 
 
 .07 
 
 100.52 
 
 1 yr. 
 
 2 
 
 
 201.04 
 
 2 yrs. 
 
 50.26 
 
 6 mo. 
 
 8.37 
 
 1 mo. 
 
 2.79 
 
 10 da. 
 
 .28 
 
 Ida. 
 
 262.74 
 
 9. What is the interest of $150 at 5% for 1 yr., 11 mo., 
 
 and 21 da.? 
 
 Solution: Having found the interest for 1 mo. 
 
 150 by dividing the interest for 1 yr. by 12, multiply by 10 
 
 ■ 05 by moving each figure of 1 mo's. interest one order to 
 
 7.50 1 yr. the left. In finding the interest for 21 da., multiply 
 
 . 62 1 mo. one month's interest by 7, placing the result one order 
 
 6.25 10 mo. to the right; for 21 da. is / of one month; i. e. divide 
 
 . 43 21 da . by 10 and multiply by 7. 
 14.80 
 
 The interest for any number of days that is a multiple of three 
 can thus be obtained. 18 = &, 12 = j^, etc. 
 
INTEREST 71 
 
 By this general rule, every figure of the partial interest 
 is expressed in a compact body with the results ready for 
 addition. 
 
 Find the interest of: 
 
 10. $842 6% 1 yr. 7 mo. 12 da. 
 
 11. 
 
 $365 
 
 4% 
 
 3 yrs. 
 
 8 mo. 
 
 15 da. 
 
 12. 
 
 $1,216 
 
 5% 
 
 1 yr. 
 
 8 mo. 
 
 17 da, 
 
 13. 
 
 $450 
 
 8% 
 
 2 yrs. 
 
 3 mo. 
 
 14 da. 
 
 14. 
 
 $720 
 
 6% 
 
 1 yr. 
 
 1 mo. 
 
 9 da. 
 
 15. $486 3% 2 yrs. 5 mo. 27 da. 
 
 MISCELLANEOUS PROBLEMS 
 
 1. If wheat weighed 2,760 lbs. when put in the bin and 
 in the spring weighed 2,484 lbs., what per cent of the for- 
 mer weight did it decrease? If worth 92c. per bushel in the 
 fall, for how much per bushel must it be sold in the spring to 
 give a 6% profit for holding? 
 
 2. I was offered $400 cash for a team of horses, but 
 decided to send them to a Chicago sale stable. They were 
 sold after two weeks for $500. Charges: $25 freight, $1 
 per day each for board, commission 4%. Which bargain 
 was the better and how much? 
 
 3. Bought a bunch of. pigs for $1,400 at 7c. a pound. 
 I kept them 5 months and fed them 2,500 bu. of corn at 40c. 
 per bushel and $200 worth of other feed. They gained 125% 
 in weight and were sold at 8c. a pound. If I borrowed the 
 purchase money at 8%, what was my profit in the trans- 
 action? 
 
 4. In example 3, what was my per cent of gain, the feed 
 being paid for when pigs were sold? 
 
 5. A note of $245 dated January 12, 1912, with interest 
 at 6%, was paid July 18, 1914. How much was due at the 
 
72 AGRICULTURAL ARITHMETIC 
 
 latter date, no interest having been previously paid? Write 
 the note if payable to John Doe signed by Richard Graham. 
 
 6. Capitalize the value of a hen that for 2 years pro- 
 duces $1.25 per year at an annual expense of 65c. 
 
 Suggestion: If she is worth 50c. for meat, add that sum to get 
 her true value. 
 
 7. Capitalize the value of a cow that produces 12,020 
 lbs. of milk testing at 4.2%, butter-fat worth 30c, for 5 yrs. 
 at an annual expense of $83.50, worth $65 for beef. 
 
 8. Suppose in example 7 the cow averages 6,000 lbs. of 
 3% milk, at an expense of $55 for feed, what is she worth? 
 
 RATIO AND PROPORTION 
 
 The ratio of one quantity to another is the quotient obtained 
 by dividing the first by the second. 
 
 The ratio of 2 to 1 is 2; of 9 to 3 is 3; of 3 to 4 is f , of 7 to 
 8 is |; 1 in. to 4 in. is \; 7 yds to 9 yds. is f ; of 6 bu. to 3 bu. 
 is 2. 
 
 A proportion is a statement of equality of two ratios; as, 
 f = |, the ratio of 6 to 3 = the ratio of 8 to 4. 
 
 The extremes of a proportion are the first and last terms; 
 the means are the second and third terms. 
 
 In the proportion 1r , ,,' = „„ , '' 8 ft. and 36 yds. 
 12 ft. 36 yds., 
 
 are the extremes and 12 ft. and 24 yds. are the means. 
 
 The extremes may be used in computation as 8 and 36, the means 
 12 and 24. The proportion ' = . numerically is the 
 
 same as & = §}. 
 
 An important working principle of proportion is: The 
 product of the means equals the product of the extremes. 
 
 1. If two quantities have the ratio of 5 to 7; two other 
 quantities have the same ratio, and the first term of the 
 second ratio is 15, what is the other term? 
 
RATIO AND PROPORTION 73 
 
 is 
 
 2. Find x in the proportion f = 
 
 a 7 X 15 „, 
 
 Solution: x = — = — = 21. 
 o 
 
 Find x in the proportions: -fa = T V; in ^ = ^; in £ = -J-f-; 
 
 in 
 
 1 9 
 X • 
 
 A proportion may be stated 3 : 12 = 4 : 6 and is read 
 
 3 is to 4 as 4 is to 6. 
 
 Find x in the proportions 3 : 7 = 9 : x; § = 4; 6 : 7 = 9 : x; 
 2.-% '. o = x : otj-. 
 
 Cement, sand, and gravel are frequently mixed in the 
 proportion of 1, 3, and 7 by bulk. If a bag of cement is 
 used', what amount of sand and gravel should be mixed with 
 it? Assume a bag of cement =1 cubic foot. If mixed in 
 the proportion of 1, 3, and 5, how many yards of sand and 
 gravel should be used to 2 barrels of cement? (1 bbl. = 
 
 4 bags.) 
 
 Principle I : The ratio of any two surfaces equals the ratio 
 of their areas or the products of their dimensions. 
 
 3. Two rectangles are 4 by 12 and 3 by 8. What is 
 their ratio? 
 
 Solution: Their areas are respectively 48 and 24 .•. the ratio of 
 
 48 4 X 12 
 
 the first rectangle to the second is — - = 2 or = 2. 
 
 24 o X o 
 
 4. What is the ratio of two squares whose sides are 
 respectively 4 and 6? 4 and 2? 1 and 4? 
 
 Principle II: Circles have the ratio of the squares of 
 their radii. 
 
 5. One circle is 4 times the area of the other. If the 
 radius of the smaller circle is one, what is the radius of the 
 other? 
 
 Solution: j=-j;* ! =4, x = 2 (page 51). 
 
74 AGRICULTURAL ARITHMETIC 
 
 6. If the areas of two circles are as 1 to 9, what is the 
 
 ratio of the radii? 
 
 Suggestion: Use 1 for the radius of the first and x for radius of 
 the second. 
 
 7. If the radii of two circles are in the ratio of 1 to 4; 
 compare their areas. If radii are in ratio of 1 to 3. 
 
 8. The radii of two circles are as 1 to 2. Compare 
 their areas. 
 
 9. Two circles have radii 5 and 7. Determine their 
 ratio. 
 
 10. If one circle is twice another, compare their radii. 
 
 2 2** 
 
 Solution: The ratio of their areas is =•.'.=- = =-= .'. x z = 2 
 
 111^ 
 
 .-. x = VT = 1.414 + 
 
 Principle III. The ratio of any two solids equals the ratio 
 of their volumes or the ratio of the products of their dimensions. 
 
 11. The edges of two cubes are 1 ft. and 2 ft. respect- 
 ively. How do their volumes compare? If the edges are 
 1 and 3 ft., compare their volumes. 
 
 12. The radii and altitude of one cylinder are each one 
 half those of another; compare their volumes. 
 
 Suggestion: Use 1 and 2 for their respective dimensions and 
 compare their volumes. (Page 60.) 
 
 13. If the edges of two cubes are in the ratio of 1 to 4; 
 of 1 to 5; of 1 to 6; compare their volumes. 
 
 MISCELLANEOUS PROBLEMS 
 
 14. One pine log has twice the diameter and three 
 times the length of another. The smaller produces 60 ft. 
 of lumber. How much will the other produce, if sawed into 
 the same dimension lumber? 
 
 15. One silo has twice the diameter and twice the alti- 
 tude of another. How does the first compare with the 
 second in capacity? 
 
RATIO AND PROPORTION 75 
 
 16. If the smaller contains 1,056 cu. ft., what is the 
 capacity of the other? 
 
 17. The area of a circle is 81f sq. ft. What is its 
 radius? 
 
 18. The ratio of two numbers is as 5 to 7 and their 
 sum is 96. What are the numbers? 
 
 Suggestion: One is /, and the other & of 96. Why? 
 
 19. Two men are to divide a crop of corn, 1,500 bu., 
 so that one will have one half as much as the other. How 
 many bushels should each have? 
 
 20. A sale of cattle for $2,842 is to be divided so that 
 one partner will have § as much as the other. What is the 
 share of each? 
 
 Suggestion: What is the ratio of their shares? 
 
 21. How shall I divide a 48-in. evener or doubletree 
 so that one horse will draw one half as much as the other? 
 
 Suggestion: The amount of draft is in inverse ratio to the 
 length of the arms of the doubletree. 
 
 22. How divide it so that one horse will draw f as 
 much as the other? To which end should the weaker horse 
 be hitched? 
 
 23. If one apple is 2 in. in diameter and another is 3 in., 
 how many times as large as the first is the second? 
 
 24. One cylindrical tank has a radius of 7' and depth of 
 2£', another a radius of 5' and depth of 3£'. What is the 
 ratio of the first to the second? 
 
 25. The price of 2£-in. oranges, pulp measure, is 40 
 cents per dozen; of 3-in. oranges is 50 cents. If of the same 
 quality, which is better to buy? What is the real value 
 of the larger oranges at the rate of the smaller? Of the 
 smaller at the rate of the larger? 
 
PART TWO 
 
 CHAPTER I 
 FARM CROPS 
 
 The first great business of the farmer is to raise crops. 
 The world must be fed and clothed, and it is he who pro- 
 vides the bulk of the raw material which is manufactured 
 into food stuffs and clothing. Without the farmer commerce 
 would cease. From the ever increasing city population 
 comes the cry for bread even more urgent. There will 
 always be cities, and thus the farmer will always find a mar- 
 ket for his farm produce. The successful farmer is an 
 artisan who skillfully combines the art and science of farm- 
 ing with business and enterprise. 
 
 ORAL PROBLEMS 
 
 1. A man produced 425 bushels of oats from five 
 acres. What was the yield per acre? 
 
 2. Another man produced 156 bushels from 13 acres. 
 What was the yield per acre? 
 
 3. In your state what is the weight of 
 
 (a) 6 bushels of shelled corn? 
 
 (b) 5 bushels of corn on cob? 
 
 (c) 2 bushels of oats? 
 
 (d) 9 bushels of potatoes? 
 
 (e) 7 bushels of wheat? 
 
 (f) 5 bushels of rye? 
 
 (g) 6 bushels of onions? 
 
 4. 350 pounds of cured corn on the cob is equivalent 
 to how many bushels of shelled corn? (See page 242.) 
 
78 
 
 AGRICULTURAL ARITHMETIC 
 
 5. How many acres will 45 bushels of rice plant, when 
 sowing is done at the rate of 1£ bushels per acre? 
 
 6. At the rate of 10 bushels per acre, how many bushels 
 of seed potatoes will be required for f of an acre? 
 
 7. At the rate of five pecks per acre, how many bushels 
 of cowpeas are required to sow 12 acres? 
 
 8. What is the weight of 4 quarts of timothy seed in 
 your state? 
 
 9. Seeding 12 pounds of red clover seed per acre is 
 equivalent to seeding how many quarts per acre? 
 
 Figure 1. — Corn, — The Greatest Crop in the United States. If one has never 
 taken any interest in plants, one should study corn. Some of the won- 
 ders of creation will be revealed. 
 
 Courtesy of Wis. Agr'l Exp. Ass'n. 
 
FARM CROPS 79 
 
 10. Hay that averages 30 pounds per square rod yields 
 how many tons per acre? 
 
 11. How many more pounds in 2 bushels of potatoes 
 than in 2 bushels of oats? 
 
 12. 3 bushels of shelled corn is equivalent in weight 
 to how many bushels of oats? 
 
 13. 33| per cent of f of a bushel of barley in Wisconsin 
 is equal in weight to what part of a bushel of beans? 
 
 WRITTEN PROBLEMS 
 
 1. According to government statistics, 105,820,000 
 acres of corn were raised in the United States in 1913, pro- 
 ducing 2,446,988,000 bushels. What was the average yield 
 of corn per acre in the United States for that year? 
 
 2. What was the average yield of corn in your state for 
 that year? What is the average yield on your farm or in 
 your community? 
 
 3. Official statistics show that the grand total corn 
 crop of the world for the year 1913 was approximately 
 3,605,442,000 bushels. What per cent of the world's crop 
 was produced in the United States during that year? 
 
 4. How much of the nation's wealth was represented 
 by its corn crop December 1, 1913, when the average farm 
 price per bushel was 69.1 cents? 
 
 5. Statistics show that the average acre wheat yield 
 of the United States for the nine years 1905 to 1913, inclu- 
 sive, is only 14.6 bushels and that of Germany for the same 
 period is 30.7 bushels. The average wheat yield of Germany 
 is what per cent of that of the United States? 
 
 6. The population of the United States in 1910 was 
 92 millions; in 1890 it was 63 millions. What was the per 
 cent of increase in the twenty years? 
 
80 
 
 AGRICULTURAL ARITHMETIC 
 
 (a) The rural population in 1890 was 
 63.9%; in 1910 it was 53.7%. 
 What was the per cent of decrease 
 in rural population during these two 
 decades? 
 
 7. The average annual acreage of corn, 
 wheat, oats, barley, rye, and buckwheat during 
 the five years from 1906 to 1910, inclusive, was 
 210,935,000 acres; for the five years from 1886 
 to 1890, inclusive, it was 144,141,000 acres. Did 
 the per cent of acreage increase keep pace with 
 the per cent of increase in population during that 
 period? Can this per cent of acreage increase 
 continue very long? Why? 
 
 8. From the data given in Figure 2 de- 
 termine the decrease in the average number of 
 bushels of corn and other grains exported annually 
 from the United States from the banner five-year 
 period, 1896 to 1900 inclusive, to the average for 
 the five years from 1906 to 1910 inclusive. 
 
 9. When we consider the facts brought out 
 in problems 6, 7 and 8, what greater responsibil- 
 ity is placed upon the farmers of this country? 
 
 10. The estimated production of cotton in 
 China in 1912 is 4,000,000 bales. This quantity 
 is 29.18% of the total production in continental 
 United States for the same year. How many 
 bales of cotton were produced in the United 
 States in 1912? 
 
 11. Texas produces more cotton than any 
 other state. The total crop for the United States in 1912 
 was 188% more than that produced by Texas alone. How 
 many bales of cotton did Texas produce in 1912? 
 
 < OQ < 
 
 Figure 2. — De- 
 crease in Ex- 
 portation of 
 Grain from the 
 United States. 
 
FARM CROPS 81 
 
 12. A grain farmer sold 30 tons, or f of his wheat crop, 
 at 90 cents per bushel, and the remainder for 91.5 cents 
 per bushel. How much did he get for his crop? 
 
 13. "A" produced from 8 acres, 1,680 pounds of clover 
 seed, or 2\ times more than his neighbor from an equal 
 area. What was the yield in bushels per acre of clover seed 
 on each of the two farms? 
 
 14. An acre of corn carefully planted should average 
 at least 3 stalks to the hill which are generally planted 44 
 inches apart each way. 
 
 (a) How many square feet are allowed for each 
 hill? Draw a diagram to illustrate this. 
 
 (b) How many hills does one acre of this corn 
 contain? 
 
 (c) A quart of average dent corn contains about . 
 2,500 kernels. How many quarts of seed corn 
 are required to plant this acre? (Allow a little 
 extra for overrun.) How many pounds? 
 
 (d) At this rate how many bushels of corn are 
 required to plant 30 acres? How many 
 pounds? 
 
 15. If the corn were planted in drills 3 feet apart and 
 averaging one stalk per foot in the row, how much corn 
 would be required to plant an acre? How many quarts? 
 Pounds? 
 
 16. Good average seed corn contains from six to seven 
 hundred kernels per ear suitable for planting, tips and butts 
 excluded. How many ears would be required to plant an 
 acre described in problem 14? 
 
 17. How many ears would be required to plant an acre 
 described in problem 15? 
 
 18. Illinois Station demonstrated that on rich soil the 
 greatest amount of total digestible food substances (pro- 
 
82 AGRICULTURAL ARITHMETIC 
 
 tein, carbohydrates and fats) were secured per acre in grow- 
 ing dent corn when the kernels were planted 3 inches apart 
 in the row and the rows 3 feet 8 inches apart. 
 
 (a) How many kernels are required to plant an 
 acre of such corn? 
 
 (b) About how many quarts? Pounds? 
 
 (c) About how many ears of good seed corn? 
 
 19. 4.8 tons of stover were produced per acre in the 
 test described in problem 18, or 3.6 pounds of stover for 
 each pound of shelled corn. How many bushels of shelled 
 corn were produced per acre? 
 
 20. A farmer growing dent corn on rich soil found he 
 could get best results by planting 4 kernels in the hill and 
 the hills placed 3 feet 8 inches apart each way. He secured 
 74.8 bushels of shelled corn per acre. The ratio of field 
 
 "cured stover to corn on cob was 1.3 to 1. 
 
 (a) How much seed corn was required to plant 
 an acre of this corn? 
 
 (b) How many tons of stover were produced per 
 acre? 
 
 21. In one gram of good alfalfa seed are 475 seeds. 
 How many seeds are sown per square foot when 20 pounds 
 are evenly distributed over one acre? (See page 246.) 
 
 Fifteen plants per square foot may be regarded an excellent 
 "catch" and 6 to 9 per square foot a good "producing stand." What 
 becomes of the rest of the seed planted? Should less seed be sown per 
 acre? 
 
 22. Shelled corn at 70 cents per bushel is equivalent 
 to how much per ton? 
 
 23. A farmer received $600 for 16 tons of oats. How 
 much did he get per bushel? 
 
 24. A man secured barley at $25 per ton when the 
 market quotation was 65 cents per bushel. Did he give 
 more or less per ton as compared with market quotation, 
 and how much? 
 
FARM CROPS 
 
 83 
 
 Figure 3. — A Good Crop of Alfalfa. Leguminous crops of all kinds not only 
 - furnish abundant supply of protein and mineral matter for animal needs, 
 but help to maintain the fertility of the soil by gathering nitrogen from 
 the air. 
 
 25. 10,500 pounds of corn on cob were placed in a crib 
 when husked. After a year's time 119.25 bushels were taken 
 from the crib as the total content. What was the per cent 
 shrinkage of the ear corn? 
 
 (a) When the corn is worth 40 cents at time of 
 cribbing, what should it be worth per bushel 
 when removed to compensate for the shrinkage? 
 
 26. 100 pounds of potatoes containing 17% starch 
 would yield approximately 1.3 gallons of denatured alcohol. 
 How many quarts would one bushel yield? 
 
 (a) How many gallons would one acre yield when 
 220 bushels are harvested averaging 20% 
 starch? 
 
 27. Cabbage plants are usually set in rows about 30 
 inches apart and 13 to 18 inches apart in the row. How 
 many tons would an acre yield when a perfect stand is 
 secured, if plants are 15 inches apart in the row, and each 
 head averages 4 pounds? 
 
84 AGRICULTURAL ARITHMETIC 
 
 (a) Estimate without use of pencil the yield of 
 cabbage per acre when the rows are 30 inches 
 apart; the plants were set 20 inches apart in 
 the rows, but every third plant on an average 
 failed to grow. The heads averaged 4 pounds. 
 Check your estimate. 
 
 (b) Make another estimate for conditions same as 
 in (a) only that the heads averaged 3 pounds. 
 Check your estimate. 
 
 28. Estimate the yield of shelled corn per acre when 
 corn is standing ripe, is planted in hills 3 feet 8 inches each 
 way, and one hill out of seven is missing. The stalks stand- 
 ing average 3 one-pound ears per hill. Deduct 30% for 
 moisture. 
 
 Points of Interest: 
 
 1. The farmers feed the world. 
 
 2. The United States with her expanses of "new" soil 
 falls far below Germany in production of wheat per acre. 
 
 3. Better agriculture is a national necessity. 
 
 Name other points of practical value brought out in 
 Chapter I. 
 
CHAPTER II 
 FARM ANIMALS 
 
 As far back as we can go in human history we find that 
 domestic animals have played an important role in the 
 development of civilization. Man and the cow, as it were, 
 seem to have been rocked in the same cradle. The 
 economic importance of domestic animals ascends to large 
 proportions when we consider their use for food, clothing 
 and labor. 
 
 Figure 4. — A Good Type of a Draft Horse (Clydesdale). 
 
 is never reared by the careless. 
 
 An animal like this 
 
AGRICULTURAL ARITHMETIC 
 
 WRITTEN PROBLEMS 
 
 1. The 1910 census reports the following numbers 
 and values as applied to domestic animals on the farms in 
 the United States for that year: 
 
 Class of Stock 
 
 Number 
 
 Value 
 
 Horses 
 
 19,833,000 
 
 $2,083,588,000 
 
 Cattle 
 
 61,804,000 
 
 1,499,524,000 
 
 Mules 
 
 4,210,000 
 
 525,392,000 
 
 Swine 
 
 58,186,000 
 
 399,338,000 
 
 Sheep 
 
 52,448,000 
 
 232,842,000 
 
 What is the total number and value of these farm animals? 
 
 2. If 10 men can each count silver dollars at the rate 
 
 of 60 per minute, how many years would it take them, 
 
 Figure 5. — A Good Type of Percheron Mare. 
 
FARM ANIMALS 87 
 
 working 8 hours per day and 300 days in a year, to count 
 a number of dollars equal to the amount invested in these 
 farm animals? 
 
 3. What was the average value of cattle per head? : 
 
 4. Compare the average value of horses per head with 
 that of mules. 
 
 5. What value was placed on swine per head? On 
 sheep? 
 
 6. What is the value per head of these various farm 
 animals in your community? 
 
 7. In 1911, nearly 272,000 car loads of live stock were 
 received in the Union Stock Yards of Chicago. What was 
 the average number of carloads received per day? 
 
 8. The average person in the United States eats about 
 180 pounds of meat a year. What is the average amount 
 eaten per day? 
 
 9. The champion Jersey cow (1914), (Sophie 19th of 
 Hood Farm), produced 999.14 pounds of butter-fat in 365 
 days. Her milk tested 5.69% fat. 
 
 (a) How many pounds of milk did this cow pro- 
 duce in a year? Average per day? 
 
 (b) How many pounds of butter testing 85% but- 
 ter-fat? Average per day? 
 
 10. The champion Ayrshire cow (1914), (Auchenbrain 
 Brown Kate 4th), produced in one year enough butter-fat 
 to make 1,070.6 pounds of butter, containing 85.7% butter- 
 fat. Her milk tested 3.99% fat. How many pounds of 
 milk were produced during the year's test? What was the 
 average per day? 
 
 11. The once champion Holstein cow of the world, 
 Banostine Belle De-Kol (May 1, 1914) produced 27,404 
 
AGRICULTURAL ARITHMETIC 
 
 Figure 6. — Auchenbrain Brown Kate IV. A former world's champion Ayrshire. 
 (See problem 10, page 87.) 
 
 pounds of milk in one year, testing 3.86% butter-fat. De- 
 termine the value of the milk produced at 6 cents per quart.* 
 
 12. May Rilma (Guernsey) produced in one year 
 19,673 pounds of milk, testing 5.45% butter-fat. How 
 many more pounds of butter-fat were produced by this cow 
 than by the champion Holstein? 
 
 13. Murne Cowan, the champion Guernsey cow of the 
 world and for a time the champion of all breeds, completed 
 her record February, 1915, after producing 24,008 pounds 
 of milk, testing 4.57% in 365 days. 
 
 (a) Determine the value of the milk produced at 
 6 cents per quart. 
 
 (b) How many pounds of milk did she produce 
 per day on an average? Quarts per day? 
 
 *1,000 pounds of milk = 465 quarts. 
 
FARM ANIMALS 
 
 89 
 
 (c) How many more pounds of butter-fat did this 
 champion cow produce than May Rilma, the 
 world's champion before her? 
 
 14. When Finderne Holingen Fayne, a 3-year-old Hol- 
 stein cow, finished her record March, 1915, she won the 
 world's championship* of all the breeds. She produced 
 24,612.8 pounds of milk yielding 1,116.05 pounds of butter- 
 fat. 
 
 (a) Determine the average test of this cow's milk. 
 
 (b) How many gallons of milk did this cow produce 
 per day on an average? 
 
 15. Compare the average daily production of milk of 
 May Rilma, Murne Cowan and Finderne Holingen Fayne. 
 Average daily production of butter-fat. Yearly production 
 of milk and butter-fat. 
 
 Figure 7. — Murne Cowan. The champion Guernsey cow of the world. (See 
 problem 13, page 88.) 
 
 *See page 228, problem 27. 
 
90 AGRICULTURAL ARITHMETIC 
 
 16. Tilly Alcartra (Holstein) produced 30,452.6 pounds 
 of milk in 365 days — the most milk ever produced by one cow 
 in one year. She produced 1,189 pounds of butter containing 
 80% butter-fat. 
 
 (a) What was the average amount of milk pro- 
 duced in a day? Pounds? Quarts? 
 
 (b) What was the average test of this cow's milk? 
 
 17. How many pounds of milk do the cows in your 
 neighborhood or on your farm produce daily? Per year? 
 
 (a) How many such cows would be required to 
 produce as many pounds of butter-fat in one 
 year as were produced by Finderne Holingen 
 Fayne? 
 
CHAPTER III 
 FEEDS AND FEEDING PRACTICE 
 
 To eliminate guessing in feeding, the farmer must have a 
 clear understanding of the needs of the various animals on 
 his farm, and he must know the comparative feeding value 
 of the many feeds. All this demands study. Food value of 
 feeds is determined by their composition which is usually 
 expressed by the following substances: water, ash*, protein, 
 crude fiber, nitrogen-free extract and fat. Nitrogen-free 
 extract includes sugars, starches and gums. The group- 
 term carbohydrates is commonly used to include the nitro- 
 gen-free extract and crude fiber. 
 
 ORAL PROBLEMS 
 
 1. Dent corn contains 10.5% water. What per cent 
 dry matter does it contain? 
 
 2. Oats contain 9.2% water. How many pounds of 
 dry matter in 100 pounds? 
 
 3. Cowpea hay consists of 9.7% water. How many 
 pounds of water in one ton? 
 
 4. Fresh pasture grass is 80% water. How many 
 pounds of water are consumed when a cow eats 70 pounds 
 of grass a day? 
 
 5. If a bushel of wheat were burned, 1.9% of it would 
 remain as ash (ashes). How many pounds of ash would 
 be left? 
 
 6. Thirteen and six tenths per cent is the ash content 
 of alfalfa leaves. How much ash is contained in 60 pounds 
 of leaves? 
 
 *Ash is commonly referred to as mineral or inorganic matter. 
 91 
 
92 AGRICULTURAL ARITHMETIC 
 
 7. Wheat bran analyzes 6.3% ash. How many pounds 
 of mineral matter are contained in 1,000 pounds of bran? 
 
 8. Cottonseed meal consists of 6.4% ash. How many 
 pounds of inorganic matter are contained in i of a ton? 
 
 9. Cottonseed meal contains 8.2% fat. How many 
 pounds of oil in 80 pounds of cottonseed meal? 
 
 10. Corn stover contains 5.9% protein. How many 
 pounds of this flesh-producing substance are contained in 
 half a ton of corn stover? 
 
 11. 14.9% of the weight of alfalfa hay is protein. How 
 many pounds of this flesh and milk-producing substance are 
 contained in 1,000 pounds? 
 
 12. Oil meal (linseed meal) analyzes 7.5% fat. How 
 many pounds of fat in 2 tons? 
 
 13. Oat straw contains 36.3% crude fiber and alfalfa 
 hay 28.3%. How many more pounds of crude fiber, or 
 woody fiber, are contained in 200 pounds of oat straw than 
 in the same amount of alfalfa hay? 
 
 14. Oat straw contains 40.8% nitrogen-free extract, 
 and alfalfa hay 37.3%. How many pounds of carbohydrates* 
 are contained in 100 pounds of oat straw? In 100 pounds 
 of alfalfa hay? 
 
 WRITTEN PROBLEMS 
 
 1. By consulting table on "Composition of Common 
 Feeds" in the appendix (page 235), determine the number of 
 pounds of ash, or mineral matter, contained in 2 tons each 
 of dent corn, alfalfa hay, red clover hay and oat straw. 
 
 2. How many more pounds of crude protein are con- 
 tained in 3 tons of alfalfa hay than in the same amount of 
 timothy hay? 
 
 ♦Carbohydrates ia the sum of the crude fiber and nitrogen-free extract. (See 
 introduction to Chapter III.) 
 
FEEDS AND FEEDING 93 
 
 3. Only 3 pounds of crude protein out of 100 pounds 
 of timothy hay can be digested by farm animals, while 10.6 
 pounds of protein in every 100 pounds of alfalfa are digesti- 
 ble. (See page 247.) What per cent is the amount of 
 digestible protein contained in one ton of alfalfa hay of that 
 in a ton of timothy hay? Which would you rather feed to a 
 dairy cow, alfalfa or timothy? Why? What per cent of 
 the total crude protein in alfalfa hay is digestible? In 
 timothy? 
 
 4. Which contains more digestible protein, one ton of 
 alfalfa hay or one ton of red clover hay? How much more? 
 
 5. How many times is the digestible protein con- 
 tained in 4 tons of cottonseed meal that in the same amount 
 of gluten feed?. 
 
 6. A farmer bought 20 bushels of shelled dent corn. 
 
 (a) How many pounds of. crude fiber did he buy? 
 
 (b) How many pounds of nitrogen-free extract 
 did he buy? 
 
 (c) How many pounds of carbohydrates? 
 
 (d) How many pounds of digestible carbohydrates 
 are contained in this amount of corn? (See 
 page 247.) 
 
 7. What is the difference in the amounts of digestible 
 carbohydrates contained in 5 tons each of oat straw and 
 timothy hay? How would good oat straw do for winter 
 horse feed? 
 
 8. How many acres of alfalfa yielding 5 tons of hay per 
 acre per year are required to produce as much digestible 
 protein as is produced by 30 acres of timothy yielding 2 
 tons of hay per acre? 
 
 9. Five tons of corn silage contain as much digestible 
 carbohydrates as is contained in how many tons of sugar 
 beets? 
 
94 
 
 AGRICULTURAL ARITHMETIC 
 
 10. How many pounds of fat, or oil, are contained in 
 25 bushels of shelled corn? How many pounds of digestible 
 fat? 
 
 11. How many pounds of nutrients* are contained in 
 1,000 bushels of oats? Digestible nutrients? 
 
 <£j$ii$fflsBEa ^^HBMESB^k 
 
 
 
 . ■ ■ 
 
 ■A 
 
 
 
 '-■ ' 
 
 
 
 .J£gBB&$l£: 
 
 
 
 
 
 
 ^^^V^^£^§^^£^^^§^^ 
 
 
 
 Figure 8. — A Well Bred and Thrifty Family. Successful management of 
 swine depends largely upon the man behind the swill cart. 
 
 12. Ten tons each of wheat bran and cottonseed meal 
 are piled separately in two piles. Which pile of feed contains 
 the greater amount of nutrients? How much more? 
 
 13. Compare the amount of nutrients in 3 tons of alfalfa 
 hay with that contained in 3 tons of wheat bran. 
 
 14. Determine the amount of digestible nutrients in 
 one ton each of alfalfa hay, wheat bran and cottonseed 
 meal. Place these three feeds in the order of the amount 
 of digestible nutrients they contain. 
 
 * "Nutrients — a term commonly used, when feeds are considered, to include 
 crude protein, carbohydrates and fats; although air, water and mineral matter might 
 likewise be so termed. 
 
FEEDS AND FEEDING 95 
 
 15. A man purchased 15 tons of alfalfa hay at $10 per 
 ton. Would he have purchased as much digestible crude 
 protein, carbohydrates and fats had he purchased the same 
 amount of good sweet clover hay at the same price? 
 
 16. A dairyman went out to buy some hay for his cows. 
 He had a choice between 7 tons of good alfalfa hay for $107 
 and 11 tons of good red clover hay for $110. He purchased 
 the hay which gave him the larger amount of total crude 
 protein to the dollar. Which hay did he purchase and how 
 much more crude protein did he get for every dollar spent? 
 
 (a) From the standpoint of digestible crude pro- 
 tein, did this dairyman purchase wisely? What 
 was the difference in amount of digestible 
 crude protein purchased for one dollar? 
 
 (b) Could he have purchased more total nutrients 
 had he purchased the other hay? The differ- 
 ence is in favor of which hay? 
 
 (c) Some feeding tests have demonstrated that 
 one pound of alfalfa hay, in feeding dairy cattle, 
 is equal to 1.6 pounds of red clover hay in pro- 
 ducing results. From this standpoint which 
 would have been the cheaper hay? How 
 much cheaper? 
 
 .17. When alsike clover is worth $8 aton, what ought 
 crimson clover to be worth on the basis of digestible crude 
 protein? 
 
 18. 1,000 pounds of shelled corn contain ^ of a pound 
 of lime* and 3 pounds of phosphorus. The same amount 
 of alfalfa hay contains 43.1 pounds of lime and 2.4 pounds of 
 phosphorus. How many pounds of bone-building material 
 (lime and phosphorus) are contained in 400 bushels of corn 
 and 5 tons of alfalfa hay? 
 
 ^Equivalent in quicklime. 
 
96 AGRICULTURAL ARITHMETIC 
 
 19. How much alfalfa hay must be fed in order to feed 
 84 pounds of digestible crude protein? How many pounds 
 of timothy? 
 
 20. How many bushels of barley must be fed in order 
 to feed 326.5 pounds of digestible carbohydrates? How 
 many bushels of corn? Would ground or rolled barley with 
 corn make a good feed for fattening hogs?* Why? 
 
 21. How many tons of mangels are required to equal 
 the amount of digestible nutrients contained in 2 tons each 
 of alfalfa and Johnson grass hay? What is the great value 
 of mangels and other roots in feeding, especially when no 
 silage is fed? 
 
 22. Cornell Station demonstrated that 1 pound of dry 
 matter in mangels is equal to 1 pound of dry matter in grain 
 in feeding dairy cows. 
 
 (a) One pound of dry matter in mangels is equiva- 
 lent to how many pounds of mangel roots? 
 
 (b) One pound of dry matter in dent corn is equiva- 
 lent to how many pounds of the corn grain? 
 
 (c) One ton of mangels in feeding dairy cows is 
 equal to how many pounds of shelled corn in 
 producing results? How many bushels? 
 
 23. A farmer had 25 cows and fed them an average of 
 30 pounds of corn silage for 270 days. 
 
 (a) How many tons of silage did he feed during 
 that time? 
 
 (b) How many acres were required to produce this 
 silage when the average yield was 12 tons of 
 silage corn per acre? 
 
 *Barley alone has about 10% less value than corn for fattening swine. It 
 should never be fed separately, but always in combination with corn, wheat, 
 Toots, alfalfa, etc. 
 
FEEDS AND FEEDING 97 
 
 (c) When a cubic foot of silage in a silo, 2 days 
 after filling, averages 40 pounds, what is the 
 minimum cubic content of. a silo required to 
 contain this amount of feed? 
 
 (d) The silo was cylindrical with an inside diam- 
 eter of 14 feet. What was the depth of the 
 silage 2 days after filling? About how high 
 was the silo, inside measurement? (See page 
 63.) 
 
 (e) About how high ought a 16-foot silo to be to 
 contain this feed? 
 
 (f) Which one is the better shape for the above 
 herd? What dimensions would you suggest?* 
 
 24. A farm of 120 acres produced all the feed fed to a 
 herd of 30 dairy cows. The average daily ration consisted 
 of 35 pounds of corn silage fed for 11 months, 6 pounds of 
 alfalfa hay fed for 8 months, 5 pounds of clover hay fed for 
 8 months, 8 pounds of ground oats and barley (equal parts 
 by weight) fed for 9 months. 
 
 How many acres were required to produce the amount 
 of feed necessary when the following yields were obtained: 
 corn, 14 tons of green corn; alfalfa, 4.5 tons (3 cuttings); 
 red clover, 2 tons (2 cuttings); barley, 30 bushels; oats, 50 
 bushels per acre? 
 
 25. 850 lambs were fed 112 days on a ration consisting 
 of 2 pounds of mixed grains, 1.4 pounds of corn silage, 1.6 
 pounds of alfalfa hay. The average gain per lamb during 
 this period was 46.04 pounds. 
 
 (a) What was the total number of tons of each 
 feed fed? 
 
 ♦When silage ia left exposed to the air for any length of time it spoils. To 
 prevent spoiling in the silo during the time in which it is being used it is necessary 
 to remove a layer of silage to a depth of not less than 1 ^ inches daily. 
 
98 AGRICULTURAL ARITHMETIC 
 
 (b) What was the average amount of each feed 
 consumed per lamb? 
 
 (c) What was the average daily gain for the whole 
 flock? Average per lamb? 
 
 (d) How many pounds of feed were necessary to 
 produce 100 pounds of gain? 
 
 (e) If alfalfa was worth $10, corn silage $3, and 
 mixed grains $20 per ton, what did it cost 
 in feed to produce a pound of gain? 
 
 26. A man had 20 cows. He gave them plenty of good 
 roughage* and fed each cow as many pounds of concentratesf 
 daily as she gave pounds of butter-fat in a week — a simple 
 rule followed by the Wisconsin Station. 
 
 The cows produced as follows: 5 averaged 15 pounds of 
 milk each daily, testing 4% butter-fat; 6 averaged 20 pounds 
 each, testing 3.8%; 5 averaged 22 pounds each, testing 4%; 
 3 produced 20 pounds each of milk daily, testing 4.2%; and 
 one cow averaged 35 pounds of milk daily, testing 3.8% 
 butter-fat. 
 
 (a) Determine the number of pounds of concen- 
 trates fed each cow daily. 
 
 (b) How many pounds of concentrates were re- 
 quired to feed the herd for 30 days? 
 
 27. A feeder plans to produce at least 10 pounds of 
 pork per bushel of shelled corn. He has 150 pigs averaging 
 110 pounds, and plans to market them when they average 
 225 pounds. 
 
 (a) On this basis how many bushels of shelled 
 corn would be required to fit these pigs for 
 the market? 
 
 *Roughage includes the coarser feeding stuffs, such as hay, corn silage, roots, 
 etc. 
 
 ^Concentrates are feeding stuffs of condensed nature, such as grains, bran, 
 - cottonseed meal, etc. 
 
FEEDS AND FEEDING 99 
 
 (b) This amount of shelled corn is equivalent to 
 how many tons of cured ear corn? 
 
 (c) What would be the average amount of shelled 
 corn consumed per pig? 
 
 Facts in Feeding: 
 
 1. There is a wide variation in the total digestible pro- 
 tein, carbohydrates and fats contained in feeds. 
 
 2. Feeds containing much protein are best for flesh 
 and nerve building and for milk production. 
 
 3. In knowing the composition of feeds and the needs 
 of animals, the farmer can raise the kind of crops, or buy 
 the feeds, to meet his particular needs in feeding. 
 
 7— 
 
CHAPTER IV 
 THE THEORY OF FEEDING 
 
 A great deal of study has been given to animal feeding. 
 Numerous experiments have been made the world over 
 determining the value of feeds and combinations of different 
 feeds for various animals in all stages of growth and for all 
 purposes and uses to which domestic animals are put. These 
 tests have demonstrated that best results are obtained 
 through intelligent and regular feeding. To do this the 
 feeder must know the particular needs of the animals he is 
 feeding and what feeds he can use to best advantage to meet 
 those needs. 
 
 WRITTEN PROBLEMS 
 
 1. The heat value of digestible fat is 2.25 times that 
 of digestible carbohydrates. 75 pounds of digestible fat in 
 a certain amount of feed is equivalent in heat value to how 
 many pounds of digestible carbohydrates? 
 
 2. 8.25 pounds of digestible fat in corn silage is equiva- 
 lent to how many pounds of digestible carbohydrates? 
 
 (a) One hundred' pounds of barley contain 66.8 
 pounds of digestible carbohydrates and 1.6 
 pounds of digestible fat. Express the total 
 amount of digestible carbohydrates and fat in 
 terms of digestible carbohydrates. 
 Solution: 
 
 Step 1. The heat value of fat is 2.25 times that of carbohydrates, 
 hence the digestible carbohydrate equivalent of 1.6 pounds of fat is 1.6 
 X 2.25, or 3.6 pounds of digestible carbohydrates. 
 
 Step 2. The sum of 66.8 and 3.6, or 70.4 pounds, is the total 
 amount of digestible carbohydrates and fat expressed in terms of 
 digestible carbohydrates. 
 
 100 
 
THEORY OF FEEDING 101 
 
 3. 100 pounds of germ oil meal contain 42.6 pounds of 
 digestible carbohydrates* and 10.4 pounds of digestible fat. 
 Express the total amount of carbohydrates and fat in terms 
 of digestible carbohydrates; i. e., the equivalent of digest- 
 ible carbohydrates. 
 
 4. Germ oil meal contains 16.5% digestible crude pro- 
 tein. Determine the ratio between the digestible crude 
 protein and the digestible carbohydrates equivalent in this 
 feed; in other words, for every pound of digestible crude 
 protein how many pounds of combined digestible carbo- 
 hydrates and fats? (Fat reduced to equivalent in carbohy- 
 drates.) 
 
 The result expresses the nutritive ratio of germ oil meal. 
 
 Nutritive ratio of any feed, or ration, is the ratio be- 
 tween the digestible crude protein and the combined digest- 
 ible carbohydrates and fat — the amount of fat being first 
 reduced to its equivalent in digestible carbohydrates. 
 
 The first factor in a nutritive ratio is always 1; the 
 second factor is determined by dividing the total amount 
 of digestible carbohydrates and fat (expressed in equivalent 
 of digestible carbohydrates) by the amount of digestible 
 crude protein. 
 
 5. What is the nutritive ratio of oat straw? 
 
 Solution: . 
 
 100 pounds of oat straw contain 1.0 pound of digestible crude 
 protein, 42.6 pounds of digestible carbohydrates and 0.9 pound of 
 digestible fat. 
 
 Heat Diges. Diges. crude 
 Fat equiv. carb. protein 
 
 [(0.9 X 2.25) + 42.6] -h 1.0 = 44.6 (The second factor of the ratio.) 
 
 Diges. carb. equiv. 
 
 The nutritive ratio of oat straw is, therefore, 1 to 44.6; or it may 
 be expressed as 1 : 44.6 or ?£& which means one pound of digestible 
 crude protein to 44.6 pounds of combined digestible carbohydrates and 
 fat. 
 
 *One hundred pounds is taken when the amount of feed is not expressed; 
 otherwise the total amount of digestible nutrients is determined from the amount 
 of feed fed. 
 
 ^Digestible carbohydrates includes digestible nitrogen-free extract and di- 
 gestible crude fiber combined. 
 
102 AGRICULTURAL ARITHMETIC 
 
 6. Determine the nutritive ratio of oats. 
 
 7. What is the nutritive ratio of crimson clover hay? 
 
 8. Compare the nutritive ratio of soy beans with that of 
 shelled dent corn. 
 
 Which contains the more digestible crude protein in 
 proportion to combined digestible carbohydrates and fat? 
 
 A feed or ration is said to have a narrow nutritive ratio 
 when it contains much digestible crude protein in pro- 
 portion to combined carbohydrates and fat; and a wide 
 nutritive ratio when it contains a small amount of protein 
 to carbohydrates and fats. The second factor in a narrow 
 nutritive ratio, therefore, is a small number; in a wide nutri- 
 tive ratio it is a comparatively large number. 
 
 9. Which has the narrower nutritive ratio, shelled 
 corn or gluten meal? 
 
 10. What is the nutritive ratio of wheat bran? Is it 
 wide or narrow? Compare this nutritive ratio with that 
 of alfalfa hay. 
 
 (a) Determine the nutritive ratio of the following 
 ration fed to a lot of range lambs at the Okla- 
 homa Experiment Station* with as good results 
 as compared with an alfalfa ration. (The 
 lambs were being fattened and were given all 
 they could consume.) 
 
 Cowpea hay 1.5 pounds 
 Cornmeal 1.6 pounds 
 Solution: 
 
 Step 1. ' Table on page 247 shows cowpea hay to contain digestible 
 nutrients as follows, — 13.1% of protein, 33.7% of carbohydrates, 1.0% 
 of fat.- Cornmeal contains 6.9% of protein, 69.0% of carbohydrates, 
 3.5% of fat. 
 
 SUp 2. 
 
 1.5 X .131 = 0.196 lb. digestible proteins fed in the cowpea hay. 
 1.5 X .337 = 0.505 lb. digestible carbohydrates fed in the cowpea 
 hay. 
 
 "Oklahoma Bulletin 78. 
 
THEORY OF FEEDING. 103 
 
 1.5 X .01 = 0.015 lb. digestible fat fed in the cowpea hay. 
 
 1.6 X .069 = 0.11 lb. digestible protein fed in the cornmeal. 
 
 1.6 X .69 = 1.104 lbs. digestible carbohydrates fed in the cornmeal. 
 1.6 X .035 = 0.056 lb. digestible fat fed in the cornmeal. 
 
 SUp 3. 
 
 [(0.015 + 0.056) X 2.25] + (0.505 + 1.104) = 1.77 lbs. (carbohy- 
 fat equivalent carbohydrates 
 
 drates equivalent, or combined digestible carbohydrates and fat). 
 
 Step 4. 
 
 1.77 -T- (0.196 + 0.11) = 5.8 (second factor in nutritive ratio) 
 protein 
 
 The nutritive ratio of the above ration is, therefore, 1 to 5.8, or 
 1:5.8. Is this ratio wide or narrow? 
 
 .11. A farmer fed the following fair ration to his dairy 
 cows: 22 pounds of red clover hay; 8 pounds of ground dent 
 corn. 
 
 (a) Determine the total digestible crude protein 
 fed in this ration. 
 
 (b) How many pounds of digestible carbohydrates 
 were fed daily to each cow? 
 
 (c) How many pounds of digestible fat were fed? 
 
 (d) What is the nutritive ratio of this ration? 
 
 12. Horses weighing 1,300 pounds at hard work were 
 
 fed on an average daily ration consisting of: 
 
 16 pounds of timothy hay. 
 10 pounds of Oats. 
 
 5 pounds of corn (cracked). 
 
 1 pound of oil meal (linseed meal). 
 
 1 pound of wheat bran. 
 
 (a) Tabulate the amount of each digestible nu- 
 trient fed daily in each feed. 
 
 (b^ Determine the total amount of each of the 
 
 digestible nutrients fed in this ration, 
 (c) What is the nutritive ratio of this ration? 
 
 13. According to Wolff feeding standards, horses 
 weighing 1,300 pounds and at heavy work should receive 
 
104 AGRICULTURAL ARITHMETIC 
 
 daily 34 pounds of dry matter*, 3.3 pounds of digestible 
 crude protein, 17.7 pounds of digestible carbohydrates and 
 1.05 pounds of digestible fat. How does this feeding stand- 
 ard compare with what was fed in problem 12? The horses 
 in problem 12 each lost 50 pounds in four weeks. Why? 
 
 (a) Would adding 3 pounds of alfalfa hay, 5 pounds 
 of oats and 3 pounds of corn to the ration in 
 problem 12 make it near a balanced ration?t 
 
 14. A dairy cow weighing 1,000 pounds and giving 22 
 pounds of milk daily should receive 29 pounds of dry matter*, 
 2.5 pounds of digestible crude protein, 13 pounds of digest- 
 ible carbohydrates and 0.5 pound of digestible fat. (Wolff 
 feeding standards.) 
 
 (a) What is the nutritive ratio of a ration satis- 
 fying these requirements? 
 
 (b) According to this standard, did the farmer in 
 problem 11 feed a balanced ration — assuming 
 his cows weighed 1,000 pounds and averaged 
 22 pounds of milk daily? 
 
 15. The following ration may be regarded as a good one 
 for a dairy cow: 
 
 35 pounds of corn silage. 
 12 pounds of alfalfa hay. 
 
 3 pounds of ground corn. 
 
 1 pound of cottonseed meal. 
 
 (a) Tabulate the amount of each digestible nutrient 
 in each of the feeds fed daily in this ration. 
 
 (b) How many pounds of roughage in this ration? 
 How many pounds of concentrates? 
 
 *Dry matter is only an indication of the bulk of a feed or a ration. 
 
 tA balanced ration is the feed or combination of feeds furnishing the several 
 nutrients in such proportion and amount as will properly and without excess of 
 any nutrient nourish a given animal for 24 hours. — Feeds and Feeding. — HENRY 
 &■ MORRISON. 
 
THEORY OF FEEDING 105 
 
 (c) In actual practice this was regarded as a well- 
 balanced ration for cows giving 30 pounds of 
 milk daily testing 4% fat. Determine its 
 nutritive ratio. 
 
 16. The Haecker standard (Minnesota) requires that 
 a 1,000-pound cow should receive daily 0.7 pound crude 
 protein, 7.0 pounds carbohydrates and 0.1 pound of fat 
 (all digestible) to maintain her, independent of the milk 
 she produces. For each pound of 4% milk produced daily 
 there should be added 0.054 pound crude protein, 0.24 
 pound carbohydrates and 0.021 pound of fat (all digest- 
 ible). (See page 236.) 
 
 (a) Determine the total amount of each nutrient 
 (digestible), required to feed a 1,000 -pound 
 dairy cow giving 30 pounds of 4% milk daily 
 according to this feeding standard. 
 
 (b) Determine the nutritive ratio of a ration sat- 
 isfying these requirements. 
 
 (c) How nearly does the ration in problem 1 5 satisfy 
 these requirements according to the Haecker 
 standard, (a) from standpoint of nutrients 
 required and (b) from standpoint of nutri- 
 tive ratio? 
 
 17. A dairy cow is to be fed 30 pounds of corn silage and 
 8 pounds of red clover hay daily. She produces each day 
 30 pounds of milk testing 5% butter-fat. According to the 
 Wisconsin Station rule, this cow should receive as many 
 pounds of concentrates daily as she gives pounds of butter-fat 
 in a week. Fill out this ration with gluten feed and ground 
 corn, mixed in proportion of 3 parts to 2 parts by weight 
 respectively. 
 
106 
 
 AGRICULTURAL ARITHMETIC 
 
 (a) How many pounds of each of these concen- 
 trates are required to satisfy this feeding rule 
 and condition? 
 
 (b) Determine the nutritive ratio of this ration. 
 
 (c) Assuming this cow to weigh 1,000 pounds, 
 what would the Haecker feeding standard re- 
 quire in digestible nutrients when to the main- 
 tenance requirement of the cow must be added 
 daily 0.060 pound of digestible crude protein, 
 0.28 pound of digestible carbohydrates, and 
 0.024 pound of digestible fat for each pound 
 of 5% milk produced? (See problem 16 or 
 page 236 for Haecker's maintenance* require- 
 ment.) 
 
 (d) What should the nutritive ratio of the ration 
 be according to the above Haecker standard? 
 
 18. A ration containing oats, 5 parts; corn, 4 parts; and 
 wheat bran, 1 part, fed to yearling wethers gave excellent 
 results as compared with other combinations tried. The 
 average amount of grain consumed per head daily was 1.54 
 pounds. 
 
 &"-~mmm 
 
 ~:.i- .- .: 
 
 Figure 9.— A Champion Fen of Berkshirea. 
 
 ^Maintenance ration is one that furnishes a sufficient amount of nutrients 
 but no more than is required to maintain a given animal at rest, so that it will 
 not gain or lose in weight. — HENRY AND MORRISON, Feeds and Feeding. 
 
THEORY OF FEEDING 
 
 107 
 
 (a) Determine the amount of each concentrate 
 fed per head daily. 
 
 (b) How many pounds of grain* were required to 
 produce one pound of gain when the average 
 weekly gain per head was 2.57 pounds? 
 
 (c) Determine the nutritive ratio of the grains fed. 
 19. The Wolff Feeding Standard for Fattening Swine. 
 
 Requirement per day per 1,000 pounds live weight: 
 
 
 Dry 
 matter 
 
 Digestible Nutrients 
 
 
 Crude 
 protein 
 
 Carbohy- 
 drates 
 
 Fat 
 
 Nutritive 
 ratio 
 
 First period 
 
 36 lbs. 
 
 4.5 lbs. 
 
 25.0 lbs. 
 
 0.7 lb. 
 
 
 Second period 
 
 32 lbs. 
 
 4.0 lbs. 
 
 24.0 lbs. 
 
 0.5 lb. 
 
 
 Third period 
 
 25 lbs, 
 
 2.7 lbs. 
 
 18.0 lbs. 
 
 0.4 lb. 
 
 
 (a) Determine the nutritive ratio for each of the 
 feeding periods. What do these nutritive 
 ratios mean with reference to kinds of feed to 
 be fed in each period? 
 
 (b) What kinds of feeds may be used to narrow up 
 a ration or a nutritive ratio? 
 
 (c) What kinds of feeds will widen a ration or a 
 nutritive ratio? 
 
 (d) Turn to table on page 247 and make out a 
 list of feeds containing large amounts of pro- 
 tein. Another list containing those rich in 
 carbohydrates. 
 
 (e) How much corn and barley would you feed a 
 pig daily at the beginning of the fattening 
 period (first period), when the pigs average 
 125 pounds? Compound a balanced ration. 
 
 *Crain — commonly used to mean concentrates. 
 
108 AGRICULTURAL ARITHMETIC 
 
 Solution: 
 
 Stepl. For each 1,000 lbs. of live weight, or to 8 pigs, must be 
 fed daily 36 lbs. of dry matter, 4.5 lbs. of protein, 25 lbs. of carbohy- 
 drates, and .7 lb. of fat. (All digestible.) 
 
 Step 2. 
 
 20 lbs. of each grain will be taken for a trial. 
 
 20 X .895 = 17.9 lbs. of dry matter in the dent corn. 
 
 20 X .075 = 1.5 lbs. of protein in the dent corn. 
 
 20 X .907 = 18.1 lbs. of dry matter in the barley. 
 
 20 X .09 = 1.8 lbs. of protein in the barley. 
 
 17.9 + 18.1 = 36.0 lbs. of dry matter, which indicates enough bulk. 
 
 1.5 + 1.8 = 3.3 lbs. of protein. (Not enough.) 
 
 Step 3. 
 
 For second trial, 10 lbs. of corn, 30 lbs. of barley. 
 
 This combination supplies 36.2 lbs. of dry matter, 3.45 lbs. of 
 protein, 26.8 lbs. of carbohydrates and .94 lb. of fat. (Not sufficient 
 protein, though a trifle too much combined carbohydrates and fat.) 
 
 Step 4. 
 
 12 lbs. of corn, 25 lbs. of barley, and 2% lbs. of tankage meets 
 the requirement very well, supplying 36 lbs. of dry matter, 4.55 lbs. 
 of protein, 24.8 lbs. of carbohydrates, and 1.2 lbs. of fat, (all digestible). 
 
 Step 5. 
 
 Daily ration for one pig will be ]/% of these amounts, or 1J^ lbs. 
 of corn (shelled), 3J^ lbs. of barley (ground o r soaked before feeding), 
 and about ft lb. of tankage. 
 
 20. Figure out a ration for fattening swine in the second 
 period. As a trial, feed 1 part of wheat middlings to 2 parts 
 corn. Approach as near the above Wolff feeding stand- 
 ard as possible. 
 
 (a) How much of each feed is required per 1,000 
 pounds of live weight daily? 
 
 (b) When the pigs average 145 pounds, how much 
 of each feed is required per hog per day? 
 Total per hog per day? Amount per hog at 
 each feeding when fed 3 times per day? 
 
 (c) How many pounds of each feed are required 
 for 50 pigs for 3 weeks? Total' amount? 
 
 21. Compare the nutritive ratio of timothy and alfalfa 
 hay. Which hay would balance up a corn and corn silage 
 ration in feeding dairy cows? 
 
THEORY OF FEEDING 109 
 
 22. Illinois Station concluded that in feeding dairy cows 
 alfalfa hay was worth $10.86 per ton more than timothy. 
 On this basis how much more are 6 acres of alfalfa worth 
 than 6 acres of timothy when alfalfa yields on an average 
 1.6 tons per acre in each of 3 cuttings and timothy yields on 
 an average \ x /% tons and sells for $9.50 per ton? 
 
 23. Compare the nutritive ratio of the following two 
 rations fed to dairy cows: 
 
 No. 1 No. 2 
 
 Red clover hay 6 lbs. Red clover hay 6 lbs. 
 
 Corn silage 30 lbs. Corn silage 30 lbs. 
 
 Cornmeal 6 lbs. Cornmeal 6 lbs. 
 
 Alfalfa 8 lbs. Wheat bran 8 lbs. 
 
 (a) These cows averaged about 16 pounds of milk 
 daily. Wolff feeding standards show that 
 these cows should receive daily about 2 pounds 
 crude protein, 11 pounds carbohydrates, 0.4 
 pound fat (all digestible). Compute the 
 nutritive ratio of this standard and compare 
 with the nutritive ratio of the two rations. 
 
 24. The results obtained from feeding the two rations 
 in problem 23 showed that 8 pounds of alfalfa were equal to 
 8 pounds of wheat bran. 
 
 (a) On this basis, 12J-£ tons of alfalfa hay cocked 
 up in a field is equivalent to how many piles 
 of bran, each pile containing 40 pounds? 
 
 (b) What would be the value of each pile when 
 wheat bran costs $25 per ton? 
 
 25. Work out a balanced ration for a cow on your farm, 
 using feeds available, (a) according to the Wisconsin Station 
 rule. (See page 105) ; (b) according to Wolff feeding stand- 
 ard. (See table on page 244); (c) according to Haecker 
 feeding standard. (See table on page 236.) 
 
110 AGRICULTURAL ARITHMETIC 
 
 Hints on Feeding: 
 
 1. It is impossible to formulate an iron-clad rule in feed- 
 ing. 
 
 2. Food value and palatability of feeds and the individ- 
 uality of the animal fed are important factors to consider in 
 feeding. 
 
 3. The stockman can gain much help in feeding by 
 studying feeding standards and scientific feeding tests. 
 
CHAPTER V 
 THE DAIRY 
 
 The dairy cow is one of the most useful of our farm 
 animals. When she is carefully studied and cared for she 
 becomes a constant source of profit. The products of the 
 dairy, milk, butter and cheese, form an important part of 
 human food. The successful dairyman knows the business 
 end of dairying so that he may be able to reduce milk to its 
 equivalent of butter-fat*, butter and cheese, and to balance 
 milk and its products over against dollars. and cents. 
 
 ORAL PROBLEMS 
 
 1. A cow gives 35 pounds of milk per day. At this 
 rate how much will she give in a month? 
 
 2. How many pounds of butter-fat in 125 pounds of 
 milk testing 3% fat? Testing 5%? 
 
 3. How many pounds of butter-fat in 300 pounds of 
 milk testing 4.5% butter-fat? 
 
 4. 8.2 pounds of butter-fat are contained in 200 pounds 
 of milk. What is the test? 
 
 5. If 500 pounds of milk contain 19 pounds of butter- 
 fat, what does the milk test? 
 
 6. Three fourths of a pound of butter-fat per day for 
 an average for 300 days- may be considered a good average 
 yield for a dairy cow. How many pounds of fat does this 
 amount to? 
 
 7. What is the average butter-fat production of the 
 cows on your farm or in your* neighborhood? 
 
 *Butter-fat, milk-fat, is the fat of milk. 
 Ill 
 
112 AGRICULTURAL ARITHMETIC 
 
 8. Colantha 4th's Johanna's seven-day record (Hol- 
 stein) was 651.7 pounds of milk. What was the average 
 per day? 
 
 9. A cow gives 200 pounds of milk in 10 days. How 
 many quarts does she average daily? (For short, regard 
 25 pounds = 3 gallons.) 
 
 10. When it costs 21 cents to produce 1 pound of butter- 
 fat in 4% milk, what does it cost per quart of milk? 
 
 11. 8 t% pounds of butter is how much more than 5£ 
 pounds? 
 
 12. Many dairy farmers aim to have all their mature 
 cows produce a pound of butter-fat, on an average, for every 
 day in the year. To do this a cow producing 4% milk must 
 give on an average how many pounds of milk daily? How 
 many quarts daily? 
 
 WRITTEN PROBLEMS 
 
 1. The number of dairy cows reported on the farms in 
 the United States in the twelfth United States census, taken 
 in 1900, was 17,135,633, which was 16.9% less than was 
 reported in 1910. How many dairy cows reported in 1910? 
 
 2. The 1900 census reported 7,728,583,349 gallons of 
 milk produced in the United States. How much did that 
 average for every man, woman and child per year? Per day? 
 (See page 227, problem 24(a).) 
 
 3. In 100 pounds of average milk are contained 87.1 
 pounds of water and 4.9 pounds of milk sugar. 
 
 (a) How many gallons of water are contained in 
 2,500 pounds of milk? 
 
 (b) How many pounds of milk sugar? 
 
THE DAIRY 
 
 113 
 
 Figure 10. — The Original Bab- 
 cock Tester. 
 
 4 . According to actual tests, 
 it costs on an average 15.8 cents 
 to produce one pound of fat in 
 Guernsey milk testing 5.2% but- 
 ter-fat. Determine the cost per 
 quart* of milk. 
 
 5. Other tests have«shown 
 that it costs on an average 21.5 
 cents per pound of fat in Hol- 
 stein milk testing 3.43% butter- 
 fat. Determine the cost on this 
 basis to produce a quart of milk. 
 
 6.- K P. Pontiac Lass (Hol- 
 stein) produced 585.5 pounds of 
 milk in 7 days testing 6.03% 
 butter-fat. 
 
 (a) 
 (b) 
 
 (c) 
 
 How many pounds of butter-fat were produced? 
 
 How many pounds of butter is that equivalent 
 to according to the rule of the Holstein-Friesian 
 Association; viz., pounds of butter-fat Xl£ = 
 pounds of butter? 
 
 This estimated amount of butter contains what 
 per cent of butter-fat? 
 
 7. May Rilma (Guernsey) produced 19,673 pounds of 
 milk in 365 days testing on an average 5.45%. 
 
 (a) How many pounds of butter-fat were produced 
 in that year? 
 
 (b) What was the average daily production of 
 pounds of milk? Of butter-fat? Quarts of 
 milk? (1,000 pounds of milk = 465 quarts.) 
 
 *One gallon of milk weighs 8.59 pounds. 
 
 8— 
 
114 AGRICULTURAL ARITHMETIC 
 
 (c) How many pounds of butter would the total 
 amount of butter-fat be equivalent to accord- 
 ing to the rule of the Association of American 
 Agricultural Colleges; viz., pounds of butter-fat 
 X 1£ = pounds of butter? 
 
 (d) According to creamery experience 1 pound of 
 butter-fat in milk produces 1.15 pounds of 
 butter. At 35 cents per pound, determine 
 the value of the butter produced by this cow 
 in one year. 
 
 (e) How many average cows in your neighborhood 
 are required to produce as much butter-fat 
 in one day as was produced by MayRilma in 
 one day? 
 
 8. When one pound of butter-fat in milk produces 1.15 
 pounds of butter, what per cent of butter-fat does that butter 
 contain? 
 
 9. According to this rule, when 100 pounds of milk 
 will produce 4 pounds of butter, what does the milk test? 
 (Assuming no loss of butter-fat.) 
 
 10. A dairyman fed a ration consisting of 40 pounds of 
 corn silage, 15 pounds of clover hay, 3 pounds of ground corn 
 and one pound of cottonseed meal to his cows averaging 22 
 pounds of milk daily. When corn silage was worth $3 per ton, 
 clover hay $10, ground corn $1 per hundred and cottonseed 
 meal $30 per ton, what was the cost of feed per quart of milk? 
 
 11. A man sold 650 pounds of cream* and was credited 
 with 227£ pounds of butter-fat. What was the per cent of 
 butter-fat in the cream sold, or what per cent cream did he 
 sell? 
 
 *Cream is that portion of milk rich in milk-fat which rises to the surface of 
 milk on standing, or is separated from it by centrifugal force, and contains not 
 less than 18% of milk-fat. 
 
THE DAIRY 115 
 
 12. 850 pounds of 30% cream is equivalent to how many 
 pounds of butter-fat? 
 
 (a) The butter-fat contained in 1,250 pounds of 
 4% milk is equivalent to how many pounds of 
 30% cream? 
 
 Solution: 1250 X .04 = 50 lbs. fat in the milk. 50 lbs. of fat 
 must be .3 or .30 of the amount of 30% cream, hence 50 -s- .3 = 166| 
 lbs. of 30% cream. 
 
 Proof: 166| X .30 = 50 lbs. of fat. 
 
 13. How many pounds of 30% cream may be taken 
 from 500 pounds of milk testing 4.5% fat, assuming no loss 
 of butter-fat in separating? (See page 114, footnote.) 
 
 14. How many pounds of 35% cream may be taken 
 from 1,500 pounds of milk testing 5.6% fat, assuming 1^ 
 pounds of butter-fat is lost in separating? 
 
 (a) How many pounds of skim milk were drawn 
 from the separator? 
 
 (b) Determine the per cent of fat in the skim milk. 
 
 15. 360 pounds of 25% cream were taken from how 
 many pounds of milk testing 4% fat when 2 pounds of fat 
 were lost in skimming? 
 
 (a) How many pounds of skim milk were obtained? 
 
 (b) What per cent fat did it contain? 
 
 16. . At a small creamery 2,930 pounds of skim milk 
 testing 0.05% fat were drawn from 3,500 pounds of whole 
 milk received in one day. The cream obtained tested 25% 
 fat and produced 166 pounds of butter. 
 
 (a) How many pounds of cream were obtained? 
 
 (b) What was the average test of the whole milk 
 received? 
 
 (c) What was the per cent of overrun* in the 
 butter? 
 
 *Overrun in butter is the amount of water, casein and salt incorporated in 
 the butter-fat in making butter. Per cent overrun is determined as follows: 
 
 [(Pounds of butter made — pounds of butter-fat received) + pounds of butter-fat 
 received] X 100 = per cent of overrun. 
 
116 
 
 AGRICULTURAL ARITHMETIC 
 
 17. 5,460 pounds of milk testing on an average 3.6% 
 butter-fat were taken in at a creamery as a daily average for 
 one month. The average overrun in butter-making was 
 18.2%. 
 
 (a) How many pounds of butter were made during 
 that month? 
 
 (b) When the farmers were paid on an average 
 $1.20 per hundred for their milk; i. e., for the* 
 butter-fat contained in 100 pounds of milk, 
 and 36 cents per pound was realized for the 
 butter, what was the creameryman's net 
 receipts? 
 
 18. 75 pounds of milk testing 4.8% fat were mixed with 
 160 pounds of skim milk testirg 0.02% fat. Determine the 
 per cent butter-fat in the resulting mixture. 
 
 19. A milk customer wanted 20 pounds of 25% cream. 
 The milkman gave him a mixture of 5 pounds of 4% milk 
 and 15 pounds of 30% cream. Did the milkman give his 
 customer what he wanted? 
 
 20. It is desirable at times to standardize milk or cream 
 by using milks or creams containing different amounts of 
 butter-fat. The Pearson 
 
 rule is the simplest to 
 follow; viz., draw a rec- 
 tangle with two diago- 
 nals as shown in Fig. 11, 
 place the tests of the 
 milks or creams to be 
 mixed at the left hand 
 corners (25% and 4%, 
 Figure 11); in center 
 place the per cent butter- 
 fat desired in the mix- 
 
 3 Us. 
 
 fibs, 
 
 Figure 11.— The Pearsdn Method in Stand- 
 ardizing Milk or Cream. 
 
THE DAIRY 117 
 
 ture (20%). At the right-hand corners place the differences 
 between the two numbers- in line with these corners. The 
 number at the upper right-hand corner (16) represents the 
 number of pounds of milk or cream to use containing the 
 per cent butter-fat indicated in the upper left-hand corner; 
 and the number at the lower right-hand corner (5) repre- 
 sents the number of pounds of milk or cream to use con- 
 taining the per cent fat indicated in the lower left-hand 
 corner. 
 
 (a) How many pounds each of 25% cream and 4% 
 milk are required to make 20% cream? 
 
 (b) Prove the rule. 
 
 (c) When 8 pounds of 25% cream and 2.5 pounds 
 of 4% milk are mixed what per cent cream is 
 obtained? 
 
 (d) Suppose 8 pounds of 25% cream were mixed 
 with 15 pounds 4% milk. Would the mixture 
 be cream? 
 
 21. How many pounds each of 30% cream and 3.5% 
 milk are required to make 18% cream? Prove your answer. 
 
 22. How many pounds each of 6.5% milk and 18% 
 cream are required to make a mixture testing 12% fat? 
 
 23. How many pounds each of 3.5% and 6.3% milks 
 are required to produce 5% milk? 
 
 (a) How many pounds of each milk would be 
 required to make 100 pounds of 5% milk? 
 
 24. A dairyman had 20% cream and milk testing 3.8%. 
 In what proportion must he mix his cream and milk to make 
 5% milk? How many pounds of each must he use to bottle 
 150 quarts of the 5% milk? (See footnote, page 113.) 
 
 25. How many pounds of milk testing 4.8% butter-fat 
 are required to produce one pound of butter? (See problem 
 7 (d) page 114.) 
 
118 AGRICULTURAL ARITHMETIC 
 
 26. A cow gives in one milking 29 pounds of milk testing 
 6% fat. How many pounds of 85% butter does that amount 
 of milk represent? How many quarts of milk per pound of 
 butter? 
 
 The approximate amount of cured cheddar* cheese that 
 can be made from 100 pounds of milk may be determined as 
 follows: 
 
 Yield of cured cheddar cheese = per cent of butter-fat in 
 milk X 2.6. 
 
 27. A dairyman has a herd of 25 cows. The average 
 daily yield per cow is 30 pounds of 4.2% milk. 
 
 (a) How many pounds of cured cheddar cheese 
 may be made from the milk produced from this 
 herd in 30 days? 
 
 (b) How many pounds of butter? 
 
 (c) What is the difference in value when cheese 
 is worth 18 cents and butter 35 cents per 
 pound? When cheese sells for 13 cents per 
 pound and butter for 30 cents? 
 
 28. In one month a private creamery took in on an 
 average 10,500 pounds of milk daily. The farmers took 
 back home daily 8,752 pounds of skim milk testing 0.034% 
 fat, and 1,315 pounds of buttermilk testing 0.304% butter- 
 fat. The cream separated tested 25% fat. 
 
 (a) How many pounds of 25% cream were sep- 
 arated daily? 
 
 (b) What was the daily loss of butter-fat in the 
 skim milk? In the buttermilk? The total 
 loss? Total loss per month? 
 
 (c) How many pounds of fat were taken in at the 
 creamery daily? 
 
 *Cheddar cheese is the common American cheese, sometimes called American 
 cheddar. It receives its name from Cheddar, England, a small town noted for 
 its cheese. ' 
 
THE DAIRY 119 
 
 (d) What was the average test of the milk received? 
 
 (e) How many pounds of butter-fat were churned 
 into butter? 
 
 (f) How many pounds of butter were made daily 
 when the butter produced contained 13.5% of 
 water, 3% salt and 1% casein? 
 
 (g) What was the per cent of overrun in this butter? 
 (See footnote on page 115.) 
 
 (h) During that month the owner of the creamery 
 
 paid the farmers 35 cents for every pound of 
 
 butter-fat taken, in at the creamery, which was 
 
 the same price he received per pound of butter. 
 
 What was his net income from this creamery 
 
 per day? Per month? 
 
 29. Suppose the above creamery, instead of taking in 
 
 the whole milk, took in the amount of 25% cream daily as 
 
 it was separated from the milk and made the same kind 
 
 and amount of butter. 
 
 (a) Determine the per cent of overrun. 
 
 (b) Explain this difference. 
 
 (c) > In which case would the owner of the factory 
 likely receive the greater profits? Explain. 
 
 Summary of Facts: 
 
 State the important facts brought out in the preceding 
 problems. 
 
CHAPTER VI 
 THE SOIL 
 
 The soil, for the most part, is material resulting from the 
 weathering of rocks, hence it is underlaid by rocks wherever 
 it is found. This comparatively thin layer covering the 
 earth as a soft blanket has supported myriads of plants and 
 animals for millions of years. It is the foundation of all 
 agriculture. A few facts and figures concerning its physical 
 character should be of interest to us all. 
 
 PROBLEMS 
 
 1. How many square feet of soil in one acre? 
 
 2. An acre of dry, sandy soil 7 inches deep weighs in 
 round numbers 2,500,000 pounds; a clay loam or silt loam 
 about 2,000,000 pounds; a muck about 1,000,000; and a 
 
 Figure 12. — A Good Drainage Ditch. The first thing to do in reclaiming a 
 marsh is to drain it. This ditch is deep enough to permit drain tile to 
 empty into it. 
 
 120 
 
THE SOIL 121 
 
 dry, raw peat 350,000 pounds. What is the weight of a 
 cubic foot of dry sand? Of clay or silt loam? Of muck? 
 Of peat? 
 
 3. An acre of dry, "heavy"* clay loam 10 inches deep 
 weighs how many tons less than an acre 10 inches deep of a 
 "light" sandy soil? 
 
 4. It took 107 cubic inches of water to saturate one 
 gallon jar full of dry clay loam soil and 79 cubic inches to 
 saturate the same volume of dry sand. What is the approx- 
 imate per cent of pore space in the clay loam? In the sand? 
 
 5. It was found that a fine clay had 56% pore space, 
 and a coarse sand 40%. What per cent of pore space has 
 the clay compared with the sand? 
 
 6. An examination of a fine sandy loam showed that 
 it consisted of 59.4 parts, by weight, of sandf; 33.2 parts of 
 silt; and 7.4 parts of clay. How many ounces each of sand, 
 silt and clay are contained in two pounds of this class of soil? 
 
 7. A sample of silt loam soil was found by a mechanical 
 analysis to consist of 3.8 pounds of coarse and fine sand; 
 15.7 pounds of silt; and 5.5 pounds of clay. This class of 
 soil consists of what per cent of sand? Of silt? Of clay? 
 
 8. How many soil particles (clay), each measuring 
 0.0005 of a millimeter in diameter, will be required to span 
 one linear inch? 
 
 9. How many fewer particles would it require to span 
 a linear inch if they were silt particles measuring .01 of a 
 millimeter in diameter? 
 
 10. Weigh out an amount of dry soil equal in weight 
 to a new nickel. Divide this amount of soil into five equal 
 
 *"Heavy" soil — a soil hard to work, like clay. 
 "Light" soil — a soil easy to work, like sand. 
 tSand — the ooarser soil particles. 
 Silt — the medium sized soil particles. 
 Clay — the finest of soil particles. 
 
122 
 
 AGRICULTURAL ARITHMETIC 
 
 portions. Each portion weighs one gram. What part of 
 an average thimbleful does one gram of dry soil make? 
 
 11. A scientist was able to estimate 70,500,000 soil 
 bacteria in one gram of a rich garden soil, and only 50,000 
 in the same amount of poor soil. How many more bacteria 
 were contained in the gram of rich soil? The rich soil con- 
 tained how many times the number of bacteria found in the 
 poor soil? 
 
 12. One gram of soil near the surface of a fertile field 
 was found to contain 56,250,000 soil bacteria, and only 
 15,000 were found in the same quantity of soil taken 3 feet 
 
 Figure 13. — One Way of Breaking a Marsh. After drainage cultivation is the 
 next step in reclaiming a marsh. The wild sod must be subdued and the 
 deficient mineral elements supplied. 
 
 below the surface. What per cent of bacteria is in the 
 surface soil compared with that which is found in the 
 subsoil? 
 
 13. The total amount of surface area of the soil par- 
 ticles in one pound of coarse sand, the particles of which 
 measure one millimeter in diameter, is estimated at 11 square 
 
THE SOIL 123 
 
 feet. How many square rods of surface are contained in 
 one pound of fine clay when the particles measure .001 of a 
 millimeter in diameter? 
 
 14. 7J pounds of water were slowly poured over each of 
 four 6-pound samples of dry soil — sand, clay, loam, muck 
 (marsh soil) and a mixture of equal parts of sand and peat. 
 All samples were held in funnels having cloth strainers at 
 the bottoms. 6y pounds of water were caught as drippings 
 from the sand; 4 pounds, 12.8 ounces, from the clay loam; 
 two drops from the muck; and 5.3 pounds from the mixture 
 of sand and peat. Express in per cent of the dry weight of 
 soil the water-holding capacity of each kind of soil used. 
 
 A Few Facts Worth Remembering: 
 
 1. A "light" soil is by weight the heaviest soil. 
 
 2. A "heavy" soil is more porous than sand. 
 
 3. The amount of sand, silt and clay a soil contains 
 determines its class name; i. e., sand, silt loam, clay loam, 
 etc. 
 
 4. Most soils teem with bacteria and other organisms. 
 
 5. Sand has the lowest water-holding capacity of any 
 soil. 
 
 6. Adding organic matter and thus increasing the humus 
 in a soil increases its water-holding capacity. How was this 
 shown? 
 
CHAPTER VII 
 
 WHAT CROPS REQUIRE 
 
 At first thought it is hard to perceive the fact that grow- 
 ing crops require other things besides water, sunlight and 
 air. Indeed, it was thought at one time that a tree was 
 nothing more than water that had undergone a mysterious 
 change in the soil. Scientists have demonstrated that 
 plants also require nitrogen, phosphorus, potassium, calcium, 
 magnesium and other elements. 
 
 ORAL PROBLEMS 
 
 1. A cubic foot of sand weighed 110 pounds. What 
 is the weight of 432 cubic inches of that sand? 
 
 2. A cubic foot of dry, heavy clay soil was found to 
 weigh 75 pounds. What is the weight of 216 cubic inches 
 of that soil? 
 
 3. Approximately \, by weight, of the atmosphere is 
 nitrogen gas. A column of atmosphere with base one inch 
 square contains how many pounds of this element? 
 
 4. King found that it requires 576.5 pounds of water to 
 produce one pound of dry matter in clover. How much 
 water was required to produce 15 pounds of green clover 
 containing 60% water? 
 
 5. King further found that it requires 464 pounds of 
 water to produce 1 pound of dry matter in barley. How 
 many pounds of water were removed from the soil when 22 
 bushels of barley were raised, representing 2,000 pounds of 
 dry matter? How many tons? How many gallons?* 
 
 *1 cubic foot of water = 62H pounds approximately. 
 121 
 
WHAT CROPS REQUIRE 
 
 125 
 
 6. A 65-bushel com crop per acre removes from the soil 
 about 18.06 pounds of phosphorus. How many pounds of 
 this element would be removed from the soil by 9 acres of 
 such corn? 
 
 7. One ton of alfalfa hay contains 4.7 pounds of phos- 
 phorus. How many pounds of phosphorus are removed 
 from the soil by 8 acres, averaging 5 tons per acre? 
 
 8. One ton of sugar beets contains, on an average, 5.2 
 pounds of nitrogen, 0.7 pound of phosphorus and 5.3 pounds 
 of potassium. How many pounds of each of these elements 
 are lost from the soil when 15 tons of sugar beets are sold? 
 
 
 
 
 ^|. « g| 
 
 , t 
 
 
 
 
 
 
 
 "*?.". 
 
 
 
 g, ^-*W-"^EJ^%W-.. W.«di : 
 
 
 
 
 4i?fi»a 
 
 
 
 
 ; 
 
 ,.-_J 
 
 
 
 BB^^; / ''-' : ''-" 
 
 
 m 
 
 
 
 gjP-jH&at 
 
 
 
 Figure 14. — Preparing a Seed Bed. Good plowing ia an accomplishment not 
 only to be proud of, but which has its reward in better crops. 
 
 WRITTEN PROBLEMS 
 
 1. All plants take carbon from the air through their 
 leaves in the form of carbon dioxide gas. 45% of the corn 
 kernel is carbon. How many pounds of carbon in 20 bushels 
 of shelled corn? 
 
126 AGRICULTURAL ARITHMETIC 
 
 (a) How many pounds of soft coal will this amount 
 of carbon be equivalent to when soft coal is 
 about ■& carbon? 
 
 2. King found that it requires, on an average, 271 
 pounds of water to produce one pound of dry matter in corn. 
 How many acre inches* of water are necessary to produce a 
 75-bushel crop of dent corn? 
 
 (a) Good corn will produce 1$ times as much 
 field-cured stoverf as ear corn. How many 
 tons of stover are produced per acre in growing 
 this 75-bushel corn crop? 
 
 (b) Very dry stover contains, on an average, 9.4% 
 moisture, dent corn 10.5%, and cobs 10% 
 moisture. How many pounds of dry matter 
 will be produced in this crop? 
 
 (c) How many tons of water does it require to 
 cover one acre one inch deep? 
 
 (d) How many acre inchest of water are necessary 
 to grow the 75-bushel corn crop? What is the 
 average amount of rainfall in your section or 
 state? Is all this rainfall used by crops? 
 
 (e) How many square rods are required to produce 
 one bushel of this corn? Measure off this 
 area. 
 
 (f) How many barrels of water must be furnished 
 by the soil on this area of land to produce one 
 bushel of corn? 
 
 3. According to King it requires 504 pounds of water to 
 produce one pound of dry matter in the oat crop. How many 
 inches of water are required to produce a 50-bushel oat crop 
 
 ♦Follow steps in problems a, b, c, d. 
 tStover — corn stalk with ears removed. 
 
 tAn acre inch of water is that amount of water required to cover an acre one 
 inch deep. 
 
WHAT CROPS REQUIRE 127 
 
 when the ratio of straw to grain is 3 to 1? (Consult table 
 on page 247 for dry matter.) 
 
 4. A silt loam 21 inches deep underlaid by coarse 
 gravel was found to contain 22% of water in the spring. 
 (Per cent based on dry weight of soil.) A good catch of 
 red clover was secured on this land the previous year. It 
 requires 576 pounds of water to produce one pound of dry 
 matter in clover. Assuming that 8.4% of the moisture in 
 the soil cannot be used by the clover, how many inches of 
 rainfall will be necessary to produce a two-ton hay crop, 
 assuming further that 40% of the rainfall is lost in surface 
 runoff and leaching? 
 
 5. How much water will be supplied by the gravelly 
 subsoil in the previous problem? 
 
 6. Two fields were lying side by side. A crop of oats 
 was just threshed from one, and the other, which was well 
 cultivated, was growing a fine crop of corn. It was found 
 that the soil in the cultivated field eight inches deep contained 
 15.3% of moisture, or 70% more than in the other field. 
 What per cent of moisture was contained in the soil in the 
 stubble field? Explain this difference. 
 
 7. The average annual rainfall of some parts of the 
 dry-farm territory in the United States is 15 inches, or 
 61.4% less than that of Ohio. What is the average annual 
 rainfall of Ohio? 
 
 8. A farmer raised 30 acres of corn averaging 65 bushels 
 of shelled corn per acre, and 24 acres of timothy hay yielding 
 H tons per acre. How many pounds of nitrogen were 
 removed from his soil by these crops? (See page 240.) 
 
 9. How many pounds of phosphorus were removed 
 from the soil by the crops in problem 8? How many pounds 
 of potassium? (See page 240.) 
 
128 
 
 AGRICULTURAL ARITHMETIC 
 
 10. What per cent of the total 
 amount of potassium taken from the 
 soil by a 15-bushel flax crop and a 
 30-bushel wheat crop is contained in 
 the stalks and straw? 
 
 11. In Figure 15 determine the 
 number of pounds of plant-food ele- 
 ments represented by each \ inch of 
 the heavy lines. 
 
 12. By a similar diagram as shown 
 in Figure 15 show the comparative 
 amounts of nitrogen, phosphorus, and 
 potassium removed from one acre by 
 a 1,500-pound tobacco crop. (Use the 
 same scale.) 
 
 (a) A man sold his oats 
 from 20 acres which 
 averaged 60 bushels 
 per acre. What frac- 
 tional part of the to- 
 tal amount of phos- 
 IC ■ ^ phorus taken from 
 
 <; H the soil by the oat 
 
 {^-H crop did he sell? 
 
 q I Solution: 
 
 j£ ■ Step 1. 
 
 **• H A 50-bushel oat crop removes approxi- 
 
 mately 8 lbs. of phosphorus in straw and 
 grain. (Table, page 240.) 
 
 A 60-bushel crop will remove approxi- 
 mately £ as much more, or 9.6 lbs. per acre. 
 9.6 X 20 = 192 lbs. phosphorus re- 
 moved from the soil by 20 acres of 60-bushel oat crop. 
 Step 2. 
 
 68.7% of the phosphorus in the total oat crop (grain and straw) is 
 contained in the grain (table, page 240); hence the amount of phos- 
 phorus the man sold in his grain would be 196 X 68.7%, or 132 lbs. 
 
 C 
 
 5 
 
 i 
 
 Figure 15. — Plant-Food 
 Elements Removed by a 
 75-bushel Corn Crop. 109.5 
 lbs. nitrogen, 20.S lbs. 
 phosphorus, and 63 lbs. 
 potassium. 
 
WHAT CROPS REQUIRE 129 
 
 132 lbs. is H of 192, the total amount of phosphorus removed by 
 his oat crop. 
 
 The man sold, therefore, {& of the total amount of phosphorus 
 taken from the soil by his 60-bushel oat crop. 
 
 13. A grain farmer sold his entire 20-acre flax crop 
 yielding 15 bushels per acre and 50 acres of wheat averaging 
 18 bushels per acre. The straw of each crop was returned 
 to the soil. What per cent of the total amount of phosphorus 
 removed from the soil by these crops did he sell? 
 
 14. It is difficult to determine how much nitrogen 
 leguminous plants, like clover, alfalfa, etc., take from the 
 soil reserve and how much is secured from the air indirectly 
 through organisms in nodules on their roots. From some 
 experiments, however, it is reasonable to assume, under 
 average normal conditions, that about £ of the total amount 
 of nitrogen contained in the clover and alfalfa plants (roots 
 and all) is taken from the soil reserve. The rest is taken 
 from the air through the organisms in the root nodules. 
 A third of this total amount of nitrogen is left in the soil in ■ 
 the roots and stubble when the crop is cut for hay. The 
 nitrogen contained in the hay, therefore, may be regarded 
 as the amount "fixed" or taken from the air by the organ- 
 isms in the nodules; and what is left in the roots and stubble 
 will offset the amount taken from the soil reserve. Raising 
 clover and cutting it for hay, therefore, does not -increase 
 the nitrogen content of the soil growing the clover. 
 
 (a) When the nitrogen content of the clover crop 
 (roots and all) is equal to 120 pounds per acre, 
 how many pounds are contained in the roots 
 and stubble? In the portion cut for hay? 
 How many pounds of nitrogen were taken 
 from the soil reserve? From the air by means 
 of organisms? 
 
130 AGRICULTURAL ARITHMETIC 
 
 15. "A" thought he could enrich a 10-acre field of silt 
 loam soil he owned by just growing red clover. The follow- 
 ing year, being a favorable year, he cut 1-f tons of hay per 
 acre. He sold the entire crop. Was his soil enriched or 
 further depleted in phosphorus, potassium and calcium, and 
 how much? 
 
 16. Is it likely that Mr. "A" in problem 15 increased 
 the nitrogen content of his soil by his method? How many 
 pounds of nitrogen did he sell? .What may this amount of 
 nitrogen sold represent with respect to the quantity of 
 nitrogen "fixed," or taken from the air by the clover crop? 
 
 (a) At 15 cents per pound what was the value of 
 the nitrogen sold? 
 
 (b) If the clover crop were plowed under as a 
 green manuring crop, each acre would have 
 been enriched by how many pounds of nitro- 
 gen? 
 
 17. How much nitrogen, phosphorus and potassium 
 are removed per acre by a 25-ton cabbage crop, a 200-bushel 
 potato crop and a 15-ton sugar beet crop, in a three-year 
 rotation? (See table, page 240.) 
 
 18. Determine the amount of phosphorus removed from 
 one acre by a 2-ton red clover crop, a 15-bushel wheat crop 
 and a 20-bushel pea crop, in a three-year rotation. 
 
 19. Compare the amount of calcium removed from the 
 soil by 6 acres of alfalfa, averaging 5 tons of hay per acre 
 per season, and 12 acres of medium red clover, averaging 
 2 tons per acre, with that removed by 25 acres of corn, 
 averaging 50 bushels of shelled corn per acre, and 20 acres 
 of oats, averaging 40 bushels per acre. 
 
 20. A field was cropped 60 years, as follows: 30 crops 
 of wheat, averaging 15 bushels per acre; 20 crops of oats, 
 
WHAT CROPS REQUIRE 
 
 131 
 
 averaging 30 bushels per acre; 5 crops of corn, averaging 40 
 bushels per acre; and 5 crops of red clover, averaging l^tons 
 per acre. 
 
 (a) How many pounds of phosphorus ,and potas- 
 sium were removed per acre during these 60 
 years of cropping? How many pounds of 
 nitrogen? 
 
 Figure 16. — A Field of Harvested Grain. The soil has given up a part ■ of 
 itself in producing this crop. All crops remove fertility from the soil. 
 
 Important Truths: 
 
 1. Water is of the greatest importance to plants. It is 
 used by them in tremendous amounts; hence, soil moisture 
 and rainfall should be conserved for use by growing crops. 
 
 2. All crops take nitrogen and mineral elements from 
 the soil. 
 
 3. Most of the phosphorus required by plants is stored 
 in the seeds, or grain; and most of the potassium used is 
 found in the straw and stalks. 
 
CHAPTER VIII 
 
 THE SOIL A GREAT STOREHOUSE 
 
 The soil not only affords a firm support for the aerial 
 portion of plants, but it also supplies them with substances 
 necessary for growth and development. The great work 
 of plants is to manufacture these raw materials absorbed into 
 food for animals. Thus the nitrogen, phosphorus, potassium, 
 calcium and other mineral elements of our bodies come from 
 the soil either directly or indirectly by way of plants. It is 
 interesting to study the supply of plant food material stored 
 in the soil and how this reserve may be used. 
 
 ORAL PROBLEMS 
 
 1. It is estimated that 780 out of 1,000 parts of a 
 fertile clay loam soil is material incapable of supporting 
 plant life. What per cent of this soil is of no use to the plant 
 in supplying it with food elements? 
 
 2. A soil analyzed 2% potassium. Express this per 
 cent in the decimal form. In the fractional form. 
 
 3. A soil analyzed 0.2% nitrogen. What fractional 
 part of the weight of this soil is nitrogen? What decimal 
 is this equivalent to? 
 
 4. A soil contains ^ of one per cent of phosphorus. 
 What fractional part of the weight of the soil is this? Ex- 
 press this in decimal form. 
 
 5. 0.21% nitrogen is what fractional part of a soil? 
 Express decimally. 
 
 . 6. 0.3% potassium is what fractional part of a marsh 
 soil? Express decimally. 
 
 132 
 
SOIL A STOREHOUSE 133 
 
 7. Five tenths of one per cent of potassium is what 
 fractional part of a muck soil? 
 
 8. A soil analyzed 0.25% nitrogen, 0.11% phosphorus 
 and two and five tenths per cent potassium. What fractional 
 part of this soil is nitrogen? Phosphorus? Potassium? 
 
 9. 0.02 per cent of phosphorus is what fractional part 
 of the weight of a poor sand? What is the decimal^ form of 
 this per cent? 
 
 10. Five hundredths of one per cent of nitrogen is what 
 fractional part of a poor sand? What is the decimal form of 
 this per cent? 
 
 WRITTEN PROBLEMS 
 
 1. Some barren soils in Maryland contain 0!009% of 
 phosphorus. How many pounds of this element are con- 
 tained in one acre 7 inches deep weighing 2 million pounds? 
 
 2. Compare this amount of phosphorus with that 
 contained in a rich alluvial soil in Holland which analyzes 
 0.205 per cent of this element. (Use the same weight of 
 soil per acre, 7 inches.) 
 
 3. A soil was found to contain 2,000 pounds of nitrogen, 
 300 pounds of phosphorus and 12,000 pounds of potassium 
 in 2 million pounds of soil. What per cent of each of these 
 elements was contained in that soil? 
 
 4. A soil in Manitoba, Canada, contains 20,100 pounds 
 
 of nitrogen, 2,530 pounds of phosphorus and 17,100 pounds 
 
 of potassium in 2 million pounds of dry soil. Express the 
 
 amount of each of these elements in per cent of the weight 
 
 of soil. 
 
 A great many soil surveys and analyses are now being made by 
 the several states and the Federal government. Analyses stating the 
 per cents of the important elements found in the soils mean but little 
 to the average person if he has no standard for comparison. The fol- 
 
134 
 
 AGRICULTURAL ARITHMETIC 
 
 lowing little table based on many analyses will help the student to 
 judge soils of similar classes when only the chemical analyses are 
 given. These figures may be used as basis for comparison. 
 
 Soil 
 A fertile clay loam 
 
 or silt loam 
 A poor sand 
 An average peat 
 
 Nitrogen 
 
 0.25 per cent 
 0.08 per cent 
 3.00 per cent 
 
 Phosphorus 
 
 0.1 per cent 
 0.02 per cent 
 0.12 per cent 
 
 Potassium 
 
 2.0 per cent 
 0.5 per cent 
 0.3 per cent 
 
 5. What may be regarded as the standard amount in 
 pounds per acre 7 inches deep, of nitrogen, phosphorus and 
 potassium in a fertile clay loam or silt loam? (See page 120.) 
 
 6. Construct a table similar to the one above and 
 determine the actual number of pounds of each of the ele- 
 ments in each of the three soils per acre 7 inches. (Assume 
 weights of soils as given on page 120.) Do all soils contain 
 the same*amount of each of the different plant-food elements? 
 
 Figure 17. — A Greenhouse Fertilizer Test. This is one of the ways in which 
 the needs of a soil are determined. This soil needs phosphorus. N 
 stands for nitrogen, F for phosphorus and K for potassium. 
 
 7. In comparing the per cents in the table above, what 
 is the ratio of the nitrogen in the peat and that in the fertile 
 clay or silt loam? What per cent is this ratio equivalent to? 
 
SOIL A STOREHOUSE 
 
 135 
 
 8. In comparing the actual number of pounds in an 
 acre seven inches deep, what is the ratio of the nitrogen in 
 the peat and that in the clay loam? What per cent? 
 
 9. In buying land, which is safer to consider, the per 
 cent or the actual number of pounds per acre seven inches of 
 the plant-food elements in soils? Explain. 
 
 10. A cropped and unproductive silt loam was analyzed 
 and found to contain 0.13% nitrogen and 0.043% phosphorus. 
 How many pounds of each of these elements must be supplied 
 to this soil per acre to equal that contained in the fertile silt 
 loam given in table above? 
 
 Soils Are Not Inexhaustible: 
 
 A few scientists believe that soils are inexhaustible and 
 are able to produce crops for untold ages to come. It must 
 be remembered that every crop removed from the land 
 reduces the fertility of the soil. The stern reality is that 
 soils do "wear" out and fail to produce the profitable crops 
 that they once did. 
 
 11. How many 50-bushel oat crops can be supplied by 
 the amount of nitrogen contained in an acre 7 inches of sand 
 which analyzes 0.091 per cent nitrogen? 
 
 Figure 18. — Making Corn Grow by Supplying Deficient Plant-Food Ele- 
 ments. For two years corn refused to 'grow on this marsh even though 
 well drained. This shows how the trouble was analyzed. 
 
136 AGRICULTURAL ARITHMETIC 
 
 12. Eleven crops of corn from one acre, averaging 65 
 bushels of shelled corn, removes as much potassium from the 
 soil as is contained in one acre of a certain peat soil f of a foot 
 deep. What is the per cent of potassium in this peat? 
 
 13. A rich silt loam weighing 2,100,000 pounds per acre 
 eight inches (dry weight) contains 0.12% phosphorus. 
 How many cotton crops averaging 500 pounds of lint per 
 acre will the phosphorus in this soil supply? 
 
 14. A soil in Wisconsin was cropped 60 years, largely 
 to wheat during the first 10 or 12 years; then to corn and 
 oats and some clover. The land is now in a depleted con- 
 dition and contains only 0.061% phosphorus. The virgin 
 soil analyzes 0.123% phosphorus. What was the average 
 annual loss of this element per acre seven inches deep? 
 Assume soil weight to be 2 million pounds. 
 
 Important Truths: 
 
 1. Only a small part of a soil is capable of supplying 
 crops with plant-food elements. 
 
 2. The most important elements — nitrogen and phos- 
 phorus — are contained in the soil only in small amounts. 
 
 3. Every crop harvested reduces the amount of the 
 plant-food elements in the soil. 
 
 4. The amount of one, two or three of the important 
 elements in some soils has been reduced to such an extent 
 by continuous cropping that these soils are no longer regarded 
 as productive. 
 
CHAPTER IX 
 
 THE BALANCE SHEET OF SOIL FERTILITY 
 
 A good business man keeps a strict account of all trans- 
 actions he makes with losses and gains. This is the only 
 way he knows for a certainty whether or not he is contracting 
 debt or reaping a profit. Every farmer should have some 
 idea as to what the gain or loss may be in the fertility of his 
 soil brought about by his method of farming. A balance 
 sheet of soil fertility worked out by the following plan will 
 give a fair idea of the approximate exchange of fertility on 
 an average farm under average normal soil conditions, when 
 the manure produced is given the best of care and all straw 
 and other material like uneaten shredded corn stalks are 
 used for bedding. 
 
 1. The amount of nitrogen, phosphorus and potassium 
 contained in any crop sold, except the nitrogen in legumes* 
 sold, to be considered lost to the soil. (Consult table, 
 page 247, for fertility contained in crops and feeds.) 
 
 2. When manure produced is well cared for, lossesf 
 sustained when feeds are fed may be reasonably considered 
 as follows : 
 
 (a) Nitrogen about 40 per cent loss. 
 
 (b) Phosphorus about 20 per cent loss. 
 
 (c) Potassium about 20 per cent loss. 
 
 3. The fertility elements in feeds purchased minus the 
 loss in feeding to be regarded as gain to the soil. 
 
 *See explanatory note page 129 prob. 14. 
 
 fLoss of fertility in the feeding transaction includes the amount of the fer- 
 tility elements retained by the animals and the unavoidable loss in handling the 
 manure. 
 
 137 
 
138 
 
 AGRICULTURAL ARITHMETIC 
 
 4. The nitrogen in clover and alfalfa hay fed on the 
 farm minus the loss in feeding to be regarded as gain to 
 the soil. 
 
 5. The nitrogen in clover and alfalfa hay sold to be 
 regarded neither loss nor gain to the soil. 
 
 ORAL PROBLEMS 
 
 Figure 19. — Exchange of Fertility in Farming. Diagram showing how the soil 
 may gain or lose in fertility. Assuming good care given the manure pro- 
 duced, the loss of fertility, when the feeds are fed, is about 40% of the 
 nitrogen, 20% of the phosphorus, and 20% of the potassium. 
 
 1. In 1,000 pounds of oats (grain) are contained 19.8 
 pounds of nitrogen, 3.5 pounds of phosphorus and 4.6 pounds 
 of potassium. How many pounds of each element are lost 
 to the soil when 3 tons of grain are sold? 
 
BALANCE SHEET OF SOIL FERTILITY 139 
 
 2. One ton of alfalfa contains on an average 4.7 pounds 
 of phosphorus and 37 pounds of potassium. How many 
 pounds of each of these elements are lost to the soil when 7 
 tons of alfalfa hay are sold off the farm? Any nitrogen lost 
 from the soil? 
 
 3. A man raised and fed 1,000 pounds of oats (grain). 
 How many pounds of nitrogen, phosphorus and potassium 
 were lost in the feeding transaction? How many pounds 
 regained in the manure? 
 
 4. Suppose the 1,000 pounds of oats were purchased 
 and fed. Determine the amount of nitrogen, phosphorus 
 and potassium gained through this transaction. 
 
 5. In 1,000 pounds of wheat bran are contained 25.6 
 pounds of nitrogen, 12.8 pounds of phosphorus, and 13.4 
 pounds of potassium. When 5 tons are fed what is the 
 approximate gain of the fertility elements? 
 
 6. A farmer raised and fed 10 tons of red clover hay. 
 When 1,000 pounds of hay contain 20.5 pounds of nitrogen 
 and 1.7 pounds of phosphorus, determine the amount of phos- 
 phorus lost from the soil, assuming all manure is well cared ' 
 for and returned to the soil. How many pounds of nitrogen 
 gained? 
 
 WRITTEN PROBLEMS 
 
 1. A dairyman purchased 20 tons of red clover hay. 
 This he fed to his stock. All manure produced was hauled 
 directly to the fields. 
 
 (a) Determine the amount of nitrogen, phosphorus 
 and potassium purchased in the hay. (See 
 table, page 247.) 
 
 (b) Determine the amount of each of these fer- 
 tility elements lost in the feeding transaction. 
 
140 AGRICULTURAL ARITHMETIC 
 
 (c) Determine the amount of each of these ele- 
 ments added to the soil on his farm through 
 this transaction. 
 
 2. Determine the amount of each of the fertility ele- 
 ments lost from the soil on the farm from which the 20 tons 
 of clover hay were purchased. 
 
 3. A man exchanged 12 tons of timothy hay for 8 tons 
 of wheat bran which he fed to his stock. Did the soil on 
 his farm gain or lose in nitrogen, phosphorus and potassium, 
 and how much? Assume good care given to the manure 
 produced. 
 
 4. A small land-holder had 2 five-acre fields. One 
 produced seven and one half tons of red clover hay and the 
 other was kept in pasture. The hay was all fed without 
 grain and all the manure produced from feeding this hay was 
 hauled directly to the pasture field. 
 
 (a) How many pounds each of nitrogen and phos- 
 phorus were taken from the soil in the clover 
 field? (See page 129, problem 14.) 
 
 (b) How many pounds of each of these two ele- 
 ments were added to the pasture field in the 
 form of manure which was produced from 
 feeding the clover hay? 
 
 (c) Determine the gain or loss of nitrogen and 
 phosphorus to the soil in the clover field if all 
 the manure were applied to that field instead. 
 
 (d) If the entire clover crop were plowed under, 
 what would the balance sheet of fertility for 
 that clover field show regarding the gain or 
 loss of nitrogen and phosphorus? (See page 
 129.) 
 
BALANCE SHEET OF SOIL FERTILITY 141 
 
 5. Ten tons of corn silage, 5 tons of red clover hay and 
 100 bushels of oats were fed mixed stock. The manure 
 produced was well cared for. 
 
 (a) Determine the amount of nitrogen, phosphorus 
 and potassium contained in the feeds fed. 
 
 (b) About how many pounds of each of these three 
 elements would the manure produced from 
 feeding these feeds contain? 
 
 (c) About what per cent of the fertility contained 
 in these feeds may be recovered in the manure? 
 (Assume that losses occur as given on page 137.) 
 
 6. A man had 30 acres of good pasture and pastured it 
 to cows and some young stock. Assuming that each acre 
 yielded as much green feed as is equivalent to 1\ tons of 
 mixed grass hay, and that 30 per cent of the nitrogen and 
 20 per cent of the phosphorus in the grass eaten is retained 
 by the animals, determine the total loss of nitrogen and 
 phosphorus to the soil in that field. 
 
 7. A farm produces annually 20 acres of alfalfa aver- 
 aging 5 tons of hay per acre per season, and 30 acres of red 
 clover averaging 2 tons of hay per acre. All the hay is fed 
 on the farm, and all manure produced is well cared for. 
 The soil on this farm is enriched by how many pounds of 
 nitrogen annually through the growing and feeding of the 
 alfalfa and clover? 
 
 (a) How many pounds of phosphorus are lost 
 annually from the soil on the farm through 
 this feeding transaction? 
 
 8. A farmer purchased 10 tons of cottonseed meal and 
 5 tons of oil meal (linseed meal). 
 
 (a) In the feeding of these feeds, how many pounds 
 of nitrogen and phosphorus would be lost, 
 assuming a 40% loss of nitrogen and 20% loss 
 of phosphorus? 
 
142 AGRICULTURAL ARITHMETIC 
 
 (b) If the manure were well cared for, what would 
 be the gain of nitrogen and phosphorus to the 
 soil through this purchase and feeding trans- 
 action? 
 
 9. About how many tons of wheat bran or cottonseed 
 meal must be fed during the time that 15 tons of alfalfa hay 
 and 40 tons of corn silage are fed in order that the total 
 phosphorus content of the manure produced may be equal 
 to that contained in the hay and silage fed? 
 
 10. A successful dairyman possessing a farm of 80 acres 
 raises on the average the following crops: 
 
 Crops 
 
 Acres 
 
 Average Yield per Acre 
 
 Corn 
 
 10 
 
 20 tons silage. 
 
 Oats 
 
 28 
 
 75 bushels. 
 
 Hay (red clover) 
 
 10 
 
 4 tons (2 cuttings). 
 
 Pasture 
 
 30 
 
 Equivalent to 1J tons of mixed 
 grass hay. 
 
 He sells all the oats except 300 bushels which he feeds. 
 All other crops are fed on the farm, and all straw is used for 
 bedding. All corn is made into silage. 
 
 He buys and feeds each year 20 tons of wheat bran and 
 10 tons of gluten feed. 
 
 All manure is hauled directly to the field when possible. 
 All milk is sold*. 
 
 (a) Determine the gain or loss of nitrogen in the 
 soil on this farm by use of a balance sheet like 
 the one below. Assume nitrogen loss on pas- 
 ture to be 30%. 
 
 *When Bkim milk ia fed on the farm the fertility in the. milk fed regained in 
 the manure should be regarded as gain. 
 
BALANCE SHEET OF SOIL FERTILITY 143 
 
 Nitrogen Balance Sheet 
 
 Crops 
 
 and 
 
 Feeds 
 
 Loss of nitro- 
 gen in feeding 
 (clover ex- 
 cepted) 
 
 Loss of nitro- 
 gen in sale of 
 crops 
 
 Nitrogen gained 
 through feeding 
 clover and pur- 
 chased feeds 
 
 Oats, — grain 
 
 
 
 
 Corn, — silage 
 
 
 
 
 Hay, (red clover).. . . 
 
 
 
 
 
 
 
 
 
 
 Wheat bran 
 
 
 
 
 
 
 
 
 
 
 Totals 
 
 
 
 
 
 
 Total loss and gain. . . 
 
 
 
 Loss or gain ........ 
 
 
 
 It is safe to assume that very little or no fertility is lost 
 from the straw used for bedding. 
 
 Fertility lost through leaching from the soil and washing 
 is not taken into account. 
 
144 
 
 AGRICULTURAL ARITHMETIC 
 
 (b) Construct a "Phosphorus Balance Sheet" and 
 determine the loss or gain of phosphorus to 
 the soil on this farm, — as follows: 
 
 
 Phosphorus Balance Sheet 
 
 
 Crops 
 
 and 
 
 Feeds 
 
 Loss of phos- 
 phorus in 
 feeding 
 
 Loss of phos- 
 phorus in sale 
 of crops 
 
 Phosphorus gain- 
 ed through pur- 
 chased feeds 
 
 Oats, — grain 
 
 
 
 
 Corn silage 
 
 
 
 
 
 
 Hay, — clover 
 
 
 
 
 
 
 
 
 
 
 Wheat bran 
 
 
 
 
 Gluten feed 
 
 
 
 
 Totals 
 
 
 
 
 
 
 Total loss and gain. . . 
 
 
 
 Loss or gain 
 
 
 
 N. B. Assume no loss of phosphorus from the straw used 
 for bedding. 
 
 * Assume 20% loss of phosphorus. 
 
BALANCE SHEET OF SOIL FERTILITY 145 
 
 (c) If no feeds were purchased and no oats sold, 
 but fed, what would the records show regarding 
 the loss and gain of nitrogen and phosphorus? 
 How many pounds of rock phosphate analyz- 
 ing 13% phosphorus must be mixed with the 
 manure to offset this loss of phosphorus? 
 
 (d) Suppose all the crops raised were fed and half 
 the amount of each kind of feed purchased; 
 determine the loss or gain of nitrogen and 
 phosphorus. 
 
 (e) Determine the loss or gain of potassium on 
 this dairy farm in a similar manner. 
 
 (f) Suppose this dairyman took poor care of the 
 manure produced and allowed it to accumulate 
 in the barnyard, thus subjecting it to heavy 
 losses through leaching. Assuming under such 
 conditions that the total loss of fertility in the 
 feeding transaction to be 60% for nitrogen, 
 37% for phosphorus and 60% for potassium, 
 determine the annual gain or loss of nitrogen, 
 phosphorus and potassium to or from the soil 
 on his farm. 
 
 11. Assuming good care given all manure produced, 
 determine the approximate loss or gain of nitrogen and 
 phosphorus to the soil on a farm when the following feeds 
 were raised and fed on the farm: 100 tons of corn silage, 45 
 tons of red clover hay, 15 tons of timothy hay, 4 tons of oat 
 straw, 300 bushels of corn, 333 tons of stover (shredded — 
 § of which was used for bedding and absorbent), 1,000 
 bushels of oats and 30 acres of pasture* equivalent to f of 
 a ton of mixed grass hay per acre. 
 
 Feeds purchased, 5 tons wheat bran, 3 tons tankage. 
 
 ♦Assume 30% loss of nitrogen and 20% loss of phosphorus in pasturing. 
 10— 
 
146 AGRICULTURAL ARITHMETIC 
 
 Crops sold, 100 tons of cabbage. 
 
 Eight acres of red clover (second growth equivalent to 
 f of a ton of hay per acre) were plowed under. 
 
 12. Make out a phosphorus balance sheet for a farm 
 with which you are familiar. 
 
 Fertility Truths: 
 
 1. Grain farming without the use of fertilizers and 
 legumes reduces the fertility of the soil most rapidly. 
 
 2. Even on a good dairy farm where no feeds or ferti- 
 lizers are purchased, the phosphorus supply of the soil is 
 reduced year by year. 
 
 3. The nitrogen supply may be maintained and increased 
 through the growing and feeding of a good acreage of clover, 
 alfalfa and other legumes. 
 
 '4. To establish permanent agriculture, at least as much 
 fertility should be returned to the soil as is removed by crops. 
 
CHAPTER X 
 
 MANURE AND COMMERCIAL FERTILIZERS 
 
 Manure may be denned as a natural substance possessing 
 a direct fertilizing value. The liquid excrement from farm 
 animals contains nearly half of the nitrogen and potassium 
 voided by them and should be carefully preserved. Open 
 barnyard manure is about half as valuable as stall manure. 
 A fertilizer may be defined as a substance which simply 
 supplies one or other of the elements necessary for plant 
 growth, as sodium nitrate, rock phosphate, etc. Com- 
 mercial fertilizers may be grouped as follows: nitrogen or 
 nitrate; phosphorus or phosphate; potassium or potash; arid 
 mixed or complete fertilizers. 
 
 MANURE 
 
 ORAL PROBLEMS 
 
 1. For practical purposes one is sufficiently accurate in 
 estimating one ton average, mixed barnyard manure to con- 
 tain approximately 0.5% nitrogen, 0.1% phosphorus and 
 0.4% potassium. How many pounds of each element are 
 contained in one ton? 
 
 2. What is the value of the fertility contained in one 
 ton of mixed manure when the nitrogen is worth 15 cents 
 per pound, phosphorus, 10 cents, and potassium, 6 cents per 
 pound? 
 
 3. Under average conditions, when straw is used for 
 bedding, it requires about 25 cows to produce one ton of 
 barnyard manure per day during winter or indoor feeding. 
 At this rate about how many tons will 5 cows produce in one 
 month? 
 
 147 
 
148 AGRICULTURAL ARITHMETIC 
 
 4. About one ton of manure ready for field application 
 (including bedding) is produced in one day by 50 average 
 working horses. (Amount produced when horses are at 
 work excluded.) How much manure will be produced by 
 5 horses in 10 days? 
 
 5. An average 1,300-pound horse will produce approx- 
 imately 65 pounds of manure per day (including bedding), 
 when kept in the barn most of the time. About how much 
 manure will be produced by 3 such horses in 3 months? 
 
 6. Ten 480-pound steers in a fattening pen produced 
 approximately £ of a ton of manure per day (bedding in- 
 cluded). At this rate how much would be produced by 20 
 such steers in 20 days? 
 
 7. A 150-pound pig produces about 9 pounds of manure 
 per day. At this rate now many pigs are required to pro- 
 duce one ton in one day? 
 
 8. 500 65-pound lambs in a feeding pen, during the 
 first period, produced about one ton of yard manure daily. 
 How much would have been produced by 3,000 lambs in 3 
 weeks? 
 
 9. A man keeps 5 horses and feeds 12 cows. About 
 how much barnyard manure is produced on his farm during 
 7 winter months? 
 
 10. A man applies 8 tons of manure per acre. About 
 how many acres can he cover with the manure produced on 
 his farm by 10 horses and 25 cows during 6 winter months? 
 
 WRITTEN PROBLEMS 
 1. 250 tons of mixed barnyard manure were produced 
 on a farm in one year. All manure was well cared for and 
 applied to the soil at the rate of 8 tons per acre. 
 
 (a) Determine the total amount of nitrogen, phos- 
 phorus and potassium contained in this amount 
 of manure. (See oral problem 1, page 147.) 
 
MANURE 
 
 149 
 
 (b) Determine the total value of the fertility con- 
 tained in this manure. (See oral problem 2, 
 page 147.) 
 
 (c) Determine the value of the fertility applied 
 per acre. 
 
 Average Composition of Fresh Manures- 
 Liquids and Bedding) 
 
 •(Including Solids, 
 
 Animal 
 
 Per cent 
 water 
 
 Per cent 
 nitrogen 
 
 Per cent 
 phos- 
 phorus 
 
 Per cent 
 potas- 
 sium 
 
 Fertility 
 value per ton 
 
 
 77 
 
 0.44 
 
 0.07 
 
 0.33 
 
 
 
 
 Pig 
 
 73 
 
 0.45 
 
 0.08 
 
 0.5 
 
 
 
 
 
 70 
 
 0.58 
 
 0.12 
 
 0.44 
 
 
 
 
 
 64 
 
 0.83 
 
 0.17 
 
 0.56 
 
 
 
 
 Hen 
 
 50 
 
 0.97 
 
 0.3 
 
 0.26 
 
 
 
 
 2. On the basis of 15 cents per pound for nitrogen, 10 
 cents for phosphorus and 6 cents for potassium, determine 
 the fertility value per ton of each of the manures in the above 
 table. 
 
 3. In an experiment extending over 15 years at the Ohio 
 Experiment Station, an 8-ton application of cattle manure 
 per acre once in a five-year rotation resulted in an average 
 increase per rotation of 13.55 bushels of corn and 517 pounds 
 of stover; 6.2 bushels of oats and 359 pounds of oat straw; 
 7.24 bushels of wheat and 836 pounds of wheat straw; 1,052 
 pounds of clover hay and 1,007 pounds of timothy hay. 
 When corn is worth 50 cents per bushel, oats 40 cents, wheat 
 90 cents, stover $3 per ton, straw $2 and clover and timothy 
 hay each $8, determine the real value of the manure per ton 
 from the value of the increased crop production. 
 
150 AGRICULTURAL ARITHMETIC 
 
 4. During the same time that the above test was made, 
 a 16-ton application of manure per acre in each five-year 
 rotation resulted in an average increase per acre per rota- 
 tion of 18.12 bushels of corn and 577 pounds of stover; 10.83 
 bushels of oats and 688 pounds of straw; 10.94 bushels of 
 wheat and 1,277 pounds of wheat straw; 1,977 pounds of 
 clover hay and 1,549 pounds of timothy hay. At the same 
 prices as given in problem 3— 
 
 (a) Determine the real value per ton of the manure 
 applied. 
 
 (b) Which method produced the greater value of 
 crop increase per acre? What per cent greater? 
 
 (c) Which method produced the greater value of 
 crop increase per ton of manure applied per 
 acre and how much more? 
 
 (d) What is the practical lesson to be learned from 
 results of problems 3 and 4? 
 
 5. A chemist* found a ton of fresh steer manure to con- 
 tain 10.3 pounds of nitrogen, 3.24 pounds of phosphorus and 
 8.14 pounds of potassium. After three months' exposure to 
 the weather in the barnyard he found the same manure to 
 contain 0.359% nitrogen, 0.123% phosphorus and 0.167% 
 potassium. 
 
 (a) Determine the per cent of loss of each of the 
 fertilizing elements. 
 
 (b) Determine the per cent of loss of total fertility. 
 
 (c) What is the value of the fertility lost per ton? 
 (See problem 2.) 
 
 (d) If 250 tons of manure were left exposed with 
 similar losses, what would be the value of the 
 fertility lost? 
 
 *Ohio Experiment Station Bulletin 183, page 205. 
 
MANURE - 151 
 
 6. A farmer applied sheep manure to his soil at the rate 
 of 18 loads per acre, each load averaging 1J tons. If this 
 manure has the chemical composition as given in the pre- 
 ceding table (page 149), what was the value of the fertility 
 applied per acre? 
 
 (a) If he secured a tobacco crop yielding 1,600 
 pounds of leaves per acre and sold the crop 
 for 22 cents per pound, what were the profits 
 per acre after deducting 50 per cent of the 
 value of the fertility applied? 
 
 (b) If \ of the fertility contained in the manure is 
 leached out of the soil after this heavy applica- 
 tion of manure is made, determine the value 
 of the fertility lost. 
 
 7. At the beginning of a 4-year rotation a farmer applied 
 
 20 tons of sheep manure per acre. The following crops were 
 
 grown : 
 
 1st year — corn yielding 75 bushels of shelled corn. 
 2nd year — oats yielding 65 bushels per acre. 
 3rd year — barley yielding 45 bushels per acre. 
 4th year — red clover yielding 2.5 tons per acre. 
 
 (a) Determine the amount of nitrogen, phosphorus 
 and potassium added to the soil per acre in the 
 manure. 
 
 (b) Assuming no loss of fertility through leaching 
 from the soil, determine the loss or gain of 
 fertility during these four years. Assume 
 clover to leave as much nitrogen in the soil as 
 it took out. (See table on page 240 for crop 
 requirements.) 
 
 8. A man believes he is maintaining the fertility of his 
 soil when he applies 8 tons of average mixed barnyard manure 
 per acre once in a four-year rotation, as follows: 
 
 1st year — corn (manured) — yielding 60 bushels of shelled corn. 
 2nd year — corn yielding 40 bushels of shelled corn. 
 3rd year — oats yielding 40 bushels per acre. 
 4th year — red clover yielding \\ tons per acre. 
 
152 AGRICULTURAL ARITHMETIC 
 
 (a) Assuming no loss through leaching, of the 
 fertility applied, determine the gain or loss of 
 nitrogen, phosphorus and potassium per acre 
 during these four years. 
 
 (b) Determine the loss or gain if the clover crop 
 were plowed under. 
 
 (c) When one cow produces 80 pounds of manure 
 per day and all is saved and applied to the soil, 
 about how many cows must be kept per 100 
 acres to produce enough manure in one year to 
 return to the soil the amount of fertility 
 removed by these crops? 
 
 COMMERCIAL FERTILIZERS 
 ORAL PROBLEMS 
 
 1. When rock phosphate contains 13% phosphorus and 
 costs $7.80 per ton, what is the cost of one pound of phos- 
 phorus? How many pounds of the element phosphorus to 
 the dollar? 
 
 2. When acid phosphate contains 7% phosphorus 
 (P)*, and costs $18.20 per ton, what is the cost of one pound 
 of phosphorus (P)? 
 
 3. When basic slag analyzes 8% phosphorus, how many 
 pounds of phosphorus are added to the soil when 300 pounds 
 are applied per acre? 
 
 4. Muriate of potash contains 43% potassium. How 
 many pounds of this element are added to an acre when 200 
 pounds of this potash fertilizer are applied? 
 
 5. At 55 dollars per ton what is the cost of 100 pounds 
 of sulphate of potash? 
 
 *"P" is the chemical symbol for the element phosphorus. 
 
COMMERCIAL FERTILIZERS 153 
 
 6. One and one half tons of sulphate of potash contain 
 105 pounds of potassium (K)*. What per cent "K" does 
 this potash or potassium fertilizer contain? 
 
 7. Nitrate of soda contains on an average 15% nitro- 
 gen. How many pounds of nitrogen in one ton of this nitro- 
 gen fertilizer? When nitrogen is valued at 15 cents per 
 pound what is the value of one ton of this fertilizer? 
 
 8. One hundred pounds of ammonium sulphate contain 
 20 pounds of nitrogen. What is this fertilizer worth per ton 
 when nitrogen is worth 15 cents per pound? 
 
 9. One ton of dried blood contains 280 pounds of 
 nitrogen (N)f. What per cent "N" does this fertilizer 
 contain? 
 
 10. A mixed or complete fertilizer contains \\% nitro- 
 gen (N), 4% phosphorus (P), and 3% potassium (K). How 
 many pounds of each of these fertilizing elements are con- 
 tained in two tons? 
 
 11. When unleached wood ashes contain 5% potassium 
 and 40% carbonate of lime, how many pounds of potassium 
 in 5 tons. Carbonate of lime? 
 
 WRITTEN PROBLEMS 
 
 1. How many pounds of pulverized limestone contain- 
 ing 36% calcium is required per acre to return to the soil the 
 amount of calcium removed by a 3-ton red clover crop? 
 (See table, page 240.) 
 
 2. If from a soil producing 6 tons of alfalfa hay per 
 acre per season, 200 pounds of calcium is lost by leaching 
 of water through the soil, how many pounds of pulverized 
 limestone analyzing 36% of calcium would it take to replace 
 the total loss of calcium to the soil? (See table, page 240.) 
 
 *"K" is the chemical symbol for the element potassium. 
 f'N" is the chemical symbol for the element nitrogen. 
 
154 AGRICULTURAL ARITHMETIC 
 
 3. Determine the cost of one pound of the element 
 potassium (K) in each of the following potassium fertilizers: 
 
 Muriate of potash containing 43% K, costing $46 per ton. 
 Sulphate of potash containing 42% K, costing $56 per ton. 
 Kainit containing 12% K, costing $15 per ton. 
 
 4. A pint of rock phosphate weighs about one pound. 
 How much of this phosphorus fertilizer* should be sprinkled 
 in the gutters in a barn per cow per day in order to give a 
 1,000-pound application per acre when 10 tons of manure 
 are added to an acre? (See problem 3, page 147.) 
 
 (a) This amount of rock phosphate adds an amount 
 of phosphorus equal to that amount removed 
 by how many 60-bushel oat crops? By how 
 many 6-ton alfalfa crops? 
 
 5. Manure applied to a soil low in phosphorus proves 
 an unbalanced fertilizer. How many pounds of rock phos- 
 phate containing 13% phosphorus must be mixed in a ton 
 of average mixed barnyard manure to raise the phosphorus 
 content up to 0.8%, thus making a more balanced fertilizer 
 for such a soil? (See Figure 17.) 
 
 The fertilizing constituents in a mixed or complete com- 
 mercial fertilizer are usually expressed in terms of nitrogen 
 (N), "phosphoric acid" (P 2 5 )t and "potash" (K 2 0).J A 
 "2-8-10" mixed or complete fertilizer, for example, means 
 one that contains 2% nitrogen (N), 8% "phosphoric acid" 
 (P 2 5 ) and 10% "potash" (K 2 0). 
 
 Give the meaning of the following fertilizers: "1.5-5-8," 
 "3-10-8," "1-6-3." 
 
 *To render rock phosphate more soluble and hence more available to crops it 
 is best to mix it with manure or plow it under with green-manuring crops, 
 like clover. 
 
 t"Phosphoric a^cid" is a chemical compound containing the elements phos- 
 phorus and oxygen, 2 parts of phosphorus combined with 5 parts of oxygen, hence 
 the chemical symbol '^gOs." 
 
 fPotash" is a chemical compound containing two parts of the element 
 potassium united with one part of oxygen, hence the chemical symbol "K^O," 
 
COMMERCIAL FERTILIZERS 155 
 
 Rules: 
 
 (a) To reduce "phosphoric acid" (P2O5) to the 
 element phosphorus (P), multiply the per 
 cent or amount in pounds of "phosphoric acid" 
 (P2O5) by -436. 
 
 (b) Per cent or amount in pounds of "potash" 
 (K2O) X .83 = per cent or amount of the ele- 
 ment potassium (K), respectively. 
 
 6. 33% of "phosphoric acid" (P2O5) is equivalent to 
 what per cent of the element phosphorus (P)? 
 
 7. 632 pounds of "P2O5" is equivalent to how many 
 pounds of "P"? 
 
 8. 50% "potash" (K2O) is equivalent to what per cent 
 of the element potassium (K)? 
 
 9. 350 pounds of "K 2 0" = how many pounds of "K"? 
 
 10. A firm advertises a brand of rock phosphate contain- 
 ing 13% phosphorus (P) at $8 per ton. Another firm adver- 
 tises the same kind of fertilizer under a different brand as 
 containing 32% "phosphoric acid" (P2O5) for $8.25 per ton. 
 Which brand will give the greater amount of phosphorus for 
 one dollar and how much more? 
 
 11. Four tons of a potash, or potassium, fertilizer, 
 which analyzes 16% "potash" (K2O), contains how many 
 pounds of the element potassium (K)? 
 
 12. A low grade, mixed, commercial fertilizer, contain- 
 ing 1.1% nitrogen (N), 3% phosphorus (P) and 2.5% potas- 
 sium (K), sells for $24.50 per ton. 
 
 (a) Determine the amount of each of the plant- 
 food elements in one ton. (Use no pencil.) 
 
 (b) When "N" in mixed fertilizer is valued at 20 
 cents a pound, "P" at 8 cents, and "K" at 
 6 cents, determine the fertility value per ton. 
 
156 AGRICULTURAL ARITHMETIC 
 
 (c) What is the excess of selling price over fertility 
 valuation? 
 
 13. A high grade, mixed, commercial fertilizer contain- 
 ing 3.5% nitrogen (N), 3.6% phosphorus (P) and 6.6% 
 potassium (K), sells on an average, for $36.50 per ton. 
 
 (a) Compare this fertilizer with the low grade in 
 points brought out in (a), (b), (c). Prob. 12. 
 
 (b) Is a low grade, mixed, commercial fertilizer 
 necessarily "cheap"? 
 
 14. A farmer was advised to apply a fertilizer con- 
 taining potassium and phosphorus to some drained peat* soil 
 he had. He purchased, muriate of potash containing 43% 
 potassium costing him $46 per ton, and acid phosphate con- 
 taining 7% phosphorus costing him $18 per ton. He mixed 
 the two in equal proportions and applied the mixture broad- 
 cast at the rate of 350 pounds per acre. This treatment 
 resulted in an average yield of 40 bushels of corn per acre 
 which was 7 times greater than the yield on an unfertilized 
 acre. 
 
 (a) The mixture applied contained what per cent 
 potassium (K)? Phosphorus (P)? 
 
 (b) How many pounds of each element did he 
 apply per acre? 
 
 (c) What did the application cost him per acre? 
 
 (d) What did the potassium and phosphorus cost 
 him per pound? 
 
 (e) How much of the low grade, mixed fertilizer 
 (Problem 12) would have been necessary to 
 add the same amount of potassium per acre? 
 What would the cost have been? How much 
 money was saved in "home-mixing"? 
 
 *Peat — a marsh soil consisting largely of organic matter, hence containing 
 much nitrogen and comparatively small amounts of potassium and phosphorus. 
 (See page 136, Figure 18.) 
 
COMMERCIAL FERTILIZERS 157 
 
 (f) Compare the value of the increased yield of 
 corn with the cost of the fertilizer applied. 
 
 (g) Compare the amount of fertility added with 
 that removed by the crop. (See table on 
 page 240.) 
 
 (h) Compound a mixture of these two fertilizers 
 and determine the amount necessary to add as 
 much potassium and phosphorus as is removed 
 by a 50-bushel corn crop. 
 
 15. How many pounds of nitrate of soda containing 
 15% nitrogen, acid phosphate containing 7% phosphorus, 
 and muriate of potash containing 42% potassium must be 
 used to make a ton of a mixed fertilizer analyzing 3.5% 
 nitrogen, 4% phosphorus and 6.3% potassium? 
 
 (a) How many pounds of "filler" must be used to 
 make this mixture equal a ton in weight? 
 
 (b) When nitrate of soda costs $45 per ton, acid 
 phosphate $18, and muriate of potash $46, 
 determine the cost of the fertilizers to com- 
 pound this mixture. Compare this cost with 
 that of the low grade and high grade fertilizers 
 in problems 12 and 13. 
 
 (c) Mention advantages in home-mixing of fer- 
 tilizers. ' 
 
 16. Assuming that the amount of nitrogen in alfalfa hay 
 is gathered from the air, determine the amount of fertility 
 removed from an acre in four years when the average yield 
 per season is 5.5 tons. 
 
 (a) How many pounds of muriate of potash and 
 acid phosphate would be required to return 
 to the soil the amount of phosphorus and 
 potassium removed during those 4 years? 
 
158 AGRICULTURAL ARITHMETIC 
 
 Figure 20. — Results of Manure vs. Mineral Fertilizers on Peat. On this 
 marsh soil, deficient in both phosphorus and potassium, muriate of potash 
 with rock phosphate proved: to be the more economical fertilizer treat- 
 ment. 
 
 (b) Compound a mixture of these two fertilizers 
 and determine the amount of the mixture to 
 be applied per acre as a top-dressing* to return 
 to the soil the fertility removed annually. 
 
 (c) Would you apply nitrogen fertilizer to the 
 alfalfa as a top-dressing? Why? 
 
 17. A farmer applied 18 tons of good, mixed barnyard 
 manure per .acre to some well-drained peat soil and was dis- 
 appointed with results produced the second year after appli- 
 cation. If applied to upland soil it would have continued in 
 its effect. 
 
 (a) This soil needed no nitrogen. Determine the 
 . amount of nitrogen applied per acre from 
 
 which he received very little or no returns. 
 
 (b) Determine the value of the useless nitrogen 
 applied per acre at 15 cents per pound. 
 
 (c) How many pounds of potassium and phos- 
 phorus, which no doubt produced the results, 
 were applied? Determine their value. (See 
 page 149.) 
 
 *Since the potassium and phosphorus compounds in muriate of potash and 
 acid phosphate are soluble, these fertilizers can very well be used for top-dressing. 
 
COMMERCIAL FERTILIZERS 159 
 
 (d) How many pounds of muriate of potash and 
 basic slag could have been applied per acre to 
 add the same amount of potassium and phos- 
 phorus as was added in the manure? (See 
 page 152, problems 3 and 4.) 
 
 (e) Determine the cost of the required amount of 
 muriate of potash and basic slag (problem 
 d), and compare it with the value of the manure 
 applied. (Consider basic slag costing $16 per 
 ton.) 
 
 Points Worth Remembering: 
 
 1. Barnyard manure is a valuable fertilizer and should 
 be well cared for and used intelligently. 
 
 2. The proper use of fertilizers depends largely on the 
 soil and crops to be grown. 
 
 3. Some legume should be given a place in every system 
 of crop rotation. 
 
 Name other points of practical value you have learned 
 from the problems in this chapter. 
 
CHAPTER XI 
 
 FARM MANAGEMENT 
 
 A little farming done on paper at times produces sur- 
 prising results. One progressive farmer went so far as to 
 say that the pencil is the greatest implement in promoting 
 better agriculture. It is only when a man knows the cost 
 of the feed he fed his hogs, for example, that he knows for a 
 certainty whether or not he is paid for his labor. Some farm 
 accounts have shown that some farmers have actually paid 
 a good price for the privilege of doing good, physical exercise 
 on their farms. The application of business principles is 
 required to determine losses and gains. 
 
 ORAL PROBLEMS 
 
 1. From a test made in Ohio it was found that the 
 feed-cost per chicken per year averaged 61 cents. What 
 would be the cost of feed for a flock of 80 chickens? 
 
 2. It was also found that the average labor cost per 
 chicken per year averaged 28 cents. Determine the annual 
 cost of labor for a flock of 80. 
 
 3. When interest on investment, shelter and deprecia- 
 tion amounted to 6 cents per chicken per year, determine 
 the total annual cost for the flock of 80. 
 
 4. When the average hen produces 6 dozen eggs an- 
 nually, how many are produced by a flock of 80? 
 
 5. When the eggs sell for an average of 20 cents per 
 dozen, determine the average labor income* from this flock 
 of 80. 
 
 *Labor income is profits above total costs, including interest on investment, 
 and may be compared to the wages paid to hired help. 
 
 160 
 
FARM MANAGEMENT 161 
 
 6. When one bushel of corn produces 10.5 pounds of 
 pork worth 8 cents per pound, how much is realized per 
 bushel of corn? 
 
 7. A farmer has 120 acres of tillable land on his farm on 
 which he keeps 25 cows. How many acres per cow? 
 
 8. A farm having 4 tillable acres per cow would receive 
 how many tons of manure per acre once in a four-year rota- 
 tion, when a cow produces an average of 13 tons barnyard 
 manure annually? Will this maintain fertility? 
 
 9. When 16 tons of manure are applied per acre and 
 are valued at $1.50 per ton, how many bushels of corn at 50 
 cents per bushel must be raised to offset 50% of the value 
 of manure applied? 
 
 (a) The following year how many bushels of barley 
 at 80 cents per bushel are required to offset 30%* 
 of the value of the manure applied for corn? 
 
 (b) The third year how many bushels of oats are 
 required to offset 20% of the value of the 
 manure applied for corn? 
 
 10. A man paid $200 per acre for some land. What is 
 the value of this land as an investment, when he realizes a 
 profit of $12 per acre above total cost in sale of crops? (5% 
 land rental f is "included in the cost of crop production.) 
 
 Solution: In gaming the net profit of $12 per acre, $10 per 
 acre as land rental was charged against the crops. Thus the man 
 realized at least 5 % on the money invested in the land. $12 is the labor 
 income which is equivalent to the interest on $240 at 5%. $22 consti- 
 tutes the investment returns and is equivalent to 5% on $440. As an 
 investment, therefore, each acre is equal to $440 at 5%. 
 
 11. A grain farmer, on account of poor crops, realized 
 no profits above cost of producing his crops. In his cost 
 accounting, $8.75 per acre as land rental at 5% was charged 
 against the crops. What value was placed on the land? 
 
 *S0% of the value of manure applied is commonly charged against the first 
 crop, 30% against the second, and 20% against the third crop. 
 
 fLand rental is an item of expense in cost accounting charged against crops. 
 The total cost in raising and marketing crops includes this item, which is usually 
 reckoned at 5%. This is to cover interest charges, taxes and insurance. 
 
 11— 
 
162 AGRICULTURAL ARITHMETIC 
 
 12. Excluding the land rental of 5%, the annual profits 
 above other costs realized from crop values on a rich, prairie 
 soil were $42.50. In capitalizing these profits what is the 
 land worth as an investment from the standpoint of produc- 
 ing power? 
 
 13. When it costs on an average about $14 on 100- 
 dollar land to grow an acre of corn, how many bushels must 
 be raised per acre to offset this cost when corn is worth 50 
 cents per bushel? ($5 of the cost is land rental at 5%.) 
 
 14. On this basis about how much is land worth per 
 acre as an investment* that raises 48 bushels of corn per acre 
 without increasing the cost except that of land rental, which 
 is usually considered at 5%? (Take same value for corn.) 
 
 15. In the Middle West it costs approximately $60 per 
 year to maintain a dairy cow. How many pounds of milk, 
 selling for $1.50 per hundred, must be produced to offset this 
 cost? How many pounds does this average per day when 
 lactation period is 300 days? 
 
 WRITTEN PROBLEMS 
 
 Miscellaneous: 
 1. A young farmer invested $20 in a 1,000-pound 
 portable scales. 
 
 (a) He sold his barley for 70.5 cents per bushel. 
 He weighed the contents of each load before 
 leaving the farm. The first load detected a 
 defective buyer's scales which underweighed 
 2.1 lbs. on a bushel. How much money 
 would he have lost had he not known the weight 
 of his loads, when he was credited for correct 
 weight for 800 bushels? 
 
 *See footnote, page 173. 
 
FARM MANAGEMENT 
 
 163 
 
 (b) 'Determine the per cent gained on money in- 
 vested in the scales. 
 
 (c) He fattened 20 hogs and was offered $325 as 
 a lump sum for them by a local stock buyer, 
 when hogs -like his were quoted at $7.35 per 
 hundred. He asked $352, knowing the aver- 
 age weight. The following day he took his 
 pigs to market and found that they averaged 
 253 pounds, and received $7.30 per hundred 
 pounds. 
 
 How much money did the scales save him 
 when offered sum is considered? 
 
 2. A good farmer spent half an hour in teaching his 
 hired man to make good, round shocks of barley properly 
 capped with two sheaves, each shock containing 12 bundles. 
 A rainy week resulted in little coloring of the grain other than 
 the cap sheaves, which were threshed separately. Because 
 of better color, 3^ cents more per bushel were realized for 
 his barley. He sold j^ of his crop of 15 acres averaging 40 
 bushels per acre. How much was he paid for the one half 
 hour spent in teaching his hired man how to do a small job 
 right? 
 
 Figure 21. — A Simple Method for Testing Corn. Upon muslin cloth squares 
 are drawn and numbered, upon which are laid the kernels from each ear to 
 be tested. When the tester is filled, a moist, sawdust pad, shown at left, 
 is placed on top. 
 
164 
 
 AGRICULTURAL ARITHMETIC 
 
 3. A man failing to test the greater portion of his seed 
 corn secured f of a crop as compared with the tested seed. 
 The yield of the poor corn averaged 25 bushels of shelled 
 corn per acre for 25 acres. When corn is worth 65 cents per 
 bushel, determine the cost of carelessness. 
 
 4. Sixty bushels of seed oats were given the formalin 
 seed treatment* for loose smut at a cost of 3£ cents per bushel 
 (including labor cost). Three bushels were sown per acre, 
 which yielded 65 bushels per acre, which was a 12% increase 
 over an untreated acre. Determine the percent profit on 
 cost of treatment when oats are worth 38 cents per bushel. 
 
 5. A man raised 20 acres of a high grade, pure-bred 
 oat, averaging 70 bushels per acre. He needed the oats for 
 feed, but had the opportunity to sell his crop for seed at 75 
 
 cents per bushel. 
 Doing so, he pur- 
 chased the same 
 amount of oats 
 for feed at an 
 auction sale for 
 35 cents. Allow- 
 ing 4 cents a 
 bu'shef as cost in 
 making the ex- 
 change, what 
 were his profits 
 above cost of la- 
 bor? 
 
 6. Mr. "A" 
 took good care 
 
 Figure 22.— Results of a Test. A Section. The ears of hlS grain 
 from which the weak and sterile kernels were taken should l' j m l,ioh 
 
 be thrown out. Dinaer wnicn 
 
 *One pound (pint) of formalin in 30 gallons of water for 50 bushels of oats. 
 Solution is sprayed on grain and thoroughly mixed or dipped in loosely filled 
 burlap sacks for 2 hours. 
 
FARM MANAGEMENT 
 
 165 
 
 cost him $125. He used it 14 years, each year cutting an 
 average of 50 acres of grain. Repairs averaged $4.75 per 
 year (including cash and labor repairs). The value of the 
 machine each year was inventoried as follows: 
 
 $125, 80, 75, 75, 65, 65, 60, 60, 50, 45, 40, 25, 20, 15. 
 
 (a) Determine the average yearly investment* 
 made in this machine. 
 
 (b) At 6% what is the interest on the total invest- 
 ment for the 14 years? Average yearly inter- 
 est?. 
 
 (c) Determine the average yearly cost of the ma- 
 chine. 
 
 (d) Determine the average machinery cost per 
 acre of grain per year. 
 
 7. Mr. "B" kept his grain harvester, costing the same 
 price, out of doors the year round, cutting on an average 
 
 Figure 23. — Going to the Junk Pile. What must be the depreciation of a 
 high-priced machine cared for like this? Is it a profitable investment? 
 
 Courtesy of Country Gentleman. 
 ♦Average yearly investment in this case is the sum of the yearly values 
 divided by 14. 
 
166 AGRICULTURAL ARITHMETIC 
 
 of 45 acres for 8 years. Repairs averaged $7 per year. 
 Inventory for each year showed value of machine as follows: 
 $125, 60, 50, 40, 30, 25, 20, 15. 
 
 Make the same determinations as in (a), (b), (c), (d), 
 problem 6. 
 
 8. In a test made on a heavy, silt loam, corn planted in 
 hills 3 feet 8 inches each way, thoroughly and scientifically 
 cultivated, yielded 74.8 bushels of shelled corn per acre and 
 2.03 tons of field-cured stover. Corn not cultivated, but 
 having weeds cut without stirring the soil, yielded 44.6 
 bushels per acre and 1.66 tons of stover. 
 
 (a) What was the per cent of increase in yield of 
 shelled corn? Stover? 
 
 (b) When corn is worth 55 cents per bushel and 
 stover $3 per ton, w&at was the value of the 
 increased yield, due to thorough cultivation? 
 
 9. A farmer hired cheap labor in the form of a boy at 
 75 cents per. day to cultivate a 15-acre field of young corn 
 planted in hills 3 feet 8 inches apart each way. It was 
 cultivated twice in a row with a one-horse walking cultivator 
 and cultivated both ways. 2\ acres were worked each day. 
 
 (a) What did it cost in cash for labor to cultivate 
 that corn two ways? How much per acre? 
 
 (b) The boy through carelessness covered up and 
 dug out on ends and in field, on an average, 
 one hill out of 11. Assuming that the yield 
 was reduced proportionately, what was the 
 loss in the crop, when the corn averaged 60 
 bushels of shelled corn and 3 tons of stover 
 per acre? 
 
 (c) What was the real price paid per day for hiring 
 the careless and inefficient boy when corn is 
 
FARM MANAGEMENT 167 
 
 worth 50 cents per bushel and stover $3 per 
 ton? 
 
 (d) What would have been the saving in the end 
 had the farmer hired a good man at $2 per day, 
 doing the same amount of work, but having 
 no corn destroyed? 
 
 10. A man sold a high grade Guernsey cow for $150 and 
 thought he had made a good bargain. The buyer found that 
 this cow produced 13,560 pounds of milk in 300 days, testing 
 4.5% fat. The cost of feed averaged 18 cents per day for 
 300 days. 2,437 pounds of 25% cream were sold at 25 cents 
 per quart* - 
 
 (a) Assuming the value of the manure produced 
 and the calf offset labor cost, determine the 
 net profits. Consider 6% interest on money 
 invested in the cow, feed cost, .value of skim 
 milk at 15 cents per hundred and cash returns 
 for cream. 
 
 (b) Suppose the man selling the cow sold the feed 
 required to feed that cow for $25, and put all 
 money received (including money received for 
 cow) out at 6% interest. What would have 
 been his net income from this investment per 
 year? 
 
 (c) Suppose he kept the cow and valued her at 
 $150; fed her a ration costing 16 cents per day 
 for 300 days; and sold f as much milk as was 
 produced by the buyer of the cow at $1.35 per 
 hundred. Determine gain or loss as in (a). 
 Compare results with (b). 
 
 *One gallon of 25% cream weighs approximately 8.35 pounds. 
 
168 AGRICULTURAL ARITHMETIC 
 
 Profit and Loss in Feeding: 
 
 11. A man produced 11.6 pounds of pork, on an average, 
 from each bushel of corn fed. When pork was sold for $8.30 
 per hundred pounds, what did he realize per bushel for his 
 corn? 
 
 12. A feeder produced 10 pounds of pork, on an average, 
 for each bushel of corn fed. When corn is worth 75 cents 
 and hogs sell for $6.60 per hundred weight, what does he gain 
 or lose on each 100 pounds of pork produced? 
 
 13. Over 500 feeding trials* show that pigs weighing 
 between 150 and 200 pounds consume, on an average, 5.9 
 pounds of feed per pig per day, producing, on an average, a 
 daily gain of 1.2 pounds; and pigs weighing between 300 and 
 350 pounds consume 7.5 pounds of feed per day, producing 
 a daily gain of 1.4 pounds. 
 
 (a) Determine the number of pounds of feed re- 
 quired to produce 100 pounds of gain in each 
 case. 
 
 (b) How many more pounds of pork could be pro- 
 duced in feeding the younger pigs from the 
 amount of feed required to produce 5,580 
 pounds of gain in feeding the older and heavier 
 pigs? 
 
 (c) Name other advantages in fattening younger 
 pigs. 
 
 14. A man figured that he had 12.5 tons of corn silage 
 and 8 tons of alfalfa more than he needed to supply his dairy 
 herd. He borrowed money at 6% on 9 months' time and 
 bought 3 cows at $75 per head. At the end of 9 months, 
 they were sold at $60 per head. The cows averaged 20 
 pounds of milk a day for 9 months. The cost of extra con- 
 centrates fed each cow was $14. Milk was sold at $1.35 per 
 
 *Henry and Morrison's Feeds and Feeding. 
 
FARM MANAGEMENT 169 
 
 hundred. Assuming that the value of the manure produced 
 and the calves was sufficient to offset the cost of labor, did 
 this man gain or lose in his feeding when he could have sold 
 his surplus silage for $3 and alfalfa hay for $15 per ton? 
 
 (a) How would the accounts stand, if the cows 
 each gave 30 pounds per day? 
 
 (b) At this price of milk how much should each 
 cow have averaged per day to balance accounts? 
 
 (c) If the milk were sold for 3 cents per quart* 
 in the original problem, what would have been 
 the profit or loss? 
 
 15. A man purchased 2,000 lambs averaging 60 pounds 
 for $6 per hundred pounds including freight. He fed them 
 for 90 days on pea silage, costing $1 per ton, at the rate of 
 3.5 pounds per lamb per day, and grain (screenings) which 
 cost 5 cents for every pound of gain. He sold them on the 
 Chicago market for 8 cents a pound — they averaged 84 
 pounds. Freight charges averaged 27 cents per lamb. Loss 
 from death was one per cent of the original number. Labor 
 cost was estimated at 1 cent per pound gain. 150 tons of 
 manure were produced per month. 
 
 (a) What was the daily gain per head? Per flock? 
 
 (b) Determine the total net profits — consider in- 
 terest on money invested in the lambs at 6% 
 and value of manure in the yard at $1.50 per 
 ton. 
 
 (c) What were the profits per head? 
 
 (d) At $15 per ton, how many tons of screenings 
 were fed? Average per lamb per day? 
 
 (e) How many lambs were required to consume 
 one ton of pea silage per month? 
 
 *100 lbs. of milk = 11% gallons, or 1,000 lbs. = 465 qts. 
 
170 
 
 AGRICULTURAL ARITHMETIC 
 
 16. The following is an itemized statement made by a 
 feeder who fed 20 steers: 
 
 Total 
 
 Av. per head 
 
 Number of steers bought 
 
 Weight 
 
 Number of pigs bought* 
 
 Weight 
 
 Cost of steers at 7.5c. per lb. 
 delivered 
 
 Cost of pigs at 7.5c. per lb. de- 
 livered 
 
 .20. 
 
 .10. 
 
 .1,050 pounds 
 
 . 150 pounds 
 
 Feed consumed by steers. 
 
 Cost of feed:'' Alfalfa at $16 
 
 per ton 
 
 Corn at 50c. per bushel 
 
 Cottonseed meal at $28 a ton. . 
 50 bu. ear corn for pigs at 50c. 
 
 per bushel 
 
 Freight charges in marketing 
 
 steers 
 
 Cost of yardage for steers. . . 
 
 Cost of hay on market 
 
 Commission in selling 
 
 Interest on money invested in 
 
 steers at 6% for 4 mo 
 
 Interest on money invested in 
 
 hogs at 6% for 4 mo 
 
 Selling weight of steers 
 
 Selling price of steers at 8c. . 
 
 Selling weight of hogs 
 
 Selling price of hogs at $7.50 
 
 per hundred 
 
 Manure produced from steers 
 Value of manure at $1.25 per 
 
 ton 
 
 Alfalfa 10 tons, broken 
 ear corn 18 tons, cotton- 
 seed meal 1.02 tons 
 
 .$1.00. 
 
 .$1.87. 
 . .25. 
 
 .50. 
 
 . 1,305 pounds 
 . .225 pounds 
 
 .85 tons. 
 
 (a) Complete the above itemized statement. 
 
 (b) Select the items of expense and of income and 
 determine profit and loss in this feeding ven- 
 ture. (Cost of labor was not included.) 
 
 *One pig following two Steers was considered sufficient, since the steers were 
 fed broken ear corn. 
 
FARM MANAGEMENT 
 
 171 
 
 (c) What would have been the profit or loss if no 
 pigs followed the steers? 
 
 (d) Determine the cost of feed per pound of gain 
 in the steers. 
 
 (e) Determine the total cost per pound of gain in 
 the steer-feeding venture, cost of labor ex- 
 cluded. 
 
 17. A farmer decided to test his herd of 10 cows and keep 
 a milk record to ascertain the worth of each individual. At 
 the end of the year he had on hand the following data: 
 
 Cow 
 
 Milk 
 pro- 
 duced 
 Pounds 
 
 Average 
 test 
 
 Lbs. 
 but- 
 ter-fat 
 
 Value of 
 
 fat at 
 
 30c. per 
 
 lb. 
 
 Inven- 
 tory 
 value 
 
 Cost of 
 feed 
 
 Net 
 returns 
 
 1. 
 2. 
 3. 
 4. 
 5. 
 6. 
 7. 
 8. 
 9. 
 10. 
 
 3,624 
 
 9,546 
 
 6,145 
 
 7,254.J 
 
 3,933 
 
 1,838 
 
 6,542 
 
 3,830 
 
 6,874 
 
 6,740 
 
 3.84% 
 4.5 ' 
 
 6 
 
 
 
 34 
 
 $50 
 80 
 60 
 75 
 50 
 30 
 70 
 50 
 70 
 70 
 
 60 
 50 
 55 
 44 
 43 
 50 
 45 
 52 
 52 
 
 All feeds were raised on the farm. The value of the 
 calves, skim milk and manure were regarded as sufficient to 
 offset the cost of feed during time cows were dry, and labor 
 cost throughout the year. 
 
 (a) Determine the pounds of butter-fat produced 
 by each cow. Value at 30 cents per pound. 
 
 (b) Determine net returns above cost of feed for 
 each cow. 
 
 (c) Determine the total net returns for the herd. 
 Average per cow. 
 
172 
 
 AGRICULTURAL ARITHMETIC 
 
 (d) Determine the total net returns from the four 
 best cows. Compare this amount with the 
 total net returns from the remaining six. 
 
 (e) How many cows averaging as the six poorer 
 cows would be required to produce as much 
 net profits as the four best cows? 
 
 (f) How many cows like No. 5 would be required 
 to produce net returns equal to cow No. 2? 
 
 Figure 24. — A Profitable Dairy Cow. Thia cow produced in one year 663 
 pounds of butter-fat. Above cost of feed she netted her owner $104.70. 
 
 (g) Suppose the man had sold cows N°s. 1, 5, 6, 
 and 8 at the stated values and the feed required 
 for each of these cows for the values given, and 
 invested all the money received at 6% for one 
 year, determine the total net returns in this 
 change in management of the herd. Compare 
 with results obtained, in problem (c). 
 
FARM MANAGEMENT 
 
 173 
 
 (h) Suppose the four poorest cows were sold for the 
 values given them and the money invested in 
 two cows which were fed the feeds required to 
 feed the four poor cows, and each produced, on 
 an average, 10,350 pounds of milk testing 4%. 
 Determine the net returns for the herd under 
 these conditions. Compare results with prob- 
 lems (g) and (c). 
 18. A man set out to determine the true* value of the 
 
 best and poorest cows in his herd of 12 grade Shorthorns. 
 
 With some help he got the following data covering one year: 
 
 Star 
 (best cow) 
 
 Nig 
 
 (poorest 
 
 cow) 
 
 Lactation period 
 
 Milk produced 
 
 Average butter-fat test 
 
 Amount of 25% cream sold 
 
 Value of fat in cream at 25c. per lb 
 
 Amount skim milk 
 
 Value skim milk at 15c. per hundred 
 
 Average cost of feed per day 
 
 Labor cost during lactation period 
 
 Cost of shelter 
 
 Miscellaneous expense other than interest on 
 
 investment and depreciation 
 
 Beef value 
 
 305 days 
 10,550 lbs. 
 
 4.1% 
 1,725 lbs. 
 
 295 days 
 5,810 lbs. 
 3.6% 
 834 lbs. 
 
 $ .20 
 
 18.50 
 
 2.46 
 
 3.25 
 60.00 
 
 8 .16 
 
 18.50 
 
 2.46 
 
 2.60 
 60.00 
 
 The value of the manure and the calves was supposed 
 to cover the cost of labor and feed while cows were dry. 
 
 (a) Determine the items left blank. 
 
 (b) Assuming that the maximum productive period 
 of a cow is 6 years, the profit per year should be 
 considered at 20% f of her value as capital 
 above her value for beef. Determine the real 
 value of each cow. 
 
 *True value of a cow is obtained by capitalizing her net profits. Land, for 
 example, which produces a profit of $5 per acre above total costs, excluding interest 
 on investment at 5%, is capitalized at S100 from the standpoint of producing 
 power. (See also problems 10-12 oral and page 69, Part I.) 
 
 f20% charges are made to include interest on investment, depreciation, etc. 
 Capitalizing net profits necessarily increases interest charges and depreciation, etc. 
 
174 
 
 AGRICULTURAL ARITHMETIC 
 
 Figure 25. — The Products of a Poor, a Good, and an Exceptionally Good 
 Dairy Cow. Large tubs each contain 60 lbs. butter. Many dairy cows do 
 not pay for their board. 
 
 Profit and Loss in Manuring and Fertilizing and in pro- 
 ducing Crops: 
 
 19. A farmer having some acid, silt loam, purchased 
 some pulverized limestone at $2.25 per ton. Freight charges 
 were 2.5 cents per hundred. Getting it on the land cost 75 
 cents per ton. Extra harrowing to work lime into soil cost 
 50 cents per acre. Two tons were applied per acre. The 
 following year the clover yielded 2.8 tons per acre on the 
 limed area — an increase of 63.3% over the unlimed area. 
 
 (a) Charging 50% of the cost of liming against the 
 first clover crop, determine the per cent pro- 
 fit on cost of liming when clover hay is worth 
 $9.75 per ton. 
 
 20. The plowing under of a green-manuring* crop of 
 cowpeas equivalent to 2 tons of hay per acre resulted in a 
 yield of 76 bushels of shelled corn per acre, which was an 
 increase of 55% over the yield on an area having received no 
 green-manuring. 
 
 *Green-manuring is the plowing under of green crops which were sown for 
 that purpose. 
 
FARM MANAGEMENT 175 
 
 (a) How many pounds of nitrogen were contained 
 in the cowpea crop plowed under? (See 
 page 240.) Value at 15 cents per pound? 
 
 (b) When corn is worth 65 cents per bushel, de- 
 termine value received from plowing under 
 this green-manuring crop. 
 
 21. The standard fertilizer for cotton adopted by the 
 Georgia Experiment Station is a mixture of 468 pounds of 
 acid phosphate (containing 7% phosphorus); 36 pounds of 
 muriate of potash (containing 41.5% potassium); and 130 
 pounds of nitrate of soda (containing 16% nitrogen), to be 
 applied per acre in one application by drill before planting. 
 
 (a) Determine the number of pounds of each fer- 
 tility element added per acre when this mixture 
 is applied. 
 
 (b) When muriate of potash costs S45 per ton, 
 acid phosphate $15.20- and nitrate of soda $45 
 per ton, what is the cost of this application 
 per acre? 
 
 (c) If such a treatment results in a yield of 460 
 pounds of lint per acre, which sells for 12 cents 
 per pound, determine the profits above cost 
 of fertilizers. 
 
 22. On a peat* soil in Wisconsin a fertilizer test was 
 made as indicated in the following diagram, which shows the 
 kinds and rate at which fertilizers were applied per acre : 
 
 No. 1 
 
 No. 2 
 
 No. 3 
 No. 4 
 
 Barnyard manure — 15 tons per acre 
 
 Muriate of potash — 400 lbs. per acre 
 Acid phosphate — 600 lbs, per acre 
 
 No treatment 
 
 Muriate of potash — 400 lbs. per acre 
 
 ♦See footnote concerning peat on page 156. 
 
176 AGRICULTURAL ARITHMETIC 
 
 The fertilizers were applied broadcast. 
 
 Two crops of corn were grown following the above treat- 
 ment, yielding per acre as follows: Strip No. 1, first year, 
 10.5 tons of corn fit for the silo, second year, 5.8 tons; Strip 
 No. 2, first year, 16.8 tons, second year 14.2 tons; No. 3, 
 first year, 3.25 tons, second year, 3.8 tons; No. 4, first year, 
 13.9 tons, second year, 13.4 tons. 
 
 (a) Determine the value of the fertility contained 
 in the manure applied per acre. (See page 147.) 
 
 (b) What was the value of the fertility contained 
 in the manure applied per ton of increase in 
 corn over the yield on the strip having received 
 no treatment? 
 
 (c) When manure is valued at $1.50 per ton, de- 
 termine the fertilizer cost per ton increase in 
 the corn crop. 
 
 (d) When green corn is worth $2 per ton standing, 
 what was the profit or loss in the use of manure 
 on this soil? Use both values given the 
 manure. Determine per cent profit or loss on 
 values given manure. 
 
 (e) Determine the cost of the fertilizer applica- 
 tion made on strip No. 2, when muriate of 
 potash cost $46 per ton and acid phosphate, 
 $18. 
 
 (f) What did it cost in fertilizers to produce one 
 ton increase on this strip? 
 
 (g) When green corn is worth $2 per ton, determine 
 the per cent profit realized on cost of fertilizers, 
 Strip No. 2. 
 
 (h) Determine the cost to produce one ton increase, 
 and per cent profit realized on cost of fertilizer 
 applied to strip No. 4. Compare these results 
 with strips Nos. 1 and 2. 
 
FARM MANAGEMENT 177 
 
 (i) Mention some lessons that may be learned 
 from these results. (See also Figures 18 and 20.) 
 
 (j) In comparing strips Nos. 2 and 4 which pro- 
 duced the greater amount of profits above 
 cost of fertilizers? What per cent greater 
 amount of profits? Which application, then, 
 would you recommend as being the more pro- 
 fitable fertilizer treatment for this marsh soil, 
 basing opinion on these 2-year results? Which 
 application would appeal to the man of small 
 means? To the man of ample means? 
 
 (k) If the money used to purchase fertilizers were 
 
 to be borrowed for one year at 6%, which 
 
 would be more profitable, to fertilize as on 
 
 strip No. 2 or as on No. 4? 
 
 23. According to Government data collected in Rice 
 
 County, Minnesota, to grow an acre of' oats costs, on an 
 
 average, as follows : 
 
 Operation Cost per acre 
 
 Seed value $0,962 
 
 Cleaning seed . 021 
 
 Plowing 1 .230 
 
 Dragging 0.258 
 
 Seeding 0.241 
 
 Cutting 0.380 
 
 Twine 0.377 
 
 Shocking 0. 158 
 
 Stacking 0.767 
 
 Stack-threshing (labor) . 632 
 
 Threshing (cash cost) . 865 
 
 Machinery 0.446 
 
 Land rental* 3 . 500 
 
 (a) Determine the total cost to grow an acre of 
 oats according to these figures. Cost per 
 bushel, when yield is 45 bushels. When yield 
 is 65 bushels, assuming costs remaining the 
 same. 
 
 *Land rental covers interest charges, taxes and insurance. In this case land 
 was valued at $70 per acre. 
 
 12— 
 
178 
 
 AGRICULTURAL ARITHMETIC 
 
 (b) When oats are worth 35 cents per bushel, at 
 least how many bushels must be produced per 
 acre to offset this cost? 
 
 (c) Determine the cost to grow an acre of oats 
 when land is valued at $225 per acre. Assume 
 5% to cover land rental and other costs to 
 remain the same. 
 
 (d) Suppose $10 worth of fertilizer were applied 
 per acre to this 225-dollar land at the time of 
 sowing the oats, and 50% of the cost of the 
 fertilizer be charged to the oat crop; how many- 
 bushels of oats worth 34 cents per bushel must 
 be produced per acre to offset this cost? In- 
 clude 2 tons of straw valued at $2.50 per ton. 
 
 Figure 26. — Harvest Time. What does it cost to raise a crop of grain? 
 
 (e) Suppose the fertilizer treatment mentioned 
 in problem (d) resulted in an average yield of 
 75 bushels of oats and 2 tons of straw per acre 
 — an increase of 15% over an unfertilized strip. 
 
FARM MANAGEMENT 179 
 
 Determine profit or loss in fertilizing when 
 oats are worth 35 cents per bushel and straw 
 $2.50 per ton. Would it have been profitable 
 were there a 20% increase in the crop? A 25% 
 increase? 
 
 24. "A" raises on an average 40 bushels of corn and 2 
 tons of stover per acre at a cost of $9.55 per acre on low- 
 priced land. "B" succeeds in producing on an average 75 
 bushels of corn and 3.4 tons of stover on highly fertilized and 
 high-priced land at a cost of $35.20 per acre. 
 
 (a) When corn is worth 60 cents per bushel and 
 stover $3 per ton, who realizes the greater per 
 cent profits per acre? 
 
 (b) What per cent greater profits does the one 
 realize over the other? 
 
 25. Mr. Brown invested $300 in fertilizers. He applied 
 $10 worth per acre once in a four-year rotation and received 
 a net profit as increased crop production, as total for the 4 
 years, amounting to $20 per acre above cost of fertilizer. 
 Jones, having $150, invested only $5 worth of fertilizer per 
 acre applied over an area equal to that fertilized by Brown, 
 and, in the same period of time, received a net profit in in- 
 creased crop yields amounting to $12.50 per acre above cost 
 of fertilizers. 
 
 (a) Who received the greater per cent profit on 
 fertilizer invested? 
 
 (b) Who made the more profitable investment, and 
 how much more profitable? 
 
 (c) Suppose both men borrowed the money invest- 
 ed in fertilizers at 6% for 4 years. Who then 
 would have made the greater per cent profit on 
 fertilizers? Who would have made the more 
 profitable investment? How much more pro- 
 fitable? 
 
180 AGRICULTURAL ARITHMETIC 
 
 (d) Would it have been more profitable for Jones 
 had he borrowed $300 at 6% for 4 years and 
 applied $10 worth of fertilizer per acre and 
 received a total increase for the 4 years of $15 
 per acre above cost of fertilizer? How much 
 more profits would he have gained as com- 
 pared with problem (c)? 
 26. A farmer who succeeded in growing alfalfa on acid 
 soil by liming and inoculating* figured profit and loss from 
 the following figures: 
 
 Operation Cost per acre 
 
 Cost of liming (initial cost) $5 . 84 
 
 Alfalfa seed (initial cost) 6 . 50 
 
 Inoculation (initial cost) 2 . 00 
 
 Fertilizer (Av. of 10% per yr. of value of manure applied to corn) * 3 . 00 
 
 Mowing (man and horse labor, 3 cuttings) 1 .00 
 
 Baking 0.54 
 
 Cocking. 0.60 
 
 Hauling in 3 . 00 
 
 Machinery cost 0.65 
 
 Land rental at 5% 7 . 50 
 
 Cost of preparation of seed bed was offset by value of nurse crop. 
 
 Returns: 
 
 Average yield for four years — 4.6 tons of hay per acre. 
 Average value per • ton — $16.00. 
 
 (a) Determine the average profits per acre per 
 year. 
 
 (b) Determine the value of this land as an invest- 
 ment, f growing alfalfa, when for every dollar 
 the land is worth above $150, 5 cents be charged 
 against it to cover annual land rental. 
 
 (c) When before liming! it cost $25 to produce 65 
 bushels of corn and 3 tons of stover per acre, 
 husked and shredded, determine the value of 
 
 *Liming corrects acidity. Inoculation is necessary to supply the proper 
 nodule-forming organisms. 
 
 tSee footnote on page 173. 
 
 %A crop of corn was raised on this land before the alfalfa. 
 
FARM MANAGEMENT 
 
 181 
 
 this land as an investment when it produces 
 this amount of corn, when corn is worth 65 
 cents per bushel and stover $3 per ton. For 
 every dollar the land is worth above $150, 5 
 cents should be charged against it to cover 
 annual land rental. 
 
 (d) From this point of view, is liming acid soils for 
 growing alfalfa profitable? (See Figure 27.) 
 
 (e) Suppose this soil were well supplied with lime 
 and required no lime and inoculation. De- 
 termine the value of the land as an income 
 producer when alfalfa yields 5J- tons per acre 
 per season. (Solve as in (b).) 
 
 27. Determine the cost to produce a bushel of potatoes 
 in your neighborhood. A bushel of corn. A bushel of oats. 
 
 Farm Records and Accounts: 
 
 28. The following is an account a farmer kept with his 
 hired man for four months: 
 
 John Zoellmer 
 
 Commenced work March 1st, 1909, at $28 per month 
 
 and board and room. 
 
 Mar. 
 
 10 
 
 Apr. 
 
 1 
 
 Apr. 
 
 4 
 
 May 
 
 10 
 
 June 
 
 31 
 
 Cash 
 
 Paid bill at store 
 
 Cash 
 
 Cash 
 
 Cash 
 
 $ 6 
 
 4 
 
 15 
 
 10 
 
 74 
 
 $110 
 
 Apr. 
 May 
 June 
 June 
 June 
 
 1 month's work in March. 
 1 month's work in April . . 
 1 month's work in May. . . 
 Did not work June 10-11.. 
 28 days' work in June 
 
 $28 
 28 
 28 
 
 26 
 $110 
 
 (a) Make out an imaginary account with a hired 
 man for six months. 
 
 29. The following brief account was kept by a farmer 
 with 15 acres of mixed clover and timothy hay. Average 
 for two years. 
 
182 
 
 AGRICULTURAL ARITHMETIC 
 
 Figure 27. — Success with Alfalfa Due to Lime. Nine acres in this field of 
 acid soil were limed and one acre left unlimed. The whole field was inoc- 
 ulated. Under ordinary conditions alfalfa cannot grow on acid soils un- 
 less first limed and inoculated with alfalfa nodule-forming organisms. 
 
 Account with 15 Acres of Hay (Av. for 2 years). 
 
 Charges* 
 
 25% of value of manure applied to 
 corn 
 
 Seed at 50 cents per acre 
 
 Mowing at 40 cents per acre. . . 
 
 Raking at 18 cents per acre . . . 
 
 Cocking, hauling and stacking at 
 $1.30 per acre 
 
 Machinery cost at 55c. per acre 
 
 Land rental at $7.50 per acre. . 
 
 Total costs. 
 Profits 
 
 Grand total . 
 
 $5 
 
 Credits 
 
 15 tons hay at $10.50 per ton 
 4 tons hay at 9.40 per ton 
 8 tons hay at 9.50 per ton 
 
 Total credits and receipts. . . 
 
 (a) Fold a sheet of paper to represent the left and 
 right-hand pages of an account book, rule as 
 shown above, and complete the account. 
 
 (b) Determine net profits per acre. 
 
 *In the farm account book the opening inventory and charges to an account ■ 
 occupy the entire left-hand page, and these items include cash paid out by you, 
 work done by you, or any item of value for which you are not paid. The credit 
 items, or what is received from an account, occupy the entire right-hand page, 
 and include cash paid to you, work done for you, value received for stock and 
 farm products sold, and the closing inventory. 
 
FARM MANAGEMENT 
 
 183 
 
 30. Fold a sheet of paper to represent the right and left- 
 hand pages of an account book, rule properly, and enter the ^" 
 following items under "charges" and "credits," as: 
 
 Account with 12 Acres of Potatoes. 
 
 Charges* 
 
 Credits f 
 
 1913 
 
 Item 
 
 Amount 
 
 1913 
 
 Item 
 
 Amount 
 
 May 25 
 
 
 
 
 Oct. 20 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 May 25, 1913, 135 bushels of seed potatoes at 55 cents 
 per bushel; July 10, 10 pounds of Paris green at 22 cents; 
 July 15, 7 pounds Paris green at 22 cents; 120 tons manure 
 at $1.50 per ton; plowing at $1.60 per acre; dragging at 92 
 cents per acre; planting at $2.50 per acre; cultivating, $1.70 
 per acre; hoeing, $1.00 per acre; spraying, $1.60 per acre; 
 digging, picking and hauling, $6.75 per acre; machinery cost, 
 35 cents per acre; land rental, $7.50 per acre; Oct. 20, sold 
 307 bushels at 56 cents per bushel; Oct. 25, sold 1,000 bushels 
 at 63 cents; Nov. 3, sold 350 bushels for seed at $1.25 per 
 bushel; saved for seed 150 bushels at 75 cents; kept for home 
 use 75 bushels at 60 cents; fed 38 bushels small potatoes at 
 20 cents; estimated value of manure left in the soil, $108. 
 
 (a) What were the total profits? 
 
 (b) Determine average net returns per acre. 
 
 (c) Determine the average cost per bushel. 
 
 31. Fold a sheet of paper to represent the left and right- 
 hand pages of an account book, rule as in problem 29, and 
 arrange the following summary items for a herd of 10 cows 
 under "charges" on left-hand page and "credits" on right- 
 hand page. 
 
 ♦Charges usually occupy the entire left-hand page of account book. 
 tCredits usually occupy the entire right-hand page of account book. 
 
184 AGRICULTURAL ARITHMETIC 
 
 CHARGES 
 
 Opening Inventory: Jan. 1st, 1914, 3 cows at $45 each, 
 4 at $60 each, 3 at $75 each; 10 tons alfalfa hay at- $16 per 
 ton, 10 tons straw at $3 per ton (used for bedding), 50 tons 
 corn silage at $3 per ton, f ton ground corn worth $21 per 
 ton, \ ton bran at $24 per ton and \ ton cottonseed meal at 
 $28 per ton. 
 
 March 5, bought \\ tons ground corn at $20 per ton and 
 \ ton wheat bran at $24 per ton; April 2, \ ton cottonseed 
 meal at $28 per ton; June 10, harvested 6 tons alfalfa worth 
 $16 per ton; July 21, 8 tons alfalfa hay worth $14 per ton; 
 Aug. 21, 8 tons alfalfa hay worth $16 per ton; Sept. 23, 60 tons 
 corn silage worth $3 per ton; Sept. 26, 15 tons straw at $3 
 per ton; Sept. 30, 1 ton ground corn at $22 per ton, 1 ton 
 bran at $23 and \ ton cottonseed meal at $28 per ton; Oct. 
 30, \\ tons ground corn at $22 per ton and 2 tons bran at 
 $24. Pasture for summer $25. Man and horse labor $22 
 per cow; shelter $2.45 average per cow; other expenses $3.50 
 per cow. Interest and taxes on amount invested at 6%. 
 (Take average of the two inventories* to obtain amount 
 invested.) 
 
 CREDITS 
 
 Milk was shipped to city market averaging $1.53 per 
 hundred net. 
 
 Feb. 1, 6,400 lbs. sold in Jan.; Mar. 1, 7,400 lbs. in 
 Feb.; Apr. 1, 7,500 lbs. in March; May 1, 6,800 lbs. in Apr.; 
 June 1, 6,900 lbs. in May; June 5, 2 veal calves for $14; July 
 1, 8,900 lbs. in June; Aug. 1, 6,100 lbs. in July; Sept. 1, 4,200 
 lbs. in Aug.; Oct. 1, 3,000 lbs. in Sept.; Nov. 1, 5,400 lbs. in 
 Oct.; Nov. 10, 3 veal calves, $18.50; Nov. 12, sold cow and 
 calf for $95; Dec. 1, 7,200 lbs. milk for Nov.; Dec. 7, 1 veal 
 
 *See inventory under "credits." Closing inventory is commonly used as the 
 opening inventory for the new year. 
 
FARM MANAGEMENT 185 
 
 calf, $6.25; Dec. 31, 6,400 lbs. for Dec; total milk for home 
 use 1,500 lbs. at $1.53 per hundred. Manure for year 105 
 tons at $1.50 per ton. 
 
 Closing Inventory: Dec. 31, 3 cows at $45 each, 4 at $55 
 each, 2 at $72 each; 3 calves valued at $25; 12 tons alfalfa 
 hay valued at $16 per ton; 5 tons straw at $3 per ton; 45 
 tons corn silage at $3 per ton; $ ton ground corn at $22; £ 
 ton cottonseed meal at $28 per ton. 
 
 (a) Determine total profits or loss. 
 
 (b) What was the average feed-cost per cow? 
 
 (c) Determine the total cost per cow. 
 
 (d) Determine the total average returns per cow. 
 The Farmer's Income: 
 
 32. Government statistics for 1910 show that the aver- 
 age farm in the United States consisted of 138.1 acres, 
 (75.2 acres improved land). The average gross .income per 
 farm was $980.55 and average total expenses were $340.15. 
 Determine the average net farm income. (Use no pencil.) 
 
 33. The total investment per farm was $6,443.67. 
 Determine the average labor income* per farm when interest 
 on investment was figured at 5%. 
 
 34. The average farmer pays 6% on a mortgage aver- 
 aging $1,715. How much money on the average is available 
 for purchase of stock and for family use on the average 
 farm in the United States? 
 
 35. Statisticsf gathered from Tompkins County, N. Y., 
 show that farmers on general farms averaging in size from 
 31 to 60 acres receive on ah average an annual labor income 
 of $254; while those on farms averaging from 150 to 200 
 acres receive $635. What per cent of the labor income from 
 the small farms is the labor income from the larger farms? 
 
 *Labor or managerial income means profits above total costs — total oosts in- 
 cluding interest on investment and unpaid family labor. Family labor means all 
 work done by wife and children on the farm, not including the household. 
 
 tNew York, Cornell bulletin 295, p. 414. 
 
186 AGRICULTURAL ARITHMETIC 
 
 36. Other statistics from the same county show that 
 the average labor income from $2,000 to $4,000 invested in 
 farm capital amounts to $240; while $10,000 to $15,000 in 
 capital returns an average labor income of $870. 
 
 (a) Is the increase in labor income in proportion 
 to increase in capital? 
 
 (b) Determine the average net farm income* in 
 each case. Consider interest on investment at 5%. 
 
 37. Farmers "A" and "B" each have a 120-acrefarm in 
 
 which capital is distributed as follows : 
 
 "A'' "B" 
 
 Land value $15,000 $18,000 
 
 Live stock (mostly cows) 3,500 2,300 
 
 Farm buildings 4,800 9,250 
 
 Implements and machinery 650 2,860 
 
 (a) What per cent of A's capital is invested in 
 land? B's? 
 
 (b) What per cent of A's capital is invested in 
 stock? B's? 
 
 (c) What per cent of A's capital is invested in 
 farm buildings, implements and machinery? 
 B's? 
 
 (d) A's net farm income for one year was $3,456 
 and B's $1,233. When interest on investment 
 is figured at 5%, what was A's labor income? 
 B's? 
 
 (e) What was one of the troubles in B's farming? 
 
 38. In Tompkins County, N. Y., it was found that on 
 land valued at from $21 to $40 per acre, the tenant's labor 
 income per year averaged $402 and the landlord's profit on 
 investment averaged 8.9%. On land valued at $60 and 
 above, the tenant's labor income averaged $562 and the 
 landlord's profit on investment averaged 6.4%. 
 
 ♦When labor income is the net farm income minus the interest on capital in- 
 vested, the net farm income may be easily determined. 
 
FARM MANAGEMENT 187 
 
 (a) For the tenant the higher priced land is how 
 many per cent more profitable than land of 
 less value? 
 
 (b) What per cent greater profits does the land- 
 lord realize when he rents the lower valued 
 land than when he rents land of higher value? 
 
 39. The labor income on farms averaging in size from 
 150 to 200 acres in that county averaged $635 which was 150 
 per cent more than on farms averaging in size from 30 to 60 
 acres. Determine the average labor income on the smaller 
 farms. 
 
 40. In Missouri it was found that small farms of 40 
 acres or less averaged $128 as labor income for the owner, 
 which was 69.8% less than that from farms averaging 200 
 to 400 acres. Determine the average labor income for 
 owners of the larger farms. 
 
 41. The following summary was made at a year's end 
 on a successful dairy farm consisting of 211 acres, of which 
 160 are tillable. 
 
 Capital invested: 
 
 Farm value at $60 per acre 
 
 Machinery $450.00 
 
 Horses: 
 
 3 @ $175 each 
 
 2 @ 75 each 
 
 Cows: 
 
 10 @ $85 each 
 
 6@ 75 each 
 
 5 @ 65 each 
 
 8 @ 45 each 
 
 7 @ 40 each 
 
 Other stock 320.00 
 
 Other items 116.00 
 
 Receipts (Cash): 
 
 300 bu. barley @ 75 cents 
 
 2 tons alfalfa hay @ $16 per ton 
 
 2,500 bu. potatoes @ 50 cents 
 
 259,200 lbs. milk @ $1.40 per hundred net 
 
 Stock sold 260.00 
 
 Miscellaneous 43.00 
 
188 AGRICULTURAL ARITHMETIC 
 
 Expenses: 
 
 Labor and board $1,100.00 
 
 Feeds purchased 560.00 
 
 Fertilizer 108.00 
 
 Seed 64.00 
 
 Miscellaneous 286.00 
 
 Unpaid family labor* (estimated) 300.00 
 
 (a) Determine the total amount of capital in- 
 vested in the farm. 
 
 (b) Determine the total cash receipts. 
 
 (c) Determine the total expenses. 
 
 (d) Determine the farm income. 
 
 (e) Determine the interest on investment at 5%. 
 
 (f) What was the labor or managerial income from 
 this farm? 
 
 A Few Farm Facts: 
 
 1. The successful farmer is a good business man. 
 
 2. It is necessary for a farmer to keep accounts to de- 
 termine profits, losses and labor income. 
 
 3. The farm is a poor place for the inefficient. 
 
 4. A farmer on a farm receiving a labor income receives 
 in addition the use of a house and some products for home 
 use. 
 
 5. Fewer farmers are needed to feed the world when 
 better farming is practiced. 
 
 6. What facts brought out in the preceding problems 
 especially interested you? 
 
 ♦Family labor includes labor done on the farm by the wife and children, 
 work done in household excluded. This item is usually estimated. 
 
CHAPTER XII 
 
 FARM MEASUREMENTS 
 
 The farmer frequently has occasion to map his farm, or a 
 part of it, for use in working out crop rotation systems, in 
 rearranging his fields, in building, etc. Such work can be 
 more successfully done when diagrams are drawn to some 
 scale. .In some communities very few farmers can give the 
 description of the farms they own. Almost every practical 
 farmer should carry in his mind a few rules to enable him 
 to work out the common problems in farm measurements. 
 
 ORAL PROBLEMS 
 
 1. A Congressional township consists of an area of 
 land 6 miles square and comprises 36 sections.* 
 
 (a) How many miles around a section? Around 
 a half-section? Quartern-section? 
 
 (b) How many rods square is a quarter-section? 
 How many rods long is half of a quarter-sec- 
 tion? 
 
 (c) How many rods long is a quarter of a quarter- 
 section? Wide? 
 
 (d) What part of a section is a 10-acre field? 
 
 2. A field measuring 20 X 80 rods contains how many 
 
 acres? 
 
 (a) A square field containing the same number of 
 acres would be how long and how wide? 
 
 (b) What is the difference in the number of rods 
 around the two fields? How does this fact 
 affect the fencing problem? 
 
 *36 sections may or may uot be a township. 
 189 
 
190 AGRICULTURAL ARITHMETIC 
 
 3. 450 cubic feet of settled hay make a ton. How 
 many tons of hay in a settled mow 15 X 30 X 20 feet? 
 
 4. When corn silage averages 40 pounds per cubic foot, 
 how many cubic feet are required for one and one half tons? 
 
 5. When a map is drawn on a scale of "\ inch = one 
 mile," how many miles between two cities when map meas- 
 urement is 8£ inches? 
 
 6. How long and how wide must a drawing be to 
 represent a half-section when \ inch = 10 rods? 
 
 7. How many rods around 3 contiguous quarter-sec- 
 tions? Which costs the more to fence, f of a section or a 
 section? 
 
 8. What is the area of a circle whose diameter is 2 
 inches? (See page 61, Part I.) 
 
 9. How many paces in a mile? 
 
 10. How many paces on the end of a square ten-acre 
 field are required to mark off one acre? 
 
 WRITTEN PROBLEMS 
 
 1. Draw a diagram representing a section of land. 
 Use scale | inch = 10 rods. (Review pages 33 and 34.) 
 
 (a) Locate in this section the S £ of NE \; the SE 
 $ of the NW i; and the NE £ of SW i of the 
 SW -j. How many acres in each block of land? 
 
 2. Draw a diagram to represent a Congressional town- 
 ship. Use scale 1 inch = 2 miles. 
 
 (a) Locate the S $ of Sec. 8; the NE \ of Sec. 24; 
 the W i of Sec. 32; the E \ of SE \ of Sec. 16. 
 
 3. Locate in another diagram drawn to the scale of 
 \ inch = 6 miles, a township which lies in the fifth row north 
 of the base line and in the third tier of townships east of 
 the principal meridian. 
 
FARM MEASUREMENTS 191 
 
 (a) Locate in this township sections 8, 12, 15, 25 
 and 34. 
 
 (b) Describe the location of section 16 as to town- 
 ship (T) and range (R). 
 
 4. Using the same base line and principal meridian as 
 in problem 3, locate T 6 N, R 5 E. 
 
 (a) Locate the W 1 of Sec. 28. 
 
 5. In a diagram drawn to a scale of 1 inch = 6 miles, 
 locate the NW i and the S i of Sec. 20, T 3 N, R 2 W. 
 
 6. Sketch an outline map of your state, use scale 
 1 inch = 48 miles, and locate the .base line from which "T" 
 is reckoned, and 'principal meridian from which "R" is de- 
 termined. 
 
 (a) Locate T 4 N or S, R 3 E; T 5 N, R 4 W; 
 T 6 S, R 7 E. 
 
 (b) In what counties in your state are these town- 
 ships located? 
 
 (c) Locate the township in which you live. De- 
 scribe the section in which you live, giving 
 section, number, "T," and "R." 
 
 (d) Get the description of your farm, or one with 
 which you are familiar. Locate it in the sec- 
 tion or sections; within the township; range. 
 
 7. The sides of a triangular field measure 40 rods, 36 
 rods and 20 rods, respectively. 
 
 (a) Draw a diagram — scale £ inch = 2 rods, rep- 
 resenting this field. 
 
 (b) Determine the number of acres in this field. 
 (See page 57.) 
 
 8. Another triangular field has sides measuring 30, 20 
 and 20 rods, respectively. 
 
192 
 
 AGRICULTURAL ARITHMETIC 
 
 (a) Draw a diagram of this field to a scale of £ inch 
 = 5 rods. 
 
 (b) How many acres in this field? 
 
 9. Figure 28 represents an ir- 
 regular field drawn to scale of ^ 
 inch = 5 rods. 
 
 (a) Draw a diagram of this 
 field to the scale of 2 inches = 
 5 rods and determine the number 
 of acres in it. (Page 31.) 
 
 Figure 28. 
 
 Figure 29. 
 
 The number of acres in a field having the shape of Figure 
 29 may be determined by measuring the breadth in rods in a 
 number of places at equal distances apart, as a, b, c, d, e and 
 f j add these measurements and find the average breadth; 
 then multiply by average length in rods and divide by 160. 
 
 10. The length of a field similar in shape to Figure 29 
 having crooked and irregular sides measured 30 rods; the 
 breadth at intervals of 3 rods measured 9, 12, 9, 6, 8, 12, 12, 
 8, 14, 9 and 12 rods, respectively. 
 
 (a) Draw a diagram of this field to a scale of £ inch 
 = 3 rods. 
 
 (b) How many acres in this field? 
 
FARM MEASUREMENTS 
 
 193 
 
 (c) If a fence were to be built on the boundary of 
 this field as you have drawn it, how many rods of fencing 
 would be required to inclose it? 
 
 11. Figure 30 is the 
 outline of an island 
 showing where and how 
 measurements were made 
 to determine its area. 
 How many acres in the 
 island? Diagram is 
 drawn to scale of £ inch 
 = 4 rods. 
 
 Figure 30. 
 
 12. Figure 31 is an outline of a farm, drawn to a 
 scale of 1 inch = 40 rods. This farm represents the W ■§■ 
 and a part of the NE \ of the SE \ of Section 16, T 2, R 1 W. 
 
 (a) By measurement, determine the number of 
 acres in each field. 
 
 (b) How many rods of fencing around each field? 
 (No fence along water-front in field No. 8.) 
 
 (c) How many rods of fencing are required for 
 the whole farm? 
 
 13. The following map, Figure 32, shows a tiling system 
 a farmer put in one of his fields at an average cost of 47 
 cents per rod. 
 
 (a) Determine the total cost to tile this field. 
 
 (b) The first year after tiling, this field averaged 
 70 bushels of shelled corn per acre, which was 
 a 95% increase over the average yield before 
 tiling. When corn is worth 65 cents a bushel 
 determine the per cent of net profits on cost 
 of tiling for that year. 
 
 13— 
 
194 
 
 AGRICULTURAL ARITHMETIC 
 
 Lake 
 
 Figure 31. 
 
 14. A man had a silo 35 feet deep and 16 feet in diam- 
 eter, as inside measurements, containing 15 feet of silage. 
 Assuming the silage in the bottom five feet to weigh on an 
 average 55 pounds per cubic foot, the second 5 feet, 51.4 
 pounds and the top 5 feet, 46 pounds per cubic foot, de- 
 termine the amount and value of the silage at $3.50 per ton. 
 (See page 61, on area of a circle.) 
 
 Measuring Hay: 
 
 To measure hay in mows: Volume ■*■ 420 to 500 cubic 
 feet = tons. 450 cubic feet of settled hay in a mow is 
 
FARM MEASUREMENTS 
 
 195 
 
 Figure 32. 
 
 For timothy use 420, for 
 
 commonly taken to equal a ton. 
 clover 500 cubic feet. 
 
 To measure hay in ricks*: Volume of rick -s- 515 to 
 590 cubic feet = tons. The volume (cubic feet) of a rick 
 is determined by multiplying the cross-section of a rick by 
 the length (L). The cross-section is obtained by multiply- 
 ing the "over" (0), which is the distance from the ground on 
 
 *U. S. Dept. of Agriculture, Bureau of Plant Industry, Circular 131. 
 
196 
 
 AGRICULTURAL ARITHMETIC 
 
 F-.37 
 
 Figure 33. Cross-section of Hayricks of Different Shapes. 
 
 one side of the rick over the top of the rick to the ground on 
 the other side, by the width (W), by a fraction (F), which 
 varies from .25 to .37, depending upon the height and full- 
 ness of the rick. If the rick has a cross-section similar to 1, 
 Figure 33, the fraction is .25; if like 9, it is .37. Therefore: 
 
 Volume (V), of rick = Fraction (F), X Over (0) X 
 Width (W) X Length (L); or V = FOWL. 
 
 590 cubic feet of hayf in rick or stack standing less than 
 30 days = 1 ton. 
 
 fMixture of clover and timothy. 
 
FARM MEASUREMENTS 197 
 
 580 cubic feet of hay in rick or stack standing from 30 to 
 60 days = 1 ton. 
 
 515 cubic feet of hay in rick or stack standing from 75 
 to 155 days = 1 ton. 
 
 15. A mow 18 feet wide and 36 feet long has 18 feet of 
 settled mixed hay in it. How many tons of hay does it 
 contain? How many tons if it were timothy? If it were 
 clover? 
 
 16. A hayrick is 16 feet wide, 20 feet long, and the 
 "over" is 30 feet. The end view indicates that the shape 
 of the rick is very close to No. 5 in Figure 33. How many 
 tons of hay in the rick when it has been standing about 50 
 days? 
 
 Solution: 
 
 Volume (V) -H 580 = tons. V = FOWL. 
 
 No. 5 in Figure 33 shows F equals .31. Therefore: 
 
 .31 X 30 X 16 X 20 K1Q , 
 ^tr = 5.13 tons. 
 
 29 
 
 17. A hayrick measures 14 feet wide, 24 feet long, and 
 the "over" is 34.2 feet. Inspection of the end of the rick 
 shows that it is of the type of No.- 8, Figure 33. The rick 
 has been standing 5 months. Determine the value of the 
 hay at $15 per ton. 
 
 18. A man bought a rick of hay for $65. After it had 
 been standing for 3 weeks it measured 30 feet long, 11.5 
 feet wide, with an "over" of 33 feet. The end view of the 
 rick showed that it was similar in shape to No. 6, Figure 33. 
 What price per ton was paid for the hay? 
 
 19. The "over" of a hayrick which has been standing 
 5 months measures 34 feet, the length is 40 feet, and the 
 width 16 feet. The end view of this rick indicates that it 
 has the shape of No. 9 in Figure 33. How many tons of 
 hay in the rick? 
 
198 
 
 AGRICULTURAL ARITHMETIC 
 
 V=.08HC* 
 
 V*.08HCc 
 
 Figure 34. — Diagram showing various Bhapes of round haystacks. 1, 2 and 
 3 represent upper portions of stacks; 4 and 5, lower parts. V = vol- 
 ume, H = height of part of stack measured; C 2 = circumference squared; 
 C c = circumference at top of base. 
 
 To measure hay in round, stacks: Volume (TO ■*■ 515 to 
 590 cubic feet = tons. 
 
 Haystacks are built either with a cylindrical base and 
 like outline No. 4, Figure 34, or like No. 5, Figure 34. 
 Tops may be formed like Nos. 1, 2 and 3, Figure 34. It 
 is necessary to determine the volume in cubic feet of the 
 top and bottom parts separately and add them together. 
 Formulas for determining volume ( V) of the parts are given 
 in the outlines in Figure 34. Read legend. 
 
 20. The base of a haystack is cylindrical and the top 
 is like No. 1, Figure 34. The height of the base after stand- 
 ing 3 months, is 4 feet, and the height of the top is 6 feet, 
 the circumference of base of stack is 30 feet. How many 
 tons of hay in the stack? 
 
 Solution: 
 
 Step 1. Top of stack (No. 1, Figure 34) V = .027 X 6 X 30 2 = 
 145.8 cubic feet. 
 
 Base of stack (No. 4, Figure 34) V = .08 X 4 X 30 2 = 288.0 cubic 
 feet. Total 433.8 cubic feet. 
 
 Step 2. 433.8 -f- 515 (page 195) = .84 ton. 
 
 21. A haystack has a base of the shape of outline No. 5, 
 Figure 34, and a top similar in shape to No. 2, Figure 34. 
 The height of the top, after standing 4 months, is 6 feet, 
 
FARM MEASUREMENTS 199 
 
 the circumference at the bulge is 48 feet, the "height of base 
 is 5£ feet, and the circumference at the bottom of the stack 
 is 40 feet. Determine the number of tons of hay in the 
 stack? 
 
 Solution: < 
 
 Step 1. Top of stack (No. 2, Figure 34) V = .04 X 6 X 48 2 = 552.9 
 cubic feet. 
 
 Base of stack (No. 5, Figure 34) V = .08 X 5J^ X 48 X 40 = 
 844.8 cubic feet. Total, 1,397.7 cubic feet. 
 
 Step 2. 1,397.7 -5- 515 (page 195) = 2.7 tons. 
 
 22. • The base of a haystack is like No. 4, Figure 34, 
 and the top like No. 2, Figure 34. After standing about 
 3 weeks, the height of the top portion measured 15 feet, the 
 circumference of the base measured 50.5 feet, and the height 
 of the base 12 feet. What is the stack worth at $12 per ton? 
 
 23. A haystack standing about 2 months had a base 
 10 feet high the shape of No. 5, Figure 34, measuring 60 feet 
 at the bulge and 51 feet at the bottom of the stack. The 
 top shaped like No. 1, Figure 34, was 1 1 feet high. The stack 
 was sold for $54. At what price per ton was the hay sold? 
 To determine bushels of grain, apples, corn, etc., in bins : 
 
 To determine bushels (not heaped) of grain in bin, divide 
 cubic content in inches by 2,150.4. For quick calculation 
 let 1.2 cubic feet = 1 bushel. 
 
 To determine bushels (heaped measure) of apples, potatoes, 
 ear corn, etc., in bins, divide cubic content in inches by 
 2,747.7. (I^et"1.6 cubic feet = 1 heaped bushel for quick 
 calculation.) Deduct -J from heaped bushels for bushels of 
 shelled corn. (56 lbs.) 
 
 24. How many bushels of wheat will a bin 20' X 20' 
 X 10' level full hold? 
 
 25. An old, circular water-tank 15 feet in diameter and 
 12 feet high (inside measurements) will hold how many 
 bushels of barley when level full? How does the weight of 
 this volume of barley compare with that of water? 
 
200 AGRICULTURAL ARITHMETIC 
 
 26. A bin 9' X 9' X 8' level full of oats was sold for 
 $20 per ton. How many bushels were sold and how much 
 per bushel was received for them? 
 
 27. How many bushels of potatoes in a bin 20' X 30' 
 X 6' level full? 
 
 28. A crib 30 feet long, 8 feet wide, and 8 feet high is 
 f full of ear corn. When shelled corn is worth 74 cents per 
 bushel, determine the value of the corn in the crib. 
 
 29. How many heaped bushels of ear corn are contained 
 in a crib with flared sides 25 feet long, 8 feet high, 6 feet 
 wide at the bottom* and 9 feet wide at the top, when level 
 full? Equivalent in shelled corn? 
 
 30. A man built a circular cement tank 16 feet in diam- 
 eter and 3 feet deep (inside measurements). How many 
 barrels of water would this tank hold when full. 
 
 31. How many gallons in one cubic foot of water? 
 
 32. A man wanted to build a cement tank 2 feet deep and 
 12 feet long (inside measurements), to hold 50 barrels of water 
 when full. How wide must it be (inside measurement)? 
 
 33. A farmer had a field of acid soil 16 X 20 rods. He 
 planned to apply lump-lime to the plowed land at the rate 
 of 1 ton per acre. When 50 pounds of the lime are to be 
 placed in each pile, how far apart must the piles be placed? 
 
 (a) Draw a diagram of this field on the scale of 
 ^ inch = 2 rods, and show the position of the 
 piles in the field. 
 
 (b) When lump-lime increases 80 per cent in weight 
 on air-slaking, determine the amount of air- 
 slaked lime in each pile. 
 
 (c) Over how many square rods must the contents 
 of each pile be spread? 
 
 ♦Average width = )4 sum of top and bottom widths. 
 
FARM MEASUREMENTS 
 
 201 
 
 34. Having no tape, a man measured the length and 
 width of a potato patehjDy counting the revolutions made by 
 the fore wheel of his buggy. He found the measurements 
 to be 60 and 22 revolutions, respectively. The wheel 
 measured 3 feet 4 inches in diameter. How many acres 
 in the potato patch? 
 
 35. A drive-pulley 4 feet in diameter must make how 
 many revolutions per minute to give a speed of 1884 revolu- 
 tions per minute to a shaft-pulley 8 inches in diameter? 
 
 36. How many bushels of oats will a bin 10 X 9 X 8 
 feet hold when full? How many tons of oats? Tons of 
 barley? Tons of wheat? 
 
 Concrete Work: 
 
 37. When 1 part of cement is to be mixed with 2 parts 
 of sand and 4 parts of gravel or crushed stone (a 1:2:4 con- 
 crete mixture), give the number of shovelfuls of sand and 
 cement that must be added to 20 shovelfuls of gravel to make 
 this mixture. 
 
 + 
 
 ■ 
 
 Cement 1 
 
 Sand 2 
 
 Concrete 1:2:4 
 
 Figure 35. — Comparative Amounts of Ingredients for a 1:2:4 Concrete Mixture. 
 The volume of concrete resulting from the mixture is slightly greater 
 than that of the stone. The cement and sand completely fill the open 
 spaces in between the stone particles. 
 
 38. One bag of cement is equal to about £ of a cubic 
 foot. How many cubic feet of sand and gravel must be 
 used for a 2-bag batch 1:2:4 concrete mixture? For a 4-bag 
 batch 1:3:6 mixture? 
 
202 
 
 AGRICULTURAL ARITHMETIC 
 
 39. When 1 barrel* of cement is regarded as equal to 
 3.8 cubic feet, how many cubic feet of sand and crushed stone 
 must be used to make a 1 : 3 : 6 concrete mixture? 
 
 Ingredients in one cubic yard of concrete.! 
 
 Proportion by vol 
 
 1:2:4 
 
 1:2|:41 
 
 1 :2:5 
 
 1 : 2| : 5 
 
 1:3:5 
 
 1 :3:6 
 
 Bbls. cement (1 bbl. = 
 3.8 cu. ft.) 
 
 1.46 
 
 1.32 
 
 1.25 
 
 1.20 
 
 1.13 
 
 1.00 
 
 Cubic yards sand 
 
 .41 
 
 .46 
 
 .35 
 
 .42 
 
 .48 
 
 .42 
 
 Cubic yards gravel or 
 stone 
 
 .82 
 
 .83 
 
 .88 
 
 .84 
 
 .80 
 
 .84 
 
 
 
 40. By consulting the table, determine the amount of 
 cement, sand and gravel necessary to construct a wall 
 30' X 8' X 10" when a 1: 2|: 5 concrete mixture is used. 
 
 41. How many barrels of cement, and cubic yards of 
 sand and gravel are required to construct a foundation for a 
 house 34' X 30'? The foundation is to be 9 feet high on all 
 sides and 10 inches thick. An impervious wall is desired; 
 hence a 1 : 2 : 4 cement mixturef is to be used. 
 
 42. Determine the amount of material required to fill 
 10 iron pipes (for end fence-posts) 8 feet long and 6 inches 
 in diameter (inside measurement), when a 1: 2£: 5 mixture 
 is used. 
 
 43. A man wishes to build a concrete silo large enough 
 to supply silage for his herd of 25 cows to be fed on an aver- 
 age of 35 pounds daily for 9 months. When settled silage 
 averages 42 pounds per cubic foot, how large a silo must he 
 build? (See footnote page 97.) 
 
 *4 bags cement are usually equal to a barrel. 
 
 tFrom Popular Handbook for Cement and Concrete Users, Lewis & Chandler. 
 
 +A perfect concrete mixture contains enough cement to fill the empty spaces 
 in between the sand particles, and coat each grain, while the sand with its coating 
 of cement fills the voids in the aggregate and also covers each stone with a film of 
 mortar, 
 
FARM MEASUREMENTS 203 
 
 (a) When 6 feet of the silo is to be a stone founda- 
 tion, determine the concrete material required 
 to construct the portion above the stone foun- 
 dation, wall to be 6 inches thick, and a 1 : 2: 4 
 mixture used. Determine the cost of this 
 material at local prices. 
 
 Do you know: 
 
 1. How many acres in a section? 
 
 2. How many cubic feet of hay equals a ton? 
 
 3. A formula to determine the area of a circle? 
 
 4. The meaning of range (R), and township (T), in 
 land description? 
 
 5. How to determine bushels of grain or ear corn in a 
 bin? 
 
 6. What a 1 : 3 : 5 concrete mixture means? 
 
CHAPTER XIII 
 
 ORCHARD AND GARDEN 
 
 The fruit-growing industry thrives best in certain geo- 
 graphical areas, since climate is the chief determinative 
 factor. Home-grown fruits, however, can be made the rule 
 in humid climates rather than the exception when proper 
 attention is given to location, planting, varieties, pruning 
 and spraying. The farmer's orchard and garden are often 
 objects of disgrace rather than sources of profit. It has 
 been demonstrated that the expense for labor in caring for a 
 garden large enough to supply the needs of an average 
 family need not exceed $30 per year*, and the retail value 
 of the vegetables which may be grown in a carefully planned 
 and well-kept garden greatly exceeds the cost to produce 
 them. 
 
 ORAL PROBLEMS 
 
 1. How many cherry trees on 7 acres, when 170 are 
 planted per acre? When 239 are planted per acre? 
 
 2. How many apple trees on 9 acres, when 48 trees is 
 the average per acre? On 12 acres, when 84 are planted per 
 acre? 
 
 3. How many plum trees on eleven acres, when 196 
 are planted per acre? When 239 are planted per acre? 
 
 4. When 60.8 barrels of apples, on an average, were 
 harvested from half an acre, how many barrels from 5 
 acres? From 7 acres? 
 
 ♦Illinois Agricultural Experiment Station, Bulletin 105. 
 204 
 
ORCHARD AND GARDEN 205 
 
 5. When 96.8 barrels of apples were raised per acre at 
 a cost of 90 cents per barrel, what was the total cost per 
 acre? 
 
 (a) When the apples were sold for $2.90 per 
 barrel, what were the total net profits per acre? 
 
 6. When it cost, on an average, $3.80 per acre to 
 spray an orchard, what did it cost for 5£ acres? 
 
 7. In a test made in New York, a part of an orchard 
 kept in sod produced 69.2 barrels of apples per acre as an 
 average for 10 years, while 116.8 barrels were harvested per 
 acre from a tilled portion. What was the increase in yield 
 due to tillage? 
 
 (a) $126.04 was secured as average annual gross 
 returns per acre from the sod portion, and 
 $224.15 per acre from the tilled portion. 
 What was the difference in receipts in favor of 
 tillage? 
 
 (b) The average acre-cost of growing the apples 
 on sod was $51.73, and under tillage, $83.48. 
 What were the net profits in each case? In- 
 crease in net profits due to tillage? 
 
 8. 16 bushels of potatoes were harvested from a 
 garden plat 2 rods by 4 rods. At this rate what would 
 an acre yield? 
 
 9. When a crate of strawberries (16 quarts) sells for 
 $1.92, what is the selling price per quart? 
 
 10. An acre of strawberries yielding 5,120 quarts pro- 
 duces how many bushels? 
 
 11. Twelve tomato plants, occupying one square rod 
 in a garden, yielded 7 bushels of tomatoes. At this rate 
 what would the yield per acre be? Pounds per acre? 
 
206 AGRICULTURAL ARITHMETIC 
 
 12. A small garden plat of potatoes sprayed for blight 
 yielded 21 bushels, while an equal area not sprayed pro- 
 duced 12 bushels. What was the per cent increase in yield 
 due to spraying? 
 
 13. According to U. S. statistics, the average cost to 
 grow an acre of tomatoes in the South for. the canning fac- 
 tory is as follows: $5 for land rental, $3.50 for plowing 
 and fitting the land, $2.25 for cost of plants, $7 for cost of 
 fertilizers, $1.50 for setting the plants, $3 for cultivation, 
 $8 for picking and $10 for hauling to factory. Determine 
 total cost. 
 
 (a) When tomatoes yield 200 bushels per acre and 
 average $12 per ton, what are the receipts 
 per acre? Net profits? (60 lbs. per bushel.) 
 
 WRITTEN PROBLEMS 
 
 1. When pear trees are planted from 20 to 30 feet 
 each way; plums and peaches from 16 to 20 feet each way; 
 and cherries from 16 to 25 feet each way, how many trees of 
 each kind can be planted on an acre? 
 
 2. When apple trees are planted 36 feet apart each 
 way, according to the rectangular system of planting, how 
 many trees can be planted per acre? 
 
 (a) Draw a diagram to a scale of i inch = 36 feet, 
 representing a square plat large enough to 
 accommodate 36 trees according to this sys- 
 tem of planting. Place a dot for each tree, 
 and place no trees on the boundary lines. 
 
 3. If , in problem 2, a tree were planted in the center of 
 each square, how many more trees would the small plat 
 contain? 
 
ORCHARD AND GARDEN 207 
 
 (a) Draw a diagram, as in problem 2 (a), to illus- 
 trate this system of planting which is called 
 the quincunx system. 
 
 (b) How many apple trees could be planted per 
 acre in problem 2, according to this system, 
 when in actual practice about 75% more trees 
 can be planted per acre? 
 
 4. When apple trees are planted, according to the 
 rectangular system, 33 feet each way, how many trees can 
 be planted per acre. How many per acre according to the 
 quincunx system? 
 
 5. Draw a diagram the same size and to the same 
 scale as in problem 2 (a), and place the trees 36 feet apart 
 each way in three directions. How many more trees can 
 be placed on this plat than when they are placed according 
 to the rectangular system of planting? 
 
 (a) This system of planting is called the hexag- 
 onal system and gives about 15% more trees 
 per acre than the rectangular system. Can 
 you determine why this is called the hexag- 
 onal system? 
 
 (b) Determine the distance between the rows. 
 
 6. When young prune trees are planted 16£ feet apart 
 each way, how many are planted per acre, according to the 
 rectangular system? According to the quincunx system? 
 According to the hexagonal system? 
 
 7. Determine the amount of each material required 
 to make 350 gallons of Bordeaux mixture* to be used as a 
 fungicide in spraying an orchard. 
 
 *Bordequx mixture for spraying is made aa follows, according to one of the 
 methods: Dissolve 4 pounds of blue vitriol (copper sulphate) in 10 gallons of water 
 in a wooden or earthen vessel. In another vessel slake 4 pounds of good, fresh 
 quicklime in 10 gallons of water. Add enough water to make 25 gallons of each 
 solution. When mixture is wanted pour the blue vitriol solution and the milk of 
 lime slowly and at the same time into a barrel, stirring all the time. 
 
208 AGRICULTURAL ARITHMETIC 
 
 8. 360 gallons of kerosene emulsion f are to be used for 
 scale insects and plant lice. Determine the amount of 
 material required to make this amount for summer spray- 
 ing. For winter spraying. 
 
 9. In West Virginia an orchard consisting of 600 trees 
 and neglected for 17 years became almost obscure by growth 
 of brush and saplings. The new owner had it rejuvenated. 
 The total expenses and receipts for the first three years 
 were as follows: 
 
 EXPENSE 
 
 Cost of trimming and clearing up land averaged 16f 
 cents per tree. 
 
 200 loads of manure and straw at 75 cents per load. 
 
 Cost of picking and hauling apples averaged 25 cents 
 per tree. 
 
 Cost of 833 barrels at 37 cents. 
 
 RECEIPTS 
 
 1st year: Cash for sale of fruit % 400.00 
 
 2nd year: No crop 
 
 3rd year: Cash for sale of fruit 1,431.75 
 
 (a) Determine the total ■ expenses for the three 
 years. 
 
 (b) Determine the total net profits from this re- 
 juvenated orchard during the first three years. 
 
 (c) What was the average annual cost per tree? 
 Average yearly net profits? (Interest on in- 
 vestment not included in cost.) 
 
 t Kerosene Emulsion: Dissolve H pound common hard soap in one gallon of 
 hot water, add 2 gallons of kerosene and churn together until a white creamy mass 
 is formed which thickens on cooling. For summer spraying dilute to 27 gallons 
 and for winter spraying, dilute to about 13 gallons before using. 
 
ORCHARD AND GARDEN 
 
 209 
 
 (d) If the trees covered 10 acres of land valued at 
 
 $300, what was the interest realized on the 
 
 investment for the three years? If land value 
 
 was $85 per acre? 
 
 10. In New York* an account was kept with an apple 
 
 orchard for 10 years to determine profits. The annual 
 
 yield of fruit from this orchard for 10 years was as follows: 
 
 
 Per tree 
 
 Per acre 
 
 Year 
 
 
 
 
 
 
 
 
 Barreled 
 
 Culls and 
 
 
 Barreled 
 
 Culls and 
 
 
 
 apples 
 
 drops 
 
 Total yield 
 
 apples 
 
 drops 
 
 Total yield 
 
 1904 
 
 2.45 
 
 2.13 
 
 
 66.53 
 
 58.08 
 
 
 1905 
 
 1.42 
 
 0.74 
 
 
 38.59 
 
 20.12 
 
 
 1906 
 
 2.67 
 
 1.44 
 
 
 72.69 
 
 39.12 
 
 
 1907 
 
 2.41 
 
 0.88 
 
 
 65.53 
 
 23.79 
 
 
 1908 
 
 4.18 
 
 1.41 
 
 
 113.85 
 
 38.25 
 
 
 1909 
 
 2.37 
 
 1.64 
 
 
 64.63 
 
 44.57 
 
 
 1910 
 
 1.92 
 
 0.69 
 
 
 52.21 
 
 18.80 
 
 
 1911 
 
 3.41 
 
 2.19 
 
 
 92.84 
 
 59.60 
 
 
 1912 
 
 3.86 
 
 1.70 
 
 
 105.05 
 
 46.17 
 
 
 1913 
 
 4.41 
 
 1.02 
 
 
 120.00 
 
 27.62 
 
 
 Total 
 
 
 
 
 
 
 
 10 yr. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Determine the total yield of apples per tree 
 
 and per acre for each year. 
 
 Determine -the production of (1) barreled 
 
 apples, (2) culls and drops, and (3) total, for 
 
 ten years, per tree and per acre. Determine 
 
 the 10-year averages. 
 
 The barreled apples were sold for an average 
 
 of $2.61 per barrel, and the culls and drops 
 
 for an average of 72 cents per barrel. What 
 
 were the average annual receipts per tree? 
 
 Per acre? 
 
 *New York, Geneva Bulletin 376. 
 
 (a) 
 (b) 
 
 (c) 
 
210 AGRICULTURAL ARITHMETIC 
 
 11. The average cost to grow and harvest a barrel of 
 apples in problem 10 was as follows: 
 
 Interest on investment at 5% 21.4 cents 
 
 Taxes 1.2 cents 
 
 Tillage 6.3 cents 
 
 Pruning. 3.0 cents 
 
 Spraying . , 9.6 cents 
 
 Cover crops 2.3 cents 
 
 Superintending orchard 25.0 cents 
 
 Picking, packing, sorting, hauling 24.4 cents 
 
 Note: All work was hired done, and workmen fur- 
 nished their own teams and tools. 
 
 (a) What was the total cost to grow and harvest a 
 barrel of apples? Cost per acre? 
 
 (b) " The cost of barrels averaged 36 cents, and the 
 culls and drops were sold without being bar- 
 reled. Determine the average yearly net 
 profits per tree. Per acre. 
 
 (c) What was the average 10-year net income re- 
 ceived per acre from this orchard? What was 
 the annual per cent interest received on the 
 investment? 
 
 12. How much seed potatoes would be required to 
 plant a garden patch 2.5 by 6.4 rods, when ordinary planting 
 requires about 12 bushels per acre, and these are to be 
 planted | thicker than ordinary planting? 
 
 (a) The potatoes are to be treated for scab. One 
 pound (-£ quart) of formalin will make 30 gal- 
 lons of disinfecting solution,* which is enough 
 to treat 50 bushels of potatoes. Determine 
 the amount of formalin and formalin solution 
 required to treat these potatoes. 
 
 (b) How much Paris greenf would be required 
 for each application to kill potato beetles 
 
 The whole potatoes should be soaked 2 hours in the formalin solution, and 
 the seed should be planted within two or three days after treatment, 
 t Paris green may also be applied with Bordeaux mixture. 
 
Orchard and garden 211 
 
 when one pound of Paris green with 3 pounds 
 of lime in 50 gallons of water is sufficient for 
 one acre? How much lime? How much 
 water? 
 
 (c) When 5 gallons of Bordeaux mixture are re- 
 quired on this patch for one spraying for 
 blight, determine the amount of material 
 required, f Write out the full directions for 
 making this amount of Bordeaux mixture. 
 (See footnote.) 
 
 13. ' A country boy moving to town started a garden on 
 a vacant lot. He agreed to pay the taxes for the privilege 
 of selling all crops grown. He raised and sold the following. 
 
 30 lbs. of wax beans at 8c. per lb.; \ bu. late carrots at 
 50c. per bu.; 140 bunches celery at 5c. per bunch; 1,400 
 small cucumbers at 25c. per hundred; 500 medium sized cu- 
 cumbers at 35c. per hundred; 36 large cucumbers at 8c; 
 25 bunches of green onions at 6c; 3 bu. onions at 3c per lb.; 
 93 lbs. peas averaging 9c; 35 bunches radishes at 5c; 72 
 bunches salsify at 6c; 2£ bu. parsnips at 50c per bu.; 186 
 green peppers at 25c per doz.; 150 lbs. early potatoes averag- 
 ing 2.1c per lb.; 660 lbs. late potatoes at 75c per bu.; 32.5 
 lbs. string beans at 10c ; 30 summer squash at 5c ; 5 hubbard 
 squash at 15c; 35 doz. ears sweet corn at 16c per doz.; 8 
 bu. tomatoes averaging 2.1c per lb. 
 
 (a) Determine gross receipts. 
 
 (b) Taxes were $26.05; plowing, $1.25; garden 
 tools, $3.15; seeds, $2.46; miscellaneous, 96c 
 What were the net profits? 
 
 (c) What part of an acre did the boy have in 
 garden when the patch was 3-J- by 6 rods. 
 
 ■{"Bordeaux mixture for potatoes should be made by using 5 pounds each of 
 blue vitriol and quicklime. (See footnote, page 207). 
 
212 AGRICULTURAL ARITHMETIC 
 
 (d) His late potatoes occupied a plat 1.2 X 4.17 
 rods. At what rate did the potatoes yield 
 per acre? (Solve by proportion.) 
 
 (e) The onions were planted in two rows 70 feet 
 long and 16 inches apart. What would the 
 yield have been had there been an acre of 
 such onions? Returns per acre, if sold at 2.5 
 cents per pound? 
 
 Before beginning the next chapter, mention 10 facts of 
 practical value you have learned in working the problems in 
 Chapter XIII. 
 
CHAPTER XIV 
 
 HOUSEHOLD ECONOMY AND HUMAN FEEDING 
 
 The feeding of stock has long been reduced to a science, 
 but scientific human feeding has only recently received much 
 attention from the average housekeeper. The demands of 
 the human body are properly met only when the right kind 
 and amount of food are supplied. To this end an under- 
 standing of the basic principles of nutrition is necessary, 
 together with the relative nutritive values of foods. The 
 right food in right amounts for the least cost must be the 
 slogan of the efficient housekeeper. 
 
 ORAL PROBLEMS 
 
 1. When 2 cupfuls of milk are equivalent to one pint, 
 one and one half cupfuls is what part of a pint? What part 
 of a quart? Gallon? 
 
 2. When 36 cents per dozen is paid for eggs and 3 out 
 of a dozen on an average are bad, what is the real price 
 paid per dozen for good eggs? 
 
 3. 8 to 10 hen eggs are equivalent to a pound. What 
 are 10 dozen eggs worth at 20 cents per pound? 
 
 4. A tablespoon level full of butter contains one half 
 an ounce. How many tablespoonfuls of butter in 4 pounds? 
 
 5. Three teaspoonfuls of liquid equals 1 tablespoonful, 
 and 1 tablespoonful equals one half an ounce. How many 
 teaspoonfuls in 5 pounds of water? 
 
 6. The net weight of a 50-pound sack of flour is 49 
 pounds. When about 3 cupfuls of flour are required for a 
 loaf of bread, how many loaves ought a sack of flour to make? 
 (4 cupfuls of flour = 1 pound.) Value at 5 cents per loaf? 
 
214 AGRICULTURAL ARITHMETIC 
 
 7. When bacon costs 30 cents per pound, and 66f of 
 its weight is lost in cooking, what is the cost per pound of the 
 portion eaten? 
 
 8. When round steak costs 22 cents per pound and 
 about 2% is lost in cooking, what is the cost of the portion 
 consumed? 
 
 9. When round steak containing 18.4% digestible 
 protein can be purchased for 20 cents per pound, and bacon 
 containing 8.8% protein for 25 cents, how much protein can 
 be purchased for one dollar in each case? 
 
 10. Whole milk contains 3.2% digestible protein, and 
 dried beef 25.6%. How much dried beef is equivalent to 
 one quart of milk (2 pounds) in protein value? 
 
 11. Dried white beans contain 2.6% digestible mineral 
 matter (ash) and 17.5% digestible protein; apples contain 
 only 0.2% ash and 0.3% digestible protein. How many 
 pounds of apples are equivalent to 2 pounds of beans in sup- 
 plying digestible mineral matter? In supplying protein? 
 Of what value is ash and protein in nutrition? (See page 237.) 
 
 12. Cheddar cheese contains 27.7% protein and 36.8% 
 fat. Fresh pork (shoulder) 12% protein and 29.8% fat. 
 When cheese can be purchased at 18 cents per pound and 
 pork at 22 cents which is the more economical food from 
 standpoint of nutrients contained? From standpoint of 
 preparation? 
 
 WRITTEN PROBLEMS 
 
 1. According to U. S. investigators, the average annual 
 value of the food purchased and consumed by the average 
 farm* was $150.74; and that which was furnished by the 
 farm was valued at $261.35. 
 
 *In the southern states the value of food products furnished by_ the farm 
 is appreciably higher than in the northern states, owing to the long growing season 
 for vegetables, and to the fact that more meat is furnished by the farm for home 
 consumption. U. S. Farmers' Bulletin, No. 635. 
 
\ 
 
 HOUSEHOLD ECONOMY 215 
 
 (a) What per cent of the total value of the food 
 consumed by the average family was pur- 
 chased? Furnished by the farm? 
 
 (b) The average size of the families in this investi- 
 gation was 4.6 persons. What was the annual 
 consumption of food per person? 
 
 (c) It was further found that the yearly cost of 
 food per person in families of two and three 
 averaged $108; and in families of six persons 
 and over, $80. What per cent less was the 
 food-cost per person in the larger families 
 than in the smaller families? What per cent 
 greater was the food-cost per person in the 
 smaller families than in the larger? 
 
 2. When 1.1 pounds of a 4-pound chicken is refuse 
 (bones, head, etc.) what greater per cent of edible portion is 
 obtained when a 5^-pound chicken is purchased? (Assume 
 same amount of refuse.) 
 
 3. When ribs (beef) are purchased at 22 cents per pound, 
 and 20.8% is refuse, what is the price paid per pound for the 
 edible portion? 
 
 4. When round steak (medium) is purchased at 22 
 cents per pound, and 7.2% of it is refuse, what is the price 
 paid per pound for the edible portion? 
 
 5. Determine the material and total cost of making 
 
 6 loaves of slow process bread when the following recipe is 
 
 used: 
 
 3 cupfuls scalded milk 3 tablespoonfuls lard 
 
 3 cupfuls cold water 3 teaspoonfuls salt 
 
 3 tablespoonfuls sugar f cake compressed yeast 
 
 About 18 or 19 \ cupfuls flour | cupful lukewarm water 
 
 Note : Consult table, page 246, for household weights and 
 measures. 
 
216 AGRICULTURAL ARITHMETIC 
 
 6. Determine the material and total cost in making a 
 one-crust lemon pie when the following recipe is used: 
 
 1 cupful of sugar 4 tablespoonfuls of lemon juice, 
 
 2 egg yolks grated rind of one lemon 
 
 3 tablespoonfuls ol corn starch 1 teaspoonful of butter 
 
 1 cupful of boiling water 
 
 FROSTING (Meringue) 
 
 2 egg whites \ teaspoonful of lemon juice 
 
 2 tablespoonfuls of powdered sugar \ teaspoonful of vanilla 
 
 7. Determine the material and total cost of making a 
 
 Spanish chocolate layer cake, as follows: 
 
 i cupful of butter 1 teaspoonful of vanilla 
 
 1| cupfuls of sugar 2 squares of bitter chocolate 
 
 4 eggs 5 tablespoonfuls of boiling water 
 i cupful of milk If scant cupfuls of flour 
 
 4 teaspoonfuls baking powder 
 
 CARAMEL FILLING 
 
 2 cupfuls of dark brown sugar \ cupful of cream 
 
 1 cupful of white sugar \ cupful of butter 
 
 1 cupful of hot water 
 
 8. When 20 pounds of fish (mackerel) may be had at 
 15 cents per pound, determine the amount of edible portion 
 purchased. (Table on page 237.) 
 
 (a) Determine real cost of edible portion. 
 
 9. If each person in a family of 6 averages, 10 ounces 
 of potatoes (edible portion) per day, how many bushels 
 would be consumed in 3 months? 
 
 10. Ten bushels of tomatoes contain how many gallons 
 of water? (60 pounds = 1 bushel of tomatoes.) 
 
 11. Determine the total amount of edible portion and 
 digestible nutrients (protein, carbohydrates and fat) in 25 
 pounds of cabbage. 
 
 Solution: 
 
 Step 1: 
 
 15% of the cabbage is refuse, hence 85% is edible. 
 
 .85 X 25 lbs. = 21.25 lbs. edible portion in 25 lbs. 
 
HOUSEHOLD ECONOMY 217 
 
 Step 2: 
 
 .012 X 25 lbs. = .3 lb. digestible protein. 
 .046 X 25 lbs. = 1.15 lb. digestible carbohydrates. 
 .002 X 25 lbs. = ,051b. d igestible fat. 
 
 1.50 lbs. total digestible nutrients. 
 
 12. Determine the total amount of digestible nutrients 
 (protein, carbohydrates and fat) contained in 6 pounds of 
 canned salmon, 15 pounds of cabbage and 12 pounds of but- 
 ter. (Table, page 237.) 
 
 13. Compare the total amount of digestible nutrients 
 contained in 25 pounds each of apples and bananas. At 
 local prices, which fruit will furnish the greater amount of 
 nutrients for one dollar? 
 
 14. A housewife purchased 8 dozen eggs at 20 cents 
 per dozen. Would she have purchased more digestible 
 nutrients had she invested the same amount of money in 
 beef (shoulder) at 16 cents per pound? Determine the 
 difference. 
 
 15. Which food will give the most digestible nutrients 
 for $2, and how much more: veal (shoulder) at 18 cents, 
 beef (neck) at 14 cents, or limburger cheese at 20 cents per 
 pound? 
 
 16. Nine eggs (1 pound) are equivalent in digestible 
 nutrients to how many pounds of bananas? 
 
 Solution: 
 Step 1: 
 
 .127 X 16 oz. = 2.03 oz. digestible protein. 
 .088 X 16 oz. = 1.41 oz. digestible fat. 
 
 3.44 oz. = total digestible nutrients in 1 lb. of eggs. 
 Step 2: 
 
 .007 X 16 oz. = .112 oz. digestible protein. 
 
 .129 X 16 oz. = 2.064 oz. digestible carbohydrates. 
 
 .004 X 16 oz. = .064 oz. digestible fat. 
 
 2.24 oz. total digestible nutrients in 1 lb. of 
 bananas. 
 
218 
 
 AGRICULTURAL ARITHMETIC 
 
 Since 1 lb. of eggs contains 3.44 oz. of digestible nutrients and 
 bananas 2.24 oz., 1 lb. of eggs is equivalent to 1.5 lbs. of bananas in 
 total nutrients. (3.44 -r- 2.24. = 1.5+) 
 
 17. One quart of whole milk (2 pounds) is equivalent 
 
 in total digestible nutrients to how many pounds or ounces 
 
 of eggs, cheddar cheese, ham (fresh), peanuts, potatoes, 
 
 apples and wheat bread? 
 
 (a) At local prices determine the cost of the quart 
 
 of milk and the equivalents of the other foods. 
 
 Dietary Standards — ■ Daily Requirements. 
 
 Age and Work Performed 
 
 Digestible Nutrients, Pounds 
 
 Nutri- 
 tive 
 
 
 Protein 
 
 Carbohydrates 
 
 Fat 
 
 Ratio 
 
 Children 6-15 years (average).. 
 
 Boy 14-16 years 
 
 Girl 14-16 years 
 
 Man, light muscular work (At- 
 
 Man, moderate work (Atwater) 
 Man, hard work (Atwater) 
 
 0.16 
 0.22 
 0.19 
 
 0.22 
 0.28 
 0.39 
 
 0.71 
 1.00 
 0.89 
 
 0.77 
 0.99 
 1.43 
 
 0.10 
 
 0.1 
 
 0.09 
 
 0.22 
 0.28 
 0.55 
 
 1 :5.2 
 1 :5.5 
 1:5.7 
 
 1:5.7 
 1:5.8 
 1 : 6.9 
 
 18. What per cent greater amount of nutrients does a 
 man at hard muscular work require than one at light work? 
 A boy than a girl? 
 
 19. In one day a boy 15 years old ate 2 ounces of grape- 
 nuts, 1 pound of wheat bread, 1 quart of whole milk, 5 table- 
 spoonfuls of granulated sugar and £ of a pound of butter. 
 
 (a) Determine the amounts of digestible protein, 
 carbohydrates and fat. in each food eaten. 
 Total amounts. 
 
 (b) Did the boy eat enough to satisfy his require- 
 ments? Explain. 
 
 20. What per cent more digestible protein and ash has 
 whole wheat bread than white bread? 
 
HOUSEHOLD ECONOMY 219 
 
 21. What per cent more digestible fat is contained in 
 Brazil nuts than in tenderloin? 
 
 22. A bowl of bread-and-milk consisting of a pint of 
 milk and 4 ounces of wheat bread is equivalent in total 
 nutrients to how many ounces of "grape-nuts'' with 2 table- 
 spoonfuls of sugar (granulated) and 6 ounces of cream? 
 
 (a) Are "grape-nuts" as nourishing as you sup- 
 posed they were? Compare the cost of these 
 two dishes when milk costs 7 cents per quart, 
 bread 5 cents per loaf (1 pound), sugar 6 cents, 
 cream 30 cents per quart, and "grape-nuts" 
 12.4 cents per pound. If an ounce of "grape- 
 nuts" is eaten, what would you suggest in 
 addition to equal food value of the bread-and- 
 milk? 
 
 23. Of the common foods mentioned in table, page 237, 
 which will give the greatest amount of digestible protein 
 for one dollar at local prices? The greatest amount of 
 carbohydrates? The greatest amount of fat? The greatest 
 amount of total nutrients? The greatest amount of mineral 
 matter (ash), excepting the salted articles? 
 
 24. Daily Ration No. 1. Daily Ration No. 2. 
 1J lbs. of potatoes* 1 lb. of potatoes 
 
 i lb. of rice 1{ lbs. of wheat bread 
 
 l| lbs. of wheat bread i lb. of cheese (cheddar) 
 
 | lb. of butter 2 oz. of dried beef 
 
 i lb. of sugar (gran.) | lb. of pork chops 
 
 | lb. of cornmeal (corn cake) 3 tablespoonfuls of sugar (gran.) 
 
 | qt. of whole milk f qt. of milk (whole) 
 
 J lb. of pork chops J lb. of butter 
 
 \ lb. of apricots \ lb. of cabbage 
 
 | lb. of macaroni \ lb. of apricots 
 
 3 oz. of white beans 
 J lb. of bananas 
 
 Note: Pupils can work in pairs or groups of four on long 
 problems like these. 
 
 *A11 amounts refer to digestible portion. 
 
220 AGRICULTURAL ARITHMETIC 
 
 (a) Determine the amounts of digestible protein, 
 carbohydrates and fat in each article in the 
 two rations. 
 
 (b) Determine the total amounts of digestible 
 nutrients in each ration. 
 
 (c) Determine the nutritive ratio of each ration. 
 
 (d) Which is the better ration for a man at hard 
 work? Explain. Suggest ways of improving 
 ration No. 1. 
 
 (e) Determine the amount of digestible mineral 
 matter in each ration. 
 
 25. Make a list of 15 foods (salted foods excepted), 
 containing comparatively large amounts of mineral matter 
 (ash). Should we give much attention to the ash content 
 of the foods we eat? What is "rickets"? 
 
 Important Facts: 
 
 1. The most desirable foods are those which furnish 
 the most easily digestible nutrients at least cost. 
 
 2. Beans, peas and oat meal are rich in protein and are, 
 therefore, especially valuable foods. 
 
 3. Proper cooking renders foods more digestible. 
 
 4. An individual should eat foods adapted to his pe- 
 culiar requirements, and should avoid those that do not 
 "agree" with him. 
 
 5. Too much food is as bad as too little. 
 
CHAPTER XV 
 MISCELLANEOUS PROBLEMS 
 
 1. The ground floor in a round barn measures 50 feet in 
 diameter. An 18-foot silo (19 feet outside measure) is in 
 the center. How many square feet are used for stalls, 
 walks and feeding space? 
 
 2. A man paid $57.16 taxes on land including the W \ 
 of the SW i (valued at $56.25 per acre) and the NE \ of the 
 SW i (valued at $45), Sec. 15, T 5 N, R 19 E. 
 
 (a) What rate of taxes did he pay? 
 
 (b) Draw a diagram showing the location of this 
 land within the section, location of section in 
 township and relation of township with refer- 
 ence to base line and principal meridian. 
 
 3. Mr. Doe paid $27 taxes on a vacant city lot 60' X 
 120'. The rate was 1J- cents on a dollar. Determine the 
 valuation of the lot. 
 
 (a) What would the taxes be on an acre of such 
 lots? 
 
 4. Corn is planted in hills 3 feet 8 inches apart each way 
 in a square 10-acre field. How long is the longest diagonal 
 row? 
 
 Figure 36. — Types of Milk Paila. 
 221 
 
222 
 
 AGRICULTURAL ARITHMETIC 
 
 (a) If the corn were planted in a field 20 rods X 
 80 rods, how long would be the longest diag- 
 onal row? 
 
 5. Bergy determined that milk from the udder averages 
 400 bacteria per cubic centimeter (1 c.c. = .061 cubic inch). 
 After passing the strainer it contains 60,000 per c.c. How 
 many 'times more bacteria in the strained milk than that 
 from the udder? What per cent greater? 
 
 6. Milk from an ordinary cleanly, carefully kept cow 
 may have no more than from 10,000 to 50,000* bacteria per 
 c.c. at the outset; but this milk, carelessly handled, may, in 
 
 Figure 37. — Sediment From Open and Small-Top Pails. The open pail is a 
 good dirt catcher in the hands of careless milkers. This explains in 
 part why some milk is not good. 
 
 the course of one or two days, have numbers as high as 
 100,000,000 per c.c. Determine the possible increase per 
 minute in the number of bacteria per c.c. of milk. 
 
 7. Draw a diagram of your school grounds, to some 
 convenient scale showing buildings, etc. 
 
 8. A farmer applied 1,000 pounds of rock phosphate, 
 containing 13.5% phosphorus, to an acre of silt loam analyz- 
 
 *There is plenty of room in milk for these many minute organisms, when we 
 consider that a space the size of a pin head may hold 8,000,000,000 of them. 
 
MISCELLANEOUS PROBLEMS 
 
 223 
 
 ing .04% phosphorus (P). How much increase was made 
 in the per cent of phosphorus in the soil? (See problem 2, 
 page 120.) 
 
 9. A man sold $3,200 worth of fruit from 12 acres of 
 orchard. Each dollar he received for the fruit cost him 40 
 cents. What were his total net profits? Profits per acre? 
 10. A farmer finding that his fanning-mill blew away- 
 half of his oats, decided to feed the oats to his horses with- 
 out fanning them. He fed the usual allowance of 12 quarts 
 per horse per day. Out of how many pounds of digestible 
 nutrients did he cheat each horse per day? 
 
 Figure 38. — A Good Start for a Future Agriculturist. This boy has an am- 
 bition which will not only affect his pigs, but will some day extend to the 
 broad fields around him. Courtesy of Country Gentleman. 
 
 11. Suppose the boy in Figure 38 sells his two pigs when 
 
 they average 200 pounds and succeeds in "making" pork 
 
 at the rate of 10 pounds per bushel of corn. How many 
 
 bushels of corn would he feed when each pig gains 75 pounds? 
 
 (a) When the pigs cost him 6 cents per pound at 
 
 the beginning of the feeding, and he receives 
 
 $7.50 per hundred, determine his profits. 
 
224 AGRICULTURAL ARITHMETIC 
 
 (b) The corn he fed was his own raising from a 
 strip 4 rods X 7.4 rods. At what rate did 
 his corn yield per acre, if he sold the pigs wheD 
 his corn gave out? 
 
 (c) What would be his gross income resulting 
 from the raising of corn and the feeding of it? 
 At this rate what would the receipts be on the 
 basis of an acre? 
 
 (d) Would he have done better had he sold his 
 corn at 65 cents per bushel? What would he 
 realize per bushel in feeding it to his pigs? 
 
 12. Write out a formula for making one gallon of kero- 
 sene emulsion. (See footnote, page 208.) 
 
 13. Write out a formula for making 10 gallons of Bor- 
 deaux mixture. (See footnote, page 207.) 
 
 14. It costs about 12 cents to set a fence-post and put 
 on two or three wires. The average life of untreated cotton- 
 wood is 4 years. When the original cost per post is 6 cents 
 determine the total cost of setting and renewals per post 
 for 20 years. Determine the cost of setting and maintain- 
 ing a mile of fence for 20 years. Posts set a rod apart. 
 
 (a) These posts can be made to last the 20 years, 
 if treated at a cost of 15 cents per post. 
 Determine the saving. 
 
 15. The following is a bill of lumber for a certified dairy 
 barn in Kentucky: 
 
 34 sills 6 X 8 X 12J 40 joists 2 X 
 
 20 posts 5 X 5 X 12 80 joists 2 X 
 
 12 posts 6X6X7 80 rafters 2 X 
 
 14 top posts 5 X 5 X 13 80 rafters 2 X 
 
 90 streamers and 30 braces 2 X 
 
 plates 2 X 5 X 12 Siding, 3,300 feet. . . 1 X 
 
 14 girders 6 X 6 X 12 270 strips | X 
 
 16 plates 4 X 6 X 12 50 pieces cornice. ...IX 
 
 8 ties 2 X 6 X 10 Sheathing 4,000 feet 
 
 8 X 10 
 
 8 X 12 
 
 5 X 16 
 
 5X8 
 
 5 X 14 
 
 12 X 12 
 
 4 X 12 
 
 7 X 12 
 
MISCELLANEOUS PROBLEMS 225 
 
 Determine the total number of feet of lumber. (See 
 page 51.) 
 
 16. A farmer had a 12' X 36' silo (inside measurements). 
 If 1£ inches of silage should be removed from the top 
 to insure good silage, how many cows ought he to feed, 
 when 40 pounds are fed per animal per day? Calculate 
 average weight of silage at 40 pounds per cubic foot. 
 
 (a) How long will the silage last at this rate of 
 feeding when there are 33 feet of silage to be 
 fed? 
 
 17. At the West Virginia Experiment Station* a record 
 of 600 hens (Single Comb White Leghorn pullets at start) 
 was kept for one year, as follows: 
 
 Amounts and Cost of Feed Consumed 
 
 Cornmeal 3,441 lbs $42.88 
 
 Wheat Bran 5,434 lbs 71.74 
 
 Wheat Middlings 2,932 lbs 39.82 
 
 Oil Meal 950 lbs 18.72 
 
 Wheat 10,509.6 lbs 152.21 
 
 Corn 8,952 lbs 103.81 
 
 Beef Scrap 2,443 lbs^ 57.70 
 
 Green Cut Bone 178 lbs 1.33 
 
 Ensilage 1,536 lbs 3.75 
 
 Rye 1,120 lbs... 16.80 
 
 Ground Oats 337 lbs 5.73 
 
 Oyster Shell \ . . 1,510 lbs 8.30 
 
 Miea Crystal Grit 1,400 lbs 7.70 
 
 Skim Milk 205 gals 4.10 
 
 Number and Value of Eggs Produced 
 November 85| doz. @ 35c. 
 
 December 281 
 
 January 3873^ 
 
 February 481*1 
 
 March, 1-10 231& 
 
 March, 10-15 125f 
 
 March, 15-31 459| 
 
 April 73711 
 
 May 722/j 
 
 *Weat Virginia Experiment Station, Bulletin No. 115. 
 
 40c. 
 35c. 
 30c. 
 30e. 
 22c. 
 20c. 
 20c. 
 20c. 
 
 J5— 
 
a a 
 
 20c. 
 
 tt tt 
 
 25c. 
 
 a tt 
 
 25c. 
 
 tt u 
 
 25c. 
 
 it it 
 
 28c. 
 
 tt it 
 
 30c. 
 
 tt u 
 
 32c. 
 
 tt tt 
 
 35c. 
 
 it tt 
 
 40c. 
 
 226 AGRICULTURAL ARITHMETIC 
 
 June, 1-14 295H 
 
 June, 14-30 321J 
 
 July 552| 
 
 August 435ft 
 
 August 72| 
 
 September 264 
 
 September 62£ 
 
 October 46| 
 
 October 82J 
 
 (a) Determine the total cost of feed. 
 
 (b) Determine the value of the eggs for each month 
 
 and total for year. 
 
 (c) How many eggs were produced in all? Aver- 
 age per fowl? Average value of eggs per 
 fowl? 
 
 (d) The other expenses were $120 for labor, $36 
 for fowls which died, $100 for depreciation and 
 $66 for interest on investment. What were 
 the total net profits? Average per hen? 
 
 (e) How much money was represented in the 
 investment? 
 
 18. Which weighs the more and how much more; 
 10 pounds of cotton or 10 pounds of gold? (See page 245). 
 
 19. Make out a list of 15 groceries and amount of each 
 that may be purchased at local prices for the money spent 
 during a year for liquor or tobacco at the rate of 15 cents 
 per day. 
 
 20. Land purchased at $175 per acre produced a net 
 profit in crop value of $20 per acre. (Land rental of 5% 
 was charged against the crop.) What per cent income was 
 realized from this land investment? 
 
 (a) This investment is equal to how much money 
 let out at 5%? At 6%? 
 
 21. Mr. B learned that rock phosphate containing 
 about 32% "phosphoric acid," (P2O5), could be purchased 
 
Miscellaneous problems 227 
 
 for about $10 per ton (including freight). When he wanted 
 to buy some and was offered a good grade of this fertilizer 
 having a guarantee of 13.5% phosphorus (P), at $9 per ton, 
 he suspected dishonesty. 
 
 Write a letter to this man explaining that the price was 
 reasonable. (See page 155.) 
 
 22. In'a thousand parts by weight of a fertile silt loam 
 soil, are how many parts of the element phosphorus? (See 
 page 132.) If this portion were removed could any plant 
 grow in that soil? 
 
 23. Four pounds of shelled corn is equal to what deci- 
 mal part of a bushel of cured ear corn? 
 
 24. Tuberculosis, which claims more victims than any 
 other disease in the United States, was the cause of 147,200 
 deaths in 1910*. Out of 100,000 population how many died 
 of this disease during that year? (See page 79, problem 6.) 
 
 (a) In 1900 when the population of the United 
 States was 76 millions, 204 out of 100,000 
 died of tuberculosis. Determine the number 
 of deaths resulting from this cause in 1900. 
 
 (b) Determine the per cent of decrease in the 
 number of deaths resulting from this White 
 Plague between the years 1900 and 1910. 
 
 25. A soft water faucet leaked 48 drops a minute. How 
 many gallons of water leaked away during 24 hours? (A 
 drop of water weighs approximately .14 grams.) (See page 
 245.) 
 
 (a) The faucet leaked for 6 weeks before it was 
 repaired. How many barrels of water were 
 lost through leakage? 
 
 26. If cow manure were reduced to the same water 
 content as sheep manure, how would the per cents of the 
 fertility elements compare? (See page 149.) 
 
 ♦Digest of 1910, U. S. Census. 
 
228 
 
 AGRICULTURAL ARITHMETIC 
 
 Figure 39. — Finderne Pride Johanna Rue (Holstein,) once the champion butter 
 cow of the world. She completed her 365-day record, June, 1915, after 
 producing 28,403.7 lbs. milk containing 1,176.47 lbs. butter-fat. 
 
 27. When Finderne Pride Johanna Rue, (Holstein), 
 completed her 365-day record, June, 1915, she became the 
 world's champion dairy cow,* producing in one year 28,403.7 
 pounds of milk, containing 1,176.47 pounds of butter-fat. 
 What was the average test of her milk? 
 
 (a) How many more pounds of butter-fat were 
 produced by this cow than by the world's 
 champion before her? (See page 89, prob- 
 lem 14.) 
 
 (b) 1,176.47 pounds of butter-fat is equivalent to 
 how many pounds of 85% butter? (Assum- 
 ing no loss of fat.) Value of butter at 30 cents 
 per pound? Value of skim milk at 20 cents per 
 hundred? Total value of dairy products? 
 
 *Finderne Pride Johanna Rue lost her world's championship to Duchess Sky- 
 lark Ormsby, (Holstein), Nov. 1915. The latter produced 1205.09 pounds of but- 
 ter-fat in 365 days. 
 
MISCELLANEOUS PROBLEMS 229 
 
 (c) How many cows averaging 30 pounds of milk 
 testing 3.5% and milking for 270 days, are 
 required to equal the yearly production of 
 Finderne Pride Johanna Rue? 
 
 28. What per cent more digestible mineral matter has 
 whole wheat flour than white wheat flour? 
 
 29. What per cent of the protein in timothy hay is 
 digestible? In clover hay? In alfalfa hay? 
 
 30. A farmer agreed to let his son Henry have all the 
 profits from the sale of 2 acres of corn, with the understand- 
 ing that Henry pay his father $10 per acre as cost to raise 
 the corn (excluding Henry's time) and also pay 15 cents per 
 pound for the nitrogen, 10 cents for the phosphorus and 6 
 cents per pound for the potassium removed by the crop. 
 The corn yielded 75 bushels of shelled corn and 2.02 tons of 
 stover per acre. The corn was sold for 60 cents per bushel 
 and the stover for $2.75 per ton. How much money did 
 Henry realize from his transaction? 
 
 31. On one of two plots of acid silt loam were applied 
 3 tons of pulverized limestone per acre. (1914.) Red 
 clover was seeded. The following year 50.5 pounds of green 
 clover containing 76.6% water were cut from an average 
 square rod on the unlimed plot, and 69 pounds containing 
 75.5% water were cut per square rod from the limed area. 
 Determine the production of dry matter per acre on each 
 plot. What was the per cent increase due to liming? 
 
 32. Eighty acres of prairie soil rented to a large grain 
 farmer produced in 10 years the following crops: 
 
 Average Yield 
 Crops Acres per Acre 
 
 Corn 160 40 bu. 
 
 Oats 120 40 " 
 
 Wheat 180 20 " 
 
 Barley 100 35 " 
 
 Clover 80 1.5 tons 
 
 Timothy 160 2 " 
 
230 
 
 AGRICULTURAL ARITHMETIC 
 
 All crops were taken from the "80" and no fertility re- 
 turned. At the beginning of the 10-year lease the land was 
 valued at $175 per acre, and during the rental period land 
 value increased 20%. Was this increase sufficient to offset 
 the value of the fertility removed by the crops? Consider 
 value of nitrogen at 15 cents, phosphorus at 10 cents and 
 potassium at 6 cents per pound. 
 
 33. If on the "80" in problem 32, 30 acres of corn, 30 
 of oats and 20 of clover were raised in a year, yielding as 
 above, and all crops fed on the "80," and manure hauled 
 directly to the land, how much fertility (nitrogen, phos- 
 phorus and potassium) would be lost from the soil? (As- 
 sume no feeds purchased.) (See page 137.) 
 
 2"' b" Runner 
 
 Figure 40. — Frame Work of an A-Shaped Portable Hog House as Shown in 
 Fig. 41. 
 
MISCELLANEOUS PROBLEMS 
 
 231 
 
 (a) About how many 30-bushel wheat crops raised 
 and sold (straw and grain sold) would cause an 
 equal loss of phosphorus? Of nitrogen? Of 
 potassium? 
 
 34. Obtain the local prices per M feet of the materials 
 in the following bill of lumber necessary to construct an 
 A-shaped portable hog house* as shown in Figures 40 and 41, 
 and determine the total cost of the lumber. 
 
 9 pieces 1" X 12" X 16' and 11 O. G. batten 16 ft. long for roof. 
 3 pieces 2" X 6" X 16' and 11 O. G. batten 16 ft. long for roof. 
 5 pieces 1" X 12" X 14' for ends. 
 
 1 piece 2" X 4" X 10' for ridge. 
 
 2 pieces 2" X 8" X 10' for plates. 
 
 7 pieces 2"" X 4" X 16' for rafters and braces in frame. 
 
 3 pieces 2" X 6" X 8' for runners. 
 
 4 pieces 1" X 12" X 16' rough for flooring. 
 
 35. A 2-12-8 
 fertilizer costing 
 $30.50 per ton 
 was applied at 
 the rate of 1,200 
 pounds per acre 
 ^once in 3 years 
 on some long- 
 cropped, silt 
 loam soil. The 
 value of the in- 
 creased crops 
 covering 3 years 
 was estimated at $52.15 per acre. When the same fertilizer 
 was applied at the rate of 600 pounds once in 3 years on 
 the same soil, the value of the increased crops was $31.57 
 per acre. 
 
 ♦Wisconsin Experiment Station, Bulletin No. 242. 
 
 Figure 41. — A-Shaped Portable Hog House. 
 
232 AGRICULTURAL ARITHMETIC 
 
 (a) Determine the per cent of nitrogen, phos- 
 phorus and potassium in the fertilizer used. 
 (See page 154 and 155.) How many pounds 
 of each element in a ton? 
 
 (b) Determine the per cent profit* on cost of ferti- 
 lizer invested above cost of fertilizer in each 
 case. 
 
 (c) From the standpoint of available money, 
 under these conditions, which would be better 
 for a man of small means — to apply the ferti- 
 lizer at the rate of 1,200 or 600 pounds per acre? 
 For the man of ample means? 
 
 (d) Under similar conditions would it be more 
 profitable for a man of small means to borrow 
 for 2 years at 6% all money used to purchase 
 fertilisers, and apply 1,200 pounds per acre 
 than to use his own money for this purpose 
 and apply 600 pounds? 
 
 (e) Does it necessarily follow under like conditions 
 that the more fertilizers applied per acre the 
 greater the amount of profits? 
 
 *It is a well established fact that on soils which respond to fertilizers a greater 
 per cent pro&t is realized from a light application of a required fertilizer than from 
 a heavy application. 
 
APPENDIX 
 
APPENDIX 
 
 235 
 
 Average Composition of Some Common Feeds* 
 Amount in 100 Pounds 
 
 Water 
 
 Ash 
 
 Crude 
 pro- 
 tein 
 
 Carbohy-- 
 drates 
 
 Fiber 
 
 Nitro- 
 gen- 
 free 
 ex- 
 tract 
 
 Alfalfa hay 
 
 Barley (grain) 
 
 Brewers' grains (dried).. . . 
 
 Clover hay, alsike 
 
 Clover hay, crimson 
 
 Clover hay, Japan 
 
 Clover hay, red 
 
 Clover hay, sweet, (white) . 
 Clover and timothy hay 
 
 (mixed) 
 
 Corn, dent 
 
 Corn, flint 
 
 Corn meal 
 
 Corn silage (well matured) 
 Corn stover (ears removed, 
 
 very dry) 
 
 Cottonseed meal (good) 
 
 Cowpeas 
 
 Cowpeas hay 
 
 Germ oil meal (high grade) 
 Gluten feed (high grade) 
 
 Gluten meal 
 
 Johnson grass hay 
 
 Mangels 
 
 Milk, cow's (whole) 
 
 Milk, skim 
 
 Millet hay (common) . . . 
 Mixed grasses, hay ..... 
 
 Oats 
 
 Oats straw 
 
 Oil meal (old process) . . . 
 
 Potatoes 
 
 Rice meal. 
 
 Rye- 
 
 Soybeans 
 
 8.6 
 
 9.3 
 
 7.5 
 
 12.3 
 
 10.6 
 
 11.8 
 
 12.9 
 
 8.6 
 
 12.2 
 10.5 
 12.2 
 11.3 
 73.7 
 
 9.4 
 
 7.9 
 
 11.6 
 
 9.7 
 
 8.9 
 
 8.7 
 
 9.1 
 
 10.1 
 
 90.6 
 
 86.4 
 
 90.1 
 
 14.3 
 
 12.8 
 
 9.2 
 
 11.5 
 
 9.1 
 
 78.8 
 
 9.5 
 
 9.4 
 
 9.9 
 
 8.6 
 2.7 
 3.5 
 8.3 
 8.8 
 5.8 
 7.1 
 7.2 
 
 6.1 
 1.5 
 1.5 
 1.3 
 1.7 
 
 5.8 
 6.4 
 3.4 
 11.9 
 2.7 
 2.1 
 1.1 
 7.5 
 1.0 
 0.7 
 0.7 
 6.3 
 5.6 
 3.5 
 5.4 
 5.4 
 1.1 
 9.1 
 2.0 
 5.3 
 
 14.9 
 11.5 
 26.5 
 12.8 
 14.1 
 12.1 
 12.8 
 14.5 
 
 8.6 
 
 10.1 
 
 10.4 
 
 9.3 
 
 2.1 
 
 5.9 
 
 37.6 
 
 23.6 
 
 19.3 
 
 22.6 
 
 25.4 
 
 35.5 
 
 6.6 
 
 1.4 
 
 3.5 
 
 3.8 
 
 8.3 
 
 7.6 
 
 12.4 
 
 3.6 
 
 33.9 
 
 2.2 
 
 11.8 
 
 11.8 
 
 36.5 
 
 28.3 
 4.6 
 14.6 
 25.7 
 27.3 
 25.9 
 25.5 
 27.4 
 
 29.9 
 2.0 
 1.5 
 2.3 
 6.3 
 
 30.7 
 
 11.5 
 4.1 
 
 22.5 
 9.0 
 7.1 
 2.1 
 
 30.2 
 0.8 
 
 24.0 
 
 28.8 
 
 10.9 
 
 36.3 
 
 8.4 
 
 0.4 
 
 9.3 
 
 1.8 
 
 4.3 
 
 37.3 
 69.8 
 41.0 
 38.4 
 36.9 
 41.6 
 38.7 
 40.1 
 
 40.8 
 70.9 
 69.4 
 72.0 
 15.4 
 
 46.6 
 28.4 
 55.8 
 34.0 
 46.0 
 52.9 
 47.5 
 43.5 
 6.1 
 5.0 
 5.2 
 44.3 
 42.7 
 59.6 
 40.8 
 35.7 
 17.4 
 48.7 
 73.2 
 26.5 
 
 ♦Compiled from Feeds and Feeding, Henry and Morrison, 15th Edition, 1915. 
 
236 
 
 AGRICULTURAL ARITHMETIC 
 
 Average Composition of Some Common Feeds (Cont.) 
 Amount in 100 Pounds 
 
 Feeds 
 
 
 
 
 Carbohy- 
 
 Water 
 
 Ash 
 
 
 drates 
 
 
 Nitro- 
 
 
 
 Crude 
 
 
 gen- 
 
 
 
 pro- 
 
 Fiber 
 
 free 
 
 
 
 tein 
 
 
 ex- 
 tract 
 
 8.6 
 
 8.6 
 
 16.0 
 
 24.9 
 
 39.1 
 
 86.3 
 
 1.1 
 
 1.6 
 
 1.0 
 
 12.6 
 
 7.5 
 
 12.0 
 
 60.5 
 
 4.2 
 
 2.7 
 
 11.6 
 
 4.9 
 
 6.2 
 
 29.9 
 
 45.0 
 
 10.2 
 
 1.9 
 
 12.4 
 
 2.2 
 
 71.2 
 
 10.1 
 
 6.3 
 
 16.0 
 
 9.5 
 
 53.7 
 
 10.4 
 
 4.4 
 
 17.3 
 
 6.0 
 
 57.0 
 
 10.2 
 
 3.9 
 
 13.3 
 
 7.4 
 
 61.1 
 
 92.5 
 
 1.6 
 
 1.0 
 
 
 4.6 
 
 i- Fat 
 
 Soy beans hay 
 
 Sugar beets (roots) 
 
 Tankage (high grade) 
 
 Timothy hay 
 
 Wheat 
 
 Wheat bran 
 
 Wheat middlings (shorts) 
 
 Wheat screenings 
 
 Whey 
 
 2.8 
 0.1 
 13.0 
 2.5 
 2.1 
 4.4 
 4.9 
 4.1 
 0.3 
 
 Haecker's Feeding Standard for the Dairy Cow 
 
 
 
 Daily allowance of digesti- 
 ble nutrients 
 
 
 Crude 
 Protein 
 
 Carbo- 
 hydrates 
 
 Fat 
 
 For support of the 1000-lb. cow 
 
 Lbs. 
 0.7 
 
 0.047 
 0.049 
 0.054 
 0.057 
 0.060 
 0.064 
 0.067 
 
 Lbs. 
 7.0 
 
 0.20 
 0.22 
 0.24 
 0.26 
 0.28 
 0.30 
 0.32 
 
 Lbs. 
 0.1 
 
 To the allowance for support add: 
 For each lb. of 3.0 per cent milk 
 
 0.017 
 
 For each lb. of 3.5 per cent milk 
 
 0.019 
 0.021 
 
 
 0.023 
 
 
 0.024 
 
 
 0.026 
 
 
 Q.028 
 
 
 
APPENDIX 
 
 237 
 
 Digestible Nutrients in Some Common Foods* 
 
 Kind of Food 
 
 Animal Foods 
 Beef, fresh 
 
 Flank 
 
 Sirloin steak 
 
 Neck 
 
 Ribs 
 
 Round. 
 
 Shoulders and clod . . 
 
 Beef, dried 
 
 Dairy Products 
 
 Butter 
 
 Buttermilk 
 
 Cream 
 
 Cheese, cheddar 
 
 Cheese, full cream 
 
 Cheese, limburger. . . . 
 
 Cheese, Swiss 
 
 Milk, whole 
 
 Milk, skim 
 
 Eggs, hens' 
 
 Fish 
 
 Cod, salt 
 
 Mackerel (fresh) 
 
 Salmon, canned 
 
 Lobster 
 
 Oysters 
 
 Mutton 
 
 Flank 
 
 Leg, hind 
 
 Fore quarter 
 
 Hind quarter (without 
 
 tallow) 
 
 Pork 
 
 Bacon, smoked 
 
 Ham, fresh 
 
 Ham, smoked 
 
 Loin chops 
 
 Shoulder 
 
 Shoulder, smoked 
 
 Salt pork 
 
 Tenderloin 
 
 Sausage (pork) 
 
 Per cent 
 
 Refuse 
 
 10.2 f 
 12.8 
 27.6 
 20.8 
 
 7.2 
 16.4 
 
 4.7 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 11.2 
 
 24.9 
 44.7 
 
 0.0 
 61.7 
 
 0.0 
 
 9.9 
 18.4 
 21.2 
 
 17.2 
 
 7.7 
 10.7 
 13.6 
 19.7 
 12.4 
 18.2 
 
 0.0 
 
 0.0 
 
 0.0 
 
 Water 
 
 54.0 
 54.0 
 45.9 
 43.8 
 60.7 
 56.8 
 53.7 
 
 11.0 
 91.0 
 74.0 
 27.4 
 34.2 
 35.7 
 36.1 
 87.0 
 90.5 
 65.5 
 
 40.2 
 40.7 
 63.5 
 30.7 
 88.3 
 
 39.0 
 51.2 
 41.6 
 
 45.4 
 
 17.4 
 48.0 
 34.8 
 41.8 
 44.9 
 36.8 
 7.9 
 66.5 
 39.8 
 
 Per cent Digestible Nutri- 
 ents and Mineral Matter 
 (ash) 
 
 Pro- 
 tein 
 
 16.5 
 16.0 
 14.1 
 13.5 
 18.4 
 16.0 
 25.6 
 
 1.0 
 
 3.0 
 
 2.4 
 
 26.9 
 
 25.0 
 
 33.2 
 
 27.0 
 
 3.2 
 
 3.3 
 
 12.7 
 
 15.5 
 9.9 
 
 21.1 
 5.7 
 
 5.8 
 
 13.4 
 14.6 
 12.0 
 
 13.4 
 
 13.1 
 13.8 
 13.0 
 11.6 
 12.6 
 1.8 
 18.3 
 12.6 
 
 Car- 
 bohy- 
 drates 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 
 0.0 
 4.7 
 4.5 
 4.0 
 2.4 
 2.9 
 2.5 
 5.0 
 5.1 
 0.0 
 
 0.0 
 0.0 
 0.0 
 0.2 
 3.3 
 
 0.0 
 0.0 
 0.0 
 
 0.0 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 
 1.1 
 
 Fat 
 
 18.0 
 15.3 
 11.3 
 20.0 
 12.2 
 9.3 
 6.6 
 
 80.8 
 
 0.5 
 
 17.6 
 
 35.0 
 
 32.0 
 
 23.0 
 
 28.0 
 
 3.8 
 
 0.3 
 
 8.8 
 
 0.4 
 4.0 
 11.5 
 6.7 
 1.2 
 
 35.0 
 14.0 
 23.3 
 
 22.0 
 
 59.1 
 24.6 
 31.7 
 23.0 
 28.3 
 25.3 
 81.9 
 12.4 
 42.0 
 
 ♦Compiled mainly from U. S. Farmers' Bulletin No. 
 tPer cents refer to total amount purchased. 
 
238 
 
 AGRICULTURAL ARITHMETIC 
 
 Digestible Nutrients in Some Common Foods (Continued) 
 
 Kind of Food 
 
 Refuse 
 
 Per cent 
 
 Water 
 
 Per cent Digestible Nutri- 
 ents and Mineral Matter 
 (ash) 
 
 Pro- 
 tein 
 
 Car- 
 bohy- 
 drates 
 
 Fat 
 
 Poultry 
 
 Fowls 
 
 Goose 
 
 Turkey 
 
 Veal 
 
 Leg 
 
 Fore quarter 
 
 Hind quarter 
 
 Vegetable Foods 
 Bread, pastry, etc. 
 
 Cream crackers 
 
 Graham bread 
 
 Rye bread 
 
 White bread (wheat) . 
 
 Whole-wheat bread. . . 
 Breakfast foods 
 
 Corn meal 
 
 Cream of wheat 
 
 Grape nuts 
 
 Oat meal 
 
 Quaker oats. 
 
 Flour, rice, etc. 
 
 Buckwheat flour 
 
 Graham -flour 
 
 Macaroni. . .: 
 
 Popcorn, popped 
 
 Rice 
 
 Rye flour 
 
 Tapioca 
 
 Wheat flour 
 
 Whole-wheat flour 
 
 Fruits 
 
 Apples 
 
 Apples, dried 
 
 Apricots, dried 
 
 Bananas 
 
 Figs, dried 
 
 Grapes 
 
 Oranges 
 
 Raisins 
 
 25.9 
 17.6 
 22.7 
 
 14.2 
 24.5 
 20.7 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 0.0 
 
 25.0 
 
 0.0 
 
 0.0 
 
 35.0 
 
 0.0 
 
 25.0 
 
 27.0 
 
 10.0 
 
 47.1 
 38.5 
 42.4 
 
 60.1 
 54.2 
 56.2 
 
 6.8 
 35.7 
 35.7 
 35.3 
 38.4 
 
 12.5 
 8.8 
 6.2 
 7.8 
 7.8 
 
 13.6 
 11.3 
 10.3 
 4.3 
 1.2 
 12.9 
 11.4 
 12.0 
 11.4 
 
 63.3 
 28.1 
 29.4 
 48.9 
 18.8 
 58.0 
 63.4 
 13.1 
 
 13.3 
 13.0 
 15.6 
 
 15.0 
 14.6 
 15.7 
 
 8.2 
 7.5 
 7.6 
 7.8 
 8.2 
 
 7.8 
 11.1 
 10.2 
 14.2 
 12.5 
 
 5.4 
 
 11.3 
 
 11.4 
 
 9.1 
 
 6.8 
 
 5.8 
 
 0.3 
 
 9.7 
 
 11.7 
 
 0.3 
 1.4 
 3.9 
 0.7 
 3.7 
 0.9 
 0.5 
 2.0 
 
 0.0 
 0.0 
 0.0 
 
 0.0 
 0.0 
 0.0 
 
 68.3 
 50.5 
 51.6 
 52.0 
 48.3 
 
 73.9 
 74.7 
 77.7 
 64.9 
 68.4 
 
 76.1 
 70.0 
 72.6 
 77.1 
 77.4 
 77.1 
 86.2 
 73.6 
 70.5 
 
 9.7 
 59.5 
 61.9 
 12.9 
 66.8 
 13.0 
 
 7.7 
 61.7 
 
 11.7 
 28.3 
 17.5 
 
 7.5 
 5.7 
 6.3 
 
 10.9 
 1.6 
 0.5 
 1.2 
 0.8 
 
 1.7 
 5.4 
 5.5 
 6.6 
 5.6 
 
 1.1 
 2.0 
 0.8 
 4.5 
 0.3 
 0.8 
 0.1 
 0.9 
 1.8 
 
 0.3 
 2.0 
 0.9 
 0.4 
 0.3 
 1.1 
 0.1 
 2.7 
 
APPENDIX 
 
 239 
 
 Digestible Nutrients in Some Common Foods (Continued) 
 
 Kind of Food 
 
 Per cent 
 
 Refuse 
 
 Water 
 
 Per cent Digestible Nutri- 
 ents and Mineral Matter 
 (ash) 
 
 Pro- 
 tein 
 
 Car- 
 bohy- 
 drates 
 
 Fat 
 
 Min- 
 eral 
 
 Mat- 
 ter 
 
 Nuts 
 
 Almonds 
 
 Brazil nuts 
 
 Filberts 
 
 Pecans, polished 
 
 Peanuts 
 
 Walnuts, English 
 
 Vegetables 
 
 Beans, white dried 
 
 Beans, string 
 
 Beans, baked 
 
 Beets 
 
 Cabbage 
 
 Celery..' 
 
 Corn, green (edible part) 
 
 Lettuce 
 
 Onions 
 
 Peas, green, shelled. . . 
 
 Potatoes 
 
 Spinach 
 
 Tomatoes 
 
 Miscellaneous 
 
 Cocoa (drink made with 
 milk) 
 
 Coffee (drink) 
 
 Tea (drink) 
 
 Soups 
 
 Beef 
 
 Tomato 
 
 Sugar (gran.) 
 
 Potato chips 
 
 45.0 
 49.6 
 52.1 
 53.2 
 24.5 
 58.1 
 
 0.0 
 
 7.0 
 
 0.0 
 
 20.0 
 
 15.0 
 
 20.0 
 
 0.0 
 
 1.5 
 
 10.0 
 
 0.0 
 
 20.0 
 
 0.0 
 
 0.0 
 
 0.0 
 0.0 
 0.0 
 
 0.0 
 0.0 
 0.0 
 0.0 
 
 2.7 
 2.6 
 1.8 
 1.4 
 6.9 
 1.0 
 
 12.6 
 83.0 
 68.9 
 70.0 
 77.7 
 75.6 
 75.4 
 80.5 
 78.9 
 74.6 
 62.6 
 92.3 
 94.3 
 
 84.5 
 98.2 
 99.5 
 
 92.9 
 
 90.0 
 
 0.0 
 
 4.6 
 
 9.5 
 7.1 
 6.2 
 4.3 
 16.2 
 5.7 
 
 17.2 
 1.7 
 5.7 
 1.1 
 1.2 
 0.8 
 2.6 
 0.8 
 1.2 
 6.0 
 1.5 
 1.8 
 0.7 
 
 3.3 
 0.2 
 0.2 
 
 4.3 
 1.7 
 0.0 
 
 4.7 
 
 9.0 
 3.3 
 5.9 
 5.9 
 17.6 
 6.5 
 
 57.8 
 6.6 
 
 18.6 
 7.3 
 4.6 
 2.5 
 
 18.7 
 2.4 
 8.5 
 
 16.0 
 
 14.0 
 3.0 
 3.7' 
 
 5.8 
 1.3 
 0.6 
 
 1.1 
 
 5.5 
 
 98.0 
 
 42.6 
 
 27.2 
 30.3 
 28.2 
 30.0 
 26.2 
 24.0 
 
 1.6 
 0.3 
 2.3 
 0.9 
 0.2 
 0.1 
 1.0 
 0.2 
 0.3 
 0.5 
 0.1 
 0.3 
 0.4 
 
 4.2 
 0.0 
 0.0 
 
 0.4 
 
 1.0 
 
 0.0 
 
 34.7 
 
 0.8 
 1.5 
 0.8 
 0.5 
 1.1 
 0.5 
 
 2.6 
 0.5 
 1.6 
 0.7 
 0.7 
 0.6 
 0.5 
 0.6 
 0.4 
 0.8 
 0.6 
 1.6 
 0.4 
 
 0.1 
 
 0.9 
 1.1 
 0.0 
 3.1 
 
240 
 
 AGRICULTURAL ARITHMETIC 
 
 Pounds of Plant-Food Elements Removed by Crops.' 
 
 (Approximate amounts per acre annually) 
 
 Crops ' 
 
 Yield 
 per acre 
 
 Nitrogen 
 (N) 
 
 Phosphorus 
 (P) 
 
 Potassium 
 (K) 
 
 Calcium 
 (Ca) 
 
 Alfalfa hay 
 
 4 tons 
 
 (190.4) t 
 
 18.9 
 
 148.0 
 
 145.0 
 
 straw 
 
 40 bu. 
 1 ton 
 
 35.3 
 11.2 
 
 46.5 
 
 7.1 
 
 1.57 
 
 8.67 
 
 11.8 
 19.9 
 31.7 
 
 0.8 
 4.6 
 5.4 
 
 
 
 
 Buckwheat, grain. . 
 
 straw. . 
 
 Total crop 
 
 30 bu. 
 M ton 
 
 21.8 
 
 12.45 
 
 34.25 
 
 5.5 
 
 0.85 
 
 6.35 
 
 7.3 
 14.1 
 21.4 
 
 0.3 
 10.3 
 10.6 
 
 
 20 tons 
 
 120.0 
 
 20.0 
 
 144.0 
 
 36.0 
 
 
 
 Clover hay (red).. . 
 
 2 tons 
 
 (82.0) 
 
 6.8 
 
 54.0 
 
 61.6 
 
 Clover hay(Japan) 
 
 2 tons 
 
 (77.6) 
 
 18.0 
 
 68.8 
 
 
 Clover and timothy 
 (mixed) 
 
 2 tons 
 
 (55.2) 
 
 8.0 
 
 63.2 
 
 
 Clover seed 
 
 3bu. 
 
 (5.25) 
 
 1.5 
 
 2.25 
 
 3.2 
 
 Corn, grain 
 
 stover 
 
 cob 
 
 Total crop 
 
 65 bu. 
 
 l%tODS 
 
 900 lbs. 
 
 59.0 
 
 33.0 
 
 2.9 
 
 94.9 
 
 10.9 
 6.86 
 0.3 
 
 18.06 
 
 12.1 
 
 37.45 
 
 4.9 
 
 54.45 
 
 0.7 
 12.2 
 
 0.1 
 13.0 
 
 Cotton, lint 
 
 seed 
 
 stalk 
 
 Total crop 
 
 500 lbs 
 1000 lbs 
 2000 lbs 
 
 1.5 
 31.5 
 51.0 
 84.0 
 
 [0.2 
 5.5 
 9.0 
 
 14.7 
 
 2.0 
 
 9.5 
 
 29.5 
 
 41.0 
 
 
 Cowpeas, hay 
 
 2 tons 
 
 (123.6) 
 
 16.4 
 
 137.2 
 
 36.0 
 
 Flax, grain 
 
 straw 
 
 15 bu. 
 0.9 ton 
 
 30.4 
 20.7 
 51.1 
 
 5.5 
 1.5 
 7.0 
 
 6.63 
 15.66 
 22.3 
 
 2.0 
 
 9.3 
 
 11.3 
 
 
 
 
 Millet hay (common 
 
 3 tons 
 
 79.8 
 
 9.4 
 
 106.8 
 
 16.2 
 
 straw 
 
 Total crop 
 
 50 bu. 
 lj^tons 
 
 31.7 
 14.5 
 46.2 
 
 5.65 
 2.25 
 7.9 
 
 7.44 
 31.13 
 38.57 
 
 1.1 
 
 7.5 
 8.6 
 
 Peas, grain 
 
 straw 
 
 Total crop 
 
 25 bu. 
 1 % tons 
 
 54.9 
 30.3 
 
 (85.2) 
 
 5.5 
 2.5 
 8.0 
 
 12.6 
 26.3 
 38.9 
 
 2.3 
 42.3 
 44.6 
 
 Potatoes (tubers) . . 
 
 200 bu. 
 
 42.0 
 
 6.3 
 
 52.8 
 
 2.4 
 
 ♦Compiled from various sources. 
 
 1 Figures in parentheses in Nitrogen column indicate total nitrogen content of 
 the leguminous crops. See pages 127, 136. 
 
APPENDIX 
 
 241 
 
 Pounds of Plant-Food Elements Removed by Crops. (Con't.) 
 
 (Approximate amounts per acre annually) 
 
 Crops 
 
 Yield 
 per acre 
 
 Nitrogen 
 
 (N) 
 
 Phosphorus 
 
 (P) 
 
 Potassium 
 (K) 
 
 Calcium 
 
 (Ca) 
 
 Rye, grain 
 
 25 bu. 
 lj^tons 
 
 26.5 
 12.0 
 38.5 
 
 4.48 
 3.05 
 7.53 
 
 6.6 
 16.4 
 23.0 
 
 0.56 
 
 5.5 
 
 6.06 
 
 
 
 
 Soybeans, grain. . . 
 straw. . . 
 
 20 bu. 
 l^tons 
 
 70.0 
 52.7 
 
 (122.7) 
 
 7.1 
 
 5.4 
 12.5 
 
 24.6 
 32.7 
 57.3 
 
 2.2 
 31.2 
 33.4 
 
 
 
 
 Sugar beets, (roots 
 
 15 tons 
 
 78.0 
 
 10.5 
 
 79.5 
 
 9.0 
 
 
 
 Timothy hay 
 
 lj^tons 
 
 29.7 
 
 4.05 
 
 33.9 
 
 7.5 
 
 Tobacco, (leaves) . . 
 
 1500 lbs 
 
 65.0 
 
 3.3 
 
 70.5 
 
 48.0 
 
 straw 
 
 30 bu. 
 1.6 tons 
 
 35.6 
 16.0 
 51.6 
 
 6.75 
 
 1.8 
 
 8.55 
 
 7.92 
 19.64 
 27.56 
 
 0.7 
 
 • 4.2 
 
 4.9 
 
 
 
 
 Specific 
 
 Gravity of a 
 
 ?ew Substances 
 
 
 Aluminium . . . 
 
 2.58 
 
 Iron 
 
 7.93 
 
 
 21.50 
 
 
 1.54 
 
 
 11.38 
 
 Potassium .... 
 
 .87 
 
 
 
 
 Copper 
 
 8.89 
 
 Mercury 
 
 13.59 
 
 
 2.35 
 
 Gold 
 
 19.30 
 
 Nickel 
 
 8.90 
 
 
 10.50 
 
 
 
 Granite 
 
 2.60 
 
 Phosphorus. . . 
 
 .1.80 
 
 Sodium 
 
 .97 
 
 Specific gravity is the relative density or weight of any volume of 
 a substance compared with an equal volume of some other substance 
 taken as the standard or unit. Solids and liquids are compared with 
 water. 
 
 A body submerged in a liquid displaces an equal volume of the 
 liquid. A body floating in a liquid displaces an equal weight of the 
 liquid. 
 
 16— 
 
242 AGRICULTURAL ARITHMETIC 
 
 Weights Per Bushel of Grain, Seeds, etc., in Different States. 
 
 States 
 
 ffl 
 
 -*3 T3 
 c3 " 
 
 w 
 
 Q 
 
 I 
 
 Alabama 
 
 Arkansas 
 
 Arizona 
 
 California 
 
 Colorado 
 
 Connecticut 
 
 Delaware 
 
 District of Columbia. ■. . 
 
 Florida 
 
 Georgia 
 
 Idaho 
 
 Illinois 
 
 Indiana 
 
 Iowa 
 
 Kansas 
 
 Kentucky 
 
 Louisiana 
 
 Maine , 
 
 Maryland 
 
 Massachusetts 
 
 Michigan 
 
 Minnesota 
 
 Mississippi 
 
 Missouri 
 
 Montana 
 
 Nebraska 
 
 Nevada 
 
 New Hampshire 
 
 New Jersey 
 
 New Mexico 
 
 New York 
 
 North Carolina 
 
 North Dakota 
 
 Ohio 
 
 Oklahoma 
 
 Oregon 
 
 Pennsylvania 
 
 Rhode Island 
 
 14 
 
 56 
 
 14 
 
 48 
 
 70 
 
 55 
 
 55 60 
 57 60 
 
 45 
 
 54 
 
 50 
 
 55 
 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 
 45 
 
 60 
 
 60 
 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 
 48 
 
 50 
 
 64 
 
 55 
 
 57 
 
 54 
 
 45 
 
 60 
 60 
 
 60 
 
 56 
 
 70 
 
 56 
 
 60 
 
 54 
 
 44 
 
 54 
 
 45 
 
 50 
 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 60 
 
Appendix 
 
 243 
 
 'Weights Per Bushel of Grain, Seeds, etc., in Different States 
 — Continued. 
 
 States 
 
 1 
 
 PQ 
 
 I 
 
 -a 
 
 o 
 
 Ph 
 
 J3 
 
 South Carolina . 
 South Dakota. . 
 
 Tennessee 
 
 Texas 
 
 Utah 
 
 Vermont 
 
 Virginia 
 
 Washington. . . . 
 West Virginia . . 
 
 Wisconsin 
 
 Wyoming 
 
 United States . . 
 
 48 
 48 
 
 48 
 48 
 
 48 
 48 
 
 48 
 
 60 
 60 
 60 
 
 62 
 60 
 
 60 
 60 
 
 14 
 
 14 
 
 42 
 50 
 42 
 
 48 
 52 
 42 
 52 
 50 
 
 42 
 
 56 
 56 
 56 
 56 
 
 56 
 
 52 
 56 
 
 57 
 
 52 
 57 
 
 57 
 
 60 
 
 60 
 60 
 60 
 
 60 
 
 56 
 56 
 56 
 
 56 
 56 
 56 
 56 
 56 
 
 56 
 
 42 
 
 60 
 60 
 60 
 
 60 
 60 
 60 
 60 
 60 
 
 60 
 
244 AGRICULTURAL ARITHMETIC 
 
 The Wolff Feeding Standards for Farm Animals 
 
 Animal 
 
 Per day per 1000 lbs. live weight 
 
 Dry 
 
 matter 
 
 Digestible nutrients 
 
 Crude 
 protein 
 
 Carbo- 
 hydrates 
 
 Fat 
 
 Fattening cattle — 
 
 First period 
 
 Second period 
 
 Third period 
 
 Milch cows, — when yielding 
 
 16.6 lbs. of milk 
 
 22.0 lbs. of milk 
 
 27.5 lbs. of milk 
 
 Horses — 
 
 Light work 
 
 Medium work 
 
 Heavy work 
 
 Brood sows 
 
 Fattening swine — 
 
 First period 
 
 Second period 
 
 Third period, 
 
 Growing fattening swine — 
 
 Weighing 50 lbs 
 
 Weighing 100 lbs 
 
 Weighing 150 lbs 
 
 Growing sheep — mutton breeds 
 
 Weighing 60 lbs 
 
 Weighing 80 lbs 
 
 Weighing 100 lbs 
 
 Fattening sheep — 
 
 First period 
 
 Second period 
 
 30 
 30 
 
 26 
 
 27 
 29 
 32 
 
 20 
 24 
 26 
 22 
 
 36 
 32 
 25 
 
 44 
 35 
 33 
 
 26 
 26 
 
 24 
 
 30 
 
 28 
 
 2.5 
 3.0 
 2.7 
 
 2.0 
 2.5 
 3.3 
 
 1.5 
 2.0 
 
 2.5 
 2.5 
 
 4.5 
 4.0 
 2.7 
 
 7.6 
 5.0 
 4.3 
 
 4.4 
 3.5 
 3.0 
 
 3.0 
 3.5 
 
 15.0 
 14.5 
 15.0 
 
 11.0 
 13.0 
 13.0 
 
 9.5 
 11.0 
 13.3 
 15.5 
 
 25.0 
 24.0 
 18.0 
 
 28.0 
 23.1 
 22.3 
 
 15.5 
 15.0 
 14.3 
 
 15.0 
 14.5 
 
 0.5 
 0.7 
 0.7 
 
 0.4 
 0.5 
 0.8 
 
 0.4 
 0.6 
 0.8 
 0.4 
 
 0.7 
 0.5 
 0.4 
 
 1.0 
 0.8 
 0.6 
 
 0.9 
 0.7 
 0.5 
 
 0.5 
 0.6 
 
APPENDIX 245 
 
 TABLES OF WEIGHTS 
 
 Avoirdupois or Commercial Weight. 
 16 drams = 1 ounce (oz.) 
 16 ounces = 1 pound (lb.) 
 100 pounds = 1 hundredweight (cwt.) 
 2000 pounds = 1 ton (T.) 
 
 The following denominations are also used in Avoirdupois Weight — 
 
 100 lbs. nails = 1 keg 
 
 196 lbs. flour = 1 barrel 
 
 200 lbs. pork or beef = 1 barrel 
 
 230 lbs. salt (at N. Y. works) = 1 barrel 
 
 Troy or Jeweler's Weight. 
 
 24 grains = l pennyweight (pwt.) 
 
 20 pennyweight = 1 ounce 
 12 ounces = 1 pound 
 
 Metric Equivalent ofAvoirdupois Weight. 
 1 dram = 1.77 grams (g.) 
 1 ounce = 28.35 grams 
 1 pound = .4536 kilogram (K.), or kilo. 
 1 kilogram = 2.2 pounds 
 
 1 ton = .9 metric ton 
 
 UNITED STATES DRY AND LIQUID MEASURES 
 Dry Measure. 
 
 2 pints (pt.) = 1 quart (qt.) 
 8 quarts = 1 peck (pk.) 
 
 4 pecks = 1 bushel (bu.) 
 
 1 bushel = 2,150.42 cubic inches, or 1.2 cu. ft. 
 
 1 heaped bushel = 2,747.7 cubic inches, or 1.6 cu. ft. 
 
 Liquid Measure. 
 
 4 gills (gi.) = 1 pint (pt.) 
 
 2 pints = 1 quart (qt.) 
 4 quarts = 1 gallon (gal.) 
 
 1 gallon = 231 cu. in. 
 3VA gallons = 1 barrel (bbl.) 
 
 2 barrels, or 63 gallons = 1 hogshead. 
 
 In approximate values 7J^ gallons = 1 cu. ft., 4% cu. ft. = 1 
 barrel. 
 
 Metric equivalent of dry and liquid measures. 
 1 liquid quart = '.95 liter 
 
 1 quart in dry measure = 1.1 liters 
 1 liter = 1.05 liquid quarts, or .9 dry quart. 
 
 LINEAR MEASURE 
 
 12"inches (in.) = 1 foot (ft.) 5}4 yards, or 16H feet = 1 rod (rd.) 
 3 feet = 1 yard (yd.) 320 rods, or 5280 feet = 1 mile (mi.) 
 
246 AGRICULTURAL ARITHMETIC 
 
 SQUARE MEASURE 
 
 144 square inches (sq. in.) = 1 square foot (sq. ft.) 
 
 9 square feet = 1 square yard (sq. yd.) 
 
 160 square rods = 1 acre (A.) 
 
 640 acres = 1 square mile (sq. mi.), or section (sec.) 
 
 SOLID OR CUBIC MEASURE 
 
 1,728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 
 27 cubic feet = 1 cubic yard (cu. yd.) 
 
 METRIC MEASURE OF LENGTH 
 
 10 millimeters (mm.) = 1 centimeter (cm.) 
 10 centimeters = 1 decimeter (dm.) 
 
 10 decimeters = 1 meter (m.) 
 
 Metric equivalent of linear, square, and cubic measures. 
 
 1 inch (in.) 
 
 = 2.54 centimeters (cm.) 
 
 1 inch 
 
 = 25.4 millimeters (mm.) 
 
 1 yard 
 
 = .914 meter (m.) 
 
 1 meter 
 
 = 3.28 feet 
 
 1 mile 
 
 = 1.609 kilometers (km.) 
 
 1 hectare 
 
 = 2.47 acres 
 
 1 cubic centimeter (c.c.) = 0.061 cu. in. 
 
 
 MISCELLANEOUS TABLE 
 
 12 units 
 
 = 1 dozen (doz.) 
 
 12 dozen 
 
 = 1 gross 
 
 20 units 
 
 = 1 score 
 
 24 sheets 
 
 = 1 quire 
 
 20 quires 
 
 = 1 ream 
 
 4 inches 
 
 = 1 hand (used in measuring height o 
 
 3 feet 
 
 = 1 pace 
 
 4' x 4' x 8' 
 
 = 1 cord wood. 
 
 TABLE OF HOUSEHOLD MEASURES 
 
 3 teaspoonfuls (tsp.) = to 1 tablespoonful (tbsp.). 
 2 cupfuls (c.) = to 1 pint (pt.) 
 
 4 cupfuls (about) flour = 1 lb. 
 2 c. sugar (gran.) = 1 lb. 
 
 2 c. butter packed solid = 1 lb. 
 
 2 c. chopped meat = 1 lb. 
 
 2 tbsp. butter (level full) = 1 oz. 
 
 1 tbsp. sugar (level full) = 1 oz. 
 
 1 tbsp. liquid = J^ oz. 
 
 8 or 10 eggs, depending on size, = 1 lb. 
 
 The juice of 1 lemon = 3 tbsp. 
 
APPENDIX 
 
 247 
 
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 o 
 
 Ph 
 
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 ddoddodoooooooiHooooooo 
 
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 dd^^dc6^d^i>t^di-H(N'-Hoic6cdwdNd 
 
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 fin 
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 Thi^ici^Tjjc^i^^oqicoqt^copi-H-^coi-icompTjH 
 
 ^dNwdco^rHi>di>o6cDdc<io6dr-Hi--Hdoio5 
 
 OSOiOiOOOOOOOOOiOOWXOOC^Ci05COOS05010SOO 
 
 
 
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 - , •; 2 is . "S +? 
 
 S3-B &"§ I a 
 
 CJ-| 03 Sl-s K 5-J 
 
 & § m ^ & & b &"S -a 
 
 ■a-rs-g.is.a.fl-a-s os^ 
 
 ■g-a £ > s> > > > > B 
 S S £.3.2.2,2,3.2 3 
 
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 en cars += 
 
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 OOOO 
 
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 fla5o g fldOi» 
 
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248 
 
 AGRICULTURAL ARITHMETIC 
 
 O 
 
 o 
 O 
 
 ID 
 
 a 
 
 o 
 W 
 
 d 
 
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 3 "O 
 
 5 e 
 
 o 
 
 3 
 O 
 
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 N O 
 
 3 
 
 5 
 
 S 
 
 
 
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 o 
 
 3 
 *-+j 
 
 a 
 
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 §« 
 
 o 
 Hi 
 
 IS 
 
 o 
 
 
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 i-jr-jr-.M^cNipp 
 
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 PHrtrttawcocHtH 
 
APPENDIX 249 
 
 SQUARE ROOT 
 
 The extraction 'of the square root is the 23 
 
 analysis of the power into two equal factors. •*_££ 
 
 Hence it is important to see how the power 'X3 
 
 was built up. In order not to lose the identity 20X20 
 
 of the elements entering into the power, the 20 2 +2f20X3V+3 2 
 multiplication is indicated, A. Compare the 
 indicated expression with the algebraic formula, B 400 
 
 (a + b) 2 . By studying operation A, it can be "jj 
 
 seen that the square of the tens figure, or any — — 
 
 part of it, cannot lie in the first or second or- 
 ders, and it must lie in the third order or in 
 
 the third and fourth orders. If hundreds, as 200 or 900, were squared 
 in the same way, it can be seen that the significant figures are in the 
 fifth or fifth and sixth orders.; hence we point off in periods of two 
 orders. 
 
 As 20 occupies the larger part of 500, (B), we take the largest 
 square 400 from the product 529. The remainder, 129, in the main 
 consists of twice the root 2 tens, or 40 times the units figure; hence to 
 find the units figure, divide 129 by 40, or, for short, divide 12 by 4, 
 making allowance for what else is in this number. As 129 con- 
 tains 40 X 3 and also 3 X 3, it is the product of (40 + 3) X 3 or 43 X 
 3; hence annex 3 to 4 the trial divisor, multiply and subtract, if there 
 is a remainder. It can be shown by analyzing the squaring of a number 
 of three orders, as 235, that the extraction of the root is but a repetition 
 of the preceding process from this point on. If pupils have studied 
 algebra, the analysis of (a+b + c) is easier to understand, for it is » 
 general statement of the above. It is interesting to note how the 
 algebraio>statement is symbolic of the arithmetical statement. 
 
 Square roots of numbers 1 to 20 inclusive correct to three deci- 
 mal places. 
 
 Vl = l V 8 = 2.828 V15 = 3.873 
 
 V2 = 1.414 V 9 = 3 V16 = 4 
 
 V3 = 1.732 V10 = 3.162 V17=4.123 
 
 V4 = 2 VH=3.316 V18 = 4.242 
 
 V 5 = 2.236 V 12 = 3.464 V 19 = 4.359 
 V6 = 2.449 V13 = 3.605 V20=4.472 
 
 V 7 = 2.645 V 14 = 3.741 
 
INDEX 
 
 References are to pages 
 
 Acre inch of s water — Defined, 126, 
 
 footnote. 
 Alcohol from potatoes, 83, prob. 
 
 26. 
 Alfalfa- 
 Seed per sq. ft. at 20 lbs. per 
 
 acre, 82, prob. 21. 
 Value of for feed and soil im- 
 provement, 83, figure. 
 Versus timothy as protein pro- 
 ducer, 93, prob. 8, 
 Versus timothy in feeding, 109, 
 prob. 22. 
 Apples — 
 How to determine heaped 
 
 bushels in bins, 19. 
 Average cost to produce a barrel 
 of, 210, prob. 11. 
 Area — How to determine 
 Circle, 61 
 
 Irregular fields, 192 
 Island, 193 
 Rectangle, 31 
 Trapezoid, 58 
 Triangle, 57, 
 Ash— Defined, 91 
 Avoirdupois vs. Troy pounds, 
 24, prob. 18 
 
 Barley — How best to feed, 96, 
 
 footnote 
 Bordeaux mixture — 
 As a fungicide, 207, prob. 7 
 How to make, 207, footnote 
 Boys — A way to make a start, 
 
 223, figure 
 Butter-fat— Defined,.lll 
 Butter-fat — Loss of in skimming, 
 
 118, prob. 28 
 Butter — How much made from 
 fat 
 According to creamery experi- 
 ence, 114, prob. (d) 
 
 Rule of Association of American 
 Colleges, 114, prob. (c) 
 
 Rule of Holstein-Friesian Asso- 
 ciation, 113, prob. 6(b) 
 
 Cabbage — Usual method of plant- 
 ing, 83, prob. 27 
 Carbohydrates — Defined, 92, foot- 
 note 
 Cement — How much equal to a 
 bbl., 202 
 Amount required for concrete 
 mixtures, 202 
 Checking and proving problems, 15 
 Cheddar cheese — Origin of name, 
 
 118, footnote 
 Cheese — Amount made from milk, 
 
 118, prob. 26 
 Circle — Defined, 59 
 Clay — Defined, 121, footnote 
 Concentrates in feeding — Defined, 
 
 98, footnote 
 Concrete — Measurements and 
 rules, 201 
 Meaning of 1-2-4 mixture, 201, 
 
 figure 
 Mixtures — Ingredients required, 
 
 202 
 Perfect mixture — Defined, 202, 
 footnote 
 Cone — Defined, 64 
 
 How determine volume of, 64 
 Congressional township — Defined, 
 
 189, prob. 1 
 Corn — Crop of world in 1913, 
 79, probs. 1 and 3 
 Amount to plant per acre, 81 
 How to measure heaped bushels 
 and shelled corn equivalent 
 In bins, 65, footnote, 199 
 In cribs having flared sides, 
 200, prob. 29 
 Kernels in one quart, 81 
 
 251 
 
252 
 
 AGRICULTURAL ARITHMETIC 
 
 Method of planting to secure 
 most food substances, 81, 
 probs. 18 and 19 
 Profits in growing, 179 
 Ratio of stover to corn, 82, 
 
 probs. 19 and 20 
 Seed tester, 163, 164 
 Stover — Defined, 126, footnote 
 Shrinkage in cribs, 83, prob. 25 
 Silage — Weight per cu. ft. in 
 silo, 97, 194, prob. 14 
 Amount to feed to prevent 
 spoiling, 97, footnote 
 Cream — Defined, 114, footnote 
 Amounts from milks, 115, prob. 
 
 12 
 How to standardize, 116, prob. 
 
 20 
 Weight of per gallon, 167, foot- 
 note 
 Crop requirements per acre per 
 
 year, 240. 
 Cylinders — How to determine vol- 
 umes of, 02 
 
 Dairy cows — Champions, 87, 88, 
 90, 112, 228 
 Feeding of, according to — 
 Haecker Standard, 105, 
 
 prob. 16; 236 
 Wolff Standard, 244 
 Wisconsin rule, 98, prob. 
 26 
 Profit and loss in feeding, 171 
 
 to 173 
 Products of poor and good, 174 
 True value of, 67, prob. 4; 170, 
 footnote and 173 
 Decimals, 38 
 
 Multiplication of, 39 
 Division of, 40 
 Dietary Standards for man, 218 
 Dry matter in common feeds, 247 
 What it indicates in a ration, 
 104, footnote 
 
 Factoring, 16 
 
 Family labor — Defined, 188, 
 
 footnote 
 Farm capital, distribution of, 
 
 186, prob. 37 
 
 Farm management problems, 
 
 160 
 Farm measurements problems, 
 
 189 
 Farm records and accounts, 181 
 Farm summary, 187, prob. 41 
 Farmer's income, 185 
 
 On large vs. small farms, 
 186, 187 
 Fat — Heat value of, 100, prob. 1 
 Equivalent, 103, prob. 10, step 
 3 
 Feeding — Hints on, 110 
 How to secure best results, 99 
 Profit and loss in, 168 
 Feeds — Composition of common, 
 235 
 Food value of, 91 
 Fence-posts — Prolonging life of, 
 
 224, prob. 14 
 Fertility in soils, standards, 134 
 Balance sheet of fertility, 137 
 Exchange of fertility in farm- 
 ing, 138 
 Loss or gain of in feeding, 137, 
 
 138 
 Needs of soils — How deter- 
 mined, 134, 135, 158 
 Fertilizers, commercial, 147 
 Amount applied vs. per cent 
 
 profits, 232, footnote 
 High grade vs. low grade, 156, 
 
 prob. 13 
 Home mixing of, 157 
 Mixed or complete — Defined, 
 
 154 
 Nitrate fertilizers, 153 
 Phosphate fertilizers, 152 
 Potash fertilizers, 154, prob. 3 
 Practical interpretation of re- 
 sults, 231, prob. 35 
 Profit and loss in using, 174 
 Soluble, as top dressing, 158, 
 footnote 
 Fertilizing constituents in common 
 
 feeds, 247 
 Fractions, 18 
 Addition and subtraction of, 26 
 Division of, 28 
 Multiplication of, 27 
 Three problems of, 29 
 
INDEX 
 
 253 
 
 Fruit trees, three methods of 
 planting, 206, 207 
 Rate of planting per acre, 204 
 to 207 
 
 Grain — Falling off in exportation 
 from U. S., 80 
 How to determine bushels in 
 
 bins, 199 
 Increase in acreage in U. S., 79 
 Weight per bushel in different 
 states, 242 
 Green manuring — Defined, 174, 
 footnote 
 
 Hay — How to measure in mows, 
 
 194 
 
 How to measure in ricks, 195 
 
 How to measure in round stacks, 
 
 198 
 
 "Hog house — A-shaped, portable, 
 
 231, prob. 34 
 Household economy and human 
 
 feeding, 213 
 Hypotenuse — How to determine, 
 54 
 
 Inoculation — Why necessary, 180, 
 
 footnote 
 Interest, 69 
 
 Kerosene emulsion — Use of, 208, 
 prob. 8 
 How to make, 208, footnote 
 
 Labor income— Defined, 160, 185, 
 
 footnote 
 Landlord's income on large vs. 
 
 small farms, 186, 187 
 Tenant's income on large vs. 
 
 small farms, 186, 187 
 Land as an investment, 161, prob. 
 
 10 
 Land description explained, 33, 34 
 Land rental — Defined, 161, foot- 
 note 
 Least common multiple, 23 
 Lime — A method of applying lump 
 
 or quicklime, 200, prob. 33 
 Liming — Why necessary, 180, 
 
 footnote 
 
 Profits derived from, 180, 229, 
 prob. 31 
 Lumber — How to measure, 56 
 
 Machinery — How to determine 
 annual investment, 165, foot- 
 note 
 Managerial income — Denned, 185, 
 
 footnote 
 Manure — Amount produced by 
 farm animals, 147; 148 
 
 Compared with commercial fer- 
 tilizers on marsh, 158, figure 
 
 Composition of fresh, 149 
 
 Fertility contained in, 147, 
 prob. 1 
 
 Fertility lost through exposure 
 of, 150, prob. 5 
 
 Light vs. heavy applications, 
 150, prob. 4(d) 
 
 Profits derived from using, 174 
 
 Manure utilization by crops, 
 161, footnote 
 
 Value of fertility in, 147, prob. 2 
 
 Meat — Bacon vs. round steak, 
 
 cost of, 214, probs. 7, 8 and 9 
 
 Milk— Bacteria in, 222, prob. 6 
 
 Gallons in 100 pounds, 169, 
 footnote 
 
 How to standardize, 116, prob 
 20 
 
 Pails— Three types of, 221 
 
 Quarts in 1,000 pounds, 88, 
 footnote 
 
 Weight per gallon, 113, footnote 
 
 Nitrogen — Amount fixed by clover 
 and alfalfa, 128 
 
 Balance sheet for, 143 
 
 Chemical symbol for, 153, foot- 
 note 
 
 Contained in air, 124, prob. 3 
 
 Contained in common feeds, 247 
 
 Gained in feeding clover and 
 alfalfa, 138 
 
 Gained in plowing under clover 
 and alfalfa, 140 
 Nitrogen-free extract — Defined, 91 
 Nutrients — Defined, 94, footnote 
 
 Digestible contained in common 
 feeds, 247 
 
254 
 
 AGRICULTURAL ARITHMETIC 
 
 Digestible contained in human 
 
 foods, 237 
 Digestible amounts in feeds — 
 
 How to determine, 102, prob. 
 
 10 
 Digestible amounts in foods — 
 
 How to determine, 216, prob. 
 
 11 
 Total in common feeds, 235 
 Nutritive ratio — Denned, 101 
 How to determine, 101, prob. 5 
 Narrow vs. wide, 102 
 
 Oats — Cost to grow an acre of, 
 177, prob. 23 
 Formalin [treatment for smut, 
 
 164, footnote 
 Weight per bu., 242 
 Orchard — (Apple),an account with 
 an, 208 
 Profits due to tillage in, 205, 
 
 prob. 7 
 Results of rejuvenating an old, 
 208 
 Orchard and garden, 204 
 Overrun in butter — Denned and 
 how should be determined, 
 115, footnote 
 
 Paris Green with Bordeaux mix- 
 ture for potatoes, 210, prob. 
 12 
 Pearson method for standardizing 
 
 milks and creams, 116 
 Peat — Defined, 156, footnote 
 Percentage, 65 
 
 "Phosphoric acid" — Defined, sym- 
 bol for, 154, footnote 
 How to reduce to equivalent of 
 phosphorus, 155 
 Phosphorus — Amount of in some 
 soils, 134 
 Amount contained in common 
 
 feeds, 247 
 Balance sheet for, 143 
 Chemical symbol for, 152, foot- 
 note 
 Fertilizers, 152 
 
 Required for seed and grain 
 development, 131 
 
 Pigs — When most profitable to 
 
 fatten, 168 
 Plant-food elements removed by 
 
 crops; 240 
 Plant requirements, 124 
 Carbon from air through leaves, 
 
 125, prob. 1 
 Water required per pound of 
 dry matter, 124 
 Pork produced per bushel of corn, 
 
 168, probs. 11 and 12 
 Potassium — Amount of in some 
 soils, 134 
 Amount contained in common 
 
 feeds, 247 
 Chemical symbol for, 153, foot- 
 note 
 Fertilizers, 154, prob. 3 
 "Potash" — Defined, symbol for, 
 154, footnote 
 Fertilizers, 154, prob. 3 
 How to reduce to equivalent 
 of potassium, 155 
 Potatoes — Benefit from spraying 
 for blight, 206, prob. 12 
 Bordeaux mixture for, 211, foot- 
 note 
 Formalin treatment for scab, 
 
 210, prob. 12 
 How to determine bushels in 
 bins, 199 
 Poultry— Profits in, 160; 225, 
 
 prob. 17 
 Pound — Avoidupois vs. Troy, 245 
 Protein — Explained, 91 
 
 Range in land description — Ex- 
 plained, 33 
 Ratio and proportion, 72 
 Ration — Balanced, 104, footnote 
 How to figure out, 108, prob. 
 
 19(e) 
 Maintenance ration — Denned, 
 106, footnote 
 Rock phosphate — To render solu- 
 ble, 154, footnote 
 Wrong interpretation of fertility 
 value, 226, prob. 21 
 Roughage in feeding — Defined, 98, 
 footnote 
 
INDEX 
 
 255 
 
 Sand — Defined, 121, footnote 
 Section of land — Defined, 189, 
 
 prob. 1 
 Seeds — Weight per bu. in different 
 
 states, 242 
 Short cuts in working problems, 
 
 42 
 Silage — See under corn 
 Silt — Meaning of, 121, footnote 
 Soil — A great storehouse, 132 
 Bacteria in, 122, probs. 11 and 
 
 12 
 "Heavy"and "Light"— Defined, 
 
 121, footnote 
 Not inexhaustible, 135 
 Weights of per acre, 119 
 What it is, 120 
 Soil fertility— Exhaustible, 134, 
 136, prob. 14 
 Value of — Removed, should be 
 
 charged against crops, 229, 
 
 probs. 30 and 32 
 Solids— Value of rectangular, 35 
 Square root — How determined, 53 
 Steer feeding — Profit and loss in, 
 
 170 
 Survey — Rectangular, 33 
 
 Tables of weights, measures, etc., 
 
 245 246 
 Township— Defined, 33 ; 189, 
 
 prob. 1 
 Tuberculosis claims many deaths, 
 
 227, prob. 24 
 
 Water — Required by crops, 123, 
 126, 127 
 Weight of per cu. ft., 124, foot- 
 note 
 
 Wheat yield of U. S. vs. Germany, 
 79, prob. 5 
 
ANSWERS 
 
 PART I 
 
 Page 31— 1. $1,890. 2. $1,100. 3. 1,500 lbs. 4. fandf 541? and 
 722£. 6. 474 bu. 
 
 Page 32—4. $17.78. 
 
 Page 33—5. 320 acres. 6. 40, 60, 160. 7. $380.16. 8. 1,584. 
 
 9. 75,093.34. 
 
 Page 34—1. E 40, D 160, A 80. 2. $2,800. 3. $1,800. 4. $103.20. 
 43c. 
 
 Page 35—5. 160 acres. 640 rds. 2 miles. 6. 800 rds. 160 
 acres. 160 rds. 
 
 Page 36—1. 480. 2. 384 bu. 3. 169 gal. 5.3 bbls. 
 
 Page 37^4. 282 cu. ft. 32J bags. 3.57 cu. yds. sand. 5,94 
 cu. yds. gravel. 5. 5.84 bags. .65 cu. yd. sand. 1.08 cu. yds. gravel. 
 6. 28.5 bbls. 7. 582 gal. 8. 20 cords. 9. $189. 10. 20 ft. 2\ 
 cords. 11. 18. 12. ^. 13. 6". 14. 52,272 cu. ft. 
 
 Page 38.— 15. 9.9 ft. 16. 3+acres. 17. 11 and 10 tons. 19. 
 10.8 in. 20. 1 ft. 5 in. 21. 1.26 tons. 
 
 Page 40.— 10. $11.28. 11.15.58. 12. 430.05 sq. rds. 
 
 Page 41.— 7. 2.68 + acres. 8. 191.1. 600 bu. 9. 143 bbls. 
 
 10. 1,180.09. 
 
 Page 42—11. 2.3 ft. 3.8 ft. 1. 15. 2. 11.7. 3. 1.3. 
 
 Page 50—5. 1,296 bu. 6. $680.40. 
 
 Page 51—8. 20. 9. 1,920. 10. 90 and 121. 11. 14.9. 12. 113.4 
 tons. 13. $2,250. 
 
 Page 63—1. $5.76. 2. $6. 3. $16. 4. $19.20. 5. $3.60. 
 6. $20.70. 7. $4.10. 8. $20.74. 9. $40.32. 10. $15.84. 
 
 Page 54—1. 5.196 + . 29,875. 79.84 + . 3. 9.472. .799. 
 54.531. 4,242. 
 
 Page 56— 2. 1.4ft. 4.8ft. 3.6.3ft. 4.18.02ft. 5.<W- 6.7.07. 
 9. 369. 10. 1,506. 
 
 Page 57—11. 20.8—. 1. 480. 204 ft. 117 ft. 195J ft. 
 2. | acre. 
 
 Page 58—1. 6J acres. 2. 21,952. 3. 70 acres. 4. 201.8 acres. 
 338.2 acres. 355.4+ rds. 6. 247£ acres. 
 
 Page 59—1. 6^. 2. 380 and 260. 3. 70. 4. 193.2 and 446.8 
 355.4 rods. 5. 247.5. 
 
 Page 60 — 4. 11.8. 5. 3 in. nearly. 
 
 Page 61—6. 31.68 in. too large. 1. 154. 452.5. 50.2. 346.5. 
 452.5. 
 
 17- 1 
 
2 AGRICULTURAL ARITHMETIC 
 
 Page 62—2. 103.1. 658.7. 3. 346.5. 4. .314. 5. 4 times. 
 
 6. 9, 16, 25. 7. 5 in. 8. 10.7 + in. 
 
 Page 63—2. 53.9 bbls. 3. 4,620 cu. ft. 
 
 Page 64—4. 32.8 cu. ft., 2,118 cu. ft., 12.3 ft. waste. .37 of log. 
 
 5. 1,202 gal. 
 
 Page 65—1. 100 bu. approximately. 3. 109 bu. approximately. 
 Page 67—1. .54 lbs. 2. 315.67 lbs. $89.97. 5. 25.1 tons. 
 $23.04. 6. 37.5 lbs. 3,750. 
 
 Page 68—3. 43f%. 2f 4. 8%. 5. 87.7%. 
 
 Page 69—3. $25,000. 4. $276.64 + beef value. 5. $9,000. 
 
 6. $12,800. 
 
 Page 70—2. $28. 3. $28.62. 4. $259.67. 5. $153.12. 6. $45. 
 
 7. $2,861. 
 
 Page 71—10. $81.67. 11. $54.14. 12. $104.20. 13. $82.40. 
 14. $87.48. 16. $36.33. 1. 10%. $1.08. 2. The latter. $27. 
 3. $953.34. 4. 36.02%. 5. $282. 
 
 Page 72—6. $1.57. 7. $323.80. 8. $65 for beef. 
 
 Page 73—4. f, \, &. 
 
 Page 74-6. I 7. A, i- 8. \. 9. ||. 11. h &. 12. J. 
 13. <h, lis, 5 ls- 14- 720 ft. 15, 8 to 1. 
 
 Page "75— 16. 8 cu. ft. 17. 5.09 ft. 18. 40.56. 19. 500 bu. 
 1,000 bu. 20. 1,218, 1,624. 21. 16 and 32. 22. 30 ft., 18 ft. 
 23. 3 3 / g times. 24. |. 25. The latter. 69+ c. 29c. 
 
 PART II 
 
 Page 79—1. 23.12 bu. 3. 67.86%. 4. $1,690,868,708.00. 
 5. 210.27% greater. 6. 46.03% increase. 
 
 Page 80— (a). 15.96% decrease. 7. 46.34% increase. 8. 191,- 
 140,000 bu. 10. 1,370,800 bales. 11. 729,150 bales. 
 
 Page 81—12. $2,425.00. 13. 34 bu. and 1 bu. 14(a). 13.44 sq. 
 ft. (b). 3,240. (c). About 4 qts., allowing a little for overrun. 7 lbs. 
 (d). About 3| bu., or 210 lbs. 15. About 5.8 qts., or 10 lbs. 16. About 
 14 to 16. 17. About 20.5 to 24. 
 
 Page 82 — 18(a). 47,520. (b). About 19 qts., or" 33* lbs. 
 (c). Between 67 and 80. 19. 47.6 bu. 20(a). About 5 qts., 8.75 lbs., 
 or 18 to 20 ears. (b). 3.4 tons. 21. About 100. 22. $25. 23. 60c. 
 24. $2.08 less. 
 
 Page 83— 25.20.5%. 26(a). 50.3 + cents. 26. 3.12 qts. (a). 201 + 
 gal. 27. 27.8 tons. 
 
 Page 84— (a). About 14 tons. (b). About 10.5 tons. 28. 83.3 bu. 
 
ANSWERS 3 
 
 Page 86—1. 196,481,000, valued at $4,740,684,000.00. 2. 54 yrs. 
 10 mo. 12 da. 
 
 Page 87—3. $24.26. 4. Horses, $105.05; mules, $124.80. 5. Swine, 
 $6.86: sheep, $4.42. 7. 745.2. 8. 7.9 oz. 9(a). 17,559 lbs.; 48.1 lbs. 
 per day. (b). 1,175.46 lbs. 3.22 lbs. per day. 10. 23,000 lbs., 
 averaging 63 + lbs. per day. 11. $764.57. 
 
 Page 88—12. 14.4 lbs. 13(a). $669.82. (b). 65.8 lbs., or 30.6 
 qts. 
 
 Page 89— 13(c). 25 lbs. 14(a). 4.53%. (b). 7.8+ gal. 
 15. Fat per yr. Fat per day Milk per yr. 
 
 May Rilma 1073.17 lbs. 2.93 lbs. 19673 lbs. 
 
 Murne Cowan 1097.16 lbs. 3 lbs. 24008 lbs. 
 
 Finderne Holingen Fayne 1116.05 lbs. 3.05 lbs. 24612.8 lbs. 
 
 Page 90— 16(a). 83.4 lbs., or 38.8 qts. (b). 3.12%. 
 
 Page 92—1. 60 lbs., 344 lbs., 284 lbs., 216 lbs., respectively. 
 2. 522 lbs. 
 
 Page 93 — 3. 353|%. Alfalfa. Protein required for milk produc- 
 tion. 71.1%. 48.4%. 4. 60 lbs. more in alfalfa. 6. 1.48+. 6(a). 22.4 
 lbs. 6(b). 794.08 lbs. 6(c). 816.48 lbs. 6(d). 759.36 lbs. 7. Only 20 
 lbs. more in timothy. 8. 3.39 acres. 9. 5.95 tons. 
 
 Page 94—10. 70 lbs. and 64.4 respectively. 11. 27,936 lbs. and 
 20,992 lbs., respectively. 12. 420 lbs. more in cottonseed meal. 
 13. 48 lbs. more in bran. 14. 1,300 lbs. in cottonseed meal, 1,142 
 lbs. in bran, 1,010 lbs. in alfalfa. 
 
 Page 95—15. 210 lbs. more in alfalfa. 16. Clover, 6.1 lbs. (a). 
 Yes. 1.3 lbs. (b). 6,030 lbs. in favor of clover, (c). 11 tons 
 clover equivalent to 6.8 tons alfalfa. Alfalfa, 89 cents per ton cheaper, 
 or total of $3.00 cheaper. 17. $9.82. 18. 528.92 lbs. 
 
 Page 96—19. 792.4 lbs. and 1.4 tons, respectively. 20. 10.18 bu. 
 barley; 8.6 bu. dent corn. 21. 27.6 tons. 22(a). 10.64 lbs. (b). 1.12 
 lbs. (c). 3.75 bu. 23(a). 101.25 tons. (b). About 8.5 acres. 
 
 Page 97— 23(c). 5,062.5 cu. ft. (d). 32.8 ft. About 38.8 ft. if 
 silage settled 6 ft. (e). At least 30 ft. to allow for settling, (f). The 
 14-ft., since 750 lbs. must be fed daily. 24. 12.3; 4.8; 9; 22.5; 20.25 
 acres, respectively. 25(a). 95.2; 66.64; 76.16 tons, respectively. 
 
 Page 98— 25(b). 224; 156.8; 179.2 lbs., respectively, (c). 349.35 
 lbs. .411 lb. (d). 1,216.5 lbs. (e). 7.3+ cents. 26(a). 4.2; 5.3; 
 6.2; 5.9; 9.3 lbs., respectively, in each group. (b).j 3,324 lbs. 
 27(a). 1,725 bu. 
 
 Page 99— 27(b). 60| tons. (c). 11.5 bu 
 
 Page 100—1. 168| lbs. 2. 18.56 lbs. 
 
 Page 101—3. 66 lbs. 4. 1 : 4 or 1 to 4. 
 
 Page 102— «. 1:6.2. 7.1:4. 8. 1:1 .8 vs. 1 : 10.4. Soybeans. 
 9. Gluten meal (1:1.8). 10. 1:3.9 (narrow). 1:3.9 (alfalfa). 
 
 Page 103— 11(a). 2.27 lbs. (b). 14.07 lbs. (c). .76 lb. (d).l:7. 
 12(a). 
 
4 AGRICULTURAL ARITHMETIC 
 
 Lbs. Protein Lbs. Carbohydrates Lbs. Fat 
 
 Timothy 48 6.85 .192 
 
 Oats 97 5.21 .38 
 
 Com 375 3.39 .23 
 
 Oil Meal 302 0.32 .067 
 
 Wheat Bran 125 0.42 .03 
 
 12(b). 2.25 16.2. .9. 
 
 12(c). 1:8.1. 13. 1:6 vs. 1:8 (not enough nutrients). 
 29.5 lbs. dry matter (not sufficient bulk). 
 
 Page 104 — 13(a). Better. 39 lbs. dry matter; 3.3 lbs. protein; 
 
 22 lbs. carbohydrate; 1.2 lbs. fat. 14(a). 1:5.7. (b). No. Short on 
 dry matter also. 
 15(a). 
 
 Lbs. Protein Lbs. Carbohydrates Lbs. Fat 
 
 Com Silage 0.38 5.25 24 
 
 Alfalfa 1.27 4.68 .11 
 
 Corn 0.22 2.03 .14 
 
 Cottonseed Meal 0.32 0.26 .08 
 
 (b). 47 and 4 respectively. 
 
 Page 105— 15(c). 1:6.2. 16(a). 2.32 lbs. protein; 14.2 lbs. carbo- 
 hydrates; .73 lb. fat. (b). 1:6.8. (c). Very well, though 15 contains 
 less protein in particular and is a slightly narrower ration. 
 
 Page 106— 17(a). 6.3 lbs. and 4.2 lbs. respectively, (b). 1:7.5. 
 (c). 2.5 lbs. protein; 15.4 lbs. carbohydrates; .82 lb. fat. (d). 1:6.9. 
 
 Page 107— 18(a). .77 lb. .62 lb. .15 lb., respectively, (b). 4.2 
 lbs. (c). 1:7.2 19(a). 1:5.9, 1:6.3, 1 : 7, respectively. More protein 
 should be fed at beginning of fattening period, (b). Protein feeds, 
 (c). Feeds high in carbohydrates. 
 
 Page 108 — 20(a). About 13 lbs. middlings and 25 lbs. com. 
 (b). If lbs. and 3J lbs., respectively. 4f lbs. total daily. About 
 1.6 lbs. (c). 1,706.25 lbs. and 3,281.25 lbs., respectively. 4,987.5 
 lbs. 21. 1:15.2 vs. 1:3.9. Alfalfa. 
 
 Page 109—22. $500.87. 23. The same. 23(a). 1:6. 24(a). 625. 
 (b). 50c. 
 
 Page 112—1. 20,620,497. 2. 101.06 gal.; 1.1 qts. 3(a). 261.3 
 gal. (b). 122.5. 
 
 Page 113-4. 1.76c. 5.1.59c. 6(a). 35.3. (b).< 44|. (c). 80%. 
 7(a). 1,072.17. (b). 53.9; 2.93; 25+. 
 
 Page 114— 7(c). 1,250.8. (d). $431.55. 8. 86.9%. 9. 3.47%. 
 10. lfc. 11. 35% cream. 
 
 Page 115—12. 255. 13. 75. 14. 235.7. (a). 1,264.3. (b). .118% 
 15.2,300. (a). 1,940 lbs., testing .103 + %. 16(a). 570+. (b). 4.114%. 
 (c). 15.3%. 
 
 Page 116— 17(a). 6,970. (b). $543.60. 18. 1.54%. 19. No. 
 (23.5%). 
 
 Page 117— 20(a). 16 lbs. cream, 5 lbs. milk. (c). 20%. (d). No. 
 (Footnote, page 113). 21. 14.5 lbs. cream, 12 lbs. milk. 22. 6 lbs. 
 milk, 5.5 lbs. cream. 23. 1.3 lbs. 3.5% milk and 1.5 lbs. 6.3% 
 
ANSWERS 5 
 
 milk, (a). 46.4 lbs. of 3.5% milk and 53.6 lbs. of 6.3%. 24. 1.2 lbs. 
 cream to 15 lbs. milk; 23.9 lbs. cream and 298.3 lbs. milk. 25. 18.1 lbs. 
 
 Page 118—26. 2 + ; &% qts. 27(a). 2,457. (b). 1,086.75. 
 (c). $61.90 in favor of cheese; $6.62 in favor of butter. 28(a). 1,748 lbs. 
 (b). 3 lbs.; 4 lbs.; 7 lbs.; 210— lbs. (c). 440. 
 
 Page 119— 28(d). 4.19+%. (e). 433 per day. 4 lbs. lost in 
 buttermilk; 3 lbs. lost in skim milk. (f). 508.8 lbs. (g). 15.63%. 
 (h). $24.08 per day; $722.40 per mo. 29(a). 16.4% (3 lbs. less fat 
 received in form of cream than milk), (b). 3 lbs. less fat received, 
 (c). $31.50 more returns per mo. when cream is brought in instead 
 of whole milk. He pays for 3 lbs. less of cream per day, not reckoning 
 extra machinery and labor cost involved in skimming. 
 
 Page 120—1. 43,560. 2. 98 lbs.; 78 lbs.; 39 lbs.; 14 lbs. 
 
 Page 121—3. Approximately 357 tons less. 4. 46.3%. 34.2%. 
 6. 140%. 6. 19, 10.6, 2.4 oz., respectively. 7. 15.2%, 62.8%, 22%, 
 respectively. 8. 50,800. 9. 48,260 less. 
 
 Page 122—11. 70,450,000 more; 1,410 times. 12. 3,750%. 
 13. 40.4. 14. 21|, 45, 125, 36f, respectively. 
 
 Page 125—1. 504. 
 
 Page 126— 1(a). 560. 2(a). 3| tons. (b). 11,046. (c). 113.42 
 approximately, (d). 13.2— . (e). 2.13 sq. rds. (f). 152+bbls. (3). 12£ 
 inches. 
 
 Page 127—4. 8f inches. 5. None. 6. 9%. Cultivated field 
 had mulch which hindered evaporation. 7. 38.86 inches. 8. 3,559.8. 
 9. 639 and 2,447.1, respectively. 
 
 Page 128—10.70.8%. 11.6.31bs. 12. Lines to represent 65 nitrogen, 
 3.3 phosphorus and 70.5 potassium. 
 
 Page 129—13. 78.7%. 14(a). 40; 80; 40; 80. 
 
 Page 130 — 15. Further depleted by 5.95 lbs. phosphorus, 47} lbs. 
 potassium and 53.9 lbs. calcium. 16. No. 71f lbs. 16(a). $10.76. 
 (b). 71| lbs. 17. 270; 41.8; 312.3 lbs., respectively. 18. 17.475. 
 19. 1,439.1 lbs. more removed from 18 acres of alfalfa and clover than 
 from 45 acres of corn and oats. 
 
 Page 131 — 20(a). 303.9 lbs. phosphorus; 1,250J lbs. potassium; 
 1,620 lbs. nitrogen. 
 
 Page 133—1. 180. 2. 4,100 lbs. 3. .1% nitrogen; .015% phos- 
 phorus; .6% potassium. 4. 1.005% nitrogen; .1265% phosphorus; 
 .855% potassium. 
 
 Page 134—5. 5,000 lbs. nitrogen; 2,000 lbs. phosphorus; 40,000 
 lbs. potassium. 
 
 6. 
 Soil Lbs. Nitrogen Lbs. Phosphorus Lbs. Potassium 
 
 Fertile silt or clay 
 
 loam 5,000 2,000 40,000 
 
 A poor sand 2,000 500 12,500 
 
 An average peat.... 10,500 420 1,050 
 
 7. 12 times, or 1,200%. 
 
6 AGRICULTURAL ARITHMETIC 
 
 Page 135—8. 2.1. 210%. 9. Pounds, because some soils are 
 much heavier than others. 10. 2,400 lbs. nitrogen; 1,140 lbs. phos- 
 phorus. 11. Only about 49 crops. 
 
 Page 136—12. .15%. 13. 171. 14. 20| lbs. 
 
 Page 139— 1(a). 820; 68; 540 lbs., respectively, (b). 328; 13.6; 
 108 lbs., respectively. 
 
 Page 140— 1(c). 492; 54.4; 432 lbs., respectively. 2. 68 lbs. 
 phosphorus; 540 lbs. potassium; no nitrogen. 3. 8.2 lbs. nitrogen 
 gained; 132.2 lbs. phosphorus gained; 99.2 lbs. potassium lost. 4(a). 
 No nitrogen; 25.5 lbs.' phosphorus. (b). 184.5 lbs. nitrogen; 
 20.4 lbs. phosphorus, (c). Gain of 184.5 lbs. nitrogen; loss of 5.1 lbs. 
 phosphorus, (d). Gain of 307.5 lbs. nitrogen; no increase in phos- 
 phorus. 
 
 Page 141— 5(a). 336.4; 42.1; 222 lbs., respectively, (b). 201.84; 
 33.68; 177.6 lbs., respectively, (c). 68.8%. 6. 274.5 and 23.5 lbs., 
 respectively. 7. 4,332. (a). 135.2. 8(a). 698.4 and 61.2 respectively. 
 
 Page 142— 8(b). 1,047.6 and 245 lbs., respectively. 9. 1.19 tons. 
 10(a). 50.6 lbs. gain. 
 
 Page 144— (b). 138.7 lbs. gain. 
 
 Page 145 — (c). 366.7 lbs. loss of nitrogen; 154.4 lbs. loss of phos- 
 phorus; 1,187.7 lbs. (d). 134.1 lbs. nitrogen gained; 74 lbs. phos- 
 phorus gained, (e). 524 lbs. lost. (f). 954.6 lbs. loss of nitrogen; 
 24 lbs. loss of phosphorus; 1,780| lbs. loss of potassium. 11. 1,939 
 lbs. loss of nitrogen; 123.5 lbs. loss of phosphorus. 
 
 Page 148— 1(a). 2,500; 500; 2,000 lbs., respectively. 
 
 Page 149— (b). $545. (c). $17.44. 2. $1.85; $2.11; $2.51; $3.50; 
 $3.82, respectively. 3. $3.25. 
 
 Page 150 — 4(a). $2.51. (b). The 16-ton application produced 
 54.5% greater crop value, (c). The 8-ton application, 74 cents more. 
 5(a). 30.3%; 24.07%; 58.96%, respectively, (b). 40.12%. (c). 83.4 
 cents, (d). $208.50. 
 
 Page 151—6. $94.50. (a). $304.75. (b). $31.50. 7(a). 332; 68; 
 224 lbs., respectively, (b). Gain of 75.3 lbs. nitrogen, 11 lbs. phos- 
 phorus and 3.6 lbs. potassium. 
 
 Page 152 — (a). Losses of 103 lbs. nitrogen, 23 lbs. phosphorus 
 and 91 lbs. potassium, (b). Losses of 41.5 lbs. nitrogen, 18 lbs. phos- 
 phorus and 50.5 lbs. potassium, (c). An average of between 35 and 
 48 cows. 
 
 Page 153—1. 256|. 2. 1,160. 
 
 Page 154—3. 5.34c; 6|c; 6}c, respectively. 4. About 2 qts. 
 or 4 lbs. (a). About 13. About 4. 5. 107.7. 
 
 Page 155—6. 14.38% P. 7. 275.55 lbs. P. 8. 41.5% K. 
 9. 290.5 lbs. K. 10. 1.32 lbs. P. more per $1 at $8.25. 11. 1,062.4. 
 12(a). 22 lbs. N.; 60 lbs. P.; 50 lbs. K.; (b). $12.20. 
 
 Page 156— 12(c). $12.30. 13(a). 70 lbs. N. ; 72 lbs. P.; 132 lbs. K. ; 
 $27.68; $8.82. 14(a). 21.5% K.; 3.5% P. (b). 75} lbs. K.; 12} 
 lbs. P. (c). $5.60. (d). K. 5.34c; P. 12fc. (e). 3,010 lbs.; $36.87. 
 
ANSWERS 1 
 
 824.05 saved. (The phosphorus in the mixed fertilizer is valued at 
 $7.22. Nitrogen was not needed). 
 
 Page 157 — (f). $17.50 returned from $5.60 when corn is worth 
 50c per bu. (g). 75| lbs. K. added, 33$ removed; \2\ lbs. P. added, 
 11.1 removed, (h). 97.4 lbs. muriate of potash mixed with 195.7 
 lbs. acid phosphate. 15. 466|; 1,143; 300 lbs., respectively, (a). 90£ 
 lbs. (b). $27.69, costing the least per 100 lbs. of fertility. 16. 104 
 lbs. P.; 814 lbs. K. (a). Between 1,893 and 1,938 lbs. of muriate of 
 potash; 1,485.7 lbs. acid phosphate. 
 
 Page 158— 16(b). 844.7 lbs., using 43% potash. 17(a). 180. 
 (b). $27. (c). 36 lbs. P.; 144 lbs. K.; $12.24. 
 
 Page 159 — (d). 334.8 lbs. potash and 450 lbs. basic slag, 
 (e). $11.30 compared with $39.24 as value of manure. 
 
 Page 162— 1(a). $24.68. 
 
 Page 163— (b). 123.4%. (c). $44.40, not counting cost of mar- 
 keting. 2. $17.55. 
 
 Page 164—3. $677.08. 4. 2,646.5%. 5. $504. 
 
 Page 165— (a). $57.14. (b). $48; 83.43. (c). S17.11.~~(d). 34.2c. 
 7. $45.63; $21.90; $2.74 (average yearly interest); $25.36; 56.35c. 
 
 Page 166— 8(a) ~67.7%; 22.3%. . (b). $17.72. 9(a). $9; 60c. 
 (b). 90 bu. of corn; 4.5 tons of stover, (c). $5.63. 
 
 Page 167— 9(d). $34.50. 10(a). $245.53. (b). $10.50/ (c). $65 
 net; $54.50 more than (b). 
 
 Page 168—11. 96.3c 12. 90c loss. 13. 491f and 535.7 lbs., 
 respectively, (b). 5. 14. $35.93 lost. 
 
 Page 
 
 169- 
 
 -14(a). 
 
 $73.42 
 
 gain. (b). 23.3- 
 
 -lbs. 
 
 (c). $28.64 loss. 
 
 15. .26+ lb.; 
 
 514f lbs 
 
 i. for flock of 1,980. 
 
 (b). 
 
 $208.80. (c). 10.54c. 
 
 (d). 154.4, 
 
 ; 1.7+ lbs. 
 
 (e). 19. 
 
 
 
 
 
 Page 
 
 170- 
 
 -16(b). 
 
 $117.65 profit. 
 
 
 
 
 Page 171- 
 
 -16(c). 
 
 $88.65 profit, (d). 
 
 8.74c. 
 
 (e). 10.4c. 
 
 17 (a and b). 
 
 
 
 
 
 
 
 Cow 
 
 Lbs. butter-fat 
 
 Value of fat 
 
 
 Net returns 
 
 1 
 
 
 139.2 
 
 
 $41.76 
 
 
 
 $1.24 loss 
 
 2 
 
 
 429.6 
 
 
 128.88 
 
 
 
 68.88 
 
 3 
 
 
 221.2 
 
 
 66.36 
 
 
 
 16.36 
 
 4 
 
 
 362.7 
 
 
 108.81 
 
 
 
 53.81 
 
 5 
 
 
 149.5 
 
 
 44.85 
 
 
 
 0.85 
 
 6 
 
 
 75.4 
 
 
 22.62 
 
 
 
 20.38 loss 
 
 7 
 
 
 314.0 
 
 
 94.20 
 
 
 
 44.20 
 
 8 
 
 
 127.9 
 
 
 38.37 
 
 
 
 6.63 loss 
 
 9 
 
 
 261.2 
 
 
 78.36 
 
 
 
 26.36 
 
 10 
 
 
 289.8 
 
 
 86.94 
 
 
 
 34.94 
 
 (c). $217.15; $21.72. 
 
 Page 172(a)— (d). $201.83 or $186.51 more than from the re- 
 maining six. (e). Between 13 and 14. (f). 81. (g). $440.85 or 
 $223.70 more than when these four cows were kept. 
 
8 AGRICULTURAL ARITHMETIC 
 
 Page 173— (h). $317.95. 
 18(a). 
 
 Star Nig 
 
 Value of fat in cream $107.81 $52.12 
 
 Amount skim milk 8,825 4,976 
 
 Value skim milk $13.23 $7.46 
 
 (b). Nig caused a loss of $11.18, hence she is worth more for beef. 
 $239.15, true value of Star. 
 
 Page 174— 19(a). 338.8%. 
 
 Page 176— 20(a). 123.6 lbs. valued at $18.54. (b). $17.53. 
 21(a). 32.76 lbs. P.; 14.9 lbs. K; 20.8 lbs. N. (b). $7.30. (c). $47.90. 
 
 Page 176— (a). $32.70. (b). $3.53. (c). $2.43. -(d). $4 or 
 17.7% loss when manure is valued at $1.50 per ton, and $14, or 42.8% 
 loss when valued at $2.18 per ton. (e). $14.60. (f).61c. (g). 228.08% 
 (h). 45.4c; 340.2%; $1.98 less than on strip No. 1 when manure is 
 valued at $1.50 per ton, and 15.6c. less than on strip No. 2. 
 
 Page 177— (j). Strip No. 2; $2 or 6.4% greater profits per acre 
 when other costs remain the same. (k). No. 2, $1.65 or 5.3% greater 
 profits per acre. 23(a). $9.84; 21.8c; 15.1c 
 
 Page 178— (b). 28 bu. (c). $17.59. (d). 51.73 bu. (e). 92c 
 loss, 80c profit per acre; $1.25 profit. 
 
 Page 179— 24(a). "A," 214.1% profit; "B," 56.2%. (b). "A" 
 realizes 281% more profit than "B" (per cent on fertilizer cost not 
 included). 25(a). Jones realized 250% profit as compared with 200% 
 by Brown, (b). Brown gained $225 more than Jones, other condi- 
 tions remaining the same. Jones 226% profit, Brown, 176%; Brown's 
 investment more profitable by $189.00. 
 
 Page 180— 25(d). More profitable by $75; $111. 26(a). $53.72. 
 (b). $1,074.50. (c). $525. (e). $1,321.60 when producing cost re- 
 mains the same. 
 
 Page 182— (b). $7.27. 
 
 Page 183— 30(a). $967.51. (b). $80.62. (c). 28.4c 
 
 Page 185— (a). $265.14. (b). $80.39. (c). $116.49. (d). $143. 
 32. $640.40. 33. $318.22. 34. $537.50. 35. 250%. 
 
 Page 186— 36(a). A 316|% increase in capital gives a 262.5% 
 increase in labor income, (b). $390; $1,495. 37(a). 62.6%; 55.5%. 
 (b). 14.6%; 7.1 - %. (c). 22.8%; 37.3%. (d). $2,258.50; "B" received 
 no labor or managerial income, since he realized only 3.8% profit on 
 his investment, (e). "B" has too large a proportion of "fixed" 
 capital. 
 
 Page 187 — 38(a). 39.8% more profitable, other conditions being 
 equal, (b). 26.4%. 39. $254. 40. $424. 
 
 Page 188— (a). $16,486.00. (b). $5,438.80. (c). $2,418.00. 
 (d). $3,020.00. (e). $824.30. (f). $2,195.70. 
 
 Page 189 — 1. 80; 40; 10 acres, respectively. 
 
 Page 191— 7(b). 2.25 acres. 
 
 Page 192— 8(b). 1.2 acres. 9(a). 2.8 acres. 10(b). 2.04 acres. 
 
ANSWERS 9 
 
 Page 193— 10(c). About 95 rds. (Use thread to measure). 
 11. 9.03 acres, approximately. 12(a). (1) 12.7 acres; (2) 14.76 acres; 
 (3) 11.7 acres; (4) 11.25 acres; (5) 2.5 acres; (6) 10.9 acres; (7) 10.9 
 acres; (8) About 5.7 acres, (b). (1) 205 rds.; (2) 205 rds.; (3) 190 
 rds.; (4) 163.75 rds.; (5) 80 rds.; (6) 170 rds.; (7) 170 rds.; (8) 102.5 
 rds. (c). 9333 rds. 13(a). $174.37. (b). 194.8%. 
 
 Page 194—14. $268.08. 
 
 Page 197—15. 25.92; 27.77; 23.32, respectively. 17. 7.56 tons. 
 18. $10. 19. 15.6 tons. 
 
 Page 199—22. $80.88. 23. 6.8 tons. 24. 3,214.2 bu. 26. 1,768 
 bu.; 23.8 tons more of water. 
 
 Page 200—26. 540 bu. (quick calculation); 32c. per bu. 
 27. 2,250 bu. (quick calculation). 28. $394.66 (short method). 
 29. 937.5 baskets (heaped) or 625 bu. (shelled). 30. 49.4 bbls. 
 31. 7.48 gal. 32. 8.77 ft. 33. 2 rds. apart each way. (b). 90 lbs. 
 (c). 4. 
 
 Page 201—34. 33.13 acres. 35. 314. 36. 600 bu.; 9.6 tons; 
 14.5; 18 (short method). 37. 10 of sand, 5 of cement. 38. 3J cu. 
 ft. sand, 7 cu. ft. gravel; 10J cu. ft. sand, 21 cu. ft. gravel. 
 
 Page 202—39. 11.4 cu. ft. sand, 22.8 cu. ft. stone. 40. 8.88 bbls.; 
 3.1 cu. yds.; 6,2 cu. yds., respectively. 41. 50.5 bbls. cement, 14.2 cu. 
 yds. sand, 28.4 cu. yds. gravel. 42. About 2 bags cement, 4.6 cu. 
 ft. sand and 9.2 cu ft. gravel. 43. One of at least 120-ton capacity, 
 and should be 16 ft. in diameter and about 35 ft. deep (allowing a 
 few feet for settling of silage). 
 
 Page 203 — (a). 33 bbls. cement, 11 cu. yds. sand, 22 cu. yds. 
 gravel or crushed stone. 
 
 Page 206—1. 48 to 108; 108 to 170; 69 to 170, respectively. 
 2. 33. 3. 25 more. 
 
 Page 207— 3(b). 58. 4. 40; about 70. 5(a). 3 more. (b). 31.1 
 feet. 6. 160; 280; 184, respectively. 7. 28 lbs. of each material. 
 
 Page 208 — 8. 6| lbs. soap and 26 i gal. kerosene for summer 
 spraying; 13.8 lbs. soap and 55.4 gal. kerosene for winter spraying. 
 9(a). $708.21. (b). $1,123.54. (c). $236.07; $374.51. 
 
 Page 209— (d). 12.48%; 44.06%. 10(c). $8.59; $233.76. 
 
 Page 210— 11(a). 93.2c; $108.86. (b). $3.55 per tree; $96.65 
 per acre. (c). $121.65 (net income includes interest on investment); 
 24.33%. 12. 1.6 bu. - (a). J oz. of formalin in one gallon of water, 
 (b). i*s lb. Paris green and f lb. lime and 5 gal. water. 
 
 Page 211— 12(c). \ lb. blue vitriol and \ lb. lime. 13(a). $76.55. 
 (b). $42.68. (c). i acre. 
 
 Page 212— (d). 352 bu. (e). 700.2 bu.; $997.78. 
 
 Page 216— 1(a). 36.5%; 63.4%, respectively. (b). $89.58. 
 (c). 25.9% less; 35% greater. 2. 51.7%. 3. 27.8c. 4. 24c. 
 
 Page 216—8. 11.06 lbs. at 27c. per lb. 9. 7.03 bu. 10. 67.7 gal. 
 
10 AGRICULTURAL ARITHMETIC 
 
 Page 217—12. 12.68 lbs. 13. .9 lb. more in the bananas. 
 14. yi oz. in favor of meat (11 lbs. eggs). 15. The cheese, 2.3 lbs. 
 more than beef and 3.6 lbs. more than veal. 
 
 Page 218 — 17. 1.1 lbs. or about 10 eggs, 5.6 oz. cheese, 10 oz. 
 ham, 4 oz. peanuts, 1.5 lbs. potatoes, 2.4 lbs. apples, 4 oz. wheat bread. 
 18. 95.8%; 12.8%. 19(a). Total of .16 lb. protein, 1.03 lbs. carbo- 
 hydrates, .2 lb. fat. 20. 5.1% more protein, 25% more ash. 
 
 Page 219—21. 144.3% more. 22. .06 lb. nutrients to be supplied 
 by 1 +oz. grapenuts. (a). 4fc. and 7 + c, respectively. 
 
 Page 220— 24(b). 3.1 lbs.; 2.46 lbs., respectively, (c). 1:10.9 
 and 1:7.1, respectively, (e). .06 lb. and .06, respectively. 
 
 Page 221—1. 1,680.4 sq. ft. 2 .9 + c. on a dollar. 3. $1,800.00. 
 (a). $163.35. 4. About 56.5 rds. 
 
 Page 222— (a). About 28 rds. 5. 126.5 times or 12,650% more. 
 6. Between 34,700 and 69,400 per min. 8. From .04% to .0467%. 
 
 Page 223—9. $1,920.00; $160. 10. About 3.9 lbs. 11. 15 bu. 
 11(a). $11.26. 
 
 Page 224— 11(b). 81 bu. (c). $11.26 or $61.30 per acre, 
 (d). 75c. per bu. in feeding. 12. About J oz. soap and 6 pts. kerosene 
 for summer, and about .6 oz. soap and 1.2 pts. kerosene for winter 
 spraying. 13. .8 lb. each of blue vitriol and lime. 14. $288.90. 
 (a). $182.97. 15. 17,191 ft. 
 
 Page 225—16. About 14. (a). 8.8 mo. 
 
 Page 226— 17(a). $534.59. (b). $1,458.87 total for year, 
 (c). 67,757 eggs; nearly 112 per fowl; $2.43. (d). $602.28; practically 
 $1 per fowl. (e). $1,100 at 6%. 18. Cotton, 40 oz. more. 19. $54.75 
 spent foolishly. 20. 16.4 + %. (a). $575; $479.16. 21. 2 lbs. more 
 phosphorus may be purchased per dollar at $9 per ton. 
 
 Page 227—22. One part. 23. .057. 24. 160. (a). 155,040. 
 (b). 22%. 26. 21.3 gal. (a). 3.4 bbl. 26. There would be but little 
 difference. 
 
 Page 228—27. 4.15%. (a). 60.42 lbs. (b). 1,384.08 lbs. valued 
 at $415.22; $48.96 (30% cream considered drawn); $464.18. 
 
 Page 229— (c). 4 +. 28. 100% more. 29. 48.3%; 59.3% red 
 clover; 71.1%. 30. $38.35. 31. .94 and 1.35 tons respectively; 43.6% 
 increase. 32. No, $4.81 loss per acre per year. (Clover considered 
 as not reducing nitrogen content of soil.) 
 
 Page 230 — 33. About 264 lbs. loss of nitrogen, 114 lbs. phosphorus 
 and 398 lbs. potassium. (Oat straw used for bedding.) 
 
 Page 231— (a). About 13; 5; 15. 
 
 Page 232— 35(a). 2% N., 5.2% P., 6.6% K, 40 lbs. N., 104 
 lbs. P., 132.8 lbs. K. (b). 179.5% and 245% profits, respectively, 
 (c). Man of small means can get greater per cent on his limited money 
 when he applies 600 lbs. per acre; whereas a man of wnple means 
 can realize more net returns per acre when he applies 1,200 lbs. per 
 acre. (d). $9.23 more profitable per acre (other conditions remain- 
 ing the same). 
 
Agricultural Text Books 
 
 FOR 
 
 HIGH SCHOOLS 
 
 Published by 
 
 WEBB PUBLISHING CO., ST. PAUL, MINN. 
 
 FIELD CROPS 
 
 By A. D. WILSON, Sup't of Farmers' Institutes and Extension, 
 Minnesota College of Agriculture, and C. W. WAR- 
 BURTON, Agronomist, U. S. D. A. 
 
 544 pages, 162 illustrations, cloth, $1 50 net. 
 
 This book discusses the peculiarities of each of the various classes 
 and varieties of farm crops, the handling of the soil, selections of seed, 
 and general crop management. It covers the cereals, including corn, 
 wheat, oats, rye, barley, etc.; forage crops, including hay grasses, clo- 
 ver, alfalfa, cowpeas and other legumes; how to make good meadows 
 and pastures, and the art of hay making, etc.; root crops; sugar crops; 
 fiber crops, including cotton, flax, hemp; tobacco, potatoes, in fact 
 every farm crop of any importance is discussed. The introductory 
 chapters give the general classification of farm crops and their uses and 
 relative importance, and review the subject of how plants grow. The 
 concluding chapters discuss the theory and practice of crop rotation and 
 weeds and their eradication. A list of supplementary references is giv- 
 en at the close of each chapter. The style is easy, subject matter well 
 arranged and vital, and the book is of excellent mechanical makeup 
 throughout. 
 
 AGRICULTURAL ENGINEERING 
 
 By J. B. DAVIDSON, Professor of Agricultural Engineering, 
 Iowa State College 
 
 554 pages, 342 illustrations, clotb, $1.50 net. 
 
 A volume intended primarily as a text for secondary schools of agri- 
 culture, and for colleges where only a general course can be offered. 
 The subjects discussed are so applicable to the every-day work of the 
 farm that the book will also prove of great interest and value to those 
 engaged in practical agriculture. The following subjects are given space 
 according to their importance: Agricultural Surveying, Drainage, Irri- 
 gation, Road Construction, Farm Machinery, Farm Motors, Farm 
 Structures, Farm Sanitation, and Rope Work. Each chapter is fol- 
 lowed by a set of questions for review and for thought promotion. 
 Lists of references to best books and bulletins are included. Complete 
 index. A splendid text in every detail. 
 
POPULAR FRUIT GROWING 
 
 Samuel B. Green, Late Professor of Horticulture, University 
 of Minnesota, 4th edition revised by Le Roy Cady, Associate Pro- 
 fessor of Horticulture, University of Minnesota. Cloth, 12 mo., 300 
 
 pages, illustrated $1.00 
 
 Paper 60 
 
 All phases of fruit growing are discussed in a clear and interesting 
 manner. 
 
 The book deals with every branch of the work, from the prepara- 
 tion of the soil to marketing, and explains every operation in detail. 
 The principal topics treated are Factors in Successful Fruit Growing, 
 Orchard Protection, Insects Injurious to Fruits, Diseases, Harvesting 
 and Marketing, Principles of Plant Growth, Propagation, Pome 
 Fruits, Stone Fruits, Grapes, Small Fruits and Nuts. An appendix 
 contains many formulas and other matters of importance. 
 
 This text has fully justified itself by years of popularity. It 
 has maintained its hold in many schools of high standing and over 
 a wide area solely on account of its perfect adaptability to the very 
 uses for which it was intended. 
 
 VEGETABLE GARDENING 
 
 Samuel B. Green, Late Professor of Horticulture, University of 
 Minnesota, 12th edition revised by Le Roy Cady, University of Minne- 
 sota. Cloth, 12 mo., 336 pages, illustrated $1.00 
 
 Paper .60 
 
 Vegetable Gardening presents the subject of gardening in a clear 
 and definite way. Its aim is to enable the student to understand 
 the fundamental principles involved in each practical undertaking 
 and to recognize the importance of careful, systematic and scientific 
 methods in the growing of vegetable crops. The entire plan of the 
 book is noteworthy. Each garden crop is discussed fully from seed 
 to marketing. There are also valuable treatments of Gardening 
 in General, the Home Garden, Fertilizers, Tillage, Seeds, Transplanting, 
 Development of Varieties, Glass Structures .Insects and Insect- 
 icides, Marketing, Exhibiting, etc. 
 
 The above features make this a textbook especially adapted to 
 agricultural high schools and other secondary agricultural schools 
 and for beginning classes in colleges. For years it has been regarded 
 as a leading authority. 
 
 PROBLEMS IN CARPENTRY 
 
 Louis M. Roehl, Director of Farm Mechanics, Milwaukee County 
 School of Agriculture and Domestic Science, Wauwatosa, Wis. Cloth, 
 Ty^&Yi inches, 112 pages $1-00 
 
 A Course in Practical Carpentry for Manual Training Classes 
 in Public Schools. It comprises a series of sixteen projects fully 
 illustrated and covering all the features of house constructing, including 
 practice in framing, flooring, lathing, plastering, calcimining, paint- 
 ing, varnishing, etc. 
 
 Full scale lumber is used for each project. Actual building con- 
 ditions are made the basis of the work. The course is exceptionally 
 practical and exactly adapted to industrial classes. 
 
 A plan for each project is given along with two or three different 
 views of the finished project. 
 
j Industrial W>rk/^ Boys | 
 
 A. E. PICKARD 
 
 A COMPANION VOLUME TO 
 "INDUSTRIAL WORK FOR GIRLS" 
 
 This volume is in keeping with the rapid strides that are 
 being made by industrial education. It is an up-to-date text 
 for teaching industrial work to boys in rural and graded 
 schools. With the exception of its companion volume, no 
 other book is better designed for training the hand as well as 
 the head. 
 
 CONTENTS 
 
 Chapter I— Course and Equipment. Purpose of Industrial Work, 
 Preliminary Industrial Work, Second and Third Division Work. 
 
 Chapter II^General Industrial Work. School Exercises in Weav- 
 ing, Paper Folding and Construction, Raffia and Rattan Work, 
 Modeling. 
 
 Chapter III—*Rope Work and Belt Lacing. Whipping, Crowning, 
 Splicing, Making Knots, Rope Halters, Block and Tackle Reev- 
 ing, Lacing Three-inch and Six-inch Belts. 
 
 Chapter IV— Woodwork at School. Equipment, Thirty-two Man- 
 ual Training Exercises. 
 
 Chapter V—^Home Projects in Woodwork. Equipment, Nineteen 
 Home Credit Projects, including the Making of a Work Bench, 
 Folding Ironing Table, Stepladder, Chicken Coop, Stock Rack, 
 Wagon Box, Farm Gate and Road Drag. 
 
 Chapter VI— Projects in Cement and Iron. Making Concrete 
 Walks, Floors, Posts and Building Blocks, Iron Work. 
 
 Chapter VII— Home Credit Work in Agriculture. Soil Study, Ro- 
 tation of Crops, Germination Tests, Garden Work, Weed Col- 
 lection, Insect Collection, Collection of Woods, Study of Birds 
 and Rodents, Study of Machinery, Stock and Grain Judging, 
 Tree Grafting, Strawberry Raising. 
 
 Chapter VIII— Contests and Club Work. Acre-yield Corn Plot, 
 Potato Yield Contest, Tomato Contest, Home Canning, Poultry 
 Contest, Pig Contest, Savings Banks, Keeping Accounts, In- 
 dustrial Exhibit. » 
 
 The book contains over 100 illustrations, most of which 
 are working drawings for the projects suggested. These 
 diagrams alone are worth in school or at home many, times 
 the price of the book. 
 
 12mo., about 150 pages. Illustrated. Price, 40 cents net 
 
 WEBB PUBLISHING COMPANY, 
 
 SAINT PAUL, MINN. 
 
INDUSTRIAL WORK 
 FOR GIRLS 
 
 A. E. PICKARD 
 
 AND 
 
 MARIE C. HENEGREN 
 
 A COMPANION VOLUME TO 
 "INDUSTRIAL WORK FOR BOYS" 
 
 THIS volume is in keeping with the rapid strides that are 
 being made by industrial education. It is an up-to-date 
 text for teaching industrial work to girls in rural and graded 
 schools. With the exception of its companion volume, no other 
 book is better designed for training the hand as well as the head. 
 
 =CONTENTS= 
 
 =f Chapter I— Course and Equipment. Purpose of Industrial Work, = 
 
 == Preliminary Industrial Work, Second and Third Division Work. == 
 
 H Chapter II— -General Industrial Work. School Exercises in Weav- = 
 
 = ing. Paper Folding and Construction, Raffia and. Rattan Work, EE 
 
 = Modeling. ' j=j 
 
 = Chapter III— Sewing in the Rural School. Equipmant, Classifica- £| 
 
 S tion of Stitches, School Exercises, Home Projects. = 
 
 H Chapter IV— Principles of Home Science. Food Denned, Food f| 
 
 = - Principles, Preservation of Food, Planning Menus, Purpose and = 
 
 == Principles of Cooking, Bread Making, Cake Making, Pie Making, == 
 
 U Salads, Beverages, Experiments. = 
 
 = Chapter V— The Hot Lunch. Equipment, General Directions. Sauces == 
 
 = and Thickening for Cream Soups, Suggestive Dishes, Recipes. == 
 
 M Chapter VI— Industrial Club Work. Tomato Contest, Home Can- =J 
 
 = ning. Bread Baking Contest. == 
 
 ^ Chapter VII— Home Credit Exercises. The Flower Garden, House = 
 
 = Plants, Bird Study, Bed Making, Preparing a Meal, Laying the == 
 
 Table, Serving Meals, Clearing Dining Table and Washing Dishes, = 
 
 = Fly Control, Planning the Home, Ventilation, Savings Banks, = 
 
 = Home Accounts, Industrial Exhibit. == 
 
 3 The book contains 62 illustrations, many of which are of = 
 
 |[ sewing stitches and exercises. It contains all the practical g 
 
 5 features of a modern industrial course for girls. = 
 
 = 12mo. y about ISO pages. Illustrated. Price, 40 cents net =E 
 
 | WEBB PUBLISHING COMPANY | 
 
 I SAINT PAUL, MINN. |j 
 
 llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllffl 
 
I INDUSTRIAL BOOKLETS I 
 
 I A. E. PICKARD | 
 
 = A BOOK ON BOOKLETS— TEACHING THE EXPRESSION = 
 
 = AND ILLUSTRATION OF INDUSTRIAL SUBJECTS THRU = 
 
 H COMPOSITION AND ART || 
 
 = Suitable for Grade Work in All Kinds of Schools = 
 
 = This book on booklets contains a series of outlines on Agriculture, ^ 
 
 = Horticulture, Animal Husbandry, Home Economics, and other subjects. = 
 
 m One topic — Poultry — is thoroly developed- as an example of the way in = 
 
 = which the work should be done. == 
 
 = The making of the booklets is a part of the language work. The ^ 
 
 |E character of the subjects and the interest taken in their study vitalize = 
 
 = the school life by supplying a wealth of pleasing material for discussion = 
 
 = and composition. Different important results are secured by the use = 
 
 = of these outlines. == 
 
 = This investigational method of study is the most practical and = 
 
 = pedagogical. = 
 
 = Topics of vital interest are impressed with their bearing on modern ^ 
 
 = activities of general concern, and pupils, under proper direction, ac- 3 
 
 = quire the habit of orderly and effective expression. The general char- == 
 
 s acter of the school work is elevated to a new plane. =_ 
 
 ^ 12mo., 144 pages. Illustrated. Price 40 cents net = 
 
 1 Webb Publishing Go. Saint Paul, Minn. j| 
 
 I Rural School Lunch | 
 
 1 NELLIE WING FARNSWORTH 1 
 
 3 DIRECTOR OF HOME ECONOMICS =_ 
 
 == STATE NORMAL SCHOOL, VALLEY CITY, N. D, =_ 
 
 m A TIMELY TREATMENT OF THIS TOPIC 1 
 
 3 FULL OF PRACTICAL VALUE FOR TEACHERS = 
 
 = As a help to convenience, comfort, health, vigor of mind, education |= 
 
 = and culture, this little booklet brings its offering in the hope and belief = 
 
 = that it may have a part in securing better conditions for multitudes who =_ 
 
 = need and deserve them. = 
 
 = TOPICS TREATED BY THE AUTHOR M 
 
 = Need of the School Lunch Advantages of the Lunch = 
 
 = Necessary Equipment Teachers' Special Problems = 
 
 = Methods of Maintenance Pood Study i= 
 
 = Suitable Dishes Composition of Food Stuffs = 
 
 = Management Recipes = 
 
 =i Full details are given for the installation and conduct of the rural j= 
 
 s school lunch. By means of the tables and directions a teacher can = 
 
 = easily work out the whole problem of not only the one dish but ofa i = 
 
 s= whole meal and of child nutrition in general. The plans embody the = 
 
 = extensive experience of the author. 3 
 
 = Price in paper covers, illustrated, 25 cents. == 
 
 s Webb Publishing Co. Saint Paul, Minn. g