WlLjL'S COMMERCIAL ARITHMETIC GIFT OF If M7 ms ' if fto WILL'S COMMERCIAL ARITHMETIC PRESENTING THE BEST USAGE IN MODERN BUSINESS PRACTICE BT WILLIAM R. WILL n FOR THIRTY-TWO YEARS PRINCIPAL OP THE MATHEMATICAL DEPARTMENT OF SADLER'S BRYANT AND STRATTON BUSINESS COLLEGE BALTIMORE; AND AUTHOR OF THE SADLER-ROWE SERIES OF COMMERCIAL ARITHMETICS THE GREGG PUBLISHING COMPANY NEW YORK CHICAGO SAN FRANCISCO COPYRIGHT, 1913, BY THE GREGG PUBLISHING COMPANY f -V THH-PLIMPTON-PRESS NORWOOD- MASS-U-S. A PREFACE IT has been found extremely difficult, if not impossible, for the average student of arithmetic to remember its many topical rules and explanations, and to exercise due discrimination in giving to each its local application. "Will's Commercial Arithmetic" has therefore ignored this unsatisfactory method of treatment, and has substituted the few cardinal principles which underlie all arithmetical processes, and which can be used in the solution and explanation of all problems. The learner is thus liberated from slavish adherence to mechanical rules and hackneyed explana- tions which are soon forgotten, and is properly cultured in the exercise of his reasoning faculties in the solution of problems, the effect of which is permanent. The only conceivable modifications of any given number in a problem are effected either by increasing it or by decreasing it. The law of increase or decrease is uniform in its application. Addition and multiplication are the only known numerical pro- cesses for increasing a number, the former being employed when the increase is by one or more given unequal components, and the latter when by equal components. Conversely, subtraction and division are the only known means for diminishing a num- ber, the former being employed when the decrease is by one or more given unequal components, and the latter when by equal components. Each of the four preceding processes is thus seen to be invariably limited to its own prerequisite conditions ; and the proper process to select will depend upon the particular pre- requisite conditions which are given. The intelligent solution of a problem can therefore involve nothing further than the identification of its numerical terms, a knowledge of the specific process to which these identified terms invariably belong, and the mechanical performance of that process. 314603 iv PREFACE The above method of treatment of all problems, though a departure from the traditional standards, is not an untested theory, but a thoroughly demonstrated success. It is the ripe fruitage of forty-eight years' experience in the class-room, during the last thirty-two of which the author has been engaged in teaching commercial arithmetic exclusively in one of the oldest and most reputable business colleges in the United States. To his fellow teachers who have been disappointed either in the lack of permanency or of quality in the results hitherto accom- plished by the conventional treatment of this most important study, the methods of this volume are confidently submitted as an efficient means for the accomplishment of entirely satisfactory results in the future. THE AUTHOR CONTENTS INTRODUCTION PAGE Preliminary Definitions 1 NUMERICAL EXPRESSION Reading Numbers 2 Writing Numbers 4 Roman Numerals 6 NUMERICAL OPERATIONS Addition 7 Subtraction 9 Multiplication 11 Division 17 United States Money 24 REVIEW Relation of Numbers 29 Review Examples 34 PROPERTIES OF NUMBERS Factoring 39 Least Common Dividends .... 41 Greatest Common Divisors ... 43 Cancelation 45 COMMON FRACTIONS Introduction 46 Definitions 47 General Principles 48 Reductions 49 Addition 55 Subtraction 57 Multiplication 60 Division 63 Relation of Fractions 69 Fractional Comparisons 71 Review Examples 75 DECIMAL FRACTIONS PAGE Introduction 77 Definitions 78 General Principles 79 Reading Decimals 81 Writing Decimals 81 Reductions 82 Addition 85 Subtraction 87 Multiplication 88 Division 89 Review Examples 92 COMPOUND NUMBERS Introduction 94 Definitions 95 Tables of Measures 95 Reductions Denominate Integers 100 Denominate Fractions 102 Addition 104 Subtraction 105 Multiplication 107 Division 108 Review Examples 110 Metric System Ill MEASUREMENTS General Surfaces 116 Carpeting, Paving, Roofing ... 118 Boards, Joists, etc 121 Volumes, generally 122 Bins, Cisterns, Wells 125 PERCENTAGE Introduction 129 Definitions. . 129 VI CONTENTS PERCENTAGE (Continued] PAGE Identification of Terms 130 Relation of Terms 131 To find Percentages 132 To find Rates 135 To find Bases 137 Review Examples 145 APPLICATIONS OF PERCENT- AGE Profit and Loss 148 Trade Discount 154 Ordinary Interest 161 Percentage Method 162 Aliquot Method 165 Accurate Interest 174 Relation of Terms 175 Compound Interest 181 True Discount 185 Bank Discount 189 To find Proceeds 190 To find Face of Note 193 Partial Payments United States Rule 195 Merchants' Rule 197 Commission 200 Stocks and Bonds 205 Stock Investments . 211 APPLICATIONS OF PERCENT- AGE (Continued) PA QE Exchange 214 Domestic 216 Foreign 218 Bankruptcy 221 Insurance 223 Taxes 224 Duties or Customs 225 MISCELLANEOUS Proportion Introduction 227 Simple Proportion 228 Compound Proportion 231 Equation of Payments 233 Averaging Accounts 241 Averaging Accounts Sales 247 Cash Balance 248 Partnership 252 APPENDIX Aliquot Calculations 267 Addition by Grouping 272 Cross Multiplication 274 Powers and Roots 274 Answers . . 283 COMMEECIAL ARITHMETIC INTRODUCTION 1. Arithmetic is that branch of knowledge which treats of numbers. It explains the possible relations between the numer- ical expression of one given quantity and that of another given quantity, and how those relations may be utilized to find the numerical expression of a required quantity. 2. Quantity is that property of matter which may be meas- ured, weighed, valued or counted, and which may be expressed by a number. 3. A number is the expression of a quantity by the use of figures. NOTE 1. (a) A number is said to be abstract when the kind of quantity is not expressed, but only its extent, as three, five times; (6) to be concrete when the kind of quantity is expressed as well as its extent, as five gallons, seven yards; (c) to be simple when a concrete quantity is expressed in only one denomination, as three bushels, twelve barrels; (d) to be com- pound when one concrete quantity is expressed by the use of two or more different denominations, as three pounds and nine ounces of sugar, six gallons three quarts and one pint of water; (e) to be integral when a quantity is expressed in one or more whole units, as six, nine pounds; (/) to be fractional when a quantity is expressed in one or more parts of a whole unit, as one-half, three-fourths. NOTE 2. When one number is placed in contrast with one or more other numbers it is said to be (a) like when it expresses a quantity of the same name, as five yards is a like number to eight yards; (b) and it is said to be unlike when it expresses a quantity of a different name, as eight inches is an unlike number to five gallons. 4. A unit is one thing of the quantity expressed by a number. Thus, the unit of eight yards is one yard, of twelve miles is one mile. 5. A problem in arithmetic is a statement containing numerical terms which are mutually related, and which requires READING 1 . NUMBERS that relation to be perfected by finding an omitted term, called the answer. NOTE. The separate unequal components and their total are distinctive terms of addition (if a total is required), or of subtraction (if a total is given). One equal component, the number of equal components, and the total of equal components, are distinctive terms of multiplication (if a total is required), or of division (if a total is given). These five terms are the alphabet of arithmetic. 6. The analysis of a problem is the process of separating it into its given numerical terms, ascertaining their character, discovering their relation to each other, and thus deducing the appropriate operation to find the required term which will com- plete that relation. NOTE. A completed relationship of arithmetical terms is a numerical total and the several equal or unequal numerical terms which constitute that total. 7. The solution of a problem is the process of finding the required term which will complete the relationship which exists between the given terms. NOTE. A relation is completed by finding a required component which will complete a given total, by finding a required total which will include all of its given components, or by finding the number of given equal components, which are contained in a given total. READING NUMBERS 8. Reading numbers is the mental conception or the vocal expression of the value of a written number. 9. Numbers are commonly expressed by the use of the ten following figures, used singly or in combination: 123456789 Naught One Two Three Four Five Six Seven Eight Nine The first of the above figures is called naught or cipher, and it expresses the absence of value. The remaining nine figures are called significant figures, (a) When written by itself or at the extreme right of a number, each figure expresses the value placed underneath it in the above illustration. Numbers greater than nine are expressed by a systematic combination READING NUMBERS 3 of the above figures. Thus, (6) to increase any of the above figures tenfold, place one figure at the right of it; (c) to increase it a hundredfold, place two figures at the right of it; (d) to in- crease it a thousandfold, place three figures at the right of it; etc. (e) The local value of a figure is determined by its position, in- creasing tenfold for each place that it is moved to the left; (/) and decreasing tenfold for each place that it is moved to the right, (g) The position which a figure occupies in a number, counting from the right, determines its value and is called its order. ILLUSTRATIVE EXERCISE 10. Read 4976. EXPLANATION. Commencing with the left-hand figure (4) read it and name its local value (4 thousand), as it has three figures at its right [9, d]; then read the next following figure and name its local value (9 hundred), as it has two figures at its right [9, c]; then read the next following figure and name its local value (7ty or 7 tens, ty being a corruption of ten), as it has one figure at its right [9, 6]; then read the next following and last figure (6), omitting its name (units), as it is understood [9, a]; thus making the entire expression read " four thousand nine hundred seventy-six." 11. Numbers composed of more than three figures are read in sets of three figures each, (a) The first set of three figures, count- ing from the right, is called the Units 1 Period; (b) the second set of three figures from the right is called the Thousands' Period; (c) the third set of three figures, the Millions' Period; (d) the fourth set, the Billions' Period; (e) the fifth set, the Trillions' Period; (/) the sixth, the Quadrillions' Period; etc. SUGGESTION. To read and write numbers with facility, it is necessary that the above periods be memorized until, without hesitation, they can be repeated in regular order, forward from the highest period, or backward from the lowest. EXERCISES Express the following numbers in written or spoken words : 1. 9 6. 72 9. 426 13. 7836 17. 8003 2. 14 6. 45 10. 865 14. 5492 18. 4600 3. 18 7. 97 11. 309 16. 6285 19. 3280 4. 36 8. 58 12. 200 16. 9058 20. 7025 4 WRITING NUMBERS ILLUSTRATIVE EXERCISE 12. Read 26637289Q219. EXPLANATION. To discover the name of each period in the given number and the figures which compose it, commence at the right and separate it into periods of three figures each, as follows: 26,463,728,496,219. As the right- hand period (219) is pointed off, think of its name (units); as the second period (496) is pointed off, think of its name (thousands) ; as the third period (728) is pointed off, think of its name (millions), and so continue until the left-hand period (26) is reached, which is thus found to be trillions. Then read the number expressed by the left-hand period as if it stood alone, calling its name (twenty-six trillion) ; then read the number expressed by the next following period as if it stood alone, calling its name (four hundred sixty-three billion) ; then, similarly, the next following period (seven hundred twenty-eight million); then the next period (four hundred ninety-six thou- sand); and finally the last period, omitting its name (units) as it is under- stood (two hundred nineteen). EXERCISES Express the following numbers in written or spoken words: 1. 2785 11. 678349526 21. 51836274813429 2. 16298 12. 2817462785 22. 200070600083294 3. 34276 13. 7132975138 23. 56000000400000 4. 416298 14. 231678261582 24. 290867320000078 6. 728341 16. 57281334967864 25. 689712000091863 6. 2685913 16. 1963475162834748 26. 500030007100006 7. 5873265 17. 5600328176003452 27. 8620053000007008 8. 37264897 18. 92086005027316814 28. 65000000002009 9. 52819624 19. 51378260000009162 29. 1834000000007 10006 10. 83429857 20. 60000520080076003 30. 2000009000070005 WRITING NUMBERS 13. (a) Writing numbers is the expression in figures of the value of a number when given in words. (6) Knowing the names of the several periods and their relative positions [11], any num- ber is written by commencing with its highest period and writing it as if it stood alone, then writing its next highest period as if it stood alone, and so continuing with period after period in regular order until the lowest period is written. NOTE, (a) If, after writing the highest period, any subsequent period should be found to contain fewer than three figures, supply the deficiency by WRITING NUMBERS 5 prefixing the requisite number of ciphers. (6) If an entire period is omitted, supply three ciphers for its three vacant orders. ILLUSTRATIVE EXERCISE Express in figures, twenty-seven quadrillion, nine hundred sixty-eight trillion, seventy-two million, eight thousand forty. EXPLANATION. Commencing with the highest period of the given num- ber (quadrillions), write it as if it stood alone (27 . . .); then write the next lower period (trillions) as if it stood alone (27,968 . . .) then write the next lower period (billions) and since no billions are expressed in the given number [Note, 6,13], supply three ciphers for its three vacant orders (27,968,000 . . .); then write the next lower period (millions), and as seventy-two requires only two figures to express it and each period except the highest must contain three figures [Note, a, 13], supply the deficiency by prefixing one cipher (27,968,000,- 072 . . .); then write the next lower period (thousands), and as eight requires only one figure to express it, and each period except the highest must contain three figures, supply the deficiency by prefixing two ciphers (27,968,000,072,- 008 . . .); then write the next lower period (units), and as forty requires only two figures to express it, prefix one cipher, obtaining 27,968,000,072,- 008,040 as the complete numerical expression. EXERCISES Express in figures the following numbers: 1. Six thousand, three hundred fifty-two. 2. Forty-nine thousand, eight hundred seventy-six. 3. Two hundred nine thousand, seven hundred. 4. Five hundred eight thousand, sixty-three. 6. Ninety thousand, eight. 6. Eighty-three thousand, six hundred ninety-five. 7. Twelve million, seventy-one thousand six. 8. Three million, sixteen thousand, two hundred eight. 9. Nineteen million, three hundred fifty-eight thousand forty. 10. Twenty-five million, five hundred twenty-seven. 11. Four trillion, seventy-six million, two hundred fifty- three thousand, seven. 12. Fifteen quadrillion, two hundred ninety-one trillion, five billion, seven-two thousand, sixteen. 6 ROMAN NUMERALS Express in figures the following numbers: 13. Ninety-seven quadrillion, one hundred twenty-eight million, four hundred thousand, ninety-two. 14. Six quadrillion, seven hundred trillion, eighty billion, nine hundred fifty thousand, twenty. ROMAN METHOD OF EXPRESSING NUMBERS 14. The Roman method of expressing numbers is dependent upon the following seven capital letters, used as numerals. I One V Five X Ten L Fifty c One hundred D Five hundred M One thousand NOTE 1. Placed by itself, each of the above Roman numerals expresses the value written underneath it. NOTE 2. Successive repetitions of the above numerals denote repetitions of the value which they express. Thus, II express two ones, or two; III, three ones, or three; XXX, three tens, or thirty; CC, two hundred; etc. NOTE 3. The combined value when one numeral is placed at the left of another of greater value expresses the difference between their respective values. Thus, IV expresses the difference between one and five, or four; IX expresses nine; XL, forty; XC, ninety; etc. NOTE 4. The combined value when one numeral is placed at the right of another of greater value expresses the sum of their respective values. Thus, VIII expresses eight; XXV, twenty-five; LXII, sixty-two; etc. 15. The manner in which Roman numerals are combined to express numbers is shown in the following TABLE OF ROMAN NUMERALS I One XIV Fourteen XC Ninety II Two XV Fifteen C One hundred III Three XVI Sixteen CC Two hundred IV Four XVII Seventeen ccc Three hundred V Five XVIII Eighteen cccc Four hundred VI Six XIX Nineteen D Five hundred VII Seven XX Twenty DC Six hundred VIII Eight XXX Thirty DCC Seven hundred IX Nine XL Forty DCCC Eight hundred X Ten L Fifty DCCCC Nine hundred XI Eleven LX Sixty M One thousand XII Twelve LXX Seventy MM Two thousand XIII Thirteen LXXX Eighty MMM Three thousand ADDITION 7 NOTE 1. No numeral should be repeated more than three successive times, except C which may be repeated four times, but no more. NOTE 2. Only one numeral of less value can be written at the left of another of greater value. NOTE 3. A bar placed over a Roman numeral, or a combination of such numerals, increases its value a thousandfold. Thus V expresses 5000; XXIV, 24000; CCXLV1DXII, 246512. NOTE 4. Vacant orders are not expressed as is found necessary when figures are employed. There is no Roman numeral to express naught. ILLUSTRATIVE EXERCISE 16. Express 3675 by the use of Roman numerals. EXPLANATION. First write the highest order, three thousand (MMM); then the next lower order, six hundred (MMMDC); then the next lower order, seven tens or seventy (MMMDCLXX); and finally the lowest order, five units (MMMDCLXXV). Express the following by the use of Roman numerals: 1. 12 6. 92 9. 219 13. 729 17. 1468 2. 27 6. 69 10. 188 14. 942 18. 2372 3. 16 7. 35 11. 347 15. 638 19. 1090 4. 45 8. 74 12. 406 16. 826 20. 3007 Express the following by the use of figures : 21. LXXXVI 25. DCCXXVI 29. MCXLVIII 22. CCXLIV 26. MCCLVII 30. MDCCCLV 23. DCLXIX 27. DCCCCXXI 31. MDCCCCIX 24. CCCCXII 28. MMCXXXV 32. DCCCLXXXIX ADDITION 17. Addition is the name of the process for finding the total of two or more unequal quantities. NOTE 1. The combined value of all the quantities to be added is called the sum, or total, or amount. NOTE 2. (a) The sign of addition is an erect cross (+) and is called plus. (b) When placed between quantities, it denotes that they are to be added. (c) The sign of equality is two short horizontal lines ( = ), and is read equals or equal, (d) The sign of equality denotes that the combined expression on the left of it is equal to the combined expression on the right of it. 8 ADDITION 18. Principles of Addition. 1. Only like numbers [Note 2, a, 3], or like orders [9, g], can be added. 2. The sum expresses a quantity of the same name as the quantities added. ILLUSTRATIVE EXAMPLE 19. Add 8167, 59, 374, 2742, 19, 3816, 7 and 34. SOLUTION EXPLANATION. So place the numbers to be added that figures of the same order [9, g] shall fall in the same column 81o7 [Prin. 1, 18], and draw a horizontal line beneath. Beginning 59 with the right-hand or units' column, find the sum of 4 + 7 374 +6 + 9+2+4 + 9 + 7, amounting to 48 units, or 4 tens 2742 anc * ** un its. Place the 8 units beneath the horizontal line 1 Q and under the units' column, and carry the 4 tens to the tens' column, that is, mentally add the 4 carried tens with the oolt figures in the tens' column. Thus, 4 (the carried figure) 7 + 34-i-j-i+4_f_7 + 5_|_6=31 tens, or 3 hundred 34 and 1 ten. Place the 1 ten beneath the line and under the tens> c l umn > an d carry the 3 hundred to the hundreds' 234 column, as follows: 3 (the carried figure) +8 + 7 + 3 + 1 = 22 hundred, or 2 thousand and 2 hundred. Place the 2 hun- dred beneath the line and under the hundreds' column, and carry 2 thousand to the thousands' column, as follows: 2 (the carried figure) +3 + 2 + 8 = 15 thousand. As this is the last column to be added, place the entire result beneath the line in such a manner that the 5 thousand shall fall under the thousands' column. NOTE 1. To prove addition, add each column in the opposite direction to that first employed. If the same results are obtained, the answer is pre- sumed to be correct. NOTE 2. When adding, write the carrying figure under the column from which it was obtained, as shown in the illustrative example, that it may also be included in the proof. NOTE 3. In adding columns, acquire the habit of pronouncing results without naming the figures from which they have been obtained. Thus, in adding the units' column of the illustrative solution, it is better to read 4, 11, 17, 26, 28, 32, 41, 48, than to spell 4 and 7 are 11, 11 and 6 are 17, etc. EXAMPLES FOR PRACTISE Find the sum of 1. 58, 36, 75, 92, 85, 36, 27, 48, 62, 59, 23, 37, 56, 18 2. 326, 62, 927, 35, 5268, 76, 248, 9, 628, 3253, 769 3. 8296, 573, 2894, 37, 518, 6275, 93, 235, 58, 617, 75 SUBTRACTION 9 Find the sum of 4. 5368, 7864, 3295, 48, 2834, 376, 58, 675, 49, 3278 5. 18237, 625, 34296, 5867, 928, 2678, 18326, 754 6. Three thousand two hundred ninety-five; seven hun- dred sixteen; one thousand forty-eight; thirty-five thousand ninety-seven; eight thousand three hundred five; sixty-nine thousand eight; eighty thousand four; forty thousand one hun- dred; two hundred six thousand seven hundred three; four hundred thousand five hundred twenty-eight. 20. It is frequently found necessary to add numbers which are written horizontally, or in some other way than in the usual ver- tical arrangement. To find the total of numbers thus irregularly arranged, first add all the right-hand orders; next add all the second orders from the right; next, all the third orders from the right, etc.; setting down and carrying in the usual manner. EXAMPLES FOR PRACTISE Find the total by adding horizontally : 1. 562, 375, 812, 492, 716 5. 826, 9187, 46, 3285, 768 2. 283, 826, 749, 514, 328 6. 96, 487, 3256, 7834, 279 3. 617, 482, 627, 835, 569 7. 38, 46, 298, 84, 2637, 526 4. 986, 789, 582, 796, 328 8. 3682, 597, 2876, 736, 98 SUBTRACTION 21. Subtraction is the name of the process for finding the difference between two unequal quantities. NOTE 1. (a) Subtraction involves the reverse use of the same terms as have been already considered in addition [Note 1, 17]. The names of these terms of addition are changed, however, to accord with the reverse process employed in their use. (6) The total value of all the unequal components formerly called the sum, is now distinguished as the minuend, which means the quantity to be diminished; (c) the given component (or the sum of the given components if more than one is given) is now called the subtrahend, which means quantity to be subtracted; (d) and the required component is now called the remainder, which means that (quantity) which remains. Hence, NOTE 2. (a) The minuend is the quantity to be diminished; (6) the subtrahend is the quantity to be taken from the minuend; (c) and the re- mainder or difference is what remains of the minuend after the subtrahend has been taken from it. 10 SUBTRACTION NOTE 3. (a) The sign of subtraction is a short horizontal line, as follows ( ). (6) It is called minus; (c) it means less; (d) and when placed between two numbers, it indicates that the number at its right is to be subtracted from the number at its left. 22. Principles of subtraction. 1. The difference can be found only between like numbers [Note 2, a, 3] or like orders [9, g]. 2. The difference between like numbers or like orders will also be like. 3. An increase of any order in the subtrahend is equivalent to a correspond- ing decrease of the similar order in the minuend. 4. The difference between two numbers is the difference of the several orders in the subtrahend separately taken from their respective similar orders in the minuend. ILLUSTRATIVE EXAMPLE 23. Subtract 482 from 827. SOLUTION EXPLANATION. Write the less number under the greater BO that each order [9, g] of the less number shall fall under a similar order of the greater. Commencing with the lowest order, subtract 2 units of the subtrahend from 7 units of the minuend, leaving 5 units to occupy the units' order of the remainder. Then subtract the next higher order of the subtrahend (8 tens) from the corresponding next higher order of the minuend (2 tens), and since the 8 tens of the subtrahend cannot be taken from the 2 tens in the minuend, "borrow" 1 hundred from the 8 hundred in the minuend, and add the bor- rowed hundred (or 10 tens) to the 2 tens, thus making 12 tens, and then subtract 8 tens from 12 tens, leaving 4 tens to occupy the tens' order of the remainder. Lastly, subtract the next higher order of the subtrahend (4 hundred) from the next higher order of the minuend (7 hundred), that is, 8 hundred diminished by the previously borrowed 1 hundred, obtaining 3 hundred as the hundreds' order of the remainder. NOTE. To prove subtraction, add the remainder (the required component, (Note 1, d, 21) to the subtrahend (the given component, Note 1, c, 21). The result should equal the minuend (the given total, Note 1, 6, 21). EXAMPLES FOR PRACTISE Find the difference between 1. 9426 and 4732 4. 79218 and 24835 2. 8592 and 5278 5. 121829 and 72636 3. 92816 and 37284 6. 187213 and 95847 SUBTRACTION 11 Find the difference between 7. 68329142 and 41768315 10. 918326781 and 40067007 8. 50380674 and 27300407 11. 700400305 and 22080026 9. 80000000 and 26040058 12. 602005009 and 41030070 13. Subtract three hundred forty-five thousand seven hun- dred nine from eight hundred four thousand six hundred. 14. Subtract one million eight hundred seven thousand five hundred thirty-six from twenty-five million one hundred thousand ninety. 15. A speculator bought a tract of land for $52786 and afterwards sold it for $71000. What was his gain? 16. A firm's deposits in a bank amounted to $62309 and its withdrawals to $18027. How much did the firm have re- maining in the bank? 17. If a man was born in the year 1843 and died in the year 1911, how old was he at the time of his death? 18. A house was sold for $28134 which was $2698 more than the sum paid for it. What did the house cost? 24. To save the time consumed in re-arranging the terms of subtraction, learners should acquire the art of finding the differ- ence between two numbers in whatever position they may have been written. This may be done as follows : Subtract the right- hand order of the less number from the right-hand order of the greater ; proceed in a similar manner with the second orders from the right; with the third orders from the right, etc.; until all the orders have been considered, checking off each order as it is subtracted to facilitate the proper identification of the next order to be considered. EXAMPLES FOR PRACTISE Subtracting horizontally, find the difference between 1. 92863 and 75392 4. 3289658 and 1832982 2. 57029 and 20807 5. 8050027 and 4207009 3. 80035 and 62008 6. 7200503 and 2830408 MULTIPLICATION 25. Multiplication is the name of the process for finding the total of two or more equal quantities. 12 MULTIPLICATION NOTE 1. (a) One of the considered equal quantities is called the mul- tiplicand; (6) the number of equal quantities is called the multiplier; and (c) the total of all the considered equal quantities is called the product. NOTE 2. The multiplicand and multiplier are called factors (producers) of the product. NOTE 3. (a) The sign of multiplication is an oblique cross ( X ) . (6) It is read multiplied by; (c) and when placed between numbers, it indicates that they are to be multiplied. 26. Principles of multiplication. 1. The product expresses a quantity of the same kind as the multiplicand [Prin. 2, 18]. 2. The complete product equals the product of each order of the multi- plicand separately multiplied by each order of the multiplier. 3. To multiply or divide the multiplier by any number, the multipli- cand remaining unchanged, is equivalent to multiplying or dividing the product by that number. 4. To multiply or divide the multi- plicand by any number, the multiplier remaining unchanged, is equivalent to multiplying or dividing the product by that number. MULTIPLICATION TABLE 1 2 3 4 5 6 7 8 9 10 11 12 1 2 4 6 8 10 12 14 16 18 20 22 24 2 3 6 9 12 15 18 21 24 27 30 33 36 3 4 8 12 16 20 24 28 32 36 40 44 48 4 5 10 15 20 25 30 35 40 45 50 55 60 5 6 12 18 24 30 36 42 48 54 60 66 72 6 7 14 21 28 35 42 49 56 63 70 77 84 7 8 9 16 18 24 27 32 36 40 45 50 48 56 63 64 72 72 80 88 96 8 54 81 90 99 108 9 10 20 30 40 60 70 80 90 100 110 120 10 11 22 33 44 55 66 77 88 99 110 121 132 11 12 24 36 48 60 72 84 96 108 120 132 144 12 1 2 3 4 5 6 7 8 9 10 11 12 1 MULTIPLICATION 13 ILLUSTRATIVE EXAMPLE 27. Multiply 976 by 8. SOLUTION EXPLANATION. Write the multiplier (8) under the lowest order of the multiplicand (6). Commencing with this lowest 97 6 order of the multiplicand, separately multiply it (6 units) and 8 each successive higher order (7 tens and 9 hundred) by the multiplier (8), setting down and carrying as usual. Thus, 8 times 6 units equal 48 units, or 4 tens and 8 units. Write the 8 units beneath the line and under the units' order of the multi- plicand, and carry the 4 tens to 8 times 7 tens or 56 tens, making 60 tens, or 6 hundred and tens. Write the tens beneath the line and under the tens' order of the multiplicand, and carry the 6 hundred to 8 times 9 hundred or 72 hundred, making 78 hundred. As this completes the multiplication, write this final result in full beneath the line, but in such a manner that the 8 hundred shall fall under the hundreds' order of the multiplicand. EXAMPLES FOR PRACTISE Multiply Multiply 1. 68734258 by 5 7. 84037028009 by 8 2. 27893542 by 7 8. 70040006897 by 12 3. 72486359 by 9 9. 40398002756 by 4 4. 86957423 by 2 10. 920007100008 by 6 5. 51627384 by 11 11. 53647298307 by 8 6. 91827436 by 10 12. 80043090076 by 3 ILLUSTRATIVE EXAMPLE 28. Multiply 6723 by 894. SOLUTION EXPLANATION. First, as explained in 27, multiply the successive orders of the multiplicand (6723) by the units' order of the multiplier (4) as if it stood alone, obtain- 894 ing 26892 as the first partial product. 26892 Next, similarly multiply the successive orders of the fi QrQ 7 multiplicand (6723) by the next higher order of the mul- tiplier (9 tens) as if it stood alone, obtaining 60507 tens. 53784 g p i ace t hi s result that its right-hand order (7 tens) shall 6010362 fall under the tens' order of the preceding partial product [Prin. 1, 18]. Next, similarly multiply the successive orders of the multiplicand (6723) by the hundreds' order of the multiplier (8) as if it stood alone, obtain- ing 53784 hundred. So place this result that its right-hand order (4 hundred) shall fall in the hundreds' column of the preceding products. [Prin. 1, 18.] The sum of these partial products (6010362) will be the complete product. 14 MULTIPLICATION NOTE 1. The first figure of each partial product when properly placed will fall directly under that order of the multiplier which was used to produce it. Thus, in the above solution, the figure at the right of the first partial product (26892), which is 2, should fall under that order of the multiplier (4) which produced it. The figure at the right of the second partial product (60507), which is 7, should fall under that order of the multiplier (9) which pro- duced it. The figure at the right of the third partial product (53784), which is 4, should fall under that order of the multiplier (8) which produced it. NOTE 2. One proof of multiplication is to reverse the position of the multiplicand and multiplier, and multiply. If the same final product is obtained, it may be accepted as correct. After division has been learned, another method of proof is to divide the obtained product by either of its fac- tors [Note 2, 25], and the quotient should equal its other factor. EXAMPLES FOR PRACTISE Multiply Multiply 1. 4768359 by 86 8. 683291482 by 8296 2. 96378342 by 74 9. 503800209 by 6083 3. 72839457 by 63 10. 700680005 by 4009 4. 84976523 by 29 11. 829600706 by 9206 5. 58297386 by 629 12. 920053008 by 70806 6. 72682943 by 785 13. 438007056 by 40075 7. 83795824 by 246 14. 320008062 by 32004 15. Find the total weight of 8965 barrels of flour which aver- age 196 pounds per barrel. 16. What is the cost of building 374 miles of railroad at $34- 768 per mile? 17. A man earns 135 dollars per month and spends 46 dol- lars per month. How much does he save in 9 months? 18. If the average speed of a steamer is 365 miles per day, what distance can it traverse in 137 days? 29. Continued multiplication is the successive multiplication of more than two factors. The final result of such a multiplica- tion is called a continued product. Thus, if 816 be multiplied by 6, and the resulting product by 7, the process is called a continued multiplication, and the final result, a continued product. Continued multiplication is em- ployed when an inconvenient multiplier, as 42, can be changed to two or more convenient multipliers, the product of which is equal to 42, as 6 X 7. MULTIPLICATION 15 ILLUSTRATIVE EXAMPLE Multiply 986 by 63. SOLUTION EXPLANATION. One of the factors (63) is separable into 986 two other and more convenient factors (9 X 7). Therefore, g first multiply the multiplicand (986) by one of these two factors of the multiplier (9), obtaining 8874; and multiply 8874 tn ig resu it by the remaining factor of the multiplier (7), 7 obtaining 62118 as the required product. If 63 = 7 times 9, 62118 tnen ^ ti mes 986 must equal 7 times 9 times 986. EXAMPLES FOR PRACTISE Employing continued multiplication, multiply 1. 6783 by 72 5. 758439 by 96 2. 52896 by 49 6. 4273869 by 132 3. 78348 by 54 7. 3842768 by 144 4. 925687 by 42 8. 52783 by 432 (=8X9X6) 30. If a number be multiplied by 1, the resulting product will be found equal to the multiplicand. Hence, if the same number be multiplied by ten times 1, or 10, the resulting product must be ten times the multiplicand, or that multiplicand with one cipher at the right of it [9, b] ; or if that number be multiplied by one hun- dred times 1, or 100, the resulting product must be one hundred times the multiplicand, or that multiplicand with two ciphers at the right of it [9, c], etc. Hence, to multiply any number by 1, followed by one or more ciphers, simply place as many ciphers at the right of the multiplicand as are found at the right of 1 in the multiplier. EXAMPLES FOR PRACTISE Multiply Multiply 1. 8763 by 100 4. 62875 by 1000 2. 9578 by 10000 6. 41329 by 100000 3. 75623 by 10 6. 72865 by 10000 31. If a multiplier be greater than 1, the product must be as many times the multiplicand as that multiplier is times 1. There- fore if the multiplier be 8, the product must be eight times the multiplicand, and if the multiplier be ten times 8, or 80, the result- 16 MULTIPLICATION ing product must be ten times the preceding product by 8, or eight times the multiplicand with one cipher at the right of it [9, 6]; and if the multiplier be one hundred times 8, or 800, the resulting product must be one hundred times the product by 8, or eight times the multiplicand with two ciphers at the right of it [9, c], etc. ILLUSTRATIVE EXAMPLE Multiply 876 by 9000. SOLUTION EXPLANATION. Multiply the multiplicand (876) by 9, as if it stood alone, obtaining 7884. As the true multiplier is one thousand times 9. the true product must be one thou- onnn sand times the product by 9, or 7884 with three ciphers at 7884000 the right of it [9, d]. EXAMPLES FOR PRACTISE Multiply Multiply 1. 894 by 700 4. 6792 by 57000 2. 7683 by 32000 5. 863 by 437000 3. 5789 by 64900 6. 9187 by 190000 32. If, as shown in 111. Ex., 31, the product of 876 multi- plied by 9000 is 9 times 876, or 7884, with three ciphers at the right of it (7884000), it must follow that one hundred times 876, or 87600, multiplied by 9000, must be one hundred times the former product, or the former product (7884000) with two addi- tional ciphers at the right of it, that is, 788400000. ILLUSTRATIVE EXAMPLE Multiply 27300 by 85000. EXPLANATION. So arrange the given factors that the right-hand significant figure of the multiplier (5) shall 27300 faU under the right-hand significant figure of the multipli- 85000 cand (3). Then multiply as if there were no ciphers at the right of either factor, obtaining 23205. As, however, the true multiplier is one thousand times 85, the product 2184 of 273 multiplied by the true multiplier (85000) must be 2320500000 one tnousan cl times 23205, or 23205000; and as the true multiplicand is one hundred times 273, or 27300, the true product of 27300 multiplied by 85000 must be one hundred times 23205000, or 2320500000. Hence, DIVISION 17 RULE. To multiply when either or both factors terminate hi ciphers: So place the multiplier under the multiplicand that the right-hand significant figure of the former shall fall under the right- hand significant figure of the latter, and multiply as if there were no ciphers at the right of either. At the right of the resulting product, place as many ciphers as are at the right of both factors. EXAMPLES FOR PRACTISE Multiply Multiply 1. 468000 by 7200 5. 823000 by 47000 2. 67200 by 9000 6. 12900 by 680000 3. 5286000 by 30 7. 3400000 by 90000 4. 9270000 by 2800 8. 7820000 by 5300 DIVISION 33. Division is the name of the process which is used when the total of all the considered equal quantities is given, and it is re- quired to find the value of one of those equal quantities, or the number of those equal quantities. NOTE 1. Division involves the reverse use of the same terms as have already been considered in multiplication [Notes 1 and 2, 26]. These terms of multiplication are however changed in name to accord with the reverse process employed in their use. (a) The total of all the equal quantities, for- merly called product [Note 1, c, 26], is now called dividend which means a quan- tity to be divided. Hence, one of the equal quantities and the number of such equal quantities must be the factors or producers of the dividend [Note 2, 25]. (6) Either of these two factors, if given, and used to divide the dividend, is distinguishably called the divisor which means "that by which a quantity is divided"; and (c) the remaining factor, resulting from the division, is distinguishably called the quotient, which means "the number of times one quantity is contained in another." Hence, NOTE 2. (a) The dividend is the number to be divided; (6) the divisor is the number by which the dividend is divided; and (c) the quotient is the result obtained from the division. NOTE 3. After the last quotient figure has been obtained, the undivided portion of the dividend, if any, is called the remainder. NOTE 4. (a) Division is indicated by the sign -f- which follows the dividend and precedes the divisor. (6) It is read divided by. 34. Principles of division. 1. // the divisor and dividend are concrete numbers, they must both represent units of the same 18 DIVISION name to produce the true quotient. 2. // the dividend be concrete and the divisor abstract, the quotient will be concrete and express units of the same name as the dividend. 3. The remainder, being the undivided portion of the dividend [Note. 3, 33], must express units of the same name as the dividend. 4. To multiply or divide the dividend by a number is equivalent to multiplying or dividing the quotient by the same number. 5. To multiply or divide both divisor and dividend by the same number produces no change in the quotient. 6. To multiply the divisor by a number, the dividend remaining the same, is equivalent to dividing the quotient by the same number. 7. To divide the divisor by a number, the dividend remaining the same, is equivalent to multiplying the quotient by the same number. 8. The complete quotient equals the quotient of each order of the dividend separately divided by the divisor. SHORT DIVISION 35. Divide 895 by 5. EXPLANATION. Divide the highest order of the dividend SOLUTION ( 8 hundred) by the divisor (5) to find the highest order of the 5)895 quotient, obtaining 1 hundred, and leaving an undivided re- mainder of 3 hundred in the dividend. Write the first quotient figure (1 hundred) underneath the 8 hundred of the dividend from which it was obtained. Carry the 3 undivided hundred (=30 tens) of the preceding operation to the next lower order of the dividend (9 tens), and divide the result (39 tens) by the divisor (5), obtaining 7 tens as the next lower order of the quotient, leaving 4 tens of the 39 tens as an undivided remainder. Write this second quotient figure (7 tens) underneath the order of the dividend (9 tens) from which it was obtained. Carry the 4 undivided tens (= 40 units) to the next lower order of the dividend (5 units), and divide the result (45 units) by the divisor (5), obtain- ing exactly 9 units, as the next lower order of the quotient. Write this third and final quotient figure (9 units) underneath the units' order of the dividend from which it was obtained. NOTE 1. The above method of obtaining the successive quotient figures by mental processes is called short division. Short division is em- ployed when the divisor is not greater than 12. NOTE 2. To prove division, multiply the quotient by the divisor, and to the product add the remainder, if any. The result should equal the dividend [Note 1, 33]. DIVISION EXAMPLES FOR PRACTISE 19 Divide 1. 71862496 by 4 2. 827359281 by 3 3. 952873656 by 6 4. 829673845 by 7 5. 528379186 by 2 6. 734829635 by 5 Divide 7. 48329157684 by 12 8. 25671928347 by 9 9. 76982763451 by 11 10. 82534826780 by 10 11. 50380700608 by 8 12. 90070008301 by 7 LONG DIVISION 36. Divide 74865 by 58. SOLUTION 58)74865(1290j| 58 168 116 526 522 45 EXPLANATION. Divide as many of the highest orders of the dividend as will contain the divisor (74 thousand) by the divisor (58), obtaining 1 thou- sand as the highest order of the quotient. Write 1 (thousand) as the first quotient figure, multiply the divisor (58) by this quotient figure (1 thousand) and subtract the product (58 thousand) from 74 thousand, thus finding that 16 thousand of the 74 thousand remain undivided. Carry the 16 undivided thousands ( = 160 hun- dred) of the dividend to the next lower order of the dividend (8 hundred), which is most conveniently done by "bringing down" the 8 hundred of the dividend to the right of the 16 undivided thousand, ob- taining 168 hundred as the next portion of the dividend to be divided, and 168 hundred -5- 58 equal 2 hundred, the next quotient figure. Multiply the divisor (58) by this second quotient figure (2 hundred), and subtract the product (116 hundred) from 168 hundred, thus finding that 52 hundred of the 168 hundred remain undivided. Carry the 52 undivided hundreds ( = 520 tens) of the last considered part of the dividend to the next lower order of the dividend (6 tens), by "bring- ing down" the 6 tens of the dividend to the right of the 52 hundreds, obtaining 526 tens as the next portion of the dividend to be divided; and 526 tens -5- 58 = 9 tens. Write 9 tens as the next quotient figure, multiply the divisor (58) by this third quotient figure (9 tens) and subtract the product (522 tens) from 526 tens, thus finding that 4 tens of the 526 tens remain undivided. Carry the 4 undivided tens (= 40 units) of the considered part of the dividend to the next lower order of the dividend (5 units), and divide the result (45 units) by 58, obtaining units times, thus finding that 58 are contained in 74865, 1290 times, and that 45 is the final undivided portion of the dividend. Write the divisor 58 under this undivided part of the dividend to denote this unexecuted part of the division, and place 20 DIVISION this fractional part of the quotient (ff) to the right of the integral part (1290), obtaining 1290ff as the complete quotient. NOTE 1. The above method of obtaining the successive quotient figures by written processes is called long division. Long division is usually em- ployed when the divisor is greater than 12. NOTE 2. (a) To find the first quotient figure when the divisor consists of several figures, mentally divide the first figure or the first two figures at the left of the dividend by the first figure at the left of the divisor, or by the first figure plus 1 if the second figure from the left of the divisor is greater than 5. (fc) To find each succeeding quotient figure, point off from the right of each partial dividend as many figures lacking one as there are figures in the divisor, and divide the remaining figures of the partial dividend by the left-hand figure of the divisor as in (a). NOTE 3. If, after multiplying the divisor by the last obtained quotient figure, the product is found to be greater than the partial dividend above it, the last quotient figure is too great and should be diminished by 1. NOTE 4. If, after any subtraction from a partial dividend, a remainder is obtained which is greater than the divisor, the last obtained quotient figure is incorrect, and should be increased by 1. EXAMPLES FOR PRACTISE Divide Divide 1. 876349572 by 23 10. 4378692573 by 486 2. 924183762 by 41 11. 5827394618 by 928 3. 729385647 by 63 12. 6537428719 by 4365 4. 682753725 by 37 13. 8291736451 by 72849 6. 521837649 by 58 14. 5278060072 by 603005 6. 437529183 by 79 15. 9428910003 by 412078 7. 621938726 by 48 16. 8006000342 by 890065 8. 529486312 by 67 17. 7200650004 by 27384 9. 297834263 by 325 18. 4120082005 by 728163 37. A continued division is a successive division by two or more divisors. It is usually employed when an inconvenient divisor, as 48, involving long division, can be changed to two or more convenient divisors (48 = 8 X 6), involving successive short divisions. ILLUSTRATIVE EXAMPLE Divide 6487 by 336. DIVISION 21 SOLUTION 336 = 8 X 7 X 6 8)6487 7)810 + 7 = 6)115 + 5 (X 8) = 19 + 1 Quotient EXPLANATION. If 6487 be divided by 8 or one forty-second of the true divisor, the quotient (810) will be 42 times the true quotient [Prin. 7, 34], and a remainder of 7. As this remainder is from a partial division of the true dividend it must form 7 a part of the true remainder [Prin. 3, 34]. If 810 (forty-two times the true quo- ~ tient) be divided by 42, the result will be 8 X 7) =56 the true quotient; but if 810 be divided by the second factor 7, or one-sixth of 42, the resulting quotient (115) must be 6 times the true quotient [Prin. 7, 34] and a remainder of 5. As this remainder (5) is from a partial division of one-eighth of the true dividend (810 = one- eighth of 6487), it must be one-eighth of the second part of the true re- mainder [Prin. 3, 34] and 8 times 5 or 40 will form the second part of the true remainder. As 115 is 6 times the true quotient, divide 115 by 6, the third factor, to find the true quotient (19), and a remainder of 1. As this remainder (1) is from a division of one-seventh of one-eighth of the true dividend, it must be one-seventh of one-eighth of the true remainder [Prin. 3, 34], and 7 times 8 times 1, or 56, will form another part of the true remainder. Hence, add these several parts of the true remainder, obtaining 7 + 40 + 56, or 103 as the complete remainder. SUMMARY. 1. The successive division by the factors of a divisor is equivalent to a division by that divisor. 2. The complete remain- der of a successive division by the factors of a divisor equals the first remainder -\-the product of the second remainder multiplied by the first divisor + the continued product of any succeeding remainder multiplied successively by all the divisors which precede its own divisor. PRACTISE Divide 3678251968 by 96 4826731853 by 84 2372853 by 168 (8X7X3) 7283945 by 504 (9X8X7) 5678325 by 125 (5X5X5) 9671847 by 315 (9X7X5) EXAMPLES FOR Divide 1. 896542 by 28 7. 2. 1768329 by 32 8. 3. 768296854 by 45 9. 4. 93725823 by 72 10. 5. 62834251 by 64 11. 6. 5162738429 by 54 12. 22 DIVISION 38. Since the value of an order [9, g], is decreased tenfold for each place that it is moved to the right [9, /], if all the orders of a dividend be moved one place to the right, it will be equivalent to dividing that dividend by 10; if moved two places to the right it will be equivalent to dividing that dividend by 100; if three places, by 1000, etc. Now to strike off one figure from the right of any dividend will have the effect of moving its remaining orders one place further to the right than their original position, and thus to dividing that dividend by 10; to strike off two figures will move the remaining orders two places to the right of their original position, and thus to dividing by 100, etc. It also follows that the figure or figures which are taken from the right of the dividend will still retain the same local position and value which they had in the original dividend, and must be the undivided portion of the original dividend, or the true remainder. Hence, RULE. To divide by 10, 100, 1000, etc., cut off from the right of the dividend as many figures as there are O's at the right of 1 in the divisor. The remaining figures of the dividend will constitute the quotient; and the figures cut off will be the remainder. EXAMPLES FOR PRACTISE Divide Divide 1. 67859 by 10 6. 1483762 by 1000 2. 82678 by 100 6. 7834106 by 100 3. 92567 by 1000 7. 28710015 by 10000 4. 26835 by 10000 8. 62520003 by 1000 39. If, as found in 38, when three figures are stricken from the right of a number, the remaining figures will be the true quotient of that number divided by 1000, and the figures stricken off will be the true remainder, it must follow that striking off three figures from the right of that number will be seven times the true quotient of that number divided by 7 times 1000, or 7000 [Prin. 7, 34] and, therefore, that striking off three figures from a number and divid- ing its remaining figures by 7, must be the quotient of that number divided by 7000. Further, if, as also shown in 38, the three figures stricken from the right of the dividend constitute the true remain- der from dividing by 1000, they must also form a part of the true DIVISION 23 remainder from dividing by seven times 1000 [Prin. 3, 34]; and any succeeding remainder from dividing the remaining figures of the dividend (or one-thousandth of the true dividend) by 7 will be one-thousandth of the second part of the true remainder, and hence must have three figures at its right to exhibit its true local value [9, d]. If, therefore, the three figures cut off from the right of the original dividend, which constitute one part of the true re- mainder, be placed at the right of the second remainder, the result will be the complete true remainder. RULE. When any divisor has ciphers at its right: 1. Cut off from the right of the dividend as many figures as there are O's at the right of the divisor. 2. Divide the remaining figures of the dividend by the divisor without the O's at its right, to find the true quotient. 3. Place the figures cut off from the original dividend at the right of the final remainder to obtain the complete remainder. ILLUSTRATIVE EXAMPLE Divide 93865 by 3700. EXPLANATION. 3700 = 100 X 37. To divide 93865 by the first factor (100), cut 37p))938>;)(25 Quo. off two fig 1 "" 68 from tne ri g nt of tne divi- dend, obtaining 938 as 37 times the true quotient, and 65 as part of the true remain- der [38]. Then, 185 Divide 938 by 37, obtaining the true ~~Ts65 "Rp m quotient (25) and a second remainder of 13. As this remainder (13) is from divid- ing one-hundredth of the original dividend (93865), it is one-hundredth of the true remainder [Prin. 3, 34], and 100 times 13, or 13 hundred must be the second part of the true remainder. Hence, place the first part of the true remainder (65) to the right of the second part (13 hundred), obtaining 1365 as the complete remainder. EXAMPLES FOR PRACTISE Divide Divide 1. 8736541 by 800 6. 73289653 by 218000 2. 9183762 by 6000 7. 52671839 by 493000 3. 4275918 by 400 8. 42835267 by 63000 (1000 X 9 X 7) 4. 6823475 by 7000 9. 84276834 by 4800 (100 X 8 X 6) 6. 5938764 by 4100 10. 64837291 by 350000 (10000 X 7 X 5) 24 UNITED STATES MONEY UNITED STATES MONEY 40. United States money is usually expressed in dollars and cents; but in performing calculations in United States money, it has been found convenient in certain cases to be more specific in designating its orders by naming the first figure to the right of the decimal point, dimes, and the third figure to the right of the deci- mal point, mills. NOTE 1. Scale: 10 mills = 1 cent; 10 cents = 1 dime; 10 dimes = 1 dollar. NOTE 2. In expressing the several orders of United States money, the dollars are written in accordance with 9. A point (.) is used to separate the dollars from the lower orders. The first order to the right of the point expresses dimes; the first two orders to the right of the point, considered as one number, express cents; and the third order to the right of the point expresses mills. NOTE 3. To indicate that a certain number expresses dollars, the sign $ is placed at its left; and that a certain number expresses cents, the sign is placed at its right. NOTE 4. If the cents to be expressed are less than 10, a cipher should be written after the point to denote the absence of tens of cents or dimes. Thus, 8 dollars 7 cents should be written $8.07. EXERCISES Write Write 1. 78 dollars, 19 cents 8. 16 dollars, 8 dimes 2. 267 dollars, 23 cents 9. 341 dollars, 6 mills 3. 59 dollars, 42 cents 10. 472 dollars, 2 cents 4. 167 dollars, 3 cents 11. 673 dollars, 34 cents, 7 mills 5. 5273 dollars, 8 cents 12. 9183 dollars, 7 cents, 2 mills 6. 435 dollars, 36 cents 13. 4865 dollars, 5 mills 7. 2926 dollars, 5 cents 14. 793 dollars, 1 cent, 9 mills Express the following by spoken or written words : 15. $35.42 18. $500.34 21. $86.352 24. $80.706 16. $76.28 19. $2176.09 22. $75.089 25. $625.105 17. $95.05 20. $513.06 23. $54.003 26. $532.017 41. (a) As 10 dimes, or 100 cents, or 1000 mills equal 1 dollar [Note 1, 40], dollars may be changed to dimes by annexing one UNITED STATES MONEY 25 cipher, to cents by annexing two ciphers, and to mills by annex- ing three ciphers [30]. (b) As 10 cents or 100 mills equal 1 dime, dimes may be changed to cents by annexing one cipher, and to mills by annexing two ciphers, (c) Similarly cents may be changed to mills by annexing one cipher, (d) The lower orders may be changed to dollars by pointing off one figure from the right of the dimes, two figures from the right of the cents, or three figures from the right of the mills [38], prefixing ciphers if necessary. EXERCISES Change Change 1. $748 to cents 6. 8 cents to mills 2. $57 to mills 7. 5278 cents to dollars 3. $382 to cents 8. 6175 mills to cents 4. 7 dimes to cents 9. 69183 cents to dollars 6. 3 dimes to mills 10. 5368 dimes to dollars ADDITION OF UNITED STATES MONEY 42. The processes for adding, subtracting, or multiplying United States money are the same as for abstract numbers, both being expressed upon a scale in which 10 units of any order equal 1 unit of the next higher order [compare Note 1, 40 with e and/, 9], Any expression in United States money may therefore be considered a simple integral number of the same name as its right- hand order, and added as such, and the resulting sum changed back to dollars by d, 41 . ILLUSTRATIVE EXAMPLE 43. Add $372, $631.18, $9352.605, and $43. EXPLANATION. So arrange the numbers to be added SOLUTION that the p i n t s o f each shall fall in an erect column, thus < 372 causing mills to fall under mills, cents under cents, dimes *, -, o under dimes, dollars under dollars [18]. Add and carry as in ordinary addition [19]. As the sum of the dimes 9352.605 column (6 + 1) must be 7 dimes, and the sum of the 43. dollars' column (3+2 + 1+2) must be 8 dollars, a $10398 785 point should be placed in the sum between the 7 dimes and 8 dollars to denote their respective values [Note 2, 40], that is, the point in the sum should fall directly under the column of points in the numbers added. 26 UNITED STATES MONEY EXAMPLES FOR PRACTISE 1. Add $573.25, $6187.92, $48.34, $7283.16, $8.64, and $58.25. 2. Add $589.18, $39.42, $643.28, $7.24, $826.32, and $65.58. 3. Add $82.74, $9.36, $628.15, $97.87, $527.62, and $4.50. 4. Add 3 dollars, 4 cents; 28 dollars, thirty cents; 7 dollars, 2 mills; 45 dollars, 9 mills; 19 dollars, 5 cents; 356 dollars, 8 cents. 5. Add 9 cents; 2 dollars, 18 cents; 74 cents; 3 cents, 5 mills; 4 dollars, 25 cents; 2 cents; 34 cents; 4 dollars; 15 dollars, 5 cents. SUBTRACTION OF UNITED STATES MONEY ILLUSTRATIVE EXAMPLE 44. Subtract $8, 7^ from $22, 15^ SOLUTION EXPLANATION. So place the subtrahend (8.07) that the point at the right of its dollars shall fall under the point at $22.15 the right of the dollars in the minuend (22.15), thus causing g 07 similar orders to fall in the same column. Subtract, borrow, and carry as in ordinary subtraction [23], and place a point in the remainder directly under those of the minuend and subtrahend, as already explained in pointing off the sum in 111. Ex., 43. EXAMPLES FOR PRACTISE Subtract Subtract 1. $518.25 from $838.13 5. $358, 5ff from $500 2. $28.32 from $47.157 6. $67 from $285, 80^ 3. $813 from $1826.45 7. $5, 4^f from $12, 3 mills 4. $534.68 from $913 8. $248 from $326, 5^ MULTIPLICATION OF UNITED STATES MONEY ILLUSTRATIVE EXAMPLE 45. Multiply $873.42 by 53. SOLUTION EXPLANATION. Write the multiplier under the mul- tiplicand so that the right-hand significant figure of the $873.42 multiplier shall fall under the right-hand significant figure 53 of the multiplicand. Multiply and carry as in ordinary mul- 262026 tiplication [28]. As the multiplicand is equivalent to 87342 4^710 cents, 53 times 87342 cents must also be cents [Prin. 1, 26]. Hence, change the resulting product 4629126 cents to dollars $46291.26 as shown in d, 41, that is, point off from the right of the product as many figures as are pointed off from the right of the multiplicand. DIVISION OF UNITED STATES MONEY 27 EXAMPLES FOR PRACTISE Multiply Multiply 1. $6875 by 253 6. $357 and 16^ by 72 (8 X 9) 2. $768.25 by 48 (8 X 6) 7. $213 and Si by 258 3. $27.53 by 37 8. $562, 3?f, 4 mills by 24 4. $827.50 by 468 9. $65, 9 mills by 900 6. $91 and 3fi by 53 10. $8, 15^, 7 mills by 32000 DIVISION OF UNITED STATES MONEY BY AN ABSTRACT DIVISOR 46. In division of United States money, the dividend may be regarded as a simple integer of the same name as its right-hand or lowest order. Thus, $5.26 may be regarded as 526 cents, and $2.065 as 2065 mills. Hence, if the divisor is an abstract number, the process of division can differ in no respect from ordinary division as shown in 36, and the resulting quotient must express units of the same name as the dividend [Prin. 2, 34]. ILLUSTRATIVE EXAMPLE Divide $593.95 by 47. SOLUTION EXPLANATION. Divide as in ordinary division. 4 As the dividend expresses 59395 cents, one forty- A7^p;QQ Qf^l9 A3 seventh of the dividend, or 1264 must also express ^*' cents [Prin. 2, 34]. Hence change the cents (1264) 47 to dollars by cutting off two figures from the right 123 of the quotient [41, d] obtaining $12.64. Q4 As there is no coinage in U. S. money below the cents' order, it is customary to stop the division after bringing down and dividing the cents' order 282 of the dividend, and if the final remainder be one- 175 half the divisor, or more, to increase the last figure .. ,- of the quotient by 1. Hence, as 34, the final re- mainder is more than half the divisor (47), the last 34 figure of the quotient (3) is increased by 1. NOTE. If the dividend is expressed in dollars only, and it should not be exactly divisible by the divisor, change the divi- dend to cents [41, a] by annexing two ciphers, and continue the division until these annexed ciphers have been brought down and divided. The resulting quotient will then express cents. 28 DIVISION OF UNITED STATES MONEY CAUTION. If the division is not the final operation of a solution, but is to be followed by an operation in multiplication, it will be necessary to retain the exact remainder and express it as a fraction of a cent as in 111. Ex. 36; for any omitted inconsiderable fraction of a cent if multiplied by a con- siderable multiplier may cause an error in the final answer of many integral cents. EXAMPLES FOR PRACTISE Divide Divide 1. $6783.25 by 5 5. $763 by 19 2. $786.97 by 8 6. $578 by 83 3. $428.34 by 23 7. $2783.42 by 327 4. $562.78 by 53 8. $8675 by 465 DIVISION OF UNITED STATES MONEY BY UNITED STATES MONEY 47. When both the divisor and dividend express United States money, and they are regarded as integers of the same name as their respective right-hand orders, they must first be changed to express units of the same name if their right-hand orders are dis- similar [Prin. 1, 34]. This is accomplished by annexing the neces- sary number of ciphers to that term which has the fewer figures at the right of the point, so that the right-hand orders of both the divisor and the dividend shall be equidistant from their respective points. ILLUSTRATIVE EXAMPLE Divide $6273 by $18.25. SOLUTION $18.25)$6273.00(343 f|M EXPLANATION. As the divisor _ 7 _ expresses 1825 cents, reduce the divi- dend ($6273) to cents by annexing 7980 two ciphers [Prin. 1, 34]. Divide 7300 as in ordinary division, finding that 6800 1825 cents are contained in 627300 5475 C ents 343 iff! times. 1325 RELATION OF NUMBERS 29 EXAMPLES FOR PRACTISE Divide Divide 1. $178.32 by $.03 6. $9258.63 by 18ff 2. $2735.15 by $2.83 7. $62583 by 9ff 3. $6825.42 by $17.38 8. $43897.38 by $27 4. $48578 by $52.67 9. $59687.42 by $826 5. $89264 by $128.30 10. $18273 by $145 RELATION OF NUMBERS 48. (a) All arithmetical operations include only four elemen- tary mechanical processes addition, subtraction, multiplication, and division; and the intelligent solution of any problem, however complicated, depends upon a knowledge of the conditions to which each of these processes is invariably limited. (6) The only con- ceivable manner in which a given numerical expression of a quan- tity can be operated upon is by adding another quantity to it (increasing it by addition or by multiplication), or by taking a quantity from it (decreasing it by subtraction or by division). The arithmetical law of increase or decrease is uniform in appli- cation, and may be stated as f ollows : (c) A given quantity is increased by addition if the quantity or quantities by which it is required to be augmented are unequal to the given quantity, or to each other; or (d) by multiplication if the quantity or quantities by which it is required to be augmented are equal to the given quantity and to each other, (e) The result of the increase is distinguishably called the sum if the augmenta- tion has been by unequal quantities; or (/) the product if the augmentation has been by equal quantities, (g) Each unequal numerical quantity employed in obtaining a sum is distinguishably called a component of that sum ; and (h) one of the equal numerical quantities employed in obtaining a product is called its multipli- cand, or quantity to be repeated; and (i) the number of such equal quantities is called its multiplier, or times the equal quantity is to be repeated. (k) A given quantity is decreased by recognizing it as a total of two or more quantities; and this total is distinguishably called a 30 RELATION OF NUMBERS minuend if it is the total of unequal quantities; or a dividend if it is the total of equal quantities. (m) Though all increases can be effected by addition, the process of addition is usually limited to such increases as are maole by combining unequal quantities [17], and the process of multiplication is more conveniently employed in combining equal quantities [25]. (n) While all decreases can be accomplished by subtraction, the process of subtraction is usually limited to the decrease of such quantities as express a total of unequal com- ponents [21], and the process of division is more conveniently employed in the decrease of quantities which express a total of equal components [33]. Hence, PKINCIPLES. 1. To find the total of two or more given unequal quantities, employ addition. 2. To find the total of two or more given equal quantities, employ multiplication. 3. To find one quantity of a given total of two or more unequal quantities, employ subtraction. 4. To find one quantity of a given total of two or more equal quantities, employ division. NOTE 1. It is thus seen that addition and subtraction are opposite processes pertaining to unequal quantities, and that multiplication and divi- sion are opposite processes pertaining to equal quantities; also, that addition and multiplication are different processes for obtaining a required total, the former limited to unequal and the latter to equal quantities; and that sub- traction and division are different processes for diminishing a given total, the former by one or more unequal, and the latter by one or more equal quantities. NOTE 2. It is also seen that the successive steps in the solution of any problem are: 1. Identifying its given numerical terms. 2. Ascertaining the relation of these terms to each other. 3. Perfecting that relation by per- forming the appropriate operation [Note, 7]. 49. (a) The total of two or more unequal quantities is dis- tinguishably called the sum if required, or (b) the minuend if given, (c) The total of two or more equal quantities is distin- guishably called the product if required, or (d) the dividend, if given, (e) The several unequal quantities which constitute a total are called its components if the total (sum) is required, or (/) they are called its subtrahend and its remainder if the total (minuend) is given, (g) If a total of two or more equal quantities (product) is required, one of the several equal quantities which constitute RELATION OF NUMBERS 31 the required total is distinguishably called its multiplicand factor, and (h) the number of times the equal quantity is to be repeated is called its multiplier factor, (i) If a total of two or more equal quantities (dividend) is given, that one of its two factors just explained in g and h, which is given with it, is distinguishably called its divisor; and (j) its other factor [Note 1, 33] which will then be required, is called its quotient. 50. Addition and subtraction contrasted. ILLUSTRATIVE EXAMPLES 1. J. A. Smith and S. H. Brown have formed a copartner- ship, Smith investing $5000 and Brown $7000. What is the cap- ital of the firm? 2. The capital of the firm of Smith and Brown is $12000. If Smith's investment is $5000, how much is Brown's? SOLUTIONS ( $5000 ) Components of a Components of a given minuend { . ( 7000 ) required sum Minuend, if given = $12000 = Sum, if required. EXPLANATION. The total of two or more unequal quantities is the sum if required as in Ex. 1, or the minuend if given as in Ex. 2 [a and 6, 49], Hence, In Ex. 1, add the given unequal quantities, obtaining $12000 as the required sum or capital [Prin. 1, 48]. In Ex. 2, subtract the given unequal quantity ($5000), now called the subtrahend, from the given total of two unequal quantities ($12000), now called the minuend, to obtain the remaining unequal quantity ($7000), now called the remainder [Prin. 3, 48], SUMMARY. 1. If a total of two or more unequal quantities is required, add the several unequal quantities. 2. // a total of two unequal quantities is given, diminish it by its given unequal quantity to find its remaining unequal quantity. 3. If a total of more than two unequal quantities is given, diminish it by the sum of all its unequal quantities except one, and the remainder will be its one remaining unequal quantity. 32 RELATION OF NUMBERS 51. Multiplication and division contrasted. ILLUSTRATIVE EXAMPLES 1. What is the cost of 12 pounds of butter at 35 per pound? 2. If 12 pounds of butter cost $4.20, what is the price of one pound? 3. At 35^f per pound, how many pounds of butter can be bought for $4.20? SOLUTIONS Factors of a given dividend \ " ? Factors of a required product. Dividend, if given = $4.20 = Product, if required. EXPLANATION. That factor of a product which expresses one of the considered equal quantities (35 j) is the multiplicand [Note 1, a, 25] ; and that factor which expresses the number of such equal quantities (12) is the multiplier [Note 1, b, 25] ; and the total of all the considered equal quantities ($4.20) is the product if required [Note 1, c, 25]; or the dividend if given [Note 1, 33]. Hence, In Ex. 1, multiply one of the equal quantities or multiplicand (35 f) by the number of such quantities or multiplier (12), obtaining the required total of all the equal quantities or product ($4.20). In Ex. 2, divide the given total of all the equal quantities or product ($4.20), now called the dividend, by the number of such equal quantities or its given multiplier factor (12), now called the divisor, obtaining one of the re- quired equal quantities or its multiplicand factor (35^), now called the quotient [Note 1, 33]. In Ex. 3, divide the given total of all the equal quantities or product ($4.20), now called the dividend, by one of the equal quantities or its given multiplicand factor (35^), now called the divisor, obtaining the required number of such equal quantities or multiplier factor (12), now called the quotient. SUMMARY. 1. If a product is required, multiply its two given factors. 2. If a product is given, divide it by its given factor, and the quotient will be its other factor. RELATION OF NUMBERS 33 52. Addition and multiplication contrasted. ILLUSTRATIVE EXAMPLES 1. A farm consists of three fields, the first containing 275 acres, the second 342 acres, and the third 378 acres. How many acres does the farm contain? 2. A farm consists of three fields, each containing 468 acres. How many acres does the farm contain? SOLUTIONS EXPLANATION. In each example, the same (1) (2) question is asked, but the proper process for 275 4(jg Ex. 1 is addition, because the several components n.n v ^ ^ ^ ne required total are unequal [Prin. 1, 48]; and for Ex. 2 is multiplication, because the 378 1404 acres several components of the required total are 995 acres ec * ual - t prin - 2 > 48 ^ SUMMARY. 1. Employ addition to find the total of two or more unequal quantities. 2. Employ multiplication to find the total of two or more equal quantities. NOTE. The multiplier factor, denoting the number of equal quantities is peculiar to multiplication and division which relate exclusively to equal quantities. The multiplier factor cannot have its counterpart in addition or subtraction because in each of these processes the number of unequal quan- tities is altogether fortuitous, depending upon the extent of the inequality of the several components of a given or required total. 53. Subtraction and division contrasted. ILLUSTRATIVE EXAMPLES 1. How many acres will remain after selling one field of 78 acres and another field of 65 acres from a farm of 216 acres? 2. How many fields of 72 acres each can I sell from a farm containing 186 acres, and how many acres will remain unsold? SOLUTIONS (1) (2) 78 216 72)186(2 fields _65 143 144 143 sold 73 acres unsold 42 acres unsold 34 REVIEW EXAMPLES EXPLANATION. In Ex. 1, the remainder is obtained by adding the un- equal subtrahends (78 + 65) to find the complete subtrahend (143) and sub- tracting it from the given total of all the unequal fields or minuend (216), obtaining 73 acres as the remaining unequal field [Summary, 3, 60]. In Ex. 2, multiply the equal subtrahends (72 acres) by the obtained num- ber of such subtrahends (2) or quotient [from the Latin, quot, how many] to find the complete subtrahend (144 acres); and subtract this product or total of equal subtrahends (144) from the dividend or total of equal sub- trahends plus a possible remainder (186), obtaining 42 acres as the remainder [48, k, m, ri\. SUMMARY. 1. Employ subtraction to find the remainder, if any, after diminishing a given total (minuend) by its known unequal com- ponent (subtrahend). 2. Employ division to find how many times (quotient) a given total (dividend) can be diminished by a known equal component (divisor), and what will be the final remainder, if any. REVIEW EXAMPLES Li 54. 1. What is the cost of 853 yards of cloth at $1.85 per yard? 2. If 42 barrels of flour cost $273, what will be the cost of 78 barrels of the same kind of flour? 3. What is the total cost of 28 pounds of sugar at 6 cents per pound, and 17 pounds of coffee at 23 cents per pound? A i 4. A man bought 278 bushels of wheat at $1.07 per bushel, 327 bushels of oats at 56 cents per bushel, and 129 bushels of rye at 95 cents per bushel. What was the total cost? 5. A man who had an estate of $52356 willed $25378 to his wife, $9678 to his son, $12368 to his daughter, and the remainder to an asylum. How much of the estate did the asylum receive at his death? , 4 6. A dealer bought 285 tons of coal at $6.25 per ton, and sold it for $1995. What was his profit? 7. A speculator bought two adjoining tracts of land, the first containing 78 acres at $42 per acre and the second contain- ing 94 acres at $35 per acre, expended $2537 in erecting a barn, $300 for other improvements, and then sold both tracts for $12300. What was his profit? REVIEW EXAMPLES 35 V) '8. A merchant bought 435 barrels of flour at $7.15 per barrel and sold it for $3175.50. What was his gain per barrel? 9. If I buy a house for $6500, pay $3200 on the day of purchase and $2758 on a subsequent day, how much shall I still owe? 10. If I buy 28 yards of muslin at 12 cents per yard, 30 yards of linen at 64 cents per yard, 15 yards of cloth at $1.37 per yard, and give the seller a fifty-dollar bank note, how much change should I receive? 4, *11. A dealer bought 75 barrels of potatoes and 863 bushes of corn for $774.34. If the price of the potatoes was $2.50 per barrel, what was the price per bushel of the corn? 12. I originally contracted a debt of $3468.75. How much do I now owe, having paid $1268.75 at one time and $875.36 at another time? 13. A firm's sales were as follows: Monday, $375.82; Tuesday, $468.74; Wednesday, $416.25; Thursday, $396.75; Friday, $298.16; and Saturday, $547.60. What were its average daily sales? EXPLANATION. To average or equalize two or more unequal quantities means to change them to the same number of equal quantities. Hence, the sum of these six unequal quantities when found should be regarded as the total of six equal quantities, or product [Note 1, c, 25], and 6 should be regarded as the number of such equal quantities, or multiplier factor [Note 1, 6, 25]. Hence, divide the product by its multiplier factor to obtain its multiplicand factor, or one of the six equal daily sales. 14. There were five boys whose ages were 8, 10, 12, 14 and 16 years, respectively. What was the average age of these boys? 15. The temperature of a certain locality was 75 Fahren- heit for October, 64 for November, and 56 for December. What was the average temperature for these three months? yQ 4 16. A grocer mixed 170 pounds coffee at 12 cents per pound, 105 pounds coffee at 15 cents per pound, and 84 pounds coffee at 18 cents per pound. What was the average price per pound of the mixture? 17. If it is estimated that the labor of 7 men is necessary 36 REVIEW EXAMPLES to complete a certain contract in 6 days, in how many days can one man complete it? EXPLANATION. 6 days is a multiplicand, as it expresses the equal labor estimated to be performed by each man; 9 is a multiplier, as it expresses the considered number of equal men. Hence, multiply these two factors to find the product which will express the total days' labor of one man which is performed by all 9 of the men. Problems of this character are solved upon the supposition that all laborers possess equal strength, equal skill, and equal industry. ^1&. If 35 men can complete a given amount of work in 13 days, in how many days can one man, working alone, complete the same work? 19. If one man working alone can perform a certain task in 24 days, how many days will be required for 6 men to complete the same task? 20. If 50 men can do a certain job of work in 24 days, how many days will be required for 40 men to perform the same job? ^ 21. A man was born in the year 1840 and died in the year 1911, how old was he at the time of his death? EXPLANATION. 1840 years after the birth of Christ (one given unequal component of a given sum) -f- how many years (a required second unequal component of that given sum) = 1911 years after the birth of Christ (the given sum or minuend) [Note 1, b, 21]. 22. George Washington died in the year 1799, aged 67 years. What was the year of his birth? J 23. The government of the United States was established in 1789. How many years had intervened between that event and the commencement of the Civil War in 1861? 24. William H. Taft was born in the year 1857, and was inaugurated President of the United States in the year 1909. How old was he at the time of his inauguration? 25. A, B, C and D formed a copartnership with a joint capital of $55000. If B invested twice as much as A; C, three times as much as A; and D, as much as B and C together; how much did each partner invest? REVIEW EXAMPLES 37 EXPLANATION. Twice (or 2 times) is the given multiplier of A's invest- ment to produce B's investment; 3 is the given multiplier of A's investment to produce C's investment; and 2 + 3, or 5, is the indirectly given multiplier of A's investment to produce D's investment. Therefore, 1 time A's invest- ment or A's investment + 2 times A's investment, or B's investment -f- 3 times A's investment or C's investment + 5 times A's investment or D's investment, that is, 11 times A's investment, or all the partners' investments, must equal the joint capital of the firm ($55000) or joint product. Hence, divide the joint product ($55000) by its joint multiplier (11) to find the common multiplicand or A's investment ($5000). Next multiply the obtained common multiplicand ($5000) by the respective multipliers for B's investment (2), for C's investment (3) and for D's investment (5), to find their respective investment products. NOTE. If the sum of two or more products is given, its multiplicand factor will be the primary multiplicand (from which all the other mentioned multiplicands have been successively derived) ; and its multiplier factor will be the sum of the expressed multipliers + 1, first changing the multipliers which have been expressed upon derivative multiplicands, to equivalent multipliers upon the primary multiplicand [Prin. 1, 18]. 26. A and B together have $260, B having three times as much as A. How much has each? 27. A, B, and C have $520, B having four times as much as A, and C twice as much as B. How much has each? j 28. A man bought two houses for $6275, paying $275 more for one than for the other. How much did he pay for each? EXPLANATION. $6275 is the given sum (minuend) of two equal quan- tities, and $275 an unequal quantity [Note 1, 6, 21]; that is, $6275 is $275 more than the sum of two equal quantities, each denoting the cost of the cheaper house. Hence, diminish the minuend ($6275) by its given unequal quantity ($275) to find the sum of its two remaining equal quantities ($6000). As $6000 is the total of two equal quantities, or product, divide it by its multiplier factor 2 (the number of equal quantities), obtaining its multipli- cand factor, or the cost of the cheaper house ($3000), etc. 29. A and B formed a copartnership with a capital of $13168, A investing $568 more than B. How much did each invest? 30. The firm of A and B has a capital of $12387, A's investment being $2387 less than B's. What is each partner's investment? 38 REVIEW EXAMPLES 31. A, B, and C together own property worth $17000, B's property being worth $1200 more than A's, and C's $400 less than B's. What is the worth of the property of each? 32. A man bequeathed $37025 to his wife, son, and daughter; his son receiving $625 less than his wife, and his daughter $925 more than his son. How much did each receive? 33. Two ships leave a certain harbor at the same time; one going east at the rate of 350 miles per day, and the other going west at the rate of 275 miles per day. How far apart will they be at the end of the ninth day? 34. At an election for mayor of a certain city, the total number of votes cast for three candidates was 52300. The first candidate received 34500 votes, and the second 1828 votes less than the third. How many votes did the third candidate receive? . 35. A man made five partial payments on a debt of $1565.78, and then found that he still owed $237.62. If his first payment was $128.37, his second $75.40, his third $216.15, and his fourth $375.28, what was the amount of his fifth payment? ,, 36. Two steamers start from the same place at the same time to go in the same direction. Four days after starting the faster steamer was 140 miles ahead of the slower. If the faster steamer's average speed was 375 miles per day, how far was the slower steamer from the starting point? 37. Two pedestrians lived 75 miles apart, and started from their respective homes at the same time to approach each other, one walking at the rate of four miles per hour, and the other at the rate of three miles per hour. How far apart were they after walking eight hours? , 38. A farmer exchanged a tract of land containing 275 acres, and valued at $45 per acre for another tract containing 495 acres. What was the estimated value per acre of the other tract? 39. A certain library contains 43 book cases, each case containing 12 shelves, and each shelf 78 books. How many books does the library contain? 40. A merchant examined the contents of his cash drawer FACTORING 39 and found that it contained $48. Of this sum he found $30 to be in notes, and the remainder consisted of an equal number of 50-cent coins, 25-cent coins, dimes and nickels. How many coins did he find in the cash drawer? FACTORING 55. Factoring is that process by which a number is separated into two or more integers [Note 1, e, 3] which, when multiplied, will produce that number. NOTE 1. Each of the integers into which a number is separated as above described, is called a factor of that number. NOTE 2. When a number cannot be separated into integral factors other than itself or 1, it is called a prime number; and when it can be so separated, it is called a composite number. NOTE 3. The factors of a number are also distinguished as prime factors when they are not separable into minor integral factors; and as composite factors when they are so separable. 56. An exact divisor of a number is one which will divide it without a remainder. NOTE. Reversely, one number is said to be divisible by another when it contains that other a certain number of times without a remainder. 57. Any number is divisible by itself and 1; and it will be divisible: (a) By 2, if it ends with a cipher, or with a figure which is divisible by 2. Thus, 480 and 7818 are divisible by 2; the former because it ends with a cipher, the latter because it ends with 8 which is divisible by 2. (6) By 3, if the sum of all its figures is divisible by 3. Thus, 4791 is divisible by 3 because 4 + 7 + 9 + 1, or 21, is divisible by 3. (c) By 4, if it ends with two ciphers, or with two figures which express a number that is divisible by 4. Thus, 34700 and 72324 are divisible by 4; the former because it ends with two ciphers, the latter because it ends with two figures (24) which express a number that is divisible by 4. 40 FACTORING (d) By 5, if it ends either with a cipher or with 5. Thus, 720 and 1465 are divisible by 5; the former because it ends with a cipher, and the latter because it ends with 5. (e) By 6, if it ends with a figure which is divisible by 2 (one factor of 6), and the sum of all its figures is divisible by 3 (the other factor of 6). Thus, 1368 is divisible by 6 because it ends with a figure (8) which is divisible by 2; and 1 + 3 + 6 + 8, or 18, is divisible by 3. (/) By 8, if it ends with three ciphers, or with three figures which express a number that is divisible by 8. Thus, 276000 and 18376 are divisible by 8; the former because it ends with three ciphers, and the latter because it ends with three figures (376) which express a number that is divisible by 8. (g) By 9, if the sum of its figures is divisible by 9. Thus, 32472 is divisible by 9 because 3 + 2 -f 4 + 7 + 2, or 18, is divisible by 9. (h) By 10, if it ends with one or more ciphers. Thus, 6730 is divisible by 10 because it ends with a cipher. (i) By 12, if it ends with two figures which express a number that is divisible by 4 (one factor of 12), and the sum of all its figures is divisible by 3 (the other factor of 12). Thus, 71928 is divisible by 12 because it ends with two figures (28) which express a number that is divisible by 4, and 7 + 1 + 9 -f 2 + 8, or 27, is divisible by 3. ILLUSTRATIVE EXAMPLE 58. What are the prime factors of 630? SOLUTION EXPLANATION. As 630 ends with a cipher, divide it by its prime factor 5 [57, d\, obtaining 126 as the unfactored part of 5)630 630. As 126 ends with 6 which is divisible by 2, the entire un- 2)126 factored part is so divisible [57, a], obtaining 63 as the remain- o\ ing unfactored part of the original number. As the sum of the figures of 63 (6 + 3), or 9, is divisible by 3, the entire unfactored 3)21 part (63) is so divisible, obtaining 21 as the remaining unfactored "7 part [57, 6]. As the sum of the figures of 21 (2 + 1), or 3, is divisible by 3, the entire unfactored part (21) is so divisible, obtaining 7, a prime number. Therefore, the several prime divisors and LEAST COMMON DIVIDEND 41 the final prime quotient 7, that is, 5, 2, 3, 3, 7, are the required prime factors of 630. PROOF. The continued product of all its prime factors should equal the factored number. Thus 5X2X3X3X7 should equal 630. RULE. 1. Divide the given composite number by any prime number which will exactly divide it. 2. Divide the resulting quo- tient by any prime number which will exactly divide it; and so continue until a quotient is obtained which is a prime number. 3. The several divisors and the final quotient will be the prime factors of the given composite number. EXAMPLES FOR PRACTISE Find the prime factors of 1. 54 4. 360 7. 1260 10. 27000 13. 16632 2. 36 6. 252 8. 5292 11. 17640 14. 45000 3. 84 6. 945 9. 3375 12. 28512 15. 16380 LEAST COMMON DIVIDEND 59. A dividend of a number is one that contains it an exact number of times. Thus, 63 is a dividend of 7 because 63 contains 7 exactly nine times. 60. A common dividend of two or more numbers is one that contains each of them an exact number of times. Thus, 24 is a common dividend of 3, 6 and 8, because it contains 3 exactly eight times, 6 exactly four times, and 8 exactly three times. 61. The least common dividend of two or more numbers is the least number which contains each of them an exact number of times. 62. General principles. 1. A dividend of a number must in- clude all its prime factors, and may include other factors. 2. A common dividend of two or more numbers must include all the prime factors of each of those numbers, and may include other factors. 3. The least common dividend of two or more numbers must include all the prime factors of those numbers, and no other factor. 42 LEAST COMMON DIVIDEND ILLUSTRATIVE EXAMPLE 63. Find the least common dividend of 12, 18, and 20. SOLUTION EXPLANATION. 2 is a common prime factor of each of the given numbers, 18 20 therefore, the least common dividend must 2)6 910 contain the factor 2 at least once. 3)3~I Q~ 5 Of the unfactored part of 12 (6), of 18 (9) and of 20 (10), 2 is a common prime factor 135 of 6 and 10. Therefore the least common dividend must contain the factor 2 at least 2X2X3X3X5= 180 twice to contain 12 and 20 which include this factor twice. Of the yet unfactored part of 12 (3), of 18 (9) and of 20 (5), 3 is a common prime factor of 3 and 9. Therefore the least common dividend must contain the factor 3 at least once in order to contain 12 and 18 which include this factor at least once. The least common dividend must therefore not only contain the prime factors already found (2, 2, 3), but also the yet unfactored part of 18 (3) or it cannot contain 18, and the yet unfactored part of 20 (5) or it cannot contain 20. Therefore the least common dividend of 12, 18 and 20 must contain these necessary factors (2, 2, 3, 3 and 5) and no other factor, and the continued product of these factors is 180. NOTE 1. When no two of the given numbers have a common prime factor, the least common dividend of those numbers must be their continued product. NOTE 2. When it is seen that all of the given numbers have a common composite factor [Notes 2 and 3, 65], it will save unnecessary labor to divide at once by the composite factor. NOTE 3. When one of the given numbers of which it is required to find the least common dividend is found to be an exact divisor of another of the given numbers, it may be omitted from the solution, as all of its factors must then be included in that other number. EXAMPLES FOR PRACTISE Find the least common dividend of 1. 15, 18, 24, and 36 7. 360, 450, 540, and 630 2. 21, 49, 56, and 63 8. 150, 500, 600, and 900 3. 18, 27, 45, and 72 9. 132, 165, 231, and 297 4. 24, 36, 48, and 64 10. 5, 15, 25, 30, 45 [Note 3] 5. 27, 36, 54, and 81 11. 12, 36, 48, 72, and 96 6. 30, 45, 75, and 90 12. 2, 3, 7, 11, 13 [Note l] GREATEST COMMON DIVISOR 43 GREATEST COMMON DIVISOR 64. A common divisor of two or more numbers is an exact divisor of each of them. Thus, 7 is a common divisor of 21, 35 and 56, as 7 is an exact divisor, or common factor, of each of them. 65. The greatest common divisor of two or more numbers is the greatest exact divisor of each of them. Thus, while 2 is a common divisor of 24, 40 and 56, it is not the greatest common divisor; and while 4 is a greater common divisor of these numbers than 2, it is also not the greatest common divisor. The greatest divisor com- mon to these numbers is 8, because 8 is the greatest factor common to all of them. 66. General principles. 1. A common divisor of two or more numbers must include in itself only such prime factors as are common to those numbers. Thus, 6 is a common divisor of 30 and 42, because all the prime factors of 6 (3 X 2) are found both in 30 (3 X 2 X 5) and in 42 (3 X 2 X 7). 2. The greatest common divisor of two or more numbers must in- clude all the prime factors which are common to those numbers, and no other factor. Thus, 9 is the greatest common divisor of 45 and 63 because all the prime factors which are common to 45 (3 X 3 X 5) and to 63 (3 X 3 X 7) are also found in 9 (3 X 3). A greater composite number than 9 cannot be a common divisor of 45 and 63 because it will include a factor which is not common to 45 and 63; and a less number than 9, say 3, which contains one factor common to 45 and 63 (3 X 1), and therefore a common divisor, would not contain all the common factors of those numbers and therefore could not be their greatest common divisor. 3. The greatest common divisor of two numbers must also be a divisor (a) of their sum, (b) of their difference, (c) of any number of times their sum, (d) or of any number of times their difference. 4. The greatest common divisor of two numbers must also be (e) a divisor of any number of times each of them; or (f) of any number of times one of them and a different number of times the other; or (g) of the difference between the greater number and any number of times the kss number that can be subtracted from the greater number. 44 GREATEST COMMON DIVISOR 5. The greatest common divisor of more than two numbers can- not be greater than the greatest common divisor of the two which express the fewest units. ILLUSTRATIVE EXAMPLE 67. Find the greatest common divisor of 531 and 767. EXPLANATION. The greatest divisor of 531 SOLUTION could not also be the greatest divisor of 767 except ro-i \TT/i it be also a divisor of their difference which is 236 531)76 ' [Prin. 3, b, 66]. Therefore, the G. C. D. of 531 and 767 must 236)531(2 also be the G. C. D. of 236, 531 and 767; and this 472 G- C. D. cannot be greater than the G. C. D. of 236 and 531 [Prin. 5, 66]. Now the greatest divisor of 236 could not also be the greatest 236 divisor of 531 except it be also a divisor of the difference between 531 and twice 236, or 472, which is 59 [Prin. 4, g, 66]. Therefore, the G. C. D. of 236, 531 and 767 must also be the G. C. D. of 59, 236, 531 and 767, and this G. C. D. cannot be greater than the G. C. D. of 59 and 236 [Prin. 5, 66]. As 59 will divide itself and as it is also found to be an exact divisor of 236, 59 is the G. C. D. of 59 and 236, and as 236 is the difference between 531 and 767, 59 must also be the G. C. D. of 531 and 767 [Prin. 3, 6, 66]. RULE. 1. Divide the greater number by the less; and if there be no remainder, the less number will be the greatest common divisor. 2. Divide the last divisor by the remainder if there should be a remainder, and if the division be exact, that remainder will be the greatest common divisor. 3. // the second division be inexact, continue to divide the last divisor by the last remainder until there shall be no remainder. The last divisor will be the greatest common divisor. NOTE 1. If, at any time, 1 is obtained as a remainder, it will demonstrate that the numbers are prime to each other. NOTE 2. To find the G. C. D. of more than two numbers, first find the G. C. D. of two of them, preferably of the two which express the fewest units; then find the G. C. D. of this obtained G. C. D. and one of the remaining numbers; and thus continue until all the given numbers have been considered. The last obtained G. C. D. will be the G. C. D. of all the numbers. Note 3. If the numbers are not inconveniently great, their greatest common divisor may also be obtained by finding the continued product of such prime factors as are common to all of them. CANCELATION 45 EXAMPLES FOR PRACTISE Find the greatest common divisor of 1. 185 and 259 5. 539 and 637 9. 16767 and 29889 2. 315 and 405 6. 10625 and 14375 10. 14406 and 26411 3. 204 and 612 7. 1344 and 4416 11. 12096, 19008, 22464 4. 875 and 1375 8. 1215 and 2673 12. 704, 448, 880 CANCELATION 68. Cane elation is the process of diminishing the mechanical labor of division by casting out factors which are common to both divisor and dividend. Thus, in dividing 168 by 56, if it be noticed that 8 is a common factor of both divisor and dividend, the labor of division will be diminished by mentally dividing one-eighth of 168, or 21, by one-eighth of 56, or 7, obtaining 3 as the quotient [Prin. 5, 34]. 69. General principles. 1. To cancel any factor of a number has the same effect as dividing that number by the canceled factor. 2. To cancel the same factor, or the same combination of factors, or equivalent combinations of factors, from both divisor and dividend, does not affect the value of the quotient [Prin. 5, 34]. ILLUSTRATIVE EXAMPLE The factors of a dividend are 210, 16 and 18, and of its divisor are 27, 8 and 7. What is the quotient? SOLUTION EXPLANATION. Instead of finding the continued }0 product of 210, 16 and 18, which is 60480, and 00 2 o dividing this result by the continued product of TT ffl 27 X 8 X 7, which is 1512, obtaining 40, the me- /P X /P_ . chanical labor of the solution can be materially $7 X $ X /7 diminished by cancelation, or continued division, [37], as follows: Place the factors of the dividend above a hori- zontal line and the factors of the divisor below that line. Then cancel the common factor 9 from 27 in the divisor and from 18 in the dividend, writ- ing underneath 27 its uncanceled factor 3, and above 18 its uncanceled factor 2. Then cancel the common factor 8 from 8 in the divisor and from 16 in the dividend, and over 16 place its uncanceled factor 2. Then cancel the common factor 7 from 7 in the divisor and from 210 in the dividend, and 46 COMMON FRACTIONS over 210 place its uncanceled factor 30. Then cancel the yet uncanceled factor 3 from 3 in the divisor and from 30 in the dividend, and place over 30 its uncanceled factor 10. The product of the uncanceled factors in the dividend (10 X 2 X 2 = 40) will be the required quotient. NOTE 1. If, after the cancelation has been completed, any uncan- celed factors are still found in the divisor, the product of the uncanceled factors of the dividend should be divided by the product of the uncanceled factors of the divisor, to obtain the quotient. NOTE 2. If the product of two or more factors on one side of the hori- zontal line is seen to be equal to a single factor on the other side of that line, they should be canceled in one operation. NOTE 3. If 1 is obtained as a result of cancelation, it should be ex- pressed in the dividend, but it may be omitted from the divisor. EXAMPLES FOR PRACTISE 1. Divide 73 X 27 X 16 by 18 X 4. 2. Divide 145 X 36 X 12 by 30 X 18 X 4. 3. Divide 648 X 75 X 32 by 54 X 45 X 8. 4. Divide 180 X 63 X 24 by 48 X 18 X 9. 5. How many bushels of wheat worth $1.12 per bushel should a farmer deliver at a mill to receive 3 barrels of flour worth $5.60 per barrel? 6. A speculator offered a tract of land containing 375 acres in exchange for 250 acres of land worth $75 per acre. At what price per acre did the speculator value his land? COMMON FRACTIONS 70. INTRODUCTION. Hitherto, all the operations which have been described and illustrated have been performed upon inte- gral units [Note 1, e, 3]. It is frequently necessary, however, to operate upon a part of an integral unit, as upon the half of one integral unit (i), the third of one integral unit (J), etc.; or upon a part of a collection of integral units, as upon a third of two inte- gral units (|), an eighth of five integral units (f ), etc. The general principles already learned are unlimited in their application to all numbers, whether integral or fractional. Any perplexity attending the study of this new fractional form of numerical expression will disappear if every fraction be regarded as an unexecuted division of which its numerator is the dividend, its denominator is the divisor, and the value of the entire frac- tional expression is the quotient. Thus regarded, the general COMMON FRACTIONS 47 principles relating to division [34], so far as they apply to divi- dends, will also apply to numerators [72]; so far as they apply to divisors, will also apply to denominators [73]; and so far as they apply to quotients, will also apply to the entire fractional expression [71]. All reductions of fractions [82], therefore, are applications of the general principles of 34. 71. A fraction expresses the result of dividing one or more units into a given number of equal parts. 72. The numerator is that term of a fraction which is written above a horizontal line, and which denotes how many of the equal parts are expressed. Thus, $f expresses 5 fractional units, the value of each of which is one- eighth of a dollar. Num(b)er-ator means that which numbers. 73. The denominator is that term of a fraction which is written below a horizontal line, and which denotes the value of each fractional unit expressed by the numerator. Thus, in the fraction $f, the numerator (4) denotes the number of frac- tional units; and the denominator (5) names the value of each fractional unit as being one-fifth of a dollar. Denominator means that which denominates or names. 74. The terms of a fraction are its numerator and denomina- tor, considered separately. Thus, in |, the numerator 7 is one term and the denominator 8 is the other. Considered jointly, they constitute the fraction f . 75. A common fraction is one which expresses the division of one or more units into any number of equal parts other than such as are represented by 1 followed by one or more ciphers. Thus, |, T 5 *, *ib are common fractions; but T %, T&T, Tolhr are not. 76. A proper fraction is one in which the numerator is less than the denominator. A proper fraction Is so-called because a fraction properly means part of a whole; and if the numerator of a fraction should exceed its denominator, it would express more than a whole and could not properly be called a fraction, though for convenience it may be so indicated. 77. An improper fraction is one in which the numerator either equals or exceeds its denominator. 48 COMMON FRACTIONS Thus, |, 1 are proper fractions because the value of each is less than 1 integral unit; and |, f are improper fractions, the former because its value is exactly equal to 1 integral unit, and the latter because its value is greater than 1 integral unit. NOTE. Any integer may be expressed in the form of an improper fraction by writing 1 under it for a denominator. Thus, 5 = f; 19 = T. 78. A mixed number is an expression which is composed partly of an integer and partly of a fraction. Thus, 35 1 is suggestively called a mixed number because it is composed partly of an integer (35) and partly of a fraction (), mixed or united in one expression (35|). 79. The value of a fraction is the value expressed by con- summating the unexecuted division of its numerator by its de- nominator. Hence, the nearer a numerator of a proper fraction approximates to its denominator, the greater its value; because any increase in a dividend (num- erator), the divisor (denominator) remaining the same, must necessarily increase the value of the quotient. Prin. 4, 34, relating to an increase of the dividend (numerator) by multiplication must, though in a different manner, apply also to an increase by addition. 80. (a) The unit of a fraction is one of the integral units of which the fraction expresses a part; and (6) a fractional unit is one of the equal parts into which the unit of the fraction is divided. Thus, in the expression $f, one dollar is the unit of the fraction; and one eighth-of-a-dollar is the fractional unit. 81. General principles. 1. To multiply or to divide the num- erator by any number is equivalent to multiplying or to dividing the fraction by that number [Prin. 4, 34]. 2. To multiply the denominator by any number is equivalent to dividing the fraction by that number [Prin. 6, 34]. 3. To divide the denominator by any number is equivalent to multiplying the fraction by that number [Prin. 7, 34]. 4. To multiply or to divide both the numerator and the denomina- tor by the same number will produce no change in the value of the fraction [Prin. 5, 34]. SUMMARY, (a) Any change in the numerator by multiplication or divi- sion, the denominator remaining the same, will produce a similar change in COMMON FRACTIONS 49 the value of the fraction; (6) Any change in the denominator by multiplica- tion or division, the numerator remaining the same, will produce an opposite change in the value of the fraction; (c) A similar change in both numerator and denominator, by multiplication or division, will produce no change in the value of the fraction. REDUCTION OF FRACTIONS TO HIGHER TERMS 82. Reduction of fractions is the process of changing their form (appearance) without producing any change in their value. Thus, if it is desired to change a fraction in low terms (possessing a less numerator and denominator) to an equivalent fraction in higher terms (pos- sessing a greater numerator and denominator), the name of the process is reduction, and of the act, to reduce. ILLUSTRATIVE EXAMPLE Reduce f to 126ths. SOLUTION EXPLANATION. The required denominator (126) is 18 times the given denominator 7 (126 *- 7 = 18); l^u ' == 1* therefore, the proper numerator for the required -~ _ ..,. denominator must be 18 times the given numerator 6 (6 X 18 = 108), to maintain a value equivalent to 7 X 18 = 126 that of f [Prin. 4, 81]. RULE. Multiply both the numerator and the denominator by such a number as will change the given denominator to the required de- nominator. NOTE 1. To find the common multiplier of both terms to produce a required higher denominator, divide the required denominator (the greater) by the given denominator (the less). NOTE 2. To reduce an integer to a fraction of a required denominator, first change the integer to the form of a fraction by writing the denominator 1 under it, or by conceiving it to be so written. Thus, 5 = yj 19 = *i ', etc. EXAMPLES FOR PRACTISE Reduce Reduce Reduce 1. to25ths 5. f to216ths 9. J to 256ths 2. f to 56ths 6. f to 512ths 10. T \ to 64ths 3. f to32nds 7. } to 16ths 11. 28 to 4ths 4. \ to 64ths 8. I to 128ths 12. 176 to 16ths 50 COMMON FRACTIONS REDUCTION OF FRACTIONS TO LOWER TERMS ILLUSTRATIVE EXAMPLE 83. Reduce & 6 * to 32nds. SOLUTION EXPLANATION. First divide the given denominator _ (544) by the required denominator (32) to find the ** divisor of the given denominator (544) which will OK _._ 17 _ c produce the required denominator (32), obtaining 17. - * If the given denominator must be divided by 17 544 -r- 17 = 32 to find the required denominator, the given numerator (85) must also be divided by 17 to find the corre- sponding numerator (5) of the required denominator [Prin. 4, 81]. RULE. Divide both the numerator and the denominator by such a common divisor as will change the given denominator to the required denominator. NOTE. To find the common divisor of both terms of a fraction which will produce an equivalent fraction of the required lower denominator, divide the given denominator (the greater) by the required denominator (the less). EXAMPLES FOR PRACTISE Reduce Reduce Reduce 1. f M to 8ths 4. iji to 32nds 7. J3i to 64ths 2. -ffy to 16ths 5. jfi to halves 8. ^-f* to an integer (firsts) 3. T W to 4ths 6. -Bf to 4ths 9. Vff- to an integer REDUCTION OF FRACTIONS TO LOWEST TERMS 84. A fraction is said to be reduced to lowest terms when its numerator and denominator are prime to each other. ILLUSTRATIVE EXAMPLES Reduce (1) J|J and (2) fj& to lowest terms. SOLUTIONS (1) (2) 168 -s- 8 = 21 384 -v- 8 = 48 G. C. D. of 949 and 2336 = 73 21 *- 3 = 7_ 949 -T- 73 = 13 48 -s- 3 = 16 AnS ' 2336 -J- 73 = 32 nS ' COMMON FRACTIONS 51 EXPLANATIONS. (1) Divide both terms of Iff by any observed common factor, say 8, obtaining |J as an equivalent fraction in lower terms [Prin. 4, 81]. It will now be further observed that 3 is a common factor of both terms of ft. Hence, divide both terms of \ \ by 3, obtaining T V as an equivalent frac- tion in still lower terms. As the terms of yV are prime to each other, they are incapable of further reduction, and therefore T^ is in its lowest possible terms. (2) If the factor or factors common to both terms are not easily discover- able, as in 2T&, it will be more convenient to divide both numerator and denominator by their greatest common divisor [67]. EXAMPLES FOR PRACTISE Reduce the following fractions to lowest terms: i. jt 5. m 9. is. 2. if 6. J 10. IIH 14. 3. || 7. }}{ 11. Uii 15. Hill 4. 8. HI 12. tffi 16. Httt REDUCTION OF FRACTIONS TO APPROXIMATE LOWEST TERMS 85. INTRODUCTION. To approximate means to approach very near without exactly reaching. Approximations are frequently necessary in business calculations to secure serviceable results. The market reports of our daily papers, contain no fractional quo- tations other than halves, quarters, eighths, and sometimes, though rarely, sixteenths. In exceptional occupations which re- quire other denominators than the above, any obtained fractional results are approximated to such exceptional denominators. Hence, if the final result of a calculation contains a fraction with an impractical denominator, it is customary to reduce it to a practical denominator by approximation. ILLUSTRATIVE EXAMPLES 86. Reduce (1) H to 8ths; (2) f Jf to 4ths. FIRST SOLUTION FIRST EXPLANATION. If the division of 21 X 8 = 168 * ts numerator by its denominator expresses the value of any given fraction [79], then to divide 168 -r- 56 = 3, exactly 8 times the numerator of U (21 X 8 = 168) by the denominator of fi (168 -5- 56) must express Hence, f = I, exactly a result (exactly 3) which is exactly 8 times the value of H [Prin. 1, 81]; and this result (3) divided by 8, or f , must be the exact value of H expressed in eighths. 52 COMMON FRACTIONS SECOND SOLUTION SECOND EXPLANATION. To 413 X 4 = 1652 divide 4 times the numerator of f H (413 X 4 = 1652) by the 1652 -T- 548 = 3, approximately denominator of fit (1652 ^ 548) will produce a result (3, approxi- Hence, HI = I, approximately mately) which must be approxi- mately 4 times the value of |?f [Prin. 1, 81]. Hence, this result (3) divided by 4, or f, must be the approximate value of f if when expressed in fourths. RULE. Multiply the numerator of the given fraction by the required denominator, and divide the resulting product by the de- nominator of the given fraction, disregarding any remainder from the division if less than half the divisor, or increasing the quotient by 1 if the remainder is half the divisor, or more. The quotient will be the exact numerator if the division be exact, or the approximate num- erator if the division be inexact. Under the quotient place the required denominator. If the resulting fraction is not in lowest terms, men- tally apply Solution 1, 84. NOTE. Intermediate fractions should never be approximated as the least departure from strict accuracy near the beginning of a continued operation may terminate in a considerable error when the final result is reached. EXAMPLES FOR PRACTISE Reduce the following to lowest terms : 1. iff, not higher than 8ths 6. fji, not higher than 16ths 2. ^T, not higher than 4ths 6. if ft, not higher than 8ths 3. Iff, not higher than 16ths 7. IHI, not higher than 32nds 4. *!, not higher than 8ths 8. ||f-J-, not higher than 4ths. REDUCTION OF INTEGRAL OR MIXED NUMBERS TO IMPROPER FRACTIONS ILLUSTRATIVE EXAMPLE 87. Reduce 37f to an improper fraction. SOLUTION EXPLANATION. 8 eighths = 1, therefore, 37 X 8 = 296 eighths 37 times 8 eighths, or 296 eighths must equal 296 + 5 = 301 eighths 37 times 1 or 37 > and 37 * must e ? ual 5 eighths more than 296 eighths or 301 eighths. Hence, 37| = ^ COMMON FRACTIONS 53 PRINCIPLES. 1. One integral unit is expressed as a fraction with a given denominator by making the numerator the same as the given denominator. 2. Two or more integral units are expressed as a fraction with a given denominator by making the numerator as many times the given denominator as the integer is times 1. RULE. Multiply the integer by the denominator of the fraction, add the numerator to the product, and the result will be the numerator of the required improper fraction; under which write the given denominator. NOTE. When no particular denominator is required, an integer is re- duced to the form of an improper fraction by simply writing 1 underneath it as a denominator [Note, 77]. EXAMPLES FOR PRACTISE Reduce the following to improper fractions : 1. 9 to 16ths 5. 28* 9. 463B 13. 354flfc 2. 75to8ths 6. 53J 10. 89H 14. 2186^ 3. 61 7. 176A 11. 756J 16. 854* 4. 12f 8. 253& 12. 817| 16. 612| REDUCTION OF IMPROPER FRACTIONS TO INTEGRAL OR MIXED NUMBERS ILLUSTRATIVE EXAMPLE 88. Reduce -I 3 - to an integer or mixed number. EXPLANATION. As 8 eighths equal one integral unit SOLUTION [Prin. 1, 87], there must be as many integral units in 173 __ g = 21 5 173 eighths as 8 eighths are contained times in 173 eighths; that is, 21 integral units and 5 eighths. NOTE. The fractional part of the obtained mixed number should be reduced to lowest terms, if not already so. EXAMPLES FOR PRACTISE Reduce to integers or to mixed numbers: 1. V- 4. W 7. W- 10. MF 13. 2. H- 8 - 5. W 8. -HF 11. *V* 14. 3. V- 6. W 9. ** 12. Vff- 15. 54 COMMON FRACTIONS COMMON DENOMINATOR 89. A common denominator of two or more fractions is any one denominator to which each of them can be separately reduced. 90. The least common denominator of two or more fractions is the least denominator to which each of them can be separately reduced. NOTE. A fraction can be reduced to any higher denominator which is a dividend of its own denominator [Note 1, 82]. A common denominator of two or more fractions must therefore be any common dividend [60] of their several denominators; and their least common denominator must be the least common dividend [61] of their several denominators. REDUCTION OF FRACTIONS TO THEIR LEAST COMMON DENOMINATOR ILLUSTRATIVE EXAMPLE 91. Reduce f , f, and & to their least common denominator. SOLUTION EXPLANATION. First find the least L. C. D. of 6, 8, 12 = 24 common dividend of the several denomina- tors, as shown in 111. Ex., 63, obtaining 24 $ = (24 -r- 6) X 5 = f as the least common denominator. Then reduce each fraction to the ob- = (^ ~ o) X o = sr tained least common denominator (24) as A = (24 + 12) X 7 = sho * IU ' Ex " 82 ' MENTAL PROCESS. The required least common denominator cannot obviously be less than the greatest of the given denominators (12); and it must also contain all the factors of the remaining denominators 6 and 8 which are not found in 12. That is, 12 contains the denominator 6 in | and it contains the factor 4 of the denominator 8 in f. Hence, incorporate with 12 the remaining factor 2 of the denominator 8 in |, and 12 X 2, or 24 is the required least common denominator. RULE. 1. Find the least common dividend of the denominators of the given fractions as shown in III. Ex., 63, and this will be the required least common denominator. 2. Reduce each of the given fractions to this obtained least com- mon denominator as shown in III. Ex., 82. COMMON FRACTIONS 55 NOTE 1. If the denominators of the several fractions have no common factor, they are said to be prime to each other. The continued product of all the denominators will then be their least common denominator [Note 1, 63]. NOTE 2. If any given denominator is seen to be an exact divisor of another given denominator, it need not be considered in finding the least common denominator [Note 3, 63]. NOTE 3. Before attempting to find their least common denominator, the several fractions should be reduced to lowest terms, if not already so. EXAMPLES FOR PRACTISE Reduce to equivalent fractions of the least common denom- inator: 1. *, ! 9. A, H, A 2. i i | 10. 7, , f 3. A, , * H. it, 4, A 4- f, I, 12. |, f , * 6. i, !, A 13. I, f , T V, U 6. !, i f U. i f, f , H 7. * , A, H 15. i f , f , f 8. f , A, A 16. i 9, 5, H ADDITION OF FRACTIONS 92. Addition of fractions is the process of finding the sum of two or more unequal fractions. PRINCIPLES. 1. Only like numbers can be added [Prin. 1, 18]. Therefore the several numbers of fractional units (the several numera- tors, 72) , must express like fractional parts (similar denominators) . 2. Numerators express the number of fractional units [72] and denominators simply denote the names of those fractional units [73]. The sum of two or more fractions must therefore be the sum of their respective numerators. 3. The sum expresses a quantity of the same name as the num- bers added [Prin. 2, 18]. The sum of the several numerators must therefore express similar fractional units (must have the same denominator) as the several fractions added. 56 COMMON FRACTIONS TO FIND THE SUM OF TWO OR MORE FRACTIONS ILLUSTRATIVE EXAMPLES 1. Find the sum of J, f , and T V SOLUTION T n r> *Q Q 10 oxi .b. ^. JJ. 01 cJ, 8, LZ = 24 4 - (24 T) V 1 8 f94tV^ 8 (24ths) I = (24 -=- 8) X 3 = 9 (24ths) = (24 -i- 12) X 7 = 14 (24ths) 31 = 1 2T> EXPLANATION. First reduce the fractions to a common denom- mat r M shown m I1L Ex "> 91 ' Then add the resulting numera _ t ? rs ( 8 + 9 + 14 = 3D, and under the obtained sum (31) place the common denominator (24). As the result (ii) is an improper fraction, reduce it to a mixed number as shown in 111. x. 88. 2. Find the sum of 25J, SOLUTION L. C. D. of 4, 12, 16 = 48 251 = (48 ^ 4) X 3 = 36 (48ths) 91 fi or 1 * > Write the frac- tional part of the obtained m f ed number (n) under the column of fractions; and carry the integral part of the obtained mixed number (1) to the units' column of the Then add the integers (1, the carried figure, + 73 + 216 + 25), 6) X 11 = 33 (48ths) ~89 141 ? ^ ? 8~ integers. obtaining 315fi as the complete sum. EXAMPLES FOR PRACTISE Find the sum of + I + H 2. 3. f + i + t 4- ! + * + J 6. i + f + A 6- ! + | + T 6 * 7. A + tt + S + t 8. f + J + f + A 9. * + H + * + A 10. 11. 12. 13. 14. 15. 16. 17. 18. | + } + A + A I + f + f + f i + | + A + il 8J + 16* + 27f + 46f 73$ + 27J + 62| + 91 A 49 + 56| + 16J + 39A 628A + 9H + 28| + 63 f 19} + 17f + 26| + COMMON FRACTIONS 57 NOTE 1. It is usual, in the dry-goods business, to express all fractions of a yard in fourths. The denominator, being thus understood, is not expressed; and the numerator is written, like an exponent, at the right of the integer. 19. Add 47 1 yards, 51 2 yards, 39 1 yards, 48 3 yards, 45 2 yards, 46 yards. 20. Add 38 3 yards, 46 2 yards, 51 yards, 42 1 yards, 38 2 yards, 47 yards, 39 s yards. NOTE 2. It is customary in the grocery business to express all fractions of a pound in sixteenths, or ounces, and to omit writing the denominator understood. 21. Add 12 9 Ibs., 35 3 Ibs., 26 Ibs., 18 8 Ibs., 23 11 Ibs., 41 1 Ibs., 23 16 Ibs. 22. Add 273 5 Ibs., 286 12 Ibs., 563 8 Ibs., 725 3 Ibs., 586 13 Ibs., 75 11 Ibs. NOTE 3. In the grain business fractions of a bushel are similarly ex- pressed in GOths (pounds) for wheat, in 56ths for corn, in 32nds for oats, etc. 23. Add the following weights in wheat: 758 37 bu., 645 62 bu., 698 43 bu. 24. How much corn in 687 39 bu., 587 23 bu., 934 18 bu., 896 42 bu.? 25. How much oats in 2183 12 bu., 968 27 bu., 1472 30 bu., 5683 19 bu.? SUBTRACTION OF FRACTIONS 93. Subtraction of fractions is the process of finding the differ- ence between two unequal fractions. PRINCIPLES. 1. The difference can be found only between like numbers [Prin. 1, 22]. Therefore, the number of fractional units (the numerator) of the subtrahend must be of the same name (must have the same denominator) as the number of fractional units (the numerator) of the minuend. 2. Numerators express the number of fractional units [72] and denominators merely denote the names of those fractional units [73]. The difference between two fractions must therefore express the differ- ence between their respective numerators. 58 COMMON FRACTIONS 3. The difference between like numbers must also be like [Prin. 2, 22]. Therefore, the difference between two numbers of fractional units (two numerators) must express fractional units of the same name (the same denominator) as those of the subtrahend and min- uend. TO FIND THE DIFFERENCE BETWEEN TWO FRACTIONS ILLUSTRATIVE EXAMPLE 94. Subtract if from SOLUTION EXPLANATION. First reduce the given mixed L.C.D. of 18 and 12 = 36 number (2^) to an im- 2* = H (36 + 12) X 31 = 93(36ths) %%% fS^J^m. H (36 + 18) X 13 = 26(36ths) ^ re f uce *?*?**: v ' lent fractions of the least ?5 == **> Ans. common denominator as shown in 111. Ex., 91, obtaining ft and f f . Next subtract the less numerator (26) from the greater numerator (93), obtaining 67 as the difference between the two numerators, under which write the common denominator (36ths) to denote the name of this difference. As the result (ft) is an improper fraction, reduce to a mixed number as shown in 111. Ex., 68, obtaining Ifi. NOTE. If the mixed number contains few integral units, reduce it to the form of an improper fraction before performing the subtraction; but if it con- tains many integral units, apply the method shown in 95. EXAMPLES FOR PRACTISE Find the difference between 1. | and it 9. and J 2. and } 10. H and M 3. f and H 11. I and 1 4. f and | 12. J and \\ 5. J and 1 13. 2\ and 3i 6. 1 and | 14. If and 2 T V 7. T % and H 15. 2 T V and 8. f and f 16. 2| and 3J COMMON FRACTIONS 59 SUBTRACTION OF MIXED NUMBERS ILLUSTRATIVE EXAMPLES 95. 1. Subtract 75 from 87J. SOLUTION EXPLANATION. First find the fractional part of 87- 3(4ths) tne remam der by subtracting no fourths in the sub- __ P. , .., ^ trahend from 3 fourths in the minuend, obtaining f. Next find the integral part of the remainder by 12 f 3(4ths) subtracting the integral part of the subtrahend (75) from the integral part of the minuend (87), obtaining 12, thus making the complete remainder 12 f. 2. Subtract 216 from 325. SOLUTION EXPLANATION. As 5 sixths in the subtrahend 4 . cannot be subtracted from no sixths hi the minuend, 325 6(6ths) "borrow" 1 integral unit ( = 6 sixths) from the integral 91ft 5 ^ffitV* \ P art f tne minuend, 325. Then subtract 5 sixths of " S) the subtrahend from the borrowed 6 sixths of the 108^ l(6th) minuend, obtaining 1 sixth as the fractional part of the remainder. Next subtract the integral part of the subtrahend (216) from the integral part of the minuend (324, that is, 325 less the borrowed 1), obtaining 108 as the integral part of the remainder, thus making the complete remainder 108&. 3. Subtract 176f from 358|. SOLUTION EXPLANATION. First reduce the fractional parts of the subtrahend and .L.U.L). OI 4 and o = \& minuend to a common denominator < 12ths )- Then subtract 9 twelfths in the subtrahend from 10 twelfths in 176f (12 -5- 4) X 3 = 9(12th8) the minuend, obtaining 1 twelfth as 182 T Vj Ans. l(12th) the fractional part of the remainder, to which prefix the difference between the integers (182), obtaining 182 T V as the complete remainder. 4. Subtract 249H from 368|. EXPLANATION. SOLUTION First reduce the fractions to a com- L.C.D. of 12 and 8 = 24 mon denominator (24. 8)Xll = 33(24ths) 249H (24 . 12) X 11 = ?2 (24ths) the minuend and 11 (24ths) 22 twenty-fourths in the subtrahend. 60 COMMON FRACTIONS As 22 twenty-fourths in the subtrahend cannot be taken from 9 twenty- fourths in the minuend, borrow 1 integral unit from the integral part of the minuend (368). Add the borrowed integral unit, which equals 24 twenty- fourths to the 9 twenty-fourths, obtaining 367|| as the equivalent minuend. Subtract 22 twenty-fourths in the subtrahend from 33 twenty-fourths in the minuend, obtaining 1 1 twenty-fourths as the fractional part of the remainder. Lastly, subtract the integral part of the subtrahend (249) from the in- tegral part of the minuend (367, that is, 368 less the borrowed unit) obtaining 118 as the integral part of the remainder, thus making the complete remainder 118H- EXAMPLES FOR PRACTISE Find the difference between 1. 617 and 928f 9. 52J and 71f 2. 248f and 562 10. 341 and 63 T % 3. 325 and 7151 11. 28i| and 57^ 4. 183| and 327 12. 65| and 82 J 6. 29f and 86f 13. 24| and 46| 6. 42f and 59J 14. 35^ and 64M 7. 72i and 92H 15. 48 and 71 T \ 8. 23 T 3 * and 58f 16. 56i and 84| MULTIPLICATION OF FRACTIONS 96. Multiplication of fractions is the process of finding the product if one or both factors are fractions. NOTE. If the multiplicand denotes the value of one (entire) considered equal quantity [Note 1, a, 25], and the multiplier denotes the number of such considered quantities [Note 1, b, 25], and the product denotes the value of all such considered quantities [Note 1, c, 25], it must follow that if the multi- plier is 1, or an improper fraction equal to 1, the product must exactly equal the multiplicand; that if the multiplier is a mixed number or an improper fraction, that is, if the multiplier is greater than 1 and therefore denotes that the multiplicand is required to be taken more than one time, the product must be greater than the multiplicand; and if the multiplier is a proper fraction, or less than 1, it will denote that the multiplicand is required to be taken less than one time, and, therefore, the product must be less than the multiplicand. Hence, PRINCIPLE. Multiplication signifies increase when the multi- plier is greater than I, and decrease when the multiplier is less than 1. COMMON FRACTIONS 61 TO MULTIPLY ONE FRACTION BY ANOTHER FRACTION ILLUSTRATIVE EXAMPLES 97. Multiply (1) ! by |; (2) 16 by f ; (3) | by 12; (4) 2| by A. SOLUTION FIRST OPERATION. 8 times 3 fourths equal 24 ,*^ fourths; therefore one-ninth of 8 times 3 fourths, that is, f times f , must equal $ of ^, or ft [Prin. 3, 26] V I = II = - wmc h i n lowest terms [84] equal f . SECOND OPERATION. As the numerators 3 and 8 ^ r > have been shown in the first operation to be factors 12 of the numerator of the answer, and therefore the ^ ^ 2 factors of the dividend of an unexecuted division [79], 4 9 ~~ an( l the denominators 4 and 9 have been there shown 13 to be factors of the denominator of the answer, and, therefore, factors of the divisor of the same unexecuted division [79], and any factor common to both divisor and dividend may be canceled as shown in 111. Ex., 69, it will be found more convenient first to cancel all factors common to opposite terms as in the second operation; and then multiply the uncanceled factors of the numerators to obtain the numer- ator of the product, and the uncanceled factors of the denominators to find the denominator of the product. SOLUTIONS (2) (3) (4) 2 4 1 4 # X | - 6 p X n = V = 6f | X I = | 13 15 The explanation of Ex. 1 will also apply to Exs. 2 and 3, as will become evident by conceiving the denominator 1 to be written under each integral factor [Note, 77]. RULE. Multiply the numerators of the fractions to obtain the numerator of the product, and multiply their denominators to obtain the denominator of the product, first canceling all factors which are common to opposite terms. NOTE 1. Before multiplication, mixed numbers which express few integral units may be reduced to improper fractions [87]; and simple integers should be regarded as numerators with the denominator 1 understood [Note, 77]. 62 COMMON FRACTIONS NOTE 2. If one of the factors is a simple integer of many figures, it will usually be found more convenient to multiply the integer by the numerator of the fraction, and to divide the result by the denominator. NOTE 3. When the word "of" immediately follows a fraction, it signifies times; thus, f of f means f times f ; and 3 of a fortune means times a fortune, a fortune being called the verbal multiplicand to distinguish it from the ordi- nary numerical multiplicand. EXAMPLES FOR PRACTISE Multiply Multiply Multiply 1. * by A 6. & by 5J 11. 673 by } 2. f by | 7. 74 by | 12. f by 276 3. f by A 8. 2f by }J 13. 783 by 2i 4. T 3 * by I 9. 2J by f 14. Find f of A 5. f by 34 10. 6f by 5i 15. Find f of f of T 8 * MULTIPLICATION OF INTEGRAL AND MIXED NUMBERS ILLUSTRATIVE EXAMPLES 98. Multiply (1) 426| by 8; and (2) 675 by 29}. SOLUTIONS EXPLANATION. (1) 8 /j\ /2) times 2 thirds equal 16 g 7 thirds, or 5i Write $ as the fractional part of the 426 f ^9* product, and carry the in- 8 4)2025 tegral part (5) to the in- o/i7oi ~XM\ i tegral part of the product OU 4 (426 X 8), obtaining 3413 1 b075 ^ t^ complete product. (f X 8 = ^ = 5J) 1350 (2) First multiply 675 2008 1 1 k y the fractional part of the multiplier (f), as di- rected in Note 2, 97, obtaining 506 j. Then multiply 675 by the integral part of the multiplier (29), and add the several partial products, obtaining 2008 lj as the complete product. EXAMPLES FOR PRACTISE Multiply Multiply Multiply 1. 678| by 7 5. 816 by 8} 9. 629 by 18* 2. 295|by9 6. 295 by 6 f 10. 571 f by 35 3. 427f by 8 7. 378 by 52J 11. 826f by 42 4. 572by3f 8. 483 by 25 } 12. 382| by 85 COMMON FRACTIONS 63 MULTIPLICATION OF MIXED NUMBERS BY MIXED NUMBERS ILLUSTRATIVE EXAMPLE 99. Multiply 79f by 27 f. SOLUTION EXPLANATION. The total product must equal the product of each part of 79 f the multiplicand separately multiplied by 27 1 each part of the multiplier [Prin. 2, 26]. 3 2 I _ /9, +v \ Therefore, multiply the fractional part of ns ' the multiplicand (f) by the fractional 79 X f = 52f = 16(24ths) part O f the multiplier (f), obtaining . | X 27 = 10J = _3(24ths) Next multiply the integral part of the 79 X 7 = 553 S = I. 1 multiplicand (79) by the fractional part 79 v/ 2 = 158 ^ tne mu ltipl* er (s)> obtaining 52f. 1 Now that att of the multiplicand has been multiplied by the fractional part of 2196^, Ans. the multiplier, separately multiply the fractional and integral parts of the multiplicand (| and 79) by the integral part of the multiplier (27), and add the several partial products obtaining 2196s 1 ! as the complete product of all the orders of the multiplicand multiplied by att the orders of the multiplier. RULE. 1. Multiply the fraction of the multiplicand by the fraction of the multiplier. 2. Multiply the integer of the multipli- cand by the fraction of the multiplier. 3. Multiply the fraction of the multiplicand by the integer of the multiplier. 4. Multiply the integer of the multiplicand by the integer of the multiplier. 5. Add the several partial products. EXAMPLES FOR PRACTISE Multiply Multiply Multiply 1. 56fby36| 6. 286^ by 28 11. 5684 f by 28 A 2. 42| by 64J 7. 672| by 40 12. 9275 by 93| 3. 631 by 75| 8. 548! by 36i 13. 3246J by 74| 4. 37A by 231 9. 715| by 12f 14. 4183f by 300f 6. 51| by 18J 10. 368f by 74J 15. 958U by 673| DIVISION OF FRACTIONS 100. Division of fractions is the process of dividing when the divisor or dividend, or both of them, are fractions or mixed numbers. 64 COMMON FRACTIONS NOTE. If division is the opposite process to multiplication [Note 1, 48], Principle, 96, taken reversely, will apply to division; that is, PRINCIPLES. 1. // the divisor is greater than 1, the quotient will be less than the dividend. 2. // the divisor is less than 1, the quotient will be greater than the dividend. DIVISION OF FRACTIONS BY INTEGERS ILLUSTRATIVE EXAMPLES 101. Divide (1) f by 4; (2) f by 3; (3) | by 6; (4) 3} by 21. SOLUTIONS EXPLANATION. (1) 8 ninths divided by 4 equal /n 2 ninths [Prin. 1, 81]. For, if f expresses the quotient g _._ 4 2 of an unexecuted division [79] of which the numerator = Q, Ans. (8) is the dividend and the denominator (9) is the divisor, then to divide the dividend (8) by 4 is equiva- (%\ lent to dividing the quotient (the fraction) by 4 [Prin. 5 5 4 ' 34 >- o v/ o = ^r, Ans. (2) As the numerator (5) is not divisible by the divisor (3), as in the preceding operation, use Prin. 2 2, 81, and multiply the denominator (8) by the divisor 2 (3), obtaining ^ as the quotient. For, to multiply y $ = rv7> Ans. the divisor of an unexecuted division by 3 is equivalent to dividing the quotient (the fraction) by 3 [Prin. 6, 34]. ,. (3) As the numerator (4) is not divisible by the 1 entire divisor (6) but is divisible by one of its factors J 1 (2), divide the numerator (4) by the factor 2, as in 2 V 21 = 6' Ans. Hi. Ex. 1, and multiply the denominator (9) by the 3 remaining factor 3 (2 X 3 = 6), as in 111. Ex. 2, obtain- ing tfV [Summary, 1, 37]. (4) First reduce the small mixed number (3?) to an improper fraction (1). As the numerator of | is not divisible by the entire divisor (21) but is divisible by one of its factors (7), divide the numerator (7) by the factor 7 of the divisor (21 = 7 X 3), obtaining 1 as the new numerator and \ as the quotient, then, in succession, [Summary 1, 37] divide ^ by the remaining factor of the divisor (3) as already shown in the second 111. Ex., obtaining i as the final quotient. RULE. 1. Divide the numerator of the fractional dividend by the divisor, and let the denominator remain unchanged. Or, 2. Multiply the denominator of the fractional dividend by the divisor, and let the numerator remain unchanged. Or, COMMON FRACTIONS 65 3. Divide the numerator of the fractional dividend by the great- est factor of the divisor which will exactly divide the numerator, and multiply the denominator of the fractional dividend by the remaining factor of the divisor. NOTE. If the dividend is a mixed number which contains few integral units, it should be reduced to an improper fraction. EXAMPLES FOR PRACTISE Divide Divide Divide 1. f by 3 6. ft by 6 11. 4 by 7 2. | by 6 7. 5i by 14 12. 6f by 5 3. iJbyS 8. 6by8 13. A by 20 4. I by 16 9. 2i by 9 14. J by 28 5. | by 20 10. f by 8 15. 8J by 14 DIVISIONS OF INTEGERS BY FRACTIONS 102. Division by any number, integral or fractional, is the reverse of multiplication by the same number [Note 1, 48]; that is, 8 divided by 2 is the opposite process to 8 multiplied by 2; for in the former 8 is made one-half as great, and in the latter, 8 is inversely made twice as great. Therefore, any division by a fractional divisor must be the inverse process of multiplication by that same fractional quantity. Now, in multiplying 6 by f , it has already been seen that 6 is multiplied by the numerator (2) and divided by the denominator (3). Hence, when performing the reverse operation of dividing 6 by f , the reverse process should be followed and 6 should be multiplied by the denominator (3) and divided by the numerator (2) ; that is, in multiplying 6 by f , the statement is 6 X f [111. Ex. 2, 97]; but in dividing 6 by f, the statement should be 6 X i, or the former process with the terms of the divisor reversed. Hence, SUMMARY. 1. Division by a fractional divisor is the opposite process to multiplication by a fractional multiplier. 2. Division by any fractional divisor is the equivalent of mul- tiplication by that same fractional divisor with its terms inverted. 66 COMMON FRACTIONS ILLUSTRATIVE EXAMPLES Divide (1) 8 by ; (2) 15 by 3 J. SOLUTIONS EXPLANATION. (1) If the dividend (8) be divided g -i- - by the numerator of the divisor (4), that is, by five 5 times the true divisor (|), the resulting quotient (2) 2 g will be one-fifth of the true quotient [Prin. 6, 34]. $ X ~i = 10 Therefore, five times 2 or 10, must be the true quo- tient; and 5 times \ of 8 or f of 8, that is, 8 multiplied by f with its terms inverted (8 X I) must be the /o\ proper process for dividing 8 by f. * ' (2) The quotient of 15 divided by 25 eighths tc _i_?_ (3 1) is the same as the quotient of one-fifth of 15, or 3, 8 divided by one-fifth of 25 eighths, or f [Prin. 5, 34]; 3 n and 3 -T- f, as shown in the first solution, is equal to 5 RULE. Multiply the dividend by the fractional divisor with its terms inverted. EXAMPLES FOR PRACTISE Divide Divide Divide 1. 6 by f 5. 15 by f 9. 15 by 2 2. 15 by f 6. 2 by f 10. 2967 by f 3. 24byf 7. 19 by f 11. 6183 by 3| 4. 8byf . 8. 28 by I 12. 7458 by 8 i DIVISION OF FRACTIONS BY FRACTIONS ILLUSTRATIVE EXAMPLE 103. Divide f by f SOLUTION EXPLANATION. If 8 ninths be divided by 2, which g _i_ 2 _ is three times the true divisor (f), the quotient (f) will be one-third of the true quotient [Prin. 6, 34]. There- fore 3 times f divided by 2, that is, three times one-half $_4_ 1 of f , or X f must be the true quotient, and this is ^$~~3 = * equivalent to multiplying the dividend by the divisor 3 i with its terms inverted. COMMON FRACTIONS 67 RULE. Multiply the dividend by the fractional divisor with its terms inverted. EXAMPLES FOR PRACTISE Divide Divide Divide 1. f by f 6. 3f by f 9. f by f 2. J by f 6. 8J by f 10. 31 by 4f 3. *by 7. Aby2| 11. 11 by 21 4. 2iby| 8. A by 4} 12. 5J by 3J DIVISION OF MIXED NUMBERS BY INTEGERS ILLUSTRATIVE EXAMPLE 104. Divide 919f by 27. EXPLANATION. First divide the integral part of 27)9191 (34 A the dividend ( 919 ) y the divisor (27), obtaining 34 gj as the integral part of the quotient. Then bring down the fractional part of dividend (I) to the remainder (1), obtaining 1| or V; and 108 divide this as yet undivided part of the dividend (V) j! _ jus by the divisor (27), obtaining 7 5 j as the fractional part of the quotient, thus making the complete quotient 1> 1 5 L- x = If desired, 7 \ may be reduced to any desired de- 8 ff 72 nominator as shown in 111. Ex., 86. RULE. Separately divide the integral and the fractional parts of the dividend by the divisor. Or, proceed as shown in 105. EXAMPLES FOR PRACTISE Divide Divide Divide 1. 8375f by 15 4. 62515 by 45 7. 3265* by 372 2. 5268f by 36 5. 9286f\ by 84 8. 58293A by 173 3. 6739| by 28 6. 2480JI by 16 9. 6875| by 540 COMMON FRACTIONS DIVISION OF INTEGERS BY MIXED NUMBERS 105. Divide 896 by 27 j. SOLUTION EXPLANATION. To multiply both divisor 27-) 896 an d dividend by the same number does not o g affect the value of the quotient [Prin. 5, 34]. Therefore, multiply both divisor (27|) and divi- 221)7168(32^ 6 T dend (896) by the denominator of the fractional 663 part of the divisor (8), obtaining 221 eighths roo as the equivalent divisor and 7168 eighths as the equivalent dividend. Divide as with integers [36]. If necessary, $& can be reduced 96 to an approximate fraction of any desired 96 221 = 9 6 denominator by 86. RULE. Multiply both divisor and dividend by the denominator of the terminal fraction, and proceed with the results as in integral division. EXAMPLES FOR PRACTISE Divide Divide Divide 1. 6783by41f 4. 8729 by 328f 7. 28973 by 269* 2. 5286by37| 5. 7123 by 416$ 8. 35286 by 713t 3. 2857 by 25f 6. 9285 by 876f 9. 41627 by 697J DIVISION OF MIXED NUMBERS BY MIXED NUMBERS ILLUSTRATIVE EXAMPLE 106. Divide 684f by 47J. SOLUTION EXPLANATION. Multiply both divisor and divi- dend by the least common denominator of their 285)4108(14H terminal fractions (L. C. D. of 2 and 3 = 6), obtain- 2g5 ing 285 sixths as the equivalent divisor, and 4108 sixths as the equivalent dividend [Prin. 5, 34]. 1258 Divide as with integers [36]. 1140 If required, ilf can be reduced to an approxi- mate fraction of any desired denominator by 86. 118 -5- 285 RELATION OF COMMON FRACTIONS 69 EXAMPLES FOR PRACTISE Divide Divide Divide 1. 695| by 27| 4. 6395| by 34} 7. 3486f by 537J 2. 438^ by 18* 5. 5284A by 65J 8. 27581- by 238f 3. 741 f by 53f 6. 9172| by 24i 9. 8175| by 196^ Approximating the fractional part of the quotient to the specified de- nominator, divide 10. 475f by 73*, to 16ths 13. 7283* by 196f , to 4ths 11. 9183| by 29J, to 8ths 14. 5976J by 39J, to 64ths 12. 6198| by 16f , to 32nds 15. 2743 J by 216, to halves RELATION OF COMMON FRACTIONS 107. INTRODUCTION. The principles which govern the rela- tion of integers [48 to 53], inclusive, are unlimited in their applica- tion to all numerical expressions, whether integral or fractional. It only remains to explain the fractional form which the multi- plier now assumes, and to harmonize its definition [Note, 96] with that of an integral multiplier [Note 1, b, 25]. A number is the expression of a quantity by the use of figures [3], and may mean "how much" as well as "how many." Hence, f of a pound is as much a number of pounds as 1 pound is, or as 2 or more pounds are. The only difference between them is in the form of expres- sion; | (3 fourths) expressing a fractional number of pounds, and the others expressing integral numbers of pounds. Hence, PRINCIPLES. 1. The multiplicand denotes the value of one considered equal quantity, whether integral or fractional. 2. The multiplier denotes the integral or fractional number of considered equal quantities. 3. The product denotes the value of all the inte- gral or fractional number of considered equal quantities. NOTE. The student is cautioned against confounding the expression "fractional number of units," with "number of fractional units." In the numerical expression f , f is the fractional number of units, and the numerator 3 is the number of fractional units, or fourths. ILLUSTRATIVE EXAMPLES 1. Find the cost of f of a yard of silk at 80 cents per yard. 2. If | of a yard of silk cost 60 cents, what is the price per yard? 70 RELATION OF COMMON FRACTIONS 3. How much silk at 80 cents per yard can be bought for 60 cents? SOLUTION Factors of f 80 ff = cost of one entire yard = multiplicand 1 Factors of dividend 1 f = fractional number of yards = multiplier J product Dividend, if _ _ ^ rf fract . onal number _ product if given required EXPLANATION. In Ex. 1, as 80?f is the price of one (entire) yard, it is a multiplicand; as f expresses the fractional number of yards, it is a multiplier. Hence, multiply these two factors, obtaining 60 j as the cost of all the frac- tional number of yards, or product. In Exs. 2 and 3, the cost of all the fractional number of yards, or product (60 ff), is given. Hence, in Ex. 2, divide the given product (60 ) by its given multiplier factor (the fractional number of yards, f ) to obtain its required multiplicand factor (the price of one entire yard, 80 jf); and in Ex. 3, divide the given product (the cost of all the fractional number of yards, 60^) by its given multiplicand factor (the price of one entire yard, 80^) to obtain its required multiplier factor (the fractional number of yards, f ). 4. If a man can walk 3| miles in one hour, what part of an hour will he require to walk 2J miles? EXPLANATION. 3 miles is a multiplicand because it is the distance walked in one entire hour [Prin. 1] and 2\ miles is a product because it is the distance walked in all the required fractional number of hours [Prin. 3]. Therefore, divide the given product (!) by its given multiplicand factor ( J s a ), to obtain its required multiplier factor (| X iV = f ), or the fractional number of hours. EXAMPLES FOR PRACTISE 108. x l. What is the cost of f of a gallon of syrup bought at $J per gallon? -* 2. At $f per yard, how many yards of goods can be bought for $2i? 3. If a horse is fed f of a peck of corn twice every day, how much corn will be necessary to last 4J days? >4/ How much butter at S| per pound can be bought for $3? 6. If the speed of a steamboat is 15| miles per hour, in what time can it go a distance of 9J miles? 6. A gardener wishes to divide 2J acres of land into 16 equal flower beds. How much land will each bed contain? COMPARISON OF QUANTITIES 71 7. If a man can complete f of a task in -fy of an hour, in what time can he perform the entire task? [Explanation of Ex. 17, 54]. 8. If 8 men can complete a piece of work in f of a working day, how many working days will be required for one man to com- plete it? 9. If 28 men can do a certain piece of work in 3 J working days, how many working days will be necessary for 7 men to do the same piece of work? 10. If there are 146 days in f of a year, how many days are in || of a year? 11. The distance between two villages is 18f miles. If a trav- eler has walked f of the distance, how many miles has he yet to walk? -f 12. What is the total cost of J of a pound of cheese at $J per pound, A of a pound of butter at $f per pound, and 9f gallons of syrup at $| per gallon? -^ 13. If it cost $T 3 ff to buy f of a yard of goods, how many yards Of the same goods can be bought for $6?? 14. An estate was bequeathed to a mother and her five chil- dren, the mother to receive one-third, and the remainder to be equally divided among the children. What part of the estate did each child receive? 15. From a barrel of cider containing 49 f gallons, there were drawn 7| gallons at one time, 16| gallons at another time, 5 1 gallons at a third drawing, and 12 J gallons at a fourth. How many gallons remained in the barrel? 16. 2J times 17f yards of goods are how much less than 5? times 28 J yards of the same goods? COMPARISON OF QUANTITIES 109. INTRODUCTION, (a) In a comparison of quantities whether integral or fractional, any one of them may be arbitrar- ily assumed as the standard, as the author of the problem may direct. It is therefore necessary that the problem should indi- cate which of the considered quantities is to be accepted as the standard of comparison. The conventional form of language for arbitrarily indicating the standard is the expression "of" [Note 3, 97] placed immediately after the fractional multiplier and immedi- ately before the standard of comparison, or multiplicand. Thus, 72 COMPARISON OF QUANTITIES in the completed comparison, $2400 are f of $3600, "$3600" is the indicated standard of comparison or multiplicand (because it is made to follow " of"), and expresses the value of one con- sidered quantity [Prin. 1, 107]; is the indicated multiplier (be- cause it immediately precedes "of $3600"), and expresses the frac- tional number of times the considered standard which is expressed by the quantity compared with the standard [Prin. 2, 107] ; and $2400 is a product because it is the quantity which is placed in comparison with the standard by denoting the number of times (f ) the standard ($3600) which is expressed by $2400, the quan- tity compared with the standard. That is, " $2400 are f of $3600," simply means $240*0 are f times $3600. (6) If, however, the standard of comparison is to be indicated with the view of increasing it, the phrase "more than" should immediately follow the fractional multiplier and precede the as- sumed standard of comparison. Thus, in the completed compari- son "$840 are f more than $600," $600 is the indicated standard of comparison, or multiplicand (because it is thus pointed out by being made to follow " f more than ") , and expresses the value of one considered equal quantity; f is the indicated multiplier (because it immediately precedes "more than $600"), and expresses the fractional number of times the standard which the quantity com- pared with the standard is more than that standard; and $840 is the quantity compared with the standard, or a product, because it expresses the value ($840) of all the considered fractional num- ber of times the standard (f more than once or of the standard, that is, I of $600). Expressed in more familiar terms, "$840 are f more than $600" means, "$840 are f of (times) $600, added to $600." (c) The standard of comparison may be indicated in a way to show that it is to be decreased, in which case the indicator "less than" should follow immediately after the fractional mul- tiplier and immediately precede the indicated standard of com- parison, or multiplicand. Thus, in the completed comparison "$750 are f less than $1200, the standard of comparison, or multi- plicand, is $1200; f less than once (f) the standard, or f times the standard, is the multiplier; and $750 is the quantity compared with the standard or product. In other words, the problem means, $750 are f of (times) $1200, subtracted from $1200. PRINCIPLES. 1. The multiplicand is the value of one indicated quantity which is assumed as a standard with which some other quantity is to be compared. 2. The multiplier is the fractional part of the standard (the fractional number of times the standard) COMPARISON OF QUANTITIES 73 which is expressed by the quantity compared with the standard. 3. The product is the quantity compared with the standard (the value of all the fractional number of times the standard which is expressed by the multiplier). NOTE 1. A problem in comparison is one in which two of the three terms of a perfected comparison are given, and it is required to find the omitted third term. NOTE 2. In problems in which the numerical standard of comparison is the omitted third term, the verbal standard will have to be substituted after the indicators "of," "more than," or "less than." Thus, in the problem: "If 200 acres are f of A's farm, how many acres does A's farm contain," "A's farm" is the indicated verbal standard of comparison, or verbal multiplicand which is required to be expressed numerically; 200 acres is the quantity compared with the standard (A's farm), or the product; and f is the multi- plier or fractional number of times A's farm expressed by the product (200 acres). ILLUSTRATIVE EXAMPLES 1. What part of $200 are $80? SOLUTION EXPLANATION. $200 is the indicated stand- ard of comparison, or multiplicand; $80 is the = Stfff = 5 quantity compared with the indicated standard, or product. Hence, divide the given product ($80) by its given multiplicand factor ($200), expressing the unexecuted division as directed in Introduction, a, 70, obtaining \ as the required multiplier fac- tor, or fractional number of times (fractional part of) the standard ($200) which is expressed by $80. 2. f of how much wheat equal \ of a bushel? SOLUTION EXPLANATION. "How much wheat" is the indicated verbal standard, or multiplicand, of which it is required to find the equivalent numer- 1 _._ 4 = _}/L._ u n ical expression [Note 2, above], f is the mul- ^48 tiplier, because it denotes the fractional number 2 of times the standard (how much wheat) which is expressed by the given product ( of a bushel). Hence, divide the given product (i bu.) by its given multiplier factor (f), obtaining f bu. as the required multiplicand factor, or standard I of which equal i bu. 74 COMPARISON OF QUANTITIES 3. 280 men are J more than how many men? SOLUTION EXPLANATION. "How many g 4 men" is the indicated verbal stand- 4 more than ? = ar( j O f comparison, or multiplicand, . of which it is required to find the 7 4 equivalent numerical expression. 280 -f- 4 = WP X| = 160 men 280 men is a product, because it is the quantity compared with the required standard. f more than once the required standard, that is, f more than I of the standard, or | of the standard, is the multiplier, because it denotes the fractional num- ber of times the standard, or multiplicand, which is expressed by the prod- uct (280 men) or quantity compared with the standard. Hence, divide the given product (280 men) by its indirectly given multiplier factor (1), obtain- ing 160 men as the required multiplicand factor, or standard f of which ( times which) equal 280 men. EXAMPLES FOR PRACTISE 110.M. I paid $7200 for a farm and afterwards sold it for | of what I paid for it. How much did I receive for the farm? 2. If | of a bushel of wheat weigh 40 pounds, what is the weight of an entire bushel? 3. What part of $320 are $180? 4. A dealer sold f of his stock of merchandise for $4824. At how much was the entire stock valued? 5. A man bequeathed & of an estate worth $12352 to a daugh- ter. How much did she receive? 6. 48 bushels of corn were sold from a bin containing 56 bushels. What part of the contents of the bin was sold? 7. In a certain school were enrolled 246 boys and 318 girls. What part of the total enrolment were the girls? 8. The capital of a firm of two partners was $5600, of which one partner invested $2100. What part of the capital did the other partner invest? 9. A grazier sold $ of a flock of sheep and had 114 sheep re- maining. How many sheep did he sell? 10. A man's salary is $90 per month and his expenses $60 per month. What part of his salary does he save? > 11. A is worth $13000, or f more than B. How much is B worth? REVIEW OF FRACTIONS 75 12. A man paid of his indebtedness and then found that he still owed $7310. How much did he originally owe? REVIEW OF FRACTIONS 111. 1. What is the sum of 137f, 463f, 285J, 348 T V, 825H, 567 f, and 362J? 2. A man engaged to plow a field containing 78f acres. How many acres remained unplowed after completing f of his engage- ment? 3. What is the cost of 58f yards of goods at 37 i cents per yard? 4. A dealer sold 16j yards from a piece containing 38y 5 s yards. What part of the piece of goods did he sell? 5. The distance between two harbors is 847H miles. After a ship has sailed J of this distance, how many miles remain to complete the voyage? 6. What is the total cost of 6J pounds beef at 18 J cents per pound, 3J pounds mutton at 13 f cents per pound, and f of a pound of lard at 11 J cents per pound? 7. A farm consists of 6 fields containing respectively 68 J, 57 1, 65TW, 42|f, 52 1, and 63|f acres, a vegetable garden con- taining H of an acre, an orchard of If acres, and a lawn of 2 acres. If it was bought at the average rate of $36 per acre, how much did the farm cost? 8. If a ship completes A of its voyage on the first day, ^ on the second, \ on the third day, and is then 1300 miles from its destination, what is the total distance which it has already sailed? 9. If a railroad train moves with the average speed of 37$ miles per hour, in how many hours can it traverse 718J miles? 10. What is the cost of 685 J tons coal at $5.75 per ton? 11. /? of a day is f of what part of a day? 12. 345 gallons are what part of 460 gallons? 13. A grammar and speller together cost 92 cents, and the speller cost \ as much as the grammar. What was the cost of each? 14. A can complete a job of work in 8 days, and B in 10 days. In what time can they complete the task by working together? 76 REVIEW OF FRACTIONS EXPLANATION. If A can complete the entire job in 8 days, he can per- form I of the job in 1 day; and if B can complete the entire job in 10 days, he can perform rff of the job in 1 day. Therefore, by working together, A and B can perform i of the job plus yV of the job, or & of the job in 1 day. That is, & of the job is the multiplicand, because it expresses the value of one equal day's labor performed by both men [Prin. 1, 107]; f o of the job is the product understood, because it expresses the value of all the equal days' labor performed by both men. Hence, divide the product understood (?{}) by its given multiplicand factor (?%), obtaining 4| days as the required multiplier factor, or number of times one day's labor of both men (& of the job) which will equal the total labor of both men (?$ of the job), or the entire job. 15. A can do a piece of work in 5 days and B in 7 days. In how many days can they do the piece of work by uniting their labor? 16. A can perform a certain task in 6 days, B in 9 days, and C in 12 days. In what time can the task be completed if they work together? 17. A and B, working together, can finish a piece of work in 12 days. If A working alone can do it in 20 days, in what time can B do it without any help? 18. The greater of two numbers is 347/j and the less 265. What is the difference between them? 19. The greater of two fractions is ff, and their difference iV What is the less fraction? 20. The less of two numbers is 758 f, and the difference between them is 125. What is the greater number? 21. The total of six weighings is 936 f pounds. If five of the weighings are respectively 168 J, 271 J, 82^, 125f, and 76 J pounds, what is the sixth weighing? 22. If coffee loses yV of its weight in roasting, how much green coffee will be required to produce 375 pounds of roasted coffee? 23. A water tank has two outlets of different diameters. If only the smaller outlet is opened, the tank will be emptied in 15 minutes; and if both are opened, it will be emptied in 6 minutes. In what time will the tank be emptied if only the larger outlet be opened? 24. A merchant withdrew if of his deposit from his bank and afterwards had $510 remaining on deposit. How much did the merchant withdraw? DECIMAL FRACTIONS 77 25. If a barrel containing 28 gallons of vinegar is only f full, what part of the barrel will be filled if 7 gallons of vinegar are added? 26. The capital of a firm was $24210. If the investment of one partner was f of the investment of the other partner, what did each partner invest? 27. A man bought a house, paid 3! of the purchase money in cash and still owed $5200. What was the cost of the house? 28. Multiply 275f by 38f and divide the product by 18|. 29. Subtract the sum of 268}, 425, and 362f from the sum of 349f, 562J, 416A, and 183}. 30. A man owed $3500 and afterwards paid $1800 at one time and $300 at another time. What part of his original debt did he still owe? 31. A invested } of the capital of a firm, B }, C }, and D the remainder. If D's investment was $2782, what was A's, B's, and C's? 32. A man bought a house for $8500, paid } of the purchase money at one time, and f of the remainder at another time. How much remained unpaid? DECIMAL FRACTIONS 112. INTRODUCTION. The new form of fraction now to be con- sidered differs from that of common fractions, already studied, in this one respect, that the denominators of decimals are not usu- ally expressed, whereas in common fractions, by reason of the varying scale employed, the denominators require expression, except in a certain class of commercial fractions [Notes 1, 2, 3, 92]. The distinguishing characteristic of decimal fractions is the uni- form scale of ten employed in expressing the equal parts into which the unit of the fraction [80] is divided, instead of the arbitrary, inconstant scale by which the unit of common fractions may be divided into any number of equal parts. Reading and writing decimals are simply extensions on a descending scale of the manner of reading and writing integers. Taking the unit as the initial order, the first figure on the left of it is the tens' order, and the first figure on the right of it is the tenths' order; the second figure on the left of the units' order is the hundreds' order, and the second figure on its right is the hundredths' order, etc. While the methods already shown for operating upon common 78 DECIMAL FRACTIONS fractions are also applicable to decimal fractions, if the latter are made to assume the form of common fractions by expressing their denominators, the much more simple and convenient methods for reduction of integers, and for adding, subtracting, multiplying, and dividing integers, can be extended to decimal fractions by reason of the same decimal system of notation which is employed for expressing both. 113. A decimal fraction expresses one or more of the decimal parts of an integral unit. Thus, any fraction the denominator of which is 10, or the product of 10 taken twice as a factor (10 X 10 = 100), or the continued product of 10 taken any number of times as a factor (10 X 10 X 10 = 1000), is called a decimal fraction, or simply a decimal. 114. A decimal point is a dot (.) placed at the left of a decimal to indicate its denominator, and if a mixed decimal, to separate the fractional from the integral orders. The denominator of a decimal is understood to be 10 taken as many times as a factor as the number of places which the right-hand order of its numera- tor is located to the right of the decimal point. Thus, in .5, the numerator 5 is located one place to the right of the decimal point and therefore the denom- inator is understood to be the factor 10 taken once; in .17 the right-hand order of the numerator (7) is located two places to the right of the decimal point, and the denominator is understood to be the factor 10 taken two times; that is, 10 X 10, or 100; etc. 115. The decimal order or decimal value of any figure of the numerator is its distance from the decimal point. The nearer any order of the numerator is to the decimal point, the greater its value, because 10 will be a factor of its denominator fewer times, thus dividing the unit of the fraction into fewer equal parts, and correspondingly increasing the value of those parts; and the more distant any order of the numerator is from the decimal point, the less its value, because 10 will be a factor of its denominator a greater number of times, thus dividing the unit of the fraction into a greater number of equal parts, and correspondingly diminish- ing the value of those parts. The names of the several decimal orders and their relative values, are shown in the following. | 09 ousands 1 Etc., etc. BILLIONS 1 1 Ten-million MILLIONS Hundred-th Ten-thousa THOUSANDS Hundreds 1 | t> 7 3 8 2 4 9 1 5 1 6 4 jt -n ^ JS 4 ^ \ 2 .- ^ S S $ ""O CO S -2| DECIMAL FRACTIONS 79 TABLE OF DECIMAL VALUES I i . 1 i ii]ii 1 a a i s Y -s f i Sc G oa2 i a 'S^g^^KS^W 7 1 8639475 Integral Orders Decimal Orders EXPLANATION. In considering the above TABLE, observe 1st, that the decimal point separates the decimal orders from the integral orders; 2nd, that, commencing at the decimal point, the integral orders are enumerated from right to left, and the decimal orders from left to right; 3rd, that the names of the several decimal orders correspond with those of the integral orders which are equally distant from the units 1 order. 116. A simple decimal is one which is composed entirely of decimal orders, having no integer at its left, and no common fraction at its right; as, .175. 117. A complex decimal is one which has a common fraction at the right of its lowest decimal order; as, .286 f. NOTE. A complex decimal is said to be terminate when the common fraction at its right can be expanded to an equivalent simple decimal [116] and to be interminate when it cannot be thus expanded. 118. A mixed decimal is one which contains integral orders at the left of its decimal point; as, 28.356. 119. The unit of a decimal is the whole unit of which the decimal expresses a part; and a decimal unit is one of the equal decimal parts into which the unit of the decimal has been divided. Thus, in the expression .25 of a yard, one whole yard is the unit of the decimal, and one one-hundredth of a yard is the decimal unit. 120. PRINCIPLES. 1. The denominator of a decimal is 1 fol- lowed by a cipher for every place that its right-hand order is distant from the decimal point. 80 DECIMAL FRACTIONS Thus, in .0375 the denominator is 1 followed by four O's (10000), because its right-hand order (5) is four places to the right of the decimal point; and the decimal is read 375 ten-thousandths. 2. Adding ciphers to, or taking them from the right of a decimal, makes no change in its value. To place O's at the right of a decimal numerator is equivalent to multiply- ing both the numerator and the denominator by 10 for every annexed [30]; and to omit O's from the right of a decimal numerator is equivalent to dividing both the numerator and the denominator by 10 for every omitted [38] ; and therefore, in neither case, does it change the value of the fraction [Prin. 4, 81]. 3. Prefixing ciphers to a decimal numerator and moving the decimal point to the left of the prefixed ciphers, is equivalent to divid- ing the decimal by 10 for each cipher prefixed. To place O's on the left of a decimal numerator does not affect the value of the numerator, but increases the denominator tenfold for each so prefixed [Prin. 1], and therefore is equivalent to dividing the fraction by 10 for each prefixed cipher [Prin. 2, 81]. 4. Moving its decimal point one or more places to the right of its former position, multiplies the decimal by 10 for each place so removed. To move the decimal point to the right of its former position will bring the right-hand order of the decimal correspondingly nearer to the point, and thus diminish the number of O's in the denominator [Prin. 1] ; but to diminish the O's in the denominator is equivalent to dividing the denominator by 10 for each rejected, and therefore of correspondingly multiplying the decimal [Prin. 3, 81]. 5. Moving its decimal point one or more places to the left of its former position, divides the decimal by 10 for each place so removed. To move the decimal point to the left of its former position will throw the right-hand order of the decimal farther from the point, and thus increase the number of O's in the denominator [Prin. 1]; but to increase the O's in the denominator is equivalent to multiplying the denominator by 10 for each additional 0, and therefore of correspondingly dividing the decimal [Prin. 2, 81]. DECIMAL FRACTIONS 81 READING DECIMALS 121. Read the decimal .0289. EXPLANATION. First read the decimal as if it were an integer (two hun- dred eighty-nine); then name its denominator (1 followed by four ciphers, or 10000, Prin. 1, 120). Read the following decimals: 1. .7 7. .375 13. .06345 19. 57.816372 2. .09 8. .0375 14. .00918 20. 219.0071832 3. .005 9. .58 15. .07016 21. 8.00031678 4. .35 10. .291 16. .00053 22. 2.006521873 5. .046 11. .8273 17. .21897 23. 15.000005281 6. .673 12. .625 18. .00725 24. 7.8123768542 WRITING DECIMALS 122. Express two thousand one hundred nine hundred thou- sandths by the use of figures. EXPLANATION. The denominator (100000) contains five ciphers, and the numerator (2109) contains only four of the five places necessary for its right- hand order (9) to be removed from the decimal point to express such a denom- inator [Prin. 1, 120]. Therefore, write the decimal point and one cipher for the one lacking decimal place (.0), and then annex the numerator, obtaining .02109, in which the right-hand order (9) is located five places to the right of the point, and is thus made to express a denominator of 1 followed by five O's, or 100000. NOTE. The number of ciphers, if any, to follow the decimal point (or to be prefixed to the numerator) will always be the number of ciphers in the denominator diminished by the number of figures necessary to express the numerator. Express the following as numerical decimals : 1. Six tenths 9. 328 ten-thousandths 2. Three hundredths 10. 75 hundred-thousandths 3. Twenty-seven hundredths 11. 4768 millionths 4. Five thousandths 12. 27 ten-millionths 5. Forty-two thousandths 13. 92536 millionths 6. Nine hundredths 14. 527 hundred-thousandths 7. Sixty-three thousandths 15. 8 millionths 8. Eighteen ten-thousandths 16. 263 ten-millionths 82 DECIMAL FRACTIONS 17. Nineteen, and twenty-three thousandths 18. Forty-six, and two hundred fifty-three ten-thousandths 19. Ninety-five, and seventy-nine thousandths 20. Six, and one thousand three hundred four millionths 21. Two, and eight thousand ninety-six hundred thousandths 22. Three and four thousand seven millionths 23. Four hundred fifty-three, and twenty-eight thousand seven hundred forty-six hundred-millionths 24. Eight thousand three hundred sixty-two, and one hundred twenty-one thousand three hundred seventeen billionths 25. Nine, and eighty-seven million two hundred ninety-four thousand eight hundred seventy-one billionths REDUCTION OF COMMON FRACTIONS TO DECIMALS ILLUSTRATIVE EXAMPLE 123. Reduce T ? S to an equivalent decimal. SOLUTION EXPLANATION. The value of any fraction is the 1 fV>7 OOfW 437^ quotient from dividing its numerator by its denomina- 10;/.UUI tor [7Q j Hence> jf 10000 timeg the numerator 7 ( or 70000) be divided by the denominator (16), the result- 60 ing quotient (4375) will be 10000 times the true quo- 4g tient [Prin. 4, 34]. Therefore cut off four places from the right of the quotient (.4375) which is equivalent to dividing it by 10000 [38], thus obtaining the true 112 quotient. on If no numerator can be exactly divisible by its Q denominator except it contain all the factors of that denominator [Prin. 1, 62], and the denominator (16) contains the factor 2 four times (2X2X2X2 = 1 6), it is necessary for the numerator (7) to contain the factor 2 four times in order to be exactly divisible by the denominator (16). As annexing one cipher to the numerator is equivalent to multiplying it by 10 [30], and as 10 contains the factor 2 once (2X5 = 10), it follows that the factor 2 is introduced once for every cipher annexed to the numerator, and therefore to annex four O's to the numerator will introduce the factor 2 four times, and thus make it exactly divisible by the denominator. RULE. Annex ciphers to the numerator and divide by the denom- inator. Place a decimal point as many places from the right of the quotient as there have been ciphers annexed to the numerator. DECIMAL FRACTIONS 83 NOTE 1. The number of ciphers to be annexed to the numerator will be equal to the greatest number of factors 2 or 5 in the denominator. A common fraction in lowest terms cannot therefore be reduced to a simple decimal [116] if its denominator contains other prime factors than 2 or 5, because annexing O's to the numerator cannot introduce such other factors without which exact division of the numerator by the denominator is impossible [Prin. 1, 62]. NOTE 2. Common fractions should be in lowest terms before they are reduced to decimals. NOTE 3. Because they recur so frequently, learners should memorize the decimal equivalents of halves, fourths, and eighths. \ = .5; i = .25; | = .75; | = .125; f = .375; f = .625; J = .875. EXAMPLES FOR PRACTISE Reduce the following to decimals: 1. i 4. i 7. & 10. ft 13. ft 16. 18 T 3 * 2. f 5. 1 8. ft 11. H 14. AV 17. 465ft 3. f 6. | 9. I 12. | 16. 18. 671H NOTE 4. Inconvenient long divisions may frequently be avoided by separating the common fraction into two factors, the decimal equivalent of one of which is already known [Note 3]. Thus, if = | of V; and ^ = 2f or 2.375; therefore \- of V must equal 1 of 2.375. Reduce the following to decimals by factoring : 19. M 22. H 25. f J 28. 63H 20. H 23. ft 26. ft 29. 37ft 21. {$ 24. 27. 58A 30. 78ft NOTE 5. It is usually sufficient for the requirements of business to extend a decimal result to a certain number of decimal places, ordinarily not more than three or four; that is, as many ciphers are annexed to the numerator as there are decimal places required in the result; and when the last quotient figure has been obtained, to increase it by 1 if the final remainder is at least one-half the divisor. Reduce the following to decimals of the number of places indicated : 2 dec. pi. 3 dec. pi. 4 dec. pi. 5 dec. pi. 6 dec. pi. 31. 168A 34. 911J 37. 231* 40. 8.26 fr 43. 1.2764 A 32. 34T** 35. 26ft 38. 71*1 41. 7.348| 44. 3.296? 33. 72| 36. 49A 39. 3715 42. 9.527f 45. 2.83451 84 DECIMAL FRACTIONS REDUCTION OF DECIMALS TO COMMON FRACTIONS ILLUSTRATIVE EXAMPLES 124. Reduce .57^ and .1875 to common fractions. SECOND SOLUTION FIRST SOLUTION 1875 -^- 625 = 3 57| X 7 = 400 = 4 ' 1875 = 15555 - 625 = 16 100 X 7 = 700 7 Or, 1875= 11X8 = 15-5 = ^ 10 X 8 = 80 -5- 5 = 16 FIRST EXPLANATION. Change the complex decimal to the form of a complex common fraction by expressing its decimal denominator 100 (f^). Then multiply both terms of the obtained common fraction by the denom- inator of the complex numerator (7) to obtain an equivalent simple common fraction f$ [Prin. 4, 81]. Finally, reduce to lowest terms [84], obtaining f. SECOND EXPLANATION. Change the decimal to the form of a common fraction (rVirw). Then reduce to lowest terms [84]. Or, applying Note 3, 123, it is seen that the last three figures of the given decimal (875) equal . Hence, .1875 equal j|. Then proceed as in the first solution. NOTE. Business men do not usually take the unnecessary trouble to find the greatest common divisor as was done in the second solution; but as 5 and 2 are the only prime factors of a decimal denominator, they successively divide by 5 until that factor has been eliminated and then similarly by 2; or, to avoid many petty divisions by 2, they divide by 4 or by 8, if seen to be com- mon to both terms. RULE. Omit the decimal point and prefixed ciphers from the given decimal ; express the remaining figures in the form of a com- mon fraction by writing the denominator understood ; and reduce to lowest terms as in 84. EXAMPLES FOR PRACTISE Reduce the following to common fractions: 1. .75 6. .6875 11. .59375 . 16. 2.00875 2. .375 7. .9375 12. .00625 17. 19.66| 3. .005 8. .0025 13. 5.0375 18. 6.333J 4. .125 9. .15625 14. 17.0075 19. 9.714f 5. .0875 10. .34375 15. 8.28125 20. 16.272A DECIMAL FRACTIONS 85 APPROXIMATION OF DECIMALS TO COMMON FRACTIONS ILLUSTRATIVE EXAMPLE 125. Approximate .75173 to a common fraction with a denom- inator not greater than eighths. SOLUTION EXPLANATION. The unit of the given decimal, ex- 75173 pressed in eighths equals 8 eighths, therefore .75173 of the unit of the fraction must equal .75173 of (times) 8 eighths, or 6.01384 eighths. Hence, .75173 = f, 6.01384 = f = f approximately, or in lowest terms, RULE. Multiply the given decimal by the required denominator. Point off as many places in the product as are found in the given decimal. If the decimal part of the product equals one-half or more, increase the integral part of the product by 1. The integral part of the product will be the required numerator, under which write the given denominator. EXAMPLES FOR PRACTISE Approximate to common fractions of the required denominator : 1. .6879 to 16ths 4. .6247 to 16ths 7. .5168 to 32nds 2. .4912 to 32nds 5. .7987 to 5ths 8. .6672 to 3ds 3. .7489 to 8ths 6. .2531 to 4ths 9. .8334 to 6ths ADDITION OF DECIMALS ILLUSTRATIVE EXAMPLES 126. 1. Add .375, 2.75, .0058, 15.5 and .0625. SOLUTION EXPLANATION. Only like orders can be added [Prin. 1, 18]. Therefore arrange the decimals to be added so that their .o/O points shall fall in an upright column, thus causing orders 2.75 which are at the same distance from the decimal point and .0058 which must consequently be like [115], to fall in the same 15.5 column. Add the columns in the usual manner. Place the Ofi25 decimal point in the sum directly under the points of the num- bers added, for the sum of any column must be of the same 18.6933 order as that of the column added [Prin. 2, 18], and must there- fore be located at the same distance from the decimal point as the column added, to be properly indicated. 86 DECIMAL FRACTIONS 2. Add 19.29f, .5f, 3.06if, and 25/5, to 4 decimal places. SOLUTION iqoQoyr EXPLANATION. First expand the complex decimals [117] and mixed numbers [78] to five decimal places [Note .OC )O7 - 5^ 123]. The expression to one decimal place more than 3.06813 the required number is intended to insure accuracy in the 25.09375 four places required in the answer by obtaining the neces- ,10 nooon sar y carrying figure (2) from the fifth decimal column. "iO.U^ZOU m, jj j , rf ., ,. Ihen arrange, add, and point off, as in the preceding 48.0223, Ans. example. NOTE. If the rejected figure at the right of any answer is 5 or more, the last retained figure should be increased by 1. That is, if, in the 111. Ex., 48.02237 had been the obtained total, 48.0224 would have been a closer approximation to this exact result than 48.0223. EXAMPLES FOR PRACTISE Find the sum of 1. 37.41, 5.8967, .3562, 863.7, and 16.25. 2. 2.183, .713, .005, 43.81, 2.1875, and 32.025. 3. 682.5, 6.073, .058, 25.305, 97.1834, and 5.07168. 4. 29.38, 1.35867, 26.038, 534.7, 6.28734, and 59.0625. 6. .0635, .0072, .9, .85, 3.15825, .075, and .6. 6. 32.05i, 7.08A, 563f|, 16.07f, and 6.9 J, as a simple decimal. 7. .08U, .72f, .84i, 46 T V, 28.3 J, and A, as a simple decimal. 8. .62J, J, 42 J, .07 T 3 ir, 48 V, and .25 T f y, as a simple decimal. 9. 52.6H, 7.82A, .5|, .17f, and 61.8-f, to 3 decimal places. 10. 7t, -5J, .68A, 2.91 A-, -781, and J}, to 4 decimal places. 11. .516f, 6.8f , 12.343 T V, .6f, and f , to 5 decimal places. 12. Sixteen, and one hundred sixty-four ten thousandths; two hundred twenty-eight, and three thousand five hundred seventy-two millionths; forty-seven, and twenty-one ten-thou- sandths; eight hundred thirty-seven, and fifteen thousand one hundred ninety-three millionths. 13. Four hundred nineteen, and six thousand one hundred fifty-six hundred-thousandths; twenty-three, and eighteen thou- sand nine hundred seventy-four ten-millionths ; eight, and forty- two thousand eight hundred fifty-one millionths; one hundred thirty-five thousand six hundred eighty-seven, and sixteen hundred-thousandths. DECIMAL FRACTIONS 87 14. Twenty-three millionths; seven, and one thousand four hundred thirty-two hundred-thousandths; fifty-nine, and thirteen thousand seven hundred twenty-eight ten-millionths; eight thou- sand three hundred fifty-eight ten-thousandths; seventy-five, and eighty-six thousandths; nine thousand two hundred, and seven thousand three hundred nineteen millionths. SUBTRACTION OF DECIMALS ILLUSTRATIVE EXAMPLE 127. Subtract 68.25 3 5 2 from 183.6|, to 3 decimal places. SOLUTION EXPLANATION. First expand each decimal to four 1 83 fi8 c n -4- decimal places, or to one more place than is required, as explained in 111. Ex. 2, 126; and so arrange them that the decimal point of the subtrahend shall fall directly 115.4317 underneath that of the minuend, thus causing similar orders to fall in the same column [Prin. 1, 22]. Subtract 115.432 , Ans. in the usual manner. Place the decimal point in the result directly under those of the minuend and sub- trahend, because the difference between like orders must also be like [Prin. 2, 22], and must therefore be located at the same distance from the decimal point [115] to indicate this likeness. EXAMPLES FOR PRACTISE Find the difference between 1. 75.106 and 28.32 6. 43.007 and 18.26 2. 837.25 and 256.3845 7. 528.1586 and 271.03 3. 72.68 and 34.2478 8. 68 and .3129 4. 39.658 and 4.8 9. 42.175 and 28 6. 76.23 and .847 10. .75 and .21867 11. 53. 28 f and 27.2 A, expanded to simple decimals 12. 268.4J and 125.92|, expanded to simple decimals 13. 9.68| and 2.934|, expanded to simple decimals 14. 67.2f and 39.16|, to 3 decimal places 15. 95.13f and 27.9 F \, to 4 decimal places 16. 18J and 12J, to 2 decimal places. 17. Subtract three hundred twenty-eight, and fifty-six ten- thousandths from five hundred sixty. 18. Subtract two thousand five hundred seventy-three hun- dred-thousandths from eight-tenths. 88 DECIMAL FRACTIONS MULTIPLICATION OF DECIMALS ILLUSTRATIVE EXAMPLE 128. Multiply 9.865 by 2.45. SOLUTION EXPLANATION. If the multiplicand (9.865) be expressed g ogc in the form of an equivalent improper common fraction ' (ftM); and if the multiplier (2.45) be similarly expressed ^4o (fof); and their product obtained as common fractions by 49325 97, the numerator of the product would be the product of 39460 these numerators (9865 X 245, or 2416925), or the product 1Q730 ^ t ^ ie severa> l 01> ders of the decimal multiplicand and decimal multiplier (9.865 X 2.45 = 2416925); and the denominator 24.16925 of the product would be the product of these denominators (1000 X 100 = 100000), that is, 1 followed by as many ciphers as are found in both factors, or five ciphers [34] ; and as each cipher in the denominator requires a decimal place in the numerator to express it [Prin. 1, 120], cut off five decimal places from the right of the product (24.16925) to indicate that the denominator is 1 followed by 5 ciphers, or 100000. RULE. So place one factor under the other that the right-hand order of the lower factor shall fall under the right-hand order of the upper factor. Multiply as in integral multiplication. Point off from the right of the product as many decimal places as are found in both factors. NOTE 1. In arranging the factors for multiplication, their decimal points need not fall in an upright column, as in addition and subtraction of decimals. NOTE 2. If any obtained product has fewer figures than are required to be pointed off, prefix ciphers to supply the deficiency. NOTE 3. To multiply a decimal by 10, 100, 1000, etc., move its decimal point as many places to the right as there are ciphers in the multiplier, annex- ing ciphers to the decimal if necessary. NOTE 4. Complex decimals, if terminate [Note, 117], should be expanded to simple decimals before multiplication; and if interminate, should be mul- tiplied as shown in 98 or 99. EXAMPLES FOR PRACTISE Multiply Multiply Multiply 1. 516.78 by 43 4. 2.086 by .034 7. .685 by .0058 2. 79.832 by 7.4 5. 56.8 by .0096 8. 7.6f by 5.2 3. 56.283 by .59 6. 8.27 by 1.038 9. 6.28 A by .76 DECIMAL FRACTIONS 89 Multiply Multiply Multiply 10. 75.1ft- by .2f 14. 2.41/^by 1.6U 18. 2.34J by .16? 11. 6.8by3.6 15. 2.89| by 3.5 19. .7183 by 100 [Note 3] 12. 2.96| by 5.12i 16. .684| by 2.63 20. .62 by 1000 13. 6.73fby.913 17. .715J by 3.8| [99] 21. .7183 by 10 DIVISION OF DECIMALS ILLUSTRATIVE EXAMPLE 129. Divide 10.02677 by 2.873. SOLUTION EXPLANATION. If the dividend be expressed o 07QMA rOA77^Q AC\ m ^ e ^ orm * an e( l u i va l ent improper common fraction WW), and the divisor be similarly expressed (|ff&$), the quotient obtained as in 1 4077 common fractions [103] would be VoV&k 7 - X iff! , 1 1492 or 2 1 873z 6 i 7 o 7 o; that is, the quotient of the dividend regarded as an integer (1002677) divided 25857 by the divisor regarded as an integer (2873), 25857 obtaining 349, and this result (349) divided by 100 [38], or by 1 followed by as many ciphers as the number of decimal places in the dividend (five) exceed the number of decimal places in the divisor (three), obtaining 3.49. DEDUCTIONS. 1. If the dividend is the product of the divisor and quo- tient (Note 2, 35), then the dividend (10.02677) must contain as many deci- mal places as the divisor (2.873) and the quotient (3.49) together [Note, 128]. Therefore, if the divisor contains only three of the five decimal places found in the dividend, the quotient must contain the remaining two decimal places. Hence, subtract the number of decimal places in the divisor from the number of decimal places in the dividend to find the number of decimal places to point off in the quotient. 2. If the dividend contains as many decimal places as both the div'sor and the quotient, it must contain at least as many as the divisor. Hence, if, before division, any dividend is seen to contain fewer decimal places than its divisor, annex as many ciphers to such a dividend as will supply the deficiency. RULE. (1) See that the dividend contains at least as many decimal places as the divisor. (2) Divide as with whole numbers. (3) From the right of the quotient point off as many decimal places as the number of decimal places in the dividend exceeds the number of decimal places in the divisor. NOTE 1. If the quotient does not contain as many figures as it must have decimal places, prefix ciphers to supply the deficiency. NOTE 2. To divide a decimal by 10, 100, 1000, etc., move its decimal 90 DECIMAL FRACTIONS point as many places to the left of its former position, as there are ciphers in the divisor, prefixing ciphers to the dividend if necessary. NOTE 3. Complex decimals, if terminate [Note 117], should be expanded to simple decimals before commencing the division; and if interminate, should be divided as shown in 105 and 106. NOTE 4. To obtain an approximate quotient of a given number of deci- mal places, use as many figures after the decimal point of the dividend as will equal the number of decimal places in the divisor plus the number of decimal places required in the quotient, omitting the remaining decimal places of the dividend if it has an excess, or annexing decimal ciphers to the dividend if it has a deficiency, of decimal places. EXAMPLES FOR PRACTISE Divide Divide 1. 16.341 by 3 9. 95 by .0005 2. 138.1716 by 49.347 10. .0002512 by .628 3. 32.50676 by 56.83 11. 9.72 by .00036 4. 2175.4754 by 842 12. .001 by .000008 5. 37 by .0008 13. 1 53.8671 f by 37.8 J [Note 3] 6. .0366234 by .537 14. .5224 by .006| 7. .00003 by .005 15. 16.47H by 3.7J 8. .000219936 by 68.73 16. 11.7196| by .016| Applying Note 4, divide 17. 19.3 by 17 (to 4 dec. places) 18. 43.5782 by 3.4 (to 2 dec. places) 19. .268345 by 5.3 (to 3 dec. places) 20. 5.93472 by .26 (to 2 dec. places) 21. 617.35 by 48.3 (to 3 dec. places) 22. .18 T 5 s- by .348 (to 3 dec. places) 23. 394.8f by .417 (to 2 dec. places) 24. 58.52f by .78 (to 3 dec. places) 25. 16.7| by .18f (to 4 dec. places) 26. 7.5J by 13| (to 2 dec. places) Applying Note 2, divide 27. 283.45 by 100 31. .6| by 10 (to 3 dec. pi.) 28. 4167.8 by 10 32. 1.7| by 100 (to 6 dec. pi.) 29. 52.384 by 1000 33. 38.74 by 400 (= 100 X 4) 30. 3.4| by 100 34. 763.4 by 7000 (= 1000 X 7) DECIMAL FRACTIONS 91 WHEN THE DIVISOR IS A SIMPLE DECIMAL OF MANY PLACES 130. It is frequently necessary in commercial calculations to divide by a simple decimal of many places. This will require much unnecessary labor if the methods of 129 are employed. The following contraction is absolutely correct, and will often save more than half the labor required by the ordinary processes. ILLUSTRATIVE EXAMPLE Divide $5.86 by .621783 to 2 decimal places (cents). SOLUTION EXPLANATION. First see that the divi- 9 -U dend contains no grater or less number than four decimal places, cutting out all beyond the fourth place if it has more, or 2640 annexing ciphers to supply the deficiency if it 2487 Then divide by as many figures of the divisor (.621783) as this dividend of four 124 decimal places (5.8600) will contain, that is, 29 by .6217$$, canceling the remaining figures of the divisor as unnecessary to secure the cor- rect quotient. Dividing this adapted dividend, it is found that 58600 ten-thousandths contain 6217 ten-thousandths 9 (integral) times with a remainder of 2640 ten-thousandths. Then dropping one figure (7) from the right of the preceding divisor (62 1/), it is seen that the remainder from the preceding division (2640 ten- thousandths) contains 621 thousandths, 4 (tenths) times, with a remainder of 153 ten-thousandths. Again dropping one figure (1) from the right of the preceding divisor (.62J), it is seen that the remainder from the preceding division (153 ten- thousandths) contains 62 hundredths, 2 (hundredths) times. As the quo- tient is required to two decimal places, or to hundredths, the division is now completed. RULE. 1. First see that the dividend contains exactly four decimal places, annexing ciphers if it has a deficiency, or rejecting all beyond the fourth place if it has an excess. 2. At the first partial division, use as many of the higher orders of the divisor as the modified dividend of four decimal places can contain, and reject the remaining orders. 92 REVIEW OF DECIMALS 3. At each successive partial division, reject one additional order from the right of the immediately preceding divisor. The last partial division should be by the number expressed by the first two figures after the decimal point of the divisor. 4. //, at any stage of the division, the remainder does not contain the divisor, write a cipher as the next quotient figure, reject one more order from the right of the divisor, and proceed as before. NOTE 1. When multiplying by the successive quotient figures, always include one or two figures next adjoining the proper divisor, to secure the correct carrying figure to the expressed product. NOTE 2. If the final remainder is one-half or more of the final divisor, increase the final quotient figure by 1. NOTE 3. The above method of division will also be found useful when a divisor is the integer 1 followed by many decimal places. In this case, see that the dividend contains exactly three decimal places, instead of the four places required when the divisor is a simple decimal. The final divisor should be the two highest orders as before, but these two orders (regarded as one quantity) will now express tenths. EXAMPLES FOR PRACTISE Carrying the quotients to 2 decimal places, divide 1. $9.16 by .48736291 6. $3.167894 by .7264583 2. $12.50 by .9376284 7. $7.25 by 1.3427862 [Note 3] 3. $35.72 by .8675913 8. $45.6 by 1.2783245 4. $138.65 by .9163748 9. $25 by 1.6824357 6. $400 by .28683297 10. $18.67 by 1.2354987 REVIEW OF DECIMALS 131. The relation of mixed decimals [118] to each other the same as that of integers [48-53]; and the relation of simple decimal quantities, the same as that of proper fractions [107- 109]. 1. What is the difference between .087 times 72.5634 and the sum of 8.73, 16.035, 12.4875, and 15.3? 2. The average yield per acre of a field of wheat was 32.1875 bushels. What was the total yield if the field contained 75.0625 acres? 3. What is the sum of 15.6fV, 38.24^, 39|, 265, f , and 2.34J, to four decimal places? REVIEW OF DECIMALS 93 4. How many 32nds are in .34378? [125] 5. What is the cost of 37 barrels of vinegar, averaging 39.0625 gallons per barrel at 16 f cents per gallon? 6. The product of two decimals is 11392.5625 and one of its factors is 156.0625. What is its other factor? 7. What decimal part of a yard of cloth can be bought for $.60, at $.80 per yard? 8. How much corn in .75 of 850 bushels? 9. A and B own a tract of land, A owning .625 of the tract. What decimal part of the tract does B own? 10. A merchant sold .375 of his stock of flour. How many barrels of flour did he originally have in stock if the quantity sold was 162 barrels? 11. After paying .1875 of his debts, a merchant found that he still owes $7800. What was his original indebtedness? 12. A man had 650 yards of silk, and at various times sold 481 yards. What decimal part of the silk did he sell? 13. A merchant had a certain sum deposited in a bank. If he withdrew .35 of his deposit to buy 140 barrels of flour at $6 per barrel, how much has he still remaining in the bank? 14. A man bequeathed $28560 to his wife, son, and daughter, leaving .475 of this sum to his wife, .315 of it to his daughter, and the remainder to his son. How much did each receive? 15. A man bought a house and lot for $3500. If the lot cost .75 of what he paid for the house, how much did he pay for each? 16. A man lost .465 of his wealth in speculation, and then found that he was still worth $8774. How much was he worth before the speculation? 17. $119.48 is .45 more than how much money? * 18. $28.90 is .15 less than how much money? 19. A man has $678.60, which is .45 more than his neighbor has. How much has his neighbor? 20. Since the census of 1900, a town has lost .18 of its popu- lation. If the census of 1910 showed a population of 4428, how many inhabitants did it have at the census of 1900? 21. A and B engage in partnership, A investing $31188 and B investing $36612. What decimal part of the capital did each invest? 94 COMPOUND NUMBERS 22. A farmer sold 450 bushels of wheat to one man at 97 cents per bushel, and .36 less than that quantity to another man at 95 cents per bushel. How much did he receive from both sales? 23. A man's income is $2500 per annum and his expenses $1800 per annum. What decimal part of his income does he save? 24. A man paid $23.75 for a suit of clothes, $33.25 for an overcoat, $3.80 for a hat, and had .36 of his money remaining. How much money did he have at first? 25. A man's expenses are .78 of his income. How much does he save, if his expenses are $1404 per annum? COMPOUND NUMBERS 132. INTRODUCTION, (a) Quantities have now been consid- ered the various orders of which have been expressed upon a uniform, ascending scale of 10, commencing with the unit as the initial order, as with integers [9-13 and 17-69]. It has also been seen that this same convenient arrangement of orders can be extended conversely to fractional parts of the initial unit by express- ing them upon a uniform descending scale of 10, as with decimal fractions [112-131]. (6) In contrast with the subdivisions of the initial unit into convenient decimal fractional parts, has also been placed the more inconvenient divisions of the integral unit upon an arbi- trary changeable scale, as with common fractions [70-111], in the consideration of which increased difficulty was experienced in completing the relation of quantities thus expressed, by reason of the varying scale in which the subdivisions (denominators) of the mutually related terms were expressed. * (c) A compound number, now to be considered, is the expres- sion of an integral quantity upon the same arbitrary varying scale, as that of common fractions. So that, as common fractions are when contrasted with decimal fractions, so are compound inte- gral quantities when contrasted with simple integral quantities; the scale in passing from one denominate order to the next higher or lower order being as irregular as in passing from a common fraction of one denominator to another fraction of a higher or lower denominator, thus requiring the application of the same prin- ciples of reduction in the one as in the other. (d) The law of increase and decrease is universal in its appli- COMPOUND NUMBERS 95 cation to all quantities, in whatever form they may be expressed. The processes for completing the relation of quantities are there- fore the same for compound integers as for simple integers [48-53], and the same for common fractions [107-109] as for decimals, though the method for executing these processes may seem to dif- fer with the form in which the related terms are expressed. Any seeming difference will quickly disappear if related compound numbers or common fractions are conceived to be reduced to equivalent expressions which have a common decimal scale. 133. A denominate number is the expression of a quantity in units of weight, of measure, or of value. 134. A simple denominate number expresses a denominate quantity in units of a single denomination or name. Thus, 3 pounds (units of weight), 5 bushels (units of measure), and 6 dollars (units of value) are simple denominate numbers. 135. A compound denominate number, or, as it is usually abbreviated, a compound number, is the expression of one de- nominate quantity in two or more different denominations or subdivisional names. Thus, "5 gallons 3 quarts 1 pint" is a compound number, because it expresses one fluid quantity in units of three different denominations. MEASURES OF VALUE 136. United States money is the legal currency of the United States. Scale: 10 cents (f) = 1 dime (d.); 10 dimes or 100 cents = 1 dollar ($); 10 dollars = 1 eagle (E). For convenience, the denomination mill is frequently used to express tenths of a cent. 137. English or sterling money is the legal currency of Great Britain. Scale: 4 farthings (far.) = 1 penny (d.) ; 12 pence = 1 shilling (s); 20 shillings = 1 pound (). 138. The following Table exhibits the intrinsic value of the several units of currency of the principal nations of the world, estimated in United States money, as proclaimed by the Secretary of the United States Treasury on Jan. 1, 1912. 96 COMPOUND NUMBERS Country Legal Standard Monetary Unit Value in Terms of U. S. Gold Dollar Argentine Republic Gold Peso $ 965 Austria-Hungary . . Gold Crown . 203 Belgium Gold and silver . Franc .193 Brazil Gold Milreis 546 British America . . . Gold Dollar . 1 000 Chili Gold Peso 365 Denmark Gold Crown 268 Egypt France . . Gold Gold and silver . Pound (100 piasters) . Franc 4.943 193 German Empire. . . . Gold Mark .238 Great Britain India (British) . . Gold Gold Pound Sterling Rupee . . 4.8665 324 5 Italy Gold and silver Lira 193 Japan Gold Yen 498 Mexico . . . Gold . ...... Peso 498 Norway Gold Crown . 268 Panama Gold Balboa 1 000 Peru . . . Gold Libra . .... 48665 Philippine Islands Gold Peso . 500 Portugal Gold Milreis 1.080 - Russia Gold Ruble .515 Spain Gold and silver . Peseta 193 Sweden Gold Crown 268 Switzerland . . . Gold Franc 193 Turkey . Gold . Piaster . .044 MEASURES OF WEIGHT 139. Avoirdupois weight is the usual weight of commerce. Scale: 16 ounces (oz.) = 1 pound (lb.); 100 pounds = 1 hun- dredweight (cwt.); 20 hundredweight = 1 ton (T.). 140. Troy weight is used by jewelers in weighing gold, silver, and other precious metals. Scale: 24 grains (gr.) = 1 penny- weight (pwt.); 20 pennyweights = 1 ounce (oz.); 12 ounces = 1 pound (lb.). 141. Apothecaries' weight is used by apothecaries and physi- cians in weighing the non-liquid ingredients of prescriptions. Scale: 20 grains (gr.) = 1 scruple (3); 3 scruples = 1 dram (3); 8 drams = 1 ounce (3); 12 ounces = 1 pound (lb.). COMPOUND NUMBERS 97 142. Miscellaneous table of weights: 2240 pounds = 1 long ton; 100 pounds of grain or flour = 1 cental; 100 pounds of nails = 1 keg; 196 pounds of flour = 1 barrel; 200 pounds of salted beef, pork, or fish = 1 barrel. NOTE 1. It is customary to allow the following weights to 1 bushel: Barley 48 Ib. Ear corn 70 Ib. Potatoes 60 Ib. Beans 60 " Corn meal 50 " Rye 56 " Buckwheat .... 48 " Cotton seed ... 30 " Timothy seed ... 45 " Clover seed. . . .60 " Oats 32 " Wheat GO " Shelled corn . . .56 " Peas 60 " Wheat bran ... .20 " NOTE 2. Denominations of the same name in the Troy Scale and Apothe- caries' Scale have the same weight; but both of these differ in weight from denominations of the same name in the Avoirdupois Scale. Thus, there are 5760 Troy grains in 1 pound Troy, and 7000 Troy grains in 1 pound Avoir- dupois; and there are 480 Troy grains in a Troy ounce (5760 gr. -s- 12), but only 4371 Troy grains in 1 ounce Avoirdupois (7000 gr. -5- 16). MEASURES OF EXTENSION 143. Linear measure is used in measuring lengths, widths, heights, depths, and distances. Scale: 12 inches (in.) = 1 foot (ft.); 3 feet = 1 yard (yd.); 5J yards = 1 rod (rd.); 40 rods = 1 furlong (fur.); 8 furlongs or 320 rods = 1 mile (mi.). 144. Surveyors' linear measure is used by surveyors in measur- ing lengths of roads, boundaries of land, etc. Scale: 7.92 inches (in.) = 1 link (1.); 100 links = 1 chain (ch.); 80 chains = 1 mile (mi.). 145. Circular measure is used in measuring angles and arcs, (parts of a circle) ; and by navigators in estimating latitude and longitude. Scale: 60 seconds (") = 1 minute ('); 60 minutes = 1 degree () ; 360 degrees = 1 circle (C), or circumference of the earth. MEASURES OF SURFACE 146. Square measure is used in measuring ordinary areas, such as the surfaces of boards, of painting, plastering, papering, paving, etc., and, unofficially, of land. Scale: 144 square inches (sq. in.) = 1 square foot (sq. ft.); 9 square feet = 1 square yard COMPOUND NUMBERS (sq. yd.); 30i square yards = 1 square rod (sq. rd.); 160 square rods = 1 acre (A); 640 acres = 1 square mile (sq. mi.). 147. Surveyors' square measure is used by surveyors in measuring land. Scale: 10,000 square links (sq. 1.) = 1 square chain (sq. ch.); 10 square chains = 1 acre (A); 640 acres = 1 square mile (sq. mi.). MEASURES OF VOLUME 148. Cubic measure is used in measuring the contents or volume of bodies. Scale: 1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.); 27 cubic feet = 1 cubic yard (cu. yd.). 149. Wood measure is used in measuring cord wood. Scale: 16 cubic feet (cu. ft.) = 1 cord foot (cd. ft.); 8 cord feet or 128 cubic feet = 1 cord (cd.). MEASURES OF CAPACITY 150. Liquid measure is used in measuring ordinary liquids. Scale: 4 gills (gi.) = 1 pint (pt.); 2 pints = 1 quart (qt.); 4 quarts = 1 gallon (gal.). 151. Apothecaries' fluid measure is used by apothecaries and physicians in measuring liquid medicines. Scale: 60 minims ("I) = 1 fluiddrachm (f3); 8 fluiddrachms = 1 fluidounce (f 5); 16 fluidounces = 1 pint (0). 152. Dry measure is used in measuring fruits, vegetables, and grains. Scale: 2 pints (pt.) = 1 quart (qt.); 8 quarts = 1 peck (pk.) 4 pecks = 1 bushel (bu.). MEASURE OF TIME 153. Scale: 60 seconds (sec.) = 1 minute (min.) ; 60 minutes = 1 hour (hr.); 24 hours = 1 day (da.); 365 days = 1 common year(yr.); 366 days = Heap year; 100 years = 1 century (cent.). NOTE 1. Leap years and common years are identified as follows: 1. Every year which is not exactly divisible by 4, is a common year. 2. Every year which is exactly divisible by 4, w a leap year. 3. Centennial years (years ending in two ciphers) when divisible by 400, are leap years; and when not divisible by 400, are common years. COMPOUND NUMBERS 99 NOTE 2. The order, name, and length in days of each month are: Order Name First January Second February Third March . . Fourth April . . . Fifth May Sixth . . . June . . Length . 31 days 28 or 29 da, .31 days .30 " .31 " .30 " Order Name Length Seventh July 31 daya Eighth August 31 " Ninth September. . . 30 " Tenth October 31 " Eleventh November ... 30 " Twelfth .... December ... 31 " OTHER MEASURES 154. In enumerating certain articles of merchandise, the fol- lowing Scale is used: 12 things = 1 dozen (doz.); 12 dozen = 1 gross (gro.); 12 gross = 1 great gross (g. gro.); 6 things = 1 set; 20 things = 1 score. 155. In measuring paper, the following Scale is used: 24 sheets (sh.) = 1 quire (qu.); 20 quires = 1 ream (rm.); 2 reams = 1 bundle (bdl.); 5 bundles = 1 bale (bl.). REDUCTION OF COMPOUND DENOMINATE NUMBERS TO SIMPLE DENOMINATE NUMBERS ILLUSTRATIVE EXAMPLE 156. SOLUTION 15 da. 17 hr. 28 min. X 24 (hrs. in 1 da.) 377 (15 X 24 + 17) X60 (min. in 1 hr.) Reduce 15 da. 17 hr. 28 min. to seconds. EXPLANATION. Commence with the highest denomination (15 da.) and reduce it to the next lower denomination (hr.) as fol- lows: If 1 day = 24 hr., 15 days will equal 15 times 24 hr., or 360 hr., and 15 days 17 hr. will equal 360 hr. + 17 hr., or 377 hr. Next reduce 377 hr. to the next lower denomination of the scale (min.) as follows: If 1 hr. = 60 min., 377 hr. will equal 377 times 60 min., or 22620 min., and 377 hr. 28 min. will equal 22620 min. + 28 min., or 22648 min. Lastly, reduce 22648 min. to the next lower denomination (sec.) as follows: If 1 min. = 60 sec., 22648 min. will equal 22648 times 60 sec., or 1358880 sec., the required answer. RULE. Multiply the highest given denomination by the number of units of the next given denomination which equal one unit of that 22648 (377 X 60 + 28) X 60 (sec. in 1 min.) 1358880 seconds. 100 COMPOUND NUMBERS highest denomination, and to the result add the given units of that next given denomination. Reduce the obtained results in a similar manner to the next lower given denomination, and continue this process until all the given denominations have been considered and the required lower denomi- nation has been obtained. NOTE. It should be remembered that a multiplicand and multiplier are not so called because they have a certain position with respect to each other, but because they possess a certain characteristic relation to each other [Note 1, 26]. Thus, in the 111. Ex., 24, 60, and 60 are the successive multiplicands though placed for convenience in positions usually occupied by multipliers. EXAMPLES FOR PRACTISE Reduce 1. 39 18s. 7d. to pence 2. 73 T. 16 cwt. to pounds 3. 16 bu. 3 pk. 5 qt. to pints 4. $19 and 8 i to mills 5. 3 yr. 216 da. 12 min. to seconds 6. 18 lb. 11 pwt. to grains 7. 196 16s. 2d. to pence 8. 28 rd. 2 yd. 1 ft. to inches 9. 25 cd. 3 cd. ft. to cubic feet 10. 4 A. 25 sq. rd. 16 sq. yd. to square feet 11. 27 mi. 18 rd. 2 yd. to inches 12. 5 T. 12 cwt. 16 Ib. to ounces 13. 3 wk. 4 da. 17 hr. to seconds 14. 35 bu. 2 pk. 3 qt. to quarts 15. 65 gal. 3 qt. 1 pt. to gills 16. 73 A. 75 sq. rd. 12 sq. yd. to square inches 17. 25 cu. yd. 13 cu. ft. to cubic inches 18. How much was received for 3 bu. 2 pk. chestnuts if sold at 6 cents per quart? 19. What is the cost of 1 Ib. 7 oz. 13 pwt. of jewelers' gold at $1.05 per pennyweight? 20. How much should be received for 37 gallons of cider if retailed at 5 cents per pint? COMPOUND NUMBERS 101 REDUCTION OF SIMPLE DENOMINATE NUMBERS TO COMPOUND DENOMINATE NUMBERS ILLUSTRATIVE EXAMPLE SOLUTION 4)3470 gi. 2)857 _j_ 2 gi 4)433 + 1 pt. 108 gal. + 1 qt. 157. Reduce 3470 gills to the higher denominations. EXPLANATION. 4 gills of the given denomination (3470 gills) equal 1 pint, the next higher denomina- tion; and 3470 gi. contain 4 gi. 867 times and 2 gi. remaining. Therefore 3470 gi. = 867 pt. and 2 gi. 2 pt. of the obtained denomination (867 pt.) equa i i qt t> t h e ne xt higher denomination; and 867 Pt- contain 2 pt. 433 times and 1 pt. remaining. Therefore, 867 pt. = 433 qt. and 1 pt. 4 qt. of the last obtained denomination (433 qt.) equal 1 gal., the next higher denomination; and 433 qt. contain 4 qt. 108 times and 1 qt. remain- ing. Therefore 433 qt. = 108 gal. and 1 qt.; and 3470 gi. = 108 gal. 1 qt. 1 pt. 2 gi. RULE. Reduce the given denomination to the next higher denom- ination. Continue this process in regular order from one obtained denomination to the next higher until the highest required denom- ination is reached. NOTE, (a) Each successive divisor should be of the same denomination as its dividend; (6) each successive remainder is of the same denomination as the dividend from which it was obtained, and (c) each successive quotient is of that denomination in the scale which equals the divisor that produced it. EXAMPLES FOR PRACTISE Reduce 1. 6173 pt. to bushels 2. 3567282 sec. to days 3. 273465 in. to miles 4. 827356 sq. ft. to acres 5. 23468 far. to 's 6. 853764 oz., Avoir., to tons 7. 5295 sheets to reams 8. 7835 cu. ft. to cords 9. 2346 pwt. to Ibs, Troy 10. 3496 cd. ft. to cords Reduce 11. 7683d. to 's 12. 53867 mills to dollars 13. 6287 gi. to gallons 14. 2863452 sec. to days 15. 627853 in. to miles. 16. 7456 qt. to bushels 17. 876345 ft. to miles 18. 917345 cu. in. to cu. 19. 17832d. to 's 20. 1378 gi. to gallons. yds. 102 COMPOUND NUMBERS REDUCTION OF DENOMINATE COMMON FRACTIONS TO INTEGERS ILLUSTRATIVE EXAMPLE 158. Reduce f of a year to integers of the lower denominations. SOLUTIONS EXPLANATION. $ of a year are equivalent to $ of (times) 365 4 of 365 da. = 202 da. (and | da.) da. or 202 da. Reserving the integral part of this result (202 | of 24 hr. = 18 hr. (and f = f hr.) da>) as the first term of the re- .~ . ,, quired answer, similarly reduce I of 60 mm. = 40 mm. Hence, the fractional part (| da } to the I yr. = 202 da. 18 hr. 40 min. next lower denomination 0*-> * follows: | of a day = 1 of (times) 24 hr., or 18 f hr. Reserving the integral part of the result (18 hr.) as the second term of the required answer, re- duce the fractional part (f hr.) to the next lower denomination (min.), as follows: f of an hour equals f of (times) 60 min., or 40 min. Therefore f of a year equals 202 da. 18 hr. 40 min. EXAMPLES FOR PRACTISE Reduce to integers of the lower denominations. 1. f 3. A gal. 5. | yr. 7. % yd. 9. 2. | yr. 4. T V bu. 6. | mi. 8. A T. 10. H yd. REDUCTION OF DENOMINATE DECIMALS TO INTEGERS ILLUSTRATIVE EXAMPLE 159. Reduce .375 to integers of the lower denominations. SOLUTION p 075 EXPLANATION. .375 of a = .375 of (times) 20s., or 7.5s. Reserving the integral part of this result (7s.) as the first term of the required answer, reduce the fractional part S. 7.500 (.5s) to the next lower denomination (d.) as follows: 12 .5s = .5 of (times) 12d., or 6d. Therefore .375 = 7s.6d. d.1U) .375 = 7s. 6d. COMPOUND NUMBERS 103 EXAMPLES FOR PRACTISE Reduce to integers of the lower denominations. 1. .1875 gal. 4. .84375 gal. 7. .34375 mi. 2. .0375 yr. 5. .876 T. 8. .52375 3. .546875 bu. 6. .375 lb., Troy 9. .921875 bu. REDUCTION OF COMPOUND INTEGRAL NUMBERS TO FRACTIONS OF ANY REQUIRED HIGHER DENOMINATION ILLUSTRATIVE EXAMPLES 160. 1. Reduce 2 pk. 3 qt. to a common fraction of a bushel. SOLUTION EXPLANATION. 1 pk. = 8 qt., therefore 2 pk. 3 qt. 2 pk. 3 qt. = (8 qt. X 2) + 3, or 19 qt. = 2 times 8 qt. (16 qt.) + 3 qt., or 19 qt. 1 bu. = 4 pk., or 4 times 8 qt. = 32 qt. 1 bu. = 4 pk., and 4 pk. o u o i < r u u i =4 times 8 qt., or 32 qt. Hence, 2 bu. 3 qt. = if of a bushel If 32 ^ J l ^ *, q{ must equal A of a bu., and 2 pk. 3 qt., or 19 qt., must equal 19 times ?? of a bu., or If of a bu. 2. Reduce 17s. 6d. to a decimal of a sterling. SOLUTION EXPLANATION. 12d. = Is.; therefore 6d. must be such a 12)6 decimal part of a shilling as 6d. are of 12d., that is, j\ or .5 of a shilling [123]; and 17s. 6d. must equal 17.5 shillings. 20s. = 1; and 17.5s. must be such a decimal part of a 20) 17.5 ^ 17 - 5s - are of 20s ., that is ~^T or -875 of a . .875 Or, 17s. 6d. may be reduced to a common fraction of a as in the preceding solution, obtaining , and this result re- duced to a decimal by 123, obtaining .875. EXAMPLES FOR PRACTISE What fraction of 1. A are 2s. 6d. 2. A dollar are 12 5 mills. 3. A gallon are 2 qt. 1 pt. 4. A bushel are 1 pk. 4 qt. 5. A day are 4 hr. 48 min. 104 COMPOUND NUMBERS What decimal of 6. A gallon are 3 qt. 1 pt. 7. A are 12s. 6d. 8. A yard are 1 ft. 6 in. 9. A bushel are 1 pk. 7 qt. 1 pt. (to 4 places). 10. A day are 5 hr. 19 min. (to 3 places). ADDITION OF COMPOUND NUMBERS ILLUSTRATIVE EXAMPLE 161. Add 3 gal. 2 qt. 1 pt.; 2 gal. 1 pt.; and 3 qt. 1 pt. SOLUTION EXPLANATION. So arrange the given compound num- bers that their denominations which are similar shall fall gal. qt. pt. in the same column [Prin. 1, 18]. Then add the column 321 of the lowest given denomination (pt.), obtaining 3 pt., or 2 Q l 1 qt. 1 pt. Write 1 pt. under the column of pints, and carry 1 qt. to the column of quarts and add, obtaining 6 qt., or 1 gal. 2 qt. Write 2 qt. under the column of quarts, 6 2 an d carry 1 gal. to the column of gallons and add, obtaining 6 gal. EXAMPLES FOR PRACTISE 123 da. hr. min. sec. T. cwt. Ib. oz. s. d. far. 5 13 25 19 45 18 29 9 15 16 8 3 17 21 18 27 37 3 75 3 7 10 6 1 12 15 35 40 15 12 34 11 24 3 10 2 8 7 43 52 7 6 17 6 18 12 93 4. AddSbu. 7qt.; 3 bu. 2pk.; 7 bu. 1 pt.; and 3 pk. 5 qt. 5. Add 35 7s. 6d.; 53 8d.; 65 2 far.; 45 15s. 6. Add 16 da. 18 hr. 5 sec. ; 7 da. 3 hr. 48 min. ; 19 da. 7 hr. 50 sec. ; 12 da. 2 hr. 34 sec. ; and 2 da. 23 hr. 7. Add 19 cu. yd. 16 cu. ft. 1217 cu. in. ; 35 cu. yd. 24 cu. ft. 197 cu. in.; 2 cu. yd. 15 cu. ft. and 718 cu. in.; 13 cu. yd. 16 cu. ft. ; and 25 cu. yd. 11 cu. ft. 1235 cu. in. 8. Add ^ da.; A da.; H da.; and $ da.; reduced to com- pound integers. 9. Add .875 cd. ft.; .25 cd.; .125 cd.; reduced to compound integers. COMPOUND NUMBERS 105 SUBTRACTION OF COMPOUND NUMBERS ILLUSTRATIVE EXAMPLE 162. Subtract 15 12s. 3 far. from 48 7s. 6d. SOLUTION EXPLANATION. So place the several denom- inations of the subtrahend that they shall fall s. d. far. under sjmilar denominations of the minuend; and 48 7 6 commence the subtraction with the lowest de- 15 12 3 nomination. As 3 far. cannot be subtracted from far. 32 15s. 5d. 1 far. borrow Id. (=4 far.) from the next higher de- nomination of the minuend (6d.), and then sub- tract 3 far. from the borrowed 4 far., obtaining 1 far. as the first partial remainder. Next subtract the next higher denomination of the subtrahend (Od.) from the similar denomination of the minuend (6d. less the borrowed Id., or 5d.), obtaining 5d. as the second partial remainder. As the next higher denomination of the subtrahend (12s.) cannot be sub- tracted from the similar denomination in the minuend (7s.), borrow 1 (= 20s.) from the 48 of the minuend, and then subtract 12s. from 7s. in- creased by the borrowed 20s., or 27s., obtaining 15s. as the third partial remainder. Lastly, subtract the next higher denomination of the subtrahend (15) from the similar denomination of the minuend (48 less the borrowed 1, or 47), obtaining 32 as the final partial remainder; thus making the com- plete remainder 32 15s. 5d. 1 far. EXAMPLES FOR PRACTISE 1. Subtract 5 bu. 1 pk. 1 pt. from 12 bu. 3 qt. 2. Subtract 19 da. 18 hr. 45 min. from 45 da. 12 hr. 3. Subtract 25 cd. 95 cu. ft. from 42 cd. 37 cu. ft. 4. Subtract 167 gal. 3 qt. 2 gi. from 258 gal. 2 qt. 1 pt. 5. Subtract 235 17s. 6d. from 452 12s. 9d. 6. Subtract 15 A. 90 sq. rd. 18 sq. yd. and 5 sq. ft. from 23 A. 73 sq. rd. 24 sq. ft. 7. A grocer sold 27 gal. 3 qt. 1 pt. from a barrel of molasses containing 39 gal. 1 qt. How much molasses remained in the barrel? 8. A merchant in London owed 285 2s 6d and subse- quently made a payment of 97 12s 9d. How much of the debt remained unpaid? 106 COMPOUND NUMBERS COMPOUND SUBTRACTION OF DATES ILLUSTRATIVE EXAMPLE 163. Find the interval between Oct. 25, 1854, and July 16, 1911. SOLUTION EXPLANATION. Take the later date (July 16, 1911) as the minuend because it expresses yr. mo. da. a greater period of time, denoting the 1911th 1911 7 16 year of the Christian era, the 7th month [Note 1854 10 25 2 153 ^ and the 16th dav ' underneath which ^ft ft 9? r\ wr * te tne earlier date which expresses a less 00 yr. 8 mo. 21 da. per iod of time, denoting the 1854th year of the Christian era, the 10th month, and the 25th day. Subtract as shown in 111. Ex. 162. NOTE 1. In obtaining the interval between two dates by compound subtraction, it is customary to consider any borrowed month as equal to 30 days. NOTE 2. If both the subtrahend and minuend express the last days of their respective months, and the subtrahend month is the longer, as in Ex. 7 and 9, following, the difference between them will be 0; but if the subtrahend month is the shorter as in Ex. 10, following, the difference is found in the usual manner. The courts have decided when a male child is born on Feb. 29th of a leap year, that he has reached his majority on Feb. 28th of the 21st year thereafter; and when a note falls due on the 29th, 30th, or 31st of a month which contains less than that number of days, that it is payable on the last day of such a month. Thus, 3 months after Aug. 31 = Nov. 30, and con- versely from Aug. 31 to Nov. 30 must equal 3 months. EXAMPLES FOR PRACTISE By compound subtraction find the interval from 1. Mch. 16, 1892 to Dec. 3, 1911. 2. Oct. 28, 1886 to May 4, 1912. 3. July 2, 1875 to Feb. 12, 1912. 4. Dec. 18, 1895 to Aug. 9, 1911. 6. Nov. 30, 1845 to Feb. 8, 1910. 6. Sept. 19, 1902 to July 31, 1910. 7. Aug. 31, 1897 to June 30, 1911. [Note 2] 8. May 31, 1875 to Jan. 31, 1909. 9. Feb. 29, 1892 to Feb. 28, 1910. 10. Feb. 28, 1887 to Feb. 29, 1912. * COMPOUND NUMBERS 107 TO FIND THE EXACT NUMBER OF DAYS BETWEEN TWO DATES ILLUSTRATIVE EXAMPLE 164. How many days after Oct. 13, 1911, is Apr. 24, 1912? SOLUTION j t r\ A u EXPLANATION. Subtract the initial 18 da., remainder of October date (Qct 13) from ^ full number of 30 da., in November days m that month ( 31)) obtaining 18 31 da., in December days as the included remainder of Oct., 31 da., in January under which place in regular order the 29 da in February number of days in each of the succeeding 1 , A/TV, ^ U ^ mon ths of the interval, and the in- da., in Marcn duded dayg of the final month (24) . 24 da., included in April g^ M} obtaining 194 days. 194 da., required interval EXAMPLES FOR PRACTISE Find the exact number of days from 1. Jan. 16, 1911 to Nov. 12, 1911. 2. July 23, 1912 to Mch. 8, 1913. 3. Sept. 3, 1909 to May 31, 1910. 4. Feb. 16, 1910 to Oct. 6, 1910. 5. Dec. 31, 1911 to Aug. 12, 1912. 6. Nov. 16, 1910 to Feb. 3, 1911. 7. Mch. 26, 1911 to Dec. 20, 1911. 8. Apr. 30, 1909 to Jan. 14, 1910. 9. June 16, 1911 to Apr. 8, 1912. 10. May 19, 1910 to May 28, 1910. 11. Aug. 27, 1911 to Mch. 5, 1912. 12. Oct. 1, 1907 to July 17, 1908. MULTIPLICATION OF COMPOUND NUMBERS ILLUSTRATIVE EXAMPLE 165. Multiply 28 da. 16 hr. 34 min. by 6. SOLUTION EXPLANATION. Commencing with the da hr min lowest denomination (34 min.) successively ' ' ' multiply each denomination of the multipli- cand by the multiplier, as follows: 5. 6 times 34 min. = 204 min., or 3 hr. 172 da 3 hr. 24 min. 24 min. Write 24 min. as a partial product, 108 COMPOUND NUMBERS and carry 3 hr. to 6 times 16 hr., or 96 hr., obtaining 99 hr., or 4 da. 3 hr. Write 3 hr. as a second partial product, and carry 4 da. to 6 times 28 da., or 168 da., obtaining 172 da. as a third and final partial product; thus mak- ing the complete product equal 172 da. 3 hr. 24 min. EXAMPLES FOR PRACTISE Multiply Multiply 1. 5 gal. 2 qt. 1 pt. by 8 7. 17 mi. 48 rd. 4 yd. 2 ft. by 73 2. 24 3s. 5d. by 7 8. 53 yr. 7 mo. 23 da. by 85 3. 16 da. 21 hr. 15 min. by 9 9. 78 bu. 3 pk. 2 qt. 1 pt. by 147 4. 46 Ib. 8 oz. 12 pwt. by 6 10. 47 T. 15 cwt. 73 Ib. by 524 5. 53 bu. 2 pk. 5 qt. by 12 11. 16 da. 15 hr. 49 min. by 97 6. 273 16s. 7d. 3 far. by 97 12. 9 gal. 3qt. 1 pt. by 276 13. The distance between two towns is 42 miles. If a man walk from one towards the other for 8 hours at an average speed of 3 miles and 192 rods per hour, how far will he then be from the other town? DIVISION OF COMPOUND NUMBERS BY ABSTRACT NUMBERS ILLUSTRATIVE EXAMPLES 166. 1. Divide 35 Ib. 10. oz. 18 pwt. by 8. SOLUTION EXPLANATION. Commencing with H, irv the highest denomination (35 Ib.) suc- 8)35 Ib. 10 oz. 18 pwt. cessively divide each denomination of 4 Ib. 5 OZ. 17 pwt. 6 gr. the dividend by the divisor as follows: 35 Ib. -r- 8 = 4 Ib. and an un- divided remainder of 3 Ib. Write 4 Ib. as a partial quotient; and carry the undivided remainder (3 Ib. or 36 oz.) to the next lower denomination of the dividend (10 oz.), obtaining 46 oz. as the second partial dividend. 46 oz. -5- 8 = 5 oz. and an undivided remainder of 6 oz. Write 5 oz. as the second part of the quotient; and carry the undivided remainder (6 oz. or 120 pwt.) to the next lower denomination of the dividend (18 pwt.), obtaining 138 pwt. as the third partial dividend. 138 pwt. -f- 8 = 17 pwt. and an undivided remainder of 2 pwt. Write 17 pwt. as the third part of the quotient; and reduce the undivided remainder (2 pwt.) to the next lower denomination of the Troy scale (2 pwt. = 48 gr.). 48 gr. -T- 8 = 6 gr. and no undivided remainder. Write 6 gr. as the fourth and final part of the quotient. These several partial quotients combined make the complete quotient 4 Ib. 5 oz. 17 pwt. 6 gr. COMPOUND NUMBERS 109 2. Divide 249 10s. 5d. by 29. SOLUTION 29) 249 10s. 5d.(8 232 17 remainder X 20 (and add 10s.) EXPLANATION. When the divisor is over 12, and not a composite num- ^yjoous.^izs. ber, long division may be advan- tageously employed. gQ The processes in the present ro example are similar to those of the preceding example; but here they 2s. remainder are written because they are too X 12 (and add 5d.) unwieldy to be carried mentally. 29)29 d.(ld. 29 Complete quotient, 8 12s. Id. EXAMPLES FOR PRACTISE Divide Divide 1. 39 gal. 1 qt. 1 pt. by 5. 6. 189 9s. 4d. by 56 (= 8 X 7) 2. 78 yd. 2 ft. 8 in. by 8 7. 105 T. 4 cwt. 38 Ib. by 27 (9X3) 3. 225 18s. 6d. by 7 8. 369 gal. 1 pt. by 47 4. 155 cd. 50 cu. ft. by 9 9. 277 yr. 4 mo. 24 da. by 73 5. 287 bu. 2 pk. 4 qt. by 12 10. 930 39' 55" by 269 DIVISION OF COMPOUND NUMBERS BY COMPOUND NUMBERS ILLUSTRATIVE EXAMPLE 167. Divide 907 yd. 2 ft. 6 in. by 34 yd. 2 ft. 9 in. SOLUTION EXPLANATION. Reduce both K)7 yd. 2 ft. 6 in. = 32682 in. compound numbers to simple 34 yd. 2 ft. 9 in. = 1257 in. denominate numbers of the lowest denomination expressed 1257)32f 32(26 times in either) obtaining 32682 in. 2514 as the equivalent simple divi- 7542 dend, and 1257 in. as the equiv- alent simple divisor. Divide in the usual manner. 110 REVIEW OF COMPOUND NUMBERS EXAMPLES FOR PRACTISE 1. Divide 1000 8s. 4d. by 35 14s. 7d. 2. Divide 764 bu. by 5 bu. 3 pk. 7 qt. 3. Divide 1172 gal. 2 qt. 1 pt. by 19 gal. 3 qt. 1 pt. 4. Divide 53 da. 11 hr. 56 min. 54 sec. by 17 hr. 35 min. 18 sec. 5. Divide 183 Ib. 3 oz. 12 pwt. by 3 Ib. 9 oz. 16 pwt. 12 gr. REVIEW OF COMPOUND NUMBERS 168. 1. From a field of wheat containing 78 A., was gathered an average crop of 28 bu. 1 pk. 3 qt. per acre. How much wheat did the field produce? 2. How many spoons weighing 1 oz. 17 pwt. can be made from 3 Ib. 8 oz. 8 pwt of silver? 3. A tenant rented a house on March 18, 1909, at $22.50 per month, and occupied it until February 18, 1912. What was the total rent paid during his tenancy? 4. A man contracted a debt on June 16, 1910, and discharged it on April 23, 1911. How many days did the debt remain unpaid? 5. A coal dealer bought 58240 Ib. of coal at $5.20 per long ton and sold it at $6.10 per short ton. What was his total gain? 6. A customer bought a piece of goods at 2s. 6d. per yard, and paid 5 7s. 6d. How many yards did the piece contain? 7. A grocer bought 3 bu. 2 pk. 7 qt. of peanuts and retailed his purchase at 5^ per pint. How much did he receive? 8. What is the difference between 15 Ib. Avoir., and 25 Ib. Troy expressed in denominations of Troy Weight? 9. A charity organization divided 51 bu. 1 pk. 7 qt. of pota- toes equally between 27 needy families. How much did each family receive? 10. A gardener divided H of an acre of ground into 12 equal vegetable plats. How much space did each plat occupy, expressed integrally? 11. An importer paid 253 5s 7d. for one invoice of merchan- dise, 75 18s. 6d. for a second invoice, 148 12s. 9d. for a third, 62 9s. 3d. for a fourth, and 197 17s. 3d. for a fifth. What was his total payment for the five invoices? 12. A grocer sold 18 gal. 3 qt. 1 pt. 2 gi. of vinegar from a THE METRIC SYSTEM 111 barrel containing 42 gal. 1 qt. How much vinegar remained in the barrel? 13. A pedestrian walked 7 hours per day at an average speed of 4 mi. 37 rd. 3 yd. per hour. What distance had he traveled at the end of the ninth day's walk? 14. How much will be received for a barrel of cider contain- ing 42.375 gallons, if retailed at 3^f per pint? 15. How many barrels will be needed to ship 79 bu. of pota- toes, if the average capacity per barrel is 2 bu. 1 pk. 7 qt? 16. What is the value in United States money of 873.45 marks (Germany)? 17. Of 728.75 crowns (Sweden)? 18. Of 673 2s. 6d. (Great Britain)? 19. Of 3278 lira (Italy)? 20. Of 348.25 yen (Japan)? 21. Of 758.35 francs (France)? 22. Reduce $675.80 to French francs. 23. $528.36 to 's sterling. 24. $912.50 to Russian rubles. 25. $248.75 to Spanish peseta. THE METRIC SYSTEM 169. INTRODUCTION. The metric system is the expression of compound denominate numbers upon the decimal scale; and by its use the convenient processes incident to operations upon simple integral quantities and upon decimal fractions can be substituted in the place of the inconvenient methods necessary for operating upon compound integral quantities and upon common fractions which are expressed upon an arbitrary variable scale. The system derives its name from the meter (39.37079 inches long) which is the established standard not only for measuring distances, but also for determining the size of the other standards for weights, measures, surfaces, and volumes, by expressing them in decimal multiples or aliquots of the meter. 170. The higher denominations of the metric system are denoted by prefixing the Greek numerals deka (10 standard units), hekto (100 standard units), kilo (1000 standard units), and myria (10,000 standard units). Thus, 1 dekameter means 10 meters; 5 hektometers, 500 meters; 7 kilo- meters, 7000 meters; and 9 myriameters, 90000 meters. The metric com- pound 5 hektometers 9 dekameters and 7 meters, is written 597 M.; the meters (7) occupying the units' order, the dekameters (9) the tens' order, the hektometers (5) the hundreds' order, in accordance with the signification of their respective prefixes. 112 THE METRIC SYSTEM 171. The lower denominations of the metric system are de- noted by prefixing the Latin numerals deci (one-tenth of one standard unit), centi (one-hundredth of one standard unit), milli (one-thousandth of one standard unit). Thus, 1 decimeter means one-tenth of a meter; 3 centimeters mean 3 hundred ths of a meter; 4 millimeters mean 4 thousandths of a meter. The metric compound 2 decimeters, 4 centimeters, 3 millimeters is written .243 M., in accordance with the signification of their respective prefixes; and is usually read 243 thousandths of a meter, or 243 millimeters. 172. The meter (= 39.37079 inches) is the metric standard for measuring lengths, widths, heights, depths and distances. NOTE 1. The meter (pronounced mee'ter) is used by merchants in meas- uring dry goods, and by others in measuring short distances; the millimeter being used in measuring minute distances; and the kilometer in measuring long distances. NOTE 2. The higher and lower denominations of the meter are expressed by the use of the prefixes mentioned in 170 and 171. NOTE 3. Reductions from the common denominations of length to the metric, or the reverse, may be made by the use of the following EQUIVALENTS OF LINEAR MEASURE (Common to Metric) (Metric to Common) 1 inch = 2.54001 centimeters 1 centimeter = .3937 of an inch 1 foot = .304801 of a meter 1 decimeter = .32818 of a foot 1 yard = .914402 of a meter 1 meter = 1.093611 yards 1 rod = 5.029211 meters 1 dekameter = 1.98838 rods 1 mile = 1.60935 kilometers 1 kilometer = .62137 of a mile 173. The gram (= .03527 oz., Avoir., or 15.432 gr., Troy) is the metric standard for weights. It is the weight of a cubic volume of distilled water each edge of which is .01 of a meter. NOTE 1. The higher and lower denominations of the gram are expressed by the use of the same prefixes with the same signification as those of the meter [170, 171]. Thus, 8 hektograms mean 800 grams; 4 milligrams mean 4 thou- sandths of a gram; etc. NOTE 2. The gram is used to express the weight of minute quantities, as jewels, medicines, etc.; and the kilogram (usually abbreviated to kilo) to express ordinary weights. THE METRIC SYSTEM 113 EQUIVALENTS OF WEIGHT (Common to Metric) (Metric to Common) 1 oz., Avoir. = 28.3494 grams 1 gram = .03527 oz., Avoir. 1 oz., Troy = 31.104 grains 1 gram = .03215 oz., Troy 1 lb., Avoir. = .45359 of a kilo. 1 hektogram = 3.527 oz., Avoir. 1 lb., Troy = .37324 of a kilo. 1 kilogram = 2.20462 lb., Avoir. 174. The liter (= 1.0567 liquid quarts, or .908 of a dry quart) is the metric standard for capacities. It is the capacity of a cubic measure each edge of which is one-tenth of a meter. The liter is used to measure not only liquids, but also salt, grains, fruits, vegetables, etc. It is pronounced "lee'ter." NOTE 1. The higher and lower denominations of the liter are denoted by the prefixes mentioned in 170, 171. NOTE 2. Of liquids, the dekaliter is used to express large quantities, the liter moderate quantities, and the centiliter or milliliter minute quantities. Of grains, fruits, vegetables, the hektoliter, like the bushel, is used to express large quantities (wholesale); and the dekaliter or liter, small quantities (retail). EQUIVALENTS OF CAPACITY (Common to Metric) (Metric to Common) 1 liquid quart = .94636 of a liter 1 liter = 1.05668 liquid qts. 1 dry quart = 1.1012 liters 1 liter = .9081 of a dry qt. 1 liquid gallon = .37854 of a dekaliter 1 dekaliter = 2.6417 liquid gal. 1 peck = .881 of a dekaliter 1 dekaliter = 1.1351 pecks 1 bushel = .35239 of a hektoliter 1 hektoliter = 2.8375 bu. 175. The ar ( = 119.6034 square yards) is the metric stan- dard for measuring land. It is a square each side of which is 10 meters, and is therefore equal to 100 square meters. It is pro- nounced "air." Scale: 100 centars = 1 ar; 100 ars = 1 hektar. 176. The square meter (= 1.196 square yards) is the metric standard for measuring surfaces other than those of land. It is a square each side of which is 1 meter. Scale: 100 sq. centimeters = 1 sq. decimeter; 100 sq. deci- meters = 1 sq. meter; 100 sq. meters = 1 sq. dekameter. 114 THE METRIC SYSTEM NOTE. As in metric square measures 100 units of any order make one unit of the next higher order, it is necessary to give each order, except the high- est, two places in expressing either land surfaces [175] or ordinary surfaces [176]. Thus, 19 sq. dekameters 5 sq. meters 49 sq. decimeters 9 sq. centi- meters should be written 1905.4909 M.; and 2 hektars 7 ars 5 centars of land are written 207.05 ars. EQUIVALENTS OF SQUARE MEASURE (Common to Metric) (Metric to Common) 1 sq. inch = 6.4516 sq. centimeters 1 sq. centimeter = .155 of a sq. in. 1 sq. foot = .0929 of a sq. meter 1 sq. decimeter = .1076 of a sq. ft. 1 sq. yard = .8361 of a sq. meter 1 sq. meter = 1.1960 sq. yd. 1 sq. rod = 25.293 sq. meters 1 ar = 3.954 sq. rods 1 acre = 40.469 ars 1 hektar = 2.47104 acres 1 sq. mile = 259 hektars 1 sq. kilometer = .3861 of a sq. mi. 177. The ster (= .2759 of a cord) is the metric standard for measuring wood. It is a cube each edge of which is 1 meter. It is pronounced "stair." Scale: 10 decisters = 1 ster; 10 sters = 1 dekaster. 178. The cubic meter is the metric standard for measuring solids other than wood. Like the Ster [177], it is a cube each edge of which is 1 meter. Scale: 1000 cubic millimeters = 1 cubic centimeter; 1000 cubic centimeters = 1 cubic decimeter; 1000 cubic decimeters = 1 cubic meter. NOTE. As in metric cubic measure 1000 units of any order make 1 unit of the next higher order, it is necessary to allot three places to each order except the highest in expressing solids other than wood. In wood measure the scale is 10, and therefore only one place is required to express each order. EQUIVALENTS OF CUBIC MEASURE (Common to Metric) (Metric to Common) 1 cu. inch = 16.387 cu. centimeters 1 cu. centimeter = .061 of a cu. in. 1 cu. foot = 28.317 cu. decimeters 1 cu. decimeter = .0353 of a cu. ft. 1 cu. yard = .7645 of a cu. meter 1 cu. meter = 1.308 cu. yards 1 cord = 3.624 sters 1 ster = .2759 of a cord 179. Reductions from one metric denomination to another are effected by simply placing a decimal point at the right of the required denomination, removing the original decimal point, and THE METRIC SYSTEM 115 making the corresponding change in the abbreviation to that of the required denomination. Thus, to reduce to a higher metric denomination, remove the decimal point one place to the left if the scale of reduction is 10, two places to the left if the scale is 100, or three places to the left if the scale is 1000, for each interme- diate denomination until the required higher denomination is reached, prefix- ing O's if necessary; and to reduce to a lower metric denomination, remove the decimal point one, two, or three places to the right, according to the scale used, for each intermediate denomination until the required lower denomina- tion is reached, annexing O's if necessary. NOTE 1. When a metric quantity is expressed in figures, intermediate vacant orders must also be expressed by writing as many O's for each inter- mediate vacant order as there are ciphers in its scale. NOTE 2. Metric denominations are abbreviated by writing the initial letter of the prefix followed by the initial letter of the standard unit; and by commencing the abbreviation with a capital letter when it expresses a higher denomination than the standard unit [170], and with a "small" letter when it expresses a lower denomination than the standard unit [171]. 180. Metric quantities are added, subtracted, multiplied, and divided as ordinary decimals. NOTE. Before adding or subtracting, it is necessary that the several metric quantities should be reduced to the same denomination [179], that is, that the decimal point of each should be at the right of the same denomination, thus throwing like denominations in the same column [Prin. 1, 18]. EXAMPLES FOR PRACTISE Reduce Reduce 1. 4613.48 Sq. M. to sq. dm. 8. 268 Km. to miles. 2. 346834.57 dg. to Dg. 9. 56 Dl. of vinegar to gal. 3. 46.8173 Hg. to dg. 10. 475 HI. of wheat to bush. 4. 567.8 ds. to S. 11. 468 gal. wine to Dl. 5. 196538.24 ca. to Ha. 12. 15.2 Ib. Troy, to dg. 6. 28.246 Km. to mm. 13. 75 yd. cloth to M. 7. 341728.3 cl. to Dl. 14. 347.48 sq. yd. to Sq. M. 15. Add 468.75 M., 3.567 Dm., and 6178.35 cm. 16. Subtract 46.8 HI. from 58722.348 dl. 17. Find the cost of 9348.56 Kg. of sugar at 11 cents per kilo. 18. Find the cost of 735.48 M. cloth at 5.37 francs per M. 116 COMPUTATION OF SURFACES COMPUTATION OF SURFACES 181. A surface is the outside or visible part of a solid body, without reference to its depth. NOTE. A plane surface is one which is perfectly flat, having no undula- tions. 182. A rectangular surface is a plane surface bounded by four straight sides, each perpendicular to its adjacent side. 183. A square is a rectangular surface bounded by four equal sides. Thus, a square foot is a rectangular surface each side of which is one foot; a square yard is a rectangular surface each side of which is one yard; etc. 184. The area of a rectangular surface is the amount of space included within its four sides, expressed in square units of any convenient denomination. Thus, to find the area of a rectangular surface 1 yd. 1 ft. long and 1 yd. wide, a convenient measuring unit is one square foot. The area can then be found as follows : EXPLANATION. Fig. 1 is the f FT 3 Fr 3 rr measuring unit, each side of which is 1 foot. Fig. 2 shows ^|iso/r[ Z| 3|sq.|pT.| FIG. I FIG. 2 12 SQ FT. the number of these measuring squares in 1 row (3 sq. ft.). Fig. 3 shows the number of rows (4) . Therefore, 4 times 3 sq. ft., or 12 sq. ft. is the area of a rec- FIG. 3 tangular surface which is 1 yd. 1 ft. long, and 1 yd. wide. THE LENGTH AND BREADTH BEING GIVEN TO FIND THE AREA 185. How many sq. yd. in a floor 18 ft. long and 14 ft. 6 in. wide? EXPLANATION. A floor 18 ft. long and only one foot wide con- tains 18 sq. ft.; therefore a floor 18 sq. ft. X 14 J = 261 sq. ft. 18 ft. long and 14 ft. wide must 261 9 = 29 SQ yd contain 14| times 18 sq.ft., or 261 sq. ft., which, when reduced to sq. yd. by 167, equal 29 sq. yd. COMPUTATION OF SURFACES 117 SUMMARY, (a) The number of square measuring units in one row (lying adjacent to one dimension) is the multiplicand, (b) The number of rows (as indicated by the number of the same square measuring units lying adjacent to the other dimension) is the multiplier, (c) The area (the number of square measuring units in all the rows) is the product. NOTE. Before finding any required area, reduce the given dimensions to the same denomination if not already so expressed. EXAMPLES FOR PRACTISE Find the area of a surface which is 1. 32 ft. long and 24 ft. wide. 2. 56 yd. long and 18 ft. wide. 3. 7 yd. long and 2 ft. 6 in. wide. 4. 18 ft. 6 in. long and 12 ft. wide. 6. 34 yd. 2 ft. long and 18 yd. 1 ft. wide. 6. 35 rods long and 25 yd. 2 ft. wide. 7. 173 yd. 1 ft. long and 98 yd. 2 ft. wide. 8. 18 rd. 2 yd. long and 12 rd. 4 yd. 1 ft. wide. 9. 3 mi. 190 rd. long and 2 mi. 75 rd. wide. 10. 15 mi. long and 7 mi. 200 rd. wide. 11. 3 yd. 2 ft. 9 in. long and 2 yd. 1 ft. 6 in. wide. 12. 8 rd. 4 yd. long and 5 rd. 3 yd. 2 ft. wide. 13. How many square yards in a ceiling 28 ft. 9 in. long and 20 ft. 6 in. wide? 14. How many square yards in the walls of a room 30 ft. long, 20 ft. wide, and 8 ft. 6 in. high? 15. How many acres does a rectangular field contain which is 350 yd. 2 ft. long and 235 yd. 1 ft. 6 in. wide? 16. How many sq. ft. in the outer surface of a packing case which is 3 ft. 7 in. long, 2 ft. 9 in. wide, and 2 ft. 3 in. high? 17. How many sq. ft. of surface are in a blackboard 9 ft. 6 in. long and 4 ft. wide? 18. What is the cost of paving a cellar floor with cement at 35^ per sq. yd., if the cellar is 30 ft. long and 18 ft. wide? 19. What is the cost of paving a sidewalk 18 ft. long and 12 ft. wide at 25 cents per square foot? 118 COMPUTATION OF SURFACES TO FIND ONE DIMENSION, WHEN THE AREA AND THE OTHER DIMENSION ARE GIVEN ILLUSTRATIVE EXAMPLE 186. If a rectangular field contains 40 rods and is 22 yards wide, what is its length? EXPLANATION. The area (40 rods) is the given product [Summary, c, 40 rods= 1210 sq. yd., area 185]; and 22 (times the square yards adjoining the required width) is the 1210 -T- 22 = 55 yd., length given multiplier factor of that prod- uct [Summary, b, 186]. Hence, complete the relation of these terms of multiplication by dividing the given product (40 rods, or 1210 sq. yd.) by its given multiplier factor (22), obtaining 55 sq. yd. (adjoining the required width) as its required multiplicand fac- tor [Summary, a, 186]. As only one linear yard of each of these 55 sq. yd. is in immediate contact with the required width, the width must be 55 linear yards. NOTE. Before division, the given area should be made to express square units, and the given dimension linear units, of the same denomination as that of the required dimension. EXAMPLES FOR PRACTISE 1. If the area of a floor is 35 sq. yd. 8 sq. ft., and its length is 6 yd. 1 ft., what is its breadth? 2. If a rectangular field contains 74 acres and is 880 yd. long, how many rods is its breadth? 3. If a blackboard contains 112 sq. ft. of surface and is 4 ft. wide, what is its length? 4. How high is a room 18 ft. long and 16 ft. wide, if its four walls contain 68 sq. yd.? 5. The area of a rectangular garden is 14 sq. rd. 8 sq. yd. 4 sq. ft. 72 sq. in., and its length 4 rd. 2 yd. What is its width? CARPETING, PAVING, ROOFING ILLUSTRATIVE EXAMPLE 187. Utilizing all the material, how many yards of carpeting 2 ft. 4 in. wide, are required to cover a floor 21 ft. long and 16 ft. wide? COMPUTATION OF SURFACES 119 SOLUTION 21 sq. ft. (in one row adjoining the length of the room) X 16 (the number of rows in the width of the room) = 336 sq. ft. (total floor area to be carpeted). 3 sq. ft. (in one row adjoining the length of 1 yd. of carpet) X 2J (the number of such rows in the width of 1 yd. of carpet) = 7 sq. ft. (area of 1 yd. of carpet). Hence, 336 sq. ft. (area of floor) 4- 7 sq. ft. (area of 1 yd. of carpet) = 48 times (1 yd. of carpet), or 48 yd. of carpet. EXPLANATION. As many yards of carpet will be required as the obtained area of 1 yd. of the given carpet (7 sq. ft.) is contained times in the entire floor area to be carpeted (336 sq. ft.), or 48 times 1 yard. SUMMARY, (a) The floor area overlaid by one yard of the cover- ing material is the multiplicand, (b) The total number of yards of the covering material is the multiplier, (c) The total floor area to be overlaid is the product. NOTE 1. To find the number of flagstones, bricks, etc., to overlay a given surface, divide the area of the surface to be overlaid by the exposed area of one unit of the overlaying material. EXAMPLES FOR PRACTISE Utilizing all the material, how many yards will carpet 1. A room 26 ft. long and 20 ft. wide; carpet 2 ft. 6 in. wide? 2. A room 19 ft. 3 in. long and 16 ft. wide; carpet 2 ft. 9 in. wide? 3. A room 27 ft. 8 in. long and 23 ft. 7 in. wide; carpet 1 yd. wide? 4. A room 37 ft. 4 in. long, 25 ft. 6 in. wide; carpet 2 ft. 4 in. wide? 5. What is the cost of laying matting 2 yd. wide on the floor of a room 24 ft. long and 18 ft. wide at 45^ per yard? 6. How many bricks 8 in. long, 4 in. wide, and 2 in. thick, are necessary to cover a pavement 18 ft. long and 12 ft. wide, if the bricks are laid flatwise? 7. Laid edgewise, how many bricks are necessary to pave a stable entrance 14 ft. long and 12 ft. wide, if the bricks are 8 in. long, 4 in. wide, and 2 in. thick? 120 COMPUTATION OF SURFACES 8. If placed endwise, how many bricks are needed to cover a sidewalk 32 ft. 6 in. long and 18 ft. 4 in. wide, if the bricks are 8 in. long, 4 in. wide, and 2 in. thick? 9. How many granite blocks 8 in. long and 8 in. wide are needed to cover a yard 18 ft. long and 15 ft. wide? 10. How many pieces of turf 1 ft. long and 9 in. wide are needed to sod a plat of ground 62 ft. 3 in. long and 43 ft. wide? 11. How many yards of carpet 2 ft. 6 in. wide are needed to cover the floor of a room 27 ft. long and 20 ft. wide, if the strips are laid crosswise to the room? SOLUTION EXPLANATION. As the strips . are to be laid crosswise (across 27 ft. ^ 2? ft. = LO^r times 1 strip the i eng thof theroom), divide the or practically 11 strips. len & th of the room < 27 ft -> bv the width of the carpet (2 ft.) to find 20 ft. X 11 = 220 ft. = 73J yd. the number of required strips (practically 11 strips, the excess of the llth strip, 6 in., being either cut off before laying or turned under while laying). As the length of each strip must be the same as the width of the room, multiply the length of each strip or multiplicand (20 ft.) by the number of strips or multiplier (11) to find the total number of yards of carpet needed (220 ft., or 73 yd.). If there is no loss in matching figures, how many yards of carpet must be purchased to cover a floor 12. 26 ft. 3 in. long and 20 ft. 5 in. wide; carpet 1 yd. wide, laid crosswise? 13. 21 ft. 9 in. long and 18 ft. 8 in. wide; carpet 2 ft. 4 in. wide, laid lengthwise? 14. 18 ft. 6 in. long and 16 ft. 9 in. wide; carpet 2 ft. 6 in. wide, laid crosswise? NOTE 2. In matching figures, an allowance should be made for the wastage caused by adding to the length of each following strip as much length of carpet as must be cut off to enable it to match the adjoining edge of the first strip. Thus, in Ex. 11, if 18 in. had been allowed for wastage in matching figures, the first strip would have been 20 ft. long and the 10 remaining strips 20 ft. + 18 in. long; and the length of carpet needed would have been 20 ft. + (2H ft. X 10), or 235 ft., or 78 1 yd. COMPUTATION OF LUMBER 121 Allowing 10 inches per strip for matching figures, how many yards of carpet will be needed to cover a floor 15. 30 ft. 6 in. long and 26 ft. 3 in. wide; carpet 1 yd. wide, laid lengthwise? 16. 25 ft. 3 in. long and 18 ft. 6 in. wide; carpet 2 ft. 4 in. wide, laid crosswise? COMPUTATION OF LUMBER 188. Lumber measure is used in measuring timber after it has been sawed into boards or planks, scantling, joists, and other building material. 189. The board foot is the standard by which lumber is measured. A board foot is 1 ft. long, 1 ft. wide and 1 in. thick; or 1 sq. ft. of the face of a board which is 1 in. thick. Thus, a board 16 ft. long, 2 ft. wide, and 1 in. thick, is estimated to con- tain 16 X 2, or 32 board feet. NOTE 1. If a board is more than 1 in. thick, the number of square feet in its face should be proportionately increased in estimating the number of board feet it contains. Thus, the number of board feet in a plank 12 ft. long, 2 ft. wide, and U in. thick = 1 J times 12 X 2, or 30 bd. ft.; if U in. thick = li times 12 X 2, or 36 bd. ft.; if 2 in. thick = 2 times 12 X 2, or 48 bd. ft. NOTE 2. Hewn timber is sold by board measure, or by cubic measure, and sometimes by the "running foot." Even when sold by the running foot, its width and thickness are the governing factors in determining the price per running foot. NOTE 3. The average width of a tapering board is ascertained by meas- uring midway between its two ends, or by taking hah* the sum of the widths of its two ends. NOTE 4. In measuring two or more boards of uniform length and thick- ness, it is more convenient to multiply in one operation the uniform length by the total width of all the boards to obtain the total board feet. ILLUSTRATIVE EXAMPLE How many bd. ft. in a plank 18 ft. long, 6 in. wide, and 1? in. thick? SOLUTION 18 bd. ft. (if 1 ft. wide and 1 in. thick) X \ (of 1 ft. wide) = 9 bd. ft. 9 bd. ft. (if 1 in. thick) X li (actual thickness in inches) = 13J bd. ft. 122 COMPUTATION OF VOLUMES EXPLANATION. A plank 18 ft. long, 1 ft. wide, and 1 in. thick contains 18 bd. ft.; and if only i of a foot (6 in.) wide, contains 1 of 18 bd. ft., or 9 bd. ft.; and if li in. thick, contains U times 9 bd. ft., or 131 bd. ft. NOTE. 5. It is sometimes more convenient to multiply the length in feet by the width in inches (or 12 times the multiplier) to find the number of board feet in a 1-in. plank; and in correction to divide the resulting product by 12. EXAMPLES FOR PRACTISE Find the number of board feet in a plank which is Long Wide Thick Long Wide Thick 1. 16ft. 9 in. 1 in. 5. 18 ft. 6 in. 6 in. 2 in. 2. 18ft. 7 in. 1 in. 6. 15ft. 9 in. 7 in. 2} in. 3. 14ft. Gin. IJin. 7. 22ft. 5 in. If in. 4. 20ft. 8 in. If in. 8. 24ft. 1 ft. 3 in. 2 in. 9. How many board feet in 5 planks, each 16 ft. long and 1 \ in. thick, if their respective widths are 6 in., 7 in., 9 in., 10 in., and 11 in.? [Note 4] 10. How many board feet in a plank 18 ft. long, If in. thick, and 8 in. wide at one end and 6 in. wide at the other? [Note 3] 11. What is the cost of an invoice of 2-in. planks 18 ft. long, at $24.75 per thousand feet, if their united width is 45 ft. 7 in.? 12. What is the cost of 24 beams each 26 ft. long, 10 in. wide, and 3 in. thick, at $21.50 per thousand feet? COMPUTATION OF VOLUMES 190. By volume is meant the contents included within the bounding surfaces of a given amount of space. NOTE 1. A rectangular volume is one which is bounded by six rectangular surfaces [182]. In addition to the length and breadth of rectangular surfaces, rectangular volumes have a third dimension, thickness. NOTE 2. A cube is a rectangular volume bounded by six equal squares, called faces. All the edges of a cube are of equal length. Thus, a cubic inch is a rectangular volume each edge of which is one linear inch, and each of the six faces of which is one square inch; and a cubic foot is a rectangular volume each edge of which is one linear foot, and each face of which is one square foot. COMPUTATION OF VOLUMES 123 191. The contents or volume of a rectangular solid is the amount of space included within its six faces, expressed in cubic units of any convenient denomination. Thus, to find the contents of a rectangular body which is 5 ft. long, 4 ft. wide, and 3 ft. deep, take any convenient measuring unit, say, one cubic foot (Fig. 1) as the standard. The contents can then be found as follows: FIG. 3 I FT. I FT. FIG. 4 20 CU. FT. X 3 = 60 CU. FT. Fig. 1 is the measuring unit, which is 1 ft. long, 1 ft. wide, and 1 ft. deep, or 1 cu. ft. Fig. 2 shows the number of these cubic measuring units in one row of the upper layer of Fig. 4 (4 cu. ft.). Fig. 3 shows the number of these cubic measuring units in all the five rows of the upper layer of Fig. 4 (4 cu. ft. X 5 = 20 cu. ft.). Fig. 4 shows the number of these cubic measuring units in all three of the layers of the given body (20 cu. ft. X 3 = 60 cu. ft.). Hence, a rectangular body 5 ft. long, 4 ft. wide, and 3 ft. deep, contains 5 cu. ft. X 4 X 3, or 60 cu. ft. SUMMARY, (a) The number of cubic measuring units in one layer (adjoining one face of a rectangular body) is the multipli- cand, (b) The number of such layers is the multiplier, (c) The number of the cubic measuring units in all the layers (the contents of the rectangular body) is the product. NOTE 1. Before finding the contents of a rectangular body, its length, breadth, and thickness should be reduced to the same linear denomination, if not already so. 124 COMPUTATION OF VOLUMES ILLUSTRATIVE EXAMPLE How many cubic feet of earth are in a rectangular embankment 16 ft. 6 in. long, 10 ft. 9 in. wide, and 6 ft. high? SOLUTION 16 cu. ft. X 10f X 6 = 1064i cu. ft. EXPLANATION. The number of cu. ft. in the row adjoining the long edge of the lower section of the embankment (161 cu. ft.) multiplied by the number of such rows in the lower section (lOf ) will produce the number of cu. ft. in all the rows of the lower section (177f cu. ft.); and the number of cu. ft. in all the rows of the lower section of 1 ft. in height (177f ) multiplied by the number of such sections in 6 ft. of height (6) will produce the number of cu. ft. in all the sections, or in the entire embankment (1064 \ cu. ft.). EXAMPLES FOR PRACTISE 1. How many cubic feet of wood are in a pile 28 ft. long, 6 ft. high, and 4 ft. wide? 2. How many cubic yards of air are in a room 16 ft. long, 12 ft. wide, and 10 ft. high? 3. How many cubic feet of wood are in a squared log 18 ft. long, 9 in. wide, and 9 in. thick? 4. How many cubic yards of stone are in a rectangular pile which is 20 ft. long, 10 ft. wide, and 5 ft. high? 5. What is the cost of digging a rectangular excavation 24 ft. long, 18 ft. wide, and 12 ft. deep, at 45 i per cu. yd.? 6. What is the cost of a pile of cord wood which is 35 ft. long, 6 ft. high, and 3 ft. wide at $7.15 per cord? 7. A cellar is 18 ft. long and 16 ft. wide, and contains 96 cu. yd. What is its depth? SOLUTION 18 cu. ft. X 16 = 288 cu. ft. [Summary, a.] 27 cu. ft. X 96 = 2592 cu. ft. [Summary, c] 2592 cu. ft. ^ 288 cu. ft. = 9 times (1 ft. deep). EXPLANATION. A cellar which is 18 ft. long, 16 ft. wide, and 1 ft. deep, contains 18 cu. ft. X 16 X 1, or 288 cu. ft. Therefore a cellar 18 ft. long and 16 ft. wide to contain 96 cu. yd. or 2592 cu. ft. must be as many times 1 ft. deep as 288 cu. ft. are contained times in 2592 cu. ft., that is, 9 times 1 ft. deep. COMPUTATION OF CAPACITIES 125 NOTE. 2. If necessary, reduce the given dimensions to linear units, and the given volume to cubic units, of the same denomination, before finding the required dimension [Prin. 1, 34]. 8. A room 48 ft. long and 32 ft. wide contains 23040 cu. ft. What is its height? 9. A contractor dug a cellar 36 ft. long and 12 ft. deep, and removed 336 cu. yd. of earth. What was its width? 10. A purchaser paid $59.50 for a pile of cord wood 28 ft. long and 4 ft. wide, bought at $8.50 per cord. How high was the pile? 11. A wagon-body 12 ft. long and 4 ft. 4 in. wide contains 78 cu. ft. What is its height? 12. How long must a rectangular embankment be to contain 35 cu. yd. 5 cu. ft., if it is 6 ft. 3 in. wide, and 9 ft. 6 in. high? COMPUTATION OF CAPACITIES 192. By capacity is meant the amount of space within a con- taining vessel which enables it to hold a given quantity of water, grain, fruit, vegetables, coal, etc. NOTE 1. In finding the amount of any given quantity of grains, seeds, or berries, stricken measure is used. By stricken measure is meant the measuring vessel even full and stricken off with a rule or striker. A stricken bushel con- tains 2150.42 cu. in. NOTE 2. In finding the amount of any given quantity of coal, lime, un- shelled corn, vegetables, or the larger fruits, heaped measure is used. Heaped measure means the measuring vessel heaped in the form of a cone. A heaped bushel contains 2747.71 cu. in. 4 heaped bushels are usually considered equivalent to 5 stricken bushels. NOTE 3. A gallon, liquid measure, contains 231 cu. in.; and a gallon, dry measure (ipk.), 268.8 cu. in. EXAMPLES FOR PRACTISE 1. A rectangular bin is 6 ft. long, 5 ft. wide, and 4 ft. deep. How many bushels of grain will it hold? SOLUTION 6 cu. ft. X 5 X 4 = 120 cu. ft. 1728 cu. in. X 120 = 207360 cu. in. 207360 cu. in. ^ 2150.42 cu. in. = 96.42 + times 1 bushel. EXPLANATION. As grain is measured by stricken bushels [Note 1] divide the capacity of the bin (207360 cu. in.) by the contents of 1 stricken bushel (2150.42 cu. in.) to find how many times 1 bu. of grain the bin will hold. 126 COMPUTATION OF CAPACITIES 2. A rectangular bin is 8 ft. long, 6 ft. wide, and 3 ft. 9 in. deep. How many bushels of potatoes can it hold? [Note 2]. 3. How many bushels of oats can be stored in a rectangular bin 10 ft. 3 in. long, 4 ft. wide, and 3 ft. 6 in. deep? 4. A cart-body is 7 ft. long, 4 ft. 3 in. wide, and 2 ft. 9 in. deep. How many bushels of shelled corn can it hold? 5. A wagon-body is 10 ft. long, 4 ft. 2 in. wide, and 1 ft. 8 in. deep. How many bushels of apples can it hold? 6. How many gallons of water can a rectangular cistern hold if it is 6 ft. 9 in. long, 4 ft. 6 in. wide, and 10 ft. deep. SUGGESTION. Capacity of cistern in cu. in. -r- capacity of 1 liquid gallon in cu. in. [Note 3] = capacity of cistern in liquid gallons. 7. How many gallons of water will fill a rectangular cistern 12 ft. long, 12 ft. wide, and 4 ft. 3 in. deep? 8. How many gallons of water can be placed in a tank which is capable of holding 35 bushels of rye? 9. How many bushels of potatoes can be stored in a vessel capable of holding 265 gallons of water? 10. A cellar 30 ft. long, 18 ft. wide, and measuring 8 ft. from the street level to the bottom of the cellar, was filled by an over- flow from a neighboring stream. How many gallons of water did it contain? 11. What must be the depth of a bin 8 ft. long and 6 ft. wide, to hold 96 bushels of oats? SUGGESTION. Capacity of 1 bu. (2150.42 cu. in.) X number of bu. (96) = capacity of the bin. Then find required dimension as in solution of Ex. 7, 191. 12. What must be the width of a wagon-body 10 ft. long and 2 ft. deep, to hold 53 bu. of turnips? 13. What must be the depth of a rectangular cistern 6 ft. 4 in. long and 5 ft. 2 in. wide, to hold 2500 gallons of water? NOTE 4. A sufficiently close approximation to meet ordinary require- ments may be obtained by multiplying the capacity of the containing vessel / in cu. ft. by .8 to find its capacity in stricken bushels, or by .63 to find its capacity in heaped bushels. 14. A farmer has a rectangular bin 10 ft. long, 4 ft. wide, and COMPUTATION OF CAPACITIES 127 4 ft. deep. What is the approximate number of bushels of wheat which it can hold? 15. A man had a rectangular bin built in his cellar to hold coal. If the bin was 10 ft. long, 6 ft. wide, and 4 ft. deep, what approximate weight of coal will it hold, allowing 80 pounds to a bushel and 2240 pounds to a ton? 16. Allowing 28 bu. to a long ton of 2240 lb., how long must be a rectangular coal bin 16 ft. wide and 4 ft. deep to hold, approxi- mately, 6 tons of coal? 17. A rectangular bin is 12 ft. long, 8 ft. wide, and 4 ft. deep. What approximate number of bushels of potatoes can it hold? NOTE 5. If a crib or bin has uniformly sloping sides, its average width is one-half the sum of its top and bottom widths [Note 3, 189]. Thus, the contents of a granary 18 ft. long, 12 ft. high, 10 ft. wide at the bottom and 14 ft. wide at the top, are 18 cu. ft. X 12 X | of (10 + 14), or 2592 cu. ft. 18. How many bushels of shelled corn can be stored in a crib which is 12 ft. long, 8 ft. deep, 6 ft. wide at the bottom and 10 ft. wide at the top, using the standard of Note 1? 19. How many bushels of turnips can be stored in a bin 8 ft. long, 6 ft. deep, 4 ft. wide at the bottom and 6 ft. wide at the top, approximating by Note 4? NOTE 6. A containing vessel is said to be cylindrical if it is circular, is of uniform diameter throughout its depth, and its ends are parallel. The capacity of a cylindrical receptacle is .7854 of that of a rectangular vessel of the same depth, and having each of its two other dimensions A CYLINDER equal to the diameter of the cylindrical receptacle. 20. A cylindrical cistern is 12 ft. deep and 4 ft. in diameter. How many gallons of water can it hold? SOLUTION 12 cu. ft. X 4 X 4 = 192 cu. ft., capacity of a rectangular cistern of equal depth (12 ft.) and a length (4 ft.) and a width (4 ft.) equal to the diameter of the cylindrical cistern. Hence, 192 cu. ft. X .7854 = 150.7968 cu. ft., capacity of the given cylindrical cistern. (1728 cu. in. X 150.7968) -=- 231 cu. in. = 1128.0384 times 1 gal. [Note 3]. 21. A cylindrical cistern is 6 ft. in diameter and 14 ft. in depth. How many gallons of water is it capable of holding? 128 COMPUTATION OF CAPACITIES 22. A well is 4 ft. in diameter and the water in it is 6 ft. deep. How many gallons of water are in the well? 23. A cylindrical bin which is 7 ft. high and 3 ft. 6 in. in diameter is filled with wheat. How many bushels does it contain? 24. A circular railroad tank is 8 ft. in diameter and contains 3500 gallons of water. How deep is the water? NOTE 7. A receptacle whose sides are uniformly round but tapering, and whose ends are parallel but unequal circles, is equivalent to a cylindrical receptacle of equal depth and of a uniform diameter of one-half the sum of its two extreme diameters. Thus, such a receptacle if 8 ft. in depth, 6 ft. in diameter at one extreme and 4 ft. in diameter at the other extreme, is equal to a cylindrical vessel which is 8 ft. in depth and has a uniform diameter of (6 + 4) -r- 2, or 5 ft. 25. How many cubic feet in a log 20 ft. long, 18 in. in diameter at one end, and 12 in. in diameter at the other end? 26. How many cu. in. in a bucket which is 2 ft. in depth, 12 in. in diameter at the bottom, and 16 in. in diameter at the top? 27. How many gallons of water will a cistern hold which is 18 ft. in depth, 10 ft. in diameter at the bottom, and 6 ft. in diameter at the top? NOTE 8. Owing to the lack of uniformity in the bulge of a barrel or cask, its average diameter is approximately found by increasing its head diameter by % of the difference between its bung diameter and its head diameter if its staves are much curved, or by 3 of such difference if its staves are slightly curved. Thus, the capacity of a barrel whose length is 36 in., bung diam- eter 30 in., and head diameter 24 in. will approximately equal that of a cylinder 36 in. long and a uniform diameter of 24 in. + f of (30 in. 24 in.), or 28 in. 28. How many gallons of cider will a barrel hold which is 40 in. long, bung diameter 32 in., and head diameter 27 in? 29. What is the capacity in gallons of a cask 3 ft. 4 in. long, head diameter 24 in., bung diameter 30 in. ? 30. What is the capacity in gallons of a barrel 36 in. long, bung diameter 28 in., head diameter 21 in. ? PERCENTAGE 193. INTRODUCTION. All possible arithmetical terms [Note, 5] in all their various relations to each other, and the means for com- pleting those relations, have now been considered [48-53 for integers, and 107-109 for fractions]. The successive subjects from 109 to the end of this volume are simply applications of these comprehensive relationships to new topics which are of sufficient importance to justify special treatment. In 109, the comparison of quantities was made without refer- ence to any particular numerical form in which the result of the comparison was to be expressed. In percentage, however, the result of the comparison is limited in expression to a particular form, that of hundredths. The title " percentage " suggests this limi- tation, being derived from per, by; centum, hundred; age, that which. Hence, percentage is the name of that specific class of calculations, whether in addition, subtraction, multiplication, or division, which is by hundredths. Thus, to add .03 to .06, to sub- tract .03 from .06, to multiply 6 by .03, or to divide 6 by .03, may properly be called operations in percentage. The name percentage, however, is ordinarily restricted to problems which involve comparisons of quantities the results of which are expressed in hundredths, as in the following : $5 are how many hundredths of $25? 20 hundredths of how many dollars equal $5? 20 hun- dredths of $25 equal how many dollars? SUGGESTION. The learner is referred to a, 6 and c of INTRODUCTION, 109, for a full exposition of the general treatment of all numerical comparisons, which will serve as a suitable preparation for considering the special application of the same general principles to particular comparisons by hundredths. 194. Operations in percentage embrace all calculations which involve a specific comparison by hundredths. 195. The base is the quantity taken as the standard with which some other quantity is to be compared. NOTE. Any required comparison of a quantity must include a definite basis or standard, expressed or understood, with which the comparison is to be made. Thus, a given quantity, say $5, may be equal to, or more than, or 130 PERCENTAGE less than, according to the standard with which it may be compared. That is, $5 will be equal to, if compared with $5; it will be more than, if compared with $4; or it will be less than, if compared with $6. 196. The rate per cent, denotes what comparative part of the base, expressed in hundredths, is the quantity which is contrasted with the base. NOTE 1. The term rate when not followed by the phrase per cent, or by the symbol (%) which means "per cent." may signify any part of the standard or base. Thus, in the expressions 4 times $600, f of $600, .005 of $600, and .08 of $600, 4, f, .005, and .08 are rates; the first (4) being an integral rate per unit, the second (f) a common fractional rate per third, the third (.005) a decimal rate per thousandth, and only the fourth (.08) a rate per hundredth or per cent. NOTE 2. To make a comparison of one quantity with another taken as the standard, some graduated measuring unit is necessary. In percentage, the accepted measuring unit is 1, which represents the entire standard of com- parison and which for convenience is conceived to be graduated or scaled in hundredths. Thus, in comparing $2 with $5, the measuring standard or base is $5, represented by the measuring unit 1 (meaning 1 time $5, or the entire $5) ; and this measuring unit 1 is conceived to be divided or scaled into 100 equal parts. If $2 are compared to the standard ($5) by means of this graduated scale, it will be found to measure up to the fortieth of its one hundred gradua- tions and therefore $2 are said to be 40 hundredths or 40 % of the standard ($5). 197. The percentage is the quantity which is compared with the base. NOTE. A percentage expresses the value of a given or required number of hundredths of a certain quantity taken as the base. Thus, in the con- summated comparison "$200 are 25 % of $800," $800 is the indicated base or standard of comparison; $200 is the percentage or quantity compared with the standard; and 25 % is the rate % or result of the comparison, denoting what part $200 are of $800, when expressed in hundredths. 198. Identification of terms. In certain classes of problems, particularly in " applied percentage," there is a uniformly accepted base, and it can therefore be "understood." In an important class of percentage problems, however, there is no uniformly rec- ognized standard, and any base may be arbitrarily assumed, as the author of the problem may direct. All bases which are not understood must necessarily be indicated, (a) The conventional form for indicating arbitrary bases is the expression "of " (mean- PERCENTAGE 131 ing times) placed immediately after the rate % and immediately before the base. (6) It is sometimes necessary to indicate the base with the purpose of having it increased [109, b,] or decreased [109, c] by a given or required percentage, in which case the phrases "more than" (implying increase) or "less than" (imply- ing decrease), are substituted immediately after the rate % and before the indicated base, (c) After the base has been identified, any quantity which is placed in comparison with the indicated base with the view of ascertaining how many hundredths of the given base it is, or of what required base it is a given number of hundredths, is called a percentage. Thus, in the consummated comparisons, "$20 are 25% of $80," "$20 are 75% less than $80," the indicated base is $80, and the percentage $20; and in the consummated comparisons, "$30 are 120% of $25," "$30 are 20% more than $25," the indicated base is $25, and the per- centage $30. 199. Relation of terms. (a) The base is a multiplicand because it expresses the value of one quantity taken as the stand- ard measuring unit; (6) the rate % is a multiplier because it ex- presses the decimal number (hundredths) of standard measuring units; (c) the percentage is a product, because it expresses the value of all the considered decimal number (hundredths) of stand- ard measuring units [Prin. 1, 2 and 3, 107]. Hence, (d), if the base (multiplicand) and the rate % (multiplier) are given, multiply these factors to find the percentage (product) ; (e) if the base (multiplicand) and the percentage (product) are given, divide the given product by its given multiplicand factor to find its required multiplier factor or rate; (/) if the rate (multiplier) and the percentage (product) are given, divide the given product by its given multiplier factor to find its required multiplicand factor or base. NOTE. Though all rates are multipliers and all percentages are products, learners are cautioned against dividing any product by any multiplier to find a particular multiplicand. A product must be divided by its multiplier factor, that is, by the particular multiplier which was used to produce it, in order to obtain its multiplicand factor, that is, the particular multiplicand which was used to produce it. 200. Convenient common fractional rates should be substi- tuted for inconvenient rates %, particularly for complex rates % 132 PERCENTAGE [117]. The rates % in common use which can be advantageously changed to equivalent common fractional rates are shown in the following TABLE OF EQUIVALENTS: A 10% = A 25% = i 62J% = f 2J% = A- 12}% = | 33j% = i 66f% = | 5% = A 16f% = } 37}% = | 75% = j 61% = A 20% = i 50% = } 87}% = 1 NOTE. The tendency to change indiscriminately all rates %, whether convenient or inconvenient, to the form of common fractions is unbusinesslike. BASE AND RATE GIVEN, TO FIND PERCENTAGE ILLUSTRATIVE EXAMPLES 201. 1. Find 9% of $385.42. EXPLANATION. $385.42 is the indicated base or mul- tiplicand, being thus arbitrarily shown by the conventional $385.42 indicator "of" [198, a]. 9% is the rate or multiplier, .09 because it denotes the decimal number of times the base 34. fi87& which is to- be expressed by the required percentage, or product. Hence, multiply these two given factors of the required percentage, obtaining $34.69. 2. A man's salary is $875 per annum. If his total expenses are 65 % of his salary, how much does he save? SOLUTION 100 %, rate % of salary $875, indicated base. 65 %, rate % of expenses, X .35, rate % saved. 35 %, rate % saved. 4375 2625 $306.25, percentage saved. EXPLANATION. "His salary" is arbitrarily indicated as the verbal base by being placed after " % of," and $875 is the numerical expression of this verbal base and therefore the numerical base or multiplicand. 65 % is the rate or multiplier of the base to produce the percentage or product of total expenses, but not the rate or multiplier of the base to produce the required percentage or product "saved. If 100 % is the rate of his salary PERCENTAGE 133 [Note 2, 196] and 65 % is the rate of his total expenses, then 100 % 65 % or 35 %, must be the specific rate or multiplier of the base ($875) to produce the percentage or product saved ($306.25). 3. On a certain farm were harvested 425 bushels of wheat, and 48% more oats than wheat. How many bushels of oats were harvested? EXPLANATION. The bushels SOLUTION of wheat harvested is the arbj. 100 %, rate % of wheat. trarilv indicated verbal base, be- ,f excess of oats^ -* "S^USS 148%, rate % of oats harvested. cal expression of the bushels of wheat harvested and therefore 425 bu., indicated base. the numerical base or multipli- 1.48 rate % of required oats. cand - 48 % is the expressed vAr\f\ rate or multiplier for the excess of the yield of oats over that of the wheat, but not the specific 425 rate or multiplier for the re- 629700 bu., percentage of oats. Quired number of bushels of oats harvested. Expressing the indicated base (425 bu.) upon a scale of hundredths as shown in Note 2, 196 (100 %), the bu. of oats harvested expressed upon the same scale will be 48 % more than 100 %, or 148 %. Hence, multiply the indicated base or multiplicand (425 bu.) by the specific rate % or multiplier of the oats harvested (1.48), obtaining 629 bu. as the specific percentage or product of the oats harvested. RULE. To find any desired percentage, multiply the base by the specific rate % of the desired percentage. NOTE 1. Any given rate should be identified not only generally as a rate or multiplier, but also specifically as the rate or multiplier of the indicated base to produce a particular percentage or product. Thus, in 111. Ex. 2, 100 % is the specific rate of his salary, 65 % is the specific rate of his expenses, and 35 % is the specific rate of what is saved. In 111. Ex. 3, 100 % is the specific rate of the bu. of wheat harvested, 48 % is the rate of so much of the bu. of oats harvested as is more than the bu. of wheat harvested, and 148 % is the rate of the oats harvested. NOTE 2. Two or more rates must be expressed in terms of (upon the same graduated scale as) the indicated base before they can be added [Prin. 1, 18] or subtracted [Prin. 1, 22]. 134 PERCENTAGE EXAMPLES FOR PRACTISE 1. .07 times $800 = how much? 2. 8 % of $300 (that is, .08 times $300) = what percentage (that is, what product by hundredths)? 3. 28 % of 650 Ib. = how many pounds? 4. 84% of 850 men = how many men? 5. 66| % of 471 sheep = how many sheep? 6. 125 % of 584 gal. = how many gallons? 7. J % of $6400 = how much? 8. J % of 800 tons = how many tons? 9. 28 % of 618 2s. 6d. = what percentage? 10. Find 38 % more than 650 gallons. 11. Find 25 % more than 716 sheep. 12. Find 50 % more than 128 horses. 13. Find 20 % less than 315 days. 14. Find 33J % less than 714 miles. 15. Find 16| % less than 468 yards. 16. A grocer purchased 625 pounds of coffee and sold 64 % of his purchase. How many pounds of coffee did he sell? 17. A ranchman had 475 cattle, but lost 28 % of them in a blizzard. How many cattle did he save? 18. By installing new machinery, a factory was enabled to produce 38 % more than at first. If its former capacity was 6500 yards of cloth per week, what is its present capacity? 19. A man's taxable property is assessed at $28500, 43 % of his assessment being upon his real estate, and the remainder upon his personal property. What is the assessed value of each? 20. In a certain graded school 375 pupils are enrolled, 28% of the pupils being in the primary department, and the remainder in the intermediate. How many pupils were enrolled in the intermediate department? 21. By the census of 1910, it was found that the population of a certain city was 43 % greater than by the census of 1900. If its population in 1900 was 46800, what was it in 1910? 22. A farm contains 700 acres, 25 % of it being sown in wheat, 32% being planted in corn, and the remainder set apart for pasturage. How many acres of pasture land does the farm contain? 23. A man died, leaving an estate worth $128500, 54% of his estate having been bequeathed to his wife, 32% to his PERCENTAGE 135 children, and the remainder to an asylum. If the cost of administering upon the estate was 5% of its value, how much did the asylum receive? BASE AND PERCENTAGE GIVEN, TO FIND RATE ILLUSTRATIVE EXAMPLES 202. 1. What % of $428 are $184.04? EXPLANATION. $428 is the arbi- SOLUTION trarily indicated base or multiplicand, $184.04 -T- $428 = .43 = 43% bein S P laced immediately after " % of" (meaning hundred ths times). $184.04 is the percentage or product, it being the quantity which is placed in comparison with the indicated base to find what part of the base it is when expressed in hundredths [197]. If stated in more familiar language, the problem would read, "How many hundredths times $428 are $184.04." Hence, divide the given percentage or product ($184.04) by its given base or multiplicand factor ($428) to find its required rate or multiplier factor, carrying the division to hundredths (by so arranging the percentage that it will have two more decimal places than those of the base), obtaining .43 or 43 %. 2. A has $350 and B. $448. What % more than A has B? EXPLANATION. A's money is SOLUTION the indicated verbal base or mul- $448 $350 = $98, percentage more, tiplicand, as it is placed imme- $98.00-$350=.28,or28%(more). diatdy after % more than" [198, 61, and $350 is the numerical expression of A's money or the numerical base. $448 is a percentage or product, it being the quantity which is placed in comparison with the indicated base to find what % more than the base it is. Though B's money is a percentage, it is not the percentage or product of which it is required to find the rate or multiplier. Therefore, subtract A's money ($350) from B's money ($448) to find the percentage more ($98), or the specific percentage or product of which it is required to find the specific rate % or multiplier. Then divide the percentage or product more ($98) by its indicated base, or multiplicand factor ($350), to find its required rate or multiplier factor, carrying the division to hundredths as in 111. Ex. 1, to obtain the rate in hundredths (28 %). RULE. Divide the specific percentage or product of which it is required to find the rate, by its base or multiplicand factory 136 PERCENTAGE and the quotient carried to hundredths will be its required rate % or multiplier factor. NOTE 1. Percentages should be identified so specifically as to distinguish each percentage from all other percentages by naming its individual charac- teristic. Thus, in 111. Ex. 2, $448 and $98 are not only to be regarded as per- centages generally, but $448 as the specific percentage of B's money, and $98 as the specific percentage more that B's money is than A's. NOTE 2. To obtain a rate in the particular form of a %, the percentage should contain exactly two decimal places more than the base. If the per- centage contains fewer than the required number of decimal places, supply the deficiency by annexing decimal O's; and if the percentage contains more than the required number of decimal places, offset the excess by annexing decimal O's to the base. NOTE 3. To find the rate of a given percentage means simply to find what multiplier of the indicated base will produce the given percentage. EXAMPLES FOR PRACTISE 1. $30 are how many times $6? 2. $20 are how many hun- dredths times $80? 3. $40 are what % of (that is, how many hundredths times) $120? 4. $23.29 are what % of $68.50? 5. $88.69 are what % of $253.40? 6. What % of $872.60 are $741.71? 7. What % of $5249.50 are $3254.69? 8. What % of $528 are $1.32? 9. What % more than $68.40 are $85.50? 10. $322.02 are what % more than $268.35? 11. $275.17 are what % less than $786.20? 12. What % less than $687.50 are $522.50? 13. A grazier owned 350 sheep, and wolves destroyed 14 of them. What % of his flock did he lose? 14. A clerk's salary is $1200 per annum. If he pays $20 per month for board and room rent, $156 per annum for clothing, and $300 per annum for other expenses, what % of his salary does he save? 15. A grocer sold 1653 Ib. sugar and had 522 Ib. remaining. What % of his original stock of sugar did he sell? 16. A son inherited 920 acres from his father, and afterwards sold 138 acres. What % of his inheritance did he have remaining? 17. A man owns real estate valued at $23223.75 and personal property valued at $17650.05. What % less than the value of his real estate is the value of his personal property? PERCENTAGE 137 18. Three men formed a copartnership, A investing $11900, B $14875, and C $15725. What % of the capital of the firm did each invest? 19. The sailing distance from one seaport to another is 8750 miles. What % of the voyage remained after a vessel had sailed 21 days at an average speed of 150 miles per day? 20. A merchant engaged in business with a capital of $42500 and at the end of the first year found that he had gained $7650. What % of his original capital did he gain? 21. A man borrowed $4500 and made three partial payments thereon of $1820, $1230, and $415. What % of his original debt remained unpaid? PERCENTAGE AND RATE GIVEN, TO FIND BASE ILLUSTRATIVE EXAMPLES 203. 1. A man bought a house and paid 45 % of the purchase money in cash. At what price did he buy the house, if his cash payment was $3825? SOLUTION EXPLANATION. The purchase money is indicated as the verbal base or multiplicand $3825.00 -T- .45 = $8500 by being placed immediately after " % of." $3825 is a percentage or product because it is placed in comparison with the indicated verbal base (the purchase money). 45 % is the specific rate or multiplier for $3825 because it denotes the result of the comparison of $3825 with the indicated base ($3825 being .45 of the purchase money). Hence, divide the given percentage or product ($3825) by its specific rate or multiplier factor (.45) to find its specific base or mul- tiplicand factor ($8500). Expressed in more familiar terms, the question is "45 hundredths times what purchase money equal $3825?" 2. A has $296.80, which is 36 % less money than B has. How much money has B? SOLUTION 100 % = rate of B's money, the indicated verbal base. 36% = rate which A's money is less than B's. 64 % = specific rate of A's money, expressed in terms of B's. $296.80 (A's money) -r- .64 (specific rate of A's money) = $463.75 (B's money, or the base). 138 PERCENTAGE EXPLANATION. B's money is indicated as the required verbal base by being placed immediately after " % less than " [198, 6], $296.80 is a percentage or product because it is placed in comparison with the indicated base (B's money) ; and specifically it is the percentage of A's money. 36 % is the specific rate or multiplier of what A's money is less than B's, but not the specific rate or multiplier of the base (B's money) to produce the specific percentage or product of A's money ($296.80). To obtain the specific rate of A's money in terms of B's, let 100 % represent the indicated base, B's money [Note 2, 196]; and as A's money is 36 % less than B's, so 36 % less than 100 %, or 64 %, is the specific rate of A's money. Hence divide the specific percentage or product of A's money ($296.80) by its specific rate (the specific multiplier of B's money to produce A's, 64 %) to find the required base or multiplicand (B's money, $463.75). RULE. Divide the given percentage or product by its rate factor (the multiplier of the base which was used to produce the given percentage), and the quotient mil be its base factor (the multipli- cand which was used to produce the given percentage). EXAMPLES FOR PRACTISE 1. $63 are 7 times how many dollars? 2. 72 gallons are .09 times how many gallons? 3. 42 yd. are 6% of (.06 times) how many yards? 4. 252 tons are 45 % of what weight? 5. 28 % of how much vinegar = 150.64 gal.? 6. 24 % of what sum of money = $2039.10? 7. |% of what base = $23.82? 8. }% of how much money = $116.91? 9. $633.45 are 64 % more than how much money? 10. 24 % more than what sum of money = $928.45? 11. 52% less than what sum of money = $616.20? 12. 187 10s. are 76 % less than how much? 13. After selling 414 tons of coal, a dealer found that he had 52 % of his original stock remaining. How many tons did he have at first? 14. A merchant paid $247.25, which was 43 % of his cred- itor's claim against him. How much of the claim remained unpaid? 15. A's farm contains 783 acres of land, which is 8% more than the number of acres in B's farm. How many acres does B's farm contain? PERCENTAGE 139 16. A owns 12 % of the stock of a corporation, B 15 %, C 20 %, D 25 %, and E the remainder. What is the value of B's stock, if E's is worth $7084? \ 17. A's house is worf^i 46% more than B's farm. What is the value of B's farm, if A'a house is worth $9855? 18. A man purchased a building lot for $4800, erected a house thereon which cost $5600, expended $2050 in painting and deco- rating the house, $510 in furnishing it, and had 28 % of his money remaining. How much did he have at first? 19. By an unfortunate investment a merchant lost $16872, which was 37 % of his original capital. What was his subsequent capital? 20. A dealer sold an invoice of goods for 26% more than he paid for them, and received $818.37. How much did the dealer pay for the goods sold? COMMON APPROXIMATIONS IN PERCENTAGE 204. INTRODUCTION, (a) After obtaining a denominate per- centage by 201 or a denominate base by 203, it is customary among business men to reduce the decimal part of the result to the form of a common fraction as in 125. If no special reason exists for reducing the decimal to a common fraction of a partic- ular denominator, it is usually reduced to 8ths. After thus reducing the decimal to 8ths, the obtained fraction should be expressed in lowest terms, if not already so. (b) In some lines of business a special denominator is adopted which expresses the number of units of the next lower denomina- tion used in that kind of business which equal 1 integral unit of the obtained result. Thus, coal dealers reduce decimal parts of a result in tons to the denominator which expresses the num- ber of pounds (the next lower denomination to tons used in the coal business) which equal 1 ton, that is, to 2000ths, if short tons, or to 2240ths, if long tons; grain dealers reduce deci- mal parts of an obtained result in bushels to the denominator which expresses the number of pounds (the next lower denom- ination to bushels used in the grain business) which equal 1 bushel of that particular kind of grain, that is, to 60ths if wheat, to 56ths if corn, etc. In the same manner fractional parts of other units of merchandise are expressed with still other denom- inators. 140 PERCENTAGE ILLUSTRATIVE EXAMPLES 1. Find 8 % of 419 bu. wheat. SOLUTION 419 EXPLANATION. First find 8 % of 419 bu. by 201, Qg obtaining 33.52 bu. Retain the integral part of the result (33 bu.) as a part of the final answer, and DU. 33.52 reduce the decimal part (.52 bu.) to pounds (GOths) 60 by 125, obtaining 31 + lb., thus making the com- lb. 31 20 plete answer 33 bu. 31 lb., usually written 33 31 bu. Ans. 33 bu. 31 lb. 2. A dealer sold 127 gal. molasses which was 43 % of all the molasses he had in stock. How many gallons did he have remain- ing, reducing any decimal of a gallon to 8ths (pints)? SOLUTION 43)127.00(295| gallons, (base) EXPLANATION. Divide the speci- gg fie percentage sold (127 gal.) by the 7 specific rate % sold (.43), obtaining 295 gal. as the indicated base, or all 387 295 f gal. at first the molasses he had in stock, and a 230 127 gal. sold remainder of 15 (if gal.). Multiply . . this remainder (15) by the required 215 168 gal. remaining denominator (8ths) and divide the 15 result (120 eighths) by the original g divisor (43) obtaining 2 eighths and more than half of another eighth, or 43)120(3 approximately = practically } [86], making the com- 129 plete base 295f gal. EXAMPLES FOR PRACTISE 1. A man bought 765 bu. potatoes and sold 52 % of his pur- chase. How many bushels and pounds (GOths) did he have remaining? 2. A coal dealer bought 163 tons of coal and sold 37 % of his purchase at one time, and 45% of the remainder at another time. How many tons and pounds (2240ths) did he have finally remaining? 3. A grain dealer sold 32 % of his stock of corn and had 416 bushels remaining. How many bushels and pounds (56ths) of corn did he sell? PERCENTAGE 141 4. A merchant sold 15 % of an invoice of muslin and had 473 yards remaining. How many yards of muslin were in the invoice, expressing any resulting fraction of a yard in 4ths? 5. A man sold 785 bu. of oats, which was 62 % of his original purchase. How many bushels and pounds (32nds) did he have remaining? 6. The distance between two cities is 268 miles. If a man has traveled 53 % of the distance, how many miles and yards (1760ths) has he yet to travel? 7. A woodman delivered 173 cords of wood, which was 72% of the amount specified in his contract. How many cords and cord feet (8ths) remained to be delivered? APPROXIMATIONS OF RATES % 205. In the practical application of 202 to business calcula- tions, many difficulties will present themselves, methods for sur- mounting some of which will now be considered. (a) When results are obtained in United States money, they are expressed approximately to the nearest cent. This will fre- quently cause embarrassment to the learner. Thus, in a business calculation, 5% of $28.37, or $1.4185, would be entered in the books of account as $1.42. If occasion should afterward arise to learn what % of $28.37 are $1.42, the exact result (5%) cannot be obtained by the usual process. The insignificant remainder (.0015) when compared with the divisor SOLUTION I (28.37) would at once remind the experi- $28.37)$!. 4200(.05 enced calculator that it was caused by count- 1 4185 ing a fraction of a cent more than a half 15 (.85 of a cent) as a whole cent, thus making the dividend ($1.42) .15 of a cent greater than the exact expression, and causing the remainder of .15 of a cent when the exact quotient (5%) is found. (6) On the other hand, if in a practical calculation, it is re- quired to take 36% of $78.59, the exact result will be $28.2924, and its practical expression $28.29. If for any reason, it is subsequently necessary to learn what rate % had been em P loved to obtain this percentage, the exact result will not be secured, but only 35% with a very 4 7130 significant remainder. This significant re- 3 9295 mainder (7835) when contrasted with the 7835 divisor (7859) should remind the calculator 142 PERCENTAGE that the remainder was caused by discarding from the percentage ($28.29) a fraction of a cent less than a half (.24 of a cent), thus making the percentage lack .24 of a cent of being exactly 36% of the base, and causing the remainder (.7835) to lack .24 of a cent of exactly containing the divisor 1 more time than 35%, or 36%. (c) Again, it may be necessary to take 6^% of $49.32, of which the exact percentage is $3.0825, or, practically expressed, $3.08. If, afterward, it should be- SOLUTION III come necessary to find what % had $49 32) $3 0800 (6 1( 7 been used to obtain $3.08, the result 2 9592 would be 6% and an intermediate re- mainder (that is, neither an insignifi- ^ cant remainder as in a, nor a significant * remainder as in b, but intermediate 9664(; = f = f between the two) of .1208. The cal- 4932 culator should then at once conclude 4732 that there was a fraction of a per cent used in the rate. The fractional rates in common use are halves, fourths, eighths, and thirds. Of these, the last (thirds) will immediately suggest itself by being associated with 16, 33 and 66 in forming the frequently used commercial rates 16f %, 33J%, and 66f %. If thirds are not thus suggested as in the present instance, the remainder (1208) should be reduced to eighths, the highest multiple of the remaining commercial denomi- nators, by 86, obtaining 1 eighth and the significant remainder 4732 which lacks 200 (8 times the discarded $.0025) of being an- other eighth, making f , or in lowest terms i; thus showing that the rate used was exactly 6J%. NOTE. If in c, thirds are suggested by association with 16, 33 or 66, pupils are cautioned against accepting the probability for a certainty. The probability should be tested, and only an insignificant remainder as in Solu- tion 1, or a significant remainder as in Solution 2, can be accepted as con- clusive. The correct fractional rate will never leave an intermediate remainder. SUMMARY. In finding rates %, insignificant remainders should be disregarded; significant remainders which approximate closely to the base require the last quotient figure to be increased by 1: and intermediate remainders, when thirds are not suggested, should be reduced to eighths and the result expressed in lowest terms. NOTE. If a base expresses many thousands of dollars as in great whole- sale transactions, or millions of dollars as in many calculations of railroad companies and other great corporations, rates are usually carried to five deci- mal places, the first two places after the point expressing %, and the following three places denoting thousandths of a per cent. PERCENTAGE 143 EXAMPLES FOR PRACTISE What % of What % of 1. $374.86 are $157.44? 5. $9248 are $3479.56? 2. $968.43 are $377.69? 6. $9.65 are $4.04? 3. $758.65 are $216.22? 7. $67841.35 are $42740.05? 4. $4832.72 are $4687.74? 8. $6873.24 are $2019.01? 9. The gross earnings of a railroad for the year 1910 were $1375487.60, and its operating expenses $296785.37. What % of the gross earnings were the operating expenses (to 5 decimal places)? 10. A wholesale firm extended credits to the amount of $3468756.75 during the year 1910, of which it failed to collect $12378.40. What % of its credits did^it collect (to 5 decimal places) ? BASES UNDERSTOOD 206. INTRODUCTION. The idea expressed by the term "base" as employed in percentage differs little from that of the same word in ordinary usage. Thus, the base of any object, say a monument, is the foundation or pedestal, upon which it rests, and without which it cannot stand. So in a percentage problem, if the base is not directly indicated, it is " understood" to be the entire quantity of which the rate % represents a part, and of which the percentage is a part, and therefore that upon which both these terms depend for their value as a monument depends upon its pedestal for its stability. A base which is thus comprehensible is sometimes called a "normal base" to distinguish it from an "arbitrary base" which has no special characteristic by which it can be identified, but which depends simply upon the arbitrary choice of the author of the problem in selecting it as such. A normal base is therefore the quantity from which the percentage naturally derives its origin, that of which one would readily think as the whole while considering any part of it, that which would naturally suggest itself to the mind as the answer if the interroga- tive "of what" be placed immediately after the rate %. NOTE. If there is a possibility of the slightest doubt relative to the identity of the base, it should be indicated by placing of, more than, or less than, or words of equivalent import, immediately after the rate % and immediately before the otherwise doubtful base [198, a, b, c]. 144 PERCENTAGE EXAMPLES FOR PRACTISE 1. A man's income is $1800 per annum. If he pays 20 % for board, 15 % for clothing, and 25 % for incidentals, how much does he save? 2. A town has a population of 10500, 55 % being native born, 20 % Germans, 15 % Irish, and the remainder of other national- ities. How many of each does the town contain? 3. A man bought a house for $6500, paying $1800 in cash, and agreeing to settle the remainder in annual payments of 25 %. What was the amount of each annual payment? 4. A man died leaving property of the value of $13500. He bequeathed 45% to his wife, 35% to his only child, and the remainder to a hospital. How much did the hospital receive? 5. A farm of 105 acres is only 28 % as large as the farm adjoin- ing it. How many acres in the adjoining farm? 6. If coffee loses 15 % in roasting, how many pounds of green coffee must be bought to produce 1020 pounds when roasted? 7. A is worth $38500 and B only 47 % as much. What is B worth? 8. A man weighed 190 pounds before an attack of illness, and 133 pounds directly after the attack. What % did he lose in weight? PERCENTAGES OF INCREASE OR DECREASE 207. INTRODUCTION. No increase or decrease is possible, except there be first, as a basis for such a modification, a quantity to be augmented or diminished. Hence, in all problems of this character, the normal base always understood is the quantity to be increased or diminished, that is, the quantity immediately before such a change has been effected; and the normal percentages always understood are the increase or decrease itself, and the quantity after such a change has been made. Most of the applications of percentage from 209 to 291, in which bases are understood, depend upon this important principle in the proper selection of the base; profit being but another name for increase, loss for decrease, the various discounts (bank, true, or trade) for de- creases, etc., etc. PERCENTAGE 145 NOTE. Increase means more than, and decrease less than [198, 6]. Hence, when in doubt, the rational answer to the interrogative "more than what" placed after a rate of increase, or "less than what" placed after a rate of decrease, will suggest the base understood. EXAMPLES FOR PRACTISE 1. At the census of 1900, a city had a population of 37240, and at the census of 1910, its population was 50274. What was the % of increase? 2. During a financial depression, a clerk had his salary dimin- ished 15 %, or $300. What was his salary after the reduction? 3. By a rise in real estate, a house has increased in value 28 %, and is now worth $10880. What was its value before the rise? 4. A firm's gross sales last year were $287300, and this year $333268. What was the % of increase? 6. By introducing new machinery a factory has increased its output 16f %, and now produces 182753 yards of sheeting per week more than before. What was the capacity of the factory before installing the new machinery? 6. The capital of a firm was increased 20 % in 1910, and 15 % in 1911, the total increase during the two years being $17406. What was its capital on Jan. 1, 1910? 7. The net earnings of a corporation were diminished 25% during the financial depression of 1907, and still further dimin- ished 10 % during the year 1908, but during the year 1909, under improved conditions, they were increased 20 %. If the net earnings for 1909 were $268764.50, what were the net earnings during 1907? 8. If the net capital of a firm was $45000 on Jan. 1, 1908, and there was a net increase of 10 % per annum for four success- ive years, what was its capital on Jan. 1, 1912? REVIEW OF PERCENTAGE 208. 1. A merchant sold 43 % of his stock of millinery for $6875.35. What was the value of the remainder of his stock of millinery? 2. A and B are partners, A having invested $19161 and B $23419. What % of the capital of the firm did each partner invest? 146 PERCENTAGE 3. A cistern now contains 741 gallons of water, 24 % having leaked out. How many gallons did it originally contain? 4. A clerk spent $676.50 of his annual salary, and saved $523.50. What % of his salary did he spend? 5. If A's age is 25 % greater than B's, what % of A's age is B's? 6. A is worth $45000 and B $30000. What % more than B is A worth? What % less than A is B worth? 7. A man inherited $74500, invested 45 % of it in real estate, 25 % of it in United States bonds, and loaned the remainder to a friend. What was the sum loaned? 8. A bakery used 425 pounds of flour in making 748 loaves of bread, each loaf weighing 12 ounces. What % more than the flour did the bread weigh? 9. A owns 12 % of the stock of a corporation, B 15 %, C 20 %, D 30 %, and E the remainder. If B's investment is $3600, what is E's? 10. A man sold 40% of a tract of land at one time, 20% of the remainder at a second sale, 25 % of what still remained at a third sale, and then had 234 acres finally remaining. How many acres did the original tract contain? 11. In 1910, a merchant's sales amounted to $185920, and in 1911, his sales were $158032. What was the per cent, of decrease in 1911? 12. A certain property has depreciated $3375 in value, and is now worth $9125. What was the % of depreciation? 13. A man bought one house for $4500 and another house for 42 % more. What was the cost of both? 14. On Sept. 23, 1911, merchandise was bought for $1340 on 3 months' credit, and subject to a discount of 3 % if paid within 10 days. What is the proper payment on Oct. 1, 1911? 15. A firm's business was placed in the hands of a receiver who discovered that its total resources after deducting the costs of receivership were $28500, and that its total liabilities were $37500. What % of the firm's indebtedness can the receiver pay to the creditors? 16. Two railroads carried 8260 pounds freight at the through PERCENTAGE 147 rate of 28 cents per cwt. If one company was entitled to 37 % of the through charge, how much should the other company receive? 17. Two railroads, one 200 miles long and the other 300 miles long, carried 750 barrels flour at the through rate of 18 cents per barrel, which was to be divided pro rata between them (that is, in proportion to each road's per cent, of the total mileage). What % of the total freight charge should each road have received? What percentage? 18. Three railroads respectively 200 miles, 250 miles, and 350 miles in length, transported 28375 pounds freight at the through rate of 28 cents per cwt., which was pro-rated among them in ac- cordance with a general traffic agreement. What % of the total distance did each road carry the through freight? What per- centage of the freight collection should each road receive? 19. The total cost of an arithmetic and a geography is $2.10, and the arithmetic cost 32 % less than the geography. What is the cost of each? EXPLANATION. The required cost of the geography is indicated as the verbal base or multiplicand by being placed immediately after " % less than." $2.10 is a percentage because it is compared with (by being expressed in terms of) the indicated base. As $2.10 is the joint cost of the geography (the indi- cated base) plus the cost of the arithmetic (a percentage), its specific joint rate must be the specific rate of the geography or base (100 % of the cost of the geography) plus the specific rate of the cost of the arithmetic or percentage (100 % - 32 %, or 68 % of the cost of the geography), that is, 168 % of the cost of the geography. Hence, divide the joint percentage or product ($2.10) by its joint rate % or multiplier (168 %), obtaining $1.25 as the common base or multiplicand (the cost of the geography). Then, the cost of both ($2.10) minus the cost of the geography ($1.25) should equal the cost of the arithmetic ($.85). Or, the obtained common base or multiplicand ($1.25) multiplied by the specific rate or multiplier for the arithmetic (68 %) will produce the specific percentage or product of the arithmetic ($.85). 20. A man bought a corner-house and the house adjoining it for $7752, paying 28 % more for the corner-house than for the one adjoining it. How much did he pay for each? 21. A, B, C, and D are partners. A's investment is 20% greater than B's, C's investment is 10% less than A's, and D's 148 PERCENTAGE investment is 75% of C's. If the capital of the firm is $53579, what is each partner's investment? NOTE. If two or more bases are indicated in the same problem, accept the initial or root base as the base of the solution, and regard the remaining derivative bases as so many percentages of the root base. There can be no consolidation of two or more specific rates or multipliers except when they are expressed upon one common base or multiplicand [Prin. 1, 18]. 22. A has 25% more money than B, B 28% more than C, C 40 % less than D, and D 25 % as much as E. If all five together have $641.20, how much has each? 23. The sales of a firm were increased 20% per annum for three successive years. If the total increase during the three years was $36400, what were the sales during the second year? 24. The gross earnings of a railroad company during the year 1910 were $6781345.30, of which $2346785.90 was expended for wages, $385465.20 for repairs, $275864.20 for fuel, and $1753285.90 for fixed charges. What % of the gross earnings was each of the above charges (to five decimal places)? 25. A commission merchant sold 10% of a consignment of wheat at one sale, 30 % of the remainder at a second sale, 40 % of what still remained at a third sale, and then had 1701 bushels remaining. How many bushels of wheat were in the consignment? 26. A owned 20% of the capital stock of a steamboat com- pany, and sold 40 % of what he owned for $4800. At the same appraisement, what was the value of the entire capital stock? PROFIT AND LOSS 209. INTRODUCTION. Profit and Loss problems are such percentage problems as have special reference to gains or losses in selling. While differing in no respect from other percentage problems, they are grouped together for special treatment because of their relative commercial importance, and the consequent necessity for acquiring thorough familiarity with the general principles of percentage as applied to them, and for cultivating special skill in computing them. 210. The cost is the sum paid for the thing purchased. NOTE 1. The prime cost of merchandise is the first cost, that is, the cost before any expense connected with the purchase, such as freight from place PROFIT AND LOSS 149 of purchase, drayage, custom-house charges, commission on purchase, insur- ance, storage after purchase, etc., has been added. The gross cost is the last cost, that is, the first cost increased by all expenses incidental to the purchase from the moment of purchase to the moment of sale. NOTE 2. When the term cost is used without a qualifying word, the actual cost, or gross cost, is always understood. 211. The selling price is the sum received for the thing sold. NOTE 1. The gross selling price of merchandise is the total sum received from the sale, before any expense connected with the sale has been deducted. The net selling price is the gross selling price diminished by all incidental expenses incurred after the moment of sale, such as commission on sales, dis- counts allowed, free delivery charges, etc. NOTE 2. When the term selling price is used without a modifying adjec- tive, the actual or net selling price is understood. 212. The profit or gain is the excess of the net selling price above the gross cost. 213. The loss is the deficiency of the net selling price below the gross cost. 214. Relation of terms. 1. A gross cost is a base or multi- plicand. 2. A % of profit, of loss, or of selling price, is a rate or multiplier. 3. A profit, a loss, or a selling price, is a percentage or product. NOTE 1. Profit means increase (more than cost) and loss means decrease (less than cost). Calculations in profit or loss are therefore applications of 207. As the gross cost is uniformly increased by the profit, or diminished by the loss to find the net selling price, the gross cost is a normal base or standard of comparison, and does not require to be indicated as is necessary for arbi- trary bases. "Of gross cost" is therefore understood after every rate of profit or of loss, when no other base is arbitrarily indicated. NOTE 2. Since the selling price equals the cost plus the profit, or the cost minus the loss, the specific rate of selling price must equal the specific rate of cost (100 %) plus the specific rate of profit, or minus the specific rate of loss. [Note 1, 201.] EXAMPLES FOR PRACTISE RULE. If the cost (base) and a rate are given, multiply the cost by the specific rate of the required profit, or loss, or selling price. Goods cost 1. $57.80 and are sold at 25 % profit. Find profit. 2. $183.75 and are sold at 18 % loss. Find loss. 3. $275.80 and are sold at 28% profit. Find profit. 150 PERCENTAGE Goods cost 4. $628.50 and are sold at 20 % loss. Find loss. 5. $342.16 and are sold at 16f % profit. Find profit. 6. $486.25 and are sold at 33J % loss. Find loss. 7. $254.85 and are sold at 35 % profit. Find selling price. 8. $685.90 and are sold at 25 % profit. Find selling price. 9. $816.35 and are sold at 10 % profit. Find selling price. 10. $723.40 and are sold at 5 % loss. Find selling price. 11. $374.85 and are sold at 20 % loss. Find selling price. 12. $436.95 and are sold at 12? % loss. Find selling price. 13. $783.24 and are sold at 45 % gain. Find gain. 14. $294.35 and are sold at 18 % loss. Find loss. 15. $385.16 and are sold at 24 % gain. Find gain. 16. $416.28 and are sold at 16 % loss. Find loss. RULE. If the cost (base) and a profit or loss or selling price (percentages) are given, divide the specific percentage or product of which it is required to find the rate, by its cost or multiplicand factor, and the quotient carried to hundredths, will be its specific rate % or multiplier factor, applying a, b, or c, 205, when the division is inexact. Goods cost 17. $38.75 and are sold at $9.69 gain. Find % of gain. 18. $68.50 and are sold at $10.96 gain. Find % of gain. 19. $872.50 and are sold at $401.35 gain. Find % of gain. 20. $497.40 and are sold at $89.53 loss. Find % of loss. 21. $26.25 and are sold at $2.10 loss. Find % of loss. 22. $63.85 and are sold at $22.99 loss. Find % of loss. 23. $342.97 and are sold for $425.28. Find % of gain. 24. $682.25 and are sold for $777.77. Find % of gain. 25. $359.34 and are sold for $294.66. Find % of loss. 26. $458.37 and are sold for $401.07. Find % of loss. 27. $178.15 and are sold for $213.78. Find % of gain. 28. $728.35 and are sold for $591.78. Find % of loss. 29. $563.27 and are sold at $93.88 gain. Find % of gain. 30. $94.65 and are sold at $17.98 loss. Find % of loss. RULE. To find the cost, divide any given percentage or product of profit, of loss, or of selling price, by its specific rate or multiplier factor. 31. The gain is $48.65 and the rate of gain 25%. 32. The gain is $150.00 and the rate of gain 32 %. PROFIT AND LOSS 151 Find the cost, if 33. The loss is $346.05 and the rate of loss 45 %. 34. The loss is $168.65 and the rate of loss 20 %. 35. The selling price is $487.97 and the rate of gain 16f %. 36. The selling price is $607.88 and the rate of loss Yl\ %. 37. The selling price is $423.72 and the rate of gain 15%. 38. The selling price is $52.88 and the rate of loss 33%. 39. The selling price is $348.70 and the rate of gain 37J %. 40. The selling price is $58.65 and the rate of loss 6j %. 41. The selling price is $105.19 and the rate of gain 34%. 42. The selling price is $406.50 and the rate of loss 35%. 43. The selling price is $519.86 and the rate of gain 10%. Find gain. 44. The selling price is $130.58 and the rate of loss 15%. Find loss. 45. Merchandise was sold for $48.40. If the gain was $9.68, what was the rate % of gain? 46. Groceries were sold for $62.68. If the loss was $15.67, what was the rate % of loss? 47. A lot of hardware was sold at 25% gain. If the seller received $22.80, how much of it was profit? 48. A house was sold for $2125 by which a loss of 15% was sustained. How much less than cost was received by the seller? 49. A man bought two horses at $125 each. He afterwards sold one of them at 25 % gain and the other at 15 % gain. How much did he receive for the two horses? SUGGESTION. A common cost or multiplicand ($125) multiplied by the specific joint rate of selling price, or multiplier for the joint product (125 % + 115 %) = joint selling price or product. 60. A speculator in real estate bought three houses at $2400 each, and afterwards sold one of them at a profit of 35%, the second at a profit of 30 %, and the third at a profit of 15 %. How much did he receive for the three houses? 51. Two houses were sold for $2400 each, by which the seller gained 20 % on one house and lost 20 % on the other. Did the seller receive what he paid for the two houses? If not, did he gain or lose, and how much? 152 PERCENTAGE 62. Three farms were bought for $3500 each, and they were afterwards sold respectively at 25% gain, 30% gain, and 18% loss. How much was received for the three farms? 53. A man bought a tract of land for $4500, paid $625 addi- tional for improvements, and then sold it for $6150. What was his % of gain? 54. A farmer bought 48 acres of land for $3600, and afterwards sold 32 acres from his purchase at a profit of 25%. For what sum per acre must he sell the remainder of the purchase to net an average profit of 20 % on the entire purchase? 55. Allowing for a loss of 12 % in roasting, what should be the selling price of roasted coffee which was bought unroasted at 20 cents per pound, to net a gain of 10% [207]? 56. An instalment dealer bought furniture at $18 per set. What must he ask per set to net a profit of 25 %, after allowing 20 % for bad collections? 57. A wholesale fruit dealer bought a vessel load of water- melons numbering 12000 at $16 per hundred, and classifying them according to size and quality, sold 5000 of them at $25 per hun- dred, 3000 at $20 per hundred, 2500 at $18 per hundred, 1000 at $15 per hundred, and the remainder for $46. What was his % of profit on the entire purchase? 58. A merchant sold 30 % of a purchase at 25 % profit, 60 % of the remainder at 20 % profit, and what still remained at 10 % profit. What was his % of gain on the entire purchase? 59. To increase their capital, a firm agreed to permit one- half of the profits to remain hi the business for three successive years. The net gains during the first year were 20 %, during the second year 25%, and during the third year 10%. If the cap- ital of the firm at the close of the third year was $31185, what was its capital at the beginning of the first year? 60. A firm sells goods in its wholesale department at 12% profit, and in its retail department at 25 % advance on its whole- sale selling price. What % profit upon original cost is made in the retail department? 61. If 20% of an invoice of fruit is spoiled in shipment, at what % gain must the remainder be sold to net 10 % profit on the original cost? PROFIT AND LOSS 153 62. A dealer sold merchandise for $48.99 and gained 15%. What should have been the selling price to gain 25 %? 63. A retail grocer bought 5 barrels of flour at $6.20 per barrel, paid 25 cents drayage to have the purchase brought to his store, and afterwards sold one barrel to a customer for $7.75 and paid 25 cents drayage to have it delivered at his customer's resi- dence. What was the grocer's rate % of profit? 64. If the sales during the year were $60191, the inventory of stock on hand at end of year $9625, and the gain $7851, what were the total purchases, and the % of gain? 65. If the purchases during the year are $35798.56, the inven- tory of stock on hand at end of year $6500, and the rate of gain 12J %, what are the total sales? 66. The books of a manufacturer for the year 1911 show the following costs: productive material $200000, productive labor $300000, overhead expenses $100000. What % of the manufac- turing cost is each element of expense? [See Suggestions following.] SUGGESTIONS, (a) Manufacturing or factory cost consists of the follow- ing elements: productive labor, productive material, and "overhead" factory expense. (6) Productive or direct labor is labor which is employed directly upon the product manufactured to sell. The labor of the factory superintendent foremen, bookkeepers, clerks, porters, etc., is called non-productive or in- direct labor. (c) Productive material is material which is used directly in the articles manufactured to sell. Material used for general purposes around the factory, as repairs, etc., is called non-productive material. (d) Overhead factory expense includes non-productive material; non- productive labor; and any general factory expense as heat, light, power, etc., except such as is incident to selling the manufactured product. 67. In the manufacture of a certain make of bicycles, the pro- ductive labor was $82500, productive material $71000, and over- head expense $23000. What % of the manufacturing cost is the productive labor? Is the productive material? Is the over- head expense? 68. Factory order No. 16713 for 100 sewing machines shows the following costs: direct labor $500, productive material $600. Estimating the overhead expense at 40% of -the direct labor, what was the manufacturing cost of each sewing machine? 154 PERCENTAGE NOTE 3. In manufacturing cost calculations, though not the uniform practise, it is customary to distribute the indirect, unproductive or overhead expense among the units of product at an estimated % upon the direct labor employed in the production of such units. 69. In a certain factory, shop order No. 2519 shows consump- tion of productive materials $83.67, direct labor, $41.83. Esti- mating the overhead expense at 40% of the productive labor, what is the factory cost? 70. Shop order No. 1627 of Monumental Machine Works includes the following costs: direct labor $10248, productive mate- rial $3722, estimated overhead factory expense 25% of direct labor, estimated distributing expense, shipping, etc., 3 % on total sales. What is the selling price to net 25 % profit on factory cost? 71. Estimating 50% of factory cost for direct labor, what should be the selling price of 75 units of product at 16 f % profit, if the direct labor per unit is $48? 72. The annex for storing the raw materials of a factory with its contents and records pertaining to the same, are destroyed by fire. The general office records show stores on hand at last inventory $20000, purchases since last inventory $60000, manufactured and sold since last inventory $96000, labor and material on unfinished product in the factory at the time of the fire, $14400. The manufacturing costs are esti- mated at 50 % for direct labor, 30 % for productive material, and the remainder for overhead expense. If the manufacturer's selling price is 50% advance on factory cost, what was the value of the raw material in the annex at the time of the fire? . TRADE DISCOUNT 215. INTRODUCTION. Problems in Trade Discount are appli- cations of the general principles of percentage to calculations which have special reference to deductions made by manufacturers and jobbers and merchants upon sales of a certain magnitude to stimulate large purchases; or as a compensation to middle- men for handling merchandise to sell at a stipulated list price; or as an absolute allowance to purchasers on "time prices" if paid for at any time within a given period. 216. The gross price is the price before any deduction has been made. TRADE DISCOUNT 155 217. The trade discount is the deduction which has been made from the gross price. 218. The net price is the price after all deductions have been made. 219. Identification of terms. 1. A gross price is a base or multiplicand. 2. A % of trade discount or of net price is a rate or multiplier. 3. A trade discount or a net price is a percentage or product. NOTE 1. Discount (from dis, away; count, compute) means counting away. In taking away from a quantity, it is decreased. Hence, problems in trade discount are applications of 207, relating to decreases; and the quantity diminished (the gross price) is the normal base always understood; and the trade discount (the decrease), and the net price (the quantity after the de- crease) are percentages. NOTE 2. If two or more rates of discount are allowed, the gross price is the root base upon which the first rate of discount is computed; and the pro- ceeds of the first discount should be taken as a derivative base upon which the second rate of discount is computed; and the proceeds of the second discount should be regarded as a second derivative base upon which the third rate of discount is computed; etc., etc. GROSS PRICE GIVEN, TO FIND THE NET PRICE ILLUSTRATIVE EXAMPLE 220. The gross price is $278.56, upon which a discount of 28 %, 20 %, and 3 % is allowed. Find the net price. SOLUTION EXPLANATION. Multiply $278.56, gross price or root base ^J?^f? u C l r ^ b T ' b ^ ($278.56) by the specific rate 72, (100% 28%) O f proceeds of the first dis- 55712 count (100 % -28% = 72 %) 194992 to ^ nc * tne P rocee ds of the first discount ($200.5632). $200.5632, proceeds of 1st discount Then mu i t i p i y the obtained 40.1126, 20 % or J of $200.563 derivative base ($200.5632) by $160.4506, proceeds of 2d discount the second rate of discount < 20 4 R1V5 ^ <7 of 1*160 4^06 %== ^ obt aming$40.1126, and 4 ^ lt3 ' d % C subtract, leaving $160.4506 as $155.6371, proceeds of 3d discount the proceeds of the second dis- count. Then multiply the last obtained derivative base ($160.4506) by the third rate of discount (3 %), obtaining $4.8135, and subtract, leaving $155.64 as the proceeds of the third discount, or required net price. 156 PERCENTAGE NOTE 1. An aliquot rate is one which is an exact divisor of 100 %, as 20 % in the 111. Ex.; and a non-aliquot rate is one which is not an exact divisor of 100 %, as 28 % in the 111. Ex. Aliquot rates of discount are more conveniently handled by first finding the discount and then subtracting; but non-aliquot rates of discount, by at once multiplying the base by the specific rate of net price, as by 100 % - 28 %, or 72 %, in the 111. Ex. NOTE 2. A rate of discount consisting of a single figure, as 3 % in the 111. Ex., is best computed as follows: Commencing with the cents' order of the base, multiply it and each of its successive higher orders by the rate of discount, and write the successive orders of the product two places to the right of the orders multiplied, first multiplying the rejected orders at the right of the cents' order of the base to obtain the correct carrying figure to the retained product. NOTE 3. All results should be carried to four decimal places, to insure accuracy in the cents' order of the final result. NOTE 4. A fraction of a half or more of the lowest retained order is not usually counted in computing a discount (which is against the seller), and is counted as another unit of the lowest retained order in computing a net price (which is in favor of the seller). Not counting in the former instance is equivalent to counting in the latter. EXAMPLES FOR PRACTISE Find the net price Find the net price Gross Price Rates of Disc't Gross Price Rates of Disc't 1. $375.85 . . 25 %, 10 %, 5 % 6. $285.94 . . 32 %, 24 %, 5 % 2. $582.90 . . 25 %, 20 %, 10 % 7. $723.56 . . 25 %, 16 %, 9 % 3. $873.62 . . 10 %, 10 %, 10 % 8. $938.25 . . 16f %, 12J %, 2 % 4. $462.75 . . 20 %, 10 %, 7 % 9. $427.36 . . 10 %, 7 %, 1 % 5. $629.38 . . 10 %, 5 %, 3 % 10. $246.78 . . 5 %, 3 %, 2 % 11. Goods are invoiced at $658.90 and 25 %, 5 %, and 3 % off. What is the net cost to the purchaser? 12. Bought 2 doz. pocket knives, No. 278, at $4.50 per dozen; \ doz. butcher knives, No. 715, at $7.50 per dozen; and 3 doz. locks, No. 645, at $3.75 per dozen. What was the net cost, after allowing a discount of 20 %, 5 %, and 2 %? 13. Bought May 16, 1912, a bill of merchandise amounting to $738.65 gross and payable in 90 days, upon which a merchan- dise discount of 10 %, 10 %, and 5 % was allowed, with a time discount of 3 % if paid within 10 days. What should be the payment if made in full on May 25, 1912? TRADE DISCOUNT 157 REDUCTION OF RATES OF DISCOUNT ILLUSTRATIVE EXAMPLES 221. 1. Reduce the discount series of 20%, 10%, and 5% to an equivalent single discount. EXPLANATION. Sub- SOLUTION 100 % = gross price discount (20% of the gross 20 % = 1st discount price) from the rate of ~80 % = proceeds of 1st discount e ross price (100 %) to find , * . . ~ , ,. the rate of 1st proceeds ( T V of 80) _8 % = 2d discount (8Q % of the ^ price) 72 % = pro. of 2d discount Next subtract the (?V of 72) 3.6% = 3d discount rate of the 2d disc't tf ff 6M % - j pro. of 3d discount JJ^ J^ * { or rate of net price price ) from the pr0 ceeds 100 % - 68.4 % = 31.6 %, single discount of the 1st disc't (80 % of the gross price) to find the proceeds of the 2d disc't (72 % of the gross price). Next subtract the rate of the 3d disc't GV of 72 % of the gross price or 3.6 % of the gross price) from the proceeds of the 2d disc't (72 % of the gross price) to obtain the rate of proceeds of the last discount (68.4 % of the gross price). The difference between 100 % of the gross price and the rate of final pro- ceeds (68.4 % of the gross price), or 31.6 % of the gross price must therefore be the total of the three discounts expressed in a required single discount upon the gross price. 2. Reduce the discount series of 32 %, 24 %, and 16 % to an equivalent single discount. SOLUTION EXPLANATION. If 32 % of the gross price (100 %) is taken off, 100 % 32 % = 68 % then 68 % of the gross price will 100 % - 24 % = 76 % remain. 100 % 16 % = 84 % If 24 % of what remains of the 68 X 76 X 84 = 434112 gross price is tnen taken off > 76 % 100 % -- 43.4112 % - 56.5888 % If 16 % of what yet remains of the gross price is finally taken off, then 84 % of 76 % of 68 % of the gross price, or 43.4112 % of the gross price, will finally remain as the rate of net price. Then subtract the rate of final net price (43.4112 %) from the rate of the original gross price (100 %) to find the rate of total discount (56.5888 %). 158 PERCENTAGE NOTE 1. The solution of 111. Ex. 1 is preferable, if the rates of discount are aliquots [Note 1, 220]; but when the rates of discount are non-aliquots, the solution of Ex. 2 will be the more convenient. If the same problem con- tains both classes of rates, employ the method of Ex. 1 for aliquot rates, and of Ex. 2 for the non-aliquot rates. Note 2. The value of a series of discounts will not be changed by any variation in the order of their arrangement. Thus, a discount series of 10 %, 5 % and 3 % upon any gross price will produce the same net price as 5 %, 3 % and 10 %, or as 3 %, 10 % and 5 %; for the product of (100 10) X (100 - 5) X (100 3) will be the same, irrespective of the order in which they may be multiplied. EXAMPLES FOR PRACTISE Find the rate of net price, if the rates of discount are 1. 20 %, 10 % 4. 20 %, 20 %, 10 % 7. 35 %, 24 %, 8 % 2. 25 %, 10 %, 5 % 5. 25 %, 20 %, 10 % 8. 20 %, 15 %, 10 %, 2 % 3. 10%,5%,3% 6. 10%,10%,10% 9. 10%,5%,3%,1% Reduce the following to equivalent single discounts: 10. 25 %, 10 % 12. 20 %, 5 %, 3 % 14. 20 %, 10 %, 10 % 11. 20 %, 3 % 13. 10 %, 10 %, 5 % 15. 5 %, 3 %, 2 % TO FIND WHAT PER CENT. OF DISCOUNT WILL NET A GIVEN PRICE ILLUSTRATIVE EXAMPLE 222. What % of discount must be allowed to net $14.60, if the merchandise is marked $18.25? SOLUTION EXPLANATION. The solution of this ex- Si 8. 25 = gross price ample is an application of 202. Hence, 14.50 = net price Divide the specific percentage of which Q c _. rlioiin+ it is required to find the %, that is, the per- o.OO ui&uuunu <*Q ACAH t .7125 = rate of 2d proceeds net price or contin- (.7125 X .03) .021375 = rate of 3d discount ued multiplier fac- r . tor (.691125), found .691125 = rate of net price M ^ ^ 221 321 60 obtaining $7.60 as .691125)$5.2500($7.p^ the requ ired gross or 48379 marked price, its 4191 multiplicand factor. The short method for dividing by a 665 decimal divisor of g22 many places is fully i ^e f explained in 130. 43 = more than half of .69 The gmall in _ dex figures are placed over the lowest decimal order of the successive i 2 divisors; that is, the first divisor is .6911, the second .691, the third and 3 last .69. 160 PERCENTAGE RULE. Divide the specific percentage of net price (product) by the specific rate of net price (its multiplier factor), and the quotient will be the gross or marked price (its multiplicand factor). EXAMPLES FOR PRACTISE At what price must goods be marked to net 1. $36.50 after allowing a discount of 25 %? 2. $18.75 after allowing a discount of 15 % and 10%? 3. $62.50 after allowing a discount of 25 %, 10 %, and 7 %? 4. $213.63 after allowing a discount of 25 %, 20 %, 5 %, 3 %? 5. $85.42 after allowing a discount of 30 %, 20 %, 10 %, 4 %? 6. 25 % profit? Cost, $24.36. Discounts 10 %, 10 %, 5 %. 7. 16|% profit? Cost, $45.60. Discts., 20 %, 10 %, 3 %, 1 %. 8. 10 % loss? Cost, $16.85. Discounts, 10 %, 10 %, 4 %, 2 %. 9. 12J % loss? Cost, $9.60. Discounts, 25 %, 5 %, 5 %, 3 %. 10. What payment on May 28, 1912, will be in full of the following INVOICE BALTIMORE, May 20, 1912. Mr. R. M. BROWNING, Bought of F. A. SADLER & Co. Terms: 60 days, net; 10 days, 2 %. 3 doz Butcher Knives $4.50 ** ** 25 %, 20 %, 10 % * ** * ** $4.25 $5.75 I 7.50 2 " Pocket Knives ea. 345, 460, 624 ** ** 10 %, 10 %, 2 % * ** ** ** $6.50 $7.25 i it Hatchets ea 783 825 * ** $5.00 $6.25 $7.15 5 " Spades ea No 1 No 2 No 3 ** ** $4.50 $6.00 3 " Shovels ea. No. 1 No. 2 ** ** *** ** 50 %, 10 % ** ** ** ** ** ** Packing and Drayage 50 INTEREST 224. INTRODUCTION, (a) In the "Introduction" of 193, it was shown that Percentage is an application of the general prin- ciples which govern the relation of all quantities. These same general principles will now be shown to apply to quantities which include a hitherto unconsidered element, that of time. (6) Interest is a compensation for the use of money in the same way as rent is a compensation for the use of property, or wages for the use of personal labor. In all of these it is evident that the amount of the compensation will vary with the length of time they are used. Hence, the necessity of establishing a fixed unit of time by which to graduate the amount of compensa- tion, as wages at so many dollars per day, rent at so many dollars per month, and interest at so much % of the principal per year. (c) It is thus seen that the element of time is not a new term of percentage, but a necessary modifier of one of the three already existent terms (the rate) which is no longer absolute as before, but conditioned upon one year's use, as its specific name "rate per annum" implies. Hence, at 5% per annum, the correspond- ing rate for two years will be two times 5%, or 10%; for three years will be three times 5%, or 15%; for 73 days will be ? VV, or i of 5%, or 1%; for 146 days will be ^|f , or of 5%, or 2%, etc. 225. Interest is a compensation for the use of money. 226. The principal is the sum of money loaned, or the sum of money due but unpaid, for which a compensation is charged. NOTE. The sum of the principal and its accrued interest is called the amount. 227. The rate of interest is the % of the principal which is charged for using it one year. NOTE. The specific rate of interest for less than one year is such a part of the rate per annum, as the given time in days is part of the number of days in a year. Thus, counting 12 months of 30 days each, or 360 days to a year, as is ordinarily done in interest calculations, at 6 % per annum, it will be i of 6 %, or 1 % for of 360 days, or 60 days; and if 1 % of the principal is the rate of interest for 60 days, the rate of interest for twice 60 days, or 120 days must be twice 1 %, or 2 %; for of 60 days, or 15 days, must be | of 1 %, or \ %; and, generally, the specific rate of interest for any number of days at 6 % per annum is such a part of 1 % (the rate of interest for 60 days), as the given time in days is part of 60 days; or, stated in a form for convenient use, is equal to the quotient of the given days, regarded as hundredths, divided by 60. The 162 PERCENTAGE specific rate for less than 1 year at other rates per annum than 6 % is similarly obtained; that is, the specific rate of interest for any given time and at any given rate % per annum must equal 1 % of the given time expressed in days divided by such a part of 360 days as I % is part of the given rate per annum. Hence, if the specific rate of interest is sought for 216 days at 5 % per annum, divide 1 % of 216 da. (2.16 da.) by i of 360 da. (72 da.) obtaining 2.16 -=- 72, or .03; to find the specific rate of interest for 225 da. at 8 % per annum, divide 1 % of 225 da. (2.25 da.) by 1 of 360 da. (45 da.), obtaining 2.25 -5- 45, or .05; etc. 228. Simple interest is a compensation for the use of the original principal, or for any unpaid part of the original principal. 229. Ordinary interest is interest computed in the ordinary manner for less than one year upon the assumption that the entire year consists of 12 months of 30 days each, or of 360 days. 230. Identification of terms. 1. The principal is the base or multiplicand. 2. The specific % for the given time [Note, 227], is the rate or multiplier. 3. The interest or the amount for the given time is the percentage or product. PERCENTAGE METHOD FOR COMPUTING INTEREST AT 6% ILLUSTRATIVE EXAMPLES 231. 1. Find the interest of $685.74 at 6% for 193 days. EXPLANATION. $685.74 is a normal base understood [206] or the multiplicand [230, 1]; 6 % is the rate or mul- $685.74 tiplier, for one year's interest, but not the specific rate for jgg 193 da. At 6 % per annum of 360 da., the corresponding : - rate will be of 6 %, or 1 %, for of 360 da., or 60 da. The 205722 specific rate of interest for 193 da. must therefore be as 617166 many times 1 % of the principal as 60 da. are contained 68574 times in 193 da., or W of 1 % of the principal. As W of "^ of the P rinci P al e( l ua l *V<& of the principal, or as many thousandths of the principal as there are days, that is, | of $22.057 193 thousandths of the principal, therefore, Multiply the principal or multiplicand ($685.74) by 6 times the specific rate or multiplier (.193), obtaining 6 times the specific percentage of interest or product ($132.34782); and divide this interest by 6 to find the correct interest ($22.06-). 2. Find the interest of $417.28 for 2 yr. 228 da. at 6 % INTEREST 163 SOLUTION 2 yr. of 360 da. - 720 da. ExpLANATION . There are twice 360 da. Fraction of a yr. = 228 da. or 720 da. in the 2 complete years, and 6). 948 228 da. in the fraction of a year, or 720 da. Q ./. i r Q + 228 da. (948 da.) in the given time. S P e ' Hence, * of .948 of the principal, or .158 of the principal must be the specific rate Principal = $417.28 of interest for the given time. Therefore, Specific rate = .158 Multiply the principal or multiplicand ooo on A ($417.28) by the specific rate of interest ** or multiplier for 2 yr. 228 da. (.158) to find the specific percentage of interest or 41728 product ($65.93 +). $65.93024 RULE. Multiply the principal by one-thousandth of the time expressed in days, and divide the product by 6. NOTE 1. If the days are exactly divisible by 6, as in 111. Ex. 2, it will be found more convenient to multiply the principal by one-sixth of the days expressed in thousandths, thus minifying the multiplication and altogether avoiding the division by 6. NOTE 2. The interest at any higher or lower rate than 6 % may be found by proportionally increasing or diminishing the interest obtained by the pre- ceding process. Thus, as 7 % is I of 6 %, so the interest at 7 % must be I of the interest at 6 %; as 5 % is | of 6 %, so the interest at 5 % must be | of the interest at 6 %; etc. NOTE 3. If the time is expressed in months and days, or years, months and days, reduce it to days by allowing 360 days to each year and 30 days to each month. Though the ordinary interest for the exact time between two dates is required, if that interval contains one or more years, each year should uniformly be regarded as containing 360 days. EXAMPLES FOR PRACTISE At 6 % per annum, find the interest of 1. $748.35 for 48 da. 6. $613.28 for 324 da. 2. $813.72 for 174 da. 7. $924.13 for 7 mo. 23 da. 3. $258.37 for 253 da. 8. $416.25 for 5 mo. 16 da. 4. $521.63 for 217 da. 9. $2358.16 for 1 yr. 8 mo. 13 da. 5. $1253.18 for 119 da. 10. $758.30 for 3 yr. 10 mo. 24 da. 164 PERCENTAGE PERCENTAGE METHOD FOR COMPUTING INTEREST AT ANY PER CENT ILLUSTRATIVE EXAMPLE 232. Find the interest of $675.83 at 5% for 223 days. SOLUTION EXPLANATION. $675.83 is the principal or base (mul- tiplicand) ; and 5 % is the rate or multiplier for one year, $b75.8o k u t no ^ the specific rate of interest or multiplier for 223 da. At 5 % per annum of 360 da., the corresponding rate 202749 must be * f 5 % or 1 %> for * f 3GO da -> or 72 da - There- 13^1fifi ^ ore *ke s P ec ^ c rate f Merest or multiplier for 223 da. must be as many times 1 % of the principal as 72 da. are 135166 contained times in 223 da., or W of 1 % of the principal. 8)1507.1009 .As -W of T for of the principal equal fffo of the principal, ^ r 7^ M many hundredths of the principal as there are days, Q\-IOO that is, A of 223 hundredths of the principal, therefore, $20.9319 Multiply the principal or multiplicand ($675.83) by 72 times the specific rate or multiplier (2.23), obtaining 72 times the specific percentage of interest or product ($1507.1009); and divide by 72 (= 8 X 9) to find the correct interest ($20.93 +). RULE. Multiply the principal by one-hundredth of the time expressed in days, and divide the product by such a part of 360 as 1% is of the given rate per annum. NOTE 1. In finding the interest by the above method, it will materially lessen the mechanical labor to cancel any factor which is common to both multiplier and divisor. Thus, to find the interest at 8 % for 63 days, instead of multiplying the principal by .63 and dividing by 45 (360 -J- 8), it will be easier to multiply by $ of .63 or .07, and to divide by i of 45, or 5. NOTE 2. If the given rate per annum is not an exact divisor of 360, first find the interest at 6 %, and apply Note 2, 231. EXAMPLES FOR PRACTISE Find the interest of Find the interest of 1. $568.73 at 8 % for 97 da. 8. $372.50 at 4 \ % for 280 da. 2. $842.95 at 9 % for 265 da. 9. $563.75 at 3-| % for 297 da. 3. $673.48 at 5% for 173 da. 10. $728.34 at 7 % for 264 da. 4. $498.25 at 4 % for 82 da. 11. $815.93 at 8 % for 7 mo. 24 da. 6. $725.38 at 3 % for 141 da. 12. $425.30 at 5 % for 5 mo. 12 da. 6. $284.83 at 10 %for 327 da. 13. $785.90 at 4 % for 1 yr. 279 da. 7. $916.40atl2%for285da. 14. $893.25 at 3 % for 3 yr. 175 da. INTEREST 165 TO COMPUTE INTEREST AT 6% BY "THE ALIQUOT METHOD" 233. INTRODUCTION. The following method is a modification of the aliquot method in common use which will enable the cal- culator to verify the mechanical accuracy of the final result with- out reviewing all the figures used in obtaining that result. By this method, the aliquots are so arranged that each successive aliquot is an aliquot part of the aliquot immediately preceding it, and that the last aliquot is also an aliquot part of the initial state- ment. This arrangement will cause the verification of the few figures in the last aliquot to include the verification of all the pre- ceding aliquots; for if the last aliquot, first obtained by an un- broken succession of derivations from the initial statement, is found to be correct by being obtainable by direct derivation from the same initial statement, then the succession of aliquots from which it was originally derived must also be correct. ILLUSTRATIVE EXAMPLES 1. Find the interest of $283.75 at 6 % for 210 days. SOLUTION EXPLANATION. At the rate of 6 % I r P 61 * annum f 360 da., the correspond- Q days ing rate must be | of 6 %, or 1 % of 8 1 51 25 = " " 180 " the principal for of 360 da., or 60 da. 41 g7 _ (( n 30 tt Therefore, by means of an elon- -^"^ u 7T~1^ ~^~~ 8 ated decimal P int > cut off two fi gures J | yd 1Z - J1U from the right of the dollars, which is equivalent to taking 1 % of the prin- cipal [Prin. 5, 120], thus obtaining the initial interest for 60 da. (Note, 227.) Next find the interest for the greatest exact dividend of 60 da. (180 da.) which is found in the given number of days (210); and as 180 da. are 3 times 60 da., the interest for 180 da. must be 3 times the interest for 60 da. ($2.8375 X 3), or $8.5125. Next find the interest for the greatest aliquot part of 180 da. (the last obtained result) which can be found in the remaining 30 da. of the given time (210 da.). As 30 is itself an aliquot part of 180 (30 = | of 180), therefore of the interest for 180 da. (J of $8.5125), of $1.4187, must be the interest for 30 da. The interest for 30 da. added to the interest for 180 da. must equal the interest for 210 da. ($9.93 +). VERIFICATION. Observe if the last aliquot ($1.4187), supposed to be the interest for 30 da., is one-half of the initial statement ($2.8375) which is known 166 PERCENTAGE to be the interest for 60 da. If so, it should be accepted as correct, as the last aliquot ($1.4187) has now been obtained from two independent sources. If the interest for 30 da. is correct, then the interest for 180 da. must also be correct, as the former was originally derived from the latter. As a further precaution, add the two parts of the time (180 + 30) to see if they equal the entire time (210 da.). Lastly prove the addition by Note 1, 19. 2. Find the interest of $875.93 at 6 % for 4 yr. 285 da. EXPLANATION. If, as shown in Ex. 1, $8.7593 is the interest $8 | 75.93 = int. for 60 da. of $875.93 for 60 da., then the interest of $875.93 for 4 times 60 35 5 03 72 = ' 240 83 95 = " " 40 da., or 240 da. (the greatest div- idend of 60 da. found in 285 da.) 72 99 = 5 ' must be 4 times $8.7593, 03 72 ) $35.0372. or 72 ) Q I ~ 35 175 18 6 i * ^ r * ^ *ke m * ;erest f r 240 da. is $35.0372, the interest for of $251 | 82 98 = 4 yr. 285 da. 240 da ., or 40 da. (the greatest aliquot part of the immediately preceding 240 da. which is found in the remaining 45 da. of the given 285 da.) must be i of $35.0372, or $5.8395. If the interest for 40 da. is $5.8395, the interest for | of 40 da., or 5 da. (the greatest aliquot part of the immediately preceding 40 da., which is found in the remaining 5 da. of the originally given 285 da.) must be | of $5.8395, or $.7299. If 6 % is the rate of interest for 1 yr., 4 times 6 %, or 24 % will be the cor- responding rate for 4 yr. As the initial statement ($8.7593) is 1 % of the principal, 24 times $8.7593 ( = 4 times the initial statement, or $35.0372; and 20 times the initial statement, or $175.186) must be the interest for 4 years. The sum of the interest for these several parts of the full time is therefore the full required interest. VERIFICATION. As a precaution against a possible mechanical error observe if the interest for 5 da. ($.7299) is 12 of the initial interest for 60 da. ($8.7593). If it is, not only is it correct, but also all the preceding aliquots; for the interest for 5 days was originally obtained by successive derivations through each of the preceding aliquots, and a correct result cannot be derived from an incorrect. RULE. 1. Cut off two figures from the right of the dollars in the principal by an elongated decimal point, and regard the result as the interest for 60 days. 2. Take as many times the interest for 60 days as the greatest dividend of 60 days in the given days, is times 60 days. INTEREST 167 3. Take such an aliquot part of the last obtained interest as will equal the greatest aliquot part of the last obtained days which is found in the remaining days of the given time. 4. Thus continue until the required interest for all the days of the given time has been found. NOTE 1. If care is taken to have the terminating aliquot days also an aliquot of 60 da., it will only be necessary to test the last aliquot interest by comparison with the initial interest for 60 days, to verify not only the last result, but also all the intermediate aliquots between it and the initial state- ment. NOTE 2. If the given time contains years and days, first find the interest for the days, and verify. Then take as many times 6 times the initial state- ment (1 % of the principal) as the given years are times 1 year, and verify. At 6 % the interest for years is most conclusively verified by comparing one partial product with the other. Thus, the multiplier for 4 yr., is 4 times 6 %, or 24 %, hence, if the partial product by 4, when obtained, should be twice the partial product by 2, it will prove both partial products to be correct. NOTE 3. Should any given interval in days lack an aliquot part of 60 da. of being an exact dividend of 60 da., it will usually be more convenient to increase the multiplier by 1, and deduct the aliquot excess of interest from the resulting product. Thus, to find the interest for 117 da., first find the interest for 120 da. and diminish the result by the interest for 3 da. (*V of 120, that is, I of the interest for 120 da. written one place to the right of the orders divided). EXAMPLES FOR PRACTISE Verifying results, find the interest at 6 %. Principal Time Principal Time 1. $648.15 240 da. 10. $425.93 204 da. 2. $375.34 186 " 11. $537.52 15 3. $827.43 222 " 12. $875.40 262 " 4. $516.20 345 " 13. $647.52 25 " 5. $285.95 168 " 14. $9246.30 146 " 6. $416.62 85 " 16. $17893 78 " 7. $742.85 160 " 16. $9.50 50 " 8. $924.50 275 " 17. $68.75 2 yr. 153 da. 9. $618.80 42 " 18. $87.25 7 yr. 311 da. NOTE 4. By compound subtraction, the interval between two dates is most conveniently found as follows: From any date in the fifth month (say, May 19, 1912) to the same date in the eleventh month (Nov. 19, 1912) are 6 mo. of 168 PERCENTAGE 30 da. each or 180 da., therefore to Nov. 27, or 8 da. later, the time will be 8 da. more than 180, or 188 da., and to Nov. 13, or 6 da. earlier the time will be 6 da. less than 180, or 174 da. If the interval is greater than one year, first find the number of complete years in the given interval, and then as above described, find the time in the remaining fractional part of a year. Thus, from Sept. 25, 1908, to July 18, 1912, are 3 yr. 293 da.; that is, from Sept. 25, 1908, to the latest corresponding date in the interval (Sept. 25, 1911) are 3 years, and from Sept. 25, 1911 (9th mo. of 1911) to July 25, 1912 (7th mo. of 1912, or the equivalent of 7 + 12, or the 19th mo. of 1911) are 10 mo. of 30 da. each, or 300 da.; and to July 18, 1912, or 7 da. earlier than July 25, 1912, are 7 da. less than 300 da., or 293 da. NOTE 5. To find the exact number of days in any fractional part of a year, first find the time by the 30-day method [Note 4,] and add one day for each 31-da. mo. in the given interval and subtract two days if the interval includes the last day of February of a common year, or subtract one day if it includes the last day of February of a leap year. Finding the time by Note 4, what is the interest at 6 % 19. Of $129.60 from Sept. 18, 1910, to June 16, 1911? 20. Of $673.45 from Mch. 23, 1908, to Nov. 13, 1912? 21. Of $827.64 from Dec. 12, 1905, to May 6, 1911? 22. Of $285.96 from Oct. 31, 1903, to June 17, 1911? [Note 2, 163.] 23. Of $875.49 from May 13, 1905, to Oct. 31, 1911? [Note 2, 163.] 24. Of $5168.35 from Nov. 27, 1902, to July 12, 1912? 25. Of $87.90 from July 31, 1907, to Oct. 19, 1911? [Note 2, 163.] 26. Of $8.53 from Jan. 16, 1906, to Nov. 25, 1911? 27. Of $62.70 from June 30, 1908, to Feb. 29, 1912? [Note 2, 163.] 28. Of $85378.15 from Feb. 21, 1908, to May 17, 1912? 29. Of $73.92 from May 18, 1909, to Sept. 3, 1911? 30. Of $2354.70 from June 28, 1910, to Jan. 31, 1912? 31. Of $825.37 from Dec. 28, 1902, to Nov. 18, 1911? 32. Of $497.85 from Apr. 24, 1907, to July 18, 1911? 33. Of $745.93 from Aug. 18, 1905, to Sept. 24, 1912? Finding the exact time by Note 5, what is the interest at 6 % 34. Of $978.42 from May 23, 1911, to Dec. 16, 1911? 35. Of $427.65 from Oct. 15, 1911, to Mch. 31, 1912? INTEREST 169 36. Of $695.80 from Feb. 3, 1911, to Nov. 6, 1911? 37. Of $584.75 from June 30, 1911, to Aug. 5, 1911? 38. Of $752.35 from Dec. 15, 1911, to Dec. 22, 1911? 39. Of $859.40 from Jan. 20, 1908, to Nov. 26, 1908? 40. Of $348.59 from May 1, 1910, to Sept. 20, 1910? 41. Of $2853.60 from July 3, 1911, to Feb. 6, 1912? 42. Of $695.73 from Mch. 28, 1910, to Oct. 12, 1910? ' 43. Of $196.82 from Nov. 17, 1910, to Apr. 5, 1911? 44. Of $378.45 from Apr. 12, 1911, to Aug. 24, 1911? 45. Of $518.62 from Dec. 19, 1910, to Feb. 23, 1911? 46. Of $35.76 from Oct. 8, 1911, to May 23, 1912? 47. Of $9.20 from Sept. 29, 1907, to Aug. 6, 1908? 48. Of $18275.65 from Aug. 13, 1909, to Jan. 25, 1910? 49. Of $598.25 from May 16, 1904, to Nov. 12, 1909? 60. Of $643.50 from July 28, 1908, to Mch. 3, 1912? 61. Of $2468.75 from June 24, 1905, to Apr. 6, 1911? 62. Of $525.12 from Feb. 25, 1908, to Aug. 31, 1911? 63. Of $716.53 from Dec. 19, 1903, to May 25, 1908? TO COMPUTE INTEREST AT ANY PER CENT. BY THE 6% METHOD ILLUSTRATIVE EXAMPLE 234. Find the interest of $758.35 at 9% for 215 days? SOLUTION EXPLANATION. First find $7 | 58.35 = int. at 6 % for 60 da. the interest of the given prin- u i QT\ u cipal for the given time at 6 % 22 3 27 13 t 79 17 = " " " " 30 by 233, obtaining $27.1741. As 9 % is one-half of 6 % (3 %) 63 19 = 5 ' added to 6 %, so the interest at 17 41 = " " Qo/ " 215 " 9 % must be one-half of the in- 58 70= " " 37 " 215 " terest at 6 % ($13.5870) added to the interest at 6 % ($27.1741), $40 I 76 11 - " 9% " 215 ' or $40.76+. RULE. First find the interest at 6%, then regard whatever is over or under 6 % in the given rate, as an aliquot of 6 %, and increase or diminish the interest at 6% by the corresponding aliquot part of the interest at 6%. 170 PERCENTAGE NOTE. If the given rate is more than one aliquot of 6 % over or under 6 %, employ Note 2, 231. EXAMPLES FOR PRACTISE Verifying each result, find the interest of 1. $578.25 at 7 % for 165 da. 6. $485.60 at 4 % for 143 da. 2. $297.62 at 9% for 273 da. 7. $816.12 at 8% for 85 da. 3. $815,40 at 5% for 137 da. 8. $3487. 25 at 4J % for 118 da. 4. $765.53 at 10% for 315 da. 9. $1852.70 at 3J % for 256 da. 5. $654.25 at 3 % for 198 da. 10. $96.85 at 5J % for 174 da. 11. $5.90 at 7J % for 1 yr. 216 da. 12. $816.35 at 12% for 4 yr. 297 da. DIRECT ALIQUOT METHOD OF FINDING THE INTEREST AT ANY PER CENT. ILLUSTRATIVE EXAMPLE 235. Find the interest of $647.28 at 5 % for 205 days. EXPLANATION. At the rate SOLUTION of 5% of the principal per $6 I 47.28 = int. at 5 % for 72 da. annum of 36 <**> the c , rre - /L/ 1 j 1_ 1 * M spending rate must be -k of 5 12 9456= " " " "144" of the principal, or 1 % of the 4 31 52 = " " " " 48 " principal for & of 360 da., or 1 07 88 = " " " " 12 " 72 days. Therefore, by means g 99 = n t( u i (t of an elongated decimal point, cut off two orders from the right $18 | 42 95 = int. at 5 % for 205 da. O f the dollars, which is equiva- lent to taking 1 % of the principal [Prin. 5, 120] and thus to obtaining the interest for 72 da. at 5 % per annum [Note, 227]. Next find the interest for the greatest dividend of 72 da. (144 da.) which is found in the given number of days (205 da.) ; and as 144 da. are 2 times 72 da., the interest for 144 da. must be 2 times the interest for 72 da. ($6.4728), or $12.9456. Next find the interest for the greatest aliquot part of 144 da. (the last obtained result) which can be found in the remaining 61 da. of the given time, or for 48 da. ( J of 144 da.). If $12.9456 is the interest for 144 da., | of $12.9456 or $4.3152 must be the interest for of 144 da., or 48 da. Next find the interest for the greatest aliquot part of 48 da. (the last obtained result) which can be found in the remaining 13 da. of the given time, INTEREST 171 or for 12 da. ( of 48 da.) . If $4.3152 is the interest for 48 da., i of $4.3152, or $1.0788, must be the interest for | of 48 da., or 12 da. Next find the interest for the remaining 1 da. of the given time, which is i 1 ? of the last obtained interest for 12 da., or $0.899. Lastly add these several partial interest results, obtaining $18.43- as the complete interest. VERIFICATION. Observe if the last aliquot ($.0899) is 7 V ( = 1 of i) of the initial statement ($6.4728). If so, the last aliquot must be correct, and con- sequently all the preceding aliquots through which the last aliquot was suc- cos.-ively derived, must also be correct. As a further precaution, add the aliquot days to see if they equal 205; and prove the final addition by Note 1, 19. RULE. 1. // 360 be divided by as many times 1 as the given rate is times 1 %, the quotient will be the initial number of days for which 1% of the principal is the initial interest. 2. Take as many times the initial interest as the greatest exact dividend of the initial days in the given days, is times the initial days. 3. Take such a part of the last obtained interest as will equal the greatest aliquot part of the last obtained days which is found in the remaining days of the given time. 4. Continue the last preceding process upon each following result until all the given time has been considered. VERIFICATION. If the successive aliquot days have been so arranged that the terminating aliquot days are also an aliquot of the initial days, by observing if the last obtained aliquot interest is also derivable from the initial interest, it will not only be a verification of itself, but also of all the aliquots which precede it. NOTE 1. If the given time is expressed in years and days, first find the interest for the days, and verify. Then take as many times the initial in- terest (1 % of the principal) as will equal the rate per annum multiplied by the number of years. NOTE 2. If the given interval in days lacks an aliquot of the initial days of being 1 more time the initial days, it will usually be more convenient to in- crease the multiplier of the initial interest by 1, and deduct the excess of interest from the resulting product. Thus, to find the interest for 177 da. at 8 %, first find the interest for 180 da. (4 times 45 da.) and deduct therefrom the interest for 3 da. (-fa of 180 da.). NOTE 3. If the given rate is not an exact divisor of 360 apply 234. Or, find the interest at some convenient rate which is a divisor of 360; thus, at 3$ %, if more convenient, first find the interest at 3 % and add i of itself; or find the interest at 4 % and deduct i of itself. 172 PERCENTAGE EXAMPLES FOR PRACTISE Verifying each result, find the interest of 1. $875.90 at 9 % for 135 da. 5. $297.50 at 3 % for 165 da. 2. $462.75 at 4 % for 210 da. 6. $935.75 at 10 % for 204 da. 3. $695.82 at 5 % for 261 da. 7. $392.18 at 12 % for 316 da. 4. $715.60 at 8 % for 318 da. 8. $1586.35 at 4 J % for 205 da. Obtaining the time by the 30-day method, find the interest of 9. $3272.24 from July 20, 1902, to Feb. 17, 1908, at 4 J %. 10. $1873.16 from Jan. 26, 1908, to Sept. 20, 1911, at 10%. 11. $1384.20 from Apr. 9, 1906, to Nov. 13, 1911, at 3%. 12. $278.15 from Nov. 30, 1905, to Aug. 10, 1908, at 3J%. 13. $48.73 from Apr. 28, 1906, to Nov. 26, 1910, at 9%. Obtaining the exact days, find the interest of 14. $435.76 from Oct. 13, 1905, to July 30, 1911, at 12%. 16. $2896.37 from Aug. 19, 1897, to May 7, 1912, at 5%. 16. $9183 from June 16, 1909, to Mch. 24, 1912, at 4 %. 17. $4238.75 from May 31, 1904, to July 16, 1911, at 4J %. 18. $37.95 from Sept. 24, 1903, to Dec. 13, 1910, at 8%. INTERCHANGEABILITY OF PRINCIPAL AND TIME 236. INTRODUCTION, (a) In 232 it was seen that the princi- pal and the time in days are uniform factors in finding the inter- est at any rate per annum. As the product of any two factors will be the same in whatever order they may be arranged, it must follow that the product of the time in days multiplied by one- hundredth of the principal will be the same as that of the prin- cipal multiplied by one-hundredth of the time in days; that is, the principal and the time in days may be interchanged at the conven- ience of the calculator without affecting the correctness of the final result. Thus, the interest of $180 for 240 da. at any rate per annum will be the same as the interest $240 for 180 da. at the same rate. Advantage should be taken of this fact whenever it is seen that the principal regarded as days can be separated into fewer aliquot parts of the initial days at the given rate than can the actual time in days. (6) This interchange of principal and time may be made more frequently advantageous by amplifying the principles underlying the processes of 233 and 235. Thus, if cutting off two orders INTEREST 173 from the right of the dollars in any principal will produce the in- terest of that principal for such a part of 360 days as 1 % is of the given rate per annum, then cutting off three orders from the right of the dollars in the principal (being equivalent to taking one- tenth of the preceding initial interest) must produce the interest for one-tenth of the preceding initial days; and cutting off one order from the right of the dollars in the principal (being equiva- lent to taking ten times the same initial interest) will produce the interest for ten times the same initial days; and cutting off no order from the right of the dollars in the principal (being equiva- lent to taking one hundred times the same initial interest) will produce the interest for one hundred times the same initial days; etc.; etc. (c) Thus, if cutting off two figures from the right of the dol- lars in the principal will produce the interest of that principal at 6 % for 60 da., then cutting off three figures must produce the interest for one-tenth of 60 da., or 6 da.; cutting off one figure, for 10 times 60 da., or 600 da.; cutting oft no figure, for 100 times 60 da., or 6000 da. Similarly at 9% per annum, cutting off three orders from the right of the dollars in the principal, will produce the interest for 4 da.; cutting off two orders, for 40 da.; 1 order for 400 da.; and no order, for 4000 da.; etc.; etc. (d) To differentiate these processes, cutting off three orders is suggestively called the units' rule (3 units of days at 12 %) ; cut- ting off two orders, the tens' rule (3 tens of days) ; cutting off one order, the hundreds' rule (3 hundreds of days); and cutting off no order, the thousands' rule (3 thousands of days). NOTE 1. As all ordinary interest is computed upon the uniform basis of 360 days to a year, if the given time is expressed in years and days, the years should be reduced to days upon the same basis, even when dates are given and the exact number of days is found in the fractional part of a year in the given interval. NOTE 2. If the given principal contains cents, no advantage will be gained by interchanging, for the cents will then have to be regarded as such a fraction of a day as they are of one dollar. ILLUSTRATIVE EXAMPLES 1. Find the interest of $685.72 at 4% for 27 days. 2. Of $275.83 at 6 % for 150 days. 3. Of $5000 at 8 % for 123 days. SOLUTIONS (D (2) | $685.72 = 9 days' int. $27 | 5.83 = 600 days' int. $2 I 057 16 = 27 " " $6 I 8 95 = 150 " " 174 PERCENTAGE (3) $123 J of $123 = 13 = 4500 days' int. 666 = 500 " $136 I 67-= 5000 EXAMPLES FOR PRACTISE Employing the fewest figures, find the interest of 1. $496.75 at 6 % for 24 da. 2. $715.25 at 6 % for 54 da. 3. $563.70 at 6 % for 18 da. 4. $758.62 at 6 % for 27 da. 5. $435.80 at 6 % for 150 da. 6. $847.35 at 6 % for 200 da. 7. $6000 at 6 % for 327 da. 8. $800 at 6 % for 171 da. 9. $7500 at 6 % for 192 da. 10. $400 at 6 % for 259 da. 11. $400 at 8% for 319 da. 12. $800 at 5% for 153 da. 13. $4500 at 7% for 137 da. 14. $300 at 9 % for 273 da. 15. $600 at 4 J % for 169 da. 16. $2000 at 12 % for 317 da. 17. $1800 at 10% for 159 da. 18. $2500 at 6 % for 3 yr. 253 da. 19. $7200 at 5 % for 4 yr. 198 da. 20. $1800 at 3 % for 2 yr. 147 da. ACCURATE INTEREST 237. Accurate interest is interest computed upon the accurate basis of 365 days to a common year, or of 366 days to a leap year. NOTE 1. All disbursements of interest by the governments of the United States and of foreign nations, all calculations of interest by foreign com- mercial houses, and many computations by business houses in the United States, especially by American importers of foreign goods, are made upon the accurate basis. ILLUSTRATIVE EXAMPLE Find the accurate interest of $297.83 for 189 da. at 8 %. SOLUTION EXPLANATION. First find the in- $297.83 terest of the given principal for 1 year .08 at the given rate, obtaining $23.8264. As the accurate interest for 1 day is 3^3- of the accurate interest for 1 year, the accurate interest for 189 da. must be 189 times ^Jy, or ^f f , of the accurate interest for one year, obtained by mul- tiplying $23.8264 by the numerator (189) and dividing by the denominator (365). 23.8264 189 2144376 1906112 238264 365)4503.1896($12.3375 ACCURATE INTEREST 175 RULE. Multiply the interest for one year at the given rate per annum by the exact number of days in the given fraction of a year, and divide the result by 365 if the interval is part of a common interest year, or by 366 if part of a leap interest year. NOTE 2. If the given interval is more than one year, first find the accurate interest for the days in the fraction of a year, and then find the interest for the years, and add the two results. NOTE 3. Always obtain the accurate time [164, or Note 5, 233] between two dates when computing accurate interest. NOTE 4. The learner is cautioned against the common error of confound- ing the calendar year with the interest year. A calendar year commences with Jan. 1 and terminates with Dec. 31, both inclusive, but an interest year com- mences with the date on which the debt commences to draw interest and ter- minates at the corresponding date of the next following year. Thus, an interest-bearing debt contracted May 18, 1911, and paid July 3, 1911, has accrued interest for 46 days of the 366-day year commencing May 18, 1911, and terminating May 18, 1912; while an interest-bearing debt contracted May 18, 1912, and paid July 3, 1912, has accrued interest for 46 days of the 365-day year commencing May 18, 1912, and terminating May 18, 1913. The leap year is so called because it includes the leap day (Feb. 29); therefore any interest year which includes the leap day must be a leap interest year, though the leap day itself may not be included in the given interval of that leap interest year. EXAMPLES FOR PRACTISE Find the accurate interest of 1. $963.75 at 4 % for 82 da. 7. $296.75 at 7 % for 93 da. 2. $725.62 at 3 % for 174 da. 8. $583.92 at 4 % for 335 da. 3. $286.40 at 5 % for 230 da. 9. $857.42 at 3 J % for 252 da. 4. $437.52 at 8 % for 161 da. 10. $728.19 at 5 % for 3 yr. 87 da. 5. $582.37 at 6 % for 271 da. 11. $649.50 at 4% for 1 yr. 279 da. 6. $345.28 at 9 % for 168 da. 12. $484.32 at 4 % for 4 yr. 176 da. 13. $2983.52 from Aug. 18, 1903, to May 18, 1910, at 9%. 14. $786.25 from Mch. 14, 1907, to Nov. 28, 1907, at 10%. 15. $685.97 from Apr. 23, 1904, to Dec. 23, 1912, at 8%. 16. $925.82 from May 14, 1905, to Sept. 20, 1911, at 12%. 17. $716.25 from Jan. 3, 1909, to Feb. 18, 1912, at 3i %. 18. $1586.73 from June 19, 1895, to Mch. 17, 1900, at 7 %. 19. $6178.30 from May 23, 1910, to Aug. 16, 1913, at 5%. 20. $875.45 from Oct. 12, 1909, to July 17, 1912, at 3J %. 21. $2495.18 from Feb. 29, 1908, to Oct. 27, 1912, at 4%. 176 PERCENTAGE NOTE 5. Reduce the lower denominations of English money to a decimal of a as in 111. Ex. 2, 160, compute the interest in the usual manner, and reduce the decimal part of a in the obtained answer to integers of the lower denominations as in 111. Ex., 159. 22. Find the accurate interest of 178 15s. 6d. from May 3, 1908, to Nov. 18, 1913, at 4% per annum. 23. What is the accurate interest of 3487 2s. 6d. from Sept. 25, 1908, to June 10, 1913, at 3% per annum? 24. What is the accurate interest of 625 18s. 3d., at 5%, from Mch.21, 1910, to Sept. 6, 1913? 25. Find the accurate interest at 3| % per annum of 345 12s. 6d. from Nov. 3, 1905, to May 26, 1913. RELATION OF THE FOUR ELEMENTS OF INTEREST 238. INTRODUCTION, (a) A problem in interest is one in which three of the four elements of interest are given, and it is required to find the omitted fourth element. The four elements (principal, rate per annum, time, interest or amount) have a uni- form relation to each other; so that when any three of them are given, it is possible to complete the relation by finding the neces- sary omitted element. The most useful methods for finding the interest when it is the omitted element have already been pre- sented [231-236] . It now remains to consider problems in which the interest is not required as heretofore, but is given, together with any two of the three remaining elements (principal, rate, time), and it is required to find the omitted remaining element. The following general process is applicable to all such problems: (6) Find the interest or the amount by any one of the preceding methods which is most favored, for all the units of the given elements, and computed for 1 unit of the required element; then divide the given interest or amount for all the units of the required element by the obtained interest or amount for 1 unit of the required element, to find the number of units in the required element. SUMMARY. 1. The total interest or amount of the total prin- cipal for the total time and at the total rate % per annum, is a product. 2. The total interest or amount of any two of the three remaining elements for 1 unit of the required element, is the multiplicand. 3. The number of units in the required element is the multiplier. INTEREST 177 PRINCIPAL, INTEREST AND TIME GIVEN, TO FIND THE RATE ILLUSTRATIVE EXAMPLE 239. At what rate per annum will $758.36 produce $13.65 interest in 216 days? SOLUTION $7 I 58.36 = int. for 60 da..at 6 % 22 75 08 = " " 180 " tt tt 3 79 18 = " " 30 " tt tt 75 83 = " tt 6 tt tt tt 6)27 30 09 = " " 216 " it ft 4 55 01 = " " 216 " " 1% $4.5501)$13.6500 ( 2, practically 3 times ) 9 1002 I the int. at 1 %, or 3 % ) 4 5498 = significant remainder EXPLANATION. First find the interest of the given principal ($758.36) for the given time (216 da.) at 6 %, obtaining $27.3009 and divide by 6 to find the interest at 1 % ($4.5501). As the given interest ($13.65) is 2 times the interest at 1 %, and a very significant remainder almost equal to the divisor [206, 6], it should be regarded as practically 3 times the interest at 1 %, or 3 %. SUMMARY. The required rate % per annum will be as many times 1 %, as the given interest is times the obtained interest at 1 % of the same principal for the same time. NOTE 1. If the remainder from the final division is insignificant, follow 205, a; if significant, follow 205, b; and if intermediate, follow 205, c. In prac- tise, the division is rarely exact. NOTE 2. If the amount [Note, 226] is given, diminish it by the principal to find the interest, and then proceed as in the preceding solution. The amount at 3 % (103 %) is not 3 times the amount at 1 % (101 %). EXAMPLES FOR PRACTISE At what rate % per annum will 1. $596.85 produce $19.70 interest in 132 days? 2. $1237.29 produce $43.99 interest in 256 days? 3. $3278.45 produce $163.92 interest in 225 days? 4. $45.97 produce 62 cents interest in 108 days [205, c]? 5. $257.30 amount to $273.31 in 320 days? 6. $768.25 amount to $781.79 in 141 days [205 c]? 178 PERCENTAGE PRINCIPAL, INTEREST AND RATE GIVEN, TO FIND THE TIME ILLUSTRATIVE EXAMPLE 240. In what time will $568.75 produce $76.78 interest at 5 % per annum? SOLUTION EXPLANATION. The in- $568.75 X .05 = $28.4375, 1 yr.'s int. terest of the 8 iven principal $28.4375)$76.7800(2 years ($56875) at th ;, spe f fic rat , e V per annum (5%) equals $28.4375 for 1 year. There- 19.9050 fore it will require as many years to produce $76.78 in- terest as $28.4375 are con- tained times in $76.78, or 2 5971 50 2 yr. and HBf $ of a year. 28.4375)7165.8000(25; days As ordinar y inter est is ^ftS7 f^n computed upon the basis of 360 da. to a year, there will 1478 300 be as many days in if f f ft 1421 875 f a y r -> as it contains 360ths. c fi 4.0 en Hence, reduce the fraction of 00 4ZOU a year to 360ths by 86, ob- 28 4375 taining 251 da. and a very 27 9875, significant significant remainder almost 2 yr. 252 da., Ans. ^ to the divisor which should be counted as an- other day [205, &,] making the required time 2 yr. 252 da. SUMMARY. The required time will be as many times one year or one day, as the given interest is times the interest for one year or for one day } of the same principal and at the same rate per annum. NOTE 1. Apply 205, a, to all insignificant remainders from the final divi- sion; and 205, 6, to all significant remainders. NOTE 2. If the given interest is less than the interest for one year, divide it at once by the interest of the same principal for 1 day. EXAMPLES FOR PRACTISE In what time will 1. $958.63 produce $32.59 interest at 8 % per annum? 2. $7285.15 produce $166.65 interest at 4| % per annum? 3. $815.30 produce $14.90 interest at 7 % ^per annum? 4. $3825.75 produce $130.50 interest at 4% per annum? 5. $8270.75 amount to $10236.43 at 6 % per annum? INTEREST 179 INTEREST, TIME, AND RATE GIVEN, TO FIND THE PRINCIPAL ILLUSTRATIVE EXAMPLE 241. What principal loaned for 279 days at 8% per annum, will produce $57.50 interest? SOLUTION 2.79 -:- 45 = .062, int. of $1 at 8 % for 279 da $57.50 4- .062 = 927.42 times $1, or $927.42. EXPLANATION. 8 % per annum equals 8 cents per annum for each dollar of the required principal. If the interest of $1 for one year of 360 da. is 8 cents, the interest of $1 for | of 360 da., or for 45 da., must be | of 8 cents, or 1 cent. If the interest of $1 is 1 cent for each 45 da. of the given time, it will be as many cents for 279 days as 45 da. are contained times in 279 da., or 6.2 cents. Hence, the required principal must be as many times $1, as the given interest of the required principal ($57.50) is times the interest of $1 of principal ($.062), for the same time and at the same rate per annum, or 927.42 times $1. SUMMARY. The required principal will be as many times $1, as the given interest of the required principal is times the interest of SI for the same time and at the same rate % per annum. NOTE. To find the interest of $1 for any given time and at any given rate per annum, divide the given days regarded as cents by such a part of 360 as 1 % is of the given rate per annum. Thus, the interest of $1 for 152 da. at 9 % per annum equals 1.52 -f- (360 -r- 9) or $.038 [Note, 227]. If the given rate is not an exact divisor of 360, first find the interest of $1, at some other convenient rate which is an exact divisor of 360, and add to, or take from, the obtained result as the given rate is greater or less than the assumed rate. EXAMPLES FOR PRACTISE What principal will produce 1. $4.08 interest at 3 % in 84 days? 2. $295.58 interest at 8 % in 324 days? 3. $169.87 interest at 6% in 247 days? 4. $1524.75 interest at 9 % in 1 yr. 282 da.? 5. $5.58 interest at 5 % in 96 days? 6. $202.71 interest at 7 % in 144 days? 7. $445.73 interest at 4% in 2 yr. 315 da? 180 PERCENTAGE AMOUNT, TIME, AND RATE GIVEN TO FIND THE PRINCIPAL ILLUSTRATIVE EXAMPLE 242. What principal will amount to $349.59 in 51 days at 4 % per annum? FIRST SOLUTION $.51 -T- 90 = $.005f, int. of $1 at 4% $1 + $.005f = $1.005f, amt. of $1 at 4 % $1. 005 f) $349.59 3 3 3.017 ) 1048.770(347.62 times $1 9051 14367 12068 22990 21119 18710 18102 6080 6034 FIRST SOLUTION. The interest of $1 for 51 da. at 4 % is $.005| [Note, 241]. This, added to $1, will be the amount of $1 for the same time and at the same rate. As the given amount ($349.59) is 347.62 times the amount of $1 of principal ($1.005|) for the same time and at the same rate, so the princi- pal which produced the given amount must be 347.62 times $1, or $347.62. SECOND SOLUTION $349.59 90 $90.51)$31463.10(347.62 times $1 27153 43101 36204 68970 63357 56130 54306 18240 18102 138 SECOND SOLUTION. If it is seen that the interest of $1 for the same time and at the same rate [Note, 241] will include an in- convenient fraction, as it usually does, the second solution will be found preferable. To multiply both divisor and dividend by the same number will not affect the value of the quotient [Prin. 5, 34]. Therefore, instead of dividing the given amount ($349.59) by the amount of $1 at the same rate and for the same time ($1 + '&) as in the first solution, it will usually involve less labor to divide 90 times the given amount ($349.59 X 90 = $31463.10) COMPOUND INTEREST 181 by 90 times the amount of $1 for the given time and at the given rate ($1 X 90 + $& X 90 = 90.51). Hence, multiply the given amount by such a part of 860 as 1 % is of the given rate per annum (in the 111. Ex., by 360 -j- 4, or 90); and divide the result by the same quotient (90) increased by the days regarded as hundredths (.51), or by 90.51. SUMMARY. The required principal is as many times $1 of prin- cipal, as the given amount of the required principal is times the ob- tained amount of $1 of principal for the same time and at the same rate % per annum. NOTE. If, in using the second solution, the given rate (as 7 %, 3| %, etc.) is not an exact divisor of 360, the resulting fraction may be avoided by mul- tiplying the given amount by 360, and dividing the result by 360 increased by as many times the days of prepayment regarded as hundredths as the given rate is times 1 %. Thus, in the 111. Ex., ($349.59 X 360) -:- (360 + .5T~X4) = $347.62. That is, instead of dividing 90 times the given amount by 90 times the amount of $1, as in the second solution, by this process 4 times 90 times, or 360 times, the given amount is divided by 4 times 90 times the amount of $1 (90 X 4 -f- .51 X 4) or by 362.04 [Prin. 5, 34]. EXAMPLES FOR PRACTISE What principal will amount to 1. $628.73 in 243 days at 8% per annum? 2. $1965.37 in 162 days at 6% per annum? 3. $748.26 in 117 days at 3% per annum? 4. $3562.80 in 327 days at 7 % per annum [See Note]? 6. $523.42 in 1 yr. 257 da. at 4 J % per annum? 6. $2685.72 in 2 yr. 168 da. at 3 J % per annum? 7. What principal will amount to $986.35 on May 16, 1911, if loaned on Aug. 28, 1910, at 5% (exact days)? 8. What sum of money loaned Apr. 16, 1911 at 4%, will amount to $1628.15 on Nov. 27, 1911 (time by 30-da. method)? COMPOUND INTEREST 243. Compound interest is interest computed in the usual manner, but which, at the end of a stipulated period of time, is compounded with (that is, added to) the principal for that period, to form a new principal for the next following period. 182 PERCENTAGE ILLUSTRATIVE EXAMPLE Find the interest of $678.45 for 1 yr. 320 da., if compounded semi-annually at 6% per annum. SOLUTION EXPLANATION. At 6 % ^TO A r AC i\ P er annum > the correspond- $678.45 (first principal) ing rate is i of 6 %> or 20.3535 (3% Of $678.45) 3 %, for one-half a year. 698.8035 (first amount) l vr - = 2 half-years; 20 QfU1 (1 07 nf ftfiQK SO*}^ and 32 da> = * half year 20 - 9b41 ( ' (i of 360 da. = 180 da.) 719.7676 (second amount) and 140 da. of a fourth 21. 5930 (3% Of $719.7676) half year. Hence, 1 yr. 741.3606 (third amount) ^20 f = 3 half -y ears and 17.2984 (int. of $741.36 for 140 da.) ^secure accuracy in 758.6590 (final amount) the final result, carry all 678.45 (original principal) intermediate results to four decimal places. As hun- 80.21- (comp. int. for 1 yr. 320 da.) dredths x hundre dths = ten-thousandths, so to ob- tain the necessary four decimal places, multiply the cents' order of the principal (hundredths) by the rate % per period (3 hundredths), and write the first order of the product underneath the principal but two places to the right of the cents' order of the principal (that is, in the ten-thousandths' order). Add the obtained interest for the first period ($20.3535) to the first prin- cipal ($678.45), to find the first amount ($698.8035), which will constitute a new principal for the second half-yearly interest period. Continue this process as many times as there are complete interest periods (3 times). Then find the interest of the amount for the last complete period ($741.3606) for the fraction of the fourth period (140 da.) at 6 % per annum by 233, obtaining $17.2984, which added to the last obtained amount ($741.3606) will produce the final amount ($758.66 -). The difference between the original principal ($678.45) and the final amount ($758.66) will be the required compound interest ($80.21 ). NOTE 1. If the given time contains a fraction of a year expressed in days, regard 180 days thereof (if it contains so many) as another half-year if the compounding of interest is semi-annual; or regard each 90 days thereof as another quarter, if the compounding of interest is quarterly. If dates are given, and there is a fraction of a year, regard each successive six calendar months thereof, irrespective of the number of days they may contain, as another half-year if the compounding is semi-annual, or each three calendar months thereof as another quarter if the compounding is quarterly; and compute the time for any possible remaining fractional part of an interest period in the customary manner. COMPOUND INTEREST 183 EXAMPLES FOR PRACTISE Find the compound interest of 1. $4375.80 for 3 yr. at 6 % per annum, comp. annually. 2. $965.25 for 4 yr. at 4 % per annum, comp. half-yearly. 3. $1825.30 for 2 yr. at 5 % per annum, comp. quarterly. 4. $2469.75 for 4 yr. at 7 % per annum, comp. annually. 5. $6138.26 for 2 yr. at 8 % per annum, comp. quarterly. 6. $594.65 for 1 yr. 270 da. at 6 %, comp. half-yearly. 7. $763.40 for 1 yr. 195 da. at 4 % per annum, comp. quarterly. 8. $298.84 for 2 yr. 78 da. at 3 %, comp. half-yearly. 9. $3297 for 1 yr. 54 da. at 5% per annum, comp. quarterly. 10. $682 for 2 yr. 173 da. at 4^ % per annum, comp. annually. 11. $456.75 for 2 yr. 135 da. at 9 %, comp. half-yearly. 12. $791.62 for 1 yr. 198 da. at 8%, comp. quarterly. 13. No payment of interest having been previously made, what is the amount due Oct. 23, 1911, upon a debt of $578.60, contracted May 12, 1908, at 5 % per annum, interest to be com- pounded annually if not paid (time by compound subtraction)? 14. What payment on June 17, 1912, will discharge a debt of $1837.15, incurred Nov. 25, 1910, drawing interest at 4%, payable quarterly, if the debtor agreed to compound all deferred interest payments, and had permitted the second and fourth interest payments to lapse (time in exact days)? 15. All the interest payments having lapsed, what was the balance due Sept. 10, 1910, upon a note of $725.65, dated June 21, 1908, drawing interest at 4 J % per annum, payable semi-annually and specifying that the interest is to be compounded if not paid (time by comp. subtraction)? 16. What was the amount due Aug. 17, 1912, upon a note of $648.75, dated Feb. 17, 1911, drawing interest from date at 5% per annum, and to be compounded quarterly if not paid, if no payment had been made previous to final settlement (time by compound subtraction)? NOTE 2. Much labor in computing compound interest may be saved by the use of a complete compound amount table. The following abridgment will sufficiently illustrate its use: 184 PERCENTAGE COMPOUND AMOUNT TABLE Periods | 1% 2% 3% 4% 5% 6% Periods || 1 1.0100 000 1.0200 0000 1.0300 0000 1.0400 0000 1.0500 000 1.0600 000 1 2 1.0201 000 1.0404 0000 1.0609 0000 1.0816 0000 1.1025 000 1.1236 000" 2 3 1.0303 010 1.0612 0800 1.0927 2700 1.1248 6400 1.1576 250 1.1910 160 3 4 1.0406 040 1.0824 3216 1.1255 0881 1.1698 5856 1.2155 063 1.2624 770 4 5 1.0510 101 1.1040 8080 1.1592 7407 1.2166 5290 1.2762 816 1.3382 256 5 6 1.0615 202 1.1261 6242 1.1940 5230 1.2653 1902 1.3400 956 1.4185 191 6 7 1.0721 354 1.1486 8567 1.2298 7387 1.3159 3178 1.4071 004 1.5036 303 7 8 1.0828 567 1.1716 5938 1.2667 7008 1.3685 6905 1.4774 554 1.5938 481 8 9 1.0936 853 1.1950 9257 1.3047 7318 1.4233 1181 1.5513 282 1.6894 790 9 10 1.1046 221 1.2189 9442 1.3439 1638 1.4802 4428 1.6288 946 1.7908 477 10 11 1.1156 683 1.2433 7431 1.3842 3387 1.5394 5406 1.7103 394 1.8982 986 11 12 1.1268 250 1.2682 4179 1.4257 6089 1.6010 3222 1.7958 563 2.0121 965 12 13 1.1380 933 1.2936 0663 1.4685 3371 1.6650 7351 1.8856 491 2.1329 283 13 14 1.1494 742 1.3194 7876 1.5125 8972 1.7316 7645 1.9799 316 2.2609 040 14 15 1.1609 690 1.3458 6834 1.5579 6742 1.8009 4351 2.0789 282 2.3965 582 15 16 1.1725 786 1.3727 8570 1.6047 0644 1.8729 8125 2.1828 746 2.5403 517 16 17 1.1843 044 1.4002 4142 1.6528 4763 1.9479 0050 2.2920 183 2.6927 728 17 18 1.1961 475 1.4282 4625 1.7024 3306 2.0258 1652 2.4066 192 2.8543 392 18 19 1.2081 090 1.4568 1117 1.7535 0605 2.1068 4918 2.5269 502 3.0255 995 19 20 1.2201 900 1.4859 4740 1.8061 1123 2.1911 2314 2.6532 977 3.2071 355 20 DIRECTIONS. 1. To find the compound amount, multiply the amount of $1 for the given number of periods and at the rate % per period, as given in the above table, by the given principal. 2. To find the compound interest, deduct $1 from the amount of $1 for the given number of periods and at the rate % per period, as given in the above table, to find the compound interest of $1; and multiply by the given principal. CAUTION. The learner is warned against the error of supposing that the compound amount or compound interest at one rate per period is proportioned to the compound amount or compound interest at some other rate for the same time; that is, that the compound amount or compound interest at 7 % is one- sixth more than the compound amount or compound interest at 6 %. For, $1 + the interest of that $1, at 7 %, cannot evidently be one-sixth more than another $1 + the interest of that other $1, at 6 %; that is, that ($1 X 1.07 X 1.07 X 1.07) $1, or the compound interest of $1 at 7 %, for 3 years, can be one-sixth more than ($1 X 1.06 X 1.06 X 1.06) - $1, or the compound interest of $1 at 6 % for the same number of years. If their multiplicands TRUE DISCOUNT 185 are the same, simple products will be proportioned to their respective multi- pliers; but continued products have constantly varying multiplicands, and cannot therefore be thus proportioned. 17. Find the compound amount of $785.30 for 8 years at 6 % per annum compounded semi-annually. SOLUTION 8 years = 16 half-years; 6 % per annum = 3 % per half-year. $1.60470644 = amt. of $1 for 16 periods at 3% per period. X 785.30 = given number of dollars. $1260.175967332 = comp. amt. of $785.30 for 16 periods at 3%. Find from the table the compound 18. Amount of $968.35 for 13 yr. at 6 % per yr., comp. yearly. 19. Amount of $1873.42 for 9 yr. at 5 % per yr., comp. yearly. 20. Amount of $628.95 for 6 yr. at 8 %, comp. half-yearly. 21. Amount of $716.25 for 8 yr. at 4%, comp. half-yearly. 22. Interest of $845.73 for 3 yr. at 12 %, comp. quarterly. 23. Interest of $467.28 for 4 yr. at 4 %, comp. quarterly. 24. Interest of $2875.50 for 7 yr. at 6 %, comp. yearly. 25. Interest of $943.25 for 5 yr. at 10%, comp. half-yearly. TRUE DISCOUNT 244. INTRODUCTION. The sum to be paid in discharge of a debt will vary with the time of payment. If paid on the due date, the sum to be paid will be the debt itself, without any increase for interest or decrease for discount; if paid after it is due, the sum to be paid will be the original debt increased by the accrued inter- est thereon from its due date to the date of postpayment; and if paid before it is due, the sum to be paid will be the original debt diminished by the interest on the prepayment from the date of prepayment to the due date of the debt. That is, interest is a compensation for the use of money [224, b] . Therefore, if money is used after the stipulated time of payment, the debtor should pay compensatory interest thereon for every day during which the creditor has been deprived of the use of what was lawfully his; and if money is paid before the stipulated time, the debtor should receive compensatory interest thereon for every day dur- ing which he has deprived himself of the use of that which was 186 PERCENTAGE legally his. In either case, the question is, what sum paid at a preceding or subsequent date, is equivalent to a given sum which is legally payable at a certain fixed date; or, with particular ref- erence to a prepayment as in true discount, what sum of money placed at interest on the date of prepayment at the current rate per annum will amount to the same sum as the given debt on the date upon which that debt is due. 245. The worth of a debt is its value as expressed in the legal currency of the country in which it is estimated. 246. The present worth of a debt is its value on the day it is paid. NOTE. The true or correct value of a debt due at a certain future time and paid before its maturity, that is, the present worth of that debt by true discount, must be such a principal [226] as loaned on the day of prepayment at the current rate per annum, will amount to that debt on the day of maturity [242]; for the creditor will then have received the principal and the use (in- terest) of that principal for the days of prepayment, and if the two combined (amount) are equal to the debt at maturity, it must be an equitable discharge of that debt. 247. The future worth of a debt is its value on the day it is due. NOTE. Future refers to any time yet to come; but as here used, it is restricted to that specific future time at which a debt falls due. Hence, the future worth of a non-interest bearing debt must be the debt itself; but the future worth of an interest bearing debt must be the debt plus the accrued interest thereon from the date upon which it commenced to draw interest to its due date. 248. True discount is the deduction from the future worth of a debt which is equal to the interest upon its present worth for the days of prepayment. NOTE. True discount is so-called because it is the true or correct deduc- tion; that is, it is the compensatory interest upon the exact sum of the use of which the debtor has deprived himself by paying the debt before it was legally due. ILLUSTRATIVE EXAMPLE 249. A debt of $764.50, legally due Oct. 19, 1911, was paid June 25, 1911. What was the sum paid, allowing true discount at 6%? What was the true discount? TRUE DISCOUNT 187 SOLUTION EXPLANATION. The present solution is similar to the second June 25 to Oct. 19 = 116 da. solution of 111. Ex., 242, to which 60. $764.50 the student is referred for a full 1.16 60 explanation. 6l7l6)45870.00($750 Present Worth. Divide 60 times the per- 428 1 2 centage of future worth ($764 . 50 X 60 = $45870) by 60 times its specific rate (1 + -Vfc 6 ) X 60, 30580 or 61.16, obtaining $750 as the Q base, or required present worth. If preferable, the first solu- $764.50 - $750. = $14.50, true disct. tion of 111. Ex., 242, may be used. VERIFICATION. Find the interest of the obtained present worth ($750) for the days of prepayment (116), and add this interest ($14.50) to the present worth ($750). If the result ($764.50) equals the future worth, the solution may be regarded as mechanically correct. 250. SUMMARY. True discount is an application of 242, the present worth of a debt corresponding to the principal; the true discount, to the interest of the principal for the days of prepay- ment; the future worth, to the amount; the days of prepayment to the time; and the rate % of discount, to the rate. Hence, to find the present worth: RULE. Multiply the future worth by such a part of 860 as 1 % is part of the given rate per annum, and divide the product by the same part of 360 increased by the days of prepayment regarded as hun- dredths. NOTE 1. To avoid fractions, if the rate per annum of discount is not an aliquot part of 360, multiply the future worth by 360, and divide the resulting product by 360 increased by as many times the days of prepayment regarded as hundredths, as the given rate is times 1 %. Thus, in the 111. Ex., ($764.50 X 360) -5- (360 + 1.16 X 6) = $750 [Note, 242]. NOTE 2. To find the present worth of an interest-bearing debt, first find its future worth (amount due at maturity, Note, 247) ; and then proceed as with other future worths. NOTE 3. If the time of prepayment is expressed in years or months, reduce to days upon the basis of 360 days to each year, or of 30 days to each month. If dates are given, find the days of prepayment in accordance with local usage. 188 PERCENTAGE EXAMPLES FOR PRACTISE Find the present worth and the true discount of 1. $968.75 at 6 %, if paid 49 days before maturity. 2. $723.46 at 5 %, if paid 127 days before maturity. 3. $2965.80 at 4 } %, if paid 213 days before maturity. 4. $7816.35 at 8%, if paid 53 days before maturity. 6. $527.90 at 7%, if paid 165 da. before maturity [Note 1]. 6. $845.12 at 3? %, if paid 87 days before maturity. 7. $693.50 at 3 %, paid 2 yr. 139 da. before maturity [Note 3]. 8. $378.65 at 4%, paid 1 yr. 173 da. before maturity. 9. Find the true discount on $1945.80 at 5 %, if paid 98 days before it is due. 10. What is the true discount on a debt of $635.70, at 6 %, if paid 253 days before maturity? 11. A house was bought for $3600, payable one-half cash, and the remainder in 6 months without interest. What equivalent single cash payment can be made on date of purchase, allowing true discount on the deferred payment at 5% per annum? 12. The same kind of broadcloth can be bought from A at $3.50 per yard and 2 months' credit, and from B at $3.58 per yd. and 6 months' credit. Of whom will it be more advantageous to buy; and how much will be gained by purchasing 95 yards at the more favorable of these terms, allowing true discount at 6 % per annum? 13. Merchandise amounting to $726.85 was bought on May 23, 1911, at 90 days, upon which a trade discount of 25%, 20%, and 10 % was allowed. What cash payment on July 7, 1911, will be in full payment of the purchase, allowing true discount at 4J % per annum, and computing the exact time in days? 14. The same brand of flour can be bought from one dealer at $6.25, and 4 months' credit, and from another dealer at $6.20 cash. How much per barrel will be saved by purchasing at the more advantageous of these terms, allowing true discount at 5 % per annum? 15. A debt of $1785.90, bearing interest at 5% from Apr. 17, 1911, and legally due Nov. 13, 1911, was paid Aug. 6, 1911. BANK DISCOUNT 189 What was the sum paid, allowing true discount at 6 %, and com- puting the exact time in days? 16. Merchandise amounting to $948.60 was bought on 3 months' credit. What payment made 1 mo. 12 da. after the date of purchase will discharge the debt, allowing true discount at 4 %? BANK DISCOUNT 251. Bank discount is a deduction from the future worth of a debt which is equal to the interest upon that future worth for the number of days it is paid before maturity. NOTE. As true discount is the difference between the present worth of a debt (the principal in true discount, 248) and its future worth (the principal in bank discount), so the difference between the two discounts must be the interest upon the difference between the present worth and the future worth, that is, the interest upon the true discount. 252. The maturity of a debt is the date upon which it is legally due. NOTE 1. By the law of a few States, if a debt is expressed in the form of a note or draft, 3 days, called days of grace, are added to the nominal time (time named upon the face of a note or draft) in determining its maturity. The nominal maturity of a note or draft is therefore the expiration of the time named upon its face; and the legal maturity, if grace is allowed, will be 3 days thereafter. If no grace is allowed, as in most States, the two maturities fall upon the same date. NOTE 2. In finding the maturity of commercial paper which is drawn payable a given number of months after date, if it should fall nominally due on the 29th, 30th, or 31st of a month which contains fewer than that number of days, it will be nominally due on the last day of such a month; and if grace is allowed, it will be legally due on the third day of the next following month [Note 2, 163]. 253. The term of discount is the number of days between the date of discount and the date of legal maturity. NOTE. In computing the term of discount, the custom of many cities is to include within the interval both the day of discount and the day of maturity, which is equivalent to calculating discount for 1 day more than the time as ordinarily computed. 254. The proceeds of a note or draft is its cash value on the day upon which it is discounted. 190 PERCENTAGE NOTE. The proceeds is what remains of the future worth after deducting the interest thereon for the days of prepayment. 255. Identification of terms. 1. The sum due at maturity is the principal (multiplicand). 2. The specific rate of discount or of proceeds is the rate (multiplier). 3. The discount or the proceeds is the percentage (product). NOTE. As the proceeds of a debt equals the sum due at maturity minus the interest thereon for the days of prepayment [Note, 264], so the specific rate of proceeds must equal the rate of the sum due at maturity or base (100 %) minus the specific rate of interest for the days of prepayment at the given rate of interest per annum [Note, 227]. TO FIND THE BANK DISCOUNT OR THE PROCEEDS ILLUSTRATIVE EXAMPLE 256. A note of $875.90, dated May 24, 1912, legally payable 90 days after date, was discounted on July 5, 1912, at 6 %. Find the date of maturity, the term of discount, the discount, and the proceeds. SOLUTION May 24 + 90 days = Aug. 22, maturity. July 5 to Aug. 22 = 48 da., term of discount. $875.90 = discount for 6 da. at 6 %. 007 2 = discount for 48 da. at 6 %. $875.90 - $7.01 = $868.89, proceeds EXPLANATION. 90 days after the date of the note (May 24) will require the 7 remaining days of May + the 30 da. of June + the 31 da. of July ( = 68 da.), and 22 da. of Aug., thus making Aug. 22, the expiration of the 90th day, or the date of maturity. From July 5 (the date of discount) to Aug. 22 inclusive (the date of maturity) = 48 da. (the days of prepayment, or term of discount). Interest of the sum due at maturity (face of note, $875.90) for the time of prepayment (48 da.) equals the compensatory discount for paying the note before it is legally due ($7.01 -). The worth at maturity ($875.90) minus the discount for prepayment ($7.01) equals the proceeds or present worth of the note on the date of dis- count ($868.89). BANK DISCOUNT 191 SUMMARY. 1. Bank Discount is the interest upon the face* of a non-interest-bearing note, or upon the amount due at maturity of an interest-bearing note, or upon the future worth of any form of debt, for the number of days it is paid before it is legally due. 2. The proceeds equal the sum due at maturity minus the bank discount. *NOTE 1. The face value, or as it is commonly abbreviated, the face of a note, is the value expressed in the body of the note, exclusive of interest. NOTE 2. An interest-bearing note has three values: its face value, its value at maturity (future worth), and its cash value on the day it is discounted or paid (proceeds) . In non-interest-bearing notes, the face value also expresses the value at maturity or future worth. NOTE 3. In allowing interest upon interest-bearing notes, when the time of payment is expressed in months, most banks compute only 30 days to each of the expressed months; but, in computing the term of discount upon such notes, the exact number of days of prepayment is found. EXAMPLES FOR PRACTISE [To THE STUDENT. In the following examples, allow no grace and find the exact number of days in the term of discount, unless otherwise directed by the Teacher to conform with a differing local usage such as Note 1, 252 or Note, 263.] Find date of maturity, term of disc., disc., and proceeds. Face of Date of Term of Date of %of note note note disc. disc. 1. $596.75 May 17, 1911 90 days June 12, 1911 5 % 2. $842.90 Nov. 25, 1911 4 months Jan. 7, 1912 6 % 3. $285.32 Feb. 15, 1912 60 days Mch. 9, 1912 4J% 4. $1763.25 Dec. 3, 1911 3 months Feb. 18, 1912 7 % 6. $934.65 Oct. 19, 1910 30 days Oct. 23, 1910 4 % 6. $2481.12 Mch. 31, 1912 3 months Apr. 20, 1912 3 % 7. $672.85 Sept. 24, 1911 90 days Oct. 2, 1911 3J% 8. $486.50 Jan. 8, 1911 4 months Feb. 16, 1911 5 % 9. $3674.15 Apr. 20, 1912 60 days May 5, 1912 8 % 10. $398.48 June 12, 1911 6 months Aug. 8, 1911 5J% 11. $725.30 Aug. 18, 1911 90 days Sept. 3, 1911 6 % 12. $486.28 July 31, 1911 4 months Aug. 24, 1911 9 % 13. Find the proceeds of the following note, if discounted July 6, 1912, at 8%: 192 PERCENTAGE $2576. BOSTON, MASS., May 26, 1912. Three months after date, value received, I promise to pay to Fairman A. Sadler, or order, Two Thousand Five Hundred Seventy-six Dollars. R. MORTIMER BROWNING. 14. Find the proceeds of the following note, if discounted Nov. 17, 1911, at 5%: $845.75 CHICAGO, ILL., Oct. 18, 1911. Ninety days after date, value received, I promise to pay Philip A. Smith, or bearer, Eight Hundred Forty-five jW Dol- lars, with interest at 6 %. JAMES J. DAVIES. 15. Find the proceeds of the following note, if discounted May 23, 1912, at 4J %, applying Note 3: BALTIMORE, MD., March 3, 1912. Four months after date, value received, I promise to pay to Allen S. Will, or order, Seven Hundred Thirteen -ffj Dollars, with interest at 5 %. GEORGE C. ROUND. 16. Find the proceeds of the following draft, if accepted July 17, 1912, and paid Aug. 20, 1912, allowing discount at 4%: $491 T V<7 NEW ORLEANS, LA., July 13, 1912. At sixty days sight, pay to the order of Henry A. Griesemer & Co., Four Hundred Ninety-one $ Dollars, value received, and charge to the account of ~ , T7 ~ 7 GEORGE W. WHITESIDE. To BROWN BROS. & Co., BALTIMORE, MD. 17. Find the proceeds of the following draft, if discounted Sept. 18, 1911, at 4J%: $874fYir NEW YORK, Aug. 12, 1911. Ninety days after date, pay to the order of Joseph S. Sinclair & Co., Eight Hundred Seventy-four -ffo Dollars, value received, and charge to the account of EDMUND BERKELEY. To W. BANKHEAD THORNTON, ALEXANDRIA, VA. BANK DISCOUNT 193 18. On Oct. 15, 1911, merchandise to the amount of $917.65 was bought at 3 months. What payment on Nov. 8, 1911, will discharge the indebtedness, allowing bank discount at 5 %? 19. What payment on Feb. 13, 1912, will discharge a debt of $2173.85, due May 24, 1912, allowing bank discount at 6%? 20. At 5 %, what are the proceeds of a 60-day note of $468.75, dated June 17, 1911, and discounted July 10, 1911? 72. SOLUTION $647.52 72 129504 453264 TO FIND THE FACE OF A NOTE TO PRODUCE A GIVEN PROCEEDS ILLUSTRATIVE EXAMPLE 257. What must be the face of a note payable 90 days after date, which, if discounted on day of issue at 5%, will produce $647.52 as proceeds? EXPLANATION. $647.52 is a percentage of proceeds, or a product [256, 3]. 5 % is the rate of discount per annum, or a multiplier, but not the specific rate or multiplier to pro- duce $647.52 proceeds [Note, 256]. As the given percentage of proceeds .90 453264 2 ^ 647 - 52 ) equals the required face of the note or future worth minus the interest thereon for 90 days at 5 % per annum, so its specific rate or multiplier must be the specific rate of the required face or future worth (100 %) minus the specific rate of discount for 90 da. at 5 % per 360 da. [Note, 227], or 1 - -f. Hence, divide the given percentage of proceeds, or product ($647.52) by its specific rate or multiplier factor (1 - -ff ), or, clear- ing the divisor of its fraction, divide 72 times the given percentage ($647.52 X 72 = $46621.44) by 72 times its specific rate (72 times 1, or 72) - (72 times fj, or .90), that is, by 72 - .90, or 71.10, obtaining $655.72 as its multiplicand factor, or the required face or future worth. VERIFICATION. Find the interest on the obtained future worth ($655.72) for the given term of discount (90 da.), obtaining $8.20 as the discount, and subtract the discount ($8.20) from the obtained future worth 71.10)46621.44($655.7; 4266 3961 3555 4064 3555 5094 4977 1170 711 459 194 PERCENTAGE ($655.72). If the remainder equals the given proceeds ($647.52), the solution may be accepted as mechanically correct. RULE 1. To find the future worth or base in bank discount, divide the given percentage of proceeds by 100% minus the specific rate of discount for the days of prepayment [Note, 227]. Or, to avoid fractions, 2. Multiply the proceeds by such a part of 360 as 1% is part of the given rate per annum, and divide the resulting product by the same part of 860 after diminishing it by 1 % of the days of prepay- ment. NOTE. If the given rate per annum of discount is not an aliquot of 360, multiply the given proceeds by 360, and divide the resulting product by 360 diminished by as many times 1 % of the days of prepayment as the given rate per annum is times 1 %. Thus, in the 111. Ex. ($647.52 X 360) divided by 360 - (.90 X 5), or by 355.50 = $655.72. That is, instead of dividing the specific percentage of proceeds ($647.52) by its specific rate (1 --f-f ), pre- ferably divide 360 times the specific proceeds by 360 times its specific rate, or by (360 times 1) - (360 times -||). EXAMPLES FOR PRACTISE Find the face of a note which will produce 1. $648.73 proceeds, if discounted 85 days before maturity at 5 %. 2. $945.85 125 6 %. 3. $1374.50 62 8 %. 4. $500 145 4J %. 5. $6258.25 97 3 %. 6. $4000 132 7%. 7. $693.40 76 3| %. 8. $2500 231 4 %. 9. $1257.85 5 mo. 16 da 6 %. 10. I owe $695.80 payable to-day; and in settlement I give my creditor a 60-day note, the proceeds of which if discounted at bank to-day at 6%, will exactly discharge my indebtedness. What was the face of the note? 11. What must be the face of a 90-day note, which, if dis- counted on day of issue at 5 %, will produce $486.25 as proceeds? 12. What should be the face of a note payable 60 days after date, which given to a creditor is equivalent to a cash payment of $500, the rate of discount at the creditor's bank then being 4f %? UNITED STATES RULE FOR PARTIAL PAYMENTS 195 PARTIAL PAYMENTS 258. INTRODUCTION. Partial payments is the process for finding the balance due at a given date upon an interest-bearing debt when one or more partial payments thereon have been pre- viously made. There are several methods in use, only two of which have more than a local importance. The principal of the two is the method of finding the balance due at the time of each successive payment, adopted by the Supreme Court of the United States, and hence called THE UNITED STATES RULE FOR PARTIAL PAYMENTS ILLUSTRATIVE EXAMPLE A note of $2346.75, dated May 16, 1909, drawing interest from its date at 6 %, has the following indorsements: paid Oct. 23, 1909, $486.30; paid July 18, 1910, $76.50; paid Feb. 25, 1911, $500; paid June 17, 1911, $865. What is the balance due Nov. 2, 1911? SOLUTION Face of note or original principal $2346.75 Interest from date of note, May 16, 1909, to date of first pay- ment, Oct. 23, 1909 (time by 30-day method = 157 da.), is $61.41; which is deducted from the first payment ($486.30) leaving a balance of $424.89 to apply to the reduction of the principal 424.89 Balance due after making the first payment $1921.86 Interest of unpaid principal ($1921.86) from its date, Oct. 23, 1909, to the date of the next following payment, July 18, 1910 (265 da.), is $84.88, or $8.38 more than the second payment $76.50). This unpaid interest ($8.38) is carried forward to the accrued interest at the date of the third payment, to find the total interest then due. Interest of same principal ($1921.86) from the date of the last considered payment, July 18, 1910, to the date of the next following payment, Feb. 25, 1911 (217 da.), is $69.51. To this, add the unpaid interest from the last operation ($8.38), making the total unpaid interest on Feb. 25, 1911, equal $77.89, which is deducted from the third payment ($500), leaving a credit balance of $422.11 to apply to the reduction of the unpaid principal 422.11 Balance due after making the third payment $1499.75 196 PERCENTAGE Balance due after making the third payment $1499.75 Interest of new principal ($1499.75) from its date, Feb. 25, 1911, to the date of the next following payment, June 17, 1911 (112 da.), is $28, which is deducted from the fourth payment ($865), leaving a credit balance of $837 to apply to the reduc- tion of the unpaid principal 837. Balance due after making the fourth payment $ 662.75 Interest of new principal ($662.75) from its date, June 17, 1911, to the date of final settlement, Nov. 2, 1911 (135 da.), is 14.91 Balance due at date of final settlement, Nov. 2, 1911 $ 677.66 RULE. 1. From the first payment, deduct the accrued interest upon the principal to the date of said payment, and diminish the principal by the remainder, if any, to find the unpaid principal. 2. Deduct from the second payment the accrued interest upon the unpaid principal to the date of the second payment, and diminish the unpaid principal by the remainder, if any, to find how much of the interest-bearing principal is still unpaid. 3. // the interest due at the date of any payment should prove to be greater than said payment, deduct the payment from the interest and add the unpaid interest to the accrued interest for the next fol- lowing interval to find the total unpaid interest at the end of the next following interval; then proceed in the usual manner. 4. Continue this process consecutively until every payment has been considered. 5. The final balance will be the balance due at the last payment, increased by the interest thereon from the date of the last payment to the date of the final settlement. NOTE 1. By the U. S. rule, any payment is first applied to the dis- charge of the accrued interest to the date of such payment, and the remainder of the payment, if any, is then applied to the reduction of the interest-bearing principal; and any deficiency of a payment to discharge the accrued interest is carried forward to the interest for the next following interval to find the total unpaid interest at the end of the next following interval. NOTE 2. The unpaid balance of the principal at any intermediate settle- ment should never exceed the immediately preceding unpaid balance. Only unpaid interest could have caused it to be greater; and to compute interest upon such a balance would be virtually to compute interest upon the unpaid interest it contains (compound interest) to which the U. S. Rule is opposed. NOTE 3. The Supreme Court uniformly employs the 30-day method [163; Note 4, 233] in finding the time between two dates. MERCHANTS' RULE FOR PARTIAL PAYMENTS 197 EXAMPLES FOR PRACTISE 1. A note of $3750, dated Sept. 26, 1908, drawing interest from date at 6%, has the following payments indorsed thereon: May 9, 1910, $575; Jan. 17, 1911, $750. Find the balance due July 21, 1911. 2. A note of $2975, dated Aug. 20, 1909, drawing interest at 5%, was credited with the following payments: Nov. 12, 1910, $473.85; Mch. 21, 1911, $678.35; Sept. 9, 1911, $296.80. What was the balance due Dec. 23, 1911? 3. What was the balance due Apr. 12, 1912, on a note of $816.25, dated Oct. 18, 1908, and drawing interest at 4J %? Payments: June 24, 1909, $256.25; Mch. 8, 1910, $176.80; Nov. 20, 1910, $156.25; May 25, 1911, $150. 4. A note of $1285.60, dated Oct. 25, 1907, drawing interest at 6%, has the following indorsements: Mch. 13, 1909, $200; Feb. 12, 1910, $352.75; Dec. 29, 1910, $148; Apr. 20, 1911, $275, What was the balance due July 8, 1911? 6. A note of $6500, dated May 23, 1910, drawing interest at 5 %, was indorsed as follows: Nov. 17, 1910, $125; June 20, 1911, $400; Feb. 16, 1912, $100. What was the balance due Sept. 12, 1912? 6. A note of $5000, dated Feb. 17, 1910, drawing 6 % interest, was paid in full on Oct. 10, 1912. What was the final payment, if the following partial payments had been previously made : Dec. 19, 1910, $200; Sept. 16, 1911, $200; May 25, 1912, $200? 7. I have a note of $658.25, drawing interest at 6 %, and dated July 24, 1910, upon which I have received the following payments: Oct. 12, 1910, $45; Apr. 28, 1911, $10. What was the balance due Sept. 16, 1911? 8. A note of $956.25, dated Jan. 16, 1912, drawing 6 % interest, had a payment of $375.80, dated May 14, 1912, indorsed thereon. On Aug. 12, 1912, the maker of the note made a second payment of $250, and gave a new interest-bearing note for the remainder. What was the face of the new note? MERCHANTS' RULE FOR PARTIAL PAYMENTS 259. The merchants' rule is so called because it conforms to the usage of merchants in obtaining the balances of all accounts which contain past-due (that is, interest-bearing) items; as hi 198 PERCENTAGE finding the "cash balance" of any account current [299], or as in fmo5ng the balance due at a given date "per average" [296]. ILLUSTRATIVE EXAMPLE A debt of $478.30, drawing interest at 6 % from May 14, 1911, was credited with the following partial payments: July 7, 1911, $178.35; Sept. 20, 1911, $200; Oct. 25, 1911, $75.50. Find the balance due Dec. 3, 1911. SOLUTION The principal, or debt $478.30 Interest of $478.30 from May 14 to Dec. 3 (exact time, 203 da.) 16.18 Indebtedness at settlement if no payments had been made $494.48 First payment on July 7, 1911 $178.35 Interest of $178.35 from July 7, to Dec. 3 (149 da.) 4.43 Second payment on Sept. 20, 1911 200.00 Interest of $200 from Sept. 20 to Dec. 3 (74 da.) 2.47 Third payment on Oct. 25, 1911 75.50 Interest of $75.50 from Oct. 25 to Dec. 3 (39 da.) .49 461.24 Debit balance on Dec. 3 $ 33.24 RULE. 1. Increase the debt by its accrued interest from the date it commenced to draw interest to the date of final settlement. 2. Find the interest of each payment from the date of its payment to the date of final settlement. 3. Subtract the sum of the payments and their accrued interest from the sum of the debt and its accrued interest. NOTE 1. The above is simply the common practise among merchants of subtracting the total credits at any given date from the total debits at that same date, to find the debit balance at that date. EXAMPLES FOR PRACTISE Direction: In the following problems, find the exact time in days [Note 5, 233], though usage in this respect is not uniform in all cities of the United States. 1. What was the balance due Sept. 28, 1912, on a note of $792.50, dated May 20, 1912, drawing interest from date at 6 %, and containing the following indorsements: July 12, 1912, $150; Aug. 8, 1912, $200. 2. A note of $1873.25, dated Nov. 17, 1910, drawing interest from date at 5%, was credited with the following payments: MERCHANTS' RULE FOR PARTIAL PAYMENTS 199 Jan. 28, 1911, $375.80; May 16, 1911, $468.25; June 25, 1911, $268.10. What was the balance due Aug. 12, 1911? 3. Find balance due Dec. 28, 1912, on a bill of merchan- dise amounting to $398.75, purchased Feb. 16, 1912, on two months' credit, computing interest at 6% from the due date of the purchase. Payments: May 13, 1912, $85.30; July 27, 1912, $49.75; Oct. 3, 1912, $158.60. 4. At 6 %, find balance due Jan. 6, 1912, on a bill of groceries amounting to $678.35, purchased Mch. 3, 1911, on three months' credit without interest. Payments: Apr. 20, 1911, $275; July 26, 1911, $150; Nov. 16, 1911, $145.80. SUGGESTION. If payments are made before a non-interest bearing pur- chase falls due (as in the payment of $275 on April 20, 1911), first find the balance due at maturity by subtracting the prepayments plus their accrued interest to the due date (that is, plus their true discount), from the non-interest bearing purchase, to find the debit balance at maturity, which draws interest from the due date of the purchase; after which proceed as usual with the postpayments, if any. 6. At 6 %, find balance due Nov. 25, 1910, upon a purchase of $846.25 on May 18, 1910, at 90 days' credit without interest. Payments: June 10, 1910, $175.30; July 14, 1910, $200; Sept. 28, 1910, $250. 6. At 8 %, find balance due Jan. 29, 1912, upon a purchase of $328.65 on Apr. 7, 191 1 , at 4 months without interest. Payments : May 18, 1911, $75.40; June 23, 1911, $183.25; July 10, 1911, $45.80. 7. At 9 %, find balance due Feb. 20, 1912, upon a purchase of $419.25 on July 16, 1911, at 30 days without interest. Payments: Aug. 15, 1911, $175.90; Oct. 10, 1911, $85.30; Jan. 6, 1912, $65.70. 8. A purchase of $956.30 on April 13, 1910, and payable 3 months after date, had the following payments credited thereon: May 20, 1910, $200; Sept. 13, 1910, $300; Oct. 25, 1910, $150. Find balance due Dec. 13, 1910, at 6 %. 9. At 6%, find balance due June 12, 1913, upon a purchase of $715.25 on Nov. 20, 1912, at 60 days without interest. Pay- ments; Dec. 10, 1912, $158.60; Jan. 19, 1913, $175; Mch. 3, 1913, $148.65; Apr. 25, 1913, $125.90; May 1, 1913, $100. 200 PERCENTAGE COMMISSION AND BROKERAGE 260. INTRODUCTION. Commission or brokerage is the appli- cation of the principles of percentage, to any business transac- tion performed by one person, called an agent, for the benefit of another person, called a principal. Thus, it may be inconvenient as well as unnecessarily expensive to travel a long distance to the best market for buying a particular kind of merchandise, or one may not possess the requisite skill or knowledge of local condi- tions to buy or sell under the most favorable circumstances; or one may not have access to the floor of the " Exchange" or " Cham- ber of Commerce," or " Board of Trade" at which the fluctuations in prices are posted; and it may therefore be so highly desirable to secure the services of one who possesses these advantages, as to justify the payment of a commission, or remuneration, to him for the service rendered. NOTE. What has been said applies not only to merchandise, but to real estate, or stocks, or bonds, which are bought or sold by an agent; or to collections of money; or to any other authorized special mercantile service rendered by an agent to his principal. 261. Commission or brokerage is a charge made by an agent for buying or selling merchandise, or real estate, or other property, or for collecting or disbursing moneys. NOTE 1. The amount of commission will vary with the extent of the ser- vice rendered, and in buying or selling is usually, estimated at a certain % of the sum paid by the agent for the merchandise bought (its prime cost), or received by the agent for the merchandise sold (its gross proceeds). In many instances, however, it is computed at a certain price per unit, irrespec- tive of the sum paid or received for it, as grain per bushel, flour per barrel, cotton per bale, leaf tobacco per hogshead, stocks per share, small fruits per crate, etc. NOTE 2. If a consignment is sold on credit, in addition to the com- mission for selling, a certain % of the credit sale, called guaranty, is charged by the agent for assuming responsibility for its final payment. NOTE 3. Merchandise forwarded to an agent to be sold is called a con- signment, the person who sends the merchandise being called the consignor or shipper, and the agent to whom it is sent, the consignee. 262. The gross proceeds of a sale is the total sum received by the agent from the buyer of the consignment before deducting the expenses incidental to the sale. COMMISSION AND BROKERAGE 201 NOTE. After all the expenses of the sale have been deducted from the gross proceeds, the remainder is called the net proceeds of the sale. 263. The prime cost of a purchase is the sum paid by the agent to the seller thereof, and does not include any expense inci- dental to the purchase. NOTE. After the incidental expenses have been added to the original purchase money, the result is called the gross cost. 264.- Identification of terms. 1. The prime cost of a purchase, or the gross proceeds of a sale, is the base or multiplicand. 2. The specific rate of any given or required percentage of com- mission, of brokerage, of guaranty, of net proceeds of a sale,* or of gross cost of a purchase, f is the rate or multiplier. 3. The commission, the brokerage, the guaranty, the net pro- ceeds of a sale, or the gross cost of a purchase, is a percentage or product. * NOTE 1. As the net proceeds of a sale equal the gross proceeds or base minus all the expenses, so the specific rate of net proceeds must equal the rate of gross proceeds or base (100 %) minus the rate of all the expenses. t NOTE 2. As the gross cost of a purchase equals the prime cost or base plus all the expenses, so the specific rate of gross cost must equal the rate of prime cost or base (100 %) plus the rate of all the expenses. NOTE 3. If an expense incidental to a purchase or to a sale is a non- percentage, that is, if it has not been computed at a certain % of the base, as drayage, freight, storage, cooperage, advertising, etc., any net proceeds of a sale or gross cost of a purchase which includes one or more such non-percentage or non-product charges, must itself be a non-percentage or non-product, and can therefore have no specific rate or multiplier. Hence, add all the non-per- centage charges to a given non-percentage net proceeds, or subtract them from a given non-percentage gross cost, to reduce it to a percentage or product, after which apply the principles of percentage in the usual manner. EXAMPLES FOR PRACTISE 1. If the prime cost of a purchase is $625.80 and the rate of commission for buying is 3 %, what is the commission? Prime cost or base X specific rate of commission = commission. Find the commission, if the x 2. Prime cost is $378.25, and the rate of commission 3 %. ^ 3. Gross proceeds are $234.70, and the rate of commission 2 %. 4. Prime cost is $1348.62, and the rate of commission 2| %. 202 PERCENTAGE 6. Gross proceeds are $6358.20, and the rate of brokerage f %. 6. Prime cost is $5786.90, and the rate of brokerage f %. 7. If the gross proceeds of a sale are $875.30, the rate of com- mission 2 %, and the drayage $3.75, what are the net proceeds? Base (base X rate of com.) other charges = net proceeds. Find the net proceeds of a sale, if the gross proceeds are 8. $586.75, the commission 3 %, and the other expenses $6.35. 9. $2783.60, the com. 1J%, and the other expenses $37.20. 10. $8625, the brokerage T 3 ir %, and the other expenses $136.50. 11. $7594.16, the brok. f %, and the other expenses $213.75. 12. If the prime cost of a purchase is $763.42, the rate of com- mission li %, and the other charges $35.90, what is the gross cost? Base + (base X rate of com.) + other charges = gross cost. Find the gross cost of a purchase, if the prime cost is 13. $853.75, the com. If %, and the other charges $25.60. 14. $2346.50, the com. li%, and the other charges $17.25. 15. $18250.75, the brok. | %, and the other charges $362.75. 16. $9580, the brokerage f%, and the other charges $237.15. 17. If the prime cost of a purchase is $687.50, and the gross cost $701.25, what is the rate % of commission? EXPLANATION. Divide the specific percentage or product of commission ($701.25 $687.50 = $13.75, commission) by its base or multiplicand factor ($687.50), obtaining 2 % as the specific rate of com., or multiplier factor. Find the rate % of commission, if the 18. Gross proceeds are $685.75, and the commission $27.43. 19. Prime cost is $468.50 and the commission $9.37. 20. Gross pro. are $768.35, and the commission $23.05 [205, a]. 21. Prime cost is $18750, and the commission $70.31 [205, c]. 22. Gross proceeds are $672.85, and the net proceeds $645.94. 23. Prime cost is $296.45, gross cost $312.52, and drayage $1 .25. 24. What is the prime cost of a purchase, if the commission is $20.57, and the rate of commission 3 %? Given percentage -f- its specific rate = its specific base. Find the gross proceeds or the prime cost, if the 25. Commission for buying is $10.69, and the rate of com. 2 \ %. 26. Commission for selling is $27.48, and the rate of com. 3 %. COMMISSION AND BROKERAGE 203 27. Brokerage for buying is $42.50, and the rate of brok. J %. 28. Brokerage for selling is $37.60, and the rate of brok. f %. 29. Net proceeds are $947.36, and the rate of commission 3 %. 30. Net proceeds are $658.56, and the rate of commission 2 %. 31. Gross cost is $5018.75, and the rate of brokerage f %. 32. Gross cost is $972.36, and the rate of commission 1 J %. 33. Net proceeds are $624.25, the com. 3 %, the drayage $6.25, (Net proceeds + non-percentage charges) -5- specific rate = gross proceeds. 34. Net pro. are $487.25, com. 2J%, other expenses $1.50. 35. Net pro. are $855.24, com. 2 %, other charges $2.75. 36. Net pro. are $446.85, com. 4%, other charges $3.15. 37. Gross cost is $751.09, com. 3%, other charges $1.25. (Gross cost non-percentage charges) -5- specific rate = prime cost. 38. Gross cost is $269.52, commission 2 %, other expenses 75 f. 39. Gross cost is $585.45, commission 1 } %, freight $2.25. 40. Gross cost is $874.41, commission If %, drayage $7.50. 41. If the net proceeds are $376.35, and the rate of commis- sion 2J %, what is the commission? 42. If the gross cost is $425.39 and the rate of commission 3J %, what is the commission? 43. An agent received $375.18 to invest in wheat. If his rate of commission was 3 %, what was his commission? 44. An agent remitted $462.95 to his principal as the net pro- ceeds of a sale. If the agent's rate of commission was 1 J %, what were the gross proceeds of the sale? 45. An agent sold 9873 pounds of sugar at 4f cents per pound, charged If % commission for selling, and $2.90 for other expenses. What were the net proceeds of the sale? 46. An agent bought 5278 pounds bacon at 13| cents per pound, charged 2J % commission for buying, and 75 cents for drayage. What sum should his principal remit in settlement? 47. An agent received $348.25 to invest in potatoes. If his commission was 3%, how many barrels could he buy at $2.15 per barrel, and what was the unexpended balance, if any? SUGGESTION. In business practise, no fraction of a barrel is bought; nor is the agent entitled to a commission on any possible unexpended balance 204 PERCENTAGE of a remittance. Hence, divide the remittance ($348.25) by the gross cost of 1 barrel, to find the number of barrels which can be bought with the remit- tance, and regard any remainder from the division as an unexpended balance. 48. An agent sold 495 bushels wheat at $1.03 per bushel, and with the proceeds bought flour at $5.95 per barrel, the commission for buying and selling being 2 %. How many barrels of flour did the agent buy, and what was the unexpended balance? 49. An agent received $298.40 to invest in wheat. If his com- mission was 2J %, how many bushels could he buy at 98 cents per bushel? SUGGESTION. Unlike merchandise bought by the barrel, in buying grain there is no unexpended balance; but the remainder from the division should be expressed as a fraction, the denominator of which equals the number of pounds in one bushel of the particular kind of grain under consideration [Note 1, 142]. In this problem, to obtain pounds or 60ths of a bushel, mul- tiply the remainder from the division by 60, and divide the resulting product by the original divisor [86]. 50. An agent sold 75 barrels of flour at $6.25 per barrel net cash, and 50 barrels at $6.40 per barrel, payable in 3 months. His charges were 2 % for selling, \\% for guaranty, and $2.50 for other expenses. What sum should he remit to his principal in settlement? 51. At what price shall an agent be ordered to buy flour at 2 % commission that, after allowing 5 cents per barrel for freight and 2 cents per barrel for dray age, the flour can be sold at $5.94 per barrel and net 20 % profit? 52. A broker sold 5600 bushels of wheat at $1.09 per bushel, and charged \i per bushel for selling. What were the net pro- ceeds of the sale? 53. An agent sold 28356 pounds leaf tobacco at $15.60 per hundredweight, charged 1 J % commission for selling, and $7.50 for other expenses. What were the net proceeds of the sale? 54. An agent bought 18725 pounds wheat at $1.15 per bushel, charged 2% for commission, 25 cents per load of 36 bushels for drayage, and 6 cents per cwt. for freight prepaid. What sum should his principal remit to him in settlement? 55. An agent received $348.72 to invest in onions. His charges were 3 % for commission, 2 cents per barrel for drayage, STOCKS AND BONDS 205 5 cents per barrel for freight, and $7.60 for other expenses. How many barrels of onions could he buy at $1.95 per barrel, and what was the unexpended balance? SUGGESTIOX. First diminish the remittance by the sum of the charges which were not estimated per barrel, and divide the remainder by the prime cost per barrel increased by the charges which were directly or indirectly estimated per barrel. The quotient will denote the number of barrels. Re- gard any remainder from the division as an unexpended balance. 66. An agent received $427.95 to invest in apples. He bought the apples at $2.75 per barrel and charged 3 % commission for buying, 25 cents per load of 10 barrels for drayage, 6 cents per barrel for freight, and $2.95 for advertising. How many barrels of apples could the agent buy with the remittance, and what was the unexpended balance? SUGGESTION. Divide the drayage per load by the number of barrels per load to find the drayage per barrel; and proceed as in Ex. 55 to find the num- ber of barrels. Then multiply the drayage per barrel for full loads by the number of barrels which the last load lacks of being a full load, to find the extra drayage (in excess of the former estimated drayage per barrel for full loads) incident to the last incomplete load; and deduct it from what would otherwise have been the unexpended balance, to find the correct unexpended balance. 57. An agent received $485.30 to invest in corn. His charges were 2 % for commission, 40 cents per load of 50 bushels for dray- age, and 5 cents per cwt. for freight prepaid. How many bushels and pounds (56ths of a bushel) did he buy at 65 cents per bushel? 58. An agent received $472.10 to invest hi potatoes. He charged 3% commission for selling, 35 cents per load of 12 barrels for drayage, 4 cents per barrel for freight, and $2.40 for other expenses. How many barrels of potatoes could the agent buy at $1.95 per barrel, and what was the unexpended balance? STOCKS AND BONDS 265. INTRODUCTION, (a) A corporation is an association of persons which is authorized by a special legal instrument called a charter to transact business under certain limitations and obli- gations as if it were a person. (6) The stock of a corporation is so much of its capital as is represented by its shares. A share of a corporation is one of the 206 PERCENTAGE equal parts into which its capital stock is divided. As a con- venience in expressing its market value, in declaring dividends or assessments, in computing the premium or discount at which it is bought or sold, the capital stock is usually divided into as many shares of $100 each, or of some multiple or aliquot thereof, as will equal the required capital. The owner of one or more of these shares is called a stockholder, and the instrument issued by a cor- poration to each stockholder to certify how many shares he holds is called a stock certificate. (c) The bonds of a corporation are so much of its capital as is represented by its bonded indebtedness. For convenience in making interest calculations, etc., each bonded loan is usually divided into as many equal parts of $100 each, or of some mul- tiple or aliquot thereof, as will equal the amount of the loan. Those who take one or more of these equal parts of a loan are called bondholders. (d) Computations in stocks or bonds are made in accordance with the general principles of percentage, the par or face value of the stock or bond being very naturally regarded as the basis of such calculations [207]. Thus, if a corporation prospers and in consequence is paying better returns upon its par or face value than the same sum of money invested in other equally safe ways, its stock will sell at such a % of premium (increase) on its par value as will approximately equal this difference ; or if the returns on its par or face value are not so good as those of the same sum of money invested in other equally safe ways, its stock is not so desirable, and will only sell at such a % of discount (decrease) on its par value as will approximate the deficiency; etc.; etc. 266. The par value is the value specified upon the face of stocks or of bonds. 267. The market value is the value of stocks or of bonds upon the stock exchange or stock market. NOTE. If the market value exceeds the par value, stocks or bonds are said to be above par or at a premium; and if the market value is less than the par value, stocks or bonds are said to be below par or at a discount. The amount of premium or discount per share is the difference between its par value and its market value. 268. A dividend is that part of the earnings of a corporation which is divided among its stockholders. NOTE. Occasionally, though rarely, special dividends have been declared by corporations from other sources than their earnings, as from loans nego- tiated for that purpose, or from the sale of an extra issue of stock, etc. STOCKS AND BONDS 207 269. An assessment is a sum levied upon the stockholders of a corporation to meet its losses, or its extraordinary expenditures for betterments, etc. 270. A quotation is the market value of $100 par value of stocks or of bonds. NOTE. Mining stocks are mostly quoted at so much per share, without reference to their par value. In some stock exchanges, all stocks, the par value per share of which is less than $100, are thus quoted. Bonds of all denomina- tions are quoted at the market value of each $100 of their face value. 271. A stock-broker is a person whose business it is to buy or sell stocks pr bonds for others. NOTE. The charge of a broker for buying or selling stocks or bonds, is called brokerage. Brokerage is estimated at a certain % of the par value, or at so much per share. 272. Identification of terms. 1. The par value of stocks or of bonds is the base or multiplicand understood when no other base is specifically mentioned. 2. The specific rate of premium, of discount, of dividend, of assessment, of market value, of prime or gross cost when buying, or of net or gross proceeds when selling, is the rate or multiplier. 3. The premium, the discount, the dividend, the assessment, the market value, the prime or gross cost of a purchase, the net or gross proceeds of a sale, is a percentage or product. NOTE 1. A quotation is the specific rate % of market value. Hence, as stocks and bonds are bought and sold at their market value, the quotation should be regarded as the specific rate % of prime cost when buying, or the specific rate % of gross proceeds when selling. The usual symbol of a rate ( %) being omitted from the quotation, the learner is cautioned against the common error of confounding a quotation which is a rate or multiplier in its relationship to other terms, with a percentage which is a product in its relationship. NOTE 2. If there is no brokerage, the quotation or specific rate of prime cost when buying (Note 1) will also be the specific rate of gross cost or of investment; but if brokerage is included in the transaction, the rate of broker- age must be added to the quotation or specific rate of prime cost to obtain the specific rate of gross cost. Reversely, the rate of brokerage must be sub- tracted from any obtained rate of gross cost to find a required quotation or rate of prime cost. NOTE 3. If there is no brokerage, the quotation or specific rate of gross proceeds when selling (Note 1) will also be the specific rate of net proceeds; 208 PERCENTAGE but if brokerage is included in the transaction, the rate of brokerage should be deducted from the quotation or specific rate of gross proceeds to find the specific rate of net proceeds. Reversely, the rate of brokerage should be added to any obtained rate of net proceeds to find a required quotation or rate of gross proceeds. NOTE 4. The product of two factors being the same irrespective of the order of their arrangement, a favorite practise among brokers is to place the specific rate of the required percentage in the usual position of the multiplicand and to regard it as the required dollars per share; and to place one-hundredth of the par value or base in the usual position of the multiplier, and to regard it as the number of shares. EXAMPLES FOR PRACTISE [In the following examples, the par value per share is understood to be $100, when not otherwise expressed.] Find the 1. Dividend on $4700 par of stock, rate of dividend 5 %. Par value or base X specific rate of dividend = dividend. 2. Assessment on $6200 par of stock, rate of assessment 3? %. 3. Dividend on 163 shares of stock, rate of dividend 7%. 4. Premium on $1900 par of stock, rate of premium 4f %. 5. Discount on 72 shares of stock, rate of discount 12f %. Find the market value of 6. $2300 par of bonds if the rate of premium is 6J %. Par value or base X rate of market value [Note 1], = market value. 7. 59 shares of stock if the rate of discount is 16f %. 8. $2700 par of bonds if the rate of premium is 12f %. 9. 148 shares of stock if the quotation is 87 J. 10. $3700 par of stock if quoted at 18 f % discount. 11. $8600 par of bonds if quoted at 29| % premium. Including \ % brokerage, find the cost of 12. $6800 par of stock if bought at 9| % premium. Par value or base X rate of gross cost [Note 2], = gross cost. 13. 43 shares of stock if bought at 118f . 14. $9300 par of bonds if bought at 23| % premium. 15. $8500 par of stock if bought at 2J % discount. STOCKS AND BONDS 209 Including J % brokerage, find the net proceeds of 16. $1800 par of bonds if sold at 105f . Par value X rate of net proceeds [Note 3], = net proceeds. 17. 37 shares of stock if sold at 7f % discount. 18. $2500 par of bonds if sold at 109J. 19. $1200 par of stock if sold at 8f % premium. If the par value of stock is 20. $7300, the market value $7592, find the % of premium. Percentage of market value -5- par value or base = rate of market value. Rate of market value or quotation rate of par (100 %) = specific rate of premium. 21. $2800, market value $3433.50, find quotation [205, c]. 22. $5400, the market value $6277.50, find the % of premium. 23. $8200, market value $6211.50, find the % of discount. 24. $3900, premium $755.63, find % of premium [205, a]. 26. $3600, the discount $265.50, find the % of discount. If, allowing J % brokerage, the cost of > 26. 73 sh. stock is $7637.63, at what % prem. were they bought? v Gross cost -5- par value or base = rate of gross cost. Rate of gross cost rate of brok. = rate of prime cost, or quo. [Note. 2]. Quotation rate of par (100 %) = specific rate of required premium. 27. 163 shares is $14996, at what quotation were they bought? 28. 52 shares is $4329, at what % discount were they bought? 29. $8400 par of stock is $9544.50, at what % premium were they bought? If, allowing \ % brokerage, the net proceeds of 30. 63 shares are $5213.25, at what % discount were they sold? Net proceeds -H par value or base = rate of net proceeds. Rate of net pro. + rate of brok. = rate of gross pro., or quotation [Note 3]. Rate of par (100 %) rate of quo. = specific rate of required discount. 31. $3400 par of stock are $3574.25, at what quo. was it sold? 32. 92 sh. stock are $7854.50, at what % disc, were they sold? 33. 126 shares of stock are $15576.75, at what % premium were they sold? 210 . PERCENTAGE Find the par value, if the 34. Premium is $191.25, and the rate of premium 12f %. Percentage of prem. -r- specific rate of prem. = par value or base. 35. Discount is $954.75, and the rate of discount 8f %. 36. Dividend is $829.50, and the rate of dividend 3J %. 37. Assessment is $49.50, and the rate of assessment 5J %. 38. Market value is $5875.25, and the quotation 82f . 39. Market value is $9634.25, and the rate of premium 8i %. 40. Market value is $5372.25, and the rate of discount 7| %. Allowing J % brokerage, what par value of stock can be 41. Bought for $6424.75, if quoted at 103J? Quo. or rate of prime cost + rate of brok. = specific rate of gross cost. Percentage of gross cost -4- its specific rate = its par value or base. 42. Bought for $2259.75, if the quotation is 98|? ' 43. Bought for $6243.75, if the rate of premium is 38f %? 44. Bought for $2932, if the rate of discount is 8} %? 45. Sold for $10631.25, if the quotation is 78|? Quo. or rate of gross proceeds rate of brok. = rate of net proceeds. Percentage of net proceeds -5- its specific rate = its par value or base. 46. Sold for $1384.50, if the quotation is 115J? 47. Sold for $5577, if the premium is 26J %. 48. Sold for $8066.75, if the discount is 1J %? /.^Jjv^?. How many shares of stock can be bought at 2f % discount ^ 5nd sold at If % premium, to net a profit of $297, allowing J % > brokerage for buying and the same for selling? .j^X 50. What are the net proceeds of 126 shares of stock sold at 1S| % discount, brokerage f %? / 51. A gentleman bought stock at 15i % premium, brokerage \ %, and afterwards drew a dividend of $480 upon the stock wjiich he purchased. If the rate of dividend was 5 %, what was yrfie total cost of the stock? y 52. A man bought 80 shares of stock at 98 J ; and subsequently sold 36 shares at 99 J, and the remainder at 99f . Allowing \ % brokerage for buying and for selling, what was his net gain? / 53. I receive an annual income of $280 from 5 % bonds bought at 92f , brokerage f %. What was the cost of the bonds? RELATION OF INCOME TO INVESTMENT 211 54. A man invested $19857.50 in 4% bonds purchased at 84f , brokerage %. What was his annual income from the purchase? 55. An investor derives a quarterly income of $525 from 6 % bonds bought at 28J% discount, brokerage \%. What did he pay for the bonds? 56. If $18716.25 is realized as net proceeds from the sale of 13800 par of bonds, at what % premium were they sold, allowing \ % for brokerage? RELATION OF INCOME TO INVESTMENT 273. INTRODUCTION, (a) As viewed by a corporation when it proposes to distribute a dividend or to levy an assessment, etc., the par or face value of the stock would be the only unchanging base that could be used as a standard in prorating such a divi- dend or assessment among its stockholders which would apply generally to all its shares, irrespective of their constantly fluctu- ating market value on the different days of their purchase by the present holders of the stock. From the standpoint of the inves- tor, however, an entirely different standard would suggest itself in considering the expediency of an investment in a particular class of securities. The thought uppermost in his mind would very naturally be the % of income which he would realize upon his investment, rather than the % of income he would realize upon the par value. Other things being equal, a high rate of dividend on stocks or of interest on bonds, the market value of which is at a high premium, may not be so desirable as a lower rate of divi- dend on stocks or of interest on bonds the market value of which is at a lower premium, or possibly at a discount. Hence, in con- sidering the expediency of a particular purchase, investors regard the sum invested (gross cost) as the base. (b) If the investment is the contemplated base, it should be expressed by placing the phrase "on investment" immediately after the rate % which refers to such a base. Otherwise, in many instances, the par value may be misunderstood as the intended base. Thus, in the expressions "5% stocks," "6% bonds," etc., no base being expressed, they are respectively understood to refer to stocks paving 5% dividends on par value, bonds paying 6% interest on par value; but when reference is made to " stock pay- ing 5 % income on investment," it is meant that the stock referred to pays a dividend on its par value which is equivalent to 5 % on the necessary investment to purchase it. 212 PERCENTAGE (c) If, in the same relation, two rates % of income are men- tioned, one of which is understood to be on par value and the other is expressed on investment, the rate % of income on par (100) may be regarded as that many dollars of income per share (a percentage or product), and the rate % of income on invest- ment may be retained as the specific rate of that income, or multiplier, (d) Similarly, any rate % on par value is convert- ible into the same number of dollars per share of the same specific name as the rate, (e) Conversely, any result obtained in dollars per share may be accepted as so much % on par value of the same specific name. Identification of terms. 1. The indicated investment is the base or multiplicand. 2. The % of income on investment is the rate or multiplier. 3. The income is the percentage or product. EXAMPLES FOR PRACTISE At what price per share must '.. 7 % bonds be bought to realize 8 % income on investment? 7 % bonds are bonds paying $7 interest on $100 [Intro., c]. Hence, Percentage of income ($7) + its specific rate on inv. (8 %) = $87|, inv. . 6 % stock be bought to realize 10 % income on investment? . 5 % bonds be bought to produce 4 % income on investment? */4. 8 % stock be bought to yield 5 % income on investment? l/tfi 9 % bonds be bought to produce 8 % income on investment? Including f % brokerage, at what 'quotation must ^/' 6. 5 % stock be bought to yield 8 % income on investment? 5 % stock means stock paying $5 dividend per $100 share. Hence, inc. per sh. ($5) -f- rate of inc. on inv. (8 %) = inv. per sh. ($62.50). Investment or gross cost per share ($62|) brokerage per share ($|) = prime cost per sh. ($62 f ), or rate of quotation (62|) [See Introduction, e]. /7. 4 % bonds be bought to realize 5 % income on investment? 8. 6 % stock be bought to yield 8% income on investment? ,9. 7% bonds be bought to yield 10% inc. on investment? ^lO. 3 % stock be bought to yield 4 % income on investment? What % of income on investment will be realized if /ll. 4 % stock be bought at 80. 4 % stock means stock paying $4 dividend on each $100 share. As there is no brokerage, the quotation or prime cost per share also expresses the RELATION OF INCOME TO INVESTMENT 213 investment or gross cost per share ($80). Hence, income per share ($4) -s- in vestment per share ($80) = specific rate of inc. on inv. (5 %). i/12. 6 % stock is bought at 75? ^ 13. 5 % bonds are bought at 25 % premium? 14. 4J % stock is bought at 10 % discount? 15. 4 % bonds are bought at 20 % discount? 8 % stock be bought at 119J %, brokerage J? Income per sh. ($8) -f- inv. per sh. ($119 -f $|) = specific rate of income on the investment base (6|) %. 3 % bonds be bought at 74 J, brokerage i %? 18. 4 % stock be bought at 89 J, brokerage \ %? 5 % bonds be bought at 79 J, brokerage J %? '20. 4 % stock be bought at 112f , brokerage \ %? WhaJ/% stock can be bought 1/21. At 75, and yield 4 % income on investment? Inv. per sh. or base ($75) X rate % of inc. on that inv. (4 %) = income per sh. from that investment ($3), or 3 % stock [See Introduction, e]. f/22. At 120, and produce 5 % income on investment? 23. At 25 % premium and yield 8 % income on investment? 24. At 20 % discount and realize 10 % income on hi vestment? 25. At 124J and yield 4 % on investment, brokerage J %? ^r Investment per share or base ($124 f + $} = $125) X rate % of income on that investment (4 %) = income per share ($5), or 5 % stock. 26. At 74| and yield 8 % income on investment, brokerage \ %? 27. At 79f and produce 5 % income on investment, brok. \ %? 28. At 112f and realize 4 % income on investment, brok. \ %? 29. What % on investment is realized in purchasing a house and lot for $4000 which can be rented at $50 per month, allowing $100 per annum for deterioration in value and $180 per annum for taxes, repairs and other expenses? 30. What % on investment is realized in buying 9 % stock at 74|, allowing \ % for brokerage? 31. What % bonds can be bought at 87f to yield an income of 8 % on investment, brokerage \ %? 32. At what quotation must 3 % bonds be purchased to pro- duce an income of 5 % on investment, brokerage \ %? 214 PERCENTAGE 33. An investor paid $6580 for stock from which he derived an annual income of $329. What % of income upon his invest- ment did he realize? 34. Both being equally secure, which is the more profitable investment, 5% stock bought at 20% premium, or 4% stock bought at 10 % discount, and how much better? 35. A man bought 74 shares of stock at 85f , and afterwards sold the same at 86 J. What was his gain, allowing % brokerage for buying and selling? 36. An investor bought 50 shares of stock at If % premium, drew a dividend thereon of 4 %, and sold it at 98|. What was his gain, allowing f % brokerage each way? 37. A man bought 300 shares of stock at 109|, sold 90 shares at 110J, 150 shares at 11 Of, and the remainder at 11 If. Allowing i % for brokerage in each transaction, what was his net gain? EXCHANGE 274. INTRODUCTION, (a) Exchange is the giving of one thing as an equivalent for another. Therefore, when a credit in one city where it is not needed is bartered for a credit in another city where it is required, at a less cost than the expressage for safely transmitting the credit money from the one city to the other, the transaction is very appropriately termed " exchange." Thus, if A of Baltimore owes $5000 to B of New Orleans; and C of New Orleans owes $5000 to D of Baltimore, and this was known to the parties concerned, the simplest way to dispose of both debts would evidently be for A in Baltimore to pay $5000 to D of the same city, and for C of New Orleans to pay $5000 to B of that city; for this exchange of credits would involve no expressage for transmission of money from either place to the other. Any such arrangement, however, is rendered impracticable by the unlike- lihood of the persons named possessing the requisite knowledge to make such an exchange feasible. To make such a convenient disposition of credits effective, " exchange banks" are established in the principal commercial cities of the world, each exchange bank in one city having as its correspondent a similar bank in other cities, through which it can transact its outside business. Thus, an exchange bank in Baltimore can cash D's draft on C, as well as the drafts of other Baltimore firms on New Orleans, and EXCHANGE 215 mail them for collection to its correspondent bank in New Orleans, the proceeds to be held in New Orleans subject to the order of the Baltimore bank; and A of Baltimore would very naturally go to his local bank to purchase a draft of $5000 upon its New Orleans correspondent and mail it to B, discharging in this manner his indebtedness without the cost of transmitting the actual money, while the Baltimore bank is also saved the expense of expressing from New Orleans to Baltimore the proceeds of D's draft on C by utilizing it in paying A's debt in that city to B, besides receiving a fee for cashing D's draft on C, and a fee for selling to A a draft on its New Orleans correspondent in favor of B. (6) If New Orleans owes Baltimore about as much as Balti- more owes New Orleans the course of exchange between the two cities is said to be at par (that is, on an equality) ; but if New Or- leans owes Baltimore more than Baltimore owes New Orleans, the course of exchange in New Orleans will be at a premium on Baltimore, the amount of the premium varying with the amount of the preponderance of indebtedness and the consequent cost of transmitting the actual money to overcome the preponderance; and at the same time the course of exchange in Baltimore on New Orleans will be at a discount to stimulate the buying of exchange on New Orleans and thus to diminish the preponderance. The highest premium or lowest discount can be but little more than the cost of safely transmitting the sum of money mentioned on the face of the bill of exchange from the debtor city to the cred- itor city. 275. A banker's draft or bill of exchange is an order written by one bank (called the drawer of the exchange), directing another bank (called the drawee of the exchange), to pay a specified sum of money to a designated person (called the payee) or to whomso- ever the payee orders it to be paid. NOTE. When the drawer and drawee reside in the same country, the draft is called a domestic bill of exchange; and when they reside in different coun- tries, it is named a foreign bill of exchange. 276. The face or par of a bill of exchange is the sum specified therein to be paid to the payee or to his order. NOTE. If, as is usually the case, a debt is payable where the payee resides, the face of a bill of exchange in payment of that debt should be the sum due to the payee by the purchaser of the bill, the purchaser defraying the necessary premium charged, or receiving the benefit of any discount allowed. 216 PERCENTAGE DOMESTIC EXCHANGE 277. Domestic exchange is exchange effected between cities of the same country. NOTE. The method of computing domestic exchange is applicable to all countries which use the same coinage, and which must therefore express the rate of exchange directly by premium or discount on the one common coinage. 278. Identification of terms. 1. The face of a bill of exchange is the base or multiplicand. 2. The % of any required premium, discount, or market value, is a rate or multiplier. 3. The premium, the discount, or the market value, is a percentage or product. NOTE 1. The rate of market value of a sight draft equals the rate of its face value (100 %) increased by the rate % of premium, or diminished by the rate % of discount. NOTE 2. The rate of market value of a time draft equals the rate of market value of a sight draft [Note 1] diminished by the specific rate of in- terest for the given time [Note, 227]. Hence, if both are of the same face value and bought at the same rate of exchange, the cost of a time draft is as much less than that of a sight draft, as will equal the interest on its face (the bank discount) for the given time. NOTE 3. As drafts are bought and sold at their market value, their market value must also be their prime cost if bought, or their gross proceeds if sold. Hence, if brokerage is included, it should be computed upon the market value or prime cost, and added to the prime cost to find the gross cost, if purchased; or computed upon the market value or gross proceeds, and sub- tracted from the gross proceeds to find the net proceeds, if sold. NOTE 4. (a) If a debt is payable at the place in which the debtor resides, the amount of the debt should be regarded as the gross cost of the draft, that is, the creditor should assume the expense of remittance; (6) but if the debt is payable at the place in which the creditor resides, the amount of the debt should be taken as the face of the draft, that is, the purchaser of the draft (the debtor) is properly chargeable with the expense of remittance. EXAMPLES FOR PRACTISE 1. Find the cost of a sight draft of $720 at f % premium. Face of draft or base ($720) X rate or prem. (f %) = premium ($2.70). Face of draft ($720) + prem. ($2.70) = market value or cost ($722.70). 2. Find cost of a sight draft of $630 at J % premium. 3. Find market value of a sight draft of $960 at f % discount. DOMESTIC EXCHANGE 217 4. Find proceeds of a sight draft of $568.30 at J % premium. 6. Find cost of a sight draft of $275.85 at J % discount. 6. Find cost of a 30-day draft of $328; prem. f %, interest 6 %. Face of draft or base ($328) X rate or prem. (f %) = premium ($2.05). Interest of $328 for 30 days at 6 % = $1.64, discount for interest. Face ($328) + prem. ($2.05) - int. ($1.64) = cost, $328.41 [Note 4, a]. 7. Find cost of a 60-day draft of $843.75; prem. \ %, int. 6 %. 8. Find cost of a 90-day draft of $476.28; prem. f %, int. 5 %. 9. Find pro. of a 30-day draft of $624; prem. f %, int. 4%. 10. Find cost of a 60-day draft of $460; disc. \ %, int. 6%. Face of draft ($460) X rate of disc. (\ %) = discount ($1.15). Interest of $460 for 60 days at 6 % = $4.60, discount for interest. Face of draft ($460) - disc. ($1.15) - int. ($4.60) = cost ($454.25). 11. Find cost of 90-day draft of $728.16; disc. J%, int. 5%. 12. Find pro. of 30-day draft of $640; disc, f %, int. 6%. 13. Find market value of 60-day draft of $160; discount | %, interest 6%. Find the face of a sight draft which can be bought 14. For $1239.09, if the rate of exchange is % premium. Percentage of market value or product ($1239.09) -5- the specific rate of market value or multiplier [100 % + J %, Note 1] = face value or multi- plicand, $1236. [Note 4, b]. 16. For $875.27, if the rate of exchange is f % premium. 16. For $932.85, if the rate of exchange is f % discount. 17. For $1465.20, if the rate of exchange is J % premium. 18. For $716.50, if the rate of exchange is J % discount. Find the face of a time draft which can be bought for 19. $824.25, payable in 90 days; exch. f % disc., int. 6%. Rate of face (100 %) rate of exchange discount (f % = .00375) rate of int. for 90 da. at 6 % per annum (.015) = rate of market value (.98125). Percentage of market value or product ($824.25) -f- specific rate of market value or multiplier (.98125) = face value or multiplicand ($840). 20. $753.25, if payable in 30 days, exch. J% prem., int. 5%. 21. $1581.70, if payable in 60 days, exch. f % disc., int. 4J %. 22. $3500, if payable in 90 days, exch. \ % prem., int. 6 %. 218 PERCENTAGE 23. $875, if payable in 60 days, exch. J % disc., int. 4%. 24. The net proceeds of a sale were $2785.90, which the com- mission merchant remitted to his principal by a bank draft, pay- able at sight, purchased at f % discount. What was the face of the draft? [Note 4, a] 25. A St. Louis merchant bought $3258.25 worth of goods from a firm in New York payable in 60 days, and when the bill became due he remitted a bank draft to New York in settlement. What was the cost of the draft, if the course of St. Louis exchange on New York was at J % premium? [Note 4, b] FOREIGN EXCHANGE 279. Foreign exchange is exchange between different countries which also have different monetary units. 280. The rate of foreign exchange is the market value of the monetary unit of the country in which the drawee resides when it is estimated in the currency of the country in which the bill of exchange is purchased. NOTE 1. The intrinsic par of exchange is the inherent or true value of such foreign coins as constitute the monetary units of their respective coun- tries, as shown by the weight and purity of the gold or silver which they contain. In 138 will be found a TABLE showing the intrinsic par of the prin- cipal foreign monetary units when expressed in U. S. money, as proclaimed by the Secretary of the Treasury of the United States on January 1, 1912. NOTE 2. The commercial par of exchange, or the rate of foreign exchange, is the market value of foreign monetary units as determined by the balance of trade between the country of the drawer and that of the drawee [274, b]. If the balance of trade is against the country on which the bill of exchange is drawn, that is, if it has bought more from the country of the drawer than it has sold to it, the commercial par of exchange on the country of the drawee will be at a discount upon the intrinsic par; but if the balance of trade is in favor of the country on which the draft is drawn, the commercial par will be quoted at a premium upon the intrinsic par [274, b]. NOTE 3. In the United States, the rate of exchange on Great Britain is expressed by giving the market value of 1 in United States money; on France, Belgium and Switzerland by giving the market value of $1 in francs; on Ger- many by giving the market value of 4 reichsmarks in United States money; and on Holland by giving the market value of 1 guilder in United States money. Hence, FOREIGN EXCHANGE 219 IDENTIFICATION OF TERMS 281. If the rate of exchange expresses the market value of 1 foreign monetary unit in United States money. 1. The ex- change value in United States money of 1 foreign monetary unit is the multiplicand. 2. The number of foreign monetary units is the multiplier. 3. The exchange value in United States money of all the foreign monetary units is the product. 282. If the rate of exchange expresses the market value of $i in foreign monetary units. 1. The number of foreign monetary units that can be bought for $1 is the multiplicand. 2. The number of dollars to be invested in exchange is the multiplier. 3. The num- ber of foreign monetary units that can be bought with all the dollars to be invested in exchange is the product. NOTE 1. As American exchange upon Germany expresses the value in United States money of 4 German reichsmarks, the rate of exchange must be divided by 4 to obtain the multiplicand of 281. NOTE 2. Notes 2, 3 and 4, 278, apply also to foreign exchange, except that in computing interest upon time bills of exchange on Great Britain, 3 days of grace are invariably added to the nominal time. Grace is also allowed in a few American cities upon domestic time bills of exchange. TO FIND THE COST OF A FOREIGN BILL OF EXCHANGE 283. The following forms of solution are employed when the buyer of the exchange owes money which is payable where the payee resides. The face of the bill should then be the amount of the debt, and any expense incurred in buying the bill of exchange should be borne by the remitter. ILLUSTRATIVE EXAMPLES 1. What is the cost of a bill of exchange on London for 673 2s. 6d., if the rate of exchange is 4.8725? SOLUTION. 673 2s. 6d. = 673.125 [111. Ex. 2, 160]. $4.8725 is the exchange value in United States money of 1, or the multiplicand [281, 1]; 673.125 is the number of 's, or the multiplier [281, 2]; therefore, multiply these factors ($4.8725 X 673.125 = $3279.80) to find the exchange value in United States money of all the given 's, or product [281, 3]. 2. What is the cost of a bill of exchange on Berlin for 675.20 reichsmarks if the rate of exchange is 96 J? 220 PERCENTAGE SOLUTION. $.96 4- 4 = $.24i [Note. 1, 282], the exchange value of 1 reichsmark, or the multiplicand [281, 1]; 675.20 is the number of reichsmarks, or the multiplier [281, 2]. Therefore multiply these factors ($.24i X 675.20 = $162.89 -|-) to find the exchange value of all the reichsmarks, or the product [281, 3]. 3. What is the cost of a bill of exchange on Paris for 3245.80 francs, if the rate of exchange is 5.20$? SOLUTION. 5.20j is the number of francs that can be bought for $1, or the multiplicand [282, 1]; and 3245.80 is the number of francs to be bought with all the required dollars to be invested, or the product [282, 3]. There- fore divide the given product (3245.80 fr.) by its given multiplicand factor (5.20i fr.) to find its required multiplier factor (623.89 times $1, or $623.89), the number of dollars to be invested. EXAMPLES FOR PRACTISE Find the cost of a bill of exchange for 1. 416 10s., if the rate of exchange is 4.8650. 2. 725 17s. 6d., the rate of exchange being 4.8715. 3. 926 12s. 6d., if the rate of exchange is 4.8620. 4. 827.60 reichsmarks, if rate of exch. is 96 [Note 1, 282]. 5. 1927.50 reichsmarks, if the rate of exchange is 95f . 6. 3460.25 reichsmarks, if the rate of exchange is 96. 7. 4675.20 francs, if the course of exchange is 5.18|. 8. 798.40 francs, the rate of exchange being 5.19J. 9. 1425.80 francs, if the rate of exchange is 5.18|. 10. 2300 guilders, if the course of exchange is 40f. 11. 826 2s. 6d.; exch. 4.8665, brok. J% [Note 3, 278]. 12. 6350 francs, if exchange is 5.18J, and brokerage J %. TO FIND THE FACE OF A BILL OF EXCHANGE 284. The following forms of solution are applicable when the buyer of the exchange owes money which is payable where he (the buyer) resides. The total cost of the bill of exchange should then be equal to the amount of the debt, thus causing all expenses incurred in buying the bill to be borne by the payee. ILLUSTRATIVE EXAMPLES 1. I paid $925.70 for a bill of exchange on Liverpool. What was the face of the bill, the rate of exchange being 4.86? BANKRUPTCY 221 SOLUTION. $925.70 is the exchange value of all the required 's, or the product [281, 3]; $4.86 is the exchange value of 1, or the multiplicand [281, lj. Therefore divide the given product ($925.70) by its given multipli- cand factor ($4.86) to find the required number of 's, or the multiplier factor (190 9s. 51 d.). 2. I bought through a broker a bill of exchange on Belgium for which I paid him $640.80. If the rate of exchange was 5.18, and the brokerage | %, what was the face of the bill? SOLUTION. $640.80 -s- l.OOi = $640 [Note 3, 278], the number of dollars to be invested in exchange, or the multiplier [282, 2]; 5.18 is the number of francs that can be bought for $1, or the multiplicand [282, 1]. Therefore multiply these two factors (5.18 fr. X 640 = 3315.20 fr.) to find the required number of francs which can be bought with all the dollars to be invested in the bill of exchange, or the product [282, 3]. NOTE. If brokerage is included, regard the amount of the debt to be canceled by the remittance as the gross cost of the bill of exchange, and divide it by 100 % + the rate of brokerage to find the prime cost or market value of the bill [Note 3, 278]. EXAMPLES FOR PRACTISE Find the face of a bill which can be bought for 1. $1825.36, if the rate of exchange is 4.8668 to the . 2. $746.25, if the rate of exchange is 4.855 to the . 3. $825.70, if the course of exchange is 5.19J francs to $1. 4. $2580.20, exchange being 5.18J francs to $1. 5. $942.65, if exchange is 94f to 4 reichsmarks. 6. $6250.75, if exchange is 95i $. to 4 reichsmarks. 7. $675.80, if exchange is 41 \i to 1 guilder. 8. $1248.35, if exchange is 40|^f to 1 guilder. 9. $3650.25, exchange 4.8575 to 1, brokerage J %. 10. $785.90, exchange 41f ?f to 1 guilder, brokerage J%. BANKRUPTCY 285. INTRODUCTION. A firm is said to be bankrupt or insol- vent when its financial resources are not sufficient to pay its debts as they fall due. The assets of a bankrupt firm are its total available property, including the collectible debts due to it from others; and the liabilities of a bankrupt firm are the debts it owes to others. The assignee or trustee is one who is appointed to take charge of the business of a bankrupt firm for the purpose 222 PERCENTAGE of converting its assets into cash sufficient to discharge its financial obligations; or, if its total assets are insufficient to discharge its total liabilities, to distribute among its creditors such a percentage of their individual claims as the firm's total assets when con- verted into cash are of its total liabilities. This distribution of available assets by the assignee among the creditors is called a dividend. Identification of terms. 1. The total liability of the firm is the base or multiplicand. 2. The % of the total liability which is distributed among a bankrupt firm's creditors, is the rate of dividend or multiplier. 3. The sum distributed among the bankrupt firm's creditors is the percentage of dividend or product. NOTE 1. The total dividend to all the creditors will always be the net available assets, that is, the gross available assets diminished by the expenses of assignment. NOTE 2. Whatever the rate of dividend to all the creditors collectively may be, must also be the rate of dividend to each individual creditor. Hence, the total liability to each individual creditor is a base or multiplicand; the % of dividend is the rate or multiplier; and the dividend received by each individual creditor is a percentage or product. EXAMPLES 1. The following statement was rendered by the assignee of a bankrupt merchant: Assets merchandise as per inventory, $8411.15; personal accounts due the firm, $1618.45; real estate, $4500. Liabilities notes due to others, $12000; due to F. A. Sadler for merchandise, $6500; due to R. M. Browning for ser- vices rendered, $2700. If the expenses of assignment were $325.60, what dividend should F. A. Sadler and R. M. Browning receive? $8411.15 + $1618.45 + $4500 = $14529.60, total assets. Total assets ($14529.60) expenses of assignment ($325.60) = net assets ($14204). $12000 + $6500 + $2700 = $21200, total liabilities. Percentage of total net assets or product ($14204) -?- total liability, the base or multiplicand ($21200) = 67 %, the rate of dividend or multiplier. Hence, $6500 X .67 = $4355, dividend to F. A. Sadler [Note 2]. $2700 X .67 = $1809, dividend to R. M. Browning. 2. A firm failed with liabilities amounting to $33864.25 and assets amounting to $24293.09. What % of dividend will the assignee be able to declare; and what dividend should Geo. C. INSURANCE 223 Round & Co. receive whose claim for $837.50 was allowed, if the expenses of the assignment were $1265.40? 3. The available resources of a bankrupt firm were $10284.75, and its liabilities $12537.50. If the expenses of assignment were $756.25, how much should J. P. Smith & Co. receive whose claim for $1785.20 was accepted? 4. A bankrupt firm owed A $580, B $1275.50, and C $875.30. Its realizable assets were $12142.25, and its total liabilities $17856.25. If the expenses of settling the bankruptcy were 4 % of the assets, how much should A, B, and C receive? INSURANCE 286. INTRODUCTION. Insurance is a pecuniary indemnity guaranteed by one party (the insurer) to another party (the in- sured) in certain specified contingencies, such as the death of the insured (life insurance)', or a disabling accident to the insured (accident insurance) ; or against loss to property insured caused by fire (fire insurance) ; etc. The written contract between the insurer and the insured is called a policy. The premium is the sum paid to the insurer for the risk which is assumed, and is estimated at a certain % of the amount of insurance, though frequently expressed by giving the cost per $100 or per $1000 of the risk assumed. Thus, f %, 75 cents per $100, or $7.50 per $1000 are equivalent rates expressed in different forms. Identification of terms. 1. The sum insured is the base or multiplicand. 2. The % of premium is the rate or multiplier. 3. The premium is the percentage or product. NOTE 1. The product of the base and the rate will be the same, irre- spective of the order of their arrangement. Hence, for convenience, when the rate of insurance is expressed at so much per $100 or per $1000, the usual position of the base and rate is reversed, and the rate per $100 or per $1000 is taken as the multiplicand, and the number of hundreds or of thousands of dollars is taken as the multiplier. Thus, the premium for $8000 of risk at $6.25 per $1000, would be computed at 8 times $6.25, or $50. NOTE 2. Fire and marine policies contain an "average clause" under which, if the property is only partially insured and is totally destroyed within the period of insurance, the insurer is held responsible for the total sum in- 224 PERCENTAGE sured; but if the property is only partially destroyed, the insurer is only liable for such a part of the loss, as the sum for which the property is insured is of the full value of the property. EXAMPLES 1. A man insured his house for $6800, the rate of insurance being f %. What premium did he annually pay? Sum insured or base (6800) X rate of prem. (f %) = prem. ($42.50). 2. A house is insured at J %. What was the sum insured if the premium paid is $26.25? 3. A house is insured for $8200 and the premium paid is $71.75. What is the rate of insurance? 4. A house is insured for $7500. If the rate of insurance is | %, what premium did he annually pay? 5. A ship was insured for $75300, the rate of insurance being f %. What was the premium paid? 6. A man had his life insured for $9000 at the annual rate of $35.50 per $1000. What annual premium does he pay? [Note 1] 7. A building valued at $12000 and insured for $8000 was damaged by fire. If the appraisement of the damage was $2400, what sum was received from the insurer in settlement? [Note 2] 8. A stock of merchandise valued at $40000 was insured in one company for $20000 and in a second company for $10000. If the stock was damaged by fire and water at an appraisement of $6000, what proportion of the loss should be paid by each company? 9. A steamboat valued at $120000 was insured in one com- pany for $20000, in a second company for $30000, and in a third company for $40000. If a loss of $15000 was sustained, how much insurance should the owners receive from each company. TAXES 287. A tax is an assessment levied for public purposes upon the person (poll tax or capitation tax), upon the income (income tax), or upon the property (property tax), of the inhabitants of a country. Identification of terms. 1. The assessed value of the property, DUTIES OR CUSTOMS 225 or the amount of income when over a prescribed minimum, is the base or multiplicand. 2. The rate of taxation is the rate or mul- tiplier. 3. The tax is a percentage or product. NOTE. If the rate of taxation is expressed at so much on $100, the com- putation may be made as explained in Note 1, 286; or it may be reduced to an equivalent rate % (cents on $1) by moving the decimal point of the rate on $100 two places to the left. Thus, $1.45 on $100 = $.0145 on $1, or 1.45 %. EXAMPLES What is the tax, if the assessed value of the property is 1. $8500 and the rate of taxation $1.32 on $100? $1.32 on $100 = 1.32 % [Note]. Hence, assessed value or multiplicand ($8500) X specific rate of taxation or multiplier (.0132) = percentage of tax or product ($112.20). 2. $16200 and the rate of taxation %? 3. $9500 and the rate of taxation $1.65 on $100? 4. $12300 and the rate 15 mills on $1? 6. $8700 and the rate of taxation 1J%? 6. $23600 and the rate $1.72 on $100? 7. $17200 and the rate 8 mills on the dollar? 8. My real and personal property are assessed at $5300. What tax will I be required to pay if the rate of property tax is 75 cents on $100, and my poll tax is $1? 9. If the assessed value of the taxable property of a county is $13578200 and the sum required by taxation is $204538, what should be the rate of taxation on $100, if there are 1730 resi- dents who are subject to a poll-tax of 50 cents? 10. If the assessed value of real estate in a town is $758200, and of personal property $215000, and the sum required by taxa- tion is $14469, and 4% of the tax upon personal property is estimated to be uncollectible, what should be the rate of taxation expressed in mills on $1? DUTIES OR CUSTOMS 288. INTRODUCTION. Duties or customs are taxes levied by a nation upon goods imported from foreign countries. By the Payne-Aldrich Tariff Act of 1909, these duties on some articles are fixed at a certain per cent, of their market value at the place 226 PERCENTAGE of purchase, including the value of the container or covering in which the goods are shipped unless such container or covering is otherwise specially subject to duty under some other provision of that tariff act; and on other articles are fixed upon the weight or measure of imported goods without reference to their value. When levied at a certain per cent, of their value, duties are called ad valorem; and when levied at so much per yard, per pound, per gallon, etc., they are called specific. Sometimes, upon the same article, both an ad valorem and a specific duty may be levied; as jute matting 15 cents per square yard and 30% ad valorem. The ports at which duties are collected are called ports of entry; the public building in which business relating to customs is trans- acted is termed a custom house; and the chief official in charge of the customs is named the collector of customs. IDENTIFICATION OF TERMS Ad valorem duties. 1. The market value at the place of pur- chase, reduced to United States money, is the base or multiplicand. 2. The % of ad valorem duty is the rate or multiplier. 3. The ad valorem duty is the percentage or product. Specific duties. 1. The specific duty levied on 1 unit of the quantity subject to such duty, is the multiplicand. 2. The number of units subject to such duty is the multiplier. 3. The total specific duty is the product. NOTE 1. In reducing foreign monetary units to United States money, use Table 138; and when estimating the ad valorem duty, omit any fraction of a dollar from the base if less than 50 cents, and count it as an additional dollar if 50 cents or more. NOTE 2. In reducing tons to pounds, or the reverse, estimate 2240 pounds to a ton. EXAMPLES FOR PRACTISE Find the ad valorem duty upon foreign goods invoiced at 1. 260 17s. 6d., allowing 5 % for leakage, rate of duty 40 %. 260 17s. 6d. = 260.875 [160]. $4.8665 X 260.875 = $1269.548 + [138]. 5 % of $1269.548 = $63.477, allowance for leakage. $1269.548 - $63.477 = $1206.071, practically $1206 [Note 1], dutiable value. $1206 X .40 = $482.40, ad valorem duty. 2. 12768.70 marks, breakage 10 %, duty 30 %. 3. 9586.20 francs, leakage 5 %, duty 45 %. 4. 948 2s. 6d., duty 60 %. SIMPLE PROPORTION 227 5. 17294.40 milreis, duty 25 %. 6. 23742 yen, duty 35 %. Find the specific duty upon an importation of 7. 213 tons white lead at 2J cents per pound. 8. 345 gallons cod-liver oil at 15 cents per gallon. 9. 600 yds. matting, 2 yds. wide, at 3J cents per sq. yd. 10. What is the duty upon 2400 yards carpet which is 30 inches wide, if invoiced at 12s. 6d. per yard, and subject to an ad valorem duty of 40%, and to a specific duty of 28 cents per square yard? 11. What is the duty on 350 gross lead pencils invoiced at 32 francs per gross, if the ad valorem duty is 25 % and the specific duty 45 cents per gross? 12. What is the duty upon an importation of 3700 yards of 27-inch dress goods if invoiced at 7s. 6d. per yard, and subject to an ad valorem duty of 50 % and a specific duty of 8 cents per square yard? SIMPLE PROPORTION 289. INTRODUCTION. Proportion is the name of that process by which a quantity (the required answer) is found which will have the same relation to a similar given quantity (expressing similar units to the required answer), as the given quantity of which the answer is required has to the similar given quantity of which the same specific answer is known. Thus, if 40 bushels of corn cost $30 and it is required to find the proportionate cost of 80 bushels of the same kind of corn, the required cost must have the same relation to the given cost ($30) as the given quantity of which it is required to find the cost (80 bu.) has to the similar given quantity (40 bu.) of which the cost is known; and as 80 bushels of which the cost is required are 2 times 40 bushels of which the cost is known, so the required cost must be 2 times the known cost ($30), or $60. The preceding process is appropri- ately called a proportion because the incomplete couplet of similar quantities of which the required answer is one term ($30 to ?) is proportional to (have the same relation to each other as) another couplet of similar quantities of which both terms are given (40 bu. to 80 bu.). NOTE 1. The two terms of a proportion which are of the same name are called a couplet of that proportion, and every problem in proportion must con- 228 PROPORTION tain at least two couplets (one complete couplet and one incomplete couplet). The first or left-hand term of each couplet is distinguishably called its ante- cedent (going before); and the second or right-hand term, its consequent (fol- lowing after). The relation which the antecedent of any couplet has to its consequent is called the ratio (the relation) of that couplet. NOTE 2. A problem in simple proportion is one in which a complete couplet of similar terms is given which have a certain relation to each other; and in which one term of a second, incomplete couplet of other similar terms is given and it is required to find the remaining term of that couplet which shall have the same relation to its given term, as the corresponding terms of the complete couplet have to each other. NOTE 3. A direct proportion is one in which the required answer will be proportionately increased as the term of which the answer is required, is increased; or decreased as that term is decreased. Thus, if the required result is wages of a given number of men, the greater the number of men employed, the more the wages, and the fewer the number of men employed, the less the wages; or if the cost of a given number of articles is required, the more the number of articles bought, the greater the cost, and the fewer the number of articles, the less the cost, etc. NOTE 4. An inverse proportion is one in which the required answer will be proportionately diminished, as the term of which the answer is required is increased; or increased as the term of which the answer is required, is de- creased. Thus, if the required answer is days in which a given task can be performed by a given number of men, it is evident that the greater the number of men employed upon the task, the fewer will be the number of days to finish it; and the fewer the number of men employed, the greater the number of days to complete it. ILLUSTRATIVE EXAMPLES 1. If 56 pounds of butter cost $11.76, what will 32 pounds of the same kind of butter cost? SOLUTION EXPLANATION. The required cost of 32 Ib. has g ^ the same relation to the given cost of 56 Ib. ($11.76) Gi v #9 and 1$ the fewer will be the days required to complete the 3 wall. Hence, 12 masons (* or f of 9 masons) can 1 inversely build the wall in the inverse of | of 8 days, that is, in f of 8 days, or 6 days. RULE. 1. Compare the given term of which the answer is re- quired with the given term which is similar to it in name and con- cerning which an affirmation is made which is similar in name to the required answer; and express the result of the comparison in the form of a common fraction, inverting its terms when the proportion is seen to be inverse [Note 4]. 2. Multiply the given term which is of the same specific name as the required answer by the obtained fraction. EXAMPLES FOR PRACTISE 1. If 200 pounds of bacon cost $24, what will 475 pounds of the same kind of bacon cost? 2. If 575 yards of cloth cost $477.25, what will 340 yards of the same cloth cost? 3. If $1125 are required weekly to pay 75 operatives in a factory, how much will be required weekly if 40 additional oper- atives are employed at the same wages? 4. How many pounds of beef will feed 320 men if 90 pounds are sufficient to feed 80 men? 5. If 75 sheep of a flock were sold for $292.50, how many sheep from the same flock should be sold at the same rate for $1053? 6. What will 642 tons of coal cost if 49 tons of the same coal can be bought for $303.80? 7. If 52 pounds of ham are bought for $12.48, what should be the cost of 75 Ib. 12 oz. of the same kind of ham? SUGGESTION. If one term of a problem in proportion is a compound number, it should first be reduced to a simple denominate number by 156 or 160; and both it and the term with which it is to be compared should express similar denominate units. 230 PROPORTION 8. If $520 will produce $10.40 interest, how much interest will $820 produce if loaned for the same time and at the same rate? SUGGESTION. All of the given terms of a proportion may be of the same general name without necessarily being of the same specific name. Thus, in Ex. 8, $520 and $820 are specifically dollars of principal; and $10.40 and the required answer are specifically dollars of interest. 9. What principal will produce $17.82 interest if $3784 will produce $104.06 interest at the same rate per annum and in the same time? 10. If a certain principal will produce $19.98 interest in 296 days, in what time will the same principal produce $10.26 interest if loaned at the same rate per annum? 11. A house 30 feet high casts a shadow 18 feet long. How high is a steeple which casts a shadow 75 feet long at the same time in the day? 12. If a tree 50 feet high casts a shadow of 10 feet, how long a shadow will be cast at the same moment by another tree which is 40 feet high? 13. If 30 men can complete a given task in 40 days, how many men are necessary to complete it in 50 days? 14. In how many days can 75 men build a bridge, if it has been estimated that 125 men can build it in 18 days? 15. If 175 pounds of beef are sufficient to feed a gang of laborers for 5 days, how much beef will be necessary to feed them for 15 days? 16. What should be the tax on a house assessed at $4500, if a house in the same town which is assessed at $6000 is taxed $75? 17. A and B are partners in business, the former investing $5000 and the latter $7000. The net gains during a certain inter- val were $3600, which were divided between them in proportion to their respective investments. How much of the gain should each receive? 18. A clothier sold a suit of clothes at a profit of $4.80 and gained 20%. What would have been his % of gain if he had sold the suit at a profit of $6? COMPOUND PROPORTION 231 19. A merchant sold goods for $644 and gained 15 %. What would have been his % of gain if he had sold them for $756? 20. A dealer sold merchandise for $45 and gained 25%. At what price must he have sold the merchandise to gain 30 %? COMPOUND PROPORTION 290. INTRODUCTION. A problem in compound proportion is one in which an answer is required which depends upon more than one couplet of similar terms, and which therefore requires as many separate comparisons as there are given couplets upon which the required answer depends. Each separate comparison is made in the same manner as in simple proportion; and the resultant of the several comparisons (a compound fraction) when reduced to a simple fraction, will express what proportion of the given term which is of the same specific name as the required answer, must be taken to obtain the required answer. ILLUSTRATIVE EXAMPLE If 72 men can dig a sewer which is 240 yards long, 12 feet wide, and 9 feet deep, in 18 days, how many men will be required to dig in similar soil a sewer which is 360 yards long, 15 feet wide, and 24 feet deep, in 12 days? SOLUTION EXPLANATION. If 72 men are required to dig a sewer 240 yards long, the other dimensions being equal, it will require ff wa t <** e l ual > * wil1 ^J require V of 72 men to dig a sewer 24 ft. deep. If 72 men are required to dig a sewer in 18 days, the dimensions of each sewer being the same, it will inversely require {I of 72 men to dig it in 12 days [Note 4, 289]. Each of the above separate comparisons shows what proportion of 72 men would be required if there were no other element entering into the comparison; therefore if the results of the several separate comparisons are combined into 232 PROPORTION a single expression, the resultant compound fraction (If of If of 2 g 4 of |l of 72 men), or simplified by reduction, -^ of 72 men, or 540 men will be the re- quired number of men to dig a sewer 360 yd. long, 15 ft. wide, 24 ft. deep, in 12 da., if all the elements which enter into the comparison are jointly considered. RULE. 1. Separately compare each given term upon which the required answer depends with the corresponding given term of the same name, of which an affirmation is made which is similar in name to the required answer; and express the result of each separate comparison in the form of a common fraction. 2. Combine each separate fraction which expresses a simple pro- portion when it is separately considered into a compound fraction which will express the compound proportion when all the terms are jointly considered; and reduce the resultant compound fraction to a simple fraction in lowest terms. 3. Multiply the given term which is of the same specific name as the required answer by the obtained simple fraction. EXAMPLES FOR PRACTISE 1. If 15 men can make 30000 shingles in 5 days, how many shingles can 25 men make in 9 days? 2. If 12 men can earn $270 in 9 working days, how much can 28 men earn in 5 working days? 3. If $864 loaned for 126 days at 5 % per annum will produce $15.12 interest, what principal loaned at 8% for 54 days will produce $5.10 interest? 4. If a gain of $68 is made by selling merchandise which cost $340 at 20 % profit, what will be the proportionate gain from selling merchandise which cost $720 at 25 % profit? 5. If by walking 4 miles an hour for 9 hours per day a pedes- trian can walk a certain distance in 10 days, in how many days can he walk the same distance at the rate of 3 miles an hour for 8 hours a day? 6. If a contractor charged $336 for digging a cellar 54 ft. long, 35 ft. wide, and 12 ft. deep, what should be his proportion- ate charge for digging another cellar 45 ft. long, 30 ft. wide, and 9 ft. deep? EQUATION OF PAYMENTS 291. INTRODUCTION, (a) Equation, as a general term, means the averaging or equalizing of two or more unequal quantities [Ex. 13, 54]. (6) When applied to paj^ments, it is the process of finding one date at which a single payment of the total of two or more debts can be made which will be equal to paying each sep- arate debt at its own separate due date, (c) When applied to terms of credit it is the process of finding one uniform term of credit for all the items of a purchase or a sale which is the equivalent of the different terms of credit at which the several items were sep- arately purchased or sold. (d) The equivalent date at which two or more payments, due at different dates, can be simultaneously made is called the aver- age or equated date; and (e) the uniform term of credit which is the equivalent of two or more different terms of credit is called the average or equated term of credit. TO FIND THE EQUATED DATE OF PAYMENT WHEN THE DATES OF SALE ARE EQUAL AND THE TERMS OF CREDIT UNEQUAL ILLUSTRATIVE EXAMPLE 292. A merchant sold goods on June 16, 1911, as follows: $600 at 90 days, $800 at 30 days, and $700 at 60 days. What is the equated date for the payment of the sum of the three items? SOLUTION 1911 The use of The use of June 16 $600 for 90 da. = $54000 for 1 da. 800 " 30 " = 24000 " 700 " 60 " = 42000 " " The use of $2100 for (?) days = $120000 " $120000 -r- $2100 = 57 times 1 da., or 57 days. June 16, 1911 + 57 da. = Aug. 12, 1911. EXPLANATION. If the purchaser had been required to pay the sum ot the three items ($2100) on the day of purchase (June 16), he would have been deprived, as per terms of purchase, of the use of $600 for 90 da., or its equiva- lent of 90 times $600, or $54000 for 1 da.; and of the use of $800 for 30 da., or its equivalent of 30 times $800, or $24000 for 1 da. ; and of the use of $700 for 60 da., or its equivalent of 60 times $700, or $42000 for 1 da.; or a total 234 EQUATION OF PAYMENTS deprivation of the use of $54000 + $24000 + $42000, or $120000 for 1 da. Hence the equitable time for the payment of $600 + $800 + $700, or $2100, must be as many days after June 16 as will offset this deprivation. As the total purchase ($2100) kept 57 days will as nearly as possible equal this de- privation (for $2100 X 57 = $119700, which is as near a product to $120000 as is possible with an integral multiplier), so 57 da. after June 16, 1911, or Aug. 12, 1911, must as nearly as possible be the equated date of payment. RULE. Multiply each item by its term of credit, and divide the sum of the resulting products by the sum of the items, to find the equated term of credit. Add the equated term of credit to the date of purchase to find the equated due date. NOTE 1. As the day is the lowest unit of time considered in terms of credit, if the division is not exact, disregard the remainder if less than half the divisor, or regard the remainder as another day if one-half the divisor or more. NOTE 2. If the terms of credit are expressed in months, and the division is not exact, multiply the remainder by 30 and divide the resulting product by the original divisor, to reduce the remainder to days or SOths of a month [86], applying Note 1 to the final remainder. NOTE 3. If all the items of sale are already equal, they may be dis- regarded; and the unequal terms of sale will then need only to be averaged [Ex. 13, 54]; but if the items are also unequal, the products of the items by their respective terms of credit must be averaged. EXAMPLES FOR PRACTISE Find the equated term of credit of the following: (1) (2) (3) $500 at 3 months $800 at 2 months $525 at 3 months 700 " 2 " 200 " 4 " 680 " 2 400 " 4 " 900 " 1 " 720 " 4 " 600 " 1 " 450 " 3 " 815 " 1 " Find the equated date of the following: (4) (5) 1912 1911 Apr 16 $650 at 30 days Oct. 23 $620 at 90 days " 470 " 90 " " 400 " 30 " " 520 " 60 " " 750 " 60 " " 350 " 90 " " 800 " 90 " EQUATION OF PAYMENTS 235 6. A house was sold for $6000, one-third payable in cash, one-third in 6 months, and the remainder in 12 months, without interest. What is the equated time for paying the entire pur- chase money? [Note 3] 7. On Mch. 23, 1911, goods were sold for $645 at 60 days, for $415 at 30 days, and for $500 cash. What is the equated date for the payment of the entire bill? 8. On June 18, 1910, goods were sold to a customer as follows: $650 for cash, $800 at 60 days, and $400 for cash. At what time can the three items be paid without loss to either buyer or seller? TO FIND THE EQUATED DATE OF PAYMENT WHEN THE DATES OF SALE ARE UNEQUAL AND THE TERMS OF SALE ARE EQUAL ILLUSTRATIVE EXAMPLE 293. William Garig bought of George W. Keller merchandise as follows: Aug. 24, 1912, $862; Sept. 16, 1912, $547; Oct. 28, 1912, $485. If no credit was allowed, what was the equated date for paying the sum of the three items? FIRST SOLUTION Assumed date, July 31,. 1912 Days from 1912 assumed Use for 1 day of date Auz 24 $862 X 24 - \ 3448 = 862 X 4 units ' 1 1724 = 862 X 2 tens h 1ft *A7 V A7 i 3829 = 547 X 7 units Sept. 16 547 X47== = 547 x 4 teng Oct 28 485 X 89 - 4365 = 485 X 9 units ' {3880 = 485 X 8 tens Use of $1894 for (?) da. = $89562 used for 1 da. $89562 -v- $1894 = 47 + times 1 da. after assumed date. Hence, 47 da. after July 31, or Sept. 16 = equated date. EXPLANATION. For convenience, assume as the required equated date, the last day of the month immediately preceding the earliest month in the 236 EQUATION OF PAYMENTS account (July 31); note the extent of the error resulting from this assumption, and make the proper correction. Thus, if the first item ($862) were paid on this assumed date (July 31), though not due until Aug. 24, or 24 days later, the purchaser would be de- prived of the use of $862 for 24 da., or the equivalent of the use of 24 times $862 for 1 da. If the second item ($547) were paid on this assumed date (July 31), though not due until Sept. 16, or 47 da. later, the purchaser would be deprived of the use of $547 for 47 days, or of the equivalent use of 47 times $547 for 1 da. If the third item ($485) were paid on this assumed date (July 31), though not due until Oct. 28, or 89 da. later, the purchaser would be deprived of the use of $485 for 89 da., or of the equivalent use of 89 times $485 for 1 da. Hence, to pay the sum of the three items ($1894) on the assumed date (July 31) would cause a total deprivation to the purchaser equivalent to the use of $89562 for 1 da. ; and to prevent this inequitable deprivation, the sum of the three items should be paid as many days after July 31 as will offset this loss, that is, as the total purchase ($1894) if kept 47 da. will equal the total deprivation ($89562 used for 1 da.), for $1894 X 47 = 89018, which is as near that deprivation as is possible to obtain with an integral multiplier, so 47 da. after July 31, or Sept. 16, must as nearly as possible be the equated date. ABRIDGED SOLUTION Assumed date, July 31, 1912 Days from Hundreds 1912 assumed of dollars date for 1 da. A,^ o/L QRO v 9/L f 34 = (8 hundred X 4 units)+2 hundred to carry 1 172 = (86 tens X 2 tens) and nothing to carry 7 _ _ ( 38 = (5 hundred X 7 units) +3 hundred to carry ( 219 = (54 tens X 4 tens) + 3 hundred to carry n , Oft xoc v QQ f 44 = (4 hundred X 9 units) +8 hundred to carry (388 = (48 tens X 8 tens) + 4 hundred to carry Use of $1894 for (?) da. = $89500 used for 1 day. $89500 + $1894 = 47 + times 1 day after July 31, or Sept. 16. EXPLANATION. Both of the above solutions are similar in principle; but in the abridged solution much unnecessary labor is avoided by omitting the multiplication of such orders in the items as will not affect the final result; that is, by actually multiplying only such orders in the items as will produce hundreds of dollars for 1 day (mentally multiplying one or two of the next adjoining rejected orders of each item to obtain the correct carrying figure in hundreds, counting half a hundred or more as an additional hundred). The use of (interest of) less than half a hundred dollars for 1 day is less than our lowest unit of coinage (Iff), and does not therefore require consideration. EQUATION OF PAYMENTS 237 NOTE 1. The orders which, when multiplied, will produce results in hundreds of dollars for 1 day, are Hundreds' order of the items X units' order of the days. Tens' order of the items X tens' order of the days. Units' order of the items X hundreds' order of the days. After obtaining the sum of the products, annex two O's to make that sum express its true value in even hundreds of dollars, and then proceed as in the first solution. RULE. Assume as the equated date the last day of the month immediately preceding the earliest month in the account. Multiply each item by the interval in days between the assumed date and the date of such item. Divide the sum of the obtained products by the sum of the items to find the number of days to count forward from the assumed equated date to obtain the correct equated date. EXAMPLES FOR PRACTISE Find the equated due date of the following cash sales: (1) (2) (3) 1910, Apr. 20 $800 1911, Mch. 6 $416 1912, July 3 $582 May 13 750 May 19 300 Aug. 21 430 June 28 675 June 6 52ti Oct. 12 618 (4) (5) (6) 1911, May 23 $275 1910, Oct. 16 $275 1912, Feb. 18 $468 July 14 356 Nov. 28 350 Apr. 7 500 Aug. 10 400 Dec. 2 196 May 23 260 Sept. 21 540 Dec. 23 248 June 19 592 NOTE 2. If all the items have a uniform term of credit, and it is expressed in days, first find by the preceding process the equated time in days exclusive of the uniform term of credit; then add the uniform term of credit to the result to find the equated time inclusive of the term of credit; lastly find the equated date in the usual manner. (7) (8) 1910, May 23 $563 at 90 da. 1911, July 14 $680 at 60 da. June 10 625 " Aug. 25 560 Aug. 2 718 " Oct. 13 725 NOTE 3. If the uniform term of credit is expressed in months, first find the equated date exclusive of the months of uniform credit; then count for- ward from this obtained date, as many calendar months as are contained in the uniform term of credit. 238 EQUATION OF PAYMENTS 0) (10) 1912, Jan. 17 $368 at 2 mo. 1910, June 26 $680 at 3 mo. Mch. 8 415 " Aug. 10 753 May 23 298 " Sept. 28 560 (ID (12) 1911, Apr. 24 $250 at 30 da. 1912, Sept. 29 $175 at 1 mo. June 11 625 " Oct. 18 256 July 21 700 " Nov. 10 340 " Aug. 14 290 " Dec. 6 294 TO FIND THE EQUATED DATE OF PAYMENT WHEN THE DATES OF SALE AND THE TERMS OF CREDIT ARE BOTH UNEQUAL ILLUSTRATIVE EXAMPLE 294. Find the equated date for paying the total of the fol- lowing sales: 1912 July 28, $652.35 at 60 days; Sept. 10, $471.20 at 30 days; Oct. 13, $537.63 at 90 days. FIRST SOLUTION Assumed date, June 30, 1912 1912 Use for 1 day of * AA J A*0 Q* V $ 5 5219 = < 652 X 8 UnltS ) + 3 July 28 at 60 da. $652.35 X 8 J 5218g = (652 3 x g 471 90 V 1 09 942 = (471 X 2 Unlts) + t0 Sept. 10 at 30 da. 471.20 ( 2688 = (537 X 5 units) + 3 to carry Oct. 13 at 90 da. 537.63 X 195 = 48387 = (537.6 X 9 tens) X 3 to carry _ ( 53763 = (537.63 X 1 hundred) Use of $1661.18 for (?) da. = $210307 used for 1 day. $210307 -T- $1661 = 127 times 1 day after assumed date. 127 days after June 30, 1911 = Nov. 4, 1912, equated date. EXPLANATION. Assume as the equated date the last day of the month which immediately precedes the earliest month found in the account to be averaged, calculate the extent of the error if the assumption proves wrong, and make the proper correction, as follows: If the purchaser had paid the first item ($652.35) on June 30, though not due until 60 da. after July 28, or 88 da. after June 30, he would have been de- EQUATION OF PAYMENTS 239 prived, as per terms of sale, of the use of $652.35 for 88 da., which would have been equivalent to an error of the interest on 88 times $652.35 for 1 da. If the second item ($471.20) had been paid on June 30, though not due until 30 da. after Sept. 10, or 102 da. after June 30, the error would have been equivalent to the interest on 102 times $471.20 for 1 da. If the third item ($537.63) had been paid on June 30, though not due until 90 da. after Oct. 13, or 195 da. after June 30, the error would have been equal to the interest on 195 times $537.63 for 1 da. Therefore, if the sum of the three items ($1661.18) had been paid on the assumed date (June 30) instead of on their respective due dates, the total error would have been the sum of the three ascertained errors, or the interest on $210307 for 1 da.; and to correct this error the sum of the three items ($1661.18) should be kept by the purchaser as many days after June 30 as would be equivalent to the use of $210307 for 1 da. As the sum of the products ($210307) divided by the sum of the items ($1661.18) equals 127 -, so $1661.18 kept 127 da. after the assumed date will offset this loss; and 127 days after June 30, or Nov. 4, 1912, must be the required date upon which the sum of the three purchases ($1661.18) should be paid without involving the loss of the use of money by either purchaser or seller. ABRIDGED SOLUTION Assumed date, June 30, 1912 Hundreds 1912 July 28 at 60 d, S652.35 X 88 Sept. !0 at 30 da. 47L20X102 - % + J ( 27 (5 hund. X 5) + 2 Oct. 13 at 90 da. 537.63 X 195 = \ 484 (53 tens X 9) + 7 _ ( 538 (537 units X 1) + 1 Use of $1661.18 for ? da. = $210300 used for 1 day. $210300 -f- 1661 = 127 times 1 da. after assumed date. June 30, 1912 + 127 da. = Nov. 4, 1912, equated date. THE EXPLANATION of the abridged solution is similar to that of the first solution preceding it. They differ only in the method of performing the multiplications. The contracted method, involving no products of less value than hundreds of dollars for 1 day, is fully explained in the abridged solution of 111. Ex., 293 and Note 1 following it. 240 EQUATION OF PAYMENTS RULE. Add the term of credit of each item to the multiplier for that item as obtained by Rule 293; and continue in accordance with that Rule. NOTE 1. When dividing the sum of the products by the sum of the items, omit the cents from the divisor if less than 50, or increase the dollars by $1 if the cents are 50 or more. EXAMPLES FOR PRACTISE Find the equated date of payment of the following: (1) (2) 1910, Apr. 6 $600 at 90 da. 1911, Mch. 14 $750 at 60 da. May 18 725 " 30 " Apr. 25 825 " 90 " June 23 480 " 60 " June 13 690 " 30 " (3) (4) 1912, Feb. 17 $482.35 at 60 da. 1911, Sept. 3 $248. 15 at 90 da. Apr. 29 527.60 " 30 " Oct. 10 361.78 " 60 " June 2 368.45 " 90 " Nov. 25 419.95 " 30 " July 13 416.25 " 30 " Dec. 20 195.80 " 60 " (5) (6) 1909, July 21 $682.40 at 30 da. 1912, May 31 $425.89 at 60 da. Sept. 12 546.15 " 90 " June 14 516.75 " 90 " Nov. 25 725.87 " 60 " Aug. 3 398.78 " 60 " Dec. 4 498.78 " 90 " Sept. 20 465.85 " 30 " NOTE 2. If a term of credit is expressed in months, it has been found most convenient in practise to proceed as follows: If goods are bought on May 16 at 3 months, it is readily seen that 3 months after May 16 will include the end of May (31st), the end of June, and the end of July (31st), or 2 Slsts 2 in all; hence write a small 2 over that term of credit (3 mo.), and when it is reached in the equation, it will be identified as 92 da., that is, as 3 months of 30 da. each, plus the increase for the two 31sts. Or, if an item at 2 months is dated Jan. 4, 1910, the term of credit will include the end of Jan. (31st), the end of Feb. (28th), a net minusage of 1, hence write 1 over that term -l of credit (2 mo.), and count it as 60 da. less 1. Or, if the item is at 4 months, and is dated Nov. 24, 1911, the term of credit will include the end of Nov., the end of Dec. (31st) the end of Jan. (31st), and the end of Feb. (29th), or a l net plusage of 1, expressed 4 months, and counted 120 da. plus 1 ; etc. AVERAGING ACCOUNTS 241 (7) (8) 1912, Apr. 16 $468.75 at 3 mo. 1911, July 12 $728.15 at 2 mo. June 12 529.60 " 1 " Sept. 3 465.70 " 4 " July 15 387.90 " 2 " Oct. 18 598.67 " 1 " Aug. 28 625.75 " 4 " Nov. 23 615.80 " 3 " (9) (10) 1911, May 25 $568.35 at 3 mo. 1912, Jan. 18 $396.75 at 2 mo. June 2 275.50 " 60 da. Feb. 12 285.67 " 3 " Aug. 17 348.25 " 2 mo. Apr. 18 2968.52 " 60 da. Sept. 13 465.72 " 30 da. May 10 98.76 " 30 " AVERAGING ACCOUNTS 295. INTRODUCTION. The process of averaging an account is an application of the principles of equation of payments to an account which has both debit and credit items. The method common to both is to assume a date of payment, to ascertain the resulting error by comparison of the due date of each item with the assumed common due date for all the items, and to make the proper correction. When all the items were debits, it was seen that the assumption of a date of payment earlier than that of the several items would uniformly result in a loss to the debtor. In averaging; an account, however, which has both debit and credit entries, the resulting error from assuming a date of settlement earlier than that of the several items will only indicate a loss to the purchaser with respect to the debit entries; and will show a gain to him with regard to all credit entries. The net gain or net loss of the entire account must therefore be the determining fac- tor in finding the proper correction of the assumed date. 296. Averaging an account is finding the proper time for paying the balance of an account which includes both debit and credit entries. NOTE. The average date is that date at which the balance of an account should be paid to avoid the loss of the use of money by either party concerned in the account. Hence, if the balance of an account is paid before the average date, it is usual to diminish the balance by the compensatory discount thereon for the days of prepayment ; and if paid after the average date, to increase the balance by the compensatory interest thereon for the days of postpayment. 242 AVERAGING ACCOUNTS TO AVERAGE AN ACCOUNT WITH ITEMS ON BOTH SIDES ILLUSTRATIVE EXAMPLE Find the equated date for paying the balance of the following account : Dr. J. P. SMITH Cr. 1911 May 23 To Mdse. $768 42 1911 July 12 By Cash $694 40 July 16 M 537 80 Aug. 16 M 478 75 Aug. 27 M 659 35 Sept. 25 M 354 80 FIRST SOLUTION Assumed date, April 30, 1911 May 23 $768.42 X 23 = 2305 _ 15368 Julyl6537.80X77=| 3 ^ Aug. 27 659.35X119 = 5934 $137547 $1965.57 1527.95 $437.62, balance of items ( 9083 July 12 $694.40 X 73 = ( 4ot)Uo ( ^8*30 Aug. 16 478.75X108= | 4 ^ ( 2838 Sept. 25 354.80X148= ] 14192 (35480 $1527.95 $154906 137547 bal. of products, $17359 $17359 -^ $438 = 40 - times 1 da. before Apr. 30, or Mch. 21. EXPLANATION. Assuming April 30, 1911, as the date for paying all the debit items, it is found by 111. Solution, 293, that if Smith had paid the three debit items on the assumed date instead of their respective due dates, he would have lost the use of (the interest on) $137547 for 1 da.; and considering the credit items upon the same assumption that the debits were due (and should therefore have been paid) on April 30, it is found that Smith would have gained the use of $694.40 for 73 da., of $478.75 for 108 da., and of $354.80 for 148 da., or a total gain equivalent to the use of $154906 for 1 da. Hence, the net gain to Smith upon this assumption must be equivalent to the use of AVERAGING ACCOUNTS 243 $154906 (gain) - $137547 floss), O r to the net use of $17359 for 1 da. Smith should therefore pay the balance of his account ($437.62) as many days before the assumed date (April 30) as will offset this gain. As Smith's net gain ($17359) divided by Smith's balance ($437.62) equals 40 -, so 40 da. before April 30, or March 21, 1911, must be the equated date upon which Smith should pay this balance without loss of the use of money (of interest) by his creditor or himself. ABRIDGED SOLUTION Assumed date, April 30, 1911 ( 2*^ ( 21 May 23 $768.42 X 23 = < ^ July 12 $694.40 X 73 = | ( 38 ( July 16 537.80 X 77 = 3 * Aug. 16 478.75 X 108 - Aug. 27 659.35 X 119 = j f g ,5. Q .,_ ( ( I Ad Dtjpl/. 6U Ocrx.OU A. IrtO *\ $1965.57 $1375 ^355 1527.95 $1527.95 $1549 $437.62 1375 $174 $17400 + $438 = 40 times 1 da. before Apr. 30, or Mch. 21. RULE. Assume as the equated date, the last day of the month which immediately precedes the earliest month in the account. Mul- tiply each item by the interval in days between the assumed date and the due date of such item; and divide the balance of the resulting products by the balance of the items to find the error in the assumed date. Count the error in days forward from the assumed date if both balances are on the same side, or backward from the assumed date if the balances are on opposite sides. NOTE 1. The equated date for paying the balance of an account does not necessarily mean the date at which it must be paid, for it may be impossible for the debtor at that date to know that it would be so payable, as in the pre- ceding 111. Ex., in which the equated date is prior to the date of any of the purchases, perhaps even prior to the debtor's intention to make any purchase whatever. In other instances, the debtor may have the requisite knowledge of the equated date but may not possess the means on the equated date with which to pay his indebtedness. In either case, the equated date is the proper time, if practicable, for paying the balance of the account without loss of in- 244 AVERAGING ACCOUNTS terest to either debtor or creditor. If for any reason, the balance is paid before the equated date, the debtor is usually allowed discount on said balance for the days of prepayment, and if paid after the equated date, the creditor will usually claim interest upon said balance for the days of postpayment. Dr. (No. 1) EXAMPLES FOR PRACTISE WARREN H. SADLER Cr. 1910 Oct. 25 To Mdse. $675 1910 Nov. 2 By Cash $24900 Dr. (No. 2) GEO. C. ROUND Cr. 1911 Mch. 18 To Mdse. $385 90 1911 Apr. 8 By Cash $246 75 May 24 14 472 75 June 16 u 187 35 Dr. (No. 3) E. BENSEL & Co. Cr. 1910 Apr. 17 To Mdse. $678 35 1910 May 6 By Cash $263 50 June 12 594 20 July 18 345 75 July 24 u 725 45 Aug. 10 416 80 Dr. (No. 4) HENRY ALEXANDER Cr. 1912 July 6 To Mdse. $416 72 1912 Aug. 8 By Cash $327 90 Aug. 29 tt 532 50 Sept. 20 ii 615 25 Oct. 14 it 374 65 Nov. 12 tt 285 45 Dr. (No. 6) H. A. GRIESEMER Cr. 1911 May 13 To Mdse. $256 15 1911 June 21 By Cash $243 60 July 16 476 35 Aug. 17 a 185 75 Aug. 30 348 50 Sept. 12 372 50 AVERAGING ACCOUNTS 245 Dr. (No. 6) A. H. ATWOOD Cr. 1910 June 3 ToMdse., 60 da. $392 70 1910 July 13 By Cash $243 GO July 28 30 " 527 98 Aug. 25 " Note, 60 da. 185 75 Sept. 19 net 476 87 Oct. 10 " Cash 372 50 Dr. (No. 7) FAIRMAN A. SADLER Cr. 1912 Feb. 23 To Mdse., 90 da. $485 90 1912 Mch. 4 By Cash $275 80 Apr. 17 60 " 372 65 May 16 " Note, 60 da. 300 00 May 26 " 30 " 427 25 July 8 (( on 250 00 Dr. (No. 8) R. M. BROWNING Cr. 1911 Oct. 24 Dec. 18 1912 Jan. 20 To Mdse., net 90 da. " 60 " $628 473 532 50 85 05 1911 Nov. 12 1912 Jan. 21 Feb. 18 By Note, 90 da. " " 60 " " Cash $352 450 275 75 00 30 Dr. (No. 9) H. C. REITZ Cr. 1910 Mch. 16 To Mdse., 3 mo. $575 80 1910 Apr. 28 By Note, 1 mo. $384 45 May 19 net 618 35 June 6 " Cash 256 85 June 24 2 mo. 487 15 July 22 " Note, 3 mo. 416 75 Dr. (No. 10) W. M. CUTCHIN . Cr. 1911 Jan. 18 To Mdse., 2 mo. $394 67 1911 Feb. 23 By Note, 4 mo. $284 62 Feb. 20 t( ^ 472 50 Mch. 29 (t 2 " 316 25 Apr. 7 " 3 " 528 75 May 10 " " 3 " 472 38 246 AVERAGING ACCOUNTS Dr. (No. 11) C. L. REINDOLLAR Cr. 1912 Feb. 29 To Mdse., 60 da. $826 45 1912 Mch. 27 By Cash $375 90 Mch. 18 Net 416 20 Apr. 12 " Note, 90 da. 462 75 May 20 " 3 mo. 752 68 June 30 " Cash 400 00 June 13 90 da. 528 34 July 18 " Note, 2 mo. 275 15 12. On June 24, 1911, H. M. Lee bought merchandise for $753.25 and 90 days' credit. If Lee made a cash payment of $500 on July 3, 1911, what is the equated date for paying the remainder? NOTE 2. If part of a debt is paid before it is due, many merchants permit the debtor to postpone payment of the remainder for a proportionate time after it is due. The date of payment of the remainder will be the equated date, found in the usual manner. Such an extension, however, is not enf orcible by law if the creditor objects. 13. On July 18, 1912, G. M. Hett bought merchandise for $900 and 60 days' credit; and on Aug. 3, 1912, paid $300 on account. How much did he owe on Dec. 13, 1912, at 6%? NOTE 3. It is customary to charge interest upon the balance of an account for the number of days it is paid after its equated date. 14. A has two of B's notes; one of $580, dated Mch. 24, 1911, payable 4 months after date; and the other of $624, dated Apr. 16, 1911, payable 60 days after date. On May 6, 1911, B paid $700 to A, canceled the two notes, and gave a new 30-day non-interest-bearing note for the remainder. What was the date of the new note? NOTE 4. The note must be so dated as to fall due on the equated date, at which time the unpaid balance for which it was given would have fallen due. 15. Thomas G. Hayes bought goods for $650 on Aug. 18, 1911, and 90 days' credit. On Sept. 23, 1911, he paid $175 on account; and on Oct. 5, 1911, he paid $225 and gave a 60-day interest-bearing note for the remainder. What should have been the date of the note? NOTE 5. If a note bearing interest from its date is given in settlement of the balance of an account, its date should be the equated date (Note 3). AVERAGING ACCOUNTS SALES 247 AVERAGING ACCOUNTS SALES ILLUSTRATIVE EXAMPLE 297. Find the equated date for paying the net proceeds of the following sales on commission: Sales, 1912 June 19, $285.90; July 17, $328.70; Aug. 22, $264.16. Charges, 1912 July 2, cash advanced consignor, $350; July 16, freight, $28.70; July 30, drayage, $8.25; commission, 3%. SOLUTION Assumed date, May 31, 1912 June 19 $285.90 X 19 July 17 328.70 X 47 Aug. 22 264.16 X 83 26 29 J23 131 $878.76 $428 $42800 + $879 = 49 times 1 da. May 31 + 49 da. = July 19, 1912. $878.76 X .03 = $26.36 +, com. on sales. EXPLANATION. First find the equated date of the three sales by 293, obtaining July 19, 1912, as the date for the total sales ($878.76) and for the total com- mission ($878.76 X .03 = $26.36). Next, average the account by 296, placing all the charges includ- ing the commission and its equated date, on the debit side; and plac- ing the total sales ($878.76) and its equated date on the credit side. Assumed date, June 30, 1912 1912 July 2 $350. X 2= $7 " 16 28.70 X 16 = " 30 8.25 X 30= 2 " 19 (com.) 26.36 X 19= _ $413.31 19 1912 ($79 July 19 $878.76X19=) ( OO $167 413.31 19 Bal. $465.45 $148 Bal. $14800 ^ $465 = 32 times 1 da. June 30 + 32 da. = Aug. 1, 1911, EXAMPLES FOR PRACTISE 1. An agent sold a consignment as follows: 1910 May 18, $7695.75; June 2, $5677.20; June 25, $4108.80. His charges 248 CASH BALANCE were: 1910 May 15, freight, $315; May 20, cash advanced on consignment, $3500; commission, 1J%. Find the equated date for remitting the net proceeds to his principal. 2. Find the equated date for paying the net proceeds of the following commission sales: Sales, 1911 Aug. 5, $287.35; Sept. 2, $328.49; Oct. 20, $289.30. Charges, 1911 July 28, freight, $18.25; Aug. 3, cash, $500; commission, 2%. 3. Find the net proceeds of the following commission sales, and the equated date for remitting the proceeds to the consignor: Sales, 1912 Apr. 17, $583.75; May 9, $872.65; June 12, $1075.80. Charges, 1912 Apr. 23, drayage, $12.50; May 14, cash advanced, $1200; commission, 2%. CASH BALANCE 298. INTRODUCTION. To take the balance of any account is to find the difference between its total debits and its total cred- its. To prefix "cash" to the word balance adds nothing to its significance; for all balances are payable in cash or its generally accepted equivalent. The term "cash balance," in business usage, has special reference to the difference, on the date of tak- ing the cash balance, or date of adjustment, between the present worth of all the debit items and the present worth of all the credit items. This includes consideration of the debit interest on each debit item that falls due before the date of adjustment for the num- ber of days between its due date at which it is legally payable to the date of adjustment when it is assumed to be paid; and the credit interest (discount) on each debit item that falls due after the date of adjustment for the number of days between the date of adjustment at which it is assumed as being payable, to its subsequent due date at which it is legally payable. It also in- cludes consideration of the credit interest on each credit item, if cash, for the number of days between its date at which it is actu- ally paid to the date of adjustment at which it is assumed to be paid; or if a time note or draft maturing before the date of adjust- ment, of credit interest for the number of days between its legal due date at which it is actually paid to the date of adjustment at which it is assumed to be paid; or if a time note or draft maturing after the date of adjustment, of debit interest (discount) for the number of days between the date of adjustment at which it is assumed to be paid and its legal due date at which it is actually paid. CASH BALANCE 249 299. The cash balance of an account is the difference, on the day of adjustment, between the present worth of all its debit items and the present worth of all its credit items. ILLUSTRATIVE EXAMPLE Find the cash balance of the following account on Dec. 13, 1911, at 6%. Dr. JOSHUA LEVERING in account with J. H. TYLER Cr. 1911 Days Int. Amount 1911 Days Int. Amount July 28 To Mdse., 3 mo. $268 75 July 20 By Note, 2 mo. $175 83 Aug. 12 30 da. 327 13 Sept. 2 " 60 da. 207 27 Oct. 19 Net 295 84 Oct. 28 " Cash 148 75 Dr. SOLUTION Cr. 1911 Days Int. Am'nt 1911 Days Int. Ainou July 28 To Mdse., 3 mo. 46 $2 06 $268 7f> July 20 By Note, 2 mo. 84 $2 40 $17f Aug. 12 30 da. 93 5 07 327 13 Sept. 2 " " 60 da. 42 1 15 207 Oct. 19 Net 55 2 71 295 84 Oct. 28 " Cash 46 1 14 148 Dec. 13 Bal. of Int. 4 79 Dec. 13 " Bal. of Int. 4 79 ' % 364 $9 84 $890 >1 $Q 84 $896 EXPLANATION. Write in the days' column opposite each item, the num- ber of days intervening between the due date of that item and the date of taking the cash balance (Dec. 13). Next write in the interest column, the interest on each item for the days specified in the days' column. Then write the difference between the total debit interest and the total credit interest ($4.79) on the less side, thus balancing the interest columns; and as this interest balance shows that Levering owes Tyler $4.79 more interest than Tyler owes Levering, enter it as an additional item in the amount column on the debit side of Levering's account. Lastly, write the difference between the total of the amount column on 250 CASH BALANCE the debit side and the total of the amount column on the credit side ($364.66), on the less side, as the required "cash balance" on Dec. 13, 1911. RULE. 1. Find the interest of each item of the account from its due date to the date of taking the cash balance. 2. Balance the interest columns. 3. Carry the interest balance to the main money column on the side opposite to that in which it is written in the interest column. 4. Balance the main money columns. NOTE 1. The interest of items which fall due after the date of adjusting the cash balance should be computed for the number of days between the date of adjustment and the due date of such items; and this interest should be written with red ink in the interest column on the same side as the item; but when the totals are taken, this red interest should be excluded from the total interest of the side on which it is written, and included with the total interest of the opposite side where it properly belongs. NOTE 2. The face of an interest-bearing note should be regarded as the equivalent of so much cash paid on the day it commenced to draw interest; and the face of a non-interest-bearing note should be treated as the equivalent of so much cash paid on its due date. EXAMPLES FOR PRACTISE 1. Find the cash balance of the following on Dec. 16, 1911, at 6 %. Dr. EDMUND BERKELEY in account with J. B. T. THORNTON Cr. 1911 Days Int. Amount 1911 Days Int. Amou July 18 To Balance *** ** ** $416 70 Aug. 20 By Cash *** * ** $416 Sept. 24 " Mdse. ** * ** 286 45 Sept. 28 ** * ** 157 Oct. 2 ,, ** * ** 341 60 Oct. 18 ** * ** 275 Nov. 20 ** * ** 298 40 Dec. 1 ,, ** ** 198 Dec. 16 " Bal. of Int. * ** Dec. 16 " Bal. of Int. * ** " " " % ** ** **** ** ** ** **** 2. Find the cash balance of the following on Oct. 16, 1912, at 5 %. CASH BALANCE 251 Dr. JOB. J. JANNEY in account with N. BRAWNER Cr. 1912 Days Int. Amount 1912 Days Int. Amount Mch. 16 To Mdse., 3 mo. *** * ** $298 75 Apr. 7 By Note, 90 da. ** * ** $178 25 May 28 " 60 da. ** ** 328 76 July 3 " " 2 mo. ** * ** 210 38 July 16 " Net ** * ** 265 83 Aug. 8 " Cash ** * ** 158 16 Oct. 16 " Bal. of Int. * ** Oct. 16 " Bal. of Int. * ** *** ** Oct. 16 % *** ** ** ** *** ** ** ** *** ** 3. Find the cash balance on Nov. 18, 1911, at 6%. Dr. T. C. WILL in account with W. P. JORDAN Cr. 1911 Days Int. Amount 1911 Days Int. Amount Mch. 21 To Mdse., 30 da. *** ** ** $385 112 Apr. 7 By Note, 30 da. *** * ** $193 75 May 10 " " 3 mo. *** * ** 278 50 May 20 " " 3 mo. ** * ** $216 25 July 28 " Net *** * ** 328 65 July 7 " Cash *** * ** 265 18 Nov. 18 " Bal. of Int. * ** Nov. 18 " Bal. of Int. * ** Nov. 18 % *** ** ** ** **** ** ** ** **** ** 4. Find the cash balance on Nov. 8, 1912, at 6 %. Dr. WALTER WATT in account with NEAL FAHR Cr. 1912 Days Int. Amount 1912 Days Int. Amount July 13 To Mdse., Net *** * ** $285 96 Aug. 20 By Cash ** * ** $150 35 Sept. 23 " 60 da. ** ** 346 75 Sept. 16 " Note, 30 da. ** ** 210 50 Oct. 3 " " 3 mo. ** * ** 417 30 Oct. 18 4 mo. *** * ** 168 73 Nov. 8 " Bal. of Int. * ** Nov. 8 " Bal. of Int. * ** Nov. 8 % *** ** * " **** ** * **! **** 4-* 252 PARTNERSHIP 5. Arrange the following items in the form of an account current, ruled as in the preceding examples, and find the cash balance on Sept. 5, 1912, at 5%: John King bought of J. Malone Gaulden, as follows: May 17, 1912, mdse. at 30 da., $378.25; June 28, 1912, mdse. at 4 mo., $296.52; July 10, 1912, mdse. net, $419.75; and King was credited on June 28, 1912, for his note of $198.63, at 90 da.; on July 5, 1912, for his note of $128.46 at 2 mo., and on July 20, 1912, for a cash payment of $293.75. PARTNERSHIP 300. INTRODUCTION, (a) A business partnership is a com- bination of the resources of two or more persons for the estab- lishment or the maintenance of a business enterprise. The amount of money, or property, or time, or the character of the duties to be assumed by each partner, or the manner of distributing the gains or losses, or any other necessary detail, is stipulated in an instrument of writing called a partnership agreement. Such a combination of persons is known as a firm, a house, or a company; and each person in the firm is called a partner, (b) The aggre- gate of what each partner contributes to the partnership is termed his investment; and the difference between his total investments and his total withdrawals is called his net investment, (c) The sum of what all the partners contribute is known as the capital of the firm; (d) and the difference between what all the part- ners have invested and what all the partners have withdrawn plus the net gain or minus the net loss since the last partnership adjustment, is called the firm's net capital. (e) The assets or resources of a firm are its entire possessions, including obligations due to it from others; and (/) its liabilities are the obligations due by the firm to others, (g) If a firm's re- sources exceed its liabilities, it is said to be solvent; and the excess is known as the firm's net solvency or net capital; but (h) if a firm's liabilities exceed its resources, it is said to be insolvent, and the excess is known as the firm's net insolvency at that time. (i) Due allowance being made for possible investments or withdrawals, if any, within the settlement-period, the net gain of a firm is the excess of its net capital at the end of that period over its net capital at the beginning of that period; (j) or the decrease of its net insolvency at the end of that period below its net insol- vency at the beginning; (k) or the sum of its net capital at the end of that period and its net insolvency at the beginning. PARTNERSHIP 253 (Z) Due allowance being made for investments or withdrawals, if any, within the settlement-period, the net loss of a firm is the deficiency of its net capital at the end of that period below its net capital at the beginning of that period; (ra) or the excess of its net insolvency at the end of that period above its net insolvency at the beginning; (n) or the sum of its net insolvency at the end of that period and its net capital at the beginning. (o) It is thus seen that partnership adjustments involve con- sideration of three terms: first, the financial condition at the beginning; second, the financial condition at closing; third, the net gain or net loss during the interval, (p) These three terms have a fixed relation to each other, additive or subtractive, so that if any two of them are given, the third can be readily found. (r) Of these three terms, the financial condition at the beginning of the settlement-period, or that condition as modified by subse- quent investments or withdrawals, if any, should uniformly be regarded as the natural standard of comparison, as follows: 301. To find the gain or loss of a firm, (a) Compare its financial condition (solvency or insolvency) at the end of a settlement-period with its financial condition at the beginning of that period, or with that condition as modified by subsequent investments or withdrawals, if any, within that settlement-period; and (b) if its financial condition at the end of the period is found to be better than its condition at the beginning of the period, there has been a gain; if worse, a loss; and the amount of gain or loss will be the difference between these two conditions. NOTE 1. As the gain or loss is the difference between two financial con- ditions, and any difference added to the less term will equal the greater, so, if a gain or loss is given, and the financial condition at the close of the settle- ment-period is required, it will be the gain better, or the loss worse than the net financial condition at the beginning; or if the net financial condition at the beginning is required, it will be the gain worse, or the loss better than that at the end. NOTE 2. The difference between similar conditions, that is, between one net capital and another net capital, or between one net insolvency and another net insolvency, is found by subtracting the less from the greater; but the dif- ference between opposite conditions, that is, between a net capital and a net insolvency, or vice versa, is found by addition. NOTE 3. A capital at the end of a period is better than at the beginning when it is greater, and it is worse when it is less; while an insolvency at the end of a period is better than at the beginning, when it is less, and it is worse when it is greater. 254 PARTNERSHIP NOTE 4. A financial condition may be given directly as so much capital or insolvency; or it may be given indirectly by enumerating the different resources and liabilities, or by access to the books of account from which the several resources and liabilities may be obtained. EXAMPLES FOR PRACTISE Find the gain or loss : 1. Capital at beginning $9285; capital at closing $12345. 2. Capital at beginning $15275.80; capital at closing $12685.92. 3. Insolvency at beginning $9685.75; insolvency at closing $13472.65. 4. Insolvency at beginning $18271.36; insolvency at closing $14275.36. 6. Capital at beginning $7286.50; insolvency at closing $2615.25. 6. Insolvency at beginning $4586.25; capital at closing $6718.60. Find the capital or insolvency at closing : 7. Capital at beginning $12856.25; gain during interval $4687.50. 8. Capital at beginning $7268.42; loss during interval $2178.65. 9. Insolvency at beginning $5829.65; gain during interval $6783.75. 10. Insolvency at beginning $9685.30; gain during interval $2785.90. 11. Capital at beginning $3568.20; loss during interval $6580.65. 12. Insolvency at beginning $4689.75; loss during interval $768.50. Find the capital or insolvency at beginning : 13. Capital at closing $7856.35; gain during interval $2563.75. 14. Capital at closing $3568.25; gain during interval $6528.15. 15. Capital at closing $9275.80; loss during interval $1845.60. 16. Capital at closing $1653.40; loss during interval $8215.25. PARTNERSHIP 255 at at at at closing closing closing closing $3720.65; $675.80; $1925.25; $2678.90; gain gain loss loss during during during during interval interval interval interval 17. Insolvency $1285.80. 18. Insolvency $4718.10. 19. Insolvency $658.25. 20. Insolvency $5815.60. 21. A firm commenced business with the following resources: cash, $2500; merchandise, $12500; personal accounts due the firm, $1846.25; bills receivable, $956.75. Its liabilities at the beginning were: due to others for merchandise bought, $3500; and on outstanding notes, $1256.18. At closing they had re- sources as follows : merchandise as per inventory $15265.70; cash, $780; due the firm on personal accounts, $4625.80; bills receiv- able, $1475.50; and their liabilities were: due to others on per- sonal accounts, $1725.80; and on outstanding notes, $865.70. What was the net gain or loss of the firm during the interval? 22. A and B formed a copartnership, the former investing f of the capital and the latter J, the agreement between them re- quiring the gains or losses to be divided equally. At settlement the net gains were found to be $6000, their resources at that time being merchandise on hand $12500, cash $3500, personal accounts $4200, bills receivable $1826; and their liabilities, outstanding notes against the firm $4500, unsettled bills for merchandise bought $2700. How much did each partner invest at the begin- ning? 23. A, B, and C engage hi business as partners, with a capital of $18000, A having invested A of the capital, B T 4 T , and C &. At closing, the firm's books showed a net gain of $6420, which accord- ing to agreement was to be shared equally. What was each part- ner's interest in the firm at settlement? DIVISION OF GAINS OR LOSSES AMONG PARTNERS IN PROPORTION TO THEIR ORIGINAL INVESTMENTS 302. INTRODUCTION. Unless otherwise specified in the part- nership agreement, the courts presume that the gains or losses are, by implication, to be divided equally among all the partners, 256 PARTNERSHIP irrespective of the money they may have individually invested, that not being regarded as the only source of possible gains. What a partner may lack pecuniarily may be offset by his possession of greater skill in business, greater influence over patronage in the community, or by assuming greater responsibility of management, etc. In the examples under this article, it is assumed that the partners' investments were made at commencing business and were left unchanged by subsequent investments or withdrawals, and that inequalities among the partners in other respects than money, if any, have been provided for by special allowances (sometimes improperly called salaries) to be taken from the gross gains before general distribution of the net gain is made among all the partners in proportion to their respective pecuniary investments alone. ILLUSTRATIVE EXAMPLE A and B are partners, the former investing $6000 and the latter $8000, and agreed to share the gains or losses in proportion to their respective investments. If the net gain during the first year was $5600, what was the share of each? SOLUTION EXPLANATION. Apply- $6000 + S8000 = S14000, total capital ** * a. a a a af\<\(\ relation to the total gain - < ' ($5600) as A ' s ($6000) which produced his 20 share of the gain has to the total investment ($6000 8 400 + $8000 = $14000) which X produced the total gain J4000 = $ 3200 > B s snare ($5600). As A's investment ($6000) is T YA% or f of the total investment, so A's share of the gain should be f of the total gain ($5600), and f of $5600 = $2400. Proceed in a similar manner to find B's share of the firm's gain. In veri- fication, add A's share of gain to B's, to find if the sum equals the total gain. RULE. Write one partner's investment and the total gain as a compound numerator, and the firm's capital as the denominator. Extend the resulting fraction if necessary, to express dollars and cents. NOTE. To find each partner's investment, if his share of the gain is given, write one partner's gain and the firm's capital as a compound numerator, and the total gain as the denominator. PARTNERSHIP 257 EXAMPLES FOR PRACTISE 1. A and B united in conducting a grocery business, A invest- ing $6480 and B $5670, and agreed to divide the gains or losses between them in proportion to investment. If the net gains were $3600, how much should each partner have received? 2. A, B, and C have formed a partnership, A investing $8000, B $10000 and C $12000. The net gain during the first year was $9000 which was divided among them in proportion to their respective investments. How much of the gain should each part- ner have received? 3. A and B engaged in business with a joint capital of $23275; and when the gains were adjusted, A received $2400 and B $3600. How much did each invest? [Note] 4. A, B, and C are partners, A investing $12000, B $15000 and C $18000; and agreed to share gains or losses in proportion to their investments. When the books were adjusted, it was found that the total resources of the firm were $78216 and their total liabilities $24816. What was each partner's share of the gain? DIVISION OF GAINS OR LOSSES AMONG PARTNERS IN PROPORTION TO THEIR AVERAGE INVESTMENTS 303. INTRODUCTION, (a) To divide the gains or losses among partners in proportion to their original investments as in 302 is proper only when those investments remain unchanged during the entire settlement-period. If a change is made in the original investment of one or more partners by subsequent investments or withdrawals, some modification of 302 will be necessary to effect an equitable distribution of gains or losses. If, for instance, a partner were to make an investment subsequent to his original investment and thus increase his contribution to the gain-produc- ing capital of the firm, it is evident that he should receive a corresponding increase of the firm's gain in proportion to his in- creased investment, and also that it would be as improper to pro- portion his gains for the entire year to his subsequent greater investment when his investment for the preceding portion of the year was less, as it would have been to proportion his gains, to his original less investment when his investment for a subsequent portion of the year was greater. (b) This modification of any partner's share of the gain in 258 PARTNERSHIP proportion to his modification of his original contribution to the profit producing capital of the firm is equitably effected by reduc- ing the several investments of each partner for dissimilar periods of a year to equivalent investments for similar periods of a lower unit of time than a year, as for one month, or for one day. Hence, (c) If all the partners' investments were not made for the entire settlement period, they should be reduced, by average, to equivalent investments for the longest obtainable equal period of time, as for one month or for one day; and the gains or losses should be divided among the partners in proportion to their equivalent investments for that common lower unit of time. ILLUSTRATIVE EXAMPLE A and B formed a copartnership, and agreed to share gains or losses in proportion to investment. A invested $9000 on Jan. 1, 1910; withdrew $2000 on May 1, 1910; and invested $5000 on Aug. 1, 1910. B invested $8000 on Jan. 1, 1910; invested $4000 on Apr. 1, 1910; and withdrew $3000 on Sept. 1, 1910. If the net gains during the year were $6000, what was each partner's share? SOLUTION $9000 invested for 1 yr. =12 times $9000, or $108000 for 1 mo. 5000 invested for 5 mo. = 5 times 5000, or 25000 for 1 mo. A's total investment, $133000 for 1 mo. $2000 withdrawn for 8 mo. = 8 times $2000, or 16000 for 1 mo. A's net investment, $117000 for 1 mo. $8000,invested for 1 yr. =12 times $8000, or $ 96000 for 1 mo. 4000 invested for 9 mo. = 9 times 4000, or 36000 for 1 mo. B's total investment, $132000 for 1 mo. $3000 withdrawn for 4 mo. = 4 times $3000, or 12000 for 1 mo. B's net investment, $120000 for 1 mo. $117000 + $120000 = $237000, firm's net capital for 1 mo. 39 $6000 xWm S234000 _ Q3 _ ^ ^ ^ ^ ^ 79 79 40 $6000 X W9W $240000 79 $3037.97 +, B's share of the gain 79 PARTNERSHIP 259 EXPLANATION. As the investments of the partners were not uniform throughout the year, recourse must be had to a shorter common unit of time to which the net investments of the several partners can be reduced, and in proportion to which the net gain of the firm can be divided. As the original investments and subsequent modifications were all made on the first day of some calendar month, the investments and withdrawals of each partner are capable of being reduced by average to equivalent investments and with- drawals for the similar period of one month, that is, in A's case to a net invest- ment of $117000 for 1 mo., in B's case to a net investment of $120000 for 1 mo., and with reference to the firm to an equivalent net capital of $117000 + $120000, or $237000 for 1 mo. Hence, divide the net gain of the firm ($6000) between A and B in proportion to these equivalent investments per month, as in 302 for investments per year. RULE. 1. // all the investments and withdrawals of all the part- ners have been made on the first day of some calendar month, divide the gain or loss of the firm among the partners as in 302, but in pro- portion to each partner's average net investment and to the firm's average net capital for one month. 2. // one or more investments or withdrawals of one or more part- ners have been made on different dates of a calendar month, divide the gain or loss of the firm among the partners as in 302, but in pro- portion to each partner's average net investment and to the firm's average net capital for one day. NOTE 1. To find a partner's average net investment for one month or for one day, multiply each investment or withdrawal by the number of months or days from its date to the end of the settlement-period, and subtract the sum of the resulting withdrawal products from the sum of the resulting investment products. NOTE 2. If an allowance is made to a partner for special labor, or risk, or skill, or influence, or responsibility, it should be considered as an extra share of the gain allotted to that partner for that special investment, in addition to his proportionate share of the remainder of the gain which is divided among all the partners in proportion to their respective investments in cash or its equivalent. This special share of the gain should therefore be deducted from the given net gain (except in the unusual case of the special allowance having been previously drawn and entered in the expense account), before such gain is proportioned among all the partners; and this special allowance to a partner should be added to his proportional share to find that partner's actual share of the total gain. NOTE 3. In applying Note 1, much labor will be saved by obtaining the equivalent net investments in hundreds of dollars for one month or for one day, as shown and explained in the abridged solution of 293. 260 PARTNERSHIP EXAMPLES FOR PRACTISE 1. A and B form a partnership on Jan. 1, 1910, A investing $9000 and B $6000. On June 1, A invested $3000 additional, and on Sept. 1, he withdrew $2000. On May 1, B withdrew $2000, and on Aug. 1, he invested $5000. On Jan. 1, 1911, the net gains were $4000, which were divided between the partners hi propor- tion to their respective investments. What was each partner's share? 2. A, B, and C engaged hi partnership, and agreed to share gains or losses hi proportion to average investment per month. A invested $12500 on Jan. 1, 1911; and withdrew $2500 on June 1, 1911. B invested $9000 on Jan. 1, 1911; invested $2000 on Apr. 1, 1911; and withdrew $3000 on Aug. 1, 1911. C invested $15000 on Jan. 1, 1911, and permitted it to remain unchanged throughout the year. If the net gains for the year were $7000, what was each partner's share? What was each partner's interest hi the firm after the adjustment? 3. A and B engage in business as partners on Jan. 1, 1912, A in vesting $15000 and B $18000; and agreed to share gains or losses hi proportion to average investment per month. A made a further hi vestment of $5000 on Aug. 1, 1912; and B withdrew $2000 on Mch. 1, 1912. On May 1, 1912, the firm took in C as a third partner who then hi vested $8000, and on July 1, 1912, invested an additional $6000. On Jan. 1, 1913, the capital of the firm was $65000. What was each partner's interest hi the firm on the latter date? 4. A and B are partners, having agreed to share gains or losses in proportion to average investment per day. A invested $18000 on Jan. 1, 1911; withdrew $2000 on May 16, 1911; and withdrew $1000 on Sept. 12, 1911. B invested $12000 on Jan. 1, 1911; and invested $9000 on July 23, 1911. If the net gains during the year were $6000, what was each partner's share? What was each partner's interest in the firm on Jan. 1, 1912? 5. A and B united in partnership and agreed to share gains or losses in proportion to average investment per day. A invested $8000 on Jan. 1, 1911; invested $4000 on May 19, 1911; and withdrew $2000 on Oct. 8 ; 1911. B invested $12000 on Jan. 1, PARTNERSHIP 261 1911; withdrew $2000 on Apr. 23, 1911; and withdrew $1000 on Nov. 2, 1911. What was each partner's interest in the firm on Jan. 1, 1912, if its capital at that time was $25000? 6. A and B engaged in business as partners and agreed to share gains or losses in proportion to average investment per day. A invested $12000 on Jan. 1, 1911; invested $7000 on June 20, 1911; and withdrew $4000 on Oct. 12, 1911. B invested $15000 on Jan. 1, 1911; withdrew $3000 on Apr. 29, 1911; and withdrew $3000 on Aug. 24, 1911. On Jan. 1, 1912, the books were adjusted, when it was found that the total resources of the firm were $40000, and their total liabilities $8000. What was each partner's interest in the firm on Jan. 1, 1912? WHEN EACH PARTNER'S PROPORTIONATE PART OF THE GAIN OR LOSS IS ARBITRARILY FIXED IN THE PARTNER- SHIP AGREEMENT, AND HIS INVESTMENT IS REQUIRED TO BE IN PROPORTION TO HIS STIPULATED SHARE OF THE GAINS OR LOSSES 304. INTRODUCTION, (a) A third method of distributing gains or losses among partners, and in many respects the most equitable, is to assign to each partner a fixed part of the gain, irrespective of his pecuniary investment. There are many pos- sible forms of investment besides that of money. Any contribu- tion of a partner to a business which will add materially to its prosperity, such as superior business talent, greater social influ- ence, excess of time or labor, disproportionate responsibility in management, etc., are properly investments, and frequently the most important elements in acquiring gains. If one partner invests twice as much money as another partner, that of itself should not be a sufficient reason for giving him twice as much of the gain, except he also invest twice as much time, twice as much responsibility of management, and, generally, twice as much of all the contributory elements in the acquisition of the firm's gains. (6) A mere difference in pecuniary investment can readily be adjusted between two partners if they are equal in all other respects, by giving to each partner one-half the firm's net gains, and requiring the partner who invested less than one-half the cap- ital to pay interest upon his deficiency at the market rate per annum, to the other partner who must have invested correspond- ingly more than one-half the capital, since both together must have contributed two-halves or the entire capital. (c) If, however, the partners are unequal in other respects 262 PARTNERSHIP than the money which they have invested, that is, if A of the firm of A and B, is, in all those qualities which put together consti- tute a successful business man, twice as capable as B, but finan- cially is deficient, then A should equitably receive twice as much of the firm's net gains as B, or two-thirds to B's one-third; and as the proper measure of the use of money is the interest thereon at the current rate per annum for the time it is used, A should pav to B interest on any deficiency of his investment below two- thirds of the entire capital. Conversely B should receive one- third of the firm's net gain and receive interest on the excess of his financial investment above one-third of the entire capital; for B's excess of investment above one-third of the firm's capital must exactly balance A's deficiency below two-thirds of the firm's capital, as the firm's capital, or what both have invested, must equal three-thirds of itself. (d) The allowance of interest by one partner who is deficient in his required average investment to another partner who is in excess, is most conveniently accomplished by crediting each part- ner with the interest on his net average investment; and debiting the firm's interest account, or its practical equivalent of subtract- ing the total interest on the several partners' net average investments from the firm's net gains; and proportioning the remainder of the net gain among the partners in accordance with the partnership agreement. ILLUSTRATIVE EXAMPLE A and B engaged in business as partners on Jan. 1, 1911, the agreement between them specifying that A, by reason of his su- perior business ability, should receive -f- of the net gain of the business, and B the remaining f; and that each should be account- able for investing as many sevenths of the net capital as he was to receive sevenths of the net gain, and be required to pay interest at 6 % upon as much of his required number of sevenths of the net capital as he failed to invest, and be entitled to receive interest at the same rate per annum on his excess if he invested more than his stipulated number of sevenths. At the end of the year the net gain was found to be $4291. A invested $12000 on Jan. 1, 1911; withdrew $4000 on Apr. 16, 1911; invested $7000 on Aug. 25, 1911; and withdrew $3000 on Oct. 8, 1911. B invested $15000 on Jan. 1, 1911; withdrew $2000 on Mch. 17, 1911; withdrew $1500 on July 5, 1911; and invested $5000 on Sept. 3, 1911. What was each partner's interest in the firm at the end of the year? PARTNERSHIP 263 SOLUTION Int. of $12000 from Jan. 1 to Dec. 31, inclusive (1 yr.) $720. Int. of $7000 from Aug. 25 to Dec. 31, inclusive (128 da.) 149.33+ Interest on A's total investments $869.33 Int. of $4000 from Apr. 16 to Dec. 31, inclusive (259 da.) $172.67- Int. of $3000 from Oct. 8 to Dec. 31, inclusive (84 da.) 42. Interest on A's total withdrawals $214.67 $869.33 - $214.67 = $654.66, int. on A's net average investment. Int. of $15000 from Jan. 1 to Dec. 31, inclusive (1 yr.) $900. Int. of $5000 from Sept. 3 to Dec. 31, inclusive (119 da.) 99.17- Interest on B's total investments $999.17 Int. of $2000 from Mch. 17 to Dec. 31, inclusive (289 da.) $ 96.33+ Int. of $1500 from July 5 to Dec. 31, inclusive (179 da.) 44.75 Interest on B's total withdrawals $141.08 $999.17 - $141.08 = $858.09, int. on B's net average investment. $654.66, obtained int. on A's net average investment. 858.09, obtained int. on B's net average investment. $1512.75, interest on firm's net average capital. Firm's net gain ($4291) interest on firm's net average capi- tal ($1512.75) = firm's gain to be proportioned among the part- ners as per agreement ($2778.25). [See Introduction, d.] $ of $2778.25 = $1587.57 (+ A's int. on net av. inv.) = A's gain f of $2778.25 = $1190.68 (+ B's int. on net av. inv.) = B's gain ($12000 + $7000) - ($4000 + $3000), or A's net in- vestment = $12000. A's stipulated proportion of the firm's net gain = 1587.57 A's interest on net average investment = 654.66 A's interest in the firm at end of year = $14242.23 ($15000 + $5000) - ($2000 + $1500), or B's net in- vestment = $16500. B's stipulated proportion of the firm's net gam = 1190.68 B's interest on net average investment 858.09 B's interest hi the firm at end of year = $18548.77 264 PARTNERSHIP EXPLANATION. Compute the interest upon each of A's investments and withdrawals from its date to the end of the settlement period (Dec. 31, 1911, inclusive), and subtract the total interest on his withdrawals from the total interest on his investments to find the total interest on A's net average investment ($654.66), with which A should be credited. Similarly find the interest on B's net average investment, obtaining $858.09, with which B should be credited. Deduct A's and B's special allowance for interest ($654.66 + $858.09 = $1512.75) from the net gain of the firm ($4291) to find the remainder of the net gain ($2778.25) which is to be divided between the partners as per agree- ment, f of $2778.25 ($1587.57) being credited to A, and \ of $2778.25 ($1190.68) being credited to B. The sum of either partner's net investment, the interest on his net average investment, and his stipulated proportion of the net gain after the net gain had been diminished by the interest on the firm's net average capital, will be that partner's interest in the firm at the time of adjustment. NOTE 1. If the interest on any partner's total investments is greater than the interest on his total withdrawals, the difference between these totals will be the interest on that partner's net average investment for which he should be credited; but if less, the difference between these totals will be the interest on that partner's net average insolvency for which he should be debited. NOTE 2. If the average financial condition of each of the partners for the settlement period, without an exception, is solvent, the sum of the interest on their respective net average investments will be the interest on the firm's net average capital for the same period; and if insolvent, the sum of the interest on their respective net average insolvencies will be the interest on the firm's net average insolvency. If, however, the average financial condi- tion of one or more of the partners is solvent and that of others insolvent, the difference between the ftum of all the solvent partners' interest on their net average investments and the sum of all the insolvent partners' interest on their net average insolvencies, will be the interest on the firm's net average capital if the sum of the interest on the net average solvencies is the greater, or it will be the interest on the firm's net average insolvency if the sum of the interest on the net average insolvencies is the greater. NOTE 3. Subtract the interest on the firm's net average capital from the net gain, or add it to the net loss, to find the gain or loss which is to be appor- tioned among the partners in accordance with the partnership agreement, for if the firm's "interest account" had been debited for the interest on each solvent partner's net average investment for which that partner had been credited and that "interest account" had been credited with the interest on each insolvent partner's net average insolvency for which that partner had been debited, and the firm's "interest account" had been balanced, and the obtained balance had been included in finding the firm's gain or loss, it would have correspondingly diminished the gain or increased the loss. PARTNERSHIP 265 NOTE 4. Add the interest on the firm's net average insolvency to the firm's net gain, or subtract it from the firm's net loss, to find the gain or loss which is to be apportioned among its several partners; for this interest ex- presses the net debit interest of the several partners for which the firm's "interest account" should exhibit a corresponding "credit balance." If this "credit balance" had been included in finding the firm's gain or loss, it would have correspondingly increased the firm's net gain or diminished the firm's net loss. EXAMPLES FOR PRACTISE 1. A and B engaged in business as partners on Jan. 1, 1911, A agreeing to invest | of the capital and to receive f of the net gain of the firm, and B to invest $ of the capital and to receive of the net gain. It was further agreed that interest at 8 % annum would be allowed to either partner for investing more than his stipulated number of fifths of the capital, and to be charged against any partner for any deficiency of stipulated investment. A in- vested $9000 on Jan. 1, 1911; invested $3000 on Apr. 12, 1911; withdrew $1500 on June 9, 1911; invested $2500 on Aug. 13, 1911; and withdrew $900 on Nov. 8, 1911. B's net investment was $18000, and the interest on his net average investment was $758.90. If the net gain of the firm during the year was $7500, what was each partner's interest hi the business on Jan. 1, 1912 (that is, on Dec. 31, 1911, inclusive)? 2. A and B formed a partnership on Jan. 1, 1912, and agreed to invest equally and to share equally in gains or losses, and in- terest at 9% per annum was to be allowed to either partner for any excess over stipulated investment, or to be charged against either partner for any deficiency of investment. The net gain of the firm on Jan. 1, 1913, was found to be $7600. A invested $15200 on Jan. 1, 1912; withdrew $2600 on May 23, 1912; and invested $3800 on Sept. 18, 1912. B invested $18000 on Jan. 1, 1912; invested $1600 on Mch. 17, 1912; withdrew $2000 on June 8, 1912; and invested $3000 on July 16, 1912. What was each partner's interest in the firm on Jan. 1, 1913? 3. A, B and C entered into a copartnership agreement to commence business on Jan. 1, 1911, and agreed to invest equally and to share equally in gains or losses, and to allow interest at 6 % per annum for any excess over stipulated investment, and to 266 PARTNERSHIP charge interest at the same rate for any deficiency. A invested $16000 on Jan. 1, 1911; invested $4000 on May 26, 1911; and withdrew $2400 on July 12, 1911. B invested $18600 on Jan. 1, 1911; and withdrew $1500 on Oct. 19, 1911. C invested $14000 on Jan. 1, 1911; withdrew $2000 on May 4, 1911; and invested $5000 on Aug. 26, 1911. If the net gain of the firm at the end of the year was $12600, what was each partner's interest hi the business at that time? 4. A, B and C enter into partnership to engage in business, A agreeing to invest f of the capital, B to invest f , and C f, and to receive the same proportion of the net gain of the business. It was also agreed that each partner was to be credited with interest at 10 % per annum for any excess, and to be debited with interest at the same rate for any deficiency, of stipulated investment. A invested $9000 on Jan. 1, 1912; and withdrew $500 on June 25, 1912. B invested $14000 on Jan. 1, 1912; and invested $3000 on May 14, 1912. C invested $7500 on Jan. 1, 1912; invested $3500 on July 2, 1912; and withdrew $2000 on Sept. 20, 1912. On Jan. 1, 1913, the books were adjusted to the end of the preceding day, and showed a net gain of $8730. What was each partner's in- terest hi the firm at settlement? 6. A and B entered into partnership, A agreeing to invest -ft of the capital and to receive TT of the total gain, and B to invest TT of the capital and to receive TT of the total gain, and interest at a fixed rate per annum was to be credited to either partner for any excess, or debited against him for any deficiency, of required investment. At settlement it was discovered that the net capital of the firm was $26200; that A's total interest on investments was $518.65, his total interest on withdrawals was $168.15, and his net hi vestment $15300; and that B's total interest on invest- ments was $927.83; his total interest on withdrawals $216.48; and his net investment $8700. Find each partner's interest in the firm at settlement. APPENDIX USEFUL COMMERCIAL CALCULATIONS THE ALIQUOT METHOD OF COMPUTATION 305. An aliquot of a number is any number which will exactly divide it. NOTE. An aliquot part of a given number is the fractional part which expresses the relation of the aliquot to that number. Thus, 8 is an aliquot of 32, and J is the aliquot part which 8 is of 32. The following are aliquots of one dollar (100 cents) : *} 25* =$i ILLUSTRATIVE EXAMPLE 306. What is the cost of 52 yd. muslin at 12J^f per yard? SOLUTION. \1\t = $J. Therefore at $| per yd., 8 yd. will cost $1, and 52 yd. will cost as many times $1 as 8 yd. are contained times in 52 yd., or 6 times $1 = $6.50. RULE. Divide the given quantity by the quantity which $1 can buy. EXAMPLES FOR PRACTISE Find the cost of 1. 512yd. at 12^ per yd. 7. 292 gal. at 6i i per gal. 2. 748 Ib. at 25^ per Ib. 8. 610 Ib. at 5^ per Ib. 3. 465 gal. at 20 i per gal. 9. 583 yd. at 25 i per yd. 4. 236 bu. at 50^ per bu. 10. 329 Ib. at Vl\i per Ib. 5. 726 Ib. at 16f i per Ib. 11. 837 gal. at 50?f per gal. 6. 417 yd. at 33j^ per yd. 12. 913 doz. at 20?f per doz. 268 APPENDIX ILLUSTRATIVE EXAMPLE 307. At 25* per pound, how many pounds of cheese can be bought for $9? SOLUTION. 25 = $^. Therefore at Sj per lb., $1 can buy 4 lb., and 9 times $1, or $9, can buy 9 times 4 lb., or 36 lb. RULE. Multiply the quantity which $1 can purchase by the number of dollars to be expended, expressing cents in the total cost as a fraction of a dollar. EXAMPLES FOR PRACTISE How much can be bought for 1. $375 at 20* per yd.? 5. $61.50 at 25 * per gal.? 2. $291 at 16f*pergal.? 6. $41.25 at 12* per yd.? 3. $148 at 25* per lb.? 7. $278.75 at 33 J* per lb.? 4. $86 at 50* per bu.? 8. $109.20 at 45* per doz.? 308. A price which is not itself an aliquot of $1 may frequently be resolved into two or more components which are. Thus, 7}* = 5* + 2H or $2V + \ of 15* = 10* + 5*, or SrV + } of 18f* = 12}* + 6K or $J + i of $J. 22^ = 20^f + 2H or $i + i of Si. 27}* = 25* + 2}*, or Si + T V of Si- 37^* = 25* + 12}*, or Si + } of Si- 55* = 50* + 5*, or S} + T V of S}. 62J* = 50* + 12}*, or $} + J of S}. 75* = 50* + 25* or $J + } of S}. ILLUSTRATIVE EXAMPLE 309. Find the cost of 764 yd. cloth at 37}* per yard. SOLUTION At 25* per yd., 764 yd. will cost $191 (i of $764). " 05 *n a rkf i cm EXAMPLES FOR PRACTISE Find the cost of 1. 618yd. at 18f?fperyd. 5. 640 yd. at 55 1 per yd. 2. 428 Ib. at 15?f per Ib. 6. 732 gal. at 22} fi per gal. 3. 316 bu. at 62}^ per bu. 7. 684 Ib. at $1.37} per Ib. 4. 520 gal. at 7J?f per gal. 8. 562 bu. at $1.15 per bu. NOTE 1. The above process can be advantageously applied to common fractions of such denominators as are usually employed in business. Multiply (9) $216 by T 5 *; (10) $756 by &. SOLUTION (9) (10) $216 = 16 sixteenths $756 = 32 thirty-seconds \ of 216 = 54 = 4 } of 756 = 189 = 8 " " }of 54=JL3} = 1 " j of 189 = 23 f= 1 " $67} or $67.50 = fV $212f or $212.625 = & EXAMPLES FOR PRACTISE Multiply Multiply Multiply 11. $648 by f 14. 824 Ib. by A 17. 372 gal. by & 12. $432 by T 3 * 15. 268 yd. by 18. 716 yd. by 13. $596 by A 16. 562 bu. by \\ 19. 384 bu. by If 310. A price which is not itself an aliquot of $1, may lack an aliquot of being $1. Thus, 99^ = $1 - -riv of $1. 95^f = $1 - ^ of $1 $l-*Vof$l 87}^f = $1 - iof $1 $1 -A of $1 75^ =$l-iof$l ILLUSTRATIVE EXAMPLE Find the cost of 136 yd. cloth at 75 f per yd. SOLUTION 136 yd. at $1 per yd. = $136. " " " $J " = 34 Q of $136). " " " $J " " = $102 ($136 - $34). 270 APPENDIX EXAMPLES FOR PRACTISE Find the cost of 1. 260 yd. @ 95^ 4. 473 gal. @ 99^f 7. 152 bu. @ $1.95 2. 512 Ib. @ 87 \i 5. 390 bu. @ 98^ 8. 284 bbl. @ $6.90 3. 432 gal. @ 75^ 6. 256 bbl. @ $5.75 9. 520 bbl. @ $7.80 NOTE 1. The above process can also be used when the multiplier is a common fraction and its numerator lacks an aliquot of being equal to its denominator. Multiply (10) 1264 by f ; (11) 928 by }. SOLUTION (10) (11) 1264 = 8 eighths 928 = 4 fourths \ of 1264 = 158 = 1 " i of 928 = 232 = 1 (1264 - 158) = 1106 = | (928 - 232) = 696 = f EXAMPLES FOR PRACTISE Multiply Multiply Multiply 12. 519 by | 15. 453 by 18. 293 by 4J 13. 673 by f 16. 369 by | 19. 467 by 2| 14. 527 by | 17. 258 by A 20. 625 by 3| NOTE 2. The above process will also be found useful when a multiplier lacks 12 or less of being 100, 1000, 10000, etc. 21. Multiply 386 by 997. SOLUTION EXPLANATION. For convenience, first mul- 386 X 1000 = 386000 ^ p * y ^y 1000, as explained in 30, obtaining a 38fi v o 1 1 co result which is 3 times 386 too great. Hence, deduct 3 times 386 from the result, obtaining 386 X 997 = 384842 384842 as the correct product. EXAMPLES FOR PRACTISE Multiply Multiply Multiply 22. 875 by 97 25. 52 by 995 28. 712 by 9998 23. 426 by 99 26. 628 by 994 29. 256 by 980 24. 642 by 96 27. 329 by 993 30. 7384 by 9200 TO FIND THE COST OF GRAIN WHEN THE PRICE IS PER BUSHEL AND THE QUANTITY OF GRAIN IS EXPRESSED IN POUNDS ILLUSTRATIVE EXAMPLE 311. Find the cost of 5687 Ib. oats at 49 per bushel of 32 Ibs. SOLUTION 5687 $.49 EXPLANATION. At 49?f per pound, the cost would 5H83 have been 5687 times 49^, or $2786.63; therefore at 49?$ _ per bushel of 32 pounds, this obtained cost must be 32 ^* times the correct cost. Hence, divide $2786.63 by 32, or 8)2786.63 by the factors of 32 (8 and 4) to find the correct cost. 4)348.329- $87.082 RULE. Multiply the weight in pounds by the price per bushel, and divide the product by the weight in pounds of 1 bushel of the given kind of grain. EXAMPLES FOR PRACTISE Find the cost of 1. 15207 Ib. wheat at $1.09 per bu. of 60 Ib. (10 X 6). 2. 9873 " corn " 68# " " " 56 " (8 X 7). 3. 5928 " beans " $2.10 " " " 60 " 4. 1753 " timothy seed at $2.35 per bu. of 45 Ib. (9 X 5). 5. 2874 " buckwheat at $1.25 per bu. of 48 Ib. (8 X 6). NOTE. The above process can be advantageously employed in finding the cost of any merchandise when the quantity is expressed in units of one denomination and the price is per unit of a different denomination, thus avoid- ing an awkward intermediate fraction. 6. 5368 Ib. coal at $6.75 per ton of 2000 Ib. [39]. 7. 18249 " " " $7.25 " " " " " 8. 27235 " " " $6.50 " " " " " 9. 8739 " " "$7.10 " " "2240 "(10X8X7X4). 10. 14296 " " " $7.50 " " " 2240 " 11. 3876 " ear corn at $3.25 per bbl. of 350 Ib. (10 X 7 X 5). 12. 9285 " freight at 45 per cwt. [38]. 272 APPENDIX 13. 25462 ft. lumber at $18.50 per M. 14. 12348 Ib. dressed beef at $10.15 per 100 Ib. 15. 6285 Ib. flour at $5.20 per bbl. of 196 Ib. (7 X 7 X 4). ADDITION BY GROUPING FIGURES 312. Skill in addition is an invaluable accomplishment which should be acquired by every one who wishes to follow a business career. No method of addition has so many advantages and so few disadvantages as that of adding one column at a time and, while doing so, training the eye to take within the range of its vision as many successive figures in the column as the mind may be capable of simultaneously adding. At first, the learner should attempt to group only those figures which are within the power of his present capacity, and let actual experience demonstrate the limitations of his individual ability. Efficiency in grouping will increase by exercise. 10 is the usual maximum grouping recom- mended for beginners. !} 2 ) 3^-9 4) 9 2 6V-9 10 10 6) 1 ^10 3) 1) 4J- 3) 8 4 3 2 61 3) 9 10 2 10 82 ILLUSTRATIVE EXAMPLE EXPLANATION. Successive figures in the several columns which admit of grouping into convenient combinations are here connected by a brace to enable the learner the better to com- prehend the manner in which figures may be thus massed. The higher figures as 9's or 8's, which have no correspondingly low contiguous figures, as 1's or 2's, that can be combined with them, should be added singly. Thus, add the groupings of the units' column (8, 17, 27, 33, 42, 52, 62, 69), obtaining 69. Write 9 units in the units' column, and carry 6 tens to the tens' column. 6 (to carry), 15, 24, 34, 43, 53, 61, 70, 80, obtaining 80 tens, or 8 hundreds and tens. Write tens in the tens' column, and carry 8 hundreds to the hundreds' column. 8 (to carry), 16, 26, 36, 46, 55, 64, 73, 82, obtaining 82 hundreds, making the sum of the three columns equal 8209. },. ADDITION BY GROUPING FIGURES 273 EXAMPLES FOR PRACTISE (1) (2) (3) (4) (6) (6) (7) 3 2 21 456 142 326 569 5 7 62 633 357 762 413 4 1 93 291 213 691 324 9 4 35 832 431 325 741 6 3 21 165 565 443 246 2 2 58 312 252 232 632 8 5 14 421 343 586 171 7 4 63 256 224 213 514 2 9 31 432 769 732 465 6 6 58 318 132 964 922 4 2 42 271 348 143 873 5 3 75 726 622 212 221 3 7 34 193 896 747 512 (8) (9) (10) (ID (12) 2674 64287 487639 528964 8492716 3425 13612 261324 272625 1637154 4283 32291 432152 313428 1231142 1716 48453 145464 293112 3536126 9134 21432 925515 654577 4172641 2942 32125 343123 132322 2629239 7123 76562 326682 721831 7314275 1358 23128 471307 276146 2163913 2542 94319 819731 819120 1342592 3291 62643 961252 125394 4289411 4876 24356 327218 363283 4526956 9212 25191 712892 142317 2133242 2794 43737 231429 654976 3641727 6243 12453 546561 211732 5324361 2922 56214 324326 732142 2472613 4636 31925 453656 129124 1413486 1256 93172 652146 476164 6235612 5126 59213 254916 323622 2249232 1716 29561 756926 342429 7264145 8226 39343 154976 248928 2295724 274 APPENDIX CROSS-MULTIPLICATION 313. In obtaining the necessary extensions of many of the separate items in making out bills, invoices, etc., much valuable time will be saved by the following process, when the price and the quantity consist of 2 or 3 orders, as is ordinarily the case. ILLUSTRATIVE EXAMPLE Find the cost of 23 yards cloth at 45 cents per yard. SOLUTION EXPLANATION. The product of two numbers is the product A . - of each order of one number separately multiplied by each order of the other number [Prin. 2, 26]. Hence, First multiply the units' orders (3 times 5 units = 15 units, $10.35 or 1 ten and 5 units). Write 5 units as the units' order of the product, and carry 1 ten. Second, multiply each tens' order by its opposite units' order (3 times 4 tens + 5 times 2 tens + 1 ten to carry = 23 tens, or 2 hundreds and 3 tens). Write 3 tens as the tens' order of the product, and carry 2 hundreds. Third, multiply the tens' orders (4 tens X 2 tens = 8 hundreds) + 2 carried hundreds = 10 hundreds; thus making the complete product 1035 cents, or $10.35. EXAMPLES FOR PRACTISE Multiply Multiply Multiply Multiply 1. 48 by 37 4. 86 by 63 7. 27 by 35 10. 73 by 54 2. 53 by 74 5. 58 by 24 8. 65 by 23 11. 36 by 27 3. 29 by 82 6. 95 by 32 9. 42 by 67 12. 52 by 73 POWERS AND ROOTS 314. A power of a number is the product of that number if taken two or more times as a factor. NOTE 1. The second power or square of 3 is the product of 3 taken twice as a factor (3 X 3), or 9; the third power or cube of 2 is the product of 2 taken three times as a factor (2 X 2 X 2), or 8; the fourth power of 5 is the product of 5 taken four times as a factor (5 X 5 X 5 X 5), or 625; etc. NOTE 2. The exponent of a power is a small figure placed diagonally above a number to denote how many times it is to be taken as a factor. Thus S 2 expresses the square of 5; 7 3 the cube of 7; 8 6 the fifth power of 8; etc. 315. The root of a power is one of the number of equal fac- tors of which that power is the product. POWERS AND ROOTS 275 NOTE 1. The square root of a power is one of its two equal factors; the cube root is one of its three equal factors; the fourth root is one of its four equal factors; etc. NOTE 2. The radical sign V~ is usually written at the left of a power to indicate that a root of it is to be taken; and a small figure called an index is written within the angle of the radical sign to show what root is to be taken. When no index is written, the index 2 is understood. Thus, VT^ indicates the square root of 16; -^512 the cube root of 512; v'ST the fourth root of 81. 316. Principles. 1. //, commencing at the right of any given power of an integral number, it is divided by a separatrix into sections of as many figures each as there are units in its exponent, the number of such sections, counting a possibly incomplete section at the extreme left as another section, will denote the number of orders in its root: and the enumeration of those sections will indicate the relative value of their corresponding root orders. 2. (a) The highest or left-hand section contains the given power of the highest root order: (b) any excess of the highest section over the given power of the highest root order, written at the left of the next following section, will approximately express the value of the given power of the second highest root order and its involved relations with the highest root order: (c) any excess of the highest two sections over the given power of the highest two root orders, written at the left of the next following section, will approximately express the value of the given power of the third highest root order and its involved relations with the highest two root orders; etc. ILLUSTRATIVE EXAMPLES 317. What is the square root of 4624? SOLUTION 46|24(6 tens 70 2 = 4900 6 2 = 36 g 60 2 = 3600 130)1024(y units 10)1300 910 Average square per unit = 130 Tl4 V4624 = 6 tens and 8 units, or 68. EXPLANATION, (a) Commence at the right of the given power and divide it by a separatrix into sections of two figures each, thus finding that the required root contains two orders, tens and units (Prin. 1, 316). 276 APPENDIX (6) The greatest number the square of which is contained in the left-hand or tens' section (46) is 6 tens. Write 6 tens as the highest order of the re- quired square root, subtract its square (6 2 = 36) from its corresponding sec- tion (46), to the remainder (10) bring down the next following section (24) and the result (1024) will approximately express the square of the next following or units' order of the required root + its involved relations with the tens' order of the root (6, Prin. 2, 316). (c) The required square root is thus found to be more than 6 tens (or 60 units if expressed in terms of the next lower root order), and less than 7 tens (or 70 units if similarly expressed); and if the difference between the square of 60 units (60 2 = 3600) and the square of 70 units (70 2 = 4900), or 1300, be divided by the difference between their respective roots (70 60), or 10, the quotient (130) will be the average square of the ten consecutive numbers between 60 and 70; and as the difference is so slight as to be prac- tically negligible, it (130) can be accepted tentatively as the average square of as many of these consecutive numbers as may be included in the next following or units' order of the required root. (d) Divide the square of the units' order of the required root and its involved relations as found in b (1024) by the assumed average square per unit and its involved relations as found in c (130) to find tentatively the number of units in the units' order (7 units and a significant remainder, practically 8 units). Hence, 6 tens and 8 units, or 68, is the required square root. NOTE. In the above solution, the average square of the ten consecutive squares between 60 2 and 70 2 (130) is slightly greater than the average square of the eight consecutive squares between 60 2 (3600) and the square of the true root (68 2 = 4624) which is 128 (4624 - 3600 -^ 8). Hence, the final re- mainder (114) from dividing by the assumed average square (130) lacks (130-128) X 8, or 16 of containing the tentative divisor (130) exactly 8 times. 318. What is the cube root of 105823817? SOLUTION 105|823|817(4 hundreds 50 3 = 125000 4 3 = 64 7 40 3 = 64000 6100)41 823(J5 tens 10)61000 36600 Average cube per unit = 6100 5223 105823 480 3 = 110592000 47 3 = 103 823 3 470 3 = 103823000 676900)2 000 817$ units 10)6769000 1 353 800 Average cube per unit = 676900 647 017 Hence, s/105823817 = 4 hundreds, 7 tens, 3 units, or 473. POWERS AND ROOTS 277 EXPLANATION, (a) Commence at the right of the given cube and divide it by a separatrix into sections of three figures each, thus finding that the required root contains three orders, hundreds, tens, and units (Prin. 1, 316). (6) The greatest number the cube of which is contained in the left-hand or hundreds' section (105) is 4 hundred. Write it (4 hundred) as the highest order of the required root (a, Prin. 2, 316), subtract its cube (4 s = 64) from its corresponding section (105), to the remainder (41) bring down the next fol- lowing section (823), and the result (41823) will approximately express the cube of the next following or tens' order of the required root + its involved relations with the hundreds' order of the root (b, Prin. 2, 316). (c) It is now found that the required cube root is greater than" 4 hundred (or 40 tens if expressed in terms of the next lower root order) and less than 5 hundred (or 50 tens if similarly expressed) . If therefore the difference between the cube of 40 tens (40* = 64000) and the cube of 50 tens (50 = 125000), or 61000, be divided by the difference between their respective roots (50 - 40), or 10, the quotient (6100) will be the average cube of the ten consecutive numbers between 40 and 50; and as the difference is so slight as to be prac- tically negligible, this result (6100) can be accepted tentatively as the average cube of as many of these consecutive numbers as may be included in the next following or tens' order of the required root. (d) Divide the cube of the tens' order of the required root and its involved relations, as found in b (41823) by the assumed average cube per unit and its involved relations as found in c (6100) to find tentatively the number of units in the tens' order (6 tens and a significant remainder, practically 7 tens). (e) The required root is now found to be greater than 47 tens (or 470 units if expressed in terms of the next lower root order) and less than 48 tens (or 480 units if similarly expressed). If therefore the difference between 470 (103823000) and 480 1 (110592000), or 6769000, be divided by the difference between their respective roots (480 470), or 10, the quotient (676900) will be the average cube of the ten consecutive numbers between 470 and 480; which may also be accepted tentatively as the average cube of as many of these consecutive numbers as are included in the next following or units' order of the required root. (/) Subtract the cube of the already obtained root orders (47* = 103823) from their corresponding sections (105823), to the remainder (2000) bring down the next following section (817), and the result (2000817) will approximately express the cube of the next following or units' order of the required root + its involved relations with the preceding root orders (c, Prin. 2, 316). (g) Divide the cube of the units' order of the required root and its involved relations as obtained in / (2000817) by the assumed average cube per unit in the units' order and its involved relations as obtained in e (676900) to find tentatively the number of units in the units' order (2 units and a significant remainder, or 3 units). Hence, 4 hundreds, 7 tens, and 3 units, or 473, is the required cube root. 278 APPENDIX 319. In the following solutions, the average power per unit is more briefly obtained by finding the difference between the given power of the root orders already found and the given power of those root orders increased by 1 without expressing them in terms of the next lower root order, and annexing to this difference as many ciphers lacking 1 as there are units hi the exponent of the given power. Much unnecessary labor may also be avoided in the suc- cessive divisions to find the next following root order by using only the highest two orders of the successive divisors, and by omitting from the right of their corresponding dividends as many figures as have been rejected from the right of their respective divisors. The processes being identical for all roots, no further explanations are considered necessary. If, however, the learner should fail to understand any step hi the following solutions, he can refer to the corresponding step in the preceding explanations for the required information. ILLUSTRATIVE EXAMPLES 1. What is the fifth root of 3796375994368? SOLUTION 379|63759|94368(3 hundred 3 5 =243 2 78|10000)136|63759(; tens 4 5 = 1024 78 3 5 = 243 58 Average power per unit = 7810000 37963759 32 5 = 33554432 g 33 5 = 39135393 55|809610000)440|932794368(/f units 32 5 = 33554432 Average power per unit = 55809610000 50 N/3796375994368 = 3 hundred, 2 tens, 8 units, or 328. 2. What is the fourth root of 552114385936? POWERS AND ROOTS 279 SOLUTION 5521|1438|5936(8 hundred 8 4 = 4096 6 24|65000)142|51438(P tens 9 4 = 6561 123 8 4 = 4096 19 Average power per unit = 2465000 55211438 86 4 = 54700816 87 4 = 57289761 25|88945000)51 106225936(2 units 86 4 = 54700816 51 Average power per unit = 2588945000 < H, 4i M, II (16.) A, A, A, 41 (is. Article 92 (Pages 56-57) (1.) 1A (20 1H (30 1A (40 1H (60 If (6.) 1H (70 (8.) 1A (90 2H4 (100 2444 (HO 2|f (12.) 2U (13.) (14.) 502f (15.) 255^ (16.) 162A (17.) 789 T 4 JJ (18.) (19.) 278 1 yd. (20.) 303 3 yd. (21.) 180 15 Ib. (22.) 251 1 4 Ib. (23. 2103 12 bu. (24.) 3106 10 bu. (26.) 10308 24 bu. Article 94 (Page 58) (1.) A (20 A (30 H (40 A (60 41 (60 I (7) it (8.) (9.) A (100 H (HO f (120 f (13.) H (14.) H (160 II (16.) Article 95 (Page 60) (1.) 311| (2.) 3134 (30 390f (4.) 143$ (5.) 57& (6.) (7.) 204 (80 35|| (9.) 18I| (10.) 2844 (11.) 28|f (12.) (13.) 22H (140 29|| (15.) 22| (16.) 28 Article 97 (Page 62) (1.) I (20 & (30 A (40 rV (60 24 (6.) 3 (7.) 4 (8.) 24 (9.) 14 (10.) 35 (11.) 504f (12.) 184 (13.) 1761f (14.) 4 (16.) i ANSWERS 287 Article 98 (Page 62) (1.) 4751| (2.) 2660f (3.) 342U (4.) 2073* (6.) 1966f (7.) 19986f (8.) 12437^ (9.) 11846| (11.) 347281 (12.) 32538 (6.) 7140 (10.) 20008 (1.) (6.) 8200*V (11.) 1609381 H (16.) 645883H (1.) A (2-) (8.) | (9.) i (16.) A Article 99 (Page 63) (2.) 2757f (3 ) 4804f| (4.) (8 ) 19895^ (9.) 9062^ (120 864966 T 5 j (13.) 241458A Article 101 (Page 65) (5.) 941 (10.) 27554f (14.) 1258300 i (3.) (10.) A A (4.) (11.) |f (6.) (12.) (60 (13.) (1.) 8 (8 ) 32 Article 102 (Page 66) (2.) 24 (3.) 27 (4.) 10 (6.) 19* (6.) 2\ (9.) 6 (10.) 5192* (11.) 1766f (12.) 904 (70 I (14.) A (7.) 44* (1.) (2.) H (3.) 3f (4.) 44 (9.) 1A (10.) f (11.) | (12.) Article 103 (Page 67) (6.) 6 (6.) 11J (7.) A (8.) * (1.) 568H (6.) (2.) 146m (7.) Article 104 (Page 67) (3.) 240,% (4.) 138fft (8.) 336HH (9. (5.) 110JJJ Article 105 (Page 68) (1.) 163^V (2.) 140JH (3.) HO^V (4.) (6.) 104 JH (7.) 107HII (8.) 49HH (9.) (6.) (6.) 378 (11.) 31 Article 106 (Page 69) (3.) (8.) (13.) 37 (4.) (9.) 41H"ft" (14.) 151A Article 108 (Pages 70-71) (1.) $M (2.) 6 yd. (3.) 2 pk. (4.) 8 Ib. (6.) (7.) M hr. (8.) 3 da. (9.) 13 da. (10.) 95 da. (12.) $8 T A* (130 23 T % yd. (14.) A (160 7J gal. (10.) 6 (16.) 1 hr. (6.) A A. (11.) 7^ mi. (16.) 108A yd. Article 110 (Pages 74-75) (1.) $6300 (7.) H (8 ) (2.) 60 Ib. (3.) A 0.) 171 sheep (4.) $5628 (6.) $2316 (6.) f (10.) * (11.) $7800 (12.) $18275 288 ANSWERS Article 111 (Pages 75-77) (1.) 2990H (20 26 j\ acres (3.) $21.83ff (4.) J- (6.) 21HH mi. (6.) $1.831 (7.) $12710A (8.) 1900 mi. (9.) 19^ hr. (10.) $3943.78i (11.) T V da. (12.) ff (13.) grammar $.69; speller $.23 (16.) 2-Hda. (16.) 2Hda. (17.) 30 da. (18.) 81U (19.) U (20.) 884 (21.) 212 T V Ib. (22.) 400 Ib. (23.) 10 min. (24.) $450 (25.) f (26.) $13450; $10760 (27.) $12800 (28.) 568i (29.) 454^ (30.) f (31.) A $4280; B $3210; C $2568 (32.) $2125 Article 123 (Page 83) (1.) .5 (2.) .75 (3.) .625 (4.) .25 (5) .875 (6.) .375 (7.) .3125 (8.) .28125 (9.) .125 (10.) .140625 (11.) .6875 (12.) .8 (13.) .36 (14.) .136 (16.) .203125 (16.) 18.1875 (17.) 465.21875 (18.) 67.49609375 (19.) .71875 (20.) .328125 (21.) .9375 (22.) .34375 (23.) .109375 (24.) .8125 (26.) .546875 (26.) .03125 (27.) 58.4375 (28.) 63.53125 (29.) 37.046875 (30.) 78.0625 (31.) 168.44- (32.) 34.06- (33.) 72.72- (34.) 91.688- (36.) 26.094- (36.) 49.016- (37.) 23.4688- (38.) 71.7656+ (39.) 37.8125 (40.) 8.26563- (41.) 7.34867- (42.) 9.52722+ (43.) 1.276431+ (44.) 3.296429- (45.) 2.834538- (1.) f )) 16 T 3 T (2.) f (9.) A (15.) 8 Article 124 (Page 84) (3.) rir (4.) | (5.) -; ff (6.) (10.) H (ID if (12.) riff (13.) (16.) 2 T fo (17.) 19 (18.) 6* (7.) H (14.) (19.) 9* (1.) (8.) (2.) ) (9.) I Article 125 (Page 85) (3.) f (4.) f (5.) | Article 126 (Pages 86-87) (1.) 923.6129 (2.) 80.9235 (3.) 816.19108 (6.) 5.65395 (6.) 626.109375 (7.) 76.250375 (9.) 123.079- (10.) 12.5576+ (11.) 20.676- (13.) 136137.1064684 (14.) 9341.9448348 (6.) i (7.) (4.) 656.82651 (8.) 92.603845 (12.) 1128.037265 Article 127 (Page 87) (1.) 46.786 (6.) 75.383 (9.) 14.175 (13.) 6.7535625 (17.) 231.9944 (1.) 22221.54 (6.) .54528 (9.) 4.774225 (13.) 6.15470625 (17.) 2.7500^ (21.) 7.183 (2.) 580.8655 (6.) 24.747 (10.) .53133 (14.) 28.103 + (18.) .77427 (3.) 38.4322 (7.) 257.1286 (11.) 26.0675 (16.) 67.2320- Article 128 (Pages 88-89) (2.) 590.7568 (6.) 8.58426 (10.) 20.660234375 (14.) 4.02636 (18.) .3847^ (3.) 33.20697 (7.) .003973 (11.) 24.768 (15.) 10.138^ (19.) 71.83 (4.) 34.858 (8.) 67.6871 (12.) 142.50125 (16.) 5.83 + (4.) .070924 (8.) 39.845 (12.) 15.181809375 (16.) 1.801 Hi (20.) 620 ANSWERS 289 Article 129 (Page 90) (1.) 5.447 (5.) 46250 (9.) 190000 (13.) 4.0625 (17.) 1.1353- (21.) 12.782- (25.) 90.0671 + (29.) .052384 (33.) .0969- (1.) $18.80- (6.) $1394.54 (9.) $14.86- (2.) 2.8 (6.) .0682 (10.) .0004 (14.) (18.) (22.) (26.) (30.) (34.) 78.36 12.82- .526 + .55 .034f .1091- (3.) .572 (7.) .006 (11.) 27000 (15.) 4.36 (19.) .051- (23.) 946.92 + (27.) 2.8345 (31.) .067- Article 130 (Page 92) (2.) $13.33 + (6.) $4.36+ (10.) $15.11 + (3.) $41.17+ (7.) $5.40- (4.) 2.5837 (8.) .0000032 (12.) 125 (16.) 726.8 (20.) 22.83- (24.) 75.035 + (28.) 416.78 (32.) .017833 + (4.) $151.30+ (8.) $35.67+ Article 131 (Pages 92-94) (1.) 46.2394842 (2.) 2866.44921875 bu. (3.) 361.39525 (4.) H (5.) $240.885+ (6.) 73 (7.) .75 yd. (8.) 637.5 bu. (9.) .375 (10.) 432 brls. (11.) $9600 (12.) .74 (13.) $1560 (14.) wife $13566; daughter $8996.40; son $5997.60 (15.) house $2000; lot $1500 (16.) $16400 (17.) $82.40 (18.) $34 (19.) $468 (20.) 5400 inhabi- tants (21.) A .46; B .54 (22.) $710.10 (23.) .28 (24.) $95 (25.) $396 Article 156 (Page 100) (1.) 9583 d. (2.) 147600 Ib. (3.) 1082 pt. (4.) 19080 m. 113271120 sec. (6.) 103944 gr. (7.) 47234 d. (8.) 5628 in. 3248 cu. ft. (10.) 26091396 sq. in. (11.) 1714356 in. (12.) 211456 oz. (13.) 2221200 sec. (14.) 1139 qt. (15.) 2108 gi. (16.) 460858572 sq. in. (17.) 1188864 cu. in. (18.) $6.72 (19.) $412.65 (20.) $14.80 a) Article 157 (Page 101) (1.) 96 bu. 1 pk. 6 qt. 1 pt. (2.) 41 da. 6 hr. 54 min. 42 sec. (3.) 4 mi. 101 rd. 2 ft. 3 in. (4.) 18 A. 158 sq. rd. 28 sq. yd. 8 sq. ft. 72 sq. in. (5.) 24 8s. lid. (6.) 26 T. 13 cwt. 60 Ib. 4 oz. (7.) 11 rm. 15 sh. (8.) 61 cd. 1 cd. ft. 11 cu. ft. (9.) 9 Ib. 9 oz. 6 pwt. (10.) 437 cd. (11.) 32 3d. (12.) $53 86 7 m. (13.) 196 gal. 1 qt. 1 pt. 3 gi. (14.) 33 da. 3 hr. 24 min. 12 sec. (15.) 9 mi. 290 rd. 5 yd. 1 ft. 1 in. (16.) 233 bu. (17.) 165 mi. 311 rd. 4 yd. 1 ft. 6 in. (18.) 19 cu. yd. 17 cu. ft. 1505 cu. in. (19.) 74 6s. (20.) 43 gal. 2 gi. Article 158 (Page 102) (1.) 17s. 6d. (2.) 273 da. 18 hr. (3.) 1 pt. 2 gi. 136 da. 21 hr. (6.) 142 rd. 1 yd. 8 in. 3 cwt. 75 Ib. (9.) 13s. 9d. (10.) 2 ft. 9 in. (4.) 2 pk. 2 qt. (7.) 2 ft. 6 in. Article 159 (Page 103) 1.) 1 pt. 2 gi. (4.) 3 qt. 3 gi. (8.) 10s. 5.7d. (2.) 13 da. 16 hr. 30 min. (3.) 2 pk. 1 qt. 1 pt. (5.) 17 cwt. 52 Ib. (6.) 4 oz. 10 pwt. (7.) 110 rd. (9.) 3 pk. 5 qt. 1 pt. 290 ANSWERS Article 160 (Pages 103-104) (!.)! (2.) $| (3.) f gal. (4.) f bu. (6.) * da. (6.) .875 gal. (7.) .625 (8.) -5yd. (9.) .4844- bu. (10.) .222- da. Article 161 (Page 104) (1.) 44 da. 10 hr. 3 min. 18 sec. (2.) 106 T. 56 Ib. 13 oz. (3.) 66 3s. lid. 1 far. (4.) 16 bu. 2 pk. 4 qt. 1 pt. (6.) 199 3s. 2d. 2 far. (6.) 58 da. 5 hr. 49 min. 29 sec. (7.) 97 cu. yd. 2 cu. ft. 1639 cu. in. (8.) 2 da. 10 hr. 56 min. (9.) 3 cd. ft. 14 cu. ft. Article 162 (Page 105) (1.) 6 bu. 3 pk. 2 qt. 1 pt. (2.) 25 da. 17 hr. 15 min. (3.) 16 cd. 70 cu. ft. (4.) 90 gal. 3 qt. 2 gi. (6.) 216 15s. 3d. (6.) 7 A. 142 sq. rd. 12 sq. yd. 21 sq. ft. 36 sq. in. (7.) 11 gal. 1 qt. 1 pt. (8.) 187 9s. 9d. Article 163 (Page 106) (1.) 19 yr. 8 mo. 17 da. (2.) 25 yr. 6 mo. 6 da. (3.) 36 yr. 7 mo. 10 da. (4.) 15 yr. 7 mo. 21 da. (6.) 64 yr. 2 mo. 8 da. (6.) 7 yr. 10 mo. 12 da. (7.) 13 yr. 10 mo. (8.) 33 yr. 8 mo. (9.) 18 yr. (10.) 25 yr. 1 da. Article 164 (Page 107) (1.) 300 da. (2.) 228 da. (3.) 270 da. (4.) 232 da. (5.) 225 da. (6.) 79 da. (7.) 269 da. (8.) 259 da. (9.) 297 da. (10.) 9 da. (11.) 191 da. (12.) 290 da. Article 165 (Page 108) (1.) 45 gal. (2.) 169 3s. lid. ( 3.) 151 da. 23 hr. 15 min. (4.) 280 ft>. 3 oz. 12 pwt. (5.) 643 bu. 3 pk. 4 qt. (6.) 26561 14s. 7d. (7.) 1252 mi. 45 rd. 5 ft. 6 in. (8.) 4560 yr. 5 da. (9.) 11587 bu. 2 pk. 7 qt. 1 pt. (10.) 25040 T. 2 cwt. 52 Ib. (11.) 1615 da. 22 hr. 13 min. (12.) 2725 gal. 2 qt. (13.) 13 mi. 64 rd. Article 166 (Page 109) (1.) 7 gal. 3 qt. 1 pt. (2.) 9 yd. 2 ft. 7 in. (3.) 32 5s. 6d. (4.) 17 cd. 34 cu. ft. (5.) 23 bu. 3 pk. 7 qt. (6.) 3 7s. 8d. (7.) 3 T. 17 cwt. 94 Ib. (8.) 7 gal. 3 qt. 3f gi. (9.) 3 yr. 9 mo. 18 da. (10.) 3 27' 35" Article 167 (Page 110) (1.) 28 times (2.) 128 times (3.) 59 times (4.) 73 times (5.) 48 times Article 168 (Pages^O-lll) (1.) 2210 bu. 3 pk. 2 qt. (2.) 24 spoons (3.) $787.50 (4.) 311 da. (5.) $42.43+ (6.) 43 yd. (7.) $11.90 (8.) 6 Ib. 9 oz. 5 pwt. (9.) 1 bu. 3 pk. 5 qt. (10.) 10 sq. rd. 25 sq. yd. 1 sq. ft. 126 sq. in. (11.) 738 3s. 4d. (12.) 23 gal. 1 qt. 2 gi. (13.) 259 mi. 125 rd. 2 yd. (14.) $10.17 (16.)32brl. (16.) $207.88+ (17.) $195.31- (18.) $3275.76+ (19.) $629.05+ (20.) $173.43- (21.) $146.36+ (22.) 3501.55+ fr. (23.) 108 11s. 5+d. (24.) 1771.84+ rubles (25.) 1288.86+ pesetas. ANSWERS 291 Article 180 (Page 115) (1.) 461348 sq. dm. (2.) 3468.3457 Dg. (3.) 46817.3 dg. (4.) 56.78 S. (5.) 19.653824 Ha. (6.) 28246000 mm. (7.) 341.7283 Dl. (8.) 166.52716 mi. (9.) 147.9352 gal. (10.) 1347.8125 bu. (11.) 177.15672 Dl. (12.) 56732.48 dg. (13.) 68.58015 M. (14.) 290.528028 Sq. M. (16.) 566.2035 M. (16.) 11.922348 HI. (17.) $1028.34+ (18.) 3949.53- fr. Article 185 (Page 117) (1.) 85 sq. yd. 3 sq. ft. (2.) 336 sq. yd. (3.) 5 sq. yd. 7 sq. ft. 72 sq. in. (4.) 24 sq. yd. 6 sq. ft. (5.) 635 sq. yd. 5 sq. ft. (6.) 4940 sq. yd. 7 sq. ft. 72 sq. in. (7.) 3 A. 85 sq. rd. 10 sq. yd. 8 sq. ft. 108 sq. in. (8.) 1 A. 74 sq. rd. 25 sq. yd. 1 sq. ft. 72 sq. in. (9.) 8 sq. mi. 19 A. 10 sq. rd. (10.) 114 sq. mi. 240 A. (11.) 9 sq. yd. 7 sq. ft. 18 sq. in. (12.) 49 sq. rd. 13 sq. yd. 6 sq. ft. 108 sq. in. (13.) 65.486^ sq. yd. (14.) 94 sq. yd. (16.) 17 A. 9 sq. rd. 29.75 sq. yd. (16.) 48 sq. ft. 10 sq. in. (17.) 38 sq. ft. (18.) $21 (19.) $54 Article 186 (Page 118) (1.) 5 yd. 2 ft. (2.) 74 rd. (3.) 28 ft. (4.) 9 ft. (5.) 3 rd. 1 yd. 5 in. Article 187 (Pages 119-121) (1.) 69 yd. 1 ft. (2.) 37 yd. 1 ft. (3.) 72 yd. 1 ft. 6 in. (4.) 136 yd. (5.) $10.80 (6.) 972 bricks (7.) 1512 bricks (8.) 10725 bricks (9.) 607* blocks (10.) 3569 pieces (12.) 61 yd. 9 in. (13.) 58 yd. (14.) 44 yd. 2 ft. (16.) 93 yd. 2 ft. 2 in. (16.) 70 yd. 1 ft. 10 in. Article 189 (Page 122) (1.) 9 bd. ft. (2.) HH bd. ft. (3.) 10 bd. ft. (4.) 23 bd. ft. (5.) 18* bd. ft. (6.) 20ff bd. ft. (7.) 11H bd. ft. (8.) 60 bd. ft. (9.) 86 bd. ft. (10.) 18| bd. ft. (11.) $40.61+ (12.) $33.54 Article 191 (Pages 124-125) (1.) 672 cu. ft. (2.) 71 cu. yd. 3 cu. ft. (3.) 10 cu. ft. 216 cu. in. (4.) 37 cu. yd. 1 cu. ft. (5.) $86.40 (6.) $35.19+ (8.) 15 ft. (9.) 21 ft. (10.) 8 ft. (11.) 1 ft. 6 in. (12.) 15 ft. 11+ in. Article 192 (Pages 126-128) (2.) 113.20- bu. (3.) 115.31+ bu. (4.) 65.74+ bu. (5.) 43.67+ bu. (6.) 2272.21- gal. (7.) 4578.08- gal. (8.) 325.82+ gal. (9.) 22.28- bu. (10.) 32315.84+ gal. (11.) 2 ft. 5.87- in. (12.) 4 ft. 2.57- in. (13.) 10 ft. 2.56- in. (14.) 128 bu. (15.) 5 T. 896 Ib. (16.) 4 ft. 2 in. (17.) 241.92 bu. (18.) 617.14- bu. (19.) 151.2 bu. (21.) 2961.1008 gal. (22.) 564.0192 gal. (23.) 54.12- bu. (24.) 9ft. 3.70- in. (25.) 24cu.ft. 939.6 cu. in. (26.) 3694.5216 cu. in. (27.) 6768.2304 gal. (28.) 125.1351 gal. (29.) 106.624 gal. (30.) 80.6344 gal. Article 201 (Page 134) (1.) $56 (2.) $24 (3.) 182 Ib. (4.) 714 men (6.) 314 sheep (6.) 730 gal. (7.) $8 (8.) 6 tons (9.) 173 Is. 6 d. (10.) 897 gal. (11.) 895 292 ANSWERS sheep (12.) 192 horses (13.) 252 da. (14.) 476 mi. (16.) 390 yd. (16.) 400 Ib. (17.) 342 cattle (18.) 8970 yd. (19.) real est. $12255; pers. prop. $16245 (20.) 270 pupils (21.) 66924 inhab. (22.) 301 acres (23.) $11565 Article 202 (Pages 136-137) (1.) 5 times (2.) .25 (3.) 33% (4.) 34% (6.) 35% (6.) 85% (7.) 62% (8.) \% (9.) 25% more (10.) 20% more (11.) 65% less (12.) 24% less (13.) 4% (14.) 42% (15.) 76% (16.) 85% (17.) 24% (18.) A 28%; B35%; C 37% (19.) 64% (20.) 18% (21.) 23% Article 203 (Pages 138-139) (1.) $9 (2.) 800 gal. (3.) 700 yd. (4.) 560 tons (6.) 538 gal. (6.) $8496.25 (7.) $6352 (8.) $46764 (9.) $386.25 (10.) $748.75 (11.) $1283.75 (12.) 781 5s. (13.) 862.5 tons (14.) $327.75 (15.) 725 acres (16.) $3795 (17.) $6750 (18.) $18000 (19.) $45600 (20.) $649.50 Article 204 (Pages 140-141) (1.) 367 bu. 12 Ib. (2.) 56 T. 1074 Ib. (3.) 195 bu. 43 Ib. (4.) 556| yd. (5.) 481 bu. 4 Ib. (6.) 125 mi. 1690 yd. (7.) 67 cd. 2 cd. ft. Article 206 (Page 143) (1.) 42% (2.) 39% (3.) 28|% (4.) 97% (5.) 37f% (6.) 41|% (7.) 63% (8.) 29f% (9.) 21.577% (10.) 99.643% Article 206 (Page 144) (1.) $720 (2.) 5775 natives; 2100 Germans; 1575 Irish; 1050 others (3.) $1175 (4.) $2700 (5.) 375 acres (6.) 1200 Ib. (7.) $18095 (8.) 30% Article 207 (Page 145) (1.) 35% (2.) $1700 (3.) $8500 (4.) 16% (5.) 1096518 yd. (6.) $45805.26 (7.) $248856.02 (8). $65884.50 Article 208 (Pages 145-148) (1.) $15989.19 (2.) A 45%; B 55% (3.) 975 gal. (4.) 56f% (5.) 80% (6.) A 50% more than B; B 33|% less than A (7.) $22350 (8.) 32% (9.) $5520 (10.) 650 acres (11.) 15% (12.) 27% (13.) $10890 (14.) $1299.80 (15.) 76% (16.) $14.57 (17.) 40%; 60%; $54; $81 (18.) 25%; 3H%; 43f%; $19.86; $24.83; $34.76 (20.) adj. house $3400; corner house $4352 (21.) A $15720; B $13100; C $14148; D $10611 (22.) A $84; B $67.20; C$52.50; D $87.50; E $350 (23.) $72000 (24.) wages 34.606%; repairs 5.684%; fuel 4.068%; fixed charges 25.855% (25.) 4500 bu. (26.) $60000 Article 214 (Pages 149-154) (1.) $14.45 (2.) $33.08 (3.) $77.22 (4.) $125.70 (5.) $57.03 (6.) $162.08 (7.) $344.05 (8.) $857.38 (9.) $897.99 (10.) $687.23 (11.) $299.88 (12.) $382.33 (13.) $352.46 (14.) $52.98 (15.) $92.44 (16.) $66.60 (17.) 25% (18.) 16% (19.) 46% (20.) 18% (21.) 8% (22.) 36% (23.) 24% (24.) 14% (25.) 18% (26.) 12*% (27.) 20% ANSWERS 293 (28.)18f% (29.) 16f% (30.) 19% (31.) $194.60 (32.) $468.75 (33.) $769 (34.) $843.25 (36.) $418.26 (36.) $694.72 (37.) $368.45 (38.) $79.32 (39.) $253.60 (40.) $62.56 (41.) $78.50 (42.) $625.40 (43.) $47.26 (44.) $23.04 (46.) 25% (46.) 20% (47.) $4.56 (48.) $375 (49.) $300 (60.) $9120 (61.) net loss $200 (62.) $1179.50 (53.) 20% (64.) $82.50 (55.) 25^ (66.) $28.13- (67.) 30% (68.) 18.7% (69.) $24000 (61.) 37^% (62.) $53.25 (63.) 15% (65.) $32960.88 16^ exp 13 + . ,,_., _, ., (64.) total purchases $61965; gain DU.OO (66.) prod. mat. 33^%; prod, labor 50%; overhead (67.) prod, labor 47 %; prod. mat. 40+%; overhead exp. (68.) $13 (69.) $142.23 (70.) $21304.12 (71.) $8400 (72.) $55400 Article 220 (Page 156) (1.) $241.01 (6.) $140.39 (11.) $455.38 (2.) $314.77 (7.) $414.82 (12.) $17.88 (3.) $636.87 (8.) $670.46 (13.) $551.34 (4.) $309.86 (9.) $354.12 (5.) $521.98 (10.) $222.86 Article 221 (Page 158) (1.) .72 (2.) .64125 (3.) .82935 (4.) .576 (7.) .45448 (8.) .59976 (9.) .8210565 (10.) .325 (13.) .2305 (14.) .352 (15.) .09693 (6.) (11.) .54 .224 (6.) .729 (12.) .6628 (1.) 5% (7.) 159 (2.) 20% (8.) 5% Article 222 (Page 159) (3.) 10% (4.) 12i% (5.) 30% (6.) 16f% Article 223 (Page 160) (l. (6. $48.67 $39.57 (2.) $24.51 (7.) $76.94 (3.) $99.56 (8.) $19.90 (4.) $386.38 (9.) $12.79 Article 231 (Page 163) (1.) $5.99 (6.) $33.12 (1.) $12.26 (6.) $25.87 (11.) $42.43 (2.) (7.) $23.60 $35.89 (3.) $10.89 (8.) $11.52 (4.) $18.87 (9.) $240.93 Article 232 (Page 164) (2.) $55.85 (3.) $16.18 (4.) $4.54 (7.) $87.06 (8.) $13.04 (9.) $16.28 (12.) $9.57 (13.) $55.80 (14.) $93.42 (6.) $176.55 (10.) $96.08 (6.) $24.85 (10.) $177.44 (5.) $8.52 (10.) $37.39 (1.) (6.) (11.) (16.) (21.) (26.) (31.) (36.) (41.) (46.) $25.93 $5.90 $1.34 $.08 $268.16 $3 $440.20 $32.01 $103.68 $1.36 (2.) (7.) (12.) (17.) (22.) (27.) (32.) (37.) (42.) (47.) (61.) Article $11.64 $19.81 $38.23 $10 $130.87 $13.79 $126.45 $3.51 $22.96 $.48 $858.30 233 (3.) (8.) (13.) (18.) (23.) (28.) (33.) (38.) (43.) (48.) (62.) (Pages 167-169) $30.61 (4.) $29.68 $42.37 (9.) $4.33 $2.70 (14.) $224.99 $41.17 (19.) $5.79 $339.69 (24.) $2984.72 $21714.51 (29.) $10.16 $317.77 (34.) $33.76 $.88 (39.) $44.55 $4.56 (44.) $8.45 $502.58 (49.) $197.42 $110.89 (63.) $190.84 (6.) (10.) (16.) (20.) (25.) (30.) (36.) (40.) (46.) (50.) $8.01 $14.48 $232.61 $147.04 $22.24 $224.87 $11.97 $8.25 $5.70 $139.32 294 ANSWERS (1.) $18.55 (6.) $7.72 (11.) $.71 Article 234 (Page 170) (2.) $20.31 (3.) $15.52 (4.) $66.98 (7.) $15.42 (12.) $472.67 (3.) $15.52 (8.) $51.44 (9.) $461.12 (5.) $10.80 (10.) $2.57 (1.) $29.56 (6.) $53.03 (11.) $232.31 (16.) $1022.37 Article 235 (Page 172) (2.) $10.80 (7.) $41.31 (12.) $26.23 (17.) $1359.58 (3.) $25.22 (8.) $40.65 (13.) $20.08 (18.) $21.93 (4.) $50.57 (9.) $820.92 (14.) $303.58 (5.) $4.09 (10.) $683.70 (15.) $684.67 (1.) $1.99 (6.) $28.25 (11.) $28.36 (16.) $211.33 Article 236 (Page 174) (2.) $6.44 (7.) $327 (12.) $17 (17.) $79.50 (3.) $1.69 (8.) $22.80 (13.) $119.88 (18.) $555.42 (4.) $3.41 (9.) $240 (14.) $20.48 (19.) $1638 (6.) $10.90 (10.) $17.27 (15.) $12.68 (20.) $130.05 Article 237 (Pages 175-176) (1.) $8.66 (2.) $10.38 (3.) $9.02 (4.) $15.44 (5.) $25.94 (6.) $14.30 (7.) $5.29 (8.) $24.12 (9.) $20.72 (10.) $117.91 (11.) $51.57 (12.) $86.83 (13.) $1811.94 (14.) $55.64 (15.) $475.71- (16.) $705.75 (17.) $78.36 (18.) $526.75 (19.) $998.68 (20.) $846.39 (21.) $465.13 (22.) 39 13s. 2.88 far. (23.) 492 8s. 1.05 far. (24.) 108 7s. 6d. 2.09 far. (25.) 91 8s. 9d. 1.55 far. (1.) 9% Article 239 (Page 177) (2.) 5% (3.) 5% (4.) 4*% (6.) 7% (6.) Article 240 (Page 178) (1.) 153 da. (2.) 183 da. (3.) 94 da. (4.) 307 da. (5.) 3 yr. 346 da. Article 241 (Page 179) (1.) $582.86 (2.) $4105.28 (3.) $4126.40 (4.) $9500 (5.) $418.50 (6.) $7239.64 (7.) $3875.91 (1.) (6.) $596.52 $2472.28 (2.) $1913.70 (7.) $951.85 Article 242 (Page 181) (4.) $3349.81 (3.) $741.03 (8.) $1589.13 (5.) $485.94 Article 243 (Pages 183-185) (1.) $83^85 (2.) $165.69 (3.) $190.72 (4.) $767.59 (5.) $1053.69 (6.) $64.89 (7.) $48.31 (8.) $20.40 (9.) $193.95 (10.) $78.87 (11.) $106.32 (12.) $103.44 (13.) $684.78 (14.) $1878.87 (15.) $801.03 (16.) $698.95 (18.) $2065.42 (19.) $2906.29 (20.) $1006.97 (21.) $9832.58 (22.) $360.08 (23.) $80.64 (24.) $1448.19 (25.) $593.20 ANSWERS 295 Article 250 (Pages 188-189) (1.) $960.90; $7.85 (2.) $710.92; $12.54 (3.) $2888.88; $76.92 (4.) $7725.36; $90.99 (6.) $511.49; $16.41 (6.) $838.03; $7.09 (7.) $647.17; $46.33 (8.) $346.99; $31.66 (9.) $26.13 (10.) $25.72 (11.) $3556.10 (12.) Of A; gain $.98 (13.) $390.30 (14.) $.05 (15.) $1808.16 (16.) $943.57 Article 256 (Pages 191-193) (1.) Aug. 15, 1911; 64 da.; $5.30; $591.45 (2.) Mar. 25, 1912; 78 da.; $10.96; $831.94 (3.) Apr. 15, 1912; 37 da.; $1.32; $284 (4.) Mar. 3, 1912; 14 da.; $4.80; $1758.45 (5.) Nov. 18, 1910; 26 da.; $2.70; $931.95 (6.) June 30, 1912; 79 da.; $16.33; $2464.79 (7.) Dec. 23, 1911; 82 da.; $4.60; $668.25 (8.) May 8, 1911; 81 da.; $5.47; $481.03 (9.) June 19, 1912; 45 da.; $36.74; $3637.41 (10.) Dec. 12, 1911; 126 da.; $7.67; $390.81 (11.) Nov. 16, 1911; 74 da.; $8.95; $716.35 (12.) Nov. 30, 1911; 98 da.; $11.91; $474.37 (13.) $2546.81 (14.) $851.29 (16.) $721.72 (16.) $489.83 (17.) $868.56 (18.) $908.98 (19.) $2137.26 (20.) $466.34 Article 257 (Page 194) (1.) $656.48 (2.) $965.97 (3.) $1393.70 (4.) $509.23 (5.) $6309.25 (6.) $4105.37 (7.) $698.56 (8.) $2565.86 (9.) $1293.64 (10.) $702.83 (11.) $492.41 (12.) $503.78 Article 258 (Page 197) (1.) $3025.70 (2.) $1830.70 (3.) $146.12 (4.) $551.71 (6.) $6612.62 (6.) $5194.17 (7.) $646.42 (8.) $358.05 Article 259 (Pages 198-199) (1.) $456.15 (2.) $813.09 (3.) $115.30 (4.) $114.70 (6.) $223.35 (6.) $21.54 (7.) $100.27 (8.) $317.96 (9.) $10.96 Article 264 (Pages 201-205) (1.) $18.77 (2.) $11.35 (3.) $4.69 (4.) $33.72 (6.) $23.84 (6.) $36.17 (7.) $854.04 (8.) $562.80 (9.) $2704.65 (10.) $8472.33 (11.) $7313.96 (12.) $808.86 (13.) $893.22 (14.) $2393.08 (16.) $18636.31 (16.) $9889 (18.) 4% (19.) 2% (20.) 3% (21.) f% (22.) 4% (23.) 5% (24.) $685.67 (26.) $427.60 (26.) $916 (27.) $17000 (28.) $6016 (29.) $976.66 (30.) $672 (31.) $5000 (32.) $957.99 (33.) $650 (34.) $500 (35.) $875.50 (36.) $468.75 (37.) $728 (38.) $263.50 (39.) $576 (40.) $852 (41.) $9.65 (42.) $13.39 (43.) $10.93 (44.) $7.05 (46.) $422.02 (46.) $739.66 (47.) 157 brls.; 57 bal. (48.) 82 brls.; $1.99 bal. (49.)297 bu. 4 Ib. (60.) $765.68 (61.) $4.78 (62.) $6097 (63.) $4360.75 (54.) $379.57 (55.) $164 brls.; 25?< bal. (56.) 145 brls.; $1.84 bal. (67.) 694 bu. 12 Ib. (68.) 226 brls.; 9 bal. Article 272 (Pages 208-211) (1.) $235 y (2.) $217^ (3.) $114F (4.) $83.13 (5.) $909 (6.) $2443.75 / (7.) $4933.88 (8./ $3040.88 (9.) $12894.50 (10.) $3006.25 (11.) $11169.25 // (12.) $7463 (I/.) $5095.50 (14.) $11532 (15.) $8319.38 /16.) $1890 V / (17.) $3422.50x/ (18.) $2734.38 (19.) $1303.50 (20.) 4%V (21.) 122f S 296 ANSWERS (23.) 24J% (24.) 19|% (26.) 7f% (26.) 4J% * (27.) 91| (29.) 13i% (30.) 17|% (31.) 5J% (32.) 14*% (34.) $1500 (36.) $11400 (36.) $23700 (37.) $900 (38.) $7100 (40.) $5800 (41.) $6200 (42.) $2300 (43.) $4500 (45.) $13500 (46.) $1200 (47.) $4400 (48.) $8200 (49.) $7200 (61.) $11076 (62.) $36.50 (63.) $5194 (54.) $940 (22.) 161% (28.) 161% (33.) 23-1% (39.) $8900 (44.) $3200 (60.) $10237.50 (56.) $24937.50 Article 273 (Pages 212-214) (1.) $87.50H2.) $66 (3.) $125 (4.) $160 (5.) $112.50 (6.)62f (7.) 79 (8.) 74 (9.) 69| (10.) 741 (11.) 5% (12.) 8% (13.) 4% (14.) 5 (15.) 5% (16.) 6f% (17.) 4% (18.) 4|% (19.) 6*% (20.) 4 (21.) 3% (22.) 6% (23.) 10% (24.) 8% (26.) 5% (26.) 6% (27.) 4 (28.) 4i% (29.) 8% (30.) 12% (31.) 7% (32.) 59| (33.) 5% (34.)4% stock yields T \% more on inv. (35.) $46.25 (36.) $62.50 (37.) $258.75 (2.) $631.58 (7.) $837.42 (12.) $636.96 (17.) $1457.91 (22.) $3535.35 (1.) $2026.27 (6.) $459.59 (9.) $274.79 Article 278 (Pages 216-218) (3.) $954 (4.) $569.01 (8.) $472.12 (9.) $622.70 (13.) $158.20 (15.) $872 (18.) $721.91 (20.) $754.51 (23.) $881.98 (24.) $2796.39 Article 283 (Page 220) (2.) $3536.10 (6.) $831.54 (10.) $928.63 (3.) $4505.25 (7.) $901.90 (11.) $4025.36 (6.) $274.47 (11.) $715.42 (16.) $938.72 (21.) $1603.75 (25.) $3266.40 (4.) $198.62 (8.) $153.80 (12.) $1226.81 Article 284 (Page 221) (1.) 375 Is. 3d. (2.) 153 14s. Ifd. (3.) 4287.45 fr. (4.) 13384.79 fr. (6.) 3974.28 marks (6.) 26249.87 marks (7.) 1638.30 guilders (8.) 3054.07 guilders (9.) 750 10s. 6|d. (10.) 1894.72 guilders Article 285 (Pages 222-223) (2.) 68%; $569.50 (3.) 76%; $1356.75 (4.) 65.28%; A $378.62; B $832.65; C $571.40 Article 286 (Page 224) (3.) |% (4.) $65.63 (6.) $564,75 (6.) $319.50 (2.) $3500 (7.) $1600 (2.) $141.75 (7.) $137.60 (2.) $820.50 (6.) $4138.40 (10.) $3480 (8.) $3000; $1500 (9.) $2500; $3750; $5000 Article 287 (Page 225) (3.) $156.75 (4.) $184.50 (6.) $108.75 (6.) $405.92 (8.) $40.75 (9.) $1.50 (10.) 15 mills Article 288 (Pages 226-227) (3.) $791.10 (7.) $11928 (11.) $698 (4.) $2768.40 (8.) $51.75 (12.) $3598 (5.) $2360.75 (9.) $42 ANSWERS 297 Article 289 (Pages 229-231) (1.) $57 (2.) $282.20 (3.) $1725 (4.) 360 Ib. (5.) 270 sheep (6.) $3980.40 (7.) $18.18 (8.) $16.40 (9.) $648 (10.) 152 da. (11.) 125 ft. (12.) 8 ft. (13.) 24 men (14.) 30 da. (16.) 525 Ib. (16.) $56.25 (17.) A $1500; B $2100 (18.) 25% (19.) 35% (20.) $46.80 Article 290 (Page 232) (1.) 90000 shingles (2.) $350 (3.) $425 (4.) $180 (5.) 15 da. (6.) $180 Article 292 (Pages 234-235) (1.) 2 mo. 10 da. (2.) 1 mo. 29 da. (3.) 2 mo. 13 da. (4.) June 18, 1912 (5.) Jan. 3, 1912 (6.) 6 mo. (7.) Apr. 24, 1911 (8.) July 14, 1910 Article 293 (Pages 237-238) (1.) May 19, 1910 (2.) May 2, 1911 (3.) Aug. 23, 1912 (4.) Aug. 4, 1911 (5.) Nov. 24, 1910 (6.) Apr. 25, 1912 (7.) Sept. 23, 1910 (8.) Oct. 28, 1911 (9.) May 12, 1912 (10.) Nov. 8, 1910 (11.) July 30, 1911 (12.) Dec. 5, 1912 Article 294 (Pages 240-241) (1.) July 11, 1910 (2.) June 27, 1911 (3.) June 23, 1912 (4.) Dec. 24, 1911 (5.) Dec. 10, 1909 (6.) Sept. 16, 1912 (7.) Sept. 16, 1912 (8.) Dec. 2, 1911 (9.) Sept. 15, 1911 (10.) June 4, 1912 Article 296 (Pages 244-246) (1.) Oct. 20, 1910 (2.) Apr. 10, 1911 (3.) May 7, 1910 (4.) Sept. 25, 1912 (5.) Apr. 27, 1911 ' (6.) Aug. 2, 1910 (7.) June 14, 1912 (8.) Nov. 11, 1911 (9.) May 3, 1910 (10.) Sept. 17, 1911 (11.) June 21, 1912 (12.) Feb. 29, 1912 (13.) $606.60 (14.) Aug. 25, 1911 (16.) Jan. 30, 1912 Article 297 (Pages 247-248) (1.) June 5, 1910 (2.) Oct. 20, 1911 (3.) Net pro. $1269.06; May 22, 1912 Article 299 (Pages 250-252) 1.) $301.47 (2.) $353.40 (3.) $326.90 (4.) $521.41 6.) $477.74 Article 301 (Pages 254-255) (1.) $3060 gain (2.) $2589.88 loss (3.) $3786.90 loss (4.) $3996 gain (5.) $9901.75 loss (6.) $11304.85 gain (7.) $17543.75 cap. (8.) $5089.77 cap. (9.) $954.10 cap. (10.) $6899.40 insolv. (11.) $3012.45 insolv. (12.) $5458.25 insolv. (13.) $5292.60 cap. (14.) $2959.90 insolv. (15.) $11121.40 cap. (16.) $9868.65 cap. (17.) $5006.45 insolv. (18.) $4042.30 cap. (19.) $1267 insolv. (20.) $3136.70 cap. (21.) $6508.68 (22.) A's $5884; B's $2942 (23.) A's $9640; B's $8140; C's $6640 ANSWERS Article 302 (Page 257) (1.) A's $1920; B's $1680 (2.) A's $2400; B's $3000; C's $3600 (3.) A's $9310; B's $13965 (4.) A's $2240; B's $2800; C's $3360 Article 303 (Pages 260-261) (1.) A's $2396.04; B's $1603.96 (2.) A's gain $2190.08; B's gain $1834.71; C's gain $2975.21; A's present int. $12190.08; B's present int. $9834.71; C's present int. $17975.21 (3.) A's $26137.72; B's $21868.26; C's $16994.01 (4.) A's gain $3043.87; A's present int. $18043.87; B's gain $2956.13; B's present int. $23956.13 (6.) A's $12935.37; B's $12064.63 (6.) A's $19437.35; B's $12562.65 Article 304 (Pages 265-266) (1.) A's $16502.33; B's $21097.67 (2.) A's $19981.95; B's $24618.05 (3.) A's $21837.05; B's $21397.60; C's $21065.35 (4.) A's $11770.85; B's $20390.33; C's $11068.82 (6.) A's $16271.31; B's $9928.69 Article 306 (Page 267) (1.) $64 (6.) $139 (11.) $418.50 $187 (3.) $93 $18.25 (8.) $3050 (12.) $182.60 (4.) $118 (9.) $145.75 (5.) $121 (10.) $41.13 (1.) 1875yd. (6.) 246 gal. Article 307 (Page 268) (2.) 1746 gal. (3.) 592 Ib. (4.) 172 bu. (6.) 330 yd. (7.) 836i Ib. (8.) 243f doz. (1.) $115.875 (6.) $352 (11.) $405 (16.) 142f yd. (19.) 624 bu. Article 309 (Page 269) (2.) $64.20 (6.) $164.70 (12.) $81 (16.) 386f bu. (3.) $197.50 (7.) $940.50 13.) $93.125 17.) 52 T \ gal. (4.) $39 (8.) $646.30 (14.) 463^1b. (18.) 425| yd. Article 310 (Page 270) (1.) $247 (6.) $382.20 (9.) $4056 (16.) 362^ (19.) 1284i (24.) 61632 (28.) 7118576 (2.) $448 (6.) $1472 (12.) 346 (16.) 307^ (20.) 2375 (26.) 51740 (29.) 250880 (3.) $324 (7.) $296.40 (13.) 504| (17.) 232 (22.) 84875 (26.) 624232 (30.) 67932800 (4.) $468.27 (8.) $1959.60 (14.) 461* (18.) 1428| (23.) 42174 (27.) 326697 Article 311 (Pages 271-272) (1.) $276.26 (2.) $119.89 (3.) $207.48 (4.) $91.55 (5.) $74.84 (6.) $18.12 (7.) $66.15 (8.) $88.51 (9.) $27.70 (10.) $47.87 (11.) $35.99 (12.) $41.78 (13.) $471.05 (14.) $1253.32 (15.) $166.74 ANSWERS 299 Article 312 (Page 273) (1.) 64 (2.) 55 (3.) 607 (4.) 5306 (5.) 529 bl. (6.) 6376 (7.) 6603 (8.) 85495 (9.) 884017 (10.) 9589235 (11.) 7762236 (12.) 74366307 Article 313 (Page 274) (1.) 1776 (2.) 3922 (3.) 2378 (4.) 5418 (5.) 1392 (6.) 3040 (7.) 945 (8.) 1495 (9.) 2814 (10.) 3942 (11.) 972 (12.) 3796 Article 319 (Pages 281-282) (1.) 81 (2.) 784 (3.) 26244 (4.) 119025 (5.) 512 (6.) 3375 (7.) 14886936 (8.) 147197952 (9.) 625 (10.) 371293 (11.) 279936 (12.) 262144 (13.) ft (14.) & (15.) 39^ (16.) 144t| (17.) .07776 (18.) .00531441 (19.) 2.097152 (20.) 188.2384 (21.) 74 (22.) 364 (23.) 139 (24.) 781 (26.) 412 (26.) 9103 (27.) 8643 (28.) 3816 (29.) 48937 (30.) 62534 (31.) 93 (32.) 752 (33.) 246 (34.) 999 (35.) 562 (36.) 832 (37.) 2106 (38.) 9372 (39.) 1001 (40.) 3475 (41.) 73 (42.) 642 (43.) 82 (44.) 643 (45.) 4 ft. 6.43+ in. (46.) 15 ft. 10.653+ in. (47.) 60.16+ ft. (48.) 7 cu. ft. 71 cu. in. (49.) 67 sq. rd. 13 sq. yd. 2 sq. ft. 72 sq. in. (50.) 8.56+ ft. (61.) 32 yd. long; 24 yd. wide (62.) 129 ft. long; 43 ft. wide (63.) 48 ft. wide; 32 ft. long (64.) 522 ft. long; 87 ft. wide; 87 ft. high (65.) 228 ft. long; 57 ft. wide; 57 ft. deep (66.) 32 ft. long; 32 ft. high; 32 ft. wide THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. JUM29 1933 AUG 13 1934 10 1937 184J MAR 3Mar'5E DS REC'D DEC 6 - 1959 LD 21-50rn-l,'33 YC 24884 UNIVERSITY OF CALIFORNIA LIBRARY