in.rv OF T! :^iTY EDUCA!DI015 ITM] "^ Z^^ ?, J^ ^-J Digitized by tine Internet Arciiive in 2008 with funding from Microsoft Corporation http://www.arcliive.org/details/completearithmetOOmacvricli COMPLETE AEITHMETIC, ORAL AKD WEITTE]Sr. PART SECOND, BY MALCOLM MacVIOAR, Ph.D., LL.D., FBINCIFAIi STATB KOBMAL SCHOOL, POTSDAM, N. Y. PUBLISHED BY TAINTOR BROTHERS, MERRILL & CO., NEW YORK. ^ Copyright by TAINTOR BROTHERS, MERRILL & CO., 1878. Electrotyped by Smith & MoDougal, 83 Beekman St., N. Y. -^flK^ ^^^SV^^ QA 10, E.aCK!. , ♦-*s^| PREFACE. JJt^^^^ •B ■«r :y:77.CT '^ ff" ^^^^ THE aim of the author in the preparation of this work may be stated as follows : 1. To present each subject in arithmetic in such a manner as to lead the pupil by means of preparatory steps and proposi- tions which he is required to examine for himself, to gain clear perceptions of the elements necessary to enable him to grasp as a reality the more complex and complete processes. 2. To present, wherever it can be done, each process object- ively, so that the truth under discussion is exhibited to the eye and thus sharply defined in the mind. 3. To give such a systematic drill on oral and written exer- cises and review and test questions as will fix permanently in the mind the principles and processes of numbers with their applications in practical business. 4. To arrange the pupil's work in arithmetic in such a manr ner that he will not fail to acquire such a knowledge of princi- ples and facts, and to receive such mental discipline, as will fit him properly for the study of the higher mathematics. The intelligent and experienced teacher can readily deter- mine by an examination of the work how well the author has succeeded in accomplishing his aim. Si5770ri7 PREP ACE. Special attention is invited to the method of presentation given in the teacher's edition. This is arranged at the begin- ning of each subject, just where it is required, and contains definite and full instructions regarding the order in which the subject should be presented, the points ^at require special attention and illustration, the kind of illustrations that should be used, a method for drill exercise, additional oral exercises where required for the teacher's use, and such other instructions as are necessary to form a complete guide to the teacher in the discussion and presentation of each subject. The plan adopted of having a separate teacher's edition avoids entirely the injurious course usually pursued of cum- bering the pupil's book with hints and suggestions which are intended strictly for the teacher. Attention is also invited to the Properties of Numbers, Great- est Common Divisor, Fractions, Decimals, Compound Num- bers, Business Arithmetic, Ratio and Proportion, Alligation, and Square and Cube Root, with the belief that the treatment will be found new and an improvement upon former methods. The author acknowledges with pleasure his indebtedness to Prof. D. H. Mac Vicar, LL.D., Montreal, for valuable aid rendered in the preparation of the work, and to Charles D. McLean, A. M., Principal of the State Normal and Training School, at Brockport, N. Y., for valuable suggestions on several subjects. M. MacVICAR. Potsdam, September ^ 1877. l^r^i^r^ ^^ /J^NS^ BEVIEW OF PART FIBST. PAGE Notation and Numeration. 1 Addition 4 Subtraction 5 Multiplication 7 Division 11 Properties of Numbers. ... 16 Exact Division 16 Prime Numbers 19 Factoring 20 Cancellation 31 Greatest Common Divisor.. 22 Least Common Multiple — 25 Fractions 29 Complex Fractions 40 Decimal Fractions 46 Denominate Numbers 59 Metric System 75 part second, Business Arithmetic 79 Aliquot Parts 82 Business Problems 85 Applications 100 Profit and Loss 101 Commission 103 Insurance 105 Stocks 107 Taxes Ill Duties or Customs 113 Review and Test Questions. 114 Interest 115 FAGS Method by Aliquot Parts. 117 Method by Six Per Cent.. 119 Method by Decimals 121 Exact Interest 122 Compound Interest 127 Interest Tables 120 Annual Interest 131 Partial Payments 1 32 Discount 136 Bank Discount 138 Exchange 141 Domestic Exchange 142 Foreign Exchange 146 Equation of Payments 151 Review and Test Questions. 160 Ratio 161 Proportion 169 Simple Proportion 170 Compound Proportion 174 Partnership 177 Alligation Medial 181 Alligation Alternate 182 Involution 187 Evolution 189 Progressions 205 Arithmetical Progression. 206 Geometrical Progression.. 208 Annuities 210 Mensuration 213 Review and Test Examples. 227 Answers 243 ♦>♦»♦■ REVIEW OF PART FIRST.* ,<»wfi ^^^S^^^'g)ij/ NOTATION AND NUMERATION, DEFINITIONS. 11. A Unit is a single thing, or group of single things, regarded as one ; as, one ox, one yard, one ten, one hundred. 12. Units are of two hinds — Mathematical and Common. A mathematical unit is a single thing which has a fixed value ; as, one yard, one quart, one hour, one ten, A common unit is a single thing which has no fixed value ; as, one house, one tree, one garden, one farm. * Note. — The first 78 pages of this part contains so much of the matter in Part First as is necessary for a thorough review of each subject, in- cluding all the tables of Compound Numbers. For convenience in making references, the Articles retained are numbered the same as in Part First. Hence the numbers of the Articles are not consecutive. 2 NOTATION AND NUMERATION. 13. A Number is a unit, or collection of units ; as, one man, three houses, four, six hundred. Observe, the number is "the how many" and is represented by what- ever answers the question, How many ? Thus in the expression seven j&rdB, seven represents the number. 14. The Unit of a Number is one of the things numbered. Thus, the unit of eight bushels is one bushel, of five boys is one boy, of nine is one. 15. A Concrete Number is a number which is applied to objects that are named ; as four chairs, ten hells. 16. An Abstract Number is a number which is not applied to any named objects ; as nine, five, thirteen. 17. Like Numbers are such as have the same unit. Thus, four windows and eleven windows are like numbers, eight and ten, three hundred and seven hundred. 18. Unlike Numbers are such as have different units. Thus, twelve yards and five days are unlike numbers, also six cents and nine minutes, 19. Figures are characters used to express numbers. 20. The Value of a figure is the number whicii it represents. 21. The Simple or Absolute Value of a figure is the number it represents when standing alone, as 8. 22. The Local or Representative Value of a figure is the number it represents in consequence of the place it occupies. Thus, in 66 the 6 in the second place from the right represents a num- ber ten times as great as the 6 in the first place. 2.3. Notation is the method of writing numbers by means of figures or letters. NOTATION AND NUMERATION. 3 24. Numeratioii is the method of reading numbers which are expressed by figures or letters. 35. A Scale in Arithmetic is a succession of mathematical units which increase or decrease in value according to a fixed order. 26. A Decimal Scale is one in which the fixed order of increase or decrease is uniformly ten. This is the scale used in expressing numbers by figures. • 27. Arithinetic is the Science of Numbers and the Art of Computation. REVIEW AND TEST QUESTIONS. 31. Study carefully and answer each of the following questions : 1. Define a scale. A decimal scale. 2. How many figures are required to express numbers in the decimal scale, and why ? 3. Explain the use of the cipher, and illustrate by examples, 4. State reasons why a scale is necessary in expressing numbers. 5. Explain the use of each of the three elements— /^?/re."?, place, and comma — in expressing numbers. 6. What is meant by the simple or absolute value of figures ? What by the local or representative value? 7. How is the local value of a figure affected by changing it from the Jirst to the tliird place in a number ? 8. How by changing a figure from the second to the fourth ? From the fourth to the ninth ? 9. Explain how the names of numbers from twelve to twenty are formed. From twenty to nine hundred ninety. 10. What is meant by a period of figures ? 11. Explain how the name for each order in any period is formed. 12. State the name of the right-hand order in each of the first six periods, commencing with units. 13. State the two things mentioned in (9) which must be observed when writing large numbers. 14. Give a rule for reading numbers ; also for writing numbers. 4 ADDITION, ADDITIOlSr. 50. Addition is the process of uniting two or more numbers into one number. 51. uMdends are the numbers added. 52. The Sum or Amount is the number found by addition. 53. The Process of Addition consists in forming units of the same order into groups of ten, so as to express their amount in terms of a higher order. 54. The Sign of Addition is +, and is read plus. When placed between two or more numbers, thus, 8 + 3 + 6 + 3 + 9, it means that they are to be added. 55. The Sif/n of Equality is =, and is read equals, or equal to ; thus, 9 + 4 = 13 is read, nine plus four equals thirteen, 5G. Principles. — /. Only numhers of the same denom- ination and units of the same order can he added. II. The sum is of the same denomination as the addends. III. The whole is equal to the sum of all the parts, REVIEW AND TEST QUESTIONS. 57. 1. Define Addition, Addends, and Sum or Amount. 2. Name each step in the process of Addition. 3. Why place the numbers, preparatory to adding, units under units, tens under tens, etc.? 4. Why commence adding with the units' column ? 5. What objections to adding the columns in an irregular order? Illustrate by an example. 6. Construct, and explain the use of the addition table. 7. IIow many combinations in the table, and how found? 8. Explain carrying in addition. What objection to the use of the word? SUBTRACTION. 5 9. Define counting and illustrate by an example. 10. Write five examples illustrating the general problem of addition, " Given all the parts to find the whole." 11. State the diflerence between the addition of objects and the addi- tion of numbers. 12. Show how addition is performed by using the addition table. 13. What is meant by the denomination of a number? What by units of the same order ? 14. Show by analysis that in adding numbers of two or more places, the orders are treated as independent of each other. SUBTEACTION. 70, Subtraction is the process of finding the difference between two numbers. 71« The Minuend is the greater of two numbers whose difference is to be found. 112 • The Subtrahend is the smaller of two numbers whose difference is to be found. 73. The Difference or Hetnainder is the result obtained by subtraction. 74. The Ih^ocess of Subtraction consists in com- paring two numbers, and resolving the greater into two parts, one of which is equal to the less and the other to the differ- ence of the numbers. 75. The Sign of Subtraction is — , and is called minus. When placed between two numbers it indicates that their difference is to be found ; thus, 14 — 6 is read, 14 minus 6, and means that the dif- ference between 14 and 6 is to be found. 76. JParentheses ( ) denote that the numbers inclosed between them are to be considered as one number. 77. A Vinculum affects numbers in the same manner as parentheses. Thus, 19 + (13— 5), or 19 + 13—5 signifies that the difference between 13 and 5 is to b'e added to 19. 6 SUBTRACTION. 78. Prtn'Ciples. — /. Only like Jiurtihers and units of the same order ean he subtracted. II. The minuend is the sum of the subtrahend and. difference, or the minuend is the whole of luhich the subtrahend and difference are the parts. III. An equal increase or decrease of the minuend and subtrahend does not change the difference. KEVIEW AND TEST QUESTIONS. *79. 1. Define the process of subtraction. Illustrate each step by an example. 2. Explain how subtraction should be performed when an order in the subtrahend is greater than the corresponding order in the minuend. Illustrate by an example. 3. Indicate the difference between the subtraction of numbers and the subtraction of objects. 4. When is the result in subtraction a remainder, and when a differ- ence? 5. Show that so far as the process with numbers is concerned, the result is always a difference. 6. Prepare four original examples under each of the following prob- lems and explain the method of solution : Prob. I.— Given the whole and one of the parts to find the other part. Prob. II. — Given the sum of four numbers and three of them to find the fourth. 7. Construct a Subtraction Table. 8. Define counting by subtraction. 9. Show that counting by addition, when we add a number larger than one, necessarily involves counting by subtraction. 10. What is the difference between the meaning of denomination and orders of units ? 11. State Principle III and illustrate its meaning by an example. 12. Show that the difference between 63 and 9 is the same as the difference between (63 + 10) and (9 + 10). 13. Show that 28 can be subtracted from 92, without analyzing the minuend as in (64), by adding 10 to each number. 14. What must be added to each number, to subtract 275 from 829 without analyzing the minuend as in ((>4:)? 15. What is meant by borrowing and canning in subtraction ? MULTIPLICATION. 7 MULTIPLIOATIOIT. IZIjUSTMATION of l*JtOCESS. 92. Step II. — To multiply by using the parts of the multiplier. 1. The multiplier may be made into any desired parts, and tlie mul- tiplicand taken separately the number of times expressed by each part. The sum of the products thus found is the required product. Thus, to find 9 times 12 we may take 4 times 13 which are 48, then 5 times 12 which are 60. 4 times 12 plus 5 times 13 are 9 times 13 ; hence, 48 plus 60, or 108, are 9 times 13. 2. When we multiply by one of the equal parts of the multiplier, we find one of the equal parts of the required product. Hence, by multi- plying the part thus found by the number of such parts, we find the required product. For example, to find 12 times 64 we may proceed thus : (1.) ANALYSIS. (2.) 64x4 = 256^ 64 64x 4. = 26e>=d times 256 4 64x4 = 256 3 256 64x12 = 768 3 768 (1.) Observe, that 12 = 4 -f- 4 -f- 4 ; hence, 4 is one of the 3 equal parts of 12. (2.) That 64 is taken 12 times by taking it 4 times -I- 4 times -^ 4 times, as shown in the analysis. (3.) That 4 times 64, or 256, is one of the 3 equal parts of 12 times 64. Hence, multiplying 256 by 3 gives 12 times 64, or 768. 3. In multiplying by 20, 30, and so on up to 90, we invariably multi- ply by 10, one of the equal parts of these numbers, and then by the number of such parts. For example, to multiply 43 by 30, we take 10 times 43, or 430, and multiply this product by 3 ; 430 x 3 = 1290, which is 30 times 43. We multiply in the same manner by 300, 300, etc., 2000, 3000, etc. f MULTIPLICATION. 93. Prob. II. — To multiply by a number containing only one order of units. 1. Multiply 347 by 500. (1.) ANALYSIS. (2.) Firststep, 347x100= 34700 347 Second step, 34700 X 5 = 173500 500 173500 Explanation. — 500 is equal to 5 times 100 ; hence, by taking 347, as in first step, 100 times, 5 times this result, or 5 times 34700, as shown in second step, will make 500 times 347. Hence 173500 is 500 times 347. In practice we multiply first by the significant figure, and annex to the product as many ciphers as there are ciphers in the multiplier, as shown in (2). 96. Prob. III. — To multiply by a number containing two or more orders of units. 1. Multiply 539 by 374. (1.) analysis. (2.) -~\ 5 3 9 Multiplicand. 3 7 4 Multiplier. 539X 4= 2156 Ist partial product. 539X374=<5539X 70= 37 7 30 2d partial product. 539x300 = 161700 3d partial product. 2 015 8 6 Whole product. Explanation. — 1. The multiplier, 874, is analyzed into the parts 4. 70, and 300, according to (92). 2. The multiplicand, 539, is taken first 4 times = 2156 (86) ; then 70 times = 37730 (93) ; then 300 times = 161700 (93). 3. 4 times + 70 times + 300 times are equal to 374 times ; hence the sum of the partial products, 2156, 37780, and 161700, is equal to 374 times 539 = 201586. 4. Observe, that in practice we arrange the partial products as shown in (2), omitting the ciphers at the right, and placing the first significant figure of each product under the order to which it belongs. MULTIPLICATION. SF DEFINITIONS. 100. Multiplication is the process of taking one number as many times as there are units in another. 101. The Multiplicand is the number taken, or mul- tiplied. 102. The Multiplier is the number which denotes how many times the multiplicand is taken. 103. The I^roduct is the result obtained by multipli- cation. 104. A Partial Product is the result obtained by multiplying by one order of units in the multiplier, or by any part of the multiplier. 105. The Total or Whole Product is the sum of all the partial products. 106. The Process of Multiplication consists, /rs^, in finding partial products by using the memorized results of the Multiplication Table; second, in uniting these partial products by addition into a total product. 107. A Factor is one of the equal parts of a number. Thus, 13 is composed of six 2's, four 3's, three 4's, or two 6's ; hence, 2, 3, 4, and 6 are factors of 12. The multiplicand and multiplier are factors of the product. Thus, 37 X 25 = 925. The product 925 is composed of twenty-five 37's, or thirtyse'oen 25's. Hence, both 37 and 25 are equal parts or factors of 925. 108. The Sign of Multiplication is x, and is read times, or miiUipUed hy. When placed between two numbers, it denotes that either is to be multiplied by the other. Thus, 8x6 shows that 8 is to be taken 6 times, or that 6 is to be taken 8 times ; hence it may be read either 8 times 6 or 6 times 8. 10 MULTIPLICATION. 109. Principles. — /. Tlxe multiplicand may he either an abstract or concrete nunvber. II. The multiplier is always an abstract number. III. Tize product is of the same denomination as the multiplicand. KEVIEW AND TEST QUESTIONS. 110. 1. Define Multiplication, Multiplicand, Multiplier, and Product. 2. What is meant by Partial Product ? Illustrate by an example. 3. Define Factor, and illustrate by examples. 4. What are the factors of 6 ? 14 ? 15 ? 9 ? 20 ? 24 ? 25 ? 27 ? 32? 10? 30? 50? and 70? 5. Show that the multiplicand and multiplier are factors of the product. 6. What must the denomination of the product always be, and why ? 7. Explain the process in each of the following cases and illustrate by examples : I. To multiply by numbers less than 10. II. To multiply by 10, 100, 1000, and so on. III. To multiply by one order of units. XV. To multiply by two or more orders of units. V. To multiply by the factors of a number (92 — ^2). . 8. Give a rule for the third, fourth, and fifth cases. 9. Give a rule for the shortest method of- working examples where both the multiplicand and multiplier have one or more ciphers on the right. 10. Show how multiplication may be performed by addition. 11. Explain the construction of the Multiplication Table, and illus- trate its use in multiplying. 12. Why may the ciphers be omitted at the right of partial products ? 13. Why commence multiplying the units' order in the multi- plicand first, then the tens', and so on? Illustrate your answer by an example. 14. Multiply 8795 by 629, multiplying first by the tens, then by the hundreds, and last by the units. DIVISION. 11 15. Multiply 3573 by 483, commencing with the thousands of the multiplicand and hundreds of the multiplier. 16. Show that hundreds multiplied by hundreds will give ten thou- sands \u. the product. 17. Multiplying thousands by thousands, what order will the pro- duct be? 18. Name at sight the lowest order which each of the following examples will give in the product : (1.) 8000 X 8000 ; 2000000 x 3000 ; 5000000000 x 7000. (2.) 40000 X 20000 ; 7000000 x 4000000. 19. What orders in 3928 can be multiplied by each order in 473, and not have any order in the product less than thousands ? DIVISION. ILLUSTRATION OF PROCESS. 119. Prob. I. — To divide any number by any divisor not greater tlian 12. 1. Divide 986 by 4. Explanation. — Follow the analysis and notice each step in the process; thus, 1. We commence by dividing the higher order of units. We know that 9, the figure expressing hundreds, contains twice the divisor 4, and 1 remaining. Hence 900 contains, according to ( 1 1 7 — 2), 200 times the divisor 4, and 100 remaining. We multiply the divisor 4 by 200, and subtract the product 800 from 986, leaving 186 of the dividend yet to be divided. ANALYSIS. 2. We know that 18, the number 4)986(200 expressed by the two left-hand 4x200 = 800 40 figures of the undivided dividend, contains 4 times 4, and 2 remaining. -'- " " Hence 18 tens, or 180, contains, 4x40 =160 6 according to (117—2), 40 times 4, 4x6 =24 n n o 4. p 2 ^^^ 20 remaining. We multiply ^ the divisor 4 by 40, and subtract the product 160 from 186, leaving 2 26 yet to be divided. 12 DI VISION, 3. We know that 26 contains 6 times 4, and 2 remaining, wliich is less than the divisor ; hence the division is completed. 4. We have now found that there are 200 fours in 800, 40 fours in 160^, and 6 fours in 26, and 2 remaining ; and we know that 800 + 160 + 26 = 986. Hence 986 contains (200 + 40 + 6) or 246 fours, and 2 remaining. The remainder is placed over the divisor and written after the quotient ; thus, 246|. SHORT AND Ij O N G DIVISION C O M P A. It E D , 121. Compare carefully the following forms of writing the work in division : (1- rOEM USED roR Two Bteps in the 4 ) EXPLANATION. process written. ) 986 ( 200 (2.) LONG DIVISION. One step written. 4 ) 986 ( 246 (3.) SHOBT DIVISION. Entirely mental. 4)986 4x200= 800 40 8 ■"2461 186 6 18 4x 40 = 160 246 16 26 26 4x 6 = 24 24 To divide any number by any given Explanation. — 1. We find how many times the divisor is contained in the fewest of the left-hand figures of the dividend which will contain it. 59 is contained 3 times in 215, with a remainder 38, hence, according to (115—1), it is contained 300 times in 21500, with a remainder 3800. 2. We annex the figure in the next lower order of the dividend to the remainder of the previous division, and divide the number thus found by the divisor. 2 tens annexed to 380 tens make 382 tens. 129. Prob. II.— To divisor. 1. Divide 21524 by 59, 59)21524(364 177 382 354 284 236 48 DIVISION. 13 59 is contained 6 times in 383, with a remainder 28 ; hence, according to (1 15 — 1), it is contained 60 times in 3820, with a remainder 280. 3. We annex the next lower figure and proceed as before. 137. Prob. III. — To divide by using the factors of the divisor. Ex. 1. Divide 315 by 35. 5)315 Explanation.— 1. The divisor 35 = 7 fives. 7 nvesyol fives ^- ^^^^^"^^ *l^° ^^^ ^^ ^' ''^ ^"^ *^'* J I — ^ 315 = 63>«5. (138.) 3. The 63 fives contain 9 times 7 fives; hence 315 contains 9 times 7 fives or 9 times 35. Ex. 2. Divide 359 by 24. 2 1^ 5^ 3 twos I 1 7 9 twos and 1 remaining = 1 4 (3 twos) I 5 9 (3 tivos) and 2 twos remaining = 4 Quotient, 1 4 and 3 (3 twos) remaining = 18 True remainder, 2 3 Explanation.— 1. The divisor 24 = 4x3x2 = 4 (3 twos). 2. Dividing 359 by 2, we find that 359=179 twos and 1 unit remaining. 3. Dividing 179 twos by 3 twos, we find that 179 twos — 59 (3 twos) and 2 twos remaining. 4. Dividing 59 (3 twoi) by 4 (3 twos), we find that 59 (3 two^ contain 4 (3 twos) 14 times and 3 (3 twos) remaining. Hence 359, which is equal to 59 (3 twos) and 2 twos + 1, contains 4 (3 twoi), or 24, 14 times, and 3 (3 twos) + 2 twos + 1, or 23, remaining. 142. Prob. IV. — To divide v^^hen the divisor con- sists of only one order of units. 1. Divide 8736 by 500. Explanation.— 1. We divide ^ ^ ft 7 I ^ r ^^^* ^^ *^® factor 100. This is ^ Lz2A2^ done by cutting off 36, the units 1 7 and 236 remaining. and tens at the right of the divi- dend. 2. We divide the quotient, 87 hundreds, by the factor 5, which gives a quotient of 17 and 2 hundred remaining, which added to 36 gives 236, the true remainder. 14 DIVISION, DEFINITIONS. 144. Division is the process of finding how many times one number is contained in another. 145. The Dividend is the number divided. 146. The Divisor is the number by which the dividend is divided. 147. The Quotient is the result obtained by division. 148. The Remainder is the part of the dividend left after the division is performed. 149. A Partial Dividend is any part of the dividend which is divided in one operation. 150. A Partial Quotient is any part of the quotient which expresses the number of times the divisor is contained in a partial dividend. 151. The Process of Division consists, first, in find- ing the partial quotients by means of memorized results ; second, in multiplying the divisor by the partial quotients to find the partial dividends ; tMrd, in subtracting the partial dividends from the part of the dividend that remains un- divided to find the part yet to be divided. 153. Short Division is that form of division in which no step of the process is written. 153. Long Division is that form of division in which the third step of the process is written. 154. The S iff ft of Division is -^, and is read divided by. When placed between two numbers, it denotes that the number before it is to be divided by the number after it; thus, 28 -h 7 is read, 28 divided by 7. Division is also expressed by placing the dividend above the divisor, Avitli a short horizontal line between them ; thus, ^^- is read, 35 divided by 5. DIVISION. 15 155. Prin^ciples. — /. Tlie dividend and divisor must he nuinbers of the same denomination. II. The denomination of the quotient is determined "by the nature of the problem solved. III. The remainder is of the same denomination as the dividend. REVIEW AND TEST QUESTIONS. 156. 1. Define Division, and illustrate each step in the process by- examples. 2. Explain and illustrate by examples Partial Dividend, Partial Quotient, and Remainder. 3. Prepare two examples illustrating each of the f ollowhig problems : I. Given all the parts, to find the whole. II. Given the whole and one of the parts, to find the other part. III. Given one of the equal parts and the number of parts, to find the whole. rV. Given the whole and the size of one of the parts, to find the number of parts. V. Given the whole and the number of equal parts, to find the size of one of the parts. 4. Show that 45 can be expressed as nines, as jives, as threes. 5. What is meant by true remainder, and how found ? 6. Explain division by factors. Illustrate by an example. 7. Why cut off as many figures at the right of the dividend as there are ciphers at the right of the divisor ? Illustrate by an example. 8. Give a rule for dividing by a number with one or more ciphers at the right. Illastrate the steps in the process by an example. 9. Explain the difference between Long and Short Division, and show that the process in both cases is performed mentally. 10. Illustrate each of the following problems by three examples : VI. Given the final quotient of a continued division, the true remainder, and the several divisors, to find the dividend. VII. Given the product of a continued multiplication and the several multipliers, to find the multiplicand. VIII. Given the sum and the difference of two numbers, to find the numbers. X6 PROPERTIES OF NUMBERS. PEOPEETIES OF NUMBERS. 163. An Integer is a number that expresses how many there are in a collection of wJioIe things. Thus, 8 yards, 12 Louses, 32 dollars. 164. An Exact JDivisor is a number that will divide another number without a remainder. Thus, 3 or 5 is an exact divisor of 15. All numbers with reference to exact divisors are either prime or composite. 165. A JPrime Number is a number that has no exact divisor besides 1 and itself. Thus, 1, 3, 5, 7, 11, 13, etc., are prime numbers. 166. A Composite JVtimber is a number that has other exact divisors besides 1 and itself. Thus, 6 is divisible by either 2 or 3 ; hence is composite. 167. A I^rime I>ivisor is a prime number used as a divisor. Thus, in 35 -^- 7, 7 is a prime divisor. 168. A Composite Divisor is a composite number used as a divisor. Thus, in 18 -i- 6, 6 is a composite divisor. EXACT DIVISION". 169. The following tests of exact division should be care- fully studied and fixed in the memory for future use. Piiop. I. — A divisor of any number is a divisor ofanynum- her of times that numher. Thus, 12 = 3 fours. Hence, 12 x 6 = 3 fours x 6 = 18 fours. But 18 fours are divisible by 4. Hence, 12 x 6, or 72, is divisible by 4. EXACT DIVISION. 17 Prop. II.— ^ divisor of each of two or more numhers is a divisor of their simi. Thus, 5 is a divisor of 10 and 30 ; that is, 10 = 2 fives and 30 = 6 fives. Hence, 10 + 30 = 2 fives + 6 fives = 8 fives. But 8 fives are divisible by 5. Hence, 5 Ls a divisor of the sum of 10 and 30. Prop. III. — A divisor of each of two numiers is a divisor of their differetice. Thus, 3 is a divisor of ^7 and 15 ; that is, 27 = 9 threes and 15 = 5 threes. Hence, 27—15 = 9 threes — 5 threes = 4 threes. But 4 threes are divisible hj 3. Hence 3 is a divisor of the difference between 27 and 15. Prop. IV. — A7iy numher ending with a cipher is divisible hy the divisors of 10, viz,, 2 and 5. Thus, 370 = 37 times 10. Hence is divisible by 2 and 5, the divisors of 10, according to Prop. I. Prop. V. — A7iy ^lumber is divisible by either of the divisors of 10, when its right-hand figure is divisible by the same. Thus, 498 = 490 + 8. Each of these parts is divisible by 2. Hence the number 498 is divisible by 2, according to Prop. II, In the same way it may be shown that 495 is divisible by 5. Prop. VI. — Any number endi^ig with two ciphers is divisible by the divisors of 100, viz., 2, 4, 5, 10, 20, 25, and 50. Thus, 8900 — 89 times 100. Hence is divisible by any of the divisors of 100, according to Prop. I. Prop. VII. — Any number is divisible by any one of the divisors of 100, when the number expressed by its two right- hand figures is divisible by the same. Thus, 4975 = 4900 + 75. Any divisor of 100 is a divisor of 4900 (Prop, VI). Hence, any divisor of 100 which will divide 75 is a divisor of 4975 (Prop. II). Prop. VIII. — Any numher ending with three ciphers is divisible by the divisors of 1000, viz., 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, and 500. Thus, 83000 = 83 times 1000. Hence is divisible by any of the divi- Bors of 1000, according to Prop. I. 18 EXACT DIVISION. Prop, IX. — Any number is divisible by any one of the divi- sors of 1000, when the number expressed by its three right-hand figures is divisible by the same. Thus, 92625 = 92000 + 625. Any divisor of 1000 is a divisor of 92000 (Prop. VIII). Hence, any divisor of 1000 which will divide 625 is a divi- sor of 92625 (Prop. II). Prop. X. — Any number is divisible, by 9, if the sum of its digits is divisible by 9. This proposition may be shown thus : (1.) 486 = 400 + 80 + 6. (2.) 100 = 99 + 1 =: 11 nines + 1. Hence, 400 = 44 nines + 4, and is divisible by 9 with a remainder 4. (3.) 10 = 9 + 1 r= 1 nine + l. Hence, 80 = 8 nines + 8, and is divisible by 9 with a remainder 8. (4.) From the foregoing it follows that 400 + 80 + 6, or 486, is divisible by 9 with a remainder 4 + 8 + 6, the sum of the digits. Hence, if the Bum of the digits is divisible by 9, the number 486 is divisible by 9 (Prop. II). Prop. XI. — Any number is divisible by 3, if the sum of its digits is divisible by 3. This proposition is shown in the same manner as Prop. X ; as 3 divides 10, 100, 1000, etc., with a remainder 1 in each case. Prop. XII. — A7iy number is divisible by 11, if the difference of the sums of the digits in the odd and even places is zero or is divisible by 11. This may be shown thus : (1.) 4928 = 4000 + 900 + 20 + 8. (2.) 1000 = 91 elevens — 1. Hence, 4000 = 364 elevens — 4. (3.) 100 = 9 elevens + 1. Hence, 900 = 81 elevens + 9. (4.) 10 = 1 eleven — 1. Hence, 20 = 2 elevens — 2. (5.) From the foregoing it follows that 4928 = 364 elevens + 81 elevens + 2 elevens— 4 + 9-2 + 8. But - 4 + 9 - 2 + 8 = 11. Hence, 4928 = 364 elevens + 81 elevens + 2 elevens + 1 eleven = 448 elevens, and is therefore divisible by 11. The same course of reasoning applies where the difference is minus or zero. PRIME NUMBERS. J.9 PRIME NUMBEES. PRErAltATORY fMOrOSITlONS. 171. Prop. I. — All even numbers are divisible by 2, and consequently all even mimbers, exce]^t 2, are composite. Hence, in finding the prime numbers, we cancel as composite all even numbers except 2. Thus, 3, 4, 5, 0, 7, $, 9, X^, 11, It, and so on. Prop. II. — Each number i7i the series of odd numbers is 2 greater than the number immediately ^preceding it. Thus, the numbers left after cancelling the even numbers are 3 5 7 9 11 13, and so on. 3 3 + 2 5 + 2 7 + 2 9 + 2 11 + 2 Prop. III. — In the series of odd numbers, every third number from 3 is divisible by 3, every fifth number from 5 is divisible by 5, and so on ivith each number in the series. This proposition may be shown thus : According to Prop. II, the series of odd numbers increase by 2's. Hence the third number from 3 is found by adding 2 three times, thus : 3 5 7 9 3 3+2 3+2+2 3+2+2+2 From this it will be seen that 9, the third number from 3, is composed of 3, plus 3 twos, and is divisible by 3 (Prop. II) ; and so with the third number from 9, and so on. By the same course of reasoning, each fifth number in the series, counting from 5, may be shown to be divisible by 5 ; and so with any other number in the series ; hence the following method of finding the prime numbers. ILLZrSTIiATION OF P It O C E S S . 172. Prob.— To find all the Prime Numbers from 1 to any given number. Find all the prime numbers from 1 to 63. 30 FA CTORING, 1 3 5 7 9 3 11 13 15 17 19 21 23 25 27 3 6 3 1 6 3 9 29 31 33 35 37 39 41 3 11 5 1 3 13 43 45 47 49 51 53 55 3 5 9 16 7 3 17 6 11 57 59 61 63 3 19 3 7 9 31 Explanation. — 1. Arrange tlie series of odd numbers in lines, at con- venient distances from each otlier, as sliown in illustration. 3. Write 3 under every third number from 3, 5 under ewevj fifth num- ber from 5, 7 under every seventh number from 7, and so on with each of the other numbers. 3. The terms under which the numbers are written are composite, and the numbers written under are their factors, according to Prop. III. All the remaining numbers are prime. Hence all the prime numbers from 1 to 63 are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 39, 31, 37, 41, 43, 47, 53, 59, 61. FAOTOEIISTG, 175. A Factor is one of the equal parts of a number, or one of its exact divisors. Thus, 15 is composed of 3 fives or 5 threes ; hence, 5 and 3 are factors of 15. 176. A JPrifne Factor is a prime number which is a factor of a given number. Thus, 5 is a prime factor of 30. 177. A Co^nposite Factor is a composite number which is a factor of a given number. Thus, 6 is a composite factor of 34. 178. Factoring is the process of resolving a composite number into its factors. CANCELLATION. 21 1*79. An JExjwnent is a small figure placed at the right of a number and a little above, to show how many times the number is used as a factor. Thus, 3^ = 3x3x3x3x3. The 5 at the right of 3 denotes that the 3 is used 5 times as a factor. 180. A Cofunion Factor is a number that is a factor of each of two or more numbers. Thus, 3 is a factor of 6, 9, 12, and 15 ; hence is a common factor. 181. The Greatest Common Factor is the greatest number that is a factor of each of two or more numbers. Thus, 4 is the greatest number that is a factor of 8 and also of 12. Hence 4 is the greatest common factor of 8 and 12. IIjZUSTItA TIO N OF P B O C E S S . 182. Find the prime factors of 462. 2)462 Explanation.— 1. We observe that the number o \~9V7 462 is divisible by 2, the smallest prime number. I — ! — Hence we divide by 2. 7)77 2. We observe that the first quotient, 231, is divis- -, 1 ible by 3, which is a prime number. Hence we divide by 3. 8. We observe that the second quotient, 77, is divisible by 7, which is a prime number. Hence we divide by 7. 4. The third quotient, 11, is a prime number. Hence the prime fac- tors of 462 are 2, 3, 7, and 11 ; that is, 462 = 2 x 3 x 7 x 11. OAE^OELLATION. PB E PAIt ATORT P R O P O S I TJ O N 8, 185. Prop. I. — Rejecting a factor from a number divides the number by that factor. Thus, 72 = 24 x 3. Hence, rejecting the factor 3 from 72, we have 24, the quotient of 72 divided by 3. Prop. II. — Dividing both dividend and divisor by the same number does not change the quotient. 22 GREATEST COMMON DIVISOR. Thus, 60^12 = 20 threes -f- 4 threes == 5. Observe that the unit three, in 20 threes -s- 4 threes, does not in any way affect the size of the quotient ; therefore, it may be rejected and the quotient will not be changed. Hence, dividing both the dividend 60 and the divisor 12 by 3 does not change the quotient. ILLUSTRATION OF P H O C E 8 8 , 186. Ex. 1. Divide 462 by 42. 6 ) 462 77 Explanation.— We divide both the n \ ~^ ^^^ ~w" ^^ -'■-'■• divisor and dividend by 6. According to Prop. II, the quotient is not changed. Hence, 77-^7 = 462-^42 = 11. Ex. 2. Divide 65 x 24 x 55 by 39 x 15 x 35. •Hi' Explanation.— 1. We divide any factor in the dividend by any num- ber that will divide a factor in the divisor. Thus, 65 in the dividend and 15 in the divisor are divided each by 5. In the same manner, 55 and 35, 13 and 39, 24 and 3 are divided. The remaining factors, 8 and 11, in the dividend are prime to each of the remaining factors in the divisor. Hence, no further division can be performed. 2. We divide the product of 8 and 11, the remaining factors in the dividend, by the product of 3 and 7, the remaining factors in the divisor, and find as a quotient 4/y> which, according to (185 — II), is equal to the quotient of 65 x 24 x 55 divided by 39 x 15 x 35. All similar cases may be treated in the same manner. GREATEST OOMMOE DIVISOR. 190. A Common Divisor is a number that is an exact divisor of each of two or more numbers. Thus, 5 is a divisor of 10, 15, and 20. 1.^ 8 11 8 X 11 88 $0 X 10 X $$ 3^7 3x7 ~ 21 GREATEST COMMON DIVTSOR. 23 191. The Greatest Connnon Divisor is the great- est number that is an exact divisor of each of two or more numbers. Thus, 3 is the greatest exact divisor of each of the numbers 6 and 15. Hence 3 is their greatest common divisor. 192. Numbers are prime to each other when they have no common divisor besides 1 ; thus, 8, 9, 25. METHOD BY FACTOEING. ritEBAItATORT PROPOSITION, 193. Illustrate the following proposition by examples. Tlie greatest common divisor is the product of the prime faO' tors that are common to all the given numbers ; thus, 42= 7x2x3= 7 sixes; 66 = 11 X 2 X 3 = 11 sixes. 7 and 11 being prime to each other, 6 must be the greatest common divisor of 7 sixes and 11 sixes. But 6 is the product of 2 and 3, the com- mon prime factors ; hence the greatest common divisor of 42 and 64 is the product of their common prime factors. ILLUSTRATION OF PROCESS. 194. Prob. I.— To find the Greatest Common Divisor of two or more numbers by factoring". Find the greatest common divisor of 98, 70, and 154. (1.) (2.) 2 ) 98 2)^70 2 )154 2 )98 70 154 7 )49 7 )35 7 )77 Or, 7 )49 35 77 "7 5 11 7 5 11 2x7 = greatest common divisor. M GREATEST COMMON DIVISOR. Explanation. — 1. We resolve each of tlie numbers into their prime factors, as shown in (1) or (2). 2. We observe that 2 and 7 are the only prime factors common to all the numbers. Hence the product of 2 and 7, or 14, according to (193), is the greatest common divisor of 98, 70, and 1 54. The greatest common divisor of any two or more numbers is found in the same manner. METHOD BY DIVISION. mErAItATORT rJiOrOSlTIONS. 107. Let the two following propositions be carefully studied and illustrated by other examples, before attempting to find the greatest common divisor by this method. Prop. I. — The greatest commoii divisor of two numbers is the greatest common divisor of the smaller number and their difference. Thus, 3 is the greatest common divisor of 15 and 37. Hence, 15 = 5 threes and 27 = 9 threes ; and 9 threes — 5 threes == 4 threes. But 9 and 5 are prime to each other ; hence, 4 and 5 must be prime to each other, for if not, their common divisor will divide their sum, accord- ing to (169—11), and be a common divisor of 9 and 5. Therefore, 3 is the greatest common divisor of 5 threes and 4 threes, or of 15 and 12. Hence, the greatest common divisor of two numbers is the greatest common divisor of the smaller number and their difference. Pkop. II. — The greatest common divisor of two numbers is the greatest common divisor of the smaller number and the rejnainder after the division of the greater by the less. This proposition may be illustrated thus : 22 — 6 = 16 1- Subtract 6 from 22, then from the differ- ■j^g g __ ^Q ence, 16, etc., until a remainder less than 6 is ^^ r, . obtained. 2. Observe that the number of times 6 has been subtracted is the quotient of 22 divided by 6, and hence that the remainder, 4, is the remainder after the division of 22 by 6. 3. According to Prop. 1, the greatest common divisor of 22 and G is LEAST COMMON MULTIPLE. 25 the greatest common divisor of tlieir diflference, 16, and 6. It is also, according to the same Proposition, the greatest common divisor of 10 and 6, and of 4 and 6. But 4 is the remainder after division and 6 the smaller number. Hence the greatest common divisor of 22 and G is the greatest common divisor of the smaller numher and the remainder after division. ILLVSTRJiTION OF 1* It O C i: S S , 198. Pkob. II. — To find the Greatest Common Divisor of two or more numbers by continued division. Find the greatest common divisor of 28 and 176. 2 8)170(6 Explanation.— 1. We divide 176 1 g Q by 28 and find 8 for a remainder ; then we divide 28 by 8, and find 4 for a re- o ) /i 8 ( o mainder ; then we divide 8 by 4, and 2 4 find for a remainder. 2 \ « /" 2 ^* -According to Prop. II, the great- ' ^ est common divisor of 28 and 176 is the _ same as the greatest common divisor of 28 and 8, also of 8 and 4. But 4 is the greatest common divisor of 8 and 4. Hence 4 is the greatest common divisor of 28 and 176. LEAST COMMON MULTIPLE. rnEJ^AItATOltY PROPOSITIONS. 211. Prop. I. — A multiple of a numher contains as a factor each prime factor of the number as many times as it eiiters into the numher. Thus, 60, which is a multiple of 12, contains 5 times 12, or 5 times 2x2x3, the prime factors of 12. Hence, each of the prime factors of 12 enters as a factor into 60 as many times as it enters into 12. Prop. II. — Tlie least common multiple of ttco or more given numhers must contain, as a factor, each prime factor in those numhers the greatest numher of times that it enters into any one of them. 2 26 LEAST COMMON MULTIPLE, Thus, 12 = 2 X 2 X 3, and 9=3x3. The prime factors in 12 and 9 are 2 and 3. A multiple of 12, according to Prop. I, must contain 2 as a factor twice and 3 once. A multiple of 9, according to the same propo- sition, must contain 3 as a factor twice. Hence a number which is a multiple of both 12 and 9 must contain 2 as a factor twice and 3 twice, which is equal to 2 x 2 x 3 x 3 = 36. Hence 36 is the least common mul- tiple of 12 and 9. DEFINITIONS. 212. A Multiple of a number is a number that is exactly divisible by the given number. Thus, 24 is divisible by 8 ; hence, 24 is a multiple of 8. 213. A Common Multifile of two or more numbers is a number that is exactly divisible by each of them. Thus, 36 is divisible by each of the numbers 4, 9, and 12 ; hence, 36 . is a common multiple of 4, 9, and 12. 214. The Least Common Multiple of two or more numbers is the least number that is exactly divisible by each of them. Thus, 24 is the least number that is divisible by each of the numbers 6 and 8 ; hence, 24 is the least common multiple of 6 and 8. METHOD BY FACTORING. ILLUSTRATION OF 1* R O C E S S . 215. Prob. I. — To find, by factoring, the least com- mon multiple of two or more numbers. Find the least common multiple of 18, 24, 15, and 35. Explanation. — 1. We observe that 3 is a factor of 18, 24, and 15. Dividing these numbers by 3, wo write the quotients with 35, in the second line. 2. Observing that 2 is a factor of G and 8, we divide as before, and fin*' the third line of numbers. Dividing 3 18 24 15 35 2 G 8 5 35 5 3 4 5 35 3 4 7 LEAST C03IM0X MULTIPLE. 2tt by 5, we find the fourth line of numbers, which are prime to each other ; hence cannot be further divided. 3. Observe the divisors 3, 2, and 5 are all the factors that are common to any two or more of the given numbers, and the quotients 3, 4, and 7 are the factors that belong each only to one number. Therefore the divisors and quotients together contain each of the prime factors of 18, 24, 15, and 35 as many times as it enters into any one of these numbers. Thus, the divisors 3 and 2, with the quotient 3, are the i)rime factors of 18 ; and so with the other numbers. Hence, according to (211 — II), the continued product of the divisors 3, 2, and 5, and the quotients 3, 4, and 7, which is equal to 2520, is the least common multiple of 18, 24, 15, and 35. METHOD BY GKEATEST COMMON DIVISOR. ILLUSTRATION OF PROCESS, 218. Pkob. 1 1. — To find, by using the greatest com- mon divisor, the least common multiple of two or more numbers. Find the least common multiple of 195 and 255. Explanation. — 1. We find the greatest common divisor of 195 and 255, wliicli is 15. 2. The greatest common divisor, 15, according to (19J5), contains all tlie prime factors that are common to 195 and 255. Dividing each of tliese numbers by 15, we find tlie factors that are not common, namely, 13 and 17. 3. The common divisor 15 and the quotient 13 contain all the prime factors of 195, and the common divisor 15 and the quotient 17 contain all the prime factors of 255. Hence, according to (211 — II), the continued product of the common divisor 15 and the quotients 13 and 17, which is 3315, is the least com- mon multiple of 195 and 255. The least common multiple of any two numbers is found in the same manner ; and of three or more by finding first the least common multiple of two of them, then finding the least common multiple of the multiple thus found and the third number, and so on. f8 PROPERTIES OF NUMBERS. REVIEW AND TEST QUESTIONS. 223. 1. Define Prime Number, Composite Number, and Exact Divisor, and illustrate each by an example. 2. What is meant by an Odd Number ? An Even Number ? 3. Show that if an even number is divisible by an odd num- ber, the quotient must be even. 4. Name the prime numbers from 1 to 40. 5. Why are all even numbers except 2 composite ? G. State how you would show, in the series of odd numbers, that every fifth number from 5 is divisible by 5. 7. What is a Factor ? A Prime Factor ? 8. What are the prime factors of 81 ? Of 64 ? Of 125 ? 9. Show that rejecting the same factor from the divisor and dividend does not change the quotient. 10. Explain Cancellation, and illustrate by an example. 11. Give reasons for calling an exact divisor a measure. 12. What is a Common Measure ? The Greatest Common Measure ? Illustrate each answer by an example. 13. Show that the greatest common divisor of 42 and 114 is the greatest common divisor of 42 and the remainder after the division of 114 by 42. 14. Explain the rule for finding the greatest common divisor by factoring ; by division. 15. Why must we finally get a common divisor if the greater of two numbers be divided by the less, and the divisor by the remainder, and so on ? 16. What is a Multiple ? The Least Common Multiple ? 17. Explain how the Least Common Multiple of two or more numbers is found by using their greatest common divisor. 18. Prove that a number is divisible by 9 when the sum of its digits is divisible by 9. 19. Prove that a number is divisible by 11 when the differ- ence of the sums of the digits in the odd and even places is zero. REDUCTION OF FRACTIONS, 29 FEACTIONS. 225. A Fractional Unit is one of the equal parts of anything regarded as a whole. 226. A Fraction is one or more of the equal parts of anything regarded as a whole. 227. The Unit of a Fraction is the unit or whole which is considered as divided into equal parts. 228. The Numerator is the number above the dividing line in the expression of a fraction, and indicates how many equal parts are in the fraction. 229. The Denominator is the number below the dividing line in the expression of a fraction, and indicates how many eq^ial parts are in the whole. 230. The Terms of a fraction are the numerator and denominator. 231. Taken together, the terms of a fraction are called a Fraction^ or Fractional I^iunher, 232. Hence, the word Fraction means one or more of the equal parts of anything, or the expression that denotes one or more of the equal parts of anything. PRINCIPLES OF REDUCTION 235. Illustrate the following principles with examples : Piii;n". I. — Tlie numerator and denominator of a frac- tion represent, each, parts of the same size ; thus, 3 4 7 5 (1.) ^^^^ (2.) Observe in illustration (1) the denominator 7 represents the whole or 7 sevenths, and the numerator 3 represents 3 sevenths ; in illustration (2), the denominator represents ^fifths, and the numerator Affihs. Hence the numerator and denominator of a fraction represent parts of the same size. 30 REDUCTION OF FRACTIONS, Prin", IL — Multiplying both the terms of a fraction hy the same number does not change the value of the fraction; thus, 2 _ ? X 4_^ 3 "" 3 X 4 "■ 12 Be particular to observe in the illustration that the amount expressed by the 2 in the numerator or the 3 in the denommator of | is not changed by making each part into 4 equal parts ; therefore f and -f^ express, each, the same amount of the same whole. Hence, multiplying the numerator and denominator by the same num- ber means, so far as the real fraction is concerned, dividing the equal parts in each into as many equal parts as there are units in the nmnber hy which they are multiplied. Prin. hi. — Dividing both terms of a fraction by the same nurnber does not change the value of the fraction ; thus, ^ -^ 3 _ 3 12 -^ 3 ~ 4 The amount expressed by the 9 in the numerator or the 12 m the de- nominator of /^ is not changed by putting every 3 parts into one, as will be seen from the illustration. Hence -^^ and f express each the same amount of the same whole, and dividing the numerator and denominator by the same number means putting as many parts in each into one as there are units in the number by which they are divided. DEFmiTIONS. 336. The Value of a fraction is the amount which it represents. 237. deduction is the process of changing the terms of a fraction without altering its vakie. 238. A fraction is reduced to Higher Terms when its numerator and denominator are expressed by larger numbers. Thus, i = A- REDUCTION OF FB ACTIONS. 31 239. A fraction is reduced to Lower Terms when its numerator and denominator are expressed by smaller numbers. Thus, A = f 240. A fraction is expressed in its Loivest Terms when its numerator and denominator are 2^ rime to each other. Thus, in I, the numerator and denominator 4 and 9 are prime to each other ; hence the fraction is expressed in its lowest terms. 241. A Common Denominator is a denominator that belongs to two or more fractions. 242. The Least Cominon I^enominator of two or more fractions is the least denominator to which they can all be reduced. 243. A ^roxier Ft^action is one whose numerator is less than the denominator ; as f , ^. 244. An Imiiroper Fraction is one whose numerator is equal to, or greater than, the denominator ; as |, |. 245. A Mixed dumber is a number composed of an integer and a fraction ; as 4f , I84. ILl^USTJtATION OF P It O C E S 8 , 246. Prob. I. — To reduce a whole or mixed number to an improper fraction. 1. Reduce 3f equal lines to fifths. = IS fifths ^ ~ Explanation, — Each whole line is equal to 5 fifths, as shown in the illustration; 3 lines must therefore be equal to 15 fift?is. 15 fifths + 3 fifths = 18 fifths. Hence in 3| lines there are Y of a line. 33 REDUCTION OF FRACTIONS. 249. Prob. II. — To reduce an improper fraction to an integer or a mixed number. 1. Keduce ^fourths of a line to whole lines. f = 9 -^ 4 = 2i Explanation. — A whole line is composed of 4 fourths. Hence, to make tlie 9 fourtJis of a line into whole lines, we put every four parts into one, as shown in the illustration, or divide the 9 by 4, which gives 2 wholes and 1 of the fourths remaining. 352. Prob. III. — To reduce a fraction to higher terms. 1. Keduce f of a line to twelfths. 2 __ 2 X 4:_^ 3 ~ 3 X 4 ~ 12 Explanation. — 1. To make a whole, which is already in thirds, into 12 equal parts, each third must be made mto four equal parts, 2. The numerator of the given fraction expresses 2 thirds, and the denominator 3 thirds ; making each third in both into four equal parts (ti45 — II), as shown in the illustration, the new numerator and denom- inator will each contain 4 times as many parts as in the given fraction. Hence, | of a line is reduced to twelfths by multiplying both numera- tor and denominator by 4. lower 255. Pr OB. IV.— To reduce a fraction t< terms. Keduce A of a given line to fourths. 9 12 -^3 ^3 = 3 4 ., .^ «■■ ■ MMM MM m — M. MM ^M ADDITION OF FRACTIONS. 33 Explanation. — 1. To make into 4 ^qual parts or fourths a whole, which is already in 12 equal parts, or twelfths, every 3 of the 12 parts must be put into one. 2. The numerator of the given fraction expresses 9 twelfths, and the denominator 12 twelfths; putting every 3 twelfths into one, in both (235 — III), as shown in the illustration, the new numerator and denominator will each contain one third as many parts as in the given fraction. Hence, ^-^ of a line is reduced to fourths by dividing both numerator and denominator by 3. 258. Prob. V. — To change fractions to equivalent ones having a common denominator. 1. Reduce | and f of a line to fractions haying a common denominator. (1-) (2.) Explanation.— 1. We find the least common multiple of the denom- inators 3 and 4, which is 12. 8. We reduce each of the fractions to twelfths (252 — III), as shown in illustrations (1) and (2). ADDITION AND SUBTRACTION. PJtEP AR ATOMY 1* B O P O S I T I O N S . 261. Prop. I. — Fractional units of the same kind, that are fractions of the same whole, are added or subtracted in the same manner as integral units. Thus, I of a yard can be added to or subtracted from f of a yard, because they are each fifths of one yard. But | of a yard cannot be added to or subtracted from | of a day. 2 3 . "" 2x4 8 3x4 12 ^■1^ B^^H ■ HBIH BiBHH ■■^iH H^HM m^t^ 3 4 ~ 3x3 9 4 X 3 ~ 12 ^— ■— ■ ^mm m ^^ ^^ ^^ ^^ «M_ M»M __« M«M 4 32 9~72 5 60 6 ""72 155 ■ ~ 72 7 63 8~72. 34 SUBTRACTION OF FRACTIONS. Prop. II. — Fractions expressed in different fractional units rmist he changed to equivalent fractions having the same frac- tional unit, before they can he added or subtracted. 263. Prob. I. — To find the sum of any two or more given fractions. 1. Find the sum of f + |- + |. Explanation. — 1. We reduce the fractions to tlie satae fractional unit, by reducing them to their least common denominator, which is 72 ^ -^ = m- (258-V). 2. We find the sum of the numer- ators, 155, and write it over the com- mon denominator, 72, and reduce VV- to 2H. The sum of any number of fractions may be found in the same manner. 366. Prob. I.— To find the diflferenee of any two given fractions. 1. Find the difference between f and -^. 7 5 21 10 11 Explanation. — 1. We reduce o 12 ^^ 24 24 ^^^ 24 *^® given fractions to their least common denominator, which is 24, 2. We find the difference of the numerators, 21 and 10, and write it over the common denominator, giving ^, the required difference. 2. Find the difference between 35 1 and 16|. 35f = 35-^^ Explanation.— 1. We reduce the | and f to ■^Q3 -- IQJL. their least common denominator. — ~ 2. ^\ cannot be taken from /^ ; hence, we in- l^H" crease the -^% by f| or 1, taken from the 35. We now subtract -^^ from ff , leaving i|. 3. We subtract 16 from the remaining 34, leaving 18, which united with 1^ gives 18^-, the required difference. 3IULTTP LIGATION OF FRACTIONS. 35 MULTIPLICATION. PREPARATORY PROPOSITIONS. 269, The following propositions must be mastered per- fectly, to understand and explain the process in multiplica- tion and division of fractions. Prop. I. — Multiplying the numerator of a fraction, while the denominator remai^is unchanged, multiplies the fraction ; thus, 2 X 4 _ 8 5 ~" 5 Observe that since the denominator is not changed, the size of the parts remain the same. Hence the fraction | is multiplied by 4, as shown in the illustration, by multiplying the numerator by 4. Peop. IL — Dividing the denominator of a fraction ivhile the numerator remains unchanged multiplies the fraction j thus, A - ? 12 -^ 4 ~ 3 (1.) (2.) Observe that in (1) the whole is made into 1 2 equal parts. By put- ting every 4 of these parts into one, or dividing the denominator by 4, the whole, as shown in (2), is made into 3 equal parts, and each of the 2 parts in the numerator is 4 times 1 twelfth. Hence, dividing the denominator of -^^ by 4, the number of parts in the numerator remaining the same, multiplies the fraction by 4. Prop. III. — Dividing the ^lumerator of a fraction while the denominator remains unchanged divides the fraction ; thus, 6-^3 _ 2 9 "9 (l.)iof (^^^^^ . 36 MULTIPLICATION OF FRACTIONS, In (1) the numerator 6 expresses the parts taken, and one-third of these 6 parts, as shown by comparing (1) and (2), the denommator remaining the same, is one-third of the value of the fraction. Hence, the fraction f is divided by 3 by dividing the numerator by 3. Prop. IV. — Multiplying the denominatoi- of a fraction luliile the nvmerator remains unchanged divides the fraction ; thus, 5x2 ~ 10 (!•) m (^0 I", In (1) the whole is made into 5 equal parts ; multiplying the denom- inator by 2, or making each of these 5 parts into 2 equal parts, as shown in (2), the whole is made into 10 equal parts, and the 3 parts in the numerator are one-half the size they were before. Hence, multiplying the denominator of f by 2, the numerator remain- ing the same, divides the fraction by 2. ILJLUSTHATION OF I* It O C E S S . 271. Prob. I. — To multiply a fraction by an integer. 1. Multiply i by 7. Solution. — 1. According to (2G9 — I), multiplying the numerator, the denominator remaining the same, multiplies the fraction. Hence, 7 , , 4 X 7 28 „, times f IS equal to - = — = 3J. y " 2. According to (269 — II), a fraction is also multiplied by dividing the denominator. 274. Prob. II. — To find any given part of an integer, or To multiply an integer by a fraction, 1. Find I of S395. Solution.— 1. We find the \ of $395 by dividing it by 5. Hence the first step, 1395 -^ 5 = $79. 2. Since |79 is 1 fifth of $395, four times $79 will be 4 fifths. Hence the second step, $79 x 4 = $316. To avoid fractions until the final result, we multiply by the numerator first, then divide by the denominator. MULTIPLICATION OF FRACTIONS. 37 276. Prob. III. — To find any given part of a fraction, or To multiply a fraction by a fraction. Find the f of j of a given line. FIKST STEP. 1 „ 3 3 3 of 3 4 4 X 3 12 SECOND STEP. _3_ X 2 _ A — i 12 ~ 12 ~ 2 Explanation. — According to (269 — IV), a fraction is divided by multiplying its denominator. Hence we find the \ of f by multiplying the denominator 4 by 3, as shown in First Step. Having found 1 third of f, we find 2 thirds, according to (2G9 — I), by multiplying the numerator of ^^ by 2, as shown in Second Step. In practice, the Second Step is usually made the First. 379. Pkob. IV. — To multiply by a mixed number. Multiply 372 by 64. 372 Explanation. — 1. In multiplying by a mixed g g number, the multiplicand is taken separately (92) as many times as there are units in the multiplier, and '^ -^ *^ '' such a part of a time as is indicated by the fraction in 2 6 5 -^ the multiplier ; hence, 249^5 2. We multiply 372 by 6| by multiplying first by 6, * which gives the product 2232, and adding to this pro- duct f of 372, which is 265 f (2 74), giving 2497f , the product of 372 and 6f . 383. Prob. V.— To multiply when both multiplicand and multiplier are mixed numbers. Multiply 86| by 54|. (1.) s^=^^', u\=^K (2.) ^ X ^ = -a Vi^^ = 47411-J. 38 DIVISION OF FRACTIONS. Explanation. — 1. We reduce, as shown in (1), both multiplicand and multiplier to improper fractions. 3. We multiply, as shown in (2), the numerators together for the numerator of the product, and the denominators together for the denom- inators of the product (275), then reduce the resulUto a mixed number. DIVISION. IZLUSTRATION OF PROCESS. 287. Pkob. I. — To divide a fraction by an integer. 1. Divide f by 4. 8 8 -r- 4 2 Explanation. — 1. Accord- (-* •) 9 "^ 9 ~ 9 ing to (269—111), a fraction is divided by dividing the numer- (9\ ^ ' A. ^ ^ ^ *'^*^^* Hence we divide f by 4, ^''^9* 9x4 36 9 as shown in (1), by dividing the numerator 8 by 4. 2. According to (269 — IV), a fraction is divided by multiplying the denominator. Hence we divide | by 4, as shown in (2), by multiplying the denominator by 4, and reducing the result to its lowest terms. 290. Prob. II.— To divide by a fraction. 1. How many times is | of a given line contained in twice the same line ? FIRST STEP, 2 lines = -^ of a line. 5 \ = SECOND STEP. 10 3 _ 10 _ y "^ 5 - ¥ - '^=*- Explanation. — 1. We can find how many times one number is con- tained in another, only when both are of the same denomination (155). DIVISION OF FRACTIONS. 39 Hence we first reduce, as sliown in First Step, the 2 lines to 10 ffthsot a line ; the same fractional denomination as the divisor, 3 fifths. 2. The 3 fifths in the divisor, as shown in Second Step, are contained in the 10 fifths in the dividend 3 times, and 1 part remaining, which makes ^ of a time. Hence 2 equal lines contain | of one of them 3 J times. Observe the following regarding this solution : (1.) The dividend is reduced to the same fractional denomination as the divisor hy multiplying it by the denominator of the divisor ; and when reduced, the division is performed by dividing the numerator of the dividend by the numerator of the divisor. (2.) By inverting the terms of the divisor, these two operations are expressed by the sign of multiplication. Thus, 2 -4- ? = 2 x f, which means that 2 is to be multiplied by 5, and the product divided by 3. 2. How many times is ^ of a given line contained in f of it ? FIKST STEP. 2 _ 4 3 ~ 6 1 _ 3 2 ■" 6 SECOND STEP. -T- 3 6 6 6 3* ^^^^:^^^^ ^ = li Explanation. — 1. We reduce, as shown in First Step, the dividend f and the divisor i both to sixths (155 — 1). 2. We divide the f by J5 by dividing the numerator of the dividend by the numerator of the divisor. The | is contained in f , as shown in Second Step, 1\ times. Hence | is contained 1^ times in f. 391. When dividing by a fraction, we abbreviate the work by inverting the divisor, as follows : 1. In reducing the dividend and divisor to the same fractional unit, the product of the denominators is taken as the common denominator, f^' COMPLEX FRACTIONS. and each numerator is multiplied by the denominator of the other frac- tion, thus, 5 ^ 2 5x3 ^ 2x7 15 ^ 14 15 Numerator of dividend. 7 * 3~7x3 * 3x7 21 * 21"' 14 Numerator of divisor. 2. By inverting the divisor, thus, f -5- 1 = f x 4 = fl, the numerators 15 and 14 are found at once, without going through the operation of finding the common denominator. Hence the rule. Invert the terms of the divisor and proceed as in multiplication. 294. Pkob. III. — To divide when the divisor or divi- dend is a mixed number, or both. 1. Divide 48 by 4f . Explanation.— 1. We re- \^'} 4:0 -r- 4r|- = 4:0 -7- -^ ducc the divisor 4|, as shown (2.) 48-^-V- = 48 X ■^ = l^ in (1). to the improper frac- tion J^. 2. We invert the divisor, as shown in (2), according to (201), and multiply the 48 by -^, giving lOf as the quotient of 48 divided by 4|. 2. Divide ^ by 3|. Explanation. — 1. We (1.) 8f -^ 3 J = 4^ -^ -^ reduce the dividend and 3 divisor, as shown in (1), (2.) 4^ -^ ^^1- = :^ X 'A: = V = 2^ to improper fractions, giving -V- -^ V-- 2. We invert the divisor, y-, as shown in (2), according to (291), and cancel 31 in the numerator 63 and denominator 31 (186), giving y ,' or3f . Hence, 8f -5- 3| = 2f . COMPLEX FEACTIONS. 29T. Certain results are obtained by dividing the numera- tor and denominator of a fraction by a number that is not an exact divisor of each, which are fractional in form, but are not fractions according to the definition of a fraction. These fractional forms are called Complex Fractions. 298. A Complex Fraction is an expression in the form of a fraction, having a fraction in its numerator or denominator, or in both ; thus, |, ^, -|' COMPLEX FRACTIONS. 41 299. A Simple Fraction is a fraction having a whole number for its numerator and for its denominator. fbobijEms in complex fractions. 300. Prob. L— To reduce a complex fraction to a simple fraction. Eeduce 3 to a simple fraction. ^ 4f X 12 56 Explanation.— 1. We find the least 'if^^^ 'rf% y 12^^93 common multiple of the denominators of the partial fractions | and |, which is 12. 2. Multiplying both terras of the complex fraction by 12 (235—11), which is divisible by the denominators of the partial fractions, f and |, reduces each term to a whole number. 4| x 12 = 56 ; 7 J x 12 = 93. 41 Therefore ^ is equal to the simple fraction f |. 303. The three classes of complex fractions are forms of expressing three cases of division ; thus, 7. A mixed number divided by an integer. 9^. An integer divided by a mixed number. 2-|. A mixed number divided by a mixed number. Hence, when we reduce a complex fraction to a simple fraction, as directed (301), we in fact reduce the dividend and divisor to a common denominator, and reject the denominator by indicating the division of the numerator of the dividend by the numerator of the divisor ; thus, (3.) 5}--2f = ¥--f, and ^-M = «^fl = ll, the same result as obtained by the method of multiplying by the least C5ommon multiple of the denominators of the partial fractions. (!•) 5|_ 7 ~ 32 .51 (3.) n" = 32 (3.) 8|_ = 8i 42 REVIEW EXAMPLES IN FRACTIONS. 304. Pkob. II. — To reduce a fraction to any given denominator. 1. Exam2)les where the denominator of the required fraction is a factor of the denominator of the given fraction. Eeduce \\ to a fraction whose denominator is 8. 17 17 _i_ 3 5|. Explanation. — We observe tliat 8, the oT -" 24 _i_'\~~ 1^ denominator of the required fraction, is a factor of 24, the denominator of the given fraction. Hence, dividing both terms of W by 3, the other factor of 24, the 5- fraction is reduced (235 — III) to -r^, a fraction whose denominator i»8. o 2. Examples ivhere the denominator of the required fraction is not a factor of the denominator of the given fraction. Reduce -^ to a fraction whose denominator is 10. 8 8 X 10 80 Explanation. — 1. We intro- (1.) T^=T7T in^^ T^ ^^^® ^^^ given denominator 10 Id Id X 10 loU ^g ^ factor into the denominator 80 _ 80 -^ 13 _ 6-^% of y\ by multiplying, as shown ^'' 130 — 130 -^13 — 1(7 ^'^ ^^' ^"*^^ terms of the fraction by 10 (235-11). 2. The denominator 130 now contains the factors 13 and 10. Hence, dividing both terms of the fraction -^^^^^ by 13 (235— III), as shown in 6 ^- (2), the result is — ^, a fraction whose denominator is 10. REVIEW EXAMPLES. 307. 1. How many thirtieths in |, and why? In f ? 2. Reduce f , ^^j, j^, /^, and fl each to twenty-eighths. 5 3.^ 3. Reduce to a common denominator ~^, -^, and /j. 5 5x4 4. State the reason why . = — - — (209). 5. Reduce | to a fraction whose numerator is 12 ; is 20 ; is 2 ; is 3 ; is 7 (235). 6. Reduce to a common numerator f and f (252). 7. Find the sum of -«, ^■-^, |, f , and ||. REVIEW EXAMPLES IN FRACTIONS, 43 8. Find the value of (f of ^\ - j\) ^(^ + ~^. 9. If I of an estate is worth $3460, what is | of it worth ? 10. $4 is what part of $8 ? Of $12 ? Of $32 ? Of $48 ? Write the solution of this example, with reason for each step. 11. If a man can do a piece of work in 150 days, what part of it can he do in 5 days ? In 15 days ? In 25 days ? In 7| days ? In 3f days ? In 12|-days? 12. A's farm contains 120 acres and B's 380; what part of B's farm isA's? 13. 42 is f of what number ? Write the solution of this example, with reason for each step. 14. $897 is f of how many dollars ? 15. f of 76 tons of coal is j% of how many tons ? 16. A piece of cloth containing 73 yards is | of another piece. How many yards in the latter? 17. Bought a horse for $286, and sold him for I of what he cost ; how much did I lose ? 18. 84 is --^^.j of 8 times what number ? Write the s:)lution of this example, with reason for each step. 19. A has $694 in a bank, which is | of 3 times the amount B has in the same bank ; what is B's money ? 20. Two men are 86| miles apart ; when they meet, one has traveled 8 J miles more than the other ; how far has each traveled ? 21. If j\ of a farm is valued at $4732 1, what is the value of the whole farm ? 22. The less of two numbers is 432f, and their difference 123 j^'^. Find the greater number. 23. A man owning | of a saw-mill, sold f of his share for $2800 ; what was the value of the mill ? 24. What number diminished by f and f of itself leaves a remainder of 32? 25. I put * of my money in the bank and gave f of what I had left to a friend, and had still remaining $400. How much had I at first ? 26. Sold 342 bushels of wheat at $1| a bushel, and expended the amount received in buying wood at $4| a cord. How many cords of wood did I purchase ? 27. If f of 4 pounds of tea cost $2 J, how many pounds of tea can be bought for $7 i ? For $12.}? For^^^^? 44 REVIEW QUESTIONS ON FRACTIONS, 28. If 5 be added to both terms of the fraction 2, how much will its value be changed, and why ? 29. I exchanged 47f bushels of corn, at $| per bushel, for 24| bushels of wheat ; how much did the wheat cost a bushel ? 30. A can do a piece of work in 5 days, B can do the same work in 7 days ; in what time can both together do it ? 31. Bought I of 844 acres of land for | of $3584^- ; what was the price per acre? 32. A boy while fishing lost | of his line ; he then added 8 feet, which was I of what he lost; what was the length of the line at first? ^^ 33. A merchant bought a quantity of cloth for $2849 1, and sold it for y'^ of what it cost him, thereby losing $? a yard. How many yards did he purchase, and at what price per yard ? 34. A tailor having 276| yards of cloth, sold f of it at one time and f at another ; what is the value of the remainder at $3 a yard ? 35. A man sold /^ of his farm at one time, { at another, and the remainder for $180 at $45 an acre ; how many acres were there in the farm ? 36. A merchant owning || of a ship, sells ^ of his share to B, and f of the remainder to C for $600|^ ; what is the value of the ship ? KEVIEW AND TEST QUESTIONS. 308. 1. Define Fractional Unit, Numerator, Denominator, Im- proper Fraction, Reduction, Lowest Terms, Simple Fraction, Common Denominator, and Complex Fraction. 2. What is meant by the unit of a fraction ? Illustrate by an example. 3. When may | be greater than ^ ? \ than \ ? 4. State the three principles of Reduction of Fractions, and illustrate each by lines. 5. Illustrate with lines or objects each of the following propositions : I. To diminish the numerator, the denominator remaining the same, diminishes the value of the fraction. II. To increase the denominator, the numerator remaining the same, diminishes the value of the fraction. III. To increase the numerator, the denominator remaining the same, increases the value of the fraction. IV. To diminish the denominator, the numerator remaining the same, increases the value of the fraction. ^%^ ' :---=7 DECIMAL FRACTIONS. 45 6. What is meant by tlie Least Common Denominator ? 7. When the denominators of the given fractions are prime to each other, how is the Least Common Denominator found, and why ? 8. State the five problems in reduction of fractions, and illustrate each by the use of lines or objects. 9. Show that multiply- ing the denominator of a fraction by any num- ber divides the fraction by that number (269). 10. Show by the use of lines or objects the truth of the following : (1.) i of 2 equals f of 1. (3.) ^ of 5 equals | of 1. (3.) f of 1 equals ^ of 3. (4.) -f of 9 equals 4 times \ of 9. 11. To give to another person f of 14 silver dollars, how many of the dollar-pieces must you change, and what is the largest denomination of change you can use ? 12. Show by the use of objects that the quotient of 1 divided by a fraction is tlie given f motion inverted. 13. Why is it impossible to perform the operation in | + |, or in f— f, without reducing the fractions to a common denominator ? 14. Why do we invert the divisor when dividing by a fraction? Illustrate your answer by an example. 15. What objection to calling — a fraction (226) ? 16. State, and illustrate with lines or objects, each of the three classes of so-called Complex Fractions. 17. Which is the greater fraction, ^ or f J, and how much? 18. To compare the value of two or more fractions, what must be done with them, and why ? Qi 4 2i 3 5^ 2- 19. Compare ~ and - ; -^ and — ; -| and -j, and show in each case 7 8 9 10 7i o^ which is the greater fraction, and how much. 20. State the rule for working each of the following examples : (1.) 3| + 4f + 8|. (2.) (7f + 5^) - (8 - 21). (3.) 5 X I of I of 27. (4.) 8| X 5|. (5.) I X f . Explain by objects. (6.) YS-i-^, Explain by objects. 21. Illustrate by an example the application of Cancellation in multi- plication and division of fractions. 4f6 DECIMAL FRACTIONS. DECIMAL FEAOTIOE"S. DEFINITIONS. 309. A unit is separated into decimal parts when it is divided into teiiths ; thus. DECIMAL PARTS. 310. A Decimal Fractional Unit is one of the decimal parts of anything. 311. By making a whole or ^mit into decimal parts, and one of these parts into decimal parts, and so on, we obtain a series of distinct orders oi decimal fractional tmits, each -^^ of the preceding, having as denominators, respectively, 10, 100, 1000, and so on. Thus, separating a whole into decimal parts, we have, according to (246), 1 = {%; making J^ into decimal parts, we have, according to (252), ^V = TTH7 ; in tlie same manner, ^-^^ = ^-^, ^-^^o = WriViy and so on. Hence, in the series of fractional units, y^^^, j^^, j \^^, and so on, each unit is one-tenth of the preceding unit. 312. A Decimal Fraction is a fraction whose denom- inator is 10, 100, 1000, etc., or 1 with any number of ciphers annexed. Thus, -^, yj^, yjfoj ^^'^ decimal fractions. 31.3. The Decimal Sir/n (.), called the decirnal pointy is used to express a decimal fraction without writing the denominator, and to distinguish it from an integer. 315, A flecimaJ fraction expressed without writing the denominator h called simply a Decimal. Tlius, wc speak of .79 as the (Primal seventy-nine, yet we mean the decimal fraction seventy-nine hundredths. LECIMAL FRACTIONS, 47 NOTATION AND NUMERATION. 314. Prop. I. — A decimal fraction is expressed without writing the denominator by using the decimal point, and placing the numerator at the right of the period. Thus, -^-Q is expressed .7 ; {^^^ is expressed .35. 316. Prop. II. — Ciphers at the left of significant figures do not iiicrease or diminish the member expressed by these figures. Thus, 0034 is thirty-four, the same as if written 34 without the two ciphers. 317. Prop. 111.— When the fraction in a mixed number is expressed decimally, it is written after the integer, with the decimal point between them. Thus, 57 and .09 are written 57.09 ; 8 and .0034 are written 8.0034. 320. Prop. IV. — Every figure in the numerator of a deci- mal fraction represents a distinct order of decimal units. Thus, tVoV is equal to ^Wif + t^^ + nrW- But, according to (255), ■f^^j^ = ■^^, and yf ^^ = yf ^. Hence, 5, 3, and 7 each represent a distinct order of decimal fractional units, and {'^^^, or .537 may be read 5 tenths 3 hundredths and 7 thousandtlis. 321. Prop. V. — A decimal is read correctly by reading it as if it were an integer and giving the name of the right-hand order. Thus, .975 = ^"o"^ + y^^ + -n/VTr* Hence is read, nine hundred seventy-five thousandths. 1 . Observe that when there are ciphers at the left of the decimal, according to (316), they are not regarded in reading the number ; thus, .062 is read sixty-two thousandths. 2. The name of the lowest order is found, according to (314), by prefixing 1 to as many ciphers as there are figures in the decimal. For example, in .00209 there are five figures ; hence the denominator is 1 with five ciphers ; thus, 100000, read handred-thousandths. 48 DECIMAL FRACTIONS: REDUCTION. PBEPARATOltT l*JiOrOSITION8, The following preparatory propositions should be ver^ carefully studied. 325. Prop. I. — Annexing a cipher or multiplying a num- ber hy 10 introduces into the number the tivo prime factors 2 and 5, Thus, 10 being equal 2 x 5, 7 x 10 or 70 = 7 x (2 x 5). Hence a number must contain 2 and 5 as a factor at least as many times as there are ciphers annexed. 326. Prop. II.— ^ fraction in its lowest terms, whose denomi7iator contains no other prime factors than 2 or 5, can be reduced to a simple decimal. Observe that every cipher annexed to the numerator and denominator makes each divisible once by 2 and 5 (325). Hence, if the denominator of the given fraction contains no other factors except 2 and 5, by annex- ing ciphers the numerator can be made divisible by the denominator, and the fraction reduced to a decimal. Thus, I = i^^-o (235—11). Dividing both terms of the fraction by 8 (235-III), we have f^^ = ^^^ = .875. 327. Prop. III. — A fraction in its lowest terms, whose denominator contains any other prime factors than 2 or 5, can be reduced only to a complex decimal. Observe that in this case annexing ciphers to the numerator and denominator, which (325) introduces only the factors 2 and 5, cannot make the numerator divisible by the given denominator, which contains other prime factors than 2 or 5. Hence, a fraction will remain in the numerator, after di^iding the numerator and denominator by the denominator of the given fraction, however far the division may be carried. Thus, ^\ = ^l^n (235—11). Dividing both numerator and denom- inator by 31, we have = Inoo ~ -^^^H* ^ complex decimal. DECIMAL FRACTIONS. 49 328. Pkop. IV. — The same set of figures must recur in-' definitely in the same order in a complex decimal which ca?mot he reduced to a simple decimal. _ 7 700Q0 6363yV T^^^' li = IIoooo = loooo" = -^^^^A- Observe carefully the following: 1. In any division, the number of different remainders that can occur is 1 less than the number of units in the divisor. Thus, if 5 is the divisor, 4 must be the greatest remainder we can have, and 4, 3, 3, and 1 are the only possible different remainders ; hence, if the division is continued, any one of these remainders may recur. 2. Since in dividing the numerator by the denominator of the given fraction, each partial dividend is formed by annexing a cipher to the remainder of the previous division, when a remainder recurs the partial dividend must again be the same as was used when this remainder occurred before ; hence the same remainders and quotient figures must recur in the same order as at first. 3. If we stop the division at any point where the given numerator recurs as a remainder, we have the same fraction remaining in the nu- merator of the decimal as the fraction from which the decimal is derived. Thus 7_700_63t'._ 7 70000 6363W noao ■, a "' u = noooo == -mm == • ''^'^^'^' ''"'^ ^^ °°- 329. Prop. V. — The value of a fraction luhich can only he reduced to a complex decimal is expressed, nearly, as a sim- ple decimal, hy rejecting the fraction from tlie numerator. 3 27 ^ Thus, r-j- = -j^ (327). Rejecting the ^ from the numerator, we have y^^, a simple fraction, which is only y\ of -^ smaller than the 27tt given fraction y\ or -~^ • 100 Observe the following : 2. By taking a sufficient number of places in the decimal, the true value of a complex decimal can be expressed so nearly that what is re- jected is of no consequence. 3 so DECIMAL FRACTIONS. ^, 3 27272727A . . ' 11 ~ TOOOOOOOO ' ^^J®^*"^& *^® h from tlie numerator, we have ■NmoWVuj or .27272727, a simple decimal, which is only -^j of 1 hundred- millionths smaller than the given fraction. 2. The approximate value of a complex decimal which is expressed by rejecting the given fraction from its numerator is called a Circulating Becimaly because the same figure or set of figures constantly recur. 330. Prop. VI. — Diminishing ilie numerator and denom- inator hj the same fractional part of each does not change ihe value of a fraction. Be particular to master the following, as the reduction of circulating decimals to common fractions depends upon this proposition. 1. The truth of the proposition may be shown thus : ^_9^-|of 9_^-3_6_3 12 ~ 12 - i of 12 ~ 12 - 4 ~ 8 ~ 4 Observe that to diminish the numerator and denominator each by i of itself is the same as multiplying each by f . But to multiply each by I , we multiply each by 2 (235 — II), and then divide each by 3 (235— III), which does not change the value of the fraction ; hence the truth of the proposition. 2. From this proposition it follows that the value of a fraction is not changed by subtracting 1 from the denominator and the fraction itself from the numerator. 3 3 3 2?^ Thus, ^ = = ^ =-^' Observe that 1 is the I of the denominator 5, 5 — 1 4 and I is \ of the numerator 3 ; hence, the numerator and denominator being each diminished by the same fractional part, the value of the fraction is not changed. DEFINITIONS. 331. A Simple Decimal is a decimal wbose numerator is a whole number ; thus, -f^ or .93. Simple decimals are also called Mnite Decimals. 332. A Complex Decimal is a decimal whose numer- ator is a mixed number; as ~^^ or .26|. DECIMAL FRACTIONS, 51 There are two classes of complex decimals : 1. Those whose value can be expressed as a simple decimal (326), as .23^ = .235 ; .32| = .3275. 2. Those whose value cannot be expressed as a simple decimal (327), as .53^ — .53333 and so on, leaving, however far we may carry the deci- mal places, I of 1 of the lowest order unexpressed. See (328j. 333. A Circalating Decimal is an approximate value for a complex decimal which cannot he reduced to a sim- ple decimal. Thus, .0G6 is an approximate value for .666| (329). 334. A Hepetend is the figure or set of figures that are repeated in a circulating decimal. 335. A Clrculafinf/ Decimal is exiwessed by writing the repetend once. When the repetend consists of one figure, a point is placed over it; when of more than one figure, points are placed over the first and last figures; thus, .333 and so on, and .592592+ are written .3 and .592. 336. A Pare Cifculatinff Decimal is one which commences with a repetend, as .8 or .394. 337. A Mixed Circulatincf Decimal is one in which the repetend is preceded by one or more decimal places, called the finite part of the decimal, as .73 or .004725, in which .7 or .004 is the finite part. jltjTjsthation of process. 338. Prob. I. — To reduce a commoii fraction to a decimal. Reduce | to a decimal. 3 _ 3000 _ 375 Explanation. — 1. We annex 8 8000 1000 ^^ the same number of ciphers to both terms of the fraction (235—11), and divide the resulting terms by 8, the significant figure in the denom- inator which must give a decimal denominator. Hence, | expressed decimally is .375. 6'Z DECI3IAL FRACTIONS, 2. In case annexing ciphers does not make the numerator divisible (327) by the significant figures in the denominator, the number of places in the decimal can be extended indefinitely. In practice, we abbreviate the work by annexing the ciphers to the numerator only, and dividing by the denominator of the given fraction, pointing off as many decimal places in the result as there were ciphers annexed. 341. Pkob. II. — To reduce a simple decimal to a eoiiiinou fraction. Eeduce .35 to a common fraction. 35 7 Explanation. — We write the decimal with •^^ ^^ Tqq ^^ 20 ^^'^ denominator, and reduce the fraction (255) to its lowest terms 344, Prob. III. — To find tlie true value of a pure circulating- decimal. Find the true value of .72. ..72 72 72 8 EXPLANATION.-In tak- .72 = = = — = — ijiff -72 as the approximate 1 nn 1 no i qq i i x\j\j xuu X xjv j.± mlue of a given fraction, we have subtracted the given fraction from its own numerator, as shown in (329 — V). Hence, to find the true value of y^^, we must, according to (330 — VI, 2), subtract 1 from the denominator 100, which makes the denominator as many 9's as there are places in the repetend. 347. Prob. IV. — To find the trvie value of a mixed circulating decimal. Find the true value of .318. (1.) .3i8 = .3H = f = S.|. Explanation. — 1. We find, according to (344), the true value of the repetend .018, which is .0J|. Annexing this to the ,3, the finite part, we have .3^§^. the true value of .318 in the form of a complex decimal. gi ft 2. We reduce the complex decimal .3.^^, or ~^^ to a simple fraction by multiplying, according to (300), both terms of the fraction by 99, 318 giving -5^ = m = /jj. Hence the true value of .318 is ^V DECIMAL FRACTIONS, 53 Abbreviated Solution.— Observe (2.) .318 Given decimal. gi s _3 Finite part. *^^* ^^ simplifying -j^, we multiplied 315 315 — - 7_, T^oth terms by 99. Instead of raulti- ^ ^^ ^^* plying tlie 3 by 99, we may multiply by 100 and subtract 3 from the product. Hence we add tlie 18 to 300, and subtract 3 from the result, which gives us the true numerator. To abbreviate the work ; From the given decimal subtract the finite part for a numer- ator, and for a denominator write as majiy 9^s as there are figures in the repetend, with as maiiy ciphers annexed as there are figures in the finite part. ADDITION. IliljVSTJtATION OF PROCESS. 351. Find the sum of 34.8, 6.037, and 27.62. Explanation. — 1. We arrange the numbers so that units of the same order stand in the same column. 2. We reduce the decimals to a common denominator, as shown in (1), by annexing ciphers. 3. We add as in integers, placing the decimal point before the tenths in the sum. SUBTRACTION. 353. Find the difference between 83.7 and 45.392. g 3 7 Explanation. — 1. We arrange the numbers so /L ^ q Q 9 *hat units of the same order stand in the same column. '- — '- — 2. We reduce the decimals, or regard them as 3 8.308 reduced to a common denominator, and then subtract as in whole numbers. The reason of this course is the same as given in addition. The ciphers are also usually omitted. (1-) (2.) 34.800 34.8 6.037 6.037 27.620 27.62 68.457 68.457 54 DECIMAL FRACTIONS. MULTIPLICATION. 355. Multiply 3.27 by 8.3. (1.) 3.27 X 8.3 = ^ x^ 327 83_ a7141 _ ^^■> loo"" 10- 1000 -^'-^^i Explanation. — 1. Observe that 3.27 and 8.3 are mixed numbers; hence, according to (282), they are reduced before being multiplied to improper fractions, as shown in (1). 2. According to (270), ff^ x ff, as shown in (2), equals 27.141. Hence 27.141 is the product of 3.27 and 8.3. The work is abbreviated thus : (3.) q n w We observe, as shown in (2), that the product of * 3.27 and 8.3 must contain as many decimal places as . ^ • ^ there are decimal places in both numbers. Hence 9 81 ^ve multiply the numbers as if integers, as shown in 2 (5 2 6 ^^' ^^^ point off in the product as many decimal — places as there are decimal places in both numbers. 27.141 DIVISION. pmeparatout propositions, 358. Prop. I. — Wlien the divisor is greater than the divi- dend, the quotient expresses the part the dividend is of the divisor. Thus, 4 H- 6 = f = |. The quotient f expresses the part the 4 is of 6. 1. Observe that the process in examples of this kind consists in reducing the fraction formed by placing the divisor over the dividend to its lowest term^. Thus, 32 -?- 56 = f f , which reduced to its lowest terms gives :^. 2- In case the result is to be expressed decimally, the process then consists in reducing to a decimal according to (338), the fraction formed by placing the dividend over the divisor. Thus, 5 -5- 8 = |, reduced to a decimal equals .625. DECIMAL FRACTIONS, 55 359. Prop. II. — The fraction remaining after the division of one i7iteger by another expresses the part the remainder is of the divisor. Tims, 42 -r- 11 = 3i\. The divisor 11 is contained 3 times in 42 and 9 left, wliicli is 9 parts or y\ of Ihe divisor 11. Hence we say that the divisor 11 is contained 3^^^ times in 42. We express the -^j decimally by reducing it according to (338). Hence, S^^ = 3.81. 360. Prop. III. — Divisio7i is possible only when the divi- defid and divisor are both of the same denomination (155 — I). For example, f^ -^ y§^, or .3 -f- .07 is impossible until the dividend and divisor are reduced to the same fractional denomination ; thus, .3 -J- .07 = .30 -V- .07 = 4| = 4.285714. ILLUSTRATION OF P M O C E S S , 361. Ex. 1. Divide .6 by .64. (1.) .8^.64 = .60-^■. 64 (2.) 60-^64 = — = .9375 Explanation. — 1. We reduce, as shown in (1), the dividend and divisor to the same decimal unit or denomination (200). 2. We divide, according to (200), as shown in (2), the numerator 60 by the numerator 64, which gives f J. Reducing ff to a decimal (338), we have .6 -4- .64 == .9375. Ex. 2. Divide .63 by .0022. (1.) .63-^. 0022 = . 6300-T-. 0022 (2.) 6300-T-22 = 286^ = 386.36 Explanation. — 1. We reduce, as shown in (1), the dividend and divisor to the same decimal unit by annexing ciphers to the dividend (850). 2. We divide, according to (290), as shown in (2), the numerator 6300 by the numerator 22, givinpr as a quotient 286y\. 3. We reduce, according to (338), the y\ in the quotient to a decimal, giving the repetend .36. Hence, .63 -*- .0022 = 286.36. i DECIMAL FRACTIONS Ex.3. Divide 16.821 by 2.7. (1.) 16. 821 -V- 2.7 = 16 .821-T-2 (2.) 16. 821 -^ 2.700 = 16821 1000 * (3.) 2 7 1 ) 1 6 8| 21(6. 23 EXPLAI 162 duce, as 81 56 J. 700 27 00 ' 1000 isfATiON. — 1. We re- shown in (1), the dividend and divisor to the " '^ same decimal unit by annex. 5 4 ing ciphers to the divisor "~g"Y (350). 2. The dividend and divisor each express thousandths as shown in (2). Hence we reject the denominators and divide as in integers (1290). 3. Since there are ciphers at the right of the divisor, they may be cut off by cutting off the same number of figures at the right of the dividend (142). Dividing by 37, we find that it is contained 6 times in 1G8, with G remaining. 4. The 6 remaining, with the two figures cut off, make a remainder of 621 or /7V5. This is reduced to a decimal by dividing both terms by 27. Hence, as shown in (3), we continue dividing by 27 by taking down the two figures cut off. The work is ubbreyiated thus: We reduce the dividend and divisor to the same decimal unit by cut- ting off from the right of the dividend the figures that express lower decimal units than the divisor. We then divide as shown in (3), prefix- ing the remainder to the figures cut off and reducing the result to a decimal. REVIEW EXAMPLES. 364. Answers involving decimals, unless otherwise stated, are carried to four decimal places. Reduce each of the following examples to decimals : 8. U. 11 31 14. (3H-i)xl 7^ 8 12. foflf. 15. f of .3 13. 5A-5|. 8| - 4.3 10. ^. 20. Four loads of hay weighed respectively 2583.07, 3007? , 2567|, and 3074}^^ pounds ; what was the total weight ? DECIMAL FRACTIO NS. 57 22. At $1.75 per 100, what is the cost of 5384 oranges ? 24. If freight from St. Louis to New York is $.39* per 100 pounds, what is the cost of transporting 3 boxes of goods, weighing respectively 7831, 32o|, and 288| pounds? 25. A piece of broadcloth cost $195. 38 1, at f3.27 per yard. How many yards does it contain ? 26. A person having $1142.49|, wishes to buy an equal number of bushels of wheat, corn, and oats ; the wheat at |1.37, the corn at $.87|, and the oats at $.35|. How many bushels of each can he buy ? 20. A produce dealer exchanged 48| bushels oats at 39| cts. per bushel, and 13 1 barrels of apples at $3.85 a barrel, for butter at 37| cts. a pound ; how many pounds of butter did he receive ? 30. A grain merchant bought 1830 bushels of wheat at $1.25 a bushel, 570 bushels corn at 784^ cts. a bushel, and 468 bushels oats at 35f cts. a bushel. He sold the wheat at an advance of 11 \ cts. a bushel, the corn at an advance of 9| cts. a bushel, and the oats at a loss of 3 cts. a bushel. How much did he pay for the entire quantity, and what was his gain on the transaction ? EEVIEW AND TEST QUESTIONS. 365. 1. Define Decimal Unit, Decimal Fraction, Repetend, Cir- culating Decimal, Mixed Circulating Decimal, Finite Decimal, and Complex Decimal. 2. In how many ways may | be expressed as a decimal fraction, and why? 3. What effect have ciphers written at the left of an integer ? At the left of a decimal, and why in each case (316) ? 4. Show that each figure in the numerator of a decimal represents a distinct order of decimal units (320). 5. How are integral orders and decimal orders each related to the units (323) ? Illustrate your answer by lines or objects. 6. Why in reading a decimal is the lowest order the only one named ? Illustrate by examples (321). 7. Give reasons for not regarding the ciphers at the left in reading the numerator of the decimal ,000403. 8. Reduce l to a decimal, and give a reason for each step in the process. 9. When expressed decimally, how many places must J^\ give, and why ? How many must ^'^ give, and why ? 10. Illustrate by an example the reason why ^\ cannot be expressed as a simple decimal (327). 58 DECIMAL FRACTIONS. 11. State wiiat fractions can and what fractions cannot be expressed as simjile decimals (31i5(> and 3ii7). Illustrate by examples. 12. In reducing f to a complex decimal, why must the numerator 5 recur as a remainder (r528 — 1 and 2j ? 13. Show that, according to (ii35 — II and III), the value of ^| will not be changed if we diminish the numerator and denominator each by I of itself. 14. Show that multiplying 9 by I5 increases the 9 by | of itself. 15. Multiplying the numerat(n- and denominator of ^| each by 1| produces what change in the fraction, and why ? 16. Show that in diminishing the numerator of | by f and the denominator by 1 we diminish each by the same part of itself. 17. In taking .3 as the value of 3,, what fraction has been rejected from the numerator? What must be rejected from the denominator to make .3 = -J, and why V 18. Show that the true value of .81 is f J^. Give a reason for each step. 19. Explain the process of reducing a mixed circulating decimal to a fraction. Give a reason for each step. 20. How much is .33333 less than i and why? 21. How much is .571428 less than |, and why? 22. Find the sum of .73, .0049, .089, 6.58, and 9.08703, and explain each step in the process (201 — I and II). 23. If tenths are multiplied by hundredths, how many decimal places will there be in the product, and why (3155) ? 24. Show that a number is multiplied by 10 by moving the decimal point one place to the right ; by 100 by moving it two places ; by 1000 three places, and so on. 25. State a rule for pointing off the decimal places in the product of two decimals. Illustrate by an example, and give reasons for your rule. 26. Multiply 385.28 by .742, multiplying first by the 4 hundredths, then by the 7 tenths, and last by the 2 tJwusandths. 27. Why is the quotient of an integer divided by a proper fraction greater than the dividend ? 28. Show that a number is divided by 10 by moving the decimal point one place to the left ; by 100 by moving it two places ; by 1000, three places ; by 10000, four places, and so on. 29. Divide 4.9 by 1.305, and give a reason for each step in the process. Carry the decimal to three places. 30. Give a rule for division of decimals. DEX0 3IINATE NUMBERS. 59 DElsrOMIKATE l^UMBEES. DEFINITIONS. 366. A Helated Unit is a unit which has an invariable relation to one or more other units. Thus, 1 foot = 13 inches, or i of a yard ; hence, 1 foot has an invaria- ble relation to the units inch and yard, and is therefore a r dated unit. 367. A Deno^ninate dumber is a concrete number (15) whose unit (14) is a related unit. Thus, 17 yards is a denominate number, because its unit, yard, has an invariable relation to the waits, foot and inch, 1 yard making always 3 feet or 38 inches. 368. A JDenominate Fraction is a fraction of a related unit. Thus, f of a yard is a denominate fraction. 369. The Orders of related units are called Denomi" nations. Thus, yards, feet, and inches are denominations of length ; dollars, dimes, and cents are denominations of money. 370. A Coinpoiind Number consists of several num- bers expressing related denominations, written together in the order of the relation of their units, and read as one number. Thus, 23 yd. 2 ft. 9 in. is a compound number. 371* A Standard Unit is a unit established by law oc custom, from which other units of the same kind are derived. Thus, the standard unit of measures of extension is the yard. By dividing the yard into three equal parts, we obtain the umifoot ; into 36 equal parts, we obtain the unit inch ; multiplying it by 5^, we obtain the unit rod, and so on. 373. Reduction of Denominate Numbers is the process of changing their denomination without altering their value. TABLE OP UKTTS. 24: gr. = 1 pwt 20 2)wL =z 1 OZ. 12 oz. =: 1 Ih, S.2gr. = 1 carat. 60 DENOMINATE NU3IBERS. UI^ITS OF WEIGHT. 374. The Troy pound of the mint is the Standard Unit of weight. TROY WEIGHT. 1. Denominations, — Grains {gr.\ Pennyweights {pwt^, Ounces (02.), Pounds (Jb.), and Carats. 3. Equivalents.— 1 lb. = 13 oz =r 340 pwt. = 5760 gr. 3. Use, — Used in weighing gold, silver, and precious stones, and in philosophical experiments. AVOIRDUPOIS WEIGHT. TABLE OP UNITS. 1. Denominations, — Ounces {oz,\ 16 oz. = 1 lb, pounds (^6.), hundredweights (cic^.), tons (71). -^^Q jlf __ 2 c2Vf ^' -KQ'^**^^^^**^*'— 1 Ton = 20 cwt. = ^^ * -, rv ' 3000 lb. = 33000 oz. 20 CWi. = 1 T. o xr TT J • • I,- 3. iJse. — Used in weighing groceries, drugs at wholesale, and all coarse and heavy articles. 4. In the United States Custom House, and in wholesale transactions in coal and iron, 1 quarter = 38 lbs., 1 cwt. = 113 lb., 1 T. r:r 3340 lb. This is usually called the Long Ton table. APOTHECARIES' WEIGHT. TABLE OF UNITS. 1. Denominations. — Grains ^gr.), 20 gr, =zlsc,OV 3, Scruples O), Drams (3), Ounces (3), 3 3 =ldr.ov z. ^^^^^^^ ^^''•)- o ^ _ 1 ^^ ^^ z 3- JEquiva7ei,ts.-lb. 1 = 112 =zm » 3 — IOZ.OT t. =3 288 = gr. 5760. iZ OZ. = 1 to. 3 jjsf>^ — Used in medical prescriptions. 4. Medical prescriptions are usually written in Roman notation. The number is written after the symbol, and the final " i " is always written j. Thus, 3 vij is 7 ounces. DENOMINATE NUMBERS. 61 Comparative Table of Units of WeighU TROT. AVOIRDin?OIS. APOTHE( 1 pound = 5760 grains = 7000 grains = 5760 grains. 1 ounce = 480 " = 437.5 " = 480 " Table of Avoirdupois Pounds in a Bushel, as Established by Law in the States named. Wheat Indian Corn. . . Oat? Barley Buckwheat Rye Clover Seed . . . Timothy Seed. 60 60 52:56 32 32 48'48 40;50 54 56 oieo 45:4) 60 60 |60 50|56 56 35|33J;32 48|48 52 52 5656 60|60 45:45 s:^!^ 160 60 156 56 30 30 32 46 48 46 42 56 56 160 60 60 60 56:58 30 1 32 48:48 50 48 56 56 64i60 44 56 60 60 60 60 5(5 56 5656 34 32 32 36 46 47 4645 42 48 4642 5656 56 56 601 I ieo 54 50 Peas, Beans, and Potatoes are usually weighed 60 lb. to the bushel. The following are also in use : 100 lb. of Grain or Flour = 1 Cental. 200 lb. Pork or Beef = 1 Barrel. 100 " of Dry Fish = 1 Quintal. 196 " Flour = 1 Barrel. 100 " of Nails = 1 Keg. 240 " Lime = 1 Cask. 280 lb. salt at N. Y. Salt Works - 1 Barrel. PROBLEMS ON RELATED UOTTS. 375. Prob. L — To reduce a denominate or a com- pound number to a lower denomination. Reduce 23 lb. 7 oz. 9 pwt. to pennyweights. 2 3 lb. 7 oz. 9 pwt. Solution. — 1. Since 12 oz. make 1 lb,, in any -^ 2 number of pounds there are 12 times as many ounces as pounds. Hence we multiply the 23 lb. by 12, and add the 7 oz., giving 283 oz. 2. Again, since 20 pwt. make 1 oz., in any number of ounces there are 20 times as many pennyweights as ounces. Hence we multiply 2 8 3 OZ. 20 5 6 6 9 pwt. the 283 oz. by 20, and add the 9 pwt., giving 5669 pwt. 6» DE NOMINATE NU3IBERS. 378. Prob. II. — To reduce a denominate number to a compound or a higher denominate number. Eeduce 7487 so. to a compound number. 3 ) 7 4 8 7 sc. Solution.— Since 3 sc. make 1 dr., o \n A qT t , n 7487 sc. must make as many drams as ) ±±^ ar. -f- ^ SO. g jg contained times in 7487, or 2495 dr. 1 2 )311 oz. + 7 dr. +2 sc. 2 5 lb 11 07 ^' ^^"^® ^ ^^' ™^^6 1 oz., 2495 dr. must make as many ounces as 8 is con- tained times in 2495, or 311 oz. + 7 dr. 3. Since 12 oz. make 1 lb., 311 oz. must make as many pounds as 12 is contained times in 311, or 25 lb. + 11 oz. Hence, 7487 sc. are equal to the compound number 25 lb. 11 oz. 7 dr. 2 sc. 381. Prob. III. — To retluce a denominate fraction or decimal to integers of lower denominations. Keduce f of a ton to lower denominations. (1.) f T. = f of 20 cwt. = f X 20 = 14 cwt. + f cwt. (2.) f cwt. = -I of 100 lb. = f X 100 := 28 lb. + 4 lb. (3.) 4- lb. = 4 of 16 oz. = 4 X 16 = 91 oz. Solution. — Since 20 cwt. is equal 1 T., f of 20 cwt., or 14f cwt., equals f of 1 T. Hence, to reduce the f of a ton to hundredweights, we take f of 20 cwt., or multiply, as shown in (1), the f by 20, the number of hundredweights in a ton. In the same manner we reduce the f cwt. remaining to pounds, as shown in (2), and the ^ lb. remaining to ounces, as shown in (3). 384. Prob. IY. — To reduce a denominate fraction or decimal of a lower to a fraction or decimal of a higher denomination. Eeduce f of a dram to a fraction of a pound. (1.) I dr. = \oz. X I = tV oz. (2.) ^ oz. = ^ lb. X A = -ih lb. Solution.— 1. Since 8 drams = 1 ounce, 1 dram is equal I of an oz., and § of a dram is equal ^ of | oz. Hence, as shown in (1), f dr.= -^^ oz. DENOMINATE NUMBERS, 63 2. Since 12 ounces = 1 pound, 1 ounce is equal ^^ of a pound, and, as shown in (2), 5% of an ounce is equal ^q of ^^ lb., or y^^ lb. Hence, ? dr. 387. Pkob. V. — To reduce a compound number to a fraction of a higher denomination. Reduce §4 36 32 to a fraction of 1 pound. (1.) ! 4 3 6 33 = 3116; lb. 1 = 3288. (2.) iM = f I ; hence, 1 4 3 6 32 = lb. ff- Solution. — 1. Two numbers can be compared only when they are the same denomination. Hence we reduce, as shown in (1), the 3 4 3 6 32 and the lb. 1 to scruples, the lowest denomination mentioned in either number. 2. §4 36 32 being equal 3116, and lb. 1 being equal 3288, §4 36 32 is the same part of lb. 1 as 3 116 is of 3288, which is |^f, or |f. Hence §4 3 6 32 = lb. f |, or .004027. 15. Reduce 8 cwt. 3 qr. 16 lb. to the decimal of a ton. a K\-\ c Oftlh Abbreviated Solution.— Since the 16 I '. * pounds are reduced to a decimal of a quar- 4)3.64 qr. ter by reducing ^ to a decimal, we annex o /\ \ o qT __^i. two ciphers to the 16, as shown in the mar- ' — '- * gin, and divide by 25, giving .64 qr. . 4 4 5 5 T. To this result we prefix the 3 quarters, 3 64 giving 8.64 qr., which is equivalent to -^ hundredweights ; hence we divide by 4, as shown in the margin, giving .91 cwt. To the result we again prefix the 8 c^vt., giving 8.91, which is equiva- 8 91 lent to -'^ of a ton, equal .4455 T. Hence, 8 cwt. 3 qr. 16 lb. = .4455 T. 16. Reduce 8 oz. 6 dr. 2 sc. to the decimal of a pound. 17. What decimal of 24 lb. Troy is 2 lb. 8 oz. 16 pwt.? 18. 9 oz. 16 pwt. 12 gr. are what decimal of a pound? 19. Reduce 12 cwt. 2 qr. 18 lb. to the decimal of a ton. 20. What decimal of a pound are § 9 3 5 32 gr. 18 ? 21. Reduce 11 oz. 16 pwt. 20 gr. to the decimal of a pound. 22. Reduce 7 lb. 5 oz. Avoir, to a decimal of 12 lb. 5 oz. 3 pwt. Troy. 23. 1 lb, 9 oz. 8 pwt. is what part of 3 lb. Apoth, weight ? 64 DENOMINATE NUMBERS. 390. Prob. VI. — To find the sum of two or more de- nominate or compound numbers, or of two or more denominate fractions. 1. Find the sum of 7 cwt. 84 lb. 14 oz., 5 cwt. 97 lb. 8 oz., and 2 cwt. 9 lb. 15 oz. cwt. lb. oz. Solution. — 1. We write numbers of the 7 8 4 14 same denomination under each other, as shown c Q w g in the margin. jj 1^ ^' ^® ^^^ ^^ ^^ Simple Numbers, com- mencing with the lowest denomination. Thus, 15 9 2 5 15, 8, and 14 ounces make 37 ounces, equal 2 lb. 5 oz. We write the 5 oz. under the ounces and add the 2 lb. to the pounds. We proceed in the same manner with each denomination until the entire sum, 15 cwt. 92 lb. 5 oz., is found. 2. Find the sum of ^ lb., f dr., and j sc. Solution.— 1. According to (261), only fractional units of the same kind and of the same whole can be added ; hence we reduce f lb., f dr., and f sc. to integers of lower denomi- 5 3 2 3 nations (381), and then add the results, as shown in the margin. Or, 2. The given fractions may be reduced to fractions of the same de- nomination (384), and the results added according to (261), and the value of the sum expressed in integers of lower denominations according to (381). 393. Prob. VII. — To find the difference between any two denominate or compound numbers, or denominate fractions. Find the difference between 27 lb. 7 oz. 15 pwt. and 13 lb. 9 oz. 18 pwt. Solution. — 1. We write numbers of the same denomination under each other. 2. We subtract as in simple numbers. W^hen the number of any denomination of the subtra- 13 9 17 hend cannot be taken from the number of tlie oz. dr. sc. gr. -Jib. = 5 2 2 idr. = 2 8 } sc. = 15 lb. oz. pwt. 27 7 15 13 9 18 DENOMINATE NUMB ERS. G5 same denomination in the minuend, we add as in simple numbers (05 — III) one from the next higher denomination. Thus, 18 pwt. can- not be taken from 15 pwt. ; we add 1 of the 7 oz. to the 15 pwt., making 35 pwt. 18 pwt. from 35 pwt. leaves 17 pwt., which we write under the pennyweights. We proceed in the same manner with each denomination until the entire difference, 13 lb. 9 oz. 17 pwt., is found. To subtract denominate fractions, we reduce as directed in addition, and then subtract. 394. Peob. VIII. — To multiply a denominate or com- pound number by an abstract number. Multiply 18 cwt. 74 lb. 9 oz. by 6. 18 cwt. 74 lb. 9 oz. Solution.— We multiply as a in simple numbers, commencing with the lowest denomination. 5 T. 12 cwt. 47 lb. 6 oz. Thus, 6 times 9 oz. equals 54 oz. We reduce the 54 oz. to pounds (378), equal 3 lb. 6 oz. We write the 6 oz. under the ounces, and add the 3 lb. to the product of the pounds. We proceed in this manner with each denomination until the entire product, 5 T. 12 cwt. 47 lb. 6 oz., is found. 396. Prob. IX. — To divide a denominate or com- pound number by any abstract number. Divide 29 lb. 7 oz. 2 dr. by 7. lb. oz. dr. Solution.— 1. The object of the division 7)29 7 2 is to find i of the compound number. This 7 ^ I is done by finding the \ of each denomina- tion separately. Hence the process is the same as in finding one of the equal parts of a concrete number. Thus, the | of 29 lb. is 4 lb. and 1 lb. remaining. We write the 4 lb. in the quotient, and reduce the 1 lb. to ounces, which added to 7 oz. make 19 oz. We now find the } of the 19 oz., and proceed as before. 1. Divide 9 T. 15 cwt. 3 qr. 18 lb. by 2 ; by 5 ; by 8 ; by 12. 2. If 29 lb. 7 oz. 16 pwt. are made into 7 equal parts, how much will there be in each part ? 66 DENOMINATE NUMBERS. TJITITS OF LENGTH. 397. A yard is the Standard Unit in linear, surface, and solid measure. LINEAR MEASURE. TABLE OF UNITS. 1. Denominations, — Inches (in.), Feet 12 in, = 1ft, (ft.), Yards (yd.), Rods (rd.), Miles (mi.). S ft z=z 1 yd ^' Equivalents,—! mi. = 320 rd. = 5280 6'^ v'd — Ird ft. = 63360 in. ^ " * * 3. Use. — Used in measuring lines and dis- 320 rd. = 1 7iii, tances. 4. In measuring cloth, the yard is divided iato Tialves, fourtlis, eighths, &nd sixteenths. In estimating duties in the Custom House, it is divided into tenths and hundredths. Table of Special Denominations, 60 Geographic or) _ ^ n 3 ^^ Latitude on a Meridian or of 69.16 Statute Miles S ~ ^^^^ \ Longitude on the Equator. 360 Degrees = the Circumference of the Earth. 1.16 Statute Miles = 1 Geog. Mi, Used to measure distances at sea. 3 Geographical Miles= 1 League ^ Feet = 1 Fathom. Used to measure depths at sea. Used to measure tlie height of horses 4 Inches = 1 Hand. \ at the shoulder SURVEYORS' LINEAR MEASURE. TABLE OF UNITS. 1. Denominations,— Itiiiis. (1.), Rod (rd.), 7.92 in. = 1 Z. Chain (ch.). Mile (mi.). 25 Z =1 rd. ^- Equivalents,—! mi. -80 ch. = 320 rd. A */7 — 1 7 * =8000 L ^ra, — I Ctl.^ g jj^^^^ _ jj^^^ jjj measuring roads and 80 cTl. = 1 mi. boundaries of land. 4. The Unit of measure is the Ounter's Chain, vjrhich contains 100 links, equal 4 rods or 66 feet. DENOMINATE NUMBERS, 67 UE'ITS OF SUEFACE. 399. A square yard is the Standard Unit of surface measure. 400. A Surface has two dimensions — length and hreadth, 401. A Square is a jyZa/^e surface bounded by four equal lines, and having four right angles. 402. A Mectangle is any plane surface having four sides and four right angles. 403. The Unit of Measure for surfaces is usually a square, each side of which is one unit of a known length. Thus, in 14 sq. ft., tlie unit of measure is a square foot. 404. The Area of a rectangle is the surface included within its boundaries, and is expressed by the number of times it contains a given U7iit of measure. Thus, since a square yard is a surface, each side of which is 3 feet long, it can be divided into 3 rows of square feet, as shown in the diagram, each row containing 3 square feet. Hence, if 1 square foot is taken as the Unit of Measure, the area of a square yard is 3 sq. ft. X 3 = 9 sq. ft. The area of any rectangle is found in the same manner. 3 sq. ft, x 3 = -9 sq. ft. 3 feet long. SQUARE MEASURE. TABLE OF UNITS. 144 sq. in. ^ sq.ft. 30^ 6-^. yd. 100 sq. rd. 640^. 1 sq.ft. 1 sq. yd. 1 sq. rd., or P. 1 sq. mi. 1. Denortiinntioiis. — Square Inch (sq. in.). Square Yard (sq. yd.), Square Rod (sq. rd.), Acre (A.), Square Mile (sq. mi.). 2. Eqifivalents. — 1 sq. mi. = 640 A. = 102400 sq. rd. = 3097600 sq. yd.= 27878400 sq. ft. = 4014489000 sq. in. TABLE OF UNITS. 625 sq. I. IP. 16 P. z=z 1 sq. cli. 10 sq. cli. \A. 640 X ; 1 sq. mi. 36 sq. mi. : 1 Tp. 68 DENOMINATE NUMBERS. 3. Use, — Used in computing areas or surfaces. 4. Glazing and stone-cutting are estimated by the square foot ; plaster- ing, paving, painting, etc. , by the square foot or square yard ; roofing, flooring, etc., generally by the square of 100 square feet. 5. In laying shingles, one thousand, averaging 4 inches wide, and laid 5 inches to the weather, are estimated to cover a square. SURVEYOKS' SQUARE MEASURE. 1. Denotninatiotis.—Sqa&Te Link (sq. 1.), Poles (P.), Square Chain (sq. ch.), Acres (A.), Square Mile (sq. mi.), Town- ship (Tp.). 2. Equivalents, — 1 Tp.=36 sq. mi. =23040 A.=230400 sq. ch. = 3686400 P. = 2304000000 sq.l. 3. Use. — Used in computing the area of land. 4. The Unit of land measure is the acre. The measurement of a tract of land is usually recorded in square miles, acres, and hundredths of an acre. insriTS OF VOLUME. 408. A Solid or Volume has three dimensions — length, hreadtli, and thichness. 409. A Mectangular Solid is a body bounded by six rectangles called faces. 410. A Cube is a rectangular solid, bounded by six equal squares. 411. The Unit of Measure is a cube whose edge is a unit of some known length. 412. The Volume^ or Solid Contents of a body is expressed by the number of times it contains a given unit of measure. For- example, the contents "of a cubic yard is expressed as 27 cuUcfeet. DENOMINATE NUMBERS. G9 Thus, piuce each face of a cubic yard contaius 9 square feet, if a section 1 ft. thick is taken it must contain 3 times 3 cu. ft., or 9 cu. ft. , as shown in the diagram. And since the cubic yard is 3 feet thick, it must contain 3 sec tions, each containing 9 cu. ft., which is 37 cu. ft. Hence, the volume or contents of a cubic yard expressed in cubic feet, is found by taking the product of the numbers denoting its 3 dimensions in feet. The conterds of any rectangular solid is found in the same manner. ^ CUBIC MEASURE. TABLE OF UNITS. 1728 CU. in, = 1 cu.ft. 27 cu. ft. = 1 cti. yd. 1. Denominations. — Cubic Inch (cu. in.), Cubic Foot (cu. ft.), Cubic Yard (cu. yd.). 2. Eqnivnleiits.—l cu. yd. = 27 cu. ft. = 46656 cu. in. 3. Use, — Used in computing the volume or contents of solids. Table of Units for Measuring Wood and Stone. IG cu.ft. = 1 Cord Foot (cd.fi.) \ Used for raeasur- 8 cd.ft. ^^\ — I n .A( fj) \ i^g ^^^^i 128 cu. ft. ) ~ ) wood and stone. 24J cu.ft. = 1 perch {pcli.) of stone or masonry. 1 cu. yd. of earth is called a load. 1. The materials for masonry are usually estimated by the cord or percli, the work by the perch and cubic foot, also by the square foot and square yard. 2. In estimating the mason work in a building, each wall is measured on the outside, and no allowance is ordinarily made for doors, windows, and cornices, unless specified in contract. In estimating the material, the doors, windows, and cornices are deducted. 70 DENOMINATE NU3[BERS. 3. Brickwork is usually estimated by the thousand bricks. Tlie size of a brick varies thus : North River bricks are 8 in. x 3| x 2|, Philadelphia and Baltimore bricks are 8| in. x 4| x 2|, Milwaukee bricks 8| in. x 4| x 2|, and Maine bricks 7^ in. x 3f x 2|. 4. Excavations and embankments are estimated by the cubic yard. BOAKD MEASUEE. TABLE OF UNITS. 12 B. in. = 13. ft. 12 B. ft. = 1 cu. ft. 418. A Board Foot is 1 ft. long, 1 ft. wide, and 1 in. thick. Hence, 12 board feet equals 1 cu. ft. 419. A Board Inch is 1 ft. long, 1 in. wide, and 1 in. thick, or yV ^^ ^ board foot. Hence, 12 board inches equals 1 board foot. Observe carefully the following : 1. Diagram 1 represents a board (1) 4 feet long. Square loot. ^ « 4 X 2 = 8 sq. ft. or 8 B. ft. (2.) 4 feet long. where both dimensions are feet. Hence the product of the two di- mensions gives the square feet in surface (405), or the number of board feet when the lumber is not more than 1 inch thick. 2. Diagram (2) represents a board where one dimension is feet and the other inches. It is evident (418) that a board 1 foot long, 1 inch thick, and any number of inches wide, contains as many board inches as there are inches in the width. Hence the number of square feet or board feet in a board 1 inch thick is equal to the length in feet multiplied by the width in inches divided by 12, the number of board inches in a board foot. 3. In case the lumber is more than 1 inch thick, the number of board feet is equal to the number of square feet in the surface multiplied by the thickness. 16. Find the length of a stick of timber 8 in. by 10 in., which will contain 20 cu. ft. Operation.— (1728 x 20)-*-(8 x 10)=432; 432-5-12 = 36 ft., the length. 1 ft. by 9 in. °* 4x9=36B.iu.; 36B.in.-i-12=3B.ft. TABLE OF UNITS. 60 sec. = 1 771171. 60 min. = 1 hr. 24 hr. = Ida. 1 da. := 1 ivk. 365 da. = 1 common yr. 366 da. = 1 /e«;? ?^r. 100 yr. 1= 1 ceil. DE NO 311 NATE NUMBERS. 71 UNITS OF TIME. 424. The mean solar day is the Standard Unit of time. 1. Denominations. — Seconds (sec). Minutes (min.), Hours (hi*.), Days (da.), Weeks (wk.), Months (mo.), Years (yr.), Centuries (cen.). 3. There are 12 Calendar Months in a year; of these, April, June, September, and November, have 30 da. each. All the other months except February have 31 da. each. February, in common years, has 28 da., in leap years it has 29 da. 3. In computing interest, 80 days are usually considered one month. For business purposes the day begins and ends at 12 o'clock midnight. 425. The reason for commo7i and leap years will be seen from the following : The true year is the time the earth takes to go once around the mn, which is 365 days, 5 hours, 48 minutes and 49.7 seconds. Taking 385 days as a common year, the time lost in the calendar in 4 years will lack only 44 minutes and 41.2 seconds of 1 day. Hence we add 1 day to February every fourth year, making the year 366 days, or Ijeap Year, This correction is 44 min. 41.2 sec. more than should be added, amounting in 100 years to 18 hr. 37 miti. 10 sec. ; hence at the end of 100 years we omit adding a day, thus losing again 5 hr. 22 min. 50 sec, which we again correct by adding a day at the end of 400 years. How many yr., mo., da. and hr. from 6 o'clock P. M., July 19, 1862, to 6 o'clock A. M., April 9, 1876. Solution. — 1. Since the latter date denotes the greater period of time, it is the minuend, and the earlier date, the subtrahend. 13 8 19 13 2. Since each year commences with yr. mo. da. hr. 1876 4- 9 7 1862 7 19 18 n DENOMINATE NUMBERS. January, and each day with 12 o'clock midnight, 7 o'clock A. M., April 9, 1876, is the 7th hour of the 9th day of the fourth month of 1876 ; and 6 o'clock p. M., July 19, 1862, is the 18th hour of the 19th day of the 7th month of 1862. Hence the minuend and subtrahend are written as shown in the margin. 3. Considering 24 hours 1 day, 30 days 1 month, and 12 months 1 year, the subtraction is performed as in compound numbers (892), and 13 yr. 8 mo. 19 da. 13 hr. is the interval of time between the given dates. CIECULAR MEASURE. 428. A Circle is a plane figure bounded by a curved line, all points of which are equally distant from a point within called the centre. 429. A Circumference ia the line that bounds a circle. 430. A Degree is one of the 360 equal parts into which the circumference of a circle is sup- posed to be divided. 431. The degree is the Standard Unit of circular measure. 1. Denominations.— Seconds ("), Minutes 0, Degrees (°), Signs (S.), Circle (Cir.). 2. One-half of a circumference, or 180°, as shown by the figure in the margin, is called a Semi-circumference; One-fourth, or 90°, a Quad- rant; One-sixth, or 60°, a Sextant; and One- twelfth, or 30°, a Sign. 3. The length of a degree varies with the size of the circle, as will be seen by examining the foregoing diagram. 4. A degree of latitude or a degree of longitude on the Equator is 69.16 statute miles. A minute on the earth's circumference is a geograph- ical or nautical mile. TABLE OF UNITS. 60" == 1' 60' = 1° 30° =18. 12 /S'. = 1 Cir. 360° = 1 Cir. DENOMINATE NUMBERS, 73 SPECIAL UNITS. Table for Paper, 24 Sheets = 1 Quire. 30 Quires = 1 Ream. 2 Reams = 1 Bundle. 5 Bundles = 1 Bale. Table for Counting, 12 Things = 1 Dozen (doz.) 12 Dozen = 1 Gross (gro.) 12 Gross = 1 Great Gross (G. Gro.) 20 Things = 1 Score (Sc.) UI^ITS OF MO]NET UNITED STATES MONEY. 433. The dollar is States money. TABLE OF UNITS. 10 m. — 1 ct. 10 ct. = 10 d. = the Standard Unit of United Id. 1. Denominations.— Mills (m,). Cents (ct.), Dunes (d.), Dollars ($), Eagles (E.). 2. The United States coin, as fixed by the Coinage Acts of 1873 and 1878, is as follows; Goldf the double-eagle, eagle, half-eagle, quar- $10 = 1 ^. ter-eagle, three-dollar, and one-dollar; Silver, the trade-dollar, dollar, half-dollar, quarter-dollar, and dime ; Nickel, the five-cent and three-cent ; Bronze, one-cent. 3. Composition of Coins. — Gold coin contains .9 pure gold and .1 silver and copper. Silver coin contains .9 pure silver and .1 pure copper. Nickel coin contains .25 nickel and .75 copper. Bronze coin contains .95 copper and .05 zinc and tin. 4. The Trade-dottar weighs 420 grains and is designed for commercial purposes solely. The silver Dollar weighs 412^ grains. CANADA MONEY. 434. 1. Denominations. — Mills, Cents, and Dollars. These have the same nominal value as in United States Money. 2. The Coin of the Dominion of Canada is as follows : Gold, the coins in use are the sovereign and half-sovereign ; Silver, the fifty-cent, twenty-five cent, ten-cent, and five-cent pieces ; Bronze, the one-cent piece. 74 DENOMINATE NUMBERS. ENGLISH MONEY. 435. The pound sterling is the Standard Unit of Enghsh money. It is equal to $4.8665 United States money. TABLE OP UNITS. 1. Denominations, — Farthings (far,), ^ r^^^ = 1 ^. Pennies (d.), Shillings (s,), Sovereign (sov.), -j 2 ^ -\ g Pound (£), Florin (fi.), Crown (cr.). . ' a ^- "^^^ Coins in general use in Great 20 5. = -^ Britain are as follows : Gold, sovereign and ( or £1. half-sovereign ; Silver, crown, half-crown, 2 5. z=z Ifl, florin, shilling, six-penny, and three-penny ; Kg -— -^ Q,f.^ Coj>l)er, penny, half -penny, and farthing. FRENCH MONEY. 436. The silver franc is the StaJidard Unit of French money. It is equal to $.193 United States money. 1. Denominations. — Millimes (m.), Cen- TABLE OF UNITS. ^j^^^ ^^^^^^ Decimes (dc). Francs (fr.). 10 7n. z= 1 ct. 2. Equivalents,—! fr. = 10 dc. = 100 ct.= 10 ct = 1 dc. 1000 m. ' 10 dc. = 1 fr. 3. The Coin of France is as follows : Gold, 100, 40, 20, 10, and 5 francs; Silver, 5, 2, and 1 franc, and 50 and 25 centimes ; Dronze, 10, 5, 2, and 1 centime pieces. GERMAN MONEY. 437. The mark is the Standard Unit of the Mw Ger- man Empire. It is equal to 23.85 cents United States money, and is divided into 100 equal parts, one of which is called a Pfenniff. 1. The Coins of the Neio Empire are as follows : Gold, 20, 10, and 5 marks ; Silver , 2 and 1 mark ; Nickelf 10 and 5 pfennig. 2. The coins most frequently referred to in the United States are the silver Thaler, equal 74G cents, and the silver Groschen, equal 2^ cents. DENOMINATE NUMBERS. 75 THE METRIC SYSTEM. 439. The 3Iefric System of Related Units is formed according to ine decimal scale. 440. The Meter, which is 39.37079 inches long, or nearly one ^■en-millionth of the distance on the earth's surface from the equator to the pole, is the base of the system. 441. The I*rlmary or JPtincipal Units of the system are the MeteVy the Are (air), the Stere (stair), the Liter (leeter), and the Gram. All other units are multiples and sub-multiples of these. 442. The names of Multiple Units or higher denominations are formed by prefixing to the names of the iirimary units the Greek nu- merals Dekc 10), Hecto (100), Kilo (1000), and Myria (10000). 443. The names of Snh-multiple UnitSf or lower denomina- tions, are formed by prefixing to the names of the primary units the Latin numerals, Beci (yV), Centi (x^ir), and Milli (ttjW- ¥NITS OF LENGTH. 444. The Meter is the prindpal unit of length. TABLE OF UNITS. 10 Millimeters, mm. = 1 Centimeter = .3937079 in. 10 Centimeters, cm. — 1 Decimeter = 3.937079 in. 10 Decimeters, dm. — 1 Meter = 39.37079 in. 10 Meters, M. = 1 Dekameter =r 32.808992 ft. iO Dekameters, Dm,. = 1 Hectometer = 19.884237 rd. 10 Hectometers, Hm. = 1 Kilometer = .6213824 mi. 10 Kilometers, Kw.. = 1 Myriameter (Mm.) = 6.213824 mi. The meter \B used in place of one yard in measuring cloth and short distances. Long distances are usually measured by the kilometer. UNITS OF SITEFACE. 445* The Square Meter is the principal unit of surfaces. TABLE OF UNITS. 100 Sq. Millimeters, sq. mm. = 1 Sq. Centimeter = .155+ sq. in. 100 Sq. Centimeters, sq. cm. = 1 Sq. Decimeter = 15.5 + sq. in. 100 Sq. Decimeters, sq. dm. --== 1 Sq, Meter {Sq. M.) = 1.193+ sq.yd 76 DENOMINATE NUMBERS, 446. The Are, a square whose side is 10 meters, is the principal unit for measuring land. TABLE OF UNITS. 100 Centiares, ca. — 1 Are = 119.6034 sq. yd. 100 Ares, A. = \ Hectare {Ha.) = 2.47114 acres. UNITS OF VOLUME. 44*7. The Cubic Meter is the principal unit for measuring ordi- nary solids, as embankments, etc. TABLE OF UNITS. 1000 Cu. Millimeters, cu. mm. = 1 Cu. Centimeter = ,061 cu. in. 1000 Cu Centimeters, cu. cm. = 1 Cu. Decimeter = 61.026 cu. in. 1000 Cu. Decimeters, cu. dm. = 1 Cu. dieter = 35.316 cu. ft. 448. The Stere, or Cu. Meter, is the principal unit for measuring wood. TABLE OF UNITS. 10 Decisteres, (?5^. = 1 Stere — 35.316+ cu. ft. 10 Steres, St. = 1 Dekastere (Z)«^.) = 13.079+ cu. yd. UNITS OF CAPACITY. 449. The Liter is the principal unit both of Liquid and Dry Measure. It is equal to a vessel whose volume is equal to a cube whose edge is one4enth of a meter, TABLE OF UNITS. 10 Milliliters, ml. = 1 Centiliter = .6102 cu. in. = .338 fl. oz. 10 Centiliters, d. = 1 Deciliter = 6.1022 " " = .845 gill. 10 Deciliters, dl. = 1 Liter = .908 qt. = 1.0567 qt. 10 Liters, L. = 1 Dekaliter = 9.08 •* = 2.6417 gal. 10 Dekaliters, JDl. = 1 Hectoliter = 2.8372+ bu. = 26.417 " 10 Hectoliters,^;. =: 1 Kiloliter = 28.372+ " = 264.17 " 10 Kiloliters, Kl. = 1 Myrialiter =283.72+ " =2641.7 The Hectoliter is used in measuring large quantities in both liquid and dry measure. UNITS OF WEIGHT. 450. The Gram is the principal unit of weight, and is equal to the weight of a cube of distilled water whose edge is one centimeter. DENOMINATE NUMBERS, 77 TABLE OF UNITS. 10 Milligrams, 10 Centigrams, 10 Decigrams, 10 Grams, 10 Dekagrams, mg. eg. dg. a. Dg. -10 Hectograms, Hg, 10 Kilograms, Kg. 10 Myriagrams, Mg. 10 Quintals, = 1 Centigram Decigram Gram Dekagram Hectogram \ Kilogram ) } or Kilo. S Myriagram Quintal Tonneau or Ton. .15432+ gr. Troy. 1.54334+ " " 15.43248+ " " .3527 + oz. Avoir. 3.52739+ " " 2.20462+ lb. = 22.04621 + = 220.46212 + = 2204.6212 + The Kilogram or Kilo., which is little more than 2| lb. Avoir., is the common weight in trade. Heavy articles are weighed by the Ton- neau, which is 204 lb. more than a common ton. Comparative Table of Units, 1 Inch = .0254 meter. 1 Cu. foot = .2832 Hectoliter. 1 Foot = .3048 " 1 Cu. yard = .7646 Steres. 1 Yard = .9144 " 1 Cord = 3.625 Steres. 1 Mile = 1.6093 Kilometers. 1 Fl. ounce =: .02958 Liter. 1 Sq. inch = .0006452 sq. meter. 1 Gallon = 3.786 Liters. 1 Sq. foot = .0929 1 Bushel = .3524 Hectoliter. 1 Sq. yard = .8361 1 Troy grain = .0648 Gram. 1 Acre =40.47 Ares. 1 Troy lb. = .373 Kilogram. 1 Sq. mile = .259 Hectares. 1 Avoir, lb. =: .4536 Kilogram. 1 Cu, inch = .01639 Liter. 1 Ton = .9071 Touneau. EXAMPLES FOR PRACTICE 451. Reduce 1. 84 lb. Avoir, to kilograms. 2. 37 T. to tonneau. 3. 96 bu. to hectoliters. 4. 75 fl. oz. to liters. 5. 89 cu. yd. to steres. 6. 328 acres to ares. 13. If the price per gram is $. 7. 4.0975 liters to cu. in. 8. 31.7718 sq. meters to sq. yd. 9. 272.592 liters to bushels. 10. 35.808 kilograms to Troy gr. 11. 133.75 steres to cords. 12. 33.307 steres to cu. ft. what is it per grain ? 78 DEN0 3IINATE NUMBERS, 14. If the price per liter is $1.50, what is it per quart? 15. At 26.33 cents per hectoliter, what will be the cost of 157 bushels of peas ? 16. When sugar is selling at 2.168 cents per kilogram, what will be the cost of 138 lb. at the same rate? 17. Reduce 834 grams to decigrams ; to dekagrams. 18. In 84 hectoliters how many liters ? how many centiliters ? 19. A man travels at the rate of 28.279 kilometers a day. How many miles at the same rate will he travel in 45 days ? 20. If hay is sold at $18,142 per ton, what is the cost of 48 tonneau at the same rate ? 21. When a kilogram of coffee costs $1.1023, what is the cost of 148 lb. at the same rate ? EEVIEW AND TEST QUESTIONS. 4:65. 1. Define Related Unit, Denominate Number, Denominate Fraction, Denomination, and Compound Number. 2. Repeat Troy Weight and Avoirdupois Weight. 3. Reduce 9 bu. 3 pk. 5 qt. to quarts, and give a reason for each step in the process. 4. In 9 rd. 5 yd. 2 ft. how many inches, and why ? 5. Repeat Square Measure and Surveyors' Linear Measure. 6. Reduce 2345G sq. in. to a compound number, and give a reason for each step in the process. 7. Define a cube, a rectangular volume, and a cord foot. 8. Show by a diagram that the contents of a rectangle is found by multiplying together its two dimensions. 9. Define a Board Foot, a Board Inch ; and show by diagrams that there are 12 hoard feet in 1 cubic foot and 12 hoard inches in 1 board foot. 10. Reduce f of an inch to a decimal of a foot, and give a reason for each step in the process. 11. How can a pound Troy and a pound Avoirdupois be compared? 12. Reduce .84 of an oz. Troy to a decimal of an ounce Avoirdupois, and give reason for each step in the process. 13. Explain how a compomid number is reduced to a fraction or deci- mal of a higher denomination. Illustrate the abbreviated method, and give a reason for each step in the process. PART SECOND. BUSINESS ARITHME imh SHORT METHODS. 466. Practical devices for reaching results rapidly are of first importance in all business calculations. Hence the fol- lowing summary of short methods should be thoroughly mastered and applied in all future work. The exercises under each problem are designed simply to illustrate the application of the contraction. When the directions given to perform the work are not clearly understood, the references to former explanations should be carefully examined. 461'. Prob. I.— To multiply by lO, 100, 1000, etc. Move tJie decimal ^mnt in the multiplicand as many places to the right as there are ciphers in the multiplier, annexing ciphers ivhen necessary (91). Multiply the following: 1. 84 X 100. 4. 3.8097 x 10000. 7. 3426 x 1000. 2. 76 X 1000. 5. .89752 x 1000. 8. 7200 x 100000. 3. 5. 73 X 100. 6. 3.0084 x 10000. 9. 463 x 1000000. 468. Prob. II. — To multiply where there are ciphers at the right of the multiplier. Move the decimal point in the multiplicand as many places to the right as there are ciphers at the right of the multiplier, annexing ciphers when necessary, and multiply the result by the significant figures in the multiplier (93). (207) 80 BUSINESS ARITHMETIC. Multiply the following : 1. 376 X 800. 4. 836.9 x 2000. 7. 3800 x 7200. 2. 42.9 X 420. 5. 7.648 x 3200. 8. 460 x 900. 3. 500 X 700. 6. 2300 x 5000. 9. .8725 x 3600. 469. Prob. III.~To multiply by 9, 99, 999, etc. Move the decimal point in the imdtiplicand as many places to the right as there are nines in the multiplier, annexing ciphers when necessary, and subtract the given multiplicand from the result. Observe that by moving the decimal point as directed, we multiply by a number 1 greater than the given multiplier ; hence the multiplicand is subtracted from the result. To multiply by 8, 98, 998, and so on, we move the decimal point in the same manner; and subtract from the result twice the multiplicand. Perform the following multiplication : 1. 736458 X 9. 4. 53648 x 990. 7. 7364 x 998. 2. 3895 X 99. 5. 83960 x 9999. 8. 6283 x 9990. 3. 87634 X 999. 6. 26384 x 98. 9. 4397 x 998. 470. Prob. IV.— To divide by lO, lOO, lOOO, etc. Move the decimal point in the dividend as many places to the left as there are ciphers in the divisor, prefixing ciphers when necessary. Perform the division in the following : 1. 8736-r-lOO. 4. 23.97-j-lOOO. 7. .54-^100. 2. 437.2-r-lO. 5. 5.236-^100. 8. .07-7-1000. 3. 790.3-f-lOO. 6. .6934-T-lOOO. 9. 7.2-f-lOOO. 471. Prob. V. — To divide where there are ciphers at the right of the divisor. Move the decimal point in the dividend as many places to the left as there are ciphers at the right of the divisor, prefixing ciphers when necessary (140), and divide the result ly the significant figures in the divisor (143). (208) SHORT METHODS, 81 Perform the division in the following : 1. 7352-r-40. 4. 5.2-r-400. 7. 364.2-^640. 2. 523.7-r-80. 5. .96-r-120. 8. 973.5—360. 3. 329.5-^3000. 6. .08^-200. 9. 8.357-^600. 472. Peob. YI. — To multiply one fraction by another. Cancel all factors common to a numerator and a denomina- pr before multiplying (185 — II). Perform the following multiplications by canceling common factors : l-Hxff. 6. ixMx«. 11. Mx-AVxf. 2. If x^V 7. Axlf xA. 12. MMxtV^x^^. 3. tttxff. 8. AxHxff. 13. «xi|ix^V 4. ^xA- 9. fxffxA. 14. HfxttxJ. 5. exT%. 10. «fxffXT|^. 15. il^xyij^xf 473. Prob. YII. — ^To divide one fraction by another. Cancel all factors common to both numerators or common to hotli denominators before dividing (291). Or, Invert tlie divisor and cancel as directed in Prob. VL Perform the division in the following, canceling as directed : 1. H-f 5. 3*A-«. 9. i^f-Jf 2. H-^A- 6. H-if. 10. AV-tW«. 3. H -^ H- 7. AV -- ^. 11. .39 ^ .003. 4. .9 -T- .03. 8. .28 -i- .04. 12. .63 — .0027. 474. Prob. VIII. — To divide one number by another. Cancel the factors that are common to the dividend and divi- sor before dividing (185 — II). Perform the following divisions, canceling as directed : 1. 8400-J-300. 4. 62500-^2500. 7. 9999-^63. 2. 3900-7-130. 5. 3420-r-5400. 8. 32000-r-400. 3. 4635—45. 6. 89600-^800. 9. 75000—1500. ^209^ BUSINESS ARITirilETIO, ALIQUOT PARTS. 475. An Aliquot Part of a number is any number, integral or mixed, which will exactly divide it. Thus, 2, 2^, 3^, are aliquot parts of 10. 476. The aliquot parts of any number are found by divid- ing by 2, 3, 4, 5, and so on, up to 1 less than the given number. Thus, 100-^2 = 50 ; 100-^3 = 33| ; 100 -^ 4 = 25. Each of the quotients 50, 33-J, and 25, is an aliquot part of 100. 477. The character @ is followed by the price of a unit or one article. Thus, 7 cords of wood @ 14.50 means 7 cords of wood at $4.50 a cord. 478. Memorize the following aliquot parts of 100, 1000, and $1. Table of Aliquot Parts, 50 = 1' 500 = J' 50 ct. = i 33i= i 333| = I 33| ct. = i 25 = J 250 = i 25 ct. = i 20 = i 200 = i 20 ct. = -J- 16|= i ^ of 100. 166f = i ► of 1000. 16fct. = i l^= \ 142f = 1 14f ct. = 1 12i=i 125 = i 12ict. =: i 1U= i IIH = i Hi ct. = i 10 =M 100 =^ 10 ct. =-^ of $1. 479. Peob. IX. — To multiply by using aliquot parts. 1. Multiply 459 by 33|. 3 ) 45900 Explanation. — We multiply by 100 by annexing two ciphers to tlie multiplicand, or by moving the decimal 15300 point two places to the right. But 100 being equal to 8 times the multiplier 33|^, the product 45900 is 3 times as large as tho required product ; hence we divide by 3. (210) SHORT METHODS, 83 Perform the following multiplications by aliquot parts. 2. 974 X 50. 5. 234 x 333J. 8. 4.38 x 3^. 3. 35.8 X 16f. 6. 869 x llj. 9. 7.63 x 142f. 4.895x125. 7. 72xlllf. 10.58.9x250. Solve the following examples orally, by aliquot parts. 11. What cost 48 lb. butter @ 25 ct. ? @ 50 ct. ? @ 33 J ct. ? Soi.UTiON. — At $1 a pound, 48 would cost $48. Hence at 33| cts. a pound, which is ^ of $1, 48 pounds would cost i of $48, which is $16. 12. What cost 96 lb. sugar @ 12^ ct. ? @ 14f ct. ? @ 16| ct. ? 13. What is the cost of 24 bushels wheat @ $1.33 J? Solution-.— At $1 a bushel, 24 bushels cost $24 ; at 33| ct., which is I- of $1 a bushel, 24 bushels cost $8. Hence at $1.33 J- a bushel» 24 bushels cost the sum of $24 and $8, which is $32. 14. What cost 42 yards cloth @ |1.16f ? @ $2.14f ? 15. What cost 72 cords of wood @ U.l^ ? @ $3.25 ? Find the cost of the following, using aliquot parts for the cents in the price. 16. 834 bu. wheat @ I1.33J ; @ $1.50 ; @ $1.25 ; at $1.1 6f. 17. 100 tons coal @ $4.25 ; at $5.50; @ $6.12^: @ $5.33J. 18. 280 yd. cloth @$2.14f ; @ $1.12|-; @$3.25; @ $2.50. 19. 150 bbl. apples® $4.20; @$4.50; ©$4.33^. 20. 2940 bu. oats @ 33 ct. ; @ 50 ct. ; @ 25 ct. 21. 896 lb. sugar @l^•, @ 14f ; @ 16f. 22. What is the cost of 2960 yd. cloth at 37^ ct. a yard ? 25 =i of 100, hence 4 ) 2960 ExPLANATioN.-At $1 a yard, ^ 2960 yd. will cost $2960. But l?i=i of ^5, hence 2 ) 740 25 ct. is i of $1, hence i of $2960 37^ 370 which is $740, is the cost at $1110 ^^ ^*' ^ ^'^• 2. Again, 12^- ct. is the ^ of 35 ct., hence $740, the cost at 25 cts., divided by 2, gives the cost at 12i ct., which is $370. But 25 ct. + 12^ ct. = 37^ hence $740 + $370 or $1110 is the cost at 37^ ct. (211) 84 . BUSINGS ARITHMETIC. 23. 495 bu. barley @ 75 ct. ; @ 62^ ct. ; @ 87^ ct 24. 680 lb. coffee @ 37^ ct. ; @ 75 ct. ; @ 60 ct. 25. 4384 yd. cloth @ 12| ct. ; @ 15 ct. ; @ 30 ct. ; @ 35 ct. Observe, that 10 ct. = yV of 100 ct., and 5 ct. = ^ of 10 ct. 26. 870 lb. tea @ 60 ct ; @ 62^ ct. ; @ 80 ct. ; @ 87^ ct. 480, Prob. X. — To divide by using aliquot parts. 1. Divide 7258 by 33^. 72.58 Explanation.— 1. We divide by 100 by moving the g decimal point two places to the left. c).. y r/A 2. Since 100 is 3 times 33^, the given divisor, the quotient 72.58 is only ^ of the required quotient ; hence we multiply the 72.58 by 3, giving 217.74, the required quotient. Perform by aliquot parts the division in the following : 2. 8730-^3J. 5. 379.6-^-33^. 8. 460.85-^250. 3. 9764-^5. 6. 98.54-^50. 9. 90.638-T-25. 4. 8.375-r-16|. 7. 394.8 -j- 125. 10. 73096-333^. Solve the following examples orally, using aliquot parts. 11. At 33 J ct., how many yards of cloth can be bought for $4 ? Solution. — Since $1, or 100 ct., is 3 times 33| ct., we can buy 3 yards for $1. Hence for $4 dollars we can buy 4 times 3 yd., which is 12 yd. Observe, that in this solution we divide by 100 and multiply by 3, the number of times 33J, the given price, is contained in 100. Thus, $4=400 ct., 400-1-100=4, and 4 x 3=12. In the solution, the reduction of the $4 to cents is omitted, as we recognize at sight that 100 ct, or $1, is contained 4 times in $4. 12. How many yards of cloth can be bought for $8 @ 12|^ ct. ? @ 14f ct. ? @ 33i ct. ? @ 16f ct. ? @ 25 ct. ? @ 10 ct. ? @ 50 ct. ? @ 8 ct. ? @ 5 ct. ? @ 4 ct. ? 13. How many pounds of butter @ 33-^ ct. can be bought for $7? For $10? For $40? 14. How much sugar can be bought at 12|- ct. per pound for $3? For $8? For $12? For $30? For $120? (212) BUSINESS PROBLEMS. 85 Solve the following, performing the division by aliquot parts : 15. How many acres of land can be bought for $8954 at f 25 per acre? At $50 ? AtlSS^? At $125? At$16f? At 1250? 16. How many bushels of wheat can be bought for IG354 at 11.25 per bushel ? At $2.50 ? Observe, $1.25 = J of $10 and $2.50 = i of $10. Hence by moving the decimal point one place to the left, which will give the number of bu. at $10, and multiplying by 8, will give the number of bu. at $1.35. Multiplying by 4 will give the number at $2.50. 17. How many yards of cloth can be bought for $2642 at 33i ct. per yard? At 14f ct? At 25 ct.? At $3.33^? At $2.50? At$l.lH? At$1.42f? 18. What is the cost of 138 tons of hay at $12J ? At 14f ? At 16| ? At $25 ? At $13.50 ? At $15.33J ? At $17.25 ? BUSINESS PEOBLEMS, DEFINITIONS. 481. Quantity is the amount of any thing considered in a business transaction. 483. JPrice^ or Rate, is the value in money allowed for a given unit, a given number of units, or a given part of a quantity. Thus, in 74 bu. of wheat at $2 per bushel, the price is the value of a unit of the quantity ; in 8735 feet of boards at 45 ct. per 100 feet, the price is the value of 100 units. 483. When the rate is the value of a given number of units, it may be expressed as a fraction or decimal. Thus, cloth at $3 for 4 yards may be expressed as $5 per yard ; 7 for every 100 in a given number may be expressed j^^j^ or .07. Hence, f of 64 means 5 for every 8 in 64 or 5 per 8 of 64, and .08 means 8 per 100, (313) 86 BUSINESS ARITHMETIC. 484. Cost is the value in money allowed for an entire quantity. Thus, in 5 barrels of apples at $4 per barrel, $4 is tbe price, and $4x5 or .^20, the entire value of the 5 barrels, is the cost. 485. JPer Cent means Per Hundred. Thus, 8 per cent of $600 means $8 out of every $100, which is $48. Hence a given per cent is the price or rate per 100. 486. The Sign of Per Cent is %. Thus, %% is read, 8 2)er cent. Since per cent means per hundred, any J* -D A \ '^^^^ ^^*^ ^ equal to the per- 510. Prob. IX. B = -^. Read, -^ ^- -^ ^ ,> w . R { centage divided hy the rate. K-t-i-r.^^ ^-P T>j( TJie rate is equal to the per- 511. Prob. X. -R = -B- Read,-^ ^- v, ^ t. ^x , R ( centage divided by the base. i Tt— ^ T? d i ^^^(^^^^^^Q^^^^^i^^^^^Of^f^^ rt-io Tit-xTT") ~ 1 + It' { divided by 1 plus the rate. OVa. Prob. XI. ) _^ / mr 7. . 7. .7 3.^. i _. X) T> J i Thebaseiscqualtothedtffnce { 1 — R { divided by 1 minus tTie rate. 513. Refer to the problems on pages 222 to 226 inclusive, and answer the following questions regarding these formulae : 1. What is meant hy BxE, and why is P=B xR9 Illus- trate your answer by an example, giving a reason for each step. 2. Why is P -^ E equal B ? Give reasons in full for your answer. 3. If i2 is 135^, which is the greater, P or By and why ? (228) Bisrn ^? PRO Fir AND LOSS. 101 4. If R is 248^, how would you express R without the 5. Why is R equal to P -r- ^, and how must the quotient of P -7- i5 be expressed to represent R correctly ? G. What is meant by ^ ? How many times R in P (502) ? How many times 1 in P ? How many times 1 + R must there be in A and why ? 7. How many times R in P (503)? D is equal to B minus liow many times R (501) ? 8. Why is B equal to D -- {I — R)? Give reasons in full for your answer. PEOFIT AND LOSS. 514. The quantities considered in Profit and Loss corres- pond with those in Percentage thus : 1. The Cost, or Capital invested, is the Base, 2. The I^er cent of Profit or Loss is the Hate. 3. The JProfit or Loss is the Percentage. 4. The Selling Price when equal the Cost plus the Frojit is the Amount, when equal the Cost minus the Loss is the Uifference. EXAMPLKS FOR PRACTICES. 515. 1. A firkin of butter was bought for 119 and sold at a profit of 1Q%. What was the gain ? Formula P = B x R. Read, Profit or Loss — Cost x Bate %. Find the profit on the sale 2. Of 320 yd. cloth bought @ $1.50, sold at a gain of 11%. 3. Of 84 cd. wood bought @ $4.43^, sold at a gain of 20^. 4. Of 873 bu. wheat bought @ $1.25, sold at -a gain of 14J^. Find the loss on the sale 5. Of 180 T. coal bought @ 17.85, sold at a loss of %\%, (229) 102 BUSINESS ARITHMETIC. 6. Of 124 A. land bought @ $84.50, sold at a loss of 21|^. 7. If a farm was bought for $4860 and sold for $729 more than the cost, what was the gain per cent Formula R = P -i- B. Read, Rate % Gain = Profit -i- Cost, 8. A piece of cloth is bought at $2.85 per yard and sold at $2.10 per yard. What is the loss per cent ? Ans. 9. If f of a cord of wood is sold for f of the cost of 1 cord, what is the gain per cent ? Ans. 10. Find the selling price of a house bought at $5385.90, and sold at a gain of 18^. Formula A=B x (1 + R). Read, Selling Price =Ci?si x (1 + Rate % Oain). 11. Corn that cost 65 ct. a bushel was sold at 20^ gain. What was the selling price ? Ans. 78 ct. a bu. 12. A grocer bought 43 bu. clover seed @ 14.50, and sold it in small quantities at a gain of 40^. What was the selling price per bu. and total gain ? 13. Bought 184 barrels of flour for $1650, and sold the whole at a loss of S%. What was the selling price per batrel ? Formula D = B (1 - R). Read, Selling Price = Cost x (1 - Rate % Loss), 14. Flour was bought at $8.40 a barrel, and sold so as to lose 15^. What was the selling price? 15. 0. Baldwin bought coal at $6.25 per ton, and sold it at a loss of 18^. What was the selling price ? 16. Sold a house at a loss of $879, which was 15^ of the cost. What was the cost ? Formula B = P -j- R. Read, Cost = Profit or Loss -i- Rate %. 17. A grain merchant sold 284 barrels of flour at a loss of $674.50, which was 25^ of the cost. Wliat was the buying and selling price per barrel ? 18. A drover wished to realize on the sale of a flock of 236 sheep $531* which is 30% of the cost. At what price per head must he sell the flock? 19. Two men engaged in business, each having $4380. A (230) COMMISSION. 103 gained 33|^ and B 75^. How much was B's gain more than A's? 20. If I buy 72 head of cattle at 136 a head, and sell 33^^ of them at a gain of 18^, and the remainder at a gain of 24^, what is my gain ? 21. A grocer sells coffee that costs 13|^ cents per pound, for lOf cents a pound. What is the loss per cent ? 22. Fisk and Gould sold stock for $3300 at a profit of ^^%, What was the cost of it ? 23. A man bought 24 acres of land at 175 an acre, and sold it at a profit of 8 J^. What was his total gain ? 24. A merchant sold cloth for $3.84 a yard, and thus made 20^. What was the cost price ? 25. Bought wood at $3.25 a cord, and sold it at an average gain of 30^. What did it bring per cord ? 26. If land when sold at a loss of l^% brings 111.20 per acre, what would be the gain per cent if sold for $15.36 ? 27. Bought a barrel of syrup for $20 ; what must I charge a gallon in order to gain 20^ on the whole? COMMISSION. 516. A Commission Merchant or Agent is a per- son who transacts business for another for a percentage. 517. A Broker is a person who buys or sells stocks, bills of exchange, etc., for a percentage. 518. Commission is the amount paid a commission merchant or agent for the transaction of business. 519. lirokeraffe is the amount paid a broker for the transaction of business. 530. The Net I^roceeds of any transaction is the sum of money that is left after all expenses of commission, etc., are paid. (331) 104 BUSINUSS ARITHMETIC, 531. The quantities considered in commission correspond with those in percentage thus : 1. The amount of money invested or collected is the Base. 2. The per cent allowed for services is the Hate. 3. The Commission or Brokerage is the I*erc€ntaf/e. 4. The sum invested or collected, plus the commission, is the Aiiioimt, minus the commission is the Difference. EXAMPIiES FOR PRACTICE. 522. Let the pupil write out the formulae for each kind of examples in commission in the same manner as they are given in Profit and Loss. What is the commission or brokerage on the following : 1. The sale of 85 cords of wood @ $4.75, commission 3^% ? 2. The sale of 484 yds. cloth @ $2.15, commission 1}^ ? 3. The sale of 176 shares stocks at 187.50 a share, broker- age i% ? 4. The collection of 13462.84, commission 2^% ? What is the rate of commission on the following : 5. Selling a farm for $4800, commission 1120 ? 6. Collecting a debt of $7500, commission $350 ? 7. Selling wheat worth $1.80 a bu., commission 4 ct. a bushel ? What is the amount of the sale in the following : 8. The commission is $360, rate of commission 2 J^ ? 9. The brokerage is $754.85, rate of brokerage 1^% ? 10. The commission is $26.86, rate of commission If ^ ? Find the amount of the sales in the following: Observe, that the commission is pn the amount of the sales. Hence the formula for finding the amount of the sales when the net proceeds are given is (506) Amount of sales = Wet proceeds ^ (1 — Bate %). 11. Net proceeds, $8360 ; rate of commission, 3^%. (232) INSURANCE, 105 12. Net proceeds, $3G40 ; rate of commission, |^. 13. Net proceeds, $1850; rate of commission, f^. Find the amount to be invested in the following : Observe, that when an agent is to deduct his commission from the amount of money in his hand the formula is (500) Sum invested = Amount in hand -r- (1 + Rate % ). 14. Amount in hand, $3401.01 ; rate of commission, 3J^. 15. Eemittance was $393.17 ; rate of commission, 2f^. 16. Amount in hand, $606.43; rate of commission, IJ^. 17. A lawyer collects bills amounting to $492; what is his commission at 5^ ? ^7i5. $24.60. 18. An agent sold 824 barrels of beef, averaging 202 J lb. each at 9 cents a pound ; what was his commission at 2 J^ ? 19. A merchant has sent me $582.40 to invest in apples, at $5 a barrel ; how many can I buy, commission being 4^ ? 20. I have remitted $1120 to my correspondent in Lynn to invest in shares, after deducting his commission of 1^% ; what is his commission ? 21. An auctioneer sold goods at auction for $13825, and others at a private sale for $12050; what was his commission at J^? Ans. $129.3750. 22. A man sends $6897.12 to his agent in New Orleans, requesting him to invest in cotton after deducting his com- mission of 2^; what was the amount invested? INSUEANCE. 533. Insurance is a contract which binds one party to indemnify another against possible loss or damage. It is of two kinds : insurance on property and insurance on life, 524. The Policy is the written contract made between the parties. 535. The I^remium is the percentage paid for insurance. (233) 106 BUSIJV^ESS ARITHMETIC. 526. The quantities considered in insurance correspond with those in percentage ; thus, 1. The amount insured is the Base, 2. The jicr cent of premium is the Hate, 3. The premium is the Percentage, EXAMPLES FOR PRACTICE. 537. Let the pupil write out the formulae as in Profit and Loss. 1. What is the premium on a policy for $3500, at 3^ ? 2. My house is insured for $7250 ; what is the yearly pre- mium, at 2f ^ ? 3. Justus Weston's house is insured for $3250 at 3|- per cent, his furniture for $945 at 1} per cent, and his bam for $1220 at 1^ per cent ; what is the amount of premium on the whole property ? 4. A factory is insured for $27430, and the premium is $685.75 ; what is the rate of insurance ? 5. The Pacific Mills of Lawrence, worth $28000, being insured for f their value, v/ere destroyed by fire; at 2f per cent, what is the actual loss of the insurance company ? 6. The premium on a house, afc f per cent, is $40 ; what is the sum insured ? 7. It costs me $72 annually to keep my house insured for $18000 ; what is the rate ? 8. What must be paid to insure from Boston to New Or- leans a ship valued at $37600, at | of 1^ ? 9. A cargo of 800 bundles of hay, worth $4.80 a bundle, is insured at 1^% on J of its full value. If the cargo be destroyed, how much will the owner lose ? 10. My dwelling-house is insured for $4800 at Y/o ; ^ij f i^r- niture, library, etc., for $2500 at ^% ; my horses, cattle, etc., for $3900 at Y/o > ^^^^ ^ carriage manufactory, including machinery, for $4700 at 1J%'. What is my annual premium? (234) T0CK8, 107 STOCKS. 528. A Corporation is a body of individuals or com- pany authorized by law to transact business as one person. 529. The Capital Stock is the money contributed and employed by the company or corporation to carry on its business. Tlie term stock is also used to denote Government and State bonds, etc. 530. A Share is one of the equal parts into which the mpital stock is divided. 531. A Certificate of Stocky or Scripy is a paper issued by a corporation, securing to the holder a given num- ber of shares of the capital stock. 532. The Par Value of stock is the sum for which the Bcrip or certificate is issued. 533. The Market Value of stock is the price per share for which it can be sold. 534. The Premium^ Discount^ and Brokerage are always computed on t\\Q par value of the stock. 535. The Wet Earnings are the moneys left after deducting all expenses, losses, and interest upon borrowed capital. 536. A Bond is a written instrument, securing the pay- ment of a sum of money at or before a specified time. 537. A Coupon is a certificate of interest attached to a bond, which is cut off and delivered to the payor when the interest is discharged. 538. U". S. Bonds may be regarded as of two classes : those payable at a fixed date, and those payable at any time between two fixed dates, at the option of the government. (235) 108 BUSINESS ARITHMETIC, 539. In commercial language, the two classes of TJ. S. bonds are distinguished from each other thus : (1.) U, S, O's, bonds payable at a fixed time. (3.) U. S. 6's 5-20, bonds payable, at the option of the Govern- ment, at any time from 5 to 20 years from their date. EXAMPLES FOR PRACTICE. 540. Let the pupil write out the formula for each class of examples, as shown in Profit and Loss : 1. Find the cost of 120 shares N. Y. Central stock, the market value of which is 108, brokerage \%. Solution. -Since 1 share cost 108% +1%, or 108|% of $100 = 108|, the cost of 120 shares will be $108| x 120 = $13020. 2. What is the market value of 86 shares in the Salem and Lowell Eailroad, at d^% premium, brokerage 1% ? 3. Find the cost of 95 shares bank stock, at Q% premium, brokerage |^. 4. How many shares of Erie Eailroad stock at 8^ discount can be bought for 17030, brokerage ^% ? Solution.— Since 1 share cost 100%— 8% +^%, or 92i% of $100 = $92.50, as many shares can be bought as $92.50 are contained times in $7020, which is 76. How many shares of stock can be bought 5. For $10092, at a premium of b%, brokerage \%? 6. For $13428, at a discount of 7^, brokerage \% ? 7. For $16830, at a premium of ^%, brokerage J^? , 8. What sum must be invested in stocks at 112, paying 9^, to obtain a yearly income of $1260 ? Solution. — Since $9 is the annual income on 1 share, the number of shares must be equal $1260-^$9, or 140 shares, and 140 shares at $112 a share amount to $15680, the required investment. Find the investment for the following: 9. Income $2660, stock purchased at 105J. yielding 7^. 10. Income $1800, stock purchased at 109|, yielding 12^. (236) STOCKS. 109 11. Income $3900, stock purchased at 92, yielding 6^. 12. What must be paid for stocks yielding 1% dividends, that 10%' may be realized annually from the investment ? Solution. — Since $7, the annual income on 1 share, must be 10% of the cost of 1 share, yV of $7, or 70 ct. , is 1 % . Hence 100% , or 70 ct. x 100 = $70, is the amount that must be paid for the stock. What must be paid for stocks yielding 13. b% dividends* to obtain an annual income of 8^ ? 14. 1% dividends to obtain an annual income of 12^ ? 15. 9^ dividends to obtain an annual income of 7^ ? 16. How much currency can be bought for 1350 in gold, when the latter is at 12;^ premium? Solution.— Since $1 in gold is worth $1.12 in currency, $350 in gold are equal $1.13 x 350 = $392. How much currency can be bought 17. For $780 in gold, when it is at a premium of 9^? 18. For 1^396 in gold, when it is at a premium of 13-J^ ? 19. For 1520 in gold, when it is at a premium of 12^^? 20. How much is 1507.50 in currency worth in gold, the latter being at a premium of 12^%? Solution. — Since $1 of gold is equal to $1.12| in currency, $507.50 in currency must be worth as many dollars in gold as $1.12| is contained times in $507.50, which is $451. llj. How much gold can be bought 21. For $1053.17 currency, when gold is at a premium of 9i7; ? 22. For $317.47 currency, when gold is at a premium ofllf;^;? 23. For $418.14 currency, when gold is at a premium of 13f ^ ? 24. Bought 80 shares in Boston and Maine Railroad, at a discount of 2|^, and sold the same at an advance of 12^; what did I gain ? Ans, $1160. (237) 110 S USINES S ARITHMETIC, 25. An agent sells 415 barrels of flour, at $6 a barrel, com- mission b%, and invests the proceeds in 'stocks of the Suffolk Bank, Boston, at 17|-^ discount, brokerage i%; how many shares did he buy ? 2G. Bought 84 shares in Michigan Southern Eailroad, at 1% discount, and sold them at Q>\% advance; what was my profit, the brokerage in buying and selling being \ per cent ? 27. Bought bonds at 70^^, bearing 4:\% interest; what is the rate of income ? Ans. (j%. 28. I invest $2397.50 in Empire Iron Foundry stock, whose shares, worth $50 each, are sold at ^43.50, brokerage J^; what annual income shall I derive, the stock yielding 1% ? 29. 0. E. Bonney sold $6000 Pacific Eailroad 6's at 107, and with a part of the proceeds bought St. Lawrence County bonds at 90, yielding ^% dividends sufficient to give an annual income of $180 ; how much has he left ? 30. What rate of income can be derived from money invested in the stock of a company paying a semi-annual dividend of b%, purchased at 84J^, brokerage ^% ? 31. What must I pay for bonds yielding 4^^ annually, that my investment may pay Q% ? 32. What must be paid for stocks paying 5 per cent, that the investment may return %% ? 33. How much more is $1400 gold worth than 11515 cur- rency, when gold is 112^? 34. A man bought a farm, giving a note for $3400, payable in gold in 5 years; at the expiration of the time gold was 175/^: what did his farm cost in currency? 35. I invested $785.40 of currency in gold when it waswortli 115|;^ ; what amount of gold did I purchase ? 36. How much gold at a jDremium of {)^% can be purchased for $876.90 currency ? For $85.50 ? For $136.80 ? 37. What is the difference in the value of $800 in gold and $900 in currency, when gold is at a premium of 13|^^ ? (238) TAXES, 111 TAXES. 541, A Tax is a sum of money assessed upon a person or property, for any public purpose. 54^. Property is of two kinds: Real Estate, such as liouses and lands ; Personal Property, such as merchandise, cash, furniture, ships, notes, bonds, mortgages, etc. 543. Taxes are of two kinds : Property Tax, which is assessed upon taxable property according to its estimated yalue ; Poll Tax, which is a sum assessed without regard to property upon each male citizen liable to taxation. 544. An Assessment Eoll is a schedule or list which con- tains the names and the taxable value of the property of all persons subject to a given tax. 545. The Hate of Property Tax is the rate per cent on the valuation of the property. 546. An Assessor is an officer appointed to prepare the Assessment Roll and apportion to each person his tax. Method of A2>portloning a Tax, 54*7. 1. Tlie Assessor determines by a personal examination the taxable value of the real estate and personal property of each p6rson subject to tlie tax, and fills an Assessment Roll, thus : ASSESSMENT KOLL. NAMES. KEAL ESTATE. PERSONAL PROPERTY. TOTAL PROPERTY. POLLS. AMOUNT OF TAX. L. Henry, W. Mann, P. Duncan, R. Storey, $6984 8095 9709 0093 $1863 1983 2300 1975 $8846 10078 12009 8057 1 1 1 1 $45.48 51.64 61.295 41.585 Totals. 30880 8120 39000 4 $200. (239) 112 BUSIIiUSS ARITHMETIC, 2. If the amount to be raised on this Assessment Roll is $195, the collector's fees 2|%, and the poll tax $1.25 i)er poll, the Assessor, or party authorized to do so, would proceed to apportion the tax thus : (1.) Since the collector is paid 2|% of the whole tax for collecting, the $195 is 97 1% of the assessment which must be made. Hence, $195 -4- .97^ = $200, the whole tax. (2.) Since each poll pays $1.25, the 4 polls will pay $5, and the amount which must be assessed on the property is $200 — $5 = $195. (3.) Since $195 are to be assessed on $39000, the whole amount of property, the rate per dollar is $195 -4- $39000 = .005, or 5 mills. (4) Multiplying each man's taxable property by .005 will give his property tax, to which we add the poll tax ; hence, J. Henry's tax = $8846 x .005 + $1.25 = $45.48. . W.Mann's tax = $10078 x .005 + $1.25 = $51.64. P. Duncan's tax = $12009 x .005 + $1.25 =: $61,295. R. Storey's tax = $8067 x .005 + $1.25 = $41,585. (5.) These results are now inserted in the blank under "Amount of Tax," and the Assessment Roll is thus completed ready for the collector. EXAMPIiES FOR PRACTICE. 548, Prepare an Assessment Roll ready for the collector for each of the following : 1. Net tax to be raised $1930, collector's fee Z^%, poll tax $2.50 per poll. Property taxed. — A, real estate $10800, personal property $3200 ; B, real estate $9600, personal property $5200 ; C, personal property $4200; D, real estate $12800, personal property $4000; E, real estate $20000, personal property $6200 ; Polls without property 35. 2. Net tax to be assessed $2387.50, collector's fee A\%, poll tax $1.25 per poll. Property taxed. — A, real estate $9700, personal property $5000; B, real estate $14600, personal property $5400; 0, real estate $8900, personal property $3100 ; D, real estate $40000, personal property $12000 ; E, real estate $21600, per- sonal property $3700 ; Polls without property 11. (240) DUTIES AND CUSTOMS, 113 DUTIES OR CUSTOMS. 549. Duties or Ctistoms are taxes levied by the gov- ernment upon imported goods. 550. A Specific Duty is a certain sum imposed upon an article without regard to its value. 551. An Ad Valorem Duty is a per cent assessed upon the value of an article in the country from which it is brought. 552. A Tariff is a schedule giving the rates of duties fixed by law. 553. The following deductions or allowances are made before computing specific duties : 1. Tare. — An allowance for the box, cask, bag, etc., containing the merchandise. 2. Leakage, — ^An allowance for waste of liquors imported in casks or barrels. 8. Brealcage, — An allowance for loss of liquors imported in bottles EXAMPIiES FOR PRACTICES. 554. 1. "What is the duty on 420 boxes of raisins, each containing 40 pounds, bought for 8 cents a pound, at 20 per cent ad valorem ? 2. Imported 21 barrels of wine, each containing 31 gallons ; 2% being allowed for leakage, what is the duty at 40 cents per gallon ? 3. A merchant imported from Havana 100 boxes oranges @ $2.25 per box ; 75 hogsheads of molasses, each containing G3 gal., @ 23 cents per gal. ; 50 hogsheads of sugar, each containing 340 lb., @ 6 cents per lb. The duty on the molasses was 25^, on the sugar 30^, and on the oranges 20^. What was the duty on the whole ? (341) 114 BUSINES S ARITHMETIC, 4. What is the duty on 320 yards of cloth, invoiced at 11.15 per yard, at 20% ad valorem ? 5. At 12% ad valorem, what i» the duty on 100 barrels of kerosene, invoiced at $.18 a gallon, 2% leakage ? EEVIEW AND TEST QUESTIONS. 555. 1. When a fraction is to be divided by a fraction, why can the factors that are common to the denominators of the dividend and divisor be cancelled ? 2. How does moving the decimal point one or more places to the left or right affect a number, and why ? 3. Show that multiplying by 1000 and subtracting three times the multiplicand from the product is the same as multi- plying by 997 ? 4. Define Base, Percentage, Amount and Difference. 5. When the amount and rate per cent is given to find the base, why add the rate expressed decimally to 1 and divide by the result ? 6. Eepresent the quantities by letters and write a formula for solving each of the following problems (508). I. Given, the Cost and the Profit^ to find the rate per cent profit. II. Given, the rate per cent profit and the selling price, to find the buying price, in. Given, the amount of money sent to an agent to purchase goods and the rate per cent commission, to find the amount of the purchase. rV. Given, the rate at which stocks can be purchased, to find how much can be secured for a given sum. V. Given, the rate at which stocks can be purchased and the rate per cent of dividend, to find the rate per cent of income on the investment. VI. Given, the premium on gold, to find how much can be purchased for a given sum in currency. (242) -•-•s^l INTEREST ifo^*^ ^ ^ ^^ DEFINITIONS. 55G. Interest is a sum paid for the use of money. Thus, I owe Wm. Henry $200, which he allows me to use for one year after it is due. At the end of the year I pay him the $200 and $14 for its use. The $14 is called the Interest and the $200 the Principal. 557. Principal is a sum of money for the use of which interest is paid. 558. Rate of Interest is the number of units of any denomination of money paid for the use of 100 units of the same denomination for one year or some j^ven interval of time. 559. The Amount is the sum of the principal and interest. 560. Simxple Interest is interest which falls due when the principal is paid, or when a partial payment is made. 561. Legal Interest is interest reckoned at the rata per cent Jixcd by law. 562. Usury is interest reckoned at a higher rate than is allowed by law. 563. The following table gives the legal rates of interest in the different States. Where two rates are given, any rate between these limits is allowed, if specified in tenting. When no rate is named in a paper involving interest, the legal or lowest rate is always understood. (243) 116 B USINUSS ARITHMETIC, STATES. BATE 5^. STATES. RATEjg. STATES. RATE %. STATES. RATE^. Ala Ark Arizona.. Cal Conn Colo Dakota... Del D.C Flor Geo Idaho.... I 10 10 7 10 I 6 8 7 10 Any Any Any Any Any 10 Any 10 Ill Ind Iowa Kan Ken La Maine.... Md Mass Mich Minn.... Miss 6 6 6 7 6 5 6 6 6 7 7 6 10 10 12 10 8 Any Any 10 12 10 Mo Montana . N.H N.J N.Y N.C Neb Nevada . Ohio Oregon,.. Penn R.I 6 10 & 7 7 6 10 10 6 10 6 6 10 8 15 An' 8 12 7 Any S. C Tenn Texas.... Utah Vt Va W.Va.... W. T Wis Wy 7 6 8 10 6 6 6 10 7 12 Any 10 12 Any 12 Any 10 The legal rate for England and France is 5^ ; for Canada and Ireland, Q%, 564. PROB. I. — To find tlie simple interest of any- given sum for one or more years. 1. Find the interest on $384 for 5 years, at 7^. Solution. — 1. Since the interest of $100 for one year is |7, the interest of $1 for one year is $.07. Hence the interest of $1 for 5 years is $.07x5 = $.35. 2. Since the interest of $1 for 5 yr. is $ 35, the interest of $384 for the , same time must be 384 times $.35, or $134.40. Hence the following 565. Rule.* — I. Find the intei^est of $1 at the giiien rate for the given time, and multiply this result hy the numher of dollars in the given pHneipal. II. To -find the amount add the interest and principal. EXAMPIi FOR PRACTICE 566. Find the interest on the following orally: 1. $800 for 2 years at 4^. 2. $1200 for 3 years at Z%, 3. $200 for 5 years at Q>%. 4. $600 for 4 years at b%. 5. $90 for 2 years at 7^. 6. $70 for 4 years at %%. 7. $400 for 8 years at h%, 8. $100 for 12 years at 9^. 9. e$600 for 7 years at 10^ 10. $1000 for 5 years at S%, 11. $20 for 3 years at ^%. 12. $500 for 5 years at b% (244) SIMPLE INTEREST, 117 Find the interest on the following : 13. 1245.36 for 3 years at 1%. 20. $375.84 for 3 years at ^%. 14. $784.25 for .9 years at 4^. 21. $293.50 for 6 years at ^%, 15. $836.95 for 2 years at ^%. 22. $899.00 for 12 years at 7f ^. 16. $795.86 for 7 years at \%, 23. $600.80 for 9 years at ^%. 17. $896.84 for 3J years at 2f^. 24. $50.84 for 5 years at 1^. 1 8. $28.95 for 1^ years at ^%. 25. $95. 60 for | of a yr. at 1^%, 19. $414.14 for 4 years at ^%, 26. $262.62 for 6 years at ^%, METHOD BY ALIQUOT PARTS. 567. Prob. XL— To find the interest on any sum at any rate for years, months^ and days by aliquot i>arts. 1. In business transactions involving interest, 30 days are usually considered one month, and 12 months 1 year. Hence the interest for days and months may be found according to (499), by regarding the time as a compound number; thus, Find the interest and amount of $840 for 2 yr. 7 mo. 20 da., at 7^. $840 Principal. .07 Rate of Interest. 6 mo.= J of 1 yr., hence 2) 68.80 Interest for 1 yr. 2 117.60 Interest for 2 yr. 1 mo. = J of 6 mo., hence 6) 29.40 « " 6 mo. 15da.= ^of lmo.,hence2) 4.90 « " 1 mo. 5da.=^ofl5da.hence3) 2.45 " " 15 da. 81f « " 5 da. $155.1 6f " " 2 yr. 7 mo. 20 da. 840.00 Principal. $995.16| Amount for 2 yr. 7 mo. 20 da 568. The interest, by the method of aliquot parts, is usually found by finding first the interest of $1 for the given (245) 118 BUSINESS ARITH3IETIC, time, and multiplying the given principal by the decimal expressing the interest of $1 ; thus, Find the interest of 1680 for 4 yr. 9 mo. 1 5 da. at 8^. 1. We first find tlie interest of $1 for tlie given time thus : 8 ct. = Int. of $1 for 1 yr., 8 ct. x 4 == Int. for 4 yr. = 33 ct. 6 mo. = ^ of 1 yr., hence, -|- of 8 ct. = " « 6 mo. — 4 ct. 3 mo. = i of 6 mo., " ^ of 4 ct. = " "3 mo. = 2 ct. 15 da. = i of 3 mo., " i of 2 ct. = " " 15 da. = .03| m. Hence the interest on $1 for 4 yr. 9 mo. 15 da. — $.383^-. 2. The decimal .383^ expresses the part of $1 which is the interest of $1 for the given time at the given rate. Hence, $680 x .383|=$260.66|, is the interest of $680 for 4 yr. 9 mo. 15 da., at 8% ; hence the following 569. Rule. — I. Find by aliquot parts the interest of $1 for the given rate and time. II. Multiply the principal hy the decimal expressing the interest for $1, and the product will he the required interest. III. To find the Amount, add the interest to the principal. EXAMPIiES FOR PRACTICE. 570. Find the interest 1. Of 1284 for 3 yr. 8 mo. 12 da. at 6% ; at 8|^. 2. Of 1500.40 for 2 yr. 10 mo. 18 da. at 7% ; at 9^. 3. Of $296.85 for 4 yr. 11 mo. 24 da. at S% ; at 5^. 4. Of $860 for 1 yr. 7 mo. 27 da. at 4.-}% ; at 7|-^. 6. Of 12940.75 for 3 yr. 11 mo. 17 da. at 7% ; at di%. 6. Find the amount of $250.70 for 2 yr. 28 da. at 8%. 7. Find the amount of $38.90 for 3 yr. 13 da. at 9%. 8. A man invested $795 at S% for 4 yr. 8 mo. 13 da. How much was the amount of principal and interest ? 9. Paid a debt of 1384.60, which was upon interest for 11 mo. 16 da. at 7%. What was the amount of the payment? 10. Find the amount of $1000 for 9 yr. 11 mo. 29 da. at 7%. (246) SIMPLE INTEREST, 119 METHOD BY SIX PER CENT. PJREP ARJ^TOItY STEPS. 571. Step I. — To find the interest for any number of months at 6%. 1. Since the interest of $1 for 12 months, or 1 yr., at Q%y is 6 cents, the interest for two months, which is \ of 12 months, must be 1 cent, or yj^ part of the principal. 2. Since the interest for 2 months is yott of the principal, the interest for any number of months will be as many times yj(j- of the principal as 2 is contained times in the given num- ber of months. Hence the following 573. EuLE. — /. Move the decimal point in the pj^n- cipal TWO PLACES to the left (470), prefixing ciphers, if necessary. II. Multiply this result hy one-half the number of months. Or, Multiply -^^-^ of the principal hy the number of months and divide the result by 2. EXAMPLES FOR PRACTICE. 573. Find the interest at 6^ 1. Of $890 for 8 mo. 6. Of 1398 for 1 yr. 6 mo. =18 mo. 2. Of 1973.50 for 10 mo. 7. Of $750 for 2 yr. 8 mo. 3. Of $486.80 for 18 mo. 8. Of $186 for 4 yr. 2 mo. 4. Of $364.40 for 7 mo. 9. Of $268 for 2 yr. 6 mo. 5. Of $432.90 for 13 mo. 10. Of $873 for 1 yr. 11 mo. 574. Step II. — To find the interest for any number of days at 6%. 1. Since the interest of $1 for 2 months at 6% is 1 cent, the interest for 1 month, or 30 days, must be | cent or 5 mills. And since 6 days are \ of 30 days, the interest for 6 days must be I of 5 mills, or 1 mill, which is j^q^ of the principal. (247) 120 BUSINESS ARITH31ETIC. 2. Since the interest for 6 days is -^^-^ of the principal, the interest for any number of days will be as many times yqwo of the principal as 6 is contained times in the given number of days. Hence the following 51i^, KuLE. — /. Move the deeimal -point in the prin- cipal THREE PLACES to the left (470), prefixing ciphers, if necessary. II. Multiply this result hy one-sixth the numher of days. Or, Multiply y^^^ of the principal hy the number of days and divide the result by 6. EXAMPLES FOR PRACTICE. 511G, Find the interest at (j% 1. Of $790 for 12 da. 6. Of 1584 for 19 da. 2. Of 1384 for 24 da. 7. Of $730 for 22 da. 3. Of 1850 for 15 da. 8. Of $809 for 28 da. 4. Of $935 for 27 da. 9. Of $396 for 17 da. 5. Of $580 for 16 da. 10. Of $840 for 14 da. 577. Prob. III. — To find the interest on any snm at any rate for years, months, and days, by the six per cent method. Find the interest of $542 for 4 years 9 months 17 days at 8 per cent. Solution.— 1. The interest of $542 for 4 years at 6^, according to (5<>4), is $542 X .06 X 4 = $130.08. 2. The interest for 9 months, according to (571), is j^^ of $543 or $5.42 multiplied by 9, and this product divided by 2 = $24.39. 3. The interest for 17 days, according to (574), is t^Vtt ^^ $542 or $.542 multiplied by 17, and this product divided by 6 = $1,535 + . Hence $130.08 + $24.39 + $1.54 = $156.01, the interest of $542 for 4 years 9 months and 17 days. 4. Having found the interest of $542 at 0%, to find the interest at 8^ we have 8% - 6% + 2%, and 2% is i of 6%. Hence, $156.01 + ^ of $156.01 = $208,013, the interest of $542 at 8% for 4 yr. 9 mo. 17 da. (248) SIMPLE INTEREST, 121 EXAMPLES FOR PRACTICE. 518. Find the interest by the 6% method 1. Of $384.96 for 2 yr. 8 mo. 12 da. afc G%; at 9%; at S%. 2. Of $890.70 for 4 yr. 10 mo. 15 da. at 7% ; at 10^; at 4^. 3. Of $280.60 for 11 mo. 27 da. at S%; at 4:%; at 7%. 4. Of $480 for 2 yr. 7 mo. 15 da. at 9%; at 12^ ; at 4J^. 6. Of $890 for 9 mo. 13 da. at 6^%; at 8^;^ ; at 9^%, METHOD BY DECIMALS. 579, In this method the time is regarded as a compound number, and the months and days expressed as a decimal of a year. When the principal is a small sum, suflBcient accuracy will be secured by carrying the decimal to three places ; but when a large sum, a greater number of decimal places should be taken. 580. Peob. IV. — To find the interest on any sum at any rate for years, months, and days, by decimals. What is the amount of $450 for 5 yr. 7 mo. 16 da. at 6^? Explanation.— 1. We express, accord- ing to (389—15), the days and months as a decimal of a year, as shown in (1). 2. We find the interest on $450, the given principal, for 1 year, which is $27, as shown in (3). 3. Since $37 is the interest on $450 for 1 year, the interest for 5.637 years is 5.637 times $37, which is $151,929, as sliowu in(l). 4. The amount is equal to the principal plus the interest (559); hence, $151.93 + $450 = $601.93 is the amount. Hence the following (249) (1.) 30 ) 16 da. (2.) $450 12 ) 7.533 mo. .06 5.627 yr. 27 $27.00 39 389 112 54 $151,929 Interest. 450 $601.93 Amount 1^ B USIiYSSS ARITHMETIC, 581. EuLE. — /. To find the interest, multiply the prineipal by the rate, and this product by the tiine, expressed in years and decimals of a year. II. To find the amount, add the interest to the prin- cipal. EXAMPLES FOR PRACTICE. 582. Find the interest by the decimal method 1. Of $290 for 1 yr. 8 mo. 12 da. at b% ; at 8^; at 7%- 2. Of $374.05 for 2 yr. 9 mo. 15 da. at 0% ; at 9% ; at 4^. 3. Of $790.80 for 5 yr. 3 mo. 7 da. at 7% ; at 11^; at 3%. 4. Of $460.90 for 3 yr. 5 mo. 13 da. at Gi% ; at 8^% ; at 3|^. 6. Of $700 for 11 mo. 27 da. at ^% ; at 7i^; at 2^%. 6. Of $580.40 for 17 da. at 6^%; at 9^^ ; at 5i^. 7. Of $890 for 7 yr. 19 da. at 6% ; at 8^ ; at 5^. EXACT INTEREST. 583. In the foregoing methods of reckoning interest the year is regarded as 360 days, which is 5 days less than a com- mon year, and 6 days less than a leap year ; hence, the interest when found for a part of a year is incorrect. Thus, if the interest of $100 is $7 for a common year or 365 days, the interest for 75 days at the same rate must be //g of $7 ; but by the fore- going method //^ of $7 is taken as the interest, -.vhich is too great. Observe, that in using //j instead of //;,, the denominator is dimin- ished ^f g^ = j\ part of itself, and consequently (330) the result is ^^ part of itself too great. Hence, when interest is calculated by the foregoing methods, it must be diminished by -^^ of itself for a common year, and for like reasons g^ of itself for a leap year. To find the exact interest we have the following : 584. Rule. — /. Find the interest for the ^iven num- ber of years (563). //. Find the exact number of days in the given months and days, and take such a part of the interest (250) SIMPLE INTEREST. 123 of the principal for one year, as the whole number of days is of 365 days. Or, Find the interest for the given months and days hy either of the foregoing methods, then subtract /j- part of itself for a comjnon year, and gV f(^^ ^ leap year. III. Add the result to the interest for the given num- ber of years. EXAMPLES FOR PRACTICE. 5S^. Find the exact interest by both rules 1. Of 1836 for 84 da. at 6^. 5. Of $2300 for 7 da. at ^%. 2. Of $2G0 for 55 da. at 8^, 6. Of $120 for 133 da. at ^%. 3. Of $690 for 25 da. at 7^. 7. Of $380.50 for 93 da. at 6f ^. 4. Of $985 for 13 da. at 9^. 8. Of $260.80 for 17 da. at 12^. 9. Required the exact interest of $385.75 at 7^, from Jan- uary 15, 1875, to Aug 23 following. 10. 'What is the difference between the exact interest of $896 at 7^ from January 11, 1872, to November 19, 1876, and the interest reckoned by the six per cent method ? 11. A note for $360.80, bearing interest at 8^, was given March 1st, 1873, and is due August 23, 1876. How much will be required to pay the note when due ? 12. What is the exact interest of $586.90 from March 13 to October 23 of the same year, at 1% ? 586. Prob. V. — To find the principal when the inter- est, time, and rate are g^iven. Obserneyihaii the interest of any principal for a given time at a given rate, is the interest of $1 taken (503) as many times as there are dollars in the principal ; hence, the following 58?. Rule. — Divide the given interest by the interest of $1 for the given time at the given rate. (351) 124 BUSINUSS ARITHMETIC. EXAMPLES FOR PRACTICE. 588. 1. What sum of money will gain $110.25 in 3 yr. 9 mo. at 11% ? Solution. — Tlie interest of $1 for 3 yr. 9 mo. at 7%, is $.2625. Now fsince $.2625 is the interest of $1 for the given time at the given rate, $110.25 is the interest of as many dollars for the same time and rate as $.2625 is contained times in $110.25. Hence $110.25 ^ .2625 = $420, the required principal. What principal or sum of money 2. Will gain $95,456 in 3 yr. 8 mo. 25 da. at 7%? 3. Will gain $63,488 in 2 yr. 9 mo. 16 da. at S% ? 4. Will gain $106,611 in 3 yr. 6 mo. 18 da. at 6^% ? 6. Will gain $235,609 in 4 yr. 7 mo. 24 da at 9% ? 6. Will gain $30,636 in 1 yr. 9 mo. 18 da. at 5}^? 7. Will gain $74,221 in 2 yr. 3 mo. 9 da. at '7^% ? 589. Prob. VI.— To find the principal when the amount, time, and rate are given. Observe, that the amount is the principal plus the interest, and that the interest contains the interest (564) of $1 as many times as there are dollars in the principal ; consequently the amount must contain (500) $1 plus the interest of $1 for the given time at the given rate as many times as there are dollars in the principal ; hence, the following 590. EuLE. — Divide the amount hy the amount of $1 for the given time at the given rate. EXAMPIiES FOR PRACTICB. 591, 1. What sum of money will amount to $290.50 in 2 yr. 8 mo. 12 da. at 6^ ? Solution. —The amount of $1 for 2 yr. 8 mo. 12 da. at 6% is $1,162. Now since $1,162 is the amount of $1 for the given time at the given rate, $290.50 is the amount of as many dollars as $1,162 is contained times in it Hence, $290.50 -f- $1 162 — $250, is the required principal. (252) SIMPLE INTEREST. 125 2. What principal will amount to 1310.60 in 3 yr. 5 mo. 9 da. at b% ? 3. What is the interest for 1 yr. 7 mo. 13 da. on a sum of money which in this time amounts to $487.65, at 7% ? 4. What sum of money at 10^ will amount to $436.02 in 4 yr. 8 mo. 23 da. 5. At S% a certain principal in 2 yr. 9 mo. 6 da. amounted to $699.82. Find the principal and the interest. 592. Prob. VII.— To find the rate when the princi- pal, interest and time are given. Observe, that the given interest must be as many times 1% oi the given principal for the given time as there are units in the rate ; hence the following 593. Rule. — Divide the given interest hy the interest of the given principal for the given time at 1 per cent. SXAMPIiES FOR PRACTICE. 594. 1. At what rate will $260 gain $45.50 in 2 yr. 6 mo.? Solution.— The interest of $260 for 2 yr. 6 mo. at 1% is $6.50. Now Bince $6.50 is 1 % of $260 for the given time, $45.50 is as many per cent as $6.50 is contained times in $45.50 ; hence, $45.50 -i- $6.50 = 7, is the required rate. At what rate per cent 2. Will $732 gain $99,674 in 2 yr. 3 mo. 7 da. ? 3. AVill $524 gain $206.63 in 5 yr. 7 mo. 18 da. ? 4. Will $873 gain $132.89 in 1 yr. 10 mo. 25 da.? 5. Will $395.80 gain $53,873 in 2 yr. 8 mo. 20 da.? 6. Will $908.50 gain $325,422 in 4 yr. 2 mo. 17 da. ? 7. A man purchased a house for $3486, which rents for $418.32. What rate per cent does he make on the invest- ment ? 8. Which is the better investment and what rate per cent (253) 126 BUSINESS ARITHMETIC. per annum, ^4360 which yields in 5 years 11635, or $3860 which yields in 9 years 12692.45 ? 9. At what rate per cent per annum will a sum of money double itself in 7 years ? Solution.— Since in 100 years at 1 % any sum doubles itself, to double itself in 7 years the rate per cent must be as many times Ifo as T is con- tained times in 100, which is 14f . Hence, etc. ] 0. At what rate per cent per annum will any sum double itself in 4, 8, 9, 12, and 25 years respectively ? 11. At what rate per cent per annum will any sum triple or quadruple itself in 6, 9, 14, and 18 years respectively ? 12. Invested $3648 in a business that yields 11659.84 in 5 years. What per cent annual interest did I receive on my investment ? 595. Prob. VIII. — To find the time when the princi- pal, interest, and rate are given. Observe, that the interest is found (580) by multiplying the interest of the given principal for 1 year at the given rate by the time expressed in years ; hence the following 596. EuLE. — /. Divide the given interest hy the inter- est of the given principal for 1 year at the given rate. II. Reduce (579), when called for, fractions of a year to months and days, BXAMPIiBS FOR PRACTICS. 597. 1. In what time will S350 gain $63 at 8,^ ? Solution.— The interest of $350 for 1 yr. at 8% is $28. Now since $28 is the interest of $350 at 8% for 1 year, it will take as many years to gain $63 as $28 is contained times in $63 ; hence $63 h- $28 = 2\ jt., or 2 yr. 3 mo., the required time. In what time will 2. $80 gain $36 at 7i%? 6. $477 gain $152.64 at 12^? 3. $460 gain $80.50 at 5% ? 6. $600 gain $301,392 at 7^%? 4. $260 gain $98.80 at 8% ? 7. $385 gain $214.72 at S^^ ? (254) COMPOUND INTEREST. 127 8. My total gain on an investment of $860 at 1% per annum, is $455.70. How long has the investment been made ? 9. How long will it take any sum of money to double itself at 1% per annum ? Solution. — At 100% any sum will double itself in 1 year ; hence to double itself at 7/^ it will require as many years as 7% is contained times in 100%, which is 14f. Hence, etc. Observe, that to find how long it will take to triple, quadruple, etc., any sum, we must take 200%, 300%, etc. 10. How long will it take any sum of money at b%, 8%, 6 J^, or 9% per annum to double itself? To triple itself, etc. ? 11. At 7% the interest of $480 is equal to 5 times the prin- cipal, How long has the money been on interest ? COMPOUND INTEREST. 598. Compound Interest is interest upon principal and interest united, at given intervals of time. Observe, that the interest may be made a part of the principal, or compounded at any interval of time agreed upon ; as, annually, semi-an- nually, quarterly, etc. 599. Prob. IX. — To find the compound interest on any sum for any given time. Find the compound interest of $850 for 2 yr. 6 mo. at 6^. $850 Prin. for Ist yr. EXPLANATION.— Since at 6% the amount IQQ is 1.06 of the principal, we multiply $850, 7 the principal for the first year, by 1.06, giv- $901 Prin. for 2d yr. i^g ^901^ the amount at the end of the first 1.06 year, which forms the principal for the ^955.00 Prin. for 6 mo. *^^^^ >^^^^' ^^ *^^ ^^^® manner we find ^ „^ $955.06, the amount at the end of the 1_L second year which forms the principal for $983.71 Total amount. the 6 months. 1850 Given Prin. 2. Since 6% for one year is dfo for 6 months, we multiply $955.06, the principal $133.71 Compound Int. fo^ ^^^^ g months, by 1.03, which gives the total amount at the end of the 2 years 6 months. (255) 12S BUSINUSS ARITHMETIC. 3. From the total amount we subtract $850, the given principal, which gives $133.71, the compound interest of $850 for 2 years 6 months at 6^. Hence the following 600. EuLE. — I. Find the amount of the principal for the first interval of time at the end of which interest is due, and make it the principal for the second interval. II. Find the amount of this principal for the second interval of time, and so continue for each successive interval and fraction of an interval, if any. III. Subtract the given principal from the last amount and the remainder will he the compound interest, CXAMPLCS FOR PRACTICE. 601. 1. What is the compound interest of $650 for 3 years, at 7^, payable annually? 2. Find the amount of $870 for 2 years at 6^ compound interest. 3. Find tlie compound interest of 1380.80 for 1 year at 8^, interest payable quarterly. 4. What is the amount of $1500 for 2 years 9 months at 8^ compound interest, payable annually ? 5. What is the amount of $600 for 1 year 9 months at h% compound interest, payable quarterly ? 6. What is the difference in the simple interest and com- pound interest of $480 for 4 yr. and 6 mo. at 7%' ? 7. What is the annual income from an investment of $2860 at 7^ compound interest, payable quarterly ? 8. What will be the compound interest at the end of 2 yr. 5 mo. on a note for $600 at 7^, payable semi-annually ? 9. A man invests $3750 for 3 years at 7^ compound interest, payable semi-annually, and the same amount for the same time at '^\% simple interest. Which will yield the greater amount of interest at the end of the time, and how much ? (256) INTEREST TABLES, 120 INTEREST TABLES. 602. Interest, both simple and compound, is now almost invariably reckoned by means of tables, which give the interest or amount of SI at different rates for years, months, and days. The following illustrate the nature and use of such tables. Table showing the simple interest of $1 at 6, 7, and 8% for years, months, and days. Years, 1 6%. 7/.. .08 Years^ 4 6^. 7/. 8%. .06 .07 .34 .28 .32 3 .12 .14 .16 5 .30 .35 .40 3 .18 21 .24 6 .36 .42 .48 Months, 1 Months. 7 .005 .00583 .00666 .035 .04083 .04666 2 .01 .01166 .01833 1 8 .04 .04666 .05333 3 .015 .01750 .02000! 9 .045 .05250 .06000 4 .02 .03333 .02660 10 .05 .05833 .06666 5 .025 .02916 .03333 11 .055 .06416 .07333 6 Days, 1 .03 .03500 .04000 Days, 16 .00016 00019 .00032 1 — ■ .00366 .00311 .00355 2 .00033 .00038 .00044 17 .00283 .00330 .00377 3 .00050 .00058 .00066 18 .00300 .00350 .00400 4 .00060 .00077 .00088 19 .00316 .00369 .C0422 5 .00083 .00007 .00111 20 .00333 .00388 .00444 6 .00100 .00116 .00133 i 31 .00350 .00408 .00160 7 .00116 .00136 .00155 32 .00366 .00427 .00488 8 .00133 .00155 .00177 33 .00383 .00447 .00511 9 .00150 .00175 .00200 24 .00400 .00466 .00533 10 .00166 .00194 .00223 25 .00416 .00486 .00555 11 .00183 .00213 .00344 36 .00433 .00505 .00577 12 .00200 .00333 .00266 37 .00450 .00525 .00600 13 .00216 .00252 .00388 38 .00466 .00544 .00633 14 .00233 .00272 .00311 39 .00483 .00563 .00644 15 .00250 .00291 .00333 (257) 130 B USINES S ARITHMETIC, Method of using the Simple Interest Table, 603. Find the interest of $250 for 5 yr. 9 mo. 18 da. at 7;^. 1. We find the interest of %\\ j f^^^ ^^*"'^^<^ ^^ *^^1^ ^°^ ^ y^' for,.,^^t^ \-\^^ ^^ ;; ;; :;^- Interest of $1 for 5 yr. 9 mo. 18 da is .406 of $1. 2. Since the interest of $1 for 5 yr. 9 mo. 18 da. is .406 of $1, the interest of $350 for the same time is .406 of $350. Hence, $350 x .406 = $101.50, the required interest. EXAMPLES FOR PRACTICE. 604. Find by using the table the interest, at ^%, of 1. $860 for 3 yr. 7 mo. 23 da. 4. $325.86 for 5 yr. 13 da. 2. $438 for 5 yr. 11 mo. 19 da. 5. $796.50 for 11 mo. 28 da. 3. $283 for 6 yr. 8 mo. 27 da. 6. $395.75 for 3 yr. 7 mo. Table sliowing tlie amount of $1 at 6, 7, and 8% compound interest from 1 to 12 years. YRS. 6%. 7%. 8%. YRS. 7 6%. 7%. 8f., 1 1.060000 1.070000 1.080000 1.503630 1.605781 1.713824 2 1.123000 1.144900 1.166400 1 8 1.593848 1.718180 1.850930 3 1.191016 1.235043 1.269712 9 1.089479 1.838459 1.999005 4 1.362477 1.310796 1.360489 10 1.790848 1.967151 3.158935 5 1.338226 1.402552 1.469328 11 1.898299 2.104852 3.331639 6 1.418519 1.500730 1.586874 13 2.013197 2.252192 3.518170 3Iethod of using the Compound Interest Table, 605. Find the compound interest of $2800 for 7 years at G^. 1. The amount of $1 for 7 years at 6% in the tahle is 1.50303. 3. Since the amount of $1 for 7 years is 1 .50363, the amount of $3800 for the same time must be 3800 times $1.50363 = $1.50303 x 2800 r:^ $4310.164. Hence, $4210. 164 -$3800 = $1410. 161, the required interest. (258) ANNUAL INTEREST, 131 BXAMPIiES FOR PRACTICE. 606. Find by using the table the compound interest of 1. $500 for 9 yr. at 1%, 8. 1384.50 for 8 yr. at Q%, 2. 82000 for 5 yr. at 8^. 9. 8400 for 4 yr. 1 mo. at 1%. 3. 8870 for 11 yr. at Q%. 10. $900 for 6 yr. 3 mo. at 8^. 4. 83800 for 7 yr. at 8^. 11. 8690 for 12 yr. 8 mo. at G^. 5. 82500 for 3 J yr. at 6^. 12. $4000 for 9 yr. 2 mo. at 1%. 6. 8640 for 4J yr. at 8^. 13. 83900 for 4 yr. 3 mo. at Q%, 7. $285 for 9J yr. at !%• 14. 8600 for 11 yr. G mo. at 8^. ANNUAL INTEREST. 60T. Annual Interest is simple interest on the prin- cipal, and each year's interest remaining unpaid. Annual interest is allowed on promissory notes and other contracts which contain the words, " interest payable annually if the interest remains unpaid." 608. Prob. X. — To find the annual interest on a promissory note or contract. What is the interest on a note for 8600 at 7^ at the end of 3 yr. 6 mo., interest payable annually, but remaining unpaid. Solution. — 1. At 7^ the payment of interest on $600 due at the end of each year is $42, and the simple interest for 3 yr. 6 mo, is $147. 2. The first payment of $42 of interest is due at the end of the first year and must bear simple interest for 2 yr. 6 mo. The second payment is due at the end of the second year and must bear simple interest for 1 y r. 6 mo, , and the third payment being due at the end of the third year must bear interest for 6 mo. Hence, there is simple interest on $42 for 2 yr. 6 mo. + 1 yr. 6 mo. + 6 mo. = 4 yr. 6 mo. , and the interest of the $42 for this time at 7 % is $13.23. 3. The simple interest on $600 being $147, and the simple interest on the interest remaining unpaid being $13.23, the total interest on the note at the end of the given time is $160.23. (259) Id2 BUSINESS ARITHMETIC, EXAMPIiES FOR PRACTICE. 609. 1. How much interest is due at the end of 4 yr. 9 mo. on a note for $460 at G%, interest payable annually, but remaining unpaid ? 2. Wilbur H. Reynolds has J. G. MacVicar's note dated July 29, 1876, for $800, interest payable annually ; what will be due November 29, 1880, at 7^ ? 3. Find the amount of $780 at 11% annual interest for 5i yr. 4. What is the difference between the annual interest and the compound interest of $1800 for 7 yr. at 7^ ? 5. What is the annual interest of $830 for 4 yr. 9 mo. at S% ? 6. What is the difference in the simple, annual, and com- pound interest of $790 for 5 years at S% ? PARTIAL PAYMENTS. 610. A Promissory Note is a written promise to pay a sum of money at a specified time or on demand. The Face of a note is the sum of money made payable by it. The Maker or Drawer of a note is the person who signs the note. The Payee is the person to whom or to whose order the money is paid. An Indorser is a person who signs his name on the back of the note, and thus makes himself responsible for its payment. 611. A Negotiable JSFote is a note made payable to the bearer, or to some person's order. When a note is so written it can be bought and sold in the same manner as any other property. 612. A Partial Payment is a payment in part of a note, bond, or other obligation. 613. An Indorsement is a written acknowledgment of a partial payment, placed on the back of a note, bond, etc, stating the time and amount of the same. (260) PARTIAL PAYMENTS, 133 MERCANTILE RULE. 614. The method of reckoning partial payments known as the Mercantile Rule is very commonly used in computing interest on notes and amounts running for a year or less. The rule is as follows : 6J 5. Rule. — I. Find the amount of the note or debt from the time it begins to bear interest, and of each paynieiit until the date of settlement. II. Subtract the sum of the amounts of payments from the amount of the note or debt; the remainder will be the balance due. Observe, that an accurate application of the rule requires that the exact interest should be found according to (683). EXAMPLES FOR PRACTICE, 616. . 1. $900. Potsdam, N. Y., Sept. M, 1876. On demand I promise to pay Henry Watkins, or order, nine hundred dollars with interest, value received. Warrei^ Mankt. Indorsed as follows: Oct. 18th, 1876, $150; Dec. 22d, 1876, $200 ; March 15th, 1877, $300. What is due on the note July 19th, 1877 ? 2. A note for $600 bearing interest at 8^ from July 1st, 1874, was paid May 16th, 1875. The indorsements were: July 12th, 1874, $185; Sept. 15,1874, $76; Jan. 13, 1875, $230 ; and March 2, 1875, $115. What was due on the note at the time of payment ? 3. An account amounting to $485 was due Sept. 3, 1875, and was not settled until Aug. 15, 1876. The payments made upon it were: $125, Dec. 4, 1875; $84, Jan. 17, 1876; $95, June 23, 1876. What was due at the time of settlement, allowing interest at 1% ? (361) 134 B USINESS ARITHMETIC, 4. 1250. Ogdensburg, N. Y., Mardi 25, 1876. Ninety-eight days after date I promise to pay E. D. Brooks, or order, two hundred fifty dollars with interest, value received. Silas Jones. Indorsements : 187, April 12, 1876 ; $48, May 9. What is to pay when the note is due ? UNITED STATES EULE. 617. The United States courts have adopted the following for reckoning the interest on partial payments : 618. Rule. — I. Find the aviount of the given prin- cipal to the time of the first payment ; if the payment equals or exceeds the interest then due, subtract it from the avvount obtained and regard the rem^ainder as the new principal. II. If the payment is less than the interest due, find the amount of the given pHncipal to a time when the sum of the payments equals or exceeds the interest then due, and subtract the sum of the payments from this amount, and regard the remainder as the new prin- cipal. III. Proceed with this new pHncipal and with each succeeding principal in the same manner. 619. The method of applying the above rule will be seen from the following example : 1. A note for ^900, dated Syracuse, Jan. 5th, 1876, and paid Dec. 20th, 1876, had endorsed upon it the following payments: Feb. 23d, 1876, $40; April 26th, $6; July 19th, 1876, $70. How much was the payment Dec. 20th, 1876, interest at 7^ ? (262) PARTIAL PAY31ENTS, 135 SOLUTION. First Step. 1. The first principal is the face of the note ....... $900 2. We find the interest from the date of the note to the first payment, Feb. 23, 1876 (49 da.), at 7% 8.43 Amount $908.43 5. The first payment, $40, being greater than the interest then due, is subtracted from the amount 40.00 Second prindpcU .... $868.43 Second Step, 1. The second principal is the remainder after subtracting the first payment from the amount at that date .... $868.43 2. The interest on $868.43, from Feb. 23 to Apr. 26, 1876 (63 da.), is $10,463 3. This interest being greater than the second pay- ment ($6), we find the interest on $868.55 from April 26 to July 19, 1876, (84 da.), which is 13.951 Interest from first to third payment . $24,414 24.414 Amount $892,844 4. Tlie sum of the second and third payments being greater than the interest due, we subtract it from the amount . 76 Third principal .... $816,844 Third Step. We find the interest on $816,844, from July 19 to Dec. 20, 1876 (154 da.), which is 24.057 Payment due Dec. 20, 1876 $840.90 In the above example, the interest has been reckoned according to (583) ; in the following, 360 days have been regarded as a year. EXAMPLES FOR PRACTICE. 2. A note for $1630 at S% interest was dated March 18, 1872, and was paid Aug. 13, 1875. The following sums were endorsed upon it: $160, Feb. 12, 1873; $48, March 7, 18 H; and $350, Aug. 25, 1874. How much was paid Aug. 13? (263) 136 BUSINESS ARITHMETIC. 3. A mortgage for ^^3500 was dated Aug. 24, 1873. It had endorsed upon it the following payments : May 17, 1874, $89 ; Sept. 12, 1874, $635; March 4, 1875, $420. How much was due upon it Feb. 9, 1876, interest at 11% ? 4. What was the last payment on a note for $1000 at 7^, which was dated Jan. 7, 1876, and paid Dec. 26, 1876, endorsed as follows : April 12, 1876, $16 ; July 10, 1876, $250 ; and Oct. 26, 1876, $370 ? ' 5. A mortgage for $4600, dated Leavenworth, Kansas, Sept. 25, 1871, had endorsed upon it: $400, June 23, 1872; $125, Aug. 3, 1873 ; $580, May 7, 1874 ; and $86, Mar. 5, 1875. How much was due upon it Sept. 25, 1875, interest at 11^ ? DISCOUNT. 620. Discount is a deduction made for any reason from an account, debt, price of goods, and the like, or for the interest of money advanced upon a bill or note due at a future date. 631. The Present Worth of a note, debt, or other obliga- tion, payable at a future time without interest, is such a sum as, being placed at interest at a legal rate, will amount to the given sum when it becomes due. 632. True Discount is a difference between any sum of money payable at a future time and its present ivorth. 633. Prob. XL — To find the present worth of any sum. Find the present worth of a debt of $890, due in 2 yr. 6 mo. without interest, allowing %% discount. Solution.— Since $1 placed at interest for 2 yr. 6 mo. at 8% amounts to $1.20, the present worth (021) of $1.20, due in 3 yr.6 mo., is $1. Hence the present worth of $890, which is due ■^^ithout interest in 2 yr. 6 mo., must contain as many dollars as $1.20 is contained times in $890 = $741.66. Observe, that this problem is an application of (506, Prob. XI). (864) PRESENT WORTH, 137 EXAMPIiES FOR PRACTICE. 624. What is the present worth 1. Of $800 at 6^, due in 6 mo. ? At 8^, due in 9 mo. ? 2. Of $360 at 7^, due in 2 yr, ? At b%, due in 8 mo. ? 3. Of $490 at %%, due in 42 da. ? At 1%, due in 128 da. ? What is the true discount 4. Of $580 at t%, due in 90 da. ? At 8^^, due in 4 yr. 17 da.? 5. Of $260 at %\%, due in 120 da.? At 9^, due in 2 yr. 25 da.? 6. Of $860 at 7^, due in 93 da. ? At 12^, due in 3 yr. 19 da.? 7. What is the true discount at ^% on a debt of 13200, due in 2 yr. 5 mo. and 24 da. ? X 8. Sold my farm for $3800 cash and a mortgage for $6500 running for 3 years without interest. The use of money being worth 1% per annum, what is the cash value of the farm? 9. What is the difference between the interest and true dis- count at 1% of $460, due 8 months hence ? 10. A man is offered a house for $4800 cash, or for $5250 payable in 2 yr. 6 mo. without interest. If he accepts the former, how much will he lose when money is worth 8^ ? 11. Which is more profitable, and how much, to buy wood at $4.50 a cord cash, or at $4.66 payable in 9 months without interest, money being worth 8;^? X 12. A merchant buys $2645.50 worth of goods on 3 mo. credit, but is offered 3^ discount for cash. Which is the better bargain, and how much, when money is at 1% per annum ? 13. A grain merchant sold 2400 bu. of wheat for $3600, for which he took a note at 4 mo. without interest. What was the cash price per bushel, when money is at 6^ ? (265) 138 BUSIIfESS ARITHMETIC, BANK DISCOUNT. 625. ^anlz Discount is the interest on the amount of a note at maturity, computed from the date the note is discounted to the date of maturity. 1. Observe that when a note bears no interest, its amount at maturity is its face. 2. When the time of a note is given in months, calendar months are understood ; if the month in which the note falls due has no day cor- responding to the date of the note, then the note is due on the last day of that month. 3. In computing bank discount, it is customary to reckon the time in days. 4. When a note becomes due on Sunday or a legal holiday, it must be paid on the day previous. 636. JDcujs of Grace are three days usually allowed by law for the payment of a note after the expiration of the time specified in it. 627. The MaUirity of a note is the expiration of the time for which it is made, including days of grace. 628. The I^poceeds or Avails of a note is the sum left after deducting the discount. 629. A Protest is a declaration in writing by a Notary Public, giving legal notice to the maker and endorsers of a note of its non-payinent. 630. Prob. XII.— To find the bank discount and proceeds of a note for any g:iven rate and time. Observe, that the bank discount is the interest on the amount of the note at maturity, computed from the time the note is discounted to the date of maturity ; and the proceeds is the amount of the note at maturity, minus the bank discount. Hence the following 631. EuLE. — J. Fmcl the amount of the note at ma- tioj^ity, compute the interest upon this sum from the date of discounting the note to the date of Wjaturity ; the result is the hanh discount. (266) BANK DISCOUNT, 139 II. Subtract the hank discount from the amount of the note at maturity ; the remainder is the proceeds. EXAl^IPIiES FOR PRACTICE. 632. What are the bank discount and proceeds of a note 1. Of ^280 for 3 mo. 15 da. at 1% ? For 6 mo. 9 da. at S% ? 2. Of 1790 for 154 da. at 6% ? For 2 mo. 12 da. at 7% ? 3. Of $1600 for 80 da. at 7% ? For 140 da. at 8^% ? 4. What is the difference between the ba7ik and true dis- count on a note of $1000 at 7%, payable in 90 days ? 5. Valuing my horse at $212, I sold him and took a note for $235 payable in 60 days, which I discounted at the bank. How much did I gain on the transaction ? ^ 6. A man bought 130 acres of land at $16 per acre. He paid for the land by discounting a note at the bank for $2140.37 for 90 da. at 6^. How much cash has he left? Find the date of mattirityj the time, and th« proceeds of the following notes : 7. $480.90. RocHESTEK, N. Y., Mar. 15, 1876. Seventy days after date I promise to pay to the order of N. L. Sage, four hundred eighty t% dollars for value received. Discounted Mar. 29. DuNCAi^ MacVicar. 8. $590. Potsdam, N. Y, Map 18, 1876. Three months after date I promise to pay to the order of Wm. Flint, five hundred ninety dollars, for value received. Discounted June 2. Peter Henderson. 9. $1600. Rome, N. Y., Jan. 19, 1876. Seven months after date we jointly and severally agree to pay James Richards, or order, one thousand six hundred dol- lars at the National Bank, Potsdam, N. Y., value received. Discounted May 23. Robert Button, James Jackson. (267) 140 BUSINESS ARITHMETIC. 633. Prob. XIII.— To find the face of a note when the proceeds, time, and rate are given. Observe, that the proceeds is the face of the note minus the interest on it for the given time and rate, and consequently that the proceeds must contain $1 minus the interest of $1 for the given time and rate as many times as there are dollars in the face of the note. Hence the following 634. Rule. — Divide the given proceeds hy the pro- ceeds of $1 for the given time and rate ; the quotient is the face of the note. fiXAMPLES FOR PRACTICE. 635. What must be the face of a note which will give 1. For 3 mo. 17 da., at Q%, I860 proceeds? $290? $530.80? 2. For 90 da., at 7^, $450 proceeds ? $186.25 ? $97.32 ? 3. For 73 da., at $8^, $234.60 proceeds? $1800? $506.94? v4. "What must be the face of a note for 80 days, at t%, on which I can raise at a bank $472.86 ? 6. The avails of a note for 50 days when discounted at a bank were $350.80 ; what was the face of the note ? 6. How much must I make my note at a bank for 40 da., at 7^, to pay a debt of $296.40 ? 7. A merchant paid a bill of goods amounting to $2850 by discounting three notes at a bank at '7%, the proceeds of each paying one-third of the bill ; the time of the first note was 60 days, of the second 90 days, and of the third 154 days. What was i\iQ face of each note ? 8. For what sum must I draw my note March 23, 1876, for 90 days, so that when discounted at 7^ on May 1 the pro- ceeds may be $490 ? 9. Settled a bill of $2380 by giving my note for $890 at 30 days, bearing interest, and another note at 90 days, which when discounted at 1% will settle the balance. What is the face of the latter note ? (268) '^^ t EXCHANGE 636. Exchange is a method of paying debts or other obligations at a distance without transmitting the money. Thus, a merchant in Chicago desiring to pay a debt of $1800 in New York, pays a bank in Chicago $1800, plus a small per cent for their trouble, and obtains an order for this amount on a bank in New York, which he remits to his creditor, who receives the money from the New York bank. Exchange between places in the same country is called Inland or Domestic Exchange, and between different countries Foreign Exchavge. 637. A Draft or Bill of Exchange is a written order for the payment of money at a specified time, drawn in one place and payable in another. 1. The Drawer of a bill or draft is the person who signs it; the Drawee, the person directed to pay it ; the Payee, the person to whom the money is directed to be paid ; the Jndorser, the person who transfers his right to a bill or draft by indorsing it ; and the Holder, the person who has legal possession of it. 2. A Sight Draft or Bill is one which requires payment to be made when presented to the payor. 3. A Time Draft or BiU is one which requires payment to be made at a specified time after date, or after sight or being presented to the pay(yr. Three days of graco are usually allowed on bills of exchange. 4. The Acceptance of a bill or draft is the agreement of the party on whom it is drawn to pay it at maturity. This is indicated by writing the word "Accepted" across the face of the bill and signing it. When a bill is protested for non-acceptance, the drawer is bound to pay it immediately. 5. Foreign bills of exchange are usually drawn in duplicate or tripli- cate, and sent by different conveyances, to provide against miscarriage, each copy being valid until the bill is paid. (269) 142 BUSIJVUSS ARITHMETIC. 638. The Par of Exchange is the relative value of the coins of two countries. Thus, the par of exchange between the United States and England is the number of gold dollars, the standard unit of United States money, which is equal to a pound sterling, the standard unit of English money. Hence $4.8665 = £1 is the par of exchange. DOMESTIC EXCHANGE. 639. Domestic JExchange is a method of paying debts or other obligations at distant places in the same coun- try, without transmitting the money. Fortns of Sight and Time Drafts, Third National Bank op Rochester, ) $890. Rochester, N. Y., May 4, 1876. \ At sight, pay to the order of Chas. D. McLean, eight hun- dred ninety dollars. William Roberts, Cashier. To the Seventh National Bank, ) New York, N. Y. \ This is the usual form of a draft drawn by one bank upon another. 12700. ' Syracuse, N. Y., July 25, 1876. At fifteen days sight, pay to the order of Taintor Brothers & Co., tioo thousand seven hundred dollars, vahce received, and charge the same to the account of ^ ^ Stewart & Co. To the Tenth National Bank, ) New York, N. Y. \ 1. This is the usual form of a draft drawn by a firm or individual upon a bank. It may also be made payable at a given time after date. 2. All time drafts should be presented for acceptance as soon as received. When the cashier writes the word "Accepted," with the date of acceptance across the face, and signs his name, the bank is responsible for the payment of the draft when due. (370) DOMESTIC EXCHANGE. 143 METHODS OF DOMESTIC EXCHANGE. 640. First Method. — Tlie party desiring to Jransmit mo7iey, purchases a draft for the amount at a banic, and sends it by mail to its destination. Observe carefully the following: 1. Banks can sell drafts only upon others in which they have deposits in money or equivalent security. Hence banks throughout the country, in order to give them this facility, have such deposits at centres of trade, such as New York, Boston, Chicago, etc. 2. A Bank Draft will usually be purchased by banks in any part of the country, in case the person offering it is fully identified as the party to whom the draft is payable. Hence, a debt or other liability may be discharged at any place by a draft on a New York bank. 3. A draft may be made payable to the person to whom it is sent, or to the person buying it. In the latter case the person buying it must write on the back " Pay to the order of " (name of party to whom it is Bent), and sign his own name. Second Method. — /. The party desiring to transmit money, deposits the amotmt in a bank and takes a certificate of deptosit, tvhich he sends as by first method. Or, II. If he has a deposit already in a bank, subject to his check or order, it is customary to send his check, certified to be good by the cashier of the bank. This method, in either of these forms, is ordinarily followed in making payments at a distance by persons in New York and other large centres of trade. Banks in such places have no deposits in cities and villages throughout the country, and hence do not sell drafts. Certificates of deposits and certified checks are. purchased by banks in the same manner as bank drafts. Third Method. — The party desiring to transmit money, obtains a Post Office order for the amount and remits it as before. As the amount that can be included in one Post Ofiice order is limited, this method is restricted in its application. It is usually employed in remitting small sums of money. (371) 144 BUSIN^BSS ARITHME TIC, Fourth Method. — The party desiring to transmit money, makes a draft or order for the amount upon apartyoivinghim, at the place tvhere the money is to be sent, and remits this as previously directed. 1. By this method one person is snid to draw upon another. Such drafts should be presented for payment as soon as received, and if not paid or accepted should be protested for non-payment immediately. 2. Large business firms have deposits in banks at business centres, and credit with other business firms ; hence, their drafts are used by them- selves and others the same as bank drafts. 641. The Premium or Discount on a draft depends chiefly on the condition of trade between the place where it is pur- chased and the place on which it is made. Thus, for example, merchants and other business men at Buffalo con- tract more obligations in New York, for which they pay by draft, than New York business men contract in Buffalo ; consequently, banks at Buffalo must actually send money to New York by Express or other conveyance. Hence, for the expense thus incurred and other trouble in handling the money, a small premium is charged at Buffalo on New York drafts. EXAMPIiES FOR PRACTICE. 642. 1. What is the cost of a sight draft for $2400, at |^ premium ? Solution.— Cost = $2400 + |% of $2400 = $2416. 2. What is the cost of a draft for -^3200, at \% premium ? Solution.— Cost = $3300 + i% of $3200 = $3204. Find the cost of sight draft 3. For $834, premium 2%. 6. For $1500, discount ^%. 4. For $6300, premium ^%. 7. For $384.50, discount ^%. 5. For $132.80, premium 1%. 8. For $295.20, discount 1|^. 9. The cost of a sight draft purchased at 1^% premium is $493.29 ; what is the face of the draft? Solution.— At 1|% premium, $1 of the face of the draft cost $1,015. Hence the face of the draft is as many dollars as $1,015 is contained times in $493.29, which is $486. (372) DOMESTIC EXCHANGE. 145 Find the face of a. draft which cost 10. $575.41, premium 2f^. 13. $819.88, discount J^. 11. $731.70, premium 1^%, 14. $273,847, discount \%. 12. $483.20, premium, f^. 15. $315.65, discount If ^. 16. What is the cost of a draft for $400, payable in 3 mo., premium IJ^, the bank allowing interest at 4^ until the draft is paid ? Solution.— A sight draft for $400, at \\% premium, costs $406, but the bank allows interest at 4% on the face, $400, for 3 mo., which is $4. Hence the draft will cost Find the cost of drafts , 17. For $700, premium \%, time 60 da., interest at Z%, 18. For $1600, premium IJ^, time 50 da., interest at 4^. 19. For $2460, discount f^, time 90 da., interest at ^%. 20. For $1800, discount 1%, time 30 da., interest at b%, 21. A merchant in Albany wishing to pay a debt of $498.48 in Chicago, sends a draft on New York, exchange on New York being at \% premium in Chicago ; what did he pay for the draft ? Solution. — Tlie draft cashed in Chicago commands a premium of ^% on its face. The man requires, therefore, to purchase a draft whose face plus ^% of it equals $498.48. Hence, according to (506 — 5), the amount paid, or face of the draft, is $498.48 -f- 1.005 = 22. Exchange being at 98} {1^% discount), what is the cost of a draft, time 4 mo. , interest at 6% ? 23. The face of a draft which was purchased at- 1^% pre- mium is $2500, the time 40 da., rate of interest allowed 4^; what was its cost ? 24. My agent in Detroit sold a consignment of goods for $8260, commission on the sale 2^%. He remitted the pro- ceeds by draft on New York, at a premium of i%. What is the amount remitted ? (273) 146 BUSINESS ARITHMETIC, FOREIGN EXOHAISTGE. 643. Foreign Exchange is a method of paying debts or other obligations in foreign countries without transmitting the money. OUerce, that foreign exchange is based upon the fact that different countries exchange products, securities, etc., with each other. Thus, the United States sells wheat, etc., to England, and England in return sells manufactured goods, etc., to the United States. Hence, par- ties in each country hecome indebted to parties in the other. For this reason, a merchant in the United States can pay for goods purchased in England hy buying an order upon a firm in England which is indebted to a firm in the United States. Form of a Bill or Set of Exchange, £400. New York, July 13, 1876. At sight of this First of Exchange (second and third of the sa7ne date and tenor unpaid), pay to the order of E. D. Blakeslee Four Hundred Pounds Sterling, for value received, and charge the same to the account of Williams, Beown" & Co. To Martin, Williams & Co., London. The person purchasing the exchange receives three bills, which he sends by different mails to avoid miscarriage. When one has been received and paid, the others are void. The above is the form of the first bill. In the Second Bill the word *• First " is used instead of " Second," and the parenthesis reads, " First and Tliird of the same date and tenor unpaid." A similar change is made in the Third Bill. 644. Exchange with Europe is conducted chiefly through prominent financial centres, as London, Paris, Berlin, Antwerp, Amsterdam, etc. 645. Quotations are the published rates at which bills of exchange, stocks, bonds, etc., are bought and sold in the money market from day to day. (274) FOREIGN EXCHANGE, 147 These quotations give the market gold value in United States money of one or more units of the foreign coin. Thus, quotations on London give the value of £1 sterling in dollars ; on Paris, Antwerp, and Geneva, the value of $1 in francs ; on Hamburg, Berlin, Bremen, and Frankfort, the value of 4 marks in cents; on Amsterdam, the value of a guilder in cents. 646. The following table gives the par of exchange, or gold value of foreign monetary units, as published by the Secretary of the Treasury, January 1, 1876 : TABLE OF PAR OF EXCHAN"GE. COUNTBIES. MONETARY UNIT. STANDARD. VATUE IN U. S. MONET. Austria Florin Silver .45, 3 .11), 3 .96, 5 .54, 5 .96, 5 $1.00 .91, 8 .91, 3 .26, 8 .91, 8 4.97, 4 .19, 3 4.86, 6i .19. 3" .23, 8 .09, 7 .43, 6 .19, 3 1.00 .99, 8 .38, 5 .26, 8 .91, 8 1.08 .73, 4 1.00 .19, 3 .26, 8 .19, 3 .^2, 9 .11, 8 ,04, 3 .91, 8 Belt^ium Bolivia Franc Gold and silver. Gold and silver. Gold Gold Dollar Brazil Milreis of 1000 reis. Peso Bogota Canada Dollar Gold : Central America. Chili Dollar Silver Peso Gold Denmark Ecuador Crown Gold Dollar .... Silver Effvpt Pound of 100 piasters . . Franc. . Pound sterling Drachma Gold France Gold and silver. Gold Great Britain. . . . Greece Gold and silver. Gold German Empire . Japan Mark Yen Gold India Rupee of 10 annas Lira . . Silver Italy Gold and silver. Gold Silver. Liberia Dollar Mexico Dollar Netherlands Norway Florin Gold and silver. Gold Crown Peru Dollar Silver . . Portugal Russia Sandwich Islands Spain Milreis of 1000 reis Rouble of 100 copecks. . Dollar Peseta of 100 centimes . Crown Franc Gold Silver Gold Gold and silver. Gold Sweden Switzerland Tripoli Gold and silver. Silver Mahbub of 20 piasters. . Piaster of 16 caroubs. . . Piaster Tunis Silver Turkey Gold U. S. of Colombia Peso..- Silver (375) 148 BUSINESS ARITHMETIC, METHODS OF DIRECT EXCHANGE. 647. Direct Exchange is a method of making pay- ments in a foreign country at the quoted rate of exchange with that country. First Method. — The person desiring to traiismit the money purchases a Set of Exchange for the amoiuit on the country to tuhich the money is to be sent, and forwards the three hills ly different 7nails or routes to their destination. Second Method. — Hie person desiring to transmit the money instructs his creditor in the foreig^i country to draw upon him, that is, to sell a set of exchange upon him, which he pays in his ow7i country when prese^ited, EXAMPLES FOR PRACTICES. 648. 1. What is the cost in currency of a bill of exchange on Liverpool for £285 9s. 6d., exchange being quoted at $4.88, and gold at 1.12, brokerage i% ? Solution. — 1. We reduce £285 9s. 6d. = £285.475 the 9s. 6d. to a decimal of £1. $4.88 X 285.475 = 11393.118 Hence £285 9s. 6d. = £285.475. $1.1225 X 1393.118= 1563.77+ ^- Since £1 = $4. 88, £285.475 must be equal $4.88x285.475 = $1393.118, the gold value of the bill without brokerage. 3. Since %\ gold is equal $1.12 currency, and the brokerage is ^%, the cost of $1 gold in currency is $1.1225. Hence the bill cost in cur- rency $1.1225 X 1393.118 = $1563.77 + . What is tlie cost of a bill on 2. London for £436 8s. 3d., sterling at 4.84|, brokerage ^% ? 3. Paris for 4500 francs at .198, brokerage ^% ? 4. Geneva, Switzerland, for 8690 francs at .189 ? 5. Antiverp for 4000 francs at .175, in currency, gold at 1.09? 6. Amsterdam for 8400 guilders at 41 J, brokerage ^% ? 7. Frankfort for 2500 marks, quoted at .974? (276) FOREIGN EXCHANGE. 149 8. A merchant in Boston instructed his agent at Berlin to draw on him for a bill of goods of 43000 marks, exchange at 24J, gold being at 1.08|, brokerage \% ; what did the mer- chant pay in currency for the goods ? METHODS OF INDIRECT EXCHANGE. 649. Indirect Exchange is a method of making payments in a foreign country by taking advantage of the rate of exchange between that country and one or more other countries. Observe carefully the following : 1. The advantage of indirect over direct exchange under certain finan- cial conditions which sometimes, owing to various causes, exist between different countries, may be shown as follows : Suppose exchange in New York to be at par on London, but on Paris at 17 cents for 1 franc, and at Paris on London at 24 francs for £1. With these conditions, a bill on London for £100 will cost in New York $486.65 ; but a bill on London for £100 will cost in Paris 24 francs x 100 = 2400 francs, and a bill on Paris for 2400 francs will cost in New York 17 cents x 2400 = $408. Hence £100 can be sent from New York to London by direct exchange for $486.65, and by indirect exchange or through Paris for $408, giving an advantage of $480.65 — $408 = $78.65 in favor of the latter method. 2. The process of computing indirect exchange is called Arbitration of Exchange. When there is only one intermediate place, it is called Simple Arbitration ; when there are two or more intermediate places, it is called Compound Arbitration. Either of the following methods may be pursued: First Mettlob.— The person desiring to transmit the money may huy a hill of exchange for the amount on an intermediate jdace^ which he sends to his agent at that place ivith instruc- tions to huy a hill with the proceeds on the place to which the money is to he sent, atid to for tear d it to the proper party. This is called the method hy remittance. (377) 150 BUSINESS ARITHMETIC, Second Method. — The person desiring tosend the money instructs his creditor to draw for the amount on his agent at an intermediate place, and his agent to draiu upon him for the same amount. This is called the method by drawing. Third Method. — TJie person desiring to send the money instructs his agent at an intermediate place to draio upon him for the amount, and buy a hill on the place to which the money is to he sent, and forward it to the proper party. This is called the method hy drawing and remitting. These methods are equally applicable when the exchange is made through two or more interniediate places, and the solu- tion of examples under each is only an application of compound numbers and business. Probs. VIII, IX, X, and XI. BXAMPIiES FOR PRACTICE, 650. 1. Exchange in New York on London is 4.83, and on Paris in London is 244- ; what is the cost of transmitting 63994 francs to Paris through London ? Solution.— 1. We find tlie cost of a bill of exchange in London for 63994 francs. Since 24^ francs = £1, 63994 -4- 24i is equal the number of £ in 63994 francs, which is £2612. 2. We find the cost of a bill of exchange in New York for £2613. Since £1 = $4.83, the bill must cost $4.83 x 2612 = |12615.96. 2. A merchant in !N"ew York wishes to pay a debt in Berlin of 7000 marks. He finds he can buy exchange on Berlin at .25, and on Paris at .18, and in Paris on Berlin at 1 murk for 1.15 francs. Will he gain or lose by remitting by indirect excliange, and how much ? 3. "What will be the cost to remit 4800 guilders from New York to Amsterdam through Paris and London, exchange being quoted as follows: at New York on Paris, .]n B. Whitney. Or, 1876. Mar. 17 May 10 Aug. 7 Tomdse. . . . " " at 4 mo. ** « at 2 mo. $400 880 540 1876. Apr. 13 June 15 • By cash .... '♦ draft at 30 da. $300 450 (284) EQUATION OF P ATME NTS. I57 Before examining the following solution, study carefvUy the three propositions under (658). SOLUTION BY PRODUCT METHOD. Due. Ami. Days. Products. Mar. 17. 400 x 204 = 81600 Sept. 10. 380 X 27 = 10260 Oct. 7. 540 Paid. Ami. Days, Products. Apr. 13. 200 X 177 = 35400 July 15. 250 X 84 = 21000 450 56400 Total debt, $1320 $91860 Amt. whose Int. for 1 da. is due to Creditor. Total paid, 450 56400 Amt; whose Int for 1 da. is due to Debtor. Balance, $870 $35460 Bai. whose Int. for 1 day is due to Creditor. Explanation.— Assuming Oct. 7, the latest maturity on either side of the account, as the date of settlement, the creditor is entitled to interest on each item of the debit side, and the debtor on each item of the credit side to this date (658). Hence, we find, according to (059 — 3), the amount whose interest for 1 day both creditor and debtor are entitled to Oct. 7. 2. The creditor being entitled to the most interest, we subtract the amount whose interest for 1 day tlie debtor is entitled to from the cred- itor's amount, leaving $3540, the amount whose interest for 1 day the creditor is still entitled to receive. 3. We find the sum of the debit and credit items, and subtract the latter from the former, leaving $870 yet unpaid. This, with 68 cents, the interest on $3540, is the amount equitably due Oct. 7, equal f 870.68. 4. According to (658-4), $3540 ^ $870 = 40«&, the number of days previous to Oct. 7 when the debt can be discharged by paying the bal- ance, $870, in cash, or by a note bearing interest. Hence the equated time of paying the balance is Aug. 27. The following points regarding the foregoing solution should be carefully studied : 1. In the given example, the sum of the debit is greater than the sum of the credit items ; consequently the balance on the account is due to the creditor. But the balance of interest being also due him, it is evi- dent that to settle the account equitably he should be paid the $870 before the assumed date of settlement. Hence the equated time of pay- ing the balance must be before Oct. 7. 2. Had the balance of interest been on the credit side, it is evident the debtor would be entitled to keep the balance on the account until the (285) 158 BUSINESS ARITHMETIC. interest upon it would be equal the interest due him. Hence the equated time of paying the balance would be after Oct. 7. 3. Had the balance of the account been on the credit side, the creditor would be overpaid, and hence the balance would be due to the debtor. Now in case the balance of interest is also on the credit side and due to the debtor, it is evident that to settle the account equitably the debtor should be paid the amount of the balance before the assumed date of settlement. Hence the equated time would be before Oct. 7, In case the balance of interest is on the debtor side, it is evident that while the creditor has been overpaid on the account, he is entitled to a. balance of interest, and consequently should keep the amount he has been overpaid until the interest upon it would be equal to the interest due him. Hence the equated time would be after Oct 7. 4. The interest method given (0«59) can be used to advantage in find- ing the equated time when the time is long between the maturity of the items and the assumed date of settlement. In case this method of solu- tion is adopted, the foregoing conditions are equally applicable From these illustrations we obtain the following 663. Rule. — I. Find the maturity of each item on the debit and credit side of the account. II. Assume as the date of settlei7ient the latest ma- turity on either side of the account, and find the numr- her of days from this date to the maturity of each, on hoth sides of the account. III. Multiply each debit and credit item hy the num* her of days from its maturity to the date of settlement, and divide the balance of the debit and credit products hy the balance of the debit and credit items ; the quo- tient is the numher of days the ecfuated tirne is from the assumed date of settlement. IT. In case the balance of items and balance of interest are both on the same side of the account, siib- tract this number of days from the assumed date of settlement, but add it in case theij are on opposite sides; the result is the equated time. (286) EQUATION OF PAYMENTS. 1^9 EXAMPIiES FOR PRACTICE. 664. 1. Find the face of a note and the date from which it must bear interest to settle equitably the following account : Dr. James Hai^d m acct. tvith P. Anstead. Cr. 1876. Jan. 7 May 11 June 6 To mdse. on 3 mo. *' " " 2 mo. " " " 5 mo. 1430 390 570 1876. Mar. 15 May 17 Aug. 9 By draft at 90 da, " cash .... " mdse. on 30 da. 1500 280 400 2. Equate the following account, and find the cash payment Dec. 7, 187G: Cr. Dr. William Hekdersoi^". 1876. Mar. 23 May 16 Aug. 7 To mdse. on 45 da. " 25 da. " 35 da. §^? 1876. Apr. 16 June 25 July 13 By cash .... " mdse. on 30 da. " draft at 60 da. $490 650 200 3. Find the equated time of paying the balance on the fol- lowing account : Dr. Hugh MacVicar. Or. 1876. 1876. Jan. 13 To mdse. on 60 da. $840 Feb. 15 By note at 60 da. $700 Mar. 24 " " •* 40 da. 580 Apr. 17 " cash .... 460 June 7 " " " 4mo. 360 June 9 " draft at 30 da. 1150 July 14 " " " 80da. 730 4. I purchased of Wm. Rodgers, March 10, 187G, $930 worth of goods ; June 23, $680; and paid, April 3, 1870 cash, and gave a note May 24 on 30 days for %500. What must be the date of a note bearing interest that will equitably settle the balance ? (287) 160 BUSINESS ARITHMETIC, REVIEW AND TEST QUESTION'S. 665. 1. Define Simple, Compound, and Annual Interest. 2. Illustrate by an example every step in the six per cent, method. 3. Show that 1%% may be used as conveniently as 6^, and write a rule for finding the interest for months by this method. 4. Explain the method of finding the exact interest of any sum for any given time. Give reasons for each step in the process. 5. Show by an example the difference between true and hank discount. Give reasons for your answer. 6. Explain the method of finding ih.Q 2)resent worth. 7. Explain how the face of a note is found when the pro- ceeds are given. Illustrate each step in the process. 8. Define Exchange, and state the difference between Do- mestic and Foreign Exchange. 9. State the difference in the three bills in a Set of Exchange. 10. What is meant by Par of Exchange ? 11. State the various methods of Domestic Exchange, and illustrate each by an example. 12. Illustrate the method of finding the cost of a draft when exchange is at a discount and brokerage allowed. Give reasons for each step. 13. State the methods of Foreign Exchange. 14. Illustrate by an example the difference between Direct and Indirect exchange. 15. Define Equation of Payments, an Account, Equated Time, and Term of Credit. 16. Illustrate the Interest Method of finding the Equated Time when there are but debit items. 17. State when and why you count forward from the assumed date of settlement to find the equated time. (288) [ RATIO PREPARATORY PROPOSITIONS, 666. Two numbers are compared and their relation detet' mined hy dividing the first hy the second. For example, the relation of $8 to $4 is determined thus, |8-f-$4 = 3. Observe, the quotient 3 indicates that for every (me dollar in the $4, there are two dollars in the $8. Be particular to observe the following: 1. When the greater of two numbers is compared with the less, the relation of the numbers is- expressed either by the relation of an integer or of a mixed number to the unit 1, that is, by an improper fraction whose denominator is 1. Thus, 20 compared with 4 gives 20 -i- 4 = 5 ; that is, for every 1 in the 4 there are 5 in the 30. Hence the relation of 30 to 4 is that of the integer 5 to the unit 1, expressed fractionally thus, f . Again, 39 compared with 4 gives 29 -^- 4 = 7^ ; that is, for every 1 in the 4 there are 7^ in 29. Hence, the relation of 29 to 4 is that of the mixed number 7^ to the unit 1. 2. When the less of two numbers is compared with the greater, the relation is expressed by a proper fraction. Thus, 6 compared with 14 gives 6 -5- 14 = ^^ = | (255) ; that is, for every 3 in the 6 there is a 7 in the 14. Hence, the relation of 6 to 14 is that of 3 to 7, expressed fractionally thus, f . Observe, that the relation in this case may be expressed, if desired, as that of the unit 1 to a mixed number. Thus, 6 -J- 14 = ^ = ^ '^ (255) ; that is, the relation of 6 to 14 is that of the unit 1 to 3^-, (289) 163 BUSINESS ARITUMETIC. EXAMPIiES FOR PRACTICES. 667. Find orally the relation 1. Of 24 to 3. 5. Of 113 to 9. 9. Of 42 to 77. 2. Of 56 to 8. 6. Of 25 to 100. 10. Of 85 to 9. 3. Of 7G to 4. 7. Of 16 to 48. 11. Of 10 to 1000. 4. Of 38 to 5. 8. Of 13 to 90. 12. Of 75 to 300. 668. Prop. II. — No numhers can it compared lut those which are of the same denominatio7i. Thus we can compare $8 witli $3, and 7 inches with 3 inches, but we cannot compare $8 with 2 inches (155—1). Observe carefully the following : 1. Denominate numbers must be reduced to the lowest denomination named, before they can be compared. For example, to compare 1 yd. 2 ft. with 1 ft. 3 in., both numbers must be reduced to inches. Thus, 1 yd. 2 ft. =60 in., 1 ft. 3 in. = 15 in., and GO in. h- 15 in. = 4 ; hence, 1 yd. 2 ft. are 4 times 1 ft. 3 in. 2. Fractions must be reduced to ihQ ^'amQ fractional denom- ination before they can be compared. For example, to compare 3| lb. with | oz. we must first reduce the 3| lb. to oz., then reduce both members to the same fractional unit. " Thus, (1) U lb. = 56 oz. ; (2) 56 oz. = if^ oz. ; (3) ^^ oz. -^ ^ oz, = i-f = 45 (290) ; hence, the relation 3^ lb. to | oz. is that of 45 to 1. BXAMPLES FOR PRACTICE. 669. Find orally the relation 1. Of 4 yd. to 2 ft. 4.' Of | to |. 7. Of 1^ pk. to 3 bu. 2. Of $2 to 25 ct. 5. Of I to 1 J. 8. Of 3 cd. to 6 cd. ft. 3. Of 2i gal. to I qt. 6. Of | oz. to 2 lb. 9. Of If to 2J. Find the relation 10. Of 105 to 28. 13. Of $364 to |4|. 11. Of 6 1 to f 14. Of 2 yd. If ft. to| in. 12. Of 9-J bu. to ^ pk. 15. Of 1 i pt. to 2| gal. (290) EAT 10. 163 DEFINITIONS. 670. A Hatio is a fraction which expresses the relatioTi which the first of two numbers of the same denomination has to the second. Thus the relation of $6 to $15 is expressed by f ; that is, 16 is I of $15, or for every $2 in |6 there are $5 in $15. In like manner the relation of $12 to $10 is expressed by |. 671. The Special Sign of Ratio is a colon (:). Thus 4 : 7 denotes that 4 and 7 express the ratio 4^ ; hence, 4 : 7 and f are two ways of expressing the same thing. The fractional form being the more convenient, should be used in preference to the form with the colon. 672. The Terms of a Hatio are the numerator and denominator of the fraction that expresses the relation between the quantities compared. The first term or numerator is called the Antecedent, the second term or denominator is called the Consequent, 673. A Simple Ratio is a ratio in which each term is a single integer. Thus 9 : 3, or f , is a simple ratio. 674. A Compound Ratio is a ratio whose terms are formed by multiplying together the corresponding terms of two or more simple ratios. Thus, multiplying together the corresponding terms of the simple ratios 7 : 3 and 5 : 2, we have the compound ratio 5x7:3x2 = 35: 6, , » ,. ,,7 5 7x5 35 or expressed fractionally - x - = ~ — - = — . 3 2 o y. 4i D Observe, that when the multiplication of the corresponding terms is performed, the compound ratio is reduced to a simple ratio. 675. The Reciprocal of a number is 1 divided by that number. Thus, the reciprocal of 8 is 1 ~ 8 = ^. (291) 164 'li^sT^^uss arithmetic. 676. The Heciprocal of a Matio is 1 divided by the ratio. Thus, the ratio of 7 to 4 is 7 : 4 or |, and its reciprocal is 1 -^ J = 4, according to (291). Hence the reciprocal of a ratio is the ratio invert- ed, or the consequent divided by the antecedent. GHH. A Matio is in its Shnplest Terms when the antecedent and consequent are prime to each other. 6*78. The MMuction of a Matio is the process of changing its terms without changing the relation they express. Thus |, f , §, each express the same relation. PKOBLEMS ON KATIO. 679. Since every ratio is either a proper or improper fraction, the principles of reduction discussed in (235) apply to the reduction of ratios. The wording of the principles must be shghtly modified thus: Pein. I. — The terms of a ratio must each represent units of the same hind, Prin". II. — Multiplying both terms of a ratio by the same number does not change the value of the ratio, Prin. III. — Dividing both terms of a ratio by the same number does not change the value of the ratio. For the illustration of these principles refer to (235). 680. Prob. I. — To find the ratio between two given numbers. Ex. 1. Find the ratio of $56 to $84. Solution. — Since, according to (66G), two numbers are compared by dividing the first by the second, we divide $56 by $84, giving $56 -^ $84 = If ; that is, $56 is || of $84. Hence the ratio of (292) RATIO, • 165 Ex. 2. Find the ratio of 1 yd. 2 ft. to 1 ft. 3 in. Solution. — 1. Since, according to (668), only numbers of the same denomination can be compared, we reduce botli terms to inches, giving GO in. and 15 in. 2, Dividing 60 in. by 15 in. we have 60 in. -f- 15 in. = 4 ; that is, 60 in. is 4 times 15 in. Hence the ratio of 1 yd. 2 ft. to 1 ft. 3 in. is f. EXAMPLES FOR PRACTICE. 681. Find tlie ratio 1. Of $512 to 1256. 3. Of 982 da. to 2946 da. 2. Of 143 yd. to 365 yd. 4. Of 73 A. to 365 A. 5. Of £41 5s. 6d. to £2 3s. 6d. 6. Of 20 T. 6 cwt. 93 lb. to 25 cwt. 43 lb. 5 oz. 683. Pbob. II. — To reduce a ratio to its simplest terms. Reduce the ratio ^ to its simplest terms. Solution.— Since, according to (679 — III), the ralue of the ratio J/ is not changed by dividing? both terms by tlie same number, we divide the antecedent 15 and the consequent 9 by 3, their greatest common divisor, giving -^ \ o = 6- ^^^ having divided 15 and 9 by their greatest common divisor, the quotients 5 and 3 must be prime to each other. Hence (677) f are the simplest terms of the ratio -^/. E^XAMPIiES FOR PRACTICES. 683. Reduce to its simplest terms 1. The ratio 6 : 9. 4. The ratio fff. 2. The ratio 21 : 56. 5. The ratio 65 : 85. 3. The ratio ^. 6. The ratio 195 : 39. Express in its simplest terms the ratio (see 668) 7. Of 96 T. to 56 T. 9. Of 8s. 9d. to £1. 8. Of f ft. to 2 yd. 10. Of 3 pk. 5 qt. to 1 bu. 2 pk. (293) 166 • BUSINESS ARITHMETIC. 684. Prob. III. — To find a number that has a given ratio to a given number. How many dollars are | of $72 ? Solution. — The fraction | denotes the ratio of the required number to $73 ; namely, for every $8 in $72 there are $5 in the required number. Consequently we divide the $72 by $8, and multiply $5 by the quotient. Hence, first step, $72 -^ $8 = 9 ; second step $5x9 = $45, the required number. Observe, that this problem is the same as Prob. VIII, 501, and Prob. II, 274. Compare this solution with the solution in each of these problems. SXAMPIiES FOR PRACTICE. 685. Solve and explain each of the following examples, regarding the fraction in every case as a ratio. 1. How many days are -^ of 360 days ? 2. A man owning a farm of 243 acres, sold -^ of it ; how many acres did he sell ? 3. James has $'i'96 and John has f as much ; how much has John ? 4. A man's capital is $4500, and he gains -^ of his capital ; how much does he gain ? 5. Mr. Jones has a quantity of flour worth $3140 ; part of it being damaged he sells the whole for -J-f of its value; how much does he receive for it ? 686. Prob. IV. — To find a number to which a given number has a given ratio. $42 are J of how many dollars ? Solution.— The fraction | denotes the ratio of $42 to the required number ; namely, for every $7 in $42 there are $4 in the required number. Consequently we divide the $42 by $7 and multiply $4 by the quotient. Hence, first step, $42 -^ $7 = G ; second step, $4x0 = $24, the required number. Observe, that this problem is the same as Prob. IX, 502. Compare the solutions and notice the points of diflference. (394) RATIOi • IW EXAMPLES FOR PRACTICE. 687. Solve and explain each of the following examples, regarding the fraction in every case as a ratio. 1. 96 acres are f| of how many acres ? 2. I received $75, which is f of my wages ; how much is still due ? 3. James attended school 117 days, or -f^ of the term ; how many days in the term ? 4. Sold my house for $2150, which was |^ of what I paid for it ; how much did I lose ? 5. Henry reviewed 249 lines of Latin, or | of the term's work ; how many hnes did he read during the term ? 6. 48 cd. 3 cd. ft. of wood is y\ of what I bought ; how much did I buy ? 7. Mr. Smith's expenses are J of his income. He spends $1500 per year ; what is his income ? 8. 4 gal. 3 qt. 1 pt. are ^ of how many gallons ? 9. A merchant sells a piece of cloth at a profit of 12.50, which is -^ of what it cost him; how much did he pay for it ? 688. Prob. V. — To find a number to which a given number has the same ratio that two other given num- bers have to each other. To how many dollars have $18 the same ratio that 6 yd. have to 15 yd.? Solution.— 1. We find by (680—1) the ratio of 6 yd. to 15 yd., which is j^j = I, according to (077). 2. Since | denotes the ratio of the $18 to the required number, the $18 must be the antecedent ; hence we have, according to (086), first step, $18 -4- $2 = 9 ; second step, $5x9 = $45, the required number. Observe, that in this problem we have the antecedent of a ratio given to find the consequent. In the foUov/ing we have the consequent given to find the antecedent. (295) 168 BUSINESS ARITH3IETIC, 689. Prob. VI. — To find a number that has the same ratio to a given number that two other given numbers have to each other. How many acres have the same ratio to 12 acres that $56 have to $84 ? Solution.— 1. We find by (680— I) the ratio of $56 to $84, which is tf = I, according to (677). 3. Since f denotes the ratio of the required number to 12 acres, the 13 acres must be the consequent ; hence we liave, according to (684), first step, 13 acr. -r- 3 acr. = 4 ; second step, 3 acr. x 4 = 8 acres, the required number. EXAMPLES FOR PRACTICE. 690. The following are applications of Prob. V and VL 1. If 12 bu. of wheat cost $15, what will 42 bu. cost? Ecgarding the solution of examples of this kind, observe that the price or rate per unit is assumed to be the same for each of the quantities given. Thus, since the 12 bu. cost $15. the price per bushel or unit is $1.25, and the example asks for the cost of 43 bu. at this price per bushel. Consequently whatever part the 13 bu. are of 43 bu., the $15. the cost of 13 bu., must be the same part of the cost of 43 bu. Hence we 'find the ratio of 13 bu. to 43 bu. and solve the example by Prob. V. 2. What will 16 cords of wood cost, if 2 cords cost $9 ? 3. If a man earn $18 in 2 weeks, how much will he earn in 52 weeks ? 4. If 24 bu. of wheat cost $18, what will 36 bu. cost ? 5. If 24 cords of wood cost $60, what will 40 cords cost ? 6. Bought 170 pounds of butter for $51; what would 680 pounds cost, at the same price ? 7. Two numbers are to each other as 10 to 15, and the less number is 329 ; what is the greater? 8. At the rate of 16 yards for $7, how many yards of cloth can be bought for $100 ? (296) 1 . '^y ^ ^ .., 1 '-M^ [[PROPORTIONl] |^i^ DEFINITIONS. 691. A Proportion is an equality of ratios, the terms of the ratios being expressed. Thus the ratio f is equal to the ratio ^f ; hence f = ^f is a propor- tion, and is read, The ratio of 3 to 5 is equal to the ratio of 12 to 20, or 3 is to 5 as 12 is to 20. 692. The equality of two ratios constituting a proportion is indicated either by a double colon (: :) or by the sign (=). Thus, I = ^y, or 3 : 4 = 9 : 12, or 3 : 4 : : 9 : 12. 693. A Simple JProportion is an expression of the equality of two simple ratios. Thus, t\ = f f » or 8 : 12 : : 3 : 48, or 8 : 12 = 3 : 8 is a shnple pro- portion. Hence a simple proportion contains four terms. 694. A Compound Proportion is an expression of the equality of a compound (674) and a simple ratio (673). Thus, « i K [ : : 48 : 60, or f X I = f§, is a compound proportion. It is read, The ratio of 2 into 6 is to 3 into 5 as 48 is to 60. 695. A Proportional is a number used as a term in a proportion. Thus in the simple proportion 2:5 : : 6 : 15 the numbers 2, 5, 6, and 15 are its terms ; hence, each one of these numbers is called a propor- tional, and the four numbers together are called proportionals. (397) 170 BUSINESS ARITHMETIC, 696. A Mean Proportional is a number that is the Consequent of one and the Antecedent of the other of the two ratios forming a proportion. Thus in the proportion 4 : 8 : : 8 : 16, the number 8 is the consequent of the first ratio and the antecedent of the second ; hence is a mean proportioned. 697. The Antecedents of a proportion are the first and third terms, and the Consequents are the second and fourth terms. 698. The Extremes of a proportion are its first and fourth terms, and the Means are its second and third terms, SIMPLE PROPORTIOK. PRJEPARATOUT STEPS, 699. The following preparatory steps should be perfectly- mastered before applying proportion in the solution of problems. The solution of each example under Step I should be given in full, as shown in (688 and 689), aud Step II and III should be illustrated by the pupil, .in the manner shown, by a number of examples. 700. Step I. — Find by Prob. V and VI, in ratio, the missing term in the following proportions : The required term is represented by the letter x. 1. 6 : 42 : : 5 : ic. 4. 5 bu. 2 pk. : 3 pk. : : a: : 4 bn. 2. 24 : 60 : : a; : 15. 5. 2 yd. : 8 in. : : a; : 3 ft. 4 in. 3. 84 : a; : : 21 : 68. 6. a; : £3 2s. : : 49 T. : 18 cwt. Step II. — Show that the product of the extremes of apro^ portion is equal to the product of the means. Thus the proportion 3 : 3 : : G : 9 expressed fractionally gives f = f . (298) SIMPLE PROPORTION. 171 Now if both terms of this equality be multiplied by 3 and by 9, the consequents of the given ratios, the equality is not changed ; hence, 2jr9 >^ ^ 62^x_9^ Cancelling (186) the factor 3 in the left-hand 3 9 term and 9 in the right-hand term we have 3x9 = 6x3. But 2 and 9 are the extremes of the proportion and 6 and 3 are the means ; hence the truth of the proposition. Step III. — Sliow thai, since the product of the extremes is equal to the product of the means, any term of a proportion can he found ivhen the other three are known. Thus in the proportion 3 : a; : : 9 : 15 we have known the two extremes 3 and 15 and the mean 9. But by Step II, 3 x 15, or 45, is equal to 9 times the required mean ; hence 45 -f- 9 = 5, the required mean. In the same manner any one of the terms may be found ; hence the truth of the proposition. Find by this method the missing term in the following : $13 : a; : : 5 yd. : 3 yd. 128 bii. : 3 pk. : : a; : $1.25. 64 cwt. : X :: $120 : $15. Solution by Simple Proportion, 701. The quantities considered in problems that occur in practical business are so related that when certain conditions are assumed as invariable, they form ratios that must be equal to each other, and hence can be stated as a proportion thus. If 4 yd. of cloth cost $10, what will 18 yd. cost ? Observe, that in this example the price per yard is assumed to be invariable, that is, the price is the same in both cases ; consequently whatever part the 4 yd. are of the 18 yd., the $10 are the same part of the cost of the 18 yd., hence the ratio of the 4 yd. to the 18 yd. is equal the ratio of the $10 to the required cost, giving the proportion 4 yd. : 18 yd. : : $10 : %x, (299) 1. 14 : 3 : : a; : 12. 4 2. tc : 24 : : 7 : 8. 5, 3. 27 : iz; : : 9 : 5. 6. 172 BUSINESS ARITH3IETIC, EXAItIPr.ES FOR PRACTICE. 702. Examine carefully the following proportions and state what must be considered in each case as invariable, and why, in order that the proportion may be correct. 1. 8. The number ' The number ("The cost 1 f The cost of units is to of units -«J ^^*^^ \isto in the bought in bought in first second one case - another case- I case J case. The ■ r The The ■ r The Principal . Principal .. interest in interest in in one 1 in another the first 1 the second case J I case J case J I case. The number ' The number ' The ' The of men of men number number that can do : . that can do , , of days , of days a piece of the same the the work in work in second first one case . another case J . work . . work. Wliy is the second ratio of this proportion made the ratio of the number of days the second work to the number of days the first work ? Illustrate this arrangement of the terms of the ratio by other examples. In solving examples by simple proportion, the following course should be pursued : 7. Represent the required term hj x, and maTce it the last extreme or consequent of the seco?id ratio in the jjroportion. II. Find the term in the example that is of the same denom- ination as the required term, and maJce it the second mean or the antecedent of the second ratio of the proportion. III. Determine, by inspecting carefully the conditio7is given in the example, whether x, the required term of the ratio now expiressed, must be greater or less than the given term^ (300) SIMPLE PROPORTION. 173 IV. If X, the required term of the ratio expressed, must he greater than the given term, make the greater of the remaining terms in the example the consequent of the first ratio of the proportion ; if less, make it the antecedent. V. When the proportion is stated^ find the required term either as shown in (688) or in (689). Observe, that in either way of finding the required term, any factor that is common to the given extreme and either of the given means should be cancelled, as shown in (186). 4. How maoy bushels of wheat would- be required to make 39 barrels of flour, if 15 bushels will make 3 barrels? 5. If 77 pounds of sugar cost $8.25, what will 84 pounds cost? 6. I raised 245 bushels of corn on 7 acres of land; how many bushels grow on 2 acres ? 7. If 6 men put up 73 feet of fence in 3 days, how many feet will they put up in 33 days ?• 8. What will 168 pounds of salt cost, if 3J^ pounds cost 37i cents ? 9. If 25 cwt. of iron cost $84.50, what will 24f cwt. cost ? 10. Paid $2225 for 18 cows, and sold them for $2675; what should I gain on 120 cows at the same rate ? 11. If 5 lb. 10 oz. of tea cost $5.25 ; w^hat will 7 lb. 8 oz. cost ? 12. If a piece of cloth containing 18 yards is worth $10.80, what are 4 yards of it worth ? 13. My horse can travel 2 mi. 107 rd. in 20 minutes ; how far can lie travel in 2 hr. 20 min. 14. If 18 gal. 3 qt. 1 pt. of water leaks out of a cistern in 4 hours, how much will leak out in 36 hours ? 15. Bought 28 yards of cloth for $20; what price per yard would give me a gain of $7.50 on the whole ? 16. If I lend a man $69.60 for 8|- months, how long should he lend me $17.40 to counterbalance it? 17. My annual income on U. S. 6^'s is $337.50 when gold is at 112^ ; what would it be if gold were at 125 ? (301) in BUSINESS ARITHMETIC, COMPOUND PROPORTION. PRE1*A.RATOJtY STEPS. 703, Step I. — A compound ratio is reduced to a simple one hy m^ultiplying the antecedents together for an antecedent and the consequents for a consequent (6*74). i 6 • 7 ) Thus the compound ratio ) j^ 1 o f ^^ reduced to a simple ratio by multiplying the antecedents 6 and 4 together, and the consequents 7 and 3. Expressing the ratios fractionally we "have f x f = |f = f (68^). Observe, that any factor that is common to any antecedent and consequent may he cancelled before the terms are multiplied. Reduce the following compound ratios to simple ratios in their simplest terms, 35 L16J Step II. — A compound proportion is reduced to a simple proportion by reducing the compound ratio to a simple ratio. Thus, in the compound proportion ] « .\£ [ : : 24 : 18, the com- pound ratio f X f is equal the simple ratio f ; substituting this in tlm proportion for the compound ratio we have the simple proportion 4 : 3 : : 24 : 18. Observe, that when a compound proportion is reduced to a simpl proportion, the missing term is found according to (688), o. (689) 1. \ r ^ • 25-^ 15 : 18 28 : 50 . 3 : 7, rl6 : 9- 27: 15 28 ^ 8 Find the missing term in the following : ( 24 : 15 ) ( 48 : 20 1. •< 7 : 16 [ : : 40 : a;. 2. •] 3 ( 25 : 21 ) (6 (302) \- : 28 : a;. COMPOUND PROPORTION. 175 Solution by Compound Proportion, 704. The following preparatory propositions should be carefully studied and the course indicated observed in solving l)roblems involving compound proportion. Prop. I. — There are one or more conditions in every example involving proportion, wliicli must he regarded as invariable in order that a solution may he given, thus If 9 horses can subsist on 50 bu. of oats for 20 days, how long can 6 horses subsist on 70 bu. In this example there are two conditions that must be considered as invariable in order to give a solution : 1. The fact that each horse subsists on the same quantity of oats each day. 3. The fact that each bushel of oats contains the same amount of food. Prop. II. — To solve a proUem involving a compound pro- portion, the effect of each ratio, ivhich forms the compound ratio, on the required term must he considered separately, thus : If 5 men can build 40 yards of a fence in 12 days, how many yards can 8 men build in 9 days. 1. We observe that the invariahle co7iditions in this example are (1.) That each man in both cases does the same a/mount of work in the same time. (2.) That the same amount of work is required in each case to bvHd one yard of the fence. 2. We determine by examining the problem how the re- quired term is affected by the relations of the given term, thus : (1.) We observe that the 5 men in 12 days can build 40 yards. Now since each man can build the same extent of the fence in one" day, it is evident that if the 8 men work 12 days the same as the 5 men, the 40 yards built by the 5 men in 13 days must have the (303) 176 BUSTNUSS ARITHMETIC, same ratio to the number of yards that can be built by the 8 men in 12 days as 5 men have to 8 men ; hence the proportion 5 men : 8 men : : 40 yards : x yards. This proportion will give the number of yards the 8 men can build in 12 days. (2.) We now observe that the 8 men work only 9 days ; and since they can do the same amount of work each day, the work done in 12 days must have the same ratio to the work they can do in 9 days that 12 days have to 9 days. Hence we have the compound proportion 5 men : 8 men ) .^ , •, 13 days: 9 days [ = ^ 40 yards : 0= sards. We find from this proportion, according to (703 — II), that the 8 men can build 48 yards of fence in 9 days. EXAMPLES FOR PRACTICJB. 705. 1. If it cost 188 to hire 12 horses for 5 days, what will it cost to hire 10 horses for 18 days ? 2. If 12 men can saw 45 cords of wood in 3 days, working 9 hours a day, how much can 4 men saw in 18 days, working 12 hours a day ? 3. If 28 horses consume 240 bushels of corn in 112 days, how many bushels will 12 horses consume in 196 days ?. 4. When the charge for candying 20 centals of grain 50 miles is $4.50, what is the charge for carrying 40 centals 100 miles? 5. The average cost of keeping 25 soldiers 1 year is $3000 ; what would it cost to keep 139 soldiers 7 years ? 6. If 1 pound of thread makes 3 yards of linen, IJ yard wide, how many pounds would make 45 yards of linen, 1 yard wide ? 7. G4 men dig a ditch 72 feet long, 4 feet wide, and 2 feet deep, in 8 days ;. how long a ditch, 2 J feet wide and IJ feet deep, can 96 men dig in 60 days ? 8. If it requires 8400 yd. of cloth 1^ yd. wide to clothe 3500 soldiers, how many yards | wide will clothe 6720 ? (304) - *>e H^^ ^SJ^r ^g)^ j ^ PARTNERSHIP DEFINITIONS. 706. A I^artnevshii) is an association of two or more persons for the transaction of business. The persons associated are called partners, and the Association is called a Company, Firm, or House. 707. The Capital is the money or other property invest- ed in the business. The Capital is also called the Investment or Jdnt-stock of the Company. 708. The Assets or Effects of a Company are the property of all kinds belonging to it, together with all the amounts due to it. 709. The lAahilities of a company are its debts. PBEPARATOBT P B O P O 8 1 T I O If 8 . 710. Prop. I. — 77ie profits and the losses of a com' pany are divided among the partners, according to the value of each marCs investment at the time the division is made. Observe carefully the following regarding this proposition: Since the use of money or property is itself value, it is evident that the value of an investment at any time after it is made, depends first upon (305) 178 BUSITTESS ARITHMETIC. the amount invested, second on the length of the time the investment has been made, and third the rate of interest. Thus the value of an investment of $500 at the time it is made is just $500 ; but at the end of 9 years, reckoning its use to be worth 7 % per annum, its mine will be |500 + $315 = $815. Prop. II. — Tlie value of any investment made for a given number of intervals of time, can he represented by another investment made for one interval of time. Thus, for example, the value of an investment of $40 for 5 months at any given rate of interest is the same as the value of 5 times $40, or $200, for one month. SXAMPLBS FOR practice:. 711. Find the value at simple interest 1. Of $800 invested 4 years at 6^ per annum. 2. Of 1350 invested 2 yr. 3 mo. at 7;^ per annum. 3. Of $2860 invested 19 months at S% per annum. Solve the following by applying (710 — II). 4. An investment of 1200 for 6 months is equal in value to what investment for 4 months? 5. A man invests 1600 for 9 months, $700 for 3 months, and $300 for 7 months, each at tlie same rate of interest. What sum can he invest for 4 months at the given rate of interest, to be equal in value to the three investments ? IZZZrSTRATION OF PROCESS . 713. Prob. I. — To apportion grains or losses when each partner's capital is invested the same length of time. Observe, that when each partner's capital is used for the same length of time, it is evident that his share of the gain or loss must be the same fraction of the whole gain or loss that his capital is of the whole capital. Hence, examples under this problem may be solved— (306) PARTNERSHIP, 179 1. By Proportion thus : The whole \ {Each man's \ ( Whole \ ( Each capital y : ] capital y :: Xgain or> : \man'sgain invested ) ( invested ) ( loss ) \ or loss, II. By Percentage thus : Find loliat per cent (504) the whole gain or loss is of the whole capital invested, and take the same per cent of each man's investment as his share of the gain or loss, III. By Fractions thus : Find what fractional part each man's investment is of the imhole capital invested, and take the same fractional part of the gain or loss as each man's share of the gain or loss, EXAMPLES FOR PRACTICE. 713. 1. Three men, A, B, and C, form a company ; A puts in $G000; B $4000; and C $5600; they gain $4320; what is each man's share ? 2. A man failing in business owes A $9600, B $7000, and C $5400, and his available property amounts to $5460 ; what is each man's share of the property ? 3. Three men agree to liquidate a church debt of $7890, each paying in proportion to his property ; A's property is valued at $6470, B'sat $3780, and C's at $7890; what portion of the debt does each man pay ? 4. A building worth $28500 is insured in the ^tna for $3200, in the Home for $4200, and in the Mutual for $0500; it having been partially destroyed, the damage is set at $10500; what should each company pay ? 5. The sum of $2600 is to be divided among four school districts in proportion to the number of scholars in each ; in the first there are 108, in the second 84, in the third 72, in the fourth 48 ; what part should each receive ? (307) 180 BUSINESS ARITHMETIC. 714. Prob. II. — To apportion gains or losses when each partner's capital is invested dift'erent lengths of time. Observe carefully the following : 1. According to (710 — II) we can find for eacli partner an amount whose value invested one interval of time is equal to the value of his capital for the given intervals of time. 2. Having found this we can, by adding these amounts, find an amount whose value invested one interval of time is equal to the total value of the whole capital invested. When this is done it is evident that each man's share of the gain or loss must be the same fraction of the whole gain or loss that the mlue of his investment is of the total value of the whole capital Invested. Hence the problem from this point can be solved by either of the three methods given un(Jer Prob. I (712). EXAMPLES FOR PRACTICES. 715. 1. A and B engage in business ; A puts in $1120 for 5 months and B $480 for 8 months ; they gain 1354 ; what is each man's share of the gain ? 2. Three men hire a pasture for 1136.50 ; A puts in 16 cows for 8 weeks, B puts in 6 cows for 12 weeks, and C the same number for 8 weeks; what should each man pay ? 3. The joint capital of a company was 17800, which was doubled at the end of the year. A put in ^ for 9 mo., B ^ for 8 mo., and C the remainder for 1 year. What is each one's stock at the end of the year ? 4. Jan. 1, 1875, three persons began business. A put in $1200, B put in $500 and May 1 $800 more, C put in $700 and July 1 $400 more ; at the end of the year the profits were $875 ; how shall it be divided? 5. A and B formed a partnership Jan. 1, 1876. A put in $6000 and at the end of 3 mo. $900 more, and at the end of 10 mo. drew out $300 ; B put in $9000 and 8 mo. after $1500 more, and drew out $500 Dec. 1 ; at the end of the year the aet profits were $8900. Find the share of each. (308) alligation] f^^s^^ ALLIGATION MEDIAL. 716. Alligation Medial is the process of finding the mean or average price or quality of a mixture composed of several ingredients of different prices or qualities. BXAMPIiSS FOR PRACTICE. 717. 1. A grocer mixed 7 lb. of coffee worth 30 ct. a pound witli 4 lb. @ 25 ct. and 10 lb. @ 32 ; in order that he may neither gain or lose, at what price must he sell the mixture ? 7 lb. @ 30 ct. ~ $2.10 Solution.— 1. Since the value of 4 lb. @ 25 ct. = 1.00 G^^ kind of coffee is not changed 10 lb! @ 32 ct = 3^20 ^^ °''^'''^' ^^ ^^^ *^'^ ""^^^^ ""^ *^'® ■ entire mixture by finding the value KL lb. = ^o.oO of each kind at the given price, and A/. Of) _i_ 0-1 OA pj- taking the sum of these values as shown in illustration. 2. Having found that the 21 lb. of coifee are worth at the given prices |6.30, it is evident that to realize this amount from the sale of the 21 lb. at a uniform price per pound, he must get for each pound ^^^ of $0.30 ; hence, .$6.30 -t- 21 = 30 cents, the selling price of the mixture. 2. A wine merchant mixes 2 gallons of wine worth 11.20 a gallon with 4 gallons Avorth $1.40 a gallon, 4 gallons worth $.90 and 8 gallons worth 8.80 a gallon ; what is the mixture worth per gallon ? (309) 182 BUSINESS ARITHMETIC. 3. A grocer mixes 48 lb. of sugar at 17 ct. a pound with 58 lb. at 13 ct. and 94 lb. at 11 ct.; what is a pound of the mixture worth ? 4. A goldsmith melts together 6 ounces of gold 22 carats fine, 30 ounces 20 carats fine, and 12 ounces 14 carats fine ; how many carats fine is the mixture ? 5. A merchant purchased 60 gallons of molasses at 30 ct. per gallon and 40 gallons at 25 cents, which he mixed with 8 gallons of water. He sold the entire mixture so as to gain 20 per cent on the original cost ; what was his selling price per gallon ? ALLIGATION ALTERNATE. 718. Alligation Alternate is the process of finding the proportional quantities of ingredients of different prices or quahties that must be used to form any required mixture, when the price or quality of the mixture is given. PREPARA.TORT PROPOSITIONS, 719. Prop. I. — In forming any mixture, it is assumed that the value of the entire mixture must he equal to the aggregate value of its ingredients at their given prices. Thus, if 10 pounds of tea at 45 ct. and 5 pounds at 60 ct. be mixed, the value of the mixture must be the value of the 10 pounds plus the value of the 5 pounds at the given prices, which is equal $4.50 + $3.00 = $7.50. Hence there is neither gain or loss in fonning a mixture. Prop. II. — The price of a mixture must he less than the highest and greater than the lowest price of any ingredient used informing the mixture. Thus, if sugar at 10 ct. and at 15 ct, per pound be mixed, it is evident the price of the mixture must be less than 15 cents and greater than 10 cents ; that is, it must be some price between 10 and 15 cents. (310) ALLIGATION, \%% ILLVSTBATION OF rRQCESS. 720. If tea at 56 ct., 60 ct, 75 ct., and 90 ct. per pound be mixed and sold at 66 ct. per pound, how much of each kind of tea can be put in the mixture ? First Step in Solution, We find the gain or loss on one unit of each ingredient thus : . . (66 ct. — 56 ct. = 10 ct. gain. ^ ' I 66 ct. — 60 ct. = 6 ct. gain. 75 ct. — 66 ct. =: 9 ct. loss. 90 ct. — QQ ct. = 24 ct. loss. (^.) { Second Step in Solution, We now take an ingredient on which there is a gain, and one on which there is a loss, and ascertain how much of each must be put in the mix- ture to make the gain and loss equal ; thus : Producing Gain. Gained and Lost. Pboditcing Loss. (1.) 9 lb. at 10 ct. per lb. gain. = 90 ct. = 10 lb. at 9 ct. per lb. loss. (2.) 4 lb. at 24 ct. per lb. gain. = 24 Ct. = 1 lb. at 24 ct. per lb. loss. Hence the mixture must contain 9 lb. at 56 cts. per pound, 10 lb. at 75 ct. per pound, 4 lb. at 60 ct. per pound, and 1 lb. at 9 ct. per pound. 731. Observe carefully the following: 1. The gain and loss on any two ingredients may be balanced by assuming any amount as the sum gained and lost. Thus, instead of taking 90 cents, as in (1) in the above solution, as the amount gained and lost, we might take 360 cents ; and dividing 360 cents by 10 cents would give 36, the number of pounds of 56 ct. tea that would gain this sum. Again, dividing 360 cents by 9 cents would give 40, the number of pounds of 75 ct. tea that would lose this sum. 2. To obtain integral proportional parts the amount assumed must be a multiple of the gain and loss on one unit of the ingredients balanced, and to obtain the least integral propor- tional parts it must be the least common multiple. (311) 184 BUSINJSSS ARITHMETIC. 3. When a number of ingredients are given on which there is a gain and also on which there is a loss, they may be balanced with each other in several ways ; hence a series of different mixtures may be formed as follows: Taking the foregoing example we have A Second Mixture thus: Prodtjcing Gain. Gained and Lost. Producing Loss. (1.) 24 lb. at 10 Ct. per lb. gain. — 240 ct. = 10 lb. at 24 Ct. per lb. loss. (2.) 9 lb. at C ct. per lb. gain. :zz 54 ct. = G lb. at 9 ct. per lb. loss. Hence the mixture is composed of 24 lb. @ 56 ct., 9 lb. @, 60 ct., 10 lb. @ 90 ct, and 6 lb. @ 75 ct. A Tliird Mixture thus: Pboducing Gain. Gained and Lost. Pboducing Loss. (1.) 9 lb. at 10 ct. per lb. gain. == 90 Ct. = 10 lb. at 9 ct. per lb. loss. (2.) 24 lb. at 10 ct. per lb. gain. = 240 ct. = 10 lb. at 24 ct. per lb. loss. (3.) 9 lb. at Get. per lb. gain. =: 54 ct. = 6 lb. at 9 Ct. per lb. loss. Observe, that in (1) and (2) we have balanced the loss on the 75 ct, and 90 ct. tea by the gain on the 56 ct. tea ; hence we have 9 lb. + 24 lb. , or 33 lb. of the 56 ct. tea in the mixture. Observe, also, that in (3) we have balanced the ga.in on the 60 ct. tea by a loss on the 75 ct. tea ; hence we have 10 lb. + 6 lb,, or 16 lb. of the 75 ct. tea in the mixture. Hence the mixture is composed of 33 lb. @ 56 ct., 9 lb. @ 60 ct., 16 lb. @ 75 ct., and 10 lb. @ 90 ct. 4. Mixtures may be formed as follows : /. Take any pair of ingredients, one giving a gain and the other a loss, and find the gain and loss on one unit of each, 11. Assume the least common multiple of the gain and loss on one unit as the amount gained and lost, hy putting the tiuo ingredients in the mixture. IIL Divide the amount thus assumed hy the gain and then by the loss on one unit ; the results will be respectively the (313) ALLIG ATION, 185 number of units of each i7igredient that must be in the mixture that the gain and loss may balance each other. I V. Proceed in the same manner with other ingredients j the results ivill be the proportional parts, EXAMPLES FOR PRACTICK. 732. 1. How much sugar at 10, 9, 7, and 5 ct. will pro- duce a mixture worth 8 cents a pound ? 2. A man wishes to mix sufficient water with molasses worth 40 cents a gallon to make the mixture worth 24 cents a gallon ; what amount must he take of each ? 3. A jeweller has gold 16, 18, 22, and 24 carats fine; how much of each must he use to form gold 20 carats fine ? 4. A merchant desires to mix flour worth $6, $7|-, and $10 a barrel so as to sell the mixture at $9 ; what proportion of each kind can he use ? 6. A farmer has wheat worth 40, 55, 80, and 90 cents a bushel ; how many bushels of each must be mixed with 290 @40 ct. to form a mixture worth 70 cents a bushel? Examples like this where the quantity of one or more ingredients is limited may be solved thus: First, we find the gain or loss on one unit as in (730). Second, we balance the whole gain or loss on an ingredient where the quantity is limited, by using any ingredient giving an opposite result thus : Producing Gain. Gained and Lost. Producing Loss. (1.) 270 bu. at 30 ct. per bu. gain. = $81.00 = 405 bu. at 20 ct. per ba. loss. (2.) 2bu.at loct. perba.gam.= .30= 3 bu.at 10 ct.per bu.los8. 272 bu. + 408 bu. = 680 bu. in mixture. Observe, the gain on the 370 bu. may be balanced with the other ingredient that produces a loss, or with both ingredients that produce a loss, and these may be put in the mixture in different proportions ; hence a series of different mixtures may thus be formed. (313) 186 BUSINESS ARITHMETIC. 6. A merchant having good flour worth $7, $9, and $12 a barrel, and 240 barrels of a poorer quality worth $5 a barrel, wishes to sell enough of each kind to realize an average price of 110 a barrel on the entire quantity sold. How many bar- rels of each kind can he sell ? 7. I wish to mix vinegar worth 18, 21, and 27 cents a gallon with 8 gallons of water, making a mixture worth 25 cents a gallon ; how much of each kind of vinegar can I use ? 8. A man bought a lot of sheep at an average price of $2 apiece. He paid for 50 of them $2.50 per head, and for the rest $1.50, $1.75, and $3.25 per head; how many sheep could there be in the lot at each price ? 9. A milkman mixes milk worth 8 cents a quart with water, making 24 quarts worth 6 cents a quart ; how much water did he use ^ Examples like this, where the quantity of the mixture is limited, may be solved thus : Solution.— 1. We find, according to (720), the smallest proportional parts that can be used, namely, 3 quarts of milk and 1 quart of water, making a mixture of 4 quarts. 2. Now, since in 4 qt. of the mixture there are 3 qt. of milk and 1 qt. of water, in 24 qt. there must be as many times 3 qt. of milk and 1 qt. of water as 4 qt. are contained times in 24 qt. Consequently we have as the first step 24 qt. -f- 4 qt. = 6, second step 3 qt. x 6 = 18 qt. and 1 qt. x 6 = G qt. Hence in 24 qt. of the mixture there are 18 qt. of milk and 6 qt. of water. 10. A grocer has four kinds of coffee worth 20, 25, 35, and 40 cents a pound, from which he fills an order for 135 pounds worth 32 cents a pound ; how may he form the mixture ? 11. A jeweler melts together gold 14, 18, and 24 carats fine, so as to make 240 oz. 22 carats fine ; how much Of each kind did it require ? 12. I wish to fill an order for 224 lb. of sugar at 12 cents, by forming a mixture from 8, 10, and IG cent sugar ; how much of each must I take ? (314) 3- -"-^^^ DEFINITIONS. 723. A Power of a number is either the number itself or the product obtained by taking the number two or more times as a factor. Thus 25 is the product of 5 x 5 or of 5 taken twice as a factor ; hence 25 is a power of 5. 724. An Exponent is a number written at the right and a Uttle above a number to indicate : (1.) The number of times the given number is taken as a factor. Thus in 7^ the 3 indicates that the 7 is taken 3 times as a factor ; hence 73 = 7x7x7 = 343. (2.) The degree of the powor or the order of the power with reference to the other powers of the given number. Thus, in 5"* the 4 indicates that the given power is the fourth power of 5, and hence there are three powers of 5 below 5^ ; namely, 5, 5"^ and 5^. 735. The Square of a number is its second power, so called because in finding the superficial contents of a given square we take the second power of the number of linear units in one of its sides (404). 736. The Cube of a number is its third poiuer, so called because in findmg the cubic contents of a given cube we take the tliird poiver of the number of linear units in one of its edges (413). 737. Involution is the process of finding any required power of a given number. (315) 188 BUSIITJSSS ARITHMETIC, PROBLEMS IN INVOLUTION. 728. Pkob. I. — To find any power of any given number. 1. Find the fourth power of 17. Solution. — Since according to (721) the fourth power of 17 is the product of 17 taken as a factor 4 times, we have 17 x 17 x 17 x 17=83521, the required power. 2. Find the second power of 48. Of 65. Of 432. 3. Find the square of 294. Of 386. Of 497. Of 253. 4. Find the cube of 63. Of 25. Of 76. Of 392. 5. Find the third power of 4. Of |. Of ^. Of .8. Observe, any power of a fraction is found by involving each of its terms separately to the required power (267). Find the required power of the following : 6. 2372. 8. (iJ)3. 10. (.25)4. 12. (.7f)2. 14. (.005J)8. 7. 45^. 9. (^)^ 11. (.3|)3. 13. (.l^^y. 15. .03022. 739. Prob. II.— To find the exponent of the pro- duct of two or more powers of a given number. 1. Find the exponent of product of 7^ and 7^. Solution.— Since 7^ = 7 x 7 x 7 and 7^ = 7 x 7, the product of 7' and 72 must be (7 X 7 X 7) X (7 X 7), or 7 taken as a factor as many times as the sum of the exponents 3 and 2. Hence to find the exponent of the pro- duct of two or more powers of a given number, we take the sum of the given exponents. Find the exponent of the product 2. Of 35* X 353. 4. Of 182 X 181 6. Of 23^ x 235. 3. Of (1)5 X (1)2. 5. Of (i)7 X (i)«. 7. Of (^Y X i^y. 8. Of (74)2. Observe, (7^f = 7* x 7* = 7*^2 :,:, 78. Hence the required exponent is the product of the given exponents. 9. Of (123)4. 10. Of (96)5. 11. Of (168)8. 12. Of [(iYY. (316) £y^^ [e V O L. U T I O N]] ^(^ .» ,— ^^„— ^ .-. ®> DEFINITIONS. 730. A Itoot of a number is either the number itself- op one of the equal factors into which it can be resolved. Thus, since 7 x 7 = 49, the factor 7 is a root of 49. - 731. The Second or Square Hoot is one of the two equal factors of a number. Thus, 5 is the square root of 35. 732. The Tliird or Cube Hoot is one of the three equal factors of a number. Thus, 2 is the cube root of 8. 733. The Madical or Hoot Sign is ^, or ^fractional exponent. Wlien the sign, |/, is used, the degree or name of the root is indicated by a small figure written over the sign ; when the fractional exponent is used, the denominator indicates the name of the root ; thus, ^9 or 9" indicates that the second or square root is to be found. Y^27 or 27^ indicates that the third or cube root is to be found. Any required root is expressed in the same manner. The index is usually omitted when the square root is required. 734. A Perfect Power is a number whose exact root can be found. 735. An Imperfect Power is a number whose exact root cannot be found. The indicated root of an imperfect power is called a surd ; thus ^/^. 736. Evolution is the process of finding the roots of numbers. (317) 190 BUSIjyUSS ARITHMETIC, SQUARE ROOT. PBEPABATOltT rJtOPOSITIONS, 737. Prop. I. — Any perfect second poioer may he represented to the eye by a square, and the number of units in the side of such square luill represent the second or square ROOT of the given power. For example, if 25 is the given power, we can suppose the number represents 25 small squares and arrange them thus : 1. Since 25 = 5 x 5, we can arrange the 25 squares 5 in a row, making 5 rows, and hence forming a square as shown in the illustration. 2. Since the side of the square is 5 units, it represents the square root of 25, the given power; hence the truth of the proposition. 738. Prop. II. — Any number being given, by supposing it to represent small squares, we can find by arranging these squares in a large square the largest perfect second power the given number contains, and hence its square root. For example, if we take 83 as the given number and sup- pose it to represent 83 small squares, we can proceed, thus : '"1 ;.[■■, ■ .1 _■„ ■1 _!!: L. (1) (2) 1. We can take any number of the 83 squares, as 36, that we know will form a perfect square (Prop. I), and arrange them in a square, as shown in (I), leav- ing 47 of the 83 squares yet to be disposed of. 2. We can now place a row of squares on two adjacent sides of the square in (1) and a square in the corner, and still have a perfect square as shown in (2\ 3. Observe, that in putting one row of small squares r:i each of two adjacent sides of the square first formed, we must use twice as many squares as there are units in the side of the square. 4. Now since it takes twice 6 or 12 squares to put one row on each of two adjacent sides, we can put on as many rows as (318) I I I I I W EVOLUTION-. 191 6xr (3) =18. 3^=9. 1 1 ■ 1 1 1 ---1 a f - ^-:- ^1: tj-- c a 12 is contained times in 47, the number of squares remaining. Hence we can put on 3 rows as shown in (3) and have 11 squares still remaining. 5. Again, having put 3 rows of squares on each of two adjacent sides, it takes 3 x 3 or 9 squares to fill the comer thus formed, as shown in (3), leaving only 2 of the 1 1 squares. Hence, the square in (3) represents the greatest perfect power in 83, namely 81 ; and 9, the number of units in its side, represents the square root of 81. 6. Now cbserw that the length of the side of the square in (3) is 6 + 3 units, and that the number of (Small squares may be represented in terms of 6 + 3 ; thus, (1.) (6 + 3)2 = Q^-]-^^-\-Uvice 6x3 = 36 + 9 + 36 = 81. Again, suppose 5 units had been taken as the side of the first square, the number of small squares would be represented thus : (2.) (5 + 4)2 ^ 52 + 42 + ^mce 5 x4 z= 25 + 16 + 40 = 81. In the same manner it may be shown that the square of the sum of any two numbers expressed in terms of the numbers, is the square of each of the numbers plus twice i\\e\t product. Hence the square of any number may be expressed in terms of its tens and units ; thus 57 = 50 + 7 ; hence (3.) 572 = (50 + 7)2 = bOf^+'l^+tivice 50 x 7 = 3249. Tliis may also be shown by actual multiplication. Thus, in multiply- ing 57 by 57 we have, first, 57 x7 = 7x7 + 50x7 = 7^ + 50 x7; we have, second, 57 x 50 = 50 x 7 + 50 x 50 = 50 x 7 + 50^ ; hence, 57^ = 50^ + 7'^ + twice 50 X 7. Find, by constructing a diagram as above, the square root of each of the following : Observe, that when the number is large enough to give tens in the root, we can take as the side of the first square we construct the greatest number of tens whose square can be taken out of the given number. 1. Of 144. 4. Of 529. 7. Of 1125. 10. Of 1054. 2. Of 196. 5. Of 729. 8. Of 584. 11. Of 2760. 3. Of 289. 6. Of 1089. 9. Of 793. 12. Of 3832. (319) 192 BUSINESS ARITH3IETIC. 739, Prop. III. — Tlie square of any mimher must con- tain twice as many figures as the number, or twice as many less one. This proposition may be shown thus : 1. Observe, the square of either of the digits 1, 2, 3, is expressed by one figure, and the square of either of the digits 4, 5, 6, 7, 8, 9, is expressed by two figures; thus, 2x3=4, 3x3 = 9, and 4x4 = 16, 5 X 5 = 25, and so on. 2. Since 10 x 10 = 100, it is evident the square of any number of tens must have two ciphers at the right ; thus, 20' = 20 x 20 = 400. Now since the square of either of the digits 1, 2, 3, is expressed by one figure, if we have 1, 2, or 3 tens, the square of the number must be expressed by 3 figures ; that is, one figure less than twice as many as are required to express the number. Again, since the square of either of the digits 4, 5, 6, 7, 8, 9, is expressed by two figures, if we have 4, 5, 6, 7, 8, or 9 tens, the square of the number must contain four figures ; that is, twice as many figures as are required to express the number. Hence it is evident that, in the square of a number, the square of the tens must occupy the third or the third and fourth place. By the same method it may be shown that the square of hundreds must occupy the fifth or the fifth and sixth places, the square of thousands the seventh or the seventh and eighth places, and so on ; hence the truth of the proposition. From this proposition we have the following conclusions : 740. /. If any numler he separated into periods of two figures each, beginning with the units place, the number of periods will be equal to the number of places in the square root of the greatest perfect power luhich the given number contains. 11. In the square of any number the square of the units are found in the units and tens place, the square of the tens in the hundreds and thousands place, the square of the hun- dreds in the tens and hundreds of thousands place, and so on. (320) EVOL UTION, 193 IZZUSTJtATIOK OF PROCESS, 741. 1. Find the square root of 225. (c) 10x5=50. (^5- =25. 1st Step. 2d Step. (a) 10^=100. 225(10 102 = 10x10= 100 i(l) T. divisor 10 x 2=20)125 ( 5 root 15 Explanation. — 1. We observe, as shown in (a), that 1 ten is the largest num- ber of tens whose square is contained in 225. Hence in 1st step we subtract 10^= 100 from 225, leaving 125. 2. Having formed a square whose side is 10 units, we observe, as shown in (&) and (<•), that it will take twke ten to put one row on two adjacent sides. Hence the Trial Divisor is 10 x 2=20. 3. We observe that 20 is contained 6 times in 125, but if we add 6 units to the side of the square {a) we will not have enough left for the corner (d), hence we add 5 units. 4. Having added 5 units to the side of the square (a), we observe, as shown in (b) and (c), that it requires t^cice 10 or 20 multiplied by 5 plus 5 X 5, as shown in (d), to complete the square ; hence (2) in 2d step. Solution with every Operation Indicated^ 743. 2. Find the square root of 466489. FmsT Step. 600x600 466489 ( 600 360000 80 Second Step, 80x80: 6400 ) ~ 102400 683 required root Thibd Step, r (1) Trial divisor 600 x 2=1200)106489 •i (1200x80=96000/ ((2)] |(1) 3 !(2)| ^(1) Trial divisor m)x2=\Zm) 4089 • \_ i 1360x3=4080/ 3x3= r Explanation.— 1. We place a point over every second figure beginning with the units, and thus find, according to (740), that the root must have three places. Hence the first figure of the roofc expresses hundreds. (321) 194 BUSINESS ARITHMETIC, 2. We observe that the square of 600 is the greatest second power of hundreds contained in 466489. Hence in the first step we subtract 600 X 600 = 360000 from 466489, leaving 103489. 3. We now double the 600, the root found, for a trial divisor, according to (741 — 2). Dividing 106489 by 1200 we find, according to (741 — 2), that we can add 80 to the root. For this addition we use, as shown in (2), second step, 1200 x 80 = 90000 and 80 x 80 =6400 (741—3), making in all 102400. Subtracting 102400 from 106489, we have still remaining 4089. 4. We again double 680, the root found, for a trial divisor, according to (741 — 2), and proceed in the same manner as before, as shown in third step. •743, Contracted Solution of the foregoing Example. 466489 ( 683 First Step. 6x6= 36 1064 „ ^ I (1) 6x2= 12 Second Step. { ^ ' \ (2) 128 X 8 = 1024 THIRD Step, j^ 68x2 = 136) 4089 1(2) 1363x3 = 4089 Explanation. — 1. Obseru, in the first step we know that the square 600 must occupy the fifth and sixth place (738). Hence the ciphers are omitted. 3. Observe, that in (1), second step, we use 6 instead of 600, thus dividing the divisor by 100; hence we reject the tens and units from the right of the dividend (142). 3. Observe, also, in (2), second step, we imite in one three operations. Instead of multiplying 12 by 80, the part of the root found by dividing 1064 by 12, we multiply the 12 first by 10 by annexing the 8 to it (91 ), and having annexed the 8 we multiply the result by 8, which gives us the product of 12 by 80, plus the square of 8. But the square of 8, written, as it is, in the third and fourth place, is the square of 80. Hence by annexing the 8 and writing the result as we do, we have united in oTie three operations ; thus, 128 x 8 = 12 x 80 + 80 x 80. (322) EVOLUTION, 195 From these illnstrations we have the following 744. EuLE. — /. Separate the nurriber into periods of two figures each, by placing a point over every second figure, beginning with the units figure. II. Find the greatest square in the left-hand period and place its root on the right. Subtract this square froiiv the period and annex to the reinainder the next period for a dividend. III. Double the part of the root found for a trial divisor, and find how many times this divisor is con- tained in the dividend, omitting the right-hand figure. Annex the quotient thus found both to the root and to the divisor. Multiply the divisor thus completed by the figure of the root last obtained, and subtract the product froin the dividend. IV. If there are more periods, continue the operation in the same manner as before. In appl3^ng this rule be particular to observe 1. When there is a remainder after the last period has been used, annex periods of ciphers and continue the root to as many decimal places as may be required. 2. We separate a number into periods of two figures by beginning at the units place and proceeding to the left if the number is an integer, and to the right if a decimal, and to the right and left if both. 3. Mixed numbers and fractions are reduced to decimals before extracting the root. But in case the numerator and the denom- inator are perfect powers, or the denominator alone, the root may bo more readily formed by extracting the root of each term separately, rru . /49 1/49 7 , /35 ^35 V35 so on. Extract the square root 1. OfVW- 3. Of HI. 5. Of J||. 7. Of^^. 2. Of^. 4. OffH- 6. Oft^. 8. Of^ftV- (323) 196' BUSINESS ARITHMETIC. EXAMPLES FOR PRACTICE. 745. Extract the square root 1. Of 4096. 7. Of^^. 13. Of 137641. 2. Of 3481. 8. Offifi^. 14. Of 4160.25. 3. Of 2809. 9. Of ttf f 15. Of 768427.56. 4. Of 7569. 10. Of .0225. 16. Of 28022.76. 5. Of 8649. 11. Of .2304. 17. Of 57.1536. 6. Of 9216. 12. Of .5776. 18. Of 474.8041. Find the square root to three decimal places 19. Of 32. 22. Of .93. 25. Of 14.7. 28. Of J. 20. Of 59. 23. Of .8. 26. Of 86.2. 29. Of ^V 21. Of 7. 24. Of .375. 27. Of 5.973. 30. Of A- Perform the operations indicated in the following : 31. \/6889 — 'v/I024. 34. 76796^-^^2136. 32. V2209 + a/225. 35. \/558009-^(TjJ:r)"^. 33. VfHl X a/2209. 36. (f f f|)^ X 131376^. 37. What is the length of a square floor containing 9025 square feet of lumber ? 38. A square garden contains 237169 square foot; how many feet in one of its sides ? 39. How mony yards in one of the equal sides of a square acre? 40. An orchard containing 9216 trees is planted in the form of a square, eacli tree an equal distance from another ; how many trees in each row ? 41. A triangular field contains 1966.24 P. What is the length of one side of a square field of equal area ? 42. Find the square root of 2, of 5, and of 11, to 4 decimal places. 43. Find the square root of 4, ^, and of |}, to 3 decimal places. (324) VOL UTION. m CUBE ROOT. PJREPAJtATOIiT PROPOSITIONS, 46. Prop. I. — A ny perfect third power may he repre- sented to the eye hy a cube, mid the number of units in the side of such cube will represent the third or cube root of the given power. Represent to the eye by a cube 343. (1) 1. We can suppose the number 343 to repre- sent small cubes, and we can take 2 or more of these cubes and arrange tliem in a row, as shown in (1). 2. Having formed a row of 5 cubes, as shown in (1), we can arrange 6 of these rows side by side, as shown in (3), forming a square slab containing 5x5 small cubes, or as many small ("^) cubes as the square of the number of unUs in the side of the slab. 3. Placing 5 such slabs together, as shown in (3), we form a cube. Now, since each slab contains 5x5 small cubes, and since 5 slabs are placed together, the cube in (3) contains 5 x 5 x 5, or 125 small cubes, and hence represents the tldvd power 125, and each edge of the cube ^^^ represents to the eye 5, the cube root of 125. We have now remaining yet to be disposed of 843—125, or 218 small cubes. 4. Now, ohserce, that to enlarge the cube in (3) so that it may contain the 343 small cubes, we must build the same number of tiers of small cubes upon each of three adjacent sides, as shown in (4). Obnerve, also, that a slab of small cubes to cover one side of the cube in (3) must contain 5x5 or 25 small cubes, as shown in (4), or as many small cubes as the square of the number of units in one edge of the cube in (3). Hence, to find the number of cubes necessary to put one slab on each of three sides of the cube in (3), we nmltiply the square of its edge by 3 giving 5' X 3 = 5 X 5 X 3 = 75 small cubes. (325) 198 B USINESS ARITHMETIC. Examine careful 5. Having found that 75 small cubes will put one tier on each of three adjacent sides of the cube in (3), we divide 218, the number of small cubes yet remaining, by 75, and find how many such tiers we can form. Thus, 218-^75 = 2 and 68 remaining. Hence we can put 2 tiers on each of three adjacent sides, as shown in (5), and have G8 small cubes remaining. G. Now, observe, that to complete this cube we must fill each of the thj'ee corners formed by building on three adjacent sides. (6) and obserce that to fill one of these three corners we require as many small cubes as is expressed by the square of the number of tiers added, multiplied by the number oi units in the side of the cube to which the addition is made. Hence we require 2^ x 5 or 20 small cubes. And to fill the three corners we require 3 times 2"^ x 5 or 60, leaving 68—60 or 8 of the small cubes. 7. Examine again (5) and (6) and observe that when the three comers are filled we require to complete the cube as ^~^ shown in (7), another cube whose side contains as many units as there are units added to the side of the cube on which we have built. Consequently we require 2^ or 2 x 2 x 2 = 8 small cubes. Hence we have formed a cube containing 343 small cubes, and any one of its edges repre- sents to the eye 5 + 2 or 7 units, the cube root of 343. From these illustrations it will be seen that the Bteps in finding the cube root of 343 may be stated thus : We assume that 343 represents small cubes, and take 5 as the length of the side of a large cube formed from these. Hence we subtract the cube of 5 = f 1. We observe it takes 5^ x 3 = 75 to put one tier on three adjacent sides. Hence we can put on 2. We have now found that we can add 2 units to the side of the cube. Hence to add this we require (1) For the 3 sides of the cube 5^ X 2 X 3 = 150 ^ (2) For the 3 corners thus formed 2^x5x3= 60 >= 218 . (3) For the cube in the comer last formed 2^= 8 J Hence the cube root of 343 is 5 + 2 = 7. (326) FmsT Step. Second Step. 343 125 75)218(2 E VOL UTION. 199 747. Ohserm, that the number of small cubes in the cube (7) in the foregoinr: illustrations, are expressed in terms of 5 + 3 ; namely, the num- ber of units in the side of the first cube formed, plus the number of tiers added in enlarging this cube ; thus : (5 + 2)^ 53+52x2x3 + 22x5x3 + 21 In this manner it may be shown that the cube of the sum of any two numbers is equal to the cube of each number, plus 3 times the square of the^r*^ multiplied by the second number, plus 3 times the square of the second multiplied by the^r«^ number. Hence the cube of any number may be expressed in terms of its tens and units ; thus, 74 = 70 + 4 ; hence, (70 + 4)3 =703 + 3 times 70^ x 4 + 3 times 4^ x 70 + 43 = 405224. Solve each- of the following examples, by applying the fore- going illustrations : 1. Find the side of a cube which contains 729 small cubes, taking 6 units as the side of the first cube formed. 2. Take 20 units as the side of the first cube formed, and find the side of the cube that contains 15625 cubic units. 3. How many must be added to 9 that the sum may be the cube root of 4096? Of 2197? Of 2744? 4. Find the cube root of 1368. Of 3405. Of 2231. Of 5832. 5. Express the cube of 83 in terms of 80 + 3. 6. Express the cube of 54, of 72, of 95, of 123, of 274, in terms of the tens and units of each number. (327) 200 BUSINESS ARITHMETIC, 748. Pkop. II. — The cube of any member must contain three times as many places as the number, or three times as many less one or two places. This proposition may be shown thus : 1. Observe, 1^ = 1, 2^ = 9, 3^ = 27, ¥ = 64, 5^ = 125, and 9^ = 729 ; hence the cube of 1 and 2 is expressed each by one figure, the cube of 3 and 4 each by two figures, and any number from 5 to 9 inclusive each by three figures. 2. Observe, also, that for every cipher at the right of a number there must (91) be three ciphers at the right of its cube; thus, 10^ = 1,000, 100"^ = 1,000,000. Hence the cube of tens can occupy no place lower than thousands, the cube of hundreds no place lower than mUlicns, and so on with higher orders. 8. From the foregoing we have the following : (1.) Since the cube of 1 or 2 contains one figure, the cube of 1 or 2 tens must contain /6>wr places; of 1 or 2 hundreds^ seven places, and so on with higher orders. (2.) Since the cube of 3 or 4 contains two figures, the cube of 3 or 4 tens must contain jive places ; of 3 or 4 hundreds, eight places, and so on with higlier orders. (3.) Since the cube of any number from 5 to 9 inclusive contains three places, the cube of any number of tens from 5 to 9 tens inclusive must contain six places ; of hundreds, from 5 to 9 hundred inclasive, nine places, and so on with higher orders ; hence the truth of the propos'tion. Hence also the following : 749. /. If any number be separated into periods of three figures each, beginning with the units place, the number of periods will be equal to the number of places in the cube root of the greatest perfect third power which the given mtmhcr contains. II. The cube of units contains no order higher than hundreds. III. TJie cube of tens contains no order lower than thousands nor higher than hundred thousands, the cube of hundreds no order lower than millions nor higher than hundred millions^ and so on with higher orders. (328) EVOLUTION, 201 IZrUaTJtATION OF mOC ESS. 150. Solution with every Operation Indicated, Find the cube root of 92345408. 92345408 ( 400 First Step. 400^=400 x 400 x 400 = 64000000 Second Step. Third Step. (1) Trial divisor 400- x 3=480000 ) 28345408 ( 50 4002 X 50x3 = 24000000 \ (2) ^ 50-' X 400 X 3 = 3000000 V = 27125000 503= 135000 ) (1) Trial divisor 450^ x 3=607500 ) 1220408 ( 2 ) 450^x3x2 =1215000) Moot 462 (2) -J 2* X 450x3= 5400^ 1220408 2»= 8 J Explanation. — 1. We place a period over every third figure begin- ning with the units, and thus find, according to (749), that the root must have three places. Hence the first figure of the root expresses hundreds. 2. We observe that 400 is the greatest number whose cube is contained in the given number. Subtracting 400^ = 64000000 from 92345408, we have 28345408 remaining. 3. We find a trial divisor, according to (746 — 4), by taking 3 times the square of 400, as shown in (1), second step. Dividing by this divisor, according to (746 — 5), we find we can add 50 to the root already found. Observe, the root now found is 400 + 50, and that according to (747), (400 + 50)3 = 4003 + 4002 x 50 x 3 + 50' x 400 x 3 + 503. We have already subtracted 4003 = 640000 from the given number. Hence we have only now to subtract, 400' X 50 X 3 + 502 x 400 X 3 + 503 ^ 27125000, as shown in (2), second step, leaving 1220408. 5. We find another trial divisor and proceed in the same manner to find the next figure of the root, as shown in the third step. (329) 202 BUSINESS ARITHMETIC. 751. Contracted Solution of the foregoing Example. First Step. Second Step. 4^ = 4x4x4 = (1) Trial divisor 40' x 3 = r 402 X 5 X 3 = 24000 ) (3) <5-'x40x3= 3000 [ = ( 53 = 125 3 92345403 ( 452 G4 4800 ) 28345 27125 Third Step. ( (1) Trial divisor 450^ xS = 6075( ] C 450-' X 2 x 3 := 1215000 ) ((2) ^2^x450x3= 5400> = ; (1) Trial divisor 450^ x 3 ( 23 = s) 607500 ) 1220408 1220408 Explanation.— 1. Observe, in the first step, we know that the cube of 400 must occupy the seventh and eighth places (749— III). Hence the ciphers are omitted. 2. Observe, also, that no part of the cube of hundreds and tens is found below thousands (749 — HI). We therefore, in finding the number of tens in the root, disregard, as shown in second step, the right-hand period in the given number, and consider the hundreds and tens in the root as tens and units respectively. Hence, in general, whatever number of places there are in the root, we disregard, in finding any figure, as many periods at the right of the given number as there are places in the root at the right of the figure we are finding, and consider the part of the root found as tens, and the figure we are finding as units, and proceed accordingly. From these illustrations we have the following : *li52. Rule. — I. Separate the nmnherinto periods of • three figures each, hy placing a point over every third figure, beginning with the units figure. II. Find the greatest cube in the left-hand period, and place its root on the right. Subtract this cube from the period and annex to the remainder the next period for a dividend. III. Divide this dividend by the tHal divisor, which is 3 times the square of the root already found, con- (330) EVOLUTION, 203 sidered as tens} the quotient is the next figure of the root. IV. Subtract from the dividend S times the square of the root before found, considered as tens, multiplied by the figure last found, plus 3 times the square of the figure last found, multiplied by the root before found, plus the cube of the figure last found, and to the re- mainder annex the next period, if any, for a new dividend. V. If there are more figures iiv the root, find in the Same manner trial divisors and proceed as before. In applying this rule be particular to observe : 1. In dividing by the Trial Bicisor the quotient may be larger than the required figure in the root, on account of the addition to be made, as shown in (T4C$ — 0) mcond step. In such case try a figure 1 less than the quotient found. 2. When there is a remainder after the last period has been used, annex periods of ciphers and continue the root to as many decimal places as may be required. 3. We separate a number into periods of three figures by beginning at the units place and proceeding to the left if the number is nn integer, and to the right if a decimal, and to the right and left if both. 4. Mixed numbers and fractions are reduced to decimals before extracting the root. But in case the mimerator and the denominator are perfect third powers, or the denominator alone, the root may be more readily found by extracting the root of each term separately. EXAMPIiES FOR PRACTICE. 753. Find the cube root of 9. fiff. 13. 24137569. 10- t¥^- 14. 47245881. 11. 250047. 15. tAtVW 12. 438976. 16. 113.379904. 17. Find, to two decimal places, the cube root of 11. Of 36. Of 84. Of 235. Of^. Oi^-i^. Of 75.4. Of 6.7. (331) 1. 216. 5. 4096. 2. 729. 6. 10648, 3. 1331. 7. 6859. 4. 2197. 8. A4V 304 BUSINESS ARITHMETIC. 18. Find to three decimal places the cube root of 3. Of 7. Of .5. Of .04. Of .009. Of 2.06. 19. Find the sixth root of 4096. Observe, the sixth root may be found by extracting first the square root, then the cube root of the result. For example, y^4096 = 64 ; hence, 4096=64 x 64. Now, if we extract the cube root of 64 we will have one of the three equal factors of 64, and hence one of the six equal factors or sixth root of 4096. Thus, ^^64 =4 ; hence, 64 = 4x4x4. But we found by extracting it3 square root that 4096 = 64 x 64, and now by extracting the cube root that 64= 4x4x4; consequently we know that 4096 =(4 x 4 x 4) x (4 x 4 x 4). Hence 4 is the required sixth root of 4096. In this manner, it is evident, we can find any root whose index con- tains no other factor than 2 or 3. 20. Find the sixth root of 2565726409. 21. Find the eighth root of 43046721. 22. What is the fourth root of 34012224? 23. What is the ninth root of 134217728? 24. A pond contains 84604519 cubic feet of water; what must be the length of the side of a cubical reservoir which will exactly contain the same quantity ? 25. What is the length of the inner edge of a cubical cistern that contains 2079 gal. of water? ^6. How many square feet in the surface of a cube whose Tolume is 16777216 cubic inches? 27. A pile of cord wood is 256 ft. long, 8 ft. high, and 16 ft. wide; what would be the length of the side of a cubical pile containing the same quantity of wood ? 28. What is the length of the inner edge of a cubical bin that contains 3550 bushels ? 29. What are the dimensions of a cube whose volume is equal to 82881856 cubic feet? 30. What is the length in feet of the side of a cubical reservoir which contains 1221187^ pounds avoirdupois, pure water? (332) 1 ,. -^r m> ^ ,., 1 JT I PROGRESSIONS f0- DEFINITIONS. 754. A ^Progression is a series of numbers so related, that each number in the series may be found in the same manner, from the number immediately preceding it. ^55, An Arithmetical Progression is a series of numbers, which increases or decreases in such a manner that the difference between any Uvo consecutive numbers is constant. Thus, 3, 7, 11, 15, 19, 23. 115^, A Geometrical Progression is a series of numbers, which increase or decrease in such a manner that the ratio between any tiuo consecutive numbers is constant. Thus, 5, 10, 20, 40, 80, is a geometrical progression. ^511. The Terms of a progression are the numbers of which it consists. The First and Last Terms are called the Extremes and the intervening terms the Means, ^5S. The Common or Constant JDifference of an arithmetical progression is the difference between any two consecutive terms. 159. The Common or Constant Matio or Multi- plier of a geometrical progression is the quotient obtained by dividing any term by the preceding one. 760. An Ascending or Increasing Progression is one in which each term is greater than the preceding one. 761. A Descending or Decreasing Progression is one in which each term is less than the preceding one. (333) 206 BUSINESS ARITHMETIC. ARITHMETICAL PROGEESSION. 762. There are five quantities considered in Arithmetical Progression, which, for convenience in expressing rules, we denote by letters, thus : 1. A represents the First Term of a progression. 2. JL represents tlie Last Term. 3. D represents tlie Constant or Common Difference, 4. N represents the Numler of Terms. 5. S represents the Sum of oM the Terms. 763. Any three of these quantities being given, the other two may be found. This may be shown thus : Taking 7 as the first term of an increasing series, and 5 the constant diflference, the series may be written in two forms ; thus : 1st Term. Sd Term. (1) 7 13 (3) 7 + (5) 7 + (5+5) Jith Term. 23, and so on. 7 + (5 + 5 + 5) Observe, in (3), each term is composed of the fi/rst term 7 plus as many times the constant difference 5 as the number of the term less 1. Thus, for example, the ninth term in this series would be 7 + 5 x (9— 1)=47. Hence, from the manner in which each term is composed, we have the following forfnulae or rules : \, A = L-I>x {N^ 1). Read, i = ^ + i> X {N- 1). Read .{ The first term is equal to the last term., minus the common difference multiplied by the number qf terms less 1. The last term is equal to the first term, plus the common difference mul- tiplied by the number of terms less 1. 3. D = L-A N-1' Read, (334) The common difference is equal to the last term, minus the first term divided by the number of terms l«is L PROGRESSION. 207 jf^ __ ^ r TJie number of terms is equal to the 4. JV= — fz — + 1. "Read, -l last term, minus the first term divided ■^ [ l^the common difference, plus 1. Obsefve, that in a decreasing series, the first term is the largest and the last term the smallest in the series. Hence, to make the above for- mulae apply to a decreasing series, we must place L where A. is, and A. where L is, and read the formulae acccordingly. 764. To show how to find the sum of a series let (1.) 4 7 10 13 16 19 be an arithmetical series. (2.) 19 16 13 10 7 4 be the same series reversed. (3.) 23 -f 23 + 23 + 23 + 23 -{- 23 =twice the sum of the terms. Now, observe, that in (3), which is equal to twice the sum of the series, each term is equal to the first term plus the last term ; hence, The sum of the terms of an arithmetical series is equal to one-half of the sum of the first and last term^ multiplied by Vie number (f terms. S = loi{A + L)x N, Read, ■ EXAMPLES FOR PRACTICE. 765. 1. The first term of an arithmetical progression is 4, the common difference 2 ; what is the 12th term ? 2. The first round of an upright ladder is 12 inches from the ground, and the nineteenth 246 inches; how far apart are the rounds ? 3. The tenth term of an arithmetical progression is 190, the common difference 20 ; what is the first term ? 4. Weston traveled 14 miles the first day, increasing 4 miles each day; how far did he travel the 15th day, and how many miles did he travel in all the first 12 days ? 5. The amount of $^6^} for 'i^,years at simple interest was $486 ; what was the yearly interest ? ^. The first term of an arithmetical series of 100 terms is 150, and the last term 1338; what is the common difference ? (335) SOS BUSINUSS ARITHMETIC. 7. What is the sum of the first 1000 numbers in their natural order ? 8. A merchant bought 16 pieces of cloth, giving 10 cents for the first and $12.10 for the last, the several prices form an arithmetical series ; find the cost of the cloth ? 9. A man set out on a journey, going 6 miles the first day, increasing the distance 4 miles each day. The last day he went 50 miles ; how long and how far did he travel ? 10. How many less strokes are made daily by a clock which strikes the hours from 1 to 12, than by one which strikes from 1 to 24. GEOMETRICAL PROGRESSIOK 766. There are five quantities considered in geometrical progression, which we denote by letters in the same manner as in arithmetical progression ; thus : \. A = First Term. 2. i = Last Term. 3. £ = Constant Ratio. ^. N= Number of Tenns. 5. 8 = the Sum of all the terms. 767. Any three of these quantities being given, the. other two may be found. This may be shown thus : Taking 3 as the first term and 2 as the constant ratio or multiplier, the series may be written in three forms' ; thus : l8t Term. M Term. 3d Term. 4th Term. Sth Term. (1.) 8 6 12 24 48 (2.) 8 3x2 3x(2x2) 3x (2x2x2) 3x (2x2x2x2) (3.) 3 3x2 3x2* ^ 3x2^ 3x2* Observe, in (3), each term is composed of the first term, 8, multiplied by the constant multiplier 2, raised to the power indicated by the number of the term less 1. Thus, for example, the seventh term would be3x2'-» = 3x2« = 192. (336) PROG RE8SI0N, 209 Hence, from tlie manner in wliicli each term is composed, we have the following formulae or rules : L { The first term is egual to the last term, divided 1, A. =— _.. Read, -j by the constant mvltiplier raised to the power -K"" I indicated by the number of terms less 1. iThe last term is equal to the first term, multi- plied by the constant mvltiplier raised to the power indicated by the number of terms less 1. r The constant multiplier is equal to the root, _ _y_n— i/i -^ , I whose index is indicated by th^ numi)er of terms 6. M— y -g^ Iteaa, J ^^^^ ^^^ ^ ^^ quotient of the last term divided I by the first. {The number of terms less one is equal to the exponent of the power to which the common multiplier micst be raised to be equal to tlie quotient qf the last term divided by the first. 768. To show how to find the sum of a geometrical series, we take a series whose common multiplier is known ; thus : >Sf = 5 + 15 4- 45 -f 135 + 405. Multiplying each term in this series by 3, the common multiplier, we will have 3 times the sum. (1.) iS^x 3=5 X 3 + 15x3-1-45x3 + 135x3 + 405x3, or {2.) Sx3=15 +45 +135 +405 +405x3. Subtracting the sum of the series from this result as expressed in (2), we have, ;Sfx3 = 15 + 45 + 135 + 405+405x3 8 = 5 + 15 + 45 + 135 + 405 8x2 = 405x3-5 Now, observe, in this remainder S -x 2 ib S x (JB — 1), and 405 x 3 is X X li, and 5 is A. Hence, Sx{It — l) = Lx M — A. And since ^ — 1 times the Sum is equal to i x R — A, we have, (The mm of a geometrical seiies is equal to the difference, between the last term multiplied __ by the ratio and the first term, divided by the I ratio minus 1. (337) ^10 BUSINESS ARITHMETIC. EXAMPIil^S FOR PRACTICE. 769. 1. The first term of a geometrical progression is 3, the ratio 4 ; what is the 8th term ? 2. The first term of a geometrical progression is 1, and the ratio 2 ; what is the 12th term ? 3. The extremes are 4 and 2916, and the ratio 3 ; what is the number of terms ? 4. The extremes of a geometrical progression are 2 and 1458, and the ratio 3; what is the sum of all the terms ? 5. The first term is 3, the seventeenth 196608 ; what is the sum of all the terms ? 6. A man traveled 6 days ; the first day he went 5 miles and doubled "the distance each day; his last day's ride was 160 miles; how far did he travel? 7. Supposing an engine should start at a speed of 3 miles an hour, and the speed could be doubled each hour until it equalled 96 miles, how far would it have moved in all, and how many hours would it be in motion ? 8. The first term of a geometrical progression is 4, the 7th term is 2916 ; what is the ratio and the sum of the series ? ANNUITIES. 770. An Annuity is a fixed sum of money, payable annually, or at the end of any equal periods of time. 771. The Amount or Final Value of annuity is the sum of all the payments, each payment being increased by its interest from the time it is due until the annuity ceases. 772. The JPresent Worth of an annuity is such a sum of money as will amount, at the given rate per cent, in the given time, to the Amount or Final Value of the annuity. (338) PROGRESSION, 211 773. An Annuity at Simple Interest forms an arithmetical progression whose common difference is the interest on the given annuity for one interval of time. Thus an annuity of $400 for 4 years, at 7% simple interest, gives the following progression : 1st Term. 2d Term. 3d Term. ^th Term. (1.) $400 $400 + ($28) $400 + ($38 + $28) $400 + ($28 + $28 + $28), or (2.) $400 $428 $456 Observe, there is no interest on the last payment ; hence it forms the 1st Term. The payment before the last bears one year's interest, hence forms the 2d Term ; and so on with the other terms. Hence all problems in annuities at simple interest are solved by arithmetical progression. 774. An Annuity at Compound Interest forms a geometrical progression whose common multiplier is repre- sented by the amount of $1 for one interval of time. Thus an annuity of $300 for 4 years, at 6% compound interest, gives the following progression : 1st Term. 2d Term. 3d Term. 4th Term. $300 X 1.06 $300 X 1.06 x 1.06 $300 x 1.06 x 1.06 x 1.06. Observe carefully the following : (1.) The last payment bears no interest, and hence forms the 1st Term of the progression. (2.) The payment before the last, when not paid until the annuity ceases, bears interest for one year ; hence its amount is $300 x 1.06 and forms the 2d Term. (3.) The second payment before the last, bears interest when the annuity ceases, for two years ; hence its amount at compound interest is $300 X 1.06, the amount for one year, multiplied by 1.06, equal $300 x 1.06 X 1.06, and forms the od Term, and so on with other terms. Hence all problems in annuities at compound interest are solved by geometrical progression. (339) 212 BUSINESS ARITHMETIC, SXAMPIiES FOR PRACTICE. 775. 1. What is the amount of an annuity of $200 for 6 years at 11% simple interest ? 2. A father deposits $150 annually for the benefit of his son, beginning with his 12th birthday; what will be the amount of the annuity on his 21st birthday, allowing simple interest at 6% ? 3. What is the present worth of an annuity of $600 for 5 years at 8^, simple interest ? 4. What is the amount of an annuity of $400 for 4 years at 7%, compound interest ? 5. What is the present worth of an annuity of $100 for G years at 6%, compound interest ? 6. What is the present worth of an annuity of $700 at S%, simple interest, for 10 years ? 7. What is the amount of an annuity of $500 at '7%, com- pound interest, for 12 years? 8. What is the present worth of an annuity of $350 for 9 years at 6^, compound interest ? This example and the four following should be solved by applying the formulae for geometrical progression on page 337. 10. At what rate ^ will $1000 amount to $1500.73 in 6 years, compound interest ? 11. The amount of a certain sum of money for 12 years, at 7^ compound interest, was $1126.096; what was the original sum ? 12. What sum at compound interest 8 years, at 7^ will amount to $4295.465 ? 13. In how many years will $20 amount to $23.82032, at 6^ compound interest? (340) saf/yr'^ggyf^^. .5^ [mensuration] -^^L^2jj±^ ^l^wp^ GENERAL DEFINITIONS. 776. A Line is that whicli has only length. 777. A Straight Line is a line which has the same direction at every point. 778. A Curved Line is a line which changes its direction at every point. 779. Parallel Lines are lines which have the same direction. 780. An Angle is the opening between two lines which meet in a common point, called the vertex. Angles are of three kinds, thus : (1) (2) (3)^ (4) u > 1 1 4 ..c )i ^ OUme Angle. 'A HORIZ TwoRi ONTAl.. ' A ght Angles. On c e Eight Angle. Acu r — i te Angle. 781. When a line meets another line, making, as shown in (1), two equal angles, each angle is a Right Angle, and the lines are said to be perpendicular to each other. 782. An Obtuse Angle, as shown in (3), is greater than a right angle, and an Acute Angle, as shown in (4), is less than a right angle. Angles are read by using letters, the letter at the vertex being always read in the middle. Thus, in (2), we read, the angle BAG or GAB. 783. A Plane is a surface such that if any two points in it be joined by a straight line, every point of that line will be in the surface. (341) 214 BUSINESS ARITHMETIC. 784. A Plane Figure is a plane bounded either by straight or curved lines, or by one curved line. 185. A Polygon is a plane figure bounded by straight lines. It is named by the number of sides in its boundary ; thus : Trigon. Tetragon. Pentagon. Hexagon, and so on. Observe, that a regular polygon is one that has all its sides and all its angles equal, and that the Base of a polygon is the side on which it stands. ^S(y, A Trigon is a ^8 > 28 ' ^^' /r 13 . 30. A. A. _Z. 7. 3if. ^- FT- 9. $5272 jV 12. f. i.?. 49. i4- $1614|. 15. 40| tons, ie. 121 1 yards. 11. $63 1. i^. 2H. 19. %im\. 20. 47i-;irai.; 38^| mi- 21. $8113 rV 22. 556,%. 23. $10500. 2J^. 504. ;?5. $1800. 26. 123 1 cords. ;^7. 4i|f pounds ; Sj^'V pounds ; f I pounds. j?5. Increased by 4i. 29. %imh 30. 2}4 days. ol- S65HfM. ^^. 15 feet. 33. 1196|fayds.; %%i-^. 34. $289||. 35. 13.V.I acres. ^u. ; 458j\ ^u. 7. 51 rd. 8. 217 A. 9. 285 lb. 10. 395 yd. 11, 476 bu. Art. 498. {. 30 yd. ?. 37 T. ; 3. 24T8rV T. 38| bu. ; 40 bu. Jt. 96 ft. 5. 483 yd. 6. 89. 7. 53 cd. 5. 9 ft. Art. 499. 7. $585.50. 8, $120.25. 5. $5075. 10. $847.09|. 11. $40455. Art. 500. ^. f ; f; t;i. 7 71 8. i8f. 9. 4t^V. 2.^. II, or .68. 13. i,"||. ^5. H;*. ANSWERS. 24 Art. 601. 22. 550. U. $4000. 15. $384,048+. 8. 42 ; 84. ^e?. 10 yd. 16. $596. 329; 905|. 2I^. 8. Art. 515. 17. $24.60. S. 20.4 ; 77.76 ; ^5. $35500. 1. $3.04. i. 42f. 11. $1000. 12. $3653.70 + . ^.9. $191.80. €0. $36. i^. $6.40. 13. $1863.97+. 29. $3720. ^1. 87^. i5. $2000. U- $3286. J(?. Uif %. 24G ANSWERS. 51. 175. 82. |62i. 53. $53." 54. $5950. 56. $680. ^6. $800,821+ ; $78.^ 82+ ; $124,931 + . 57. $8. Art. 554. 1. $268 80. J?. $255,192. 5. $622. 68f. 4. $73.60. 6. $66.6792. Art. 566. 13. $51.5256. U- $282.33. 15. $8.3695. 16. $41.78265. 17. $86.3208 + . IS. $1.6559 + . 19. $13.25248. SO. $107.1144. 81. S85.115. 52. $827.08. 83. $462,616. ^4' $42362. 85. $5,736. S6. $97.1694. Art. 570. 1, $63,048; $89,318. 5. $113.1074; $145.4238. 5. $118.3442; $73,965 + . /f. $64.1775; $106.9625. 6. $815,976+; $1078.254 + , 6. $292.3719. 7. $49,529 + . 8. $1094.096. 9. $410,475 + . 10. |1699.b0 + . Art. 573. 1. $35.84. 2. $48,675. 3. S43.812. J,. $12,754 5. $28.1885. G. $35.82. 7. $120. 8. $46.50. 9. $40.20. 10. $100,395. Art. 576. 1. $1.58. 2. $1,536. 3. $2,125. 4. $4.2075. $1.54f. $1.849i $2.67f. $3.7754. $1,122. 6. 6. 7. 8 9. 10. $1.96. Art. 578. 1. $62.36352; $93.54528 ; $83.15136. 2. $303.9513; $434.216 ; $173.6875. S. $22.2609 ; $11.1304 ; $19.4783. 4. $113.40; .'^151.20; $56.70. 5. $45.4765; $57.7202; $66.4656. Art. 582. 1. $24.65; $39.44; $34.51. 2. $62.6533 $93.9800; $41.7689. 3. $291,695; $458,378; $120.Om. 4. $103.44004 $131.2892 ; $57.6877. ■>. $65.9458; $50.3270 ; $19.0895. >. $1.85; $2.6037 ; $l.r3074. y. $376.6183 ; $502.1577; $313.8486. Art. 585. t. $11.5436. $3.1342. $3.3082. $3.1574. $3.3945. $3.6073. $6.54407. $1.4576. $16.2754. $.8939. $461,193. 12. $25.2125. Art. 588. 2. $364.9937 + . 3. $283,992 + . 4. $462,019. 5. $562,984. G. $296. 7. $434,994. Art. 591. 2. $264,998+. 3. $49,652 + . 4. $295,996 + . 5. $572,996 + . Art. 594. 2. Qfc. 3. 7%. 4. Sfo. 5.5%. 6. 8^% nearly. 7. 12%. 5. ^% better 2d. 0. 14|%, 10. 25% I 12^%. ]nfo ; S^% ; 4%. 11. 33i%; 50%; 22|%;33^%; 14f %; 21f %; 111%; 16|%. Art. 597. 1. 2 yr. 3 mo. 2. 6 yr. 3. 3 yr. 6 mo. 4. 4 yr. 9 mo. 5. 2 yr. 8 mo. G. 5 vr. 7 mo. 6 da. 7. 6 yr. 4 mo. 24 da. 8. 7 yr. 6 mo. 25+ da. 9. 14f yr. 10. 20 yr. ; 12iyr.; 15^yr.; ll|^yr.;40yr.; 25yr. ; 'SOU yr. ; 22| yr. 11. 71|yr. Art. 601. 1. $146.27795. 2. $977,532. 3. $31390106+. 4. $1854.576. 5. !5^6o4.5102 + . 6. $20.0034. 7. $205,516 + . 8. $108,595 + . 9. $44.0824 +gr. at comp. int. Art. 604. 1. 1219.558. 2. $183.0183. 3. $133.55053. 4. $11 4.8721673^ 5. $55 4364. C. $99.2659725, AJVSWURS. 247 Art. 606. 1. $469.53704. ^. $1230.2528. 3. $781.52013. 4. $2755.3006. 5. $506.8663. 6. $348.1372193. 7. $357.399443. 8. $328.3345. 9. $145.7:>8068. 20. $556.75033. 11. $753.052567. li\ 83439.63075. 13. $1097.5152. 14. $854.943736. Art. 609. 1. $146,004. ^. $1071.41i. 3. $1138.075. 4- $33.1858. 5. $363,308. 6. $50.56 dif. between Sim. and Annual Int. ; $4,309 dif. between An. and Com. Int ; $51,769+ dif. be- tween Sim. and Com. Int. Art. 616. 1. $283.46}. £. $11,254. 3. $302,793. 4- $117,942. Art. 619. e, $1491.49 + . 3. $2891.5^7. 4. $420,293. 6. $5434.651 + . Art. 624. i. $776,699+; $754,717 + . e. $315,789+ ; $348,387 + . 5. $485,468+ ; $478.10+. 4 $9 975 + . $148,456. 5. $5,513 ; $40.83 + . 6. $15,275; $230.57R. 7. $530,367. 8. $9171.90 + . 9. $.957. 10. $435. 11. $.103 more profit- able at $4.66. 1^. 3d $33 865 better. 13. $1.47 + . Art. 632. 1. $5.88 Bk. Dis. ; $374.13 Proceeds ; $11,941 Bk. Dis. ; $368.05i Proceeds. 2. $30.6711 Bk. Dis. ; $769.3381, Proc'ds; $11.530|Bk. Dis. ; $778.479.V Proc'ds. 3. $35.83f Bk. Dis. ; $1574. 17f Proc'ds $54.033f Bk. Dis. $1545.9771 Proc'ds 4. $.884. 5. $30.13U. 6. $37,194. 7. Due May 37; 59 da. Time ; $475,383+ Proc'ds 8. Due Aug. 16 ; 75 da. Time ; $581.3951 Proc'ds. 9. Due Aug. 22 ; 91 da. Time ; $1571.681 Proc'ds. Art. 635. 1. $876,061 + ; $295,415 + ; $540713. ^. $458,387 ; 4. $480,616. 5. $354,453. 6. $398,899 + . 7. $961,781, 1st; $967,495, 3d; $979,914, 3d. 8. $495,363. 9. $1517.440. Art. 643. 1. $3416. $99,113. $238.63 ; $1830.923; $515,648. 3. $850.68. 4. $6331.50. 5. $133,796. 6. $1491. 7. $382.3038*. 8. $391,141. 9. $486. 10. $560. 11. $730. 12. $480. 13. $824 U. $375.50. 15. $321. 16. $403. 17. $698.25. 18. $1615.11*-. 19. $2415.925. ^0. $1779. 21. $496. 22. 97TV'%,or3U%dia 23. $3536.38^. 24. $8013.43 + . Art. G48. 2. $2134.99065 + . 3. $81)3.22?. 4. $1(>42.41. 6. $763. 6. $3469. 33k 7. $609,375. 8. $11456.8125. Art. 650. 1. $12615.96. 2. $301 gain by Ind. 3. $124852+ less by Ind. 4. 3011.58+ marks. 248 ANSWERS. Art. 661. 1. July 1, 1876. ^. Dec. 27, 1876. 3. April 30, 1876. 4. Oct. It), 1876. ^. Sept. 3, 1876. C. July 21, 1877. 7. Oct. 19, 1877. 5. 60 da. i?. Feb. 5. Art. 664. 1. $210, Face of note ; Due Dec. 12, 1876. ^. $100 due; Dec. 7, 1876, equat- ed time. S, Apr. 6, 1877. 4. March 1, 1876. Art. 681. i. f. 14» "• 3TTK- 4. //s- 5. «gW. 6. Y- Art. 683. 5. C. y. 7. V- s.h 9. jV Art. 685. /. 150 da. S. 45 A. ;.'. $597. 4. $2100. 5. $2386.40. Art. 687. 1. 104 A. e. $50. ^. 130 da. 4. $fi78|f. 5. 415 lines. 6. 177 cd. 3 cd. ft. 7. $2000. <9. 17 gal. 3 qt. 1 pt. 9. $13.50. Art. 690» 2. $72. 5. $468. ^. $27. 5. $100. 6. $204. 7. 403|. 5. 228f yd. Art. 700. 1. 35. ^. 6. S. 272. 4. 29^ bu. 5. 10 yd. C. £168 15s. 6f€L 1. 56. 2. 21. ^. 15. 5. 213^. 510, B's stock ; $7150, C's stock. 4. P35.10 + , A's profit ; ^288.56 + , B's profit ; $251.32 + , C's profit. 6. $3666.06 + , A's share ; $5238.93 + , B's share. Art. 717. 5. $1.00. 5, $.13^V. 4. 18 J carats. 6. $.311. Art. 722. 1. 1, 3, 2, 1 lb. 5. 3 gal. of mo. to 3 gal. of water. 5. 1 part of each. 4. 1, 2. and 6 bbl. 6. 2, 2, 604, and 240 bbl. 7. 2, 1, and 109 gal. 8. 50, 50, 5, and 1 10. 18, 27, 63, and 27 lb. 11. '30, 30, and 180 oz. 12. 44|,89t,and89|lb. Art. 728. 8. 2304: 4225,186624. 5. 86436; 148996; 247009 ; 64009. 4. 250047 ; 15625 ; 438976; 60236288. .512. 6. 56169. 7. 4100625. O 2197 Q 289 10. .00390625. 11. .039304 13. .00028561. U. .000000166375. 15. .00091204. Art. 729. 2. 7. 3. 7. 4. 9. 5. 15. 6. 12. 7. 13. 5. 12. iO. 30. 11. 24. ii?. 12. Art. 745. 1. 64. 7. If. 2. 59. 5. If. 3. 53. 5. If. 4. 87. 10. .15. 5. 93. 11. .48. G. 96. i;^. .76. 13. 371. i4. 64.5. 15. 876.6. i6. 167.4. i7. 7.56. IS. 21.79. i9 5.656+. 20. 7.681 +. 21. 2.645+. j^^. .964 + . 23. .894+. ^4. .612 + . 25. 3.834+. ;?e. 9.284+. ^7. 2.443+. 28. .881+. ;?«?. .346 + . 30. .404 + . ,?i. 51. ^^. Q\ S3. 33587. J4. 6. 35. 2656. 56'. 354.906 + . 57. 95 ft. 38. 487 ft. 39. 69.57+ yd. <^. 96 trees. 41. 44.342+ rd. 4^. 1.4142; 2.2360; 3.3166. 43. .654;. 852; .735. Art. 753. 1. 6. 9. M. ^. 9. m j.f. 5. 11. i7. 63. 4. 13. i^. 76. 5. 16. 13. 289. 6. 22. 14. 361. 7. 19. i5. .y^. .-au^'.i*i»t- -b