IjItlBllllllIf ■■■■ HBBfl^^H ^SSiSISHtMili'ti' .» M*3*f y/f *M&? & * Digitized by the Internet Archive in 2007 with funding from Microsoft Corporation http://www.archive.org/details/commercialarithmOOwillrich WILLIAMS & ROGERS SERIES COMMERCIAL ARITHMETIC BY OSCAR F. WILLIAMS, A. M. HEW YORK . : • CINCINNATI . : • CHICAGO AMERICAN BOOK COMPANY ^ Entered according" to Act of Congress, in the year By WILLIAMS & ROGERS, |In the Office of the Librarian of Congress, at Washington, D. C Copyright, 1894, BY WILLIAMS & ROGERS. W. P. 12 PREFACE. TVrO school text-book is used now-a-days in consequence of prefatory compli- *~ ments paid it by its authors or publishers. In this day of general enlight- enment, teachers understand the necessities of their classes, and, as a rule, need no advice as to what or how to teach. Since the wants of American schools and the ideas of American teachers are various, a variety of text-books upon every topic of school instruction is required, and with the hope and belief that the contents of this volume will more nearly meet the necessities of some schools and the ideas of some teachers, than any of the several good books upon the subject of Arithmetic now in print, the work is respectfully submitted by THE AUTHORS. 541^71 CONTENTS. SIMPLE NUMBERS. PAGE Definitions .. 1 Signs . 2 Abbreviations and Contractions _ _ _ 3 Notation and Numeration 3 Arabic Method of Notation 4 French System of Numeration 4 English System of Numeration 5 Roman Notation and Numeration 7 Addition _ .. _ 8 Addition Table 9 Group Method of Addition 11 Horizontal Addition „_ 11 Subtraction .... „.. 16 Subtraction Table 17 Multiplication _ 20 Multiplication Table 21 Division 28 Long Division 34 Average 37 Complement 37 Factors and Factoring 38 Divisors... 39 Multiples _ 41 Cancellation _ 43 COMMON FRACTIONS. Definitions. ..... 45 Reduction of Fractions 46 Addition of Fractions .__„_„ 49 Subtraction of Fractions. 51 Multiplication of Fractions ..... . 54 Division of Fractions 57 Complex Fractions 61 Miscellaneous Examples 61 Definitions... ... ........ 66 Numeration of Decimals.... 67 Notation of Decimals., 68 Reduction of Decimals 70 Circulating Decimals . 72 DECIMALS. Addition of Decimals 73 Subtraction of Decimals 75 Multiplication of Decimals 75 Division of Decimals 76 Miscellaneous Examples 80 UNITED STATES MONEY. Definitions .... 81 United States Coins 82 United States Paper Money 83 Reduction of United States Money 84 Addition of United States Money 84 Subtraction of United States Money... 84 Multiplication of United States Money. . 85 Division of United States Money ... 86 Definitions , ANALYSIS. 87 | Examples. 87 SPECIAL APPLICATIONS. Definitions 89 Aliquot Parts 89 Instructions for Practice with Aliquot Parts 90 Miscellaneous Contractions 90 Instructions for Finding Quantity 91 Miscellaneous Contractions 92 Bills, Statements, and Inventories 100 Miscellaneous Examples 106 CONTENTS. Vll DENOMINATE NUMBERS Definitions 108 Measures of Time. .••..... 108 Reduction of Time........ .... 110 Addition of Time Ill Subtraction of Time 112 Circular Measure ........ 113 Latitude, Longitude, and Time 113 Standard Time 114 English Money 116 Reduction of English Money 117 Measures of Weight ... 119 Troy Weight 119 Reduction of Denominate Numbers ... 119 Addition of Denominate Numbers 122 Subtraction of Denominate Numbers ._ 123 Multiplication of Denominate Numbers 123 Division of Denominate Numbers 123 Avoirdupois Weight 125 Table of Avoirdupois Pounds per Bushel 126 Additional Tables 127 Apothecaries' Weight 128 Comparative Table of Weights .... 128 Measures of Capacity. 129 Dry Measure ................... ... 129 Liquid Measure 130 Comparative Table of Liquid and Dry Measures 130 Measures of Extension.. , . 131 Linear Measure 131 Special Table, Linear Measure 132? Square Measure. - 13$ Involution 137 Evolution 137 Square Root .. .. 137 Surveyors' Long Measure , ...... 141 Surveyors' Square Measure 142= Cubic Measure 14& Table Special Cubic Measures 144 Producers' and Dealers* Approximate Rules 146 Hay Measurements 147 Cube Root 147 Duodecimals . 151 Miscell aneous Measurements 151 The Metric System 155 FrenchMoney 157 German Money.... ..... .».*.......... 157 Miscellaneous Examples 158 PERCENTAGE. Definitions 160 To find the Percentage, the Base and Rate being given ... 162 To find the Base, the Percentage and Rate being given ... ... 163 To find the Rate, the Percentage and Base being given _ 164 To find the Amount Per Cent., the Rate being given 165 To find the Difference Per Cent., the Rate being given 166 To find the Amount, the Base and Rate being given 166 To find the Difference, the Base and Rate being given 167 To find the Base, the Amount or Differ- ence, and the Rate being given 168 Review of the Principles of Percentage 169 Profit and Loss 173 To find the Profit or Loss, the Cost and Rate being given 173 To find the Cost, the Gain or Loss, and the Rate of Gain, or Loss being given 174 To find the Rate of Profit or Loss, the Cost and the Profit or Loss being given 17t To find the Cost, the Selling Price and the Rate Per Cent, of Profit or Loss being given Vfi Review of the Principles of Profit and Loss , „ ITS Trade Discount !.8fc To find the Selling Price, the List Price and Discount Series being given 18& To find the Price at which Goods must be Marked to Insure a Given Per Cent, of Profit or Loss, the Cost and Discount Series being given„_. 184 To find a Simple Equivalent Per Cent, of Discount, a Discount Series being given 186 Storage... „. 187 To find the Simple Average Storage.... 187 To find the Charge for Storage with Credits 188 To find, the Storage when Charges Vary 190 Vlll CONTENTS. PAGE Commission ._ _ 191 To find the Commission, the Cost or Sell- ing Price and Per Cent, of Commis- sion being given 192 To find the Investment or Gross Sales, the Commission and Per Cent, of Commission being given 192 To find the Investment and Commission, when both are included in a Remit- tance by the Principal 192 Custom-House Business 197 To find Specific Duty 198 To find Ad Valorem Duty 199 Taxes _._ 201 To find Property Tax .... 201 To find a General Tax 202 Insurance __ 204 To find the Cost of Insurance.. 206 To find the Amount Insured, the Pre- mium and Per Cent, of Premium being given _ 206 Personal Insurance 208 Interest 209 Six Per Cent. Method 210 To find the Interest on Any Sum of Money, at Other Rates than 6 Per Cent 210 To find the Interest, the Principal, Rate, and Time being given 211 To find the Principal, the Interest, Rate, and Time being given 214 To find the Principal, the Amount, Rate, and Time being given 214 To find the Rate, the Principal, Interest, and Time being given „ 215 To find the Time, the Principal, Interest, and Rate being given 216 Short Methods for Finding Interest 216 To find Interest for Days, at 6 Per Cent., 360 Day Basis 217 Stocks and Bonds 273 To Find the Dividend on Stocks, the Capital Stock and Rate Per Cent, of Dividend being given _ _ 274 To Find the Rate Per Cent, of Dividend, the Capital Stock and Net Earnings being given _ 275 To Find the Par Value, the Premium or Discount being given 276 To Find the Market Value, the Premium or Discount being given 277 To Find the Rate Per Cent, of Invest- ment, the Cost and Dividend being given _ 277 Exchange 280 PAGB To find Interest at Other Rates than 6 Per Cent., 360 Day Basis 217 To find Interest for Days at 6 Per Cent , 365 Day Basis 220 Periodic Interest _ 221 To find Periodic Interest 221 Compound Interest 222 To find Compound Interest _ 223 Compound Interest Table 224 True Discount _ 230 To find the Present Worth of a Debt... 230 Bank Discount 233 General Remarks on Commercial Paper, 234 To find the Discount and Proceeds of a Note 236 To find the Face of a Note 238 Partial Payments 239 United States Rule 240 Merchants' Rule.... s 241 Equation op Accounts 243 When the Items are all Debits, or all Credits, and have no Terms of Credit 244 When the Items have Different Dates and the Same or Different Terms of Credit 248 When an Account has Both Debits and Credits 250 Ratio 259 Proportion 260 Simple Proportion .. 260 Compound Proportion 261 Partnership 263 To Divide the Gain or Loss, when Each Partner's Investment has been Em- ployed for the Same Period of Time, 264 To Divide the Gain or Loss, According to the Amount of Capital Invested and the Time it is Employed 266 To Find the Cost of Domestic Exchange 283 To Find the Cost of Exchange by Ex- press M. O., U. S. M. O 283 To Find the Proceeds of Commercial Paper Drawn at Sight 284 To Find the Proceeds of Commercial Paper Drawn on Time 284 Foreign Exchange... 285 To Find the Cost of a Foreign Bill of Exchange, the Face of the Bill and the Rate of Exchange being given.. 289 To Find the Face of a Foreign Bill of Exchange, the Amount ~and Rate . being given 289 Answers 291 COMMERCIAL ARITHMETIC. DEFINITIONS. 1. Arithmetic is the Science of Numbers and the Art of Computation. 2. A Unit is a single thing. 3. A Number is a unit or a collection of units. 4:. The Unit of a number is one of the collection of units forming the number; thus, the unit of 5 is 1; of 17 dollars, 1 dollar; of 30 pupils, 1 pupil. 5. An Integer is a whole or entire number. 6. An Even Number is one that can be exactly divided by 2; as, 6, 8, 44. 7. An Odd Number is one that cannot be exactly divided by 2; as, 5, 9, 23. 8. A Composite Number is one that can be resolved or separated into factors; as, 4 = 2 X 2; 12 = 3 X 2 X 2. 9. A Prime Number is one that cannot be resolved or separated into factors, being divisible only by itself and unity; as, 1, 2, 3, 5, 7, 19, 83. 10. An Abstract Number is one used without reference to any particular thing or quantity; as, 3, 11, 24. 11. A Concrete Number is one used with reference to some particular thing or quantity; as, 3 dollars, 11 men, 24 cords of wood. 12. A Compound Denominate, or Compound Number, is a concrete number expressed by two or more orders of units; as, 3 dollars and 11 cents; 5 pounds, 2 ounces and 15 pennyweights. 13. Like Numbers are such as have the same unit value; as, 5, 14, 37; or, 5 men, 14 men, 37 men; or, if denominate, the same kind of quantity; as, 5 hours 14 minutes 37 seconds. 14. Unlike Numbers are such as have different unit values; as, 11, 16 days, 265 dollars, 5 pounds, 4 yards. 15. Ratio is the comparison of magnitudes. It is of two kinds; aritlimeticaX and geometrical, 16. Arithmetical Ratio expresses a difference. 17. Geometrical Ratio expresses a quotient. 18. A Problem in Arithmetic is a question to be solved; its analysis, the logical statement of its conditions and of the steps required for its solution. SIGNS. \ \IQ.. The Conclusion of the analysis is called the answer, or result. 20. A Rule is an outline of the steps to be taken in a solution. SIGNS. 21. A Sign is a character used to express a relation of terms or to indicate an operation to be performed. The following are the principal and most useful arithmetical signs: 22. The Sign of Addition is a perpendicular cross, +. It is called Plus, and indicates that the numbers between which it is placed are to be added ; thus, 5 + 4 indicates that 4 is to be added to 5. 23. The Sign of Subtraction is a short horizontal line, — . It is called Minus, and indicates, when placed between two numbers, that the value of the number on its right is to be taken from the value of the number on its left ; thus, 8 — 3, indicates that 3 is to be subtracted from 8. 24. The Sign of Multiplication is an oblique cross, x. It indicates that the numbers between which it is placed are to be multiplied together; thus, 7X9, indicates that the value of 7 is to be taken 9 times. 25. The Common Sign of Division is a short horizontal line with a point above and one below, -r-. It indicates a comparison of numbers to determine a quotient, it being understood that the number at the left of the sign is to be divided by the one at its right ; thus, 20 -*• 5, indicates that 20 is to be divided by 5. 26. The Sign of Ratio is the colon, : ; it also indicates division. 27. The Sign of Equality is two short horizontal lines, =. It is read equals, or, is equal to, and indicates that the numbers, or expressions, between which it is placed are equal to each other; thus, 2 + 2 = 4. 28. The Signs of Aggregation are the parenthesis, ( ), brackets, [ \ brace, { j, and vinculum, . They indicate that the quantities included within, or connected by them, are to be taken together and subjected to the same operation. 29. The Index, or Power Sign, is a small figure placed at the right of and above another figure. It indicates that the number over which it is placed is to be taken as a factor a number of times equal to the numerical value of the index. Thus 4 2 indicates that 4 is' to be taken twice as a factor, or multiplied by itself once; 4 3 indicates that 4 is to be used three times as a factor. 4 2 is read 4 squared; 4 3 is read 4 cubed; also, the second power of 4; the third power of 4. 30. The Root, or Radical Sign, is the character, V> it is the opposite of the index, or power sign. When there is no figure in the opening, it indicates that the quantity over which the sign is placed is to be separated into two equal factors, or its square root taken. A figure placed in the opening indicates the number of equal factors required, or the root to be extracted ; as, %/~^, %/~Ts. ABBREVIATIONS AND CONTEACTIONS. 3 31. The Dollar Sign is the character, $. 32. The Cent Sign is the character, o «— < 00 § .2 .2 s K H H w H § 2, 5 7 3 1 8 1, •a a P o o *C — a> ^ 9 6 Each period of the higher orders has also six places. |1 tS<£ 7 I 2. Remark. — The English system of Numeration being of no practical value to pupils in the schools of the United States, it will not be hereafter referred to. 41. The Arabic method of notation is based upon the following General Principles. — 1. The removal of any figure one place toward the left multiplies its value by ten ; two places, by one hundred ; three places, by one thousand, etc. . 2. The removal of any figure one place toward the right divides its value by ten ; two places, by one hundred, etc. 8. A cipher placed after a significant figure multiplies it by ten ; two ciphers so placed, multiplies it by one hundred, etc. 6 DOTATION AND » I AlERATION. 42. Write and read : 1. Nine units of the first order. 2. Five units of the first order and two units of the second order. 8. Eight units of the first order, three of the second, and one of the first. J/.. Four units of the fourth order, nine of the third, and two of the second. 5. Two units of the fifth order, nine of the third, and seven of the first. 6. One unit of the sixth order, nine of the fourth, six of the second, and eight of the first. 7. Seven units of the seventh order and seven of the first. 8. Six units of the eighth order, four of the sixth, seven of the fourth, and one of the second. 9. One unit of the ninth order, two of the eighth, three of the seventh, four of the sixth, five of the fifth, six of the fourth, seven of 'the third, eight of the second, and nine of the first. 43. Express by figures the following numbers: 1. Sixty-four. 2. One hundred forty-eight. 3. One thousand four hundred six. J.. Twenty thousand twenty-one. 5. Three hundred sixty-five thousand. 6. Eighty million forty-two. 7. Ninety thousand nine hundred. 8. Fifty million fifty-one 9. Eighty-seven billion seven thousand twelve. 10. Ninety-seven million ninety-seven thousand ninety-seven. 11. Twenty-one million twenty-five. 12. Sixteen billion sixteen million sixteen. IS. Six hundred eighty-nine thousand nine hundred seven. 111.. Nineteen billion five hundred forty-one million eleven thousand eleven. 15. Twenty-seven quintillion eighty-one quadrillion two trillion seven hundred sixty billion one million two. 44. Point off into periods, numerate, and read the following numbers: 1. 386. 8. 77010016. 15. 987000460000. 2. 1341. 9. 200020. 16. 27510304050. 3. 12406. 10. 1400246780. 17. 11002200330044. 1 79001. 11. 2100211. 18. 2234567890. 5. 872463. 12. 5860092. 19. 46122555003. 6. 901008. 13. 34307061. 20. 621438001246709. 7. 4681005. u. 100010001000. 21. 12345325596001503 45. Write in words an d read the following numb Brs: 1. 920. 5. 50115. • 9. 1406250. 2. 1146. 6. 100000. 10. 54790207. 3. 3070. 7. 521469. 11. 1021714. If. 61036. 8. 201012. 12. 579607359. NOTATION" AND NUMERATION. < ROMAN NOTATION AND NUMERATION. 46. By combining, according to certain principles, the letters used in this method of writing numbers, any number can be expressed. Principles. — 1. Repeating a letter repeats its value. Thus, I = one, II = two, X = ten, XX = twenty. 2. If a letter of any value is annexed to one of greater value, the sum of the two values is indicated; if a letter of any value is prefixed to one of greater value, the difference of their values is indicated. Thus, XI denotes X + I = eleven, IX denotes X — I = nine. S. A dash — placed over a letter multiplies its value by one thousand. Thus, V = five, V= five thousa nd, CD = four hundred, CD = four hundred thousand, LXVII = sixty-seven, LXVI I = sixty-seven thousand. Table of Komau Numerals with Arabic Equivalents. I, 1 xi r, 12. L, 50. II, 2. XIII, 13. LX, 60. III, 3. XIV, 14. LXX, 70. IV, 4. XV, 15. LXXX, 80. v, 5. XVI, 16. XC, 90. VI, 6. XVII, 17. c, 100. VII, 7. XVIII, 18. cc, 200. VIII, 8. XIX, 19. ccc, 300. IX, 9. XX, 20. CD, 400. x, 10. XXX, 30. D, 500. XI, 11. XL, 40. DC, 600. 47. Eead the following express sions: CM MC XIX. LXXIX. XCII. CCXVII. DCV. ] XXVII ] XXIX. DCCX. MC DXCII. CLX. ( MXXV. MD CCLVI. J DCC, 700. DCCC, 800. CM, 900. M, 1000. MM, 2000. X, 10000. L, 50000. o, 100000 D, 500000 M, 1000000 DLXX. DCCXLV. MDC. MDCCCLXXXVIIL 48. 42. 111. 666. 1125. 7000. 11451. 997. 56104. 3001. Express by the Roman system the following numbers: 7454. 8709. 62550. 1620. 399. 25406. 48250. 3700. 2865. 1629. 1889. 460012. 3658. 175400. 1761. 1887. 1000000. 20000. 45450. 19015. 1111. 6057. 3113. 90055. 805000. 365. 1515. 6059. 21021. 4888. 90909. 5168. 1890. 1775. 1893. 1900. ADDITION, ADDITION, 49. Addition is the process of combining several numbers into one equiva lent number. 50. The Sum or Amount is the result obtained by the addition of two 01 more numbers. 51. The Sign of Addition is +, and is called Plus, which signifies more. "When placed between two numbers or combinations of numbers, it indicates their addition; as, 5 -{- 2 is read 5 plus 2, and shows that 5 and 2 are to be added. 52. The Sign of Equality is = . When placed between two numbers or combinations of numbers, it indicates that there is no difference in their value; thus, 5 + 2 = 7, is read 5 plus 2 equals 7, and indicates that the value of 7 equals the value of the sum of the numbers at the left of the sign of equality. 53. Carrying the Tens is the process of reserving the tens and adding them with the next column. 54. Principles. — 1. Only like numbers and like unit orders can be added one to another. 2. The sum or amount contains as many units as all the numbers added. 8. The sum or amount is the same in zahatever order the numbers be added. 55. Addition is the Reverse of Subtraction and may be proved by it; as, 5 + 2 = 7. Now if 7 be diminished by 5, the result will be 2, while if 7 be diminished by 2, the result will be 5. 56. Numbers are written for addition either in vertical or horizontal order. 57. General Rules. — 1. If the sum, of two numbers and one of the numbers be given, the unknown number may be found by taking the given number from the sum. 2. If the sum of several numbers and all of the numbers bat one be given, the unknown number may be found by subtracting the sum of those given from the sum of all the numbers. Notes to Teacher. — 1. Classes should have frequent and extended drill in rapid mental addition. 2. The following table is given simply to facilitate class drill, preparatory to work in rapid addition. ADDITION. 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 16 17 18 19 8 + 8 + Addition 12 = 20 13 = 21 14 = 22 15 = 23 16 = 24 17 = 25 18 = 26 19 = 27 20 = 28 21 = 29 22 = 30 23 = 31 24 = 32 25 = 33 9 = 18 10 = 19 11 = 20 12 = 21 13 = 22 14 = 23 15 = 24 16 = 25 17 = 26 18 = 27 19 = 28 20. = 29 21 = 30 22 = 31 23 = 32 24 = 33 25 = 34 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 20 11 = 21 12 = 22 13 = 23 14 = 24 15 = 25 16 = 26 17 = 27 18 = 28 19 = 29 20 = 30 21 = 31 Table for Class Drill. 10 + 22 = 32 I 13 + 22 = 35 10 + 23 = 33 ' 13 + 23 = 36 10 + 24 = 34 13 + 24 = 37 10 + 25 = 35 13 + 25 = 38 11 + 11 11 + 12 11 + 13 11 + 14 11 + 15 11 + 16 11 + 17 11 + 18 11 + 19 11 + 20 11+21 11 + 22 11 + 23 11 + 24 11 + 25 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 12 + 12 12 + 13 12 + 14 12 + 15 12 + 16 12 + 17 12 + 18 12 + 19 12 + 20 12 + 21 12 + 22 12 + 23 12 + 24 12 + 25 24 25 26 27 28 29 30 31 32 33 34 35 36 37 13 + 13 13 + 14 13 + 15 13 + 16 13 + 17 13 + 18 13 + 19 13 + 20 13 + 21 26 27 28 29 30 31 32 33 34 4 + 14 4 + 15 4+16 4 + 17 4 + 18 4 + 19 4 + 20 4 + 21 4 + 22 4+23 4 + 24 4 + 25 28 29 30 31 32 33 34 35 36 37 38 39 5 + 15 5 + 16 5 + 17 5 + 18 5 + 19 5 + 20 5 + 21 5 + 22 5 + 23 5 + 24 5 + 25 30 31 32 33 34 35 36 37 38 39 4C 6 + 16 6 + 17 6 + 18 6 + 19 6 + 20 6 + 21 6 + 22 6 + 23 6 + 24 6 + 25 32 33 34 35 36 37 38 39 40 41 7 + 17 7 + 18 7 + 19 7 + 20 7 + 21 34 35 36 37 38 17 + 22 s 39 17 + 23 = 40 17 + 24 = 41 17 + 25 = 42 18 18 18 18 18 18 18 18 + 18 = 36 + 19 = 37 + 20 = 38 + 21 = 39 + 22 = 40 + 23 = 41 + 24 = 42 + 25 = 43 19 19 19 19 19 19 19 + 19 = 38 + 20 =- 39 + 21 = 40 + 22 = 41 + 23 = 42 + 24 = 43 + 25 = 44 20 20 20 20 20 20 + 20 = 40 + 21 = 41 + 22 = 42 + 23 = 43 + 24 = 44 + 25 = 45 21 21 21 21 21 + 21 = 42 + 22 = 43 + 23 = 44 + 24 = 45 + 25 = 46 22 22 22 22 + 22 = 44 + 23 = 45 + 24 = 46 + 25 = 47 23 23 23 + 23 = 46 + 24 = 47 + 25 = 48 24 24 + 24 = 48 + 25 = 49 25 + 25 = 5C ADDITION. MENTAL EXERCISES. 58. Add (1.) (*.) (A) (A) (5.) (6.) (7.) (*) 1 5 1 2 2 11 12 25 2 6 3 4 5 14 14 50 3 7 5 6 4 13 18 15 4 8 7 8 8 15 15 25 5 9 9 10 6 12 1G 10 6 10 11 12 12 18 20 25 7 11 13 14 10 17 14 15 8 12 15 16 14 20 16 10 59. There are three methods of addition in common use, viz. ; the Elementary method, the Result method, and the Group method. Kemarks. — 1. These methods of addition are recommended to be taught in their order to pupils in elementary work; the first, as soon as mastered, should be abandoned for the second, and the second in its turn, when mastered, abandoned for the third. 2. Daily drill in the third method is urgently advised with all pupils during the entire period of their study of Arithmetic. Too much importance can scarcely be attached to this suggestion. 60. The Elementary Method of Addition. Example.— Add 32, 71, 25, 48, 90, 12, and 63. ■operation. Explanation.— Having arranged the numbers so that units of like order* 32 71 stand directly under each other, begin with the last figure in the right-hand, or units' column, and add upward as follows : 3 and 2 are 5, 5 and 8 are 13, 13 and 5 are 18, 18 and 1 are 19, 19 and 2 are 21. Having thus obtained the sum, 25 place the 1 beneath the line, in units' column, and treat the 2 as a part of the 48. second, or tens' column, which add upward as before; thus, 2 and 6 are 8, 90 8 and 1 are 9, 9 and 9 are 18, 18 and 4 are 22, 22 and 2 are 24, 24 and 7 are 31, 22 31 and 3 are 34. Having obtained the sum, write it in full at the left of th€ £„ figure 1 before written, and the result is 341, the numerical expression of the sum of the numbers added. 341 T° Prove. — Add the columns downward ; if the two results agree, the work is presumed to be correct. 61. The Result Method of Addition. Example.— Add 32, 71, 25, 48, 90, 12, and 63. OPERATION. 32 7 Explanation. — Beginning as before, with the lower figure in units' column, name the result only of each successive addition, thus: 3, 5, 13, 18, 19, 21 ; then, ' as before, write the 1 beneath the line in units' column and carrying the 2 to 0q tens' column as a part of it, add upward, thus : 2, 8, 9, 18, 22, 24, 31, 34 ; as -. 2 before, write 34 at the left, and the result is 341, the same as before. 63 To Prove.— Add the columns downward. "341 ADDITION. 11 62. The Group Method of Addition. Example.—.?. Add 32, 71, 25, 48, 90, 12, and 63. operation. Explanation. — Treat the same numbers thus: add upward; 3, 13, 21; 32 n grouping 2 and 8 for 10 to add to 3, making 13, and 5, 1, and 2 for 8, to add to iy-1 (o 13, making 21. Having written the 1 beneath the line, in units' place, carry the f 2 or 2 tens, to its column, and again add ; 2, 8, 18, 24, 34 ; grouping 2 and 6 ^ ) for 8, 9 and 1 for 10, 4 and 2 for 6, and 7 and 3 for 10 j then write the result 48 \ in full as before. 90 V 10 To Prove. — Review the first column by adding downward ; 8, 18, 21 ; 12 ) grouping 2, 1, and 5 for 8, 8 and 2 for 10, to add for 18, and to this add the 63 remaining figure 3, for 21, the same result as before. Then review the second column by adding downward ; 10, 16, 26, 34 ; grouping 3 and 7 for 10, 2 and 4 341 for 6, 9 and 1 for 10, and 6 and 2, for 8, with the same result. Example.— & Add 3417, 2140, 439, 7164, 1538, 5046, 6116, 8735, 971, 4880, 1263, 9270, 192, and 634. bPERATiON. Explanation. — Beginning with the lower unit figure add upward; 10, 15, 35, 55, grouping 4, 2, 3, and 1 for 10, which added to 5 gives 15 ; grouping 6, 6, and 8 for 20 to add to 15 obtaining 35 ; and grouping 4, 9, and 7 for 20 to add to 35 for 55 the result. Write the units' figure 5 in its place, and carrying the tens' figure 5 to its column proceed thus : 8, 24, 38, 48, 56, 66, 70, grouping the 5 carried and 3 for 8 ; 9 and 7 for 16 to add to 8 for 24 ; 6 and 8 for 14 to add to 20 24 to make 38 ; 7 and 3 for 10, making 48 ; 1, 4, and 3 for 8, making 56 ; 6, 3, and 1 for 10, making 66, to which we add the 4 for 70, the result. Write the 34171 2140 439 7164 1538 5046 6116 20 8735 cipher of the 70 at the left of the unit figure already written beneath the line and carrying the 7 to the third, or hundreds' column group as before; 16, 26, 36, 48, 58, grouping upward thus : 7, 6, 1, 2 = 16 ; 2, 8 = 10 ; 9, 1 = 10 ; 7, 5 = 12 ; v 1, 4, 1, 4 = 10. Write the 8 in hundreds' column, and carrying the 5 to thou- sands' column, group 15, 27, 39, 49, 51. 5, 9, 1 = 15 ; 4, 8 = 12 ; 6, 5, 1 = 12 ; 7, 3 = 10, and adding 2 write the result, 51, at the left of the figures before written, thus obtaining 51805 the numerical expression of the sum of the num- 51805 k ers added. Prove by adding downward, grouping as illustrated above. 971 4880 1263 9270 192 634 Remark. — Practice in grouping will lead to great proficiency, and after the pupil becomes somewhat skilled, he should be encouraged to skip about somewhat along the column, in order to select those numbers which can be most conveniently grouped. Ordinarily thorough drill in the addition table will greatly assist in grouping, and multiples of the nine digits can be added with ease. Except with very bright pupils, groups greater than 25 are not to be recommended. HORIZONTAL ADDITION. 63. Numbers when written in horizontal order, as in invoices and other business forms, may be added without being re-written in vertical columns. Remarks.— 1. In adding numbers written horizontally, more care is requisite that the units added shall be of like order, and greater certainty of correctness can be had by adding first from left to right, and then from right to left. 2. The group method may be employed with equal advantage where numbers are written horizontally. 12 ADDITION. MENTAL EXERCISES. 64:. Add from left to right, and review from right to left. 1. 5, 3, 6, 1, 8, 2, 7, 9, 4. 2. 21, 56, 12, 93, 47, 60, 17. 3. 66, 29, 5, 14, 71, 19, 2, 11. h. 149, 865, 73, 40, 5, 13, 502. 6. 365, 10, 88, 46, 200, 175, 95. 6. 15, 23, 36, 18, 25, 53, 92. 7. 11, 85, 315, 125, 111, 206. 8. 8, 42, 87, 20, 112, 108, 94, 128. 9. 61, 400, 1, 126, 25, 440. 10. 25, 50, 511, 3, 209, 8, 804. WRITTEN EXERCISES. 65. Copy, and add from left to right; review from right to left, preserving results. 1. 510, 297, 69, 841, 638, 203, 40, 7, 700, 28, 9. 2. 1260, 2700, 408, 9206, 51, 7240, 27, 1620. 3. 8809, 1492, 1000, 20, 1, 504, 6620, 7596, 10. 4. 50000, 20000, 8900, 21050, 47800, 14090. 5. 76030, 20500, 38037, 69000, 81, 107, 2, .19975. 6. 346211, 218040, 173508, 973200, 701001, 555555. 7. 604000, 181523, 51, 19406, 200, 309, 5, 2, 8000. 8. 2463911, 7054133, 4444044, 1371005, 6090400. 9. 8500500, 1035660, 5000000, 2987400, 7020319. 10. 416, 49, 2, 7967400, 8i, 307, 21021, 190200, 40, 3. Remark. — Horizontal addition is rarely practiced with numbers containing more than four or five figures. It may sometimes be employed to advantage in adding dollars and cents; in such cases it is best to omit the dollar sign; as, for $5.25 write 5.25. 66. Copy and add horizontally; review and preserve results. 1. 5.25, 8.17, 11.40, 1.82, 16.02, 90.70. 2. 146.24, 9.11, 210.10, 46.98, 5.50, 108.12, 4.75. 3. 26.53, 92, 5.71, 108.97, 29.33, 150, 46.07, 19, 76. It.. 231.45, 50, 75, 19.78, 40, 50, 63, 100. 5. 63, 51, 87, 25, 75, 18, .09, 95, 1.25, 6. 6. 278.19, 105.29, 80.50, 19.93, 52, 1. 7. 29.30, 403, 51, 73, 1.14, 90, 300, 1.25. 8. 1.13, 9.25, 14, 27.16, 5.01, 8, 25, 1.75. 9. 87.50, 125, 36.21, 9.90, 14.75, 16, 25.25. 10. 117.82, 7.71, 19.03, 15, 49.55, 87.08. 11. 5.40, 88, 35, 90, 112.50, 45.95, 111.50. 12. 100, 79.22, 50.08, 2.25, 7.75, 10, 3, 8.24. 13. 216.24, 92, 15, .06, 138.50, 2.38, 9.25. Remark. — The teacher may give other examples of the same kind ; he will find extensive drill in such work of great value to all grades of pupils, in developing accuracy and rapidity. EXAMPLES FOR PRACTICE. 67. 1. A grocer's sales were, for Monday, $241; Tuesday, $306; Wednesday, $523 ; Thursday, $438 ; Friday, $497 ; on Saturday his sales amounted to $27 more than the sales of the first three days of the week. What were his total sales during the week? ADDITION. 13 2. A planter shipped eleven bales of cotton, weighing respectively 492, 504 ? 523, 487, 490, 500, 516, 499, 512, 511, and 496 pounds. What was the aggregate weight of the shipment ? 3. A portable saw-mill cut lumber for the six working days of a week, as follows : On Monday, 5116 feet ; Tuesday, 4900 feet ; Wednesday, 5750 feet • Thursday, 6100 feet ; Friday, 4580 feet ; and on Saturday, 6754 feet. What amount of lumber did the mill cut during the week? Jf. Find the sum of four units of the second order and five of the first; eight of the fifth, three of the third, and nine of the second; seven of the sixth, one of the fifth, and two of the third; one of the eighth, nine of the third, seven of the second, and six of the first; four of the fifth, three of the fourth, and nine of the third; fi>e of the tenth, one of the ninth, four of the seventh, eight of the third, two of the second, and one of the first. 5. Find the sum of sixty-nine thousand five hundred seven, one thousand six hundred twenty-two, one hundred fifty-six thousand seventy-six, ninety-nine thousand nineteen, forty-one million eighty-seven thousand five, three hundred twenty-five million sixteen thousand eight hundred eighty-eight, six billion ninety-one million four thousand two hundred fifty-six. 6. The British House of Lords was, in 1884, comprised of 4 princes, 23 dukes, 19 marquises, 139 earls, 32 viscounts, 26 bishops, and 272 barons. How many members in all ? 7. In 1883, there arrived and settled in the United States, immigrants: Germans, 192,000; English, 100,200 ; Canadians, 65,100 ; Irish, 64,400; Scandinavians, 52,200; Italians, 32,500; miscellaneous, 92,700. What was the total number of immigrants ? 8. The British national debt in March, 1883, was : Consols, £699,053,100 ; Bank debts, £13,645,900; Annuities, £27,570,900; Exchequer Bills, £8,754,400; Treasury Bills, £5,431,000 ; Savings Banks, £1,804,400 ; and the local debt, £163,501,000. What was the total debt in pounds sterling ? 9. In 1866, the U. S. collected as revenue from Customs, $179,046,651.58; from Internal Eevenue, $309,226,813.42; from Direct Taxes, $1,974,754.12; from the Public Lands, $665,031.03 ; and from other sources, $29,036,314.23. What was the total government revenue collected that year? 10. The British government collected as revenue in 1882 : From Customs, £19,300,000; from Excise, £27,230,000; from Stamps, £11,145,000; from Land Tax, £2,775,000; from Income Tax, £11,662,000; from Post Office, £7,150,000, from Telegraphs, £1,650,000; from Crown Lands, £380,000; from Interest,, £1,180,000 ; from miscellaneous sources, £4,725,000. What was the total reve- nue of the British government for that year? 11. The dwarf, Borowlaski, was only 39 inches in height ; Tom Thumb, 31 ; Mrs. Tom Thumb, 32 ; Che-Mah, of China, 25 ; Lucia Zarate, of Mexico, 20 ; and Gen. Mite, 21. What was the combined height of the six? 12. The firm of Davis & Drake own land valued at $39,750; lumber, $68,125; notes, $21,700 ; book accounts, $17,291 ; machinery, $13,250 ; cash in bank, $14,238 ; cash on hand, $4,232. What is the property value of the firm ? 14 ADDITION. 13. In 1880, there were women workers in the United States as follows: artists, 2,061; authors, 320 ; barbers, 2,902 ; dressmakers, 281,928 ; journalists, 288 ; lawyers, 75 ; musicians, 13,181 ; physicians, 2,432 ; preachers, 165 ; printers, 3,456; tailors, 52,098; teachers, 154,375. How many women workers in all ? Remark. — The three following problems can be properly used by the teacher for drill in group adding. H. The population of the United States, by the census of 1880, was as folio ws: Ala., 1,262,505 Alaska, 30,000 Ariz., 40,440 Ark., 802,525 Cal., 864,694 Colo., 194,327 Conn., 622,700 Dak., 135,177 Del., 146,608 D. C, 177,624 Fla., 209,493 Ga.,.__ ..1,542,180 Idaho, 32,610 111., ..3,077,871 Ind., ..1,978,301 Ind.T., 70,000 Kans., 996,096 Ky., 1,684,690 La., 939,946 Me., 648,936 Md., _ 934,943 Mass., 1,783,085 Mich.......... 1,636, 937 Minn.,... 780,773 Miss., 1,131,597 Mo., 2,168,380 Mont. T., 39,159 Nebr.,._ 452,402 Nev., 62,266 N. H., 346,991 N. J., 1,131,116 K Mex., 119,565 M". Y., 5,082,871 N. C, 1,399,750 Ohio, 3,108,062 Oregon,.. 174,768 Pa., 4,282,891 E. I., 276,531 S.C., 995,577 Tenn., 1,542,359 Tex., ..1,591,749 Utah, 143,963 Vt.,_ 332,286 Va.,.____ 1,512,565 Wash. T., 75,116 W. Va., 618,457 Wis., 1,315,497 Wyo. T. 20,789 Iowa, 1,624,615 What was the total population ? 15. The area of the United States, in square miles, is as follows : Ala., 51,540 Alaska, .531,409 Ariz., 112,920 Ark., 53,045 Cal., 155,980 Colo., 103,645 Conn., 4,845 Dak., ...147,700 Del., 1,960 D. C, 60 Fla., 54,240 Ga., _. 58,980 Idaho,. 84,290 111.,... 56,000 Ind., 35,910 Ind. T., 69,830 Iowa, 55,475 What is the total area ? Kans., 81,700 Ky., .... 4,000 La.,._ 45,420 Me., -. 29,895 Md., 9,860 Mass., 8,040 Mich., --.. 57,430 Minn., 79,205 Miss., 40,340 Mo., 68,735 Mont. T., 145,310 Xebr., 76,185 Nev., 109,740 N.H., 9,005 N. J., 7,455 N. Mex., 122,460 N.Y., 47,620 K C, 48,580 Ohio, 40,760 Oregon, 94,560 Pa., 44,985 E. L, 1,085 S. C, .. 30,170 Tenn., .-.. .41,750 Tex., 262,290 Utah, - 82,190 Vt, 9,135 Va., 40,125 Wash.T., 66,880 W. Va., 24,645 Wis.,.. 54,450 Wyo. T., 97,575 ADDITION. 15 16. For State tax of 1888, the several Counties of the State of New York were assessed as follows : Oneida, $58,146,279 Onondaga, 63,265,536 Ontario, 29,389,870 Orange, 42,953,974 Orleans, 14,816,445 Oswego, 23,655,679 Otsego,... 22,544,650 Putnam, .___ 7,483,530 Queens, 44,464,675 Rensselaer, . _ _ 60, 545, 955 Rockland, 13,394,485 Richmond, 12,271,105 Saratoga, . 23,189,435 Schenectady, 12, 772,451 Schoharie, 10,297,219 Schuyler, 7,248,620 Seneca, 15,347,372 St. Lawrence, 24,476,678 Steuben, 22,776,074 Suffolk, 17,262,646 Sullivan, 5,427,300 Tioga, 12,084,525 Tompkins, ._ 15,450,670 Ulster, 25,443,000 Warren, 6,555,175. Washington,.... 22,501,173 Wayne, 25,404,569 Westchester, 82,375,217 Wyoming, 14,922,986 Yates, '. 12,721,716 Albany, ....$86,606,307 Alleghany, -- 14,395,123 Broome,.* 21,383,568 Cattaraugus, - 16, 050, 985 Cayuga, 30,631,548 Chautauqua, 25, 649, 740 Chemung, 18,718,275 Chenango, 1 7,982,340 Clinton, 9,766,255 Columbia, 29,984,129 Cortland, 11,108,469 Delaware, 13,921,534 Dutchess, 44,532,280 Erie, 127,763,104 Essex, 10,515,260 Fulton, 8,383,735 Franklin, 8,026,235 Genesee, 21,384,810 Greene, 13,760,299 Hamilton, 1,157,600 Herkimer, 23,739,092 Jefferson, 23, 638,204 Kings, 342,116,976 Lewis, '..:.. 9,039,285 Livingston, 25,395,180 Madison, 19,797,535 Monroe, 85,964,190 Montgomery, 23,877,638 New York, 1,500,550,825 Niagara, 26,097,826 What was the total assessed value of the State that year ? 16 SUBTRACTION. SUBTRACTION. 68. Subtraction is the process of finding the difference between two numbers. 69. The Subtrahend is the number to be subtracted. 70. The Minuend is the number from which the Subtrahend is to be subtracted. 71. The Difference or Remainder is the result obtained by subtracting one number from another. 72. The Sign of Subtraction is — . It is called Minus and signifies less. When the sign of subtraction is placed between two numbers it indicates that the number placed after it is to be taken from the one before it. 73. The Complement of a Number is the difference between it and a unit of the next higher order. Thus the complement of 7 is 3, because 1 ten, the unit of the next higher order, diminished by 7 = 3. Again, the complement of 36 is 64, because the unit of the next higher order, 1 hundred, or 100, diminished by 36 = 64. 74. Principles. — 1. Only like numbers and units of the same order can be subtracted, one from the other. 2. The sum of the subtrahend and the remainder must be equal to the minuend, 75. General Relation of Terms in Subtraction. I. The Minuend — the Subtrahend = the Remainder. II. The Minuend — the Eemainder = the Subtrahend. III. The Subtrahend + the Eemainder = the Minuend. 76. General Rules. — 1. If the minuend and subtrahend be given, the remainder may be found by subtracting the subtrahend from the minuend. 2. If the minuend and remainder be given, the subtrahend may be found by subtracting the remainder from the minuend. 3. If the remainder and subtrahend be given, the minuend may be found by adding the remainder to the subtrahend. 77. To Prove Subtraction. — Add the remainder to the subtrahend; if the sum equals the minuend, the work is correct. SUBTRACTION. 17 Subtraction Table. 78. Find the difference, mentally, between 21 and 16 24 and 5 21 tt 17 24 n 6 21 a 18 24 tf 7 22 and 3 , 24 a 8 22 tt 4 24 iC 9 22 tt 5 24 (t 10 22 tt 6 24 it 11 22 tt 7 24 it 12 22 tt 8 24 tt 13 22 tt 9 24 it 14 22 tt 10 24 tt 15 22 tt 11 24 tt 16 22 tt 12 24 it 17 22 it 13 24 tt 18 22 ee 14 24 it 19 22 ee 15 24 tt 20 22 ee tt 16 17 24 tt 21 22 25 and 3 22 ft 18 25 tt 4 22 ft 19 25 tt 5 23 and 4 25 tt 6 23 tt 5 25 ee ' 7 23 (I 6 25 te 8 23 ee 7 25 ee 9 23 (C 8 25 ee 10 23 ee 9 25 ee 11 23 ee 10 25 ee 12 23 ee 11 25 ee 13 23 ee 12 25 ee 14 23 ee 13 25 ee 15 23 ee 14 25 t. 16 23 ee 15 25 ee 17 23 ee 16 25 ee 18 25 ee 19 23 ee 17 25 ee 20 23 ee 18 25 ee 21 23 ee 19 25 ee 22 Remarks.— 1. Frequent and thorough use of the Subtraction Table will result in great facility in all operations in this subject, and will also aid in additions and rapid work in arithmetical computations in general. 2. The above table, like the one in addition, is given for the teacher's reference, to time and labor in rapid mental exercises. 3 18 SUBTRACTION. 79. When any Figure in the Minuend is Less than the Corresponding Figure in the Subtrahend. Example. — From 435 take 176. operation. Explanation. — It is readily observed that the units figure 6 of the subtrahend cannot be taken from the corresponding figure of the minu- 43o Minuend. en( j . therefore analyze the minuend, and transform it into 4 hundreds, 176 Subtrahend. 2 tens, IS units; then from the 15 units take the 6 units of the subtrahend, obtaining 9 units as a remainder, which write as the units of the result j 259 Remainder, having reduced one of the tens of the minuend to units, we have only 2 tens remaining in the tens' column of the minuend, and since this is numerically less than the tens' figure in the subtrahend, transform as before, and read the 4 hundreds and 2 tens as 3 hundreds and 12 tens; then taking the 7 tens of the subtrahend from the 12 tens thus produced, write the remaining 5 tens for the second or tens' figure in the result; having taken 1 from the hundreds' column, we have 3 remaining in that column, from which take the 1 hundred of the subtrahend 7, obtaining 2 as the third or hundreds' figure of the result. Thus we conclude that 176 subtracted from 435, leaves a remainder of 259. Remarks. — 1. This process is called "borrowing tens," as each left-hand order is tenfold greater than the order at its right. 2. Having mastered the theory, the ordinary and most convenient method for practice is to leave the minuend figure in its original form and, when borrowing is necessary, add 1 to the succeeding subtrahend figure. Again, apply this method to the example; OPERATION. Explanation. — Subtract 6 from 15 leaving 9, which write in units' column ; T£l (adding 1 to 7) subtract 8 from 13 leaving 5, which write in its column; (adding : 1 to 1) subtract 2 from 4 leaving 2, which write in its column. 25fc Rule. — I. So write the numbers to he subtracted, that units of the same order stand in the same vertical line. II. Begin at the right and subtract each figure of the subtrahend from the corresponding figure of the minuend. When it is necessary, trans- form, or borrow ten, and -mentally add one to the next subtrahend figure. III. Write results in their proper order. EXAMPLES FOR PRACTICE. 80. -?. From 1524 take 911. *. From 3128 take 1519. 5. From 4055 take 2033. 4. From 27410 take 13520. 6. From 80500 take 30500. 6. From 123706 take 59341. 7. From 520200 take 368977. 8. From 80090 take 23084. 9. From 3406268 take 1998765. 10. From 303005 take 89700. 11. From 2046 take 1597. 12. From 40509300 take 9619475. 13. In Germany there are 2,436,000 land owners, and in France 3,226,000. How many more in France than in Germany? U. A dealer bought 1,732 sheep and sold to A 51, B 147, C 34, D 1000, and to 32 the remainder. How many did E purchase? SUBTRACTION. 19 15. A. farmer raised 1,?30 bushels of wheat, 958 of barley, 1,275 of oats, and 1,762 vi corn. If he keep for seed and feed, 116 bushels of wheat, 84 of barley, 600 of oats, and 1,150 of corn, how many bushels of grain will he have left to sell? 16. The equatorial diameter of the earth is 41847194 feet, and the polar diameter 41707.308 feet. How many feet greater is the equatorial than the polar diameter ? 17. If the sailing distance from New York to Queenstown be 2890 miles, how far from the latter port will a steamer be after running 1296 miles from the port of New York ? 18. Texas contains 274356 square miles and New York 47156 square miles. How many times may the area of New York be taken from the area of Texas and what number of square miles will remain ? 19. The area of Brazil is 3956000 square miles and of the United States 3026504 square miles. How many square miles greater is Brazil than the United States ? 20. A man bought a farm for 3250 dollars. He built a house on it at a cost of 1850 dollars, fences costing 416 dollars and then sold it for 7500 dollars. "What was his gain ? 21. I bought 23,240 acres of Dakota land, and sold at times 1000, 320, 520, 640, 3200, 2520, 160, and 1920 acres. How many acres had I remaining ? 22. During a five years' partnership a firm gained $123,475. If the gain the first year was $11,425 ; the second, $9,500 ; the third as much as the first and second, less $1,120; the fourth equal, to the second and third; how much must have been gained the fifth year? 23. The cost of my lot was $1,750. I paid for mason work on my house, $1,210; for carpenter work, $5,145; for plumbing, $985; for decorating, $1,650; for painting, $625 ; for grading, sodding, and fencing grounds, $590. The interest on outlays to date of sale was $315. I then sold the property at a loss of $20, receiving cash $6,000, and a note for the remainder. What was the face of the note? 21f.. My book-keeper's salary is $1,450 per year. If he requires for his rent, $365; for personal expenses, $170; and for support of his family, $775; what amount will he have left at the end of the year? 25. A Boston bicyclist journeying to San Francisco, distant 3,432 miles, ran the first week of six days, an average of 77 miles per day; the second week, 92 miles ; the third, 84 miles ; the fourth, 106 miles; the fifth, 95 miles, and reached his destination at the end of the sixth week. How many miles did he run the last week? 20 MULTIPLICATION. MULTIPLICATION. 81. Multiplication is the process of taking one of two numoers as many times as there are units in the other. 82. One of the numbers is called the Multiplicand and the other the Multiplier. The numbers are also called Factors of the product. 83. The Multiplicand is the factor multiplied. 84. The Multiplier is the factor by which the multiplicand is multiplied. 85. The Product is the result obtained by multiplying one number by another. 86. The Factors of a number are such numbers as will, when multiplied together, produce the given number. 87. A Continued Product is the result obtained by multiplying several factors together. 88. The Sign of Multiplication is an oblique cross, x . It is read "times," or "multiplied by" and indicates that the numbers between which it is placed are to be multiplied together, or their product obtained. Thus, 5x2 is read 5 times 2, or, 5 multiplied by 2. Remarks. — 1. In practice, the multiplier is regarded as an abstract number and the multipli- cand as a concrete number ; but as the resulting product is the same whichever factor is used as a multiplier, the above relation is recognized only in explanations of work done. 2. Where the multiplicand is concrete, the product will be concrete and of the same denomination as the multiplicand. 89. Multiplication is a short method of performing addition, and like addi- tion may be proved by subtraction. Thus, 2x2 = 4; that is, two taken twice as a factor = 4, or, 2 added to 2 = 4. We prove this by subtraction, 2 from 4 leaves 2. Again, 6 X 7 = 42; that is, 6 taken seven times as a factor = 42, or, seven 6's added == 42 ; this may be proved by subtracting seven 6's in succession from 42, when nothing remains. ' 90. General Rules. — 1. If the multiplicand and multiplier be given, the product may be found by multiplying those factors together. 2. If the product and multiplier be given, the multiplicand may be found by dividing the product by the multiplier. 3. If the product and multiplicand be given, the multiplier may be found by dividing the product by the multiplicand. J},. If the product of two numbers and one of the numbers be given, the other may be found by dividing the vroduct by the number given. M.ULTIPLICATIOX, ^ Multiplication Table. to IO 4- to CO to to — to © ^ X •'i c: 0« F> eo 10 — © © X -1 CI 0' 4- zz to UA 14. o 4=* 00 4=* Ci -'- -u. CO X 00 08 8 BO ■ X Of to CO Ci OS 4* O o co Ol eo tO 88 X — X — -1 to OS X o: — C-. 5 Or 9S Of CO GO — P 09 d CO to to X to 4- to © o 1— I to GO 4*. 0* tO ot to zz 01 o © Or zz CD O? eo o X Of * -7 Or -7 CO 01 Or Ol CI O' Or P Si 4- © CO Of 5 ro OI to CI 1— I Or O ©t © Of o lifc 09 X 03 to ro ■£ o GO Q to 60 08 80 3 GO -7 GO to O. Ol Ol Of — 4^ X to CO CO o to 4- t— «■ GO to © -01 -3 Of 01 35 OS Or 4* 4- — CO 08 80 to C~ Or BO X 80 GO 4- -7 -3J CO Ol So Or Ol CO 4^ to CO Of to GO to -— * 4=^ ^1 0D to o © to to X -> 08 ci X 39 s oJ b9 4^ 5 <0 to 4- 80 X J. X CI •3J to 4^ OJ — X — o CO to K Ci GC © to to Of ?o i— >. SB to o SO GO X 80 c: -J 99 to Or CO — 4- zl Ol iO s -7 p X 80 89 •- zz © -7 to Ol Or — Of z: a to -7 © I— © to Of o to © to co 5 JO fO o to o £0 ■-o X O § co CO Or s 4^ o ZZ CO to CO P © o --o S X CI -41 CI Ol d Of o o; ZZ zz to o © p-l to Of iO 01 to a CO tO to to CO ri ri q — 80 X X 08 Or Or 4^ 4- co 66 to to •— _» o 80 X GO -7 ~3 01 g 4^ 4^ oo o: to to — 10 CO o o ?0 X 00 69 01 ?o 8 ?o Or CO — to to X b9 lO CO ao to X S a X Or Ol 4^ 4> CO to to o O X 80 01 't -5 to Ci o fe Ol to 4^ to 08 CO to Of 00 M to tO CO CO X 5 o: S ?0 — to zz 4- ?o to )0 ~ X as Or X to co. Or 01 rfx CO CO o i— ^ — CO -7 X Ci Of Of to 2 to Ci CO tffc OS Of o CO 55 06 CO to bo OS O X re co 4- to X s to a ?o OT to ?o W X to to 4- to P CO o X to Ol X Of 4i- 4*> O to Ol to CO X X -3 © Or Ol 4^ to GO 4* Of CO Of i as o» 5 Or s - X Or o Or Or 4- co ro ro Ol to to Or X 3 O Of Of o BO Or ro CO Or CO CO -7 Of Ol 5 4^ Or CO o Of _l 4>* o o i zz OS X 60 Or to o: 08 to zz 5 4- to X GO to ro en to CO to to to S X 80 to -7 Oi CI O to -X to CO o: GO O oi GO CO to © V— 1 -J 4*> to Of 4- o -x as 4>> • i— i- GO ^J. -3 O Or 00 o: 01 CO ?o GO Of 01 X Of i— ^ CO 4» •a 00 Of © co *0 oo OS 0B CO X so 08 5 CO to CO ro -p. CO 08 to c X to to Or to to zz 4- IO 01 CO GO GO O Ol to 4- — to 01 1— k. X co o -4{ to Of 4- CO Ci — © Of Or 03 4^ CO -7 4^ 00 80 80 X o^ C5 to to SO CO 4- X Or Ol to — » o *-o GO :: Or 5 Of 80 CO Or to to to CO ^- i-*- o X o Ol X to 01 t—k- o Of X 4^ CI OO 4^ to to to Of Of o eg - to O — X z: — CO zz Or to CO c: CO 00 .0 X 08 tZ Ol to to ►u to to o 80 X -7 Ol Or — o: to r— k o GO X Ol Ol 4x 4^ to to to co Of Of Or to Ol ro CO Oj 08 4x £>■ GO OS CO O co — S S X 09 Of iO (O s <0 -7 Ol Of CO CO o ?0 -7 X 4^ Ol X 1— » 80 to 01 BS 4^ Ci to CO to 4* CS o o Ol Of q to Ol O GO 4*. o OT 08 4- ZZ to 4- O X CO X 4- SO a o zz 55 08 CO M to to X £». rf^ o to 0B •O' to Ol X rf> 4^ to © 89 Ol to 4^ GO to 4* to 0* ro en 5 5 oj Or Or Or O Of Of to o Of o — -I Or Si © ba Or 4'.- CO CO 00 Of CO Or CO BO to Or 00 5 CO to to -7 Of Of o IO to Or *0 o CO' 5 Or CO to Of © CO -3 Of Of o to Of to o* 4- 10 CO to to to to M © ^ at -i ft en ■i- CO to £ © © X -1 © 0"f 4- Ci to - Note. — It will be of great advantage to the student to fully master the above table. Any delay caused by following this suggestion will be offset by time gained in subsequent wo»-k ; such mastery will so increase the rapidity of work in business applications as to greatly lessen the labor of accounting. 52 MULTIPLICATION. 91. Multiply EXAMPLES FOR ORAL DRILL. 1. 42 by 3. 11. 102 by 4. 21. 144 by 10. SI. 595 by 13. 2. 31 by 4. 12. 511 by 8. 22. 52 by 11. 82. 70 by 22. 8. 27 by 2. IS. 125 by 6. 28. 45 by 13. S3. 90 by 25. 4. 60 by 5. u. 340 by 2. 24- 201 by 15. 84. 150 by 23. 5. 51 by 4. 15. 416 by 3. 25. 65 by 20. 85. 118 by 11. 5. 75 by 6. 16. 99 by 7. 26. 411 by 17. 86. 906 by 15. 7. 91 by 2. 17. 133 by 9. 27. 932 by 12. 37. 450 by 19. 81 » °y the divisor, 37, and obtain 2 as the first figure of the quotient; then subtract from 81 the product of 2 X 37, or 74, obtaining 7 as a remainder ; to this remainder annex 4, the succeeding figure of the dividend, which gives 74 as the next partial dividend; the divisor is contained in this dividend twice, or 2 times, giving 2 as the next or second quotient figure ; sub- tracting the product of 2 X 37 from 74, nothing remains; then bring down 3, the next figure of the dividend and as it is less than the divisor, place a in the quotient ; next bring down 7, the remaining figure of the dividend which gives 37 as the last partial dividend ; the divisor is contained in this dividend once, or 1 time ; writing this 1 as the final figure of the quotient and subtracting the last partial product from the last partial dividend nothing remains, and the quotient, 2201, is the result of dividing 81437 by 37. Rule* — 1. Write the divisor at the left of the dividend with a curved line between them, and another line at the right of the dividend to sep- arate it from the quotient when found. II. From the left of the dividend select the least number of figures that will contain the divisor one or more times, and divide. Write the quotient figure thus obtained at the right of the dividend, multiply the divisor by this quotient figure and subtract the product from the partial dividend used. To the remainder annex the succeeding figure of the dividend and divide as before; so continue until the last partial product has been subtracted from the last partial dividend. If there be a remainder place it over the divisor with a line between, and write the resulting fraction as a part of the quotient. Proof. — Multiply the divisor by the quotient, and to the product add the bemaindeb if there be any; the result should equal the dividend. DIVISION. EXAMPLES IN LONG DIVISION. 127 1. 1728 by 48. 11. 115680 by 155. 21. 2. 2025 by 135. 12. 29410 by 251. 22. S. 625 by 125. 13. 666666 by 2144. 23. 4. 1920 by 160. u. 93462007 by 1525. «*. 5. 2268 by 45. 15. 5005C0500 by 1888. 25. 6. 106295 by 28. 16. 21416009 by 5407. 26. 7. 52467 by 109. 17. 11460250 by 999. 27. 8. 4762 by 367. 18. 87629000 by 11181. 28. 9. 250000 by 793. 19. 20405701 by 820006. 29. 10. 87524 by 31. . 20. 72109904 by 72109. 30. 35 375735212 by 20812. 26800001 by 909125. 104690955 by 5642. 9000716002 by 1776, 250252500 by 1562. 5087910041 by 508791. 3641694611 by 72853. 111222333456 by 370054. 9876543210 by 12345. 210631890048 by 840263. MISCELLANEOUS EXAMPLES IN LONG DIVISION. 128. 1. In 1880 the total number of persons engaged in all occupations in the United States was 17392099, of which 7670493 were engaged in agriculture; how many times greater is the whole number of workers than those engaged in agriculture ? 2. The 2515 miles of canal in the United States cost $170028636 ; what was the average cost per mile? 3. If an elephant produces 120 lb. of ivory and the manufactories of Sheffield consume yearly 483000 lb., how many elephants must be killed each year to Bupply the Sheffield market alone ? If. In 1880 there were in attendance in the 177100 public schools of the United States 9705100 pupils ; what was the average number in attendance in each school ? 5. During the financial crisis of 1857, 7200 business houses in the United States failed for an aggregate of 111 million dollars ; what was the average insolvency ? 6. Dan. Lambert, at the age of 40, weighed 739 lb. ; if his weight at birth was 13 lb., what was his average yearly increase of weight ? 7. Between 1871 and 1884 the Kimberly diamond field of 9 acres produced 75 million dollars worth of diamonds; what average value per acre was produced each year ? Each month ? 8. A bottle thrown overboard into the Pacific Ocean was picked up 455 days later, 6700 miles distant from where it was thrown; what average distance did it float per day ? 9. The great bell of Moscow weighs 202 tons of 2240 lb. each; if 77 parts of the metal of which it is composed are copper and the remaining 23 parts tin, how many pounds of each metal does the bell contain ? 10. The log of the yacht Wanderer in circumnavigating the globe in 1880-82, showed 48490 miles run in 280 days actual running time; what was the average miles run per day ? 11. An Iowa firm manufactures daily, from 5 tons of paper, 1600 barrels, of 6 lb. weight each ; what number of barrels can be made, at this rate, from 10750 lb. of paper ? 36 DIVISION. 12. On the planet Neptune 60127 days make one year. A year on Neptune equals how many common years on the earth ? 13. For the year ending September 30, 1887, the exchanges at the Clearing House at New York amounted to $34872848786, and those of the 36 remaining important cities, $17253855702. What was the average of the exchanges per month at the New York Clearing House ? "What was the average per month of the 36 remaining Clearing Houses ? llf.. The Spanish Armada, sent in 1588, by Phillip II. of Spain for the intended conquest of England, comprised 132 ships with 34054 seamen and soldiers. What was the average number with each ship ? 15. In 1885 the total loans of the National Banks of Chicago and St. Louis were $55171842, while those of the National Banks of New York city were $236823598. How many times greater was the amount loaned by the banks of New York than by the banks of the other two cities named ? 16. The aggregate height above sea level of the 8 highest mountains of the earth, is 174173 feet. What is the average height in miles of 5280 feet each ? 17. During the year 1854, 50 banks of New York city made exchanges through the Clearing House to the amount of $5750455987 ; and in the year 1887, 64 banks made exchanges to the amount of $34872848786. Find the average clearings of each bank for each of the two years quoted. 18. The Kingdom of Belgium averages 480 inhabitants per square mile and the United States averages only 14. How many more times densely peopled is Belgium than the United States ? 19. The National Banks of St Louis in 1885 made loans to tne amount of $9182417, while those of Chicago made, during the same year, loans to the amount of $45989425. How many times greater were the loans of the banks of Chicago than those of St. Louis ? 20. The total cost of the railroads of the U. S. in 1880 was $5425772550. If the average cost per mile was $62522, how many miles had there been built ? 21. In 1880 the total railroad freight of the United States was 290897395 tons, of which 42003504 tons was grain and 89622899 tons was coal. How many times greater was the whole freight than that of coal alone? How many times greater than that of grain alone ? 22. The total expenditures of the railroads of the United States in the year 1880, were $541950795, and their net income was $119344596. How many times greater were the expenditures than the net income ? AVERAGE. 3? AVERAGE. 129. The Average of several numerical terms is the quotient obtained by dividing their sum by the number of terms taken. Thus, the average of 32, 40, 56, 16, 72, 24, 70, and 66, is 47, because 8 times 47 = 376, which is the sum of the numbers taken. 130. An average may be fractional ; as 33f is the average of 59, 43, 21, 10, and 35, because the sum of these five numbers equals 5 times 33-f. Remark. — The average numerical value of fractions, either common or decimal, may be obtained by dividing the sum of all such fractional expressions by the number of such expressions taken. Rule. — Divide the sum of the terms by the number of terms used. EXAMPLES FOR PRACTICE. 131. Find the average of the following groups of numbers and prove the results : 1. 20, 24, 52, and 88. | 8. 71, 46, 200, 11, 93, 51, and 17. 2. 32, 72, 56, 108, and 144. | £ 5, 28, 19, 72, 40, 85, 106, 29, and 54. 5. A man walked during six days of a week, 41, 47, 36, 54, 60, and 44 miles respectively. How many miles did he average per day? 6. A merchant sold during the 12 months of a year, goods in amounts as follows: $14216, $10008, $11051, $11097, $18241, $16900, $13754, $12291, $9267, $12935, $14901, and $20518. What were his average sales per month? 7. An errand boy earned on Monday 73^, Tuesday 91^, Wednesday 49^, Thurs- day 67^, Friday 81^, and Saturday 95^. What were his average earnings per day for the week? COMPLEMENT. 132. The Complement of a number is the difference between such number and a unit of the next higher order; thus, the complement of 6 is 4, because 4 is the difference between 6 and 10, or 1 ten, a unit of the next higher order than 6. Again, the complement of 83 is 17, because 17 is the difference between 83 and 100, or 1 hundred, a unit of the next higher order than 83. Again, the complement of 209 is 791, because their sum is equal to 1000. EXAMPLES FOR PRACTICE. 133. Find the complement of each of the following numbers, and prove and explain results : 1. 36. a. 71. * 8. 115. 6. 81. 1 7 - 1249. 1 9 - 28763 4- 704. 6. 258. 1 *• 1094. 1 10. 82041 38 FACTORS AND FACTORING. FACTORS AND FACTORING. 134. Factors are such numbers as multiplied together will produce a required number ; as 3 and 4, also 3, 2, and 2 are factors of 12 ; 3 and 15, also 5 and 9 are factors of 45 135. A Prime Number is one that cannot be resolved into two or more factors ; or, it is a number exactly divisible only by itself and unity; thus, 2, 3, 5, 7, 11, and 13, are prime numbers. 2 is the only even number that is prime. 136. A Composite Number is one that can be resolved into'factors. 137. A Prime Factor is a prime number used as a factor. To aid the pupil in determining the prime factors of a composite number we give the following Table of Prime Numbers from 1 to lOOO. 1 59 139 233 337 439 557 653 769 SS3 2 61 149 239 347 443 563 659 773 8S7 3 67 151 241 349 449 569 661 787 907 5 71 157 251 353 457 571 673 797 911 7 73 163 257 359 461 577 677 809 919 11 79 167 263 367 463 587 683 Sll 929 13 83 173 269 373 467 593 691 821 937 17 89 179 271 379 479 599 701 823 941 19 97 181 277 383 487 601 709 827 947 23 101 191 281 389 491 607 719 829 953 21) 103 193 283 397 499 613 727 839 967 31 107 197 29? 401 503 617 733 853 971 37 109 199 307 409 509 619 739 857 977 41 113 211 311 419 521 631 743 859 983 43 127 223 313 421 523 641 751 863 991 47 131 227 317 431 541 643 757 877 997 53 137 229 331 433 547 647 761 8S1 Remark.— The pupil can with little labor memoi !ze the prime numbers from fto 100 FACTORS AND FACTORING. 69 138. To Find the Prime Factors of a Composite Number. Example. — Find the prime factors of 4290. OPERATION. 5 ) 4290 Explanation.— Observe that the given number ends with a cipher, hence is exactly divisible by the prime number 5, by which divide it; next, observe that ' the quotient ends with an even number, and is, therefore, exactly divisible by 3 ) 429 2, so divide by 2; then observe that 3 will exactly divide the quotient 429; 11 ) 143 divide by it, obtaining 143, which divide by 11, obtaining 13, which divided by — — -" itself, gives a quotient of 1. All the divisors being prime numbers they together J__. constitute the prime factors of 4290. 1 Rule. — Divide by any prime number that is exactly contained in the dividend; divide the resulting quotient in the same, manner ; and con- tinue' this until the final quotient is 1- The prime divisors will be all the prime factors of the dividend. EXAMPLES FOR PRACTICE. 139. Kesolve 1. 27 into its prime factors. 2. 117 into its prime factors. 3. 165 into its prime factors. Jf. 93 into its prime factors. 5. 2376 into its prime factors. 6. 1 050 into its prime factors. 7. 144 into its prime factors. 8. 15025 into its prime factors. 9. 22464 into its prime factors. 10. 881790 into its prime factors. DIVISORS. 140. An Exact Divisor of a number is one which will divide it without a remainder, or which gives a whole number as a quotient ; thus, 5 is an exact divisor of 15, 3 of 12, and 2 of 4. 141. 1. Any number is divisible by itself and 1. 2. Any even number is divisible by 2. 3. Any number ending with 5 or is divisible by 5. h. Any number ending with is divisible by 10. 5. An even number is not an exact divisor of an odd number. 6. A composite number is an exact divisor of any number when all its factors are divisors of the same number. 142. A Common Divisor of two or more numbers is one that will exactly divide all the numbers considered; thus 3 is a common divisor of 6, 9, 12, and 15; also 7 is a common divisor of 14, 28, 35, and 49. 143. The Greatest Common Divisor of two or more numbers is the greatest number that is exactly contained in all of them, or that will divide each of them without a remainder. 144. Numbers having no common divisor, or factor, are said to be prime to each other. 40 DIVISORS. 145. To Find the Greatest Common Divisor. I. When the numbers are readily factored. Example. — Find the greatest common divisor of 10, 15, and 35. operation. Explanation. — By inspection find that the prime number 5 is an ^ \ 20 15 35 exact divisor of each of the numbers given; using it as a divisor, gives as quotients 2, 3, and 7; these being prime numbers have no 2 — 3 — 7 common divisor, therefore 5 is a common divisor of the numbers 10, 15, and 35, and as it is the greatest number that will exactly divide them it must be their greatest common divisor. Remark. — When it is determined by inspection that any composite number will exactly divide all the numbers of which we wish to obtain the greatest common divisor, such com- posite number may wisely be used as a divisor. II. When numbers are less readily factored. Example.— Find the greatest common divisor of 140, 210, 350, 420, and 630. operation. Explanation. — To prevent confusion, sepa ft x 1;in ~.,~ oe ~ 4nri nn ~ rate the numbers by a short dash. Observe that 2)140 — 210 — 350 — 420 — 630 m n ,. . A , * .. . ... J_ 2 will exactly divide each of the numbers, like g \ 170 2Q5 275 210 315 wise that 5 and 7 will exactly divide the successive - — — — — quotients ; therefore divide by 2, 5, and 7 ; then 7 ) 14 — 21 — 35 — 42 — 63 observe that the remaining quotients, 2, 3, 5, 6, and — 9 have no common divisor ; hence the divisors 2 — 3 — 5 — 6 — 9 2, 5, and 7 are all factors of the greatest common divisor, which is 70. Rule. — I. Write the numbers in a horizontal line, separating them by a dash. II. Divide by any number that will exactly divide all the numbers given, and so continue until the quotients have no common divisor. ILL Multiply together the divisors for the Greatest Common Divisor- Remark. — When factors cannot be readily determined by inspection the numbers may be resolved into their prime factors. The product of all the common factors of all the giveD numbers will be the greatest common divisor. EXAMPLES FOR PRACTICE. 146. Find the greatest common divisor of 252, 630, 1134, and 1456. 2150, 600, 3650. 1000, and 5001 302, 453, 755, 1057, and 1661. 126, 441, 567, 693, and 1071. 210, 350, 280, 840, and 1260. 200,325, 525, 350, and 675. 1. 22, 55, and 99. 7. 2. 24, 36, 60, and 96. 8. 3. 32, 48, 80, 112, and 144. 9. *. 54, 72, 90, 126, 180, and 216. 10. 6. 104, 156, 260, 364, and 572. 11. 6. 135, 450, 315, and 585. 12. MULTIPLES. 147. When no Common Factor can be Determined by Inspection. Example. — What is the greatest common divisor of 182 and 858. OPERATION. Explanation. — Draw two vertical lines and write the numbers on t iyoQ the "£ bt and left * Then divide 858 b J 183 > and write the quotient, 4, between the lines; then divide 182 by the remainder, 130, and write the quotient, 1, between the lines; next divide 130 by 52 and write the quotient, 2, as before; next divide 52 by 26 and write the quotient as before. As there is nothing now remaining the last divisor, 26, is the greatest common divisor of the given numbers. 182 130 52 52 858 728 130 104 26 Remarks. — 1. The greatest common divisor of several numbers which cannot be factored, may be obtained by taking any two of them and applying the above formula; then the divisor thus obtained and one of the remaining numbers, and so on until the last. If 1 be the final result they have no common divisor; if any number greater than 1, that number must be the greatest common divisor of all the given numbers. 2. The only practical use of the Greatest Common Divisor is in the reduction of a common fraction to its lowest terms; we thus find a number that will affect such reduction by a division of the terms but- once. Rule. — Divide the greater number by th"< less, the divisor by the remainder, and so continue until nothing remains, ■ The last divisor will be the Greatest Common Divisor. EXAMPLES FOR PRACTICE. 148. Find the greatest common divisor of 1. 316 and 664. J. 1377 and 1581, 2. 96 and 216. 6. 92 and 124. 8. 1226 and 2722. 7. 679 and 1869. A- 1649 and 5423. 8. 2047 and 3013, 9. 231 and 273. 10. 1179 and 1703. 11. 1888 and 1425. 12. 1900 and 1375. MULTIPLES. 149. A Multiple is a number exactly divisible by a given number; as, 12 is a multiple of 6. 150. A Common Multiple is a number exactly divisible by two or moie given numbers; as, 12 is a common multiple of 6, 3, and 2. 151. The Least Common Multiple of two or more numbers is the least number exactly divisible by each of them; as, 36 is the least common multiple of 18, 9, 6, 4, 3, and 12. 152. Principles. — 1. The product of two or more numbers, or any number of times their product, must be a common multiple of the numbers. 2. Two or more numbers may have any number of common multiples. 8. A multiple of a number must contain all the prime factors of that number. 4. The common multiple of several numbers must contain all the factors of all the numbers, 6. Tlie least common multiple of two or more numbers is the least number that will contain all the prime factors of the numbers given. 42 MULTIPLES. 153, To Find the Least Common Multiple of Two or More Numbers Example. — Find the least common multiple of 12, 16, 63, and 90. Explanation.— By factoring, find the prime factors of 12 which are 2, 2, and 3. " 16 " 2, 2, 2, and 2. " " - '* 63 " 3, 3, and 7. M " " 90 " 3, 3, 2, and 5. Since no number less than 90 can be divided by 90, it is evident that the least common multiple cannot be less than that number ; hence it must contain 3, 3, 2, and 5, the factors of 90 ; including with these another 2, gives all the factors of 12; two more 2's all the factors of 16 , and if 7 be included, all the factors of 63 are obtained ; hence the product of the factors 3, 3, 2, 5, 2, 2, 2, 2, and 7, or 5040 must be the least common multiple of the numbers 12, 16, 63, and 90. The method of determining the least common multiple by formula given below, will be found convenient. Example. — Find the least common multiple of the numbers 12, 16, 63, and 90. Write the numbers in a horizontal line to obviate confusion, and separate them by a dash. operation. . Explanation.— First divide by 2 ; 63 not being divisible by 2 2 ) 12 16 63 90 bring it to the lower line and divide again by 2; neither 63 nor 45 being divisible by 2, bring both to the lower, or quotient line. 2 ) 6 8 63 45 Next divide by 3; 4 not being divisible by 3, bring it to the quo- 3)3 4. 63 1 5 t * ent * me aQ d divide again by 3; the remaining numbers 4, 7, and 5 being prime to each other, are to be taken, together with the 3 ) 1 — 4 — 21 — 15 prime divisors 2, 2, 3, and 3, as factors of the least common raul- ~ A „ ~T tiple; their product is 5040, the same as before found. Remarks— 1. This principle has a practical value only in determining the least common denominator of common fractions, and is even then rarely used. 2. Where one of the numbers given is a factor of another, reject the smaller. 3. When it is observed that any composite number is exactly contained in all the numbers given, divide by such composite number rather than by its prime factors; the operation will thus be shortened. Rule. — t Write the number.* in a horizontal line, separating them by a dash. If Divide by any factor common to all the numbers, or by any prime factor of any two or more of them. In the same manner divide the quotients obtained, and continue until the quotients are prime to each other. III. The product of the divisors and prime remainders is the Least Common Multiple. 154. Greatest Common Divisor and Least Common Multiple Compared. I. The greatest common divisor is the product of all the prime factors common to all the numbers. II. The least common multiple is the product of all the prime factors of all the numbers. CANCELLATION. 43 EXAMPLES FOR PRACTICE. 155. Find the least common multiple of 1. 12, 20, and 32. 2. 25, 90, and 225. 8. 6, 16, 26, and 36. 4. 42, 210, 56, and 35. 5. 5, 30, 24, and 28. 6. 11, 32, 216, and 66. CANCELLATION, 7. 18, 80, 99, and 120. 8. 2, 3, 4, 5, 6, 7, and 8. 9. 21, 72, 24, and 30. 156. Cancellation is the omission of the same factor from terms sustaining to each other the relation of dividend and divisor. It is used for the purpose of saving labor in division, and is an application of the principle already given, that dividing both dividend and divisor by the same number will not alter the quotient; thus { may be read 2 divided 4; divide both terms by 2 and the result is 1 divided by 2, or -J. 2 X 27 Again, may be read 2 times 27, divided by 4 times 18 ; rejecting the 4 X 18 factor 2 from the 2 in the dividend and from the 4 of the divisor, also the factor % x %t z 3 9 from the 27 of the dividend and the 18 of the divisor, gives = , Mx«« 2X2 or f , or 3 divided by 4, as a final quotient. The correctness of this result is easily proved by factoring the dividend and 2x27 2x9x3 divisor, thus : = , then rejecting 2 and 9 from both terms, 4X18 2X2X9X2 or cancelling, obtain = f Ans. 2X2X0X2 157. We may supplement the former definition thus: The rejection of equiva- lents of factors from terms sustaining to each other the relation of dividend and divisor, is cancellation. Example. — What is the quotient of 3x2x28x5x7x51 divided by 6 X 11 X 4 X 7 X 35 X 17 ? operation. Explanation.— Cancel 6 from the divisor and 2 v v 2fA v £ v tf v SU^ 3x2 fTOm the dividend ; 4X7 from the divisor o x p x w x ? a /< x y* = ^ and 2g from the divideDd; the 35 from the divisor $ X 11 X 4 X t X W X It and 5 X 7 from the dividend; the 17 from the divisor and the 51 from the dividend, leaving 3 in the dividend, and 11 in the divisor; the quotient is T 8 T . Remark. — This principle can be put to frequent and valuable use in a great variety of business computations. Rule. — I. Write the divisor below the dividend zuith a line separating tikem. n. Cancel from the dividend and divisor all factors common to both ; then divide the product of the remaining factors of the dividend by the product of the remaining factors of the divisor. 44 CANCELLATION. EXAMPLES FOR PRACTICE. 158. 1. Determine by cancellation the quotient of5x9x2xl3x40x6 divided by 8 X 3 X 7 X 26. 2. Determine by cancellation the quotient of 64 X 25 X 3 X 15 divided by 45 X 12 X 4 X 11 X 36. In like manner, 8. Divide 210 X 9x78x5x23x10x36 by 13x144x40x3x27x5x400. 4. Divide 38 X 4 X 55 X 9 X 32 X 30 by 12 X 11 X 3 X 16 X 19 x 5. 5. Divide 51 X 7 X 9 X 27 X 40 X 54 by 63 X 17 X 9 x 200. 6. Divide 24 X 25 X 26 X 27 by 2 X 4 X 5 X 9 X 13. 7. Divide 2x3x4x5x6x7x8x9 by 23 X 45 X 67 X 89. 8. Divide the product of the numbers 98, 76, 54, and 32 by the product of the numbers 9, 8, 7, 6, 5, 4, 3, and 2. 9. Divide the product of 33, 4, 42, 9, 5, and 60 by the product of 7, 15, 12, and 11. 10. Divide the product of 416,216, and 810 by the product of 135, 52, 24, and 5. 11. How many bushels of potatoes at 60^ per bushel will pay for 450 lb. of sugar at 6^ per pound ? 12. A farmer traded 4 hogs weighing 325 lb. each, at 6$ per pound, for sugar at 5^ per pound. How many entire barrels of 312 lb. each should the farmer receive ? IS. I bought 18 car loads of apples of 216 barrels each, each barrel containing 3 bushels at 60^ per bushel, and paid for the same in woolen cloth. If each bale of cloth contained 600 yd. at 30 cents per yard, how many bales and how many odd yards did I deliver ? ££• How many yards of cloth at 15^ per yard should be given for 9 barrels of pork of 200 lb. each, at 6^ per pound ? 15. A hunter traded 6 dozen coon-skins at 40^ each, for powder at 750 per lb. How many 5 lb. cans of powder should he receive ? 16. How many pieces of cloth of 45 yd. each, should be received for 5 baskets of eggs, each basket containing 21 dozens at 18^ per dozen, if the cloth be valued at 8^ per yard ? 17. How many quarter sections of Kansas prairie land valued at $9 per acre, should be received for 80 cattle worth $78 per head ? Remark. — A section of land, in the United States, contains 640 acres. 18. How many years' work of 12 months of 26 days each, must be given for a farm of 112 acres at $78 per acre, if labor be worth $2 yer day? 19. A farmer exchanged 3 loads of oats, each load containing 27 sacks of 2 bushels each, worth 33^ per bushel, for flour at 6^ per pound. At 196 lb. per barrel, how many barrels should he have received ? 20. How many sections of Texas prairie land at $8 per acre should be given or an Ohio farm of 272 acres at $45 per acre P FRACTIONS. 45 FRACTIONS. 159. A Fraction is one or more of the equal parts of a unit. If a unit be divided into 3 equal parts, one of the parts is called one-third and is written £ , two of the parts are called two-thirds and are written f. 160. A Fractional Unit is one of the equal parts into which the number or thing is divided. |, \, -£, are fractional units. 161. The Numerator is the number above the line; it numerates, or num- bers the parts, and is a dividend. 162. The Denominator is the number below the line; it denominates, or names the value, or size, of the parts showing the number of parts into which the unit has been divided. It is a divisor. 163. The Terms of a fraction are the numerator and. denominator, taken together. 164. The Value of a fraction is the quotient of the numerator divided by the denominator. 165. Fractions are distinguished as Common Fractions and Decimal Frac- tions; and common fractions are either proper or improper. 166. A Common Fraction is one expressed by two numbers, one written above the other, with a line between. 167. A Proper Fraction is one whose value is less than 1, the numerator being less than the denominator. -J, f , \, f , -fa, ^ are proper fractions. 168. An Improper Fraction is one whose numerator is either equal to or greater than its denominator; its value is equal to or greater than 1. f, f , f , -|, V> W> H are improper fractions. 169. A Mixed Number is an entire or whole number and a fraction united. &J, 5f, 9|, 14f|, 107£J are mixed numbers. 170. A Complex Fraction is one having a fraction for its numerator or denominator, or for both of its terms. As a fraction indicates a division to be performed, a complex fraction indicates a division oi fractions to be performed, i is a complex fraction and indicates that £ is to be divided by f ; 5 * the expression is read £ -4- f ; — , and X are also complex fractions. I 8 Principles. — 1. Multiplying the numerator multiplies the fraction; dividing the numerator divides the fraction. 2. Multiplying the denominator divides the fraction; dividing the denominator multiplies the fraction. 3. Multiplying or dividing both terms of a fraction by the same number does not change the value of the fraction. 4:6 REDUCTION OF FRACTIONS. REDUCTION OF FRACTIONS. 171. To Reduce a Whole Number to a Fractional Form. Example. — Reduce 3 to a fraction the denominator of which is 7. Explanation.— The fractional unit having 7 for a denominator is \ ; and since 1 unit equals 7 sevenths, 3 units which are 3 times 1 unit must equal 3 times 7 sevenths, or 21 sev- enths ; therefore, 3 = V • Rule. — Multiply the whole number by the required denominator, and place the product over the denominator for a numerator. EXAMPLES FOR PRACTICE. 172. 1. Reduce 5 to a fraction the denominator of which will be 4. 2. Reduce 7 to a fraction the denominator of which will be 9. S. Reduce 4 to a fraction the denominator of which will be 13. 4. Reduce 3 to a fraction the denominator of which will be 8. 5. Reduce 8 to a fraction the denominator of which will be 12. 6. Reduce 15 to a fraction the denominator of which will be 10, 7. Reduce 14 to a fraction the denominator of which will be 5. 8. Reduce 27 to a fraction the denominator of which will be 11. * 9. Reduce 416 to a fraction the denominator of which will be 23. JO. Reduce 1125 to a fraction the denominator of which will be 57. 173. To Reduce a Mixed Number to an Improper Fraction. Example. — Reduce 5f to an improper fraction. Explanation. — Since 1 unit is equal to 3 thirds, 5 units, which are 5 times 1 unit, must be equal to 5 times 3 thirds, or 15 thirds; and 15 thirds plus 2 thirds equals 17 thirds; there- fore, 5J=y. Rule. — Multiply the whole number by the denominator of the fraction, to the product add the numerator, and place the sum over the denom- inator. EXAMPLES for practice. 174. Reduce 1. 3J to an improper fraction. 2. 7| to an improper fraction. 3. 10| to an improper fraction, 4. 434 to an improper fraction. 5. 16 T ^ to an improper fraction. 6. 78f to an improper fraction. 7. 26^ to an improper fraction. 8. 170^ to an improper fraction. 9. 1040^5- to an improper fraction. 10. 968 T \ to an improper fraction. 175. To Reduce an Improper Fraction to a Whole or Mixed Number. Example. — Reduce ^ 3 to a whole or mixed number. Explanation. — Since 4 fourths make 1 unit, 23 fourths will make as many units as 4 is contained times in 23, or 5 times with a remainder of 3, or three-fourths; therefore, 2 T 8 = 5f . KEDUCTION OF FRACTIONS. 47 Rule. — Divide the numerator by the denominator, place the remain- er, if any, over the denominator, and annex the fraction thus found to the entire -part of the quotient EXAMPLES FOR PRACTICE. 176. Keduce 1. V to a whole or mixed number. 2. *f to a whole or mixed number. 8. 2g 3 to a whole or mixed number. -4. ff to a whole or mixed number. 5. ^ to a whole or mixed number. 6. %ff to a whole or mixed number. 7. J |f 1 to awhcle or mixed number 8. |f to a whole or mixed number. 9. Yg 5 3 2 to a whole cr mixed number. 10. 1 ±-t-§ to a whole or mixed number. 177. To Reduce a Fraction to its Lowest Terms. Example. — Reduce -^ to its lowest terms. Explanation. — By applying the principles of factoring, change the form of the fraction Jg to — ~ ; then by cancellation reject the 2 and 3 from the numerator, and the same 2X3X3 factors from the denominator, leaving 1 for the new numerator and 3 for the new denom- inator; the resulting fraction is £. Or, observe that 6 is a factor of both the terms and that £ is the result of dividing both the terms by 6. Rules. — 1, Divide both terms of the fraction by their greatest common divisor. Or, '2. Divide both terms of the fraction by any common factor, and con- tinue the operation with the resulting fractions until they have no com- mon divisor. Remarks. — 1. When the terms of a fraction have no common factor, the fraction is in its simplest form, or its lowest terms. 2. If both terms of a fraction be divided by their greatest common divisor the fraction will be reduced to its lowest terms. This is the only use in practical arithmetic of the theory of the greatest common divisor. EXAMPLES FOR PRACTICE. 178. 1. Reduce -f-J- to its lowest terms. 2. Reduce -J^- to its lowest terms. 8. Reduce -§-§- to its lowest terms. 4» Reduce -^f to its lowest terms. 5. Reduce -^ to its lowest terms. 6. Reduce -£■£% to its lowest terms. 7. Reduce T 8 ^° 8 to its lowest terms. 8. Reduce -^fl t t° its lowest terms. 9. Reduce ££|# to its lowest terms. 10. Reduce -J|-g- to its lowest terms. 179. To Reduce a Fraction to Higher Terms. Example. — Reduce 4 to a fraction the denominator of which is 21. Explanation. — Since 7 is contained in 21 three times, the given fraction may be reduced to a fraction whose denominator is 21, by multiplying both of its terms by 3; multiplying 7x3 gives |f, the required result. This operation does not alter the value of the given fraction. Rule. — Divide the required denominator by the denominator of the given fraction and multiply the numerator by the quotient thus obtained; write the product over the required denominate- 48 REDUCTION" OF FRACTIONS. EXAMPLES FOE PRACTICE. 180. 1. Reduce f to a fraction the denominator of which is 15. 2. Reduce | to a fraction the denominator of which is 36. 3. Reduce -f- to a fraction the denominator of which is 42. 4. Reduce f to a fraction the denominator of which is 32. 5, Reduce -^ to a fraction the denominator of which is 88. 6, Reduce T \ to a fraction the denominator of which is 52. 7. Reduce -^ to a fraction the denominator of which is 115. 8, Reduce -£$ to a fraction the denominator of which is 128. 9, Reduce -^ to a fraction the denominator of which is 192. 10, Reduce \\ to a fraction the denominator of which is 147. 181. To Reduce Fractions to Equivalent Fractions Having a Common Dexxom inator. Example. — Reduce §, -J, -f, -^ to equivalent fractions having a common denominator. Explanation. — The product of the denominators 3, 2, 5, 7, = 210, and this number is exactly divisible by each of the several denominators; hence each of the given fractions may be reduced to an equivalent having 210 for a denominator; the desired result is then accom- plished, as 210 is a denominator common to all the given fractions; £ = ff$ , -£ = Mtt> I = Mf » ari A 4 120 ana T _ $f$. Rule. — Multiply together the denominators of the given fractions for a common denominator. Multiply each numerator by all the denomina- tors except its own and write the several results in turn over the common denominator. Remark. — Where one or more of the given denominators are factors of the others, th^ smaller may be rejected. EXAMPLES for practice. 182. Reduce to equivalent fractions having a common denominator: 1* i, h h and f. & 1, A, |, f , i, and A- & A> h i, 2> -ft, h and -f. 4. «, 5, i, #, i, ^ and |. A U, W> A, 38i, 23*, and 12. 6, 4f, 5fc 13i|, 6, and llf. A 2i, 7|-,llf, 23 T V, 1, |, and5. *• *i *> ffc 8, i|, 1, 14f , and J. 10. A, i> «> tVo, ih t, and 20. 183. To Reduce Fractions to Equivalent Fractions Having the Least Com- mon Denominator. The Least Common Denominator of two or more fractions is the least denom- inator to which they can all be reduced, and must be the least common multiple of the given denominators. Example. — Reduce \, f, f, -J, T 7 5 , and f to equivalent fractions having the least common denominator. operation. . Explanation.— Find the least 3, 2, ) 9 — 15 — 4 -J- = -f^. |- = -jSJL.. common multiple of the given de- ~~g jj 2 2 __ i2o i _ _ 9Q nominators for the least common ¥—180- 3 — 180- denominator, which is 180. Then 3X2X3X5^2 = 180. -fr = tW- i = Hi- by Art. 179, reduce each of the given fractions to a fraction whose denominator is 180. ADDITION OF FRACTIONS. 49 Rule. — I. Find the least common multiple of the given denominators. II. Divide this multiple by the denominator of each of the given frac- tions, and multiply its numerator by the quotient thus obtained. Remarks. — 1. The pupil should do as much of this work as possible by inspection. 2. Mixed numbers should be reduced to improper fractions before applying the rule. EXAMPLES FOR PRACTICE. 184. Reduce to equivalent fractions having the least common denominator: J. iii and |. *. h i> h h and f A i, A?> I, 2|, and ■&. * A> i> h h 2, 5i, and $. 5 > h h A> A> i> ¥> and 5 - 6. ft, i, 11, 7f, If, 9, and 1. 7. 23 J, 14f, % 5|, and |. 8. 17, 2f, 14}, 8|, 3^, and 5. & .*,*,£*,*,*,*,*, and -ft. ^. ^i^fi8,A,31,12f,and^. « tt,W,*,i±,i,H,W>W,H>ii. ^- «, if, V, ii 3, 2}, 4ft, 2, ft. ADDITION OF FRACTIONS. 185. To Add Fractions having a Common Denominator. Example. — Find the sum of {,, J, f , £, and £. OPERATION. V =2f Explanation. — As the given fractions have a com' mon denominator, their sum may be found by adding the numerators and placing the result 19, over the common denominator ; the simplest form of this sum is found by application of Art. 175. Rule. — Add the numerators and place the sum over the common denominator ; if the result be an improper fraction reduce it to a whole or mixed number EXAMPLES FOR PRACTICE. 186. Add t f, f, \, 4, \, and f ■*. if , A, A «, A and -&. * A, H» A, H, «, tt, and H. I A A, A A A H> and ff * A A A H, A tt, and f i *• ft, if > A A, «, if, and ft 7. «, A M> Ih A and A. A AAA«,«,H,4f,and||. ft A tt, A A, A, ib U> and If *>* «, A, **, «,«,«,«, and «, 187. To Add Fractions not having a Common Denominator. Example. — What is the sum of f, -f, and }. Explanation. — Since the given fractions are not of the same unit value, reduce them to a common denominator (Art. 181), and writing their equivalents below, add their numerators, and place the sum over the common denom- inator; reduce this result to an improper fraction. Rule. — Reduce the fractions to a common denominator, or if desired, to their least common denominator; add the resulting numerators < place the sum obtained over the common denominator and reduce the fraction 4 OPERATION. 50 ADDITION OF FRACTIONS. EXAMPLES FOB PRACTICE. 188. Add t i, h h h h h and f . ^ H> A, 4, i *, *, and fr A A, «> f > 14* 4, i, and |f. -*. 4> 4 > A, *, A> and f. & i, i, h h h A> and J. & tt, *, A, 4, i, i, «, and &. 7. +, f, AV> At «, i, 4, and i. & ih A, i A, H, f > fc and f, fc f, 1, h A. A. i and A- ■#. £, 4, «, i, *, 4, A, and .&. 189. To Add Mixed Numbers. Example. — Find the sum of 24, -J, 4, and •§-. OPERATION. 8$ + | + 4 + f = Explanation. — "Write the expressions in a hori zontal line; then change such of the expressions as are in mixed or entire form to fractional equivalents, and place them, together with the simple fractions, in a line below. Next find by inspection that 90 is the least common multiple of the denominators, or the least common denominator of the expressions; then apply Art. 179, add, and reduce results. For convenience the fractions may be written in a vertical line and only the fractional parts of the expressions reduced ; then adding the integers and the fractions separately, unite the results. OPERATION. 2 4 45 4 * 70 I 36 m Explanation. — Separate, mentally, or by a vertical line, the integers from the fractions. By inspection reduce the fractions to equivalents hav- ing the common denominator 90 ; now, keeping this in mind, write only the numerators 45, 70, and 36; the sum of these is 151, which placed over the common denominator in the form of a fraction, gives ^-, reducible to lf£; this added to the integers gives 7fJ, the sum as before found. Rules. — I* Reduce mixed numbers and integers to common fractional forms and then to common denominators. Add their numerators, place the result over the common denominator in the form of a fraction, and reduce to simplest form. Or, 2. Find the sum of the integers and the fractional expressions sep- arately, and add the results. EXAMPLES FOR PRACTICE. 190. Add t 4, *, f, 4, h 3, *, «, and 11. 6. *. f, 2*, I, 4, 6, 4, 1, A, and 4J. 7. 9. i, * , A. H, 5, H , h 74, and f. 8. I 2, 5f, |, 11H, h 14, 204, and 4. 9. 5- h 3, 6|, |, 19, 754, A, and ■&. 10. 3f, 10, 214, 42i, 84£, and 168^. 6**, &m, 50 U> *Hh and 86«. 59gf, 103ft, 554, 400, and 96H- 103A* H 9 ttib 297^, and 188J£ 334, 15, 124, 6±, 25, and 16$. Remark. — In invoices of cloth, &c, account of fractional parts is made only in quarters and merely the numerators are written; as, 5 8 = 5f, 3 1 = 3£, 13 3 = 13f. etc. SUBTRACTION" OP FRACTIONS. 52 Miscellaneous Examples in Addition of Fractions. EXAMPLES FOR MENTAL, PRACTICE. 191. What is the sum of 1. i, |, 4, and }.. 2. f, }, }, and £ »5. i, f, }, andf 4. }, |, }, and f. & f, f, fc and f . ft *,A>t>andA- ?• }* f , |, f, I* and ||. & i, h h h A> and 4. A }, I, h h A> and if. ^. 1, *, A, «, h i > ^d |f ^. },},},*,},},}>*,}, and}. EXAMPLES FOR WRITTEN PRACTICE, 192. Add i. 130f, 69}, 600-&, 2044*, and 46. 2. 80i, 2J, 5f, 17, 41*, 83^, and 14|. A 284-, 85^, 60.fr, 400, 20}, and 11. 4. 900, 47£, 3}, 4, 29}, 06^, and 4. 5. 16|, 334, 66}, 884, 100, and 116f. 6. 18}, 65}, 161}, 67}, 23}, and 75. 7. The six fields of a farm measure respectively, 10, 124, 19^, 26^^, 30^ and 2 J acres. How many acres in the farm ? £. Ten sheep weighed as follows : 90-^, 1104, 89}, 100, 106}, 101^, 96, 99 r 113f, and 198^- lb. respectively. What was their aggregate weight ? 9. A farmer sold 3G04 pounds of pork, 167} lb. of turkey, 24l£ lb. of chicken, 690-j-f lb. of butter, 475 lb. of lard, a cow's hide weighing 97} lb., 71|- lb. of tallow, and three quarters of beef weighing respectively, 161^, 187-£> and 190 lb. How many pounds in all had he to deliver ? 10. For 341f bushels of wheat I received $375J, For 597} bushels of barley I received $5004, For 1120| bushels of oats I received $619}, For 316J bushels of buckwheat I received $200f£, For 250 bushels of beans I received $525 T V, For 1386| bushels of potatoes I received $755|4, For 1050J bushels of apples I received $301^, For 6304 bushels of turnips I received $63^. How many bushels did I sell and what sum was received for all ? SUBTRACTION OF FRACTIONS. 193. To Subtract Fractions having a Common Denominator. Example. — Subtract } from -f. operation. Explanation. — Since the fractions have a common denominator, their difference may be found by taking the numerator 3 from the numerator * * f " 5, and placing the difference 2, over their common denominator 7. Rule. — Subtract the numerator of the subtrahend from that of the minuend, and place the difference over the common denominator. Remark. — A proper fraction may be subtracted from 1 by writing the difference between its numerator and denominator over the denominator. Results should always be reduced to- their lowest terms. Improper fractions may be treated the same as if proper. 52 SUBTRACTION" OF FRACTIONS. EXAMPLES FOR MENTAL PRACTICEo 194. What is the difference between X * and |. 2. |and|. 8. ff and^. 4- a 5. I 5. 1 and T V and |. and -J. 7. y and §. & f| and H- 9. f T and|f 10, ff and T 2 T . 11. 1 and ff . 1& 1 and -&. 13, i^ and ft. 14, V and i. 15, 1 and fJ. EXAMPLES FOR WRITTEN PRACTICE. 195. Subtract 1, ft from |f. 2. ^fromiff. 8, Jjfromft 4. iKfromfff. 5, || from 1. 0. -Hr from ff. 7. ||f from iff 8, ^from^. 9, \ 10, ■ 11. 12, - |f from |f . flfcfromH*. W from W- || from SV 3 . 15 u 15 16 . || from Vr 9 - . «* from ft* ■Hf from 1. ^5- from 3. 196. To Subtract Fractions not having a Common Denominator. Example. — From J take f. operation. Explanation. — As the denominators indicate the kind of parts, and o __ o _ only like things can be taken the one from the other, it follows that beforq * ~~~ * "" the subtraction can be performed, the fractions must be reduced to a. H — A == A- common denominator ; then, the difference between the resulting numera* tors, placed over the common denominator gives ^ as a result. Hale.— Reduce the given fractions to equivalent fractions having a> common denominator. Subtract the numerator of the subtrahend from the numerator of the minuend, and write the result over the common denominator. Remark. — Improper fractions may be treated in like manner. EXAMPLES FOR MENTAL PRACTICE. 197. What is the difference between 1. i and | ? ? f and 8, £ and f ? 198. Subtract 1. f from |. 2. f from £ . 8. •& from f . 4. |f from V- I and | ? ■Arandi? i and | ? 7. f and f| ? £ iand|? 9, Hand A? 10. IX 12. T V and | ? « and || ? Hand ft P EXAMPLES FOR WRITTEN PRACTICE. From 5. | take |. 5. H tak ef. 7. y take V- 8. ff take |. Find the difference between 9, || and f . 10. V and 1. 11. V and ^ 12. || and V- 15. u. 15. 16. ff- and |. A andV- | and f | and |. 199 To Subtract Mixed Numbers. Example.— From 1G£ take 11 J. OPERATION. 16| - 111 = ISA -HAS 16A = 15«, 4H Explanation. — Reduce the fractions to a common denominator Observing that the T 9 ^ of the subtrahend is greater than the ^ of the minuend, take 1 from the 16 of the minuend, reduce it to twelfths (^f ), and adding it to the •& obtain f § ; from this take the fV and the fractional remainder is found to be || Having taken 1 from the 16 in the minuend, there remains 15 from which to take the 11 of the subtrahend ; therefore the integral remainder is 4, and the entire result 4||. Remark.-Iu case the minuend is integral subtract 1 and reduce it to a fractional form of the required denominator. SUBTRACTION OF FRACTIONS. 53 Rule.— Write the subtrahend underneath the minuend. Reduce the fractional parts to like denominators. Subtract fractional and integral parts separately and unite the results. Remark. — In case the lower fraction be greater than the upper, take 1 from the upper whole number, reduce it and add to the upper fraction ; from this sum take the lower fraction. EXAMPLES FOR MENTAL PRACTICE. 200, What is the difference between 1, 6| and 2| ? 2. 5f and Z\ ? S. 12 1 and 3| ? 4. 17^andA? 3i and 5£ ? 8| and lift- ? 14| and 21i-|? 6| and 14ft * 9. 10. 11. 12. 3i and 12* ? 17 T V and 5-J- ? 21iandllf ? 9^and23f£? IS. 17^ and 22^ f U. 12f and3|£? 15. 113£andl4f ? 16. 215 1 Vand45|i EXAMPLES FOR WRITTEN PRACTICE. 201. From 1. 4J take l\%. U. 18^ take h\. S. 79£take49|. 4. 104ft " 84f Subtract 5. 9^ from 11. 6. 20from56£J. 7. 17ft from 80ft. 8. 41-J-J- from 5 Of . Find the difference between 9. 240f and 89^-. 10. 210|fand250. 11. 200 and lftf ^. um " mi- ls. 117ftand57f. U. 95ft and 183ft. 15. 1050H- and 2020§. 7(5. 2016f and 2503f§. EXAMPLES REQUIRING THE USE OF THE PRECEDING EXPLANATIONS. s. 4. 6. 6. 18. 14. IS. 16. 17. 18. 19. SO. 21. From the sum of f and -J take -§. j- and £ take \\. £ and f take |f. J and 2f take 4. ft and 9f take 3-§. 5 j and 4* take 9. 202. From the sum of Subtract the sum of 1. -I and I take 4. 7. ft and V from 14f. 8. 8± and 2f from 20. P. f and 5 from 11 J. 10. 18f and 15ft from 100ft. 11. 20ft and 1 5ft from 40ft. 12. 201-fr and 87f from 304£ Take the sum of f and f from the sum of 2| and |. Take the sum of 3£ and y from the sum of 4 and 8^-. Take the sum of 20 and 14ft from the sum of 18f and 19J. Take the sum of 28ft and 60J from the sum of 50J and 40£. Take the sum of 100| and 28f from the sum of 66^ and 69ft. Take the sum of 13ft and 4 an <* 100. 27. From the sum of 80f, 70£, 60$, and 1 take the difference between 1} and 101. 28. From the sum of ■£, f , f , f , -§-, -f, |, |, and -^ take the difference between V and 2. £0. Having $1302f in bank I drew checks for $2044, $189^, $640i, 182-J, $20, I30f, and $100. How much remained to my credit in the bank ? 80. A town owing $38246|, paid, in '85, $9304f ; in '86, $12000^- ; in '87, $4250£ ; and in '88 the remainder. How much was the payment of 1888 ? 31. If I pay $3500 for a house, $346| for repairs, $1126f for furniture, $400£ for carpets and curtains, and sell the entire property for $5000, how much will I lose ? MULTIPLICATION OF FRACTIONS. 203. To Multiply a Fraction by a Whole Number. Example. — Multiply £ by 4. OPERATION I. Explanation — Since the numerator is the divi- 4x4—. :Li-LZ — ig = ]4 = IX dend, the fraction may be multiplied by multiplying 8 the numerator 3 by the multiplier 4 ; the product is *5 8 , which reduced gives If, or l£. Or, since the operation II. denominator 8, is the divisor, the fraction may be 3 multiplied by dividing this divisor 8, by the multi- * * = o~T~Z = "* = ' P^ er ^» wnicn wil l gi ye tne reduced form, | = 1-J. This introduces the principle of cancellation into 0T 9 4 X 4 = •§■ = 1-J. fractional operations. For examples containing concrete numbers, reason as follows : Example. — If one pound of wool costs £ of a dollar, what will be the cost of 21 pounds ? Explanation.— Since one pound costs •§ of a dollar, 21 pounds which are 21 times 1 pound will cost 21 times f of a dollar, or 71 dollars. Rule. — Multiply the numerator, or divide the denominator, by the whole number. Remark. — To economize time and space divide the denominator or cancel when it can be done, as the numbers to be treated are thus put into simpler form. MULTIPLICATION OF FRACTION'S. EXAMPLES FOR MENTAL PRACTICE. 204. What is the product of 1. f multiplied by 3 ? 6. 2. \ multiplied by 2 ? 7. 8. | multiplied by 4 ? 8. 4. ft multiplied by 3 ? 9. 5. ft multiplied by 7 ? I 10 e | multiplied by 3 ? W multiplied by 5 ? | multiplied by 8 ? -ft multiplied by 15? ft multiplied by 6 ? 11. 12. IS. 14. is. 55 •J- multiplied by 12 ? •§■ multiplied by 5 ? | multiplied by 6 ? -^-multiplied by 30? ft multiplied by 32? 205. Multiply 1. ft by 85. 2. if by 8. 3. y by 12. EXAMPLES FOR WRITTEN PRACTICE. 5. 6. 7. 8. V by 16. V b y n - *-§- 4 by 40. V bv 28. 9. 10. 11. 12. «# by 10. A b y *i- H 9 by 20. 13. 14. 15. 16. M by 115. W by 49. A^byl05. H by 156. Remark. — It is sometimes desirable to reduce the whole number to fractional form by placing 1 for its denominator. 206. To Multiply a Whole Number by a Fraction. Example.— Multiply 6 by ft. Explanations.— 1st. If the multiplicand 6, be multiplied 3-J. by 7, the numerator of the fraction ^, the product 42 will be 12 times too large, because the multiplier was not 7, but one- twelfth of 7; hence this product must be divided by 12, which gives f| = 8& = 3£. 2d. Since 6 and T V are the factors of the product, and as it matters not which term is multiplied, reverse the order of the factors and proceed as in Art. 208. Operation L 6X^ = 41 = 3^ = Operation IL 2 Operation IIL 3d. Place 1 as a denominator for the multiplicand, then cancel and reduce. Rule. — Multiply the whole number by the numerator of the fraction, and divide the product by the denominator. Cancel when possible. EXAMPLES FOR MENTAL PRACTICE. 207. What is the product of 1. 5 multiplied by f . 2. 7 multiplied by f. 3. 6 multiplied by f . 4. 9 multiplied by ^-£. 5. 14 multiplied by f. 6. 22 multiplied by ft. 7. 40 multiplied by f. 8. 9 multiplied by §. 9. 4 multiplied by \$. 10. 7 multiplied by f|. 11. 12. 13. 14. 15. 4. 208. Multiply. 81 by f 56 by i. 61 by § . 105 by ^ EXAMPLES FOR WRITTEN PRACTICE. 5. 6. 7. 8. 19 by V*. 22 by iff 240 by «. 8byM- 9. 27 by £ J0. 210 by ft. IX 48 by if. 12. 91 by ||. 15 multiplied by £. 21 multiplied by fj. 12 multiplied by f. 18 multiplied by f. 42 multiplied by f c i& 71 by if ££. 203 by ^. 75. 415 by V- 70. 672 by |. Remark. — Entire or mixed number can be treated with facility by reducing them to fractional forms and cancelling when possible. 56 MULTIPLICATION OF FRACTIONS. 5 by f • 209. To Multiply a Fraction by a Fraction Example. — 1: (Abstract). Multiply - 2 - Operation. |X$ ( 3 times 4, f= •< or ( I of 3. Explanation. — The multiplier, f, is equal to 3 times | or A of 3. By application of 203, multiply § by 3, which gives |, which must be seven times too large, since the multiplier was not 3, but one-seventh of 3; hence, the correct result will be obtained by dividing the product, §. by 7, which gives ^ 6 r . Example. — 2. (Concrete). If a pound of tea costs £ of a dollar, what will § of a pound cost ? Explanation.— If a pound of tea costs f of a dollar, £ of a pound will cost & of £ of a dollar, or T 4 - of a dollar; if £ of a pound costs T 4 r of a dollar, £ which is 2 times £, must cost 2 times T 4 F of a dollar, or T \ of a dollar. Example. — 3. If a yard of cloth costs If (or ^) dollars, what will $ of a yd, cost? Explanation. — If 1 yd. costs */ dollars, f, of a yd. will cost | of ^ 5 , or f of a dollar; and if £ of a yd. costs f of a dollar, £ which is 4 times \, must cost 4 times f , or § = l£ dollars. Remarks — 1. Observe that the numerator of the product is the product of the numerators of the factors, and that the denominator of the product is the product of the denominators of the factors. 2. This will apply to the product of any fractions, proper or improper, or to the product of continued fractions. Rule. — I. Cancel all equivalent factors from the numerators and denominators. II. Multiply together the remaining numerators for the numerator of the product, and the remaining denominators for the denominator of the product. EXAMPLES FOR MENTAL PRACTICE (ABSTRACT). 210. Find the product of 1. \ and f 0. f and f. S. \ and £. £ I and |. 5. f and $. 6. | and ^. 7. ft and*. * *by*. Multiply 10. ft by*. 11. Hbyf ^. «by*. EXAMPLES FOR MENTAL PRACTICE (CONCRETE). 211. j?. What will be the cost of § of a pound of tea, if the cost of a pound be f of a dollar? 0. I bought £ of an acre of land and sold £ of my purchase. What part of an acre did I sell? S. What will be the cost of -J of J of a cord of wood at $6 per cord? 4. A girl having $ of a yard of ribbon used ■§- of what she had. What part ,of a yard had she left. 5. John was given f of a farm and James f as much. What part had James? 6. If i of a stock were lost by fire and the remainder sold at f of its cost, what part of the first cost was received. DIVISION OF FRACTIONS. 5? 7. Divide 20 into two parts, one of which shall be $ of the other, t 8. So divide $150 between two persons that one may have \ of the whole more than the other. 9. Tea costing f of a dollar per pound is sold for f of its cost. J<\>r what price per pound is the tea sold? 10. Coffee costing T 5 ¥ of a dollar per pound is sold for -| of its cost. What price is obtained f Oi the coffee f 11. What is ■§■ of § of a yard of cloth worth, if the entire yard costs § of a dollar? 12. If a pound of steak costs T 3 ¥ of a dollar, what will \ of \ of a pound cost? 13. After paying -J a dollar for a pound of nuts, I sold f of | of my purchase at the same rate. How much did I receive ? llf. From a gallon of oil f of f of a gallon leaked away. What part of a gallon was left? EXAMPLES FOR "WRITTEN PRACTICE. 212. Multiply together * M> A, «> and f i. 2. 3|, |, |, 21, and llf. 3. f, |, 8|, 11 T %, V, and 20. £ 12f, 16*, 21|, 60f, and ft. 5. 5f, 8, 11^, Hi, 25f, and 6. & 91, 19f, 29f , 39£, and 49. 7. 1, f , i, h h h f , and 10. 8. 206*, 187|, 40^, and 51*. P. 500, 186^-, 63f , 41, 19|, and 4 10. 23 T V, 16|, 8|, and 12|. « *>*,«>■¥> V, and if- •*& h h h h ft, and 60. 13. 13^, 21f , 12, 4, 24, and 3£. U. H, 5f,5, ^ H, and 1. 75. What will be the cost of ?■£ tons of hay at f of lo| dollars per ton? 16. Having bought f of a farm of 106J acres, I sold f of my purchase. How many acres do I sell? * 17. I bought a house for 12165-J and sold it for -f| of its cost. How many dollars did I lose? 18. If 7£ barrels of flour be consumed by a family in ten months, how many barrels would fifteen such families consume in 13-J- months ? 19. Having bought 2f tons of coal at f of 16f dollars per ton, I gave in pay- ment a twenty dollar bill. How much change should I have received ? 20. If 17J cords of wood are bought at % of 13f dollars per cord, and sold at \ of 9-J- dollars per cord, what is the gain or loss? DIVISION OF FRACTIONS. 213. To Divide a Fraction by a Whole Number. Example — 1. Divide \ by 2. operation. Explanation.— By the General Principles of Fractions, dividing the num- __ erator (dividend) divides the fraction; hence, divide the numerator 4 of the » "^ •■ fraction £ by 2, and the quotient is f . 58 DIVISION OF FRACTIONS. .Example. — 2. If a pound of tea costs f of a dollar what will \ of a pound cost ? operation. Explanation.— If one pound costs f of a dollar, \ of a pound will cost £ of |, or | of a dollar. Observe that in this operation the mul- |~2 = f Xl = | tiplier \ is the reciprocal of the divisor 2, or 1 has been written under the divisor as a denominator, the divisor inverted and the work per- formed as in multiplication. Example.— 3. Divide f by 7. Explanation. — By the General Principles of Fractions, if we multiply the denominator we divide the fraction. Therefore, § -r- 7 = $ z . Rule. — Divide the numerator or multiply the denominator by the whole number. Remarks. — 1. Divide the numerator if it be divisible, as the numbers will thus be made smaller. 2. If the dividend be a mixed number and the divisor an integer, it is not necessary to reduce the dividend to an improper fraction; divide the integral part of the dividend by the divisor, and if there be a remainder from such division, reduce it to a fraction of the same denomination as the fractional part of the dividend, add it to this fractional part and divide as before shown. Example. Divide 2815f by 8. Explanation. — (Short Method). — Write as in Short Division and divide; 8 is contained in 2815£, 351 times with a remainder of 7f not divided; reduce the 7 to 8 ) 2815§ thirds and to the result add the £ making y, which divide by 8, obtaining §f as the fractional part of the quotient; annex this to the integral part which gives 351|f. OPERATION. 351f| EXAMPLES FOR MENTAL PKACTICEo 214. What is the quotient of 1. •f divided by 3 ? 5. 2. |f divided by 4 ? 6. s. 4t divided by 9 T 7. 4. £f divided by 8 ? 8. 9. 10. 11. 12. \ divided by 4 ? \\ divided by 7? |4 divided by 18 ? f| divided by 22 ? if divided by 5 ? ff- divided by 7 ? iff- divided by 15 ? ff divided by 13 ? 18. If a pound of powder costs | of a dollar, what will £ of a pound cost ? 14. Having {% of a yard of cloth, I divided it into 7 equal pieces. How much cloth was there in each piece ? 15. If f-J- of a farm be grain land, and evenly divided into 3 fields, what pari of the farm will each field contain ? EXAMPLES FOR WRITTEN PRACTICE. 215. Divide 1. «by9. 2. 2^ by 17. $- mbylS. 4. 3Hby21. 5. ^by8. 9. 308 T V by 40. 13. 205fby6. 6. 211 by 5. 10. 1000^ by 41. U. 185 £ by 9. 7. 41f by 11. 11. 16fby5. 15. 112J by 8, 8. 2096§ by 21. 12. 108J by 25. 16. 321J by 6. DIVISION" OF FRACTIONS. 59 216. To Divide a Whole Number by a Fraction. Example. — Into how many pieces f of a yard each, may 5 yards of ribbon be cut? operation. Explanation.— Since 5 yd. equals x ¥ 5 yd. they may be cut into as 5 -f- f = many pieces, each containing f yd. as f is contained times in */ which is tf -5- § = 7|. U times. Remark. — Since the denominator names or tells the kinds, or value of the parts taken, when fractions are reduced to the same denomination, or to equivalent fractions having a common denominator, their numerators compare as whole numbers. We may consequently ignore the denominators. Rules. — 1. Multiply the denominator of the fraction by the whole number, and divide the result by the numerator. Or, 2. Reduce the whole number to a fraction of the same denomination as the divisor, and divide the numerator of the dividend by that of the divisor. EXAMPLES FOK MENTAL PRACTICE. 217. Divide 1. tfbyf 6. 14 by ^. 11. 31byy. 16. 24by|. *. Hbyf. 7. 20by^. 12. 50 by y. 17. 30by^. 3. 20 by if 8. 51 by «. IS. 21by£. IS. 12 by J. I 86 by*. 9. 39bytt- U. 60 by f . 19. 33 by \i. 5. 101 by i$. 10. 25byV- 15. 18byf. . 20. 15 by \. EXAMPLES FOR WRITTEN PRACTICE. 218. 1. If f of an acre of land sell for 45 dollars, what will an acre sell for at the same rate ? 2. A farm of 471 acres is divided into shares of 94^- acres each. How many shares are there ? 3. A church collection of 232 dollars was divided among poor families to each of which was given 5f dollars. How many families shared the bounty ? If.. When potatoes are worth £ of a dollar per bushel and apples f of a dollar per bushel, how many bushels of potatoes will pay for a load of apples measuring 30 bushels ? 5. A woman buys f of a cord of wood worth 6f dollars per cord and pays for it in work at f of a dollar per day. How many days must she work to make full payment ? 6. A dealer paid -f of 15f dollars for f of 14£ Cords of wood. What was the cost per cord ? 7. If -j^- of a farm of 67-j- acres be divided into 63 village lots, what part of an acre will each lot contain ? 8. 1760 bushels of wheat, 2100 bushels of barley, 2758 bushels of oats, and 696 bushels of beans were put into sacks ; those for the wheat contained each 2£ bushels, for the barley 2-J bushels, for the oats 2f bushels, and for the beans 1£ bushels. How many sacks in all were required ? 60 DIVISION OF FRACTIONS. 219. Tc Divide a Fraction by a Fraction. Example. — Divide •§ by 4* Operation. 4 times \ or $of4 1 + 4 = Explanation. — The divisor, 4, is equal to 4 times ^, or \ of 4. Applying the explanation of Art. 213, and dividing the dividend, f , by 4, gives ^ as a quotient ; but since the given divisor was -J- of 4, and the divisor used was 4, a number 7 times too great, / a , the quotient obtained, is 7 times too small; to correct this error multiply -£q by 7, obtaining f J as an answer. Ob- serve that 21 (the numerator of the quotient), is obtained by multiplying the numerator of the divid- end by the denominator of the iiviso*\ and that 20 (the denominator of the quotient), is obtained by multiplying the denominator of the dividend by the numerator of the divisor, or by ejecting a cross multiplication as shown by the connecting or tracing lines in the operation. =1*. Rules. — 1. Multiply the numerator of the dividend by the denomina- tor of the divisor for the numerator of the quotient, and multi-ply the denominator of the dividend by the numerator of the divisor for the denominator of the quotient; Or, 2. Invert the terms of the divisor and proceed as in multiplication of fractions. Remark. — Reduce mixed numbers to improper fractions before applying the rule. 220. Divide examples fok mental practice. byf. 8 fbyf- ibyf. *byf A byf a ic, 11. n. 1 by I fby«. I byf. IS. u. 15. 16. ilbyf *by«. «byf ttbyf. 221. Divide EXAMPLES FOK WRITTEN PRACTICE. 1- WbyH. *• Wbyff, *•■ T%by^. I 17. 18. «by«. Vby^- «byff. it by*. 5- «Jby«. 9. WbyW- #• 6- «byW- 1°- Vbyi i*. 7- ««by|f 11. AbyV- -^ 5 - *. ifbyW- n - ttbyVo'- *?• If a boy earns * of a dollar in a day, how long will it take him to earn $15f ? How many fields of 9f acres each can be made from a farm containing « by A- 125-J- acres ? 19. If a wheelman runs 93| miles per day, how long time will he require to run 1167i miles ? 20. If 12f acres produce 982£ bu. of corn, how many bu« will 15£ acres produce ? 21. If ten men cut 132£ cords of wood in six days, how many cords can eighteen men cut in twenty-one days ? 22. If a man bought 1150f bushels of wheat with § of his money, how many bushels could he have bought had all his money been invested ? 28. After traveling -fc of the distance between two cities, a pedestrian finds that there are 101-J miles still before him. How far apart are the cities ? COMPLEX FRACTIONS. 61 COMPLEX FRACTIONS. 222. A fraction is complex when either or both of its terms are fractional. 5 . Tiiu« - is a complex fraction and is read 5 -r- f ; it indicates that 5 is to be divided by f . - is read -f -f- 8 and indicates what is thus expressed. 1 is read 8 J ^ -f- -§- and indicates what is thus expressed. Remark. — The entire subject of complex fractions will, on account of its lack of practical value, be dismissed with the full illustration of one example of each of two forms. f Example 1. — What is the value of -? t i Operation: | =f -5-f =| X * = f* =?*£• Example 2. — What is the value of - — „ — ? * ** Operation: |^{= (I X V) -*- (i X «) = | X V X*X» = tt=4A. MISCELLANEOUS EXAMPLES IN FRACTIONS. 223. i. Erom the sum of -f and 5f , take the difference between 17£ and 21. #. How much will remain after the product of f , ^, 2, £$, and 3£ is taken from 10£. $. Divide into six equal parts the product of lly^ multiplied by 3}. 4. Eind the remainder after subtracting the product of 3f, -|, 7-f, 5, f, and 1, from the product of 3, f, f, 7, •£, 5, f , and 14. J. An estate is so divided among A, B, and C, that A gets f , B T 3 ^, and C the remainder, which was $4200. What was the amount of the estate ? 6. My bank deposit is $5605, which is 4|- times the amount in my purse. How much money have I in all ? 7. If 14 bu. of apples can be bought for $3|, how many bushels can be bought for $| ? 8. A woman having $1, gave f of it for coffee at 33£^ per pound. How many pounds did she buy ? 9. Having bought f of a ship, I sold f of my share for $12000. What was the value of the ship at that rate ? 10. If the ingredients are -fa sulphur, -fa saltpeter, and f charcoal, what is the number of pounds of each, in 2154^ T pounds of gunpowder ? 11. What must be the amount of an estate which, if divided into three parts, the first will be double the second, the second double the third, and the differ- ence between the second and the third be $7500 ? 12. Having paid $115 for a watch and chain, I discover that the cost of the chain was only ^ of the cost of the watch. What was the cost of each ? 62 "MISCELLANEOUS EXAMPLES IN FRACTIONS. IS. I gave two 20-dollar gold coins to a dealer, of whom I bought 2 cords of wood at 5f dollars per cord, and 3 \ tons of coal at 6| dollars per ton. How much change should I have received. lJf. A and B, working equally, can mow a meadow in 10 days of 9 hours per day. In how many days of 12 hours can A alone do the work ? 15. An estate valued at $120p00 was so distributed that A received T 2 -, B fa of the estate more than A, as much as A and B together less $6000, and tjro Q s charities the remainder in equal parts. How much did each charity receive r\jt^ 7 16. Brown owned fa of a stock of goods, f of which was destroyed by fire and ( M ' ^ of the remainder so damaged by water that it was sold at half its cost. If the uninjured goods when sold at cost brought $10800, what must have been the amount of Brown's loss ? 17. A grocer bought a cask of molasses containing 65| gal., from which he sold at one time i of it, at another -J- of it, at another 5 gal. less than J of what remained, and the remainder was sold with the cask for 20-J- dollars. If the value of the cask was one dollar, at what price per gallon was the last sale made? £\* 18. A painter worked 17-J- days, and after expending f °f his wages for board, had $15 left. How much did he earn per day ? *\!%l+ 19. A farmer having 650 bu. of wheat, kept jfor his own use 52f bu. less than^* J, sold to his neighbors for seed 45-J- bu. more than -J, and marketed the remain- der at 80^ per bushel. How much money was received from the market sales ? 20. Of a journey of 160 miles, a walker accomplished | of the distance the first day, \, less 15f miles, the second day, fa, plus 4-J miles, the third day, and finished his journey on the two following days by traveling fifteen hours each day. What must have been his average distance per hour for those two days ? 21. A mechanic worked 21-| days, and after paying his board with f of his earnings, had 66% dollars left. How much did he earn per day ? 22. So place a sum of money that -J- of it shall be in the first package, § in the second, fa in the third, and the remainder, which is $550, in the fourth package. What amount of money will be required ? 4 \^> >^ 23. If \ the trees of an orchard are apple, \ peach, \ pear, fa plum, and. the remaining 21 trees cherry, how many trees in all ? 2Jf.. John's weight is £■ as much as mine, and Ben's is f of John's. What is my weight if John is 15 pounds heavier than Ben ? • 25. If 12 boys earn $54 in a week, how much will 15 boys earn in the same time at the same rate ? 26. A, B, and C rented a pasture for $37. A put in 3 cows for 4 months, B, 5 for 6 months, and C, 8 for 4 months. How much ought each to pay ? 27. Henry, when asked his age, replied, " If 7-J- years be added to 13£ years, the sum will represent -J of my age." How old was he ? 28. Silas, Harvey, and Eobert have together $2210. Silas has 2-J- times as £* much as Harvey, who has £ as much as Eobert. How many dollars has each ? ^ 29. Theodore's age is 7£ years, and Herbert's 9f years ; three times the sum \ x i of their ages is 8 years more than the age of their mother, who is 5£ years younger than their father. What is the united age of the parents ? MISCELLANEOUS EXAMPLES IN" FRACTION'S. 63 30. A farmer sold two cows for $75, receiving for one only -J as much as for the other. What was the price of each ? 31. After selling 45 turkeys, a dealer had % of his stock remaining. How ma ay had he at first ? 32. If 8 horses consume 4£ bushels of oats in 3-j- days, how many bushels will 12 horses consume in the same time ? t jH 33. A and B can do a piece of work in 10 days, which A alone can do in .8 days. In what time can B alone do the work ? $4, John and Calvin have agreed to build a wall for $86. If Calvin can work only $ as fast as John, how shall the money be divided ? 85, A flagstaff stands -if of its length above and 7-J ft. below the surface of the ground. What is the length of the staff ? ik 36. What is the length of a pole that stands -f in the mud, f in the water, and 25^ feet above the water ? 37. A colt and cow cost $124. If the colt cost $4 more than three times tha cost of the cow, what was the cost of each ? 38. What is the hour when the time past noon equals \ of the time to mid- night ? 39. A tree 84 ft. high was so broken in a storm that the part standing was £ the length of the part broken. How many feet were standing ? 2 £, 40. A farmer has -| of his sheep in one pasture, f in another, and the remain- der of his flock, 72 sheep*, in the third pasture. How many sheep had he ? 41. For a horse and carriage I paid $540. What was the cost of each, if the cost of the carriage was \\ times the cost of the horse ? >\\>-^-v^/ Jft Calvin is 8^- years old, Leo 6 J years less than three times as old as Cal- vin, and John's age is 3 year,s more than the sum of the ages of Calvin and Leo. What is John's age ? % fc'JV J+3. Peter can do a piece of work in 12 days and Charles in 15 days. How many days will be required for its completion, if both join in the work ? \> p 44- If A can do a piece of work in 21 days, B in 18 days, and C in 15 days, in how many days can the three working together perform the work ? Remark. — In the above and similar examples, reason in general as follows: If A can build a wall in 4 days, he can build £ of it in 1 day; and if B can build the same wall in 5 days, he can build A of it in 1 day. Since in 1 day A can build i of the wall, and B \ of it, the two can, if they work together, build in 1 day the sum of £ and i or fa of it; and since they can together do -^ in 1 day, it will take them as many days to do the whole work, or §f, as -fa is contained times in §#, or 2|. 45 John and his father have joint work, which they can do working together " in 25 days. If it require 60 days for John working alone to complete the work/ how many days will it require for the father to complete it ? U -y M-7 46 A man and boy can in 16 days complete a job that can be done by the man / ^ alone in 21 days. How long would it take the boy alone to complete the work? ^1 f 47. Smith said to Brown, "f of my money is equal to -§- of yours, and the ti*-iO difference between your money and mine is $30." How much money had each? j) 48. Izaak Walton having lost | of his trolling line, added 65 ft., when he ^Ifr found it was just £ of its original length. What was its length at first ? fl&l 64 MISCELLANEOUS EXAMPLES IN FRACTIONS. 49. A cistern sprung a leak by which f of its contents ran out, but during the same time f as much ran in. What part of the cistern was filled ? '$|i-> / 50. A dog pursuing a rabbit which has 32 rods the start, runs 11 rods while A the rabbit runs but 9. How far must the dog run before he can overtake the rabbit ? St A cistern has two faucets, by the larger of which it can be emptied in 24 x\ minutes, and by the smaller in 36 minutes. If both be opened at once, what j length of time will be required to empty the cistern ? 52. Ben and John bought a cocoanut for 8 cents, of which Ben paid 5$ and John 3^. Henry offered 8$ for one-third of the cocoanut, which offer was accepted, each taking and eating one-third of it. How should Ben and John divide the 8$ received from Henry ? 53. There are 108 bu. of corn in two bins, and in one of the bins there are 12 bushels less than one-half as many bushels as in the other. How many bushels in each ? 54. At what time between one and two o'clock will the hour and minute hands of a clock be together ? 55. At what time between 6 and 7 ? 56. At what time between 9 and 10 ? 57. At what time between 10 and 11 ? 58. Nick bought a basket of oranges at the rate of 3 for 2 cents, and gained 50^ by selling them at the rate of 2 for 3 cents. How many oranges did he buy? 59. If you buy 60 lemons at the rate of 6 for 10 cents, and twice as many more , ^ at the rate of 5 for 8 cents, and sell the entire lot at the rate of 3 for 4 cents, ^ will you gain or lose, and how much ? 60. So divide $15,000 among A, B, C, and D, that their portions shall be to each other as 1, 2, 3, and 4. What is the portion of each. 61. I wish to line the carpet of a room that is 7-J yd. long and of yd. wide, with duck f of a yd. wide, How many yards of duck will be required if it shrink ■£$ in- length and ¥ V in width ? 62. A and B are engaged to perform a certain work for $35 T 5 ^g-. It is sup- posed that A does i more work than B, and they are to be paid proportionally. How much should each receive ? 63. A tank has an inlet by which it can be filled in 10 hours, and an outlet by which when filled it can be emptied in 6 hours. If both inlet and outlet be opened when the tank is full, in what time will it be emptied ? 64. A cistern has two faucets, by the larger of which its contents may be emptied in 12 minutes and by the smaller in 15 minutes; the cistern being full, the smaller faucet is left open for 6 minutes, after which both are opened. How long before the cistern will be emptied ? 65. A man being asked his age replied, " My mother was born in 1800 and my father in 1801 ; the sum of their ages at the time of my birth was two and one* third times tri} age in 1846." How old was the man in 1880 ? 66. Three men dig a well for $36. If during the time A and B working together could do \ of the work, B and together could do f of the work, and A and together could do f of the work, how should the money be divided? MISCELLANEOUS EXAMPLES IN" FRACTIONS. 65 67. Brown and Smith have joint work for 16 days. In any given time Brown does only -f as much work as Smith. How many days would each working alone require to complete the work ? If they work together, how should the $45 paid for the work be divided ? 68. Coe, Hall, Tell, and Lee have a contract to dig a ditch which Coe can dig in 35 days, Hall in 45 days, Tell in 50 days, and Lee in 60 days. How long will it take all together to do the work ? If $100 be paid, for the work and all join till it is completed, how much should each get ? 69. A and B have joint work for 21 days, but B can in a day do only f as much as A; Lifter B has worked alone for 3 days and A for 5 days, they unite and complete the work. How many days will they require ? If $75 be paid for the work, what part of it should each receive ? 70. An estate was left to A, B, and C, so that A's part was £ of the whole increased by a sum equal to -^ of C's part; B's was £ of the whole increased by a sum equal to -J of C's part: and to C was given the remainder, which was $700 less than B's share. What was the value of the estate and of each one's share ? 71. Hill, Mann, and Benton have joint work for 36 days, for which they are to receive $200. If Hill can do only -| as much as Mann and Benton does twice as much as Mann, in how many days could each working alone complete the work? 72* How long would it take Hill and Mann ? 73. How long would it take Hill and Benton ? 7 J/.. How long would it take Mann and Benton ? 75. If all work together until the job is completed, how should the money be divided ? 76. A, B, C, and D, liaving joint work for 30 days, A begins and works alone for 2 days, when he is joined by B; after the two have worked together for 3 days, they are joined by C; the three work together for 4 days, when D joins them, and all working together complete the work. If A can do but -J as much as D, B f as much as A, and Q ^ as much as B, how long would each alone require to do the entire work. 77. How long would it take A and^B ; A and C ; A and D ? 78. How long would it take A, B,$ind C ; A, C, and D ; B, C, and D ? 79. How long after D began did it take for all to do it ? 80. If $300 was paid for the work and the men worked according to conditions given in Example 76, how should the money be divided ? 66 DECIMALS. DECIMALS. 224. A Decimal Fraction or a Decimal is a fraction having for its denominator ten or some power of ten ; as 10, 100, 1000, 10000. It expresses one or more of the decimal divisions of a unit. 225. Decimals may be expressed in the same form as common fractions; that is, with the denominator written. Practically, however, this is never done. Remark. — The two points of difference between common and decimal fractions are, 1. The denominator of a common fraction is always written, while that of a decimal is only indicated. 2. The denominator of a common fraction may be any number, while that of a decimal must be 10 or some power of 10. 226. The Decimal Point ( . ) is a period, and is used to limit the value of of a decimal expression, and to determine the denominator; in this latter rela- tion it takes the place of the unit 1 of the denominator when fully w r ritten; as, in the decimal expression .3, read 3 tenths, the decimal point considered as 1 and placed before a cipher, represents the order of its units, and shows that the indicated denominator is 10. Remark — When the decimal point is used to separate the integral from the fractional part in mixed decimals, or dollars and cents in decimal currency, it is called a separatrix. 227. Decimals are either pure or mixed. 228. A Pure Decimal corresponds to a proper fraction, the value being less than the unit 1 ; as, .3, .17, .206, .5191. A Mixed Decimal corresponds to an improper fraction, the value being greater than the unit 1 ; as, 17.4, 5.192, 32.3017. 229. The Yalue of a Decimal is computed from the decimal point, and the orders have the same scale as integers. A removal of the decimal point one place to the right, multiplies the expression by ten ; removing it two places, by 100 ; three places, by 1000, and so on. A removal of the decimal point one place to the left, divides the expression by 10 ; two places, by 100 ; three places, by 1000, and so on. 230. From the above it will be observed that if a cipher be placed between the numerical expression of the decimal and the point, the expression being thereby removed one place further from the point, will be divided by 10. But as the value of the decimal expression is computed from the point to the right, it follows that one or more ciphers, placed after the decimal, will not alter its value. -f% is expressed decimally .3 ; a cipher annexed to the decimal gives .30 = -^\ two ciphers annexed gives .300 = -jVoV By this it will be observed that the expressions, though unlike in form are of equal value. Each of the expressions .5, .50, .500, .5000, .50000, .500000, is equal to -J. NUMERATION OF DECIMALS. 67 231. Principles. — 1. Decimals increase in value from right to left, and decrease from left to right, in a tenfold ratio. '2. A decimal should contain as many places as there ivould be ciphers in its denominator if written, the decimal point representing the unit 1 of such denominator. S. The value of any decimal figure depends upon its place from the decimal 'point. 4- Prefixing a cipher to a decimal decreases its value the same as dividing it uy ten. 5, Annexing one or more ciphers to a decimal does not alter its value. NUMERATION OF DECIMALS. 232. For Notation and Numeration of Decimals we begin with the decimal point as a simple separatrix ; in the integral expression the first place to the left is units (corresponding to the decimal point in the decimal expression), the 2d tens, the 3d hundreds, etc., while from the separatrix to the right we have in order (the point standing for units), tenths, hundredths, thousandths, etc. 233. The Order of a Decimal may be found by numerating either from right to left or from left to right, only let it be remembered that the decimal point stands in the position of the unit 1 of the decimal denominator. The order of a decimal may usually be determined by inspection, if the fact to bo drawn from the following illustration be observed. If .35 be numerated from the right as in integers, the point is in hundreds place; hence, read 35 hundredths; in .1463 the point is in ten-thousands' place, read 1463 ten-thous- andths; in .014065 the point is in millions place, and is read 14065 millionths 234. The value of a decimal may be determined by the same numeration as that employed in integers. The relation of orders in a mixed decimal is clearly shown by the following Table. 00 § 03 •a p CO* co r5 go' ■d — 1 CO CO* p .2 1 a* o o 00 •9 p 03 co P O CO* Q 3 CO a 2 CO n co* •a CO a 03 CO P O "Z2 CO i od M a o P 1 o - "P "P £ "c3 a cu ■p S 3 p o 8 13 2 s ■s — p p d to p 5 C •p PI P O p CO p O w s -P P P to p B ad EH CO* P CD g •p a P 32 1 5 0) ■p p p B p CD 6 "P P P W H § w H H H H s S H W H Eh W S Eh ffl ■d A £ ,a ,a* ,P . -*-i • . s ,d •B .a' ,p" £1 35 53 <*! ■p "p CO CO "P m S3 C5 3d i> «D lO "^ CO CM l-H tH c* -* io «o £- 00 1 1 1 1 1 X 1 1 1 • 1 1 1 1 1 1 1 1 The Integral Part. The Fractional Part. The above number is read 111 million 111 thousand 111, and 11 million 111 thousand 111 hundred-millionths. Remark. —It is better, in reading mixed decimals, to connect the integral and fractional parts by and; as, 2.5 read 2 and 5 tenths; 17.016, 17 and 16 thousandths. 68 NOTATION OF DECIMALS. Rule. — L Numerate from the decimal point, to determine the denom- inator. II. Read the decimal as a whole number, and give to it the denom- ination of the right-hand figure. EXAMPLES FOR PRACTICE. 235. L Bead .297, .1471, .20442, .56007. 2. Bead .105, .6931, .214698, .4003755. 3. Bead. 19005, .3050408, .690004003. > 4. Bead .2, .20, .200, .2000, .20000, .200000. 5. Bead 18.3, 29.75, 460.215, 80.03465. e. Bead 270.01, 5960.030506, 8205.506007 t 7. Bead 10002.200001, 38960041.100086341. 8. Bead 27000.000027, 8100081.81000081. 9. Bead 1001001. 1000100001, 9003009.009009. 10. Bead 39864125.86954769, 919101.01919 • 11. Bead 50000000.00000050, 1000.1000. 12. Bead 123456.10203040506, 801.00801. IS. Bead 46000046.004600046. H Bead 37538651.0352615093. IS. Bead 45316255.83715632650. * 236. Bead the following decimals: t .206. 9. 5320.008641. 17. 2000.00020002. 2. 1.423. 10. 6000.58302. # 18. 564636.002616. 3. 7.005. 11. 9001.00901. 19. 202020.20202. 4. 19.11035 • 12. 340006.583. 20. 21212121.51210021. 5. 170.2093 IS. 75075.07507. 21. 30560078.0124861. 6. 1050.0501. Ik 560.00020201. 22. 503760.2000463. 7. 300.003. 15. 53200.56931. 23. 37564.03060507. S. 1000.0001. 16. 214600.086005. 24. 10023580021. 1809010724 NOTATION OF DECIMALS. 237. The doubt which often arises in the mind of the pupil as to how a decimal should be written, may be entirely dispelled by keeping in mind the following facts: 1st. That they are fractions. 2nd. That both terms should be written or indicated. 3rd. That the denominator of any decimal (if written) would be 1, with as many ciphers to the right as the decimal contains places. 4th. When the numerator (or decimal) does not contain as many places as the denominator (if written) would contain ciphers, prefix ciphers to make the number of places equal. NOTATION OF DECIMALS. 69 Example. — Write as a decimal three-tenths. Explanation. — Observe that in writing three-tenths as a common fraction, the mental operation is as follows: after writing 3, the numerator, you ask yourself 3 what? the answei is, 3 tenths; then the ten is written below as a denominator, thus obtaining -fa. Now reason in the same way regarding the decimal, and after writing 3, the numerator, ask yourself 3 what? and answer, 3 tenths; and indicate it by placing before the three, a decimal point, which rep resents the 1 of the decimal denominator; notice that the 3 occ'upies one place corresponding to the one cipher in the denominator. Again, express decimally 416 thousandths. Explanation. — "Write the 416 and ask what? answer, thousandths, which is determined by numerating from the right; units, tenths, hundredths, and (the point answering to the figure 1 of the denominator) thousandths; then place the point. Remark. — By extending and developing this method of writing decimals, the pupil can in a few minutes master the entire matter, so that he can write any decimal as readily and with as great certainty as if it were a whole number. Rule. — I. Write the decimal the same as a whole number, prefixing ciphers when necessary, to give to each figure its true local value. IL Place the decimal point before the left-hand figure of the decimal. EXAMPLES FOR PRACTICE. 238. Express by figures the following decimals: 1. Twenty-six thousandths. 2. Twenty-seven hundredths 3. Sixteen ten-thousandths. £. Pour hundredths. 5. Twenty-two hundred-thousandths. 6. Five and seven tenths. 7. Eighty-three and five hundred four ten-thousandths. 8. Seven hundred ten and two hundred forty-three hundred thousandths. 9. Five hundred and five hundredths. 10. Forty-five and forty-six thousandths. 11. One thousand one and one hundred ten-thousandths. 12. One thousand eight hundred ninety and ninety thousandths. IS. Eight hundred fifty and five hundredths. llf. Ten hundred and ten hundredths. ADDITIONAL EXERCISES. 239. Write as decimals 1. Eleven and one hundred seven thousandths. 2. Fifteen and fourteen ten-thousandths. 3. Seven hundred twenty-six millionths. 4. Eleven hundred six and twelve ten-thousandths. 5. Sixteen hundred and sixteen hundredths. 6. Ten million and ten millionths. 7. Three hundred and sixty-five hundredths. 70 REDUCTION OF DECIMALS. 8. 9. 10. 11. 12. IS. U. 15. 16. 17. 18. 19. 20. 21. 22. 25. Twenty-five thousand four hundred and eleven hundredths. Twenty-one and fifteen thousand fifteen ten-millionths. Eighteen thousand eighteen ten- billion ths. Five hundred thousandths. Five hundred-thousandths. Nine hundred millionths. Nine hundred-million ths. Fifty-four million, fifty-four thousand, fifty-four and fifty-four million fifty thousand fifty-four ten-billionths. One hundred three thousand five hundred eighty-seven thousandths. Sixty-four thousand sixty-four hundredths. Two million six hundred four thousand two hundred-thousandths. Nine billion nineteen million twenty-nine thousand thirty-nine millionths. Seventy-seven tenths. Eighty-seven thousand one hundredths. Four hundred seventy-nine million twenty seven thousand four and ninety-nine thousand four ten-billionths. Seventy trillion and seven trillionths. Eleven hundred and eleven ten- thousandths. Three thousand one billionths. One thousand three millionths. One hundred-thousand eleven ten-millionths. Six hundred five hundred-millionths. Eighteen hundred ninety and eighteen hundred ninety hundred-billionths. 240. Write as decimals the following: i- -th- 000 "• lOGOO'd* *• looo* °' 10 00000"* 7 5 04 '• '1 ooooo- 8 2 146900 3 "• 10 000 • Q 56 9738 05 * 7 * 10 • 10. W^Mftftr- 77 3325481 J.J.. 100 12. ■* 0# To oo • •*•*• loo 0* Jfi L5JJJLJL5. * u * 10 o • 16. 17. 18. 19. 3 5 7 1000000000' 2 19_16.j0 4_8.01 10000 00000* *>• 1 WAW 1 - REDUCTION OF DECIMALS. 241. To Reduce Decimals to a Common Denominator. Example 1.— Reduce .021, .64, .03705, .5, .17272538, to equivalent decimals- having the least common denominator. operation. _ ' ;. Explanation. — Since the decimal having the greatest number of decimal .02100000 pi aces jg hundred-millionths its denominator is the least common denominator . 64000000 of the given expressions; this highest decimal contains 8 places, and by adding 5 .03705000 places, or ciphers, to the first, 6 to the second, 3 to the third, and 7 to the fourth, .50000000 all are reduced to 8 places, or to hundred-millionths, which is the least common •17272538 denominator of the given expressions. REDUCTION - OF DECIMALS. 71 Rule. By annexing ciphers make the number of decimal places equal. Remarks— 1. Decimals, like other fractions, can be neither added nor subtracted until reduced to a common denominator; but the scale in decimals being in the uniform ratio of ten, it is only- necessary to write decimals for addition or subtraction so that the decimal points are in the same vertical line ; the columns will then be of the same orders of units ; in other words the decimals will be practically reduced to a common denominator. 2 The denominator of that expression containing the highest number of places is the least common denominator of the decimals; therefore the least common denominator may in all cases be determined by inspection, and decimals reduced to their least common denominator by simply supplying decimal ciphers until all have the same number of places. 3. In practice, however, this is never done, being rendered unnecessary by observing to write decimals so that the points stand under each other. Example 2. — Reduce .7, .23, .18765, and .175 to a common denominator. Operation. Explanation. — As shown in the preceding operation, the effect of reducing • 7 decimals to a common denominator by annexing ciphers, is to cause the decimal .23 points to fall in the same vertical column. Since annexing ciphers to decimals . 18765 does not alter their value, omit the ciphers and write the decimals so that the .175 points are in the same vertical column. Rule. — Write the expressions so that the decimal points will stand in the same vertical line. Remark. — This Rule applies equally to Pure and to Mixed Decimals. EXAMPLES FOK PRACTICE. 242. 1. Reduce .26, .423, 7.05, .56931 to their least common denominator. 2. Reduce 21.18, .20463, 4636.02 to their least common denominator 8. Reduce 56 hundredths, 75 millionths, 3 tenths, and 41 thousandths to theu least common denominator. Jf.. Reduce 2.36, .0005, .1, .62053, and 15.2 to their least common denom- inator. 6. Reduce 19.0043, 3.87, 38.7 and .387 to their least common denominator 243. To Reduce Decimals to Common Fractions. Ifc has already been demonstrated 1st. That Decimals are fractions. 2d. That their denominators are merely indicated, and that the denominator may be expressed by writing 1, with as many ciphers at its right as the decimal contains places. Example. — Reduce .17 to a common fraction. Operation. Explanation. — Since the decimal contains two places, its indicated denom- .17 — -^L, inator must be 100. Rule. — Omit the decimal point and write for a denominator 1 with as many ciphers as the decimal contains places. Remark. — Mixed Decimals may be reduced in a similar manner. 72 CIRCULATING DECIMALS. EXAMPLES FOR PRACTICE. 244. — Reduce to fractions in their lowest terms 1. .3 5. . 4625 9. .42504 IS. .114608 2. .63 6. .2244 10. .28828 U. .315264 S. .105 7. .1878 11. .08004 IS. .2000534 4. .372 8. .1900 12. .24042 16. .983004752 Reduce to an ordinary mixed number 17. 5.16 £0. 3005.1258 28. 1234500.0012345 18. 13.205 21. 1600.0016 24. 6540000.0002607 19. 117.602 22. 1000000.00000001 25. 188900.00001888 245. To Reduce a Common Fraction to a Decimal. Example. — Reduce -f to an equivalent decimal. First Operation. Explanation.— From the definition of decimals, observe that the ■| X § = yV = • 6. denominator must be 10 or some power of 10, and that f may be reduced to a fraction the denominator of which is 10 by multiplying both its terms by 2. Tc change this fraction to an equivalent decimal, omit the denominator, and place a decimal point before the numerator. Explanation.— Place a decimal point and cipher after the num- erator 3. This does not alter its value, though in form it becomes 3.0 = thirty tenths; and since this numerator is a dividend and the divisor is 5, divide 3.0 (thirty-tenths) by 5, and obtain .6 (six-tenths), as A result, an equivalent in decimal form as required. Second Operation. 5 )3.0 .6 Rule. — Place a decimal point and ciphers at the right of the numerator; divide by the denominator, and from the right of the quotient point off for decimals as many places as there have been ciphers annexed. EXAMPLES FOR PRACTICE. 246. Reduce to equivalent decimals * TV «■ «• 11. }f. 16. H- «• «• «. a- 7. Afr **• *■ 17. «. **. m- s. U- 8. tf. is- *h- 18. rfv ®$' AV 4. A- »• T**- u. U- 19. $h. **• «• A H. 10. rtn I 5 . irfinr' 20. ft &5. dhr- CIRCULATING DECIMALS. 247. Certain common fractions, as -J, J , f , and -^ cannot be reduced to an equivalent decimal, because the denominator (divisor) is not an exact divisor of any power of 10. Such expressions cannot be reduced to exact decimal forms and are termed repeating, or circulating decimals ; if used in the decimal form they are followed by the sign + to indicate inexactness. The repeated part is called a repetend ; as, .3333+ is called the repetend 3 ; .171717+ is called the repetend 17 ; .206206+ is called the repetend 206. ADDITION OF DECIMALS. 73 248. To Express the Exact Value of a Repetend. The exact value of any repetend is a common fraction, the numerator of which is the repetend and the denominator as many 9's as the repetend contains places; thus .333+ = f. .171717+ = Jf .206206+ = fff. Kuie. — Take the repetend for the numerator of a common fraction, and for its denominator write as many 9's as the repetend has orders of units. EXAMPLES FOR PRACTICE. 249. Express the exact value of the following repetends 1. .2222+ I 3. .232323+ I 5. .613613613 + 2. .7777+ I 4. .105105105+ 6. .201120112011 + / // / Remarks. — 1. Limiting marks are sometimes used ; as, 234234 ; they are, however, of no importance. 2. In business, final results are carried to three places, the fourth being rejected if less than one-half, but if one-half or more than one-half, 1 is added. 3. In interest rates or other multipliers, it is generally safest to use a common fractional equivalent ADDITION OF DECIMALS. 250. Example.— Add .7, 2.43, .865, 11.5, 113.2075, and 200.00165. Operation. • • Explanation. — Since by the decimal system numbers increase in ***° value from right to left in ten-fold ratio, and the decimal point .obo separates integral from fractional orders, observe to write decimals 11QOAWK so ^ a * the points fall in the same vertical line, as units of the same %n^niL order will thus fall in the same column; the result of the addition is d then obtained in the same manner as in simple numbers. 328.70415 Remake. — As before shown, the decimals added could be reduced to a common denomina- tor, but this being practically accomplished by the order in which they are written, the actual reduction by supplying ciphers is entirely unnecessary. Rule* — Write the decimals so that the points mill fall in the same vertical line. Add as in whole numbers, and place the point in the sum directly below the points in the numbers added. EXAMPLES FOR PRACTICE. 251. 1. Add 4, .37, 2.46, 19.301, and 103.21. 2. Add 3.04, 25.001, .67, .2146, and 819.256. 3. Add 30.1257, 605.2146, 1000.864532, and 16.25694. 4. Add 896.111, 9530.216753, 1111.230004, and 1100.960005. 6. Add 265.4203, 1129.000111, 8.005, .0060008, and 1200.12000014. 6. Add 8046.0012, 250.0000001, 311.00555, and 81.0081001. 7. Add 11000.4604, 7652.0000004, 5000.500005, and 365.50053004. 74 ADDITION OF DECIMALS. 8. Add 14.0000864, .0096, 250.4, 700.0007, 1000.00000001, 563.3001468, 20.2001, 10000.001001 and 896.707075. 9. Find the sum of seventeen and forty-six ten-thousandths, eighty-three and one thousand four millionths, five hundred two and seventy-five hundred- thousandths, three thousand eleven and three hundred eleven thousandths, one million six hundred six and one ten-thousandths. 10. Add fifty-six thousand twelve and one thousand twenty millionths, six and ninety-seven million five billionths, one thousand five hundred seventy-nine and twenty-six thousand twenty-one hundred-thousandths. 11. Add one and one thousandths, ten and eleven hundred-thousandths, one hundred ten and nine millionths, eleven hundred eleven and ninety-nine billionths, one thousand eight hundred ninety and ninety-seven hundred- billionths, seven millions and seven hundred-thousandths. 12. A farmer having 315.625 acres of land, added at different times by pur- chase, 505.85 acres, 115.75 acres, 469.2 acres and 220.9 acres. How many acres had he in all ? 13. What is the sum of 16.5 acres, 21.125 acres, 86.0625 acres 111.45 acres, 216.05 acres, 37£ acres, 426-jL acres, 89$ acres, and 13^ acres ? lJk What is the number of bushels in ten bins of 93.625 bu., 111.025 bu., 306.005 bu., 81-H- bu., 193f bu., 200f bu., 300.0625 bu., 125£ bu., 250* bu., and 136 T ^- bu. respectively ? 15. I bought ten bales of cloth as follows : 321, 41|£, 39^, 461, 29 J, 38^, 431, 41-^, 42-J-J-, and 40.625 yd. respectively. How many yards in my purchase? Remark. — In invoices of goods only fourths are usually counted, and these are written as follows : 3 1 = 3£, 15 3 = 15£, 12' 2 = 12|. By the omission of the denominator time is saved. In additions, find the sum of the small figures first as so many fourths, reduce to units and carry as in other addition. 16. Add 21 1 , 54 2 , 17 1 , 30 3 , 46 1 , 61 2 , 80 1 , 39 3 , and 24 a . 17. Add 121 1 , 97 3 , 46 2 , 111 3 , 43, 71 2 , 86 3 , 50 1 , 103 3 , 72 1 , 71 3 , and 50. 252. To Add Repetends. Remark. — In addition of repetends, bear in mind their equivalents ; thus, in adding .6-f to .3 -f- remember that the value of the first is f , and of the second f ; their sum is f, or 1. In all examples in addition of repetends, before beginning the operation, continue the repetends sc that all have the same number of places, and in the right-hand column add each 9 as 10 253. Add .333333 + .171717+ ,306306 + ,811357+ EXAMPLES FOR PRACTICE. ,11111 + ,77777+ , x 46207' .55555 + ,33333 + .2222 + .3333+ .8787+ .0101 + .3467+ 4. .561561561 + .202202202 + .333333333 + /504300542' .306306306 + 5. Find the sum of the following expressions : 105.333 +, 86.1919+, 53.1031C3+, 17.66+, 204.77+, 29.11+, 815.201201+ and 73.11081108 +. 6. Add .66 +, 1.2121 +, 50.55 +, 89.99 +, 2046.33 +, 38.22 +, 106. 77+* 1593.44 +, 11.230230 +, 528.60916091 + and 1102.300300 +. . MULTIPLICATION OF DECIMALS. 75 SUBTRACTION OF DECIMALS. 254. Example. — Subtract .17 from .56. Operation. «« Explanation. — For reasons heretofore explained, place the subtrahend below the minuend, so that the decimal points shall fall in the same vertical line. Subtract as in simple numbers, and place the point in the remainder below the points in the terms above. Remarks. — 1. In case the number of decimal places of the subtrahend be greater than those of the minuend, consider decimal ciphers as annexed to the minuend, and subtract as before. 2. Mixed decimals may be treated in the same manner. Rule. — Write the terms in decimal order and subtract as in integers, placing the point in the remainder below the points in the other terms. EXAMPLES FOR PRACTICE. 255. Subtract 1. .573 from .985. 2. .13823 from .668. S. .8627 from 1.549. 4. 1.232 from 6.7584. 5. .754352 from 2.3. 6. 46.2906 from 100.52. 7. 3491. 5 from 4246. 1005. 8. .0001 from 10000.1. 9. 24.6852 from 25. 10. 286.111 from 500. 000825. 11. .09 from. 900. 12. 250. 98754 from 386. 245. MULTIPLICATION OF DECIMALS. 256. Example.— Multiply .17 by .5. First Operation. Explanation. -Write .17 as ^ and .5 as T %, and apply the rule for multiplication of common fractions. Multiplying * 17 = ToV (Com. frac'l form.) thege f rac ti nal equivalents, obtain yf^ as the common frac- • 5 = -f-Q " " " tional expression of the product; by Art. 245, this maybe iVo X A- = To to = -085. written .085 as the decimal expression of the product, or the product required. Remark. — Observe that the denominator of the product is, as in other fractions, the product of the denominators ; also that the denominator of the multiplicand contains two ciphers, or two places, and that of the multiplier one cipher, or one place; these taken together contain three ciphers, or three places, the same number of ciphers, or places, as are found in the product. Then by applying the theories of decimals already explained, the expression is changed to decimal form. Second Operation. Explanation. — Write and multiply the expressions as in .17 whole numbers. Since the numerator is 17 hundredths and 5 the denominator 5 tenths, the product must be 85 thou- sandths. Hence, change the product 85 to 85 thousandths by Qgg prefixing a cipher and a decimal point, thus: .085. Rule. — Multiply as hi whole numbers; then, from the right of the product, point off for decimals a number of places equal to the number in both factors, prefixing ciphers if needed to obtain the required number. 76 DIVISION" OF DECIMALS. EXAMPLES FOR PRACTICE. 257. Multiply 1. .78 by. 7. 2. .123 by .16. 8. 1.45 by .875. 4. 26.08 by 1.53. 25000 by .000025. 8.76 by .100. 716.0025 by 10.1005 7000 by .007. . 5. 2085.109 by 11.256. 9. 6. 1000.87 by 4621. 5. 10. 7. 10000 by. 0001. 11. 8. 300 by. 03. 12. IS. 1000000 by .0000001. | Z& 3400000.0081 by 81.000034. 15. What will be the cost of 187.0625 acres of land at $108.08 per acre ? 16. I sold 14.4 bales of cloth of 61.625 yd. each, at 8.60J- per yd. How much did I receive ? 17. What will be the cost of 5. 75 cases of paper, the average weight of which is 403.625 pounds, at $.40375 per pound ? 18. From 10.85 acres of wheat a farmer harvested 31.875 bushels per acre, and sold his crop at $.9725 per bushel. How much was received for the crop ? Remarks.— 1. The contraction of multiplication of decimals by restricting the number of places to appear in the product, is not deemed of sufficient practical importance to justify presentation. 2. As has been previously explained, decimal expressions, either pure or mixed, may be multiplied by 10 or by any power of 10, by removing the point as many places to the right as the multiplier contains ciphers. In such cases annex ciphers to the multiplicand if there is not already a sufficient number of decimal places. DIVISION OF DECIMALS. 258. Example.— Divide .085 by .17. First Operation. .085=-^. "i o o. o ~^~ Ttnr = Second Operation. .17). 085 (.5 85 00 Explanation.— Since .085 = yf^ and .17 = T W, proceed as in Division of Common Fractions ; that is, invert the terms of the divisor and multiply. Observe now, that cancelling 17 and 100 from opposite terms of the fractional multiplicand and multiplier there is left only the factor 5 for the numerator and the factor 10 for ■the denominator of the quotient, or the fraction T 5 ^ = .5. Explanation. — Divide as in whole numbers. The divid- end has 3 decimal places ; the divisor has 2 decimal places ; the dividend having one more decimal place than the divisor, point off one place from the right of the quotient. Remark,— It will be seen from the first operation that the number of decimal places of the divisor cancels, or offsets, the same number in the dividend. If the number of places in the terms be equal, it is obvious that the quotient will be a whole number. Rule. — I. Wheiv needed, annex ciphers to the dividend to mahe its places equal in number to those of the divisor. II. Divide as in integers, and, from the right of the quotient, point ofj for decimals as many places as the number of places in the dividend exceeds those in the divisor. DIVISION" OF DECIMALS. 77 259. Decimals may be readily divided if, in connection with the above explanations, attention be given to the following Suggestions. — 1. Do not commence the division until the number of decimal places in the dividend is at least equal to the number of decimal places in the divisor. Supply any deficiency in the dividend by annexing ciphers. 2. If the divisor and dividend have the same number of decimal places, the quotient obtained, to the limit of the dividend as given, will be a whole number. 3. If the number of decimal places in the dividend be greater than the number of decimal places in the divisor, point off from the right of the quotient for decimals, a number of places equal to such excess, prefixing ciphers to the quotient if necessary. 4. If after division there be a remainder, ciphers may be annexed to it and the division continued to exactness, or to the discovery of a repetend, or to the two or three places ordinarily demanded in business computations. All such added ciphers should be considered as parts of the dividend. Remarks. — 1. Inasmuch as the main difficulty experienced by pupils with decimals is found in division, and as that difficulty increases when the principles of decimals are applied to practice in percentage, it is advised that most thorough and repeated drill in division of decimals be given to all grades of pupils in all stages of class work. 2. From pleasant experience in teaching this subject, it is suggested that ten or more examples be grouped as a single exercise, and so arranged that the numerical quotient be the same for all. The pupil thus relieved from effort to determine this feature of the quotient, finds the requirement narrowed down to the placing of the decimal point, and soon fully ters all difficulty. EXAMPLES FOR PRACTICE. 260. Divide i. .625 by 2.5. 2. 15.25 by .05. 8. 1100 by 4.4. I 9.5 by 19. 5. 9.5 by 190. 6. .95 by .019. 7. 36.5 by .073. 8. 250 by .0626. -5-1 = ? -^.1 = ? -*- .01 = ? 10 -T- .1 = ? L0 -5- .01 = ? + l«f -J-.l = ? -*- .01 = ? 4- .001 = t -r-10= ? 9. 1750 by. 875. 17. 17.5 by 17500. 10. 3.6 by 1800. 18. .44 by .00011. 11. .005 by 200. 19. 10000 by. 00001. 12. 27.465 by .00015. 20. .001 by 1000. 18. 1396. 875* by 250. 21. 1.6 by .064. 14. 131300 by. 025. 22. 6400 by .0000016. 15. 62.5 by 1.25. 28. .0081 by .054. 16. .00875 by 125. 2^. 1860 by. 000031. (26.) (27.) U10=? .22 - -11 = ? 1 -*- 100 = ? 2.2 - 7- .011 = ? .1 -T- 1000 = ? .220 -^ 11000 = ? .001 -T- 100 = ? .022 -*• 110 BB ? .0001 ~r .1 = ? .00022 -T- 11000 = ? 100 -T- .00001 = ? 2.2 - t- .000011 = ? 1000 ~ .01 = ? 2200 -f- .00011 = ? .00001 ~- 1000 = ? .022 ~ 110000 = ? 10 ~ 100000 = ? .0000022 ~ 1100000 = ? 10000 -J- .0001 = ? 220000 -T- .000022 = ? Vb GREATEST COMMON DIVISOR OF DECIMALS. {28.) 1.6 -*■ 2.5 = ? 160 -T- .25 = ? .0016 -*- 250 = ? 16 ^ .00025 - ? 160 -*- 250000 = ? 16000 •* .000025 = ? .0016 ~ .00025 rr ? .000016 ~ 2500000 = ? 1600 ~ .00025 = ? 1600000 H-. 00000025=? Find the sum of the quotients. (31.) 440 -T- 1.1 = ? .00044 ^ 1100 = ? 4400 -f- .11 = ? 440 4- .0011 = ? .0044 ^ 110000 = ? 44000000 -T- 1100000=? 4400000 -5- .000011 = ? 44000 -*- .011 a= ? .00000044 ~- 110000 as ? 4400 -4- .00011 = ? . Find the sum of the Quotients. (29.) 6.25 -5- 2.5 = ? 62.5 -*- .025 = ? 6250 -f- .0025 = ? .0625 -*- 250 = ? .00625 ~ .00025 s= ? 6.25 -f- 25000 = ? .0000625 -4- .00025 = ? 625000 ■¥ .0000025 = ? .0000625 -f- 2500000 as ? 625 4- .0000025= ? Find the sum of the quotients. (32.) .375 -*- 1250 = ? 375 ~ .0125 = ? .0375 -I- 12.5 = ? 37.5 ■*■ .000125 = ? 37500 -*- .00125 = ? 3.75 -*- 1250000 = ? .00375 -4- 125000 = ? .0000375 4- .125 = ? 3750000 ~ .000125 = ? .000375 -4- 12500 = ? Find the sum of the quotients. (30.) 2.o -4- 625 = ? .025 -4- 62.5 = ? .0025 -f- 6250 = ? .00025 -4- .62b = ? .000025 -4- .000625 = ? .0000025 -4- 62500 = ? 2500 -4- .0625 = ? 2500000 -4- .0000625 = ? .00025 -4- 6250 = ? .000025 -4- 6250000 = ? Find the sum of the quotients. (33.) 2.25 ■* .015 = ? 225 -4- 1500 = ? .0225 + 150 = ? .00225 -4- .015 = ? 2250 -4- .0015 = ? 22500 -4- 15000000 = ? .000225 -4- .00015 = ? .0000225 ~ 1500000 = ? 2.25 ~- .000015 = ? 22500000 -f- .00015 = ? Find the sum of the quotients. Remark. — Any decimal may be divided by 1 with any number of ciphers annexed, as 10, 100, 1000, 10000, by removing the decimal point as many places to the left as the divisor contains ciphers. THE GREATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE OF FRACTIONS, COMMON AND DECIMAL. 261. All explanations given in finding either the Greatest Common Divisor or Least Common Multiple of integers apply equally to fractions, common or decimal. 262. To Find the Greatest Common Divisor of a set of Common Fractions. Example. — What is the Greatest Common Divisor of ■£, f, and f ? Explanation. — First reduce the given fractions to a common denominator and obtain as a result, Operation. 5 ) 15 — 20 — 25 3— 4 — ", ft. f a> II I then arrange the numerators of the resulting fractions in a horizontal line. Pro- ceeding as by previous explanations find the Greatest Common Divisor of the numbers to be 5 ; but since these numbers are numerators of fractions whose common denominator is 30, and 30 is the Least Common Multiple of this common denominator, the Greatest Common Divisor of the given fractions must be 5 -*- 30, or -fo = J. Notice that the numerator of the resulting | is the Greatest Com- mon Divisor of the numerators, and that the denominator 6 is the Least Common Multiple of the denominators, of the gi^en fractions. LEAST COMMON" MULTIPLE OF DECIMALS. 79 Rule. — Write a fraction the numerator of ivhich shall be the Greatest Common Divisor of the numerators of the given fractions, and the denom- inator the Least Common Multiple of the denomi?iators of the given fractions. 263. To Find the Least Common Multiple of a set of Common Fractions. Example. —Find the Least Common Multiple of f , {-, and y 3 ^. Operation. Explanation. — Reduce the given fractions to a common de nominator as before, and obtain §£, £g, £{j ; the Least Common 2 ) 24: oO 18 Multiple of the numerators is found to be 1800 ; but the terms o \ i o ax _ q were not 24, 50, and 18, but f$, §#, and |§, and 60 is the Greatest ' '_ Common Divisor of 60, the common denominator, therefore the ^ 25 3 Least Common Multiple is not 1800, but i§"|*, or Y; therefore 30 is the Least Common Multiple of the given fractions. Observe that the numerator of - 3 T ° is the Least Common Multiple of the numerators, and the denom- inator of the - 3 T ° is the Greatest Common Divisor of the denominators, of the given fractions. Rule. — Write a fraction the numerator of which shall be the Least Common Multi-pie of the numerators of the given fractions, and the denominator of which shall be the Greatest Common Divisor of the denom- inators of the given fractions. 264. To Find the Greatest Common Divisor of a set of Decimal Fractions. Example. — Find the Greatest Common Divisor of .5, .25, and .375. Operation. k \ ErtA o»A n„ r Explanation. — Reduce the expressions to equivalents hav- ' ' ing a common denominator, obtaining .500, .250, .375. For 5 1 100 — ~0 7*5 convenience omit the decimal points, find the Greatest Com- mon Divisor of the numerators, and obtain 125. Since 500, 5 \ 20 10 15 ^^' anc * 375 were not whole numbers, but .500, .250, and .375, the result is not 125, but .125. 4_ 2— 3 Rule. — Reduce the expressions to the same decimal order, then write the Greatest Common Divisor of the expressions as a whole number, and make it of the decimal order common to all. 265. To Find the Least Common Multiple of a set of Decimal Fractions. Example. — Find the Least Common Multiple of A, .12, and .416. Operation. 4 ) 40u — 720 — 416 Explanation. — Reduce to decimals of the same order, ob- * ~" taining .400, . 720, and .416 ; find the Least Common Multiple of j ^^ ^"^ 1^'* the numerators, which is 93600. But since the expressions were not integers, but thousandths, the result is 93600 thou sandths, or 93.600 = 93.6, the Least Common Multiple. 5 ) 25 — 45 — 26 5— 9—26 80 MISCELLANEOUS EXAMPLES IN" DECIMALS. Rule. — Treat the expressions as integers and obtain their Least Com- mon Multiple; then make it of the same decimal order as that one of the given decimals which has the greatest number of decimal places. Remark.— These illustrations, when presented before a class, may properly be combined. MISCELLANEOUS EXAMPLES IN DECIMALS. 266. 1. Add 51.01, 8.1006, 67.00102/14.5, 1750.5072003, 100.0010041. 2. Add 137 thousandths, 41 hundredths, 13 millionths, 5011 ten-millionths, 608 ten-thousandths, 200600 hundred-millionths. S. Beduce J-jj- to a decimal fraction. J/.. Reduce .015025 to a common fraction. 5. Divide 38.462 by 10000. 6. From 3006.01 take 889.01546. 7. From 540.123 take the sum of 81.625, 126.0972, 45.001, and 100.1002. 8. If 60-J bushels of corn cost $26,785, how much will 17.65 bushels cost. 9. Take the sum of nineteen millionths, five and two ten-thousandths, and sixty, from one hundred six and three tenths. 10. Multiply the sum of sixty-five and one hundred seven millionths, by the product of nine hundred millionths and one hundred twenty and seventeen hundredths. 11. From one billion take two billionths. 12. From six and fifty-hundredths take five and sixty hundredths. IS. Divide nine hundred sixteen and two thousand four millionths by sixteen ten thousandths. llf.. Find the cost of 11.6 bales of cloth, each bale containing 61f yards, at $1.54 per yard. 15. What is the cost of six barrels of sugar, weighing 301, 314, 297, 309, 313, and 315 pounds respectively, at 6-J0 per pound ? 16. How many tons of phosphate, at $34.88 per ton, will pay for 296.48 bushels of beans, at $1.25 per bushel ? 17. A contractor received $354.06 for excavating a cellar, at 35^ per cubic yard. How many yards of earth were removed ? 18. If a wheelman travels 10.3 hours per day, how many days will be required for him to travel 558.0025 miles, at the rate of 7.88 miles per hour ? 19. A teacher's salary is $1500 per annum. If he pays $650.50 for board, $119.25 for books, $31.85 for other literature, $63.40 for charity, $209.25 for clotbes, $109.90 for traveling expenses, and $41.27 for incidental expenses, how much of his salary has he left ? 20. I sold a lumberman 381.25 pounds of butter at $.2875 per pound, 2468.375 pounds of cheese at $.114 per pound, and 2356.5 pounds of dressed beef at 8.07-J per pound, and received pay in lumber at $23 .125 per thousand feet. How many thousand feet of lumber should I have received ? UNITED STATES MOtfEY. 81 UNITED STATES MONEY. 267. United States Money is the legal currency of the United States, adopted in 1786 and changed by various Acts of Congress since that date ; it is sometimes called Federal Money. 268. Money is the measure of value. 269. Legal Tender is the term applied to such money as may be legally offered in the payment of debts. 270. Bullion is pure gold or silver in bars, or ingots, and " bullion value " is the value of such metal, which varies from coin value only by the charges for coinage made by the mint. 271. Coin is the standard money of the mints, its value being established by law. 272. Currency is coin, treasury notes, bank-bills, or any substitute for money, in circulation as a medium of trade. • 273. A Decimal Currency is a currency whose denominations increase and decrease by the decimal scale. United States money is a decimal currency. .274. The Dollar is. the unit of United States money. Dollars are written as integers, with the sign ($) prefixed ; the lower denominations are written aa decimals, dimes being tenths, cents hundredths, and mills thousandths of a dollar. Thus, 15 dollars; 1 dime, 5 cents, 5 mills, is written $15,155. In business records and papers, cents are often written as fractions of a dollar ; the half-cent is expressed either as a fraction (|), or as 5 mills. Thus, $15.75 may be written $15-^; 12-J- cents, $.12|, or 1.125. 275. The denominations and scale of United States money are shown in the following Table. 10 mills = 1 cent (c. or ct.). 10 dimes = 1 dollar ($). 10 cents = 1 dime (d.). 10 dollars = 1 eagle (E.). Scale.— Descending, 10, 10, 10, 10. Ascending, 10, 10, 10, 10. Remarks. — 1. The scale being a decimal one, all operations in United States money are performed the same as with common decimal expressions. 2. The Dime is a coin, but its name is never used in reading United States money. The Mill is not coined; it is used only as a decimal of the cent, which is the smallest money of the mint and the smallest recognized in business. 6 UNITED STATES MONEY, Obverse. Reverse. ° hvmc ' ReVerse - Obverse. Coins of the United States. Reverse- UNITED STATES MONEY. 8b UNITED STATES COINS. 276. The Coins of the United States, authorized by various Acts of Con- gress, are of gold, silver, copper-nickel, and bronze. 277. The Gold Coins of the United States are as follows . 1. The Double Eagle; value, $20 ; weight, 516 Troy grains. 2. The Eagle; value, $10 ; weight, 258 Troy grains. 3. The Half-Eagle; value, $5 ; weight, 129 Troy grains. Jf. The Three Dollar piece ; value, $3 ; weight, 77.4 Troy grains. 5. The Quarter-Eagle; value, $2.50; weight, 64.5 Troy grains. 6. The One Dollar piece ; value, $1 ; weight, 25.8 Troy grains. Remarks.— 1. All United States gold coins are made of T 9 7 pure gold, and ^ alloy of copper and silver, the alloy being used to toughen the metal so as to reduce the loss from abrasion. The alloy used is never more than T V part silver. 2. United States gold coins of standard weight are legal tender for all debts. 278. The Silver Coins of the United States are as follows : 1. The Dollar; value, $1.00 ; weight, 412.5 Troy grains. 2. The Half-Dollar; value, 50^; weight, 192.9 Troy grains. 8. The Quarter- Dollar ; value, 2o I/. To find the quantity when the price of one is 20{2. dollars, considered as quantity, by 5. 5. To find the quantity when the price of one is 16f 0. dollars, considered as quantity, by 6. 6. To find the quantity "when the price of one is 12-J^. dollars, considered as quantity, by 8. 7. To find the quantity when the price of one is 10jz5. Rule. — Annex a cipher to the dollars, and consider as quantity. 8. To find the quantity when the price of one is 8% per y& 71 yd., at 50^ per yd. 947 yd., at 3^ per yd. Remark. — Quarters are often written thus : 5 8 = 5$; 17 1 = 17£ ; ll 8 = llf ; this method is not used with other fractions. llf.. Find the total cost of the following : 83 2 yd., at 6^ per yd. 127 2 yd., at 16-|0 per yd. 71 3 yd., at 8-^ per yd. 1000 yd., at If?* per yd. 230* yd., at 25^ per yd. 547 3 yd., at 5^-per yd. 28551 yd., at 20$ per yd. 872 2 yd., at 50^ per yd. 624 yd., at 2|0 per yd. 305. To Find the Cost when the Price is given and the Quantity is a Multiple, or an Aliquot part, of 100 or 1000. Remark. — When the quantity is in even hundreds or thousands, find the cost by multiply- ing the price, expressed as a decimal, by the number of hundreds or thousands. For parts of hundreds or thousands, add equivalent fractional parts. Example 1.— Find the cost of 100 yards of cloth, at 54£0 per yard. Explanation.— Since 1 yard costs $ .5425, 100 yards will cost 100 times as much, or $54. 25. • Example £.— Find the cost of 300 yards, at 17-^ per yard. Explanation.— Since 1 yard costs $ .175, 100 yards will cost $17.50, and 300 yards, or 3 times 100 yards, will cost 3 times $17.50, or $52.50. Example 8.— Find the cost of 1000 yards, at 83f^ per yard. Explanation.— Since 1 yd. costs $ .8375, 1000 yd. will cost 1000 times $ .8375, or $837.50. 94 EXAMPLES FOR PRACTICE IN SPECIAL APPLICATIONS. Example Jf. — Find the cost of 75 yards, at $1.37^ per yard. Explanation. — Since 1 yard costs $1.37&, 100 yards will cost $137.50 ; 75 yards will cost | less than $137.50, or $103,125, or $103.13. . Example 5. — Find the cost of 250 yards, at $1.75 per yard. Explanation.— Since 1 yard costs $1.75, 1000 yards will cost $1750 : 250 yards, or \ of 1000 yards, will cost $ of $1750, or $437.50. Example 6. Parts of Parts. — What will be the cost of 142fi bushels of wheat, at S1.37-J- per bushel ? Explanation.— At $1 per bushel, 1426 bushels will cost. _ _ $1426.00 At J = 250 per bushel, 1426 bushels will cost. _ 356.50 At \ — (| of {), or 12£0 per bushel, 1426 bushels will cost 178.25 At $1. 37^ per bushel, 1426 bushels will co3t _ . $1960. 75 Example 7. — What is the cost of 824 yards of cloth, at $1.75 per yard ? Explanation.— At $1 per yard, 824 yards will cost $824.00 At £ = 50^ per yard, 824 yards will cost. 412.00 At J = 25^ per yard, 824 yards will cost 206. 00 At $1. 75 per yard, 824 yards will cost $1442. 00 Example 8. — At 550 per lb., what will be the cost of 146 lb. of gunpowder ? Explanation.— At $1 per lb. , 146 lb. will cost $146 .00 At \ = 50^ per lb., 146 lb. will cost 73.00 At -fa = ( T V of \\ or 50 per lb., 146 lb. will cost 7.30 At 55^ per lb. , 146 lb. will cost. $80.30 EXAMPLES FOR WRITTEN PRACTICE. 306. Find the cost of 1. 200 lb., @ 37-i^. 2. ■ 700 lb., @ 51J0. 5. 15Olb.,@14f0. 4* 250 lb., @ 21§0. o. 1000 1b., @ $1.12^. 6. 750 lb. @ 81f0. 7. 1250 lb., @ $2.62£. 8. 400 1b., @ 95-J0. 9. 250 lb., @ 9f0. 10. 75 lb., @ 6O|0. 11. 125 lb., @ 27-^0. 12. 1100 1b., @ $1.42£. IS. 500 lb., @ 37|0. U. 1500 1b., @ 18f0. 15. 16. 17. 18. 19. 20. 21. 300 1b., @41f0. 3000 lb., @ 12J0- 2500 lb., @ $1.10. 25 lb., @ $1.85. 150 lb., @ 33£0. 75 1b., @$1.15. 125 lb., @ 11.25. EXAMPLES FOR MENTAL PRACTICE. Remark. — All extensions in the following examples should be made mentally, the pupl writing only the cost of each item for footing. 307. 1. Find the total cost of 516 lb., at 100 per lb. 484 lb., at 50 per lb. 1000 lb., at 7^0 per lb. 2500 1b., at 80 per lb. 3000 lb., at 110 per lb. 216 1b., at 1210 per lb. 1120 lb., at 500 per lb. 818 lb., at 250 per lb. 1400 lb., at 200 per lb. 3811b., at 400 per lb. 1095 lb., at 33£0 per lb, 125 lb., at 60 per lb. 711 lb., at 300 per lb. 97 1b., at 8# per lb. 150 lb., at 6£0 per lb. EXAMPLES FOR MENTAL PRACTICE. 95 2. Find the total cost of 686 yd., at 150 per yd. 2140 yd., at 50 per yd. 853 yd., at 100 per yd. 246 yd., at 200 per yd. 398 yd., at 300 per yd. 450 yd., at 33£0 per yd. 3. Find the total cost of 1400 1b., at 40 per lb. 2168 1b., at 3^0 per lb. 7000 1b., at 50 per lb. 246 2 lb., at 610 per lb. 596 3 lb., at 8^0 per lb. 1252 « lb., at 100 per lb. i. Find the total cost of 525? yd., at 80 per yd. 11T8 1 yd., at 90 per yd. 2853 3 yd., at 100 per yd. 1400 yd., at 6^0 per yd. 5. Find the total cost of 83 2 yd., at 550 per yd. 71 3 yd., at 750 per yd. 107 1 yd., at 500 per yd. 23O 3 yd.,at66|0peryd. 1753 2 yd., at 250 per yd. 46 yd., at 150 per yd. 6. Find the total cost of 629 1 yd., at 3J0 per yd. 1140 2 yd., at 50 per yd. 594 3 yd., at 6^0 per yd. 3469 yd., at 8^0 per yd. 1291 2 yd., at 110 per yd. 593 3 yd.,atl2|0peryd. 297 yd., at 250 per yd. 1100 yd., at 430 per yd. 1200 yd., at 28^0 per yd. 298 yd., at 500 per yd. 931 yd., at 250 per yd. 1315 yd., at 33^0 per yd. 936 2 lb., at 1240 per lb. 2143 1 lb., at 150 per lb. 546 1b., at 110 per lb. 2980 1b., at 16|0 per lb. 593 1b., at 130 per lb. 1250 2 lb., at 6^0 per lb. 367 yd., at $1.25 per yd. 28 2 yd., at$2.50peryd. 577 * yd., at 550 per yd. 315 yd., at 750 per yd. 127 2 yd.,at 16f0peryd. 500 yd., at I8f per yd. 269 3 yd., at 12^0 per yd. 2960 1 yd., at 900 per yd. 183 2 yd., at 8^0 per yd. 2375 3 yd., at 100 per yd. 250 yd., at 13f0 per yd. 400 yd., at 15|0 per yd. 756 3 yd.,atl6f0peryd. 1375 1 yd., at 200 per yd. 1741 yd., at 250 per yd. 9O6 3 yd.,at66|0peryd. 800 yd., at 130 per yd. 959 yd., at 16|0 per yd. 1000 yd., at 19f per yd. 2000 yd., at 21|0 per yd. 606 yd., at 1240 per yd. 150 yd., ar 250 per yd. 291 lb., at 500 per lb, 1437 2 lb., at 250 per lb. 1978 3 lb.,at33J0perlb. 844i ib. ? at 750 per lb. 930 1b., at 66|0perlb. 675 2 lb., at 12^0 per lb. 2100 yd., at 750 per yd. 146 yd., at 250 per yd. 500yd.,at$.37|peryd. 1000 yd., at $1.87^ per yd 2005 3 yd., at 6i0 per yd. 1000 yd., at 83f0 per yd. 250 yd., at 27^0 per yd. 931 2 yd., at 33^0 per yd. 768 3 yd.,at$ 1.25peryd. 1756iyd.,at$1.12|peryd. 5548 2 yd., at 140 per yd. 1250 yd., at 7|0 per yd. 300 yd., at 23|0 per yd. 500 yd., at 4l£0 per yd. 186 yd., at 160 per yd. 175 yd., at 150 per yd. 308. To Find the Quantity when the Price is an Aliquot Part of $1. Example 1. — If oats cost 33^0 per bushel, how many bushels can be bought for $54 ? Explanation. —Since 1 bushel costs 33^, or i of $1, 3 bushels can be bought for $1 ; and if $1 will buy 3 bushels, $54 will buy 54 times 3 bushels, or 162 bushels. Example 2. — If a yard of cloth costs 66|0, how many yards will $84 buy ? Explanation. — Since the price is | of itself less than $1 per yard, the number of yards will be | greater than the number of dollars expended ; | of 84 = 42; 84 -f- 42 = 126, or 1 26 yards. 96 EXAMPLES FOR PRACTICE. Example 3.— At 87^ per bushel, how many bushels of wheat can be bought for $1267 ? Explanation.— Since the price is + of itself less than $1 per bushel, the number of bushels will be i greater than the number of dollars expended; \ of 1267 = 181; 181 -}- 1267 = 1448, or 1448 bushels. Remark.— Application of the principle of reciprocals can profitably be introduced at this point; the reasoning will be the same as in the examples given above. Example 4.— At 66f ^ per yard, how many yards of cloth can be bought for $84? Explanation.— 6%* = $£ ; write its reciprocal, f , and multiply by $84. Example 5.— At 75^ per yard, how many yards of cloth can be bought for $84? Explanation.— 750 = $f ; write its reciprocal, f, and multiply by $84. Example 6.— At 87-J^ per yard, how many yards of cloth can be bought for $84? Explanation.— 87i^ = $|; write its reciprocal, f, and multiply by $84. Rules. — 1. Multiply the cost by the quantity that can be bought for $1. Or, 2. Add to the cost (as quantity) such a part of itself as the price lacks of being $1. EXAMPLES FOB PRACTICE. 309. 1. If 1 lb. of candy can be bought for 25^, how many pounds can be bought for $5.75 ? 2. At 33£^ per yard, how many yards of cloth will $1542.50 buy ? 3. A boy expended $1 for almonds, at 16f ^ per lb. How many pounds did he buy ? 4. At 75^ per yard, how many yards of cloth can be bought for $572.40 ? 5. If I invest $175.30 in eggs, at 20^ per doz., how many dozens do I purchase? 6. A farmer sold 26£ bu. buckwheat, at 87-^ per bu., and took his pay in sugar at 6£^ per lb. How many pounds should he have received ? 7. A gardener exchanged 132 qt. of berries, at 8J^ per qt., and 75 doz. corn, at 12-J0 per doz., for cloth at 25^ per yd. How many yards did he receive ? 8. If I exchange 1920 acres of wild land, at $7.50 per acre, for an improved farm at $125 per acre, what should be the number of acres in my farm ? 9. A farmer gave 8f cwt. of pork, at $7.50 per cwt., 15 bu. of beans, at $3.25 per bu., and 46£ bu. of oats, at 33£^ per bu., for 28 yd. of dress silk, at ^L.25 per yd., and 52£ yd. of delaine, at 16f

. Millet, weighing 1768 lb., at $1 per bushel of 45 lb. 5. Oats, weighing 2255 lb., at $ .35 per bushel of 32 lb. 6. Buckwheat, weighing 2172 lb., at $ .60 per bushel of 48 lb. 7. Beans, weighing 2761 lb., at $1.25 per bushel of 62 lb. 8. Peas, weighing 2500 lb., at $1.40 per bushel of 60 lb. 9. Hungarian grass seed, weighing 3146 lb., at $2.50 per bushel of 45 lb. 10. Red top grass seed, weighing 2059 lb., at $ .90 per bushel of 14 lb. 11. Timothy seed, weighing 2677 lb., at $2 per bushel of 44 lb. 12. Kentucky blue grass seed, weighing 2266 lb., at $1.50 per bushel of 14 lb. 18. Clover seed, weighing 2941 lb., at $5.10 per bushel of 45 lb. U. Flax seed, weighing 2727 lb., at $2.25 per bushel of 56 lb. 15. Castor beans, weighing 3050 lb. , at $3 per bushel of 46 lb. 16. Potatoes, weighing 2;>99 lb., at $.65 per bushel of 60 lb. 17. Turnips, weighing 2100 lb., at $ .30 per bushel of 56 lb. 18. Apples, weighing 2701 lb., at $ .25 per bushel of 56 lb. 19. Sweet potatoes, weighing 3349 lb., at $1 per bushel of 55 lb. 20. Onions, weighing 2021 lb., at $ .85 per bushel of 57 lb. 21. "Rye, weighing 1367 lb., at $ .64 per bushel of 56 lb. 22. The products of a farm were ten loads each of wheat, barley, corn, oats, and potatoes. The wheat sold at $1.12 per bushel of 60 lb., the barley at 85^ per bushel of 48 lb., corn at 70^ per bushel of 58 lb., oats at 32^ per bushel of 32 lb., and potatoes at 62^ per bushel of 60 lb. The loads of wheat weighed respectively 2585, 2640, 2721, 2594, 3063, 3354, 3145, 2720, 2938, and 2890 lb.; the barley 2163, 2487, 2225, 3004, 3121, 2742, 2907, 2525, 3140, and 3082 lb.; the corn 3100, 3126, 3097, 3040, 2872, 2950, 2777, 2981, 2547, and 2939 lb.; the oats 1973, 2946, 2172, 3148, 2500, 1951, 2031, 2997, 3005, and 2775 lb.; the potatoes 2846, 2891, 2805, 2863, -2984, 2901, 3046, 3280, 33y5, and 2584 lb. How much was received from the five products? Remark. — Add each ten loads, and compute bushels but once for each product. 100 BILLS, STATEMENTS, AND INVENTORIES. BILLS, STATEMENTS, AND INVENTORIES. 319. A Bill is a written statement in detail of articles sold or services rendered. Remark. — A Bill should state the names of both parties, the terms of credit, the name, quantity, and price of each item, and the entire amount. The Bill is said to be receipted when the words " Received Payment," or " Paid * and the creditor's signature, have been written at the bottom. 320. An Invoice is a written description of merchandise sold, or shipped to be sold on account of the shipper. Remark 1.— The terms Invoice and Bill are now used interchangeably; formerly the term Invoice was applied only to written statements of merchandise shipped to be sold for the owner. 2. An Invoice should bear the date of the sale or shipment, the special distinguishing marks, if any, upon the goods, the names of seller and buyer, or consignor and consignee, the items, prices, footing, discounts, if any, terms of sale, and manner of shipment. 321. A Statement is based upon itemized bills previously rendered, and is a written exhibit of the sum of the items charged in each of the bills, including also the dates on which the several bills were rendered. 322. An Inventory is an itemized schedule of the property possessed by an individual, firm, or corporation, and not shown by the regular books of account; or it may include all of the property possessed by an individual, firm, or corpo- ration, such as book accounts, notes, cash, merchandise., etc., and also the debts due by the individual, firm, or corporation. This, however, is generally called a statement of the business. Remark. — An inventory is usually made upon the event of taking off a balance sheet, of a change in the business, of the admission of a partner, of the issue of stock, or, in case of embarrassment or insolvency, for examination by creditors, together with the other resources and liabilities of the business. 323. Contractions and Abbreviations used in Business. Al First Quality. Acct. Account. Agt. Agent. Amt. Amount. Bal. Balance. BU. or Bar. Barrel. Bdl. Bundle. Blh. Black. % Bill of Lading. Bot. Bought. Bro. Brother. Bu. Bushel. Bx. Box. Cd. Cord. $ Cent. Ct. Cent. Chgd. Charged. Co. Company. C. 0. D. Collect on Delivery. Com. Commission. Con. Consignment. Cr, Creditor. Cwt. Hundred weight. Dft. Draft. Dis. Discount. Do. or ditto. The same. Doz. Dozen. Dr. Debtor. Ea. Each. E. & 0. E. Errors and Omissions Excepted. Exch. Exchange. Fol. Folio or page. FrH. Freight. Ft. Foot. Gal. Gallon. Gr. Gross. Guar. Guaranteed. Hhd. Hogshead. i. e. That is. In. Inch. Ins. Insurance. Jr. Junior. Lb. Pound. BILLS. ,101 Mdse. Merchandise. Mem. Memorandum. . Messrs. Gentlemen or Sirs. Mr. Mister. Mrs. Mistress. N. B. Take notice. Net. Without discount. No. Number. Oz. Ounce. P. oy p. Page. Pp. or pp. Pages. PayH. Payment. Pd. Paid. Per. By, or by the. Pkg. Package. P. 0. Post Office. Pr. Pair. Pc. Piece. Qr. Quarter. Remark. — In abbreviating measures of capacity, weight, distance, or time, it is unnecessary to add an * for the plural. 324. Time Abbreviations and Contractions used in Business. Received. Receipt. Railroad. Schooner. Shipment. Str. Steamer. Sunds. Sundries. Super. Superfine. Wt. Weight. Yd. Yard. Rec'd. RecH. R.R. Schr. Ship't. Jan. or Jarfy. January. Feb. or FeVy. February. Mar. March. Apr. April. Aug. August. Sept. September. Oct. October. Nov. November. Dec. December. Mo. Month. Yr. Year. Inst. Present month. Prox. Next month. Tilt. Last month. Cent. Century. d. Day. h. Hour. m. Minute. sec. Second. wk. Week. 325. Signs and Symbols in Common Use. @ At; as, at a price. j( Number. /$ By, or by the. <*fc Account. % V Care of. Check mark. Per cent, or dredths. Hun- New account. Old account. By, in surface measures. BILLS. 326. Find the footing of each of the following bills : (1.) John" R. Knox, 153 Pearl St., City, Knoxville, Tenn., Dec. 31, 1888. Bought of CULVER & CASS. sacks Cream Flour 950 bii. Potatoes ._ 800 bu. Sweet Potatoes 900 lb. Ginger ... _ 220 lb. Jap. Tea _ 550 lb. 0. H. Tea 750 gal. Syrup _. 450 bu. Onions _ $1 lb. Crackers 110 Paid, 2 85 1 60 45 44 1 10 75 90 50 44 * Culver & Cass, Per Cass. 102. BILLS. (2.) Folio 246. Saginaw, Mich., Sept. 1, 1888. McGraw & Sage, Tonawanda, N. Y., To WALLACE W. WESTON, Dr. Terms, Sight Draft without notice after ninety days; 5£ if paid within 60 days. 26416 ft. Clear Pine 28.00 per M. 146250 ft. Pine Plank ,_ 23.50 per M. 81275 ft. Clapboards 25.00 per M. 11670 Cedar Posts 7.00 per C. 71300 Shingles "A" 4.10 per M. 56200 ft. Pine Timber 21.00 per M. 111224 Cedar R. R. Ties 34.50 per C. 91050 ft. Flooring 27.50 per M. *5508 Shingles " B," 3. 60 per M. 31000 Barn Boards 15. 75 per M. (3.) Ole Paulsen & Bro., Detroit, Mich., Folio 41. Sales Bk. 219. Terms cash. Worcester, Mass., May 15, 1888. To FRANK DRAKE & SON, Dr. Case. J 19 8 5 8 31 if ? it 21 15 12 24 21 Pieces Bleached Cotton, 41 2 46 3 41 1 45 2 44 44* 47 1 45 3 42 42 3 43 3 431 47 : 44 44 2 Pieces Muslin, 37 1 32 3 33 35 3 34* 32 35 2 33 3 37 38 1 38* 36 Pieces Delaine*, 39 40 2 41i 393 38 2 40 42 3 44* 42 Pieces Windsor Prints, 21 3 27 3 25 3 28 26 22 2 24 25 32 31 2 28 24i 25 27 2 22 28 1 24*22 21 2 26 24 31 2 32 22 Pieces Merrimac Prints, 28 1 32 34 3 28 2 26 24* 22 2 24 2 26 2 24 26i 33 28 2 34 27 1 30 32 3 24 30 2 31 30 2 No.Yd Price. Items. Amou m n$ 16£ H except cen- tennial years, are leap years. II. All centennial years divisible by 400 are leap years. Table. 60 seconds (sec. ) =1 minute .„ mm, 60 minutes = 1 hour _„ . . . hr. 24 hours == 1 day „ da, 7 days = 1 week „ wk. 4 toeehs = 1 lunar month .......... mo. 30 days = 1 commercial month » „ mo. 565 days --= 1 common year ...... yr, 566 days = Heap year... . yr. 12 calendar months = 1 civil year - •. _ yr, 10 years =s 1 decade 100 years s= T 'century C. Scale, descending, 12, 30, 24, 60, 60; ascending, 60, 60, 24, 30, 12. Remark. — In most business transactions 30 days are considered a month, and twelve such months a year. 110 REDUCTION OF TIME. ►46. The Calendar Months are as follows: 1st. January (Jan.) having 31 2nd. February (Feb.) " 28 3rd. March (Mar.) " 31 4th. April (Apr.) " 30 5th. May (May) " 31 6th. June (June) " 30 days. -29 " 7th. July (July) 8th. August (Aug., 9th. September (Sept.) 10th. October (Oct.) 11th. November (Nov.) 12th. December (Dec.) having 31 days. 31 30 31 30 31 347. The year begins with the first day, or First, of January, and is divided into four seasons of three months each. 348. The Seasons are Winter, Spring, Summer, and Autumn, or Fall, The Winter months are December, January, and February. The Spring months are March, April, and May. The Summer months are June, July, and August. The Autumn months are September, October, and November. Remark. — The ancient Roman year began with March 1, and thus September, Octobt November, and December ranked, as their Latin derivation indicates, as the 7th, 8th, 9th, an} 10th months respectively of the Roman year. REDUCTION OF TIME. 349. The reduction of expressions of time from higher to lower denomina- tions, or the reverse, may be accomplished in the same manner as the reduction of United States money heretofore explained, the only difference being that the scale in the latter. is uniform, while that in the former is varying. 350. To Reduce Time from Higher to Lower Denominations. Example, 12 3yr. 36 mo. 7 mo. 43 mo. 30 1290 da. 11 da. 1301 da. 24 31224 hr. 7hr. 31231 hr. , — Reduce 3 yr. 7 mo. 11 da. 7 hr. Operation. 7 mo. 11 d. 7 hr. 25m. 38 sec. Operation Continued. 31231 hr. 60 1873860 m. 25 m. 1873885 m. 60 112433100 sec. 38 sec. 112433138 sec. 25 m. 38 sec. to seconds. Explanation. — Since one year equals 12 months, 3 years equal 36 months, and 7 months added gives 43 months; since one month equals 30 days, 43 months equal 1290 days, and 11 days added gives 1301 days; since one day equals 24 hours, 1301 days equal 31224 hours, and 7 hours added gives 31231 hours; since one hour equals 60 minutes, 31231 hours equal 1873860 minutes, and 25 min- utes added gives 1873885 minutes; since one minute equals 60 seconds, 1873885 minutes equal 112433100 seconds, and 38 seconds added gives 112433138 seconds. Remark. — The reduction descending of any compound denominate number can be accom- plished as above, by observing the scale of the table to which it belongs. ADDITION OF TIME. Ill Rule. — Beginning with the highest, multiply the units of each denom- ination by the number in the scale required to reduce it to the denom- ination next lower; add the units, if any, of such lower denomination, and so continue from the given to the required denomination. 351. To Reduce Time from Lower to Higher Denominations, Example. — Reduce 112 433138 seconds to years. Operation. 60 ) 112433138 sec. Explanation.— Divide the given 60 ) 1873885 min. + 38 sec. seconds by 60 < to reduce t0 minutes ; the minutes thus obtained, by 60, to 24 ) 31231 hr. + 25 min. reduce to hours; the hours by 24, to 30 ) 1301 da. + 7 hr. reduce to days: the days by 30, to reduce to months, and the months by 12, to reduce to years. 12)43 mo. +11 da. 3 yr. +7 mo. 112433138 sec. = 3 yr. 7 mo. 11 da. 7 hr. 25 min. 38 sec. Rule. — Divide the given units by that number in the scale which will reduce them to units of the next higher denomination, and so continue from the given to the required denomination. Any remainder obtained will be of the same denomination as the dividend from which it arises. ADDITION OF TIME. 352. To Add Time. Time expressions may be added as simple numbers, if only it be observed that the scale from the lowest to the higest order is 60, 60, 24, 30, and 12. The highest denomination in common use is the year. Example.— Add 41 yr. 8 mo. 22 da. 19 hr. 27 min. 14 sec, and 5 yr. 6 mo. 11 da. 10 hr. 50 min. 56 sec. Explanation. — Arrange the numbers so that those of the same denomination stand in the same ver- tical line. Then begin with the 47 yr. 3 mo. 4da. 6 hr. 18 min. 10 sec. lowest denomination, which is sec- onds, and add: 14 seconds plus 56 seconds equals 70 seconds, equals 1 minute plus 10 seconds; write the 10 underneath the column of seconds, and carry the 1 to the next column; 27 minutes plus 50 minutes equals 77 minutes, and 77 minutes plus 1 minute (to carry) equals 78 minutes, equals 1 hour plus 18 minutes; write and carry as before; 19 hours plus 10 hours equals 29 hours, and 29 hours plus 1 hour (to carry) equals 30 hours, equals 1 day plus 6 hours; 22 days plus 11 days equals 33 days, and 33 days plus 1 day (to carry) equals 34 days, equals 1 month plus 4 days; 8 months plus 6 months equals 14 months, and 14 months plus 1 month (to carry) equals 15 months, equals 1 year plus 3 months; 41 years plus 5 years equals 46 years, and 46 years plus 1 year (to carry) equals 47 years. Rule. — Add as in abstract numbers, and reduce according to the table of Time. Operation. 41 yr. 8 mo. 22 da. 19 hr. 27 min. 14 sec. 5yr. 6 mo. 11 da. 10 hr. 50 min. 56 sec. 8 F- 1 mo. 18 da. 5 y r - 4 mo. 21 da. 112 SUBTRACTION OF TIME. SUBTRACTION OF TIME. 353. Difference in time is found in two ways: 1st. By counting the actual number of days from the given to the require^ date. Thus, the number of days between May 13 and September 7 is 117, count- ing 18 days left in May, 30 for June, 31 for July, 31 for August, and the 7 of September. 2d. By Compound Subtraction. Subtraction in either simple or compound numbers is really the same, except that in the latter a varying scale is employed. That is, it may, and usually does, involve a transformation in either case. This will always be required unless the several minuend terms, or orders are each equal to or greater than the corresponding subtrahend term. 354. To Find the Difference in Time by Compound Subtraction. Example. — Subtract 5 yr. 4 mo. 21 da. from 8 yr. 1 mo. 18 da. Explanation. — Write the numbers so that those of the same denomination stand in the same column. Then begin with the lowest denomination to subtract. Since 21 days can- not be subtracted from 18 days, transform, or borrow one from 2 yr. 8 mo. 27 da. the next denomination; 1 month = 30 days, and 18 days added = 48 days; 48 days — 21 days = 27 days, which write underneath the column of days; the 1 month having been borrowed from the minuend, there are no months remaining from which to sub- tract the 4 months in the subtrahend, hence, borrow one from the next denomination; 12 months — 4 months = 8 months, which write underneath the column of months; there now remains 7 years from which to subtract; 7 years — 5 years = 2 years, which write underneath the column of years. This completes the operation, giving a remainder of 2 years, 8 months, and 27 days. Rule. — Subtract as in abstract numbers, observing the varying scale. EXAMPLES FOR PRACTICE. Remark. — In the following examples, the difference in time should be found by compound subtraction, unless it be otherwise stated. 355. i. Eeduce 27051 seconds to minutes. 2. Eeduce 83129 seconds to hours and minutes. 8. Reduce 610251 seconds to higher denominations. J/.. How many years, months, days, hours, and minutes, in 749520360 seconds? 5. How many hours from half -past three o'clock p. m. Oct. 13, 1888, to noon on the fourth day of July, 1889? 6. A note entitled to 93 days' time was dated Oct. 13, 1888. Counting actual time, on what day should it be paid? 7. How many days between Nov. 3, 1890, and Mar. 1, 1900? 8. A mortgage dated July 2, 1888, was paid Sept. 14, 1891. How many days did it run? 9. How long does a note run if dated Sept. 22, 1887, and paid Aug. 31, 1888? 10. How much time will a man gain for labor in 60 years, by rising 45 minutes earlier each day, beginning Jan. 1, 1888. LATITUDE, LONGITUDE, AND TIME. 113 11. How many more minutes in the eleven years "before Jan. 1, 1890, than in the eleven years after that date ? 12, How many seconds of difference in the time of one solar year and 12 lunar months of 29 da. 12 hr. 44 min. and 3 sec. each ? CIRCULAR MEASURE. 356. Circular Measure is used in surveying, navigation, astronomy, and geography; for reckoning latitude and longitude, determining location of places and vessels, and in computing differences of time. 357. Every circle, great or small, is divisible into four equal parts; these parts are called quadrants, and are divisible into ninety equal parts, each of which is called a degree; every circle, therefore, may be divided into 360 equal parts, called degrees. REMAKK.-^-The divisions into twelfths called signs, and into sixths called sextants, are ir occasional use. Table. 60 seconds (") = 1 minute ('). | 30 degrees = 1 sign (8.) 60 minutes = 1 degree (°). | 12 signs or 360° = 1 circle (C.) « , j descending, 12, 30, 60, 60; or, 360, 60, 60. ^ caie ' ( ascending, 60, 60, 30, 12; or, 60, 60, 360. Remark. — Minutes of the earth's circumference are called nautical or geographic mile**. EXAMPLES FOR PRACTICE. 358. i. Eeduce 2154/ to degrees. Reduce 87406" to degrees, minutes, and seconds. Reduce 330581" to higher denominations. How many seconds in a circle? 5. How many minutes in 2 8. 21° 47'? 6. How many seconds in 1 8. 27° 8' 57'? Reduce 8162 geographic miles to degrees. 8. How many geographic miles in the circumference of the earth? 9. By two different observations the position of a ship was shown to have changed 519 geographic miles. How much was her change in degrees and minutes? LATITUDE, LONGITUDE, AND TIME. 359. Latitude is distance north or south from the equator. A place is said to be in north latitude if north of the equator; and to be in south latitude if south of the equator. 360. Longitude is distance east or west from any given starting point oi meridian. A place is* said to be in west longitude if west of the given meridian- and to be in east longitude if east of the given meridian. 8 114 LATITUDE, LONGITUDE, AND TIME. 361. Since every circle may be divided into 360 equal parts, or degree^ ana the sun appears to pass from east to west around the earth, or through 360° of longitude, once in every 24 hours, it will pass through -fa of 360°, or 15° of longi- tude, in 1 hour; through 1° of longitude in -^ of 1 hour, or 4 minutes; and through 1' of longtitude in -gV of 4 minutes, or 4 seconds. Table. 360° of longitude = 24 hours or 1 day of time, da. 15°" " = 1 hour of time, hr. 1° " " = 4 minutes " min. 1' " " = 4 seconds (t sec. Remark. — Standard Time. — Previous to 1883 there were fifty-three different time- standards in use by the railroads of the United States, and as these standards were based on the local time of the principal cities which served as the center of operations of the different roads, they were a constant source of annoyance and trouble, both to the railroads and to the traveling public. To obviate this difficulty the principal railroads of the United States and Canada adopted, in 1883, what is known as the "Standard Time System." This system divides the United States and Canada into four sections or time-belts, each covering 15° of longitude, 7£° of which are east and 7£° west of the governing or standard meridian, and the time throughout each belt is the same as the astronomical or local time of the governing meridian of that belt. The governing meridians are the 75th, the 90th, the 105th and the 120th west of Greenwich, and as these meridians are just 15° apart, there is a difference in time of exactly one hour between any one of them and the one next on the east, or the one next on the west; the standard meridian next on the east being one hour faster, and the one next on the west one hour slower. The time of the 75th meridian, which is about 4 minutes slower than New York time and about 1 minute faster than Philadelphia time, is called " Eastern Time," and when it is astronomical noon on this meridian it is noon on every railroad clock from Portland, Me., to Buffalo and Pittsburg, and from Quebec to Charleston. The time of the 90th meridian, one hour slower than "Eastern Time," and 9 minutes slower than Chicago time, is known as " Central Time," and all roads operated in the second belt are run by " Central Time." The time of the 105th meridian, one hour slower than " Central Time," is distinguished as " Moun- tain Time." Time in the fourth belt, which is governed by the 120th meridian, and extends to the Pacific coast, is called "Pacific Time;" it is one hour behind "Mountain Time," two behind " Central Time," and three behind " Eastern Time." The changes from one time- standard to another are made at the termini of roads, or at well-known points of departure, and where they are attended with the least inconvenience and danger. As this system has produced satisfactory results and has been adopted by most of the principal cities for local use, it is probable that the business of the whole country will, before many years, be regulated by standard railroad time. 362. To Find the Difference in Time, when the Difference in Longitude is given. Example. —If the difference in longitude of two places be 9° 15', what must be their difference in time ? Operation. Explanation. — Since each minute of distance equals 4 seconds of go , -^~ / time, 15 minutes of distance will equal 15 times 4 seconds, or 60 seconds, which equals one minute of time. And since each degree of distance I equals 4 minutes of time, 9 degrees will equal 9 times 4 minutes, or 36 37 min. sec. minutes; adding the one minute obtained above, gives 37 minutes as the required result. LATITUDE, LONGITUDE, AND TIME. 1 f 5- II ill e. — Multiply the units of distance by J/,, and reduce according to- the table of Time. EXAMPLES FOR PRACTICE. Remark. — Examples under this topic will be restricted to variations of solar time. 363. 1. Cincinnati is 84° 24', and San Francisco 122°, west longitude. What is their difference in time? 2. New York is 74° 1% and Halifax 63° 36', west longitude. Find their difference in time. 3. St. Petersburg is 30° 19' east, and St. Louis 90° 15' west longitude When it is noon at St. Petersburg, what is the time at St. Louis. Kemakk. — If one place be east and the other west of the given meridian, to find their difference in longitude, add their respective distances from the meridian taken. U. The longitude of the City of Mexico is 99° 5', and that of Boston 71° 3', west longitude. Find their difference in time. 5. If on leaving London, 0° 0' of longitude, my watch, keeping correct time, indicates 46 minutes, 15 seconds, after 3 P. M., what time should it indicate on my arrival at Astoria, Oregon, 124° west, where it is then noon? 364. To Find the Difference of Longitude, when the Difference in Time is Given. Example. — The difference in time between two places is 2 hours, 19 minutes, and 48 seconds. What is their difference of longitude? Operation. Explanation. — 2 hours, 19 minutes, and 2 hr. 19 min. 48 sec. = 139 min. 48 sec. 48 seconds equal 139 minutes and 48 seconds; since each 4 minutes of time equal 1 degree 4 ) 139 mm. 48 sec. of distance, 139 minutes and 48 seconds equal 34° + (3 min. 48 sec.) 34 degrees, with 3 minutes and 48 seconds, or q ™;^ aq o™ ooq nnn 228 seconds, remainder; and since each 4 sec- 6 mm. 48 sec. = /J28 sec. _ ,. . , „, i ,. nnn , onds ol time equal V of distance, 228 seconds 4 ) 228 s ec. equal 57' of distance. Therefore, if the dif- 57' ference in time between two points be 2 2 hr. 19 min. 48 sec. = 34° 57'. hours, 19 minutes, and 48 seconds, their dif- ference in longitude will be 34° 57'. Rule. — Reduce the difference in time to minutes and seconds, and divide by 4> the quotient will be the difference of longitude, in degrees., minutes, and seconds. examples for practice. 365. 1. What is the difference in the longitude of New York and San Fran- cisco, their difference of time being 3 hr. 11 min. 56 sec. 2. The longitude of Sitka is 135° 18' west. What is the longitude of the city of Jerusalem if, when it is 9 o'clock and 5 minutes A. m. at Sitka, it is 27 minutes and 4 seconds after 8 p. m. in Jerusalem? 8. Find the difference in latitude of Chicago, situated 41° 54' north, and: Valparaiso, 33° 4' south. 116 REDUCTION OF ENGLISH MOHEY. 4. What is the latitude of Washington, if it be 61° 46' 20' north of Rio Janeiro, and the latter place be 24° 54' south latitude? 6. When it is 20-J minutes after noon at Washington, it is 21 min. 26 sec before noon at Chicago, 87° 30' west. What is the longtitude of Washington f MISCELLANEOUS EXAMPLES. 366. 1. A messenger leaves the Greenwich Observatory, westward bound, at noon, Dec. 31, and by a uniform rate of speed encircles the globe in 24 hours. Where is he at the end of the old year? & Suppose the messenger be eastward bound, at what point will he meet the new year? 5. When it is 20 minutes past 10 A. M. at Cape Horn, 68° west, what is the time at Cape of Good Hope, 18° 19' east? Jf, When it is noon at London, what is the time at St. Augustine, 81° 35' west? At Berlin, 13° 30' east? At New Orleans, 90° west? At Sidney, 152° 20' east? At Paris, 20° 20' 22^' east? At New York, 74° 3' west? ENGLISH MONEY. 367. English or Sterling Money is the legal currency of Great Britain. Table. 4 farthings ( far. ) = 1 penny d. 12 pence = 1 shilling s. 20 shillings HlP^r Scale, ascending, 4, 12, 20; descending, 20, 12, 4. 368. The standard unit is the pound sterling, the value of which, in United States money, is shown, together with the other coins, in the following Comparative Table. The farthing = \ % cent. i The shilling = 24 \ cents. The penny = 2^ cents. The pound =$4.8665. Remark. — The farthing is but little used, except as a fractional part of the penny. COINS OF GREAT BRITAIN. 369. The gold coins are the sovereign and the half-sovereign. The silver coins are the crown (equal to 5 shillings), the half-crown, the florin (equal to 2 shillings), the shilling, the six-penny and three-penny pieces. The copper coins are the penny, the half -penny, and the farthing. The guinea (equal to 21 shillings) and the half -guinea are in use, but are no ionger coined. REDUCTION" OF ENGLISH MONEY. 117 REDUCTION OF ENGLISH MONEY. 370. To Reduce English Money from Lower to Higher Denominations. Example. — Eeduce 13206 farthings to units of higher denominations. Operation. Explanation. — Since 4 farthings equal one penny, 13206 far- things equal as many pence as 4 is contained times in 13206, or 4 ) 13206 far. 3301, p i us 2 remainder, equal 3301 pence, 2 farthings; since 12 12 ) 3301 d. + 2 far. pence equal 1 shilling, 3301 pence equal 275 shillings, plus 1 \ 97K j_ 1 a Penny; since 20 shillings equal 1 pound, 275 shillings equal 13 ^OJ^75 s. + 1 d. pounds, plus 15 shillings. Therefore, 13206 farthings equal £13, £13 + 15 s. 15s. Id. 2 far. 13206 far. = £ 13, 15 s. 1 d. 2 far. Rule.— Divide by the units in the scale from the given to the required denomination. EXAMPLES FOR PRACTICE. 371. Reduce 1. 5124 s. to pounds. 3. 13042 d. to pounds. 11916 far. to shillings. h 187409 far. to higher denominations. 20 180 s. Operation Continued. 4 s. 2211 d. 184 s. 4 12 8844 far. 2208 d. 2 far. 3d. 8846. 372. To Reduce English Money from Higher to Lower Denominations. Example. — How many farthings in £9, 4 s. 3 d. 2 far.? Operation. £ 9, 4 s. 3 d. 2 far. Explanation. — Since 1 pound equals 20 shillings, 9 pounds equal 180 shillings, and 180 shillings, plus 4 shillings, equal 184 shillings; since 1 shilling equals 12 pence, 184 shillings equal 2208 pence, and 2208 pence, plus 3 pence, equal 2211 pence; since 1 penny equals 4 farthings, 2211 pence equal 8844 farthings, and 8844 farthings, plus 2 farthings, equal 8846 farthings. Therefore, £9, 4 s. 3 d. 2 far. = 8846 far. 8846. 2211 d. Rule. — Multiply by the units in the scale from the given to the required ^nomination . EXAMPLES FOR PRACTICE. 373. 1. How many pence in £27? 2. How many farthings in 19 s. lid.? S. How many pence in £161, 17 s. 9 d. ? 4. Reduce £41, 1 s. 10 d. 2 far. to farthings. 5. How many farthings in £13, 15 s. 1 d. 2 far. ? 374. To Reduce English Money to Equivalents in United States Currency. Example. — Reduce £15, 3 s. 7 d. 2 far. to dollars and cents. First Explanation.— Since £1 equals $4.8665, £15 equal $72.9975; since 1 shilling equals 24£ cents, 3 shillings equal $.73; since 1 penny equals 2^ cents, 7 pence equal $.1414 ; since 1 farthing equals f# cent, 2 farthings equal $.0101. Therefore, £15, 3 s. 7 d. 2 far. = $73.8789, or $73.88. 118 REDUCTION OF ENGLISH MONEY. Second Explanation. — Call each 2 shillings y 1 ^ of a pound, then 3 shillings equal £.15; call the pence and farthings, reduced to farthings, so many y^u of a pound, then 7 pence, plus 2 farthings, equal 30 farthings, equal £.030; to these add the £15, and the result is £15.18. And, since £1 equals $4.8065, £15.18 equal 15.18 times $4.8665, or $73.88, as before found. Rules. — -Z. Multiply each of the orders of Sterling money by its equiva- lent in United States currency, and add the results. Or, 2. Reduce the Sterling expression to pounds and decimals of a pound by calling each 2 shillings tV of a pound, and the pence and farthings, reduced to farthings, so many roW of a pound; multiply the pounds and decimals of a pound thus obtained by J^.8665, and the product will be the answer in dollars and cents. Remark. — This is exact to within ^ of the part represented by the pence and farthings. EXAMPLES FOR PRACTICE. 375, Reduce to equivalents in United States money 1. £71, 19 s. 5 d. and 3 far. 2. £108, 11 d. and 1 far. 3. £13057, 10 s. and 4 d. 4. £3, 1 s. 9 d. and 2 far. 5. £ 11, 3 s. 1 d. 1 far. 376. To Reduce United States Money to Sterling equivalents. Example. — Reduce $5164.28 to equivalents in English money. Operation. 4.8665 ) 5164.28 £1061 + £.189 rem. 20 Explanation.— Since $4.8665 equal £1, $5164.28 equal £1061.189; multiply the decimal by the units 3 s. -h .78 s. rem. in the scale, 20, 12, 4, in order, pointing off as in 12 decimals, and obtain 3 s. 9 d. 1 far., which, added to — — ., the £1061, equals £1061, 3 s. 9 d. 4 far. 9 d. + 36 d. rem. H 4 1 far. +.44 far. Rule. — Divide the expression of decimal currency by 4.8665, and the integers of the quotient will be pounds Sterling; reduce the decimal of the quotient, if any, by multiplying by the lower units in the scale. EXAMPLES FOR PRACTICE. 377. 1. Reduce $185 to equivalents in English money. 2. Reduce $308.50 to equivalents in English money. 3. Reduce $2500 to equivalents in English money. Jf. Reduce $3658.21 to equivalents in English money. 5. Reduce $110085.75 to equivalents in English money. REDUCTION OF DENOMINATE NUMBERS. 119 MEASURES OF WEIGHT. 378.. Weight is the measure of gravity, and is of three kinds, distinguished from each other by their uses, viz : Troy weiglit, with units of pounds, ounces, pennyweights, and grains, used for weighing precious metals. Avoirdupois tveight, with units of tons, hundred weights, pounds, ounces, and drams, used for weighing products and general merchandise. Apothecaries' weight, with units of pounds, ounces, drams, scruples, and grains, used by druggists. TROY WEIGHT. 379. The Troy pound is the standard of weight, and is equal to 22.7944 cubic inches of pure water, at its greatest density. The grains of the other weights are the same as the Troy grains Table. 24 grains ( gr. ) = 1 pennyweight pwt. 20 pennyweights = 1 ounce oz. 12 ounces — 1 pound lb. S j j descending, 12, 20, 24. 1 lb. = 5760 grains. ( ascending, 24, 20, 12. 1 oz. = 480 grains. REDUCTION OF DENOMINATE NUMBERS. 380. To Reduce Denominate Numbers from Higher to Lower Denominations. Example — Reduce G lb. 11 oz. 15 pwt. 21 gr., Troy, to grains. Operation. First Explanation. — Since 1 pound equals 12 6 lb. 11 oz. 15 pwt. 21 gr. ounces, 6 pounds equal 72 ounces, and 72 ounces plus l* v 11 ounces equal 83 ounces ; since 1 ounce equals 20 pennyweights, 83 ounces equal 1660 pennyweights, and 1660 pennyweights plus 15 pennyweights equal 1675 pennyweights; since 1 pennyweight equals 24 grains, 1675 pennyweights equal 40200 grains, plus 21 grains equal 40221 grains. Therefore, 6 lb. 11 oz. 15 pwt. 21 gr. Troy, = 40221 gr. 72 oz. 11 oz. 83 oz. 20 1660 15 pwt. pwt. 1675 24 pwt. 40200 21 gr. g r - Second Explanation.— Since 1 pound equals 5760 grains, 6 pounds equal 34560 grains; since 1 ounce equals 480 grains, 11 ounces equal 5280 grains ; since 1 pen- nyweight equals 24 grains, 15 pennyweights equal 360 grains; to these add the 21 grains, and the entire sum is 40221 gr. 40aal grains " Remark. — A thorough knowledge of the unit equivalents, together with readiness in the use of the multiplication table, renders the second form much the shorter of the two methods. 120 REDUCTION" OF DENOMINATE NUMBERS. Rules. — 1. Multiply the units of the highest denomination given by that number in the scale which will reduce it to the denomination next lower, and add the units of that lower denomination; continue in this manner until the required denomination is reached. Or, 2. Multiply the units of each denomination by the number of units of the desired equivalent which it takes to make one of that denomination, and add the products thus obtained. 381. To Reduce Denominate Numbers from Lower to Higher Denominations. Example. — Reduce 40221 gr., Troy, to higher denominations. First Operation. -^ _. Explanation. — Since 24 grains equal 1 pennyweight, 24 ) 40221 gr. 40221 grains equal 1675 pennyweights, plus 21 grains; since 20 ) 1675 DWt + 21 2T 2 ^ pennyweights equal 1 ounce, 1675 pennyweights equal 83 ounces, plus 15 pennyweights; since 12 ounces equal 12 ) 83 oz. + 15 pwt. 1 pound, 83 ounces equal 6 pounds, plus 11 ounces. There- 7 ,, fore, 40221 gr., Troy, = 6 lb. 11 oz. 15 pwt. 21 gr. ID. — p J.JL OZ. 40221 gr., Troy, = 6 lb. 11 oz. 15 pwt. 21 gr. Second Operation. 5760 ) 40221 gr. ( 6 lb. Explanation.— Since 5760 grains equal 1 pound, 40221 34560 grains equal 6 pounds, plus 5661 grains; since 480 grains 480 "I 5661 OT M 1 07 eo x ua l 1 ounce, 5661 grains equal 11 ounces, plus 381 grains; 5280 * since 24 grains eqnal 1 pennyweight 881 grains equal 15 pennyweights, plus 21 grains. Therefore, 40221 gr. = 6 lb. 24 ) 381 gr. ( 15 pwt. n oz> 15 pwt 2 1 gr., as before found. 360 ~21gr. 40221 gr., Troy, = 6 lb. 11 oz. 15 pwt. 21 gr. Remark. — The first form is advised for practice, as the operations may usually be per- formed by short division. Example 2. — Eeduce 11426 gr., Troy, to higher denominations. Explanation. — Divide the given number by 24, the integers of the quotient by 20, the integers of the new quotient by 12. Rules. — 1. Divide by the successive units in the scale. Or, 2. Divide by the unit equivalents of each of the higher denominations. 382. To Reduce Denominate Fractions from a Higher to a Lower Denomination, Example. — Eeduce y^g-g- lb., Troy, to the fraction of a pennyweight. First Operation. 16 1 6 x V X 2 T ° — fill = It- Explanation.— tc 7 ^ of a pound equals Tlr Vu of the 12 ounces in 1 pound, or T f£^ ounces; jffo of an ounce Second Operation. e 1 ua,s tH» of the 20 pennyweights in 1 ounce, or tfjfc 4 which equals \\ pennyweights. Therefore, y^ lb.. *&* X V X V = ff pwt. Troy> = || pwt . &* 31 "Rule,— Multiply the fraction by the units in the scale, from the given to the required denomination. REDUCTION OF DENOMINATE NUMBERS. 121 38?». To Reduce a Denominate Fraction from a Lower to a Higher Denomi- nation Example. — Reduce \ of a grain to the fraction of a pound, Troy. Operation. ,_ m , . -■ . . _ Explanation. — £ of a grain equals f of ^ of a pen- fX-g^X ^X^ = irirnr* ny weiglit ; f of £f of a pennyweight equals f of & of i * ^ of an ounce ; § of ^ of ^ of an ounce equals | of ^ of ^ of -^ of a pound, or - 14 | 0ff of a pound. * gr. = T4irn7 lb., Troy. Rule. — Divide by the units in the scale, from the given to the required * denomination. 384. To Reduce Denominate Fractions to Integers of Lower Denominations. Example. — Reduce fa of a pound, Troy, to integers of lower denominations. Operation. • _ __ Explanation. — r 3 ^ of a pound equals y 8 ^ of the 12 ounces •^X^f T z * 4 • i n a pound, or f £ ounces, which reduced gives 2£ ounces ; * £ of an ounce equals \ of the 20 pennyweights in an ounce, 4- X V 1 = 5 "Dwt or 5 pennyweights. Therefore, •& of a pound, Troy, equals 2 ounces, 5 pennyweights. fa lb., Troy, = 2 oz. 5 pwt. Rule. — Multiply the denominate fraction by the unit next. lower in the scale, and if the product be an improper fraction reduce it to a whole or mixed number. 385. To Reduce a Compound Denominate Number to a Fraction of a Higher Denomination. Example. — Reduce 7 oz. 5 pwt. 9 gr. to the fraction of a pound, Troy. Operation. First Explanation. — Since 1 ounce equals 20 pennyweights, -, R , q 7 ounces equal 140 pennyweights; 140 pennyweights plus 5 pen- " ' ° nyweights equals 145 pennyweights ; since 1 pennyweight equals ™ .24 grains 145 pennyweights equal 8480 grains; 8480 grains plus 9 140 pwt. grains equals 3489 grains; since 1 pound equals 5760 grains, 3489 5 pwt. grains equal ff f $ of a pound. 145 pwt. Second Explanation. — Since 1 ounce equals 480 grains, 7 o a ounces equal 3360 grains; since 1 pennyweight equals 24 grains, « 5 pennyweights equal 120 grains; 3360 grains, plus 120 grains, plus 3480 gr. 9 g ra i ns equal 3489 grains; since 1 pound, Troy, equals 5760 grains, 9 gr. 3489 grains equal f ||f of a pound. Therefore, 7 ounces, 5 pen- 3489 gr. = 44|4 lb. nyweights, 9 grains, equal f|f$ of * pound, Troy. Rule. — Reduce the compound denominate number to its lowest denomi- nation for a numerator, and a unit to the same denomination for a denominator; the fraction thus formed is the answer sought 122 ADDITION" OF DENOMINATE NUMBERS. 386. To Reduce a Denominate Decimal to Units of Lower Denominations. Example.— Reduce .865 of a pound, Troy, to integers of lower denominations. Operation. .865 lb. 22 Explanation. — .865 of a pound equals .865 A of the 12 ounces in 1 pound, or 10.38 ounces; 10. 38p oz. >38 of an ounce equals 3 8 of the 20 pennyweights "0 in 1 ounce, or 7.6 pennyweights; .6 of a penny- 7.60 pwt weight equals .6 of the 24 grains in 1 penny- 24 weight, or 14.4 grains. Therefore, .865 of a ■ pound, Troy, equals 10 ounces, 7 pennyweights, 1^.4 gr. 14.4 grains. .865 lb., Troy, = 10 oz. 7 pwt. 14.4 gr. 'Rule.— Multiply the decimal by that unit in the scale which will reduce it to units of the next lower denomination, and in the product point off -as in decimals. Proceed in like manner with all decimal remainders. 387 To Reduce Denominate Numbers to Decimals of a Higher Denomination. Example. — Reduce 8 oz. 3 pwt. 15 gr. to the decimal of a pound, Troy. Operation. 24 ) 15 gr. - — - & Explanation.— Since 24 grains equal 1 pennyweight, 15 • 62o grains equal ^| or . 625 of a pennyweight ; 3 pennyweights plus 3. pwt. .625 pennyweights equal 3.625 pennyweights; since 20 penny- H0 ) 3 625 pwt. weights equal 1 ounce, 3.625 pennyweights equal .18125 of an — '- ounce, and 8 ounces plus .18125 of an ounce equal 8.18125 .lol/tfD oz. ounces; since 12 ounces equal 1 pound, 8.18125 ounces equal 8. oz. .68177 T Vof a pound. Therefore, 8 oz. 3 pwt. 15 gr.=.68177i 12 ) 8.18125o z. lb > Tr °y- .68177 T Vlb. Rule. - Divide the lowest denomination given by the number in the scale next higher, and to the quotient add the integers of the next higher denomination. So continue to divide by all the successive orders of units in the scale. ADDITION OF DENOMINATE NUMBERS. 388. Example.— Find the sum of 2 lb. 5 oz. 13 pwt. 4 gr., 17 lb. 11 oz. 18 pwt. 20 gr., and 9 lb. 9 oz. 6 pwt. 15 gr. Explanation. — Since each of the given expressions is a compound number of the same class, and they all have the same varying scale, their addition may be performed the same as in simple numbers; in reducing the sum of each column from a lower to a higher order, observe the units in the ascending scale. 30 1b. 2oz. 18 pwt. 15 gr. Rule.— 1 - Write the numbers of the same unit value in the same column. II. Beginning with the lowest denomination, add as in simple numbers, •and reduce to higher denominations according to the scale. Operation, lb. oz. pwt. gr. 2 5 13 4c 17 11 18 20 9 9 6 15 389. Example. — Subl 6gr. Operation. 11. 23 02. 4 pwt. 17 gr. 6 11 7 13 9 11 lb. 9 oz. 3 pwt. 21 gr. MULTIPLICATION" OF DENOMINATE NUMBERS. 123 . SUBTRACTION OF DENOMINATE NUMBERS. -Subtract 11 lb. 7 oz. 13 pwt. 9 gr. from 23 lb. 4 oz. 17 pwt. Explanation. — Subtract as in simple numbers. If a subtrahend term be numerically greater than the cor- responding minuend term, borrow 1 from the next higher minuend term, reduce it to equivalent units in the denom- ination next lower, add them to the minuend units, and from their sum take the subtrahend units. Rule.— Write the numbers as for simple subtraction ; take each subtra- hend term from its corresponding minuend 'term for a remainder. In case any subtrahend term be greaier than the minuend term, borrow 1 as in simple subtraction, and reduce it to the denomination required. MULTIPLICATION OF DENOMINATE NUMBERS. 390. Example.— Each of five bars of silver weighed 16 lb. 3 oz. 10 pwt. 21 gr. What was the total weight? Explanation. — Multiply 21 grains by 5 and obtain 105 Operation. grains, which reduce to pennyweights by dividing by 24, lb. oz. pwt. ar. an( ^ 0Dt ain 4 pennyweights, with a remainder of 9 grains; jg o -in * oi* multiply 10 pennyweights by 5, add the 4 pennyweights, and reduce to ounces by dividing by 20, obtaining 2 ounces, 1_ 14 pennyweights; multiply 3 ounces by 5, add the 2 ounces 81 lb. 5 OZ. 14 pwt. 9 gr. and divide by 12, obtaining 1 pound, 5 ounces; multiply 16 pounds by 5, add the 1 pound and obtain 81 pounds. Bale. — Beginning with the lowest denomination, multiply each in succession, and reduce the product to higher denominations by the scale. Remarks. — 1. In order that the pupil may have all problems under each denominate subject given together, and so make an exhaustive study separately of each, it has seemed proper to include all of the reductions under a typical subject, that of Troy Weight, and hereafter, as may be needed, reference will be made to such reductions. 2. The teacher will appreciate the above change, as each subject will thus be made to include enough work for a lesson, and the confusion often arising from giving in the same lesson several tables, with varying scales, may be avoided. DIVISION OF DENOMINATE NUMBERS. 391. Example. — If 7 lb. 7 oz. 12 pwt. 18 gr. of silver be made into 6 plates of equal weight, what will be the weight of each? Operation. Explanation.— One plate will weigh \ as much as lb. oz. pwt. ar 6 plates. Write the dividend and divisor as in short divi- g\ iv » -.a io s io Q . Divide 7 pounds by 6, obtaining a quotient of 1 — pound and an undivided remainder of 1 pound; reduce 1 lb. 3 oz. 5 pwt. 11 gr. this remainder to ounces (12) and add to the 7 ounces of the dividend, obtaining 19 ounces, which divide by 6, obtaining 3 ounces and an undivided 124 COMPOUND DENOMINATE DIVISION. remainder of 1 ounce; reduce this remainder to pennyweights (20) and add to the 12 penny, weights of the dividend, obtaining 32 pennyweights, which divide by 6, obtaining 5 penny- weights and an undivided remainder of 2 pennyweights; reduce this remainder to grains (48) and add to the 18 grains of the dividend, obtaining 66 grains, which divide by 6, obtaining 11 grains, and thus completing the division. Therefore, the weight of each plate will be 1 pound, 3 ounces, 5 pennyweights, 11 grains. Rule.— Write the terms as in short division; divide as in integers and reduce remainders \ if any, to next lower orders by the scale. Remarks. — 1. Should the highest dividend order not contain the divisor, reduce its units to the order next lower, and so proceed to the end. 2. The above and like divisions may be accomplished by the reduction of the denominate expressions to the lowest order in its scale, then effecting the division and afterwards reducing the quotient to higher denominations. COMPOUND DENOMINATE DIVISION. 392. Example. — How many plates, each weighing 1 lb. 3 oz. 5 pwt. 11 gr., can be made from 7 lb. 7 oz. 12 pwt. 18 gr. of silver? Explanation. — Reduce each of the given Operation. expressions to its equivalent in grains. Since 1 lb 3 oz 5 pwt 11 gr = 7331 gr one plate weighs 7331 grains, and the weight of 7 lb. 7 oz! 12 pwt. 18 gr. = 43986 gr. tJ ! e f silver * be " sed \ 48986 T*? ™ immy , QQQ JT Q1 _ & plates can be made as the weight of one plate, 4dy«b gr. -f- 7ddl gr. = b 7331 grains> ig contained times in the 43tf86 grains to be so used, or 6 plates. Rule.— Reduce the dividend and divisor to the same denomination, and divide as in simple numbers. EXAMPLES FOR PRACTICE. 393. 1. Reduce 31 lb. 10 oz. 13 pwt. to pennyweights. 2. How many grains in 27 lb. 17 pwt. 20 gr. ? S. How many pounds, ounces, and pennyweights in 23051 gr. ? Jf. Reduce 30297 grains to higher denominations. 5. Reduce -fa of a pound to grains. 6. ^yVo °f a pound is what part of a pennyweight ? 7. /o of a grain is what fraction of an ounce? S. Reduce -fa of a pennyweight to the fraction of a pound. 9. Reduce -^ of a pound to lower denominations 10. Reduce § of an ounce to lower denominations. 11. Reduce 9 oz. 1 pwt. 21 gr. to the fraction of a pound. 12. What fraction of a pound equals 11 oz. 11 pwt. 18 gr. ? 18. What is the value in lower denominations of . 6425 lb. ? Of.. Find the equivalents in lower denominations of .905 oz. ? 15. 3 oz. 11 pwt. 12 gr. is what decimal of a pound ? 16. Reduce 17 pwt. 12 gr. to the decimal of an ounce. 17. Add 236 lb. 4 oz. 15 pwt., 83 lb. 11 oz. 21 gr., 46 lb. 16 pwt., 105 lb. 9 oz. 11 gr. AVOIRDUPOIS WEIGHT. 125 18. What is the sum of 16 lb. 16 pwt. 16 gr., 100 lb. 1 oz. 5 pwt. 20 gr., 76 lb. 7 oz. 6 pwt. 13 gr., 19 lb. 2 oz. 10 pwt. 20 gr.? 19. Find the equivalents in lower denominations of .1425 oz. 20. 1 pwt. 15 gr. is what decimal of a pound ? 21. Subtract 41 lb. 11 oz. 6 pwt. 18 gr. from 50 lb. 2 oz. 22. What is the difference between 19 lb. 9 oz. 11 pwt. and 11 oz. 16 pwt. 22 gr. ? 23. What will be the cost of 15 gold chains, each weighing 1 lb. 3 oz. 18 pwt. 18 gr., at 7 ! i CO CO i 'JO o CO .New York 00 ©* i 00 i O i ^ CO i i rH .i CO i JO 1 CO ia > jo C O CD O JO O O COJOCOJOJO"*CO©1 Minnesota 00 I i • CQ < O 00 00 -^ ii i "^ i c© ©a C4 ! CO . JO ©3 • i CO CO I ' JO i o . CO Michigan 00 i • « ©* • O 00 00 rtf i • • ^ i CO < • ■ ! o i CO Kentucky 00 © © ^ ©* ' O i "OC0O»0OO •^cO00r-lJO"«tfCD©*C0JO''*JOCOC0JOCOJOJO" • 1 TJH i CO ; i JO i © I CO ad H O § g Barley, - Beans,. Bituminous Coal, . . Blue Grass Seed,... Buckwheat, Castor Beans, f!lrwpr SppH Dried Apples, ----- » > i c 1 1 1 > ; 1 Hemp Seed, Indian Corn, Indian Corn in Ear, Oats, Onions, Potatoes, Rye, Salt. Timothy Seed, Wheat, Whp.nt, P»ran . ■ EXAMPLES FOR PRACTICE. Additional Table of Weights of Products, As usually given, but varied by the laws of different States : 127 Apples, green, 56 lb., per bushel. Charcoal, 22 lb. per bushel. Hungarian Grass Seed, 45 lb. per bushel. Malt, 381b. per bushel. Millet, 45 lb. per bushel. Mineral Coal, 80 lb. per bushel. Peas, . . -60 lb. per bushel. Potatoes, sweet, 55 lb. per bushel, Red Top Grass Seed, . .14 lb. per bushel. Turnips, , . . 56 lb. per bushel. Table of Gross Weights for Freighting'. Ale and Beer, 330 1b. per barrel. Apples, 150 lb. per barrel- Beef (200 lb. net),_330lb. per barrel. Cider, 400 lb. per barrel. Corn Meal, 200 lb. per barrel. Eggs, 180 lb. per barrel. Fish, 300 1b. per barrel. Flour ( 196 lb. net ), 200 lb. per barrel. High wines, 400 lb. per barrel. Lime, 230 lb. per barrel. Oil, 400 lb. per barrel. Pork (200 lb. net),_330 lb. per barrel Potatoes, 180 lb. per barrel. Salt, 3001b. per barrel. Vinegar, 400 lb. per barrel. Whiskey, 400 lb. per barreL Estimates on Lumber, Wood, Etc., for Freighting. Pine, Hemlock, and Poplar, seasoned, per M, 3000 lb. Black Walnut, Ash, Maple, and Cherry, per M, . . . 4000 lb. Oak and Hickory, per M, 5000 lb. Soft wood,' dry, per cord, 3000 lb. Hard wood, dry, per cord, 3500 lb. Remark. — For unseasoned lumber, add one-third. Brick, common, each, 4 lb. Brick, fire, each, 6 lb. Sand, cubic yard, 3000 lb. Gravel, cubic yard, 3200 lb. Stone, cubic yard, 4000 lb. Remark. — For assistance in the solution of the following examples, the pupil is referred td the explanations and rules under Troy Weight. 3. 4. 5. 6. 7. 8. 9. 10. EXAMPLES FOR PRACTICE. i. How many pounds Avoirdupois in 17 T. 6 cwt. 69 lb.? Reduce 31275 lb. Avoirdupois to higher denominations, f of a ton Avoirdupois equals how many pounds ? Reduce -^ of a cwt. Avoirdupois to ounces. Reduce .3842 of a ton Avoirdupois to lower denominations. How many Avoirdupois pounds in . 625 of a ton ? 17 cwt. 72 lb. 4 oz. Avoirdupois is what fraction of a ton ? Reduce 51 lb. 12 oz. Avoirdupois to the fraction of a hundred- weight. What decimal part of a hundred- weight is 24 lb. 2 oz. Avoirdupois ? Reduce 19 cwt. 99 lb. 15 oz. Avoirdupois to the decimal of a ton. 128 apothecaries' weight. 11. What is the sum of 7 T. 4 cwt. 78 lb. 5 oz., 3 T. 17 cwt. 19 lb. 11 oz., 5 T. 18 cwt. 96 lb., 13 T. 1 cwt. 11 oz. ? 12. A farmer sold 4 loads of hay, weighing respectively 1 T. 2 cwt. 14 lb., 19 cwt. 90 lb., 1 T. 3 cwt. 97 lb., 1 T. 5 cwt., and received for it $16 per ton. How much did he receive? 13. Six loads of lime weighed 13 T. 15 cwt. 4 lb. What was their average weight? APOTHECARIES' WEIGHT. 396. Apothecaries 5 Weight is used by druggists in retailing, and by apothecaries in mixing medicines. Table. 20 grains = 1 scruple sc. 3 scruples = 1 dram dr. 8 drams == 1 ounce oz. 12 ounces == 1 pound lb. Scale, descending, 12, 8, 3, 20 ; ascending, 20, 3, 8, 12. Remarks. — 1. The pound, ounce, and grain are the same as in Troy weight. The only difference between these weights is in the subdivisions of the ounce. 2. Drugs and medicines are sold at wholesale by Avoirdupois weight. EXAMPLES FOR PRACTICE. Remark. — For assistance, refer to rules and explanations under Troy Weight. 397. 1. Reduce 5128 sc. to higher denominations. 2. How many drams in 61 lb. 5 oz. ? 3. 10 oz. 1 dr. 1 sc. 15 gr. equal what fraction of a pound ? 4. Reduce .955 of a pound to lower denominations. 5. How many scruples in -fa of a pound ? 6. Add 6^5- lb., 7 A oz -> 3 f dr - and 2 i sc - 7. Find the sum of ^ lb., 7 oz., 7 dr., 1 sc. and 16 gr. 8. From 21 lb. 5 oz. 3 dr. 1 sc. 11 gr., take 14 lb. 1 oz. 7 dr. 19 gr. 9. What is the difference between 16 lb. 1 oz. 4 dr. 2 sc. 12 gr., and 9^ lb.? 10. In compounding six cases of medicine, an apothecary used for each 2 lb. 7 oz. 6 dr. 18 gr. What was the aggregate weight ? 11. If 19 lb. 4 oz. 7 dr. 1 sc. 5 gr. be divided into 21 packages of equal weight, what will be the weight of each ? Comparative Table of Weights. Troy. Apothecaries.' Avoirdupois. 1 pound = 5760 grains = 5760 grains = 7000 grains. 1 ounce '= 480 grains =3 480 grains = 437.5 grains. 175 pounds = 175 pounds = 144 pounds. Questions. — 1. Which is heavier, a pound Troy or a pound Avoirdupois ? 2. Which is heavier, an ounce Troy or an ounce Avoirdupois ? MEASURES OF CAPACITY. 129 Remarks. — 1. A cubic foot of water weighs 62£ lb. or 1000 oz. Avoirdupois. 2. In weighing diamonds and gems, the unit generally employed is the carat, which is about 3.2 Troy grains. 3. The term carat is also used to express the fineness of gold, 24 carats fine being pure; thus 18 carat gold = f pure EXAMPLES FOR PRACTICE. 398. 1. A dealer bought 131 lb. 5 oz. of drugs by Avoirdupois weight, at $G.25 per pound, and retailed them at 5, how much less did he receive than if he had measured by dry measure when buying and by liquid measure when selling ? MEASURES OF EXTENSION. 404. Extension is that which has one or more of the dimensions, length, breadth, and thickness; it may therefore be a line, a surface, or a solid. 405. A Line has only one dimension — length. Remarks.— 1. The United States Standard of linear, surface, and solid measure, is the yard of 3 feet, or 36 inches. 2. The standard, prescribed at "Washington, has been fixed with the greatest precision. It was determined by a brass rod, or pendulum, which vibrates seconds'va. a vacuum at the sea level, in the latitude of London, Eng., and in a temperature of 62° Fahrenheit. This pendulum is divided into 391393 equal parts, and 360000 of these parts constitute a yard. 406. A Surface or Area has two dimensions — length and breadth. 407. A Solid has three dimensions — length, breadth, and thickness. LINEAR MEASURE. 408. Linear or Long Measure is used in measuring lengths and distances. Table. 12 inches (in.) = 1 foot ft. 3 feet =1 yard yd. 5£ yards, or 16| feet = 1 rod rd. 320 rods = 1 statute mile mi. Scale, descending, 320, 5| 3, 12 ; ascending, 12, 3, 5i, 320. 1 Mile = 320 rods, or 5280 feet, or 63360 inches. 132 SQUAKE MEASUKE. Special Table. 3 geographic miles = 1 League, used f ol measuring distances at sea. 60 geographic miles or 69.16 statute miles = 1 Degree of Latitude on a meridian, or Longitude on the equa- tor. 360 degrees = Equatorial circumfer- ence of the earth. 1 geographic mile = 1 Knot, used to determine the speed of vessels. -J of an inch = 1 Size, applied to shoes. 18 inches = 1 Cubit. 3.3 feet = 1 Pace. 5 paces = 1 Eod, 4 inches == 1 Hand, used to measure the height of animals. 6 feet = 1 Fathom, used to measure depths at sea. 1.152§ statute miles = 1 Geographic or Nautical mile, used for measuring distances at sea. Remarks.— 1. In civil engineering, and at the Custom Houses, the foot and inch are divided into tenths, hundredths, and thousandths. 2. The yard is divided into halves, quarters, eighths, and sixteenths, for measuring goods sold by the yard. 3. The furlong of 40 rods is little used. 4. Degrees are of unequal length; those of latitude varying from 68.72 miles at the Equator to 69.34 miles in the polar regions. The average length, 69.16 miles, is the standard adopted by the United States Coast Survey. 5. A degree of longitude is 69.16 statute miles at the equator, but decreases gradually toward the poles, where it is 0. EXAMPLES FOR PRACTICE. Remark. — For assistance refer to Rules under Troy Weight. 409. 1. Eeduce 2 mi. 1 rd. 7 ft. to inches. 2. Reduce 2501877 inches to higher denominations. 8. What part of a mile is T 6 ¥ of a foot ? Jf. Reduce f of a mile to integers of lower denominations. 5. What fraction of a rod is 11 ft. 2 in. ? 6. Reduce .542 of a mile to integers of lower denominations. 7. Reduce 285 rd. 7 ft. 4 in. to the decimal of a mile. 8. A wheelman ran 71 mi. 246 rd. 1 yd. 2 ft. 6 in. in the forenoon, and 20 mi. 10 rd. 8 in. less in the afternoon. What distance did he run in the entire day ? 9. If a yacht makes an average of 227 mi. 227 rd. 2 yd. 2 ft. 2 in. per day, for the seven days of a week, what distance will be passed ? 10. If the Servia steams 2905 mi. in six days, what is her average rate per day? SQUARE MEASURE. 410. Square Measure is used for computing the surface of land, floors, boards, walls, roofs, etc. 411. The Area of a figure is the quantity of surface it contains. 412. An Angle is the difference in the direc- tion of two lines proceeding from a common point Angle. called the vertex. SQUARE MEASURE. 133 Two Right Angles. 413. A Right ,4ngle is the angle formed when one straight line meets another so as to make the adjacent angles equal. The lines form- ing the angles are said to be perpe?idicular to each other. E and F are right angles, and the lines A B and D are perpendicular to each other. Rectangle. inches. a Contents or c Area. Six b square inches. d 3X2 in. =6sq. iu. 3 feet. i|. ONE SQUARE ! FOOT jONE 414. A Rectangle is a plane or flat surface, having four straight sides and four square corners, or four right angles. 415. The Contents or Area of any surface having a uniform length and a uniform breadth is found by multiplying the length by the breadth. In the accompanying diagram, in which the angles (a, b, c, d), are all right angles, and the corners all square corners, the area is 6 square inches, and is found by multiplying 2 inches by 3 inches. SQUARE YARD ^"'VJ L 416. A Square is a figure bounded by four equal lines, and having four right angles. Remark. — A square inch is a square, each side of which is 1 inch. A square foot is a square, each side of which is 1 foot. A square yard is a square, each side of which is 1 yard. 3 X 3 ft. = 9 sq. ft. = 1 sq. yd. Table of Square Measure. 144 square inches (sq. in.) = 1 square foot sq. ft. 9 square feet = 1 square yard sq. yd. 30i square yards, or ) _ - , , 272£ square feet ____ ( l square rod sq. rd. 160 square rods _ _ = 1 acre A. 640 acres == 1 square mile sq. mi. 36 square miles (6 miles square), = 1 township Tp. Scale, descending, 36, 640, 160, 30£, 9, 144; ascending, 144, 9, 30J, 160, 640, 36. Remark. — All the units of square measure, except the acre, are derived by squaring the corresponding units of linear measure; as, a square foot is a surface one foot square; a square rod is a surface 1 rod or 16£ feet square; a square mile is a surface 1 mile or 320 rods square. 134 SQUARE MEASURE. 417. The Unit of Land Measure is the acre, equal to 208.71ft x 208.71ft.- Remarks. — 1, In sections of the United States where the original grants were from France; the arpent, a French unit of surface, equal to about | of an acre, is still sometimes used. 2. The Rood, equal to 40 square rods, is but little used 418. Dimension stuff is sold by board measure. 419. The Unit of Board Measure is a square foot surface, one inch thick, called a board foot. 420. To Find the Number of Board Feet in a Board. Rule —Multiply the length in feet by the width in inches, and divide by 12; the quotient will be the number of square feet. Remark. — If the*board tapers evenly, find the mean or average width, by adding the width of the two ends, and dividing by 2. 421. To Find the Number of Board Feet in Timbers or Planks. Rule.— Multiply the length in feet by the product of the width and thickness in inches, and divide by 12. 422. To Find the Number of Squares in a Floor or Roof. Remakk. — In flooring, roofing, slating, etc., the square, or 100 square feet, is used as a unit of measure. Rule,— Point off two decimal places from the right of the number of surface feet- 423. To Find the Number of Yards of Carpeting that Would be Required to Cover a Floor. Rule.— I. Divide one of the dimensions of the floor by 3, add the wastage, if any, and the result will be the length, in yards, of 1 strip of the carpet. IL Divide the other dimension by the width of the carpet, and the quotient will be the number of strips it will take to cover the floor. HI. Multiply the length of each strip by the number of strips, and the product will be the number of yards required. Remakk. — In carpeting and papering, it is usually necessary to allow for certain waste in matching the figures of patterns, and often carpets may be laid with less waste one way of the room than the other. Dealers charge for all goods furnished, regardless of the waste. EXAMPLES FOR PRACTICE. Remark. — For assistance refer to rules under Troy Weight. 424. 1. Reduce 5 A. 110 sq. rd. 7 sq. ft. to square inches. 2. Reduce 4 sq. mi. 527 A. 105| sq. rd. to square feet. 3. Reduce .1754 of a square mile to lower denominations. SQUARE MEASURE. 135 4. Reduce fj of an acre to lower denominations. 5. What fraction of a square mile is T \ of a square foot? 6. What decimal part of an acre is 150 sq. rd. 3 sq. yd. 7 sq. ft. 100 sq. in. ? 7. From .6375 of an acre take \\ of a square rod. 8. To the sum of \, §, and -fa of an acre, add .0055 of a square mile. 9. How many squares in a roof, each side of which is 26 X CO feet? 10. How many yards of carpet, 1 yard wide, will be required to cover a floor 10.5 yd. long by 6 yd. wide, if no allowance be made for matching ? 11. How many feet in 8 boards, each 15 ft. long, 9 in. wide, and 1 in. thick? 12. How many feet in 15 boards, each 16 ft. long and 1 in. thick, the boards being 13 in. wide at one end and 10 in. at the other? IS. How many acres in a square field, each side of which is 64 rods in length? 14-. 'What will be the cost of a tract of land 508 rd. long and 1350 rd. wide, at $25 per acre? 15. A field 87-J- rd. wide and 240 rd. long, produced 27} bu. of wheat to the acre. What was the crop worth, at 90^ per bushel? 16. A farm in the form of a rectangle is 75 rd. wide; if the area is 167.5 A., how long is the farm? 17. I wish to build a shed which will cover f of an acre of land. If the width of the shed is 42 ft., what must be its length? 18. 17.75 bu. of timothy seed is sown on land, at the rate of 6 lb. per acre. What is the area thus seeded? 19. What is the difference between a square rod and a rod square? 20. What is the difference between two square rods and two rods square? 21. A square yard will make how many surfaces 5 in. by 9 in. ? 22. How many acres of flooring in a six-story block 160 ft. by 220 ft. ? 23. A rectangular field containing 16^ A. is 45 rd. wide. What is its length? 24. How many fields, each of 10 A. 56 sq. rd. 21 sq. yd. 5 sq. ft. and 28 sq. in., can be formed from a farm containing 124 A. 40 sq. rd. 16 sq. yd. 8 sq. ft. 48 sq. in.? 25. How many acres in a road 17206 ft. long and 66 ft. wide? 26. What will be the cost, at $3.50 per M, of the shingles for a roof 26 ft. wide and 110 ft. long, if the shingles are 6 in. wide and 4 inches of their length be exposed to the weather? 27. A hall 7-J ft. wide and 19J ft. long is covered with oil cloth, at 65j£ per sq. yd. How much did it cost? 28. If a farm of 106 A. 94J sq. rd. is divided equally into 11 fields, what will be the area of each field ? 29. Reduce 240689740 sq. in. to higher denominations. 80. How many rods of fence will enclose 1 60 A. of land lying in the form of a square ? 81. How many additional rods will divide the farm into four fields of equal area ? 136 SQUARE MEASURE. 32. *IIow many yards of brussels carpeting, f of a yard wide, laid length- wise of the room, will be required to cover a room 22 ft. by 17 ft. 4 in., if the waste in matching be 6 in. on each strip ? Remark. — When the width of the room is not exactly divisible by the width of the carpet, drop the fraction in the quotient and add 1 to the whole number. The waste in such cases is either cut off or turned under in laying. 33. What will it cost, at %1$ per sq. yd., to plaster the sides and ceiling of a room 24 ft. by 31£ ft. and 10-j- ft. high, if one-sixth of the surface of the sides is taken up by doors and windows ? 3 If. A street 4975 ft. long and 40 ft. wide was paved with Trinidad asphaltum, at $2.65 per square yard. What was the cost ? 35. A skating rink, 204 ft, by 196| ft., was floored with 2 in. plank, at $23.50 per M. What was the cost of the lumber ? 36. What will be the cost of the carpet border for a room 16| ft. by 21 ft., if the price be 62-J-^ per yard? 37. How many single rolls of paper, 8 yd. long and 18 in. wide, will it take to cover the ceiling of a room 56 ft. long and 27 ft. 4* in. wide, if there be no waste in matching ? Remark. — When no allowance is made for waste in matching, divide the surface to be papered by the number of square feet in one roll of the paper. 38. How many yards of carpeting, f of a yard wide, will be required to carpet a room 32 ft. long and 25 ft. wide, if the lengths of carpet are laid crosswise of the room, and 8 inches is lost on each length in matching the pattern ? How many yards if the lengths are laid lengthwise and 6 in. is lost in matching ? If the carpet is laid in the most economical way, what will be the cost, at $2.55 pei yard ? 39. How many sheets of tin, 20 in. by 14 in., will be required to cover a roof 60.5 ft. wide and 15G.25 ft. long ? Jfi. What is the difference between four square feet and four feet square ? J/.1. What will it cost, at $1.15 per yard, to carpet a flight of stairs 11 ft, 4 in. high, the tread of each stair being 10 in. and the riser 8 in.? J$. How many shingles, averaging 4 in. wide and laid 5 in. to the weather, will cover the roof of a barn, one side of the roof being 24 ft. wide and the other 42 ft. wide, the length of the barn being 60 ft. ? ]$. Divide an acre of land into 8 equal sized lots, the length of each of which shall be twice its frontage. What will be the dimensions of each lot ? 44-* How many granite blocks, 12 in. by 18 in., will be required to pave a mile of roadway 42 ft. in width ? Jfi. What will be the cost, at 20^ per sq. yd., for plastering the ceiling and walls of a room 22 ft. wide, 65 ft. long, and 15 ft. high, allowance being made for 8 doors 4 ft. 6 in. wide by 11 ft. 6 in. high, and 10 windows each 42 in. wide by 8 ft. high 3 Ji-6. I wish to floor and ceil a room 27J yd. long and 15 yd. 2 ft. wide, with matched pine. What will be the cost of the material, at $26.40 per M ? SQUARE ROOT. 137 INVOLUTION. 425. A Power of a number is the product arising from multiplying a number by itself, or repeating it several times as a factor. 426. A Perfect Power is a number that can be exactly produced by the involution of some number as a root; thus, 64 and 16 are perfect powers, because 8x8 =64, and 2 X 2 X 2 x 2 = 16. 427. The Square of a number is its second power, 428. The Cube of a number is its third power. 429. Involution is the process of finding any power of a number; and a number is said to be involved or raised to a power, when any power of it is found. EXAMPLES FOR PRACTICE. 430. 1. What is the square of 1 ? 2. What is the square of 3 ? 3. What is the square of 4 ? 4. What is the square of 5 ? 5. What is the square of 9 ? 6. What is the square of 10 ? ?. What is the square of 99 ? 8. What is the square of 256 ? Remark. — From the solution of the above examples the pupil will observe: 1st. That the square of any number expressed by one figure cannot contain less than 1 nor more than 2 places. 2d. That the addition of 1 place to any number will add 2 places to its square- EVOLUTION. 431. Evolution is the process of extracting the root of a number considered as a power. It is the reverse of Involution, and each may be proved by the other. 432. A Root of a number is one of the equal factors which, multiplied together, will produce the given number; as, 4 x 4 x 4 = 64; 4 is the root from which the number 64 is produced. SQUARE ROOT. 433. The Square Root of a Number is such a number as, multiplied by itself, will produce the required number. 434. The operation of finding one of the two equal factors of a square, or product, is called extracting the square root. Remark. — The square root of any number, then, is one of its two equal factors, the given number being considered a product. 435. In practical operations, a surface and one of its dimensions being given, the wanting dimension is found by dividing the surface by the given dimension . 138 SQUARE ROOT. The accompanying diagram is a square 14 feet by 14 feet. Its square feet, or area, is by inspection found to be made up of: 1st. The tens of 14, the number representing the length of one side, or 10 squared = 100 square feet, as shown by the square within the angles a, b, c, d. 2d. Two times the product of the tens by the units of the same number, or 2 (10 X 4) = 80 square feet, as shown by the surfaces within the angles e, f, g y h, and i, j, k, I. 3d. The square of the units, 4 feet, or the product of 4 ft. by 4 ft. rs 16 square feet, as shown by the square within the angles w, x, y, z. Hence, a square 14 feet on each side will contain 10 x 10 = 100 square feet. 2 (10 X 4) = 80 square feet. 4x4 =16 square feet. 196 square feet. Or, the square of 14 is made up of or equals the square of 10, plus twice the product of 10 by 4, plus the square 4, the number to be squared. 436. General Principles. — The square of any number composed of two or more figures is equal to the square of the tens, plus twice the product of .the tens multiplied by the units, plus the square of the units. a d e Fi b c f G I I N X J K X Y 437. Units and Squares Compared. tJivrrs. Squares. Remark. — Squaring the numbers from 1 to 10 inclusive, shows: V = 2 a = 3 a = 4* = 5* = 1 4 9 16 25 36 49 r = 8* = 64 9* ss 81 10* S3 100 1st. That the square of any number will contain at least one place, or one order of units. 2d. That the square of no number represented by a single figure will contain more than two places. If the number of which the square root is sought be separated into periods of two figures each, beginning at the right, the number of periods and partial periods so made will represent the number of unit orders in the root. Therefore, the square of any num- ber will contain twice as many places, or one less than twice as many, as its root. 3d. Where the product of the left hand figure multiplied by itself is not greater than 9, then the square will contain one less than twice as many places as the root. 438. Operation. t u 6.25 )2 5 4= 400 Example. — Find the square root of 625. Explanation. — The number consists of one full and one partial period; hence its root will contain £200 places — tens and units. The given number, 625, must be the product of the root to be extracted multiplied by itself; therefore, the first figure of the root, which will be the highest order of units in that root, must be obtained from the first left hand period, or highest order of units in the given number. Hence, the first or tens figure of the root will be the square root of the greatest perfect square in 6. 6 coming between 4, the square of 3, and 9, the square of 3, its root must be 2 tens with a remainder. Subtracting »0 X 2 = 40 5 45 225 225 rem. rem. SQUARE ROOT. 139 from the 6 hundreds or 6, the square of 2 (tens) = 400 or 4, gives 225 as a remainder. Having now taken away the square of the tens, the remainder, 225, must be equal to 2 times the tens multiplied by the square of the units, plus the square of the units. Since the tens are 2 or 20, twice the tens = 40. Observe, therefore, that 225 must equal 40 times the units of the root, together with the square of such units. If, then, 225 be divided by 40, the quotient, 5, will nearly, if not exactly represent the units of the root sought. Using 40, then, as a trial divisor, the second, or unit figure of the root is found to be 5. The term, twice the tens multiplied by the units, is equal to 2 (20 X 5), or 200, and the units, or 5, squared = 25; the sum of these wanting terms, or 225, is the remainder, or what is left after taking from the power the square of the first figure of the root. Therefore, the square root of 625 is 25. Rule. — I. Beginning at the right, separate the given number into periods of two places each. II. Take the square root of the greatest perfect square contained in the left hand period for the first root figure; subtract its square from the left hand period, and to the remainder bring down the next period. III. Divide the number thus obtained, exclusive of its units, by twice the root figure already found for a second quotient, or root figure; place this figure at the right of the root figure before found, and also at the right of the divisor; multiply the divisor thus formed by the new root figure, subtract the result from the dividend, and to the remainder bHng down the next period, and so proceed till the last period has been brought down, considering the entire root already found as so many tens, in determinijxg subsequent root figures- Remarks. — 1. Whenever a divisor is greater. than the dividend, place a cipher in the root and also at the right of the divisor; bring down another period and proceed as before. 2. When the root of a mixed decimal is required, form the periods from the decimal point right and left, and if necessary supply a decimal cipher to make the decimal periods of two places each. 3. A root may be carried to any number of decimal places by the use of decimal periods. 4. Any root of a common fraction may be obtained by extracting the root of the numerator for a numerator of the root, and the root of the denominator for the denominator of the root. 5. To find a root, decimally expressed, of any common fraction, reduce such common frac- tion to a decimal, and extract the root to any number of places. EXAMPLES FOK PRACTICE, 439. 1. Find the square root of 196. 2. Find the square root of 225. 3. Find the square root of 144. 4. Find the square root of 576. 5. Find the square root of 1225. 6. Find the square root of 5625. 7. Find the square root of 42436. 8. Find the square root of 125.44. 9. Find the square root of 50.2681. 10. Find the square root of 482, carried to three decimal places. 11. Find the square root of 25.8, carried to two decimal places. 140 SQUARE ROOT. 12. Find the square root of 106.413, carried to four decimal places. IS. What is the square root of -fo ? 14. What is the square root of f -f- ? 15. What is the square root, decimally expressed, of -f|, carried to three decimal places ? 16. What is the square root, decimally expressed, of £jf-J, carried to two decimal places? 17. What is the square root of 30368921, carried to one decimal place. 18. What is the square root of 4698920043, carried to two decimal places. 440. A Triangle is a plane figure having three sides and three angles. 441. The Base is the side on which the triangle stands; as, a, c. 442. The Perpendicular is the side forming a right angle with the base; as, a, b, in fig. S. 443. The Hypothenuse is the side opposite the TRIANGLE. ^j. angle; ^ ^ ^ fo fig> g # Fig. T. is a triangle, having angles at a, b, c. Fig. a n 444. A Bight-angled Triangle is a triangle havin b a right angle. Fig. S is a right-angled triangle, the angle at b being a right angle. The line a b is the Perpendicular; the line b c is the Base; the line ac is the Hypothenuse. Remark. — It is a geometrical conclusion that the square formed on the hypothenuse is equal to the sum of the squares formed on the base and b u * c the perpendicular RIGHT-ANGLED TRIANGLE. 445. To find the hypothenuse, when the base and perpendicular are given. Rule. — To the square of the base add the square of the perpendicular, and extract the square root of their sum. To find the base, when the hypothenuse and perpendicular are given. Kule. — From the square of the hypothenuse take the square of the perpen- dicular, and extract the square root of the remainder. To find the perpendicular, when the hypothenuse and base are given. Rule. — Take the square of the base from the square of the hypothenuse, and extract the square root of the remainder. EXAMPLES FOR PRACTICE. 446. 1. The base of a figure is 6 ft. and the perpendicular 8 ft. Find the hypothenuse. 2. The perpendicular is 17.5 ft. and the base is 46.6 ft. Find the hypoth- enuse to three decimal places. SURVEYOR'S LONG MEASURE. 141 8. The hypothenuse is 110 ft. and the base is 19.5 ft. 51nd the perpendic- ular to two decimal places. 4. The hypothenuse is 86 ft. and the base is equal to the perpendicular. Find both of the wanting terms to two decimal places. 5. The hypothenuse is 127 ft. and the base is equal to - 1 of the perpendicular. Find wanting terms to three decimal places. Remarks. — 1. Observe, in example 4, that the square root of i the square of the hypothenuse is equal to the base; and in example 5, that the square root of \ of the square of the hypothenuse is equal to the base. 2. Carry all roots to two decimal places. 6. What is the length of one side of a square field, the area of which is one acre? 7. How many feet of fence will enclose a square field containing five acres? 8 I wish to lay out ten acres in the form of a square. "What must be its frontage in feet and inches? 9. What is the distance from the top of a perpendicular flag-staff 105 ft. high to a point 4 rods from the base and on a level with it? 10. What is the width of a street in which a ladder 60 ft. long can be so placed that it will reach the eaves of a building 40 ft. high on one side of the street, and of another building 50 ft. high on the opposite side of the street? 11. What length of line will reach from the lower corner to the opposite upper corner of a room 64 ft. long, 27 ft. wide, and 2 1 ft. high ? 12. If a farm be one mile square, how far is it diagonally across from corner to corner? Find the answer in rods, feet, and inches. 13. How many rods of fence will enclose a square field containing 20 acres? 14-. A farm of 180 acres is in the form of a rectangle, the length of which is twice its width. How many rods of fence will enclose it? 15. What will be the base line of a farm of 136 A. 40 sq. rd. if it is in the form of a right-angled triangle, with the base equal to the perpendicular? SURVEYOR'S LONG MEASURE. 447. The Unit of measure used by land surveyors is Gunter's Chain, 4 rods, or 66 feet, in length, and consisting of 100 links. Remakk. — Rods are seldom used in Surveyor's Measure, it being customary to give distances in chains and links or hundreths. Table. 7.92 inches ssllmk' .... 1. 25 links =1 rod rd. 4 rods, or 66 feet _ _ . = 1 chain _ . ch. 80 chains, or 320 rods = 1 mile mi. Scale, descending, 80, 4, 25, 7.92; ascending, 7.92, 25, 4, 80. 142 SURVEYOR'S square measure. EXAMPLES FOR PRACTICE. 448. t Reduce 3 mi. 27 ch. 19 1. 4 in. to inches. #. Reduce 14841 1. to higher denominations. 8. Reduce || of a chain to lower denominations. 4. Reduce .953 of a mile to links. 5. A lot having a frontage of 4 rods contains \ of an acre. What is its depth in chains, links, and inches? 6. A field 37 ch. 42 1. long, and 30 ch. 21 1. wide, will require how many feet of fence to enclose it? 7. How many rods of fence wire will enclose a farm 21 ch. 50 1. long and 18 ch. 60 1. wide, if the fence be made 6 wires high ? 8. A garden is 307f feet long and 250-^- feet wide. What is the girt, in chains, links, and inches, of a wall surrounding it ? 9. An errand boy goes from his starting point east 33 ch. 50 1. 3 in., thence north 14 ch. 90 1. 2 in., and returns. How many full steps of 2 feet 4 inches did he take, and what was the remaining distance in inches ? SURVEYORS SQUARE MEASURE. 449. The Unit of land measure is the acre. N. 6 miles. Table. 625 square links (sq. L) =s 1 square rod sq. rd. 16 square rods =1 square chain. . sq. ch. 10 square chains, or) =lacre ._._ A . 160 square rods j 640 acres =1 square mile _ . - sq. mi. Remakk. — In surveying United States lands, a selected North and South line is surveyed as a Principal Meridian, and an East and West line, intersecting this, is surveyed as a Base Line. From these, other lines are run at right angles, six miles apart, which divide the territory into Townships six miles square. The surf aoe of the earth being convex, these merid- ians converge slightly. The townships and sections are, therefore, not perfectly rectangular; thus is cre- ated the necessity for occasional offsets called Cor- rection Lines. Each township (Tp.) is divided into 36 equal squares of 1 square mile each, as shown in the first diagram. These squares are called sections (Sec), and are divided into halves and quarters; each quarter-section, 160 acres, is in turn divided into halves, or lots of 80 acres, and quarter or half lots of 40 acres each, as shown in the second diagram. The row of townships running north and south is called a Mange; the townships in each range are numbered north and south from the base line, and Township. the ranges numbered east and west from the principal * 6 ' 5 4 3 2 1 rr 8 9 10 11 12 18 17 16 15 14 13 19 20 21 22 23 24 30 29 28 27 26 25 31 32 33 34 35 36 CUBIC MEASURE. 143 & N. 1 Mile. N. % Section. 320 Acres. S. W. X Sec. 160 A. W. S. Sec. &8 80 A. N. E. H of S. E. 40 A S. E. H of S. E. A. S. Section. meridian. The numbering of the sections in every township is as in the township diagram given, and the corners of all quarter-sections are permanently marked by monuments of stone or wood, and a description of each monument and its location (sur- roundings) made in the field notes of the surveyor. The advantages of the United States survey over all others are. 1st, its official character and uni- formity; and 2d, its simplicity. Any one having a sectional map of the United States may place a pencil point upon any described land, thus knowing abso- lutely its exact location. For example, Sec. 26, Tp. 24, N. of Range 8, E. of the 5th Principal Meridian, describes a section in the 24th tier of townships north of the base line, and 8th range east of the fifth principal meridian. EXAMPLES FOR PRACTICE. 450. -Z. Make a diagram of a township, and locate S. \ of Sec. 21, and mark its acreage. 2, Make a diagram of a township, and locate S. E. £ of Sec. 16, and mark its acreage. S. Make a diagram of a township, and locate N. W. £ of S. W. i of Sec. 12, and mark its acreage. 4. Make a diagram of a township, and locate Sees. 35, 26, and E. | of 27, and mark their acreage. 1 Ft. < A room 28 ft. long, 18 ft. 'wide, and 12 ft. high, will store how many cords of wood ? 15. How many cords of wood in a pile 108 ft. long, 7 ft. 9 in. high, and 6 ft. wide ? 16. ' From a pile of wood 71 ft. 6 in. long, 9 ft. 4 in. wide, and 6 ft. 8 in. high, 21f cords were sold. What was the length of the pile remaining ? 17. At $4. 75 per cord, what will it cost to fill with wood a shed 34 ft. long, 18 ft. wide, and 10 ft. high ? 18. What is the weight of a block of granite 11 ft. 3 in. long, 3 ft. 6 in. thick, and 8 ft. 4 in. wide, if it weighs 166 lb. per cubic foot ? 19. What is the weight of a white oak timber 15 in. square and 40 ft. long, if the weight per cubic foot be 72.5 lb.? 20. How many cubes 1 in. on each edge can be cut from a cubic yard of wood, if no allowance be made for waste by sawing ? 21. Find the contents of a cube, each edge of which is 2 yd. 7-J in. 22. How many perches of masonry in a wall 7-J ft. high and 2 ft. thick, enclosing a yard 12J rods long and 9| rods wide ? How many bricks will be required, and if bricks cost $6.50 per M and laying them cost $1.60 per M, what will be the cost of the wall ? 10 146 producers' and dealers' approximate rules. 23. What is the volume of a rectangular solid 11 ft. long, 4-J- ft. wide, and 4 ft. high ? 24. A cask holding 25 6^ gal. of water will hold how many bushels of wheat? PRODUCERS' AND DEALERS' APPROXIMATE RULES. 463, To find the contents of a bin or elevator in bushels, stricken measure. Rule. — Multiply the cubic feet by .8, and add 1 bushel for each 300, or in that 'proportion. To find the contents of a bin or crib in bushels, by ^heaped measure. Rule. — Multiply the cubic feet by .63. Remark. — If the crib flare, take the mean width. To find the number of shelled bushels in a space occupied by unshelled corn. Rule.; — Divide the cubic inches by 384-0, or multiply the cubic feet by .Jfi. To find the dimensions of a bin to hold a certain number of bushels. Rule. — To the number of bushels add one-fourth of itself, and the sum will be the cubic feet required, to within one three-hundredth part. To find the exact number of stricken bushels in a bin. Rule. — Divide the cubic inches by 2150.42. To find the exact number of heaped bushels in a bin. Rule. — Divide the cubic inches by 2747. 71. To find the capacity of circular tanks, cisterns, etc. Rule. — The square of the diameter, multiplied by the depth in feet, will give the number of cylindrical feet. Multiply by 5% for gallons, or multiply by .1865 for barrels. Remark. — In tanks or casks having bilge, find the mean diameter by taking one-half of the sum of the diameters at the head and bilge. To find the number of perches of masonry in a wall, of 24f cubic feet in a perch. Rule. — Multiply the cubic feet by .0404. To find the number of perches of masonry in a wall, of 16£ cubic feet in a perch. Rule. — Multiply the cubic feet by .0606. Remark. — The above is correct within y^ part. In large contracts add ^ of 1%. Example. — How many perches, of 24f cu. ft. each, in a wall 150 ft. long, 50 ft. high, and 2 ft. thick? Explanation.— Short Method.— -150 X 50 X 2 = 15000; 15000 x .0404 = 606; add y^, or .606 = 606.606. Extended Method.— 150 X 50 X 2 = 15000; 15000 -*- 24.75 = 606.6, same as before. Same example, perch of 16£ cu. ft. Explanation.— Short Method.— 150 X 50 X 2 = 15000; 15000 X .0606 = 909; add jfa, = .9; 909 + .9 = 909.9. To find the number of cubic feet in a log. Rule. — Divide the average diameter in inches by 3, square the quotient, multiply by the length of the log in feet, and divide by 36. CUBE ROOT. 147 To find the number of feet, board measure, in a log. Rule. — Multiply the cubic feet, as above obtained, by 9. HAY MEASUREMENTS. 464. Few products are so difficult of accurate measurement as hay, owing to the pressure, or the want of it, in packing, time of settling, volume in bulk, and freedom from obstruction in packing. Plainly, the larger (higher) the stack, or mow, and the greater the foreign weight in compress, the more compact it will be. 465. The accepted measurements are of three kinds: 1st. To find the weight of hay in a load or shed loft, unpressed. Rule. — Allow S4.0 cubic feet for a ton. 2d. To find the weight in common hay barn, or small (low) stack. Rule.— Allow Jf05 cubic feet for a ton. 3d. To find the weight in mow bases in barns, compressed with gram, and in butts of large stacks of timothy hay. Rule. — Allow 82^ cubic feet for a ton, CUBE ROOT. 466. The Cube or Third Power of a number, is the product of three equal factors. 467. The Cube Root of a number is one of the three equal factors the 'product of which represents the cube. Thus, a cubic foot = 12 X 1$ X 12, or 1728 cubic inches, the product of its length, breadth, and thickness; and since 12 is one of the three equal factors of 1728, it must be its cube root. 468. The operation of finding one of the equal factors of a cube is called extracting the cube root. 469. As shown in the explanation of extracting the -square root, the first point to be settled in extracting any root is the relative number of unit orders or places in the number and its root. 470. Units and Cubes Compared. Remark. — From this comparison may be inferred the following: 1st. The cube of any number expressed by a single figure cannot have less than one nor more than three places or unit orders. 2d. Each place added to a number will add three places to its cube. 3d. If a number be separated into periods of three figures each, begin- ning at the right hand, the number of places in the root will equal the number of periods and partial periods if there are any. 471. To help in understanding the cube root, first form a cube and thus ascertain its component parts or elements. Take 57 as the number to be cubed. NITS Cubes l 3 = 1 2 s ;= 8 3 s = 27 4 3 = 64 5 3 = 125 6 3 = 216 7 3 = 343 . 8 3 = 512 9 3 — 729 10 s — 1000 18 CUBE BOOT. Opekation 50 + 7 50 + 7 (50 X 7) + 7 2 50 2 -+- (50 X 7) 50 a + 2 (50 X 7) + 7 2 50 + 7 (50 2 X 7) + 2 (50 X 7 2 ) + 7 3 50 3 + 2 (50* X7) + (50x r) Explanation. —Cubing 57, we have 57 X 57 X 57 = 185193; or, separating 57 into its tens and units gives o tens or 50 + 7 units; or, 50 + 7. Cube the given number, by using it in this form three times as a factor, and the result is 185193. 50 s 3 (50 2 X 7) + 3 (50 X 7 2 ) + V = 185193 472. From this result observe that 57 s = the cube of the tens, plus three times the square of the tens multiplied by the units, plus three times the tens, multiplied by the square of the units, plus the cube of the units; or that the cube of any number made up of tens and units = t 3 + 3t 2 u + 3 t u 2 + u 3 , which for the purpose of reference we will call Formula (a). And if all orders above simple units are considered tens, Formula (a) will apply to the cube of any number. 473. To assist in understanding the operation of extracting the cube root, observe the forms and dimensions of the illustrative blocks, and the relation of each to the other in the formation of the complete cube. Operation t 3 + 3t 2 u + 3tu 2 u s .= 185.193 [5 7 t 3 = 125 _._or 125000 3 1 2 u + 3 1 u 2 + u 3 = 60193 = rem. t = 50 t 2 = 2500 3 t 2 = 7500 3 t = 150 3 1 2 + 3 t = 7650 jrial divisor. 3 t 2 u = 52500 3tu 2 = 7350 u 3 = 343 3t 2 u + 3tu 2 +u 3 = 60193. Explanation.— Since the block (A) is a cube, the number representing the length of its side will be its cube-root. The given number consists of two periods of three figures each, therefore its cube root will contain two places, tens and units. Since the given number is a product of its root taken three times as a factor, the first figure, or highest order of the root, must be obtained from the first left hand period, or highest order of the power; therefore, find first the greatest cube in 185; since 185 comes between 125 (the cube of 5) and 216 (the cube of 6) the tens of the root must be 5 plus a certain remainder; therefore, write 5 in the root as its tens figure. Subtracting the cube of the root figure thus found (5 tens, or 50 ) 3 = 125000, by taking 125 from the left hand period, 185, and so obviate the necessity of writing the ciphers ; to this remainder bring down the next, or right hand period, 193, thus obtaining as the entire remainder 60193. Referring to Formula (a), observe that, having subtracted from the given number the cube of its tens (t 3 ), the remainder, 60193, must be equal to 3 t 2 u + 3tu 2 -fu 3 CUBE ROOT. 149 If a cube (B), 50 inches in length on each side, i3 formed, its contents will equal 125000 cubic inches, and it will be shown that the remaining 00193 cubic inches are to be so added to cube (C) that it will retain its cubical form. In order to do this, equal ad- ditions must be made to three adjacent sides, and these three sides, being each 50 inches in length and 50 inches in width, the addition to each of them in surface, or area, is 50 /s 50, or 50 2 , and on the three sides, 3 (50 2 ), or 3 t 2 , as in the squares (C). It will also be observed that three oblong blocks (D) will be required to fill out the vacancies in the edges, and also the small cube (E), to fill out the corner. Since each of the oblong blocks has a length of 5 tens, or 50, inches, the three will have a length of 3 X 50 inches, or 3 t. Observe, now, the surface to be added to cube (B), in order to include in its con- tents the 60193 remaining cubic inches, has been nearly, but not exactly obtained ; and since cubic contents divided by surface measurements must give units of length, the thickness of the three squares (C), and of the three oblong pieces (D), will be de- termined by dividing G0193 by the surface of the three squares, plus the surface of the three oblong blocks, or by 3 t 2 -f- 3 t ; this division may give a quotient too large owing to the omission in the divi- sor of the small square in the corner; hence such surface measure taken as a divisor, may with pro- priety, be called a trial divisor. So using it, 7 is obtained as the second, or unit figure of the root. Assuming this 7 to be the thickness of the three square blocks (C), and both the hight and thickness of the three oblong blocks (D), gives for the solid contents of the three square blocks (C), 52500, and for the solid contents of the three oblong blocks (D), 7350, or 3 t 2 u + 3 t u 2 = 59850 ; and by reference to the Formula (a), observe that the only term or ele- ment required to complete the cube of (t -J- u) is the cube of the units (u 3 ). Now, by reference to the illustrative blocks, observe that by placing the small cube (E) in its place in the corner, the cube is complete. And since (E) has been found to contain 7 X 7 X 7, or 343 cubic inches, add this to the sum of 3 t 2 u 3 tu 2 u 3 , and obtain 3 t 8 u -f 3 t u 2 -f- u 3 = 60193; and if to this t 3 , or 1250C0 is added, the result is t 3 -f- 3 t 2 u -+- 3 t u 2 + u 3 = 185193, Formula (a) ; then subtracting 60193 from the remainder, 60193, nothing remains. This proves that the cube root of 185193 is 57. By t he operation is also proved the correctness of Form- ula (a) : The cube of any number equals the cube of its tens, plus three times the square of its tens mul tiplied by its units, plus three times its tens multiplied by the square of its units, plus the cube of its units. 150 EXAMPLES FOR PRACTICE. Rule. — I. Beginning at the right, separate the given number into periods of three figures each. II. Take for the first root figure the cube root of the greatest perfect cube in the left hand period; subtract its cube from this left hand period, and to the remainder bring down the next period. III. Divide this remainder, using as a trial divisor three times the square of the root figure already found, so obtaining the second or units figure of the root; next, subtract from the remainder three times the square of the tens multiplied by the units, plus three times the tens multiplied by the square of the units, plus the cube of the units. Remakks.— 1. In examples of more periods than two, proceed as above, and after two root figures are found, treat both as tens for finding the third root figure. For finding subsequent root figures, treat all those found as so many tens. 2. In case the remainder, at any time after bringing down the next period, be less than the trial divisor, place a cipher in the root and proceed as before. 3. Should the cube root of a mixed decimal be required, form periods from the decimal point right and left. If the decimal be pure, point off from the decimal point to the right, and if need be annex decimal ciphers to make periods full. 4. To obtain approximate roots of imperfect cubes, to any desired degree of exactness, annex and use decimal periods. 5. The cube root of a common fraction is the cube root of its numerator divided by the cube root of its denominator. 6. The cube root of any common fraction may be found to any desired degree of exactness, either by extracting the root of its terms separately (adding decimal periods if need be) or by first reducing the common fraction to a decimal and then extracting the root. 7. The 4th root can be obtained by extracting the square root of the square root. 8. The 6th root is obtained by taking the cube root of the square root, or the square root of the cube root. EXAMPLES FOR PRACTICE. 474. 1. What is the cube root of 1728 ? 2. What is the cube root of 15625 ? 3. What is the cube root of 110592 ? Jf. What is the cube root of 65939264 ? 5. Find the cube root of 2146, to three decimal places. 6. Find the cube root of 119204, to two decimal places. 7. Find the cube root of 46982, to one decimal place. 8. Find the cube root of -££$. 9. Find the cube root of t VbtV 10. Find the cube root of ^HUfihr. 11. Find the cube root of \\\, U> one decimal place. 12. Find the cube root of ■}■$$■$, to two decimal places. 13. Find the cube root of 25.416237, to two decimal places. IJf. Find the cube root of 3496.25, to three decimal places. 15. Fimi the cube root of .4106, to three decimal places. 16. Find the decimal equivalent of the cube root of \\, to two decimal places, by reducing the fraction to a decimal of six places and extracting the root of the decimal. MISCELLANEOUS MEASUREMENTS. 151 17. What must be the highfc of a cubical bin that will hold 1000 bu. of wheat? 18. The width and hight of a crib of unshelled corn are equal, and each is one-third of its length. If the contents of the crib are 7465 bushels, what is its length ? 19. If the hight of an oat bin is twice its width, and its length is three and one- half times its hight, what must be its dimensions, if the bin holds 1750 bushels? 20. A cubical cistern contains 630 barrels. How deep is if ? 21. A square cistern, the capacity of which is 420 barrels, has a depth equal to only one-half its width. Find its dimensions. DUODECIMALS. 475. Duodecimals are denominate fractions of either linear, square, or cubic measure. They are found by successive divisions of the unit by 12, and are added, subtracted, multiplied, and divided in the same manner as compound numbers, though they may be treated as fractions, 12 being the uniform denom- inator. The scale is uniformly 12. 476. The Unit of measure in Duodecimals is the foot. Its first division by 12 gives primes ( ' ); primes divided by 12 give seconds ( " ), seconds divided by 12 give thirds ( '" ), and so on. Remark. — Duodecimals are but little used. MISCELLANEOUS MEASUREMENTS. 477. A Triangle is a plane figure bounded by three straight lines. 478 To find the area of a triangle, the base and hight being given. Rule. — Multiply the base by one-half the hight. To find the area of a triangle, when the three sides are given. Rule. — Find one-half of the sum of the three sides; from this subtract each side separately; multijily together the four results thus obtained, and extract the square root of the product. To find the area of any plane figure, the opposite sides of which ar3 equal and parallel. Rule. — Multiply the base by the perpendicular hight. To find the area of a plane figure, whose opposite sides are parallel but of Unequal length. Rule. — Obtain the average length, and multiply by the per- pendicular hight. 479. A Circle is a plane figure bounded by a curved line, every part of which is equally distant from a point within called the center. 480. The Circumference of a circle is the curved line c^cle. bounding it. 152 MISCELLANEOUS MEASUREMENTS. 481. The Diameter of a circle is a straight line passing through the center and terminating in the circumference. 482. The Radius of a circle is a straight line passing from the center to any point of the circumference. 483. To find the circumference of a circle, the diameter being given. Rule. — Multiply , the diameter by 3.1^16. To find the diameter of a circle, the circumference being given. Rule. — Divide the circumference by 3.14.16. To find the area of a circle, the circumference and diameter being given. Rule. — Multiply the circumference by the diameter, and divide tJie product by 4. To find the side of a square equal in area to a given circle. Rule. — Multiply flie circumference by .2821. To find the area of a square that can be inscribed within a given circle. Rule. — Multiply the square of the radius by 2, and extract the square root of the result. 484. A Cylinder is a circular body of uniform diameter, the ends of which are parallel circles. Remark. — The convex surface of a cylinder is equal to the surface of a rectangular body, the length and hight of which are equal to the circumference and hight of the cylinder. See the figure, A, B, C, D, back of the cylinder in the acompanying diagram. CYLINDER AND RECTANGLE. 485. To find the surface or area of a cylinder. Rule. — Multiply the cir- cumference by the hight. To find the contents of a cylinder. Rule. — Multiply the area of the base by the hight. A 486. A Pyramid is a solid, the base of which has three or more equal sides, terminating in a point called a vertex. 487. A Cone is a solid which has a circular base, its convex surface ter- minating in a point called a vertex. PYRAMID. CONE. 488. To find the surface of a regular pyramid or cone. Rule.— Multiply the perimeter or circumference of the base, by one-half the slant hight. To find the contents of a pyramid or cone. Utile.— Multiply the area of the, base by one-third the perpendicular hight. EXAMPLES FOR PRACTICE. 153 489. A Sphere is a solid bounded by a curved surface, all points of which are equally distant from a A point within called the center. 490. The Diameter of a sphere is a line drawn through its center, terminating each way at the surface. 491. To find the surface of a sphere. Rule.— Multiply the square of its diameter by 3. 11/.16. To find the volume of a sphere. Rule.— Multiply the cube of the diameter by .5236. To find how large a cube may be cut from any given sphere, or may bo inscribed within it. Rule. — Divide the square of the diameter of the sphere by 3, and extract the square root of the quotient; the root thus found will be the-, length of one side of the cube. To gauge or measure the capacity of a cask. Rule. — Multiply the square of the mean diameter in inches by the length in inches, and this product by .0034- the result ivill be the capacity in gallons. Remark.— In case the cask is only partly full, stand it on end, find the mean diameter of the part filled, multiply its square by the hight, and that product by .0034. EXAMPLES FOR PRACTICE. Remark. — In giving one example under each of the several preceding rules in measure- ments, the object is as much for reference as for practice in solving. 492 1. How many square feet in the gable end of a house 24 ft. wide and 6 ft 6 in. high ? 2. Find the numoer of square yards in a triangular sail, the sides of which arc 36 ft., 45 ft., and 48 ft. respectively. 3. How many acres in a rectangular field 108 rods long and 48 rods wide ? 4. A farm stretches across an entire section, being 200 rods wide on the west line and 160 rods wide on on the east line. How many acres in the farm ? 5 How many feet of fence will inclose a circular pond 82.5 ft. in diameter ? 6. What is the diameter of a circle, the circumference of which is 90 rods ? 7. The diameter of a circular park is 50 rods. How many acres does the park cover ? 8. What is the side of a square having an area equal to that of a circle 100 ft. in diameter ? 9. What is the largest square timber that can be hewn from a log 42 inches in diameter ? 10. What will be the cost of a sheet-iron smoke-stack 40 ft. high and 2 ft. in diameter, at 15^ per square foot ? 11. Find the capacity in gallons of a tank 14 ft. deep and 18 ft. m diameter? 12. A pyramid has a triangular base 3 ft. on each side, and a slant hight of of 10 ft. Find the number of square feet in its surface. 154 TABLES AND CUSTOMS IN THE PAPER AND BOOK TRADE. 13. A tent is in the form of a cone; if its slant hight is 16 ft. and its base circumference 30 ft., how many square yards of duck were used in making it ? lJf. How many square inches of leather will cover a foot ball 8 in. in diameter? 15. How many cubic feet in the contents of a globe 4 ft. in diameter ? 16. The diameter of the earth is 7901 miles, and that of the planet Jupiter 85390 miles. How many spheres like the earth are equal to Jupiter ? 17. What will be the length of the largest cube that can be cut from a sphere 7901 miles in diameter ? 18. A cask 28 in. at each end, and 34 in. at the bilge, is 3 ft. long. How many gallons of water will it hold ? 19. If a cask 24 inches at the chime, 30 inches at the bung and 3 feet long, is £ full, how many more gallons may be put into it ? TABLES AND CUSTOMS IN THE PAPER AND BOOK TRADE. 493. Paper in the stationery trade is sold by the following Table. 24- sheets I =1 quire. 20 quires = 1 ream. 2 reams = 1 bundle. 5 bundles = 1 bale. A bale contains 200 quires, or 4800 sheets. Remakks. — 1. In copying, a folio is usually 100 words. 2. In type-setting, an em is the square of the body of a type, used as a unit by which to measure the amount of printed matter on a page. 494. Books are sometimes classified by their size, or the number of pages in a sheet. Name. Sheet folded into. Pages. Folio, ._ _._ 2 leaves, 4 Quarto, 4to. 4 leaves, 8 Octavo, 8vo 8 leaves, 16 Duodecimo, 12mo 12 leaves, 24 16mo 16 leaves, 32 18mo 18 leaves, 36 24mo 24 leaves, 48 32mo 32 leaves, 64 Table for Counting. 12 units = 1 dozen. 20 units = 1 score. 12 dozen =-. 1 gross. 12 gross = 1 great gross. 104J feet square = J of an acre. 147 T V feet square = -J of an acre. 208 T V feet square = 1 acre. Table for Land and. Lot Measures. 10 rods X 16 rods = 1 acre. 8 rods X 20 rods =; 1 acre. 40 yards X 121 yards = 1 acre. THE METRIC SYSTEM. 155 THE METRIC SYSTEM. 495. The Metric System is a decimal system of denominate numbers. It is in use in nearly all the European States, in South America, Mexico, and Egypt. It is also used somewhat in Asia, and is authorized by law in the United States; but its use here is so limited as to justify only a reference to it, and the presentation of its unit equivalents in our weights and measures, as a reference for interested parties. 496. The Unit' of Length and basis of the system is the Meter = 39.37+ inches, being one ten-millionth of the distance from the equator to the pole. The unit of area is the Ar (A.); the unit of solidity is the Ster (S.); the unit of weight is the Gram (G-.); the unit of capacity is the Liter (L.). Higher denom- inations are called Dek'a (10), Hek'to (100), Kii'o (1000), and Myr'ia (10000). Lower orders are called Dec'i (tenths), Cen'ti (hundredths), Mil'li (thousandths). Metric Linear Table. 10 mil'li-me'ters {mm) == 1 cen'ti-me'ter cm = yj-g- M. 10 cen'ti-me'ters 10 dec'i-me'ters 10 me'ters 10 dek'a-ine'ters 10 hek'to-me'ters 10 kil'o-me'ters = 1 dec'i-meter dm = ^ M. = 1 Meter M. == 1 dek'a-me'ter Dm = 10 M. = 1 hek'to-me'ter Hm = 100 M. = 1 kil'o-me'ter Km = 1000 M. = 1 myr'ia-me'ter Mm = 10000 M. Remarks. — 1. All tables are formed in a similar manner. 2. In naming units, abbreviations are commonly used. 3. The system being on a decimal scale, the full mastery of the names of the higher and lower denominations, with unit equivalents, will be sufficient for practical use. 497. An Act of Congress requires all reductions from the Metric to the common system, or the reverse, to be made according to the following Tables of Equivalents. Linear Measure. 1 inch = 2.54 centimeters. 1 foot = .3048 of a meter. 1 yard = .9144 of a meter. 1 rod = 5.029 meters. 1 mile = 1.6093 kilometers. 1 centimeter = .3937 of an inch. 1 decimeter = .328 of a foot. 1 meter = 1.0936 yards. 1 dekameter = 1.9884 rods. 1 kilometer = .62137 of a mile. Square Measure 1 sq. inch = 6.452 sq. centimeters. 1 sq. foot = .0929 of a sq. meter. 1 sq. yard = .8361 of a sq. meter. 1 sq. rod = 25.293 of a sq. meter. 1 acre = 40.47 ars. 1 sq. mile = 259 hektars. 1 sq. centimeter = .155 of a sq. inch. 1 sq. decimeter = . 1076 of a sq. foot. 1 sq. meter =1.196 sq. yards. 1 ar = 3.954 sq. rods. 1 hektar = 2.471 acres. 1 sq. kilometer = .3861 of a sq. mile. 156 MOXEY OF THE GERMAN EMPIRE. Cubic Measure 1 cu. inch = 10.387 cu. centimeter. 1 cu. foot = 28.317 cu. decimeter. 1 cu. yard = . 7645 of a cu. meter. 1 cord = 3.624 ster. 1 cu. centimeter = .061 of a ou. inch. 1 cu. decimeter = .0353 of acu. foot. 1 cu. meter = 1.308 cu. yard. 1 ster = .2759 of a cord. Measures of Capacity. 1 liquid quart = . 9463 of a liter. 1 dry quart == 1.101 liter. 1 liquid gallon = .3785 of a dekaliter. 1 peck = .881 of a dekaliter. 1 bushel = .2524 of a hektoliter. 1 liter = 1.0567 liquid quarts. 1 liter = . 908 of a dry quart. 1 dekaliter = 2.6417 liquid gallons. 1 dekaliter = 1.135 pecks. 1 hektoliter = 2.8375 bushels. Measures of Weight. 1 grain, Troy — .0648 of a gram. 1 ounce, Avoir. = 28.35 gram. 1 ounce, Troy = 31.104 grams. 1 pound, Avoir. = .4536 of a kilogram. 1 pound, Troy = .3732 of a kilogram. 1 ton (short) = . 9072 of a tonneau. 1 gram = .03527 of an ounce, Avoir. 1 gram s= .03215 of an ounce, Troy # 1 gram = 15.432 grains, Troy. 1 kilogram = 2.2046 pounds, Avoir. 1 kilogram = 2.679 pounds, Troy. 1 tonneau = 1.1023 tons (short' Remark. — Metric quantities of any unit are read like ordinary decimals. FRENCH MONEY. 498. The Legal Currency of France is decimal, its unit being the silver Franc. 499. The French coins are as follows: 100 francs, 40 francs, 20 francs, 10 francs, 5 francs. Gold f 5 francs, Silver \ 2 francs, f 1 franc. {10 centimes, 1 centime. 10 millimes (m.) 10 centimes 10 decimes Table. 1 centime (ct.) == $.00193. 1 decime (dc.) = .0193. 1 Franc (fr.) = .193. MONEY OF THE GERMAN EMPIRE. 500. The Unit is the Mark = $ .2385 United States money, into 100 pfennigs (pennies). The silver Thaler = $. 726 United States money. 501. The German coins are: ( 20 marks, ( 20 marks, Gold < 10 marks, Silver i 1 mark, Nickel It is divided 5 marks. ( 20 pfennigs. j 10 pfennigs, ( 5 pfennigs. EXAMPLES IN DENOMINATE NUMBERS 157 MISCELLANEOUS EXAMPLES. 502. 1. What is the value, in English money, of $1750 in United States gold coin ? 2. It required 12 yr. 6 mo. 1 da. to build the Brooklyn bridge. If it was completed July 4, 1882, when was its construction begun ? 3. What is the board measure of 7 planks, each 16 ft. long, 15 in. wide, and 3 in. thick ? 4. How many acres of land can be bought for $25000, if a square foot costs 25^- ? 5. A cellar is 24 ft. square inside of the wall, which is 9 ft. high, and 2 ft- thick. How many perches of 16-J- cu. ft. each does the wall contain? Remark. — Sometimes 24f cubic feet are reckoned as a perch, but this is rarely done by contractors or architects; girt measurements are taken. 6. How many shingles, 4 inches wide, laid 6 inches to the weather, would be required to cover the roof of a barn 60 ft. long and 24 ft. wide on each side? 7. The highest chimney in the world is at Port Dundas, Scotland, it being 450 ft. high. How many rods in hight is it ? 8. The Italian Government pays out yearly $2140000 to 32590 monks and nuns. What is the average sum received by each ? 9. What will be the cost of the plank, at $18 per M, that will cover a floor 24 ft. by 13 ft., if the plank is 2£ inches in thickness ? 10. A farm having 225 rods fronting the road, is 95 rods wide at one end and 72. 5 rods at the other. How many acres does the farm contain ? 11. If the capacity of a cask is 64^ wine gallons, how many quarts of berries will it hold? 12. A bird can fly 1° in 1 hr. 10 m. 12 sec. At that rate, in what time can it encircle the earth ? 18. What will be the cost in Paris of a cargo of 38500 bu. United States wheat, at 10 fr. 60 cent, per hektoliter? 11/.. How many francs are equal to $275. 15. The largest shipping lock in the world is at Cardiff, it being 600 ft. long, 80 ft. wide, and 32 ft. deep. What is its capacity in barrels ? 16. When it is noon at the point of your observation, what is the time at a point 1500 statute miles due south-west ? 17. If your coal costs $5. 60 per ton, and you use 65 lb. per day, what will be the expense of your fire for the months of the winter of 1891-2 ? 18. How many barrels of water in a cistern 12.5 ft. long, 10 ft. wide, and 7.5 ft. deep ? 19. A carriage wheel 4 ft. 3 in. in diameter will make how many revolutions in going 62.5 miles ? 20. If Wm. H. Vanderbilt died worth two hundred millions of dollars, in what length of time could his fortune, in silver dollars, be counted by one person, counting 60 per minute and working 10 hours per day for 365 days each year ? k21. What will be the cost of 10 sticks 2 in. by 4 in., 10 sticks 2 in. by 6 in., sticks 4 in. by 4 in., and 10 sticks 2 in. by 10 in., if the sticks are each 16 ft. 158 EXAMPLES IN DENOMINATE NUMBERS. 22. How many yards of Axminster carpeting, f of a yard in width, and laid lengthwise of the room, will be required to coyer a floor 21f ft. long and 18f ft. wide, making no allowance for waste in matching the design ? 28. How many tons of 324 cu. ft. each, in a mow of hay 36 ft. 3 in. long, 18 ft. 10 in. wide, and 13 ft. 6 in. high ? 2Jf.. Two astronomers, located at different points, observed at the same instant of time an eclipse of the moon, one seeing it five minutes after 9 p. m., local time, and the other five minutes before midnight. How many degrees of longi- tude separated the observers ? 25. How many Avoirdupois pounds in 10 myriagrams 4 kilograms. 26. If the sun is 93 millions of miles from the earth, and a cannon ball travels nine miles per minute, at what time would a ball fired from the earth at one minute after 3 o'clock p. m., Dec. 25, 1889, reach the sun at that rate ? 27. How many francs are equal to £425 ? 28. Allowing 305 sq. ft. for doors and windows, what will be the cost, at 400 per square yard, of plastering the ceiling and walls of a room 45 ft. long, 35£ ft. wide, and 12 ft. 3 in. high ? 29. How many German marks are equal to $1500 United States money ? SO. A pile of wood built 10 ft. high and 22 ft. wide must be how long to contain 125 cd. ? 81. The main centennial building at Philadelphia in 1876 was 1880 ft. long and 464 ft. wide. What was its area in acres, square rods, and square feet ? 82. Eeduce 47 mi. 216 rd. 11 ft. 5 in. to metric units. 88. If £2 4 s. 6 d. is paid for a coat and vest, and the coat costs 4 s. more than twice as much as the vest, what is the cost of each, in United States money? 84. From .001 of a section, plus .01 of an acre, take .001 of a quarter section, plus .01 of a square rod. 85. A grocer bought 12 bu. of chestnuts, at $3.50 per bushel dry measure, and sold them at 15^ per quart liquid measure. Did he gain or lose, and how much ? 86. How many dollars are equal to 2150 francs ? 37. How many square feet of sheet lead will be required to line a tank 7 ft. in diameter and 1 2 ft. deep ? 88. If bricks cost $5.50 per M, what will be the cost of the brick for a wall 12 ft. high and 3 ft. thick, enclosing an acre of land 10 rd. wide and 16 rd. long ? 89. The gold coin of the commercial world suffers each year a loss of one ton by wear or abrasion. What is the value, in United States gold dollars, of the loss thus sustained? JfO. Eeduce 250 hektars to common units. 41. What will be the cost, at $16.00 per M, of a tapering board 18 ft. long, and 9 in. wide at one end and 16^- in. wide at the other ? 42. A German immigrant having 1000 thalers and 500 marks, exchanges them for United States money. How many dollars should he receive ? JjS. The hight, width, and length of a shed are equal. What are its dimen- sions, if it will contain 125 cords of wood ? EXAMPLES IN DENOMINATE NUMBERS. ]59 44- A train of 45 cars of Lehigh coal averages, by the long ton, 23 T. 7 cwt 3 qr. 10 lb. per car. What is the value of the coal, at $5.25 per short ton ? 45. How many feet of lumber in a box 6f ft. long, 5 J ft. wide, and 3^ ft. deep, inside measurements given, and lumber 1 inch in thickness ? 46. What will be the cost of carpeting f yd. wide, and lining $ yd. wide, to cover a room 24 ft. long and 20 ft. wide, if the strips of carpet are laid the long way of the room and there is a waste of 9 inches at one end in matching, also an allowance of 10$ in width and 6$ in length for shrinkage of the lining, the carpet selling at $2.25 per yd., and the lining at 300 per yd. 47. A pile of wood 56 meters long, 18^ meters wide, and 3f meters high, was sold at $6 per cord. How much was received for it ? 48. A farmer filled a bin 9 ft. wide, 12 ft. long, and 7-J ft. deep, with wheat grown from a field yielding 32£ bu. per acre. How long was the field, if its width was 50 rods ? 49. Seasoned pine in freighting is estimated to weigh 3000 lb. per M, and green oak 5000 lb. per M. How much freight must I pay, at $1 per ton, on a car load of 3265 ft. of pine and 3795 ft. of oak ? 50. How long is the side of the largest cube that can be cut from a spherical snow ball 5 ft. in diameter ? 51. Glenn's California reaper will in 12 hours cut, thresh, winnow, and put into bags, 30 A. of wheat. How many days, of 15 working hours each, will it require .to harvest and thresh the wheat of a field 125 rods wide and 240 rods long ? 52. An ounce of gold can be so beaten as to cover 146 sq. ft. What weight of gold would be required for a sheet which will cover an acre of ground ? 53. A farmer having 1240 bu. of corn in the ear to store in two rail cribs, builds each 9 ft. square on the inside. If one is built 10 ft. high and filled, how high must the other be built to hold the remainder ? 54. The Hercules ditcher, of Michigan, removes 750 cu. yd. of earth per hour. In how many days, of 12 working hours each, can it dig a ditch 7 miles in length, 8 ft. in depth, 24 ft. wide at the surface, and 16 ft. at the bottom ? 55. If a car carrying 20 tons of freight is with its couplings 42 ft. long, what would be the length of a train carrying Vanderbilt's two hundred millions of dollars, if it is all in standard silver dollars, and any fractional part of a car load is rejected ? 160 * PEKCKNTAGfi. PERCENTAGE. 503. Percentage is a term applied to computing by hundredths. 504. The Elements of Percentage are, the Base, the Rate, the Amount Pel Cent., the Difference Per Cent., the Percentage, the Amount, and the Difference. 505. The Base is the number upon which the percentage is computed. 506. The Kate Per Cent, denotes how many hundredths of the base are ta be taken, and is usually expressed as a decimal. 507. Per Cent, is an abbreviation of the Latin words per centum, signifying by the hundred, or a certain number of each one hundred parts. 508. The Sign, #, is used to denote per cent. 509. The Bate may be expressed as a part in a common fractional form, as J ; in the form of an extended decimal, as .01625 = lf$ ; but only when expressed in hundredths can it with strict propriety be considered a rate per cent. Thus, .12, .06, .15|, .05f, are each a rate per cent. 510. To read per cent. , call the first two places per cent. , and the added places, if any, fractions of' 1 per cent.; as, .2125 read as 21 and one-fourth per cent. 511. To express per cent, as a common fraction, write the per cent, for a numerator and 100 for a denominator, and reduce; thus, 25$ = -^ ss \. 512. To change a common fraction to an equivalent per cent., apply the decimal explanation, Art. 245. Divide the numerator by the denominator, and give the quotient at least two decimal places. 513. Every rate per cent., being as many hundredths, requires at least two decimals places; hence, if the per cent, be less than 10, a cipher must be prefixed to the figure denoting it; thus, 2$ = .02. 514. The Amount Per Cent, is 100 per cent, increased by the rate, or 1 plus the rate. 515. The Difference Per Cent, is 100 percent, diminished by the rate, or 1 minus the rate. Remark.— Where the rate per cent, is the equivalent of a common fraction, use in solution whichever is most convenient. 516. The Percentage is the sum obtained by multiplying the base by the rate, 517. The Amount is the sum of the base and percentage. 518. The Difference. is the remainder after deducting the percentage from the base. PERCENTAGE. 161 519. The Base is either an abstract or denominate number; the rate per cent, is always abstract, and the percentage, amount, and difference are always like the base. Remarks. — 1. In all operations where a decimal rate is used, too great care cannot be taken to express all decimal terms with exactness. 2. As the greater pai^of commercial calculations are based upon percentage, the importance of a thorough mastery of its principles will be readily perceived. 520. Since per cent, is any number of hundredths, it may be expressed eithei* as a decimal or as a common fraction, and the table of aliquot parts can be used with little variation and to great advantage in many operations in percentage* Hence, the rules given under Special Applications may be applied in this subject. Table. 11 Decimal. i 3om . Frac. Lowest Terms. 1 per cent. S3 .01 = ToT reducible to y^-. 2 per cent. 35 .02 = To~o~ reducible to -fa. 3 per cent. = .03 = ToT reducible to yf-g-. 4 per cent. = .04 = Too" reducible to fa. 5 per cent. = .05 = 5 ToT reducible to fa. 6 per cent. = .06 = Too" reducible to -^j-. 7 per cent. = .07 = TOO" reducible to yj^. 8 per cent. S3 .08 = 8 ToT reducible to -fa. 9 per cent. S3 .09 = Too" reducible to yg-g-. 10 per cent. = .10 = TO" reducible to fa. 12 per cent. ss .12 = tVSt reducible to -fa. 14 per cent. = .14 = iA reducible to -fa. 16 per cent. = .16 = tW reducible to -fa. 20 per cent. = .20 = T% reducible to \. 25 per cent. = .25 33 -h\ reducible to £. 30 per cent. = .30 = tft reducible to fa. 50 per cent. = .50 = *V reducible to -J. 75. per cent. S3 .75 . = TOO reducible to f. 100 per cent. = 1.00 = m reducible to 1. 125 per cent. = 1.25 = m reducible to £ = 1£. 150 per cent. = 1.50 = 1 50 TTT reducible to f = 1-J-. 1J- per cent. S3 .0125 = ioooo reducible to -g^. If per cent. = .0166| S3 30000 reducible to -g^. 2-J per cent. = .025 = 1000 reducible to -fa. 3£ per cent. = .033£ S3 drtfr reducible to -fa. 6£ per cent. =3 .0625 = •625 100 reducible to fa. Si per cent. =S .0833£ = JLtJLiL. Wo o o reducible to fa. 12^- per cent. 33 .125 S3 t¥A reducible to -J. 16f per cent. S3 .166f = "3~oiro reducible to •}. 33J per cent. = .333^ = an reducible to £. 62^- per cent. S3 .625 = WW reducible to f. 66J per cent. S3 ,66f S3 m reducible to f. 874 P er cent. 3= .875 = m reducible to $-. 162 EXAMPLES IX PERCENTAGE. 521. The relation between the elements of Percentage is such, that by the application of the General Principles of Multiplication and Division, if any two of the elements, except amount per cent, and difference per cent., are given, the other three may be found. 522. To find the Percentage, the Base and Rate being given. Example. — What is 25$ of $468 ? First Explanation.— 25 per cent, equals .25; therefore, Operation. 25 per cent of $468 equalg $468 multiplied by ^ equals $468 = base. $117. .25 = rate per cent. Second Explanation.— $468 is 100 per cent, of itself; ttl^OO = percentage. *fj?Z ^ P« r ^nt equals i of 100 per cent., 25 per cent 1 & of $468 will be i of that sum, or $117. Rules. — 1. Multiply the base by the rate expressed decimally. Or, 2, Take such a part of the base as the number expressing the rate is part of 1. Remark. — When the rate is an aliquot part of 100, the percentage may be found by taking a like part of the base; thus, for 10# take fa for 25$ take i, for 33£# take £, etc. Formula. — Percentage = Base X Rate. EXAMPLES FOR MENTAL, PRACTICE. 523. What is 1. 5 per cent, of 100 ? 2. 12 per cent, of 600 ? 8. 15 per cent, of 800 ? Jh 20 per cent, of 500 ? 5. 25 per cent, of 1200 ? 6. 33£ per cent, of & ? 7. 25 per cent, of 1440 ? 8. 8 per cent, of 450 ? 9. 50 per cent, of 680 ? examples for written practice. 524. 1, A man owning 250 acres of land, sold 20$ at one time, and 25$ of the remainder at another time. How many acres did he have left ? 2. If a ranchman having 5450 sheep, lost 20$ by a storm and afterwards sold 20$ of those remaining, how many sheep did he sell? 8. A collector deposited $13500 in coin, and 12£$ more than that amount in bank bills. What was the total of his deposit? £ Find 11$$ of 1680 lb. of wool. 5. Find 16|$ of 12 lb. 3 oz. of silver. 6. From a charge of $675, made for a bill of goods, 8$ was deducted. What was the net amount of the bill ? 7. If 526 barrels of salt were bought for $1.10 per bar., and sold at an advance of 15$, what was gained ? 8. Two men, each having $12500, made investments, from which one gained 15$, and the other lost 35$. How much did each then have? 9. How much greater is 12£$ of $1550, than 7£$ of $2150 ? 10. Having raised 1240 bushels of wheat, a farmer used 5$ of it for seed and 5$ for bread; he then sold to one man 10$ and to another 25$ of what remained. How many bushels had he left ? EXAMPLES IN" PERCENTAGE. 163 11. Having $75000 to invest, a gentleman bought United States bonds with 33£$ of his money, a home with 20$, and invested the remainder equally in farm lands and manufacturing stock. How much did he pay for the farm lands ? 12. I owed John Smith $1750, and paid at one time 20$ of the debt, at another time 35$ of the remainder, and at another time 25$ of what then re- mained unpaid. How much of the debt did I still owe ? 13. A capitalist owning f of a coal mine, sold 32^-$ of his share for $65000. At that rate, what was the entire mine worth ? llf. A jobber having bought 2160 bags of coffee, sold at one time 8£$, at another 25$ of what remained, and at a third sale 15$ of what still remained. Find the value of what was left, at $18 per bag. 15. Of a farm containing a half section of land, 15$ was in wheat, 32$ in oats, 5$ in potatoes, and the remainder devoted equally to orchard, corn, beans, and pasture. How many acres were in pasture ? 16. A farmer having 156 sheep to shear, agreed to pay for their shearing 4$ of the sum received for their wool. If the fleeces averaged 7-j- lb. and sold for 30^ per pound, how much was paid for shearing ? 17. A speculator having $41820, invested 50$ of it in oil, on which he lost 16f$ ; the remainder he invested in cotton, which he sold at 9$ below cost. How much was received from both sales ? IS. A trader bought 12 mustangs for $400, and after selling 25$ of the num- ber at a gain of 50$, and 33£$ of those remaining at a gain of 12-£$, sold those still on hand at $30 per head. Did he gain or lose, and how much ? 525. To find the Base, the Percentage and Rate being given. Remark. — Since the base multiplied by the rate produces the percentage, percentage must be a product; if, therefore, it is divided by either factor, the quotient will be the other factor. Example. — By selling 4$ of a stock of goods, a merchant realized $644. What was the value of the entire stock ? Operation. Rate. Percentage. Explanation.— If the value of 4 per cent, is $644, the value of .04 ) 644 00 * per cent ' wil1 be $ 161 5 and if tne value of 1 per cent, is $161, the : value of 100 per cent, will be $16100. 16100 base. Rule. — Divide the percentage by the rate, expressed decimally. Formula. — Base =s Percentage ■*■ Rate. EXAMPLES FOR MENTAL PRACTICE. 526. 1. 846 = 6$ of what number ? 2. 2150 = 10$ of what number ? S. 543 = 5$ of what number ? 4. 219 = 33£$ of what number ? 5. 150 = |$ of what number ? 6. A man sold 25$ of his farm for $2120. How much was the farm worth at that rate ? 164 EXAMPLES IIS T PERCENTAGE. 7. What is the value of a house renting for $360 per year, if the rent equals 9$ of its value ? 8. How many acres in a farm of which 12.5 acres is but 5$. 9. Of what sum is $36 but 33£$ ? EXAMPLES FOR WRITTEN PRACTICE. 527. -7. A planter sold 76 bales of cotton, which was 19$ of his crop. How many bales did he raise ? 2. I paid $123.48, which was 16f$ of a debt. What amount did I owe ? 8. A lady paid for millinery, $17.50; for shoes, $11.40; for jewelry, 113.80; for furs, $78.55; and had expended but 15$ of her money. How many dollars had she at first ? 4. A clerk's present salary of $520 per year is only 75$ of what he formerly received. How much was formerly paid him ? 6. . A grocer, after increasing his stock to the amount of $6448, found that the new purchase was but 16$ of the old stock on hand. What was the value of his old stock ? 6. The owner of 68$ of a mine, received $91510 from the sale of 25^ of his share. Find the value of the entire mine at that rate ? 7. A, B, C, and D are partners; A furnished 15$ of the capital, B 25$, 42$, and D $16200. What was the capital of the firm ? 8. A Wyoming ranchman lost 1684 cattle during a blizzard. How many had he at first, if his loss was only 1|$ of his herd ? 9. The population of a county increased 22$ in ten years. If the births exceeded the deaths by 2166, and the county received 13234 immigrants during the time, what must have been its population before the increase ? 10. A speculator»owned a quarter interest in a mill, and sold one-quarter of his part for $11250. What was the mill worth, on that basis of value ? 528. To find the Rate, the Percentage and Base being given. Remark. — The percentage is a product, the base being one of its factors. Example. — What per cent, of 480 is 120 ? First Operation. 4.80 ) 120.00 ( 25 times. First Explanation.— Since 480 is 100 per cent, of itself, 1 960 per cent, of 480 would be T £ 7 part of it, or 4.80; and since 4.80 is 1 per cent, of 480, 120 would be as many times 1 per cent. as 4.80 is contained times in 120, which is 25 times; and 25 2400 1$ — .01 2400 times t per cent> = 25 VQT ceQt /la .25 = 25$. aqt^TwhT^T-ZW Second ExPLANATiON.-Since the percentage is a product 4SU } l^SU. UU { . <£0 — AZJo» of the bage and rate> the quotient obtained by dividing the per- _ centage by the base will be the rate. Or, 120 is \\l, or ± 2400 of 480: and since 480 is 100 per cent, of itself, 120, which is 2400 £ of 480, must be \ of 100 per cent., or 25 per cent. EXAMPLES" lis PERCENTAGE. 165 Rule.— Divide the percentage by the base, carrying the quotient to two decimal places. Formula. —Rate = Percentage -4- Base. EXAMPLES FOR MENTAL PRACTICE. 529. What per cent, is i. 25 of 125 ? 4. 12£ of 100 ? 7. 37J of 150 ? ft 40 of 160 ? 5. 15 of 45 ? 8, 200 of 10 ? S. 18 of 36 ? 6. 125 of 1000 ? 9. 120 of 4 ? EXAMPLES FOR WRITTEN PRACTICE. 530. 1. From a herd of 1184 cattle, 296 were sold. What per cent, was sold? 2. R. G. Dun & Co. charged $21 for collecting an account of $600. What rate was charged ? 3. Sold f of a stock of goods for what the entire stock cost. What was my rate of gain ? 4. What per cent, of 12 lb. 8 oz. is 2 lb. 8 oz., Avoirdupois ? 5. From a half section, 120 acres were sold, and afterwards 80 acres more. What per cent, was sold ? ' 6. Of a stock of 800 bushels of potatoes, 240 bushels were sold at one time, and 135 bushels at another. What per cent, was still unsold ? 7. A merchant failed, owing $27984, his assets amounting to $16090.80. What per cent, of his debts can he pay ? 8. At a normal school there were enrolled 855 male pupils and only 185 female pupils. What per cent, more were the male than the female pupils ? 9. A girl having $5.40, expended $1.35- for gloves, 45^ for flowers, and one- half of the remainder for a pair of slippers. What per cent, of her money had she left ? 10. From a cask of lard of 314 lb., 78.5 lb. were sold at one time, and 25$ of the remainder at another. What per cent, of the whole remained unsold ? 11. Of a regiment of men entering battle, 1040 strong, only 260 came out unhurt, £ of the remainder having been killed. What per cent, of the whole were killed ? 531. To find the Amount Per Cent, the Rate being given. Example. — If the rate be 7$, what is the amount per cent. ? Operation Explanation. — Since the amount per cent. !0(w — 1 __ a 11T1 ';f (definition, page 160), is always 100 per cent, in- ** ~ creased by the rate, we may find it by adding 7$ = .07 = rate 1( )0 per ^ ;ent ? or 1? t0 the per cent gi ven< Hence, TZZ^ 7TZ , ^ the rate is 7 per cent., the amount per cent, will .07^ = 1.07 = amount per cent. be m per ^ Rule.—«4£foZ the rate to the unit 1, Formula. — Amount Per Cent. = 1 + Eate. 166 EXAMPLES 12* PERCENTAGE. EXAMPLES FOR MENTAL PRACTICE. 532. 1. If the rate be 10$, what will be the amount per cent.? 2. If the rate be 75$, what will be the amount per cent. ? 3, If the rate be 110$, what will be the amount per cent. ? 4-. Find the amount per cent., if the rate per cent, be 16|? 5. Find the amount per cent., if the rate per cent, be 87-j- ? EXAMPLES FOR WRITTEN PRACTICE. 533. i. Goods costing $1400 were sold for $1470. Find the amount per cent, of the selling price ? 2. Last month I sold $2750 worth of coffee, while the previous month I sold $3000 worth. What was the amount per cent, of my sales for the previous month as compared with those of the last month ? 3. If tea costing 62|^ per pound sell at 87-^, what amount per cent, do the sales show as compared with the cost ? 534. To find the Difference Per Cent., the Rate being given. Example. — If the rate be 5$, what is the difference per cent.? ° PERATI0N ' Explanation.— Since the difference per cent, (definition, ei — fkK — a U2U page 160 )' k equaI to 10 ° per cent > or *» less the rate > if we take ^ ~ •OP = ra * e * the given rate, 5 per cent., from 100 per cent., the remainder, 95$ = .95 = dif. fo. 95 per cent., will be the answer required. Rule.— Subtract the rate from the unit 1. Formula. — Difference Per Cent. = 1 — Rate. EXAMPLES FOR MENTAL PRACTICE. 535. 1, If the rate be 15$, what is the difference per cent. ? 2. If the rate be 37-J$, what is the difference per cent. ? 3. If the rate be -J$, what is the difference per cent. ? Jh If the rate be 3^$, what is the difference per cent. ? 5. If the rate be 70$, what is the difference per cent. ? EXAMPLES FOR WRITTEN PRACTICE. 536. 1. The pupils of a school are reduced in number from 112 to 80. What per cent, is the present -of the former attendance ? 2. Walter, having 48 marbles, gave Henry 15. What per cent, had he left ? 3. Find the difference per cent. , if the rate equals i of f . 537. To find the Amount, the Base and Rate being given. Example. — What is the amount of 550 increased by 8$ of itself ? Operation Explanation. — The amount equals base plus percentage (de 5 a« — I G ' finition, page 160). The base is 550 and 8 per cent, of 550 equals • ° rate. ^ ^ e percentage ; therefore the amount must equal 550 plus 44, 44. 00 = per cent, or 594; or, since 550 equals 100 per cent, of itself, an increase of 8 550 = base. per cent, would give 108 per cent, of the original number; and TTI 108 per cent, of 550, or 1.08 times 550 equals 594. 594 = amount EXAMPLES IN" PERCENTAGE. 167 Rules.— 1. Find the percentage and add it to the base. Or, 2, Multiply the base by 1 plus the rate- Formula. — Amount = Base + Percentage. EXAMPLES FOR MENTAL PRACTICE. 538. 1. If the base is 1500, and the rate 10$, what is the amount ? 2. If the base is 1356, and the rate 25$, what is the amount ? 3. The base is 440 and the rate 5$; find the amount. 4-. The base is 1000 and the rate 18$; find the amount. 5. The base is 252 and the rate 10$; find the amount. 6. The base is 2150 and the rate 20$; find the amount. 7. The base is 630 and the rate 33£$; find the amount. S. The base is 546 and the rate 16|$; find the amount. P. The base is 200 and the rate 125$; find the amount. EXAMPLES FOR "WRITTEN PRACTICE. 539. 1. What amount will be received for a house costing $13500, if it is sold at a gain of 7-J$ ? 2. A bought two horses for $180 each, and sold one at a gain of 20$ and the other at a gain of 33£$. How much did he receive for both ? 3. A section of Kansas prairie was bought at $12.50 per acre, and sold at an advance of 40$. How much was received for it ? Jf. What is the amount of 768 increased by 25$ of J of itself ? 5. What is the amount of $3144 increased by f of 16f$ of itself ? 6. If the base is $864.88 and the rate 3£$-of f of itself, what is the amount ? 540. To find the Difference, the Base and Rate being given. Example. — What remains after diminishing 450 by 10$ of itself ? Operation. 100$ = 450 = base. l(w — .10 = rate. Explanation. — Since 100 per cent, oi tne , number equals 450, 10 per cent, of it will equal 90$ dif. $ 45.00 = percentage. 45; and 450 minus 45 equals 405. Or, since 100 450 base ^ ei cent * e( l uals 45 0> *° P er cent - less tban 1 ^ ) 45 percentage. P er cent «> or 90 per cent, will equal 405. 405 difference. Rules.— 1. Find the percentage and subtract it from the base. Or, 2. Multiply the base by 1 minus the rate. Formula. — Difference = Base — Percentage. EXAMPLES FOR MENTAL, PRACTICE. 541. 1. If from a brood of 15 chickens 20$ are lost, how many will remain? 2. What number will remain if 225 is diminished by 33£$ of itself ? S. If the base is 1050 and the rate 10$, what is the difference ? 168 EXAMPLES m PERCENTAGE. 4* 816, less 25$ of itself, equals what number ? 5. 1440, less 16f $ of itself, equals what number ? 6. 800, less 37J$ of itself, equals what number ? 7. 40, less 87|$ of itself, equals what number ? 8. A boy having 648 ft. of kite string, lost 12-J$ of it. How many feet had he remaining ? EXAMPLES FOR WRITTEN PRACTICE. 542. 1. A speculator lost 35$ of | of $16250. How much did he lose? 2. A planter having 616 acres in rice, lost \ of 33£$ of his planting by flood. How many acres had he left for harvest ? 8. Brown deposited $1147 in a savings bank, and his son deposited $1$ less. How much was deposited by both ? J/.. An agent earned $250 in May, 15$ less in June, and 20$ less in July than in June. What was the amount earned for the three months ? 543. To find the Base, the Amount, or Difference, and the Rate being given. Example (first illustration). — What number, increased by 15$ of itself, amounts to 345 ? Operation. = 1.00 15$ = .15 Amount. Base. Explanation. — Since the number must be 100 ^% ~ !; ;; per cent, of itself, if it has been increased 15 per cent., 345 must be 115 per cent, of that number; 115$ amt. $ 1.15 ) 345.00 ( 300 if 115 per cent, is 345, 1 per cent, must be ^ of 345 345, or 3; and 100 per cent, will be 100 times 3, or _ 00 . m Example (second illustration). — What number, diminished by 35$ of itself, equals 975 ? Operation. 100$ = 1.00 35$ = .35 Diff> Base Explanation.— If the number be diminished by 35 per cent, of itself, there will be remaining but 65$ dif. $ = .65 ) 975.00 ( 1500 65 per cent, of itself; and if 65 per cent, of the 65 number be 975, 1 per cent, must be -£% of 975, or 325 15; and if 1 per cent, be 15, 100 per cent, must be 325 1500. 00 Rules.— 1. Divide the amount by 1 plus the rate. Or, 2. Divide the difference by 1 minus the rate. Formulas. — 1. Base = Amount -f- Amount Per Cent. 2. Base = Difference -t- Difference Per Cent. EXAMPUES FOR MENTAL PRACTICE. 544. 1. If the amount is 750 and the rate 25$, what is the base ? 2. What number, increased by 10$ of itself, amounts to 440 ? S. After 75$ of a number had been added to it, the amount was 525. What was the number ? REVIEW OF THE PRINCIPLES OF PERCENTAGE. 169 4. After selling 30$ of his apples, a boy had 70 left. How many had he at first ? 5. I lost $600 by a bankrupt, who paid only 85$ of his indebtedness. What was the full amount of my claim ? EXAMPLES FOR WRITTEN PRACTICE. 545. 1. A builder gained 15$ by selling a house for $1150. What was its cost ? 2. Sold 945 tubs of butter for $5113, and thereby gained 20$. How much did the butter cost per tub ? 8. The income from a tenement house is $6042 this year, which is 24$ less than it was last year. How much was it last year ? 4. A liveryman paid $180 for a horse, which was 40$ less than he paid for a carriage. How much did he pay for both ? 5. A drover gained 16f $ on 33 head of cattle sold for $4081. What was the average cost per head ? 6. Smith sold two horses for $1500 each, gaining 25$ on one, and losing 25$ on the other. What did the horses cost him ? 7. After paying 35$ of his debts, a man finds that the remainder can be paid with $19500. What was his entire indebtedness ? 8. A boat load of wheat was so damaged that it was sold for $8500, which was 15$ less than its original value. What was its value before it was damaged? 9. The attendance of pupils at a school during May was 954, which was 6$ more than attended during April, and this was 80$ more than attended during February. What was the attendance for February ? 10. Which is better, to invest in a house that will rent for $30 per month, at 6$ on its value, or to invest the same amount in a farm that in two years will bring $7000 ? How much better in the two years ? REVIEW OF THE PRINCIPLES OF PERCENTAGE. 546. 1. To find the percentage, the base and rate being given. Rule. — Multiply the base by the rate expressed decimally. 2. To find the base, the percentage and rate being given. Rule. — Divide the percentage by the rate expressed decimally. 3. To find the rate, the percentage and base being given. Rule. — Divide the percentage by the base, carrying the quotient to two decimal places. 4. To find the amount per cent., the rate being given. Rule. — Add the rate to the unit 1. 5. To find the difference per cent., the rate being given. Rule. — Subtract the rate from the unit 1. 6. To find the amount, the base and rate being given.. Rules. — 1. Mul- tiply the base by the rate, and to the product add the base. Or, 2. Multiply the base by 100 per cent, plus the rate. 170 EXAMPLES FOR PRACTICE IN PERCENTAGE. 7. To find the difference, the base and rate being given. Rules. — Multiply the base by the rate, and subtract the product from the base. Or, Multiply the base by 100 per cent, minus the rate. 8. To find the base, the amount and rate being given. Rule. — Divide the amount by 100 per cent, plus the rate. 9. To find the base, the difference and rate being given. Rule. — Divide the difference by 100 per cent, minus the rate. 547. Percentage is applied to two classes of problems: First, to those in which time is not an element; as, Profib and Loss, Com- mission, Brokerage, Insurance, Taxes, Customs or Duties, and Trade Discounts. Second, to those in which time enters as an element; as, Interest, Bank Dis- count, True Discount, Equation of Accounts, and Exchange. Remark. — The pupil should be drilled in the formulas and rules of simple or abstract Percentage as above, and in their application to problems in applied Percentage to follow. MISCELLANEOUS EXAMPLES FOR PRACTICE. 548. 1. At the battle of Waterloo, of the 145000 combatants, 51000 were either killed or wounded. What per cent, were uninjured ? 2. The pressure on a steam boiler was 61.2 lb., after it had been reduced 10$. What was it before the reduction ? 3. A pupil in examination answered correctly 56 questions, which was 20$ less than the number asked him. What should be his average, on a basis of 100? J/.. By assessing a tax of f $, $175000 was raised in a county. What amount of property was taxed ? 5. A benevolent lady gave $10500 to three charities; to the first she gave $2500, to the second $4500, and to the third the remainder. What per cent, did each receive ? 6. On attaining his majority, a son finds his age is 62-£$ less than the age of his father. Find the sum of their ages ? 7. If 8$ of B's money equals 24$ of C's, how much has C, if B has $324 ? 8. A farmer bought a horse, a mule, and a cow, for $385. The mule cost 15$ less than the horse, and the cost of the cow was 7|$ of that of the horse. What was the cost of each ? 9. A creditor, after collecting 21f$ of a claim, lost the remainder, which was $3918.75. What was the sum collected ? 10. A woman weaving a rag carpet used 185$ more weight of rags than of warp. How many pounds of each in a bale of carpet weighing 96£ pounds ? 11. The sum paid for two watches was $384, and 75$ of the sum paid for one equalled 105$ of the sum paid for the other. Find the price of each. 12. If A has 35$ more money than B, and B has 25$ more than C, how much has C, if A has $192 ? 13. If a gain of $4755 was taken out of a business at 1* d 12. Having bought 75 barrels of apples for $187.50, I sold them at a loss of 20$. How much did I receive per barrel ? IS. A sells a steam tug to B, gaining 12£$, and B sells it to C for $4130, and makes a profit of 18$. How much did the tug cost A ? llf.. What per cent, is lost on an article that is sold for two-thirds of its cost? 15. A farmer, after selling 1760 barrels of apples, had 20$ of his crop left. How many barrels had he at first? 16. I lost 25$ of a consignment of berries. At what per cent, of profit must the remainder be sold, in order that I may gain 10$ on the whole ? v\-V^ 17. A Texas farm of 160 acres was bought at $15 per acre; $354 were paid for fencing, $480 for breaking, $626 for a house, and $220 for a barn. At what price per acre must it be sold, to realize a net profit of 25$ on the investment ? 18. King sold his wheel at 33^$ gain, and with the money bought another, which he sold at a loss of 25$, receiving therefor $120. Did he gain or lose, and how much ? 19. What per cent, more is -J than £ ? 20. Cloth, bought at $4 per yard, must be marked at what price in order that the seller may make a reduction of 10$ from the asking price and still gain 12-J$ on the cost ? 21. If 25$ of the selling price is gain, what is. the per cent, of gain ? 22. I sell | of a stock of goods for $27, thereby losing 20$. For what must I sell the remainder, to make a profit of 20$ on the whole ? p ^ 23. If 30$ of a farm sold at 33£$ gain, and 30$ of the remainder at 15$ gain, how much was the total gain, if the remainder was sold at cost for $7350 ? 21/.. What per cent, of cost is realized on goods marked 25$ advance and sold at 20$ off from the marked price ? 180 MISCELLANEOUS EXAMPLES IN PROFIT AND LOSS. 25. A banker Dought a mortgage at 7£$ less than its face value, and sold it for 3$ v more than its face value, thereby gaining $981. 75. What was the face value of the mortgage ? 26. At what price should damaged goods ba marked to lose 25$, the first cost having been 36 30. What price must be asked for 1000 pounds of coffee, costing 18^ per pound, in order that the seller may deduct 10$ from the asking price for bad debts, allow 16f$ for loss in roasting, and still gain 20$ on the cost? '/S 31. B and C each invested an equal amount of money in business; B gained 12-J$ on his investment, and C lost $5275 ; C's money was then 42$ of B's. How many dollars did each invest ? 32. A trader lost 33J$ on 20$ of an investment, and gamed 12^$ on the remainder, thus realizing a net gain of $1000, Had he gained 20$ on J-, and lost 25$ on the remainder, what would have been his net profit ? 33. A manufacturing company's per cent, of gain on a self-binder was 25$ less than that of the general agent; the general agent's profit was 20$, he thereby gaining* $25.30. What did it cost to make the machine ? j>\\^ 3J+. Of a cargo of 8000 bushels of oats, costing 35^ per bushel, 25^ was destroyed by fire. What per cent, will be gained or lost, if the remainder of the oats are sold at 45^ per bushel ? 35 For what must hay be sold per ton, to gain 16|$, if, by selling it at $18 per ton, there is a gain of 25$ ? \V 36. Jones sold \ of a stock of goods at cost, i at a gain of 35$, -J at a loss of 25$, and ^ at a gain of 10$. At what per cent, of its cost must he sell the remainder to net cost on the whole ? 37. After a carriage had been used two years, it was sold for $5 less than it cost, the seller thereby sustaining a loss of 3J$ of the selling price. How much was the first cost of the carriage ? 38. If oranges cost $1.80 per hundred, at what price must they be marked to ensure a gain of 20$, and make allowance for 28$ decay, and 25$ bad debts in selling ? v 39. Having paid 40^ per pound for tea, at what retail price must it be marked, that I may allow 12|$ for bad debts and gam 40$ on the cost ? Jfi. Six wheel-rakes were sold for $21 each; three of them at a gain of 20$, and the others at a loss of 20$. What was the net gain or loss ? Jf.1. A stock of goods is marked 22-J-$ advance on cost, but becoming damaged, is sold at 20$ discount on the marked price, whereby a loss of $1186.40 is sus- tained. What was the cost of the goods ? MISCELLANEOUS EXAMPLES DST PROFIT AND LOSS. 181 4%, My retail price of Axminster carpet is $3.50 per yard, by which I gain 25$. If I sell at wholesale, at a discount of 25$ from the retail price, how much do I receive per yard. What is my per cent, of gain or loss, and how much is my actual gain or loss by selling 1000 yards at wholesale ? 43. If the loss equalled -J of the selling price, what was the per cent, of 44. A grocer bought 200 quarts of berries, at 11^ per quart, and 150 quarts of cherries, at 6^ per quart. Having sold the cherries at a loss of 30$, for how much per quart must he sell the berries, to gain 15$ on the whole ? 45. A sells two horses to B at an advance of 16f$, B sells them to at an advance of 25$, and C sells them to D for $735, thereby making a profit of 20$. How much did A pay for the horses ?^^° 46. Having bought 48 pounds of coffee, at the rate of 3-J- pounds for 91^, and 84 pounds more at the rate of 7 pounds for $1.26, I sold the lot at the rate of 9 pounds for $1.53. What was my per cent, of gain or loss ? 47. By selling at a loss of 6 or base, is $60, the discount to be deducted will be 40 per cent, of $60, or $24 ; and the net price will | 60 _ cos k be $60 minus $24, which equals $36. 24 ss discount. $ 36 = net selling price. Example (second illustration). — The list price of a threshing machine is $900. What is the net price, if a discount series of 25$, 20$, and 10$ is allowed ? Operation. $900 = list price. 235 = 25$, or \ = 1st discount. Explanation.— From the list price take the $675 = rem. after 1st discount. nrst discount, and make each remainder the base 135 = 20$, or 4 = 2d discount. for the succeedin S discount. The last remainder will be the net price. $540 = rem. after 2d discount. 54 =s 10$, or ^ = 3d discount. $486 s= rem. after 3d discount, or net price. Remark. — In like manner treat any series of discounts. 184 EXAMPLES IN TRADE DISCOUNT. Rule. — Deduct the first discount from the list price, and each subse* quent discount from each successive remainder. EXAMPLES FOR PRACTICE. 575. -?. What is the selling price per dozen of hats, listed at $36, and discounted 20$ and 15$ ? 2. Find the net price of a ton of fence wire, listed at 9^ per pound, and sold at 70$ and 5$ off. 3. Find the net cost to the purchaser of a bill of goods invoiced at $1100, from which discounts of 20$ and 25$ were allowed. 4* An invoice of silk amounting to $12000 was sold Sept. 21, 1888, at a discount of 25$, 20$, and 12|$, with a further discount of 10$ to be allowed if paid within 30 days. How much cash will pay the bill Oct. 15, 1888 ? 5. Having bought merchandise at 25$ and 15$ discount from the list price of $1500, I sell it at 15$, 15$, and 10$ from the same list price. Do I gain or lose, and how much ? 6. A wholesale dealer offers cloth at $2.40 per yard, subject to a discount of 25$, 20$, 10$, and 5$. How many yards can be bought for $246.24 ? 7. "What is the net cost of a bill of goods invoiced at $2150, and sold at a discount of 15$, 10$, 5$, and 3$ ? 8. Three drummers, A, B and C, offer me the same grade of goods at the same list price. A offers to discount 25$ and 15$; B 20$ and 20$; and C 15$, 15$, and 10$. With which will it be most advantageous for me to deal, and how much would I save from a list price of .$200 ? Remarks. — 1. It is often convenient in finding the net price to multiply the list price by 1 minus the first discount, the remainder by 1 minus the next, and so on. 2. The order in which the discounts of any series are considered is not material, a series of 25, 15, and 10 being the same as one of 15, 10, and 25, or of 10, 25, and 15. 576. To find the Price at which Goods must be Marked to Insure a Given Per Cent, of Profit or Loss, the Cost and Discount Series being given. Example (first illustration). — Having bought goods for $105, at what price must they be marked to allow a discount of 25$, and still make a profit of 10$ ? Explanation. — The cost, $105, is 100 per Operation cent, of itself ; the rate of discount to be al- &105 = the cost lowed is 25 per cent.; 100 per cent, minus 25 " ' * . per cent., or 75 per cent., is the per cent, which ♦ - 1Q = $ to be gamed. the prfce t0 be insured is of the price t0 ^ $10.50 as gain to be insured. asked. And if 10 per cent, must be insured, 105.00 = cost. the goods must actually bring 10 per cent., or r777~T7 ,,. , i „ n $10.50 more than cost, or $115.50. And since $115.50 selling price to be insured. T . , .. - OK '*•*». i * ^ ° r a deduction of 25 per cent, is to be made from 75 ^ $115 50 ^ e ^king price, the selling price, $115.50, will —^ — be only 75 per cent, of the asking price. $154. asking price. Therefore, divide $115.50 by .75, and the qucv tient, $154, will be the asking price. EXAMPLES IN TEADE DISCOUNT. 185 Example (second illustration). — A seal sacque cost a manufacturer $240. At what price must it be marked, that a discount series of 25$, 20$, and 20$ may be allowed, and he still make a profit of 30$ ? Opebation. $240 = cost or base. $1.00 = $ of price realized. .30 = $ to be gained. .25 = $ of 1st discount. 72 = gain to be insured. .75 = $ of price to be received Sn 240 = cost. order to gain 30$ and allow $312 = price to be received. 1st discount. .75 ) $312.00 $1.00 = $ of price. — 7777 , . • • j .20 = $ of 2d discount. $416 = asking price in order to pay $240, gain .80 = $ of price to be received in 30$, and allow a dis- order to gain 30$ and dis- count of 25$. count 25$ and 20$. .80 ) $416.00 $1.00 = $ of price. „>_,^ , . . . , .20 as $ of 3d discount. $o20 = asking price in order to pay $240, gain 30$, and .80 = $ of price to be received in discount 25$ and 20$. order to gain 30$ and dis- count 25$, 20$, and 20$. .80 ) $520.00 $650 = asking price in order to gain 30$ and allow the full discount series. Rule. — Add to the cost the gain required, and divide consecutively by 1 minus each of the rates in the discount series. EXAMPLES FOR* PRACTICE. 577. -?. What must be the asking price of a watch, costing $18, that 33-J$ may be gained, after allowing the purchaser a discount of 20$ ? 2. Having bought an invoice of lawn mowers at $15 each, I desire to so mark them that I may gain 20$, and still discount 25$ and 20$ to my customers. At what price must each be marked ? 8. Having paid $8800 for a stock of goods, what price must be asked for it, in order to gain $1100 and allow 12-J$ and 10$ discount ? 4. After buying velvet at $5 per yard, I so marked it as to allow discounts of 25$, 20$, and 16|$ from the marked price, and yet so sell it as to lose but 10$ on my purchase. At what price per yard was the velvet marked ? 5. The cost of manufacturing silk hats being $36 per dozen, how must they be marked, that a gain of 16f $ may be realized by the manufacturer, after allow- ing discounts to the trade of 20$ and 12-J$? 6. If a carriage be marked 33£$ above cost, what per cent, of discount can be allowed from the marked price and realize cost? 7. If the list price of an article is 25$ advance on the cost, what other per cent, of discount than 10$ must be allowed, to net 10$ gain bv sale ? 186 EXAMPLES IN TRADE DISCOUNT. 8. A merchant purchasing a bill of goods was allowed discounts from the list price of 15$, 10$, 10$, and 6$. If the total discount allowed was $352.81, what must have been the asking price of the goods ? 578. To find a Single Equivalent Per Cent, of Discount, a Discount Series being given. Example. — "What single rate of discount is equal to the series 25$, 20$, 10$, and 5$ ? Operation. $1000 = assumed list price or base. 250 = 1st discount. $750 == 1st rem. or 2d base. 150 = 2d discount. Explanation.— Assume $1000 as the list * price, and successively deduct the discounts as $(300 = 2d rem. or 3d base. by the series, and compare the result with the 60 = 3d discount. base assumed. $540 = 3d rem. or 4th base. 27 = 4th discount. $513 == 4th or last rem. or net price. $1000 = list price, or base. 513 = net price. $487 = total discount on $1000, which, divided by 1000, gives 48 T V, the per cent, of discount equivalent to the given series. Rule. — From $1000 as a list price, or base, take the discounts in order; subtract the final remainder from the base taken, and the result will be the total discount ; then point off from its right three places for decimals, and the expression thus obtained will be the equivalent per cent, of dis- count required. Remakk.— This is the usual method, and it is more convenient for business men than to compute the net price for each sale through a series of discounts. EXAMPLES FOR FKACTICE. 579. 1. Find a single discount equivalent to a series of 10$ and 10$. 2. Find a single discount equivalent to a series of 25$, 15$, and 5$. 8. Find a single discount equivalent to a series of 30$, 20$, 10$, and 3$. Jf. Goods were sold 25$, 35$, 20$, and 15$ off ; what single discount would have insured the same net price ? 5. What is the difference between a single discount of 50$ and a series of 20$, 20$, and 10$ ? 6. What per cent, of the list price will be obtained for goods sold at a discount of 35$, 20$, 15$, 10$, and 5$ ? 7. From a list price, I discounted 30$, 25$, 20$, 15$, 12£$, 10$, and 5$. What per cent, better for the purchaser would a single discount of 75$ have been ? STOfiAGB. 187 STORAGE. 580. Storage is a provision made for keeping goods in a warehouse for a time agreed upon, or for an indefinite time, subject to accepted conditions. 581. The term storage is used also to designate the charges for keeping the goods stored. 582. Rates of storage may be fixed by agreement of the parties to the con- tract, but are often regulated by Boards of Trade, Chambers of Commerce, Associations of Warehousemen, and by legislative enactment. 583. Storage Charges may be made at a fixed price per package or bushel, or at a fixed sum for a term or terms ; they may be made for a term of days or months ; but usually, if the goods stored are taken out before the storage time expires, the charge made is for the full time. 584. The rates of storage often vary for grains, or goods of different grades or values, and also on account of different modes of shipment. Remarks. — Storage Receipts, especially of grains, are frequently bought and sold under the name of "Warehouse Receipts "or "Elevator Receipts," as representing so much value by current market reports. 585. Cash Storage is a term applied to cases in which the payment of charges is made on each withdrawal or shipment, at the time of such withdrawal or shipment, notwithstanding the fact that the owner may still have goods of the same kind in store at the warehouse. 586. Credit Storage is a term applied to cases in which sundry deposits or consignments are received, from which sundry withdrawals or shipments are made, and all charges adjusted at the time of final withdrawal. Remarks. — 1. When deposits or consignments, and withdrawals or shipments, are made at different times, credit is to be given for the amount of each deposit or consignment, from the date to its final withdrawal or shipment, and credit given to the owner or consignor for each withdrawal or shipment, from date up to the time of settlement. 2. In the private bonded warehouses of the United States, goods may be taken out at any time, in quantities not less than an entire package, or, if in bulk of not less than 1 ton, by the payment of duties, storage, and labor charges. The storage charges are computed for periods of one month each, a fractional part of a month being counted the same as a full month. 1 3. Drovers sometimes hire cattle fed on account, entering and withdrawing them as circum- stances require; such accounts are closed in the same manner as are those for storage. 587. To find the Simple Average Cash Storage. Example. — There was received at a storage warehouse : Oct. 11, 300 bar. apples; Oct. 30, 250 bar. potatoes; Nov. 13, 200 bar. apples; Nov. 20, 60 bar. quinces ; Nov. 28, 280 bar. apples. The merchandise was all delivered Dec. 2. 188 STORAGE. If the contract specified that the rate of storage was 5$ per barrel for a period of 30 days average storage, what was the storage bill ? Operation. The storage of 300 bar. for 52 da. = the storage of 1 bar. for 15600 da. The storage of 250 bar. .or 33 da. = the storage of 1 bar. for 8250 da. The storage of 200 bar. for 19 da. = the storage of 1 bar. for 3800 da. . The storage of 60 bar. for 12 da. == the storage of 1 bar. for 720 da. The storage of 280 bar. for 4 da. = the storage of 1 bar. for 1120 da. The total storage = the storage of 1 bar. for 29490 da. Or, 983 periods of 30 days each; $.05 X 983 s= $49.15, storage bill. Explanation. — The 300 barrels constituting the first deposit or delivery were stored from Oct. 11 to Dec. 2, or for 52 days; the storage of 300 bar. for 52 days equals the storage of 1 barrel for 15600 days; the storage of 250 barrels for 33 days equals the storage of 1 barrel for 8250 days; that of 200 barrels for 19 days equals 1 barrel for 3800 days; that of 60 barrels for 12 days equals 1 barrel for 720 days; that of 280 barrels for 4 days equals 1 barrel for 1120 days. The total storage was equal to that of 1 barrel for 29490 days, or for 983 storage terms or periods of 30 days each. Since the storage charge was 5^ per barrel for each average period of 30 days, the charge would amount to $.05 X 983, or $49.15. Rule. — Multiply the number of articles of each receipt by the number of days between the time of their deposit and withdrawal ; divide the sum of these products by the number of days in the storage period, and multiply the quotient by the charge per period. EXAMPLES FOR PRACTICE. 588. 1* There was received at a warehouse: May 30, 4000 bu. wheat; June 5, 2600 bu. oats; June 24, 3500 bu. barley; July 18, 5000 bu. corn. If all of this was shipped July 20, what was the storage bill, the charge being 1-J^ per bushel per term of 30 days average storage ? 2. A farmer received for pasture: April 30, 12 head of cattle; May 15, 14 head of cattle; May 23, 27 head of cattle; June 9, 5 head of cattle; June 30, 8 head of cattle; July 16, 40 head of cattle. All were delivered July 25, and the charges were 75^ per head for each week of 7 days average pasturage. How much was his bill ? 589. To find the Charge for Storage with Credits. Example. — The storage charges being 2$ per barrel for a month of 30 days average, what will be the bill in the following transaction ? Received. July 19, 100 bar.; July 31, 240 bar.; Sept. 8, 360 bar. Delivered. Aug. 15, 300 bar. ; Sept. 12, 200 bar., Oct. 1, 200 bar. EXAMPLES IN" STORAGE. 189 Operation. From July 19 to July 31 = 12 da. ; 100 bar. stored for 12 da. = 1 bar. stored for 1200 da. July 3t 240 bar. received. From July 31 to Aug. 15 = 15 da.; 340 bar. stored for 15 da. = 1 bar. stored for 5100 da. Aug. 15 300 bar. delivered. From Aug. 15 to Sept. 8 = 24 da. ; 40 bar. remaining for 24 da. = 1 bar. stored for 960 da. Sept. 8 360 bar. received. From Sept. 8 to Sept. 12 = 4 da. ; 400 bar. stored for 4 da. = 1 bar. stored for 1600 da. Sept. 12 200 bar. delivered. From Sept. 12 to Oct. 1 = 19 da.; 200 bar. remaining for 19 da. = 1 bar. stored for 3800 da. Oct. 1 200 bar. delivered. 000 Total = 1 bar. stored for 12660 da., Or, 422 terms of 30 da. each; $.02 X 422 = $8.44, total storage bill. Explanation. — 100 barrels were stored for 12 days, when 240 barrels were added; these 340 barrels were stored 15 days, when 300 barrels were withdrawn; the 40 remaining barrels were stored 24 days, when 360 barrels were added; these 400 barrels were stored 4 days, when 200 barrels were withdrawn; the remaining 200 barrels were stored 19 days and then withdrawn. The total storage thus equalled that of 1 barrel for 12660 days, or for 422 terms of 30 days each; and since the charge for 1 term is $ .02, for 422 terms it would be 422 times $ .02, or $8.44, the total amount of the bill. Rule, — I. Multiply the number of articles first received by the num- ber of days between the date of their receipt and the date of the next receipt or delivery ; add the number of articles of such next receipt, or subtract the number of such delivery, as the case may be, and so pro- ceed to the time of final delivery. II.— Divide the aggregate storage by the number of days in the storage term, and multiply the quotient by the storage charge per term. EXAMPLES FOR PRACTICE. 590. -7. What will be the storage charge, at 4|^ per barrel, for a term of thirty days average, in the following transaction ? Received. Delivered. Feb. 8, 180 bar. flour. Mar. 1, 100 bar. apples. Feb. 27, 100 bar. apples. Mar. 28, 190 bar. flour. Mar. 8, 60 bar. potatoes. Apr. 15, 60 bar. potatoes. Mar. 13, 300 bar. flour. Apr. " 60 bar. flour. Apr. 29, 230 bar. flour. 2. A drover hired pasture of a farmer, agreeing to pay $4.20 per head of stock pastured for each average term of 30 days. What was the amount of the bill, the receipts and deliveries being as follows ? Received. Delivered. June 15, June 27, July 5, July 29, July 31, 21 head of cattle. 20 head of cattle. 15 head of cattle. 40 head of cattle. 40 head of cattle. July 1, 30 head of cattle. July 20, 15 head of cattle. July 30, 15 head of cattle. Aug. 21, the remainder. 190 EXAMPLES IN STORAGE. 591. To find the Storage where Charges Vary. Example. — At a warehouse there was received and delivered flour, as follows: Delivered. Jan. 23, 250 bar. Mar. 1, 400 bar. Received. Jan. 3, 150 bar. Jan. 20, 200 bar. Feb. 1, 300 bar. The storage charge on the above was, 50 per barrel for the first 10 days or part thereof, and 30 per barrel for each subsequent period of 10 days or part thereof. What sum must be paid in settlement? Date. Receipts and Deliveries. Jan. 3, received 150 bar. " 20, " 200 " Operation. Bate. Storage. " 23, 350 delivered 250 Feb. 1, 100 received 300 ft Mar. 1, 400 delivered 400 <( ec 66 in store. 150 bar. stored 20 days, or 2 terms, 80 = $12.00 100 " 3 " 1 term, 50 = 5.00 remainder. in store. 100 bar. stored 40 days, or 4 terms, 140 — $14.00 300 " 28 " 3 " 110 = 33.00 Total storage, = $64.00 Explanation. — Of the 250 barrels delivered Jan. 23, 150 barrels had been in store since Jan. 3, 20 days or 2 terms, and the charge was 5 cents plus 3 cents, or 8 cents per barrel, which equals $12 storage. The remaining 100 barrels of the delivery of Jan. 23, had been in store only since Jan. 20, 3 days or 1 term, at 5 cents per barrel, equal to $5 storage. Of the 400 barrels delivered Mar. 1, 100 barrels had been in store since Jan. 20, 40 days or 4 terms, at 5 cents plus 3 cents plus 3 cents plus 3 cents, or 14 cents per barrel, equal to $14 storage; while the remaining 300 barrels had been in store since Feb. 1, 28 days or 3 terms, at 5 cents plus 3 cents plus 3 cents, or 11 cents per barrel, equal to $33 storage. By addition, the total storage is found to be $64. Rule. — Multiply the number of articles of each delivery by the charge for the term or terms stored, and add the products so obtained. EXAMPLE FOR PRACTICE. The receipts and deliveries of goods at a storage warehouse were as Delivered. Sept. 20, 100 bar. . 1. T he receipts am s: Received. Sept. 2, 100 bar. Sept. 25, 200 bar. Oct. 19, 350 bar. Oct. 31, 150 bar. Nov. 7, 200 bar. Sept. Oct. Oct. Oct. 30, 10, 20, 30, 100 bar. 100 bar. 100 bar. 100 bar. the remainder. Nov. 20, The contract required the payment of 60 per barrel for the present term of 30 days or fraction thereof, and 30 per barrel for each subsequent term of 30 days or fraction thereof. Find the storage bill. UUMillbsiOJU. 191 COMMISSION. 593. An Agent is a person who transacts business for another; as, the purchase or sale of merchandise or real estate, collecting or investing money, etc. 594. An agent who receives goods to be sold is sometimes called a factor or commission merchant; one employed to buy or sell stocks or bonds, or to nego- tiate money securities, is called a broker. 595. Commission is an allowance made to agents or commission merchants for transacting business. It is usually a percentage of the money involved in the transaction, although sonfetimes it is computed at a certain price per bale, bushel, barrel, etc. 596. The Agent's Commission for selling is computed on the gross pro- ceeds, and for purchasing on the prime cost. 597. The Principal is the person for whom the business is transacted. 598. A Consignment is a shipment of goods from one party to another, to be sold on account of the shipper, or on joint account of the shipper and the consignee. The shipper is called the Consignor, and the one to whom the goods are shipped is called the Consignee. 599. Guaranty is a per cent, charged by an agent for assuming the risk of loss from sales made by him on credit, or for giving a pledge of the grade of goods bought; it is computed the same as are commission charges. 600. The Gross Proceeds of a sale or collection is the total amount received by the agent before deducting commission or other charges. * 601. The Net Proceeds is what remains after all charges have been deducted. Remakes.— Charges may be for commission, guaranty, freight, inspection, cartage, storage, or any other outlay incident to the sale. 602. An Account Sales is a statement in detail rendered by a Consignee to his Consignor, showing the sales of the consignment, all of the charges or expenses attending the same, and the net proceeds. 603. An Account Purchase is a detailed statement made by a purchasing agent to his principal, having the quantity, grade and price of goods bought on his account, all the expenses incident to the purchase, and the gross amount of the purchase. 604. Commission compares with Abstract Percentage, as follows: The Prime Cost or Gross Selling Price = Base. The Rate Per Cent, of Commission = Rate. 192 COMMISSION". The Commission for either buying or selling, or for guaranty of quality or credit = Percentage. The remittance to Purchasing Agent, including both Commission and Invest- ment = Amount. The Selling Price, minus the Commission = Difference. 605. — To find the Commission, the Cost or Selling Price and Per Cent, of Commission being given. Example. — How much commission will be due an agent who buys $8000 worth of coal, on a commission of 5$? Operation. Explanation.— Since the rate of commission $8000 = investment or base. \ s 5 per cent., the whole commission due the agent .05 = per cent, of commission. will be 5 per cent, of the investment, $8000, or $400 = commission or percentage. $ Remark. — In case of sales, proceed in like manner, treating the selling price as the base. Rule.— Multiply the cost or selling price by the rate per cent, of commission. Formula. — Commission = Cost or Selling Price X Rate per cent, of Commission. EXAMPLES FOR PRACTICE. 606. 1. An agent sold a house and lot for $6000, and charged 3$ for his services. How much was the commission? 2. Having agreed to pay an agent 3$ for all purchases made by him, how much will be due him, if he buys for me goods costing $2500? 3. If an agent's charges are 2$, how much commission will he earn by selling property valued at $12500? J/.. I owned one-half of a stock of goods sold by an agent for $10000. If the agent charged b\er pound, the consignor received as net proceeds $853. 74. What was the per cent, of commission charged for selling, if the charges for storage and insurance amounted to $51.60? EXAMPLES IN COMMISSION". . 195 11. Find the gross proceeds of a sale made by an agent charging %^f> for com- / mission, 5$ for guaranty, $17.65 for cartage, $11.40 for storage, and $3.25 for insurance, if the net proceeds remitted amount to $1714.10. 12. A Milwaukee agent received $83195.28, with instructions to invest one- half of it in wheat, at 800 per bushel, and the balance, less all commissions, in wool, at 200 per pound. If his commission for buying the wheat is 2$, and that for buying the wool is 5$, how many pounds of wool will he buy, and what will be the amount of his commissions? IS. I sent $3402.77 to my Atlanta agent for the purchase of sweet potatoes, at $1.60 per barrel; his charges were, for commission, 2£#; guaranty, 3$; dray- age, 10 per barrel; and freight, $200. How many barrels did he buy, and how much unexpended money was left in his hands to my credit? llf.. A Texas buyer shipped 33000 lb. of coarse wool to a Boston agent to be sold on commission, and gave instructions for the net proceeds to be invested in leather. If the agent sold the wool at 180 per pound, on a commission of 2$, and charged 10$ for the purchase and guaranty of grade of the leather, what was the amount of his commissions? 15. I received from Duluth a cargo of 1 6000 bu. of wheat, which I sold at $1. 10 per bushel, on a commission of 4$; by the consignor's instructions I invested the net proceeds in a hardware stock, for which I charged 5$ commission. What was the total commission, and how much was invested in hardware? 16. Having sent a New Orleans agent $1835.46 to be invested in sugar, after allowing 3$ on the investment for his commission, I received 32400 pounds of sugar. What price per pound did it cost the agent? 17. An agent in Providence received $828 to invest in prints, after deducting his commission of Z\i<>. If he paid 7|0 per yard for the prints, how many yards did he buy? 18. The fees of the general agent of an insurance company are 5$ on all sums received, and 5$ additional on all sums remaining in his hands at the end of the year, after all losses and the expenses of his office are paid. He receives during the year $117416.25, paid losses to the amount of $91140.50, and the expenses of his office were $3267.70. Find his total fees. 19. An agent sold on commission 81 self-binders, at $140 each, and 113 mowers, at $65 each, remitting $10224.90 to his principal. Find the rate of commission. 20. A commission merchant received a consignment of 660 bales of cotton, of an average weight of 510 pounds, which he sold at 12-|0 per pound, on a commission of 3$, charging 100 per bale for cartage. He invested for the con- signor $9416.20 in bacon, charging 5$ for buying, and remitted cash to balance consignor's account. How much was the cash remittance? 21. An agent received $4325, to invest in mess-pork, at $16 per barrel, after deducting his purchasing commission of 4$. If the charges for incidentals were >^$ $81.40, besides cartage of 750 per load of 8 barrels, how many barrels did he buy, and what unexpended balance does he place to the credit of his principal? ¥ 196 MISCELLANEOUS EXAMPLES IN" COMMISSION". 22. A street-car company bought 35 horse-cars through a Troy agent, at $850 each. If the freight was $17.50 on each car, and the agent's commission 3\% for purchasing, what was the total cost to the company? 3>l/ ^ 9 ^'l ] & 23. I received from Day & Son, of Chicago, a ship load of corn, which I sold for 60 *) 26. The holder of a doubtful claim of $850, handed it to an agent for collec- tion, agreeing that, for every dollar sent him by the agent, the agent might keep for himself 20^. The agent succeeded in collecting but 80$ of the debt. How much did the agent remit, how much commission did he receive, and what was his per cent, of commission? ((**+* fH>%- G»-+~ 'if ^73 — ff»*x /£y^ 27. I remitted $10500 to a Duluth agent to be invested in wheat, allowing him a commission of 3$ for investing. The agent paid 95^ per bushel for the wheat, and charged me \\$ a bushel per month for storage. At the end of 4 months the agent sold the wheat at $1.10 per bushel, on a commission of 5$. If I paid $350 for the use of the money, did I gain or lose by the operation, and how much? 28. My Memphis agent sends me an account purchase of 350 bales of cotton, averaging 480 pounds each, bought at 15^ per pound, on a commission of 2|$. His charges, other than for commission, were: freight advanced, $126.50; cartage, $53.25; and insurance, $13.75. What sum should I remit to pay the account?- 29. A Charleston factor received from Cincinnati a consignment of corn, which he sold at 75^ per bushel, on a commission of 5$; and by instructions of the consignor invested the net proceeds in cotton, at 20^ per pound, charging 3$ for buying, and 3$ additional for guaranty of quality. If the total amount of the agent's commission and guaranty was $1640, how many bushels of corn were received? SO. ^ Buffalo brewer remitted $21500 to a Toronto commission merchant, with instructions to invest 40$ of it in barley, and the remainder, less all charges, in hops. The agent paid 60^ per bushel for barley, and 20^ per pound for hops, charging 2$ for buying the barley, 3$ for buying the hops, and 5$ for guaran- teeing the quality of each purchase. If his incidental charges were $187.50, what quantity of each product did he buy, and what was the amount of his commission? CUSTOM-HOUSE BUSINESS. 197 CUSTOM-HOUSE BUSINESS. 612. Duties, or Customs, are taxes levied by the Government on imported goods, for revenue purposes and for the protection of home industry. 613. Duties are of two kinds, ad valorem and specific. 614. An Ad Yalorem Duty is a certain per cent, assessed or levied on the actual cost of th° goods in the country from which they are imported, as shown by the invoice. 615. A Specific Duty is a tax assessed or levied upon the number, weight, or measure of goods, regardless of their value; as, a fixed sum per bale, ton, barrel, etc. Remark. — Upon certain goods both specific and ad valorem duties are levied. 616. A Custom-House is an office established by the Government for the transaction of business relating to duties, and for the entry and clearance of vessels. Remark.— 1. The ports at which custom-houses are established are called ports of entry. 2. The waters and shores of the United States are divided into collection districts, in each of which there is a port of entry, which is also a port of delivery; other ports than those of entry may be specified as ports of delivery. Duties are paid, and entries and clearances made, at ports of entry only; but after vessels have been properly entered, their cargoes may be •discharged at any port of delivery. 617. An Invoice, or Manifest, is a written account of the particular goods sent to the purchaser or factor, with the actual cost, or value, of such goods, made out in the currency of the country from which they are imported. Remarks — All invoices are made out in the weights, measures, etc., of the place from which the goods are imported. , 618. A Tariff is a schedule of goods, and the rates of import duties imposed by law on the same. 619. The Free List includes classes of goods that are exempt from duty. 620. Tonnage is a tax levied upon a vessel independent of its cargo, for the privilege of coming into a port of entry. 621. Duties are collected at the port of entry by a custom-house officer appointed by the United States Government, and known as the Collector of the Port. Under him are deputy collectors, appraisers, weighers, gaugers, etc. 622. The Collector of the Port supervises all entries and papers pertain- ing to them; estimates all duties, receives all moneys, and employs all weighers, gaugers, etc. 62?». Before estimating specific duties, allowances are made; these allowances are called Tare, Leakage, Breakage, etc. 198 CUSTOM-HOUSE BUSINESS. 624. Tare is an allowance made for the box, bag, crate, or other covering of the goods. 625. Leakage, determined by gauging, is an allowance made for waste of liquids imported in barrels or casks. 626. Breakage is an allowance made for loss of liquids imported in bottles. 627. Gross Weight is the weight before any allowances are made. 628. Net Weight is the weight after all allowances have been made. Remarks. — 1. The ton used at the United States Custom-Houses is of 2240 avoirdupois pounds. 2. Duties are not computed on fractions of a dollar; if the cents in the invoice are less than 50, they are rejected; if 50 or more, they are counted as a dollar. 629. The Naval Officer^ appointed only at the more important ports, receives copies of all manifests, countersigns all documents issued by the Col- lector, and certifies his estimates and accounts. 630. The Surveyor superintends the employees of the Collector, and revises entries and permits. 631. The Appraiser examines imported articles, and determines their duti- able value and also the rate of duty to be charged. 632. The Store-keeper has charge of the warehouse. Remarks. — 1. Warehousing is depositing imported goods in a government or bonded ware- house. 2. A bonded warehouse is used for storing goods on which the duties have not been paid. 3. Goods may be withdrawn from a bonded warehouse for export, without the payment of the duties. If goods on which the duty has been paid are exported, the amount of duty so paid is refunded; the sum so refunded is called a drawback. 4. Smuggling is bringing foreign goods into the country without paying the required duty. This is done either by not entering them at a Custom-House, or by showing less than their real value in the invoice. It is a crime, for the prosecution and punishment of which stringent laws are enacted. 5. Many merchants employ a Custom-House Broker, one familiar with the laws, to enter goods for them. 633. To find Specific Duty. Example. — What is the specific duty on 140 casks of alcohol, of 60 gallons each, at lh

^ hicb , on , _ ? , gives 7980 gallons, on which to charge duty. _4U) gal. — leakage. Since the duty fa 15 centg per gallon> for 7980 7980 = net quantity. gallons it will be $1197. $.15 == duty per gallon. $1197.00 = specific duty. EXAMPLES IN - CUSTOM-HOUSE BUSINESS. 199 Bule. — Multiply the net quantity by the duty per single article of the hind or class considered. EXAMPLES FOR PRACTICE. 634. 1. "What is the duty on 60 packages of figs, each of 16 lb. weight, at 24-'/ per pound, tare 5$? 2. Find the duty, at 75^ per ton, on an invoice of 897120 lb. of bituminous coal. 3. If the duty on plate glass is 25^ per square foot, how much will be charged on an importation of 200 boxes, each containing 20 plates 24 X 48 in. in size ? Jf.. Find the duty, at $2 per dozen, on 40 doz. bottles of wine imported from Lyons, if an allowance of 10$ is made for breakage. 5. If the duty is 650 per cubic foot, what amount must be paid on an impor- tation of 6 blocks of marble, each 10 ft. long, 3 ft. wide, 2 ft. high ? 6. After being allowed 10$ for leakage, a wine merchant paid $864 duty, at $2 per gallon, on 12 casks of wine. How many gallons did each cask originally contain ? 7. Find the duty on 1500 dozen empty bottles, breakage 4$, and rate of duty 10^ per dozen. 635. Applying the terms of Percentage to Ad Valorem Duties, we observe the following : The net Value, or Quantity == the Base. The Eate Per Cent. Ad Valorem = the Rate. The Duty = the Percentage. 636. To find Ad Valorem Duty. Example. — What is the ad valorem duty, at 35$, on 90 boxes of brass rivets, 25 lb. per box, invoiced at 120 per pound, tare being 6 lb. per box ? Operation. 90 X 25 lb. = 2250 lb. gross weight. 90 x 6 lb. = 540 1b. tare. 1710 lb. net weight. 1710 Explanation.— Find the net weight and $.12 = cost per pound value as the base; multiply by the rate of $205.20 = net value. duty * .35 = i of duty. 1025 615 $71.75 = duty. Remark. — The cents in the net value, being less than 50, are rejected. Rule. — Multiply the value, after all deductions are made, by the per cent, of duty assessed. 200 EXAMPLES IN CUSTOM-HOUSE BUSINESS. EXAMPLES FOR PRACTICE. 637. 1. What is the duty, at 50$, upon a consignment ot 650 dozen kid gloves, invoiced at 90 francs per dozen ? 2. An importation of English crockery was invoiced at £896, 5s. 6d. Fin<2 the duty, at 40$. 8. If an importation is invoiced at 17450 francs, what will be the duty, at 35$? 4. If the duty on sperm oil is 20$, what will it amount to in an importation of 600 barrels, of 42 gallons each, invoiced at 45^ per gallon, 3|$ being allowed for leakage ? 5. I received by steamer Kaglan, from Liverpool, the following invoice of goods : 768 yd. velvet, invoiced at £1 12s. per yd. ; 2150 yd. lace, invoiced at 3s. 4d. per yd. ; 1200 yd. broadcloth, invoiced at 15s. per yd. ; 3520 yd. carpet, invoiced at lis. 6d. per yd. If the duty on the velvet was 60$, on the lace and broadcloth 35$, and on the carpet 50$, how much was the total duty to be paid? MISCELLANEOUS EXAMPLES. 638. -?. What is the duty on 1000 yd. of brussels carpet, 27 in. wide, invoiced at 6s. 9d. per yd; duty 44^ per square yard specific, and 35$ ad valorem ? 2. If the duty on flannel is 29^ per pound specific, and 35$ ad valorem, how much must be paid on an invoice of 2150 yd., weighing 420 lb., and valued in Canada, whence it was imported, at 75^ per yard ? 8. Find the duty on 3 dozen clocks, invoiced at $21.50 each, and 6 dozen watches, invoiced at $35 each, if the ad valorem duty was 35$ on the clocks, and 25$ on the watches. Jf. How much duty must be paid on an importation of 27640 lb. of wool, invoiced at £1497 10s. 4d., if the rate of duty is 10^ per pound specific; and 11$ ad valorem? 5. I imported from Canada 7240 bushels of barley, and 17-J tons of hay, invoiced at $9.50 per ton. What amount of duties had I to pay, at 10^ per bushel on the barley and 20$ on the hay? 6. A merchant imported 300 pieces of three-ply carpet, each piece containing 75 sq. yd., invoiced at 3s. 6d. per square yard, upon which he paid a duty of 17# per square yard specific, and 35$ ad valorem. What was the total amount of duty paid ? 7. An invoice of woolen cloth, imported from England, was valued at £956 6s. If its weight was 684 lb., how much was the duty, at 50^ per pound specific, and 35$ ad valorem ? 8. Find the duty on 50 cases of tobacco, each weighing 60 lb., and 50000 Havana cigars weighing 550 lb., invoiced at $75 per M, the duty being 50^ per pound specific on the tobacco, and $2.50 per pound specific and 25$ ad valorem on the cigars. TAXES. 201 TAXES. 639. Taxes are sums of money levied on persons, property, or products, for any public purpose. 640. Capitation or Poll Taxes are levied at a certain amount for each person or head of legal voters not exempt by law. 641. Property Tax is a tax assessed or levied upon property, at a given rate per cent, of the valuation. 642. Property is of two kinds: Personal and Eeal. 643. Personal Property is movable property; as, merchandise, ships, cattle, money, stocks, mortgages, etc. 644. Real Property or Real Estate consists of immovable property; as, houses and lands. 645. Assessors are public or government officers, who appraise the value of property to be taxed, and apportion the taxes pro rata; that is, in proportion to the value of each man's property. 646. Collectors are public or government officers, who collect taxes. Kemakk. — Taxes are generally assessed and made payable in money, but in "road taxes " they may be made payable in "day's work." 647. The terms of Percentage, applied to Taxes, are: The Valuation = the Base. The tax on $1.00 = the Rate. The Sum to be raised = the Percentage. The Sum, minus the Collector's fees, or commission = the Difference. 648. To find a Property Tax. Example. — The rate of taxation in the city of Des Moines, Iowa, is lf#. What amount of tax must a person pay, whose personal property is valued at $17500, and who owns real estate assessed at 124900 ? Operation. &17500 Explanation. — Since his total valuation was $42400, and the rate of taxation If per cent., his tax would be If 24900 per cent of $42400, or $742. $42400 X .01f = $742. Rule. — Multiply the total assessed value by the rate per cent, of taxation. EXAMPLES FOB PRACTICE. 649. 1. Mr. B. owns personal property assessed at $7140, and real estate assessed at $11100, in a village in which he is taxed one-half of one per cent. Find the amount of his tax. 202 EXAMPLES IN" TAXES. & A person having lands valued at $7500, $12250 in money, and a stock of goods worth $6000, pays tax on all at the rate of ]-£#. Find his total tax. 650. To find a General Tax. Example. — A tax of $2505 is to be assessed upon the village of Livonia ; the valuation of the taxable property is $G00000, and there are 324 polls, to be assessed $1.25 each. What will be the tax on a dollar, and how much will be the tax uf Mr. Scott, whose property is valued at $12500, and who pays for 2 polls. Operation. $1.25 X 324 = $405, amount of poll tax. $2505 — $405 = $2100, amount of property tax. $2100 ~- $600000 = .0035, rate of taxation. $12500 X .0035 = $43.75, Mr. Scott's property tax. $43.75 + $2.50 (2 polls) = $46.25, Mr. Scott's total tax. Explanation. — Since $2505, the amount to be raised, includes both the poll and property tax, if $405, the poll tax, is subtracted from this amount, the remainder, $2100, will be the Percentage, or sum to be assessed on the Base, or entire property. Divide this Percentage by this Base, and the quotient will be the rate of tax assessed, 3£ mills on the dollar. Multiply $12500, the assessed valuation of Mr. Scott's property, by .0035, the per cent, expressed deci- mally, and the result, $43.75, is his property tax; adding to this $2.50, the tax on two polls, gives $46.25, his entire tax. Rule. — From, the sum to be raised, deduct the poll taoc, if any; divide the remainder by the total assessment, and multiply the assessment of each individual by the quotient; add to the product the amount of poll taoc to be paid. EXAMPLES FOR PRACTICE. 651. 1. A tax of $125000 is levied on a city, the assessed valuation of which is $15000000. What is the rate of taxation, and what amount of tax will a rjerson have to pay whose property is valued at $7500 ? ■2. If a tax of $120 is assessed on a mill valued at $24000, what is the valua- tion of a residence that is taxed $17.75 at the same rate ? 3. The per cent, of tax assessed for state purposes is |$, for county ■£#, and for city 1-J-^. What will be the amount of my tax, on property assessed at $21500 I Jf.. The tax assessed upon a town is $20914.80; the town contains 2580 polls, taxed $. 62-J each, and has a real estate valuation of $4062000, and a valuation of personal property to the amount of $227400. Find the rate of taxation, and C's tax, who pays for 4 polls, and whose property is assessed at $15000. Remark.— In certain States, the common schools are supported by a tax or rate bill made out ©n the basis of the total attendance. 5. My son and daughter each attended school 214 days, and the expense, including teacher's wages and incidentals, was paid by a rate bill. How much must I pay, if the teacher's wages amounted to $440, fuel and repairs $101.50, and janitor's fees $74.75, and the total number of day's attendance was 7460? EXAMPLES IX TAXES. 203 6. For the year 1888 the rates of taxation in the State of New York were as follows: 'Schools, 1.085 mills; general purposes, 1.475 mills; new capitol, .6 of a mill; other purposes, .34 of a mill. What was the total rate of taxation, and how much was raised by Livingston County, the valuation of which, as fixed by the State board of equalization, was $25395180? How much did said county raise for school purposes? 7. The cost of maintaining the public schools of a city during the year 1888, was $112000, and the taxable property of the city was $44800000. How many mills on a dollar must be assessed for school purposes? If 10$ of the tax assessed cannot be collected, how many mills on a dollar must then be assessed ? 8. A tax of $13943.20 is assessed upon a town containing 800 taxable polls; the real estate is valued at $2708000, and the personal property at $151600. If the polls be taxed $1.25 each, what will be the rate of property taxation, and what will be the tax of Peter Parley, who pays for three polls, and has real and personal estate valued at $23750? 9. In a school district, the valuation of the taxable property is $752400, and it is proposed to repair the school house and ornament the grounds, at an expense of $5000. If old material sells for $673.70, what will be the rate per cent, of taxation, and what will be B's tax, whose property was valued at $9400? 10. The assessed value of a town is, on real estate, $1197500, and on personal property, $432500. A poll tax of $.50 per head is assessed on each of 1870 persons. The town votes to raise $8000 for schools, $1500 for highways, $1500 for salaries, $1000 for support of poor, and $310 for contingent expenses. How much tax will a milling company have to pay, on a mill valued at $46500, and stock at $19750? 11. The total assessed value of a town, real and personal, is $630000, and the town expenses are $3913.95. How much tax must be collected to provide for town expenses and allow 3$ for collecting? If the same town contains 310 polls, taxed $1.50 each, what will be the rate of taxation, and how much will be the tax of a man who pays for two polls and owns property assessed at $14500 ? 12. The assessed valuation of the real estate of a county is $1910887, of the personal property, $921073, and it has 4564 inhabitants subject to a poll tax. The year's expenses are: for schools, $8400; interest, $6850; highways, $7560; salaries, $5150; and contingent expenses, $13675. If the poll tax was $1.50, and the revenue from fairs and licenses $6200, what tax must be levied on a dollar to meet expenses and provide a sinking fund of $7000? 204 IKSUBAlfCB. INSURANCE. 652. Insurance is indemnity secured against loss or damage. It is of two kinds: Property Insurance and Perso7ial Insurance. 653. Property Insurance includes: 1. Fire Insurance, or indemnity for loss of or damage to property by fire. 2. Marine Insurance, or indemnity for loss of or damage to a ship or its cargo, by any specified casualty, at sea or on inland waters. S. Lire Stock Insurance, or indemnity for loss of or damage to horses, cattle, etc., from lightning or other casualty. 654. The Insured Party is usually the owner of the property insured, but may be any person having a financial insurable interest in the property. 655. The Insuring Parties are called Insurers or Underwriters, and are usually incorporated companies. 656. Insurance Companies are distinguished by the way in which they are organized; as Stock Insurance Companies, Mutual Insurance Companies. 657. A Stock Insurance Company is one whose capital has been con- tributed and is owned by stockholders, who share the profits and are liable for the losses. 658. A Mutual Insurance Company is one in which the profits and losses are shared by the insured parties. Remarks.— 1. Some companies combine the features of both stock and mutual companies, and are called MlMd Companies. 2. In mixed companies, all profits above a limited dividend to the stockholders are divided among the policy-holders. 659. Transit Insurance refers to risks taken on goods being transported from place to place, either by rail or water or both. 660. The Policy is the contract between the insurance company and the person whose property is insured, and contains a description of the insured property, the amount of the insurance, and the conditions under which the risk is taken. 661. The Premium is the consideration in the contract, or the sum paid for insurance. 662. The Term of Insurance is the period of time for which the risk k taken, or the property insured. Remarks. — 1. Premium rates are usually given as so much per $100 of the sum insured, and depend upon the nature of the risk and the length of time for which the policy is issued; insurance is usually effected for a year or a term of years. 2. Short Bates are for terms less than one year. 3. It is usual to make an added charge for the policy. 4. Insurance is frequently effected upon plate glass, the acts of employees, etc. INSURANCE. 205 663. An Insurance Agent is one who acts for an insurance company, in obtaining insurance, collecting premiums, adjusting losses, reinsuring, etc. 664. An Insurance Broker is a person who negotiates insurance for others, for which he receives a brokerage from the company taking the risk; he is con- sidered, however, an agent of the insured, not of the company. Remakk. — A Floating Policy is one which covers goods stored in different places, and gen- erally such as are moved from place to place in process of manufacture. 665. Losses may be total or partial, 666. Fire Insurance Losses are usually adjusted by the insurance company paying the full amount of the loss, provided such loss does not exceed the sum insured; if the policy, however, contains the "average clause," the payment made is such proportion of the loss as the amount of insurance bears to the total value of the property. 667. When a loss occurs to a vessel, the insurance company pays only such a proportion of the loss as the policy is of the entire value of the vessel. 668. It is an established rule in marine insurance, that insurers shall be allowed one-third for the superior value of the new material, as sails, masts, etc., used in repair of damage; that is, "one-third off new for old." Remark. — Marine policies usually contain the "average clause." 669. In case a policy is terminated at the request of the insured, he is charged the " short rate " premium; if, however, it be terminated at the option of the company, the lower long rate will be charged, and the company refund the premium for the unexpired time of the policy. 670. A Y alued or Closed Policy is the ordinary form, and contains a fixed valuation of the thing insured. 671. An Open Policy is one upon which additional insurances may be entered at any time from port to port, at rates and under conditions agreed upon, 672. Policies on Cargoes are issued for a certain voyage, and on vessels for a voyage or for a specified time. 673. Salvage is an allowance made to those rendering voluntary aid in saving vessels or cargoes from marine casualties. Remabks. — 1. Insurance companies usually reserve the privilege of rebuilding, replacing, or repairing damaged property. 2. Insurance policies ordinarily state that the loss, if becoming a charge upon the company, will be paid 30 days or 60 days after due notice and proof of loss. If not then paid, the amount of the claim becomes interest-bearing. 674. The computations in Property Insurance are performed the same as in Percentage, and the terms compare as follows: The Amount Insured = the Base. TheEate fo of Premium = the Rate. The Premium = the Percentage. JtfUb EXAMPLES IN INSURANCE. 675. To find the Cost of Insurance. Example. — The mixed stock in a country store is insured for $750. What is the cost of insurance for one year, at \tyf, premium, if $1.25 is charged for the policy? Operation. *i?k/\ » • j Explanation.— Since the amount insured is $foU. z= amount insured. ., , , ., . . , .. the base, and the per cent, of premium the rate, •° 15 = fo of premium. if the amounl ^ e multiplied by the rate, the $11.25 = premium. product, $11.25, will be the premium; adding 1.25 = cost of policy. $1.25, the cost of the policy, gives the full cost, ■ «I12 50 $12.50 = full cost of insurance. Rule. — Multiply the amount of insurance by the rate per cent, of premium, and add extra charges, if any. 676. To find the Amount Insured, the Premium and Per Cent, of Premium being given. Example. — I paid $141.50 to insure a stock of goods for three months. If the charge for the policy was $1.50, and the rate of premium ■£#, for what amount was the policy issued? Operation. Explanation — Smce $141.50 was the full $141.50 = full cost. cost or p r e m i um p i us the charge of $1.50 for 1*50 = cost of policy. the policy, the premium must have been $140; $140 = premium. an( l since the rate of premium was $ per cent., U = .00875 = decimal rate. if $140 is divided b ^ * P er cent " the . How much premium did each company receive? What was the gain or loss of the German? What per cent, of premium did it receive for the part of the risk not re-insured? PERSONAL INSURANCE. 678. Personal Insurance is the insurance of persons. It includes: 1. Life Insurance, or indemnity for loss of life. 2. Accident Insurance., or indemnity for loss from disability occasioned by accident. 8. Health Insurance, or indemnity for loss occasioned by sickness. 679. Policies of Life Insurance are usually either Life Policies or Endowment Policies. 680. A Life Policy stipulates to pay to the beneficiaries named in it a fixed sum of money on the death of the insured. 681. An Endowment Policy guarantees the payment of a fixed sum of money at a specified time, or at death, if the death occurs before the specified time. 682. Life insurance companies are known as Stock, Mutual, Mixed, and Co- Operative. 683. Losses sustained by Stock and Mixed companies are paid either from "reserve funds" or by assessment on the stockholders; those sustained by Mutual and Co-Operative companies are paid by pro-rata or fixed contributions of the policy holders. Remakks. — 1. The money may be made payable to any one named by the insured; if made payable to himself, at his death it becomes a part of his estate and is liable for his debts, if payable to another, that other cannot be deprived of the benefit of the insurance, either by the will of the person taking out the insurance, or by his creditors. 2. A person may insure his life in as many companies as he pleases, and to any amount. 3. Any one having an insurable interest in the life of another, may take out, hold, and be benefited by a policy of insurance upon the life of the other; or he may take out a policy in his own name, and then assign it to any creditor or to any one having an insurable interest. 4. The practical workings of life insurance are fully set forth in documents in general circu- lation, and all matters of premiums to be paid, cash value of policies surrendered, and manner of becoming insured, are determined from such documents, rendering it unnecessary to require the solution of problems under life insurance, INTEREST. 209 INTEREST. 684. Interest is a compensation paid for the use of money. 685. The Principal is the money for the use of which interest is paid. 686. The Amount is the sum of the principal and interest. 687. The Time is the period during which the principal bears interest. 688. Interest is reckoned at a certain per cent, of the principal. It is therefore a Per Cent, of which the Base is the Principal. 689. The Rate of Interest is the annual rate per cent. 690. Interest differs from the preceding applications of Percentage only by introducing time as an element, in connection with the rate per cent. The Principal = the Base. The Per Cent, per Annum = the Kate. The Interest = the Percentage. The Sum of the Principal and Interest = the Amount. 691. Legal Interest is interest according to the maximum rate fixed by law. 692. Usury is interest taken at a rate higher than the law allows. 693. Simple Interest is interest on the principal only, for the whole time of the loan or credit; and this is generally understood by the term interest. 694. Annual, Semi - Annual, or other Periodic Interest, is interest computed at a specified rate for a year, half-year or other designated period. 695. Compound Interest is interest computed on the amount at regular intervals. Remarks. — 1. The payment of periodic interest, if specified in a contract, may usually be enforced; and if not paid when due, becomes simple interest bearing, and is not usury. 2. Neither the paying nor the receiving of compound interest is usury; but its payment cannot ordinarily be enforced, even though it is mentioned in the contract. 696. Accrued Interest is interest accumulated on account of any obliga- tion, due or not due. 697. Common Interest is interest computed on a basis of 360 days for a year. Remarks. — 1. This method is generally employed by business men, and in some states has received the sanction of law. 2. In reckoning interest by this method, it is customary to consider a year to be 12 months, and a month to be 30 days. Statement. — July 22, 1887, at the annual convention of the Business Educators' Associa- tion of America, then in session at Milwaukee, Wis., the following resolution was unanimously 14 2J0 SIX PER CENT. METHOD. adopted : Resolved, That, as business educators, we uniformly teach interest and discount on a 360-day basis, finding time by compound subtraction, and calling each month thirty days, except where the day of the minuend time be thirty-one, when it shall be so counted. Remark. — In computing interest for short periods of time, it is customary to take the exact number of days. 698. Exact Interest is interest computed for the exact time in days, and regarding the days as 365ths of a year. This method is nsed by the United States Government and by some merchants and bankers; but as it is inconvenient unless interest tables are used, it is not generally adopted. Remarks. — 1. Exact interest, for any period of time expressed in days, may be obtained by subtracting T ^ part from the common interest lor that period of time. 2. Common interest may be obtained from exact interest by adding thereto fa Dart of itself. 699. For convenience, the rate of interest should always be expressed deci- mally; the rules governing the multiplication and division of decimals may then be applied to any product or quotient arising from the use of the decimal rate. Remarks. — 1. In many of the States a legal rate of interest is established, to save dispute and contention in cases of contracts in which no rate of interest is agreed upon by the parties; still the laws sanction an interest rate higher than the fixed legal rate, if such rate be agreed upon by the parties; in a few of the States, any rate, if agreed upon, is thus made legal. 2. When no particular rate of interest is named in a contract containing a general interest clause, as " with interest" or " with use," the legal rate of the place where the contract is made Is understood. S. Debts of all kinds bear interest after they become due, but not before, unless specified SIX PER CENT. METHOD. 700. The following method of computing interest is based upon time as usually reckoned; i. e., 12 months of 30 days each, or 360 days for a year, and is called the Six Per Cent. Method. It is convenient for use in all cases where time is not given in days, as for years and months, or for years, months, and days, and where exact interest is not required. Should the rate be any other than six per cent., the change can be easily made. It is a common method of computing interest. Six Per Cent. Method. $1.00 in 1 yr., at 6$, will produce $.06 interest. $1.00 in \ yr., or 2 mo., at 6$, will produce $.01 interest. $1.00 in 1 mo., or 30 da., at 6$, will produce $ .005 interest. $1.00 in 6 da., or \ mo., at 6$, will produce $.001 interest. $1.00 in 1 da., at 6$, will produce $.000£ interest. 701. To find the Interest on Any Sum of Money, at Other Rates than 6 per cent. : 1. To find the interest at 7$. Rule. — To the interest at 6$ add one-sixth of itself. 2. To find the interest at 7£$. Rule. — To the interest at 6$ add one-fourth of itself. EXAMPLES IN" INTEREST. 21} 8. To find the interest at 8$. Eule. — To the interest at 6$ add one-third of itself. 4. To find the interest at 9$. Rule. — To the interest at 6$ add one-half of itself. 5. To find the interest at 10$. Rule. — Divide the interest at 6$ by 6, and remove the decimal point one place to the right. 6. To find the interest at 12$. Rule. — Multiply the interest at 60 4 for 2 months is $ .01, for 8 months it will be $ .04; since a a a ^2 da. = .002 tne interest on $1 for 6 da. is $.001, for 12 days it will be $.002; therefore the interest on $1, at G per cent., Int. on $1 for 3 yr. ^ for the full timGj is $ 223; and the interest on $ 550 will 8 mo. 12 da. = $ . 222 De 550 times the interest on $1, or the product of the prin- $550 X .222 = $122.10. cipal and the rate for the given time, which is $122.10. Rule. — Multiply the principal by the decimal expressing the interest of one dollar for the full time. EXAMPLES FOB PRACTICE. 703. 1. Find the interest on $900, for 4 yr. 1. mo. 6 da., at 7#. *^'* Explanation. — Find the interest at 6$, and add to it one-sixth of itself. 2. What is the interest on $400, for 1 yr. 7 mo. 2 da., at 7|$ ? *Y Explanation. — Find the interest at 6%, and add to it one-fourth of itself. yo ■8. What is the interest on $150, for 6 yr. 3 mo. 18 da., at 8$ ?^ Explanation. — To the interest at 6% add one-third of itself. 1 V 4. Compute the interest on $1200, foi 3 yr. 4 mo. 15 da., at 9#. * Explanation. — To the interest at 6% add one-half of itself. .0 o. Find the interest, at 10$, on $840, for 5 yr. 5 mo. 9 da. \A Explanation. — Divide the interest at 6% by 6, to obtain the interest at 1%, and remove the decimal point 1 place to the right. 212 EXAMPLES IN" INTEREST. 6. What is the interest, at 12$, on $366, for 2 yr. 11 mo. 27 da. ? Explanation. — Multiply the interest at 6? by 2%. 7. Find the interest on $1800, for 6 yr. 9 mo. 25 da., at 5|$. (*"P ' Explanation. — From the interest at 6£ subtract one- twelfth of itself. 8. Compute the interest, at 5$, on $1000, for 11 yr. 4 mo. 24 da. /? D ' Explanation.— From the interest at 6% subtract one-sixth of itself. 9. What is the interest, at 4J#, on $1100, for 6 yr. 6 mo. 6 da.? jy? Explanation. — From the interest at 6# subtract one-fourth of itself. „ h 10. What is the interest, at 4$, on $1350, for 9 yr. 8 mo. 12 da.? P* Explanation. — From the interest at 6% subtract one-third of itself. 11. Find the interest, on $546, for 6 yr. 2 mo. 24 da., at df . y ^ Explanation. — Divide the interest at 6% by 2. Remakes.— 1. Interest at any other rate, entire or fractional, can be found by a general application of the methods above explained. 2. When the mills of a result are 5 or more, add 1 cent; if less than 5, reject them. 12. v Compute the interest on $752.50, for 4 yr. 11 mo. 9 da., at 6$. ^ IS. Compute the interest on $3560, for 9 yr. 10 mo., at 8$. *•/* 14. Compute the interest on $1540, for 9 mo. 20 da., at 6$. ^ jy 15. Compute the interest on $610.15, for 7 yr. 11 da., at 7$. ** 0,> 4 r 16. Compute the interest on $1116, for 3 yr. 11 mo. 11 da., at 5$. •» ' 17. i Compute the interest on $17500, for 2 yr. 1 mo. 10 da., at 4J#. ,lp 18. Compute the interest on $350.40, for 5 yr. 5 mo., at 7$. 19. Compute the interest on $2400, for 7 yr. 1 mo. 19 da., at 10£#. 20. Find the interest on $1450, from Aug. 12, 1882, to Nov. 10, 1890, at 6#. 21. What is the amount of $610, at 8$, for 3 yr. 8 mo. 21 da.? Explanation. — The Principal plus the Interest equals the Amount. 22. Find the amount due after 1 yr. 10 mo. 20 da., on a 6$ loan of $1941.50. 28. On the lfttli of September, 1884, 1 borrowed $3500, at H interest. How much will settle the loan Jan. 1, 1890? &4. My note, given 2 yrs. 9 mo. 27 da. ago, for $875.25, and bearing 4$ interest, is due to-day. What is the amount due? 25. A manufacturer marks a carriage with two prices; the one for a credit of 6 months on sales, and the other for cash. If the cash price was $750, and money was worth 8$, what should be the credit price? 26. Borrowed $2750 July 16, 1887, at 5$ interest, and on the same day loaned it at 71$ interest. If full settlement is made Jan. 4, 1889, how much will be gained? 27. On goods bought for $4500, on 6 months credit, I was offered 5$ off for cash. If money was worth 6$, how much did I lose by accepting the credit? 28. A man sold his farm for $16000; the terms were, $4000 cash on delivery, $5000 in 9 months, $3000 in 1 year and six months, and the remainder in 2 years from date of purchase, with 6$ interest on all deferred payments. What was the total amount paid? EXAMPLES IK INTEREST. 213 29. May 16th I bought 300 barrels of flour, at $7 per barrel; July 28tli Isold 50 barrels, at $8 per barrel; Oct. 30th, 100 barrels,, at $6.75 per barrel; and Feb. 13th following, the remainder, at $7.80 per barrel. Allowing interest at 6$, what was my gain? 80. John Doe bought bills of dry goods as follows: May 3, $250; July 1, $1125; Sept. 14, $450; Oct. 31, $150; Dec. 1st. $680; and on Dec. 21st, he paid in full, with 60 interest. What was the amount of his payment ? 81. On March 25, I sold five bills of goods, for amounts as follows: $1046.81, $952.40, $173.50, $1250, and $718.25; and on the first day of the following December I received payment in full, with interest at 6$. What was the amount received? 82. A firm bought goods on credit, and agreed to pay 7$ interest on each purchase from its date; Oct. 6, 1887, goods were bought to the amount of $268 ; Dec. 31, 1887, to the amount of $765.80; Feb. 29, 1888, to the amount of $600; Apr. 1, 1888, to the amount of $325.25. If full settlement was made Aug." 25, 1888, how much cash was paid. Remark. — In the following examples, given for teacher's use in class drill, the interest on each separate principal should be computed to its nearest cent; the sum of ^he results so obtained will be the answer sought. Find the amount of interest at 6$, by the six per cent, method, On $680, for 2 yr. 6 mo. 10 da. On $1895, for 1 yr 7 mo. 7 da. On $468, for 5 yr. 5 mo. 1 da. On $1000, for 11 yr. 1 mo. 20 da. On $645, for 4 yr. 4 mo. 5 da. On $500, for 3 yr. 1 mo. 27 da. On $895, for 5 yr. 11 mo. 11 da. On $1650, for 1 yr. 10 mo. 23 da. On $1463, for 9 yr. 1 mo. 9 da. On $365, for 4 yr. 1 mo. 25 da. 81/.. Find the amount of interest, by the six per cent, method, On $538, for 6 yr. 6 mo. 6 da., at 90. On $1200, for 7 yr. 4 mo. 27 da., at 100. On $350, for 3 yr. 7 mo. 18 da., at 60. On $586.50, for 2 yr. 9 mo. 15 da., at 70. On $1345, for 5 yr. 4 mo. 1 da., at 80. 85. Find the amount of interest, by the six per cent, method, On $550, for 4 yr. 6 mo. 21 da.,. at 60. On $2100, for 1 yr. 11 mo. 3 da., at 7$. On $750, for 8 yr. 8 mo. 8 da., at 80. On $1200, for 3 yr. 3 mo. 1 da., at 7£0. On $1500, for 7 yr. 7 mo. 9 da., at 90. On $675, for 5 yr. 5 mo. 25 da., at 100. On $1000, for 11 yr. 11 mo. 11 da.,at 50. On $2500, for 1 yr. 1 mo. 1 da., at 4^0. On $300, for 2 yr. 2 mo. 2 da., at 40. On $990, for 4 yr. 4 mo. 6 da., at 30. Find the amount of interest, by the six per cent, method, On $250, for 3 yr. 4 mo. 29 da., at 80. On $967.25, for 7 yr. mo. 27 da., at 60. On $1305. 09, for 1 yr. 11 mo. 7 da. , at 70. On $1255.84, for 9 mo. 1 da., at 100. On $316.75, for 5 yr. 11 mo. da. , at 4^0. On $2100.50, for 1 yr. 1 mo. 19 da., at 90. On $3546.81, for 5 yr. mo. 5 da., at 30. On $1867, for 2 yr. mo. 2 da., at 7|0. On $266.60, for 7 yr. 7 mo. 5 da., at 50. On $1120. 95, for 4 yr. 4 mo. da., at 40. On $1000, for 5 yr. 6 mo. 7 da., at 80. On $1743, for 2 yr. 3 mo. 6 da., at 60. 214 EXAMPLES IN INTEREST. 704. To find the Principal, the Interest, Rate, and Time being given. Example.— What principal, in 3 years and 2 months, at 60, will gain $47.50 interest ? Operation. Explanation. — Since $1 in 3 years, at $.18 == int. of $1, at 60, for 3 yr. 6 P er cent -> wiu S ain $- 18 interest, and in 2 .01 = int. of $1, at 60, for 2 mo. ™ nths .*'<* * nter + est > {t wil j JJ * e ^ ven ' / time gain $.19 interest; and if $1 will in $.19 = int. of $1, at 60, for 3 yr. 2 mo. the given time gain $.19 interest, the prin- $47.50 interest - .19 = $250, principal. ^ that will in the given time gain $47.50 x x interest must be as many times $1 as $.1& is contained times in $47.50, or $250 ; therefore $250 is the principal which will, in 3 yr. 2 mo., at 6%, gain $47.50 interest. Rule.— Divide the given interest by the interest of one dollar for the given time and rate. Remark. —Whenever the divisor contains a fraction not reducible to a decimal, as in case of some fractional or odd ratio per cent., it is better that the fractional form be retained. Before division in such cases, multiply both divisor and dividend by the denominator of the fractional divisor; the relative value of the terms will not be changed, and greater exactness will be secured in the result. EXAMPLES FOR PRACTICE 705. 1. What principal, at 70, will gain $154 in 6 yr. 4 mo. 24 da. ? 2. What sum of money, loaned at 4|0, for 7 yr. 11 mo. 15 da., will gain $1468.21 interest ? l/oH.^I 3. What sum of money, invested at 5^0, will in 7 yr. 1 mo. 1 da. produce $131.50 interest ? 3 3 j,^ Jf. A money lender received $221.68 interest on a sum loaned at 80 July 17, 1885, and paid Oct. 11, 1888. What was the sum loaned ? * fff 3 { > w, £f . £j 7/0/ 5. A dealer who clears 12J0 annually on his investment, is forced by ill health to give up his business; he lends his money at 70, by which his income is reduced $1512.50. How much had he invested in his business ? 6. How many dollars must I put at interest, at 90, Jan. 27, 1889, so that on the 18th of Nov., 1895, $506.27 interest will be due? X \ rfL ', £$f } 6 ,> 2 706. To find the Principal, the Amount, Rate, and Time being given. Example. — What principal, at 60, will, in 4 yr. 6 mo. 15 da., amount to $2372.25? Opekation. Explanation. — Since a principal of *i owok » *. AA . ., .. $1 will, in the given time, amount to $1.2725 = amount of $1.00 for the time. L ow , K u Jm « • • i * uxo uiuic. |i # 2725, it will require a principal of as $2372.25 -S- 1.2725 = $1864.24. principal many times $1 to amount t0 $2 372.25 as $1.2725 is contained times in $2372.25, or $1864.24. Rule. — Divide the amount by the amount of 1 dollar for the given time and rate. EXAMPLES IN" INTEREST. 215 EXAMPLES FOR PRACTICE. 707. 1. What sum, put at interest at 7$ for 5 yr. 11 mo. 3 da., will amount to $630.90? 2. A boy is now 15 years old. How much must be invested for him, at 7-§-# simple interest, that he may have $15000 when he becomes of age ? 3. What sum, put at interest June 1, 1888, at 7$, will amount to $687.50 July 1, 1890 ? 4. What sum of money, put at interest to-day at 5$, will amount to $1031.25 in 7 mo. 15 da. ? 5. What principal will amount to $308.34 in 11 mo. 9 da., at 6^ ? 6. A man loaned a sum of money to a friend from June 13 to Dec. 1, at 7$, when he received $763.28 in full payment. How much was loaned ? 7. Owing a debt of $2146. 18, due in 1 yr. 7 mo. 18 da., I deposited in a bank, allowing me 6^ interest, a sum sufficient to cancel my debt when due. Find the sum deposited. 708. To find the Rate Per Cent., the Principal, Interest, and Time being given. Example. — At what rate per cent, must $750 be loaned, for 2 yr. 5 mo. 6 da., to gain $164.25 interest ? Opekation. Explanation. — The principal will gain a-io oer -j. £ dbwKA £ j-r. j.' j. iv $18.25 interest in the given time at 1 per $18.25 = int. of $750 for the time at 1%. *ZI . , .. . .. . .. . * nfi4 2^^182^-qnrQ/ Cent ; m ° rder tbat ll maj m the glven time ^104.^o -v- *18.^5 — 9 or 90. gain $164>25> the rate must be as many times 1 per cent, as $18.25 is contained times in $164.25, or 9 per cent. Rule. — Divide the given interest by the interest on the given principal for the given time, at 1 per cent. Remark. — When the amount, interest, and time are given, to find the rate per cent., subtract the interest from the amount, thus finding the principal, then proceed as by the above rule. EXAMPLES FOE PRACTICE. 709. 1. If I pay $518.75 interest on $1250, for 5 yr. 6 mo. 12 da., what is the rate per cent. ? 2. At what rate wonld $710, in 3 yr. 5 mo. 20 da., produce $172.56 interest ? 3. At what rate would $4187.50 amount to $4738.68, in 1 yr. 11 mo. 12 da. ? 4. If $1200 amounts to $2135.80 in 12 yr. 11 mo. 29 da., what is the rate per cent. ? 5. A lady deposited in a savings bank $3750, on which she received $93.75 interest semi-annually. What per cent, of interest did she receive on her money? 6. A debt of $480, with interest from August 24, 1886, to Dec. 18, 1888, amounted to $546. 72. What was the rate per cent, of interest ? 7. To satisfy a debt of $1216.80, that had been on interest for 4 yr. 4 mo. 21 da., I gave my check for $1751.18. What was the rate per cent, of interest? 216 SHORT METHODS FOR FINDING INTEREST. 710. To find the Time, the Principal, Interest, and Rate being given. Example. — In what time will $540 gain $74.52 interest, at 6$ ? Operation. Explanation.— Since in 1 year $540 will, at $32.40 = int. on $540 for 1 yr., at 6#. 6 *** cent., gain $32.40 interest, it will require $74.52 — 32 40 = 2 3 as man y vears for & t0 g ain $74.52 interest as ' j "_ 9 o ' * $32.40 is contained times in $74.52, or 2.3 years; 4.6 x l yr. — Z.6 years. ^ findj by the mle for the reduction of a denomi . .3 yr. X 12 = 3. 6 months. nate decimal, that 2.3 years equals 2 yr. 3 mo. .6 mo. X 30 = 18 days. 18 da. 2 yr. 3 mo. 18 da. Remark. — When by inspection it is apparent that the time is less than a year, divide the given interest by the interest on the principal for the highest apparent unit of time; the quotient will be in units of the order taken, which reduce as above. Rule. — Divide the given interest by the interest on the principal for 1 year, at the given rate per cent. Remark. — When the amount, interest, and rate are given to find the time, subtract the interest from the amount, thus finding the principal, and* proceed as above. EXAMPLES FOR PRACTICE. 711. t How long will it take $360 to gain $53.64, at S#. 2. How long should I keep $466.25, at 8$, to have it amount to $610.48 ? 3. A debt of $1650 was paid, with h\**A $1725, for 57 days, at 9$. *k ■ $125, for 55 days, at 6$. » $3741.85, for 6 days, at 7&< $5178, for 9 days, at 9$. ^ ' $732, for 11 days, at 6$. N ^ $1174.51, for 42 days, at 8£ ^ \ EXAMPLES FOR PRACTICE. 219 23 2o, 26 27 28 29 30 31 32 S3 H 35 $340, for 70 days, at 100. \fl , $1478, for 80 days, at 60. . $2150, for 96 days, at 4^0.^ ' v $1200, for 53 days, at 60. 81500, for 87 days, at 70. $420, for 41 days, at 50. i $360, for 81 days, at 60. $2347.50, for 18 days, at 70. $1112.49, for 25 days, at 80. -' $1300, for 13 days, at 60. $17000, for 3 days, at 5*0. iV $195.50, for 33 days, at 100. $1050, for 43 days, at 70. %<%%. $1560, for 44 days, at 7&J $180, for 47 days, at 60. ^ V 36. v $120, for 49 days, at 90. 37. $60, for 50 days, at 50. - 38. 12500 X .06 = $150 = 1 yr. int. after it was duej f he second year > g Remaining unpaid S*J V 'l m °* \\ ^ interest ' $150 ' was not paid until for periods of j yr * m0, ' 1 yr - 2 ma 15 da - after [t was due » ( 2 mo. 15 da. an d the third year's interest, $150. Interest of $150. for 3 yr. 7 mo. 15 da. = $ 32.63 was not P aid Ulltil 2 mo. 15 da. 3 yr. simple interest on principal = 450.00 after U was due ' the ^^gate of -,<*-,>, • • i o-irt!- the time for which interest should 2 mo. 15 da. interest on principal = 31.25 , . n , . A r r be computed on one year s mterest, Total interest due = $513.88 $150, is 3 yr. 7 mo. 15 da., and its interest for that time is $32.63. Adding to this the interest of the principal for the full time, $481.25, gives $513.88, the amount of interest due. Rule.— To the simple interest on the principal for the full time, add the interest on one period's interest for the aggregate of time for which the payments of interest were deferred. EXAMPLES FOR PRACTICE. 729. 1. What is the annual interest of $1260, payments due semi-annually from May 21, 1884, to Nov. 9, 1888, at 7$, no interest having been paid? 2. What is the annual interest of $3416.50, payments due quarterly from Jan. 15, 1882, to Sept. 6, 1889, at 5$, no interest having been paid? S. Find the amount of interest due at the end of 4 yr. 9 mo. on a note for $1155, at 6/o, interest payable annually, but remaining unpaid. It. On a note of $1750, dated Aug. 1, 1882, given with interest payable annually at 10$, the first three payments were made when due. How much remained unpaid, debt and interest, Jan. 1, 1889? 5. Find the amount due Oct. 11, 1891, on a debt of $11000 under date of July 5, 1888, bearing 4-J$ interest, payable quarterly, notes for the quarterly interest having been given and nothing paid until final settlement. COMPOUND INTEREST. 730. Compound Interest is the interest on the principal and on the unpaid interest after it becomes due. 731. The Simple Interest may be added to the principal annually, semi- annually, quarterly, or for other agreed periods; when done, interest is said to be compounded annually, quarterly, etc., as the case may be. 732. General Rule. — Find the amount of the principal and interest for the first period, and make that the principal for the second period, and so proceed to the time of settlement. COMPOUND INTEREST. 223 Remarks, — 1 If the time contains fractional parts of a period, as months and days, find the amount due for the full periods, and to this add its interest for the months and days. 2. Compound interest is not recoverable by law, but a creditor may receive it if tendered, "without incurring the penalty of usury; a new obligation may be taken at the maturity of a compound interest claim, for the amount so shown to be due, and such new obligation will be valid and binding. 733. To find the Compound Interest, when the Principal, Kate, and Time of Computing it are given. Example. — Find the interest of $750, for 3 yr. 8 mo. 15 da., at 6% if interest be compounded annually. Operation. Explanation. — Since the interest a.. __ , , p , is to be compounded annually, the ?45 _ int. lor 1st yr. amount due at the end of the first $795 = amt. at end of 1st yr. year> which is $ 795) will be the basis $47.70 = int. for 2d yr. of the interest for the second year; - $842. 70 = amt. at end of 2d yr. or $ 768 - If tne face of dboivi ^o *«v/>o Aino /?o ± j- the debt is $871.68, and its present worth $871.68 — $768 = $103.68, true discount. . , A „J .. \ ,. *L . .,. , is only $768, the true discount will be $871.68 minus $768, or $103.68. Rule. — Divide the amount of the debt, at Us maturity, by one dollar ■plus its interest for the given time and rate, and the quotient will be the present worth; subtract the present worth from the amount, and the remainder will be the true discount. EXAMPLES FOR PRACTICE. 74:4. 1. What is the present worth of $661.50, payable in 3 yr. 9 mo. } discounting at 6$? 2. Find the present worth and true discount of a debt of $138.50, due in 5 yr. 6 mo. 18 da., if money is worth 7$ per annum. 8. Find the present worth of a debt of $1750, $1000 of which is due in 9 mo. and the remainder in 15 mo., money being worth 6$ per annum. EXAMPLES IN" TRUE DISCOUNT. 231 4- "Which is greater, and how much, the interest, or the true discount on $516, due in 1 yr. 8 mo. s if money is worth 10$ per annum? 5. Which is better, and how much, to buy flour at $6. 75 per barrel on 6 months time, or to pay $6 cash, money being worth 6$? 6. When money is worth 5$ per annum, which is preferable, to sell a house for $20,000 cash, or $21,000 due in one year ? 7. A farmer offered to sell a pair of horses for $420 cash, or for $475 due in 15 months without interest. If money is worth 8$ per annum, how much would the buyer gain or lose by accepting the latter offer ? 8. If money is worth 6$, what cash offer will be equivalent to an offer of $1546 for a bill of goods on 90 days credit ? 9. An agent paid $840 cash for a traction engine, and after holding it in stock for one year, sold it for $933.80, on eight months' credit. If money is worth 6$, what was his actual gain ? 10. A stock of moquette carpeting, bought at $1.95 per yard, on 8 months' credit, was sold on the date of purchase for $1.80 per yard, cash. If money was worth 6$ per annum, what per cent, of gain or loss did the seller realize ? 11. Marian is now fifteen months old. How much money must be invested for her, at 6$ simple interest, that she may have $15000 of principal and interest when she celebrates her eighteenth birthday? 12. A thresher is offered a new machine for $480 cash, $500 on 2 months credit, or $525 on 1 year's credit. Which offer is the most advantageous for- him, and how much better is it than the next best, with money worth 7-J$? 13. After carrying a stock of silk for 4 months, I sold it at an advance of 30$ on first cost, extending to the purchaser a credit of one year without interest. If money is worth 5$ per annum, what was my per cent, of profit or loss ? 11/.. Having bought a house for $5048 cash, I at once sold it for $7000, to be paid in 18 months without interest. If money is worth 8$ per annum, did I gain or lose, and how much ? 15. Goods to the amount of $510 were sold on 6 months' credit. If the selling price was $30 less than the goods cost, and money is worth 6$ per annum, how much was the loss and the per cent, of loss ? 16. How much must be discounted for the present payment of a debt of $8741.50, $2000 of which is on credit for 5 months; $3000 for 8 months, and the remainder for 15 months, money being worth 10$ per annum ? 17. What amount of goods, bought on 6 months time or 5$ off for cash, must be purchased, in order that they may be sold for $4180, and net the pur- chaser 10$ profit, he paying cash and getting the agreed discount off? 18. A dealer bought grain to the amount of $2700, on 4 months' credit, and immediately sold it at an advance of 10$. If from the proceeds of the sale he paid the present worth of his debt at a rate of discount of 8$ per annum, how much did he gain ? 19. A merchant bought a bill of goods for $2150, on 6 months' credit, and the seller offered to discount the bill 5$ for cash. If money is worth 7-J$ per annum, how much would the merchant gain by accepting the seller's offer. 232 EXAMPLES IN TRUE DISCOUNT. 20. The asking price of a hardware stock is $5460, on which a trade discount of 25$, 15$, and 10$ is offered, and a credit of 90 days on the selling price. If money is worth 5|$, what should be discounted for the payment of the bill ten days after its purchase ? 21. A merchant sold a bill of goods for $1800, payable without interest in three equal payments, in 3 months, 6 months, and 9 months respectively. If money is worth 5$ per annum, how much cash Would be required for full settle- ment on the date of purchase ? 22. A stationer bought a stock worth $768, at a discount of 25$ on the amount of his bill, and 4$ on the remainder for cash payment. He at once sold the stock on 4 months' time, at 10$ in advance of the price at which it was billed to him. How much will the stationer gain if his purchaser discount his bill on the date of purchase by true present worth, at the rate of 7$ per annum ? 23. I sold my farm for $10,000, the terms being one-fifth cash, and the remainder in four equal semi-annual payments, with simple interest at 5$ on each from date; three months later the purchaser settled in full by paying with cash the present worth of the deferred payments, on a basis of 10$ per annum for the use of the money. How much cash did I receive in all ? 2J/.. What amount of goods, bought on 4 months' time, 10$ off if paid in 1 month, 5$ off if paid in 2 months, must be purchased, in order that they may be sold for $11480, and -J the stock net a profit of 15$ and the remainder a •profit of 20$ to the purchaser, if he cashes his purchase within 1 month and gets the agreed discount off ? BANK DISCOUNT. 233 BANK DISCOUNT. 745. A Bank is a corporation chartered by law for the receiving and loaning of money, for facilitating its transmission from one place to another by means d)f checks, drafts, or bills of exchange, and, in case of banks of issue, for furnishing a paper circulation. Remark. — Some banks perform only a part of the functions above mentioned. 746. Negotiable Paper commonly includes all orders and promises for the payment of money, the property interest in which may be negotiated or trans- ferred by indorsement and delivery, or by either of those acts. 747. Bank Discount is a deduction from the sum due upon a negotiable paper at its maturity, for the cashing or buying of such paper before it becomes due. 748. The Proceeds of a Note or other negotiable paper is the part paid to the one discounting it, and is equal to the face of the note, less the discount. Remark. — In true discount, the present worth is taken as the principal ; in lank discount the future worth is taken as the principal. 749. The Face of a Note is the sum for which it is given. 750. The Discount may be a fixed sum, but is usually the interest at the legal rate, and taken in advance. 751. The Time in bank discount is always the number of days from the date of discounting to the date of maturity. 752. The Term of Discount is the time the note has to run after being discounted. Remark. — Bank discount is usually reckoned on a basis of 360 days for a year. 753. A Promissory Note is a written, or partly written and partly printed, agreement to pay a certain sum of money, either on demand or at a specified time. Remark. — In general, notes discounted at banks do not bear interest. If the note be interest- bearing, the discount will be reckoned on and deducted from the amount due at maturity. 754. Days of Grace are the three days usually allowed by law for the payment of a note, after the expiration of the time specified in the note. 755. The Maturity of a note is the expiration of the days of grace; a note is due at maturity. Remarks. — 1. Notes containing an interest clause will bear interest from date to maturity, unless other time be specified. 2. Non-interest bearing notes become interest bearing if not paid at maturity. 3. The maturity of a note or draft is indicated by using a short vertical line, with the date on which the note or draft is nominally due on the left, and the date of maturity on the right; thus. Oct. 21 /24. 234 GENERAL REMARKS ON" COMMERCIAL PAPER. 756. The Yalue of a note at its maturity is its face, if it does not bear interest; if the note is given with interest, its value at maturity is the face plus the interest for the time and grace. Kemakks.—I.* Grace is given on all negotiable time paper unless " without grace " be specified. 2. In some States, as Minnesota, Pennsylvania, and others, drafts drawn payable at sight are entitled to days of grace, and should be accepted in the same form as time drafts; while in such States drafts payable on demand have no days of grace, and like the sight drafts of most of the States, are dishonored if not paid on demand. Other States, as New Jersey and Pennsylvania, have statutory requirements as to the phraseology of the note; as to include the phrase "with* out defalcation or discount," etc. In such matters State laws should be observed. 757. Notes given for months, have their maturity determined by adding to their date the full months, regardless of the number of days thereby included, and also the three days of grace. 758. Notes given for days have their maturity determined by counting on from their date the expressed time, plus three days of grace. This is done regardless of the number of months compassed by the days so counted. 759. hi some States, the bank custom is to take discount for both the day of discount and the day of maturity, which is excessive. Remarks. — 1. In general, the laws of the different States provide that, if a note matures on Sunday, it shall be paid on Saturday; if Saturday be a legal holiday, then the note shall be paid on Friday; but the laws of different States vary, and should be carefully studied and lully observed, in order to hold contingent parties responsible. 2. Notes maturing on a legal holiday must be paid on the day previous, if the legal holiday occurs on Monday, payment must be made on the preceding Saturday. 760. Banks, in many of the larger cities, loan money on collateral securities, such as stocks, bonds, warehouse receipts, etc. Such loans, being made payable on demand, or on one day's notice, are termed " call loans" or " demand loans." On such the interest is usually paid at the end of the time. Remark. — Variations in practice among banks, and at the same bank with different patrons, are very common and subject to no rule of law. GENERAL REMARKS ON COMMERCIAL PAPER. 761. Commercial, or Negotiable Paper, includes promissory notes, drafts, or bills of exchange, checks, and bank bills, warehouse receipts, and certain other evidences of indebtedness; but notes and time drafts are the only two kinds entering largely into the operations of bank discount. 762. If there is no' admixture of fraud in the transaction, any negotiable paper may be bought and sold at any price agreed upon by the parties, and the purchaser thus have full right of recovery. 763. The purchaser of a negotiable paper is protected in nis right of recovery of its amount against all original and contingent parties thereto, if he can show three conditions : 1st. That he gave value for the paper. *Days of grace have been abolished by statute in California, Connecticut, Georgia, Idaho, Illinois, Montana, New Jersey, New York, North Dakota, Oregon, Pennsylvania, Utah, Vermont. Wisconsin. GENERAL REMARKS Otf COMMERCIAL PAPER. 235 zd. That he bought it before its maturity. 3d. That he did not, at the time of its purchase, know of the existence of any claim or condition affecting its validity. 764. Indorsements are made on notes for three purposes: 1st. To secure their payment. 2d. To effect their transfer. 3d. To make a memorandum of a partial payment. 765. Persons indorsing for security or transfer are liable for the payment of the paper indorsed, unless the holder of the paper fails to demand payment of its maker at maturity, and, in case of its non-payment, gives the indorser or indors- ers, within a reasonable time, notice of its dishonor by the maker. 766. If the dishonored paper be foreign — i. e., the parties to it being of different states or countries — to hold contingent parties, a formal notarial pro- test, mailed to the indorsers, is required by the laws of most States; but a verbal or other informal notice of dishonor is sufficient if the paper is domestic. 767. No demand notice or protest is necessary to hold the maker; he, being a principal debtor, is only released from his obligation by the outlawing of the note, or by his payment of it. 768. A Protest is a written, or partly written and partly printed, statement, made by a notary public, giving legal notice to the maker and indorsers of a note of its non-payment. 769. The laws governing negotiable paper are not uniform throughout the United States, and a careful observance of the laws of all the States wherein one does business is necessary to avoid risks of loss. 770. It is lawful to compute and take interest for all three of the days of grace, although the debtor may thus lose the interest for one or two days by the fact that the note matures on Sunday or on a legal holiday. 771. Interest charges for time of transfer of notes to distant places for demand, and for the return of the remittance therefor, is a matter wholly of custom with banks, as is also an added charge or fee for services in relation to such demand and remittance. 772. Patrons of good standing at banks are often given credit for the face of interest bearing notes discounted. 773. When a note is discounted at a bank, the payee indorses it, thus making it payable to the bank; both maker and payee are then responsible to the bank for its payment. 774. Indorsements for transfer are at the same time indorsements for surety, unless made "without recourse." 775. Negotiable papers may be transferred: 1st. By indorsement in full — i. e. 9 by the payee writing on the bacK of the note, in substance, as follows : "Pay to the order of John Doe, Richard Roe" 236 EXAMPLES IN BANK DISCOUNT. ( payee ). In which event Doe becomes the legal owner of the note, and possesses a right to receive payment on it, or, in case of its non-payment at maturitv, to sue and recover from the maker; and if he follows the statute law of the State as to demand and notice, he may recover, either jointly or severally, from eithei the maker, or from Roe, the indorser, as such indorser is also a surety. 2d. By indorsement "in blank" — L e., by the payee writing, on the back of the note, simply his name. After this is done, the holder is presmned to be the owner, and he may, in case of default, recover by suit from the maker; and il he observes the requirement of the law of the place, he may also hold the indorser, as such indorser becomes a surety. 3d. By indorsement " without recourse " — i. e., by the payee writing, on the back of the note, in substance, " Pay John Doe, or order, without recourse to me, Richard Eoe" (payee); or by writing simply " Without recourse, Eichard Eoe." A note so indorsed is fully transferred from the payee, but he rests under no obligation as to its payment. 776. The corresponding terms of Bank Discount and Percentage are as follows: The Face of the note = the Base. The Eate Per Cent. = the Eate. The Bank Discount = the Percentage. 777.— To find the Discount and Proceeds, the Face of a Note, Time, and Eate Per Cent, of Discount, being given. Example. — Find the bank discount and proceeds of a note for $580, due in 63 days, at 6$. Explanation. — The bank discount of a note being its inter- Operation. est f or fa e t j me pi us g race> anc i t ne proceeds being the face of $580.00 = face. a note minus the bank discount, it is only necessary to compute a Q9 __ a\„ f or AQ Q* a the interest on the face for the full time to obtain the discount, and to subtract such discount from the face to find the proceeds; $573.91 = proceeds. thus, $6.09 being the discount, $580, minus $6.09, equals $573.91, proceeds. Rule.— Compute the interest for the time and rate, for the bank discount; and subtract this bank discount from the face of the note, to find the pro- ceeds. Remark. — If the note is on interest, find the discount on the amount of the note at maturity. EXAMPLES FOR PRACTICE. 778. 1. Find the bank discount and proceeds of a note for $750, due in 90 days, at 5#. #. Find the bank discount and proceeds of a note for $286.50, due in 30 days, at 7#. S. "Pmd the bank discount and proceeds of a note for $1325, due in 60 days, at 10/. EXAMPLES IK BASTE DISCOUNT. 237 4.. What is the discount on a note for $1000, discounted at a bank for 23 days, at 7# ? 5. What are the proceeds of a 90-day note for $1000, discounted at a bank at Ufa ? 6. I paid in cash $950 for an engine, and sold it the same day for $975, taking a 60-day note, which I discounted at a bank at 8$. What was my gain or loss ? 7. Find the bank discount and proceeds of a note for $1240, dated Sept. 3, 1888, payable in 4 months, with interest at 6$, and discounted Nov. 1,-1888, at the same rate. 8. What are the proceeds of a note for $1750, due in 63 days, bearing interest at 10$, and discounted at a bank at the same rate ? 9. Find the maturity, term of discount, and proceeds of the following note: $286.00. Buffalo, NY., Oct. 25, 1888. Three months after date, I promise to pay to the order of Smith & Bro., Two Hundred Eighty-six Dollars, at the Erie County National Bank. Value received. THOMAS BROWN, JR. Discounted Jan. 1, 1889, at 6$. 10. Find the maturity, term of discount, and proceeds of the following uote: $800.00. Cleveland, 0., Jan. SI, 1888. One month after date, without grace, we promise to pay to the order of Hale & Bly, Eight Hundred Dollars, with interest at 5 per cent. Value received. HART & GOLE. Discounted Feb. 10, 1888, at 10$. 11. Find the maturity, term of discount, and proceeds of the following note. $660.90. Albany, N. Y., May 5, 1888. Ninety days after date, I promise to pay to the order of H. H. Douglas, Six Hundred Sixty and ffo Dollars, with interest. Value received. CLAYTON 8. MEYERS. Discounted June 1, 1888, at 5$. 12. Find the maturity, term of discount, and proceeds of the following note: $2400.00. St. Paul, Minn., Aug. SI, 1888. Six months after date, we promise to pay to the order of John W. Bell, Two Hwusand Four Hundred Dollars, with interest at 8 per cent, after one month. Value received. OLIVER & JONES. Discounted Sept. 5, 1888, at 8$. Remarks. — 1. If discount be required on a basis of 365 days for the year, compute the discount first on a basis of 360 days, and from the discount so obtained, subtract fa of itself. 2. The following three examples are to be worked on a discount basis of 365 days. 13. Paul Harmon's bank account is overdrawn $3596.11 ; he now discounts, at 6$ : a 90-day note for $450 ; a 60-day note for $1754.81 ; a 30-day note for $851.95 ; a 20-day note for $345.25 ; a 10-day note for $100; proceeds of all to his credit at the bank. What is the condition of his bank account after he receives these credits ? 238 EXAMPLES IN BANK DISCOUNT. lJh Swick & Sons' bank account is overdrawn $11546.19; they now discount, at 6/e : a 90-day note for $3975.21; a 60-day note for $5514.25; a 30-day note for $1546.19; a 20-day note for $2546.85; proceeds of all to their credit at the bank. What is the condition of their bank account after they receive credit as above? IS. Philo Perkins & Co.'s bank account is overdrawn $12,916.47 ; they now discount, at 6$: a 90-day note for $2428.40; a 60-day note for $6311.25; a 30-day note for $1120.50; a 20-day note for $4500; a 10-day note for $1550.50; Pro- ceeds of all to their credit at the bank. What is the condition of their bank account after they receive the above credits ? 779. To find the Face of a Note, the Proceeds, Time, and Rate Per Cent, of Discount, being given. Example. — What must be the face of a note, payable in 60 days, that, when discounted at 6$ ? the proceeds may be $573.91 ? Operation. Explanation. — If the discount of $1, at *iaa4! - , 4? db-i 6 per cent., for 63 days, is $.0105, the pro- $1.00 = face of note of $1. *, - ' s .. / :, , ' . ceeds of $1 of the note would be $1 minus • Qr05 = dls * of note of v 1 * .0105, or $.9895 ; and if the proceeds of $1 $ .9895 = proceeds of note of $1. are $.9895, it would require as many dollars face of note to give $573.91 proceeds as $573.91 -f- .9895 = $580, face required. $.9895 are contained times in $573.91, or $580. Rule. — Divide the proceeds of the note by the proceeds of one dollar for the given rate and time. Remark. — If the note be interest-bearing, find the proceeds of one dollar of such note, and proceed as above. EXAMPLES FOR PRACTICE. 780. t What must be the face of a 90-day note that will give $315.04 pro- ceeds, when discounted at 6$ ? 2. What face of a 30-day note, discounted at 7$, will give $1241.98 proceeds ? 3. Wishing to borrow $900 of a bank, for what sum must my 90-day note be drawn, to obtain the required amount, discount being at 10$ ? 4. Having bought goods to the amount of $2431.80 cash, I gave my 60-day note in settlement. If discount be at 7$, what should have been the face of the note ? 5. What must be the face of a note dated Aug. 16, 1888, and payable 6 months after date, that when discounted at a bank Oct. 1, 1888, at 6$, it will bring $2100.55 proceeds ? 6. A note dated Sept. 1, 1888, payable in 90 days, with interest at 7% was discounted 21 days after date, at 10$. If the proceeds were $690.42, what must have been the face ? 7. You have $328.40 to your credit at the bank; you give your check for $936.20, after which you discount a 30-day note for $425.40, proceeds to your credit at the bank; you also discount a 90-day note made by H. C. Davis, pro- ceeds to your credit; you now find yourself indebted to the bank $12.37. If discount be at 6$ what must have been the' face of the Davis note ? PARTIAL PAYMENTS* 239 PARTIAL PAYMENTS. 781. A Partial Payment is a part payment of the amount of a note, mortgage, or other obligation existing at the time such payment is made. 782. Part payments, or payments, are usually acknowledged, and should always be by indorsement on the back of the note or other obligation, but some- times special receipts are given for the sums paid. Indorsements should give date and state amount paid; they are then equivalent to receipts. 783. Partial payments may apply to obligations, either before or after their maturity. 784. A debtor, his attorney, or other authorized agent, may make a payment either partial or in full of any obligation, and such payment may be received and receipted for by the creditor, his attorney, authorized agent, or even by one not authorized, if such a person occupies his place and is so apparently his agent as to deceive a debtor making a payment in good faith. 785. Various rules are in use for finding the balance due on claims on which partial payments have been made; but only the United States Pule and the Merchants' Rule have more than local application. 786. The United States Rule is very generally used. It has the sanction of the law, being the rule adopted by the Supreme Court of the United States, and has been adopted by most of the States. Remarks. — 1. It was held by the Supreme Court of the United States, in its decision adopting or making the above-mentioned rule, that the payment should first be applied to cancel the interest ; that what is left, if anything, after paying the interest, should be used to diminish the principal. In case the payment is not large enough to cancel the interest, it fails of its object, and is to be passed as directed by the rule. 2. If at the time of the making of a partial payment of a debt, the debtor renew his obliga- tion by taking up the old note or bond, and giving a new one bearing interest for the unpaid part of his debt, no taint of usury can be shown affecting the validity of the new note, even though it may be clearly shown that a payment credited was less than the interest due at the time such payment was made. 787. Principles. — 1. Payments must le applied, first, to the discharge of accrued interest, and then the remainder, if any, toward the discharge of the principal. 2. Only unpaid principal can draw interest. 788. The Merchants' Rule is used by most banks and business houses, where computations are on short time obligations, as such rule is regarded as the most convenient for business purposes. Remark. — The merchants' rule is varied in its use by different creditors, and hence is rather more an agreement, founded upon custom or otherwise, between debtor and creditor as to mode of settlement, than a strict rule of law. 240 EXAMPLES US' PARTIAL PAYMENTS. 789. United States Rule for Partial Payments. Remark. — Settlements by this rule are made as follows: Example. — A note, the face of which was $3600, bearing interest at 6$, was given Oct. 17, 1884, and settled Feb. 14, 1889. Find the balance due, the following payments having been made: Mar. 3, 1885, $600; Oct. 25, 1886, $1000; Dec. 6, 1888, $2400. Operation and Explanation. Remark. — Find the time by compound subtraction. Face of note _.- $8600.00 Interest to date of first payment (4 mo. 16 da.) .... w .... . 81.00 Amount of principal and interest at time of first payment — .... ............... $3681. 60 First payment (of Mar. 3, 1885) 600.00 Remainder after deducting first payment.. $3081.60 Interest to date of second payment (1 yr. 7 mo. 22 da.) 304.05 Amount due at time of second payment ...................... $3385.65 Second payment (of Oct. 25, 1886) 1000.00 Remainder after deducting second payment $2385.65 Interest to date of third payment (2 yr. 1 mo. 11 da.) - 302.58 Amount due at time of third payment -..—......................... ... $2688.23 Third payment (of Dec. 6, 1888) 2400.00 Remainder after deducting third payment $288.23 Interest to time of settlement (2 mo. 8 da.) 3.27 Balance due at time of settlement (Feb. 14, 1889) $291.50 Rule. — Find the amount of the principal to the time when the pay- ment, or the sum of the payments, shall equal or exceed the interest then due; from this amount deduct the payment or payments made; and with the remainder as a new principal, proceed as before, to the time of settlement. Second Operation and Explanation Remark. — Find the time by compound inverse subtraction. 1884—10—17) =date of note. Face of note $3600.00 >• = 4 mo. 16 da. = time first payment, which was ...... $600.00 ) Interest on principal to time of first payment 81.60 518.40 ) Remainder after deducting proceeds of first payment $3081.60 . y = 1 yr. 7 mo. 22 da. = time to second payment which was $1000.00 ) Interest on new principal to date of second payment .. 804.05 695.95 J Remainder after deducting proceeds of second payment $2385.65 J. = 2 yr. 1 mo. 11 da.=time to third payment which was $2400.00 ) Interest on new principal to date of third payment 802.58 2097.42 1888—12— 6 x Remainder after deducting proceeds of third payment $288.23 > = 2 mo. 18 da. = time to settlement. 1889— 2 14 ) I nterest OD new principal to settlement 3.27 Balance due at time of settlement - $291.50 Rule. — Find the interest on the principal to the time when the pay- ment, or the sum of the payments, shall equal or exceed the interest then due. Deduct the interest from the payment or payments made, and this difference subtract from the principal. Treat the remainder thus found as a new principal, with which proceed as before, to the time of settlement. EXAMPLES IN" PARTIAL PAYMENTS. 24} EXAMPLES FOR PRACTICE. 790. 1. On a loan of $2000, made Mar. 19, 1884, and bearing 6$ interest, payments were made as follows: Nov. 1, 1885, $500; May 3, 1887, $700; Feb. 1, 1888, $1000. How much will be required for settlement in full, Mar. 2, 1888? 2. Oct. 1, 1885, a note for $1000 was given, payable in 4 years, with. 6$ interest. A payment of $50 was made 1 yr. from date; a payment of $250 was made 1 yr. 6 mo. from date; a payment of $224 was made 2 yr. from date; a pay- ment of $20 was made 2 yr. 8 mo. from date; a payment of $110 was made 2 yr. 10 mo. from date. How much remained due at the maturty of the note? 3. On a claim for $3000, dated Aug. 12, 1885, and bearing interest at 7#, payments were made as follows: Dec. 15, 1885, $30; Apr. 1, 1886, $550; Jan. 20, 1887, $85; June 12, 1887, $1651.50. How much was due May 30, 1888 ? 4. I gave a mortgage for $10000, May 9, 1881, bearing 6$ interest, and made thereon the following payments: Sept. 19, 1881, $500 ; Jan. 1, 1883, $500 ; Apr. 25, 1883, $4000; Oct. 15, 1885, $4000; May 1, 1888, $3525. How much was due at final settlement, June 2, 1888 ? 5. The following note was settled Oct. 13, 1888; a payment of $25 having been made Feb. 15, 1887; one of $300, July 12, 1887; and one of $200, Apr. 1, 1888. If money be worth 8$, how much was due at final settlement? $585.50. Elmira, N. Y., Aug. 1, 1886. Six months after date, I promise to pay to James H. Kingsbury, or order, Five Hundred Mqhty-five and ■&& Dollars, value received. SIMEON G. FREEMAN, 6. On a mortgage for $5500, dated Aug. 13, 1882, and bearing 6$ interest, the following payments were made: Jan. 1, 1883, $100; Mar. 2, 1883, $25; Aug. 13, 1885, $2500; Dec. 19, 1887, $2500; Mar. 1, 1889, $500. How much was required for full settlement, Mar. 11, 1889 ? 7. On the following note payments were endorsed as follows: Nor. 3, 1886, $50; Mar. 16, 1887, $50; Oct. 1, 1887, $50; Dec. 30, 1887, $1000j Apr. 1, 1888, $625. How much was due, if paid in full May 8, 1888, money being worth 6#? $1600.00. Dayton, Ohio, Apr. 1, 1886. Three years after date, I promise to pay to the order of Silas Hopkins, One Thousand Six Hundred Dollars, value received, with use. PETER 8. BRYANT. 8. On the following note indorsements were made as follows: Aug. 1, 1883, $350; Nov. 3, 1883, $1000; Mar. 20, 1886, $600; Mar. 31, 1885, $2500; Deo. 11, 1888, $2000. What was the balance due Jan. 30, 1889 ? $6500.00. Clricago, 111, Mar. 19, 1882. On demand, we promise to pay to the order of Ames & Adams, Six Thousand Five Hundred Dollars, with interest at 6 per cent. Value received. HURD <& HOUGHTON 791. Merchants' Rule for Partial Payments. Example.— Find the balance due Oct. 13, 1888, on a note for $1500, dated July 1, 1887, bearing 6$ interest, and on which the following payments had been made: Oct. 1, 1887, $300; Feb. 12, 1888, $420; June 13, 1888, $700. 16 B42 EXAMPLES IX PARTIAL PAYMENTS. Operation and Explanation Remark. — Find the time by compound subtraction. Face of note, dated July 1, 1887 $1500.00 Interest to date of settlement (1 yr. 3 mo. 12 da,) 115.50 Amount of note at date of settlement $1615.50 First payment (of Oct. 1, 1887) $300.00 Interest of first payment to date of settlement (1 yr. 12 da.) .....;.. 18.60 Second payment (of Feb. 12, 1888) . 420.00 Interest on second payment to date of settlement (8 mo. 1 da,)........ . 16.87 Third payment (of June 13, 1888) 700.00 Interest on third payment to date of settlement (4 mo.) 14.00 Total amount of the payments $1469.47 Balance due $146.03 Rule. — Find the amount of the principal to the time of settlement; also find the amount of each payment, from the time it was made to the time of settlement; subtract the sum of the payments fj*om the amount of the principal debt; the remainder will be the balajtce due. Remark. — This rule is mainly used in case of short notes or business accounts. EXAMPES FOR PRACTICE. 792. 1. What is the balance due, Apr. 27, 1889, on a note for $1050, dated Jan. 24, 1888, bearing 7$ interest, if the following indorsements were made thereon: July 1, 1888, $150; Oct. 15, 1888, $400; Jan. 21, 1889, $300; Mar. 27, 1889, $60. 2. Find the balance due at the maturity of the following note, payments having been made as follows: Apr. 1, 1888, $500; Aug. 25, 1888, $1250; Nov. 3, 1888, $240; Dec. 30, 1888, $300; Feb. 1, 1889, $200. $3000.00. St. Louis , Mo., Dec. S, 1887. Eighteen months after date, I promise to pay to the order of Ezra R. Andrews, Three Thousand Dollars, with interest at 5 per cent. Value received. GEO. J. BEATER. S. How much was due at the maturity of the following note, payments hav- ing been made as follows: Sept. 11, 1888, $75; Sept. 19, 1888, $225; Sept. 26, 1888, $159; Oct. 1, 1888, $155. $650.00. Wichita, Kan., Sept. 6, 1888. Thirty days after date, 1 promise to pay to Gideon Piatt & Co., Six Hundred and Fifty Dollars, with interest at 10 per cent., without gram. Value received. BENJ. F. COLEMAN. 4. Find the balance due on the following note, payments having been made as follows: May 28, 1888, $255.50; June 13, 1888, $168.41; Aug. 31, 1888, $50; Oct. 30, 1888, $500; Nov. 1, 1888, $684.25. $2150.00. Denver, Colo., May 1, 1888. Six months after date, we promise to pay to the order of Wm. H. Sanford, Two Thousand One Hundred Fifty Dollars, with interest at 8 per cent. Value received. MARTIN F. RIQNET t RICHARD M. PECK. EQUATION OF ACCOUNTS. 243 EQUATION OF ACCOUNTS. 793. Equation of Accounts, or Equation of Payments (called also Averaging Accounts or Averaging Payments), is the process of finding the date on which a single payment can be made of two or more debts falling due at different dates, or when the balance of an account having both debits and credits can be paid without loss of interest to either party. 794. Accounts having entries on but one side, either debit or credit, are appropriately called simple accounts, and the process of equating such accounts may be called Simple Equation. 795. Accounts having both debit and credit items may likewise be called compound accounts, and the process of equating such accounts may be called Compound Equation. 796. The Average Date of Payment, or Due Date, is the date on which such payment or settlement may be equitably made; called also the Equated Time, 797. The Focal Date is any assumed date of settlement, with which the dates of the several accounts are compared for the purpose of finding the average time or due date. Remarks. — 1. Any date conceivable may be taken as a focal date, and interest may be computed at any rate per cent., and either on a common cr exact basis, without varying the result; providing only that the dates of all items be compared with such focal date, and uniformity in rate and manner of computing interest be observed throughout. 2. In practice it is vastly better to observe a simple method, by assuming the latest date in the account as a focal date, computing all interest at 6^ by the snort method, on a 360 day basis. 3. The importance of uniformity, simplicity, accuracy, and rapidity in the equation of payments and accounts is such as to justify the use and repetition of the above suggestions as a General Hule.— In all equations, extend time if credit or time paper be involved; select the latest date as a focal date, find actual time irv days, and compute interest at 6 per cent., on a 360 day basis. 798. The Term of Credit is the time to elapse before a debt becomes due; if given in days, it is counted on from the date of purchase or sale the exact number of days of the term; if given in months, it is counted on the number of months, regardless of the number of days thus included. Remarks. — 1. Book accounts bear legal interest after they become due, and notes, even if not containing an interest clause, bear interest after maturity. 2. The importance of a thorough knowledge of both the theory and practice of Equation of Accounts, on the part of bookkeepers and accountants, can hardly be over-rated, as much of this class of work is to be found in every wholesale and commission business. 799. The equity of a settlement of an account by equation rests in the fact that, by a review of such account, one of the parties owes the other a balance to 244 EQUATION OF ACCOUNTS. which certain interest should be added or from which certain interest (discount) should be subtracted. 800. To find the Equated Time, when the Items are all Debits or all Credits and have no Terms of Credit. Example.— When does the (face) amount of the following account become due by equation? Peter Dunn, Directions.— 1. Take Nov. 1 as the focal date. To Eobt. S. Campbell, Br. 2 ' Find the exact time in da ^ s from the date ot w £oo eac h item to the focal date. a /* m- tii-j A ™ 8 Com P ute interest at 6 per cent. , 360 day basis, Sept. 5. To Mdse $ 60 on each item for its time. "26. " " _ 100 4. Find the total of interest Oct. 8. " " 200 5. Divide the total interest by the interest on the Nov. 1. " " 120 face amoiint for one day; the quotient will be the average time in days. $480 6. Count back from the focal date the number of days average time thus found Remark. — Compute interest by rules on page 217 Operation. Explanation.— Assume Nov. 1 as a focal date, 1888. Items. Time. Int. and reason as follows: If, on Nov. 1, Dunn pays Sept. 5. $60X57 = 1.57 Campbell the $120 due on that day, there will be no " 26. 100 X 36 = .60 interest charged, because that item was paid when it Oct. 8. 200 X 24 = .80 became due. If, on Nov 1, Dunn pays Campbell Nov. 1. 1-0 X 0= 00 the $200 that has been due since Oct 8, he should pay • — — interest also for the 24 days between Oct. 8, when $480 $1.97 that item became due, and Nov. 1, when, as we have Int. on $480 for 1 day = $ .08 assumed, it was paid, or he should pay, or be charged $1.97 -7- .08 = 24f or 25 days with > $- 80 interest If > on Nov. 1, Dunn pays the the average time; 25 days back fi , du ? f pt 2 ^ h + e ^>uld pay interest also for the •p XT i * n \ ys between Se P t 26 and Nov - 1, or $ . 60. If, on Irom JNov. 1 is Oct. 7. Nov> 1? Dunn pays the $60 due Sept. 5, he should pay interest also on that item from its date to Nov. 1, or for 57 days, or $ .57 Now, on Nov. 1, Dunn owes Campbell not only the $480, the total face amount of the debt, but also $1 97 interest; and if a cash balance were required Nov. 1, Dunn would owe $481.97. But the question was not, what is the cash balance due Nov. 1, but when was the $480, the face amount of the account, due; that is, from what date should such face amount draw interest, in order that neither party gain or lose. Now observe that we have the principal, $480, the interest as found, $1.97, and the rate as assumed and used, 6 per cent., to find the time. The interest on $480 for 1 day is $ 08. Since it takes the principal 1 day to accumulate $ .08, it must have taken it as many days to accumulate $1.97 — or the account was due as many days back from Nov. 1, the focal date — as $ 08 is contained times in $1.97, or 25 days. Count back 25 days from Nov. 1, 1888, and obtain Oct. 7, 1888, the equated date of payment, or the date on which Dunn could pay Campbell $480, the face of the debt, without loss of interest to either party. Again: the same example solved, when assuming Sept. 5, the earliest date, as a focal date, or by the discount method. Remark. — Explanations like the following are based upon a settlement of accounts, none of which are due at the date of settlement or adjustment, as in case of the giving of an interest bearing note or bond for the equitable amount due, or for anticipating the payments of debts thus requiring a cash balance. Sept. 5. $ GO X = $.00 " 26. 100x21= .35 Oct. 8. 200 X 33= 1.10 Nov. 1. 120 X 57 = 1.14 EQUATION OF ACCOUNTS. 245 Operation. Explanation.— Assume the earliest date (Sept. 5) Items. Time, pisct. as the focal date, and reason as follows: If, on Sept. 5, Dunn pays the $00 due on that date' he will neither have to pay interest on it nor be allowed discount; but if, on Sept. 5, he pays the $100 due Sept. 26, he should be allowed discount oa that item for the 21 days between Sept. 5 and Sept. 26, or $ .35 discount. $480 $2.59 i f> on Sept 5 ne pays ihe ^00 not due until Oct. 8, he should be allowed discount on that item for the 33 days between Sept. 5 and Oct. 8, or $1.10 discount; and if, on Sept. 5, he pays the $120 not due until Nov. 1, he should be allowed discount on that item for the 57 days between Sept. 5 and Nov. 1, or $1.14 discount. There- fore, assuming Sept. 5 as the date of settlement, Dunn does not owe on that date the face amount of the account, but such amount, $480, less the amount of the above discounts, $2.59, or really a cash balance of $480, minus $2.59, or $477.41. But the question is not, what was the cash balance Sept. 5, but on what date would the payment of the face amount, $480, have been equitable? We have thus a condition similar to that found in the first operation, viz.: the principal, $480, the rate, 6 per cent., and the discount (interest) given, to find the time; and, as before, divide the discount by the discount on the principal for 1 day, and the quotient, 32, will be the average time in days. And reason, in conclusion, that from Sept. 5 Dunn is entitled to retain the face amount of his debt, $480, for 32 days, or until it has accumulated $2.59 interest in his hands; or, in other words, in equity, he should pay such amount 32 days after Sept. 5, or Oct. 7. Again: same example, explained with an intermediate date (Oct. 1) assumed as a focal date. Opebation. Interest on $60 from Sept. 5 to Oct. 1, 26 days = $.26 . Interest on $100 from Sept. 26 to Oct. 1, 5 days = .0833 + Total interest, -' - - $.3433 + Discount on $200 from Oct. 8 back to Oct. 1, 7 days == $.2333 + Discount on $120 from Nov. 1 back to Oct. 1, 31 days = .62 Total discount, ... - 8.8533 + .8533-1 .3433+ = $.51, excess of discount. $.51 ^ .08 — 6 days. Oct. 14-6 days = Oct. 7. Explanation. — Assume Oct. 1 as the focal date, and reason as follows: If, on Oct. 1, Dunn pays the $60 due Sept. 5, he should also pay interest on that item for the 26 days between Sept. 5, when it became due, and Oct. 1, when it was (assumed to have been) paid, or he should pay or be charged with $.26 interest. If, on Oct. 1, he pays the $100 due on Sept. 20, he should also pay interest on that item for the 5 days between Sept. 26, when it became due, and Oct. 1, when it was (assumed to have been) paid, or he should pay $ .0833+ interest; thus we have a total interest charge against him of $.3433+ on the two items of his account not paid until after they were due. But if, on Oct. 1, he pays the $200 not due until Oct. 8, he should be allowed a discount for the 7 days between Oct. 8, when it became due, and Oct. 1, when it was paid, or he should be allowed a discount of $ .2333+ on that item; and if, on Oct. 1, he pays the $120 not due until Nov. 1, he should be allowed a discount on that item for the 31 days between Nov. 1, when it became due, and Oct. 1, when it was paid, or he should be allowed a discount of $ .62. Thus we have a total discount to be allowed him of $.8533+ off from the two items of his account which he paid before they were due. The difference between the amount of interest charged to him, $ .3433+, and the amount of discount for which he is given credit, $.8533+, is $.51, an excess of discount, showing that at the date assumed 246 EQUATION OF ACCOUNTS. (Oct. 1) he does not owe the face amount of the account, $480, but $480, the face amount, lesj $.51 discount, or only $479.49, which sum is the cash balance due on that date (Oct. 1). But since, as before, the question is not as to the cash balance, but is the date on which equitable settlement could have been effected by the payment of the face amount of the account, $480, we have, as before, the principal, rate, and discount (interest) given, to find the time. Divide the discount, $.51 by $.08, and find Dunn to be entitled to withhold or delay the payment of the $480 until it accumulates $ .51 interest (discount) in his hands, or that he keep the $480 for 6 days after Oct. 1, thereby in equity paying it on Oct. 7, as already twice shown. Remarks. — 1. The above explanation is given in. addition to the former two, in order to illustrate that any date may be used as a focal date, and for the object of aiding the teacher in imparting to the pupil a full understanding of the underlying principles involved, and it gives added assurance that the solutions before given led to a correct result. Yet no one of them, nor all of them taken together, can be accepted as being anything beyond assurances. They are not proofs. 2. If settlement on Oct. 7 be equitable, the interest on such of the accounts as fall due before that date must be offset or balanced by the discount (interest) of such of the accounts as fall due after that date, to within less than one-half of the interest (discount) of the face amount of the account for one day; otherwise the due date as determined would be proven wrong. Pkoof. — Oct. 7 as a focal date. Explanation.— Assume Oct, 7 as a focal date, and reason as follows. If, on Oct. 7, Dunn pays the $60 due Sept. 5, he should pay interest also on that item for the 32 days between Sept. 5 and Oct. 7, or $ .32 interest; and if, on Oct. 7, he pays the $100 due Sept. 26, he should pay interest on that bill for the 11 days between Sept. 26 and Oct. 7, or $.1833-J- interest; being thus charged $ .5033+ interest on the two items not paid until after they were due. But if, on Oct. 7, be pays the $200 not due until Oct. 8, he should be allowed a discount on that item for the 1 day between Oct. $ . 03 7 and Oct 8, or $ .0333+ discount; and if, on Oct. 7, he pays the $120 not due until Nov. 1, he should be allowed a discount on that item for the 25 days between Oct. 7 and Nov. 1, or $.50 discount; being thus allowed a total discount of $ .5333+ for the pre-payment of the items of the account coming due after Oct. 7. The difference between the amount of the interest on the items of the account falling due before Oct. 7, from their respective dates down to Oct. 7, and the amount of the discounts on the items of the account coming due after Oct. 7 from their respective dates back to Oct. 7, is only $.03, or is less than one-half the interest (or discount) on the face amount of the account for one day, thus proving Oct. 7 to be the date on which the payment of the face amount of the account, $480, will effect an equitable settlement between Dunn and Campbell. Rule.— I. Select the latest date as a focal date; find the time in days from the date of each item of the account to the focal date, and compute the interest on each of the respective items for its time as found. II. Divide the amount or sum of the interest on the items by the inter- est on the face amount of the. account or items for one day ; the quotient will be the number of days average time. III. Count bach from the focal date the number of days so found ; the date thus reached will be the due date of the face amount of the account or the date on which such face amount could be paid without loss to either party. Operation. Days to Oct. 7. Interest. Sept. 5, $ 60.. ...32... .1.32 Sept. 26, 100.. ...11... . .1833 + .5033 + Discount. Oct. 8, $200.. .. 1.... $.0333 + Nov. 1, 120 .. ..25.... .50 $.5333 + .5033 + EXAMPLES IX EQUATION" OF ACCOUNTS. 247 Remarks. — 1. In finding the average time of credit in days, fractions of a day of one-half or greater are counted as a full day; fractions less than one-half are rejected 2. In business, odd days, odd cents, and even odd dollars, are often rejected in the interest calculations in equating the time, it being correctly reasoned that, in the long run, any losses or gains thereby shown would fairly balance; and therefore business men, so settling, may cut off as they please. But for class-work, exact money, exact time, and interest computed to four decimal places, should be required. 3. Any date between the extremes, or within the account, may be taken as a focal date, the only question involved being a balance of the interest or discount ; but, except for illustrative purposes by the teacher, or test exercises for advanced pupils, the selection of any date except the latest for a focal date is not recommended. 4. The selection of the latest date saves one interest computation, and removes the objection often raised in case an earlier or the earliest date be chosen, that an account is not likely to have been settled before it was made. 5. The product method of equating accounts, often used, and in many cases capable of producing correct results, is not recommended, because: First. It is much more difficult to comprehend than the interest method. Second. It usually involves a greater number of figures. Third. By it, a cash balance, often desirable, is only obtainable by an additional operation, and with difficulty and perplexity. Fourth. Equation of accounts having debit and credit items is impossible by that method, in case, as frequently happens, the face amounts of the two sides chance to be equal; i. e. } the debtor having paid the face amount of his obligation; while there may still be an important balance of interest or discount, which can be readily adjusted if the interest method be used. Fifth. A book-keeper, equating by the interest method, can readily exhibit to his employer the equity conditions of an excess of interest or discount, even though the employer be unfa- miliar with the formal work of the equation. Suggestion to the Teacher. — Place on the blackboard, as an example, an account with a dozen or more items, having different dates, and each for a simple amount, and so assign the example that each pupil may have a different focal date from which to work; then require each pupil to prove his result and withhold the announcement until called for. Such exercises will stimulate the pupils to accuracy and speed in their work, and will result in imparting a very thorough knowledge of the subject. EXAMPLES FOB PRACTICE. 801. When are the following accounts due by equation: Remark. — The teacher should require that each result be proved. Oct. Warren Pease, To Calvin Gray, Dr. ToMdse $ 75 " " 50 " " _ 80 " " 120 " " 40 Oct. 1, 6, 14, 25, 31, Norman Colby, To Seth Stevens & Sons, Dr. 1, To Mdse $300 5, " " 150 11, " " 120 16, " " 200 28, " " 100 30, " " 180 Aug. 7, " 25, Sept. 30, Oct. 12, Parker H. Goodwin, To Perkin3 & Hawley, Dr, To Mdse $200.00 " " 180.55 " « 35.60 " " 100.00 " " 50.25 Dec. 3, " " £ Wm. P. Dugan, 1889# To Godfrey, Son & Co., Dr. To Mdse $200 " 100 " _. 100 " " 300 ". " 300 " 200 Jan. Feb. 6, 1, " 27, Apr. 3, " 20, ei 27 248 EQUATION" OF ACCOUNTS. O. Gerald, Jones & Co., 1888. io oamuei oi mm, ur. Oct. 13, To 3 ..._$500.00 Nov. r ee a _._ 821.75 " 28, '6 ee ee ... 150.00 Dec. 17, ee _.. 205.25 " 30, ee ee ... 33.00 1889. Jan. 12, ee ee ... 300.00 6. Theodore Stanley, 1887. To Paul Fleming, Dr. Nov. 6, To Mdse $500 " 28, a ee 200 Dec. 17, ee ee 150 " 29, ee ee 150 1888. Jan. 13, ee ee 300 " 30, ee ee 100 Feb. 11, ee ee 200 Mar. 31, ee ee 200 7. Felix Peterson & Bro., 1888. To Paul Paulson & Co., Dr. Dec. 1, To Mdse._ $1500 " 16, " " 2000 Jan. 19, Mar. 1, " 21, 7000 500 1000 8. Philip Darling, To Jacob V. Hall, Dr. To Mdse $300 " " 150 1887. Oct. 6, " 31, Nov. 17, Dec. 1, 1888. Jan. 20, Feb. 16, Mar. 3, Apr. 6, 150 450 300 600 300 300 802. To find the Equated Time, when the Items have Different Dates, and the Same or Different Terms of Credit. Example (requiring time extension). — When does the face amount of the following account become due by equation ? • John Price, 1888. To Volney Clark, Dr. Sept. 14, To Mdse., 1 mo $1000 5 mo 500 60 da 700 30 da 200 2 mo 600 If the time for the payment of each of the several items of the above account be extended for the term of credit indicated, the account will stand as follows: John Price, Directions. — 1. Assume the latest date as a focal date. 2. Star the focal date to distinguish it. 3. Observe general directions for example on page 244. ee 30, ee ee Nov. 10, ee ee ee 29, ee ee Dec. 31, ee ee To Volney Clark, Dr. 1888, Oct. 14... $1000 1889, Feb. 28._ 500 1889, Jan. 9 700 1888, Dec. 29 200 1889, Feb. 28 600 Rule.— I. Extend the time of credit of such items as are sold on credit. II. Select the latest date as a focal date, and find the interest on each item from its maturity date to the focal date. III. Divide the aggregate of interest thus found by the interest on the face amount of the account for one day ; the quotient ivill be the time in days to be counted bach from the focal date to determine the due date or average date. EXAMPLES IK" EQUATION" OE ACCOUNTS. EXAMPLES FOR PRACTICE. 249 803. On what dates are the face amounts of the following accounts due by equation ? Remarks. — 1. Extend the time, by adding the term of credit to the date of each item, before proceeding wjith the work. 2. Should two or more items mature on the same date, their sum may be found, and one computation of interest serve for all. Sept 12, To] lldsc a 30, (6 ee Nov. 10, << ee a 29, it ee Dec. 31, SC it 1. Herbert G. Williams, 1888. To Brewster & Brewster, Dr. Aug. 15. To Mdse., 2 mo., $180 " 29, " "' " 300 Sept. 20, " " " 200 Oct. 4, " " u 120 Nov. 1, u " M 100 Samuel S. Sloan, To A. D. Wilton, Dr. To Mdse., 1 mo., $1000 5 mo., 500 60 da., 700 30 da., 200 2 mo., 300 & H. C. Colvin, 1888. To Jas. Fowler, Dr. Nov, 3, To Mdse., 30 da., $550 " 23, " " " ... 800 Dec. 1, " " " 90 " 28, " " " 210 1889. Jan. 11, " u " 600 " 31, " « " 300 Jf. Oliver H. Brown, 1888. To Stephen Brackett, Dr. Oct. 3, To Mdse., 30 da.,__.$ 319.50 " 31, " " 4 mo., ... 750.00 Dec. 1, " " 280.50 " 31, " " 2 mo., ... 400.00 1889. Feb. 3, " *■ 60 da.,... 250.50 " 28, " " 216.75 Mar. 12, " " 80.25 Apr. 30, " " 1 mo.,... 150.00 5. T. L. King & Son, 1888. To Groves & Co., Dr. Jan. 30, To Mdse., 1 mo., $ 300 Feb. 28, " " 60 da., 300 Mar. 25, " " 2 mo., 1200 June 29, " " 30 da., . 1500 Jan. ee Feb. Mar. June 20, July 8, John Jennings, To Richard Smith, Dr. To Mdse., ._•_ $ 50 1 mo., 100 2 mo., 600 200 3 mo., 120 1 mo., 300 17, 31, 9, 3, Feb 29, 14, 25, 1, 31, June 24, Mar. ee May Porter Cass & Sons, To Phelps Bros., Dr. 19, To Mdse., 60 da., ..$519.22 60 " .. 211.50 30 " .. 120.00 30 " .. 181.75 2 mo., .. 80.00 1 " .. 69.78 3 " - 127.75 a ee ee ee ee ee a a a a ee 8. H. B. Spencer & Co., 1888. To Wood, Son & Co., Dr Sept. 14, To Mdse., 1 mo., $ 1000 « 30, " " 5 mo., 500 Nov. 10, " " 60 da., 700 " 29, " " 30 da., 200 Dec. 31, " " 2 mo., 600 Jan. 30, " Feb. 28, " Mar. 25, " June 29, " 1 " 60 da., 300 300 1200 1500 250 EQUATION* OF ACCOUNTS. 804. To find the Equated Time, when an Account has both Debits and Credits. Example. — What is the balance of the following account, and when due hy equation. Dr. James B . Gkeene. Or. 1889. Jan. Feb. 15 28 To Mdse., 600 300 I 1889. Feb. 1 Mar. 1 31 By Cash, © © © © CO CO Directions. — 1. Select the latest date as a focal date. 2. Find the time from the date (maturity) of each item to the focal date. 3. Compute the interest on each item for its time. 4. By addition, determine the sum of the interest on each side. 5. Find the difference between the Dr. and Cr. interest for an interest balance. 6. Divide this interest balance by the interest on the balance of the account for one day. 1889. Jan. 15, $600. " Feb. 28, 300. Total Dr., $900. Operation. Dr. 75 days to focal date = $7.50, interest. 31 " " " " = 1.55, " 05, total Dr. interest. Cr. 1889. Feb. 1, $300. " * Mar. 31, 300. $2.90, total Cr. interest. 58 days to focal date = $2.90, interest. " u " « =z 00, " Total Cr., $600. Dr. balance, $300. Interest on $300 for 1 day = $.05. $9.05 — $2.90 = $6.15, excess Dr. interest. $6.15 ~ $ .05 = 123, or 123 days equated time. 123 days lack from Mar. 31, 1889, gives Nov. 28, 1888. * Focal date. Explanation.— Assume the latest date, Mar. 31, as a focal date, and reason as follows: If, on Mar. 31 , Greene receives credit for the $300 paid on that day, he should not receive credit for any interest, because the money was paid on the day it fell due ; but if, on Mar. 31, he receives credit for the $300 that he paid Feb. 1, he should receive credit also for the interest on that payment for the 58 days between Feb. 1, when he paid it, and Mar. 31, when he received credit tor it, or he should be credited for $2.90 interest ; and if there were no debits or charges against him, he would be entitled, Mar. 31, 1889, to a net credit of $602.90, as a cash balance in his favor. But we have the debit of the account to be considered, as follows : If, on Mar. 31, Greene be charged with $300, the value of Mdse. sold to him Feb. 28, he should also be charged with its interest for the 31 days between Feb. 28 and Mar. 31, because he did not pay for the Mdse. when the amount of it was due; or he should, on this item, be charged $1.55 interest ; and if, on Mar. 31, he be charged with $600, the value of Mdse. sold him Jan. 15, he should also be charged with its interest for the 75 days between Jan. 15 and Mar. 31, because he did not pay for the Mdse. when the amount of it was due; or he should, on this item, be charged $7.50 interest, thus being charged a total of $9.05 interest, and showing his total debt to be $909.05 on Mar. 31, in case he had received no credit for payments made. But since he had received credit for payments amounting to $600, and for interest thereon amounting to $2.90, his debt, on Mar. 31, was not $900, as the sum of the items charged, plus EQUATION" OF ACCOUNTS. 25J 1 9.05, the sum of the interest charged, but was $900, the sum charged, less $600, the sum credited, or only $300 of principal debt or charge unpaid, and $9.05, less $2.90, or $6.15, interest balance due. And if the cash balance due was required, it would thus be found to be $306.15. But the question is not concerning the cash balance due Mar. 31, 1889,. but on what date was the $300 balance of account due by equation ? And to determine this, proceed as in the earlier explanation of this subject; having given the principal (balance of account), $300, interest (balance), $6.15, and rate, to find the time. Divide the balance of interest by the interest on the balance for 1 day, and find the time to be 123 days, and reason in conclusion that, since on Mar. 31, Greene owed not only the $300, but also $6.15 interest, he had at that date been owing the $300 for a time sufficient to enable it to accumulate $6.15 interest, or for 123 days; and if he had, on Mar. 31, 1889, been owing the $300 for 123 days, that debt must have been due by equation 123 days prior to Mar. 31, 1889, or since Nov. 28, 1888. For reference, and to give assurance of the correctness of the above conclusion, the same example is taken and solved with the earliest date assumed as a focal date. Operation. Dr. 1889. * Jan. 15, $600. days to focal date = 00, discount. Feb. 28, 300. U " " " "= $2.20, total Dr. discount. Total Dr., $900. Or. 1889. Feb. 1, $300. 17 days to focal date = % .85, discount " Mar. 31, 300. 75 " u " " = 3.75 , Total Cr., $600. $4.60, total Cr. discount. 2. 20, total Dr. discount. Dr. balance, $300. $2.40, excess Cr. discount. Interest or discount on $300 for 1 day, = .05. $2.40 -f- $.05 = 48 = number of day3 equated time. 48 days hack from Jan. 15, 1889, gives Nov. 28, 1888. . * Focal date. Explanation.— Assume Jan. 15, 1889, the earliest date, as a focal date, ana reason as follows: If, on Jan. 15, Greene pays the $600, the value of Mdse. bought on that day, he pays his debt when due, and should neither be charged with interest nor credited with discount; but if, on Jan. 15, he pays the $300 not due until Feb. 28, he should be credited with discount on that item for the 44 days between Jan. 15, when he paid it, and Feb. 28, when it becomes due ; or he should be credited with $2. 20 discount for the pre-payment of this item. Thus we find that, on Jan. 15, he did not owe the $900, the face amount of his debt, but only $900, the face, less $2. 20 discount. If there were no credits to be considered, he would, on Jan. 15, 1889, owe $897.80 as a cash balance. But we have to consider the Cr. of his account, and do so as follows: If, on Jan. 15, he be credited for the $300 not paid until Feb. 1, he should be charged discount on that sum for the 17 days between Jan. 15, when he received credit for its payment, and Feb. 1, when such payment was actually made, or he should be charged discount on this item of $ .85 ; and if, on Jan. 15, he receives credit for the $300, the payment not made until Mar. 81, he should be charged discount on this item for the 75 days between Jan. 15, when he received credit for its payment, and Mar. 31, when it was actually paid, or he should be charged discount on this item of $3.75; thus we find that, on Jan. 15, he should have received credit for the sum of his payments, $600, less the sum of the discount, $4.60, charged against him, or for $595.40 as a cash balance; or 252 EQUATION OF ACCOUNTS. that, on Jan. 15, he owed $300 and stood charged with discount balance of the difference between $4.60 and $2.20, or $2.40 ; in other words that, on Jan. 15, 1889, he not only owed the $300, balance of items, but also the $2.40 balance of discount, or had been owing the $300 for a length of time sufficient to enable that sum to accumulate $2.40 in the creditor's hands. "We have thus, as before found, the principal, interest (discount), and rate given to find the time; and divide the interest (discount) balance, $2.40, by the discount on the balance of the account for 1 day, and find that, on Jan. 15, 1889, Greene had been owing the $300 for 48 days. Counting back 48 days from Jan. 15, 1889, find, as before, the balance, $300, to have been due by equation Nov. 28, 1888. Remarks.— 1. "While the result, being the same in both the foregoing operations, gives assurance of the correctness of both, it is assurance only, it is not proof. 2. If the conclusions drawn from the above explanations be correct, and the balance be due Nov. 28, 1888, as found, then the sum of the discount of the Dr. items from their respective dates back to Nov. 28, 1888, must be offset or balanced by the sum of the discount of the Cr. items from their respective dates back to Nov. 28, 1888, to within less than one-half of the discount of the balance, $300, for 1 day, or to within less than $.02£. Proof. — Take the example as above explained, and assume Nov. 28, 1888, as a focal date. Operation. Dr. 1889. Jan. 15, $600. 48 days back to focal date = $4.80, discount. " Feb. 28, 300. 92 " " " " " = 4.60, " $9.40, total Dr. discount. Cr. 1889. Feb. 1, $300. 65 days back to focal date = $3.25, discount. " Mar. 31, 300. 123 " " " " " = 6.15, " $9.40, total Cr. discount. Focal date, Nov. 28, 1888. Explanation.— Assume Nov. 28, 1888, as a focal date, and compute the discount on each item of the account for the time between the date of such item and the focal date, and find that the total of the Dr. discount exactly balances the total of the Cr. discount. Hence it is proved that the balance of the account considered was due by equation Nov. 28, 1888, as twice shown. Remarks. — 1. In case a cash balance is required at any given date, it may be ascertained in either of the two following ways: First, compute interest or discount on all debit items; add the interest to these items, or subtract the discount from them; find the total amount of these items for a total debit; in the same way find the total amount of the credit items for a total credit; then by subtraction, find the balance due. Second, find the date on which the face balance of the account is due by equation, then if the due date comes before the actual date of settlement, add to the face balance of the account the interest upon it from the due date to the actual date of settlement; or if the due date comes after the actua 1 date of settlement, subtract the discount on the face balance for the same time. 2. After the due date is determined, the rate of interest or discount allowed should be deter- mined by the law of the place, but it may be determined by agreement of the parties. LocaJ interest and usury laws would prevail in disputed cases. 3. In proving the equation of accounts, the equitable settlement of which is found to come at a date within the account or between its extreme dates, the difference between the interest and discount of the Dr. items from their respective dates to the due date (by equation) must be offset or balanced by the difference between the interest and discount of the Cr. items, from their re spective dates to the due date, within one-half of the interest or discount on the balance for one day EQUATION OF ACCOUNTS. 253 805. Example. — What is the balance of the following account, and when is it due by equation ? Dr. Feb. a 1 10 To Mdse., (Student's Ledger.) Charles S. Williams. 1886 600 Feb. 19 1800 a 28 Mar. 6 By Cash, Or. 300 300 300 Operation. Dr. June 19, 1886, focal date. 1886. Feb. 1, $ 600. 138 days to focal date = $13.80, interest, « Feb. 10, 1800. 129 " «! " " = 38.70, " Total Dr., $2400. Cr. $52.50, total Dr. interest, 1886. Feb. 19, $300. 120 days to focal date = $6.00, interest. « Feb. 28, 300. Ill " " " " = 5.55, " Mar. 6 ; 300. 105 « * " " = 5.25, «. Total Cr., $900. $16.80, total Cr. interest. $52.50, Dr. interest. Dr. balance $1500. • 16.80, Cr. interest. Interest of $1500 for 1 day = $ .25. $35.70, excess Dr. interest. $35.70 ~ $.25 = 142| = 143 days equated time. 143 days back from June 19 = Jan. 27, 1886. Remark. — Since debit and credit accounts are accounts wherein both debtor and creditor are represented by certain purchases (debts) and payments, and since the items constituting the Dr. on the Ledger of one of the parties would constitute the Or. on the Ledger of the other party, and vice versa, it follows that an account equated from both these views must show like conclusions; i. e., the above account reversed, so that its Cr. shall appear a Dr., and its Dr. appear a Cr., and equated from any date as a focal date, must show the same conclusion as before. Example. — Same as Def ore, reversed, and with May 1 assumed as a focal date. Dr. 1886. Feb. 19 a 28 Mur. 6 To Cash, (Charles S. Williams' Ledger.) "Student." Or. 1886. 300 Feb. 1 300 (4 10 300 By Mdse., 600 1800 254 EQUATION OF ACCOUNTS. May 1, 1886, focal date. 1886. Feb. 19, $300 " Feb. 28, 300 • Mar. 6, 300 Total Dr., $900. 1886. Feb. 1, $ 600. " Feb. 10, 1800. Operation. Dr. 71 days to focal date 62 " " " " 56 " " M " = $3.55, = 3.10, = 2.80, interest. ee Or. $9.45, total Dr. interest. 89 80 days to focal date U ti €< « = $ 8.90 3 = 24.00, , interest. ee Total Cr., $2400. $32.90, total Cr. interest. 9.45, total Dr. interest. Cr. balance, $1500. $23.45 Interest of $1500 for 1 day = $.25. $23.45 -J- $.25 = 93f = 94 days equated time. 94 days back from May 1, 1886 = Jan. 27, 1886, as before found. Example ( same as first illustrated ). — Proof, assuming Jan. 27, 1886, as a focal date. Operation. Dr. 1886. Feb. 1, $ 600. 5 days after focal date = $ .50, discount. " Feb. 10, 1800. 26 " " ' " " = 4.20 , 4.70, total Dr. discount. Cr. 1886. Feb. 19, $300. 23 days after focal date = $1.15, discount. ee Feb. 28, 300. 32 ee ee ee ee = 1.60, ee ee Mar. 6, 300. 38 ee €€ cc ee = 1.90, ee $4.65, total Cr. discount. Cr. balance, $1500. $4.70, total Dr. discount. Discount on $1500 fori day, $.25. 4.65, " Cr. $.05, difference. Explanation. — The difference between the Dr. discount and the Cr. discount is 5 cents, or -fa = i of the discount on the $1500 balance for 1 day, or less than one-half of 1 day's discount, thus proving the balance to have been due since Jan. 27, 1886, as determined by both the former operations, and rendering an explanation which could be made in the usual form quite unnecessary. Rule. — Find the face balance of the account, and also the excess of interest from the latest date as a focal date. If the balance of account and excess of interest be on the same side, date bach; if on opposite sides, date forward. EXAMPLES IN EQUATION OF ACCOUNTS. 255 EXAMPLES FOR PRACTICE. 806. 1. When is the balance of the following account due by equation ? Dr. Frank H. Barnard. Or. 1887, Jan. 15 Feb. 28 To Mdse., 600 300 1887. Feb. Mar. 1 By Cash, 31 " " 300 300 2. What is the balance of the following account, and when due by equation ? Dr. Benj. F. Hawkins. Or. 1887. 1887. Jan. 14 To Mdse., 600 Jan. 20 By Cash, 1000 Feb. 28 3 a a a a 300 500, Feb. 10 a a 700 tt 15 a a 600] 3. If money be worth 7$ per annlim, what was the cash balance due on the following account July 1, 1887 ? Dr Victor E. Brown & Co. Cr. 1887. Jan. 31 Mar. 30 To Mdse., 1887. 450 Jan. 2 450 Feb. 13 Mar. 29 By Mdse., " Cash, " Mdse., 600 300 300 4. What was the cash balance due on the following account Jan. 1, 1889, if money be worth 8$ per annum ? Dr. Henry J. Sanford & Bro. Cr. 1888. Aug. 4 To: Md Sept. 1 a a Oct. 31 a a Dec. 3 a a 4 mo., 1888. 200 Oct. 1 400 Nov. 1 600 Dec. 1 300 1889. Jan. 1 Feb. 1 Mar. 1 By Cash, 150 150 150 150 150 150 5, Find the balance of the following account, and when due by equation. Dr. Louis K. Gould. Cr 1888. Sept. 21 Oct. 5 - a 30 Dec. 18 1889. Jan. 31 Feb. 28 To Mdse., 1 mo., " " 60 da., " " 30 da., " " 60 da., 1 mo., 1888. 100 Nov. 1 150 a 28 116 50 Dec. 31 251 45 1889. Jan. 15 80 75 Mar. 1 100 10[ By Cash, " Mdse., 1 mo., " " 2 mo., " Cash, 75 110 120 175 200 50 256 EXAMPLES IN EQUATION OF ACCOUNT^. 6. What is the balance of the following account, and when due by equation : Dr. Keed & Co. Cr. 1888. 1888. June 14 To Mdse., 300 July 1 By Cash, 100 a 29 « Cash, 150 Aug. 1 a a 100 Aug. 4 " Mdse., 200 Sept. 1 a 66 100 Oct. 31 " Cash, 100 Oct. 1889. Jan. 1 1 a it " Mdse., 100 450 Remark. — Interest may be computed on one of the four similar Cr. items for the aggregate of their days. 7. When is the balance of the following account due by equation ? Dr. King & Sherwood. Cr. Nov. Dec. 1889. Jan. Mar. 3 31 11 4 To Mdse., 1888. 750 Dec. 20 1000 1889. Jan. 1 600 Feb. 1 150 May 3 By Cash, " Mdse., tt a " Cash, 500 500 1500 500 8. When is the balance of the following account due by equation ? Dr, Samuel Peck & Son". Or. 1887. 1887. Mar. 3 To Mdse., 60 Apr. 1 By Cash, 150 Apr. 24 a a 100 June 1 it a 150 May 1 tt tt 150 Aug. 1 6( (t 150 a 30 tt a 90 Oct. 1 <( 66 90 Aug. 17 tt €€ 200 9. Find, 1st, the balance of the following account; 2d, when due by equation. Dr. Walter L. Parker. Cr. 1888. 1888. May 11 To Mdse., 2 mo., 108 40 June 1 By Cash, 124 27 July 1 " " 30 da., 225 Oct. 31 " 4 mo. note (no Aug. 31 66 tt 280 80 interest), 167 91 Oct. 1 66 ft 137 50 Dec. 1 " Cash, 305 05 10. Find, 1st, when the following account is due by equation ; 2d, the cash balance due Jan. 1, 1888, if money be worth 5/ per annum. Prove the result. Dr. John Montgomery & Co. Cr. 1887. 1888 Dec. 15 To Mdse., 200 Jan. 2 By Cash, 300 tt 28 " " » mo., 300 Mar. 1 " 60-da. note (no 1888. interest), 150 Jan. 14 " " 30 da., 300 EXAMPLES Itf EQUATION OF ACCOUNTS. 25? Remark.— In case a negotiable paper is given, its maturity is determined in the usuai way, by adding to its express time three days of grace. If the paper bear interest, its value is equivalent to its face as cash at its date; while if the paper be non-interest bearing, its value is equivalent to cash at its full maturity. 11. Find, 1st, the balance of the following account; 2d, when due by equation; 3d, cash balance due Jan. 1, 1888, if money be worth 6$ per annum. Prove the result. Dr. R. E. Rogers & Bro. Cr. 1887. May June July 14 3 31 To Mdse. , 1 mo., 60 da., 2 mo., 300 200 400 1887. May July 1888. Jan. 31 15 1 By 2-mo. note interest), " 30-da. note, interest. " Cash, (no on 240 150 100 12. Find, 1st, the balance of the following account; 2d, when due by equation; 3d, the cash balance due Jan. 1, 1888, if money be worth 10$ per anr am. Prove the result. Dr. King, Son & Co. Cr. 1887. Oct. 1 Nov. 3 Dec. 14 1888. Jan. 15 To Mdse., 1 mo., " " 2 mo., * « 60 da,, 1887. 150 Nov. 1 150 Dec. 1 300 1888. 300 Feb. 15 By Cash, " 3-mo. accpt. (no interest), " Cash, IS. When is the balance of the following account due by equation ? 200 200 200 Dr. Spaulding & Co. Cr. 1888. 1888. Oct. 15 To Mdse., 30 da., 278 50 Nov. 20 By Cash, 210 a 31 " " 4 mo., 147 50 Dec. 31 " 2-mo. accpt. (no 1889. interest), 175 Jan. 5 <( €4 160 25 1889. Feb. 14 " • 60 da. 311 50 Mar. 1 u 60-da. note, on interest, 220 50 14-. Find, 1st, the balance of the following account; 2d, when due by equation; 3d, the cash balance due Mar. 1, 1889, if money be worth 5/ per annum. Prove the result. Dr. Abraham Bradley. Cr. 1888, Aug. 31 i-ept. 5 Oct. 31 Dec. 19 1889. Jan. 1 By Mdse., 1 mo., " " 60 da., •* 4 mo., " 30 da., a it (( tt, 17 w 1 mo., 1888, 150 Oct. 2 200 600 (6 30 150 Dec. 1 100 1889. Jan. 25 By 30-da. note (no interest), " Cash, " 60-da. note, on interest, *' 1-mo. accpt. (no interest), 100 200 300 500 258 EXAMPLES IN" EQUATION OF ACCOUNTS. 15. Find, 1st, the balance of the following account ; 2d, when due by equa- tion ; 3d, the cash balance due Apr. 1, 1889, if money be worth 7$ per annum. Prove the result. Dr. Lee & Powers. CV. 1888. 1889. Sept. 9 To Mdse., 600 Jan. 1 By Cash, 500 Oct. 1 " " 2 mo., 300 Mar. 16 " 2-mo. note, on Dec. 13 " " 1 mo., 150 interest, 100 1889. Apr. 30 :. Cr. $15600.00 2324.00 $12000.00 1249.70 1491.01 $14740.71 3183.29 $17924.00 $17924.09 Dr. Grain. Or. $14382.50 438.20 1249.70 $16070.40 1491.01 1491.01 $15600.00 1128.42 2324.00 $19052.42 $17924.00, Brown's debit. 14740.71, Brown's credit. $19052.42 $19052.42, sales of grain. 16070.40, purchases of grain. $3183.29, excess received by Brown, or the amount due from Brown to Hart. 2 ) 2982.02, n et gain of firm. 1491.01, net gain of each. Explanation. — Credit Brown for the $12000 cash sent by him to Hart, and debit Hart for the same amount. Credit Hart for the $14382.50 paid by him for grain, and debit Grain for the same amount. Credit Grain for $15600, the price received by Brown for the 40 car loads of corn, and debit Brown for the same amount. Credit Hart for the $438.20 expenses paid by him, and debit Grain for the same amount, as an element of its cost. Credit Brown for the $1,249.70 freight paid, and debit Grain for the same amount as an added element of its cost. Now under the dissolution agreement, debit Hart for $1128.42, the inventory value of the grain taken by him, and credit Grain for the same amount, as having virtually been sold to Hart. Debit Brown for $2324, the inventory value of the oats taken by him, and credit Grain for that amount, as having virtually been sold to Brown. Having now disposed of all the grain, the difference between its cost, Dr., and the returns from its sales, Cr., will show the gain or loss. Foot the debits, and find the total cost to have been $16070.40; foot the credits, and find the total receipts from sales to have been $19052.42, showing a net gain of the difference, or $2982.02, one-half of which, or $1491.01, should go to the credit of each partner. Debit Grain for Hart's one-half of the gain, $1491.01, and credit Hart for the same amount, to which he is entitled by the partnership agreement; and for like reasons, debit Grain for $1491.01, as Brown's one-half of the gain, and credit Brown for the same amount, as his one-half of the gain, and find that while Brown is entitled, as shown by his credits, to only $14740.71, he has actually received, as shown by his debits, $17924, or that he has received the difference $3,183.29, more than he is entitled to receive. Also find that while Hart is entitled, as shown by his credits, to receive $16311.71, he has actually received, as shown by his debits, only $13128.42, or that he has received the difference, $3183.29, less than is due him. If then, Brown pays the excess, $3183.29, that he has received, over to Hart, the accounts of both, as well as the Grain account, will be in balance, and the obtained results will be shown as follows: 1st. Net gain, $2982.02. 2d. Net gain of each, $1491.01. 3d. Brown owes Hart $3183.29. 2. Hopkins and Hawley formed a partnership Sept. 1, 1886, for two years, and agreed that the gains or losses in the business should, on settlement, be adjusted- according to the average investment. Sept. 1, 1886, Hopkins invested $6250, and Hawley invested $4500. Three months later each invested $1750. On Mar. 1, 1888, Hopkins drew out $3000, and Hawley invested $2000. How should a gain of $9400 be divided ? 3. Three boys bought a watermelon for 24^, of which price Charles paid 9^, John 8^ and Walter 7^. Ralph offered 24^ for one-quarter of the melon, which offer was accepted and the melon divided. How should the 24^ received from Ralph be divided among the other three boys? •70 EXAMPLES IN" PARTNERSHIP. 4. At the time of closing business, the resources of a firm were: Cash, $931.50; Mdse., per inventory, $13196.25; notes and accounts due it, $8154; interest on same, $211.50; real estate, $11150. The firm owed, on its notes, acceptances and bills outstanding, $7142, and interest on the same, $348.50; and there was an unpaid mortgage on the real estate of $2500, with interest accrued thereon of $88.50. If the invested capital was $22500, what was the net solvency or net insolvency of the firm at closing, and how much has been the net gain or net loss ? 5. Gray, Snyder and Dillon entered into partnership with equal investments, and agreed that, in case no withdrawals of capital were made, and no added investments made by either, they should share the gains or losses equally; but in case either party increased or diminished his investment, the gains or losses should be shared according to average investment. At the end of 6 months Gray withdrew $2000, and Snyder $3000, and Dillon invested $5000. Three months later Gray invested $1000, and Snyder and Dillon each withdrew $1500. At the end of the year they dissolved the partnership, having as total resources, $51000; total liabilities, $16500. ISio interest account having been kept, what was the present worth of each at closing, and what was the gain of each, the whole gain being $6900 ? 6. Phelps, Rogers, and Wilder enter into partnership for five years. Phelps invested $10000; Rogers, $20000; and Wilder, $30000. At the end of each year Phelps withdrew $1000; Rogers, $1600; and Wilder, $1800. Upon final settle- ment, the value of the partnership property was $57200. How much of this sum should each receive? 7. Apr. 1, 1884, Smith and Jones commenced business as partners, Smith investing $8000, and Jones $6000; six months later each increased his investment $1500; and on Jan. 1, 1885, Brown was admitted as a partner with an investment of $2400. On Oct. 1, 1885, each partner drew out $1500; on Apr. 1, 1886, Smith and Jones each drew out $1000, and Brown invested $6000. On Jan. 1, 1889, it was found that a net gain of $37500 has been realized. What was the share of each? If by agreement Smith, at final settlement, was to be allowed $1200 per year for keeping the books of the concern, what was the present worth of each ? 8. Burke, Brace, and Baldwin became partners, each investing $15000, and each to have one-third of the gains or sustain one-third of the losses. Burke withdrew $2100 during the time of the partnership, Brace $1800, and Baldwin $2000. At close of business their resources were: Cash, $3540; Mdse., $14785; notes, acceptances, and accounts receivable, exclusive of partner's accounts, $16250; real estate, $28500. They owed on their outstanding notes $8125, and on sundry personal accounts $1950. Find the present worth of each partner at closing. 9. Parsons and Briggs became partners Apr. 1, 1887, under an agreement that each should be allowed 6$ simple interest on all investments, and that, on final settlement, Briggs should be allowed 10$ of the net gains, before other division, for superintending the business, but that otherwise the gains and losses be divided in proportion to average investment. Apr. 1, 1887, Parsons invested $18000, and Briggs $4000; Jan. 1, 1888, Parsons withdrew $5000, and Briggs EXAMPLES IN PARTNERSHIP. 271 invested $3000; Aug. 1, 1888, Briggs withdrew #1500; Dec. 1, 1888, the partners agreed upon a dissolution of the partnership, having resources and liabilities as follows: Liabilities. Notes and acceptances ._ $6520. 00 O utstanding accounts 21246. 50 Rent due 1200.00 Resources. Cash on hand and in bank $ 1101.05 Accounts receivable 1 6405. 50 Bills receivable 2550. 00 Int. accumulated on same 287.41 Mdse. per inventory 9716.55 If, of the accounts receivable, only 80$ prove collectible, what has been the net gain or loss? What has been the gain or loss of each partner? What is the firm's net insolvency at dissolution? What is the net insolvency of each? 10. Bradley and Maben became partners July 1, 1885, under a 3-year's contract which provided that Bradley should have $1500 each year for superintending sales, and that Maben should have $1000 each year for keeping the books of the concern, and that these salaries should be adjusted at the end of each year and before other apportionment. of gains or losses was made. July 1, 1885 each invested $12500. Six months later each increased his investment $5000. July 1, 1886, Bradley drew out $3600, and Maben drew out $3000. Oct. 1, 1886, Bradley withdrew $1000 and Maben invested $2000. July 1, 1887, each drew out $1500. At the expiration of the time of the contract the resources exceeded all liabilities $47280. What was the gain of each, and the present worth of each ? 11. Clark, Wilkin and Ames bought a section of Kansas land for $6400, of which Clark paid $1600, Wilkin $2000, and Ames the remainder. Wheeler offered $4000 for one-fourth of the land; the offer was accepted, and each of the four had set apart a quarter-section for his exclusive use. How shall the money received from Wheeler be divided ? 12. A, B, and C, formed a copartnership for 2 years, investing equal sums, with the agreement that each shall receive interest at the rate of 6$ on all sums invested, be charged interest at the same rate on all sums withdrawn, and the gains or losses shown on final settlement be apportioned according to average net investment. Three months after the formation of the partnership A drew out $1200, and six months later B and C each drew out $1000, and A invested $6000; at the end of the first year each drew out $500. On closing the affairs of the firm, the following statement was made: net gain, $15000; present worth, $75000. What was the original investment of each? What was the present worth of each at the time of dissolution? What w T as each partner's share of the gain? IS. A and B became partners for one year; A investing -f of the capital, and B |; the agreement being that the gains or losses shall be apportioned accord- ing to average net investment, and that each partner be allowed 6$ interest per annum on all investments, and be charged interest at that rate on all sums withdrawn. At the end of the year the firm had as resources: Mdse., per inventory, $21460; real estate, $15000; cash, $1950; bills receivable, $13146.50; interest accrued on the same, $519.25; accounts due it, $11218.50; 272 EXAMPLES IN PARTNERSHIP. store furniture, $1320; delivery wagons and horses, $2100. The liabilities were: mortgage on real estate, $7000; interest on same accrued, $210; notes outstand- ing $26950; interest accrued on same, $811.75. The firm owes Barnes, Clay & Co., of Boston, $33560. It is found that 33 J per cent, of the accounts due the firm are uncollectible. If the firm's losses during the year have been $12000, how much was invested by each partner ? What is the present worth or net insolvency of the firm, and of each partner, at closing ? \ lJf.. Clay and Hard commenced business Nov. 1, 1883, with the following resources: Clay invested cash $1 0000 Store, valued at 12000 Marble fixtures, valued at 1500 Hard invested Mdse., valued at. _ $13500 Cash _.. _ 3000 Good will of trade, valued at__. 7500 The firm assumed an outstanding mortgage on the store of $6000, and a note made by Hard for $3000, and due without interest July 1, 1884. Jan. 1, 1884, each partner withdrew $300; May 1 , 1S86, Clay withdrew $2000, and Hard invested the same amount. Jan. 1, 1887, Dunn was admitted to the partnership, with a cash investment of $4500. Nov. 1, 1887, each partner invested $1000; and on Nov. 1, 1888, the partners agreed upon a dissolution, the following being shown from the ledger of the firm: Liabilities. Notes and acceptances $3825.00 Interest on notes 114.60 Balance of mortgage unpaid- . 2150.00 Taxes on store, due 75. 40 Due Hard for keeping the books 5000. 00 It was agreed that Hard should, at the time of dissolution, be allowed $1000 per year for keeping the books of the concern. If no interest account was kept and the gains or losses be apportioned according to average investment, what are the net resources of the firm at closing ? What has been the net gain or loss ? What has been the gain or loss of each parter ? What h the present worth of each at closing ? Mdse. , per inventory $48450. 50 Cash 10918.20 Accounts receivable 23416. 80 Real estate 15000.00 Movable fixtures and sundries, 3114. 50 APPENDIX. APPENDIX. STOCKS AND BONDS. 843. Stocks is a term applied to shares in the capital stock of banking, insurance, railroad and other incorporated or joint stock companies. 844. A Stock Certificate is a written or printed instrument of a Joint Stock Company or Corporation, signed by the officers of the company, certifying that the holder of the certificate is the owner of a certain number of shares of its capital stock. A share represents simply a certain component part of the capital stock, which is usually divided into shares of $25, $50, $100, $1000. The Stock Certificate represents the number ot shares specified therein. The Capital Stock of a company is the sum of all the shares issued, at their par value. The Par Value of stock is the sum for which stock is issued. The Market Value is the 3um for which stock can be sold. Stocks are at par when they sell for the value written on their face. Stocks are below par when worth less than their face value, and above par when worth more than their face value. 845. The Preferred Stock of a corporation is stock on which dividends are payable before those on the original shares or common stock. Preferred Stock is usually issued to take up certain floating indebtedness of a corporation, and agreed dividends are declared at certain intervals out of the net earnings, and before any dividend can be declared on the common stock. Such stock is usually issued upon the reor, ganization of railroads and consolidated joint stock companies. 846. A Bond is a written or printed obligation of government, Joint Stock Company or Corporation. It is conditioned to pay a certain sum of money at a specified time and at an agreed or fixed rate of interest, payable at regular intervals. Bonds of business corporations are usually secured by mortgage on their real estate. Municipa, Bonds are issued by a vote of the people or their representatives, and for the payment of which a Sinking Fund is accumulated by a yearly rate per cent, levied on all the real property within the limits of the municipality. 847. Government Bonds are bonds issued by the general government. Their names are usually derived from the interest they bear and the time when due; as 4-Twenties, 4i-'91's, TJ. S. Cur. 6V97, etc. 848. A Coupon Bond is one with certificates attached showing amount, date of interest and when due. When paid the coupons are detached and canceled as vouchers. 18 274 appendix. The interest coupons on government coupon bonds are payable to the bearer, and will be cashed by any bank or banker in the United States. Coupon bonds may be converted into registered bonds of the same issue. 849. A Registered bond is one which is payable to the owner as registered in the books of the corporation or government issuing it. Kegistered bonds can be transferred only by assignment and registry on the books of the corporation. The interest on registered bonds is paid by checks, which are made to the order of the regis tered owner and sent to him by mail. The checks for interest on registered government bonds, when properly endorsed, will be cashed by any bank or banker. 850. A Corporation is a fictitious person. It consists of several natural persons, who in the name of the corporation are authorized by law to transact business. The instrument which defines the rights and duties of a corporation is called a Charter. It is issued by government, under seal. 851. Stock Quotations are the public prices or rate per share that stock sells for. 852. A Dividend is a certain profit divided among stockholders. Dividends and assessments are declared at a certain per cent, on the capital stock. Divi dends are declared yearly, semi-annually and quarterly. 853. An Assessment is a sum levied upon the stockholders of a corporation to make up its losses, etc. 854. Premium is the per cent, profit of stock over 100$, or its par value. Discount is the per cent, loss of stock below 100$, or par value. 855. To Find the Dividend on Stocks, the Capital Stock and Rate Per Cent of Dividend being given. Example. — The Wilson Manufacturing Co., of Trenton, N". J., has a net profit of $17812.50 for the year, to be adjusted. Its capital stock is $250000, divided into 2500 shares of $100 each, and the directors have declared a dividend of 6ifc. How much will the dividend be, and how much will the sinking fund be after the dividend is transferred? also, how much of the dividend will be due 0, B. Henry, who owns 15 shares? Operation. Explanation. — Since the divi ~ .. , *o-aaaa dend is a certain per cent, of the Capital _ =$2o0000. capital stock, 5££ dividend repre- Rate of dividend = 5^. sen ts 5$% of $250000 or $13750. Net earnings =$17812.50. As the net earnings are $17812. 50, Dividend = .05* X $250000 = $13750. and the dividends are $13750, Sinking fund = $17812.50 - $13750 = $4062.50. * ber ? ™ H ™ for *° sinkl ^ Kim*.!, ft ^ fund, the difference between the C. B. Henry's dividend. two amounts; and $17812.50 — , T , ., K -, * unA $13750 = $4062.50. Since C. B. Value 15 shares = $1500. * , Henry has 15 shares, the parvalue h\i of $1500 = $82.50. of which ig j lgoo ^ his dividend will be h\% of $1500 = $82.50. APPEXDIX. 275 Rule.— I- To find the total dividend: Multiply the par value of the capital stock by the rate per cent, of dividend. II. To find the dividend due a stockholder: Multiply the par value of one share by the number of shares owned, and multiply this product by the rate per cent, of dividend. EXAMPLES FOR PRACTICE. 856. 1. The Citizens Gas Co., of Rochester, N. Y., declares a dividend of 7^/ on its capital stock of $300000. Find the total dividend and the amount due Martin Bishop, who owns 25 shares of the par value of $50 each. 2. A bank with a capital of $200000 declares a dividend of 4|#. Find the amount of dividend, and B's dividend who owns 12 shares, of the par value of $100 each. 3. A Boston railroad company having a capital stock of $3500000, divided into 35000 shares, declares a semi-annual dividend of 2f#. How much is the total dividend, and how much is due A, who holds a certificate for 16 shares? Jf. The Johnson Machine Co., of Cincinnati, 0., declared a quarterly dividend of lf$ on its capital stock of $150000. Find the total dividend due stockholders, and C's portion, who owns 50 shares of the par value of $100 each. 5. A manufacturing concern has a capital stock of $200000. Its net earnings for the year are $28512.75. Pursuant to its charter 3$ of the net earnings is set aside as a Sinking Fund to cover losses, 8$ is declared for dividends, and the remainder is transferred to an account called Undivided Profits, from which to declare subsequent dividends or to meet current expenses. Find the amount carried to dividend account, sinking fund account, and undivided profits account. 857. To Find the Rate Per Cent, of Dividend, the Capital Stock and Net Earnings being given. Example. — An insurance company having a capital stock of $500000, has net earnings of $45813.50. After setting aside 1-J$ for a sinking fund, how great an even per cent, dividend may be declared out of the remainder? Operation. Capital stock =s $500000. 00 Net earnings = 45813.50 Explanation. — Since the rate per cent, of divi- Sinking fund l£4ol*ib.d0 tQ be get agide frQm thg net earniDg8 for a sinking \% of capital = $5000 fund, which equals $687.20, then $45813.50— $687.20 $45126.30 -f- $5000 = 9 times, = $45126.30, the remainder from which to declare Plus $126.30 undivided profits. dividends. As \% of the capital stock = $5000, there Rate dividend = 9$ can ^ e as £ reat an even per cent, dividend declared ^ as $5000 is contained an even number of times in Q- l • #» i w _. a rqv on $45126.30; and $15126.30 -4- $5000 = 9 times, with a binkmg fund, 1£$ — $ 687.20 remainder of $126.30; therefore the rate of dividend Dividend, 9$ = 45000.00 i s g& and undivided profits = $126.30. Undivided profits = 126.30 Net earnings = $45813.50 276 APPENDIX. Rule.— From the net earnings deduct the amount to be set aside for the sinking fund, if any, and divide the remainder by 1% of the capital stock; the quotient will represent the rate per cent, of dividend. EXAMPLES FOR PRACTICE. 858. 1. If the capital is $1500000, and the net earnings are $52500, what rate per cent, of dividend may be declared? 2. A railroad company having a capital of $400000 has net earnings of $33500 to divide among its shareholders. What is the greatest even per cent, of dividend that may be declared? 3. A gas company having 2000 shares, at the par value of $50 each, has net earnings of $18500. What even rate per cent, of dividend may be declared? J/.. On Jan. 1st, 1894, a manufacturing company issued 2000 shares of its capital stock at the par value of $100 each. At the end of the year the net earn- ings were $16850. What even rate per cent, of dividend was declared, and how much remained for undivided profits? 5. The Excelsior Bicycle Co.'s loss and gain accounts show the following for the year: Gross earnings, $29518.75; expenses, $9618.75. Its capital stock being $200000, what even per cent, of dividend may be declared, and what will be the amount of undivided profits, if 1$ is first set aside as a sinking fund? 859. To Find the Par Value, the Premium or Discount being given. Example. — Sold Mutual Gas Co. stock for $18000, at a gain of 12£#. Find the par value of the stock. Operation. 100 $ = par value. Explanation.— The par value is 100$ and the gain, 12 1( £ = premium. 12£#, is the premium, therefore, 112£# = sale of $18000. - — - . ■ ' As 112£# = $18000, the unit of measure, or 1% = 112i$ = sale, $18000. , $i 8 ooo -*- 112£ = $160; and 100<£, or par value, = 100 X 1 $ = $160 unit of measure. $160 = $16000. 100 $ = $16000, par value. Rule.— Divide the sale by 100% plus the rate of premium, or minus the rate of discount ; the quotient will represent 1% of the par value, which multiplied by 100, will represent 100%, or the par value. When the net gain or the net loss, and the per cent, of gain or loss are given, divide the gain or loss by the rate and multiply the quotient by 100. When dividends or assessments and the amount of the sale are given, proceed as under the rule. EXAMPLES FOR PRACTICE. 860. Find the par value. 1. $16640 sale, 4$ premium. 2. $24000 sale, 4=/ discount. 3. $17510 sale, 3$ premium. 4. $82616 sale, 8$ loss. 5. $17940 sale, 8$ loss. 6. $750 assessment, 5$ rate. 7. $L450 dividend, 2$ rate. 8. $640 discount, 4$ rate. 9. $845 loss, 2ifc rate. 10. $648 gain, 3$ rate. APPENDIX. 277 861. To Find the Market Value, the Premium or Discount being given. Example. — Find the market value of 150 shares Vermont Central Railroad stock, par value $100 each, quoted at 93-^$. Operation. Par value = 150 x $100 = $15000. 100$ = par value. 93-j-$ = market value. ,93^ X $15000 = $14025 market value. Explanation. — The par value of 1 share is $100. The par value of 150 shares equals 150 X $100 = $15000. 100# = the par value, and the market value = 93^ of the par value, therefore .93£ X $15000 = $14025, market value. Rule. - Multiply the par value by 100% plus the rate per cent, of prem- ium or minus the rate per cent, of discount. The product will be the market value. 862. Find the market value of the following stock of the par value of $100 per share. 1. 50 shares quoted at 92f. 2. 75 shares quoted at 31|. S. 80 shares quoted at 75f. 4. 100 shares quoted at 43-^. 5. 200 shares quoted at 72. 6. 58 shares quoted at 62£. 7. 150 shares quoted at 85. 8. 300 shares quoted at 73. 9. 500 shares quoted at 154-J-. 10. 450 shares quoted at 112-J. 863. To Find the Rate Per Cent, of Investment, the Cost and Dividend being given. Example. — What per cent, profit is made on stock costing 80 and paying a dividend of 10$? 1* Hot Operation. = cost. = dividend. = unit of measure. -4- .8$ = 12| or 12£$. Explanation. — To find the rate of investment refers to the cost, therefore, 1% of cost = 1% of 80$, or .8%. As 1% of cost equals .8% profit on the investment, 10# profit 'will equal as many per cent, as .8% is contained times in 10$; and 10# -=- .8% = 12£, or 12i£. Proof: 121% of 80# = .12J X 80# = 10#. Rule.— Divide the rate of income by 1% of the cost or market value, quotient will represent the rate of investment. The EXAMPLES FOR PRACTICE. 864. Find the rate of investment. Kate of Income. i. \%\i. 2. 10$. 3. 33$. 4. 20$. 5. 40$. Cost. Rate of Tncome. Quotations. 75. 6. 16f$. 83£. 90. 7. 10$. 80. 110. 8. 6$. 95. 120. 9. 5$. 96. 130. 10. 15$. 120. 278 APPENDIX. MISCELLANEOUS EXAMPLES. 865. 1. What per cent profit is made on the investment if stock paying 6# dividend is bought at 120? 2. A manufacturing company having a capital of $100000 has net earnings of $6500. What rate per cent dividend may be declared? 3. A natural gas company declares a semi-annual dividend of 7£$ on its capital of $300000. What is the total dividend, and how much is due C, who owns 68 shares of the par value of $50 each? 4. How much money will be required to purchase 150 shares D. R. & B. rail- road stock, par value $100 each, quoted at 76^, brokerage -J#? 5. I paid $4440 for Kansas City & Pacific Railroad stock at 74. How many shares at $50 each did I buy? 6. H. M. Barton bought through a broker, at \ brokerage, the following shares of railroad stock of the par value of $100 each: 75 shares Mobile & Ohio, at 65g; 100 shares Milwaukee & St. Paul, at 123; 80 shares Louisville & Nash- ville, at 113i; 20 shares K Y. Central, at 123±. Find the total cost. 7. A broker bought for a customer, at -J# brokerage, 800 shares New Jersey Central Railroad stock, par value $100, at a total cost to his customer of $100100. Find the market quotation and brokerage. 8. A business man owns 18 shares of gas stock of the par value of $100 per share, upon which he was paid $99 as a dividend. If the capital stock was $150000, what was the total dividend paid to stockholders? 9. A street railway company declared a dividend of 2^ per cent. If the amount of the dividend was $16875, what was the capital stock? 10. The Silver Lake Ice Co. declares a semi-annual dividend of 2£# on its capital stock of $100000. Find the total dividend and the amount due Horace Brown who owns 75 shares of the par value of $100 each. 11. When U. S. 5's, 1907, are selling at 119^, how much must be invested in them to produce a quarterly income of $1250? 12. Brown bought bank stock paying a regular annual dividend of 8$ and realized 6$ on his investment. What did he pay for the stock? 13. I purchased through a broker $5000 Rock Island 5's at 10 H. What did the bonds cost me, the broker's charges being ■£■#? U. A father invested $5712.50 in U. S. 4's, 1907, for his son. If the bonds were purchased at 114J, what was the son's annual income from the bonds? 15. What is the cost of four $1000 U. S. 4's, 1907, reg. bonds at 114^, brok- erage -J$? 16. The net earnings of a railroad company is $1336375.48, and its capital stock, $25000000. If 6$ is set aside for a sinking fund, what even per cent, of dividend can be declared, and how much will remain for undivided profits? 17. Find the total cost of 1000 shares of American Sugar Ref., at 82; 500 shares of General Electric, at 37-J-; 75 shares Michigan Central, at 99; 10 shares Manhattan Elevated, 113f. Brokerage -J#. APPENDIX. 279 18. With the proceeds of Louisville & Nashville stock which I sold at 52£, I purchased $21000 in Rock Island 5's, at 100J. What was the par value of the stock sold, brokerage %%? 19. What is the total par value and the total market value of 50 shares Amer. Ex., at 114; 125 shares Adams Ex., at 146^; 200 shares Chicago Gas, at 73^; 150 shares N. Y. C. & St. L. 1st preferred, at 65? 20. Jones bought 100 shares of railroad stock paying a regular annual divi- dend of 4-|%, at 75. Smith bought the same number of shares of bank stock paying 8% dividend, at 125. Which made the better investment, and what per cent, better? 21. I bought Canadian Pacific stock at 6*,', which paid regular annual divi- dends of 4$. What rate per cent, of income will I receive on my investment? 22. Purchased 1000 shares of Amer. Sugar Refinery at 80-J ; 500 shares General Electric at 37-J; 125 shares of Manhattan Elevated at 113|. Sold the Sugar at 81 J-, the General Electric at 37-J, and the Manhattan at 113 J. What was the net gain or loss, brokerage \% each way, no interest? 28. Jones bought dies. & Ohio bonds at 25% below par and by so doing real- ized 8% on his investment. He also purchased Texas Pacific bonds bearing the same rate of interest at 15% below par and received on these an annual income of $1800. What did he pay for the Texas Pacific bonds? 24. I own 150 shares of Western Union Telegraph stock, for which I paid $13500. If I realize 0% annually on my investment, what is the rate per cent, of dividend? 25. A capitalist invested $5306.25 in " industrial " stock at 106, paying 9% annual dividends; $2678.13 in U. S. cur. 6's, '97, at 107; $5012.50 in express stock at 50, annual dividends 3%; and $7471.88 in railroad stock at 99 J, annual dividends 1% ; brokerage \%. How much was invested, and what was the total annual income? 280 APPENDIX. EXCHANGEo 866. Exchange is a system by which debts are paid in distant places without the transmission of money. This is done by the means of written orders called Bank Drafts, Bills of Exchange, Commercial Paper, Express Orders, Telegraphic Money Orders and Post Office Money Orders. 867. Bills of Exchange are classed as Domestic or inland, and Foreign. 868. Domestic Bills of Exchange are those payable at some place in the same country in which they are drawn. Domestic Bills are called Drafts whether drawn at sight or on time. 869. Foreign Bills of Exchange are those payable at some place in another country 870. A Place of Exchange is some great money center. The principal centers of exchange in the United States are New York, Boston, Philadelphia^ (Chicago, St. Louis, Baltimore, Cincinnati and San Francisco. Those of Europe are London Paris, Antwerp, Geneva, Amsterdam, Hamburg, Frankfort, Bremen, Berlin and Vienna- DOMESTIC EXCHANGE. 871. A Bank Draft or Domestic Bill of Exchange is an order written by one bank directing another bank to pay a specified sum of money to a third party or to his order. Following is a form of BANK DRAFT. ALLIANCE BANK. No. 76G. Kochester, N. Y., Novo 18, 1894- Pat to the Order of Williamson Publishing Co., $155.86. One Hundred Fifty-five _ ■£$% Dollars, To Ninth National Bank, ) 0. J D Barton, -\ New York. ) Cashier. Nearly all banks keep money deposited with some one bank, called a correspondent, at one or more commercial centers, against which they draw drafts to sell to their customers for remit tance to creditors; these drafts pass as cash in the section of country tributary to the commercial centers where the banks are located upon which the drafts are drawn, 872. Commercial Paper consists of Sight and Time Drafts drawn by one person or firm, called the drawer, directing a second person or firm, called the drawee, to pay to a third person or firm, called the payee, or to his order, a specified sum of money. Following is a common form of W. B. Keynolds & Co. APPENDIX. 281 COMMERCIAL PAPER. $418.50. Albany, N. Y., Dec. 1, 1894. . At one day's sight Pay to the order of M. F. Brownell & Son, Four Hundred Eighteen " and T %\ Dollars, and charge to the account of To Barker Bros., ) Newark, N. J. ) Bank drafts are drawn against cash balances on deposit, while commercial drafts are drawn to collect a debt due. 873. United States Post Office Money Orders are drawn by the post- master, or his clerk, at one office, directing the postmaster of another office to pay to the person named in his private letter of advice, the sum specified in the order. Applications for money orders must be in writing, and must state the amount of each order wanted, the name and address of the person to whom the order is to be paid, and the name and address of the remitter. Application blanks may be obtained at any money order office The maximum amount for which a single money order may be issued at an office designated as a "Money Order Office" is $100, and at an office designated as a "Limited Money Order Office," $5. When a larger sum is to be sent, additional orders must be obtained But post- masters are instructed to refuse to issue in one day to the same remitter, and in favor of the same payee, on any one post office of the fourth class, money orders amounting in the aggre- gate to more than $300, as such office might not have funds sufficient for immediate payment of any large amount. The payee who desires a money order to be paid to another person must fill in and sign the form of transfer which appears on the face of the order. More than one transfer is prohibited by law. If a money order is lost, a certificate should be obtained from both the paying and issuing postmasters that it has not been paid, and will not be paid, and the Department at Washington will issue another on application. If a money order is not collected within one year from date, it is invalid, and can be paid only by the Department at Washington on application through the issuing or paying postmaster. U. S. MONEY ORDER RATES. Domestic. For $2.50 or less 3 cents. Over$2.50to$ 5.00 5 cents. Over! 5.00 to $ 10.00 8 cents. Over $10.00 to $ 20.00 10 cents. Over $20.00 to $ 30.00. 12 cents. Over $30.00 to $ 40.00 15 cents. Over $40.00 to $ 50.00 18 cents. Over $50.00 to $ 60.00 20 cents. Over $60.00 to $ 75.00 25 cents. Over $75.00 to $100.00 30 cents. International. For $10.00 or less ..1 10 cents. Over $10.00 to $ 20.00_ _ 20 cents. Over $20.00 to $ 30.00 30 cents. Over $30.00 to $ 40.00 40 cents. Over $40.00 to $ 50.00 50 cents. Over $50.00 to $ 60.00 60 cents. Over $60.00 to $ 70.00 70 cents. Over $70.00 to $ 80.00 80 cents. Over $80.00 to $ 90.00_ 90 cents. Over $90 to $100 1 dollar. The above and following tables of rates are given to aid the student in working the examples relating to money orders in this subject. These rates, while current at the time of the writing of this chapter, are subject to change, and hence should not be regarded as authentic by the business man, until verified. 282 APPENDIX. 874, An Express Money Order is an order drawn by the agent of an express company at the office where the order is bought, directing another agent of the company, at some place designated, to pay to the person named therein a certain sum of money. Express orders are transferable by indorsement the same as checks. EXPRESS MONEY ORDER RATES. Payable in U. S., Canada or Europe. Not over 85. 00 5 cents. Over! 5.00 to $10.00 Scents. Over $10.00 to $20.00 10 cents. Over $20.00 to $30.00 -12 cents. Over $30.00 to $40.00 - - 15 cents. Over $ 40.00 to $ 50.00 18 cents. Over $ 50.00 to $ 60.00 -----20 cents. Over$ 60.00 to $ 75.00 25 cents. Over $ 75.00 to $100.00 30 cents. Over $100.00 at above rates. 875. Telegraphic Money Orders represent a system of exchange by which the remitter sends a message to the payee, directing him to call at a certain tele- graph office for the sum named therein. The telegraph agent at the sending office instructs the agent at the receiving office, to pay to the person named in the message, the sum specified, upon his personal application and proper identi- fication. Express companies also contract to transmit orders for funds by telegraph, but instead of requiring the payee to call for the funds in person, they deliver them to him at his residence or place of business. TELEGRAPHIC TRANSFER RATES. For not more than $50, 50^; $50 to $100, 1%; $100 to $200, $1.25; over $200 to $300, $1.50; over $300 to $400, $1.75; over $400 to $500, $2.00; over $500, special rates. The above rates are in addition to cost of telegraphic service, which is based upon distance and the number of words contained in the message. 876. To Find the Cost of Domestic Exchange. Example. — A hardware merchant of Toledo, Ohio, owes Morgan & Co., of Taunton, Mass., an account of $750. He buys of his banker a draft on the Commercial Bank of New York, at a cost of 10^ per hundred. How much did the draft cost ? Operation. Explanation. — The cost of the draft will be its Draft face = $750.00 face, $750, plus the cost of exchange, 750 or $750.75. Exchange — 71 X 10tf s= .75 Credit wil1 be given bv Mor S an & Co ' llowever > ^ ' for only $750, for should they deposit this draft with Cost of draft, = $750. 75 their banker, or transfer it to any one on account, the draft would be worth only its face. $750. Rule.— Add to the face of the draft, the charge for drawing same; the sum will represent the cost of the draft. Formerly domestic exchange was at a premium or discount in the city where purchased, according as the balance of trade between that city and the one on which the draft was drawn, APPENDIX. 283 was in favor of or against the former city. If the drawer city owed the drawee city, exchange on the latter would be at a premium in the former. If the balance of trade was in favor of the drawer city, the conditions of exchange would be reversed in the two places. For several years past, however, domestic exchange has been practically at par throughout this country. Bankers usually make a charge, called "exchange," for the trouble of keeping funds at commercial centers and drawing drafts against these funds, and also one, called "col- lection," for collecting drafts deposited with them payable at banks outside of the section in which the banks receiving them are located, or where they have no regular correspondents. Some bank drafts, as those drawn on New York, are current at par almost everywhere in this country. Some banks make no charge for domestic exchange to regular depositors. Instead of making remittances by exchange, merchants often send their personal checks in payment of bills. These, having to be returned for collection, command a small charge, as 150 or 200 on each check, and not a, per cent, on the amount. 877. To Find the Cost of Exchange by Express Money Orders, U. S. Money Orders and Telegraphic Money Orders. Example. — W. J. Boone & Co., of Bandon, Oregon, have bills to pay <*a follows: T. W. Brooks, Dayton, 0., $650; E. L. Greyson & Sons, Rosedale, Oregon, $48.53; Barnes & Snyder, Bolton, Mo., $48.50; and their traveling salesman, W. H. Post, is wanting $100 for expenses at Denver, Colorado. They pay the amounts by remitting as follows: T. W. Brooks and E. L. Greyson & Sons, express money orders; Barnes & Snyder, P. O. money order, and W. H. Post by telegraphic transfer in a 10-word message, the message costing 50^. Find the total cost of the remittances. Operation. Express orders: Dayton remittance, $650.00 Exchange, 1.98 = $651.98 Rosedale remittance, $ 46.53 Exchange, 18 = $ 46.71 U. S. money order: Bolton remittance, $ 48.50 Fee, .18 = $ 48.68 Telegraphic transfer: Agent's remittance, $100.00 commission, l discount. What were the proceeds? ~ ' Explanation. — As the draft is at sight, no time is con- sidered, and as the B. of L. .valued at $3000 is attached to Face of draft = $1400 the draft, that, together with the indorsement of Barnum Pis. = ^ the total dis " count on $1 00 is $.0205, and the Total dis. on $1.00 = $ .0205 proceeds are $1.00 -$.0205 =$.9795. Proceeds of $1.00 = $1.00 — $ .0205 == $ .9795 Since each $1.00 of the draft cost $800 draft costs 800 X $.9795 = $783.60 $.9795, the face of the draft will cost 800 X $ .9795 = $783.60. Rule.— Find the cost of $1.00 by deducting the bank discount for the time and the commercial discount at the given rate. Multiply the cost of $1.00 by the face of the draft, the product will represent the cost of the draft. EXAMPLES FOR PRACTICE. 880. 1. A grocer at Kingsboro, Pa., had a debt of $38.17, due in New York, which he paid' by U. S. money order. Find the cost of the order. 2. The proceeds of a 15-day draft, which was sold at \ discount, are $1943.44. What was the face of the draft, interest 6$ ? 3. Find the cost of a bank draft for $2580 on Third National Bank of New York, at the rate of 10 cents exchange for each $100. If. What is the face of a time draft which can be bought for $313.14, if pay- able 24 days after sight, discount §#, interest 5#? APPENDIX. 285 5. Gates & Son, of Memphis, drew a sight draft on Perrin & Boon, of Port- land, Me., for $8750.85, which they sold at the Cotton Exchange Bank at f$ discount. How much were the proceeds? 6. I drew a 60-day draft on one of my customers and sold it to a broker at f $ discount, receiving $1354.18 as proceeds. What was the face of the draft, inter- est 6^? 7. Jno. W. Williams, of Middlesex, Mass., remitted Janis Bros. & Co., of Milwaukee, $1750 by draft on New York, exchange 15 cents per each $100; Martin & Co., of Allentown, Pa., by American Express money order, $89.75; and Theodore Emens, $28.50, by P. 0. money order. Find the total cost of exchange. 8. An Atlanta broker bought, at £ % discount, a 90-day (after date) draft on a Nashville merchant, 4 days after the draft was drawn, and gave his check for $660.17 in payment. What was the face of the draft, interest 7$? 9. L. C. Thompson, of St. Louis, bought of Glenn & Garson a 60-day draft for $3800 on Leroy & Co., of Boston, at f # discount and less 6$ interest for the. time to run. He sent the draft to a Boston broker who sold it at -J$ discount and less 6$ interest for 54 days. If the broker's commission was»J#, what amount did he remit Thompson and what did Thompson make by the transaction? > 10. A wholesale grocer owed for an invoice of $5425.40, purchased in New York, subject to a discount of 6$ if paid within 10 days. Within the required time he discounted the bill and remitted for balance as follows: A sight draft which he bought of E. M. Brooks on Gunn & Baker for $4000, at \ discount, and a bank draft for remainder, the exchange being 10 cents for each $100. How much was required to settle the bill, and how much was gained by discounting it? FOREIGN EXCHANGE. 881. Foreign Exchange is the name given to drafts or bills of exchange drawn in one Country and payable in another. Foreign bills of exchange are usually drawn in the moneys of account of the countries in which they are payable. Thus, drafts on England, Ireland, and Scotland are drawn in pounds, shillings, and pence; on France, Switzerland, and Belgium, in francs; on Germany, in marks; on Holland, in guilders. Foreign bills of exchange are usually drawn in duplicate or triplicate of the same tenor and date, one of which being paid the others are void. Formerly it was the practice to send the different bills of a set of foreign exchange by different routes or vessels, to guard against loss or accident. The present practice, however, especially between Europe and America, where the mail service is both rapid and sure, is to send only the original bill and retain the others of the set. Foreign bills of exchange are sometimes used as a means of collecting debts due in foreign countries. The method employed is similar to that used in collecting debts by means of domestic bills. That is, the drawer leaves the bill with his local banker, who forwards it to a correspondent in the place where the drawee resides, by whom it is collected, the drawee paying the equivalent of the face of the bill in local current funds. The correspondent then remits the proceeds of the bill in funds current in the place where the draft was drawn, and when received by the local banker he pays or credits the drawer the face of the bill, less the charges for collecting the same. 286 APPENDIX. SET OF EXCHANGE. 1 2 Exchange for £100. New York, Nov. 2, 1894. Thirty days after sight of this First of Exchange (Second and Third of the same tenor and date unpaid), Pay to the order of Messrs. E. P. Reed & Co One Hundred Pounds Sterling — Value received, and charge the same to account of To Union Bank, London. No. 1305. Chrystie & Jaxney. Exchange for £100. New York, Nov. 2, 1894. Thirty days after Sight of this Second of Exchange... ll, I . (First and Third of the same tenor and date unpaid), Pay to the order of Messrs. E. P. Reed & Co _ One Hundred Pounds Sterling Value received, and charg-e the same to account of To Union Bank, London. No. 1305. Chrystie & Janney\ 3 Exchange for £100. New York, Nov, 2, 1894. Thirty days after sight of this Third of Exchange il | (First and Second of the same tenor and date unpaid), lllll Pay to the order of Messrs. E. P. Reed & Co.. One Hundred Pounds Sterling Value received, and charge the same to account of To Union Bank, London. No. 1305. Chrystie & Janney. 882. The Par of Exchange is the established value or equivalent of th< standard unit of money of one country, expressed in the standard unit of nionej of another country. It is of two kinds, intrinsic and commercial. The intrinsic par value refers to bullion value. Thus, the pound sterling of Great Britaii contains 113 grains of pure gold, and the dollar of the United States contains 23.22 grains pun gold. Since 113 grains are 4.8665 times 23.22 grains, the dollar is worth 100 cents, and th( pound is worth 486.65 cents. APPENDIX. 287 The commercial par value refers to the value of the coin or currency of one country com- pared with that of another country, as determined by its market value, or by the requirements of trade or commerce. The following quotations of the foreign moneys of account are used as the basis for com- parisons in the United States: Great Britain, £1 = $4.8665; France, 1 Franc = $.193 ; German Empire, 1 Mark = $ .238 ; Spain, 1 Peseta = $ .193 ; Italy, 1 Lira = $ .193 ; Mexico, 1 Dollar = $ .79 ; Brazil. 1 Milreis = $.546; Cuba, 1 Peso = $.932 ; Hayti, 1 Gourde = $ .965. Computations in the following examples will depend upon the quotations given in each. 883. The Commercial Kate of Exchange is the market value in one country of the drafts on another. Quotations of foreign exchange are given by means of equivalents, no reference being made to the par value. The commercial par of exchange cannot be greater or less than the intrinsic par to a point beyond the transportation charges and insurance of shipping coin or bullion from one country to the other. When the cost of exchange is greater than the intrinsic value of the coin repre- sented by its face, to a point beyond the cost of transmitting such coin to that country, gold can be exported at a profit ; when less than its intrinsic value, beyond the cost of shipment and insurance, gold can be imported at a profit. American exchange on Great Britain (Sterling exchange) is quoted by giving the exchange value of £1 in dollars and cents ; on France, Belgium, and Switzerland by giving the exchange value of $1 in francs and centimes ; on Holland by giving the exchange value of one guilder in cents ; on Germany by giving the exchange value of four reichmarks in cents. 884. Documentary Exchange is a bill drawn by a shipper upon his con- signee against merchandise shipped, accompanied by the letter of hypothecation, the bill of lading endorsed to order of payee, and the insurance certificates cov- ering the property against which the bill is drawn. 885. Cable Transfer. — Within a few years the practice has arisen of trans- ferring money to foreign countries by telegraph, or, as it is termed, "cable transfer." By cable transfer a merchant who desires to ship wheat to London can complete the trans- action in a few hours. He can ship the wheat, telegraph the fact to the consignee at London, obtain particulars concerning the conditions of the market, and, if he thinks best, have the wheat sold at once, "to arrive," and to remit the proceeds through a London banker. A bill does not appear at all in the transaction. The amount of business done in this manner has materially reduced the volume of bills in some places. In the eastern trade with London, in which competition is exceedingly keen and the margin of profit consequently small, the tele- graphic transfer system has been in use for several years. The amount of cable transfer between this country and European countries is constantly increasing. 886. A Letter of Credit is a circular letter issued by a banking house to a person who desires to travel abroad. The letter is usually addressed to the foreign correspondents of the bank issuing it, requesting them to furnish the traveler such funds as he may require up to the aggregate amount specified in the letter. When the traveler desires funds he goes to the correspondent in the city which he is then visiting and draws a draft for the amount on the correspondent mentioned in the body of the letter of credit. The draft is signed in the presence of the local correspondent, who carefully compares the signature with the one on the letter, and if found to agree, the draft is cashed, and the amount inscribed on the back of the letter. The last draft drawn is attached to the letter itself. 288 APPENDIX. The difference between a bill of exchange and a letter of credit is that the former is payable at a certain designated place, at a specified time and in one amount, while the latter is payable at several places, at different times and in variable amounts. 887. Travelers Cheques are a substitute for letters of credit and bills of exchange. They are similar in form to bank bills. They are issued for fixed printed amounts, with the equivalent of each denomination in the money of the principal European countries, and are payable to order, after being signed and countersigned by the purchaser or holder. Travelers Cheques are cashed, without discount or commission, by an extended list of banks and bankers, and are received in settlement of hotel bills by the principal hotels in Europe. Following is the form of Travelers Check issued by the American Express Company: TRAVELERS CHEQUE. THE AMERICAN EXPRESS COMPANY. Estahi,»hed 1841. CAPITAL, $19,000,000. Principal Office, 65 Broadway, N. Y. London Bankers : National Provincial Bank of England. Paris Bankers • Credit Lyonnais. No Cheque issued until its full value has been paid in. Payment is assured however long checks may remain out- standing. Cheques are issued for $10. $20, $50 and $11)0 each or foreign equivalents. This Cheque will be cashed by any of the 8000 agencies of the Company in United States or Canada, and in Europe at the Company's agencies, 35 Milk St., near Cheapside, London. E. C. : 13 Water St., Liverpool •. £ Rue Scribe, Paris ; 117 Langenstrasse, Bremen. By Credit Lyonnais, Paris, Havre, Ly- ons, Nice, Brussels, Madrid, Geneva, Constantinople, Smyrna. Alexandria, Cairo and St. Petersburg . Munster & Leinster Bank, Cork and Dublin ; North- ern Banking Co. and Ulster Bank, Bel- fast . National Bank of Scotland, Royal Bank of Scotland, and British Linen Co. Bank of Edinburg and Glasgow ; Dres- dener Bank, Berlin and Dresden : Bank fur Handel & Industrie, Frankfurt, A. M.; Bayerische Vereins Bank, Munich ; Anglo-Austrian Bank, Vienna ; Maguay, Hooker & Co., Rome, Florence, Leghorn and Pisa : Hope & Co., Amsterdam ; Land man dsbanken. Copenhagen: T. J. Heftye & Sons, Christiania: Kredit Ar- tiebolaget, Stockholm, and other Banks, Bankers. Hotels, etc , as per list. In Austria, Russia, Spain, Portugal, ete., Cheques are paid at the current rate of ex- change. the 39368 When Countersigned Below with American Express Company this Signature travelers cheque. ^^ 12. 164f f | years. Art. 146. Art. 173. Jflfti f andW. 13. $2906070732^ 1. 11, 7. . 14. 2. ¥• 6 - W- 7. 8375 5000 7560 90TT0"» ¥0~lT0"» ¥0~0~0~> and $14378213- 08A- 2. 8. 12. 8. 50. 16. 9. 151. 2. 3. ¥• A MP* 2250 ¥0~IT0~- ?£ 2571ft men. * 18. 10. 63. 4. 5. mmw, IK 416136280 5. 52. 11. 70. 5. Htp.mm #• 4&V& miles. 6. 45. 12. 25. w 77. $115009119^ and Art. 174. 9. %T000 » STFlToX $544888262^1- Page 41. 1. f. e. h*. WW. and 18. 34 T 4 T . Art. 148. 2. ¥• 7. w- WftrW, '*'• Oj l g 24 l y» 1. 4. 7. 7. 8. ¥• A HP* *Hm.^Wft¥ ft?. 86781fffff. 2. 24. 8. 23. 4- io«. Q. JL0801 HttlHfP. and M. «BWttfa. 3. 2. 9. 21. 5. W- *>• 1 W 1 - 735000 ¥3TTnro~- O22028698 * 5. 17. 10. 131. 51. 11. 1. Page 47. 2a 54000 59400 T?"2T0~> TT2TT7' 4 7025 85145 T¥2T T¥¥50"' 6. 4. 22. 25. Art. 176. HHH. WW*. Page 37. 1. 4*. 6. 2ff. and MIHP. Art. 131* Page 43. 2. 6f. 7. 22f|. Page 49. 2. 46. Art. 155* 8. 8f. 5. 1&. Art. 184. *. 82f. 1. 480. 4* 8A. A 88Ht. 2. tfo"> xf^' i?2ir» 5. 69f. 2. 450. 5. 8*. 10. um and Jfc 4. 48*. 3. 1872. Art. 178. £. 5 6 2 8 8 315 5 TO"?* 50"?» 6. 47 miles. 4. 840. 1. 2. 8. 4- 6. f 6. ^ f. 7. ff. A- A ft. if. *>• f. Iff, and ¥ Vj- tf. $13764ft. 6. 840. A &k-m f .w& 7. 760. Art. 133. 2. 64. 6. 7. 8. 9504. 7920. 840. 4- ^ftr.«f*.and iff!- *. 29. 9. 2520. Htt.fiW.itH A 885. 4. 296. Page 44. Page 48. 6. and #&. io"> i$> &> ¥0"» t$» J. 19. Art. 158. Art. 180. W.«ldW 6. 742. 2. 64f. 2. «. A If- 6. I*. **, w. w, 7. 8751. 2. &v . 2. H. 7- m. 165 810 an A 5o"> "9"o""» ana & 8906. 3. m> 8. H A dfr. 90 9. 71237. 4- 120. 4- f¥* A TF2~* 7. 42 30 2592 1422 t;stt> "T8"o""> "T¥ij"> 10. 17959. 5. 97£. 6. A. A #. V/^, and iff. 6. 90. 8. HP. W. HP, Page 39. 7. T3TT4?» Art. 183. W. W. and Art. 139. 8. 35&. 2. 108 0"» TffSO"? To~50"» w- 1. 8, 3, and 3. 9. 1080. and^V eg8 9. 1260 1680 1890 f 5¥tf» 2520, F5^ff» 2. 3, 3, and 13. 10. 86f. $. S467TJ* T¥5TT7» 2016 2100 2520) '2T2"0"» 3. 3, 5, and 11. 11. 45 bu. 19 6 5 6 10580 ¥F2~0~> ¥5^0"» 4. 3, and 31. 12. 5bbl. Hfffi and fM§>and fff!, 6. 2, 2, 2, 3, 3, 3, 13. 38 bales and 528 15120 UT5T0"- 10. t¥o-» t¥j» iff' and 11. yards. 8. 3T0~8~tf, t^0"Flff» m> ¥8-V-, tVV. €. 2, 3, 5, 5, and 7. U 720 yd. 5520 44160 82 08T7' 8208ff» *f». ¥¥. and 7. 2, 2, 2, 2, 3, and 3. 15. m «m. wwftr. t/u- 8. 5, 5, 5, 5, 5, and 5. 16. 5£ pieces. and mil 11. itio 1 ' rltfj tHo 1 * 9. 2, 2, 2, 2, 2, 2, 3, 17. 4*. 4. 84 8 160 4 22 ^~> 8S~> ¥2> 8¥> 27 72 1980 "T¥¥0""» T¥5¥» 3, 3, and 13. 18. 14. if, A. and ff. 1710 780 80 10. 2,3,5,7,13,17, 19. 4m bbL 6. if, -w-, «. w. 9 8 5 5 10 6 9 2 and 19. 20. 2ff sections. W. and W- 294 ANSWERS. 10 8 5 5 1620 3060 14 ' VVTi> ~FtfTT> SWi Art. 186. 3. 6. 8tf. A- 7. 8|. 4. e 53. Art. 300 . 4 9. 9|. 2f 10. 12*. 5. 9. 11. 9f 2f I* *A 0..3ft. 24. 8tf. 7. 7ft. 25. 99ft 8. 8ft. IS. 170f Art. 301. I. 2H- A 4. 5. 6. 7. 25. 88*. **• 1*. • 25. 1A. 26. 2*. 17. 18 da. 25. . 13. 19. 12* da. £0. 1223AV bu. 21. 834f cords. 00. 1841* bu 05. 279* miles. Page 61. . Art. 233. 2. 2ff. A 7***|. A 7**. 5. $14000. 6. $6772*f. 7. 3 bu. 8. 21b. 0. $135000. 20. 323AV. sulphur, 215 T 4 A salt., and 1615f*char. 11. $52500. 12. $80 watch, and $35 chain. Page 62. 25. $5f. 24. 15 days. 25. $19000. 16. $67837*. 27. 51 + 0. 18. $2. 25. $248.70|. #0. 3rVA mae8 ' 02. $5. £0. $13200. 28. 336 trees. 04. 1001b. 25. $67*. 06. $6, $15 and $16. 27. 62* years. 28. S $1190, H $476 and R $544 29. 90A years. Page 63. SO. $35 and $40. 31. 105. 3$. 6f bu. 55. 22* da. 34. J$47*, C$38£ 35. 40 ft. 36. 94* ft. 57. Cow $30, colt 38. 2 p. m. 3d. 36 ft. 40. 405.. 42. H $216, C $324. 42. 30f years. 43. 6* da. 4*. 5A 5 yda. 45. 42f da. 46. 67* da. 47. S $180, B $1501 48. 150 ft. Page 64. 49. H- 50. 176 rd. 52. 14f min. 50. Ben 70, John 1£ 55. 28 bu. and80bu. 54. 5A m i Q « pas* 1 o'clock. 55. 27A min. of 7 o'clock. 56. 10f£ min. of 10 o'clock. 57. 5A niin. of 11 o'clock. 58. 60. 59. 520 loss. 60. $1500, $3000, $4500, $6000. 61. 62A 3 T yards. 60. A $19.75, and B $15.80. 63. 15 hr. 64. 4 min. 65. 55 yr. 66. A $HAfr, B $14AV and $10*. Page 65. 67. B 42*, S 25|; B $16*, S $28*. 65. Hfffda; C $32f|f , H $25fff , T $22|*f , L $19AV- 69. 16f da.; A, $44 f£f;andB,$30 70. $7350. A, $2650} B, $2700; O, $2000. 71. H,216;M,129f and B, 64$ 72. 81 dr 73. 49**. 296 ANSWERS. 74- 43*. 2ft .000900. & AWfc * m"Mt 75. H,$33*;M,$55f, 14. .00000009. 10. *w&. A f*f. and B, $111*. 15. 54054054.005405- u. &y&. e. m*. 76. A, 135f da.; B, 169* da.; C, 0054. 16. 103.587. &> urn- Art. 251. 188* da.; D, 674 da. 17. 640.64. 1A 26.04002. / / 4 9 2 6 ■"*• 5 00 0000' 1. 129.341. ft, 848.1816. 77. A and B, 75* da. ; 19. 9019.029039. ■». flfflfflffi*- A 1652.461772. AWC, 78ff da.; A and D, 20. 7.7. 17. 5/ r . 4- 12638.517762. 21. 870.01. 1A 13^V A 2602.55141194. 45* da. ftft 479027004.00000- 19. 117f£*. A 8688.0148502. 7A A, B,\ and C, 5^da,;A,C, 99004. m sooovyy^. 7. 24018.46093544. 23. 70000000000000- ftl. 1600***. Page 74. and D, 36*f .000000000007. #a 1000000 24. 1100.0011. 25. .000003001, 1 A 13444.61870921. da.; B, C, and D, 88ff da. 10000060 Of* 23. 1234500 S469 9. 1004219.317454. 10. 57597.358230005, 79. 26^ da. 80. A, 77^%; 26. .001003. 27. .0100011. SOOOOOO' ft£ 6540000 11. 7003122.0011890- 9997. B, 58flMft; c, 47«m; 28. .00000605. 29. 1890.0000000189- 0. T0000000* 25. 188900 12. 1627.325 A. 1A 1017.84375 A. D, HS^Vft. iiilfiooo* 14. 1798.9425 bu. Page 69. Art. 340. Art. 246. 15. 395.8125 yd. Art. 238. 1. .25. 1. .0625. 16. 376. 1. .026. ft, .27. *. .0106. 2. .65. 17. 926 s . A .00256. A .275. Art. 253. 5. .0016, 4. 9.01. 5. 1.476. 4. .09375. A .1375. 1. .811357+. 4. .04. J. .00022. 6. .193024. 7. .00504. A .52. 7. .0525. 2. 2.23985+. A 1.7912 A 5.7. 7. 83.0504. A 710.00243. A 2146.9003. P. 56973.805. A .46875 9. .024. 4. 1.907703945. 5. 1384.4959234662- 2. 10. .1934675. 10. .0024. 9. 500. 05. 10. 45.046. 11. 1001.0100. 11. 33254.81. 12. .00001876. X*. .9375. 12. .015625. 6. 5569.3518126587* 42. 12. 1890.090. 13. 850.05. U. 1000.10. 13. 10.007. 14. .00097. 15. 15.0015. 15. .035700097. 1A .0015625 14. .8875. 15. .00024. 16. .96875. Page 75. Art. 255. 1. .412. Art. 239. 17. 219760.0801. 17. .984375. 2. .52977. 1. 11.107. 18. .046700004. 1A .028. A .6863. ft, 15.0014. 19. .00068001. 19. .308. 4. 5.5264. e 1 ci c/» i«r» A .000726. 20. 1101.10011 m .95. A l. 545648. 4. 1106.0012. £1. .94. A 54.2294. 5. 1600.16. A 10000000.000010. Page 72. Art. 244. 22. .226. *A .034^75. 7. 754.6005. ■ A 10000.0999. 9. .3148. 7. 300.65. 24, .76. £A .015625. 10. 213.889625. Page 70. *- dfo. 11. .810. A 25400.11. * 4fc Page 73. 1A 135.25746. 5. 21.0015015. & §*. Art. 249. to. .000001801a ft fflfe i. f. Page 76. 11. .500. ■• *00O» ft. f Art. 257. ?2. .00005. 1 A *. A ff. 1. .546. ANSWERS. 29? t. .01968. 5. 1.26875. 4. 39.9024. 6. 23469.986904. 6. 4625520.705. 7. 1. 8. 9. 9. .625000. 10. .87600. 11. 7231.98325125. 12. 49. tf. .1. U. 275400116.25610- 02754. 15. $20217.72. '16. $536.88. 17. $937.04. 18. $336.33. Page 77. Art. 260. 1. .25. 2. 305. 3. 250. 4. .5. £. .05. 6. 50. 7. 500. 8. 4000. P. 2000. f0. .002. 11. .000025. 22. 183100. 13. 5.5875. 14. 5252000. 25. 50. lb. .00007. 27. .001. 18. 4000. 29. 1000000000. . 20. .000001. 22. 25. 22. 4000000000. S3. .15. 24. 60000000. 25. 1. 10. 100. 100.' 1000. .1. 1. 10. 100. .01. 26. .1. .01. .0001. .00001. .001. 10000000. 100000. .00000001. .0001. 100000000. 27. .02. 200. .02. .0002. .00000002. 200000. 20000000. .0000002. .000000000002. 10000000000. Page 78. 28. 6400646464647.0- 406464000064. 29. 250252502527.75- 0500000025. 80. 40000400000.044- 800440044. 31. 400044440440.00- 0000440004. 32. 30030330000.003- 60306. S3. 150001650151.80- 1650000015. 15. 16. 17. 18. 19. 12711.8750. 10.625. 1011.6. 6.875 da. $274.58. $24.9331+. Page 80. Art. 266. 1. 1991.1198244. 2. .61032010. 8. .1625. 4. 601 Twcro"- 5. .0038462. 6. 2116.99454. 7. 187.2996. 8. $7,766. 9. 41.299781. 10. 7.029956572371- 000. 11. 999999999.99999- 9998. 12. .90. 13. 572501.2525. u. $1100.869. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12 Page 84. Art. 287. 1. 6000. 2. 11100^. 3. $2.41. A $10.44 6. $214.68. 6. $18. 7. $510. 8. $98.76. 9. 100980. 10. 3750. 11. 26530. 12. 157320. Art. 289. 1. $88.84. Page 85. $144.89. $4337.77. $221.53. $378.07. $377.29. 384. 5 i. 1088 60 . 4073 76 . 3065«b. 896 86 . 2094.15. Art. 291. $8179.88. Page 86. $4859.76. $29.52. $839.26. $33.73. Art. 293. $2284.35. $1315.63. $7.13. 950 bu. $344.73 gain. $91.10. Page 88. Art. 294. 7. 26 and 39. 8. 16^ min. aftei 3 o'clock. 9. Midnight. 10. B. 12, A. 20, C. 78. U. B. 840 and O. 450. 12. 1900. Page 93. Art. 304. 2. $143. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. $65,875. $113.75. $291.83£. $133.31£. $247.60. $63.05. $1.21£. $3.70. $67.71£. $10.50. $841.15. $1156.97. Page 94* Art. 306. 2. $75. 2. $358.75. 3. $22.13. 4. $53.50. 5. $1125. 6. $612. 7. $3281.25. 8. $382. 9. $24.06. 10. $45.56. 11. $34. 12. $1567.501 ' 13. $187.50. 14. $281.25. 15. $125.25 16. $370. 17. $2750. 18. $46.25 19. $50.25 20. $86.25. 21. $156.25. ANSWERS. Art. 307. 2. $2507.96. Page 95. 2. $332446. $4321.26. $5282.20. $8096.48. $3364.72. 1. 2. 8. 4. 5. 6. 7. 8. 9. 10. 11. 3. 4- 5. 6. 7. 8. 9. 10. 1. 2. 8. 4- 5. 6. 7. 8. 9. 10. 11. 12. Page 96. Art. 309. 23 lb. 4627.5 yd. 61b. 763.2 yd. 876.5 doz. 371 lb. 81.5 yd. 115.2 acres. 689 yd. 123 lb. $40.96. Page 97, Art. 311. 1. $21.91. $70.07. $28.35. $148.28. $31.68. $25.31. $89.10. $63.63. $35.65. Art. 313. $24.75. $14.66. $158.76. $123.18. $53. $38.75. $19.80. $100.32. $140.81. $34.78. $68296.35. $2819.31. Page 98. Art. 315. 1. $4.02. $4.71. $78.70. $13.84. $2014.46. $2016.46. $941.63. $21.06. $2.77. $19.51. $296.24. $15.77. $56.72. $66162.39. Page 99. Art. 317, $28.56. $28.08. $52.83. $39.29. $24.66. $27.15. $55.67. $58.33. $174.78. $132.36. $121.68. $242.79. $333.31. $109.57. $198.91. $28.16. $11.57. $12.06. $60.89. $30.14. $15.62. $1941.92. Page 101. Art. 326. 1. $9.03. Page 102. 2. $49954.08. 3. $191.47. Page 103. 4. $1812.31. Art. 327. 1. $46731.58. Page 104. 2. $967.31. 5. 6. 7. 8. 9. 10. 11. 12. 13. 1. 2. 8. 4- 5. 6. 7. 8. 9. 10. 11. 12. 13. 15. 16. 17. 18. 19. 8. $66.21. Page 105. Art. 328. $943.54. $57269.94. Page 106. Art. 329. $115.68. $560.50. $1528.75. $190.33. $272.35. ).14. 7. 8. 9. 10. Page 107, $524.03. $1718.01. $168.68. $322.29. Page 112. Art. 355. 1. 450 min. 51 sec. 2. 23 hr. 5 min. 29 sec. 3. 7 da. 1 hr. 30 min. 51 sec. 4. 24 yr. lmo. 5 da. 6 min. 5. 6332£ hr. 6. Jan. 14, 1889. 7. 3405 da. 8. 1169 da. 9. 11 mo. 9 da. 10. 1 yr. 10 mo. 19 da. 19* hr. Page 113. 11. 1440 min. 12. 939613.7 sec. Art. 358. 1. 35° 54'. 2. 24° 16' 46". 3. 3S. 1°49'41\ 4. 1296000. 5. 4907'. 6. 205737. 7. 136° 2'. 8. 21600'. 9. 8° 39'. Page 115. Art. 363. 1. 2 hr. 30 min. 24 sec. 2. 41 min. 40 sec. 3. 57 min. 44 sec. after 3 a. m. 4. 1 hr. 52 min. 8 sec. 5. 16 min. past 8 p. m. Art. 365. 1. 47° 59'. 2. 35° 13' E. 3. 74° 58'. Page 11G. 4. 36° 52' 20" N. 5. t 77° 1'. Art. 366. 1. 180°. 2. 180°. 3. 4 5 16 p. m. 4. 6 33 40 a. m. 6 min. of 1 p. m. 6 a. m. 10 9 20 p. m. 1 21 21£ p. m„ 3 min. 48 sec. past 7 a. m. Page 117. Art. 371. 1. £256 4 s. 2. 248 s. 3 d. 3. £54 6 s. 10 d. % £195 4s.4d.) far. Art. 373. 1. . 6480 d. 2. 956 far. 3. 38853 d. 4. 39450 far. 5. 13206 far. Page 118. Art. 375. 1. $350.26. 2. $525.80. 3. $63544.40. ANSWERS. 299 4. $15.03. 28. 73 lb. 2 oz. 8 pwt. 5. $254.49. 10. 484 mi. 53 rd. 1 5. $54.29. 19 gr. 29. $154.22. 6. $5.04. yd. 2 ft. 6 in. Art. 377. 30. $360.67 gain. Art. 400. Page 134. 1. £38 3 d. 2.4 far. 1. 928 pt. Art. 434. 2. £63 7 s. 10 d. Page 127. 2. 599 pt. 1. 35676648 sq. in. 3. £513 14 s. 3 d. Art. 395. 3. 186 bu. 3qt.2pt. 2. 134498442f sq.ft. 3 far. 1. 34669 lb. £ 7qt. lfpt. 3. 112 A. 40 sq. rd. 4. £751 14 s. 3 d. 5. £22621 2 s. 8 d. 2. 15 T. 12 cwt. 75 lb. Page 130. 261 sq. ft. 51.84 sq. in. lfar. 3. 12 cwt. 50 lb. 4. 56 lb. 4 oz. 5. 5 bu. 1 pk. 1 qt. 1 pt. Page 135. Page 124. 5. 7 cwt. 68 lb. 6 .4 oz. 0. 83 bu. 1 pk. 3 qt. lfpt. 4. 110 sq. rd. "■ '44. J. $71.75. 11. 90 ft. 12. 230 ft. 12. $72.81. 6. 7 gal. 2 qt. 1 pt. 13. 25|A. gr. 13. 2 T. 5 cwt. 84 lb. % gi- 14. $107156.25. 10. 12 pwt. 12 gr. 11. T y F ib. 12. f§9 lb. Art. 397. 7. 13 gal. 1 pt. 1 gi. £. $19.16. 15. $3277.97. 16. 357£ rd. 1. 17 lb. 9 oz. 5 dr. 1 sc. 2. 5896 dr. P. $72.58, gain. 17. 414f ft. 13. 7 oz. 14 pwt. 4.8 gr. 14- 18 pwt. 2.4 gr. 25. .297916 + lb. tf. .875 oz. Page 131. 18. 130£ A. 19. Not any. 3. T 9 x¥*lb. Art. 403. 20. 2 sq. rd. 4. 11 oz. 3 dr. 2 sc. 1. 1579y 4 y pt. 21. 28f. •8gr. 2. 134.5 + pt. gain. 22. 4f|A. 17. 472 lb. 1 oz. 12 5. 63 sc. 3. $.52 gain. 23. 58£ rd. pwt. 8 gr. 6. 6 lb. 9 oz. 6 dr. 11.5 gr. 4. $5.25, gain. 5. $7.99 less. 24. 12. 25. 26 A. 11 sq. rd. 4 Page 125. 7. 1 lb. 2 oz. 4 dr. I sc. 4 gr. Page 132. sq. yd. 5 sq. ft. 36 sq. in. IS. 211 lb. 11 oz. 19 8. 7 lb. 3 oz. 4 dr. Art. 409. 26. $60.06. pwt. 21 gr. 12 gr. 1. 127002 in. 27. $10.56. 10. 2 pwt. 20. 4 gr. 5. 6 lb. 2 oz. 6 dr. 2. 39 mi. 155 rd. 4 28. 9 A. 110 sq. rd. 20. .0067+. 1 sc. 8 gr. yd. 3 in. 12 sq. yd. 3 sq. 21. 8 lb. 2 oz. 13 pwt. 10. 15 lb. 10 oz. 5 *• "i 2 6 * s • ft. 54 sq. in. 6gr. dr. 2 sc. 8 gr. 4. 213 rd. 1yd. 2 ft. 29. 38 A. 59 sq. rd. 22. 18 lb. 9 oz. 14 11. 11 oz. 2 sc. 4& 6 in. 12 sq. yd. 5 sq. pwt. 2 gr. gr- 5. fjrd. ft. 112 sq. in. 93. $8032.50. 6. 173 rd. 2 yd. 1ft. 30. 640 rd. 24. $1924.39. Page 129. 312 in. 31. 320 rd. U. 5 oz. 2 pwt. 17 Art. 398. 7. .892+. gr. 1. "$1477.27. . 3000*. Art. 530. 1. 25*. -2. 3**. 5. 33£*. 4. 20*. 5. 62£*. 6. 53i*. 7. 57£*. 8. m%\%. 9. 33i*. 10. 56±*. 22. 25*. Page 166. Art. 532. 1. 1.10, amt. per. cent. 2. 1.75, amt. per. cent. 3. 2.10, amt. per. cent. 4. 1.16$, amt. per. cent. 5. 1.87£, amt. per. cent. Art. 533. 1. 1.05 per cent. 2. 1.09^*. 3. 1.40*. Art. 535. 1. .85, difference per cent. 2. .62*, difference per cent. 3. .99£, difference per cent. 4. .96$, difference per cent. 5. .30, difference per cent. Art. 536. 1. 71f*. 3, .60, difference per cent. Page 167. Art. 538. 1. 1650. 2. 1695. 3. 462. £ 1180. 5. 277.2. 0. 2580. . 7. 840. 5. 637. 9. 450. Art. 539. 2. $14512.50. 3. $11200. 4. 816. 5. 3537. 6. $886.50. Art. 541. 2. 12. 2. 150. 3. 945. Page 168 4. 612. 5. 1200. 6. 500. 7. 5. 8. 567 ft. Art. 542. $2843.75. 581| A. $2053.13. $632.50. Art. 544. 1. 600. 2. 400. 3. 300. Page 169. 4. 100. 5. Art. 545. 1. $1000. 2. $4.51. 3. $7950. 4. $480. 5. $106. 6. $3200. 7. $30000. 8. $10000. 9. 500 pupils. 10. Invest, in farm, $280. Page 170. Art. 548. /. 64|f*. 2. 681b. 302 ANSWERS. 3. A- 5. 6. 7. 8. 9. 10. 11. 12. 13. 80%. $2333 28H, 77 yr. 42f, and H.$200, M.$170, C. $15. S1081.25. 25 lb. warp, and 71£ lb. rags. $160, and $224. $113.78. 1£% gain. Page 171. U. $4387.50. 15. $1045.45 + . 16. 216%+. 17. 27 ¥ V%, 21|f%, andl8 T f T %. 18. 100 head. 19. 9^%. 20. 25%. 22. $6344.40. 20. 18f%. 23. $1000 loss. f4. 400. 25. 3000. *?. 77§H%. 27. 10 yd. &?. $4856.25. 29. $60250. Page 172. ,20. Grazing, 504 A.; grain, 420 A. ; timber, 936 A. 81. $192 C. 82. A $93840, and B 33. $22400. 34. Not any. 35. 7500, 9750, 6825, and 9555. 36. Clover, 450; tim- othy, 450 ; or- chard grass, 150 and 50 red top. 37. $81.20, $101.50, $182.70. 38. 16f%. 89. $22629.31. 40. $1750, $3062.50, $6125, and $8575. 41. 25600 T. 42. Wife, $21750 ; D., $10000; Y. S., $12500; and E.S., $13750. Page 173. Art. 557. 3. $750. Page 174. 4. $50. 5. $32. 6. $320. 7. $225. 8. $2100. 9. $700. Art. 558. 1. $592.50 gain. 2. $3997.50 loss. 3. $677.25 loss. 4. $9.47 gain. 5. $184.92 gain. 6. $6 gain. 7. $10.16 loss. 8. $22.97. 9. $133.59. Page 175. Art. 560. 1. $100. 2. $3500. 3. $10000. £ $4400. 5. $40. 6. $300. 7. $900. 8. $1050. 9. $1. Art. 561. 1. $57.50. 2. $500 and $625. 3. $600. •4. $700. 5. $7085.71. 6. 200 A. 7. $2750. 9. $74.94. $2707.18. $4310.74. $2040.15. $2766.55. $1969.62. $3519.75. SO. 35. 1. 1. Page 214, Art. 705. $343.75. $4099.71. ^ $337.41. $857.01. $27500. $826.23. Page 215. Art. 707. $445.94. $10344.83. $600. $1000. $291.85. $739.13. $1954.63. Art. 709. 7%. 6f%. 6%. 5%. 6%. 10%. Page 216. Art. 711. 1. 2 yr. 5 ino. 24 da. 2. 3 yr. 10 mo. 12 da. 3. April, 25, 1881. 4.. 11 mo. 5. Sep. 22, 1889. 6. 12 yr. 6 mo. Page 218. Art. 715. 1. $4.38. 2. $5.25. 3. $3.71. 4. $17.44. 5. $9.24. 6. $1.13. 7. $5.83. 8. $3.33. 9. $1.76. 10. $5.70. 11. $2.34. 12. $8.02. 13. $6.32. 14. $3.00. 15. $24.58. 16. $1.15. 17. $4.37. 18. $11.65. 15. $1.34. 20. $10.96. M.