.-> yiA^t^Z^ r, U L/f^l / A PRACTICAL COMMERCIAL ARITHMETIC DESIGNED FOR USE IN ALL SCHOOLS IX WHICH THE COMMERCIAL BRANCHES ARE TAUGHT AND AS A BOOK OF REFEREJSTCE * FOR BUSIJ^ESS MEiSr. 1889. Entered according to Act of Congress, in tlie year 1888, By WILLIAMS & ROGEES, In the Office of the Librarian of Congress, at Washington, D. C. E. R. ANDREWS, PRINTER AND BOOKBINDER RIXIH ESTER, N. Y. SRUR ^^/^^^T^^^^ 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. CONTENTS. SIMPLE NUMBERS. l-AGK Definitions 1 Signs 3 Abbreviations and Contractions 3 Notation and Numeration -. 3 Arabic Metliod of Notation 4 Frencli 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 DECIMALS. Definitions 66 Numeration of Decimals .- 67 Notation of Decimals 68 Reduction of Decimals .-. 69 Circulating Decimals 72 Addition of Decimals 73 Subtraction of Decimals 75 Multiplication of Decimals 75 Division of Decimals 76 Miscellaneous Examples 80 UNITED STATES MONEY Definitions United States Coins United States Paper ]\Ioney Reduction of United States Money 81 82 83 84 Addition of United States Monej' 84 Subtraction of United States Money 84 Multiplication of United States Money.. 85 Division of United States Money _ 86 Definitions . ANALYSIS. 87 I Examjjles . 87 SPECIAL APPLICATIONS. Definitions 89 Aliquot Parts 89 Instructions for Pr^rtice with Ali»iuot Parts DO Miscellaneous Contractions 90 Instructions for Finding Quantity 91 Miscellaneous Contractions 92 Bills, Statements, and Inventories 100 Miscellaneous E.xamples 106 CONTENTS. Vll DENOMINATE NUMBERS PAGE Definitions 108 Measures of Time 108 Reduction of Time. 110 Addition of Time 111 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 Pleasure . 132 Involution 137 Evolution ^ 137 Square Root 137 Surveyors' Long Measure 141 Surveyors' Square Measure 142 Cubic Measure ^. 143 Table Special Cubic Measures 144 Produces' and Dealers' Approximate Rules 146 Hay Measurements 147 Cube Root 147 Duodecimals 151 Miscellaneous Measurements 151 The Metric System 155 French 3Ioney 157 German Money 157 M iscellaneous 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 - 176 To find the Cost, the Selling Price and the Rate Per Cent, of Profit or Loss being given 177 Review of the Principles of Profit and Loss 178 Tkade Dtscouxt 183 To find the Selling Price, the List Price and Discount Series being given 183 To find the Price at which Goods must lie 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 Tin COXTENTS. TAGS Commission 191 To find the Commission, the Cost or Sell- ing Price and Per Cent, of Commis- sion being given 192 To find tlie 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 CcsTOM-HorsE BrsixEss 197 To find Specific Duty.. 196 To find .Vd 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 Pr<>- mium and Per Cent, of Premium being given 206 Peuso5> AL 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 21 1 To find the Rate, the Principal, Interest, and Time being given. 215 To find the Time, the Principal, Interest, and Rate being given. 216 SnoKT Methods kou Finding Interest 216 To find Interest for Days, at 6 Per Cent., 360 Day Basis.. 217 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 223 To lind Compound Interes' 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 Paysients 239 United States Rule 240 Merchants' Rule .• 241 Equation of Accounts 243 "When the Items are all Debits, or all Credits, and Lave 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 R.\.Tio 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 Answers 273 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 IT dollars, 1 dollar; of 30 pupils, 1 juipil. 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 tluit 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 tiling 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. 14r. Unlike Numbers are such as have different unit values; as, 11, 16 days, 365 dollars, 5 pounds, 4 yards. *15. Ratio is the comparison of magnitudes. It is «>f two kinds; urithmetical 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 analysts, the logical statement of its conditions and of the steps required for its solution. 2 SIGNS. 19. The Conclusion of llio analysis is called the (nm^ccr, or rcsnU. 20. A Rule is an outline of the etei)s 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 followin2: are the ])rincipal and most useful arithmetical signs: 22. The Sign of Addition is a perpendicular cross, +. It is called Plus, and indicates that the numbers betAveen 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 jjlacod between two numbers, that the value of the number on its right is to be taken from the value of tlie 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 Dirision is a short horizontal line with a point above and one below, -=-. 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 eqital 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, { }, and vinculum, . They indicate that the quantities included within, or connected by them, are to be taken together and su})jected 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^ indicates that 4 is to be taken twice as a factor, or multiplied by itself once; 4^ indicates that 4 is to be used three times as a factor. 4- is reac^ 4 squared; 4^ is read 4 cubed; also, the second jjower of 4; the third power of 4. 30. The Root, or Radical Sign, is the character, \/', 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 sei)aratcd into two equal factors, or its square root taken. A figure jdaced in the ojiening indicates the number of equal factors required, or the root to be extracted ; as, '\/T\, \/~~6' ABBREVIATIONS AND CONTKACTIONS. 3 31. The Dollar Sign is the character, %. 32. The Cent 8ig|l is the character, pacLi SOh hoh p&h m § S a .2 -■ o: o; <*.! Q o C k C *C X c .S a D 0! c H K E- 5 Mh H 6, 1 3 2. 7 4 13 a ai CO •v S o O ■c ^ « E- Oh CO •c a <& cc J3 QQ c H s «»-i o o 01 H ^ TT a a c 03 D O a ZJ JS t^ E- 0) O 6, 9 8 5, 18 KOTATION AND NUMERATION. 5 The other successive periods are called Quadrillion, Quintillion, Sextillion, Septillion, Octillion, etc. Commencing with the right figure, which is called units of the first order, or simple units, the orders of figures, or units, to the left, are called units of the second order, units of the third order, fourth, fifth, sixth, etc. ENGLISH SYSTEM OF NUMERATION. 40. There is in use a system known as the English numeration, which gives to each period after thousands, six places, or figures, instead of three us given by the French numeration. Numbers are divided into periods, and enumerated and read by the English numeration as shown by the following English Nuiueratiou Table. -% T3 en 09 33 a .2 to a o a tn 3 ■a ^ 73 o a 33 ■X a 00 a o a a 3 1146. .?. 3070. S. 77010016. 9. 200020. 10. 140024G780. 11. 2100211. 12. 5800092. 13. 34307001. U- lOOOlOOOlOUd. 6103G. iu wiirils mill read the following numlx-r I n. 50415. I 6'. 100000. 7. 521469. 8. 201012. iJ. 987000460000. iij. 27510304050. 17. 11002200330044. 18. 2234507890. 19. 40122555003. 20. 621438001240709. 21. 12345325500001503. 9. 1406250. 10. 54790207. 11. 1021714. 12. 5790r)7359. (^ a NOTATION' AND NUMERATION. 7 ROMAN NOTATION AND NUMERATION. 46. Bj combining, according to certain princiiiles, 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 tlie 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. 3. A dash — placed over a letter multiplies its value by one thousand. Thus, V= five, V= five thousand, CD = four hundred, CD = four hundred thousand, LXVII = sixty-seven, LXVII = sixty-seven thousand. Table of Roman Numerals with Arabic Equivalents. I,- 11, III, IV, V, VI, VII, VIII, IX. X, XI, 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. XII, XIII, XIY, XV, XVI, XVII, XVIII, XX, / ^ XXX, XL, 12. 13. 14. 15. 16. 17. 18. 19. 20. 30. 40. L, LX, LXX, LXXX, XC, c, CC, CCC, CD, D, DC, 50. 60. 70. 80. 90. ■ 100. 200. 300. 400. 500. 600. DCC, 700. DCCC, 800. CM, 900. M, 1000. MM, 2000. X, 10000 L, 50000. c. 100000 D, 500C00 M, 1000000 47. XCII. XXVII. XXIX. CLX. 48. 42. 111. 66G, 1125. 7000. ,11451. 997. 56104. 3001. Eead the following expressions: CCXVII. CMXIX. DCV. DCCX. CMXXV. MCLXXIX. MCDXCII. MDCCLVI. DLXX. DCCXLV. MDq. . ^DCCCLXXXVIII. Express by the Roman system the following numbers 7454. 8709. 62550. 1620. 399. "^-^5406." 48250. 3700. 2865. 1629. ^1889. "^60012. 3658. 175400. 1761. 1887. 1000000. 20000. 45450. 19015. 1111. 6057. 3113. 90055. 805000. 365. 1515. 6059. -*21021. 4888. 9()909. 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 tlie result obtained by the addition of two or 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 4-2 = T, 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. 3. The sum or amount is the same in whatever order the numbers be added. 55. Addition is the Reverse of Subtraction and may be proved by it; as, 5 + 2 = 7. Xow if 7 be diminished by 5, the result Avill be 2, while if 7 be diminished by 2, the result will be 5. 56. Numbers are written for addition either in vertical or horizotital order. 57. General Kules. — l. If the sum of two numhers and one of the numbers be given, the unknown number may be found by taking the given nuTtiber from the sum. 2. If the suTti of several nunibers and all of the numbers but 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. 6 + 6 = ij + 7 = 6 + 8 = 6 + 9 = 6 + 10 = 6 + 11 = 6 + 12 = 6 + 13 = 6 + 14 = 6 + 15 = 6 + 16 = 6 + 17 = 6 + 18 = G + 19 = 6 + 20 = 6 + 21 = 6 + 22 = 6 + 23 = 6 + 24 = 6 + 25 = 7 + 7 = 7+9 7 + 10 7 + 11 V + 12 7 + 13 7 + 14 7 + 15 7 + 16 7 + 17 7 + 18 7 + 19 7 + 20 7 + 21 7 + 22 + 23 + 24 7 7 7 + 25 = 8 + 8 = 8 + 9 = 8 + 10 = 8 + 11 = 12 13 14 15 16 ir 18 10 20 21 22 23 24 25 20 27 28 29 30 31 14 15 16 = r 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 16 ir 18 19 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + Addition 12 = 20 13 = 21 14 = 22 15 = 23 16 = 24 17 = 25 18 = 2G 19 = 27 20 = 28 21 = 29 22 = 30 23 = 31 24 — 32 25 = 33 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 9 = 18 10 = 19 11 = 20 12 = 21 13 = 22 14 = 23 15 = 24 16 = 25 17 = 20 18 = 27 19 = 28 20 = 29 21 = 30 22 = 31 23 = 32 24 = 33 25 = 34 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 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 + 32 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 14 + 14 14 + 15 14 + 16 14 + 17 14 + 18 14 + 19 14 + 20 14 + 21 14 + 22 14 + 23 14 + 24 14 + 25 28 29 30 31 32 33 34 35 36 37 38 39 15 + 15 15 + 16 15 + 17 15 + 18 15 + 19 15 + 20 15 + 21 15 + 22 15 + 23 15 + 24 15 + 25 30 31 32 33 34 35 36 37 38 39 4C 16 + 16 16 + 17 16 + 18 16 + 19 16 + 20 16 + 21 16 + 22 16 + 23 16 + 24 16 + 25 32 33 34 35 36 37 38 39 40 41 17 + 17 17 + 18 17 + 19 17 + 20 17 + 21 34 35 36 37 38 17 -^ 22 = 3^ 17 + 23 = 40 17 + 24 = 41 17 + 25 = 42 18 18 18 18 18 18 IS 18 + 18 = 36 + 19 = 37 + 20 = 3& + 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 = 4e + 24 = 47 + 25 = 4& 24 24 + 24 + 25 48 49 + 25 = 50 10 ADDITION. MENTAL KXERCISES. 58. 1 Add {2.) 5 {3.) 1 a) (5.) 11 Cr.) V2 25 2 6 3 4 • 5 14 14 50 3 ^ 5 6 4 13 18 15 -t 8 • 1^ 1 8 8' 15 15 25 5 9 9 10 6 12 10 10 6 10 11 1-2 12 18 20 25 1 11 13 1-t 10 i: 14 15 8 12 15 IG 14 20 1€ 10 59. There are three methods of addition in common use, viz. : tlie Elementary vipfhofl. the Besidt method, and the Group method. Remarks. — 1. These methods of additibn 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 03. OPERATION. ExPLANATiOK.— Having arranged the numbers so that units of like orders „.^ 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 sura, 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 upVard 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. 19 31 and 3 are 34. Having obtained the sum, write it in full at the left of the ^_ figure 1 before written, and the result is 341, the numerical expression of the '_ sum of the numbers added. Q^j To Prove. — Add the columns downward ; if the two results agree, the work is presumed to be correct. 61. The Eesult Method of Addition. Example.— Add 32, 71, 25, 48, 90, 12, and 03. OPERATIOK. 32 . 71 25 48 90 12 53 To Prove.— .\dd the columns downward. 341 ExPLAKATioN. — 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 tens' column as a part of it, add upward, thus : 2, 8, 9, 18, 22. 24, 31, 34 ; as before, write 34 at the left, and the result is 341, the same as before. ADDITION. 11 6'2. The Group Method of Addition. Example.— i. Add 32, 71, 25, 48, 90, 12, and 63. ■OPERATION. Explanation. — Treat the same numbers thus : add upward ; 3, 13, 21 ; 32 grouping 2 and 8 for 10 to add to 8, making 13, and 5, 1, and 2 for 8, to add to f.-. I o 13, making 21. Having written the 1 beneath the line, in units' place, carry the _ C 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 ; then write the result 48 ^ in full as before. 90 V 10 To Provk. — 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. Tlien 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. OPERATION. Explanation. — Beginning with the lower unit figure add upward; 10, 15, 35, 341 '^ 1 00, grouping 4, 2, 3, and 1 for 10, which added to 5 gives 15 ; grouping 6, 6, and 2140 ^ ^*^"' ^^ ^" ^^^^ ^° ^^ obtaining 35 ; and grouping 4, 9, and 7 for 20 to add to 35 439 I ^"^ ^5 ^^^^ result. Write the units' figure 5 in its place, and carrying the tens' 7164 I figure 5 to its column proceed thus : 8, 24, 38, 48, 56, 60, 70, grouping the 5 1538 1 carried and 3 for 8 ; 9 and 7 for 16 to add to 8 for 24 ; 6 and 8 for 14 to add to 5046 I 20 24 to make 38 ; 7 and 3 for 10, making 48 ; 1, 4, and 3 for 8, making 56 ; 6, 3, 6116 J iind 1 for 10, making 06, to which we add the 4 for 70, the result. Write the 8735 cipher of the 70 at the left of the unit figure already written beneath the line 971 1 and carrying the 7 to the third, or hundreds' column group as before; 16, 26, 36, 4880 48, 58, grouping upward thus : 7, 6, 1, 2 = 16 ; 2, 8 = 10 ; 9, 1 == 10 ; 7, 5 = 12 ; 1263 I 1^ 1, 4, 1, 4 = 10. Write the 8 in hundreds' column, and carrying the 5 to thou- 9270 I sands' column, group 15, 27, 39, 49, 51. 5, 9, 1 = 15 ; 4, 8 = 12 ; 6, 5, 1 = 12 ; 192 I 7. 3 = 10, and adding 2 write the result, 51, at the left of the figures before 634 J written, thus obtaining 51805 the numerical expression of the sum of the num- 51805 ^^''^^ added. • Prove by adding downward, grouping as illustrated above. Remark. — Practice in grouping will lead to great proficiencj', 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 *- X to CO CO o to 4^ X to © -J -7 C5 on x^ to CO to CS to o on CO X CO X 4- -7 -7 o OS Ol CO CS CO 4^ to 05 Ol to X to 1-1 1-1 -I o oo -J C5 C5 00 l-k y- «5 31 o on to 1^ i to 00 to o 1-1. to o CO OS 00 00 X ^7 OS to .fk Ol OS 00 4^ o 00 to to CS X w to t3 OS to o -1 OO 00 o -3 CS to On CO 1-^ CO Ol to OS -1 o 00 CO o CO 00 -7 — to OS CO Ol Ol CO OS to -J 00 o © lO JO to to o to o to o oo o -1 o CS o OT o 4^ CO o i o o o CO 'X o =; -7 o o 3 4- g to o ^ I-* to to 4^ 1 to to to CO to to o to CO 00 oo -1 -1 OS CS On Ol 4^ CO CO to ro o CO 00 CO X -7 OS OS ^? 4^ CO CO to h-1 05 o o to 00 00 lO Ci to a to on to to o to to 00 to CS to o CO to CS X Ol CS CO to to o O CO X OS 00 4^ -3 lO CS o 1^ 00 CO OS ^ to CO to VX CO lO to to C5 to E2 to o to 4^ -J to CO to to to o 00 CO On 00 to 05 CO CS CO CO o I-' o -7 4^ CO -7 X On Ol to g g cc 4- 03 CO CO CO to to CO o 00 to o to 00 o to OS OS to o? to to CO oo to to to o CO 00 to 'X Ol 4i. 4^ o to l-i CS to CO X 00 -7 o on CS 4^ to to oo 4*- k^ *« CO VI CO o CO CO CO o cc On CO to X on to to Ol Ot to »«^ o ro ro to o CO On X CS On on o CO to on O !i; CO o -7 ^. 4- CO o 7| 1^ o o CO CO (X' CO Ol to CO CO OS to to CO to ■X 00 to -3 to ro en to to to to o X CO to H-l- CS OS o 4^ to 4^ 00 to CO OS 00 o OS 4^ X CO to ^ - to o oo CO CO CO ^1 CO CO to CO OS 00 CO to -7 to to 8! to CO oo to to 4^ 1-^ X -7 )-l- -J o en CO CO CS CO o to X Ol OS 00 on 1-1 CO 4^ 1^ o CO 4^ CO OS CO 00 CO OS o CO to w ro 4^ CO o CS to (O oc to -I o to or to to CO to CS CO 00 i OS i«h to 4^ to CS i 8 -3 to OJ 4^ OS X w -J en o CO 00 CO o CO ■X o CO CO to CO to CO CO o to 00 Ol to CS CS to -3 to to 00 i i <7 on H- to 10 CO 4^ CO On -I OS en -3 CO X © © -' c -x -7 *» CS CO X On CO to § 4^ CS 10 4- C5 o o U1 on to or to oo On >*>. O i30 *- o Ol 05 CO to 4^ o 00 CO 00 1*^ CO CO CO CS CO to to 8B to CS 4i- to o to ^ ^ CO CS to CS X 4^ 4i- i CO OS J3 fe to 4- o ro 05 O o on ~l o» on on o on o? to o Ol o 1^ Ol Ol o to Ol >«>■ CO -7 on CO or o CO to on to o o to -J en to On o to to to- o on O On g to Ol o o ~7 on g lO Ci< <>< 4* CO I« K) 1- © X •>! CS ha Ki hi' © tf X •^1 © «. )^ CO to h* Note. — It will be of great advantage to tlie student to fully master the above table. Any delay caused by following this suggestion will be offset by time gained in subsequent work ; such mastery will so increase the rapidity of work in business applications as to greatly lessen the labor of accounting. 22 MCLTIPLICATIOX. 91. Multiply j'^.vAjni'i.fcs ft IK ilK. \L ]>KILI.. 1. 42 bv 3. 11. 102 by 4. 21. 144 by 1(1. SI. 595 by 13. 2. 31 by 4. n. 511 by 8. 22. 52 by" 11. S2. 70 by 22. S. 2: by 2. IS. 125 by 6. 23. 45 by 13. SS. 90 by 25. J^. 60 by 5. U- 340 by 2. 24. 201 by 15. SJ,. 150 by 23 5. 51 by 4. ^ lo. 416 by 3. 25. 65 by*20. So. 118 by 11 6. 75 by 6. m. 99 by 7. 26. 411 by IT. S6. 906 by 15 7. 91 by 2. 11. 133 by 9. 27. 932 by 12. 37. 450 by 19 8. 29 by 3. IS. 208 l/v 4. 28. 43 by' 19. 38. 375 by 14 9. 57 by 2. 19. 666 by 5. 29. Ill by 23. 39. 250 by 18 to. 95 by 5. 20. 89 by 8. SO. 207 by 22. 40. 789 by 11 417 5 2085 9*2. "When either Factor is a Number within one's thorough knowledge of the Multiplication Table. Example. — 1. Multiply 417 by 5. OPERATION. Explanation. — "Write the multiplier 5 below the unit figure of the multi- plicand, and multiph" each figure of the multiplicand by the multiplier, thus; 5 times 7 = 35, or, 3 tens -f- 5 units ; ■write the 5 imits in units' place and reserve the 3 tens to add to the next product ; next, 5 times 1 ten are 5 tens, and addmg the 3 tens reserved gives 8 tens, "which -write in tens' place ; next, 5 times 4 hundreds are 20 hundreds, or 2 thousands ; write a naught, or cipher, in the hundreds' place and the 2 iu the thousands' place, thus completing the multiplication and obtaining 2085 as the product of 417 multiplied by 5. Example.—,?. Multiply 123 by 12. Explanation. — Multiply each figure of the multiplicand by the multiplier, 12 ; 12 times 3 = 36, or, 3 tens -j- 6 units; write the 6 in units' place and reserve the 3 tens to add to the next product ; next, 12 times 2 tens are 24 tens, and adding the 3 tens reserved gives 27 tens, or 2 hundreds-)- 7 tens; write the 7 in tens' place and reserve the 2 hundreds to add to the next product; next, 12 times 1 hundred are 12 hundreds, and adding the 2 hundreds reserved gives 14 hun- dreds, or, 1 thousand 4 hundreds, which write in hundreds' and thousands' places, thus com- pleting the multiplication and obtaining 1476 as the product of 123 multiplied by 12. Rule. — Write the factors one below the other, arranged so that figures of like orders will stand in the same vertical line. Multiply each figure of the upper factor, heginjiing at the right, hy the lower factor, placing in order the last figure of the product so obtained, and carrying to tlie next product all figures except the last ; continue so doing until the last product is found, which u-riie in full. 0PER.\T10N. 123 12 1476 KXA>II'M.N >oK >IKNTAI. I'ltACTICK. 4- 6. 93. Multiply 201 by 8. 1325 by 2. IS. hll by 16. li>. 641 l)y 13. 507 by 5. 8. 2108 by li. 14. 1603 by 9. 20. 7122 i)y 5. 1001 by 12. 9. 511 by' 15. lo. 3006 by 14. 21. 984 by 8. 311 by" 6. 10. 293 b"y 12. 10. 249 by 7. 22 2260 by 12 805 by 9. 11. 1801 by 13. 17. 519 by 8. 23. 461 l)y"l4. 1203 by 8. 12. 684 by' 14. 18. 1122 in- 11. 2J,. 3542 in- 15 MULTIPLICATIOK. •->3 EXAMPLKS FOK WRITTKX I'KACTICK. 94. Multiply 1. 2168 by 9. €,. 35G142 by 18. JL o 31046 by 16. t . 2147603 by 8. 12. 3. 599 by 12. S. 15286097 by 15. 13 Jf- 2170 by 13. 0. 508093240 by 13. n 5. 50890 bv 11. JO. 6381201432 i)y 14. ]■', 99084160024 by 15. 294640205580 by 9. 6620.5380777 by 7. 897352468004 by 12. 21430206041 bv 15. Multiplicand. Multiplier. Units. Tens. Hundreds. 1251 417 352 834 2085 95. When the Multiplier consists of two or more figures. Example. — Multiiily 417 l»y 352. OPERATION. Ex PI, AX AT I OX. — Write the numbers one below the other in the 00 same unit order from the right. Then, beginning with the unit figure t . of the lower factor multipl}-; 2 times 7 units are 14 units = 1 ten -f 5c~ 4 units ; write the 4 units in units' column and add the 1 ten to the next product; next, 2 times 1 ten are 2 tens and the 1 ten added makes 3 tens, which write in tens' place ; next, 2 times 4 hundreds are 8 hundreds, which write in hundreds' place, giving 834 as the first partial product, or the product of the upper factor multiplied by the unit figure of the lower factor. Next take the tens' figure of the lower number as a multiplier; 7 taken 5 tens or 50 times = 35 tens, or 350 ; write the 5 of the number 35 in tens' column, or below the 8 tens of the first partial product, and carry the 3 of the 35 to the next product; next, 5 times 1 are 5 and the 3 to carry added make 8, which write under the 8 of the first partial product ; then, 5 times 4 are 20, which write still to the left, making the second partial product 2085 tens. Next, take the third figure, or hundreds, of the lower factor, as a multiplier; 3 times 7 hundreds are 21 hundreds ; write the 1 in the hundreds' place and reserve the 2 for the next product ; then, 3 times 1 are 3 and 2 to carry makes 5, which write in its order ; then, 3 times 4 are 12, which write still to the left, having as a product 1251. Since, however, the several figures of the factor taken as a multiplier were of successive orders of units. The first partial product 834 = 834 simple units. The second partial product 2085 = 2085 tens = 20850 " The third partial product 1251 = 1251 hundreds = 125100 " 146784 And the sum = 146784 Therefore, 146784 is found to be the product of the numbers 417 and 352. Rule. — I. Place the multiplier helow the multiplicand, the unit figures in the same vertical line. II- Beginning with the unit figure, multiply all the figures of the mul- tiplicand bij each successive figure of the multiplier, writing the first figure obtained in each partial product directly below the figure by which it was multiplied. Add the partial products. Remark. — The object of writing each succeeding partial product below and one place to the left of its predecessor, is that imits of the same grade, or order, ma}% for convenience in adding, be found in tlie same vertical line; this arrangement precludes the necessity of filling the vacant orders witli ciphers. As before shown, the arrungemeiit of factors will not vary the result ; as, 4 X 5 = 20, also 5 X 4 = 20 ; therefore, in business or school ])ractice, arrange factors in such order as to save time and space; by so doing, problems otherwise long and difficult, may be- solved Ijy mental ])rocesses. 24 MULTIPLICATION. Example.— Multiply 120000 by 7256. Explanation. — Consider the factors as reversed in order ; thus, 7256 X 120000. Theo multiply the 7256 mentally by 12, and to the product, 87072, annex four ciphers, because the 12 was not 12 simple units but 12 units of the tifth order, or tens of thousands. 9(). When one Factor is 10, 100, 1000, 10000, or 1 with any number of ciphers annexed. E.\AMPLE.— Multiply 324 by lOUO. ExPL.\NATiON. — Since there are three ciphers in tlie multiplier, annex three to the multipli- cand, 324, thus obtaining the product, 324000. Rule. — To the one factor annex as many ciphers as there are ciphers in the other factor. 97. A Composite Nuinbei* is a number that may be resolved or separated into integral factors ; or, it is a number that may be formed by multiplying together two or more numbers; thus, 12 = 4 X 3 ; or 12 = 2 x 2 X 3 ; or 4 X 3 = 12; or 2 X 2 X 3 = 12. 98. When the Multiplier is a Composite Number. Whenever it is required to find the product of numbers one or more of which is composite, the result may be obtained by using as multipliers the factors of l such composite number or numbers; thus, 6 X 4 = 24, or 6 x (2 X 2) = 24. Rule. — Separate the multiplier into its factors. Multiply the miMipli- cand by one of these factors, that product hy another factor, and so on, using in succession all the factors; the last product ivlll be the result required. Rem.ark. — Since the order in which factors are used will not vary the product, the student is recommended to seek the simplest number — the one most easily factored — as a multiplier. KXAMPLES rOK PRACTICE. 99. 1. Multijily 41 by 15, itsing as factors 3 and 5, 2. Multiply 17 by 21, using as factors 7 and 3. '3. ;Multi])ly 111 by 24, using as factors 3, 2, and 4 4. Multiply 1157 by 30, using as factors 6 and 6. o. Multiply 2019 by 45, using as factors 5, 3, and 3. 6. Multiply 8T002 by 9G, using as factors 6, 4, and 4. 7. Multiply 54235 by 144, using as factors 12 and 12. 8. Multiply 54235 by 144, using as factors 9 and 16. 9. Multiply 54235 l)y 144, using as factors 9, 4, and 4. 10. Multiply 54235 by 144, using as factors 3, 3, 2, and 8. 11. ^Multiply 54235 by 144, using as factors 3, 3, 2, 2, and 4. 12. Multiply 54235 by 144, using as factors 3, 3, 2, 2, 2, and 2. 13. Multiply 81 by 64, using as factors 8 and 8, 14. Multiply 04 Ijy 81, using as factors 9 and 9. lo. Multiply 81 by 04, using as factors of 81. au'I 9, and as factors of 64, 8 and 8. 100. Multiply 1. 1431 by 7000. e ^. 900 by 2104G. 7 S. 1969 by 54 = 9 x 6. 8 Jt. 171548 by 1500 = 15 X 100. 9 5. 1653 by 25000 = 5 X 5 X 1000. 10 MULTIPLICATION-. 25 KXAMPtKS COMBINING EtEMENTAKY PKINCIPLES PKEVIOUSI.Y EXPLAINED. 3500 by 72 = 6 X 12. 1921 by 450 = 9 X 5 X 10. 321058 by 144 = 12 x 12. 504 by 288 = 9 X 8 X 4. 1043 by 105 = 7x3x5. 11. A clerk sold 9 shirts ut 80 cents each, 2 neck-ties at 35 cents each, 10 col- lars at 25 cents each, a pair of gloves for 75 cents, and two suits of underwear at 95 cents ]}qv suit. What was the price of all ? 12. I bought 15 cows at 32 dollars per head, a pair of horses for 245 dollars, a harness for 22 dollars, and 81 sheep at 3 dollars per head ; what was the total cost of my purchases? 13. The cost of furnishing a house was, for parlor and library furniture 762 dollars, halls 150 dollars, dining room and kitchen 295 dollars, chambers 648 dollars, stoves and furnace 350 dollars, carpets and curtains 825 dollars, what was the total cost ? IJf. 14250 dollars was paid for four houses, the first costing 2750 dollars, the second 400 dollars more than the first, the third 250 dollars less than the first and second together, and the fourth the remainder. Find the cost of the fourth house ? 15. Find the difference between the continued products of 91 X 4 X 3x11x9 and 5x5x12x4x6x7. 16. Find the difference between seven units of the sixth order and the con- tinued product of 15x6x5x12x4x7x11x8x2x9. 17. A merchant having 17462 dollars to his credit in a bank, gave checks as follows: for dry goods 5416 dollars, groceries 5995 dollars, boots and shoes 1416 dollars, hardware 1850 dollars, and drew out 500 dollars for family expenses; what amount was left in the bank ? 18. Exchanged a city block valued at 35000 dollars, for a farm of 175 acres valued at 95 dollars per acre, eight horses at 110 dollars cacii, 14 cows at 28 dollars each, 225 sheep at 4 dollars each, farm machinery valued at 825 dollars, and received the balance in cash. How much cash was received ? 19. A drover bought 135 horses at an average j^rice of 115 dollars for 100 of them, and 125 dollars per head for the remainder ; he sold 25 at 100 dollars per head, twice that number at twice the price per head, and the remainder at 67 dollars per head. IIow much was gained or lost ? 20. A ranchman sold to a trader, 46 ponies at 60 dollars i)er j)air, 116 calves at 9 dollars per head, 41 cows at 35 dollars per head, and a pair of mules for 375 dollars. He received in part payment, 15 barrels of flour at 9 dollars per barrel, 11 hundred weight of bacon at 12 dollars i)er hundred weight, 4 suits of clothes at 22 dollars \k'V suit, 2 saddles at 13 dollars each, a wagon at 75 dollars, a set of furniture for 58 dollars, and the remainder in cash. What amount of cash did the trader i)ay ? ' . ' 26 MULTIPLICATION. MISCELLANEOUS EXAMPLES. 101. 1. The United States export 105,000 sewing machines yearly. If each machine does the work of 12 women, what is the value of the labor thus con- tributed by the United States to other nations each year of 306 working days, if labor be estimated at $1 per day? 2. The Union Pacific Railway is 1777 miles in lengtli, and was built at an average cost of $106,775 per mile; what was the total cost of construction ? 3. The bills issued by the U. S. Treasury for National Bank circulation, are in denominations of ^1, §2, $5, 810, $20, $50, $100, $500, and $1000. How much money has one possessing 73 bills of each denomination ? J^. The gold coins of the U. S. are in denominations of $1, $2.50, $3, $5, $10, and $20. How much money in a bag containing 365 of each of these coins ? 5. The U. S. notes — greenbacks — are of the following denominations, viz.: $1, $2, $5, $10, $20, $50, $100, 8500, $1000, $5000, and $10000. How large a debt could be paid with 7 of each of the above-named greenbacks ? 6. How many feet of wire will be required to fence a field 11 10 ft. square, with six wires on each of the four sides ? 7. What is the amount of the following bill ? 28 lb. Lard @ l\f per lb. 110 lb. Beef @ 14^' per lb. 46 bii. Salt @ 15^' per bu. 50 " Butter @ 32^ per lb. 17 " Apples @ 45^ per bu. 4 pk. Onions @, 35^ per pk. 61 lb. Pork @ 9/ per lb. 15 bu. Potatoes @ 75^' per bu. 8. Find the total cost : 4 cd. Hard Wood @ $6 per cord. 13 tons Furnace Coal @ $5 i)er ton. 11 " Soft Wood @ $3 per cord. 7- " Stove Coal @ $6 i)er tun. 9 loads Kindling @ $2 \^cv load. 2 " Cannel Coal @ $9 per ton. 9. Find the cost of 7 lb. Tea @ 65^ per lb. 9 lb. Java Coffee @ 31^ per lb. 50 " A Sugar @ 7^- per lb. 52 " Br. Sugar @ bf- per lb. 15 '' Cheese @ 13^- per lb. 25 " C Sugar @ 6^- per lb. 10. What must be paid fur llie following goods ? 7 yd. Prints @ 7^ per yd. 11 yd. Jeans @ !%(/: per yd. 61 " Sheeting @ 13^ per yd. 29 " Calico @ ^

again divide. y. // the division is not exact, write the remainder over the divisor in fractional form, and annex the result to the integral part of the quotient. EXAMPLES FOR I'KACTICE. 117. Divide 1. 646 by 2. 8. 143258 by 11. lo. 7600 by 16. 2. 945 by 3. fK 81052 by 13. 16. 240000 by 13. 3. 1124 by 4. 10. 5841226 by 14. 17. 20416201 by 15 h. 2645 by 5. 11. 90090 by 7. IS. 952451 by 17. • 5. 31562 by 8. 12. 163208 by 15- 19. 200468 by 18. 6. 60703 by 9. 13. 21406 by 8. 20. 1119306 by 10. 7. 2075 by 12. u. 51007 by 11. 21. 8476432 by 12. 118. "When the Divisor is a Composite Number. When the divisor is a composite number the operation may be simplified by usinff tlie factors of the divisor. DIVISION. 31 Example. — Divide 15552 bv 288. OPERATION. Explanation.— First resolve the number 288 into the factors 3 ) 15552 '^, S, 12. Then dividing the dividend by the factor 3 obtain 5184, — the first ciuotient; dividing this quotient, treated as a new divi- 8 ) 5184 1st quotient, dend, by the factor 8 obtain 648 as the second quotient ; again, .. dividing by the factor 12 obtain 54, the third, or final quotient, 12 ) 648 2nd " which is the quotient required. Hence 14.552 divided by 288 „ , equals 54. 04 3rd Rule. — Divide the dividend by any one of the factors, and the quotient thus obtained hy another of the factors,, and so on until all of the factors have been used as a divisor. The last quotient will be the required result. EXA31PI.KS FOK PRACTICE. 1 19. 1. Divide 216 by 72, using the factors 8 j^nd 9. 2. Divide 1100 by 55, using the factors 5 and 11. 3. Divide 5280 by 480, using the factors 4, 12, and 10. 4. Divide 31248 by 144, using the factors 12 and 12. 5. Divide 31248 l)y 144, using tlio factors 9 and 16. 6. Divide 31248 by 144, using the factors 8 and 18. 7. Divide 31248 by 144, using the faptors 8, 2, and 9. 8. Divide -31248 by 144, using the ftictors 4, 2, 3, and 6. .9. Divide 31248 by 144, using the fi^ctors 2, 2, 2, 3, 3, and 2. 10. Divide 2025 by 45, using the factors 3 and 15. 11. Divide 2025 by 45, using the factors 3, 3, and 5. 12. Divide 2025 by 45, using the factors 9 and 5. Remark. — The pupil will observe that tlie order in wiiich the factors are used, does not vary the result. 120. To find the True Remainder after Dividing by the Factors of a Com- posite Number. Example.— Divide 1347 by 105, using tlio factors 5, 3, and 7. OPERATION. Explanation.— Divide the given dividend by 3, _ . . • obtaining the quotient 209, with 2 units for a remain- ^ ' ' t niits. ^gj.. jjjg quotient 269 is composed of units equal in 3 "i or-qs's , 9 ■±. value to 5 times those of the given dividend, and may ___ * be written 2696'8 ; thfe remainder, 2, is of the same 7 ) 89"^'^ + 2''''' = 10 " "°^' value as the given dividend, and is, therefore, a part of the O'we remainder; next divide the quotient 13W5's _,_ 5i5'a _. -vg It 26y^'s by 3 obtaining 89 for a quotient and 2 for a — remainder. The units of whicli tiie quotient 89 is rue rem. composed, are equal in value to 15 times those of •'-"To 5 (juotient. tijg given dividend and may be written SQi^'s ; the remainder is 25's and equals 5 X 2, or 10 units of the given dividend, next divide by 7 which gives the quotient 12, with 5 for a remainder; the quotient 12 is composed of units equal in value to 105 times those in the given dividend and may 32 DIVISION. be written 12'"^'^ ; the remainder is 5'^'^ and equals 15 X 5, or 75 units of the given dividend. The sum of the remainders, 2 units, 2^'^ or 10 units, and 5'^'*, or 75 units, equals 87, the true remainder, and the result of the division, or the quotient, is 12 with a remainder of 87 ; or^ in another form 12x*g^. EXAMPLES EOK PRACTICE. 121. 1. Divide 1121 by 25, using as factors 5 and 5, 2. Divide 819 by -42, using as factors 3, 2, and 7. 3. Divide 1705 by 64, using as factors 8 and 8. 4. Divide 4600 by 135, using as factors 3, 5, 3, and 3. 5. Divide 22406 by 125, using as factors 5, 5, and 5. 6. Divide 53479 by 144, using as factors 12 and 12. 7. Divide 53479 by 144, using as factors 9 and 16. S. Divide 53479 by 144, using as factors 8 and 18. 9. Divide 53479 by 144, using as factors 4, 9, and 4. 10. Divide 53479 by 144, using as factors 4, 3, 3, and 4. 11. Divide 53479 by 144, using as factors 2, 2, 3, 3, 2, and 2. 12. Divide 419047 by 81, using as factors 3, 3, 3, and 3. 13. Divide 341772 by 4095, using as factors 7, 5, 9, and 13. IJf. Divide 792431 by 72, using as factors 6, 2, and 6. 15. Divide 19111 by 24, using as factors 2, 2, 2, and 3. 122. To Divide by 10, or any one of its powers. Since by the decimal system, numbers increase in value from right to left and ■decrease from left to right in a tenfold ratio, it follows tliat to cut off from the right of a number one place, divides the number by 10, two jilaces by 100, three places by luOO, etc. Rule. — From the Tight of the dividend point off as many orders of units, or places, as the divisor contains ciphers. The figure or figures so ■cut off wiU express the remainder. 123. To Divide by any multiple of 10, 100, or 1000, etc. Example.— Divide 16419 by 600. FIRST OPERATION. EXPLANATION. — 6 and 100 are factors , , of 600. First divide 16419 by 100, by l/OO ) 164/19 separating from it the last two figures, obtaining 164 as the first quotient and 19 First quotient 164 — 19, first rem. as the first remainder; next divide 164 by 6 and obtain 27 as the second, or last SECOND OPERATION. quotient, and 2 as the second, or last 6 ) 164 remainder ; multiply this remainder by 100, to obtain its true value, and to the Second quotient 27. -.2 X 100 = 200, second rem. result add the first remainderobtaining 219 rtiA ^or the true remainder. The result of Zly, true rem. ...... . .-**•«-, j the division is a quotient of 27 and a 27|^ required quotient. remainder of 219, or 27Uh DIVISION. 33 B/Ule. — From the right of the dividend separate as many figures as Hhe divisor contains ciphers; divide the figures at the left of the separa- trix by the digit or digits of the divisor, and to the remainder, if there be one, annex the figures first separated from the dividend; the result will be the true remainder. EXAMPLES FOR PRACTICE. 124-. 1. Divide 519 by 40, using as factors 4 and 10. 2. Divide 1164 by 300, using as factors 3 and 100. 5. Divide 2084 by 500, using as factors 5 and 100. Jf. Divide 90406 by 1500, using as factors 15 and 100. J. Divide 83251 by 600, using as factors 6 and 100. 6. Divide 416250 by 9000, using as factors 9 and 1000. 7. Divide 94275 by 3000, using as factors 3 and 1000. 8. Divide 730246 by 11000, using as factors 11 and 1000. 9. Divide 50640231 by 120000, using as factors 12 and 10000. 10. Divide 620974 by 41000, using as factors 41 and 1000. 11. Divide 124689011 by 5910000, using as factors 591 and 10000. 12. Divide 365021467 by 6250000, using as factors 625 and 10000. MISCELLANEOUS EXAMPLES IN SHORT DIVISION. 125. 1. A gentleman left his estate worth $618330 to be shared equally by his wife and five children; what was the sliare of each? 2. A -county containing 400000 acres is divided into 25 townships of equal area. How many acres in each township? 3. $21,735 was received from the sale of a farm at $35 per acre. How many acres did the farm contain? Jf. If a speculator pays $15730 for 715 acres of Nebraska prairie land, and sells the same for $17875, what is his gain .per acre? 6. In New York City, in February, 1882, Hazel walked 660 miles in 6 days, receiving as a prize $20000. Allowing no time for stops, what was his average distance and the average amount earned per hour? 6. Great Britain makes 330 million pins weekly, or 9 for each inhabitant ; what is the number of inhabitants? 7. The dividend is 230304561, the divisor is 15 ; find the quotient and the remainder? 8. The remainder is 7, the quotient 19023, and the dividend 247306 ; what is the divisor? 9. If 8 men can do a certain piece of work in 9 days, in how many days can 12 men do the same work? 10. I sell my village home for $3250, my store for $5000, my stock of goods for $11250, receiving in part payment $8775 cash, and for the remainder Iowa prairie land at $15 per acre; ho-w manv acres should I receive? 3 34 DIVISION. 11. The steamship Servia crosses the Atlantic from New York City to Liver- pool in 150 hours, averaging for the first 24 hours, 18 miles per hour; for the next 48 hours, 17 miles per hour; for the next 30 hours, 19 miles per hour; and for the next 12 hours, 21 miles per hour. If the entire distance be 2841 miles, what was the average distance per hour traveled for the remainder uf the time ? Remark. — Short division, though a mental process, is practicable whenever the divisor ia 35 or less, if the pupil has mastered the multiplication table as given. LONG DIVISION. 1*26. When the divisor is ;i number larger than can be treated mentally, the following method, called Long Division, is employed. Example.— Divide 81437 by 37. Explanation. — Write the terms as in short division, and place a line after the dividend to separate it from the quotient, which is now to be written at the right. Then divide the first two figures of the dividend, 81, by 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. — I. Write the divisor at the left of the dividend with a curved line between them, and another Hive 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 nuniber 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, inultiply 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 siobtracted 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. Vroof.— Multiply the divisor by the quotiext, and to the product add the REMAiNDEB if there be any; the result .should equal the dividend. OPERATION. Divisor. , Dividend. Quotient. 37 ) 81437 74 74 74 37 37 ( 2201 Remainder. 127. Divide DIVISION. EXAMPLES IX LOXG DIVISION. 35 1. 1728 by 48. 11. 115680 by 155. 21. 2. 2025 by 135. 12. 29410 by 251. 22 3. 625 by 125. 13. 666666 by 2144. 23. U- 1920 by 160. u. 93462007 by 1525. 24. 5. ;>268 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. 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 sui)ply the Slieffield market alone ? 4.. 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 Kiml)erly diamond field of 9 acres produced 75 million dollars wortli of diamonds; what average value per acre was produced each year ? Each month ? 8. A bottle thrown overboard into the Pacific Ocean was picked uj) 455 days later, 6700 miles distant from where it was thrown; wliat average distance did it float per day ? 9. The great bell of Moscow weighs 202 tons of 2240 Ih. eacli; if 77 parts of the metal of which it is composed are copper and the remaining 23 parts tin. liow 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 ? 3G DIVISION. t 12. On the planet Neptune 60127 days make one year. A year on Nei)tune equals liow many common years on the earth ? 13. For the year ending September 30, 1887, the exchanges at the Clearing House at Xew York amounted to §34872848786, and those of tlie 36 remaining important cities, $17253855702. What was the average of the exchanges per month at the Xew York Clearing House ? "Wliat was the average per month of the 36 remaining Clearing Houses ? IJf. The Spanish Armada, sent in 1588, by Phillip II. of Spain for the intended coufjuest of England, comprised 132 shijis with 34054 seamen and soldiers. What was the average number with each shij) ? 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 nuide 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 $45089425. 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. 37 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, IG, 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 33-| is the average of 59, 43, 21, 10, and 35, because the sum of these five numbers equals 5 times 33|. 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 &// the numher of terms used. KXAMFLKS FOK PKACTICK. 131. Find the average of the following groups of numbers and prove the results : 1. 20, 24, 52, and 88. i 3. 71, 46, 200, 11, 93, 51, and IT. 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, 813754, 812291, $9267, $12935, $14901, and $20518. What were his average sales per month? 7. An errand boy earned on ]\Ionday 73^', Tuesday 91^-, Wednesday 49^*, Thurs- day 67^', Friday 81^', and Saturday 95^'. What were his average earnings i)er 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 G. 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 lOdo. £XAMPI.£S FOR PRACTICE. 133. Find the complement of each of the following numbers, and ])rove and explain results : 1. 36. S. 115. 5. 81. 1 ''• 1249. y. 28763. 2. 71. 4. 704. 6. 258. 1 <^- 1094. 10. 82041. 38 PACTOES AND FACTORING. FACTORS AND FACTORING. 134. Factors art- 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. 13C. A Composite Number is one that can be resolved into factors. 137. A Prime Factor is n jirime^nwmhev used as vi factor. To aid the pujjil in determining the prime factors of a composite number we give the following Table of Prime Numbers from 1 to lOOO. 1 59 139 233 337 439 557 653 769 883 2 r,i 149 239 347 443 563 659 773 887 3 07 151 241 349 449 569 661 787 907 5 71 157 251 353 457 571 673 797 911 7 73 163 257 359 461 577 677 809 919 11 79 1G7 263 367 463 587 683 811 929 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 2!) 103 193 283 397 499 613 727 839 967 31 107 197 293 401 503 617 733 853 971 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 53 137 229 331 433 547 647 761 881 Remauk.- The pupil can with little labor memorize the prime numbers from 1 to 100. It^ FACTORS AND FACTORING. 39 138. 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 — :; — r~ exactly divisible by the prime number 5, by which divide it; next, observe that 2 J^oo ^jjg 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 ; -|^ -J 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 13 ) 13 constitute the prime factors of 4290. 1 Rule. — Divide by any prime niomber that is exactly contained in the dividend; divide the resulting qiooiient in the same manner, and con- tinue this until the final quotient is 1- The prime divisors will he aU the prime factors of the dividend. KXAMPLKS FOK PRACTICE 139. Resolve 1. 27 into its prime factors. 2. 117 into its prime factors. S. 165 into its prime factors. Jf. 93 into its prime factors. J. 2376 into its prime factors. 6'. 1050 into its prime factors. 7. 144 into its prime factors. S. 15625 into its prime factors. 9. 22464 into its prime factors. 10. 881790 into its prime factors. DIVISORS. 140. An Exact Diyisor 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. 3. Any even number is divisible by 2. 3. Any number ending with 5 or is divisible by 5. 4. 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 Avhen 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 €ach 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 ai» r \ -iQ -IS o- 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. ^>\■.^/^ «,^ o-/^ . -./^ ,..-./i rate the numbers by a short dash. Observe that 2)140 — 210 — 3o0 — 420 — 630 „ .„ ., ■.. .i u c ,u k n J_ 2 will exactly divide each of the numbers, like- g \ f-Q 2Q^ -irrx .-)]() 315 wise that 5 and 7 will exactly divide the successive quotients ; therefore divide by 2, 5, and 7 ; then T ) 14 — 21 — 35 — 42 — 105 observethattheremainingquotients, 2, 3, 5, 6, and -— 15 have no common divisor ; hence the divisors ''^ — 3 — .T — 6 — 15 2, 5, and 7 are all factors of the greatest common divisor, which is 70. Rule. — I. Write the nuvibers in a horizontaZ line, separating thein hij a dash. n. Divide by any number that mill exactly divide all the numbers given, and so continue until the quotients have no common divisor. HI. 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 givea numbers will be the greatest common divisor. EXAMPLKS FOR PRACTICE. 146. Find the greatest common divisor of 1. 22, 55, and 99. 2. 24, 36, 60, and 96. 3. 32, 48, 80, 112, and 144. A. 54, 72, 90, 126, 180, and 216. 7. 252, 630, 1134, and 1456. 8. 2150, 600, 3650, 1000, and oOO. 9. 302, 453, 755, 1057, and 1661. 10. 126, 441, 567, 693, and 1071. 5. 104, 156, 260, 364, and 572. 11. 210, 350, 280, 840, and 1260. 6. 135, 450, 315, and 585. I 12, 200,325, 525, 350, and 675. MULTIPLES. 41 147. "When no Common Factor can be Determined by Inspection Example. — What is the greatest common divisor of 182 and 858. OPERATION. 182 130 52 52 858 728 130 104 26 Explanation. — Draw two vertical lines and write the numbers on the right and left. Then divide 858 by 182, 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. 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 numher by the less, the divisor by the remainder, and to continue until nothing remains. The last divisor will be the Greatest Common Divisor. 148. 1. 316 and 664. 2. 96 and 216. 3. 1226 and 2722 4. 1649 and 5423 KXAMPLES roil PKACTICK Find the greatest common divisor of J. 1377 and 1581. 6. 92 and 124. 7. 679 and 1869. <^. 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 Commou Multiple is a number exactly divisible ])y two or nioie given numbers; as, 12 is a common multiple of 6, 3, and 2. 151. The Least Commou 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. — l. The 2^roduct of two or more numbers, or any number of tim"s their product, must be a common multiple of the numbers. 2. Two or more numbers may have any number of common multiples. 3. A multijjle of a number must contain all the prime factors of that number. 4- T/ie common midtiple of several numbers must contain all the factors of all the numbers. 5. The least common multiple of ttvo or more numbers is the least number that will contain all the prime factors of the numbers yiven. 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. " 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 tlie 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- q \ Q 4 63 15 *^®°^ ^'°*^ ^°*^ 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 mul- ~ T I I 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. — 1 Write the nunibers in a horizontal line, separating them by a dash. 11. 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. in. The product of the divisors and prime remainders is the Least Common Multiple. 154. Greatest Common Divisor and Least Common Multiple Compared. I. Tlie greatest common divisor is the product of all the pj^me factors common to all tlie 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. 3. 6, 16, 26, and 36. 4. 42, 210, 56, and 35. I 7. 18, 80, 99, and 120. 5. 5, 30, 24, and 28. I 8. 2, 3, 4, 5, 6, 7, and 8. 6. 11, 32, 216, and 66. I 9. 21, 72, 24, and 30. CANCELLATION. 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 f may be read 2 divided 4; divide both terms by 2 and the result is 1 divided by 2, or -|. 2 X 27 Again, mav be read 2 times 27, divided bv 4 times 18 ; reiecting the 4 X 18 " factor 2 from the 2 in the dividend and from the 4 of the divisor, also the factor 9 from the 27 of the dividend and the 18 of the divisor, gives =■ : — , Mxi^2 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 divisor, thus : = , then reiecting 2 and 9 from both terms, 4 X 18 2 X 2 x 9 X 2 or cancelling, obtain = f Ans. ^ X 2 X X 2 157. 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 bv 6 X 11 X 4 X 7 X 35 X 17? OPERATION. Explanation.— Cancel 6 from the divisor and $x2x2$X^XlxW3 3x2 from tlie dividend; 4x7 from tlie divisor '■ = y5j. and 28 from the dividend; the 35 from the divisor ^ XllX'ixIl XUXU and 5x7 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 j\. Remark. — This principle can be put to frequent and valuable use in a great variety of business computations. Rule. — I. Write the divisor helow the dividend iinth a line separating them. II. Cancel from the dividend and divisor all factors common tu both; then divide the product of the remaining factors of the dividend hy the product of the remaining factors of the divisor. 44 CANCELLATION. EXAMPLES FOK PRACTICE. 158. 1. Determine by cancellation the quotient of 5 x 9 x 2 x 13 x 40 x 6 divided bv 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, 3. Divide 210x 9x 78x 5 x23 X 10 X 36 by 13x144x40x3x27 X5x400. 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 2 X 3 X 4 x 5 x 6 x 7 X 8 x 9 by 23 x 45 x 67 x 89. 8. Divide the product of the numbers 98, 76, 54, and 32 l)y 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^ })er 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 ? IJ^ 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 75^ 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 878 per head ? Rem.vbk. — A section of land, in the Unitefl 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 |i78 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 jirairie land at $8 per acre should be given for ap Ohio farm of 272 acres at $45 per acre .'' FEACTIONS. 45 FRACTIONS. 159. A Fraction is one or more of the equiil 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 §. 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 jorojoer or Improper. 166. A Common Fraction is one expressed by two numbers, one written above the other, Avitli a line between. 167. A Proper Fraction is one whose value is less than 1, the numerator being less than the denomimdor. \, f, \, f, ^i, -^ ^vo. p)roper 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. |, |, \, -|, V' St ' rt ^^"6 improper fractions. 169. A Mixed Number is an entire or luhole number and n fraction united. 2i, 5f, 91, 14|3-, lOTfl 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 of fractions to be performed. ^ ^^ * complex fraction and indicates that f is to be divided by S ; ' 5 * the expression is read f s- 5 ; — , and ^ are also complex fractions. i 8 Principles. — 1. Multiplying the numerator multiplies the fraction; dividing the numerator divides the fraction. 2. Multiplyhuj the denominator divides the fraction; di aiding the denominator multiplies the fraction. 3. Multiplying or dividing both terms of a Jradioti Inj the same number does not change the value of the fraction. 46 REDUCTION OF FRACTIONS. REDUCTION OF FRACTIONS. 171. To Reduce a Whole Number to a Fractional Form. Example. — Reduce 3 to ;i fraction the denominator of wliicli is 7. Explanation.— The fractional unit having 7 for a denominator is 4 ; 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 ichole number hy the required deiwrninator, and place the product over the denominator for a numerator. EXAMPLES FOR PRACTICE. Wi. 1. Reduce 5 to a fraction the" denominator of which will be 4. J. Reduce 7 to a fraction the denominator of which will be 9. J. Reduce 4 to a fraction the denominator of Avhich will be 13. ,'/. Reduce 3 to a fraction the denominator of which will l)c 8. '). Reduce 8 to a fraction the denominator of Avhich will bo 12. 6. Reduce 15 to a fraction the denominator of which will bo 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 1)e 11. 9. Reduce 416 to a fraction the denominator of which will be 23. JO. Reduce 1125 to a fraction the denominator of which Avill 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, whicli 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, ^ = 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. KXAMPLES POK I'KAi'TICK. 174. Reduce 1. 3^ to an improper fraction. 2. 7| to an impro])er fraction. 3. 10| to an improper fraction. Jf.. 43f to an improjier fraction. J. 16^ to an improjyer fraction. 'j. 78f to an improper fraction. 7. ^^^i^ to an improper fraction. 8. 170^L. to an improper fraction. •9. KMOg'y to an im])roper fraction. ]i>. 96Sj^ 1o an improper fraction. 175. To Reduce an Improper Fraction to a Whole or Mixed Number. Ex.\MPLE. — Reduce -/ tu a whole or mixed number. Explanation. — Since 4 fourths make 1 unit, 23 fourtlis will make as mrny vmits a.s 4 is contained times in 23, or 5 times with a remainder of 3, or three-fourths; therefore,-," = 5f. REDUCTION OF FRACTIONS. 47 Rule. — Divide the numerator by the dertominator, place the remain- der, if any, over the denominator, and annex the fraction thus found to the entire part of the quotient KXAMPtES FOK PRACTICE 176. Reduce 1. 1^ to a whole or mixed number. 2. Y to a whole or mixed number. 3. ^ to a whole or mixed number. 4.. ^^ to a whole or mixed number. 5. y^^ to a whole or mixed number. 6. \y> to a whole or mixed number. 7. i||i to a whcle or mixed number. 8. |-| to a Avhole or mixed number, ^- Vs¥ ^^ ^ whole cr mixed number. 10. 1 Iff" to a whole or mi xed number. 177. To Reduce a Fraction to its Lowest Terms. Example. — Reduce y\ to its lowest terms. Explanation. — By applying the principles of factoring, change the form of the fraction 2x3 JL 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 tlie new denom- inator; the resulting fraction is i. Or, observe that 6 is a factor of both the terms and that i is the result of dividing both the terms by 6. Rules. — 1. Divide both terms of the fraction hy their greatest common ■divisor. Or, 2. Divide both terms of the fraction hy any common factor, and con- tinue the operation ivith 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 terras. 2. If both terms of a fraction be divided by their greatest common divisor the fraction will be reduced to its loirest terms. This is the only use in practical arithmetic of the theory of the greatest common divisor. EXAMPLES FOK PRACTICE. 178. 1. Reduce |^ to its lowest terms. 2. Reduce ^f to its lowest terms. 3. Reduce |-§- to its bwest terms. Jf. Reduce -J-f to its lowest terms. 0. Reduce -^-^ to its lowest terms. 6'. Reduce -^^^ to its lowest terms. 7. Reduce -^^ to its lowest terms. 8. Reduce |||| to its lowest terms. 9. Reduce ||^ to its lowest terms. 10. Reduce m to its lowest terms. 179. To Reduce a Fraction to Higher Terms. Example. — Reduce f to a fraction the denominator of which is 21. Explanation. — Since 7 is contained in 21 three times, the given fraction maybe reduced to a fraction whose denominator is 21, by multiplying both of its terms by 3; multiplying 22L? gives l\, the required result. This operation does not alter the value of the given fraction. Rule. — Divide the required denominator hy the denominator of the given fraction and multiply the niimcrator hy the quotient thus obtained; write the product over the required denominator. 48 REDl'CTIOX OF FRACTIONS. EXAMPLES FOR PRACTICE. 180. 1. Reduce f to a fraction the denominator of whicli is 15. 2. Reduce ^ to a fraction the denominator of which is 36. S. Reduce -^ to a fraction the denominator of which is 42. J^. Reduce f to a fraction the denominator of which is 32. 5. Reduce -^j to a fraction the denominator of which is 88. 6. Reduce -^ to a fraction the denominator of whicli is 52. 7. Reduce 4^ to a fraction the denominator of which is 115. S. Reduce -^ to a fraction the denominator of which is 128. 0. Reduce /|- to a fraction the denominator of which is 102. 10, Reduce l\ to a fraction the denominator of which is 147. 181. To Reduce Fractions to Equivalent Fractions Having a Common Denom- inator. Example. — Reduce f, ^, -|, 4, to equivalent fractions having a common denominator. EXPLAXA.TION. — 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; f = ^*g, | = iyg, | = \\^, «nH 4 — ISO ana , — jto- Rule. — Multiply together the deivominators of the given fractions for a common deiiominator. Multiply each jiuDveT'ator hy all the denomina- tors except its aim and irrife the several results iti turn over the common denominator. Remark. — "Where one or more of the given denominators are factors of the others, the smaller may he rejected. EXAMPLES FOR PRACTICE. 182. Reduce to e i. and -iV 3. A» h \, 2, tV. h and -|. -4. n, 5, h \h h ■i\, and -J. ^. fa, 51, t3„, 38i, 23|, and]2. 6. 44, 5i, 13if, 0, and 11 J. ' • h I, II, \h If, and \. 8. 2^, 7-2,llt, 23^V, i I, and5. 9' tl, f, fj, 8, iiil^, andf 10. A, 4, ^l, -j^^gV, H, t and 20. 1S:J. To Reduce Fractions to Equivalent Fractions Having the Least Com- mon Benominator. The Least Common Denominator of two or more fractions is the least denom- inator to which tlu'V can all Ije reduced, and must be the least common multiple of the given denominators. Example. — Reduce \, |, I, \, -^^, and ^ to equivalent fractions having the least common denominator. OPERATION. Explanation. —Find the least 3, 2, ) 9 — 15 — 4 \ = -j3g^. ^ = -jSJi^. common multiple of the given de- 3 5 2 i ^= J4-'' 4 = "-" nominators for the least common igo- L — isu- denominator, which is 180. Then 3X2X3X5X2 = 180. -j^ = ,%■ f = i-J;;. by Art. 179, reduce each of the ^ven fractions to a fraction whose denominator is 180. ADDITIOX OF FRACTIONS. 49 Rule. — I Find the least common multiple of the given denom,inators. II. Divide this inidtiple hy the denmrviaator of each of the given frac- Mons, and multiply its uum^erator hy 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 PKACTICE. 184. Eeduce to equivalent fractions having the least common denominator: 7. 23i, 14f, 7^, 5f, and f . 8. 17, 2f, 14i, 8f, 3^\, and 5. T5 3j f J ?5 "B"' T> 8j TTj ^^^ A- 1- h h h and f . 2- h h h h and |. S. h -h, "!> h ^> and -jL. 4. A, i, h h 2, H', and 4. ^. h h tV -^> h ¥. and 5. ■6. A. i^ 11' n, n, ^, and 1 9. 10. 11. 12. A, h U, \h 8, iV 3i, in, and ^. it, iV i 14, 1, il, Vr«, W, 5f, 11. il, -H, V, \h 3, 2i, 4^3^, 2, tf ADDITION OF FRACTIONS. 185. To Add Fractions having a Common Denominator. Example. — Find the sum of |,, I, |, J, and |. OPEKATION. + i + «9 + i + t = V V = 2i. 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 coinmon denominator ; if the result be an improper fraction reduce it to a whole or mixed number. EXAMPLES EOK PRACTICE. 186. Add ^. f, f, h i h and f ^- if , tV f^, H, t\, and tV- 3. is, e, -h, i-i, H, il, and H- ^- -h^ -h, A. t't, t^, H» and |4. 5- A, A, t\. H, a, 14, and f 1 6. 7. TT' fi ' 2 T ITT' and -f^. ii, ih, A, ii, ih ih ii, and If. A, ii, "jV, "3%, "2??, il, if, and |f. i^, A. ii, if, If, If, ff, and i^ 187. iTo Add Fractions not having a Common Denominator. Example. — "What is the sum of f, 4, and 4. OPERATION. Explanation.— Since the given fractions are not of the 2 , B^ I 1 same unit value, reduce them to a common denominator ' ' (Art. 181), and writing their equivalents below, add their f t "I" fr + ti ^^ f I =^ lf4' 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 50 AUDITION' OF FRACTIONS. EXA3IPL£S FOK PKACTICE. 188. Add 1- h I, h h I, \, and f. ^- H» f^' i' I' i' ^' and yV ^- -fr. ft. f. fi. i, i' and \\. 4. f , f , 1*0, \, -h> and i ^. h h h h \, -h^ and \. 6' n, i, -fo, h h h H. and -^. - h h t'«V, iVf^ if. i. h and i. <^- H. ^> h tV. iuS I. h and f ^'>. f, f, §, iS. A. i. and ^K. 10. h h ii I. I. *, t'o, and WL. 189. To Add Mixed Numbers. Example. — Find the sum of 2^, ^, 4, and f. OPERATION. 2^ + ^ + 4 + 1 = 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. 170, add, and reduce results. For convenience the fraction.? 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 + 45 4 i TO i 3C 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 Vo"i reducible to 1 J J; this added to the integers gives 7f^, the sum as before found. Rules. — 1- Reduce mixed numbers and integers to common fractional forms and then to coinmon denominators. Add their numerators, place the result over the common denoTninator 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. KXAMPL,ES FOK PRACTICE 190. Add 1. 4, 4, I, i, h 3, \, H, and 11. 2. I, 2i, \, h 6, h h ^n, and 4^. 3. I, h ^, n, 5, 3i, h n, and |. Jf. 2, 5|, -I, n|4, i, 14, 20iand 4. 5. \, 3, 64, I, 19, 75i, Vj, and -^,. 6. 7. S. 0. 10. 3t, 10, 21^, 42^:, 84i, and 168^^. m, WA, 50f^, 29H, and 86^- 591^, 103e, 5oi, 400, and 96^- 10.3^^, 1191^, 2974, and 188i|. 33^ 15, 124, Ci, 25, and l^. Remark. — In invoices of cloth, «fec., account of fractional parts is made only in quarters and merely the numerators are written; as, 5* = 5|, 3^ = 3^, 13* = 13|, etc. SUBTRACTION OF FRACTIONS. 51 Miscellaneous Examples in Addition <»f Fractions. EXAMPLES FOR MENTAL PRACTICE. 191. What is 1 1. \,h \, and 1 2. hh i, and f 3. i, i, h and f Jf. 4 S t' IT' f , and 4 5. f, f , I, and |. ^- 3j To' ?» '"^"- To* 7- i, f , f, f, I, and H. '^. i, h h h i^, and h 9- i, h h h -h, and H. ■^^- 1. i. tV, H' ^, h and If ^^- I' h Ih \h and |3. 1^- i, !,f,ii, 1,1,1, i, and f. EXAMPLES FOR "WRITTEN PRACTICE, 192. Add 1. 130f, 69§, 600^\, :3044J, and 4G. j ^. 900, 47i, Sf, 4, 29f, 06^?^, and 4. ^^ 80t^, 2^, 5f, 17, 41^, 83^, and 14f. 5. 16|, 33^, 66f, 88^, 100, and llGf. ■3. 28A, 85-jV, 60^^, 400, 20|, and 11. 1 6. 18f, 65|, 161f, 67f, 23^, and 75. 7. The six fields of a farm measure respectively, 10, 12|, 19^, 26^.^, 30^ J, and 2^ acres. How many acres in the farm ? 8. Ten sheep weighed as follows : 90yV, HOi, 89f, 100, 106|, 101^, 96, 99, 113f, and 198^ lb. respectively. What was their aggregate weight ? 9. A farmer sold 3G0| pounds of pork, 167f lb. of turkey, 241| lb. of chicken, 690{f lb. of butter, 475 lb. of lard, a cow's hide weighing 97f lb., 71|-lb. of tallow, and three quarters of Ijcef weighing respectively, 161:J-, 187-i, and 190 lb. How many pounds in all had he to deliver ? 10. For 341f bushels of wheat I received $375^, For 597| bushels of barley I received $500i, For 11204- bushels of oats I received $619f, For 316^ bushels of buckwheat I received 1200^^, For 250 bushels of beans I received 1525-j''^, For 1386^ bushels of potatoes I received $755|^, For 1050^ bushels of apples I received 1301 -j^. For 630^ bushels of turnips I received 163^^^. 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 f from |, OPERATION. Explanation. — Since the fractions have a common denominator, tiieir g ^ difiFerence may be found by taking the numerator 3 from the numerator l-f 5, and placing the difference 2, over their common denominator Rule. — Subtract the numerator of tlie subtrahend from that of the minuend, and place the difference over the coimnoih deiioDiinator. 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. 194. EXAMPUES FOR MKNTAL PRACTICK. What is the difference between 1. \ and \. 2. ^ and f . 3. W and A- 4. \\ and T*j. I r. V and f . | 76». \\ and y^. 5. I and \. 8. \l and i^. 11. 1 and if 6. 1 and i 1 V. fa and if I 12. 1 and y^^. 195. Subtract 1. fl from f|. 3. ft. from ^. EXAMPL.KS FOK WRITTEN PRACTICE. s. \^ from 1. •H ^'•oni f|. Kl from i^^. P. 11. 12. \\ from ^f tVo from ifa. SV from ^i. II from SV- 13. U- 15. 10. 13. SV '»n<^ T^- U. 4" and f i.'J. 1 and 4^. ff from ^. H^fromH^. \WiYom\. yV from 3. 196. To Subtract Fractions not having a Common Denominator. Example. — From f take |. OPERATION. Explanation. — As the denominators indicate the kind of parts, and only like things can be taken the one from tlie other, it follows that before tlie subtraction can be performed, the fractions must be reduced to a 1-1 = \» — To ^^ If'o- common denominator ; then, the difference between the resulting numera- tors, placed over the common denominator gives /^ as a result. ^u\e. — Eeduce the given fractions to equivalent fractions having a common denominator. Subtract the numerator of the subtraJiend from the mimerafor of the minuend, and uidte the result over the common denominator. Remark. — Improper fractions may be treated in like manner. EXAMPIES FOR MENTAL PRACTICE. 197. What is the diflFercneo between i. i and f ? \ Jf- i and \ ? 2. i and f ? 3. I and f ? 5. y^andi? 6. \ and f ? 9. f and J4 ? i and l"? HandtV? 10. 11. 12. yV and I ? H and If ? M and \\ ? 198. Subtract 1. I from -J. 2. I from |. 3. ^j from 4- 4. II from V- EXAMPLES FOR AVRITTEN PRACTICE. Find the difference between 9. if and |. 13. \\ and i. 10. V and 1. u. 1^ and V. 11. V and yV 15. 1 and f 12. U and V- 16. 1 and |. Fnnn 5. I take -J. 6. \^ take f . 7. V take V- <^. il take i. 199 To Subtract Mixed Numbers. Example. — From IGJ take- 11 J. OPERATION. Explanation. — Reduce the fractions to a common denominator. 16| — llf = Ob.serving that the -^^ of the subtrahend is greater than the y\ of the minuend, take 1 from the 16 of the minuend, reduce it to twelfths (|f), and adding it to the ^\ obtain \% ; from this take the -f^ 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^. Rem.\rk.— In case the minuend is integral subtract 1 and reduce it to a fractional form of the required denominator. 10,^ = 15^, i1t«V = 11-/2- SUBTRACTION OF FRACTIONS. 53 'Rule. -^ Write the subtrahend underneath the minuend. Reduce the fractional paHs 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. EXAMPIiES FOR MENTAL. PRACTICE. 200. What is the difference between 1. 2. S. 4. 6i and 2^ ? 5| and 3i ? 12 1 and 34 ? 17Hand^? 3i and 5^ ? 8f and llyV ? 14f andSlfJ? 6 J and 14|^ ? 9. 10. 11. 12. 3^ and 12^ ? 17-tV and b\ ? 21iandllf ? 9^and23H? IS. 17-iVand22^? U. 12|and3Ji? 15. 113f andUi? 16. 215 iV and 45 1? EXAMPLES FOR WRITTEN PRACTICE. 201. From 1. 4i take 1\\. 2. 18TVtake5f S. 79i take 491^. ■4. 104A " 84f Subtract 9J I from 11. 20 from 56f ^. 41 II from 50|. Find the difference between 9. 10. 11. 12. 240f and 89^. 210if and 250. 200 and 1\^\. 11444 << 5^0 13. 117^and5Tf. U. 95-1^ and 383-.V -?/>. lOoOf^ and 2020|. 10. 2016| and 2503||. EXAMPLES REQUIRING THE USE OF THE PRECEDING EXPLANATIONS. 202. From the sum of I and -J take |. I and f take \\. f and f take \\. I and 2| take 4. '^ and 9 1 take 34. 5f and 4^ take 9. 1. ;■) ^. 3. Jf- 5. 6. 13. u. 15. 10. 17. 18. 19. 20. 21. 22. 23. Take the sum of ^ Subtract the sum of 7. -j\ and y from 14f. 8. 8i and ^ from 20. 9. f and 5 from 11^. 10. 18f and IS^V from 100/,. 11. 20^V and 1 5^^ f i"om 40-gi^. 12. 201 ij and 87f from 304 i. and f from tlie sum of 24- and |. Take the sum of 3| and y from the sum of 4 and 8|. Take the sum of 20 and 14yV from the sum of 18f and 19^^. Take tlie sum of 28y5^ and GO^ from the sum of bO\ and 401. Take the sum of 100^ and 28| from the sum of 60/^ and GO^V Take the sum of 13^ and 46 from the difference between 125 and 1 1\. Take the sum of 216| and lOl^from the difference between 1000 and 875- Take the sum of 45^ and 25^ from the difference between 305f and 425^. Take the sum of 23 and 41/„ from the difference between 21^ and 93^. Take the sum of 91 and 5f from the difference between lig- and 19|. From 21G^ acres of hind, lots of 21 A, IGf A, 2Gj|- A, 41.} A, and 63f acres were sold. How many acres remained unsold ? 2Jf. A lady went shopping with 135 J in her purse. She- expended for car fare, ^ of a doll?,r; for thread, |J- of a dollar; for gloves, 'i\ dollars; for a hat, b\ dollars; for a clock, 21| dollars; and invested the remainder in linen. How much was paid for the linen? 54 MULTIPLICATION OP FRACTIONS. 25. A dealer bought u farm for $3685f , the crops for $887^, the stock for 11015^, and the utensils for $602^. He sold the entire i)roperty for $6425^%. Did he gain or lose and how much ? 26. From the difference between 280^ and 1200^ take tlie sum of 20^, 16|, 5|^, 86, UH' it SS ^iid 100. 27. From the sum of 80|, 70^, GOf, and 1 take tlie difference between IJ and 101. 28. From the sum of 4, f , ^, \, |, f, I, |, and -jV take the difference between V and 2. 29. Haying $1302| in bank I drew checks for $204|, ^ISOi^^^, $G40^, 1^82^, $20, 130^, and $100. How much remained to my credit in the bank ? 30. A town owing $38246|, paid, in '35, $9304f ; in '86, $12000^ ; in '87, 14250^ ; and in '88 the remainder. How much was the jiayment of 1888 ? 31. If I jiay $3500 for a house, $346| for repairs, $1126f for furniture, I4U0^ for carpets and curtains, and sell tlie entire property for $5000, how much will I lose ? MULTIPLICATION OF FRACTIONS. 203. To Multiply a Fraction by a Whole Number. Example. — Multiply f by 4. OPERATION I. Explanation — Since the numerator is the divi- 3 V 4 A >>^ 4 :— : ^_ll__ _- 1 2 _ ^4 _ -j^i dend, the fraction may be multiplied by multiplying 8 the numerator 3 by the multiplier 4 ; the product is ^/, which reduced gives 1|, or li. Or, since the OPERATION 11. denominator 8, is the divisor, the fraction may be 3 r, 1 , multiplied by dividing this divisor 8, by the multi- plier 4, which will give the reduced form, | = 1^. This introduces the principle of cancellation into or, -j; X ^ = 2 = li' fractional operations. X 4 = = t = U ; 8--4 * For examples containing concrete numbers, reason as follows : Example. — If one pound of wool costs f of a dollar, what will be the cost of 21 i)ounds ? Explanation.— Since one pound costs | of a dollar, 21 pounds which are 21 times 1 pound will cost 21 times | of a dollar, or 7| dollars. Bule. — Multiply the numerator, or divide the denominator, by the whole munber. Remark. — To economize time and space divide the denominator or cancel when it can be done, as the numbers to be treated are tlius put into simpler form. MULTIPLICATION OF FRACTIONS. EXAMPLES rOK MENTAt PRACTICE. 204. What is the product of 1. I multiplied by 3 ? \ 6. 2. \ multiplied by 2 ? | 7. 3. % multiplied by 4 ? \ 8. Jf. j\ multiplied by d ? 9. S. -^ multiplied by 7 ? 1 10. f multiplied by 3 ? \^ multiplied by 5 ? I multiplied by 8 ? -j\ multiplied by 15? -^^ multiplied by 6 ? 11. 12. 13. U- lo. 205. Multiply 1. ^ by 85. 2. 41 by 8. ^. V by 12. 4. V by 9. KXAMPI.ES FOR WRITTEN PRACTICE. 6. 3J by 16. V by 11. i|4 by 40. V by 28. 9. 10. 11. 12. V/ by 10. \V by 57. t\ by 21. H' by 20. 55 \ multiplied by 12 ? 1^ multiplied by 5 ? f multiplied by C ? y\ multiplied by 30? ■^ multiplied by 32? 13. II by 115. IJt. \V by 49. 15. ^?jijbyl05. 16. ^\ 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 Frattion. Example. — Multiply 6 by yV Operation I. Operation II. ■jiV X = f = 3^. Operation III. !x^ = | = 3^. 3d Explanations. — 1st. If the multiplicand 6, be multiplied by 7, the numerator of the fraction yV, 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 If = S^-V = 3*. 2d. Since 6 and -^^ 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. 203. Place 1 as a denominator for the multiplicand, theu cancel and reduce. Rule. — Multiply the ivlvole numher hij the numerator of the fraction, ■and divide the product by the denominator. Cancel when possible. EXAMPLES FOR MENTAL PRACTICE. 4. 9 multiplied by W. 5. 14 multiplied by |. 6. 207. What is the product of 1. 5 multiplied by f. 2. 7 multiplied by f 3. 6 multiplied by |. 22 multiplied by ^. 7. 40 multiplied by f. 8. 9 multiplied by |. 9. 4 multiplied by \\. 10. 7 multiplied by |f . 11. 15 multiplied by ^. 12. 21 multiplied by \\. 13. 12 multiplied by \. H. 18 multiplied by |. 15. 42 multiplied by 4- 208. Multiply. 1. 81 by f 2. 56 by |. 3. 61 by f J^. 105 by \\. EXAMPLES FOR WRITTEN PRACTICE. 19 by V/. 22 by i||. 240 by \l. 8 by li. 9. 10. 11. 12. 27 bv f 210byfi. 48 by H • 91 by i|. 13. U. 15. 16. 71 by H. 203 by ^. 415 by V. 672 by f. Remark. — Entire or mixed number can be treated with facility by reducing them to fractional forms and cancelling when possible. 56 MULTIPLICATION OF FRACTIONS. 209. To Multiply a Fraction by a Fraction. Example. — 1. (Abstract). Multiply f by 4. Operation. Explanation.— The multiplier, 2, is equal to 3 times ^ |X| (3 times ^, or 1 of 3. By application of 203, multiply ? bj- 3, which ^ =z -I or gives ?, -which must be seven times too large, since the ( 1^ of 3. multiplier -n-as not 3, but one-seventh of 3; hence, the -| X 3 = -|; -1x4- = -^. conect result will be obtained by dividing the product, 0, by 7, which gives /^ . Example. — 2. (Concrete). If a pound of tea co^s ^ of a dollar, what will ^ of a pound cost ? Explanation.— If a pound of tea costs | of a dollar, i of a pound will cost ^ of | of a dollar, or ^V of a dollar; if J^ of a pound costs ^\ of a dollar, i which is 2 times i, must cost 2 times j\ of a dollar, or -j^ of a dollar. Example. — 3. If a yard of cloth costs 1|- (or y ) dolhirs, what will | of a yd. cost ? ExPLAN.^TiON. — If 1 yd. costs y dollars, ^ of a yd. will cost ^ of 5/, orf of a dollar; and if J of a yd. costs I of a dollar, i which is 4 times |, must cost 4 times f , or § = 1^ dollars. REM.4.RKS — 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- Cujicel all equivalent factors from the numerators and denojtiinators. II. Multiply togetlier the remaining nmnerators for the numerator of the product, and the reniniiiiiig dcuomijiators for the dcnomiiiator of the product. EXAMPLKS rOK MENTAL PRACTICE (ABSTRACT). 210. Find th( 1. \ and 4. ?. -I and |. ■?. I and \. 4. \ and |. luc t of Multiply 5. 1 and I 9. ^byf 6. f and ■^. ^0. ifi^yf. 7. ii and f . 11. Mhyf. s. ■1 i^y T-v- ]2. f3-bv|. IS. U. 15. 16. Hbyl. EXAMPLE.S for mental practice (CONCRETE). 211. 1. "What will be the cost of | of a pound of tea, if the cost of a pound be f of a dollar? 2. I bouglit f of an acre of land and sold ^ of my purchase. What i)art of an acre did I sell? 3. What will be the cost of -J of f 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 | as much. What part had James? 6. If :}: of a stock were lost by fire and the remainder sold at | of its cost, what part of the first cost was received. DIVISIOX OF FRACTIONS. 57 7. Divide 20 into two parts, one of which shall be f of the other. 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 2)ound is sold for | of its cost. For what price per pound is the tea sold? 10. Coffee costing -^^ of a dollar per pound is sold for ^ of its cost. What price is obtained for the coffee? 11. What is ^ of f of a yard of cloth worth, if the entire yard costs f of a dollar? 12. If a pound of steak costs -^-^ of a dollar, what will \ of ^ of a pound cost? 13. After paying ^ a dollar for a pound of nuts, I sold -| of | of my purchase at the same rate. How much did I receive ? H. From a gallon of oil f of | of a gallon leaked away. AVhut 2)urt of a gallon was left? EXAMPI.KS FOK IVKITTKN PKACTICK. 212. Multiply together 1- iil, A, II. ^"tUi- 2. 3|, I, I, 21, and llf. 3. I, I, U, ll^V V, and 20. 4. 12f, \^, 21|, 60i, and fg. 5. 5^, 8, ll^io, 17i, 25f, and 6. 6. ^, 19|, 29f, 39|, and 49. 7- h h h i, h h I, and 10. <^. 200^, 187^, 40yV, and 51f 9. 500, 186y\, 63f, 41, 19f, and 4. 10. 23^^, 16f, 8|, and 12f 11' I, 1^, li ¥, V.andH- 1'- h h i, h -^, and 60. 13. 13^, 21f, 12, 4, 2|, and 3^. U- -R, if, 5.1^, H, audi 15. What will be the cost of 74^ tons of hay at f of 15^ dollars i)er ton? 16. Having bought |- of a farm of lOG^ acres, I sold f of my ])urchase. How many acres do I sell? 17. I bought a house for $21654^ and sold it for -^1^ of its cost. How many dollars did I lose? 18. If 71" barrels of flour be consumed by a family in ten months, hoAV many barrels would fifteen such families consume in 134 months? 19. Having bought 2^ tons of coal at -f of IGf dollars per ton, I gave in i>fly- ment a twenty dollar bill. How much change should I have received ? 20. If 17^ cords of wood are bought at ^ of 13| dollars per cord, and sold at I" of 9^ 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, dividin": the num- , erator (dividend) divides the fraction; hence, divide the numerator 4 of the * "^ ^" fraction i by 2, and the quotient is f . 58 DIVISION OF FRACTIONS. Example. — •->. cost ? OPERATION. If a jjouud of tea costs f of a dollar what will 4 of a i»ound ExPLASATiox. — If one pound costs ^ of a dollar, i of a pound will cost i of ^, or f of a dollar. Observe that in this operation the mul- . .?. 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 | by 7. Explanation. — ^By the Greneral Principles of Fractions, if we multiply the denominator we divide the fraction. Therefore, | -=- 7 = /j. Rule. — Divide the incvi£rator or jnultiply the dejwmiiiator hy tlie 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 OPERATION. 8)2815f 351|| —4. Divide 28151 by 8. Explanation. — (Short Method). — Write as in Short Division and divide; 8 is contained in 2815j, 351 times with a remainder of Tj^not divided; reduce the 7 to thirds and to the result add the ^ making -/, which divide by 8, obtaining ?,| as the fractional part of the quotient; annex this to the integral part which gives 351|J. EXAJHPI.ES FOR MEXTAX PKACTICE. 211. What is the quotient of 4 divided by 3 ? J I divided by 4 ? -,«,- divided by 9 ? ^\ divided by 8 ? 5. 6. 9. 10. 11. 12. f divided by 4 ? |4 divided by 7? 14 divided by IS ? fl^ divided by 22 ? H" divided by 5 ? \\ divided by 7 ? J Of divided by 15 ? II divided by 13 ? IS. If a pound of powder costs f of a dollar, what will ^ of a pound cost ? IJf., Having \\ of a yard of cloth, I divided it into 7 equal pieces. How much cloth was there in each piece ? 15. If 1^ of a farm be grain land, and evenly divided into 3 fields, what i)art of the farm will each field contain ? EXA3IPLKS FOK WKITTEX PKACTICE. 215. Divide 1. |^by9. 5. -A\by8. 0. 308 Jj- by 40. 13. 205|- by 6. 2. 2i^ bv 17. 6'. 21i by 5. 10. 10003Vby41. U- 185 1 by 9. s. |Hbyi2. 7. 41^ by 11. 11. 16f by 5. 15. 112^ by 8 Jf. 3ff by 21. 8. 2096 1 by 21. 12. 108^ by 25. 16. 321| by 6 DIVISION" OF FRACTIONS. 59 216. To Divide a Whole Number by a Fraction. Example. — Into liow many pieces f of a yard eaub, may 5 yards of ribbon be cut? OPERATION. Explanation. — Since 5 yd. equals ^f yd. Ibey may be cut iulo as o -^ ^ =: many pieces, each containing f yd. as | is contained times in 'j^Vhich is y ^ 2 _ 7|^ 7i 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 hy the whole nuinber, and divide the result hy the numerator. <^r, 2. Reduce the ivhole number to a fraction .of the same denoinination as the divisor, and divide the numerator of the dividend hy that of the divisor. KXAMPLES FOK MKXTAL I'KAtTICK. 217. Divide 1. 17 by f 6. UhjU. 11. 31 by V- 16. Mby|. 2. 11 by |. 7. 20by^. 12. 50 by V- 17. 30 by ^ 3. 20 by if. 8. 51 by If. 13. 21 by i 18. 12 by |. 4. 86by|. 9. 39byH. U- 60 by |. 19. 33 by H S. 101 b^' ^. 10. 25byV- 15. 18 by |. 20. 15 by |. EXAMPLES FOR WRITTEN PRACTICE. 218. 1. If i of an acre of land sell for 45 dollars, what Avill an acre sell for at the same rate ? 2. A farm of 471 acres is divided into shares of 94^ acres eacli. 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 ? 4. When potatoes are worth f of a dollar per bushel and apples -| 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 Gf dollars per cord and jiays for it in work at i of a dollar per day. How many days must she work to make full payment ? 6. A dealer paid -f of 15f dollars for | 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 2J bushels, for the barley 2| bushels, for the oats 2J bushels, and for the beans 1| bushels. How many sacks in all were required ? 60 DIVISION OF FRACTIONS. 219. To Divide a Fraction by a Fraction. Example. — Divide | by 4. Operation. i 4 times 4- or ( iof4 =^^- 4 = Explanation. — Thedivisor, i, is equal to 4 times |, or ! of 4. Appl3ing the explanation of Art. 213, and dividing the dividend, :-| , by 4, gives /o as a quotient ; but since tlie given divisor was 1 of 4, and the divisor used was 4, a number 7 times too great, -^^j, the quotient obtained, is 7 times too small; to correct this error multiplj' /j by 7, obtaining f J as an answer. Ob- serve that 21 (the numerator of the quotient), i» obUiined by multiplying the numerator of the divid- end by the denominator of the divisor, 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 effecting a cross multiplication as shown by the connecting or tracing lines in the operation. Rules. — 1. Multiply the numerator of the dividend hy the denomina- tor of the divisor for the numerator of the quotient, and multiply the denoniinator of the dividend hy the numerator of the divisor for the denominator of the quotient; Or, 2. Invert the terms of the divisor and proceed as in multi plication of fractions. Remark. — Reduce mixed numbers to improper fractions before applying the rule. • KXAMPLKS FOK MKNTAI., PKAOTK'K. 220. Divide 1. f bv |. 2. |byf. S. fbyf 4. ibyf. 5- i by f . 6. ^jbyf. 7. ^byi^. 8- iVbyf ^- iibyf. 10. |byi 11. fbyii. 12. Ibyf. IS. u. 15. 16. \i by 4. -^ by W. Hbyi H by |. 221. Divide KXAMPLKS nm. w KITTKN PKACTICK. 1- HbyH- 2- liby^f 3- iVVbyT^- 4. Hbyif. •^- fUbyii- 6- HbyW- 7. Hiibyff S. IfbyV.^ 1 9. VjO by Yi. 10. y> hi I 11' iVbyV- 12. HbySV- IS. U- 15. 16. V by A- U- by 1?. Hbyi It by ■^^. 17. If 11 boy earns ^ of a dollar in a day, how long will it take him to earn $15f ? 18. How many fields of Oi; acres each can be made from a farm containing 125 1 acres ? 19. If a wlieelman runs 93| miles jter day, bow long time will he require to run 1167^ miles? 20. If 12 J acres produce 982^ l)ii. of corn, how many bu. will 15|^ acres produce? 21. If ten men cut 1324^ cords of wood in six days, how many cords can eighteen men cut in twenty-one days ? 22. If a man bought 1150J bushels of wlieat with f of his money, how many bushels could he have bought had all his money been invested ? 23. After traveling ^ of the distance between two cities, a pedestrian finds that there are 1014 miles still before him. How far ai)art are the cities ? COMPLEX FRACTIONS. 61 COMPLEX FRACTIONS. 222. A fraction is complex when either or both of its terms are fractional. ' Thus - is a complex fraction and is read 5 ^ f ; it indicates that 5 is to be I" 5 2 divided by f . - is read | -^ 8 and indicates what is thus expressed. ~ is read ^ -=- 1^ and indicates Avhat is thus expressed. Remark. — The entire subject of complex fractions will, ou account of its lack of practical -value, be dismissed with the full illustration of one example of each of two forms. I Example 1. — What is the value of -? "t 3 Operation: | = | - f =: | x | = ti = ^tV- Example 2. — Wliat is the value of ."' of ^ * ■^' Operation: ^-.-^^ = (| x V) - (i^ X fi) - | X ^ X ^ X ^ = U^^-^,- i of i| u MISCELLANEOUS EXAMPLES IN FRACTIONS. 223. 1. From the sum of f and 5|, take the difference betweeji 17^ and 21. ^. How much will remain after the product of f , -j^, 2, ^^, and 3f is taken from 10|. 3. Divide into six equal parts the product of ll^V multiplied by 3|. 4. Find the remainder after subtracting the product of 3f, -f, 7|, 5, f , and 1, from the product of 3, f, f , 7, f, 5, f, and 14. 5. An estate is so divided among A, B, and C, that A gets |, B j^^, and C the remainder, which Avas 14200. 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 $!f ? 8. A woman having $1, gave f of it for coffee at 33^^ i)er pound. How many pounds did she buy ? 9. Having bought f of a ship, I sold | of my share for $12000. What wjis the value of the ship at that rate ? 10. If the ingredients are -^ sulphur, ^l. saltpeter, and | charcoal, what is the number of pounds of each, in 2154y\ pounds of guni)0wder ? IJ. AVhat must be the amount of an estate whic^h, if divided into three ])arts, the first Avill be double the second, the second double the tliird, 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 MISCELLAXEOUS EXAMPLES IX FKACTIOXS. 13. I gave two 20-(l()llar gold coins to :i dealer, of whom I bouglit 2 cords of wood at 5f dollars per cord, and 3} tons of coal at G| dollars per ton. ITow much change should I have received. H. 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 ^1200()0 was so distributed that A received -^j B -^ of the estate more than A, C as much as A and B together less $6000, and two charities the remainder in equal j)arts. IIow much did each charity receive ? 16. Brown owned -^ of a stock of goods, ^ of wliicli Avas destroyed hy fire and ^ 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 tlie amount of Brown's loss ? 17. A grocer bought a cask of molasses containing 65^ gal., from which he sold at one time ^ of it, at another \ of it, at anotlier 5 gal. less than \ of what remained, and the remainder was sold with the cask for 'Z0\ dollars. If the vahie of the cask was one dollar, at what jirice per gallon was the last sale made? 18. A i^ainter worked 17^ days, and after expending ^ of his wages for board, had $15 left. How much did he earn per day ? 19. A farmer having G50 bu. of wheat, kept for his own use 52f bu. less than \, sold to his neighbors for seed 454- bu. more than ^, and marketed the remain- der at 80^ per bushel. How much money was received from the market sales ? 20. Of a journey of KiO miles, a walker accomplished + of the distance the first day, \, less 15f miles, the second day, r^, plus 4| miles, the third day, and finished his journey on the two following days by traveling fifteen hours each day. Wliat must luive been his average distance per hour for those two days ? 21. A mechanic worked 21 1 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 \ of it shall be in the first package, f in the second, -^^ in the third, and the remainder, which is $550, in the fourth package. What amount of money will be required ? 23. If ^ the trees of an orchard are ai)ple, \ peach, ^ pear, ^ plum, and the remaining 21 trees cherry, how many trees in all ? 2J^. John's weight is f as much as mine, and Ben's is ^ of Jolm'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 i)asture for $37. A put in 3 cows for 4 montlis, B, 5 for fi months, and C, 8 for 4 months. How much ought each to pay ? 27. Henry, when asked liis age, replied, " If 7^ years be added to 13^^ years, the sum will represent ^ of my age." How old was he ? 28. Silas, Harvey, and Eobert have together $2210. Silas has 2^ times as much as Harvey, who has | as much as I\()l)ert. How many dollars has each ? 29. Theodore's age is 7^ years, and Herbert's 9g years ; three times the sum 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' FHAf'TIOXS. 63 30. A farmer sold two cows for 175, receiving for one only \ as much as for the other. What was the price of each ? 31. After selling 45 turkeys, a dealer had a of his stock remaining. IIow many had he at first ? 32. If 8 horses consume 4^ bushels of oats in 34- days, how many bushels will 12 horses consume in the same time ? 33. A and B can do a piece of work in 10 days, which A alone can do in 18 days. In what time can B alone do the work ? 3Ji.. John and Calvin have agreed to build a wall for $(80. If Calvin can work only \ as fast as John, how shall the money be divided ? 35. A flagstaff stands \\ of its length above and 74 ft. below the surface of the ground. "What is the length of the staff ? 36. What is the length of a pole that stands f in the mud, % in the water, and 254 feet above the Avater ? 37. A colt and cow cost $124. If the colt cost $4 more than tliree times the cost of the cow, what was the cost of each ? 38. What is the hour Avhen the time jjast noon equals 4 of the time to mid- night ? 39. A tree 84 ft. high was so broken in a storm that the part standing was f the length of the part broken. How many feet were standing ? Ji-O. A farmer has \ of his sheep in one pasture, f in another, and the remain- der of his flock, 72 sheep, in the third i:)asture. How many sheep had he ? J^l. For a horse and carriage I paid $540. What was the cost of each, if the cost of the carriage was 1^ times the cost of the horse ? JfS Calvin is 84 years old, Leo %\ years less than three times as old as Cal- vin, and John's age is 3 years more than the sum of the ages of Calvin and Leo. What is John's age ? JfS. 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 ? 44- Tf A can do a i)iece of work in 21 days, B in 18 days, and C in 15 days, in how many days can the three working together iierform the work ? Remark. — In the above and similar examples, reason in general as follows: If A can build a wall in 4 days, lie can build \ of it in 1 day; and if B can build the same wall in 5 days, he can build \ of it in 1 day. Since in 1 day A can build \ of the wall, and B ! of it, the two can, if they work together, build in 1 day the sum of \ and \ or vV of it; and since they can together do -^Q in 1 day, it will take them as many days to do the whole work, or 1%, as 4^ is contained times in f g, or 2|. Jf5 John and his father have joint work, which they can do Avorking 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 ? Jfi 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? J^!. Smith said to Brown, " f of my money is equal to \ of yours, and the difference between your money and mine is $30." How much money had each? JfS. Izaak Walton having lost -| of his trolling line, added 65 ft., when he found it was just | of its original length. What was its length at first ? 64 MISCELLANEOUS EXAMPLES IN' FRACTIONS. ^9. A cistern sprung a leak by which | of its contents ran out, but during the same time f as much ran in. What part of the cistern was filled ? 50. A dog pursuing a rabbit which has 32 rods the start, runs 11 rods while the rabbit runs but 9. How far must the dog run before he can overtake the rabbit ? 51. A cistern has two faucets, by the larger of which it can be emptied in 24 minutes, and by the smaller in 36 minutes. If both be opened at once, what 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 df. 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 tlie bins there are 12 "bushels less than one-half as many bushels as in the other. How many bushels in each ? 5^ 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 T ? 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 ■60^ by selling them at the rate of 2 for 3 cents. How many oranges did he buy? 59. If vou buv GO lemons at the rate of 6 for 10 cents, and twice as manv 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 shaU be to €ach other as 1, 2, 3, and 4. What is the j)urtion of each. 61. I wish to line the carjiet of a room that is 74^ yd. long and 5f yd. wide, with duck f of a yd. wide. How many yards of duck will be required if it shrink -^ in length and -^ in width ? 62. A and B are engaged to perform a certain work for $35 j^y. It is sup- posed that A does \ 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 emi)tied ? 6Jf. 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 ojien for G 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 my age in 1846." How old was the man in 1880 ? 66. Tiiree men dig a well for |i3G. A and B working together do ^ of the work, B and C f of the work, and A and C | of the work. How should the mouev be divided ? MISCELLANEOUS EXAMPLES IN FRACTIONS. 65 67. Brown and Smith have joint work for 16 days. In any given time Brown ^oes only | as much work as Smith. How many days would each working alone require to complete the work ? If they work together, how should the 145 paid for the Avork 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 lono- 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; after 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 fi75 be paid ior 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 -J- of the whole increased by a .sum equal to ^ of C's part; and to C was given the remainder, which was 1700 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 Teceive $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 ? 7S. How long would it take Hill and Benton ? 74. 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, having joint work for 30 days, A begins and works alone ior 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 \ as much as D, B f as much as A, and C ^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, and 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 ^iven in Example 76, how should the money be divided ? fl6 DECIMALS. DECIMALS. *2*24. A Decimal Fraction or a Decimal is u 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 mav be expressed in the same form as common fractions; that is, with the denominator written. Practically, liowever, 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 written; 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 j^/wre 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, 32.301:. 229. The Talue of a Decimal is computed from the decimal poiiit, and the orders have the same scale as integers. A removal of the decimal point one place to the right, multiplies the expression ])y ten ; removing it two places, by 100 ; three places, by 1000, and so on. A removal of the decimal ])oint one place to the left, divides the expression by 10 : iico 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 com])utccl from the point to the right, it follows that one or more ciphers, placed after the decimal, will not alter its value. ^ is expressed decimally .3 ; a cipher annexed to the decimal gives .30 = -^j; tiro ciphers annexed gives .300 = t%VV- 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 ^. XUMERATIOX OF DKCIMALS. 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 shotdd contain as many jjlaces as there would be ciphers in its denominator if ivritten, the decimal point representing the unit 1 of such denominator. S. The value of any decimal figure depends upon its jjlace from the decimal point. Jf. Prefixing a cipher to a decimal decreases its value the same as dividing it by ten. 5. Annexing one or 7nore ciphers to a decimal does not alter its value. NUMERATION OF DECIMALS. 232. For Notation and Numeration of Decimals we begin witli the decimal point as a simple separatrix ; in the integral expression the first i)lacc to the left is units (corresponding to the decimal point in tlie decimal expression), the 2d tens, the 3d hundreds, etc., while from the separatrix to the right we have in order (the j^oint standing for units), tenths, hundredths, thousandths, etc. 233. The Order of a Decimal may he 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 be drawn from the following illustration be observed. If .35 be numerated from the right as in integers, the point is in Inindreds place; hence, read 35 hundredths; in .1403 the point is in ten-thousands' jilace, read 14G3 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 mixixl decimal is clearly shown by the following Table. a .9 a 03 § i o 5 O 3 cc tt-H 3 «-> fl 5 O o O X 'I m H to s 32 ^ 00 .a V «»-. o £ t-i OJ <) t^ o u O rjl u 73 3D a a 3 a o 13 a 3 s 'o o Q 3 43 ^ ^' ■a _£J ^ T3 'd Xfi 03 'O ■ •C! Oi OO I- S iO ■^ eo c* oints fall in the same vertical line, as units of the same order will thus fall in the same column; the result of the addition is then obtained in the same manner as in simple numbers. 11.5 113.2075 200.00165 328.70415 Remark.— 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.— TFri^e the decimals so that the points luill fall in the same vertical line. Add as in whole numbers, and place the point in the sum, directly helow the points in the numhers added. EXAMPLES FOR PKACTICE. 251. 1. Add 4, .37, 2.46, 19.301, and 103.21. 2. Add 3.04, 25.001, .67, .2146, and 819.256. S. Add 30.1257, 605.2146, 1000.864532, and 16.25694. J^. Add 896.111, 9530.216753, 1111.230004, and 1100.960005. 5. 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 1-4.0000864, .0096, 250.4, 700.0007, lUOO. 00000001, 563.3001468, 20.2001, 10000.001001 and 896.707075. 9. Find the sura of seventeen and forty-six ten-thousandths, eighty-three and one tliousund four millionths, five hundred two and seventy-five hundred- tliousandths, three thousand eleven and three hundred eleven thousandtlis, one million six and six million 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 milliontlis, 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, llow many acres had he in all ? 13. "What is the sum of 16.5 acres, 21.125 acres, 86.06;'5 acres 111.45 acres, 216.05 acres, 37^ acres, 426^^ acres, 80f acres, and 13-/^ acres ? IJf. What is the number of bushels in ten bins of 93.625 bu., 111.025 bu., 306.005 bu., 81J| bu., 193| bu., 200f bu., 300.0625 bu., 125^ bu., 250i bu., and 136^"^ bu. respectively ? 15. I bought ten bales of cloth as follows : 32J, 41||, 39-5!^, 46J, 29^, 38^^, 431, 41-5^, 42||^, and 40.625 yd. respectively. How many yards in my purchase? Resiark. — In invoices of goods only fourths are usually counted, and these are written as follows : 3' = 3|, 15 ' — \iS^, 12- = 12f . By the omission of the denominator time is saved. In additions, tind the sum of the small figures first as so many fourths, reduce to units and carry as in other addition. , 16. Add 21', 54=, 17', 30', 46', 61% 80', 39% and 24\ 17. Add 121', 97S 46% 111% 43, 71% 86% 50', 103% 72', 71% and 50. 252. To Add Repetends. Remark. — In addition of repetends, bear in mind their equivalents; thus, in adding .6-j- to .3 -{- remember that the value of the first is J, and of the second J; their sum is %, or 1. In all examples in addition of repetends, before beginning the operation, continue the repetends so that all have the same number of places, and in the right-hand column add each 9 as 10 253. Add . 333333 -f .171717 + .306306 + EXAMPLES FOK I'KACTICE. .811357 + .11111 + .<«(«< + .^46207' .55555 + .33333 + .2222 + .3333 + .8787+ .0101 + .3467 + .561561561 + .202202202 + .333333333 + .'504300542' .306306306 + 5. Find the sum of the following expressions: 105.333 +, 86.1919+, 53.103103+, 17.66+, 204.77+, 29.11+, 815.201201+ and 73.11081108 +. 6. Add .66 +, 1.2121 +, 50.55 -r, 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 .50. Operation. gg 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 paint in the remainder below ~ the points in the terms above. .30 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 suhti^act as in integers, placing the -point in the remainder heloiv the points in the other teinrvs. EXAMPLES FOR PRACTICE, 255. Subtract 1. .573 from .985. 2. .13823 from .668. 3. .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. 10. 11. 24.6852 from 25. 280. Ill from 500.000625. .09 from .900. 12. 250. 98754 from 386.245. MULTIPLICATION OF DECIMALS. 256. First Operation. .17 =: -j^''^y^ (Com. frac'l form.) Example. — Multiply .17 by .5. Explanation. -Write .17 as -^^^^ and .5 as ^^5, and apply 8i_ — Too — the rule for multiplication of common fractions. Multiplying these fractional equivalents, obtain j^§g as the common frac- tional expression of the product; by Art. 245, this may be 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 tiro ciphers, or two places, and that of the multiplier o?j6' 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. Explanation. — Write and multiply the expressions as in whole numbers. Since the numerator is 17 hundredths and the denominator 5 tenths, the product must be 85 thou- sandths. Hence, change the product 85 to 85 thousandths by .085 prefixing a cipher and a decimal point, thus: .085. 'Rule,— Multiply as in whole numbers; then, from the right of the product, point off for decimals a number of places etjual to the number in both factors, prefixing ciphers if needed to obtain the rcffuired number. Second Operation. .17 .5 DIVISION OF DECIMALS. KXAMPtES FOR PKACTICK 257. Multiply .78 by .7. .123 by .16. 1.45 by .875. 26.08 by 1.53 IS. 1000000 by .0000001 9. 10. 11. 13. 25000 by .000025. 8.76 by. 100. 716.0025 by 10.1006 7000 by .007. 3-100000.0081 by 81.000034. 2085.109 by 11.256. 1000.87 by 4621.5. 10000 by. 0001. .300 by. 03. 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 $.60^ 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 ? Remakks.— 1. The contraction of multiplication of decimals by restricting the number of places to appear in the product, is not deemed of sulficient 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 = nriTo' Too — 1860 6 ■nrrs-^ It: — To— •^' Second Operation. .17). 085 (.5 85 00 Explanation.— Since .085 = y^^ and .17 = ^V, 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 ^ = .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 ob\ious that the quotient will be a whole number. Rule. — I. Wlieii needed, annex ciphers to the dividend to make its places equal in number to those of the divisor- 11. Divide as in integers, and, from the right of the quotient, point off for decimals as many places as the number of places in the dividend exceeds tJiose in the divisor. DIVISION" OF DECIMALS. 77 269. 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 teii 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 mas- ters all difficulty. EXAMPLES FOK PRACTICE. 260. Divide 1. .625 by 2.5. 2. 15.25 by .05. S. 1100 b/4.4. 4- 9.5 by 19. S. 9.5 by 190. 6. .95 by .019. /v 36.5 by .073. ^. 250 by .0625. (25.) 1 -V - 1 = ? I -^ - .1 = ? 1 -i - .01 = ? 10 -^ .1 = ? 10 ■^ .01 = ? .1 - ^ 1 = ? .1 - ^ .1 = ? .1 - ^ .01 = ? .1 -^ .001 = ? 10 = ? 9. 1750 by .875. 17. 17.5 by 17500. 10. 3.6 by 1800. 18. .44 by .00011. Jl. .005 by 200. 19. 10006 by. 00001. 12. 27.405 by .00015. 20. .001 by 1000. 13. 1396. 875' by 250. 21. 1.6 by. 064. U. 131300 l)y.V25. 6400 by .0000016. 15. 62.5 by 1.25. 23. .0081 by .054. 16. .00875 by 125. U. 1860 by .000031. (26.) [27.) 1 -4- 10 = ? .22 - - 11 = ? 1 ^ 100 =1 ? 2.2 - ~ .011 = ? .1 -^ 1000 = ? 220 ■- 11000 = ? .001 -f- 100 = ? .022 ^ 110 = ? .0001 -^ .1 = ? .00022 -=- 11000 = ? 100 -V- .00001 = ? 2.2 - f- .000011 = ? 1000 -^ .01 = ? 2200 -^ .00011 = ? .00001 ~- 1000 = ? .022 ^ 110000 = ? 10 -^ 100000 = ? .0000022 -=- 1100000 = 10000 -^ .0001 = ? 220000 H- .000022 = ? 78 GREATEST COMMON DIVISOR OF DECIMALS. {29.) 6.25 -^ 2.5 = ? 62.5 H- .025 = ? 6250 -^ .0025 = ? .0625 -^ 250 = ? .00025 ~ .00025 = ? 6.25 -4- 25000 = ? .0000025 -^ .00025 := ? 625000 ^ .0000025 = ? .0000625 ^ 2500000 = ? 625 -f- .0000025 = ? Find the sum of the quotients. (30.) 2.5 -^ 625 = ? .025 -T- 62.5 = ? .0025 -V- 6250 = ? .00025 H- .625 = ? .000025 -^ .000625 = ? .0000025 -^ 62500 = ? 2500 -^ .0625 = ? 2500000 -^ .0000625 = ? .00025 -4- 6250 = ? .000025 -- 6250000 = ? Find the sum of the quotients. (38.) 1.6 -4- 2.5 = ? 160 -4- .25 = ? .0016 -=- 250 = ? 16 -f- .00025 = ? 160 -4- 250000 = ? 16000 ^ .000025 = ? .0016 -^ .00025 = ? .000016 -^ 2500000 = ? 1600 -4- .00025 = ? 1600000 -4-. 00000025=? Find the sum of the quotients. (31.) 440 ^ 1.1 = ? .00044 -f- 1100 = ? 4400 -^ .11 = ? 440 -4- .0011 = ? .0044 ^ 110000 = ? 44000000-=- 1100000=? 4400000 -f- .000011 = ? 44000 -4- .011 = ? .00000044 H- 110000 = ? 4400 ^ .00011 ■= ? 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. (32.) (33.) .375 -r- 1250 = ? 2.25 -f- .015 = ? 375 -^ .0125 = ? 225 -^ 1500 = ? .0375 -4- 12.5 = ? .0225 -^ 150 = ? 37.5 -^ .000125 = ? .00225 -4- .015 = ? 37500 -^ .00125 = ? 2250 -- .0015 = ? 3.75 ^ 1250000 = ? 22500 -4- 15000000 = ? .00375 -^ 125000 = ? .000225 H- .00015 = ? .0000375 H- .125 = ? .0000225 -4- 1500000 = ? 3750000 -4- .000125 = ? 2.25 ^ .000015 = ? .000375 -4- 12500 = ? 22500000 -^ .00015 = ? Find the sum of the quotients. Find the sum of the quotients. THE GREATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE OF FRACTIONS, COMMON AND DECIMAL. 261. All exi^lanations 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 -j, f, and | ? Explanation. — First reduce the given fractions to a common denominator and obtain as a result, ^g, |g, |o ; 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 /^ = \. Notice that the numerator of the resulting J is the Greatest Com- mon Divisor of the numerators, and that the denominator 6 is the Least Common Multiple of the denominators, of the given fractions. Operation. 5 ) 15 — 20 — 25 3— 4— 5 LEAST COMMON MULTIPLE OF DECIMALS. 79 Rule. — Write a fraction the numerator of which shall he tJie Greatest Common Divisor of the numerators of the given fractions, and the denom- inator the Least Common Multiple of the denomUiators 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 ^V- Operation. Explanation. — Reduce the given fractions to a common de- nominator as before, and obtain fj, fg, \*^ ; the Least Common '^ ) '^■^ "^^ Multiple of the numerators is found to be 1800 ; but the terms s J ~ I were not 24, 50, and 18, but |J, H, and J|, and 60 is the Greatest ^ " '_ Common Divisor of 60, the common denominator, therefore the , i)~ ., Least Common Multiple is not 1800, but '^%%^, or 'V; therefore 30 is the Least Common Multiple of the given fractions. Observe that the numerator of -Y is the Least Common 3Iultiple of the numerators, and the denom- inator of the \" is the Greatest Common Divisor of the denominators, of the given fractions. Rule. — Write a fraction the numerator of which shall he the Least Common Multi-pie of the numerators of the given fractions, and the denominator ofivhich shall he the Greatest Common Divisor of the denom- inators of the given fractions. 204. To Find the Greatest Common Divisor of a set of Decimal Fractions. Example. — Find the Greatest Common Divisor of .5, .25, and .375. Operation. - \ -t^i\ .i-rv .-,r,- Explanation. — Reduce the exoressions to equivalents hav- ' mg a common denominator, obtammg .oOO, .2o0, .370. For - \ -1 QQ -Q ~- convenience omit the decimal points, find the Greatest Com- \ mon Divisor of the numerators, and obtain 125. Since 500, 5 \ 20 10 15 ^^*^' ^"^ 375 were not whole numbers, but .500, .250, and .375, the result is not 125, but .125. 4— 3—3 Rule. — Reduce the expressions to the same decimal order, then icrite the Greatest Common Divisor of the expressions us a whole number, and make it of th c 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 .4, .Tl, and .41 (J. Operation. 4 ) 400 < "vO 41G Explanation. — Reduce to decimals of the same order, ob- , ^^ taining .400, .720, and .416 ; find the Least Common Multiple of ^ ^^ ^^^ *^^ «Mmcraullion 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 tlie mints, its value being established Tiy 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 wliose denominations increase and decrease by the decimal scale. United States money is a decmial currency. 274. The Dollar is the unit of United States money. Dollars are written :as integers, with the sign ($) prefixed ; the lower denominations arc written as decimals, dimes being tenths, cents hundredths, and mills thousandths of a •dollar. Thus, 15 dollars, 1 dime, 5 cents, 5 mills, is written ^15.150. In business records and papers, cents are often written as fractions of a dollar ; the half-cent is expressed either as a fraction (4), or as 5 mills. Thus, Sln.tS may be written $15jVo; '^^ cents, 1.124, or 1.125. 275. The denominations and scale of United States money are shown in tlie following Table. 10 mills = 1 cent (c. or et.). 10 dimes = 1 dollar ($). 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 Ctatcs 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 Slates money. The Mill is not coined; it is used only as a decimal of the cent, which is the smallest money of the jnint and the smallest recognized in business. 6 S2 U>ilTEl> STATES MONEY, Obverse. Obverse. Keverse. Coins of the United States. UNITED STATES MOXEY. ' 83 UNITED STATES COINS. 276. The Coins of tlie United States, authorized Ijy various Acts of Con- gress, are of gold, silver, copper-nickel, and bronze. 277. The Gold Coins of the United States arc as follows . 1. The Dotible Eofjle; value, $20 ; weight, 510 Troy grains. 2. The Eagle; value, 110 ; Aveight, 258 Troy grains. 3. The Half-Eagle; value, $5 ; weight, 129 Troy grains. Jf.. The Three Dollar piece ; value, $3 ; weight, 77. -i 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 -j'jj pure gold, and j'jy 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 j'„ part silver. 2. United States gold coins of standard weight are legal tender for all debt^. 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. ■3. The Quarter- Dollar; value, 25^v weight, 9(i.45 Troy grains. Jf. The Divie; value, 10/' ; weight, 38.58 Troy grains. Remarks.— 1. The value of gold and silver coins is based mainly on their weight and fine- ness, or the amount of pure metal used. Silver coins are made of ^'^ pure silver and -i\ alloy of copper. 2. United States silver dollars are lef/al tender for all sums not otherwise provided for by contract. The smaller silver coins are legal tender for all sums not exceeding ten dollars. 279. The Copper-Mckel Coins of the United States are as follows : 1. The Five- Cent incce, called tlie nickel; weight, 77.16 Troy grains. 2. The Three-Cent piece ; weight, 30 Troy grains. Remark.— The ^(j- and 3^ coins are composed of J copper and J Nickel. 280. TIk' Bronze Coin. — The only bronze coin now issued from the mint is the one cent piece, weighing 48 Troy grains, and composed of -f^^ copper and j5_ tin and zinc. Remark.— The 5f/ and 3^ nickel coins, and the If bronze coin, are called minor coins ; and while they are legal tender for all sums not exceeding twenty-live cents, their value is not a bullion value, as in case of coins of gold and silver, but an arbitrary value fixed for commercial convenience. UNITED STATES PAPER MONEY. 281. The Paper Money of the United States consists of Treasury Xotes, Treasiirif Certifiratrs. and Xational Bauh Bills. 282. Silver Certificates.— Any holder of silver dollars, to the amount of ten dollars or more, may deposit the same with the Treasurer or Assistant Treasurer of the United States and obtain therefor Silver Certificates, which are receivable for duties, taxes, and all public debts ; and any holder of the smaller silver coins to the amount of twenty dollars, or any multiple thereof, may obtain therefor lawful money at the office of the Treasurer or of any Assistant Treasurer. 84 BEDUCTIOX OF UiJTrED STATES MONET. 283. United States Treasury Notes. — Treasury Notes, or Greenbacks, are in the same denominations as the Bills of National Banks, with the addition of those of $5,000 and $10,000 value respectively. Tiiey are legal tender for all debts except customs or duties, and interest on the public debt, and are usually receivable for these also, being convertible into coin on demand when presented in sums of fifty dollars or more. 284. National Bank Bills. — National Bank Bills are the notes issued by National Banks, under the supervision of Government, and these bills are in denominations of $1, $2, $5, $10, $20, $50, $100, $500, and $1000, and being secured by deposits of Government Bonds with the United States Treasurer, and redeemable on demand with lawful money, are usually received for all dues, but yet are not legal tender ; and a debt cannot be paid with these notes if the cred- itor states as his reason for their rejection that they are not lawful money. REDUCTION OF UNITED STATES MONEY. 285. To Reduce Dollars to Cents. Example. — Reduce 5 dollars to cents. Explanation. — Since there are 100 cents in 1 dollar, in 5 dollars there are 5 times 100 cents, •or 500 cents. Rule. — Add two ciphers to the dollars. 286. To Reduce Cents to Dollars. Example. — Eeduce 1500 cents to dollars. Explanation. — Since 100 cents make 1 doUar, there are as many dollars in 1500 cents as 100 cents is contained times in 1500 cents, or 15 times, equal to 15 dollars. Rule. — Divide the cents hy 100, hy pointing off two places from the right. KXA3LPLES FOK PRACTICE. 287. Reduce 1. 6 dollars to cents. 2. Ill dollars to cents. 3. 241 cents to dollars. 4- 1044 cents to dollars. 5. 21468 cents to dollars. ; 9. $100.98 to cents. 6. 1800 cents to dollars. \ 10. $o.T5 to cents. 7. 51000 cents to dollars. 11. $26.53 to cents. S. 9876 cents to dollars. | U. $157.32 to cents. ADDITION AND SUBTRACTION OF UNITED STATES MONEY. 288. To Add or Subtract United States Money. Rule. — Write dollars under dollars and cents under cents; then add or subtract as in simple ninnhers. EX.\MPLES FOR PRACTICE. 289. 1. Add ten dollars twenty cents, six dollars forty-eight cents, fourteen dollars twenty-six cents, eleven dollars eighty cents, and forty-six dollars ten cents. MULTIPLICATIOK" OF UNITED STATES MONEY. 85 2. Subtract seven hundred sixty-five dollars nineteen cents from nine hun- dred ten dollars eight cents. 3. A farmer sold produce as follows : wheat, for |i761.25; oats, |i38:i.40; barley, $816.09 ; buckwli£>at, |;186;92; corn, $1127.50; potatoes, $063.11 ; hay, $400.50. What were his entire sales ? Jf. A lady bought groceries to the amount of $6.85; meats, $2.11; dry goods, $31.75; carpets, $167.25; millinery, $13.57. AVhat was the total amount of her purchases ? 5. A student expends for tuition and supplies, $118.75; for board, $167.50 ; for clothes, $57.25 ; for entertainment and church, $28.42 ; for charity, $6.15. What amount does he ex])end ? 6. The ex])enses of my house are as folloAvs : for interest, $167.50 ; taxes, $103.29; repairs, $56.82; insurance, $11.35; water rent, $11.25; and gas, $27.08. What are my total expenses ? Remarks. — 1. Under some circumstances it is desirable to write United States money, expressed in dollars ami cents, without the $ sign and the decimal point, with the decimal part placed slightly above that expressing the integers or dollars ; as $5.25 may be written 5-^ ; thirteen dollars and eight cents may be written 13""*. This is advisable only where the sum of several items is to be found by horizontal addition. 2. The amoimt in each of the following examples is to be found by horizontal addition. 7. Add 15^«, 29^ 1146^ 1079«, 9^3, 81*5, 12392, 601, IS^o, and ll^. S. Add 34650, 291^5, lOO^i, 269ii, and 8093. 0. What is the sum of 21658*, 7243^ 9920^ 117«S 5005o, and 1127i* ? IQ. What is the sum of 667* S 328io, 97' ^^ goO, 20", 155i«, 1101, 28^3, and 6759 ? 11. My bills for a year are: for groceries, 283^1; meats, 135^1; miller's i)ro- ducts, 76'' 5; coal, 412 0; kindling, 45"; milk, 47^5; servant, 217; incidentals, 915*. What are my expenses ? 12. A merchant bought cottons, for 3467-5; linens, for 1326^5; woolens, for 4215'' 5 ; delaines, for 1025* 5 • brocades, for 11275 ». If all were sold for 132562 6, how much was gained ? MULTIPLICATION OF UNITED STATES MONEY. 290. To Multiply United States Money. Rule. — Multiply as in abstract decimals. Remark. — Money is a concrete expression ; therefore in critical analysis of its multiplica- tion, the money cost or price of an article is a concrete multiplicand, the number of things bought or sold is an abstract multiplier, and their product is concrete and of the denomination of the multiplicand. But since the money scale is decimal, these terms may be interchanged for convenience. exampi.es for practice. 291. 1. AVhat will be the amount of the following purchases : 147f cd. hard wood, at $5.75 per cd.; 206f cd. soft wood, at $4.25 per cd.; 4 car loads slab wood, each containg 16f cd., at $2. 75 per cd.; 816^ tons hard coal, at $5,15 ton; and 536^ tons soft coal, at $3.85 per ton ? 86 DIVISION" OF UNITED STATES MONEY. 2. Bought ilb bar. superfiue flour, at $4.85 per bar.; 355 bar. extra flour at $5. 15 per bar. ; 132 bar. rye flour, at $4.90 per bar. ; 210 bar. corn meal, at $3.70 per bar. ; and 642 sacks graham flour, at 88^ per sack. What was the total cost ? S. A retailer bought 35 overcoats, at $0.75 each; IGO black suits, at $17.25 each; 125 plaid suits, at $14.05 each; 84 jean suits, at $6.90 each; and 50 pairs trousers, at $3.15 each. Find the total cost. 4. An invoice of six pieces of gingham of 51^, 49^, 50*, 54*, 49-, and 51^ yd. respectively, was sold at $.09| per yd. What was the amount of the sale ? 5. Six men worked 19f days each, at $1.90 per day; 24:^ days each, at $1.80, 11^ days each, at $1.65: and 31f days each, at $1.25. How much was earned by all in the entire time ? 6*. A laborer received $184.55 as a balance due him for his season's work. He paid a debt of $19.25; bought 8^ yd. cloth, at $1.25 per yd. ; 2 suits of clothes, at $13.25 per suit; hosiery and gloves for $2.85; 4f tons coal, at $5.65 per ton; 2 cd. wood, at $3.90 i)er cd. ; 3 bar. flour, at $4.75 jier bar. ; 628 pounds of pork, at 6f^ per lb. : and loaned the remainder of his money. How much did he loan ? DIVISION OF UNITED STATES MONEY. •>9-2. To Divide United States Money. Rule. — Divide as in abstract decimals. EXAMPLES FOK PRACTICE. 293. i. If $11421.75 be divided equally among hve persons, what will be the share of each ? 2. B sold 18T^ acres of land at $105.25 per acre, and divided the proceeds equallv among fifteen persons. What sum did each receive ? 3. A charitable farmer gave 15f bushels of apples worth $.50 per bu., 21| bushels of potatoes worth $ .75 per bu., and 30 bushels of turnips worth $.624^ per bu.. in equal shares to six families. What was the value of each share ? j^. A dealer bought wheat at $.95 per bu., oats at $.45 per bu., and corn at $.65 per bu. He paid $332.50 for the wheat, $191.25 for the oats, and $113.75 for the corn. How many bushels did he buy in all ? o. C invested $9659.50 in coal, at $5.85 per ton; $2645.30 in sand, at $2.80 per cubic vd. ; $058.40 in lime, at $1.60 per barrel. If he sold the coal at $6.05 per ton, the sand at $2.75 per cubic yd., and the lime at $1.75 per barrel, what was the gain or loss ? 6. Having sold my mill for $17250, and 316 barrels of flour in stock at $5.15 per barrel, I invested of the proceeds, $1185.85 in furnishing a house, $1259.30 in utensils, $1582.25 in live stock, and with the remainder paid in full for a farm of 163 acres. Wiiat was the cost of the farm i)er acre ? Rem.\rk. — In case exact quotients are not obtained in division of dollars, add two decimal ciphers and continue the quotient to cents ; if not then exact add one cent if the mills be 5 or more, but if less than 5, reject the mills. ANALYSIS. 87 ANALYSIS. 294. Arithmetical Analysis is the process of solving problems inde- pendently of set rules, l)v deducing, from the terms stated, the conditions and relations required in their solution. Remark. — The general subject of Analysis will be treated only as auxiliary to the subject of Common Fractions, and the Special Applications of the Fundamental Rules. Example 1. If 5 men earn %'60 m 4 days, how many dollars will T men i-arn in 9 days ? First Explanation (^extended). — If 5 men earn $30 in 4 days, 1 man, or 1 of 5 men will cam in 4 days 1 of $30, or $6 ; and if 1 man earns $6 in 4 days, in 1 day, which is \ of 4 days, he will earn \ of $G, or $li. Then, since 1 man in 1 day earns $li, in 9 days, which are 9 times 1 day, he will earn 9 times $li, or $13^ : and if 1 man in 9 daj's earns $131, 7 men, which are 7 times 1 man, will earn 7 times $131, or $941. Second Explanation (abbreviated). — If 5 men earn $30 in 4 days, they will earn $7i in 1 day ; and if 5 men earn $7^ in 1 day, 1 man will earn ! of $7|, or $U ; since 1 man in 1 day earns $U, 7 men in 1 day will earn 7 times $li, or $10^ ; and if 7 men in 1 day earn $10|, in 9 days they will earn 9 times $10i, or $94^, the same as before found. Third Explanation {tnore abbreviated). — If 5 men in 4 days, doing 20 days' work, earn $30, $11 would equal 1 daj''s work ; 7 men in 9 days do 63 days' work, and since 1 day's work equals $1|, 63 days' work will equal $941, as before found. Example 3. If 6 men can cut 45 cords of wood iu 3 days, how many chords can 8 men cut in 9 days ? First Explanation (extended). — If 6 men cut 45 cd. in 3 days, in 1 day, which is J of 3 days, they can cut ^ of 45 cd., or 15 cd.; and if 6 men can in 1 day cut 15 cd., 1 man in 1 day can cut J of 15 cd., or 21 cd. ; since 1 man in 1 day can cut 21 cd., 8 men can in 1 daj' cut 8 times 2^ cd., or 20 cd.; and if 8 men in 1 day can cut 20 cd., in 9 days they can cut 180 cd. Second Explanation (abbreviated). — 6 men in 3 days, doing 18 days' work, cut 45 cd.; hence 21 cd. can be cut by 1 man in 1 day ; then 8 men in 9 days, doing 72 days' work, can cut 72 times 2i cd., or 180 cd., as before found. Example J. If a post 4 ft. high casts a shadow 13 ft. in length, wliat must be the height of a post that Avill cast a shadow 125 ft. in length? Explanation. — If a post 4 ft. high casts a shadow 13 ft., a post 1 ft. high would cast a shadow 3] ft. ; since a shadow 3] ft. is cast by a post 1 ft. high, a post that will cast a shadow 125 ft. in length must be as many times 1 ft. in height as 3] ft. are contained times in 125 ft., or SSx'V ft. Example 4- If the hour and minute hands of a clock are together at noon, ; what times af 4 and 5 o'clock ? at what times after noon will they again be together ? At what time between 88 ANALYSIS. Explanation. — Since the minute hand passes the hour hand 11 times in 12 hours, it will pass it the first time in i\ of 12 hours; the second time in f^ of 12 hours; the third time in y\ of 12 hours; the fourth time in y\ of 12 hours. y\ of 12 hours equals 4 hours, 21 minutes, and 49^^f seconds ; therefore the hands will he together between 4 and 5 o'clock at 21 minutes 4.9^j seconds after 4 o'clock. Remark. — Apply the same reasoning to all examples of this class. Example o. If Grace were ^ older than she is, her age would equal ^ of her grandmother's. What is the age of each, if the age of both is 87 years ? Explanation. — If Grace were l older than she is, she would be | of her present age ; and since if she were !; her present age, she would be only J as old as her grandmother, the age of grandmother must be 4 times ? or V of the age of Grace, and the age of both must be | + V" or -J* of the age of Grace ; since the age of both is 87 years, 87 years must be -\"- of the age of Grace, who must be 15 years old. If Grace's age be increased by ^ of itself, or 3 years, she will be 18 years of age ; and since her age would then be only J of grandmother's age, the age of grandmother must be 4 times 18 years, or 72 years. Example 6. A mau being asked his age, replied : "My father was born in 1805 and my mother in 1806 ; the sum of their ages at the time of my birth was two and one-third times my age in 1851." How old was the man in 1888 ? Explanation, — If the father was born in 1805 and the mother in 1806, the sum of their ages in 1851 was 91 years ; and since the sum of their ages at the time of the birth of the son was 2^ times his age in 1851, and the parents each increased in years after the son's birth as fast as he did, in 1851 the sum of their ages must have been 4|^ times the age of the son; hence the son, in 1851, was 91 years -i- 4^, or 21 years of age, and he must have been born in 1830, and in 1888 would be 58 years old. 7. The sum of two numbers is 65, and their difference is equal to ^ of the greater number. Find the two numbers. 8. How long after noon will it be when the minute hand passes the hour hand the third time? 9. How long after noon will it be when the minute hand joasses the hour hand the eleventh time? 10. A's age is 2| times the age of B, and the age of C is 2^^ times the age of both A and B. If the sum of their ages is 116 years, what is the age of each? 11. A man bought 15 bushels of barley, and 36 bushels of oats, for $38.80, and 25 bushels of barley, 18 bushels of oats, for $29.10. How much per bushel did he give for each kind of grain? 12. Charles, when asked his age, replied: " My father was born in 1843, and my mother in 1847. The sum of their ages at the time of my birth was 5 times, my age in 1887." In what year will Charles be 25 years of age ? SPECIAL APPLICATIONS. 8& SPECIAL APPLICATIONS. 295. Special Applications, as here treated, embraces the use, in the sohition of problems, of any or all explanations heretofore given, and the con- sideration of cost, price, and quantity y as being the elements of every business transaction ; it also treats of such contracted methods as may be employed in dealing with aliquot parts of the powers of 10, or of other numbers. General Rules. — 1. If the price and quantity he given, the cost may he found hy multiplying, the price hy the quantity. 2. If the cost and quantity he given, the price may he found hy divid- the cost hy the quantity. 3. If the cost and j/rice he given, the quantity may he found hy divid- ing the cost hy the price. ALIQUOT PARTS. 296. The Aliquot Parts of a number are the even parts of that number. 25, 33i, \%\, are aliquot, or even, parts of 100, Remakk — The component factors of a number must be integral, while the aliquot parts of a number may be either integral or mixed. 297. The even parts of other even parts may be called parts of parts ; as, i = i of i ; or, since 33| is a part of 100, ^ of 33^, or 11^, must be a part of the part 33^. Remakk — Full illustrations of the use of aliquot parts will follow. Those of $1, equal to 100^, being the most valuable for use, will be mainly considered. 1. 50 cents = ^ of $1 3. 33^ cents = ^ of |] 3. 25 cents = -]- of |1 4. 20 cents = -\ oi U. Aliquot Parts of One Dollar. 5. 16f cents = i of $1. 10 cents = J„ of |1. 8^ cents = ^ of $1. 9. 6^ cents = iV of $1. 10. 3| cents = ^ of $1. 11. 2i cents = ^V of U. 12. If cents = ^^ of $1. Aliquot Parts of Aliquot Parts of One Dollar. \

AKATTOx.— Since 1 yard costs f 1.37i, 100 yards will cost $137.50 ; 7.5 yanls will cost 1 less than $137.50, or $103,125, or $103.13. Example .7. — Find the cost of 250 yards, at $1.75 per yard. ExPT.A^^\TIo^•.— 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 1420 bushels of wheat, at $1.37^ per bushel ? ExPLANATiox.— At $1 per bushel. 1426 bushels will cost $1426.00 At \ = 25o per bushel, 1426 bushels wUl cost 356.50 At I = (i of j), or 12io per bushel, 1426 bushels will cost 178.25 At $1,371 per bushel, 1426 bushels will cost $1960.75 Example 7. — What is the cost of 824 yards of cloth, at $1.75 per yard ? ExPLAXATiox.— At $1 per yard, 824 yards will cost $824.00 At 1 = 50o per yard, 824 yards will cost 412.00 At i = 250 i>er yard, 824 yards will cost 206.00 At $1.75 per yard, 824 yards will cost $1442.00 Example S. — At 55^ per lb., what will l)e the cost of 14G lb. of gunpowder ? ExPLA^-ATI0^-.— At $1 per lb., 146 1b. will cost $146.00 At ^ = 50C per lb., 146 1b. will cost.. 73.00 At ^ = (iV of i), or 5/;* per lb. . 146 lb. will cost 7.30 AtSS^ per lb., 146 lb. will cost $80.30 KXAMPLES FOR WRITTKN PKACTICK. 306. Find the cost of 2U0 lb., @ 37^^. i 700 lb., @ 51i^. 150 lb., @ 14f^. 250 lb., @ 21f^. 1000 lb., @ $1.12^. 750 lb. @ 81f'/. 1250 lb., @ $2.62^. 8. 400 lb., @. 95i^. lo. 9. 250 lb., @ 9f^. 1 16. 10. 75 lb., @ 60J^-. ' 17. 11. 125 lb., @ 27^'/. j 18. 12. 1100 1b., @ $1,424. I 19. 13. 500 lb., @. 374^. j 20. U. 1500 1b., @ 18f^. ' 21. 300 lb., @41|^. 3000 lb., @ 12^^. 2500 Ik, @ 61.10. 25 lb., @ $1.85. 150 lb., @ 334^. 75 lb., @, $1.15. 125 lb.,@ $1.25. KX.VMPLKS KOK MKNTAL PKACTICK. Remark. — All extensions in the following examples should be made mentally, the pupil writing only the cost of each item for footing. 307. 1. Find the total cost of 516 lb., at \05,S4 l)uy ? Explanation. — Since the price is \ of itself less than $1 per yard, the number of yards willbei greater than the number of dollars expended; .\ of 84 =42; 84 + 42 — 126, or 126 yards. 96 EXAMPLES FOR PRACTICE. Example 3. — At 87^^ per bushel, how many bushels of wheat can be bought for *12G: ? ExTLAJNATiON-.— Since the price is i of itself less than $1 per bushel, the number of bushels will be i greater than the number of dollars expended; i of 1267 = 181; 181 -f- 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 66|^ per yard, how many yards of cloth can be bought for $84? Explanation. — 66|^ = $|; write its reciprocal, |, and multiply by $84. Example o. — At 75^ per yard, liow many yards of cloth can be bought for $84? Explanation. — 75^ = $J; write its reciprocal, |, and multiply by $84. Example 6. — At 87^^ per yard, how many yards of cloth can be bought for $84? Explanation. — 81i^ — $i; write its reciprocal, i. and multiply by $84. Rules. — 1. Multiply tJie cost hij the quantity that can he bought for $1. Or, 2. Add to the cost (as qioantity) such a paH of itself as the price lacks of being $1. EXAMPLES FOK PRACTICE. 309. 1. If 1 lb. of candy can be bought for 25^, how many pounds can be bought for $5.75 ? ~. At 33^^ per yard, how many yards of cloth will $1542.50 buy ? 3. A boy expended $1 for almonds, at 16§^ 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 8^^ per qt., and 75 doz. corn, at 12^^ 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 8 J cwt. of pork, at $7.50 per cwt., 15 bu. of beans, at $3.25 per bu., and 4Gi- bu. of oats, at 33^^ per bu., for 28 yd. of dress silk, at $1.25 per yd., and 52^ yd. of delaine, at 16|^ per yd., receiving for the remainder, cotton goods at 12^^ per yd. How many yards of cotton goods should be delivered to him ? 10. "When potatoes are worth G6|^- per bu., and turnips 25^ per bu., how many pounds of coffee, at 16|5# per lb., will 2)ay for 24 bu. of potatoes and 18 bu. of turnips ? 11. Having bought 1487 lb. A. sugar, at 6^^ per lb.; 872 lb. C. sugar, at 5^ per lb. ; 628^ lb. Y. H. tea, at 33^^ per lb. ; 522 lb. J. tea, at 25/ per lb. ; 650 lb. Rio coffee, at 12^/ per lb.; and 81 sacks of flour, at $1.25 persadk, I give in pay- ment seven one-hundred dollar bills. How much should be returned to me? EXAMPLES FOR PKACTICE. 97 310. To find the Cost of Articles Sold by the C. C stands for 100. M stands for 1000. Example. — What is the cost of 416 lb. phosphate, at $2.00 per hundred? ExPLAKATiON.— 416 Ibs. = 4.16 hundred lbs. If 1 hundred pounds cost $2.00, 4.16 hundred lb. will cost 4.16 times $2, or $8.32. Rule. — Reduce the quantity to hundreds and decimals of a hundred, by pointing off two places from the right, then multiply hy the pi'ice per C. EXAMPLES FOR PKACTICE. 311. Find the cost of 1. 1753 lb. of salt, at $1.25 per C. £. 8425 lb. of scrap iron, at $1. 10 per C. 3. 2156 lb. of fence wire, at $3.25 per C. 4, 378 fence posts, at $7. 50 per C. ■5. 3295 lb. of gitano, at $4.50 per C. 6. 905 lb. of lead, at $3.50 per C. 7. 1125 lb. of castings, at $2.25 per C. 8. 1620 handles, at $5.50 per C. 9. 509 lb. of beef, at $12.50 per C. 10. 23765 lb. of nails, at 15^ per C. 312. To Find the Cost of Articles Sold by the M. Example. — At $7.00 per M, what will be the cost of 1544 bricks? ExPLA^fATI0N. — 1544 bricks = 1.544 thousand bricks; and if one thousand bricks coat $7, 1.544 thousand bricks will cost 1.544 times $7, or $10,808 - $10.81. Rule. — Reduce the quantity to thousands and decimals of a thousand, by pointing off three places from the right, then multiply by the cost per M. EXAaiPLES rOK PRACTICE. 313. 1. What will be the cost of 1650 ft. pine lumber, at $15 per M? 2. What will be the cost of 611 ft. oak lumber, at $24 per M? 3. What will be the cost of 21 168 ft. hemlock lumber, at $7.50 per M? Jf. What will be the cost of 9475 ft. elm lumber, at $13 per M? 5. What will be the cost of 2120 ft. ash lumber, at $25 per M? 6. What will be the cost of 2768 ft. maple lumber, at $14 per M? 7. What will be the cost of 1100 ft. chestnut lumber, at $18 per M ? 8. Find the cost of 4560 ft. oak lumber, at $22 per M. 9. Find the cost of 11265 ft. spruce lumber, at $12.50 per M. 10. Find the cost of 6625 shingles, at $5.25 per M. 11. A dealer bought the season's cut of a saw mill, which was as follows: ■326475 ft. clear pine, at $25 per M; 1467250 ft. seconds, at $17.50 per M; 102500 ft. culls, at $13 per M; 890000 ft. hemlock boards, at $10.50 per M; 824650 ft. hemlock timber, at $9 per M; 552720 ft. white oak plank, at $21 per M; 75690 ft. red oak plank, at $16 per M; 101145 ft. cherry, at $35 per M. What was the amount of the purchase ? 12. For constructing a house and barn I bouglit: 46210 ft. matched pine, at $21 per M; 13516 ft. siding, at $28.50 per M; 11260 ft. chestnut, at $32 per M; 4680 ft. black walnut, at $45 per M; 928 ft. cherry, at ^^Q per M; 33725 ft. hemlock timber, at $11 per M; 58660 shingles, at $6.25 per M; 13700 brick, at -5.60 per M. What was the total cost ? 7 98 EXAMPLES FOR PRACTICE. 314. To find the Cost of Articles Sold by the Short Ton, or Ton of 2000 lb. Example. — What will be the cost of 3108 lb. of coal, at $6 per ton? ExPLAXATiOK.— 3108 lb. = 3.108 thousand lb.; since 1 ton, or 2000 lb., cost $6, i ton, or 1000 lb., -will cost i of $6, or $3; and if 1000 lb. cost $3, 3.108 thousand lb. will cost 3.108 times $3, or $9,324, or $9.32. Rule. — Diiide the price of one ton by 2, and the result wiU be the price per 1000 lb. From the right of the quantity point off 3 places, thus reducing it to thousands and decimals of a thousand. Multiply by the price per 1000 lb. EXAMPI.ES FOK PRACTICE. 315. 1. At $3 per ton, what will be the cost of 2680 lb. soft coal?' x\ At 87 per ton, what will be the cost of 1345 lb. canuel coal? 3. At $36 per ton, what will be the cost of 4372 lb. phosphate? 4. At 12.50 per ton, what will be the cost of 11075 lb. salt? 5. ' At $34.50 per ton, what will be the cost of 116780 lb. pig iron? 6. At 847.60 per ton, what will be the cost of 84725 lb. steel rails? 7. At ¥125 per ton, what will be the cost of 15066 lb. sheet copper? 8. At $4.50 per ton, what will be the cost of 9362 lb. land plaster? 9. At $2.10 per ton, what will be the cost of 2640 lb. slack lime? 10. At $35 per ton, what will be the cost of 1115 lb. giiauo? 11. What will be the freight, at $5 per ton, on four cars of Mdse. of 21780, 23055, 41200, and 32460 lb. weight respectively? 12. At $16.50 per ton, what will be the express charges on five boxes weighing respectively 186, 610, 241, 519, and 356 lb? 13. My furnace consumed, in one year, six loads of hard coal, weighing respec- tively 4125, 3960, 4305, 4440, 4055, and 3775 lb. If the coal was bought at $4.60 per ton, what did it cost to run the furnace ? IJf. A dealer stocked his yard with 17500 tons of coal, as follows: 850 tons cannel, at $7.40 per ton; 52600 lb. soft, at 82.50 per ton; 193410 lb. of egg, at $3.20 per ton, and the remainder chestnut, at $3.60 per ton. What was the value of the dealer's stock ? 316. To Find the Cost of Products of Varying Weights per BnsheL Example 1. — Required, the cost of 104 lb. of clover seed, at $6.35 per bushel of 60 lb. ExPLAXATioK. — At $6.35 per lb., the cost would be 104 times $6.35, or $660.40; but since the price was not $6.35 per lb., but $6.35 per bu. of 60 lb., the cost will be ^V of $660.40, or $11,006, or $11.01. Example 2. — Required, the cost of 100 1b. of blue grass seed, at $1.25 per bushel of 14 lb. ExPLAKATiox. — At $1.25 per lb. the cost would be $125; but since the price was not $1.25 per lb., but $1.25 per bu. of 14 lb., the cost would be ^^ of $125, or $8.93. EXAMPLES FOR PRACTICE. 99 Jinie.— Multiply the number of pounds weight by the price per bushel, and divide the product hy the number of pounds in 1 bushel. Remark.— Parts of bushels are often written in smaller figures at the right and above aa pounds. Thus 1** bu. clover seed = U^ bu. = 1 bu. 44 lb. = 104 lb. 21 1^ bu. oats = 21|| bu. = 21 bu. 12 lb. = 682 lb. 119«« bu. corn = llOfl bu. = 119 bu. 25 lb. = 7689 lb. EXAMPLES rOK PRACTICE. 317. How much should be paid for a load of 1. Wheat, weighing 2142 lb., at % .80 per bushel of GO lb. 2. Corn, weighing 2506 lb., at S.G5 per bushel of 58 lb. S. Barley, weighing 3381 lb., at $ .75 per bushel of 48 lb. Jf. Millet, weighing 1768 lb., at $1 per bushel of 45 lb. 5. Oats, weighing 2255 lb., at 1.35 per bushel of 32 lb. 6. Buckwheat, weighing 2172 lb., at 8.60 per bushel of 48 lb. 7. Beans, weighing 2761 lb., at 11.25 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. Eed top grass seed, weighing 2059 lb., at $ .90 per bushel of 14 lb. 11. Timothy seed, weighing 2677 lb., at %% per bushel of 44 lb. 12. Kentucky blue grass seed, weighing 2266 lb., at $1.50 per bushel of 14 lb. 13. Clover seed, weighing 2941 lb., at $5.10 per bushel of 45 lb. IJf. 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 2599 lb., at $.65 per bushel of 60 lb. 11. Turnips, weighing 2160 lb., at $ .30 per bushel of 56 lb. 18. Apples, Aveighing 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 Avheat, 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 629^' per bushel of 60 lb. The loads of wlieat 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, 2631, 2997, 3005, and 2775 lb,; the potatoes 2846, 2891, 2805, 2863, 2984, 2901, 3046, 3280, 3395, 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, AXD INVENTORIES. BILLS, STATEMENTS, AND INVENTORIES. 319. A Bill is a \rritten rendered. statement in detail of articles sold or setvices 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 l)e receipted when the words " Received Payment," or " Paid " and the creditor's signature, have been written at the bottom. 3*20. An IiiToiee 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. 3*22. An luTeutory 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. a. Cent. E. & 0. E. Error Acct Account. C7igd. Charged. Omissions Except Agt. Agent. Co. Company. Exch. Exchange. Amt. Amount. a 0. D. Collect on Fol. Folio or page. Bed. Balance. Delivery. Fr't. Freight. Bbl or Bar. Barrel. Com. Commission. Ft. Foot. Bdl Bundle. Con. Consignment. Gal. Gallon. Blk. Black. Cr. Creditor. Gr. Gross. /l Bill of Lading. Cwt. Hundred weight. Guar. Guaranteed Bot. Bought. Dft. Draft. Hhd. Hogshead. Bro. Brother. Dis. Discount. i. e. That is. Bu. Bushel. Do. or ditto. The same. In. Inch. Bx. Box. Doz. Dozen. Ins. Insurance. Cd. Cord. Dr. Debtor. Jr. Junior. ^ c ent. Ea. ^ Each. Lb. Pound. and BILLS. 101 Mdse. Mercliandise. P. or p. Page. Pec'd. Received. Me7n. Memorandum. Pp. or pp. Pages. Rec't. Receipt. Messrs. Gentlemen or Pat/'i. Payment. R. R. Railroad. Sirs. Pd. Paid. Schr. Schooner. Mr. Mister. Per. By, or by the. Ship't. Shi])ment. Mrs. Mistress. Pkf/. Package. Str. Steamer. N. B. Take notice. P. 0. Post Office. Sunds. Sundries. Net. Without discount. Pr. Pair. Super. Superfine. No. Number. Pc. Piece. Wt. Weight. Oz. Ounce. Qr. Quarter. Yd. Yard. Remark. — In abbreviating measures of capacity, weight, distance, or time, it is unnecessary to add an s for the plural. 324. Time Abbreviations and Contractions used in Business. Jan. or Jajiy. January. Nov. November. Ce7it. Century. Feb. or FeVy. February. Dec. December. d. Dav. Mar. Marcli. Mo. Month. It. Hour. Apr. April. Yr. Year. m. Minute. Aug. August. Inst. Present month. sec. Second. Sept. September. Prox. Next month. ick AVeek. Oct. October. Ult. Last month. 325. Signs and Symbo Is in Common Use. @ At; as, at a i)rico. "'^ Care of. New account. if Number. y" Check mark. o/ Old account. ^ By, or by the. ;» Per cent, or Hun- X By, in surface ^ Account. dredths. measures. BILLS. 326. Find the footing of each of the following bills: (1.) John R. Kxox, \rvi Pearl St., City, Knoxville, Tenn,, Dec. 31, 1888. Boufjht of CULVER & CASS. 3 sac 2 bu. i bu. 2 lb. 2 lb. 1 lb. 2 gal 4 bu. 4 lb. ks Cream Flour 95^- Potatoes 80i^' Sweet Potatoes 90^' Ginger 22^'- Jap. Tea 55^'- 0. H. Tea. 7o^- . Syrup 45^' Onions $1 Crackers 11^- Paid, 2 85 1 60 45 44 10 75 90 50 44 Culver & Cass, Per Cass. 102 Folio 246. BILLS. 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. 25508 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. 1119 15 H 5 12 Pl t 7 24 it 21 21 Pieces Bleached Cotton, 412 403 411 452 44 441 471 453 42 423 433 431 47 44 44^ Pieces Muslin, 371 323 33 353 341 32 352 333 37 381 381 36 Pieces Delaine, 39 402 411 393 3^2 40 423 44^ 42 Pieces Windsor Prmts, 213 273 253 28 26 228 24 25 32 312 28 241 25 272 22 281 24^22 21^26 24 312 32 22 Pieces Merrimac Prints, 281 32 343 282 26 24i 222 242 262 24 261 33 282 34 27i 30 323 24 302 31 302 No.i'd Price, 7^- 16^^ 5i0 Items. Amount. Remark. — Any conditions as to time of credit, manner of payment, interest on balance, or discount for prepayment, are properly placed on a bill or statement. An M of shingles is equiTalent to one thousand shingles averaging 4 inches in width. STATEMENTS. (4.) 103 Book 3, Page 308. H. H. Barnes & Co. Boston, Mass., Terms, Interest after sixty dai/s. Chicago, III,, Aug. 1, 1888. Bought of PEASE & SONS. 25 baskets Pork Loins, net 312 301 297 315 302 313 8^^- tubs Lard, 71-14 70-15 69-14 pkg. 10^ each 11^ casks Shoulders, 428-68 419-70 423-65 432-72 pkg. 90^ each 9^ bar. Mess Pork $22.50 20 casks Hams, 395-67 412-71 402-71 411-67 408-68 425-71 400-69 399-70 398-71 426-68 419-69 423-69 407-67 415-75 418-68 409-71 403-71 421-71 428-68 400-78 pkg. 75^ each 13i^- PkK. STATEMENTS. 327. Find the amount of each of the following statements : (1.) Folio 1(21. Birmingham, Ala., Jan. 1. 1889. Richmond & New Orleans Railway Co., To CLIMAX FOUNDRY CO., Dr. 1888. Nov. 4 To Bill r Bnde (. 7' (( 13 (( 18 a 21 i( 25 a 29 << 30 Dec. 3 t< 6 tt 7 (( 10 i( 15 a 20 i ( 21 ii 22 ii 25 t( 30 Please remit 590 25 375 13 1150 1560 25 2506 50 763 28 846 20 1000 12750 2634 19 9374 75 871 03 767 20 8500 76 50 1438 10 119 93 1408 27 104 STATEMENTS. (2.) Austin, Texas, Mar. 21, 1888. Geo. H. Grimes, Galveston, Texas, In acconnt with CLAUDE M. OGDEN, Dr. 1888. Jan. 15 a 20 (< 24 <( 28 Feb. 1^ ti 10 K 13 (i 18 It 20 K 22 4( 24 (< 29 Feb. 4 ii 27 Mar. 3 (< 15 To Bill rendered N. Y. Dft. Cash, Or. Balance due 275 41 315 07 798 10 176 42 215 84 193 76 505 75 97 22 108 47 214 29 307 62 184 36 3392 1200 450 275 500 2425 967 31 31 (3.) William Warren, 763 Madison St., City, lUUO. Apr. Milwaukee, Wis., June 12, 1890, Bought of HARRIS BROS. & CO. 2 2 O aJ 2 29 29 29 29 29 2 pairs Kip Boots 3. 75 2 " Ladies' Shoes 4.25 1 " Child'sShoes 1.10 1 doz. Linen Handk'ch'fs 1.80 2 Neckties _ 35^- 21 yd. Dress Silk 1.40 46" Bleached Cotton 11^ 15" Muslin.. 12^^ 5 " Broadcloth 2.25 Received Payment, Harris Bros. & Co., Per L. Harkis. Remark. — In retail business, where running accounts are kept with customers, a transcript 3f the charges, or of charges and credits, is made, giving items, dates of purchases and of pay- aaeats, and so partaking of the nature of both Statement and Bill. INVENTOKIES. INVENTORIES. 328. Find the amount of each of the following inventories (1.) Merchandise Inventory, J ax. l, 1888. 105 pc. F. A. Cambric 56 52 45 50 52 54 46 50^405 gr. Jet Buttons, pc. P. D. Goods 55 453 552 503 51 52 461 50 521 54 482 503 53 55150 pc. G. Flannel 353 40 402 403 pc. E. Lining 40 522 54 551 452 5o« pc. V. Barege 201 05 232 27 263 22 242 22 2G3 28 pc. B. H. Checks 45 52 55 41 402 513 511 53 508 46 pc. W. Prints 252 313 30 282 27 pc. A. F. Cashmere 621 653 601 G3 583 6O2 562 558 60 622 553 581603 58 55i pc. L. Gingham, 45 481 461 442 453 443 46 44 48 502 513 408 471 461 48 49 451 43 22^ 1.12i 1 50^ \ 25^' 34^- ' mt 1 2i(/- HfOMI>"ATE NUMBERS. DENOMINATE NUMBERS. 330. Denominate numbers may 1>e either simple or compound. 331. A Simple Denominate Nnmber is a unit or a eollection of units of but one denomination. 332. A Componnd Denominate Nnmber is a concrete mimher expressed in two or more different denominations; as 5 lb. 4 oz, 12 dr.; 4 yr. 7 mo. 12 da. Remark. — Compound denominate numbers are sometimes called compound numbers. 333. Componnd Numbers express divisions of time, and of the money, weights, and measures of the different countries. Remark. — Most denominate scales are varying, but the uniform decimal scale i? used throughout the metric system, and, except in Great Britain, in the money of most civilized countries. The units oi all denominate numbers are treated by the decimal scale. 334. A Denominate Fraction is a fraction expressing one or more of the equal parts of a denominate or concrete unit; as f of a ton, 4 of a yd., ^ of a gal. 335. Reduction of Denominate Numbers is the process of changing them from one denomination to another, without altering their value. It is of two kinds. Reduction Descending and Rednction Ascending. 336. Reduction Descending is the process of changing a denominate num- ber to an equivalent number of a lower denomination; as the change of barrels to an equivalent in gallons, quarts, pints, or gills. 337. Reduction Ascending is the process of changing a denominate num- ber to an efjuivalent of a higher denomination; as the change of gills to an equivalent in pints, quarts, gallons, or barrels. MEASURES OF TIME. 338. Time is the measure of duration ; its computations, being based upon planetary movements, are the same in all lands and among all peoples. 339. The Solar Day is the unit of time; it includes one revolution of the earth on its axis, and is divided into -iA: hours, counting from midnight to midnight again. 340. Noon, marked M. for Meridian, is that moment of time at which a line, called a Meridian, projected from the centre of the earth to the sun, would pass through the point of observation. 341. A. M. {Ante- Meridian) denotes the 12 hours before noon. MEASURES OF TIME. 109 342. p. M. {Post- Meridian) denotes the time between noon and the follow- ing midnight. Remarks.— 1. For astronomical calculations, the day begins at 12 o'clock noon, but for civil affairs, it begins at 12 o'clock midnight. 2. In banking business, the law fixes the end of the day at the hour appointed for closing the bank. 34-3. The Solar Year is the exact time required by the earth to make one complete revolution around the sun. It is equal to 365 days, 5 hours, 48 minutes, 49.7 seconds, nearly 365^ days. 344. The Common Year consists of 365 days for 3 successive years; and exery fourth year, except it be a centennial year, contains 366 days, one day being added for the excess of the solar year over 365 days; this day is added to tiie month of February, which then has 29 days, and the year is called LeajJ Year. The slight error still existing after this addition, is again corrected by excluding from the leap years the centennial years which are not divisible by 400. Thus 1900, 2100, 2200, while divisible by 4, are not divisible by 400, hence will not be leap years; while 2000, 2400, 2800, being divisible by 400, will be leap years. Remarks. — 1. The correction last named was made by a decree of Pope Gregory XIII., in 1685, and is known as the Gregorian calendar. It is used in all civilized countries except Russia, and is so nearly correct that an error of one daj' will not be shown for 4000 years, hence it is practically correct. 2. The calendar in general use previous to 1685 was known as the Julian calendar, having been established by Julius Caesar, 46 B. C. This calendar is still in use in Russia, and as the difference in the two calendars is now 12 days, the current date in Russia is 12 days behind that of the other civilized countries of the world; thus when it is Jan. 1 in Russia, it is Jan. 13 in all other countries. 3. The Julian and the Gregorian calendars are sometimes designated by the terms Old Style (0. S.), and New Style (N. S.) 345. Rule for Leap Years.— I- All years divisible by 4> e-vcept cen- tennial years, are leap years. n. JJl centennial years divisiUe hy 400 are leap years. Table. 60 seconds (sec. ) = 1 minute min. 60 minutes = 1 hour hr. 24 hours — 1 day da. 7 (lays = 1 loeek wh. Jf weeks = 1 lunar month mo. 30 days := 1 commercial month . . mo. 565 days = 1 common year yr. 566 days = 1 leap year yr. 12 calendar months = 1 civil year yr. 10 ijears = 1 decade 100 years = 1' 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. 7th. July (July) having 31 days. 8th. August (Aug.) " 31 - 9th. September (Sept.) " 30 '' 10th. October (Oct.) 31 " nth. November (Nov.) " 30 " 12th. December (Dec.) " 31 '' 346. The Calendar Months are as folk 1st. January (Jan.) having 31 days. 2nd. February (Feb.) " 28-29 " 3rd. March (Mar.) " 31 4th. April (Apr.) '' 30 5th. May (May) " 31 6th. June (June) " 30 347. 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 montiis are December, January, and February. The Si)ring months are March, April, and May. The Summer months are J^ine, July, and August. The Autumn months are September, October, and November. Remark.— The ancient Roman year began with March 1, and thus September, October, November, and December ranked, as their Latin derivation indicates, as the 7th, 8th, 9th, and 10th months respectively of the Roman year. REDUCTION OF TIME. 349. Tiie 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, Avhile that in the former is varying. 350. To Example, 3 vr, Operation. mo. 11 d. 7 hr. 12 30 mo. 7 mo. 43 mo. 30 1290 da. 11 da. 1301 da. 24 31224 hr. 7hr. 31231 hr. Reduce Time from Higher to Lower Denominations. — Reduce 3 yr. 7 mo. 11 da. 7 hr. 25 m. 38 sec. to seconds. Explanation. — Since one year 25m. 38 sec. 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. Operation Continued. 31231 hr. 60 1873860 m. 25 m. 1873885 m. 60 112433100 sec. 38 sec. 112433138 sec. 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. HI jj^nlg, Beginning with the highest, multiply the units of each denoiyv- ination hy the numher in the scale required to reduce it to the denoin- ination next lower; add the units, if any, of such lower denomination, and so continue from the given to the required denortvination. 351. To Reduce Time from Lower to Higher Denominations. Example.— Reduce 1124:33138 seconds to years. Opekation. " 60 ) 112433138 sec. Explanation.— Divide the given . seconds by 60, to reduce to minutes; 60 ) 1873885 mm. + 38 sec. ^^^ minutes thus obtained, by 60, to 24 ) 31231 hr. + 25 min, reduce to hours; the hours by 24, to 3oTl301 d'l + 7 hr reduce to days; the days by 30, to * ' reduce to months, and the months 12J_43 mo. + 11 da. ^ij 12, to reduce to years, 3 yr. 4- 7 mo. 112433138 sec. = 3 yr. 7 mo. 11 da. 7 hr. 25 min. 38 sec. Rule. — Divide the given units hy that numher in the scale ivhich will reduce them to units of the next higher denomination, and so continue from the given to the required denomination. Any remainder ohtained will he 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. Exp LAN A T I o N . — Arrange the 27 min. 14 sec. numbers so that those of the same 50 min 56 sec denomination stand in the same ver- ! '. tical line. Then begin with the 47 yr. 3 mo. 4da. 6 lir. 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. Hula.— Add as in ahstract numl)ers, and reduce according to the table of Time. Operation. 41 yr. 8 mo. 22 da. 10 hr. 5 yr. 6 mo. 11 da. 10 hr. 11*2 SUBTKACTION OF TIME. SUBTRACTION OF TIME. 353. Difference iu time is found in two ways: 1st. B)- counting the actual number of days from the given to the required •date. Thus, the number of days between May 13 and September 7 is 117, count- ing IS 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 mu. 'il da. from S yr. 1 mo. 18 da. Operatiok. Explaxatiox. — Write the numbers so that those of the S vr. 1 mo. IS da. same denomination stand iu the same column. Then begin -5 vr 4 mo 21 da ^^^^ ^^^ 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 sul)trahend, hence, borrow one from the next denomination; 12 months — 4 months = 8 mouths, which write underneath the column of months; there now remains 7 years from which to subtract; 7 years — 5 years = 2 years, which write imderneath the column of years. This completes the operation, giving a remainder of 2 years, 8 months, .and 27 days. Rule. — Snhtract as in abstract mnubcrs, ohseri-ing the i-arying scale. EXA3IPI.ES FOK PRACTICE. Remark. — In the following examples, the difference in time should be found by compound subtraction, unless it be otherwise stated. 355. 1. Reduce 2T051 seconds to minutes. 2. Reduce 83129 seconds to hours and minutes. 3. Reduce 610251 seconds to higher denominations. 4. How many years, months, days, hours, and minutes, in 749520360 seconds? 0. How many hours from half-past three o'clock p. m. Oct. 13, 1888, to noon on the fourth day of July, 1S89? 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. Remark. — The divisions into twelfths called signs, and into sixths called sextants, are in occasional use. Table. 60 seconds (") = 1 minute ('). 30 degrees — 1 sign {S.) 60 minutes = 1 degree {). 12 signs or 360° = 1 circle (C) q , j descending, 12, 30, 60, 60; or, 360, 60, 60. ^^^^®' \ ascending, 60, 60, 30, 12; or, 60, 60, 360. Remark. — Minutes of the earth's circumference are called nautical or geographic miles. EXAMPLES FOK PRACTICE. 358. 1. Reduce 2154' to degrees. 2. Reduce 87406" to degrees, minutes, and seconds. 3. Reduce 330581" to higher denominations. 4. How many seconds in a circle? o. How many minutes in 2 S. 21° 47'? 6. How many seconds in 1 S. 27° 8' 57"? Reduce 8162 geographic miles to degrees. How many geographic miles in the circumference of the earth? 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 b^ in south latitude if south of the equator. 360. Longitude is distance east or west from any given starting point or 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 degrees, and 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 -^ of 360°, or 15° of longi- tude, in 1 hour; through 1° of longitude in ^ oi I hour, or 4 minutes; and through 1' of longtitude in -gV of -J^ minutes, or 4 seconds. Table. 360° of longitude = '^4 hours or 1 day of time, da, 15°" " = 1 hour of time,-- hr. 1° ♦• *' = 4 minutes " -. min. 1' '• •' = 4 seconds '" ..- sec. Remark. — Standard Time. — Previous to 1883 there were fifty-three different time- staodards 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, lx)th 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 di\ides 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 Toth, 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 T5th 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 aU roads operated in the second belt are run by " Central Time." The time of the 10.5th meridian, one hour slower than " Central Time," is distinguished as " Moun- tain Time." Time in the fourth belt, which is governed by the 120lh meridian, and extends to the Pacific coast, is c-alled "Pacific Time;" it is one hoiu* 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 road,*, 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 Qo I 1 =/ 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 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, AXD TIME. 115 Rule.— Multiply the units of distance hy Jj., and reduce according to the table of Time. EXAMPLES FOR PKACTICE. Remark. — Examples under this topic will be restricted to variations of solar time. 363. 1. Cincinnati is 84° 24', and San Francisco 122°, west lonsritudc What is their difference in time? 2. New York is 74° 1', and Halifax 63° 30', west longitude. Find tlieir 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. Hemakk. — 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. Jf. 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 watcli, 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 DiflFerence 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. *? seconds equal 139 minutes and 48 seconds; . V .. _ _ . since each 4 minutes of time equal 1 degree 4 ) 139 mm. 48 sec. ^^ distance, 139 minutes and 48 seconds equal 34° + (3 min. 48 sec. ) 34 degrees, with 3 minutes and 48 seconds, or 3 min 48 sec = 2*^8 sec 228 seconds, remainder; and since each 4 sec- onds of time equal 1' 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'. Kule. — Reduce the difference in time to luinutes and seconds, and divide hy Jj. ; the quotient will he the difference of longitude, in degrees, minutes, and seconds. £XAMPI.£S 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, Avhen 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? S. Find the difference in latitude of Chicago, situated 41° 54' north, and Valparaiso, 33° 4' south. 116 REDUCTION OF ENGLISH MONEY. 4. What is the latitude of Washington, if it be 61° 46' 20' north of Kio Janeiro, and the latter place be 24° 54' south latitude? 5. When it is 20| minutes after noon at Washington, it is 21 niin. 26 sec. before noon at Chicago, 87° 30' west. What is the longtitude of Washington ? MISCELLANEOUS EXAMPLiJS. 366. i. 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? 2. Suppose the messenger be eastward bound, at what point will he meet the new year? 3. 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? 4. When it is noon at London, what is the time at St. Augustine, 81° 35' west? At Berlin, 13° 30' cast? At Xew Orleans, 90° west? At Sidney, 152° 20' east? At Paris, 20° 20' 22r east? At Xew York, 74° 3' west? ENGLISH MONEY. 367. EngHsh or Sterling Money is the legal currency of Great Britain. Table. 4 farthings ( far. ) = 1 penny d. 12 pence = 1 shilling s. 20 shillings =|l^;„3«"::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. The shilling = 24^ cents. The penny = 2^ cents. The pound = 14.8665. Remark. — The farthing is but little \ised, except as a fractional part of the penny. COINS OF GREAT BRITAIN. 369. The gold coins are tlie sovereign and the half-sovereign. Tte silyer 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-j^enny, and the farthing. The guinea (equal to 21 shillings) and the half-guinea are in xase, but are no longer coined. EEDUCTION 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 higiier denominations. Operation. Explanation.— Since 4 farthings equal one penny, 13206 far- things equal as many pence as 4 is contained times in 13206, or 4J_13306 far. 3301, plus 2 remainder, equal 3301 pence, 2 farthings; since 13 12 ) 3301 d. + 2 far. pence equal 1 shilling, 8801 pence equal 275 shillings, plus 1 9nT"97" -1- 1 r1 penny; since 20 shillings equal 1 pound, 275 shillings equal 13 _i-Jll^ ^' + -■■ ^- 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. ISillle.— Divide by the units iiv the scale from the given to tlie rcquived denomination . EXAMPLKS FOR PRACTICE. 371. Eeduce 1. 5124 s. to pounds, I 3. 13042 d. to pounds. 2. 11916 far. to shillings. I 4- 18T409 far. to higher denominations. 372. To Reduce English Money from Higher to Lower Denominations. Example. — How many farthings in £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. 2211 d. Rule. — Multiply by the units in the scale from the given to the 7'equired denomination. EXAMPLES FOR PRACTICE. 373. 1. How many pence. in £27? 2. How many farthings in 19 s. 11 d. ? 3. 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 24i cents, 3 shillings equal $.73; since 1 penny equals 2/j cents, 7 pence equal $.1414 ; since 1 farthing equals 1% cent, 2 farthings equal $.0101. Therefore, £15, 3 s, 7 d. 3 far. = $73.8789, or $73.88. Operation. £ 9, 4 s. 3 d. 2 far. 20 Operation 180 8. Continued 4 s. 2211 d. 184 s. 4 12 8844 far. 2208 d. 2 far. 3d. 8846. 118 REDUCTION OF ENGLISH MONEY. Second Expl.vn-atiox. — Call each 2 shillings -j^ of a pound, then 3 shillings equal £.15; call the pence and farthings, reduced to farthings, so many yg'j^g 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 14.8665, £15.18 equal 15.18 times $4.8665, or $73.88, as before found. Rules. — 1. Multiply each of the orders of Sterling Jtwiiey by its equiva- lent iji 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 to of a pound, and the pence antl farthings, reduced to faHhings, so inany nsW of a pound; multiply the pounds and decimals of a pound thus obtained by Ji..8665, and the product will be the answer in dollars and cents. Remark. — This is exact to within -s\^ of the part represented by the pence and farthings. KXAMPUi.S FOR PKACTICK. 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 -i d. J^. £3, 1 s. 9 d. and 2 far. 5. £11, 3 s. Id. 1 far. 376. To Reduce "United States Money to Sterling equivalents. Example. — Reduce 851G4.28 to equivalents in English money. Operation. 4.8665 ) 5104.28 £1061 + £ .189 rem. 20 3 s. + .78 s. 12 rem, 9d. + 36 d. 4 rem, Explanation.— Since $4.8665 equal £1, $5164.28 equal £1061.189; multiply the decimal by the units in the scale, 20, 12, 4, in order, pointing off as in decimals, and obtain 3 s. 9 d. 1 far., which, added to the £1061, equals £1061, 3 s. 9 d. 4 far. 1 far. + .44 far. Rule. — Divide the expression of decimal cujTency by ^.8665, and the integers of the quotient will be pounds Sterling ; reduce the decimal of the quotient, if any, by multiplying by the loicer units in tJie scale. EXAMPLKS rOK 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. 6. 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 loeight, with units of j^ounds, ounces, pennyweights, and grains, used for weighing precious metals. Avoirdupois weight, 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 2:^2.7944 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 pennyweiglits = 1 ounce oz. 12 ounces =: 1 pound lb. Scale -I descending, 12, 20, 24. | 1 lb. = 5760 grains. ~Z I ascending, 24, 20, 12. I 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. ^ yiust Explanation. — Since 1 pound equals 13 6 lb. 11 oz. 15 pwt. 21 gr. ounces, 6 pounds equal 72 ounces, and 72 ounces plus J^ 11 ounces equal 83 ounces ; since 1 ounce equals 20 72 oz. pennyweights, 83 ounces equal 1660 pennyweights, and 11 oz. 1660 pennyweights plus 15 pennyweights equal 1675 "gg Qj, pennyweights; since 1 pennyweight equals 24 grains, 2Q ' 1675 pennyweights equal 40200 grains, plus 21 grains equal 40221 grains. Therefore, 6 lb. 11 oz. 15 pwt. 1660 pwt. 21 gr. Troy, = 40221 gr. 15 pwt. Second Explanation.— Since 1 pound equals 576C ^, . 1 ' grains, 6 pounds equal 84500 grains; since 1 ounce equals 480 grains, 11 ounces equal 5280 grains ; since 1 pen- 40200 gr. nyweight equals 24 grains, 15 pennyweights equal 360 21 gr. grains; to these add the 21 grains, and the entire sum is 40221 gr. ^2^1 S*"*'°^- 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 bjf that nurtibcr 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 nuTnber of units of the desired equivalent u'liich it takes to mahe one of that denomination, and add the products thus obtained. 381. To Reduce Denominate Numbers from Lower to Higher Denominations. Example. — Reduce 40'^"^! gr., Troy, to higher denominations. First Operation. t?.- a- ^a • •, -, -i,* Explanation. — Smce 24 grains equal 1 pennyweight, 24 ) 402^1 gr. 40221 grains equal 1675 pennyweights, plus 21 grains; since •^0 'k M"^~ wt 4- 21 o-r ^^ 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- ,, , , fore, 40221 gr., Troy, = 6 lb. 11 oz. 15 pwt. 21 gr. 6 lb. + 11 oz. .. 6 . J i' 40221 gr., Troy, = 6 lb. 11 oz. 15 i)wt. 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 dsTTV^fifiT o- • M 1 o equal 1 ounce, 5661 grains equal 11 ounces, plus 381 grains; 5280 ' since 24 grains eqnal 1 pennyweight 381 grains equal 15 pennyweights, plus 21 grains. Therefore, 40221 gr. = 6 lb. 24 ) 381 gr. ( 15 pwt. n oz. 15 pwt. 21 gr., as before found. 360 21 gr. 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. — Reduce 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. — Reduce ^^Vfr l^^-? Troy, to the fraction of a pennyweight. First Operation. ^^ X V X V = ilU = If- Explanation.— t^Vtj of a PO"nd equals ^/^ of the 12 ounces in 1 pound, or ^f^^ ounces; ^fl^y of an ounce Second Operation. ^^"^^"^ tUtt of the 20 pennyweights in 1 ounce, or f|f§, 4 which equals |f pennyweights. Therefore, j/^^j lb., T^ X V X ^ = M pwt. Troy, = |^ pwt. 3 t "Rnie.— Multiply the fraction hy Ihe units in the scale, from the given to the required denomination. REDUCTION OF DENOMINATE NUMBERS. 121 383. 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. O "PER ATION Explanation.— I of a grain equals f of ^ of a pen- fX^X^XiV^ Tjhnr- nyweiglit ; | of -^ of a pennyweight equals | of ^ of 12 ^ff of ^^ ounce ; f of ^ of ^^ of an ounce equals f of ^ of ^ of yij of a pound, or xrinTT of a pound. *gr. = rrhirl^-' 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 y\ of a pound, Troy, to integers of lower denominations. Opekation. ^ Explanation. — ^\ of a pound equals -^^ of the 13 ounces •^ X V t *^^' '^t ^'^- in a pound, or \\ ounces, which reduced gives 2^ ounces ; * i of an ounce equals \ of the 20 pennyweights in an ounce, 1 ^ ^ K f or 5 pennyweights. Therefore, y^j of a pound, Troy, equals ' 2 ounces, 5 pennyweights. ^ lb., Troy, == 2 oz. 5 pwt. Rule. — Multiply the denominate fraction hy the unit next lower in the scale, and if the product he an iinproper 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, ~ - ,j. q . 7 ounces equal 140 pennyweights; 140 pennyweights plus 5 pen- ' ^ ' ° * ny weights 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 f ^|^ of a pound. 145 pwt. Second Explanation. — Since 1 ounce equals 480 grains, 7 OA ounces equal 83G0 grains; since 1 pennyweight equals 24 grains, -; 5 pennyweights equal 120 grains; 3360 grains, plus 120 grains, plus o4:hU gr. 9 grains equal 3489 grains; since 1 pound, Troy, equals 5760 grains, Q^gi"' 3489 grains equal ||§| of a pound. Therefore, 7 ounces, 5 pen- 3489 ST. = ?f g§ lb. nyweights, 9 grains, equal |lf§ of a pound, Troy. YivXe.— 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 ansiver sought. 122 ADUITIOX OF DENOMINATE NUMBERS. 386. To Beduce a Denominate Decimal to Units of Lower Denominations. Example. — Reduce .805 of a pound, Troy, to integers of lower denominations. Operation. .865 lb. 1 .^ Explanation. — .865 of a pound eqxials .865 ^ of the 12 ounces in 1 pound, or 10.38 ounces; 10.380 oz. .38of an ounce equals. 38 of the 20 penny weights ~ in 1 ounce, or 7.6 pennyweights; .6 of a penny- ■^ 60 pwt weight equals .6 of the 24 grains in 1 penny- n . weight, or 14.4 grains. Therefore, .865 of a pound, Troy, equals 10 ounces, 7 pennyweights, 14.4 gr. 14.4 grains. ,865 lb., Troy, = 10 oz. 7 pwt. 14.4 gr. Rule.— Multiply tlie decimal by tJiat unit in the scale which udll reduce it to units of the next loxver denomination, and in the product point off as in decimals. Proceed in liJce manner with all decimal remainders. 3JS 7 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 o"r 1 ° ■ Explanation. — Since 24 grains equal 1 pennyweight, 15 . 625 grains equal V\ or . 625 of a pennyweight ; 3 pennj-weights plus 3. pwt. .62o pennyweights equal 3.625 pennyweights; since 20 penny- 20 "> S 6^5 T)wt weights equal 1 ounce, 3.625 pennyweights equal .18125 of an — '- * ounce, and 8 ounces plus .18125 of an ounce equal 8.18125 .18120 oz. ounces; since 12 ounces equal 1 pound, 8.18125 ounces equal 8. oz. .68177iVof apound. Therefore, 8 oz. 3 pwt. 15 gr.=.68177i 12 ) 8.18125 o z. It)' Troy. .6817:tV1>J- YiViX^.— Divide the lowest denomination given hy the number in tJie 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 11). 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 lb. 2 oz. 18 pwt. 15 gr. Rule.— I Write the nunibers 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. pict. P-- 2 5 13 4 17 11 18 20 9 9 6 15 Ih. oz. pwf. gr. 23 4 17 6 11 1 13 y 11 lb. 9oz . 3 pwt. 21 gr MULTIPLICATION OF DENOMINATE NUMBERS. 123 SUBTRACTION OF DENOMINATE NUMBERS. 389. Example.— Subtract 11 lb. 7 oz. 13 pwt. 9 gr. from ^3 lb. -i oz. 17 pwt. 6 gr. 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. Yi\\\e.— 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 he greater than the minuend term, borrow 1 as in simple subtraction, and reduce it to the denomination required. MULTIPLICATION OF DENOMINATE NUMBERS. 300. Example. — Each of five bars of silver weighed IG 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, 21 Q2 yii^i Qj- and obtain 4 pennyweights, with a remainder of 9 grains; ,p q -irv f)-| multiply 10 pennyweights by 5, add the 4 pennyweights, and reduce to ounces bj' dividing by 20, obtaining 2 ounces, iL. 14 pennyweights; multiply 3 ounces by 5, add the 2 ounces 81 lb. 5 oz. 14 pwt. 9 gl". and divide by 12, obtaining 1 pound, 5 ounces; multiply 16 pounds by 5, add the 1 pound and obtain 81 pounds. Rule. — Beginning ivith 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 G i)late8 of equal weight, what will be the weight of each? Operation. Explanation. — One plate will weigh J as much as lb. oz. Vict. nr. Opiates. Write the dividend and divisor as in short divi- g-vi*- ^ 1.) -.o sion. Divide 7 pounds by 6, obtaining a quotient of 1 ; ;^ pound and an undivided remainder of 1 pound; reduce 1 lb. 3 oz. 5 i)wt. 1 1 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 DEXOMINATE 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. Unle.— Write the terms as in short division; divide as in integers, and reduce remainders, if any, to next lower orders hy 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 -.!,'» ' -.cT ', ,o * Anr^^^' the silver to be used is 43986 grains, as many 7 lb. 7 oz. 12 pwt. 18 gr. = 43986 gr. , , , , ^, • t. ^ i / r to to plates can be made as the weight of one plate, 43986 gr. ^ 7331 gr. = 6 7331 grains, is contained times in the 43986 grains to be so used, or 6 plates. Rule. — Reduce the dividend and divisor to the same denomination, and divide as in simple numbers. JKXAMPLKS FOK PRACTICE. 393. 1. Reduce 31 lb. 10 oz. 13 jiwt. to pennyweights. 2. How many grains in 27 lb. 17 pwt. 20 gr. ? S. How many pounds, ounces, and pennyweights in 230.51 gr. .^ J^. Reduce 30297 grains to higher denominations. 5. Reduce -^^ of a pound to grains. 6. ^tVtt of a pound is what part of a pennyweight ? 7. -^7 of a grain is what fraction of an ounce? 8. Reduce -^ of a pennyweight to the fraction of a i)()und. 9. Reduce ^ of a pound to lower denominations 10. Reduce f of an ounce to lower denominations. 11. Reduce 9 oz. 1 i)wt. 21 gr. to the fraction of a pound. 12. What fraction of a pound equals 11 oz. 11 pwt. 18 gr. ? 13. What is the value in lower denominations of .6425 lb. ? H. 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., 4() 11>. l»i pwt.. l(»o lb. 9 oz. II 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 tlie difference between 19 lb. 9 oz. 11 pwt. and 11 oz. 16 pwt. 22 gr. ? 2S. What will be th6 cost of 15 gold chains, each weighing 1 lb. 3 oz. 18 pwt. 18 gr., at 7^ per grain ? 2J^. I bought 7 lb. 7 oz. 12 pwt. 18 gr. of old gold, at 81.05 j^er pwt. AVhat was the sum paid ? 25. A manufacturer made 18 vases from 7 lb. 8 oz. 8 pwt. 18 gr. of silver. What was their average weight ? 26. If 12 rings be made from 1 lb. 8 oz. of gold, what will be the weight of each ? 27. A miner having 77 lb. 10 oz. 5 pwt. of gold dust, divided \ of it among nis laborers, and had the remainder made into chains averaging 3 oz. 3 pwt. 3 gr. of pure gold each. If he sold the chains for $52.50 each, how much did he receive for them ? 28. What is the aggregate weight of five purchases of old silver, weighing respectively 4 lb. 9 oz. 20 gr., 13 lb. 17 pwt. 22 gr., 20 lb. 1 oz. 17 pwt. 4 gr., 8 lb. 2 oz., and 27 lb. 12 pwt. 21 gr.? 29. I bought 27 lb. 11 oz. 1 gr. of old silver, and after having used 15 lb. 15 pwt. 15 gr., sold the remainder at 5^ per pwt. What quantity was sold, and how much was received for it ? 80. A goldsmith bought 3 lbs. 9 oz. 1 pwt. 16 gr. of old gold, at 80^ per pwt., and made it into pins of 40 grains weight each, which he sold at $2 apiece. How much did he gain or lose ? AVOIRDUPOIS WEIGHT. 394-. Avoirdupois Weight is used for all ordinary purposes of weighing. Table. 16 ounces = 1 pound lb. 100 pounds z= 1 hundred-weight. . cwt. 20 hundred-weight., or 2000 pounds = 1 ton T. Scale, descending, 20, 100, 16 ; ascending, 16, 100, 20. Remark. — At the United States Custom Houses, in weighinggoods on which duties are paid and to a limited extent in coal and iron mines, the long ton of 2240 pounds is still used. Long Ton Table, 16 ounces = 1 pound lb. 28 pounds = 1 quarter qr. 4 quarters, or 112 pounds = 1 hundred-weight... cwt. 20 hundred-weight, or 2240 pounds = 1 ton T. 126 AVOIRDUPOIS WEIGHT. Table of Avoirdupois Pounds per Bushel. Wojth. Territory. X X ' CD W ©» 1 CO » » CD i § Wisconsin 00 >* X X «o cj cj la CO C» eo CD CO O CO •* CO Vertnont CO IS CJ CO CO LO CO Bhode Island. .. ; ; ; g g : i Pennsylvania. .. X g '■• CO ■ ts s Oregon 01 ->* X 00 (M .": sc CO 13 -* CO CJ Minnesota. X ci ■^ X X CS> OJ « CO 10 CJ so CO CD Michigan X 5? X X CJ CI CO us CJ 00 CO CO Massachusetts. .. ^ CO eo ic » CO Maine. ■ eo CO I Louisiana. CJ eo ; CO Kentucky. X o o -* -?> -r o X — 1-t » -* «D -^ IC eo CO o Lo tJI CO c* Iowa xoo-fc-»oo-^:c«o-*»XL-t-05DOooo ■^;cx^is-*off*eoo-*io;oe30i5i.';i0'*0(?» Indiana xoO'rooo««eo»-^ox»>xc^o»-'5o > 'T o e- i-i o -*< « ci « »o -r L-; ;c ec 'T ec Lt « Tf ee Illinois - LO '* CO CJ Connecticut ; 1 -i^ ; CO 10 X S «5 5 ' California o i CJ 1.0 so -1' CO CO o O ■> a C c , C . C ' a ' C , c : 1 ' CO ) S es d & 5 5 ^ - C cS C c: u "5 JS (■ a ' > . c - _a < • c a : 5 P > cc B i. - c _ c a c a .1 . 21 a ■I c a C 1 > 1 > J 5 -- ^-^ '■: v: f- '-c c^ :: •- '^, '^- -t- i;; ;:^ ?^ v; cv ^ *^ ^ 05 EXAMPLES FOR PRACTICE. 127 Additional Table of Weights of Products, As usually given, but varied by tbe laws of different States : Apples, green, 56 lb. per bushel. Charcoal, . . - .22 lb. per bushel. Hungarian Grass Seed, 45 lb. per bushel. Malt, 38 lb. per bushel. Millet, .45 lb. per bushel. Mineral Coal, 80 lb. per bushel. Peas, 60 lb. per bushel. Potatoes, sweet, 55 lb. i)er 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),. 330 lb. per barrel. Cider, .400 lb. per barrel. Corn Meal, .200 lb. per barrel. Eggs, 180 lb. per barrel. Fish, 300 lb. 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, ...300 1b. per barrel. Vinegar, 400 lb. per barrel. Whiskey, 400 lb. per barrel. Estimates on Lumber, Wood, Etc., foi* Freighting. Pine, Hemlock, and Poplar, seasoned, per M, 3000 lb. Black AValnut, 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 to the explanations and rules under Troy Weight. KXAMPLES FOR PRACTICE. 395. i. How many pounds Avoirdupois in 17 T. 6 cwt. 69 Ih. 'i ^ Reduce 31275 lb. Avoirdupois to higlier 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. 3. 4. 5 . 6. 7. 8. ■ 9. JO. 128 apothecaries' weight. 11. Wliat is the sum of T 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 respectiyely 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. Wliat was their average weight? APOTHECARIES' WEIGHT. 396. Apothecaries' Weight is used by druggists in retailing, and by apothecaries iu 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 FOK PRACTICE. Remark. — For assistance, refer to rules and explanations under Troy "Weight. 397. 1. Eeduce 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 ? J^. Eeduce .955 of a pound to lower denominations. 5. How many scruples in -^ of a pound ? ' 6. Add 6/o lb., 7y5^ oz., 3| dr. and 2| sc. 7. Find the sum of ^-i 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 '^^^ lb.? 10. In comjjounding 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 = 480 grains = 437.5 grains. 175 pounds = 175 pounds = 144 pounds. QuESTioxs. — 1. Which is heavier, a pound Troy or a pound Avoirdupois ? 2. Which is heavier, an ounce Troy or an ounce Avoirdupois ? MEASURES OF CAPACITT. 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 «bout3.3 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 PKACTICE. 398. 1. A dealer bought 131 lb. 5 oz. of drugs by Avoirdupois Aveight, at t;G.25 per pound, and retailed them at 5^ per scruple. What was his gain ? 2. How much is gained or lost by buying 23 lb. 4 oz. of medicine by Avoirdu- pois weight, at 50^ per oz., and selling by Apothecaries weight, at 1^^ per grain? S. Reduce 5f lb.. Avoirdupois, to Troy units. • Jf.. What is the remainder after subtracting hl^^ lb. Troy from 60 lb. 10 oz. Avoirdupois ? 5. I bought by Avoirdupois weight 45 lb. 6 oz. of drugs, and from the stock sold by Apothecaries weight 29 lb. 4 oz. 3 dr. 1 sc. 10 gr. What is the remainder worth, at 75j^ per Avoirdupois ounce ? 6. Having bought \l of a pound of roots by Avoirdupois weight, I sold H of a pound by Apothecaries weight. What was the remainder worth, at 10^ per scruple ? MEASURES OF CAPACITY. 390. Dry Measure is used for measuring grains, seeds, fruit, vegetables, •etc. — all articles not liquid. 'J'ho units are pints, quarts, pecks, and bushels. Table. 2 pints (pt.) = 1 quart qt. 8 quarts = 1 peck. pk. 4 pecks = 1 bushel bu. Scale, descending, 4, 8, 2 ; ascending, 2, 8, 4. Remakks. — 1. The United States Standard Unit of Dry Measure is the bushel, which, as a •circular measure, is 18i inches in diameter and 8 inches deep, contains 2150.42 cubic inches, and is in uniform use for measuring shelled grains; while the Jieaped bushel of 2747.71 cubic inches is used for measuring apples, roots, and corn unshelled. 2. The British Imperial bushel contains 2318.19 cubic inches. The English Quarter men- tioned in prices current = 8 bu. of 70 lb. each, or 560 lb. avoirdupois = \ long ton. 3. For weights of different commodities, refer to Table, page 125. EXAMPLES EOK PKACTICE. Remark. — For assistance, refer to rules under Tkoy Weight. 400. 1. How many pints in 14^ bu. ? 2. Reduce 9 bu. 1 pk. 3 qt. 1 pt. to i)ints. S. Add 51 bu. 3 \)k. 1 pt. ; 4G bu. 2 pk. ; 37 bu. 2 (it. 1 pt. ; 51 bu. 1 pk. 7 qt. 4- From I of a bushel, take f of a peck. 9 130 LIQUID MEASURE. 5. What is the difference between Tt bu. and 2 bu. 2 pk. 2 qt, 1 pt. ? 6. A teamster's 12 loads of wheat measured 1000 bu, 1 pk. 6 qt. 1 pt. How much was the average of each load ? 7. What will be the cost, at 45^ per bushel, of 5 loads of oats, weighings respectively 2619 lb., 2554 lb., 2124 lb., 3051 lb., and 2745 lb.? LIQUID MEASURE. 401. Liquid Measure is used for measuring water, oil, milk, cider,, molasses, etc. The units are gills, pints, quarts, gallons, and barrels. Table. 4 gills (gi.) = 1 pint. pt. 2 pints = 1 quart qt. 4 quarts = 1 gallon gal. 31^ gallons = 1 barrel bar. or bbl. Scale, descending, 31^, 4, 2, 4 ; ascending, 4, 2, 4, 31^. Remarks. — 1. The standard unit of Liquid Measure is the gallon, which contains 231 cubic inches. , 2. Casks, called hogsheads, pipes, butts, tierces, tuns, etc., are indefinite measures; their capacity, being determined by gauging, is usually marked upon them. 3. In sales of oils and liquors, and in certain other cases, the barrel is also of indefinite capacity. EXAMPUES FOR PKACTICE. Remark. — For assistance refer to rules under Troy Weight. 402. 1. How many gills in 5 bar. 27 gal. 3 qt. 1 pt. of cider? 2. Reduce 31479 gi. to higher denominations. 3. .046 of a barrel equals how many gills? 4. From .895 of a barrel take 21 gal. 2 qt. 1 pt. 1 gi. 5. From a cask containing 68 gal. 4^ qt. of wine, 1.625 bar. were sold. What was the remainder worth, at 50^ per pint ? 6. A reservoir contained 896 gal. 2 qt. of water, and its contents were put into 116 kegs. What was the quantity put into each ? 7. From | of a barrel take 4 gal. 1 qt. 3 gi. 8. If 2 qt. 1 pt. 1 gi. of oil be consumed per day for the year 1890, what will be its cost for the year at 8^ per gallon ? 9. From a cask of brandy containing 69 gal. 1 pt. and costing $3.75 per gallon, one-fourth leaked out, and the remainder was sold at 20^ per gill. What was the amount of gain or loss ? Comparative Table of Dry and Liquid Measures. Cu. in. in one Cu. in. in one Cu. in. in one gallon. quart. pint. Dry Measure (^pk.)268f 67^ 33|. Liquid Measure 231 57| 28|. LINEAR MEASTJBE. 131 Remarks. — 1. A pint of water weighs about 1 pound, Avoirdupois. 2. Potatoes and grains are tisually sold to dealers and shippers by weight. 3. The beer gallon of 282 cubic inches is nearly obsolete. EXAMPLES FOR PKACTICE. 403. 1. Keduce 21 bu. 6 qt. 1 pt., dry measure, to pints, liquid measure. 2. A grocer bought 12 bu. 3 pk. 3 qt. of chestnuts by dry measure and when selling used a liquid pint, measure. How many pints did he gain by the change ? 3. A bushel of cherries, bought at 10^- per quart, dry measure, was sold at the same jmce per quart, wine measure. How much was thereby gained ? 4. A cask of cranberries, containing 5^^ bu., was bought for $15, and retailed at 10^ per quart by wine measure. What was the gain ? 5. A blundering clerk bought of a gardener 375 quarts of currants, measur- ing them by a liquid quart measure, and when selling used a dry quart measure. If he bought at 6^ per quart and sold at 7^, 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. Remakks.— 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 secomUin 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 di\men&ion&— length and breadth. 407. A Solid has three diimen&ions— length, breadth, and thickness. LINEAR MEASURE. 408. Linear or Long Measure is used in measuring lengths and distances. Table. 12 inches (in.) = i foot ft. 3 feet =1 yard yd. 5^ yards, or 1G| feet = 1 rod rd. 320 rods.. = 1 statute mile mi. Scale, descending, 320, 5^ 3, 12 ; ascending, 12, 3, 5^, 320. 1 Mile = 320 rods, or 5280 feet, or 63360 inches. 132 SQUARE MEASURE, Special Table. ^ of an inch = 1 Size, applied to ^hoes. 18 inches = 1 Cubit. 3.3 feet = 1 Pace. 5 paces = 1 Rod, 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 3 geographic miles = 1 League, used for measuring distances at sea. 00 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. 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. Deirrees are of unequal length; those of latitude varying from 68.72 miles at the Equator to 69.3-1 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. KXA3IPI.ES FOK PRACTICE. Remark. — For assistance refer to Rules under Troy Weight. 409. 1. Eeduce 2 mi. 1 rd. 7 ft. to inches. Reduce 2501877 inches to higher denominations. AVhat part of a mile is ^V of a foot ? Reduce f of a mile to integers of lower denominations. What fraction of a rod is 11 ft. 2 in. ? Reduce .542 of a mile to integers of low^er denominations. Reduce 285 rd. 7 ft. 4 in. to the decimal of a mile. 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 Sei'via 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 j)roceeding from a common point called the vertex. 2. 3. 4. o. 6. 7. 8. Angle. SQUARE MEASURE. 133 Two Right Angles. 413. A Right Angle 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 he 2)erpe7idicidar to each other. E and F are right angles, and the lines A B and C D are perpendicular to each other. Rectangle. 3 inches. Contents Six inclies. 3X2 in. =6 8q. in. 3 feet. '•one! ' ! ^°^r [QUE square: YARD 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. 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. 9 square feet = 1 square yard . sq. ft. sq. yd. 30i square yards, or ( _. , j 272i square feet ....\ - - - - i square loci sq. i a. IGO square rods = 1 acre A. 040 acres = I square mile sq. mi. 36 square miles (6 miles square), = 1 township Tp. Scale, descending, 36, 640, 160, 30^, 9, 144; ascending, 144, 9, 30^, 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 MBASFEE. 417. The Unit of Land Measure is the acre, equal to 208.71ft. x 208. 71 ft. 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 hoard measure. 419. The Unit of Board Pleasure is a square foot surface, oue inch thick, called a hoard foot. 420. To Find the Number of Board Feet in a Board. llule. — Multiply the length in feet hy the ividth in iivches, and divide by 12; the quotient jrill he the iiumher of square feet. Rkmark. — If the 'board tapers evenly, find the mean or average width, by adding the width of the two ends, and dividing by 2. 4*21. To Find the Number of Board Feet in Timbers or Planks. ^w\e.— Multiply the length in feet hy the product of the ividth and thickness in incites, and divide by 12. 422. To Find the Number of Squares in a Floor or Roof. Remark. — In flooring, roofing, slating, etc., the square, or 100 square feet, is used as a unit of measure. Rule.— Poi;/^ off two decimal places from the right of the numher 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 ivill be the length, in yards, of 1 strip of the carpet. II- Divide the other dimension by tlie width of the carpet, and the quotient will be the iiuviher of strips it will take to cover the floor. III. Multiply the length of each strip by the number of strips, and the product will be the nmnhcr of yards required. Remark. — 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. KXAMPLKS rOK I'KACTICK. Remark. — For assistance refer to rules under Troy Weight. 424. 1. Reduce 5 A. 110 sq. rd. 7 sq. ft. to square inches. ~. Eeduce 4 sq. mi. 527 A. 105^ S(|. rd. to square feet. 3. Reduce .1754 of a S(iuare mile to lower denominations. SQUARE MEASURE. 136 4.. Reduce \^ of an acre to lower denominations. 5. What fraction of a square mile is j\ of a square foot? 6. "What decimal part of an acre is 150 sq. rd. 3 sq. yd. 7 scj. ft. 100 rsq. in. V 7. From .0375 of an acre take \^ of a square rod. 8. To the sum of ^, f, and -^ of an acre, add .0055 of a square mile. 9. How many squares in a roof, each side of which is 2G x CO feet? 10. How many yards of carpet, 1 yard wide, Avill be required to cover a floor 10.5 yd. long by 6 yd. Avide, if no allowance be made for matching ? 11. IIow many feet in 8 boards, each 15 ft. long, 9 in. wide, and 1 in. thick? 12. How many feet in 15 boards, each IG ft. long and 1 in. thick, the boards being 13 in. wide at one end and 10 in. at the other? 13. How many acres in a square field, each side of which is 04 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? 13. A field 87^ rd. wide and 240 rd. long, produced 27f bu. of wheat to the 4icre. What Avas the crop Avorth, at 90^ per bushel? 16. A farm in the form of a rectangle is 75 rd. Avide; if the area is 107.5 A., hoAV long is the farm? 17. I wish to build a shed which will coA'er f of an acre of land. If the Avidth of the shed is 42 ft., what must be its length? 15. 17.75 bu. of timothy seed is sown on land, at the rate of 6 lb. per acre. What is the area thus seeded? 10. What is the difference between a square rod and a rod square? 20. What is the difference between two square rods and tAvo rods square? 21. A square yard will make how many surfaces 5 in. by 9 in. ? 22. IIow many acres of flooring in a six-story block 100 ft. by 220 ft. ? 23. A rectangular field containing 10^ A. is 45 rd. wide. What is its length? 24. How many fields, each of 10 A. 50 sq. rd. 21 sq. yd. 5 sq. ft. and 28 sq. in., can be formed from a farm containing 124 A. 40 sq. rd. 10 sq. yd. 8 sq. ft. 48 sq. in. ? 25. HoAv many acres in v^ road 17200 ft. long and 00 ft. wide? 26. AVhat Avill be the cost, at $3.50 per M, of the shingles for a roof 26 ft. Avide and 110 ft. long, if the shingles are in. Avide and 4 inches of their length be exposed to the Aveather? 27. A hall 7| ft. Avide and 19| ft. long is covered witli oil cloth, at 05^ per .sq. yd. HoAv much did it cost? 28. If a farm of 100 A. 94| sq. rd. is divided equally into 11 fields, Avhat will be the area of eacii field ? 29. Reduce 240089740 sq. in. to higher denominations. 30. HoAv many rods of fence Avill enclose 100 A. of land lying in the form of a square ? 31. IIow many additional rods aviII divide the farm into four fields of ecjual .iirea ? 136 SC^lARE MEASURE. 32. How many yards of brussels carpeting, f of a yard wide, laid length- wise of the room, will be required to cover a room 23 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. 3S. AVhat will it cost, at 21^- per sq. yd., to plaster the sides and ceiling of a room 24 ft. by 3U ft. and 10^ ft. his:li, if one-sixth of the surface of the sides is taken up by doors and windows ? 34.. 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 10^ ft. by 21 ft., if the price be G2^^' 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 Avaste 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, £ of a yard wide, Avill 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 hiid 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 per yard ? 39. How many sheets of tin, 20 in. liy 14 in., will be required to cover a roof 60,5 ft, wide and 156.25 ft. long ? 40. What is the difference between four square feet and four feet square ? 41. 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.? 42. 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. ? 43. 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., Avill be required to pave a mile of roadway 42 ft. in width ? 45. What will be the coot, at 20/' per s(|. 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 '' 46. I wish to floor and ceil a room 27^^ yd. long and 15 yd. 2 ft. wide, with matched pine. What will be the cost of the material, at |!26.40 i>er 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, G-i 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 poiver. 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. • KXAMPLKS rOIt PKACTICE. 430. 1' What is the square of 1 ? 2. What is the square of 3 ? 3. What is the square of 4 ? Jf. What is the square of 5 ? 5. What is the square of 9 ? 6. AVhat is tlie square of 10 ? 7. What is the square of 99 ? 8. What is the square of 250 ? 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 I 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 tlie given number; as, 4 x 4 x 4 = 64; 4 is the root fnmi which the number 64 is produced. SQUARE ROOT. 433. The Square Root of a Number is such a number as, multij)lied 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 extractinfj 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 diuieusious being given, the wanting dimension is found by dividing the surface ])y tlie given dimension. 138 SQUARE ROOT. The accompanying diagram is a square 14 feet by 14 foot. 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 tc7is by the nnits of the same number, or 2 (10 x 4) = 80 square feet, as shown by the surfaces within the angles e, f, g, h, and /, j, k, I. 3d. The square of the units, 4 feet, or the product of 4 ft. by 4 ft. = 16 square feet, as shown by the square within the angles w, x, y, z. 14 Et.MO Ft. & + F b. ^ 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 Priuciples. — Tlie square of any mimher composed of tivo or more fif/ures is equal to the square of the tens, plus twice the inoduct of the tens multiplied hi/ the units, plus the square of the iniits. a dl e h .A* h * oK L. b c f G II. i I w S ^ J K X Y 437 I'.vrrs, 1' = 2' — 3' = 4' = 5' = 6' = »vj t — b' = 9' = 10' = TJnits and Squares Compared. Sqiakks. Remark. — Squaring the numbers from 1 to 10 inclusive, shows: 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. "VThere 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. 1 4 9 16 25 36 49 64 81 100 438. Example. — Find the square root of 625. Operation. 0.25 )2 5 4= 400 ExPL.\NATioN. — The number consists of one full and one partial period; hence its root will contain ^?ro places — tens and units. The given number, G25, 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 2, and 9, the square of 3, its root must be 2 tens with a remainder. Subtracting 20 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 zinits 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, ticice th€ 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 niunher 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 remaUvder hring down the next period. III. Divide the Tiumher 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 bring 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 deteimining subsequent root figures- Rem.\rks. — 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. KXAMPLESi I'OK VKACTICK, 4:31). 1. Find the square root of ] 96. 2. Find the square root of 225. S. Find the square root of 144. Jf. Find the square root of 576. 5. Find the square root of 1225. 6. Find the square root of 5025. 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 tlie 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 tlie square root of 106.413, carried to four decimal places. 13. What is the square root of -j^ ? IJf. What is the square root of ff ? 15. What is the square root, decimally expressed, of ||, carried to three decimal places ? -76\ What is the square root, decimally expressed, of ^\\, carried to two decimal places? 11. 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. 4:40. A Triangle is a plane figure having three sides and three angles. 4-il. 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, h, in fig. S. 443. The Hypothenuse is the side opposite the TRIANGLE. ^.jgj^^ ^^gj^. ^g^ ^^^ ^^ jj^ ^g_ g_ Fig. T. is a triangle, having angles at a, h, c. Fig. S. 444. A Right-angled Triangle is a triangle liaving a right angle. Fig. S is a right-angled triangle, the angle at b being a right angle. The line a h is the Perpendicular; tlie line h c \s the Base; the line ac is tlie 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 the peqiendicular RIOHT-ANOLEIJ TRIANGLE. 445. To find tlie liy])Otlieiiuse, when the base and perpendicular are given. Rule. — To the square of the base add tlie square of tlie iierpendicular, and extract the square root of their sum. To find the base, when the hypothenuse and perpendicular are given. Rule. — From the square of the h^jpothcnuse 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 tlie remainder. i:XAJ»IPL,K8 FOK PRACTICE. 446. 1. The base of a figure is G 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 S. The hypothenuse is 110 ft. and the base is 19.5 ft. Find the perpendic- ular to two decimal places. Jf.. Tlie 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 ^ of the perpendicular. Find wanting terms to three decimal places. Remarks. — 1. Observe, in example 4, that the square root of l_ 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 tlie form of u 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 21 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. IIow 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. AVhat 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 QQ feet, in length, and consisting of 100 links. Remark. — Rods are seldom used in Surveyor's Measure, it being customary to give distances jn chains and links or hundreths. Table. 7.92 inches = 1 link ... 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. 448. ■3 £XA9IPI^S FOR PRACTICE. Reduce 3 mi. 27 ch. 19 1. 4 in. to inches. Reduce 14841 1. to higher denominations. 3. 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 cli. 21 1. Avide, will require li6w many feet of fence to enclose it? 7. How many rods of fence wire will enclose a farm "il ch. 50 1. long and 18 ch. 60 1. wide, if tlie fence be made 6 wires high ? 8. A garden is 307f feet long and 250| feet_wide. What is tlie 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 ? SURVEYOR'S SQUARE MEASURE. 449. The ITnit of land measure is the acre. Table. 625 square links (sq. 1.) = 1 square rod sq. rd. 16 square rods =1 square chain. . sq. ch. 10 square chains, or ) _ ^ Y 160 square rods f "~ 640 acres =1 square mile sq. mi. ' Remark. — In surveying United States lands, a selected Korth 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 surface of the earth being convex, these merid- ians converge slightly. The towu.ships 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 Range; 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 CUBIC MEASURE. 143 N. 1 Mile. g^ N. }4 Section. 320 Acres. 8. W. )i Sec. 160 A. W.^of S.E.J^ Sec. 80 A. N. B. % of S. E. 40 A. S. E. H of S. E 40 A. 8. 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 FoA PRACTICE. 450. i. 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 See. 16, and mark its acreage. 3. 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. 4 of 27, and mark their acreage. CUBIC MEASURE. -^ / / ;FOOT 3 FT. N V V - \v^ V \ V X^ \ \ \ \ \ \ 451. Cubic Measure is used in measuring solids or volume. 452. A Solid is that which lias length, breadth, and thickness ; as the walls of Ijuildings, bins of grain, timber, wood, stone, etc. 453. A Cube is a regular solid bounded by six equal square sides, ot faces ; hence its length, breadth, and thickness are equal. 454. The Measuring Unit of solids is a cube, the edge of which is a linear unit. Thus a cubic foot is a cube, each edge of Avhich is 1 foot ; a cubic yard is a cube, eacli edge of which is 1 yard. See the accom- panying diagrams. CUBIC YARD 144 CUBIC MEASIRE. 455. To Find the Volume of a Solid. Rule. — Multiply iogethei its IcngtJt, hicadth, and tlii^ikueiis. Table. 1728 cubic inches (cu. in.) = 1 cubic foot cu. ft. 27 cubic feet.. = 1 cubic yard cu. yd. 128 cubic feet . = 1 cord of wood ..cd. Special Cubic Measures. 100 cubic feet = 1 register ton (shipping). 40 cubic feet = 1 freight ton (shipping). 16^ cubic feet = 1 perch of masonry. 456. A Cord of ■wood is a pile 8 feet long, 4 feet -wide, and 4 feet high. 457. A Cord Foot g is one foot in length of sucli a pile. 458. To Find the Cubical Contents of Square Timber. Bule. — Multiply together the feet ineasurements of length, width, and depth . 459. To Carry Timbers, one person supporting an end and two others with bar. DiKFxnoNS. — Let the tiro with the bar lift at a point J the length from the end. REM.VRK. — 1. Formerly a perch of masonry was 24| cu. ft.; but the perch of 16| cu. it., which is 16^ ft. long, 1 ft. high, and 1 ft. wide, is now in general use. 2. A cubic yard of earth is called a load. 3. Mechanics estimate their work on walls by the girt, and no allowance is made for windows or doors. In estimating the amount of material required, such allowances are made. Formulas for Rectaugular Solids. Lemjtlt X Breadth x Higlit = Volume. Volume H- {Length X Breadth) = Eight. Volume -=- {Length X Sight ) = Breadth. Volume -=- (Breadth x Hight) = Length. Remark. — The three given dimensions must be expressed in units of the same denomination. 46<). To Find the Number of Bricks for a WaU ^\\U.— Multiply the cubic feet by 23%, and add %U. Rem.\rk. — For guide in purchasing material the above will be found correct for bricks 8 in. X 4 in. x 2 in., after allowing for mortar. EXAMPLES FOR PRACTICE. 145 461. To Find the Number of Perches in a Wall. Rule. — Divide the contents of the wall, in feet, hy 16H. EXAIMTPLES FOK PKACTICE. 462. -?. Reduce 468093 cu. in. to higher denominations. 3. Reduce 132 cu. yd. 11 cu. ft. 981 cu. in. to cubic inches. 3. What is the volume of a solid 8 ft. 3 in. long, 5 ft. 10 in. high, and 4 ft. •6 in. wide ? 4. How many cubic feet of air in a room 26 ft. 8 in. long, 22 ft. 6 in. wide, and 12 ft. high ? 5. How many cubic yards of earth must be removed in digging a cellar 60 ft. long, 30^ ft. wide, and 7^ ft. deep ? 6. How many perches of masonry, of 16^ feet each, in a wall 85 ft. long, 32 ft. high, and li ft. thick ? 7. Reduce -| of a cubic inch to the fraction of a cubic yard. 8. What decimal part of a cubic yard is 7 cu. ft. 108 cu. in. ? 9. What fraction of a cubic foot is 220 cu. in. ? 10. Reduce .525 of a cubic yard to lower denominations. 11. What will be the cost, at 21(^' i)er cubic yard, of excavating for a reser- voir 180 ft. long, 105 ft. 3 in. wide, and 15 ft. 9 in. deep ? 12. What will be the cost of building the walls of a block 140 ft. long, 66 ft. wide, and 57 ft. high, at $1.40 per perch of 16^ cu. ft., if the wall is 16 in. thick, and no allowance bo made for openings ? 13. How many common bricks will be required for the above wall, allowance being made for 28 windows each 3^ ft. wide and 8 ft. high, 48 windows each 3 ft. 9 in. wide and 8 ft. high, and 4 doors each 8 ft. wide and 11 ft. high ? 14-. A room 28 ft. long, 18 ft. wide, and 12 ft. high, will store how many •cords of wood ? lo. How many cords of wood in a pile 108 ft. long, 7 ft. 9 in. high, and •6 ft. wide ? 16. From a i)ilo of wood 71 ft. 6 in. long, 9 ft. 4 in. wide, and 6 ft. 8 in. bigh, 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. G in. thick, and 8 ft. 4 in. wide, if it weiglis 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^ in. 22. How many perches of masonry in a wall 7^ 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 i46 producers' and dealers' approximate rules. 23. What is the volume of a rectangular solid 11 ft. long, -i^ ft. wide, and 4 ft. high ? 2j^ a cask holding %bQ\ gal. of water will hold how many bushels of wheat?" PRODUCERS' AND DEALERS' APPROXIMATE RULES. 163, To find the contents of a bin or elevator in bushels, stricken measure. Rule. — Miiltiply the cubic feet bj/.S, 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 jlare, take the mean width. To find the number of shelled bushels in a space occupied by unshelled com^ Rule. — Divide the cubic inches by SSJfO, or multiply the cubic feet by Jf5. 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 sutn will be the cubic feet required, to loithin 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, ivill 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 Bum 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 1(J^ cubic feet in a perch. Rule. — Multiply the cubic feet by .0606. Remark.— The above is correct within ^^^^ 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 3 ft. thick? Explanation.— 5A<^< Metliod.—lbQ X 50 x 2 = 15000; 15000 x .0404 = 606; add y^, or .606 = 606.606. Extended Method.— 150 x 50 X 2 = 15000; 15000 h- 24.75 = 606.6, same as before. Same example, perch of 10^ cu. ft. Explanation. --SAorf Met/u>d.— 150 x 50 x 2 = 15000; 15000 x .0606 = 909; add j^'^^ = .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 comi)act 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. Eule. — Allow 5Jfi cubic feet for a ton. 2d. TofindthcAveight in common hay barn, or small (low) stack. Eule.— ^l/Zo?/' Jfi5 cubic feet for a ton. 3d. To find the Aveight in mow bases in barns, compressed with gram, and in butts of large stacks of timothy hay. Eule. — Alloir 32Jf. 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 = 13 X 12 X 12, or 1728 cubic inches, the product of its length, breadtli, and thickness; and since 12 is one of the three e(|ual 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 ciibe 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. XJnit^ and Cubes Compared. Remauk. — 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 ordeis. 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. 10' = 1000 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 ])e cubed. U.MTS. Cubes. 1' = 1 2* = 8 3* = 27 ■i' = 64 5' = 125 C = 21G =: 343 8* = 512 9' = 729 148 CUBE ROOT. (50* X + 2 (50' 50» 50 + 7 50 + 7 (50 X 7) + 7' + (50 X 7) 50^4- - 2 (50 X 7) + 7'^ 50 + 7 50» 7) + X7) 2 (50 X 7") + r + (50 X T) Explanation. —Cubing 57, we have 57 x 57 x 57 = 185193; or, separating 57 into its tens and units gives 5 tens or 50 -f 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' + 3 (50* X 7) + 3 (50 X 7') + 7= = 185193 472. From this result observe that 57^ = 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 iinifs = t' + 3t' u + 3 t u' + \i\ 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 u t3 + 3t2u + 3tu2 + u3=185.193 [5 7 t3 = 125 --.o r 125000 StMi + 3tu2+ u3= 60193 = rem. 3t: t = 50 t2 = 25U0 3 t2 = 7500 3t = 150 + 3 t = 7650 trial divisor. 3t2u 3tu2 u^ = 52500 7350 343 3 t2 u + 3 t u2 + u3 = 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)' = 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 Formvla (a), observe that, having subtracted from the given number the cube of its tens ( t' ), the remainder, 60193, must be equal to 3 t' u + 3tu* + u''. CUBE ROOT. U9 If a cube (B), 50 inches in length on each side, h fonned, its contents will equal 125000 cubic inches, and it will be shown that the remaining 60193 cubic inches are to be so added to cube (B) 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 siiles, being each 50 inches in length and 50 inches in width, the addition to each of them in surface, or area, is 50 ^ 50, or 50-, and on the three sides, 3 (50*), or 3 t«, 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 scjuares (C), and of the three oblong pieces (D), will be de- termined by dividing 60193 by the surface of the three squares, plus the surface of tbe three oblong blocks, or by 3 t- + 3 t ; this division may give a quotient too large owing to the omission in the di^^• 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- u-f 3 t u- = 59850 ; and by reference to the Formula (a), observe that the only term or ele- ment required to complete the cube of (t -f u) is the cube of the units (u ^). 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- u 3 tu- u^ and obtain 3 t- u -f 3 t u2 + u» = 60193; and if to this t\ or 125000 is added, the result is t» +3 t -u-f-3 t u« -|-u» = 185193, Forimdn (a); then subtracting 60193 from the remainder, 60193, nothing remains. This proves that the cube root of 185193 is 57. By the 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. Bule. — I. Beginning at the right, separate the given number into periods of three figicres 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. in. 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 muUiplied by the units, plus three times the tens multiplied by the square of the units, plus the cube of the units. Remarks. — 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 dirUor, 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 FOB PKACTICE. 4:74-. 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 65939204 ? 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 jilace. 8. Find the cube root of ^^, 9. Find the cube root of y^-^. 10. Find the cube root of -g-^^mH^. 11. Find the cube root of ^\\, to one decimal place. 12. Find the cube root of \l^^, to two decimal places. 15. Find the cube root of 25.41G23T, to two decimal places. IJf. Find the cube root of 3496.25, to three decimal places. 16. Find the cube root of .4106, to three decimal places. 16. Find the decimal equivalent of the cube root of \\, to two decimal i)lace6, by reducing tlie fraction to a decimal of six places and extracting the root of the decimal. MISCELLANEOUS MEASUREMENTS. 151 ir. What must be the liiglit 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 it ? 21. A square cistern, the capacity of which is 420 barrels, has a depth equal to onlv one-hftlf its width. Find its dimensions. DUODECIMALS. 475. Duodecimals are denominate fractious 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 g'wes prinies ( ' ); 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. KuLE. — MuUiph/ 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; multi2)ly together the four results thus obtained, and extract the square root of the product. To find the area of any plane figure, the ojiposite sides of which ara equal and parallel. Rule. — Multiply the base by the perpendicidar hight. To find the area of a plane figure, Avhose opposite sides are i)arallel but of unequal length. Rule. — Obtain the average length, and multiply by the per- piendicular hight. 479. A Circle is a plane figure bounded by a curved line, every part of which is equally distant from a i)oint within called the center. 480. The Circumference of a circle is the curved line boundinsj it. 152 MISCELLANEOUS MEASUREMENTS. 481. The Diameter of a circle is a straiglit line i)assiug through tlie center and terminating in the circumference. 482. The Radius of a circle is a straight line i)as8ing 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 hy S.lJflG. To find the diameter of a circle, the circumference being given. Rule. — Divide the circumference hy S.lJflG. To find the area of a circle, the circumference and diameter being given. Rule. — Multiply the circumference hy the diameter, and divide the product hy 4- To find the side of a square equal in area to a given circle. Rule. — Multiply the circumference hy .2821. To find the area of a square that can be inscribed within a given circle. Rule. — Mulfijily the square of the radius hy 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. Remahk. — 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 hy the hight. . To find the contents of a cylinder. Rule. — Multiply the area of the base by the hight. 486. A Pyramid is a solid, the base of Avhich 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. 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 i)yramid or cone. Rule. — Multiply the area of the base hy one-third the perpendicular hight. EXAMPLES FOR PRACTICE. 153 489. A Sphere is a solid bounded by a curved surface, all points of Avhich 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. H16. To find the volume of a sphere. Rule. — Multiply the cuhe of the diameter hy .5236. To find how large a cube may be cut from any given sphere, or may be inscribed within it. Rule. — Divide the square of the diatneter 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 .003 Jf.; the result will 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 number of square yards in a triangular sail, the sides of Avhich arc 36 ft., 45 ft., and 48 ft. respectively. 3. How many acres in a rectangular field 108 rods long and 48 rods wide ? Jf. 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. in 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 IS" THE PAPER AXD 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 Ju making it ? H. 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 T901 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 ? lU. If a cask 24 inches at the chime, 30 inches at the bung and 3 feet long, is f full, how many more gallons may be put into it ? TABLES AND CUSTOMS IN THE PAPER AND BOOK TRADE. 498. Pajier in the stationery trade is sold by the following Table. 24 sheets =1 quire. 20 quires = 1 ream. 2 reams = 1 bundle. 5 bundles = 1 bale. A bale contains 200 quires, or 4800 sheets. Remarks. — 1. In copying, & 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. 49-4. Books are sometimes classified by their size, or the number of pages in a sheet. Xame. Sheet folded into. Pa^es. Folio, 2 leaves, 4 Quarto, 4to 4 leaves, 8 Octavo, 8vo. 8 leaves, 10 Duodecimo, 12mo 12 leaves, 24 16mo 16 leaves, 32 18mo 18 leaves, 3G 24mo 24 leaves, 48 32mo. 32 leaves, 64 Table for C'ouuting:. 12 units = 1 dozen, i 12 dozen = 1 gross. 20 units = 1 score. | 12 gross = 1 great gross. Table for Land and Lot Measures. 104^ feet square = ^^ of an acre. 10 rods X 16 rods = 1 acre. 14Tyy feet square = -^ of an acre. '2Q^>-^ feet square = 1 acre. 8 rods X 20 rods = 1 acre. 40 yards X 121 yards = 1 acre. THE METRIC SYSTEM. 155 THE METRIC SYSTEM. 495. Tlic 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 Meters 39.37-1- inches, being one ten-milliontli 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), Kil'o (1000), and Myr'ia (10000). Lower orders ai-e called Dec'i (tenths), Cen'ti (hundredths), Mil'li (thousandths)- Metric Linear Table. 1 cen'ti-me'ter cm = yJ-Q M. = 1 dec'i-meter dm = ^^- M. = 1 Meter M. = 1 dek'a- me'ter Dm = 10 M. = 1 hek'to-me'ter 11 m = 100M. = 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 10 mil'li-me'ters {mm) 10 cen'ti-me'ters 10 dec'i-me'ters 10 me'ters 10 dek'a-me'ters 10 hek'to-me'ters 10 kil'o-me'ters 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. Tables of Equivalent.s. Linear Measure. 1 centimeter = .3937 of an incli. 1 decimeter = .328 of a foot. 1 meter = 1.0936 yards. 1 dekameter = 1.9884 rods. 1 kilometer = .62137 of ii mile. Square Measure. i s(i. inch = G.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 S(i. mile. 156 MOXEY OF THE GERMAN EMPIRE. Ctbic Measure. 1 cu. inch = 1G.38T 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 cu. 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 = .252-4 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 = 31.104gi-ams. 1 pound, Avoir. = .4536 of a kilogram. 1 pound, Troy = .3732 of a kilogram. 1 ton (short) = .9072 of a tonneau. 1 gi-am = .03527 of an ounce, Avoir. 1 gram = .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. I^RENCH MONEY. 498. The Legal Currency of France is decimal, its unit being the .sj/t-er Franc. 499, The French coins are as follows: Gold f 100 francs, 40 francs, 20 francs, 10 francs, 5 francs. ( 5 francs, -/ 2 Silver J 2 francs / 1 franc. Bronze f 10 centimes, 5 centimes, 2 centimes, [ 1 centime. 10 millimes (m. ) 10 centimes 10 decimes Table. = 1 centime (ct.) = $.00193. = 1 dccime (dc.) = .0193. = 1 Franc (fr.) = .193. MONEY OF THE GERMAN EMPIRE. 500. Tlie I'nit is the Mark = $.2885 United States money. It is divided into 100 pfennigs (pennies). The silver Thaler = $. 746 United States monev. 501. The German coins are: ( 20 marks. Gold \ 10 marks. Silver 20 marks, 10 marks, 5 marks. f 20 marks, ■j 1 mark, ( 20 pfennigs. Nickel P?!t""'S«' ( piejinigs. EXAMPLES IX DENOMIKATE NUMBERS. 157 MISCELI.ANEOUS KXAMPLES. 502. 1. What is the value, in English money, of $1750 in United States gold coin ? 2. It required 12 yr. C mo. 1 da. to build the Brooklyn bridge. If it wjis completed July 4, 1882, when Avas its construction begun ? 3. What is the board measure of 7 planks, each 16 ft. long, 15 in. wide, and 3 in. thick ? J^. 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 tnany perches of IG^ cu. ft. each does the wall contain? Remark. — Sometimes 24} 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 inches to the weather, would be required to cover the roof of a barn GO 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 atone 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 Avhat time can it encircle the earth ? 13. What will be the cost in Paris of a cargo of 38500 bu. United States wheat, at 10 fr. 60 cent, per hektoliter? 14. 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. GO i)er ton, and you use G5 lb. i)cr day, wliat will be the expense of your fire for the months of the winter of 1891-2 ? 18. How many barrels of Avater 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 Avill make how many revolutions in going 62.5 miles ? 20. If Wm. II. Vanderbilt died Avorth two hundred millions of dollars, in what length of time could his fortune, in silver dollars, bo counted by one person, counting GO per minute and Avorking 10 hours i)cr day for 3G5 days each year ? 21. What will be the cost of 10 sticks 2 in. by 4 in., 10 sticks 2 in. by 6 in., 10 sticks 4 in. by 4 in., and 10 sticks 2 in. by 10 in., if the sticks are each 16 ft. long and the cost is $15 per M ? 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 cover a floor 21 J ft. long and 18f ft. wide, making no allowance for waste in matching the design ? 23. 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 ? 2^. 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 40^ per square yard, of plastering the ceiling and walls of a room 45 ft. long, 354- ft. wide, and 12 ft. 3 in. high ? 29. How many German marks are equal to $1500 United States money ? 30. A pile of wood built 10 ft. high and 22 ft. wide must be how long to contain 125 cd. ? 31. The main centennial building at Philadelphia in 18T6 was 1880 ft. long and 464 ft. wide. What was its area in acres, square rods, and square feet ? 32. Reduce 47 mi. 216 rd. 11 ft. 5 in. to metric units. 33. 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? 3^. From .001 of a section, plus .01 of an acre, take .001 of a quarter section, plus .01 of a square rod. 35. A grocer bought 12 bu. of chestnuts, at §3.50 per bushel dry measure, and sold them at Ibf per quart liquid measure. Did he gain or lose, and how much ?' 36. 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 12 ft. deep ? 38. If bricks cost 15.50 per M, what will be the cost of the In'ick for a wall 12 ft. high and 3 ft. thick, enclosing an acre of land 10 rd. wide and 16 rd. long ? 39. The gold coin of the commercial world suffers each year a loss of one ton by wear or abrasion. What is tlie value, in United States gold dollars, of the loss thus sustained? JfO. Reduce 250 hektars to common units. 41. What will be the cost, at $16.00 i)er M, of a tapering board 18 ft. long, and 9 in. wide at one end and 16^ in. wide at the other ? Jf^i. A German immigrant having 1000 thalcrs and 500 marks, exchanges them for United States money. How many dollars should he receive ? Jf3. 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. J 59 J^. A train of 45 cars of Lehigh coal averages, by the long ton, 2o T. 7 cwt 3 qr. to lb. per car. What is the value of the coal, at #5.25 per short ton ? Jf5. How many feet of lumber in a box 6 J ft. long, h\ ft. wide, and 3i ft. deep, inside measurements given, and lumber 1 inch in tliickness ? J^6. What Avill be the cost of carpeting | 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 \^io in width and %ic in length for shrinkage of the lining, the carpet selling at $2.25 per yd., and the lining at 30^ per yd. 47. A pile of wood 56 meters long, 18^ meters wide, and 3| meters high, was sold at $6 per cord. How much was received for it ? U8. A farmer filled a bin 9 ft. Avide, 12 ft. long, and 7^ ft. deep, with wheat grown from a field yielding 32^ bu. per acre. How long was the field, if its width was 50 rods ? J^9. . 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 81 i)cr ton, on a car load of 3205 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. \i one is built 10 ft. high and filled, how high must the other be built to hold the remainder "^ 5Jf. 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 deptli, 24 ft. wide at the surface, and 10 ft. at the bottom ? 55. If a car carrying 20 tons of freight is with its couplings 42 ft. long, what would be tlie length of a train carrying Vanderbilt's two iuindred millions of dollars, if it is all in standard silver dollars, and any fractional part of a car load 18 rejected ? iOO PERCENTAGE. PERCENTAGE. 503. Percentage is a term ajiplied to computing by hundredths. 504. The Elements of Percentage are, the Base, the Rate, the Amount Per Cent., the Difference Per Cent., the Percentage, the Amount, and the Difference. 505. The Base is the number ujion which the percentage is computed, 506. The Rate Per Cent, denotes how many hundredths of the base are to be taken, and is usually expressed as a decimal. 507. Per Cent, is an abbreviation of the Latin words jt?er centum, signifying by the liundred, or a certain number of each one hundred parts. 508. The Sign, i, is used to denote per cent. 509. The Rate may be expressed as a part in a common fractional form, as f ; in the form of an extended decimal, as .01625 = If^ ; 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 j!?er cew^. , and the added places, if anv, 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^ = -^^ = ^. 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 phis the rate. 515. The Difference Per Cent, is 100 percent, dimmished by the ra^e, 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. ' 161 619. The Base is either an abstraot 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 part of commercial calculations are based upon percentage, the importance of a thorough mastery of its principles will be readily perceived. 520. ^'\ncQ per cent, is any number of hundredths, it may be expressed either 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 perceyitage. Hence, the rules given under Special Applications may be applied in this ^subject. Table. 1\ Decimal. Com. Frac. Lowest Terms. 1 per cent. = .01 = 10 reducible to 10 0' 2 per cent. = .02 = 10 reducible to t' 3 per cent. =z .03 = 3 Too" reducible to Tinr- 4 per cent. = .04 = xio- reducible to ^• 5 per cent. = .05 = TolT reducible to ^' 6 per cent. = .06 = ro*r reducible to ■io- 7 per cent. = .07 = tJtt reducible to lod* 8 per cent. = .08 = , 8 Too reducible to ih- 9 per cent. 1= .09 = To¥ reducible to 106* 10 per cent. = .10 = Too reducible to iV- 12 per cent. =: .12 = ^^ reducible to A. 14 per cent. = .14 = j'^ reducible to A- 16 per cent. = .16 = iVo reducible to A. 20 per cent. = .20 = 100 reducible to i- 25 per cent. = .25 = tVo reducible to i. • 30 per cent. = .30 = loo" reducible to ^. 50 per cent. = .50 = ■^A reducible to i. 75 per cent. =: .75 = 100 reducible to I. 100 per cent. = 1.00 = 100 To reducible to 1. 125 per cent. = 1.25 = 12 5. 1 reducible to i = u- 150 per cent. = 1.50 = 15 0. To reducible to i = H' li per cent. =. .0125 = To reducible to ■sV. If per cent. = .01661 = tIfoot reducible to A- 2i per cent. = .025 = ^2.5_ To reducible to iV- 3i per cent. = .033^ = -1 oo_ ^000 reducible to A. 6i per cent. = .0625 = 625 10000 reducible to iV- 8i per cent. = .0833^ = 2_5_o_0_ 70000 reducible to iV. m per cent. = .125 = JL2_5_ To reducible to I m per cent. z= .1661 = WW reducible to h 33^ per cent. = .333 J = "To reducible to h 62^ per cent. = .625 =. To reducible to l. 66f per cent. = .661 =z m reducible to f. m per cent. = .875 = A^oV reducible to h Ifi2 EXAMPLES IN PERCENTAGE. 521. Tlie ri'latiou between the eleinents of Percentage is such, tliat by the- application of the General Principles of MultiiiJieation and Division, if any two of the elements, except amount ]ier cent, and difference per cent., are given, the other three may be found. 522. To find the Percentage, the Base and Rate being given. Exam PLK.— What is 25^ of 1:468 ? First Explaxatiox. — 25 per cent, equals .25; therefore, Operation. ^5 per cent, of $468 equals $468 multiplied by .25, equals $468 = base. $117. ,25 = rate per cent. Second Explanation. — $468 is 100 per cent, of itself; r~~zr7T ^ and since 25 ner cent, equals i of 100 per cent., 25 per cent. $117.00 = percentage. ^^ ^^^^ ^,^ -^ ^ ^^ ^^^^ ^^^ ^^ ^^^^ 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 iV, for ib'i take i, for 33^-; take \, etc. Formula. — Percentage = Base X Rate. KXAMPLE.S FOK MEIfTAI. PKACTICE. 523. What is 1. 5 per cent, of 100 ? 2. 12 per cent, of 600 ? 3. 15 per cent, of 800 ? J^. 20 per cent, of 500 ? 5. 25 per cent, of 1200 ? 6. 33^ p^r 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? 5. A collector deposited 813500 in coin, and 12|^ more in bank bills. What was the total of his deposit ? J^. Find ll^f^ of 1G80 lb. of wool. J. Find 1655^ of 12 lb. 3 oz. of silver. 6. From a charge of $675, made for a bill of goods, 8j^ 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 15j^, and the other lost 35^^. How much did each then have? 9. How much greater is 12^^ of $1550, than 74^ 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 PERCEHTTAGE. 163 11. Having $75000 to invest, u gentleman bought United States bonds with 3iH^ of his money, a home with 20^^, and invested the remainder equally in farm lands and manufacturing stock. How much did he pay for tlie farm lands ? 12. I owed John Smith $1750, and paid at one time 'HOfo 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 | of a coal mine, sold 324^ of his share for $65000. At that rate, what was the entire mine worth ? IJf.. 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 Avheat, 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 74- 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 IQ'^fc ; the remainder he invested in cotton, which he sold at 9^ below cost. How much was received from both sales ? 18. A trader bought 12 mustangs for $400, and after selling 2b% 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 ^ P^'" ^^°*- ^^^^ ^^ %IQI\ and if the value of 1 per cent, is $161, the '- value of 100 per cent, will be $16100. 16100 base. 'R\\\e.— Divide the percentage hy the rate, expressed decimally. Formula.— Base = Percentage -^ Rate. KXAMPIiES FOR MENTAL, I'KAOTICE. 626. 1. 846 = Q16090.80. 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 11.35 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. -iQQ^ 1 _ unit (definition, page 160), is always 100 per cent, in- ' ' creased by the rate, we may tind it by adding 7% = . ( = rate jqq p^^ ^^^^ , or 1 , to the per cent, given. Hence, 1 A.>v ~rr^r^ J. , if the rate is 7 per cent., the amount per cent, will 107^ =: 1.07 = amount per cent. . .^- ^ ^ be 107 per cent. Rule.— ^dc? the rate to the unit 1. Formula. — Amount Per Cent. = 1 4- Rate. 106 EXAMPLES IX PERCENTAGE. EXAMPLKS FOR MENTAL, PRACTICE. 53*2. 1. If the rate be 10;e, ■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 i)er cent.? Jf. Find the amount per cent., if the rate per cent, be I65? <5. Find the amount per cent., if the rate per cent, be 8T^ ? EXAMPLES FOR WRITTEN' PRACTICE. 533. 1. Goods costing 1*14:00 were sold for $14T0. 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 tlie previous month as compared with those of the last month ? S. If tea costing ^'2\^ per pound sell at 87^^, what amount per cent, do the sales show as compared with tlie cost ? 534. To find the Difference Per Cent., the Rate being given. Example. — If the rate be b^, what is the difference per cent.? Operation. Explaxatiox. — Since the difference per cent, (definition, 100^ = 1.00 = a tinit. p^gg jgo), is equal to 100 per cent., or 1, less the rate, if we take y 8< of itself? ^' Explanation. — The amount equals base plus percentage (de- 550 = base. finition, page 160). The base is 550 and 8 per cent, of 550 equals •^" rate. 4^^ ^jj^ percentage; therefore the amount niuist equal 550 plus 44, 44.00 = per cent, or 594; or, .'Jince 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 108 per cent, of 550, uv 1.08 times 550 equals 594. 594 = amount EXAMPLES IX PEKCENTAGE. 167 Rules.— i. Find the -percentage and add it to the base. Or, ^. Multiply the base by 1 plus the rate- Formula. — Amount == Base + Percentage. EXAMPLES FOR :»IEXTA1. PKACTICE. 638. i. 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 ? S. The base is 440 and the rate 5^; find the amount. Jf.. 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. 8. The base is 546 and the rate 16|^,* find the amount. 9. The base is 200 and the rate 125^; find the amount. EXAMPr.ES FOR WKITTEX PRACTICE. 639. 1. What amount will be received for a house costing %13500, if it is sold at a gain of l^fo ? 2. A bought two horses for $180 each, and sold one at a gain ef 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 i)er acre, and sold at an advance of 40;^. How much was received for it ? 4. What is the amount of 768 increased by 25,^^ of \ 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 ? 640. To find the Dijfference, the Base and Rate being given. Example. — What remains after diminishing 450 by 10^^ of itself ? Operation. 100^ = 450 = base. \Qc^ ^ jQ _ j.j^te. Explanation. — Since 100 per cent, of the —^ 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 P^^ *^^°*' ^Q*^^!^ 450, 10 per cent, less than 100 45 percentage P^*^ cent., or 90 per cent, will equal 405. 405 difference. Rules.— i. Find the percentage and subtract it from the base. Or, £. Multiply the base by 1 minus the rate. Formula. — Difference = Base — Percentage. EXAMPLES FOR 3IENTAL PRACTICE. 641. 1. If from a brood of 15 chickens 20r» arc lost, how many will remain ? 3. What number will remain if 225 is diminished by 33^^^ of itself ? 3. If the base is 1050 and the rate 10^, what is the difference ? 168 EXAMPLES IN PERCENTAGE. 4. 816, less 25^ of itself, equals what number ? 0. 1440, less 165^ of itself, equals what number ? 6. 800, less 3T^^ 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|^ of it. How many feet had he remaining ? £XAMPI^S FOK WKITTEX PRACTICE. 542. , 1. A speculator lost 35<^ of ^ of $16250. How much did he lose? A?. A planter having 616 acres in rice, lost \ of 33^^^ of his planting by flood.. How many acres had he left for harvest ? 3. Brown deposited $1147 in a savings bank, and his son deposited 21,<^ less. How much was deposited by both ? 4. 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 ? ExPLASATiox. — Since the number must be 100 J 00,^ l.« I p^j. pgQj Qf itself, if it has been increased 15 per ^^^ = j}^ Amount. Base. (.^^^ ^^^ ^just be 115 per cent, of that number; llb^ amt. J^ 1.15 ) 345.00 ( 300 if 115 per cent, is 345, 1 per cent, must be yl^ of 345 345, or 3; and 100 per cent, will be 100 times 3, or ■~oo ^- Example (second illustration). — What number, diminished by 35^ of itself^ equals 975 ? Oper.\.tiox. 1 r.A ExpLAKATiOK. — If the number be diminished by 35 per cent, of itself, there will be remaining but 65 per cent, of itself; and if 65 per cent, of the number be 975, 1 per cent, must be ^V of 975, or 15; and if 1 per cent, be 15, 100 per cent, must be 1500. 00 Rules.— i. Divide the amount hij 1 plus the rate. Or, 2. Divide the difference hy 1 minus the rate. Formulas. — 1. Base = Amount -^ Amount Per Cent. 2. Base = Difference -=- Difference Per Cent. EXAMPLES FOK MENTAL PKACTICE. 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 ? 3. After loi of a number had been added to it, the amount was 525. What was the number ? Oper.\tiox. .00^ = 1.00 35;^ 65^ dif. = .35 Diff. I 975. 65 325 325 00 Base. (1500 REVIEW OF THE PRINCIPLES OF PERCENTAGE. 16^ ^. After selling '60^ of his apples, a boy had 70 left. IIow many had he at first ? 5. I lost $600 by a bankrupt, who paid only 85^ of his indebtedness. What was the fnll amount of my claim ? EXAMPLES FOK WKITTEN PRACTICE. 645. -?. A builder gained 15^ by selling a house for $1150. What was ita cost ? 2. Sold 945 tubs of butter for $5113, and thereby gained 20^. Ilbw much did the butter cost per tub ? 3. The income from a tenement house is $6042 this year, which is 24^ less than it Avas 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 jier 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. W^hat 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 Avas the attendance for Februtiry ? 10. Which is better, to invest in a house that Avill 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 vears ? REVIEW OF THE PRINCIPLES OF PERCENTAGE. 546. 1. To find the percentage, the base and rate being given. Kule. — Multiply the base ly the rate expressed decimally. 2. To find the base, the percentage and rate being given. Rule. — Divide the lyercentage hy the rate expressed decimally. 3. To find the rate, the percentage and base being given. Rule. — Divide the percentage hy the base, carrying the quotient to tivo decimal places. 4. To find the amount per cent., the rate being given. Rule. — Add the rate to the unit 1. 6. To find the difference per cent., the rate being given. Rule. — Subtract the rate from the unit 1. 6. To find.tlic 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 TX PERCENTAGE. 7. To find the difference, the base and rate being given. Rules. — Multiply Ihe base by the rate, and subtrai't 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 ihe 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 chisses of problems: First, to those in which time is not an element; as. Profit 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 EXABIPLES 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 lOj^. What was it before the reduction ? 3. A pupil in examination answered correctly 56 qitestions, which was 20^1^ less than the number asked him. What should be his average, on a basis of 100? 4. 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 slie gave $2500, to the second $4500, and to the tiiird the remainder. What per cent, did each receive ? 6. On attaining his majority, a son finds liis age is 62-k^ less than the age of his father. Find tlie sum of their ages ? 7. If 8^ of B's money equals 245^ of C*s, how much has C, if B has $324 ? S. A farmer bought a horse, a mule, and a cow, for $385. The mule cost 15^ less than tlie 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 war}). How many pounds of each in a bale of carpet weighing 96^ pounds ? 11. The sum paid for two watches was $384, and 75jfe of the sum paid for one equalled 105-^ of the sum paid for the other. Find the price of each. 12. If Abas 35'^ more money than B, and B has 25f^ more than C. how much has C, if A has $102 ? 13. If a gain of $4755 was taken out of a business at the end of the first year, and a loss of $3566.25 was sustained tlie second year, what was the per cent, of net gain or loss, the investment having been $63400 ? EXAMPLES FOR PRACTICE IX PERCENTAGE. 171 14. After making three of the seven equal annual payments of the face of a mortgage, I find $5850 to be still unpaid. How many dollars of principal have been paid ? 15. After the salary of a book-keeper had beenincreased. 10^, and afterwards 8^, he received $1242 a year. What Avas his salary at first ? 16. By the United States Census of 1880 the total capital invested in man- ufactures in the State of Pennsylvania was $190055904, while the amount invested in Alabama was $9098181; Arkansas, $131GG10; Delaware, $5452887; Florida, $1874125; Georgia, $10890875; Louisiana, $7151172; Mississippi, $4384492; North Carolina, $9693703; South Carolina, $6931756; Texas, $3272450. What per cent, greater was the manufacturing capital invested in Pennsylvania than in the group of the ten other States named ? 17. From an estate tlie widow received $9250, which was one-third; the remainder 'Avas divided among three children, aged respectively 15, 12, and 10 years, and they shared in proportion to their age. What per cent, of the estate did each of the children receive ? 18. A herder was asked how many cattle he had, and replied: "My herd increased last year 40^; should it increase at the same rate during this year and next, and I then buy 4 head more, I shall have double my present number." How many head of cattle had he ? 19. What per cent, of the amount, at 10^, is 10^ of the base ? 20. From a farm containing 180 A. 120 sq. rd., one-half was sold at one time, and one-half of the remainder at another time. What per cent, of the whole then remained? 21. A man drew 15f^ of his deposit from a bank, and with it paid a debt of $1119.60. What balance was left in the bank .? 22. Ibfo of f of a number is wliat per cent of | of it ? 2S. A man sold two farms for $7500 each; for one he received 25^ more than it cost, and for the other 25^ less than it cost. Did he gain or lose by the sale, and how much ? 2Jf.. What number is that which, being increased by 35;^ and 46^ of itself and 76 more, will be doubled ? 25. A ranchman, when asked how many sheep he had, replied: " If my flock increases next year 20^,' the next 25^^, and the third year 40^, I can then sell 300, and have left double my present number. How many had he ? 26. The total number of Popes up to 1888 has been 253, of whom 197 have been Italians. What per cent, of all have been of that nationality ? 27. By widening a roadway 5j^, it was made lO.V yd. wide. What was its original width ? 28. Oct. 11, 1888, A bought an engine and mill for $5250, on six months credit, or 5^ off if i)aid within 90 days, or 7^^ off if paid within 30 days. What amount was required for full settlement Nov. 7, 1888 ? 29. In settling an estate, an executor found 7^r^ of it to be invested in telegraph stock, 15f^ in railroad stock, 37|^ in city bonds, $16750 in real estate, and $7350 cash in bank. Find the total value of the estate. 172 EXAMPLES FOR PRACTICE IN PERCENTAGE. 50. A farm is composed of 20^ more grazing than grain land, and the timber is one-half of the area. How many acres of each, if, after deducting 12 acres for lawn and garden, there is left of tlie farm 18G0 acres ? 51. A has 20*^ less money than B, and B has 25't more than C. How much has C, if A has 1192 ? 32. A and B were heirs of an estate of #120000, A receiving 1 jf^ of the whole more than B. For four years thereafter the property of eacli increased at an average rate of 9j^ per annum. How much had each at the end of that time ? 33. A man owning 62^,^^ of a factory, sold 7^f^ of his share for $1050. At that rate, what was the value of the factory ? ( .18 = 18^ Explanation.— If 95 dollars of cost gain $17.10, 1 dollar of 95 cost would gain as much as 95 is contained times in 17.10, or .18, ^Q equal to 18 per cent. 760 Unle.— Divide the profit or loss by the cost. Formula. — Per Cent, of Profit or Loss = Profit or Loss -^ Cost. EXAMPI.ES FOR MENTAL PRACTICE. 563. -?. I gained $12.50 on what cost me $125. Find my rate per cent, of gain. 2. I bought a bicycle for $150, and sold it for $7.50 below cost. What per cent, did I lose? S. What per cent, is lost by selling a $5 book at 62-|^ below its cost? 4. A safe costing $380 was sold at a loss of $76. Find the loss per cent. 5. A gold pen cost $2, and after being tested and found imperfect, was sold as old gold for $1. Find the per cent, of loss. 6. What per cent, of gain is realized by buying a horse for $300, and selling it at an advance of $100? 7. Find the per cent, of gain, on a section of Dakota prairie, bought at $4 per acre, and sold at $10 per acre. 8. An Ohio river steamer costing $100000 was sold for $9500 profit. Find the per cent, of profit. 9. A Vermont manufacturer, having invested $40000, gained $8250 each year. What was his per cent, of gain per annum? EXAMPLES FOR WRITTEN PRACTICE. 564. 1. What per cent, is gained by selling an article for 2-k times its cost? 2. I bought a quantity of cloth at $1.60 per yard, and sold it at $2 per yard. What was my per cent, of gain? 3. A speculator bought wheat at 80^- per bushel, and oats at 32{i^- per bushel. If he sold the wheat at 90^' per bushel, and the oats at 40^ per bushel, on which would he make the greater per cent., and how much? 4. If a boy sells three apples for what four cost him, what per cent, does he gain? 6. Four-fifths of a stock was sold at 45^ loss, and the remainder at 225^ profit. What was the per cent, of net loss or net gain on the stock? 6'. Paper bought at $2.70 per ream, and retailed at 1^ per sheet, will yield what per cent, of profit? EXAMPLES IN PROFIT AND LOSS. 177 7. Potatoes costing $1.35 per barrel, and sold at 11.62 per barrel, will net what per cent, of gain ? 8. A wood dealer, after buying 8 car loads of mixed wood, of IG cords each, at $5 per cord, sorted it and sold 35^ of it at 1%\% gain, 35^ of it at 10^ gain, and the remainder at 20f« gain? What was his average per cent, of gain? 9. If 33^^ of a barrel of salt be sold at 33^^ profit, and the remainder be sold at cost, what per cent, of profit is realized on the whole? 10. An agent sold a sewing machine for $45.70, and thereby gained $18.28. What per cent, did he gain? 11. If \ of an article is sold for what f of it cost, what is the loss per cent.? 12. If I sell ^ of an article for what ^ of it cost, what is my rate of gain? 13. A drover, buying 125 beeves at the rate of $55 per head, and 78 at $G2.50 per head, sold the lot at a profit of $2115. What was his per cent, of gain? U. A cargo of lumber cost $3000. If ^ of it is sold for 82000, 4 of the remainder for $1250, and what is left for $420, what is the per cent, of gain or loss by the transaction? 16. Oil bought at 81^^" per barrel is sold at 86|^'. If ^^' per barrel is allowed for expenses, what must have been the investment, the gain having been $1350? 565. To find the Cost, the Selling Price and the Rate per cent, of Profit or Loss being given. Rules.— i. Divide the selling price by 1 plus the rate of gain. Or, 2. Divide the selling price by 1 minus the rate of loss. Formulas — •! ^' ^^^^ ~ ^^^^^^S ^^^ce -^ 1 + Per Cent, of Gain. ( b. Cost = Selling Price ~ 1 — Per Cent, of Loss. « EXAMPLES FOR MENTAL PRACTICE. 566. 1. A buggy was sold for $105, at a gain of 5^. What the the cost? 2. What must have been the cost of a harness sold at 40^ loss, if $24 were received for it? 3. Find the cost of making a suit of clothes, if 20^ is gained by selling it at $18. Jf. Find the cost of a watch that sold at a profit of 16S^ and brought $87.50. 6. I sold a house for 125^ profit, receiving therefor $2250. Wliat was the price paid? 6. If $15360 is realized on a stock of goods after it has been damaged 40^, what was its value before being damaged? EXAMPLES FOR WRITTEN PRACTICE. 567. 1. One of a pair of horses was sold for $180, at a loss of 12A^; the other was sold for $200, at a gain of 25^. What did the pair cost? 2. A fruit dealer, after losing 16! of his apples by frost, has 147^ barrels left. If he bought his stock at $2.50 per barrel, what was his outlay? 3. What was the original value of Calumet copper mining stock, which, when «old at a gain of 175^, brought $20625? 12 178 REVIEW OF THE PRINCIPLES OF PROFIT AXD LOSS. Jf.. A paid 6^ tax on his income. What was his income, if, after paying the tax, the remainder equalled $7050.94. 5. A dairy produced 20^ more cheese in March than in February, "What was the jiroduct for March, if that for the two months was 1980 pounds ? 6. I sold a house to A at a profit of lO't; he sold it to B. gaining \h^\ and B, by selling it to C for $6072, gained 20^ on his purchase. How much did the house cost me? REVIEW OF THE PRINCIPLES OF PROFIT AND LOSS. 568. -?. To find the gain, the cost and per cent, of gain being given. Rule. — Multiply the cost by the j^er cent, of gain. 2. To find the loss, the cost and percent, of loss being given. Rule. — Mul- tiply the cost by the per cent, of loss. 3. To find the selling price, the cost and gain being given. Rule. — Add the gain to the cost. 4. To find the selling price, tlie cost and loss being given. Rule. — Sub- tract the loss from the cost. 5. To find the cost, the gain and per cent, of gain being given. Rule. — Divide the gain by the per cent, of gain. 6. To find the cost, the lossand per cent, of loss being given. Rule. — Divide the loss by the per cent, of loss. 7. To find the selling price, the gain and per cent, of gain being given. Rule. — Divide the gain by the ])er cent, of gain, and to the quotient add the gain. 8. To find tlie selling price, the loss and per cent, of loss being given. Rule. — Divide the loss by the per cent, of loss, and froin the quotient subtract the loss. 9. To find the per cent, of gain, the gain and cost being given. Rule. — Divide the gain by the cost. 10. To find the per cent, of loss, the loss and cost being given. Rule. — Divide the loss by the cost. 11. To find the per cent, of gain, the selling price and gain being given. Rule. — Subtract the gain from the selling price and divide the gain hu the quotient. 12. To find the per cent, of loss, the selling jirice and loss being given. Rule. — Add the loss to the selling price, and divide the loss by the sum obtained. MISCELLANEOUS EXAitPLES. 569. 1. What is that sum of money of which 50^ is $19.20 more than 37^j^? 2. What amount of money must an attorney collect, in order that he may pay over to his principal $475, and retain b<^ for his services? 3. A woman is 72 years old, and 16^|<^ of her age is 25,^ of the age of her daughter. Find the daughter's age. 4. Gunpowder is made of f nitre, and the remainder of equnl parts of sulphur and charcoal. Find the per cent, of each. MISCELLANEOUS EXAMPLES IN PROFIT AND LOSS. 179 5. A milkman increased his herd of cows by a purchase of 36, which was 45^ of the whole number he then owned. How many had he before buying the last lot? 6. If I make a profit of 16f^ by selling a horse at 87.50 above cost, how much must I have advanced on the cost to have realized a profit of 25f^? 7. Two persons contributed $2100 towards a business venture, from which their part of the gain was $350. If of this gain the share of one was $70 more than that of the other, what part of the original contribution must have been made by each ? 8. How much money must be invested in notes, at 4|«^ below their face value, in order that, when sold at ?>io above their face, a profit of $225 may be realized ? 9. I bought a warehouse of Brown for 12^^ less than it cost him, and sold it for 16f^ more than it cost him, gaining thereby $963. GO. How much did I pay for the warehouse ? 10. What per cent, is gained by buying pork at $17.50 per barrel, and retail- ing it at 12^ per pound ? 11. A lady wishing to sell her piano, asked 15^ more than it cost, but finally sold it at 12.T^ less than her asking price. What did the piano cost, if by its sale she gained 155 ? 12. Having bought 75 barrels of ajiples for $187.50, I sold them at a loss of 20^. How much did I receive per barrel ? 13. 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 ? III.. 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 ? n. A Texas farm of 160 acres was bought at $15 jier acre; 8354 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 \ than f ? 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 124^ on the cost ? 21. If 25^ of the selling price is gain, what is. the per cent, of gain ? 22. I sell f 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 ? 23. If 30c^ of a farm sold at 33^^ gain, and 30f^ of the remainder at 15^ gain, how much was the total gain, if the remainder was sold at cost for $7350 ? ^^. 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 biinker bought a mortgage at 7^^ less than its face value, and sold it for Z'/t more than its face value, thereby gaining $981.75. What was the face value of the mortgage ? 26. At what jn-ice should damaged goods be marked to lose 25^, the first cost having been 36^ per yard ? 27. A man sold a carriage and gained 25^, and with the proceeds bought iinother, wliicli he sold at a profit of lO,'^, thus realizing a total gain of $75. "What did he pay for eacli ? 28. If I sell f of an acre of land for what % of it cost, what Avill be my gain or loss per cent. ? 29. 21-^^ was lost by selling an engine for $2355. How much would it have brought had it been sold at a loss of XOfc ? 30. "What price must be asked for 1000 pounds of coffee, costing 18^ per pound, in order tliat tlio seller may deduct lO,'^ from the asking price for bad debts, allow 1G|^ for loss in roasting, and still gain 20,^^ on the cost? 31. B and C each invested an equal amount of money in business; B gained 12-i^ on Ills investment, and C lost $5275 ; C's money was tlien 42^ of B's. How many dollars did eacli invest ? 32. A trader lost 33^f^ on 20^ of an investment, and gained 12^^ on the remainder, thus realizing a net gain of $1000. Had he gained 20^ on -J, and lost 2")'^ 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 tlie general agent; the general agent's profit was 20^, he thereby gaining $25.30. "What did it cost to make the machine ? 31).. Of a cargo of 8000 bushels of oats, costing 35^- per busliel, 25j^ was destroyed by fire. What per cent, will be gained or lost, if the remainder of the oats are sold at 45^ per busliel ? 35 For Avhat must hay be sold per ton, to gain 16|^ if, by selling it at $18 per ton. there is a gain of 25^ ? 30. Jones sold \ of a stock of goods at cost, \ at a gain of 35^, ^ 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 'd^'fo of tlie selling price. How much was the first cost of the carriage ? 38. li oranges cost $1.80 per liundred, at what price must tliey be marked to ensure a gain of 20^, and make allowance for 28^ decay, and 25,^ bad debts in selhng ? 39. Having paid 40^- per pound for tea, at Avhat retail price must it be marked, that I may allow 12^^ for bad debts and gain 40^ on the cost ? Jf.0. Six wheel-rakes were sold for $21 each; three of them at a gain of 20j^, and tlie others at a loss of 20je. "What was the net gain or loss ? Jf.1. A stock of goods is marked 22^^ advance on cost, but becoming damaged, is sold at 20^ discount on the marked price, whereby a loss of $1180.40 is sus- tained. "What was the cost of tlu; goods ? MISCELfcANEOUS EXAMPLES IN PROFIT AXD LOSS. 181 Ji2. 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 %o^o 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 ? JfS. If the loss equalled \ of the selling price, what was the per cent, of loss ? Ji4. 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 30fc, for how much per quart must he sell the berries, to gain 15j^ on the whole ? Jf.G. A sells two horses to B at an advance of IGf;?;, B sells them to C at an advance of 25,'?^, and C sells them to D for $735, thereby making a profit of 20^. IIow much did A pay for the horses ? Jfi. Having bought 48 pounds of coffee, at the rate of 34- i:>ounds for 91^, and 84 pounds more at the rate of 7 pouads 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 ? ' Jfl. By selling at a loss of 6^ per yard I get 87+^^ of the cost of cloth. What per cent, of the cost would I have received had I lost %

in length ? If tlie carpet be laid lengthwise of the room, and be furnished at $2.25 per yard, f of a yard wide, and the duck, before shrinking, at 200 per square yard, and a i)r()fit of IGf*^ be realized on both, Avhat Avill be the gain? 58. If I pay 83.20 for, 20 gal. of vinegar, how many gallons of water must be added, that 40,'^ profit may be realized by selling it at \b was his bill ? 589. To find the Charge for Storage with Credits. Example. — The storage charges being ■2, in order that an agent may receive a commission of $175? 2. An agent received l>306.25 for selling wheat, on a commission of \\^. What Avas the amount of the sales? 3. A collector's charges of 5^ for collecting a note amounted to $14. 10. What sum was collected? 4. A factor charged $216.80 for selling a consignment of canned fruit. If his commission was 2|^, what must have been the gross sales? 5. I paid a grain dealer \\'/o for buying corn for me, at 62^ per bushel. If his •commission amounted to $83.70, how many bushels did he buy? 6. A Mobile factor earned $99.75 by selling cotton, at 2f^ commission. How many bales, averaging 560 lb., did he sell, the price being 15^ per pound ? 6(M). To find the Investment and Commission, when Both are Included in a Remittance by the Principal. Example. — If $1050 is sent to a Saginaw agent for the purchase of salt, how much will he invest, his rate of commission being o,'^? Operation. Explanation.— For each dollar invested ^1.00 = investment. by the agent, the principal supplies the dollar .05 = commission. invested and 5 cents for the agent's services; a,-, AK ^i-, ^ i. i. ■ • ^ J! 1 therefore the agent will invest only as manv $1.05 = actual cost to principal of each , „ . ,. *^ , ^ ^. ^. . , , dollars in salt as $1 plus 5 cents, or $1.05, is dollar invested by agent. contained times in the remittance, $1050; 1.05 1.05 ) $1050.00 is contained in $1050. 1000 times; hence the $1000 sum invested in salt. investment is $1000. Rule.— Divide the remittance by 1 plus the rate per cent, of coinmission. Remarks. — 1. All computations in commission may be made by applying the principles of Percentage. 2. When a charge is made for guaranty, add the per cent, of guaranty to 1 plus the rate per ■cent, of commission, and proceed as above. Formula. — Investment = Remittance to Agent -4- 1 plus the Rate Per Cent, ■of Commission. EXAMPI.es FOK PRACTICE, 610. 1. An agent receives $12504.20, with instructions to invest in avooI. If his commission is 3c^, how many dollars worth of wool will he purchase? 2. How many pounds of wool, at 27{^ per pound, can be bought for $8424, if the agent is allowed Afo for purchasing? 8. I remitted $1306.45 to a Boston agent for the purchase of soft hats. If the agent's commission is 4,'^, and he makes an added charge of 2^ for guaranty of quality, how many dozen hats, at $8.50 per dozen, should he send me? 13 194 EXAMPLES IX COMMISSION. 4. An agent receivies $13760.80 to invest in land, after deducting his charges of 3^. What amount of commission will he receive? 5. A real estate agent, whose stated commission is 2^^, receives 38302.50 to invest in Iowa prairie, at $5,40 per acre. How many acres did he jnirchase, and and how much Avas his commission. 6. I remitted $300 to an agent for the purchase of hojis. If the agent's charges were h'fo for purchase and $6 for inspection, how many pounds, at 10^ per pound, ought he to buy? MISCEL,I.AXEOrS EXAMPLES. 611. 1. A collector obtained 75*^ of the amount of an account, and after deducting 12'?; for fees, remitted his principal $495. What was the amount of his commission? 2. A Hartford fruit dealer sent a Lockport agent $1946.70, and instructed him to buy apples at $1.40 per barrel. The agent charged 3fo for buying, and shipped the purchase to his principal in six car loads of an equal number of barrels. How many barrels did each car contain? 3. Find the per cent, of commission on 'a purchase, if the gross cost is $2048.51, the commission $87.30, the cartage 820. and other charges $1.21. 4. 11500 bushels of wheat were bought through an agent, Avho charged \f^ for buying. If the agent paid 85{# per bushel for the wheat, $762.50 freight, and $12.50 insurance, what sum should be remitted to him in full settlement? 5. A collector obtained 75;:?; of a doubtful account amounting to S1750. How much was his per cent, of commission, if, by agreement with the principal, the commission was to be 50,*^ of the net proceeds remitted? 6. A farmer received from his city agent $490 as the net proceeds of a ship- ment of butter. If the agent's commission is 3ff, delivery charges $6.80, and b^^^ charge is made for guaranty of quality to purchasers, how many pounds, at 27^ per pound, must have been sold, and how much commission was allowed? 7. An agent sold 2000 bu. Alsike clover seed, at $7.85 per bushel, on a com- mission of 5^; and 1200 bu. medium red, at $5.20 per bushel, on a commission of ^\ic\ taking the purchasers 3 month's note for the amount of the sales. If the agent charges 4< for his guaranty of the notes, what amount does he earn by the transaction? 8. An agent bought l)utter on a commission of 10^, cheese on a commission of 6^, and eggs on a commission of h'lr. If his commission for buying the butter was $21, for buying the cheese $21.60, and for buying the eggs |22, and he charges 25^ additional for guaranteeing the freshness of the eggs, what sum should the jjrincipal remit to i)ay for purchases and charges? 9. Find the pet proceeds of a sale made by an agent charging Z^i^, if inci- dental charges and commission charges were each $41.30. 10. From a consignment of 3160 jjounds of tea, sold by an agent at 30^ per 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. 105 11. Find the gross jiroceeds of a sale made by an agent charging 2^^ for com- mission, hi for guaranty, $17.G5 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 80^' per bushel, and the balance, less all commissions, in wool, at 20^ per pound. If his commission for buying the wheat is 2,<, and that for buying the wool is 5^, how many pounds of avooI Avill he buy, and Avhat Avill be the amount of his commissions? 13. I sent $3402.77 to my Atlanta agent for the purchase of sweet i)otatoes, at $1.60 per barrel; his charges were, for commission, 2^^; guaranty, 3,<; dray- age, 1^ per barrel; and freight, $200. How many barrels did he buy, and how much unexpended money was left in his hands to my credit? 11^. A Texas buyer shipped 33000 lb. of coarse wool to a Boston agent to Ijo sold on commission, and gave instructions for the net proceeds to be invested in leather. If the agent sold the wool at 18^/ per pound, on a commission of 2f^, and charged 10^ for the purchase and guaranty of grade of the leather, what a\ as the amount of his commissions? i'5. I receiA'Cd from Duluth a cargo of IGOOO bu. of wheat, which I sold at $1. 10 per bushel, on a commission of 4^6^; by the consignor's instructions I invested the net proceeds in a hardware stock, for Avhich I charged 5;^ commission. What was the total commission, and how much was invested in-iiardware? 16. Having sent a New Orleans agent' $1835. 40 to be invested in sugar, after allowing 3f^ on the investment for his commission, I received 32400 pounds of sugar. "What price per i)ound did it cost the agent? n. An agent in Providence received $828 to invest in prints, after deducting his commission of Z^'fo. If lie paid 74-^'- per yard for the prints, how many yards did he buy? 18. The fees of the general agent of an insurance comjjany arc h'^ on all sums received, and 5^ additional on all sums renuiiuing in his hands at the end of the year, after all losses and the expenses of his office are paid. He receiA^es during the year $117410.25, paid losses to the amount of $01140.50, and the expenses of his office Avere $3207.70. Find his total fees. 19. An agent sold on commission 81 self-binders, at $140 each, and 113 mowers, at $05 each, remitting $10224.90 to his principal. Find the rate of commission. 20. A commission merchant received a consignment of 600 bales of cotton, of an average weight of 510 pounds, Avhich he sold at 124^ per pound, on a commission of 3^, charging 10^' per bale for cartage. He invested for the con- signor $9410.20 in bacon, charging 5^^ for buying, and remitted cash to balance consignor's account. Hoav much Avas the cash remittance? 21. An agent received $4325, to invest in mess-pork, at $10 \wx barrel, after deducting his i)urchasing commission of A^. If the charges for incidentals were $81.40, besides cartage of 75^/ per load of 8 barrels, how nuiny barrels did he buy, and Avliat unexpended balance does he place to the credit "atiox. — 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, 3i 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. — Fvom the sum to be raised, deduct the -poll tax, if any ; divide the remainder by the total assessineiit, and multiply the assessment of each, individual by the quotient; add to the product the aitvount of pdlZ tax 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 t>erson 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 i^, for county ^^, and for citv 1|^. What will be the amount of my tax, on property assessed at $21500 ? 4. The tax assessed upon a town is $20914.80; the town contains 2580 polls, taxed $.624 each, and has a real estate valuation of $4062000, and a valuation of i)ersonal 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 en the basis of the total attendance. o. My son and daughter each attended school 214 days, and the expense, including teachers 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 janitors fees $74.75, and the total number of day's attendance was 7460? EXAMPLES IN 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, .G of a mill; other purposes, .34 of a mill. What was the total rate of taxation, and hoAv much was raised by Livingston County, the valuation of which, as fixed i)y 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, liow mauy mills on a dollar must then be assessed ? 8. A tax of $13943.20 is assessed upon a town containing 8G0 taxable i)olls; the real estate is valued at $2708000, and the personal property at $151000. If the polls be taxed $1.25 each, what will bo the rate of property taxation, and what will be the tax of Peter Parley, Avho pays for three i)olls, 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 $073.70, what will be the rate per cent, of taxation, and what will be B's tax, whose i)roperty 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 Avill 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 ? 13. 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 years expenses arc: 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 lilSUEANCE. INSURANCE. 652. Insurance is indemnity secured against loss or damage. It is of two kinds: Property Insurance and Personal Insurance. 653. Property Insurance includes: 1. Fire Insurance, or indemnity for loss of or damage to property by fire. ~. 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. 3. Live 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 Stoch 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 Mixed 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, cither by rail or water or both. 660. The Policy is the contract between the insurance company and the person whose proi)erty is insured, and contains a description of tbe 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 i)aid for insurance. 662. The Term of Insurance is tlie period of time for which the risk is 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 Rates 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 wliich he receives a brokerage from the company taking the risk; he is con- sidered, however, an agent of the insured, not of the company. Remark. — 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 imrtial. 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 tlie 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 compan)' refund the premium for the unexpired time of the policy. 670. A Talued or Closed Policy is the ordinary form, and contains a tixed 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 voluntarv aid in saving vessels or cargoes from marine casualties. Remarks. — 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 c-ompany, -will be paid 30 days or GO 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. TheRate ^/o of Premium = the Rate. The Premium = the Percentage. 206 EXAMPLES IN INSURANCE. 675. To find the Cost of Insurance. Example. — The mixed stock in a country store is insured for $7500. What is the cost of insurance for one year, at 1^^ premium, if $1.25 is charged for the policy? Opkration. 17500. = amount insured. Explanation. — Since the amount insured is the base, and the per cent, of premium the rate, •Q^'^ = r^ of premium. jf ^l^^, amount be multiplied by the rate, the $11.25 = premium. product, $11.25, will be the premium; adding 1.25 = cost of ijolicy. $1.25, the cost of the policy, gives the full cost, '' $12.50 $12.50 = full cost of insurance. Jiule.— Multiply the amount of insurance hy the rate per cent, of premium, anil 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.— Since $141.50 was the full $141.50 = full cost. (,Qgt oj. premium plus the charge of $1.50 for 1-50 = cost of policy. the policy, the premium must have been $140; $140 ;= premium. ^°*i since the rate of premium was | per cent., l^ = .00875 = decimal rate. '^ ^^^^ '•' ^^^^^^^ ^^ ^ P^"" ^^°'- *^^ Quotient, A-i Ar^ t\r\c,^~ a.i r>r\,^n -p j? T $16000, will be the facc of the policy. $140 -=- .0087o = $16000, face of policy. ^ ' ^ ^ Rule. From, the full cost of insurance, subtract the extra charges, if any; divide the remainder hy the per cent, of premium, and the quotient will he the face, of the policy. EXAMPLES rOR I'KACTICE. 677. J. How much insurance, at \\^, can be procured for $62.50? 2. A ranchman paid a premium of $75.20 for insuring f of his herd of cattle, at 60^- per $100. If the cattle were valued at $40 per head, how many had he? 3. The loss on a property was $6000, of which $2000 was insured in the Home, $3000 in the Phmnix, and $2500 in the Hartford. How much did each company contribute? Jf. If it cost $663 to insure a certain block for $44200, what will be the cost, at the same rate, to insure a block valued at $105000, if $1.50 extra be charged for the policy in the latter case? J. How much will it cost to insure a factory for $42000, at f^r, and its machinery for 816500, at \\'/<,y charge for policy and survey being $2.50? 6. A gentleman paid 835.60 per annum for insuring his house, at 2f^ on two fifths of its value. What was the value of the house? EXAMPLES IX INSURANCE. 207 7. If a store and its contents are valued at $27000, for how much must it be insured, at H^ to cover loss and premium in case of total destruction? 8. A cargo of teas, valued at 8330^)0, was insured for $18000, in a policy containing an "average clause." In case of damage to tlie amount of $21000, how much should the company pay? 9. The steamer Norseman, valued at $90000, is insured for $75000, at 2^^. What will be the actual loss to the insurance company, in case the steamer is damaged to the amount of $20000? 10. A speculator bought 2000 barrels of flour, and had it insured for 80<^. of its cost, at 34^, paying a premium of $429. At what price must he sell the flour, to make a net profit of 10«^? 11. I insured my grocery store, valued at $13500, and its contents, valued at $33000, and paid $350 for premium and policy. If the policy cost $1.25, what was the rate per cent, of premium? 12. A canal-boat load of 8400 bushels of wheat, worth 90^' i)er bushel, is insured for three-fourths of its value, at If^ premium. In case of the total destruction of the wheat, how much will the owner lose ? 13. A stock of goods, valued at $30000, was insured for 18 months, at 1\'^; at the end of 12 montlis tlic owner surrendered the policy. If the "short rate" for 6 months was 65^ per $100, what should be the return jiremium? H. For how much must a house worth $G000, and furniture worth $2000, be insured, at 1^ per cent., to cover the cost of the policy, which was $2, the amount of premium paid, and f of the value of the property? 15. A man owning | of a ship, insured f of his interest, at l\fc, and i)aid $91.50 for premium and a policy charge of $1.50. If the ship becomes damaged to the extent of $12000, how much can be recovered on the policy? 16. A schooner is valued at $10500, and has a cargo of 3500 barrels of apples, worth $2.10 per barrel. What amount of insurance must be obtained, at 'i^ii, to provide, in case of loss, for the value of the property, the premium, and $5 additional which the owner paid for survey and policy? 17. A block of stores and contents was insured for $220000, and became dam- aged by fire and water to the amount of $150000. Of the risk, $40000 was taken by the Hartford Co., $05000 by the Manhattan, $35000 by the yEtna, and the remainder was divided equally between the Piuenix and the Provident. What was the net loss of each company, if the premium paid was 1^;^? 18. The furniture in my house is estimated at one-half the value of the house. I get both insured for $7687.50 for 5 years, at 24f?;, and find that in case of total destruction the face of the policy will be full indemnity for both the property and premium. Find the value of the house. 19. A factory worth $45000 is insured, with its contents, for $62500; $30000 of the insurance is on the building. $12500 on machinery worth $20000, and $20000 on stock worth $35000. A fire occurs by Avhich the building and tlie machinery are both damaged, each to the amount of $15000, and the stock is entirely destroyed. How much is the claim against the company, if the risk is covered by an "ordinary" policy? How much if tlie i)olicy contains the "aver- age clause?" 308 PERSONAL INSURANCE. 20. The German Insurance Company insured the Field block for $105000, at 60^ per $100; but thinking the risk too great, it re-insured $40,000 in the Home, at f*?^, and $45000 more in the Mutual, at ^'i. How much premium did each company receive? What -svas the gain or loss of tiie German? "Wlnit per cent, of premium did it receive for the part of the risk not re-insured? PERSONAL INSURANCE. 678. Personal Insurance is tlie insurance of ])ersons. It includes: 1. Life Insurance, or indemnity for loss of life. 2. Accident Insurance, or indemnity for loss from disability occasioned by accident. 3. Jlealth Insurance, or indemnity for loss occasioned by sickness. 679. Policies of Life Insurance are usually either Life Policies or Endoicmeut 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 jiaid 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. Remarks. — 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. Anj' 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 anj- 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 tlie money for the use of whieli interest is paid. 686. The Anionnt is the sum t)f the princijMil and interest. 687. The Time is the jjeriod during which the principal bears interest. 688. Interest is reckoned at a certain per cent, of tlie principal. It is therefore a Per Cent, of Avhich 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 Rate. The Interest = tlie Percentage. The Sum of the Principal and Intei;est = the Amount. 691 . Legal Interest is interest according to the maximum rate fixed by law. 692. Tsury is interest taken at a rate liigher than the law allows. 693. Simple Interest is interest on tlie priiu-ipal 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. 69.5. (*ompoun(l Interest is interest computed on the amount at ri>gular intervals. Remabks. — 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. 69.7. Conimoii Interest is interest comi)uted 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 l)y 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 2]0 SIX PER CENT. METHOD. adopted : Rewired, 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-on^, when it shall be so counted. RiiMABK. — In computing interest for short periods of lime, it is customary to take the exact numl)er of days. 698. Exact luterest is iuterest computed for the exact time in days, and regarding the days as 3G5ths of a year. This method is used 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. Rkmarks. — 1. Exact interest, for any period of time expressed in days, may be obtained by subtracting -L part from the common interest for that period of time. 2. Common interest may be obtained from exact interest by adding thereto J^ part 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. Remakks. — 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; stiU 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. 3. Debts of all kinds bear interest after they become due, but not hefare, 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 300 days for a year, and is called the Six Per Cent. Metliod. 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. 11.00 in 1 yr., at G,'^, will produce $.06 interest. 11.00 in \- yr., or 3 mo., at 6^, will produce $.01 interest. $1.00 in 1 mo., or 30 da., at 6^, will produce $ .005 interest. $1.00 in G da., or ^ mo., at Gf*^, will produce $.001 interest. $1.00 in 1 da., at Q^, 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 7j^. Rule. — To the interest at 6'^^ add otie-sixth ?/ itself, 2. To find the interest at 7^^. Rule. — To the interest at 6'^ add one-fourth- of itself. EXAMPLES IN INTEREST. .?11 5. To find the interest at 8^. Rule. — To the interi'st at O'i add one-third of itself. 4. To find the interest at 9^. Rule. — To the interest at 6% add one-half of itself. 6. To find the interest at 10^. Rule. — Divide the interest at 6''fo b>i >i. and remove the decimal point one place to the right. 6. To find the interest at 13^. Rule, — Multiply the interest at 6^ by ^. 7. To find the interest at 54^<. Rule. — From the intere.^t at 64, subtract one-twelfth of itself. 8. To find the interest at b'i. R»le. — From the interest at 6i, subtract one-sixth of itself. 9. To find the interest at 44^^. Rule. — From the interest at Si, subtract one-fourth of itself. 10. To find the interest at 4^. Rule. — From the interest at H':. sul)tract one-third of itself. 11. To find the interest at 3^. Rule. — Divide the interest at H', bij '. 702. To find the Interest, the Principal, Rate, and Time being given. Example, — What is the interest on $550, at 6^', for 3 yr. 8 mo. 12 da.? Operation. Explaxatton.— Since the interest on .$1 for 1 year is Int. on |!l for 3 yr, = $ . 18 $ .OG, for 3 years it will be $ . 18; since the interest on $1 " " ''8 mo. = .04 for 2 months is $ .01, for 8 months it will be $ .04; since '•- " " 13 da. — .002 ^^^ interest on |1 for 6 da. is ,$.001, for 12 days it will , . „ be $.002; therefore the interest on $1, at 6 per cent., mt. on ^1 lor 6 yr. ^ ^^^ ^^^ j^jl ^j^^^^ j^ ^222; and the interest on $550 will 8 mo, 12 da, == $ .222 ^e 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 hij tlie decimal e.vpressiiig the interest of one dollar for the full time. EXAMPLKS FOR PRACTICE, 703. 1. Find the interest on $900, for 4 yr. 1. mo. r, da., at l^L Explanation.— Find the interest at 6;^', and add to it one-si.xth of itself. 2. What is the interest on $400, for 1 yr. 7 mo. 2 (hi., at Tij^ ? Explanation.— Find the interest at 6^', and add to it one-fourth of itself. 3. What is the interest on $150, for fi yr. 3 mo. IS da., at 8^ ? Explanation. — To the interest at Q% add one-third of itself. If. Compute the interest on $1200, for 3 yr. 4 mo. 15 da., at 9^?^. Explanation. — To the interest at 6^ add one-half of itself. 5. Find the interest, at 10<^, on $840, for 5 yr. :> mo. '.» da. Explanation. — Divide the interest at 6^ by 6, to obtain tlu* interest at \[i, 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 %% by 2%. 7. Find tlie interest on «!l800, for 6 yr. 9 mo. 25 da., at 5|^. Explanation. — Fi-om the interest at 6i subtract one-twelfth of itself. S. Compute the interest, at 5^, on $1000, for 11 yr. 4 mo. 24 da. Explanation.— From the interest at 6^ subtract one-sixth of itself. !>. What is the interest, at U^t, on $1100, for 6 yr. 6 mo. 6 da. ? Explanation. — From the interest at 6^' subtract one-fourth of itself. 10. What is the interest, at 4^, on $1350, for 9 yr. 8 mo. 12 da. ? Explanation. — From the interest at 6^ subtract one-third of itself. 11. Find the interest, on 8546, for yr. 2 mo. 24 da., at 3^. Explanation. — Divide the interest at 6'^ by 2. Remarks. — 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. Compute the interest on $752.50, for 4 yr. 11 mo. 9 da., at 6^. lo. 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^. lo. Compute the interest on $610.15, for 7 yr. 11 da., at 7^. IG. Compute the interest on $1116, for 3 yr. 11 mo. 11 da., at 5^. 17. Compute the interest on $17500, for 2 yr. 1 mo. 10 da., at 4^^. 18. Compute the interest on $350.40, for 5 yr. 5 mo., at 7^. 10. 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^. 2 J. 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. 25. On the 16th of September, 18b4, I borrowed $3500, at 8,'^ interest. How much will settle the loan Jan. 1, 1890? 24. After paying $225 cash for a horse, the purchaser at once sold him for $275, on 4 months credit. Money being worth 7;^, how much was gained? 2o. 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 ])c the credit price? 26. Borrowed $2750 July 16, 1887, at bfo interest, and on the same day loaned it at 7if« interest. If full settlement is made Jan. 4, 1889, how much will be gained? 27. On goods bouglit for $4500, on 6 months credit, I was offered 5^ off for cash. If money was worth 6^r, 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 montlis, $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 iX INTEREST. 213 29. May 16th I bought 300 barrels of flour, at *7 i)er barrel; July 28th I sold 50 barrels, at $8 per barrel; Oct. 30th, 100 l>arrels, at $6.75 i)er barrel; and Feb. 13th following, the remainder, at S7.80 per barrel. Allowing interest at 6^, what was my gain? 30. John Doe bought bills of dry goods as follows: May 3, ^250; July 1, 81125; Sept. 14, $450; Oct. 31, $150; Dec. 1st. $680; and on Dec. 21st, he paid in full, with 6fo interest. What was the amount of his payment ? 31. On March 25, I sold live bills of goods, for amounts as follows: S1046.81, 1952.40, $173.50, $1250, and $718.25; and on the first day of the following December I received payment in full, with interest at 6'r. What was the amount received? 32. 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, liow much cash was paid. Remark. — In the following examples, f^xen for teacher's use in class drill, the interest on each separate principal should be computed to its nearest cent; the sum of the results so obtained will be the answer sought. 33. 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. 3Ji.. Find the amount of interest, l)y the six jier cent, method, On $538, for 6 yr. 6 mo. 6 da., at 9;^. On $1200, for 7 yr. 4 mo. 27 da., at 10^. On $350, for 3 yr. 7 mo. 18 da., at CH. On 8586.50, for 2 yr. 9 mo. 15 da., at 7'i. On $1345, for 5 yr. 4 mo. 1 da., at 8?b. 35. Find the amount of interest, by the six per cent, method, t On $675, for 5 yr. 5 mo. 25 dS, at 10^. On $1000, for llyr. 11 mo. 11 da., at 5;^. On $2500, for 1 yr. 1 nio. 1 da., atlt^^. On $300, for 2 yr. 2 mo. 2 da., at 4^. On $990, for 4 yr. 4 mo. 6 da., at 3f^. 36. Find the amount of interest, by the six per cent, method. On $550, for 4 yr. 6 mo. 21 da., at Gfc. On $2100, for 1 yr. 11 mo. 3 da., at 7^. On $750, for 8 yr. 8 mo. 8 da., at S^. On $1200, for 3 yr. 3 mo. 1 da., at 7^^. On $1500, for 7 yr. 7 mo. 9 da., at 9^^. On $250, for 3 yr. 4 mo. 29 da., at 8,^. On $967.25, fo/7 yr. mo. 27 da., at Qfc. On $1305.09, forlyr. 11 mo. 7 da., at 7^. On $1255.84, for 9 mo. 1 da., at lOj^. On $316. 75, for 5 yr. 1 1 mo. da. , at U^. On $2100. 50, for 1 yr. 1 mo. 1 9 da. , at 9^. On $3546.81, for 5 yr. mo. 5 da., at 3^. On $1867, for 2 yr. mo. 2 da., at 7^^. On $260.60, for 7 yr. 7 mo. 5 da., at 5^. On $1120.95, for 4 yr. 4 mo. da., at 4^. On $1000, for 5 yr. 6 mo. 7 da., at S^. On $1743, for 2 yr. 3 mo. 6 da., at 6^^. • 214 EXAMPLES IN INTEREST. 704. To find the Principal, the Interest, Eate, and Time being given. Example. — What principal, in 3 years and 2 nionths, at 6^, will gain $47.50 interest ? Operation. Explanation. — Since $1 in 3 years, at $.18 = int. of $1, at G'*', for 3 yr. ^ P^^" ^e°^-' ^^" g^'° ^-^^ '^^^''^''t' ^^ in 2 ^, .^ ^A, ^ n^ £ o' months |.01 interest, it will in the civen .01 = int. of %\; at e**, for 2 mo. ,. * ^^ ■ * ' ^ •* *i •„ • ' ' ' time gain $.19 interest; and if $1 will in $.19 = int. of $l,at 6jfc, for 3 yr, 2 mo. the given time gain $.19 interest, the prin- d.1- -/^ • i i. -^c^ &.n~f\ • ^ cipal that will in the given time gain $47.5*0 $4 (.oO interest ^ .19 = $2oO, pr!ncii)al. .\ ^ ^, ^ . ^.-i * -.a ' ^ ^ interest must be as many times $1 as $.19 is contained times in $47.50, or $250 ; therefore $250 is the principal which will, in 3 yr. 2 mo., at 6'V, gain $47.50 interest. Rule. — Divide the given interest by the iivtrrest of one dollar for the given time and rate. Remark.— "Whenever the divLsor 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. KXAMPLES FOR PKACTICK 705. 1. /What principal, at 'v', "will gain $154 in 6 yr. 4 mo, 24 da.? 2.- What sum of money, loaned at 4^^, for 7 yr. 11 mo. 15 da., will gain $1468.21 interest ? 3. "What sum of money, imested at 5^^, will in 7 yr. 1 mo. 1 da. produce $131.50 interest ? Jf.. A money lender received $221.68 interest on a sum loaned at 8,*^ .July 17, 1885, and paid Oct. 11, 1888. What was the sum loaned ? 5. A dealer who clears 12^^^ annually on his investment, is forced by ill health to give up his business; he lends his money at 7^, by which his income is reduced $1512.50. How much had he invested in his business ? 6. How many dollars mitst I put at interest, at 9^, Jan. 27, 1889, .so that on the 18th#>f Xov.. 1895, $506.27 interest will be due? 706. To find the Principal, the Amount, Rate, and Time being given. Ex.v.MPLK. — What i>rincipal, at ij'ft,, will, in 4 yr. G mo. 15 da., amount to $2372.25? Operation. Explanation. — Since a principal of 4,, or/.T- t. e S.A r^t\ e l\ j.- $1 "^^^h in the given time, amount to $1.272o = amount of $1.00 for the time. ^. o-o- •. -n • • • i <• $1.272.D, It will require a principal of as $2372.25 -f- 1,2725 = $1800, principal. many times $1 to amount to $2372.25 as $1.2725 is contained times in $2372.25, or $1800. Rule. — Divide the uimmnt by tlir amount of 1 dollar for the given time and rate. EXAMPLES IX INTEREST. 215 EXAMPLES FOK PRACTICE. 707. 1- What Slim, put at interest at '7'fc 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 7^5^ 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? Jf. 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 6j^ ? 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 Eate 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 ? Operation. Explanation. — The principal will gain . ^ „ , , . , ^ . $18.25 interest in the given time at 1 per $18.25 = ]nt. of $750 for the time at 1^. cent. ; in order that it may in the given time $164.25 -7- $18.25 _ 9 or 9^. ^..^^^ $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, irvterest hy the interest on tlie given principal for the given time, at 1 per cent. Remark. — When the amoimt, interest, and time are given, to find the rate percent., subtract the interest from the amount, thus finding the principal, then proceed as by the above rule. EXAMPLES FOK PRACTICE. 709. 1. If I pay $518.75 interest on $1250, for 5 yr. mo. 12 da., what is the rate per cent. ? 2. At what rate would $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.? i. If $1200 amounts to $2135.80 in 12 yr. 11 mo. 29 da., what is the rate per cent. ? 6. 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 |i74.52 interest, at 6^ ? Opekation. Explaxatiox. — Since in 1 year $540 will, at. $32,40 = int. on $540 for 1 vr., at 6<. ^ P^'" cent., gain #32.40 interest, it will require f^A 5-> _i_ •^•) 40 — o Q ~ ' as many years for it to gain .$74.52 interest as * ^ ' ^* .-) Q~" ' f32.40 is contained times in $74.52, or 2.3 years; Z:6 X 1 yr. = :..3 years. find, by the rule 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 hy the interest on the princijxil for 1 year, at the given rate ])rr eent. 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. EXABIPLES FOR PRACTICE. 711. 1. How long will it take $360 to gain $53.64, it 6^. 2. How long should I keep $466.25, at 8fr, to have it amotint to $610.48 ? 3. A debt of $1650 was paid, with bl'/c interest, on Aug. 30, 1888, by deliver- ing a check for $2316.85. At what date was the debt contracted ? 4. How long must $612 be on interest, at 7,^^', to amount to $651.27 ? 5. On April 1, 1888, I loaned $1120, at 5^, and when the money was due I received $1202.60 in full payment. What was the date of the payment ? 6. In what time will money, bearing 8^ simple interest, double itself ? ExPLAXATiox. — In order to double itself, the interest accumulated must be equal to the principal, or be 100 per cent, of the principal. And since the principal increases 8 per cent, in one year, it will require as many years to increase 100 per cent., or to double itself, as 8 per cent, is contained times in 100 per cent., or 124, equal to 12 yr. 6 mo. SHORT METHODS FOR FINDING INTEREST. 712. To find Interest for Days, at 6 per cent., 360 day basis, or Common Interest. ExPLAXATiox. — A principal of |1 will, in 1 year, at 6 per cent., gain $.06 interest. A prin- cipal of $1 will, in I year, or 2 months, or 60 days, at 6 per cent., gain .01 interest. Since $.01 equals j-J^ of the principal, the interest on any sura of money for 60 days, at 6 per cent., can be found by pointing off two integral places from the right; and since 6 is y'g of 60, the interest for 6 days Ciin be found by pointing off three places; and since ten times 60 is 600, the interest for 600 days is ten times that for 60 days, and may be found by pointing off 1 place; and since 6000 is ten times 600, the interest for 6000 days can be found by multiplying the inter- est for 600 days by 10, or in other words, the interest for 6000 days will equal the principal; the principal thus being shown to double itself in that time at 6 per cent. This may further be proved true from either of two illustrations: EXAMPLES IN FlXDINft IXTEREST. :^17 1st. 6000 da, -^ :JGO (12 X 30) = 16|, or 16 yr. + 8 nio. 2d. lOOfc -^ 6;c = 16^, or 16 yr. + 8 mo. Hence, assuming $3136 as a principal, we form the following Table. %2136 = principal. 12.136 = interest at Qfo for 6 days. $21.36 = interest at 6^ for 60 days. $213.6 = interest at ^ for 600 days. $2136.= interest at 6^ for 6000 days. Remakks. — 1. Observe, as above stated, that the interest for 6000 days equals the principal, or that anj- sura of money will, at common interest, double itself in 6000 days. 2. Since interest is ordinarily computed on the basis of 860 days, or 12 periods of 30 days each, as illustrated above, all results will be required on that basis, unless otherwise specified. 713. — 1. To find the interest of any sum of money, at ^4, for 6 days. Bulk. — Cut off three integral ;placesfrom the right of the principal. 2. To find the interest of any sum of money, at 6^ for 60 days. Rule. — Cut off two integral places from the right of the j)rincipal. 3. To find the interest of any sum of money, at (j'fc, for 600 days. Rule. — Cut off' one integral place from the right of the principal. Jf. To find the interest of any sum of money, at 6,^, for 6000 days. Rule. — Write the interest as being equal to the 2^rincipal. Remark. — Interest is a product of which the rate and time are factors. [Formula. — Interest =r Principal X Rate X Time.] Since the rate, being a constant factor, may be ignored, it will be observed that it will make no difference if, for convenience, the term principal (in dollars), and that of time (in days), be interchanged. Illustration: The interest of 500 (dollars) for 93 (days), is the same as the interest of 93 (dollars) for 500 (days); and since 500 is ^^j of 6000, the interest required can be found by dividing 93 (dollars) by 12, which gives $7.75. Again, the interest of 150 (dollars) for S8 (days) equals the interest of 88 (dollars) for 150 (days); and since 150 is i of 600, the required interest is obtained by pointing off one place from the right of 88 (dollars), as, $8.8, and dividing the result by 4, obtaining $2.2, or $2.20, as the interest. 714. To find Interest at Other Rates than 6 per cent., 360 Day Basis. 1. To find the interest on any sum of money for 12 days, at 6 per cent. Rule. — Point off' 3 ^jlaces and multi])li/ by 2. Remakks. — 1. For any number of days divisible by 6, proceed in like manner. 2. For other rates, add or subtract fractional parts of results, as in Art. 701. 3. For odd days, add fractional parts to the result. 2. To find the interest for 18 days, at 7;k Rule. — Point off' 3 places , inultiply by 3, and to the result add one-sixth of itself. 3. To find the interest for 24 days, at o^L Rule. — Point off 3 places^ mziltiply by 4, and from the result subtract one-sixth of itself. 4. To find the interest for 36 days, at 4^^. Rule. — Point off 3 places, rmiltijily by 6, and from the result subtract one-fourth of itself. 5. To find tiie interest for 78 days, at %rth, apply the principles given in Interest. The debt corresponding to the amount; the rate per cent, agreed upon to the rate; the time intervening before the maturity of the debt, to the time; and the present worth, which is the unknown term, is the principal. 2. When payments are to be made at different times, without interest, find the present worth of each payment separately, and take their sum. 3. With debts bearing interest, and discounted at the same or at a different rate of interest, the face of the debt plus its interest as due at maturity becomes the base. 743. To find the Present Worth of a Debt. Example. — Find the present worth and true discount of a claim for $871.68, due 2 yr. 3 mo. hence, if money is worth C<^ per annum. ExPL.VNATiox. — The amount of the debt ,^ at the end of 2 yr. 3 mo. is |871.68; and Operatiox. , ^ , , . , . „ smce $1 would m that time, at 6 per cent. , ,133 = int. on $1 for 2 yr, 3 mo. at 6,<. amount to f 1.135, the present worth must $1,135 = amount " " " " be as many times $1 as $1,135 is contained $8:i,68 --- 1.135 = $768, present worth. *^™<^^ »° $871.68, or $768. If the face of *or«i />o «>iao *ir>o cfu i. J- i. the debt is $871.68, and its present worth $871.68 — $768 = $103,68, true discount. . „, ...;„ ,, ' .■ , n ., ' IS only $i68, the true discount will be $871.68 minus $768, or $103.68. IXvii^.— Divide the aitvount of the debt, at its maturity, by one dollar plus its interest for the given time and rate, and the quotient ivill be tlie present worth; subtract the present ivorth from the amount, and the remainder irill be the true discount. KXAMPLKS FOK PRACTICE, 744. 1. What is the present worth of 8661.50, 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. 3. Find the i)resent worth of a del)t of $1750, $1000 of which is due in 9 mo, and the remainder in 15 mo., money being worth 6^ per annum EXAMPLES i:^ TRUE DISCOUNT. 231 i. Which is greater, and how mucli, the interest, or the true discount on ^516, due in 1 yr. 8 mo., if money is worth 10$^ 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 Avorth 6^? 6. AVhen money is worth 5^ per annum, which is preferable, to sell a house for $30,000 cash, or $31,000 due in one year ? 7. A farmer offered to sell a pair of horses for $430 cash, or for $475 dne 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 81.95 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 G^ 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 3 months •credit, or $535 on 1 3'car's credit. Which offer is the most advantageous for him, and how much better is it than the next best, with money worth 7|^? 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 ? 14' 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, $3000 of which is on credit for 5 mouths; $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 lOj^ profit, he paying cash and getting the agreed discount off? 18. A dealer bought grain to the amount of $3700, 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 S^c per annum, how much did he gain ? 19. A merchant bought a bill of goods for $3150, on G months' credit, and the seller offered to discount the bill 5^ for cash. If money is worth 7^,^ 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 lOf^ 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 ^'fc on the remainder for cash payment. He at once sold the stock on 4 months' time, at lOj:^ 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 lOj^ 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, lO,'^ off if paid in 1 month, hi) off if paid in 2 months, must be purchased, in order that they may be sold for $11480, and \ the stock net a profit of 15^ and the remainder a profit of 20j^ to the purchaser, if he cashes his purchase within 1 month and gets the agreed discount off ? BANK DISCOUNT. 233 BANK DISCOUNT. 745. A Bauk is a corporation chartered by law for the receiving and loaning of money, for facilitating its transmission from one place to another ])y means of 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 trve discount, the present worth is taken as the principal ; in hank 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 da3's from the date of discounting to the date of maturity. 752. The Term of Discount is tlie 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 BEMARK3 ON COMMERCIAL PAPER. 756. The Talue 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. Re^iarks. — 1. Grace is given on all negotiable time paper unless " mthout grace" he speci&ed. 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. Xotes 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. In some States, the bank custom is to take discount for both the day of discount and the day of maturity, which is excessive. Rejiarks. — 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 fully 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. Sucli 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. Conimercial, 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 pajier is protected in his right of recovery of its amount agamst all original and contingent parties thereto, if he can show three conditions : 1st. That he gave value for the paper. GENERAL BEMARKS OX COMMERCIAL PAPER. 235 2d. That he bought it before its mbell, Dr. , ., . .. , , ^ . * each Item to the focal date. ^^^°- 3 Compute interest at 6 percent., 360 day basis, Sept. 5. To Mdse. ^60 on each item for its time. ''26. " " 100 4. Find the total of interest Qq^ 8. ** *' • 200 ^- I^i^'ide the total interest by the interest on the Xov 1 " '' 1"^0 ^^^ amount for one day; the quotient will be tLe 1 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. $ 60 X 57 = # .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 "^00 X 24 =: 80 became due. If, on Nov. 1, Dunn pays Campbell "V ' 1 1 '>n N/ no ^^^ ^~^^ ^^^^ ^^^ heen 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 41 Q*- _i_ OR 04.4 or 25 dav with, $.80 interest If, on Nov. 1, Dunn pays the .~ ^ ^ 11* $100 due Sept. 26, he should pay interest also for the the average time; 2o days back ^ ^^^.^ ^0^^.^^^ g^pt og ^^d Nov. 1, or $ . 60. If, on from Nov. 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 reciuired 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, 18S8, 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 bearin"- note or bond for the equitable amount due, or for anticipating the payments of debts, thus requiring a cash balance. Operation. 1888. Items. Time. Disct. Sept. 5. $ 60 X = $.00 " 26. 100x21= .35 Oct. 8. 200 X 33= 1.10 Nov. 1. 120 X 57 = 1.14 $480 $2.59 EQUATION^ OF ACCOUNTS. 245 Explanation. — Assume the earliest date (Sept. 5) as the focal date, and reason as follows: If, on Sept. 5, Dunn pays the $GOdue 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. 2G, he should be allowed discount oa that item for the 21 days between Sept. 5 and Sept. 26, or $ .35 discount. If, on Sept. 5, ue pays the |200 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. Opebatiok. 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 CM $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, . . . . $.8533 + .8533H .3433+ = $.51, excess of discount. $.51 -^' .08 = 6 days. Oct. 1 + 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 $.20 interest. If, on Oct. 1, he pays the $100 due on Sept. 26, he should also pay interest ou 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-j- 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 $ .23334- on that item; and if, ou 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 ditference 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, less $ .5] discount, or only $479.49, which .sum is the cash balance duo 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 tind Dunn to be entitled to withhold or delay the payment of the $480 until it accumulates $ .ol 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 aut/ 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 iifvolved, and it gives added assurance that the solutions before given led to a correct result. yelnot»«<' of them, nor all of them taken together, can be accepted as being anything beyond aMurances. Tliey are not proofs. 2. If settlement on Oct. 7 be equitable, the interest on such of the accounts &s fall due before that date must be offset or balanced hy 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. Proof. — Oct. 7 a.s 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 till for the 11 days between Sept. 26 and Oct. 7, or $.1833-|- interest; being thus charged $ .5033+ interest on the two items not paid until after they were due. But if, on Oct. 7, he 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 Isov. 1, or $.50 discount; being thus allowed a total discount of $ .5333+ for the pre-pajTiient of the items of the account coming due after Oct. 7. The difference lM?tween the amount of the interest on the items of the account falling due before Oct. 7, from their rcsixjctive 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 le.ss 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 accoiuit, $480, will effect an equitable .settlement between Dunn and Campbell. Rule.— I. Select tlie 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. n. Divide the anwunf or sum of tJie interest on the items hij the inter- est on the face amount of the account or items for one day ; the quotient ivill he the number of days average time. III. Count hack from the focal date the nunibcr of days so found; tJis date thus reached u-ill he the due date of the face amount of the account or the date on ichich such face amount could he paid without loss to either party. Operation. Days to Oct. 7. Interest. Sept. 5, $ 60 33... -$.32 Sept. 26, 100 11... . .1833 + .5033 + Discount. Oct. 8, $200 1 $.0333 + Kov. h 120-...25 .50 $.5333 + .50.33 + EXAMPLES IN 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 fmir 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 teas 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 ligures. 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 chancfi 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 tliough the employer be unfa- miliar with the formal work of the equation. ScGOESTiON TO THE Teacher. — Placc 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 a.ssign 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 FOR PRACTICE. 801. When iire the following uccoimts due by equation: Remark. — The teacher should require that each result be proved. 1. 1888. Oct. Oct. Warren Pease, To Calvin Gray, Ur. 1, ToMdse. .- $ ;5 6, " " 50 14, " " 80 25, " *' 120 31, " " 40 Xorman Colby, To Seth Stevens & Sons, Dr. 1, To Mdse 4300 5, " " - 150 11, " " 120 IG, '' " 200 28, " '' 100 30, " *' 180 1888. Aug. Sept. 30, Oct. 12, Parker II. Goodwin,. 7o Perkins & Ilawley, Dr. 7, To Mdse. .$200.00 " 180.55 Dec. 3, 35.60 100.00 50.25 Jan. 6, Feb. 1, a 27, Apr 3, ( i 20, . Fijid \\\v balance of the following account, and when due by equation. Dr. Louis K. Gould. Cr. 1888. Sept. 21 To Mdse. Oct. 5 a ic (< 30 a a Dec. 18 er day was a cause, the effect of which is unknown; but from the application of the logical statement of the principles of proportion (1st Cause : 1st Effect : : 2d Cause : 2d Effect), we have the statement of the example given in form as follows: 20- COMPOUND PROPORTION. IstCaust'. 1st Effect. 2d Cause. 2d Effect. And since, as before shown, the extreme (or 10 1 ^'"^ outside) terms constitute the factors of the divisor, 12 200 1^ ^ and the mean (or inside) terms constitute the fac- 8 j 6 tors of the dividend, any factor of the divisor' may be cancelled against any factor of the dividend, or vice versa. Reproducing the above statement, and effecting possible cancellations, we have: : 225 cords. 5 X 15 X 3 = 225. Remark. — All problems in proportion, simple or compound, by some called the "single rule of three" or the " double rule of three," can be solved as above. 10 n n m 15 H ^ 3 kJcamplks for practick. 823. 1. If 5 men, working G days of 12 hours per day, can cut 2-4 acres of com, how many acres of corn should 8 men cut in 5 days, if they work 10 hours per day? 2. If 6 men, working for 12 days, dig a ditch 80 rods long, how many rods of such ditch should 15 men dig in 21 days? Remark. — When any term or terms is fractional, either common or decimal in form, treat them in the usual manner, or reduce such fractions to a common denominator and compare their numerators. S. If 15 men earn $607.50 in 18 days, how much should 21 men earn in 12 days? 4. If $1600, invested in a business for 3 years, gain $900, liow much should $2150 gain in the same time. 5. If 8145.35 interest accrue on $510, at 6f/, in 4 yr. 9 mo., how much interest will accrue at the same rate and time on $1350? 6'. If 40 yards of carpet, f of a yard in widtli, Avill cover a room 18 feet long and 15 feet wide, how many yards of carpet, \ of a yard in widtli, will cover a room 35 feet long and 28 feet in width? 7. If $684, at interest for 3 yr. 3 mo. 18 da., at 5fc, accrue $112.86 interest, at what rate per cent, must $1800 be put at interest for the same time to accrue $445.50 interest? 8. If $760, put at interest at 10«s5, accrue $9.50 interest in 45 days, in how many days will $1140 accrue $17.67 interest at be entitled; or it may be obtained by application of the same form as that used to determine the gain of A, viz. : $n^ : 060 : : 3^00 : (B's part). ;? 8 30 8 X 30 = $240, B's part of the rent. Rule. — The ichdle capital is to the whole gain, as each partner s sTiare of the capital is to his share of the gain. Remarks. — 1. Should the result of the investment be a los.s, the share to be sustained by each can be determined in the same manner as above. 2. If investments are made for different periods of time, compute the investment of each partner for one period of that time, day, month, or year, then make the proportion as above. EXAMPLES IN PARTNERSHIP. 265 EXAMPLES FOK PRACTICE. 841. i. Two men boiight a mine for 120000, of which sum A paid $12500, and B paid the remainder; they afterwards sold the mine for 142000. How much of the selling price was each partner entitled to receive? 2. The condition of the business of Hadley & Hunt is as follows: Mdse. OD kand, $28240; notes and accounts due the firm, 121416.54; cash on hand, $1619.62; total liabilities of the firm, $23186.75. Hadley's investment was $9000, and Hunt's $12500. What has been the gain or loss, and what is the share of each? 3. A, B, C, and D, engaged in a business, in which D invested 88400, which was also the amount of the net gain; if A's share of the gain was $1800, B's $3000, and C's $2400, what must have been the whole capital and D's gain? 4. A, B, and C are partners, A's investment being $9600, B's $8100, and C's $7500. At the end of the year they have resources amounting to $27850, and liabilities amounting to $3150. What is the present worth of each partner at closing? 5. Four partners. A, B, C, and D, invested equal amounts, and agreed ta equally ajiportion the gains or losses. At the time of dissolution, the firm had resources to the amount of $33800, and liabilities to the amount of $51975. If the net loss was $27460, what Avas the net insolvency of each partner at the time of dissolution? What was each partner's investment ? 6. A and B were partners 1 year, each investing $3500, and agreeing to equally share the gains or sustain the losses. At the close of the year their resources were: Cash, $2650; Mdse., $3040; accounts due them, $3150. During the year, A drew out $4500, and B $5750. How much has been gained or lost? What is the solvency or insolvency of the firm? What is the present worth of each? 7. Harrison and Morton bought a section of Nebraska jirairie for $8000, Harrison paying $5000, and Morton paying the remainder. Cleveland offered them $8000 for one-third interest in the land; the offer being accepted, the land was surveyed and divided, each taking for his exclusive use one-third of it. How should Harrison and Morton divide the 88000 received from Cleveland? 8. Seaman and Sullivan entered into partnership with a joint capital of $35500, of which Seaman invested $22000. During the existence of the part- nership, each withdrew $1500, and it was agreed that no interest account should be kept, and that Seaman should receive f of the gains, and sustain the same share of the losses, if any; while Sullivan should receive f of the gains, and Sustain that share of the losses, if any. At the time of the dissolution, the resources and liabilities were as follows: Eesources. Cash $ 2050 Accounts receivable 15850 Real estate 8100 Liabilities. Notes outstanding $21500 Accounts outstanding $16500 Insurance and interest due 2000 Find the net loss of the firm, and each partner's net insolvency at closing. 2G0 EXAMPLES IN" PARTNERSHIP. Operation and Expijlnation. Total liabilities $40000. 00 Total resources 2G000.00 Net insolvency 814000.00 Seaman's f of net loss 129062.50 Sullivan's f of net loss 17437.50 Total loss $46500. 00 Proof. Seaman's net insolvency $8562.50 Sullivan's net insolvency 5437.50 Net insolvency of firm $14000.00 Seaman's investment $22000 Seaman's withdrawal 1500 Seaman's net investment $20500 Whole investment $35500 Seaman's investment 22000 Sullivan's investment $13500 Sullivan's Avithdrawal 1500 Sullivan's net investment $12000 Seaman's net investment .$20500 Sullivan's net investment 12000 Firm's net investment $32500 Firm's insolvency 14000 Firm's net loss $46500 Seaman's |- of loss, $29002.50, less liis net investment, $20500 = $8562.50, Seaman's net insolvency. Sullivan's f of loss, $17437.50, less his net investment, $12000 = $5437.50, Sullivan's net insolvency. S42. To Divide the Gain or Loss, according to the Amount of Capital Invested, and Time it is Employed. Example. — A, B, and C are partners in business; A invested $3000 for four years, B invested $5000 for three years, and C invested $4500 for two years. How should a gain of $15000 be divided? Operation and Explanation. A's investment of $3000 for 4 yr. = an investment of 83000 X 4, or $12000, for 1 yr. B's investment of $5000 for 3 yr. = an investment of $5000 x 3, or $15000, for 1 yr. O's investment of $4500 for 2 yr. = an investment of $4500 x 2, or $9000, for 1 yr. A's investment for 1 vr. = $12000 B's investment for 1 yr. = $15000 C's investment for 1 yr. = 9000 Total investment for 1 yr. = $36000 J0000 : 1?000 : , 1^000 : A' gain. t 5000 $5000 = A 's part of gain. $0000 : 1W0 : : 10000 : B's gain. n 5 1250 5 X 1250 = $6250 = B's gain. ?0000 : 1^000 : : 0000 : C's gain. n 5 750 5 X 750 = $3 r50 = C's gain. EXAMPLES IX PAKTNEBSHIP. 267 Rem AKK.— Should withdrawals of capital be made at different times, or additional invest- ments be made, follow the steps taken above: i. e., by subtracting from the whole investment for 1 year (or 1 month) the Avhole withdrawal for 1 year (or 1 month). BXAMPI.ES FOR PRACTICE. 842. 1. Three persons traded together and gained $900; A had invested in the business $1000, for C months; B had invested $T50, for 10 months; and C had invested §1200, for 5 montlis. How should the gain be divided? 2. A, B, and C were partners; A had $800 in the business for 1 year, B had f 1000 in for 9 months, and C had $2000 in for 8 months. How should a gain of $2150 be divided ? 3. Martin and Eaton were partners one year, Martin investing at first $5000, and Eaton $3000; after six months Martin drew out 83000, and Eaton invested $1500; they gained $3600. What was the gain of each, and the present worth of each, at the time of the dissohition of the partnership? 4. A, B, and C hired a pasture for 6 months for $95.10; A put in 75 sheep, and 2 months hxter took out 40; B jiut in 60 sheep, and at the end of 3 months l^ut in 45 more; C put in 200, and after 4 months took them out. What part of the rent should each pay? 5. A, B, and C were partners, with a joint capital of $18600; A's capital was invested for 6 months, B's for 10 months, and C's for 1 year; A's part of the gain was $1260, B's $1500, and C's $1200. Find how much Avas invested by each. 6. A and B engaged in the grocery business for 3 years, from March 1, 1885; on that date each invested $1600; June 1, A increased his investment $400, and B drew out $300; Jan. 1, 1886, each withdrew $1000; Jan. 1, 1887, each invested $1500. How should a gain of $7500 be divided at the time of the expiration of the partnership contract? 7. A commenced digging a ditch, and after working 6 days was joined by B, after which the two worked together 9 days, when they were joined by C. The three then worked 12 days, at the end of which time A left the job and D worked with the other two 3 days and the work was comoleted. If $92 was paid for the work, how much should each receive ? 8. July 1, 1885, A and B commenced business with a capital of $7500, fur which A furnished -| and B the remainder; May 1, 1886, B invested $1500, and A withdrew $600; Oct. 1, 1886, they admitted C as a partner, with an investment of $4500; Jan. 1, 1887, each partner invested $1000, and on Jan. 1, 1888, each partner withdrew $500. On closing business, Oct. 1, 1888, it is found that a net loss of $3000 has been sustained. Find each partner's proportion of the loss. ■9. Olsen and Thompson dissolved a three-year's partnership Aug 1, 1888, having resources of $16500, and liabilities of $2150. At first Olsen invested $2750, and Thompson $2500; at the end of tlie first year Olsen drew out $1500, and Thompson invested $3000; six months later each invested $1200. Xo interest account being kept, what has been tlie gain or loss, and the share of each partner, if apportioned according to average investments ? 268 EXAMPLES IN" PARTNERSHIP. 10. Simmons and Sawyer commenced business with $25500 capital, of which Simmons invested ^13500. It was agreed that Sawyer sliould liave $1200 a year sahiry for attending to tlie business, and that tlie net gain should be divided in proportion to investments. At the close of 1 year the partnership was dissolved, the firm having resources to the amount of $;}7500, and liabilities, otlier than for Sawyer's salary, to the amount of $4150. If neither made witlulrawals during the year, what was the interest of each partner at closing? 11. Drew, Allen, and Brackett, each invested 1^15500 in a business that gave the firm a profit of $21000 in one year. Nine months before dissolution, Drew increased his investment $3000, and Allen and Brackett each Avithdrew $3000; six months before dissolution, Allen invested $2000, and Drew and Brackett each drew out $2000; three mouths before dissolution, Brackett invested $1000, and Drew and Allen each drew out $1000. If no interest account was kept, and the gain be divided according to average investment, what is each partner's share ? 12. A and B formed a copartnership for 3 years, A investing $7200, and B investing $5400. At the end of G months A increased his investment by $1500, and B Avithdrew $900; one year before the expiration of the partnership, each withdrew $1000; and 6 months later each invested $500. The net loss was $2400. How much should be sustained by each, if sustained according to aver- age investment; and if each be credited for interest at ^^ on investments and be rliarged interest on withdrawals, what will be the present worth of each at closing ? 13. Sept. 1, 1883, Martin and Gould engaged in partnership for 5 years, Martin investing $13000, and the firm assuming his debts, amounting to $2750; Gould investing $9G00, and the firm assuming his debts, to the amount of $1050. At the end of the first year Martin withdrew $2000, and Gould invested $800. At the end of the second year Cole was admitted as an equal partner, he making au investment of $C000. One year later each drew out $1000; and six months before the partnership contract expired, each invested $2500. Sept. 1, 1888, the partnership was dissolved, at which time it was found that a net loss of $7500 has been sustained. If the loss was shared in proportion to average investment* what was the loss o^ each partner? MISCELI-ANEOUS EXAMPLES. 1. Hart, of Kansas, and Brown, of New York, form a copartnership in the grain business; Hart to make jjurchases. Brown to effect sales, and they agree to share equally the gains or losses. Brown sent Hart $12,000 cash; Hart bought grain to the value of $14,382.50, and sent Brown 40 car loads of corn, of 600 bushels each, which Brown sold at 65^ per bushel. Hart paid traveling expenses to the amount of $438.20, and Brown paid freight $1249.70. At the close of the season Hart had in his possession wheat to the value of $1128.42, and Brown had on hand 8300 bushels of oats, worth 28^ per bushel in the New York market. They then dissolved partnership, each taking the grain in his possession at the values stated. What has been the gain or loss, and how should the partners settle ? Remark. — By application of the principlea of debit and credit, as used in book-keeping, a book-keeper may with ease and certainty close np the affairs of a partnership involving any agreed division of gains or losses, interest conditions, or those of prior or subsequent insolvency. EXAMPLES IX PARTNERSHIP. 269 Dr, Hart, Or. Br !|12000.00 1128.42 $13128.42 3183.29 $16311.71 114382.50 438.20 1491.01 $16311.71 Operation. Brown. Cr. $15600.00 2324.00 $17924.00 $12000.00 1249.70 1491.01 $14740.71 3183.29 $17924.09 Dr. Grain. Cr. $14382.50 438.20 1249.70 $16070.40 1491.01 1491.01 $19052.42 $17924.00, Brown's debit. 14740.71, Browirs credit. $15600.00 1128.42 2324.00 $19052.42 $19052.42, sales of grain. 10070.40, purchases of grain. $3183.29, excess received by Brown, or the amount due from Brown to Hart. 2 ) 2982.02, net gain of firm. 1491.01, net gain of each. ExPivANATiON.— 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 oae-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 bj-^ 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 partnersliip Sept. 1, 1880, 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 $0250, 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 ? S. 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 tlie 24j^ received from Ealph be divided among the other three boys? 270 EXAMPLES IN PARTNERSHIP. Jf.. At the timo of closing business, the resources of a firm were: Cash, $931.50; Mdsc, per inventory, S13196.25; notes and accounts due it, $8154; interest on same, $211.50; real estate, $11150. Tlie firm owed, on its notes, acceptances and bills outstanding, $7142, and interest on the same, $348.50; and there was an unpaid mortgage on tlie 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 ma4e, 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 G months Gray withdrew $2000, and Snyder $3000, and Dillou invested $5000. Three months later Gray invested $1000, and Snyder and Dillon each withdrew $1500. At the end of the year they dissolved the partnershi]), having as total resources, $51000; total liabilities, $10500. No 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, Kogers, and Wilder enter into partnership for five years. Phelj)S invested $10000; Rogers, $20000; and ^Yilder, $30000. At the end of each year Phelps withdrew $1000; Rogers, $1000; and Wilder, $1800. Upon final settle- ment, tlie value of the jmrtnership 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 K\)V. 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., 114785; 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 G^ sim])le 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. 5J71 invested $3000; Aug. 1, 1888, Briggs withdrew I^ISOO; Dee. 1, 1888, tlie ])artners agreed upon a dissolution of the partnership, having resources and liabilities as follows: Liabilities. Notes and acceptances $6520.00 Outstanding accounts 21246.50 Kent due 1200.00 Resources. Cash on hand and in bank $ 1101.05 Accounts receivable ] 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 sliould 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, 1S86, 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 S1500. At the expiration of the time of the contract the resources exceeded all liabilities $47280. What was the gain of each, and the present wortii 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 G^ 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 Avas each partner's share of the gain? 13. A and B became partners for one year; A investing f of the capital, and B f ; the agreement being that the gains or losses shall be apportioned accord- ing to average net investment, and that each partner be allowed 6ffe 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; casli, $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, 12100. The liabilities were: mortgage on real estate, $7000; interest on same accrued, $210; notes outstand- ing S26950; interest accrued. on same, 1811.75. 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 ? 14. Clay and Hard commenced business Nov. 1, 1883, with the following resources: Clay invested cash $10000 | Hard invested Mdse., valued at ..$13500 Store, valued at. ..- 12000 j Cash 3000 Marble fixtures, valued at 1500 i 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. Xov. 1, 1887, each partner invested $1000; and on !N"ov. 1, 1888, the partners agreed upon a dissolution, the following being shown irom the ledger of the firm: Liabilities. Xotes 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 Besources. Mdse. , per inventory $48450. 50 Cash '- 10918.20 Accounts receivable 23416.80 Eeal estate 15000.00 Movable fixtures and sundries, 3114. 50 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 is the present worth of each at closing ? ANSWERS Page 12. Art. 64. 1. 45. .2. 306. S. 217. 4. 1647. 5. 979. 6. 262. 7. 853. 5. 599. 9. 1053. iO. 1610. Art. 65. 1. 3342. .^. 22512. 26052. 161840. 223732. 2967515. 7. 813496. S. 21423493. 9. 24543879. 10. 8179519. 1. S. 3. 4. 5. 6. 7. S. 9. 10. 11. 12. 13. Art. 66. 133.36. 530.80. 553.61. 629.23. 421.34. 536.91. 948.69. 91.30. 314.61. 296.19. 488.35. 260.'54. 473.43. Art. 67. $3102. Page 13. 5530 pounds. 33200 feet. 5114836332. 18 5. 6457434373. Page 22. (>. 515. Art. 91. 7. 599100. 1. 126. 8. £919760700. 2. 124. 9. $519949564.38. 3. 54. 10. £87197000. 4. 300. 11. 168 in. 6. 204. 12. $178586. 6. 450. 7. 182. Page 14. 8. 87. 13. 513281. 9. 114. u. 50291783. 10. 475. 15. 3501409. 11. 408. 12. 4088. Page 15. 13. 750. 16. $3361127356. u. 680. 15. 1248. Page 18. 16. 693. Art. 80. 17. 1197. 18. 832. 1. 613. 19. 3330. 2. 1609. 20. 712. 3. 2022. 21. 1440. 4. 13890. 22. 572. 5. 50000. 23. 585. G. 64365. 24. 3015. / , 151223. 25. 1300. '18»> "T8 8 > Art 14G. Art. 173. V,V.and^«,V- 1. 11, 7. 14. 1, 62 p 897 7. 3376 SOOO 7660 ?000< »009> ^SOV> ,J. 12. 8. 50. 2. 8400 5800 anf\ 9o00' iroOS' '»"" 3. 16. .9. 151. 3. 2260 »(fOO- 4- 18. 10. 63. 4. 5. If. io. «v,'A. S. V2T.^W.W(f-. 5. G. 52. 45. 11. 70. 12. 25. 3«an -T50 • Art. 174. 0. 1449000 708760 »4b;ooo-> ^rsxim' Pag e41. 1. h 6\ 5^.5. BS31500 anl\ S4SOO0'. *" 1. 4. 7. 7. 3. -V. . 54000 69400 7 4 25 0' TT5?ff> 47025 S5 145 7rffS0> 'Ji^oO- 19800 33000 7T5o ? 2 » • 18 lis 14 7 6,5 "20< "2 0"' "2 0' 50 • •*'« and ^A** 'So > "■"" 20 • 10. 11. stnri 1486000 anu -YUfso Page 49. Art. 184, I 5 T20> 815 BO-f> 4 8 10 "i"2(7) I'SO' and -1V5. _6 6 8 8 8 8 11 ^ffT' KOTi BO- 358 q„,1 8 4 6 0f> ^"11 50-f- 49S 380 13860 lS5ff'T9 8ff» 198 880 4950 a,wl 1886 i»8 0- 1080 1248 13 00 "iS60' lff«TJ' 1B«0> 13 65 3 180 8 82 156-U> 1560> fStfO and 780 18 10 6 5 3 GO' «U> BO' "b0'> so, 240 nnrlSOO eu ' ^"" "80 • 24 40 9 90 6 90 90' 90, 90 > "9«> 186 810 anr\ 90 » "90", *"^ 90 4830 8698 148.8 i80"> "190 > T^r5» 103 5 .,,1,1 180 "ISO", '*"*^ 180' il-f > 84 ' 8f » 735 272 onH 420 '8f • 1280 1880 1890 2520' 252ff, 5Tr2D» 2 18 2 10 2 3S0> 2520' 816 2205 2520' 2320' ??i" and 2?5? 18 2'3-_- 8160 2520' 2320' 2240 Q„,l 2 2 88 8520' '"•"'J 2520' i 4 8 8 15 8 8ff' 18(J> 18 0' 492 1440 18 1»0' "180"' 18d> 630 2 89 6 niiH 18 0> 18 0"' •*"" 48 18 0- 1800 17.82 1760 1980, "rtSff> T980, 2 7720 1»H0 1 « uo 1980, ..V 27720 198 0"! 1710 17S0> 9865 l98ff> 81 780 TSSO • 1760 rise 1 » HO 19 8 0' 730 80 "1980'" 10698 T5da ' 276 ANSWERS. IS. 4. 6. 6. 7. 8. 9. 10. 1. 2. 3. A. s: 6. 7. 10 6 5 1680 8 6,0 840 STOO 2500 S 7 8 1800 880 Art. 186. 10. TiTff' 1. 3. G. 31|. S. ^\. 7. 2|. 3. 4. 5. 4. 3lf- 01 9 ^. 4tV. -?^- 2i| Page 50. Art. 188. 1. 4|. J.171 1 4t5T5- Art. 190. 01 9 89 m09 .5U- 58U- 105f|. 329A. 8. 715U. 9. 709xVg. 108f. Page 51. Art. 191. 1. 'i'-h- ~- 4i- :?. 3^- S. lU- 5. 2HB- 5- ^■ 4. ^n- 10. 4^. 5. 2t|. ii. 3|f. 6. 2|§. i^. 5^V- Art. 193. i. lOoV//^. ;?. 244fJ. 3. COoJUI- 4- 1055^. 5. 42 H. €. 412H. 7. lOOH acres 8. 1104IU lb. 2644xV lb. 5693^ bu. and I3341H- Page 52. Art. 194. 1. |. 9. ^. 2. \. 10. 2^. 3. 1^. 11. A- A. |. -?-'• !!• 5. |. 13. m 6. |. 14. If. 7. H. i5. A- 5. i. Art. 195. 1. tV- ^- *• 193 ;-? 9 li -Z-J- 4. A- i4. i. |. ic. 2^15. Art. 197. 1. i 7. M- 2. h S- ^• 5. i. 9. If. ^. iV- io. f. 5. A. -Zi. ^• •5 25- >sS- Art. 198. 1. i,. 9. 3|f. - ^%. 10. 6f. h 11. 4^. A- ^~'- 5U- 5. A- i5. «>5^. 6. i. i4. 2^- StIj. 15. ^. H- ^<^- A- Page 53. Art. 200. 1. 4. 9. 9|. ^. 2f 10. 12*. 5. 9. ii. 9i 4. 17i. i^. 14^^ 5. 2i. iJ. 5A- 6. 3y«g. i4. 8it. 7. 7,^. 15. 99A. .•?. S^j. i6. ITOi. Art. 201. i. 2i|. 2. \m- 3. 30J. 4. 20. 5. in. 6. 36|f 7. 63x||7r- 5. 8f. 9. 10. 11. 12. 13. 14. 15. 150^. 39^. 198|Sf. e;ses48 59if. 87,«|. 9691. 16. 487i 1. f. • 5. 4. 5. 6. 9. 10. 11. 12. 13. 14- 15. 16. 17. 18. 19. 20. 21. •^2 23. 24. Art. 202. 31 91 1 85 lA- 6A. ^^21 125^-5. «74 3 4i6o- mi- 323 55- fi 4S1 5931^. 49tVit- 47f5 acres. 6fJ dollars. Page 54. $23HJ. gain. 660HH. n4|&i. ft 51 $34i. $12690H- ^0 i *>'i~5' Page 55. Art. 204. .1. n. 9. 4i. 2. f- 10. 1*. 3. 2i. 11. a. 4- If- 12. 9. 5. 5A. 13. 3i. 6. 2. 14. 14. rv 3j. 15. 18. 8. 14. Ai-t . 205 1. 25. 9. 344. 2. 7A- 10. 600. 3. 44. 11. m- 4. 30. 12. 297i 5. 124. 13. 265. 6. 26^. 14. 679. 7. 462| . 15. 7. 8. 94i. 16. 68. Alt . 207 . 1. 3|. 9. 3f. 2. 6i. 10. 6i. 3. 10. 11. 12. 4- 5^. 12. m. 6. 35. 13. 10. 6. 2. 14. 16. 7. 15. 15. 36. 8. 6. Art . 208. 1. 46f. 9. 63. Z. 49. 10. 427. s. 33|. 11. 45. 4- 77. 12. 84. 6. 49i. 13. 65^. 6. 2U 14. 168t»c 7. 152. 15. 6972. 8. 3|. 16. 448. Page 56. Art. 210. r 4 O 25 2. i. 10. |. 5. A. 11. If. 4. if. 12. -Nn- 6. A- 13. m- 6. y%. 14- f§J- 7. 2tV. -?5. A- *. hi 1(^- i- Art. 211. 1. $|. 4. A- 5. iA. 5. A- 5. IH. 6. ^. Page 57. 7. llf, and 8^. ANSWERS. 277 s. 9. 10. 11. VS. IS. u- 1. 2. S. 4- B. 6. 7. 8. 9. 10. 11. 12. 13. U. IS. 16. 17. 18. 19. SO. |90, and $60. $1. $i. $i. lifV- n gal- Art. 213. 1 17S9 ■'tboo- 1340713^. 10667478x^5. 80134846IH. 18786149735tV- 1160851. u- $87^. 561^ acres. / 159^t bbl. $6A. $35|U gain- Page 58. Art, 214. ^- lo: A. XT- A- 12- ^v- A- 13. If. A- Art. 215. 8 q 751 T55- ''• 'BS- ^I'oV i^. 24^i- xfr. i5. 34ii. 4i. i^. 20|i. 3f?. i5. li^V- 99|§. i6. 531 J. Page 59. Art. 217. 1. 39|. ii. 12^|. ^- 5. 6. 7. 8. 9. 10. 11. 12. 13. U. 15. 16. 17. 18. 19. 20. 19f 12. 3f. 31f. IS. 24. 51|. U. 67i. 888|. 15. 81. 20. 16. 40. 22|. 17. 300. 67/^. 18. 14f. 53^. i9. 36. 3|f. 20. 21. Art. 218. $72. 5 shares. 40 families. 24 bu. 5 da. $iH- 3089^, or 3090 sacks. Page 60. Art. 220. 2i. 10. f. U- ii. H- f i~'. 14. 16. U. T5 Art. 231. 1 6 1^- 76 T7K- 1196 809 10886 ItSTB- 19 8600 8Sd8- 351. 18 da. 13. 12i da. 1223/A bu. 8341 cords. 184U bu. 23. 279i miles. Page 61. Art. 223. 1. 2||. q 71 1 J. (35. 6'' ^4 5. 6'. 7. 5. 9. 10. 11. 12. 13. U. 15. 16. 17. IS. 19. 20. 21. 25. 26. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41- $14000. $6772^1. 3bu. 21b. $135000. 323^y5, sulphur, 215^Vff salt- • ""^ 1615fi char. $52500. $80 watch, and $35 chain. Page 62. $5|. 15 days. $19000. $67837i. 51 + ^ $2. $248,701. 3iWij J^iles. $5. $13200. 336 trees. 1001b. $67i. $6. $15 and $16. 62i years. S $1190, H $476 and R $544. 90^/5 years. Page 63. $35 and $40. 105. 6tbu. 22* da. J .$475, C $38| 40 ft. 94^ ft. Cow $30, colt $94. 2 p. M. 36 ft. 405.. H $216, C $324. 42. 43 U 45 46 4 48. 49. 50. 51. .52. 53. 54. 55. 56. 58. 59. 60. 61. 62. 63. 64. 65. 66. 68. 69. 70. 71. 30t years. 6| da. 5Ay da. 425 da. 67^ da. S $180, B $150. 150 ft. Page 64. 1 3 15- 176 rd. 14| min. Ben 7^, John \. 35 75. H,$33i;M,|55f, 11 .00000009. ' 10. 7807 5. lil- and B, %\\\\. 1.',. 54054054.005405- 11. 2001 6. ISH- 76. A, 135| da.; B, 0054. 12. 18081 Art. 251. 169^ da.; C, 188i da.; D, 16. 17. 103.587. 640.64. 13. u. ^'s 5 00 • 4926 1. 2 129.341. 848.1816. 1652.461772. 67* da. IS. 26.04002. 15. 1000267 3. 77. AandB,75ida. ; 19. 9019.029039. 16. Tf^TboojMf 4- 12638.517762. A and C, 78ff 20. 7.7. '17. 5A. 5. 2002.55141194. da.; A and D, 21. 870.01. 18. 13AV 6. 8688.0148502. 45^ da. ^ 479027004.00000- 19. ii'lU- i , 24018.46093544. 7S. A, B, and C, 99004. 20. 3005^'V- 534tda.; A,C, 23. 70000000000000- 21. 1600^^. Page 74. and D, 36JA da. ; B, C, and ;?4. .000000000007. 1100.0011. 22 1000000 1 8. 9. 13444.61870921. 10 000000 0- 1004219.317454. D, 38ff da. 20. .000003001. 23. 1234500 10. 57597.358230005 79. 20,'A- 'la. 26. .001003. 11. 7003122.0011890- 80 A, 77,1^^; 2~ .0100011. 24. 6540000 9997. B, 58,V^Vir; 28. .00000605. ToTnnSCTTTi' 12. 1627.325 A. C, 47|ftH; 29. 1890.0000000189- 25. 188900 5 9 13. 1017.84375 A. D, llSAV/j 0. 3TT50000- 14 1798.9425 bu. Page 60. Art. 338. 1. Art. 240. .25. .0106. 1. 2. Art. 246. .0625. .65. 15. 16. 17. 395.8125 yd. 376. 9262. 1. .026. 07 3. .00256. 3. .275. Art. 253. 3. i. 5. C. .lit i. .0016. .04. .00032. 5.7. 4. 5. 6. 9.01. 1.476. .193024. .00504. 4. 5. 6. 7. .09375. .1375. .52. .0525. 1. 2. 3. 4- .811357 -f. 2.23985+. 1.7912 1.9^7703945. 8. 2146.9003. 8. .46875 8. 83.0504. 710.00243. 9. 10. 56973.805. .1934675. 9. 10. .024. .0024. 5. 1384.4959234662 2. 9. 10. 11. 12. 13. u. 500.05. 45.046. 1001.0100. 1890.090. 850.05. 1000.10. Art. 239. 11. 12. IS. u. 15. 16. 17. 33254.81. .00001876. 10.007. .00097. 15.0015. .035700097. 219760.0801. 11. 12. 13. 14. 15. 16. 17. .9375. .015625. .0015625 .8875. .00024. .96875. .984375. 6. 1. 2. 3. 1 5569.3518126587- 42. Page 75. Art. 355. .412. .52977. .6863. 5.5264. 1.545648. 54.2294. 1. 11.107. IS. .046700004. 18. .028. 2. 3. 15.0014. .000726. 19. 2v. .00068001. 1101.10011. 19. 20. 21. 22. 23. u. 25. .308. .95. .94. .226. .034375. .76. .015625. 4- 5. 6. 4. 5. 6. /- 1106.0012. 1600.16. 10000000.000010. 3.65. 1. Page 72. Art. 244. -A- 8. 9. 10. 754.6005. 10000.0999. .3148. 213.889625. Page 70. 3. iTcrV- 11. .810. S. 25400.11. 4. a u u Page 73. 12. 135.25740. 9. 21.0015015. 5. u- Art. 249. Page 70. 10. .0000018018. 6. 661 1. 2 11. .500. 7_ 98 9 2. f Art. 257. 12. .00005. S. iVs- 3. II- 1. .546. ANSWERS. 279 S. .01968. 3. 1.26875. 4. 39.9024. 5. 23469.986904. 6. 4625520.705. 7. 1. <^. 9. 9. .625000. 10. .87000. 11. 7231.98325125. 12. 49. 13. .1. U. 275400116.25610 02754. lo. $20217.72. IG. $536.88. 17. $937.04. 18. $336.33. Page 77. Art. 260. 1. .25. 2. 305. 3. 250. 4. .5. 5. .05. 6. 50. 7. 500. 8. 4000. 5. 2000. i(?. .002. 11. .000025. 12. 183100. IS. 5.5875. i.#. 5252000. i5. 50. 1C>. .00007. 27. .001. 18. 4000. i». 1000000000. ^C. .000001. 21. 25. i'.^. 4000000000. 23. .15. .^.4. 60000000. ^5. 1. 10. 100. 100. 1000. .1. 1. 10. 100. -01. 2G. .1. .01. .0001. .00001. .001. 10000000. 100000. .00000001. .0001. 100000000. 27. .02. 200. .02. .0002. .00000002. 200000. 20000000. .0000002. .000000000002. 10000000000. Pajfe 78. 28. 64006464C4047.0- 4U6404000064. 2U. 250252502527.75- 2750000025. 30. 40000400000.044- 800440044. 31. 400044440440.00 0000140004. 32. 30030330000.003- 60306. 33. 150001650151.80- 1650000015. Page 80. Art. 366. 1. 1991.1198244. 2. .61032010. ■ 3. .1625. / «oi "f- TOOoO- 5. .0038462. C. 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. i.5. 12711.8755*. IG. 10.625. 17. 1011.6. IS. 6.875 da. 19. $274. .58. 20. $24.9331-f 8. 9. 10. 11. 12. Page 84. Art. 387. ooo^-. 11100^. $2.41. $10.44. $214.68. $18. $510. $98.76. 100980. 375i^. 26530. 157320. Art. 389. /. $88.84. Page 85. 2. $144.89. 3. $4337.77. .'i. $221.53. 5. $378.07. 6. $877.29. 7. 384.51. cS'. 1088* 0. .'/. 4073^ «. 10. 3065* 5. 11. 8968*. 12. 2094.15. Art. 291. 1. $8179.88. Page 86. 2. $4859.76. 3. S5669.60. 4. $29.52. o. $839.26. .S33.73. i;_ Art. 393. 1. $2284.35. $1315.63. $7.13. 950 bu. $344.73 gain. $91.10. Page 88. Art. 394. 7 26 and 39. 8 16^\ min. j 3 o'clock. 9 Midnight. 10 B. 12, A. C. 78. 11 B. 840 and 450. 12 1900. Page 93. Art. 304. 1. $143. $384. 3. $65,875. 4. $113.75. 5. $291.83i. 6. $133.31^^. 7. $247.60. 8. $63.05. 9. $1.2U. 10. $3.70. 11. $67.71^. 12. $10.50. 13. $841.15. u. $1156.97. Page 94. Art. 306. 1. $75. 2 $358.75. 3. $22.13. 4. $53.50. 5. $1125. 6. $612. 7. $3281.fi5. 8. $382. 9. $24.06. IG. $45.56. 11. $34. 12. $1567.50. 13. $187.50. 14. $281.25. 15. $125.25 IG. $370. 17. $2750. IS. $46.25 19. $50.25 20. $86.25. 21. $156.25. aftei 26. O, 280 Art. 307. 2. $2507.96. Page 95. 2. $3324.46. 3. $4321.26. 4. $5282.20. 5. $8096.48. 0. $3364.72. Page 96. Art. 309. 1. 231b. n 4627.5 yd. .'. 61b. ■'{. 763.2 yd. 5. 876.5 doz. 6. 371 lb. 7_ 81.5 yd. S. 115.2 acres. 9. 689 yd. JO. 123 lb. 11. $40.96. Page 97. Art. 311. 1. $21.91. r9 $92.68. .?. $70.07. 4. $28.35. 5. $148.28. 6. $31.68. 7. $25.31. 8. $89.10. 9. $63.63. 10. $35.65. Art. 313. 1. $24.75. 2. $14:66. 3. $158.76. 4. $123.18. 5. $53. 6. $38.75. 7. $19.80. 8. $100.32. 0. $140.81. 10. $34.78. 11. $68296.35. 12. $2819.31. Page 98. Art. 315. 1. $4.02. ANSWERS. (0 $4.71. S. $66.21. 3. $78.70. 4. $13.84. Page 105. 5. $2014.46. Art. 328. 6. $2016.46. 1. $943.54. 7_ $941.63. o $57269.94. 8. $21.06. 0. $2.77. Page 106. 10. $19.51. Art. 329. 11. $296.24. 1. $115.68. 12. 13. $15.77. $56.72. 2. $560.50. $1528.75. 14. $66162.39. 4. $190.33. Page 99. Art. 317. J. $272.35. $429.14. 1. $28.56. Page 107. a $28.08. 7. $524.03. o. $52.83. s. $1718.01. 4. $39.29. 9. $168.68. 5. 6. ' $24.66. $27.15. 10. $322.29. S. $55.67. $58.33. Page 112. 0. $174.78. Art. 355. 10. $132.36. 1. 450 min. 51 sec. 11. $121.68. ■) 23 hr. 5 min. 29 12. $242.79. sec. 13. $333.31. 3. 7 da. 1 hr. 30 u. $109.57. min. 51 sec. 15. $198.91. 4- 24 yr. Imo. 5 da. 16. $28.16. 6 min. 17. $11.57. 5. 63321 hr. IS. $12.06. 0. Jan."l4, 1889. 19. $60.89. 7. 3405 da. 20. $30.14. 8. 1169 da. 21. $15.62. 9. 11 mo. 9 da. .->-> $1941.92. Page 101. 10. 1 yr. 10 mo. 19 da. 19^ hr. Art. 326. Page 113. 1. $9.03. 11. No difference. Page 102. 12. 939613.7 sec. 2. $49954.08. Art. 358. 3. $191.52. 1. 35° 54'. 2 24' 16' 46". Page 103. 3. 3S. r 49'4r. 4- $1812.31. 4. 1296000. Art. 327. 5. 4907'. 1. $46731.53. G. 205737. 7. 136° 2'. Page 104. 8. 21600'. 2 $967.31. 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. S. 74° 58'. Page 116. 4. 36° 52' 20" N. 5. 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 2U p. m_ 3 min. 48 sec. past 7 a. m. Page 117. Art. 371. 1. £256 4 s. -'. 248 s. 3 d. 3. £54 6 s. 10 d. 4. £195 4 s. 4 d. t far. Alt. 373. 1. 6480 d. 2. 956 far. 3. 38853 d. 4. 39450 far. 5. 13206 far. Page 118. Art. 375. 1. $350.26. 2. 4525.80. 3. $63544.40. ANSWEKS. 281 4. $15.03. r,. 154.29. Art. 3 77. 1. £38 3 d. 2.4 far. 2. £63 7 8. 10 d. 3. £513 14 s. 3 d. 3 far. 4. £751 14 s. 3 d. r,. £32621 2 s. 8 d. 1 far. Page 124. Art. 393. 1. 7653 pwt. 2. 155948 gr. 3. 4 lb. 11 gr. If. 5 lb. 3 oz. 2 pwt. 9gr. 5. 432 gr. 6'. -i^ pwt. 9. 1 OZ. 13 pwt. 18 gr- it;. 12 pwt. 12 gr. 11. AVlb. 12. mib. 13. 7 oz. 14 pwt. 4.8 gr. U. 18 pwt. 2.4 gr. 15. .297616 + lb. 16. .875 oz. 17. 472 lb. 1 oz. 12 pwt. 8 gr. Page 125. IS. 211 lb. 11 oz. 19 pwt. 21 gr. 19. 2 pwt. 20. 4 gr. 20. .0067+. 21. 81b. 2 oz. 13 pwt. 6gr. 22. 18 lb. 9 oz. 14 pwt. 2 gr. 23. $8032.50. 24. $1924.39. 25. 5 oz. 2 pwt. 17 gr- 26. 1 oz. 13 pwt. 8 gr. 27. $11655. 28. 73 lb. 2 oz. 8 pwt. 19 gr. 29. $154.22. 30. $360.67 gain. Page 127. Art. 395. 1. 34669 lb. ..'. 15 T. 12 cwt. 75 lb. .i. 12 cwt. 50 lb. 4. 56 lb. 4 oz. 5. 7 cwt. 68 lb. 6 .4 oz. 6. 12 cwt 50 lb. "- 7089 T 8. llS-cwt. 9. .24125 cwt. 10. .99996875 T. Page 128. n. 30 T. 1 cwt. 94 lb. 11 oz. 12. $72.81. 13. 2 T. 5 cwt. 84 lb. Art. 397. 1. 17 lb. 9 oz. 5 dr. 1 sc. 2. 5896 dr. .'f. 11 oz. 3 dr. 2 sc. .8gr. 5. 63 sc. 6. 6 lb. 9 oz. 6 dr. 11.5 gr. ;. 1 lb. 2 oz. 4 dr. 1 sc. 4 gr. S. 7 lb. 3 oz. 4 dr. 12 gr. 9. 6 lb. 2 oz. 6 dr. 1 sc. 8 gr. 10. 15 lb. 10 oz. 5 dr. 2 sc. 8 gr. ;/. 11 oz. 2 sc. 4^i gr- Page 129. Art. 398. 1. $1477.27. 2. $2255.25. 3. 61b. lOoz. 15gr. 4. 18 lb. 6.43? oz. $254.49. $5.04. Art. 400. 1. 928 pt. 2. 599 pt. 3. 180bu. 3qt.2pt. 4. 7 qt. If pt. Page 130. .7. 5 bu. 1 pk. 1 qt. 1 pt. ',. 83 bu. 1 pk. 3qt. IfPt. ;. $184.12. Art. 403. 1. 5932 gi. ,?. 31 bbl. 7 gal. 1 pt. 3gi. V?. 651.168 gi. 4. 6 gal. 2 qt. 1.16 g'- ,J. $71.75. 6. 7 gal. 2 qt. 1 pt. V^ gi- ;. 13 gal. 1 pt. 1 gi. 5. $19.16. '.). $72.58, gain. Page 131. Art. 403. 1. 1579A pt. 2. 134.5 + pt. gain. 3. $.52 gain. 4. $5.25, gain. 5. $7.99 less. Page 132. Art. 409. J. 127002 in. 2. 39 mi. 155 rd. 4 3'd. 3 in. 3. T^tyY"?- /,. 213 rd. 1yd. 2 ft. 6 in. o. %\ rd. 6. 173 rd. 2 yd. 1ft. 812 in. 7. .892+. S. 123 mi. 162 rd. 3 yd. 1 ft. 4 in. 9. 1593 mi. 312 rd. 2 yd. 1 ft. 8 in. 10. 484 mi. 53 rd, 1 yd. 2 ft. 6 in. Page 134. Art. 424. 1. 35676648 sq. in. 2. 1344984421 sq.ft. 3. 112 A. 40sq. rd. 261 sq. ft. 51.84 sq. in. Page 135. 4. 110 sq. rd. -5- 4 4 aOS440 g sq. mi. 6. .9382+ A. 7. 101 sq. rd. 2 sq. yd. 21.6 sq. in. 8. 4 A. 83 sq. rd. 6 sq. yd. 64| sq. in. 9. 31 i squares. 10. 63 yd. 11. 90 ft. 12. 230 ft. 13. 25| A. U. $107156.25. ir,. $3277.97. 16. 357i rd. 17. 414? ft. 18. 130i A. 19. Not any. 20. 2 sq. rd. 21. 2%%. 22. 4|| A. 23. 58^ rd. 24. 12. 25. 26A.llsq. rd. 4 sq. yd. 5 sq. ft. 36 sq. in. 26. $60.06. 27. $10.56. 28. 9 A. 110 sq. yd. 3 sq. ft. 54 sq. in. 29. 38 A. 59 sq. rd. 12 sq. yd. 5 sq. ft. 112 sq. in. 30. 640 rd. 31. 320 rd. Page 136. 32. 60 yd. 33. $40.30. 34. $58594.44. 282 ANSWERS. S5. $188404. 36. $15.63. 37. 43 rolls. 3S. 128J- yd., 130 yd., and $327. 25. 39. 486U|. 40. 12 sq. ft. 41. $9.46. 4^. 28512. 43. 52.177+ ft. and 104.354+ ft. 44. 147840. 45. $74.36. 46. $204.73. Page 137. Art. 430. 1. 1. 2. 9. 3. 16. ^. 25. 5, 81. 6\ 100. 7. 9801. -s'. 65536. Page 139. Alt. 439. 1. 14. S. 15. ^. 12. 4. 24. 5. 35. 6-. 75. 7. 206. <9.. 11.2. .0. 7.09. 10. 21.954. i/. 5.07. Page 140 12. 10.3156. 13. f. 14. f. l.'>. .968 H». 10. .85+. 17. 5510.8 +. 18. 68548.66+. Art. 446. 1. 10. .<> 49.777. 5. 6. 3. 9. 10. 11. i:. 13. U. 15. Page 141. 108.25+. 60.81 +. 56.796; 113.592. 208.71 + ft. 1866.76+ ft. 660 ft. 124.03 + ft. 77.88 + ft. 72.56 + ft. 452 id., 8 10.92 + in. 226.42 + rd. 720 rd. 208.80 rd. ft. Page 142. Alt. 448. 1. 211618.48 in. S. 1 mi. 68 elf. 1 rd. 161. 3rd. 181.59.4in. 7624 1. 7ch.l41. 2.27f iu. 8927,\ ft. 1924* rd. S. 16 ch. 91 1. 5.22 in. .'/. 2738 steps, llf in. rem. Page 145. Art. 459. 1. 10 cu. yd. 1533 cu. in. 2. 6178581. 3. 8cu. yd.972cu. in. 4. 7200 cu. ft. 5. 508i cu. yd. 247 j\ pch. Trr/sTff cu. yd. cV. .2615 + cu. yd. 9. ^3^ cu. ft. 10. 14 cu. ft., 302.4 cu. in. 11. $2320.76. 12. $2656.78. 13. 650471 bricks. 14. 47i cd. 15. 39^ j cd. 16. 26 ft. 9/5 in. 17. $227.11. 18. 54468f lb. 19. 20. ■21. S. 9. 10. 11. 12. 13. 14. 15. 16. 1:. IS. 19. 4531i lb. 46656. 10 cu. yd. 20 cu. ft. 1339icu. in. 659.709+ pch.; 254013 bricks; $1169.41. Page 146. 198cu. ft. 27.52+ bu. Page 150. Art. 474. 12. 25. 48. 404. 12.898 + 49.21+. 36.1 +. .8+. .92+. 2.90 +. 15.177+. .160+. .97+. Page 151. 10 ft. 1 + in. 47 ft. 5 + in. 5ft. 4 + in.wide; 10 ft. 9 + in. high, and 37 ft. 7 + in. long. 14 ft. 2 + in. 7 ft. 7 + in. deep, and 15 ft. 2 + in. square. Page 153. Art. 492. 78 sq. ft. 85.498 sq. yd. 32| A. 360 A. 259.182 ft. 472. 68 + ft. 12.27 + A. 88. 6 + ft. 29.7+ in square. lu. $37.70. 11. 26649.9 gal. /.-. 45 sq. ft. Page 154. 13. 26| sq. yd. 14- 201.0624 sq. in. 15. 33.5104 cu. ft. 16. 1260. 9 +. 1:. 4564.2 + mi. IS. 117.6264 gal. 19. 22.3074 gal. Page 157. Art. 502. 1. £359, 12 s. 1.8. + far. -'. .Tan. 3, 1870. 3. 420 sq. ft. 4. 2.295+ A. 5. 122.18 + pch. 6. 17280 shingles. 7. 27^\ rd. 33. 1.10, ami. per. cent. 1.75, ami. per. cent. 2.10, amt. per. cent. 1.161, amt. per. cent. 1.87^, amt. per. cent. Art. 533. 1.05 per cent. 1.09,',;?. i.40r;. Art. 535. .85, difference per cent. .62^, difference per cent. .99^, difference per cent. .96 J, difference per cent. .30, difference per cent. Art. 536. 712';. 68|;f. .60, difference per cent. Page 167. Art. 538. 1650. 1695. 462. 1180. 277.2. 2580. 840. 8. 637. 9. 450. 10. Art. 539. $14512.50. $456. $11200. 816. 3537. $886.50. Art. 541. 12. 150. 945. Page 168. 4. 612. 5. 1200. 6. 500. 7. 5. 8. 567 ft. Art. 543. 1. $2843.75. 2. 581 i A. 3. $2053.13. 4. $632.50. Art. 544. 1. 600. 2. 400. ■'■ 300. Page 169. 4. 100. 5. $4000. Art. 545. $1000. $4.51. $7950. $480. $106. $3200. $30000. $10000. 500 pupils. Invest, in farm. $280. Page 170. Art. 648. AAS4r 0*25 ,. 68 lb. 284 ANSWERS. 5. 80^. 4. 123333333. 33i. 5 23H. 425, and 33J. 6. 77 yr. 7. $108. 5. H.|200,3I.$170 C. $15. 9. $1081.25. 10. 25 lb. warp, and 71i lb. rags. 11. $160, and $224. 12. $113.78. IS. \\% gain. Page 171. 1^. $-i{387.50. 15. $1045.45 +. IQ. 216:;+. 11. 21^Y<, 21M!f. ""J ij?. 100 bead. 19. &^^. ^0. 25^. 21. $6344.40. S2. ISi'Tr. 23. $1000 loss. 24. 400. f5. 3000. 26. 77|i|^. i-?. 10 yd. £8. $4856.25. 29. $60250. Page 172. 50. Grazing, 504 A. ; grain, 420 A ; timber, 936 A. 51. $192 C. 32. A $93840. and B $69360 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. j:. $81.20. $101.50, $182.70. 38. 16|?. 39. $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. 1. $9. 2. $48. J. $750. Page 174. $50. $32. $320. $225. $2100. $700. .\Tt. 558. 1. $592.50 gain. 2. $3997.50 loss. J. $677.25 loss. 4. $9.47 gain. 5. $184.92 gain. 6. $6 gain. :. $10 16 lo-ss. S. $22.97. ;/. $133.59. Page 175. Art. 560. 1. $100. „'. $3500. J. $10000. 4. $4400. o. $40. C. $300. ?. $900. 5. $1050. 5. $1. Art. 561. ;. $57.. 50. .?. $500 and $625. J. $600. .^. $700. 5. $7085.71. tJ. 200 A. r. $2750. .V. $240. [>. $5000. Page 176. Art. 563. 1. m. 2. 5^. 3. m^. 4. 20^. 5. 50;?. <;. 33i:?. ^ 7. 150^. 9. 20f^. Art. 564. 1. 150.?. •? 25^. Oats. \%i%. 33i'?. 9, $386.25. 64. $4000. J. $180. 2. $75. 3. $250. 4. $275. 5. $277.38. 21. 253 bbl., and Art. 651. 65. 100500 corn, and $9.68. 1. 1;;, $62.50. 75375 wheat. 2. $3550. 66. C, $27.18J; H, Page 196. 3. $537.50. $326.25; and /■O $31403.75. 4. 4^ mills, $70. S, $1.81+. 6. 1375.90. 23. 26250 corn, and ''. $35.35. 19200 barley. Page 184. Page 193. 24. 5^. Page 203. Art. 575. Art. 608, 25. $5044.29. 0. 3^ mills, $888 1. 2. $24.48. $.^)1.30. 7. $5000. J. $24500. 26. Remitted $566^, com. $113Jt, 83.13, S. T. $27553.77. 3. 4- 5. 6. $660. $5670. $19. 13 gain. 200 yd. ■J. $282. 4. $8672. 5. 9000 bu. 0". 50 bales. 28. 29. and rate 16i^L Loss of $280.26. $26023.50. 21071.52 bu. *8. 9. 2 J mills; 24 mills. 4i mills; $110.63 $54.05. ''• $1515.64. 30. Barley, 14333^ 10. $462 33. S. A, $72.50. Art. 610. bu. ; hops, 560- 11. $.005ff,and$85.- Paj;*' 185. J. $12140. -'. 30000 lb. 67.13 lb.; and com. $508.40. IL 17. 1.25669 /. Art. .'>~~. $30. $30. 3. 145 doz. 1. 2. Page 194. Page 199. Page 206. 0. $12571.43. 4. $400.80. Art. 6.14. .Art. 6 7 7. 4. $9. 5. 1500 A., and 1. $22.80. 1. $5000. 5. $60. $202.50. r> $300.38. •:■> 470. 6. 25f.'. r,. 17501b. 3. $8000. 3. $1600, $2400, 7. 2ff/. 4- $72. $2000. Art. 611. .'). $234. 4- $1576.50. Page 18«. J. $07.50. 6. 40 gal. 5. $523.75. S. $1000. J. 225 bbl. 1 ''• $144. 6. $4450. 286 Page 207. 22. $2161.54. ERS. Page 218. 50. $.53. 7. $27411.17. 25. $4981.67. Art. 715. Art. 716. 8. $114.54.55. 2i. $44.75. 1. $4.38. 1. $79.20. 9. $14791.67. 25. $780. 2. $5.25. 9 $29.96. 10. $16853.56. 26. $100.83. 3. $3.71. 3. $80.84. 11. ir<- 27. $96.75. 4. $17.44. 4- $104.45. 12. $1967.96. 28. $16975. 5. $9.24. 13. $60. Page 213. 6. $1.13. Page 220. U. $6093.40. 15. $3200. le. $18242.66. 29. SO. $74.94. $2707.18. S. 0. $5.83. $3.33. $1.76. 5. $73.41. Art. 718. n. H, $26522.73; SI. $4310.74. 1". $5.70. 1. $5.18. M. $43099.43; 32. $2040.15. 11. $2.34. 2 $11.53. A, $23207.39; 33. $2766.55. $1969.62. $3519.75. 12. $8.02. 3. $3.62. Phoenix, $265- 34. 13. $6.32. 4. $2.80. 22 .73 and Prov- 35. 14- $3.00. 5. $5.19. ident, $26522.73 36. $2837.92. 15. $24.58. 6. $16.81. IS. $5000. Page 214. 16. $1.15. 7_ $3.95. 19. $47500. and $3- Art. 705. 17. $4.37. 8. $5.10. 9375. ' 1. $343.75. 18. $11 65. 9. $16.95. 1 Page 208. 20. G., $630; H., $150; and M., 2. 3. 4. $4099.71. $337.41. $857.01. $27500. $826.23. 19. 20. $1.34. $10.96. Page 219. 10. 1. $4. .50. Art. 719. $191.26. $337.50. 6. 21. $6.61. 2 $90.88. $142.50 gain. 22 $19.71. .n\% Page 215. 23. $25.80. Page 221. Art. 707. 24. $10.60. 3. .$45.89. Page 211. 1. $445.94. 25. $25.38. Art. :03. 2. $10344.83. 26. $2.39. Page 222. 1. $258.30. 3. $600. 27. $4.86. .\rt. 729. 2. $47.67. 3. $75.60. 4. 5. $1000. ' $291.85. 28. 29. $8.22. $6:i8. 1. .9 $448.70. $1546.70. $366.60. $2422.30. $12726.80. 4. $364.50. 6. $739.13. 30. $2.82. 5. $457.10. Page 212. 7. 1. $1954.63. Art. 709. 31. 32. S3. $7.79. $1.79. $8.78. 3. 4. 5. 6. $131.39. 7. $675.13. 8. $570. 3. 7?. 6 3 yr. 10 mo. 12 44. $.70. Page 226. n. $1662.50. da. 45. $5.50. .\rt. 737. IS. $132.86. 3. April 25. 1881. 40. .S16.53. 1. *27>t>.75. 19. $1798.30. 4. 11 mo. 47. $11.66 2 $1020.30. 20. $717.27. 5. Sep. 22, 1889. 4S. $5.70. S. H^. 21. $791.78. G. 12 yr. 6 mo. 40. $29.17 4. S%. ANSWERS. 28? Art. 738. Page 230. 9. Jan. 28, 1889; Page 248. 1. $889.58. Art. 744. Term of Dis., 5. Nov. 15, 1888. 2. $1773.73. Page 227. 1. $540. $99.75, and $38.- 27 days; Pro., $384.71. 10. Feb. 39, 1888; 6. 7. 3. Dec. 29, 1887. Jan. 14, 1889. Jan. 11, 1888. S. $1618.33. <0. 19 da., $799.09. 4. $6386.77. 3. $1654.61. 11. Aug. 6, 1888; 66 Page 249. 5. $879.71. * days; $664.99. .Art. 803. Page 231. 12. Mar. 3, 1889; 178 1. Nov. 12, 1888. Art. 730. 1. 4 yr. 2 mo. 2. 6.716J?. 4- 5. Interest, $12.29. $.55, better to days; $3383.44. 13. Dr., $125.39. 3. 4. Dec. 21, 1888. Jan. 10, 1889. Feb. 13, 1889. 3. X3.23:?. pay cash. 5. June 13, 1888. A. $920.08. 6. No difference. Page 238. 6. May 9, 1889. 5. $20.72. 7. Loss $11.82. 14. $1900.41 to their 7. May 14, 1888. 6. $8681.12. 8. $1533.15. credit. 8. Mar. 7, 1889. 7. m. 8. 3 yr. 7 mo. 24 0. 10. $7.48. A'i'o loss. 15. $3865.30. Page 255. da. 11. $7481.30. Art. 780. Art. 806. 9. $10505.94. 12. Cashoffer,$8.37. 1. $330. 1. Nov. 28, 1886. 10. 7i yr. 13. ^ll'/o profit. 2. $1350. .? Feb. 11, 1887, 11. $6856.53. u. Guin .$1303. 3. $933.87. $300. 12. ^\\i. 15. $44.85,and8|i';. 4. $3461.96. 3. $313.18. 13. $4644.61. IC. $683.33. 5. $3150. 4. 601.73. 17. $4000. ';. $691.13. 5. Oct. 14, 1888. Page 228. IS. $340.13. 7. $175.08. Page 256. U. May 18, '89. If: $39.79. G. $100. Nov. 35, 15. Gain $875. Page 240. 1890. 16. $2728.82. 17. Grace, $7678.96; Page 232. Alt. 790. /. $199.37. 7. S. Sep. 21, 1889. Dec. 33, 1886. Mabel. $7031.- 85 ; Flora, $4333.02. 18. $299.20. 20. 21. 23. $37.73. $1756.27. $372.58. $9736.94. 2. $533.68. 3. $1100.85. 4- $4.07 9. 10. Sept. 4, 1887. Jan. 24, 1888. $348.88. 19. m',L 20. $1373.81. 24. $10855.79. Page 241. 11. Page 257. July 24, 1887, 21. $4128.37. 22. 725 M. Page 236. J. $100.53. '-■. $1550.07. 12. $431. $300, Dec. 27, 23. $2660. Art. 778. 7. $1.73. 1887, $300.43. 24. 50 years. 1. Bk. Dis., $9.38; S. $1890.50. 13. Feb. 16, 1889. 25. Herbert, $5938.- Pro., $740.63. 14. $100, May 17, '89 66; Theodore, 2 Bk. Dis., $1.67; Page 242. $98.93. $4847.73. Pro., $284.83. Art. 793. 3. Bk. Dis.. $23. 08; 1. $203.98. Page 258. Page 229. Pro., $1303.93. ..'. $640.87. 15. $400, Jan. 4, '88; 2Q. $536.95, better to ,;. $39.18. $435.23. invest in land. 27. $33884.38. Page 237. 4. $563.53. IG. Dec. 7, 1888. Page 261. 28. $1508.75. 4- Bk. Dis., $4.47. Page 247. Art. 820. 29. $13006.80. 5. Proc, $988.37. Art. 801. 1. 4. SO. Chas., $5364.99; G. Gain, $11.35. 1. Oct. 16, 1888. 2. 28, John, $4590.03; 7. Bk. Dis., $13.71; 2. Oct. 13, 1888. 3. 7. Walter, $3895.- Pro., $1251.71. ,;. Sep. 7, 1888. 4. 81. 34. S. Proc, $1749.47. 4. Mar. 30, 1889. 5. $288. 288 ANSWERS. 6. $4.30+. 7. 733i ft. S. 478 bu. 9. $2812.50. 10. 2 yr. 9 mo. 22i days. Page 262. Art. 823. J. 26JA. 350 rd. $567. $1290. $384.75. 124^ yd. 74^. S. 93 da. Page 265. Art. 841. 1. A.. $26250; and B., $15750. £. $6589.41, gain; Hadley,$2758.- 36; and Hunt, $3831.05. S. Whole capital, $58800; D's gain, $1200. 4. A, $9409.-52; B, $7939.29: and C, $7351.19. 5. N.Insol.,$4543.- 75. N. In vest., $2321.25. 6. Gained, $12090; A'sP.W.,$504- 5;B'sP.W.,$37- 95; Solv.,$8840. 7. Harrison, $7000; Morton, $1000. Page 267. .Vrt. 842. 7. A, $270.92; B, $346.15; C, $276.93. i". A, $596.53; B, $559 25 ; C. $994.22. 3. Martin's gain, $1737.93; Eaton's gain, $1862.07; Martin's P. W., $3737.93; Eaton's P. W., $6362.07. 4. A, $17.40; B, $29.70; andC, $48. 5. A's investment, $8491.30; B's in- vestment, $6065.- 22; C's invest- ment. $4043.48. 6. A, $4548.39; B, $2951.61. I 7. A. $36;B, $32;C, $20; and D, $4. S. A, $731.57; B, $1483.93; and C, $784.50. 9. Net gain, $5200; Olsen's share of net gain, $1640.- 27; Thompson's share of net gain, $3559.73. Page 268. 10. Simmons, $170- 20.59, and Saw- yer, $16329.41. 11. Drews' gain, $8058.14; Allen's gain, $6837.21 ; Bracketl's gain, $6104.65. 12. B, $838.10; and A, $1561.90. 13. Martin, $3126.- 53; Gould, $3104.46; and Cole, $1269.01. Page 269. 2. Hopkins's gain, $4873.27 ; Haw- ley's gain, $4526.- 73. 3. Charles, 12f: .John, 8j^; Walter, Page 270. 4. Net resources, $23564.25 ; net solvency, $23,564.- 25; net gain, $1064.25. 5. Investm'tof each, $9866? : Gray's gain. $2161.69 ; Snyder's gain, $1894.93; Dillon's gain, $2843.38 ; Dillon's P. W., $16210.04; Sny- der's P. W., $72- 61.60 ; Gray's P. W., $11028.35. 6-. Phelps, $8000 Rogers, $18300 Wilder, $30900, 7. Smith, $16170.43 Jones, $11990.32 Brown, $9339.25. Smith's P. W., $2- 8870.43 ; Jones's P.W., $16990.32; Brown's P. W., $16239.25. i S. Burke, $17533.33; Brace, $17833.33; Baldwin, $17633.- 33. 9. Loss, $22747.09. Briggs' loss, $5- 907.62; Parson's • loss, $16839.47; net insolvency, $2187.09; Briggs' P. W., $127.38; Parson's insolv., $2314.47. 1". 11. I l-'. 13. 14- Page 271. Bradley's gain, $7517.61 ;Maben'8 gain. $8362.39 ; Bradley's P. W., $23417.61 ; Ma- ben's P. W., $26- 362.39. Wilkins' share, $1000; Ames' share, $3000. $1744.5.24. A's P. W.$30165.84;B's P. W. $22417.09; C's P. W. $22417.- 08. A's share of gain, $6033.17 -, B's share of gain, $4483.42 ; C's share of gain, $4483.41. A's capital, $386- 5.80; B's capi- tal, $2577.20; in- solvency, $5557 ; A's insolvency, $3334.20; B's in- solvency, $2222.- 80. Page 272. Net resources, $94735; net gain, $44335; Clay's share of gain, $17932.89; Hard's share of gain, $2-i- 380.43; Dunn's share of gain, $2- 021.68. Clay's P. W. at closing, $3- 4132.89 ; Hard's P. W. at clos- ing, $53080.43 ; Dunn's P. W. at closing, $7521.68. ] D 000 878 084 3 ^^T, >5Jr,-