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SERIES OF NATIONAL SCHOOL BOO viz. :— First, Second, Third, Fourth, and Fifth Be of Lessons, an English Grammar, and an Arithmetic^ Walkingame's Arithmetic; English 'Reader; Kirkhi Lennie's, and Murray's Grammars i Canada, Mavo| Carpenter's, Cobb's, and Webster's Spelling Books, < — ALSO,— A large assortment of Works on Rhetoric, Logic, Philj phy. Chemistry, Botany, Physiology, Phrenology, i Mechanics. B., McP., & Co. have always on hand a well-'^elected 8h of the Classics, and other School Books generally used inj Province, which are supplied to Merchants and TeacU on the best Terms. A LIBERAL Discount for Cash Purchases. BOOK & JOB PRINTffli TASTEFULLY EXECUTED. BOOKS, STATIONERY, &c. &c. Binding executed in all its Branches on liberal Tei Ingliih, FreDchi and Amerieaa Paper Hangiagi. les of National Sohool Books. • Co' tllATISE ON AEITHMETIC, IN THEORY AND PRACTICE, FOR THE USE OF SCHOOLS. it ITHORISED BY THE COUNCIL OF PUBLIC INSTRUC- TION FOR UPPER CANADA. TORONTO: PUBUSHED BY BREWER, MePHAIL & C:, 46, Eno Stbbbt Eait. 1864. 6) A 163. T / !■ s i -:JO^.}i ]';'•? 'h.; :iv^j- Ml : „ . , f ^ f ; ., ; C i /- u* ii V..' i ... I;. I, :/.■■:', 0* i-'' :.Vi •>.■» >'-» . (»: ■« I,, -i PREFACE ;«A In the present edition a vast number of exercises ha?e been added, that no rule, however triflmg, might be left without so many illustrations as should serve to make it sufficiently familiar to the pupil. And when it was feared that the application of any rule to a particular class of casea might not at once suggest itself, some question calculated to remove, or diminish the difficulty has been introduced among the examplelb. A considerable spaoe is devoted to the "nature of num bers,"' and "the principles of notation and numeration;' for the teacher may rest assured, that the facility, and eveiw the success, with which subsequent parts of his instructioi> will be conveyed ito the mind of the learner, depends, in a great degree, upon an adequate acquaintance with them. Hence, to proceed without securing a perfect and practical knowledge of this part of the subject, is to retard, rather than to accelerate improvement, o A-jlf: •■ i- i.-sjytja ...' n nmui The pupil, from the very oommenceipent, must be made perfedly familiar with the terms and signs which are intro- duced. Of the great utility of technical language (aoco^ rately understood) it is almost superfluous to say. anything here.: we cannot, holv^everj forbear,, upon this oocasion, re- calling to remembrance what ia. so admirably and so effec- tively inculcated ' an the ^Easy Lessons on Reasoning." ''Even in. the common inechanioalarts,! sc^mething of a technical language is found needful frar thoM^ who are lean> PREFACI. ing or ezeroiaing them. It would be a yery great in- oonvenienoe, OTeii to a common carpenter, not to have a precise, well understood name for each of the several opera* tions he performs, such as chiselling, sawing, planing, &o., and for the several tools [or instruments] he works with. And if we had not such words as addition, subtraction, multiplication, division, &c., employed in an exactly defined sense, and also fixed rules for conducting these and other arithmetical processes, it would be a tedious and uncertain work to go through even such simple calculations as a child very soon learns to perform with perfect ease. And after all there would be a fresh difficulty in making other per- sons understand clearly the correctness of the calculations made. "You are to observe, however, that technical language and rules, if you would make them really useful, must be not only distinctly understood^ but also learned and remembered as familiarly as the alphabet, and employed constantly^ and with scrupulous exactness ; otherwise, technical language will prove an encumbrance instead of an advantage, just as a suit of clothes would be if, instead of putting them on and Vfearing them, you were to carry them about in your hand." Pa,ge 11. ^.. ,■. -I.. ,,. .'■. . i'. !,(•/» (i.'i -■,.,t>-; ■■'!'■ What is said of technical language is, at least, equally true of the signs and characters by which we still further facilitate the conveyance of our ideas on such matters as form the subject of the present work. It is much more nmple to put down a character which expresses a process, than to write the name, or description of the latter, in full. Besides, in glancing over a mathematical investigation, the mind is able, with greater ease, to connect, and understand its dif- ferent portions when they are briefly expressed by familiar signs, than when they are indicated by words which have nothing particularly calculated to aitch the eye^ and which cannot even be clearly understood without considerable attenti(ni. But it must be borne in mind, that, while such m treatise as the present, will seem eaaj and intelligible PREFACE fii very great in- not to have a le several opera- g» planing, &o., ho works with. >n, subtraction, •xactlj defined bese and other and uncertain ;ions as a child ^' And after »ng other per- le calouhitions leal language il) must be not remembered as wstantly^ and language will ^0} just as a them on and your hand." equally true er facilitate u form the mple to put aJi to write in ie mind is md its dif. >y familiar 'hich have «id which nsiderable rhile such itelligibie Inough if the signs, which it contains in almost evorj pagei as familiar as they should be, it must neceMarily appear ore or less obscure to those who have not been habituated the use of them. They arc, however, so few and so simplta, [hat there is no excuse for their not being perfectly under* toed — particularly by the teacher of arithmetic. j Should peculiar circumstances render a different arrange- »nt of the rules preferable, or make the omissicm of aiqf f them, for the present at least, advisable, the jndioioiif aster vrill never be at lost: aow to act — there maif be nstances in which the shortness of the tine, or tiie limtt^i ntelligence of the pupil, will render it necessary to confine is instruction to the more important branches. The teacher hould, if possible, make it an inviolable rule to receive answer unless accompanied by its explanation, and its eason. The references which have been subjoined to the ifferent questions, and which indicate the paragraphs where answers are chiefly to be obtained, and also those refer- ences which are scattered through the work, will, be found of considerable assistance ; for, as the most intelligent pupil iwill occasionally forget something he has learned, he may ot at once see that a certain principle is applicable to a articular case, nor even remember where he has seen it Explained. • Decimals have been treated of at the same time as integers, I because, since both of them follow precisely the same laws, when the rules i: lating to integers are fully understood, there is nothing new to be learned on the subject — particu- larly if wnat has been said with reference to numeration and notation is carefully borne in mind. Should it, however, in any case, be preferred, what relates to them can be omitted until the learner shall have made some further advance. The most useful portions of mental arithmetic have been introduced into " Practice" and the other rules with which they seemed more immediately connected. The different rules should be very cai-efuUy impressed on the mind of the learner, and when he is found to have been ▼m FRr.rAri. guiltj of any inaoouraoy, he should be made to eoneet Mm telf by repeating each part of the appropriate rule, and exemplifying it, until he perceives his error. I^- should be continually kept in view that, in a work on such a subject as arithmetic, any portion must seem difficult and obscure without a knowledge of what precedes it. The table of logarithms and article on the subject, also the table of squares and cubes, square roots and cube roots «f nmnberB, which have been introduced at the end of the work, will, it is expected, prove very acceptable to the mort •dfinoed arithoMtioian. • (' ;': ii'.i'ff . . i*s- taeorreer hm itfs rule, and I^> should be loh a subject ' and obscure subject, also id cube roots le end of the I to the more CONTENTS PAKT I. Multiplication Table, . • • • Tables of money, weights, and meaiures, Definitions, ..... Section I. — Notation and Numeration, • Arabic system of numbers, Roman notation, .... Section II. — Simple Addition, . • To prove Addition, . . . Addition of quantities contuning deoimals, Simple Subtraction, . . . To prove Subtraction, Subtraction of quantities containing decimals, Simple Multiplication, . To multiply when neither multiplicand nor mul< tiplier exceeds 12, . . . When the multiplicand exceeds 12, To preve Multiplication, . .' To multiply quantities, when there are oyptiera or decimals, . . . ^ When beth multiplicand and multiplier ex- cee(\12, . . . To prove Multiplication, To prove Multiplication, by casting out the nines. To multiply, so as to have a certain number of decimal places in the product, To multiply by a composite number, - by a number not composite. To multiply by a number consisting of nines, Simple Division, .... 1 8 12 14 18 20 88 86 37 4ft 10 49 53 67 67 60 61 66 66 67 70 72 78 74 T8 ;',' -V:-. ! ' I d CONTKNTS To divide, wlifrk the diviHor does not exceed 12, nor the dividend 12 times the divisor, . 79 When the divisor does not exceed 12, but the dividend is more than 12 times the divisor, 80 To prove Division, .... 85 To divide when the dividend, divisor, or both, contain cyphers or decimals, . 86 When the divisor exceeds 12, . 89 To prove Division, .... 94 To abbreviate the process of division, when there are decimals, ..... 96 To divide by a composite number, . . 96 To divide by a number but little leas than one expressed by unity and one or more cyphers, 96 To find the greatest common measure of numbers, 101 To find their least common multiple, . 104 Section III. — Reduction Descending, . . 107 Reduction Ascending, . . • r 109 To prove Reduction, . , * • 110 The Compound Rules, . . • .^113 Compound Addition, . • » 114 Compound Subtraction, . . . 123 Compound Multiplication, when the multiplier - does not exceed 12, . . . . 126 When the multiplier exceeds 12 and is ,,, composite, . . . # . 128 When the multiplier is the sum of compo- i eite numbers, ..... 129 When the multiplier is not composite, . 130 Compound Division, when the divisor is abstract, and does not exceed 12, . . 132 When the divisor exceeds 12 and is com- posite, ..... 134 When the divisor exceeds 12 and is not com- .; ' posite, .'. . . . ,' 134 ^ When the divisor and dividend are both ap- ^r plicate, but not of the same denomination ; or ^. more than one denomination is found in either or both, • • 139 I t "^^j 4 rONTKNT*. >ed 12, Pii(r« 2, but 79 i visor, 80 • both, 85 t 86 8« • 94 N«t;TlON IV.— Vul{j;ar Kmctions, i'riu'Au) Ixed number, 1 there A one 95 lers, 96 bers, 101 104 107 109 110 113 114 123 tlier • 126 i is • 128 po- • 129 • 130 ^tf ! • 132 m- • 134 m- • 134 P- ,,. or ^ Br I 139 To reduce an impropot To reduce pa intoj^or to a fraction, To red"oe fractionH to lower terniR, To find the valuo of a fraction in terms of a lower denomination, . . . • • To express one quantity as the fracti .n of luother, To add fractions having a common denominator. To add fractions when their denominators are different and prime to each other, To add fractions having different denominators, not all prime to each other, To reduce a mixed number to an improper frac- tion, ...... To add mixed numbers, . . . . To subtract fractions whicti have a oommon denominator, . . . . To subtract fraotions which have not a oommon denominator, . To subtract mixed numbers, or fractions from mixed numbers, .... To multiply a fraction and whole number together. To multiply one fraction by another, To multiply a fraction, or mixed number by a mixed number, To divide a fraction by a whole number, To divide a fraction by a fraction, To divide a whole number by a fraction. To divide a mixed number by a whole number or fraction, ..... To divide an 'nteger by a mixed number, To divide a fraction or mixed number by a mixed number, ..... When the divisor, dividend, or both, are com- pound, or complex fractions, . Decimal Fractions, . . . - . To reduce a vulgar fraction to a decimal or to a deoimal fraction, To reduce a decimal to a lower denomination. 1^ 138 140 142 142 143 143 144 145 140 147 148 149 150 150 152 153 154 156 156 158 158 159 159 160 162 163 168 CONTENTS. I III li il To And at once the decimal equivalent to any number of shillings, pence, &c., To find the number of shillings, &c., equivalent to a decimal, ..... Circulating Decimals, . . , .> To change a circulating decimal into its equi- valent vulgar fraction. When a vulgar fraction will give a finite decimal, The number of decimal places in a finite decimal, The number of digits in the period of a circulate, When a circulating decimal will contain a finite part, ..... Contractions in multiplication and division, de- rived from the properties of fractions, Section V. — Proportion, .... Nature of ratios, . . . , , Nature of Proportion, .... To find the arithmetical mean of two quantities. To find a fourth proportional, when the first term is unity, ..... , When the second or third term is unity, . To find the geometrical mean of two quantities, Properties of a geometrical proportion, . Rule of Simple Proportion, When the first and second terms are not of the same, or contain different denominations, When the third term contains more than one 164 denomination, ..... — — - If fractions or mixed numbers are found in any of the terms, . . • . Rule of Compound Proportion, . » . ' To abbreviate the process, . • * PART n. . Section YI. — ^Practice, . • • To find aliquot parts, . • • To find the price of one denomination, ihftk of » Ugber bting givsn, 166 (If 16« 4 167 f. 170 m 170 1 171 i 172 173 176 177 179 181 184 184 184 185 185 190 192 195 202 204 209 209 164 166 16« 167 170 176 171 172 • 173 • 176 • 177 • 179 ies, 181 rm • 184 • 184 s, 184 • 185 • 185 of 18, 190 • • 1V\ 192 in • 195 • 202 • 204 • 209 • 209 A 21] CONTENTS. To find the price of more than one lower denomi- nation, ...... To find the price of one higher denomination, . To find the price of more than one higher deno- mination, ..... Given the price of one denomination, to find that of any number of another, . * When the price of any denomination is the aliquot part of a shilling, to find the price of any num- ber of that denomination. When the price of any denomination is the ali- quot part of a pound, to find the price of any number of that denomination, . When the complement of the price, but not the price itself, is the aliquot part or parts of a pound or shilling, .... When neither the price nor its complement is the aliquot part or parts of a pound or shilling. When the price of each article is an even number of shillinii;8, to find the price of a number of artioleB, . . . . When the price is an odd number and lew than 20, . . . . To find the price of a quantity represented by a mixed ^number, . . . . Given the price per cwt., to find that of cwt., qrs., &c., ..... Given the price per pound, to find that of cwt., qTB.j occ, • • • * . Given the price per pound, to find that of a ton, Given the price per ounce, to find that of ounces, pennyweights, &c., .... Given the price per yard, to find thaii c£ * Bif}.%i qn.) &c., . . . • To find the price of acresi roods, &c., . Given the price per quart, to find that of » }a^ flifHi «1m prlM ptr qoar^ to find thai if a ta^ dtt 212^ 213 213 214 21& 21» 216 217 218 219 220 221 222 223 223 225 220 226 il * ill' \ I'i' % :l !■(! ilT CONTENTS Given the pri<^e of one article in pence, to find that of any number, . . . . Given wages per day, to find their amount pet year, • • • • • • Bills of parcels, . . . . . Tare and Tret, ..... Section VII. — Simple Interest. To find the simple interest on any sum, for a year, When the rate per cent, consists of more than one denomination. To find the interest on any sum for years, For years, months, &c., . To find the interest on any sum, for any time, at 5, 6, &c., per cent., .... — ^ When the rate, or number of years, or both, are expressed by a mixed number, ; ; To find the interest for days, at five per cent., . To find the interest for days, at any other rate, . To find the interest for months, at 6 per cent., '■• To find the interest of money left after one or m(»re payments, . ... . Given the amount, rate, and time— to find the principal, ..... Given the time, rate, and principal — to find the amount, ..... Given the amount, principal, and rate — to find the time; ..... Given the amount, principal, and time, to find ili6 rate. Compound Interest — ^gi\en the principal, rate, and time — ^to find the amount and interest, To find the present worth of any sum, . Given the principal, rate, and amount — to find the time, . . . '• : • ♦ DiBOQUBlt, K : , .± . i - . - To find discount, . . . . . ^ To ooil^le Conunimion, Insurance, BrokAttigo, 227 22b 229 233 237 238 239 239 250 251 256 257 260 261 263 ■-i-fi--' - Page >;'^H 227 'fjyH 22b 229 233 m 237 238 239 239 240 242 243 244 244 246 248 249 249 250 251 256 257 260 261 263 CONTENTS. To find wLat insurance must be paid that, if the goods are lost, both their yalue and the insur- ance paid may be recoveced, . Purchase of Stock, i ;/ :'n r < ^ Equation of Payments, . . Iection VIII. — Exchange, . . Tables of foreign money. To reduce bank to current money^ To reduce current to bank money. To reduce foreign to British money. To reduce British to foreign money, To reduce florins, &c., to pounds, &o., Flemish, To reduce pounds, &o., Flesush, to florins, &c., Simple Arbitration of Exchanges, Compound Arbitration of Exchanges, To estimate the gain or loss per cent., Profit and Loss, . . . . - To find the gain or loss per cent.. Given the cost price and gain — to tind the selling price, . . . . • . Given the gain or loss per cent., and the selling price — to find the cost price, . Simple Fellowship, . . • Compound Fellowship, . . , ■ '■ Barter, . . . . • • ^ Alligation Medial, . • * ^", Alligation Alternate, . . • ' When a given amount of the mixture is re- quired, . — When the ^unt of one ingredient is given. Section IX. — Involution, To raise a number to any power, To raise a fraction to any power, . To raise a mixed numbed to any power, . Evolution, . . . . To fiind the square root, '^ When the square contains decimals, To find the square root of a fraction, XV Poge 263 265 266 268 269 272 273 274 277 281 282 283 285 286 287 288 289 290 291 293 296 298 299 302 304 306 307 307 308 308 308 "iU MS : in«fc-^»'. ■i ; -ii 1 i m Ii' ;i ■ xyx CONTENTS. To find the square root of a mljred number, To find the cube root, .... When the cube contains decimals, To find the cube root of a fraction, . ^ To find the cube root of a mixed number, To extract any root whatever, .-. ..•..!-;;',. ■ . To find the squares and cubes, the square and cube roots of numbers, by the table, . Logarithms, ..... To find the logarithm of a given number by the table, . . . . . • To find the logarithm of a fraction, . To find the logarithm of a mixed number. To find the number ccrresponding to a given logarithm, If the given logarithm is not in the table, . To multiply numbers by means of their loga-i rithms, . ...,, . ^ •. . j. > . ;i To divide numbers by means of their logarithms, To raise a quantity to any power by means of its logarithm, . . ,.. . jxp-To evolve a quantity by means of its logarithm. Section X. — On Progression, . : jit^ '^ •' "' V; To find the sum of a series of terms in arithmeti- • •L cal progression, . . . .; In an arithmetical series given the extremes aiid number of terms — to find the common di£fer«> ence, ' . . . . • * • To find any number of arithmetical means be- tween two given numbers, . , . To find any particular term of any arithmetical SGlrlGB) • • • • • • . In an arithmetical series, given the extremes and common difference — ^tofindjUxe number of terms. )r A'V t <),*r~ — Given the sum of the series, the number of ^/;<^ terms, and one extreme — ^to find the other,; i.„-i , > To find the sum of a series of terms in geometri- ^$f oal progression, j^j^fi.^, v^o^r^ ^»J3|W^rUi* Page 313 llni 313 ili^^B 315 9 To 316 9 ^ 316 ■ to 317 ■ ini 9 c 318 Vint 319 1 t •1 t 321 i A.ni 323 1 To 323 •1 " 324 ;* 324 ■|To 325 m iro 1 ^" 326 1 Do 1 M 327 ' iTal 327 329 329 330 331 332 332 333 334 ,/» 313^ 313 315 r 316 M4.v.i.. 31 G 317 318 319 321 323 323 324 324 325 326 327 327 329 329 330 331 332 332 333 CONTENTS. XVii rage In a geometrical series, given the extremes and number of terms — to find the common ratio, 335 To find any number of geometrical means between two^uantities, .... 336 To find any particular term of a geometrical series, 336 In a geometrical series, given the extremes and common ratio — to- find the number of terms, . 337 In a geometrical series, given the common ratio, the number of terms, and one extreme — ^to find the other ..... 338 A.nnuities, ..... 340 To find the amount of a certain number of pay- ments in arrears, and the interest due on them, 340 To find the present value of an annuity, , 342 When it is in perpetuity, . . . 343 To find the value of an annuity in reversion, . 344 Position, ...... 345 Single Position, . . . . .346 Double Position, ..... 347 Miscellaneous exercises, .... 355 'able of Logarithms, • « • • • 361 Table of squares and cubes, and of square and cube roots, ..... 377 Table of the amounts of £1, at compound interest, 385 Table of the amounts of an annuity of £1, . 385 Table of the present values of an annuity of jEl, 386 Irish converted into British acres, . . 386 Value of foreign money in British, . . 386 'Sd^ Hill ill!' ;!rx *>»•■ fM _ > = .^'.*«S', fir •'^?:v«uvr,M j., n ..liur,.>'i'.t-y*~ m I -i {'tirfy ff "fn' »'.' M Ui/i/i ;;!«wfi';'/;> IT'j'fiJ ^'ll.iv- !:!;M7J'V tfia r «,i! .•i:. ;«•'%; ^ • 1 ' ' 1 ' ; ; ' > . tn '*fi. ;Wrt ii.i. f*f.-.;'i"i"i ,ir;; ;/r.. 'ui- J'i-Oi ..• ■ .i(i» ir' .- 5'/!/, .•'!;'.' a, .1, rK.-'frt fr: ■1,-irf.:- •», Kill 'lii ';.'i-'«t7 ). *^}>:^•T^^V')■^ ^a i^Y;:;'^«e fU? -i»r t-'i U- ■()t'f; ,f!i'lM»,r ( : • '^iv(t'' ■-• ^ .o;>ft{-,!lJXa' :,3rMn i.tj.j; i !>:--.l: :?;» i./M' .. ^fiii*) i,ini '^.'I'iixj ♦ . s« r,i ' o!;fi; -Uti •.„ i-;i vy.; fOl':: i:. i^nj .Uix . :jii: ;:^;,-ii' /; '■w^.'^j ^'a,}}h I / .'\ : * ' . i * J r T V .11- ^.1 ' » I . . 1. 'r» t ' \ *■; i-iii;.:' ..-a.r ;;j ;i I » .■^"'np f ' '-"■ 7') jf; 'lo .tori oiin-: ■^^z^ •] ,'.A,^, .'>sd , !. ''r, I .,„ ^i- . ;^ »fe^s,'fc.v( !E^|:J'^r i: •ijc >'i SIGNS USEP IN raiS TREATISE. !;'!! '! i I I 11 J:' I'lii !:! i II. ; ■ m::l I'i, + the sign of addition; as 5+7, or 5 to be added - to 7. -~ the sign of subtraotion ; as 4—3, or 3 to be sub- tracted from 4. X the sign of multiplication; as 8X9) or 8 to be multiplied by 9. -7- the sign of division ; as 18-r6, or 18 to be divided by 6. the vinculum, which is used to show that all the quantities united by it are to be considered as but one. Thus 4+3—7X6 means 4 to be added to 3, 7 to be taken from the sum, and 6 to be multiplied into the remainder — the latter is equivalent to the whok quantity under the vinculum. = the sign of equality j as 5+6=11, or 5 added to 6, is equal to 11. I >|, andf<|, mean that ^ is greater than |, and that I is less than f . : is the sign of ratio or relation ; thus 5 : 6, mea.us the ratio of 5 to 6, and is read 5 is to 6. : : indicates the equality of ratios ; thus, 5 : 6 : : 7 : 8, means that there is the same relation between 5 and 6 as between 7 and 8 ; and is read 5 is to 6 as 7 is to 8. ^ the radical sign. By itself, it is the sign of the square root ; as ^ 5, which is the same as 5*, the square root of 6. ^6, is the cube root of 3, or 3^ /^ 4, is the 7th root of 4, or 4^, &c. ExAMPLE.-V 8-3+7 X 4-f- 6+31 X ^ 9 -^ 10^ x 5^= 641-31, &c. may be read thus : take 3 from 8, add 7 to the difference, multiply the sum by 4, divide the product by 6, take the square root of the quotient and to it add 31, then multiply the sum by the cube root of 9, divide the product by the square root of 10, multiply the quotient by the square of 5, and the product will be equal to 641*31, &c. These signs are fmy explained in their proper places. SE. ARITHMETIC. to be added 3 to be sub- ', or 8 to be to be divided jbow that all dered as but dded to 3, 7 iiltiplied into bo the whok r 5 added to than I, and 5 : 6, means > : 6 : : 7 : 8, )en 5 and 6 as 7 is to 8. sign of the , the square '^ ^4, is 4.10^X52= idd 7 to the roduct by 6, id 31, then the product ent by the 31, &c. r places. PART I. TABLES. MULTIPLICATION TABLE. JTwice |1 are 2 2—4 18—6 — 8 — 10 — 12 — 14 — 16 — 18 — 20 11 — 22 — 24 8 times 1 are 3 2—6 3—9 4 — 12 6 — 16 6 — 18 7—21 8 — 24 9 — 27 10 — 30 11 — 33 12 — 36 4 times 1 are 4 2—8 3 ^12 4 — 16 6 — 20 6 — 24 7 — 28 8 — 32 9 — 86 10 — 40 11 —44 12 — ic 6 times 1 are 6 2 — 10 3 — 16 4 — 20 6 — 25 6 — 30 7 — 35 8 — 40 9 — 46 10 — 60 11 — 66 2 — 60 6 times 1 are 6 2 — 12 8 — 18 4 — 24 6 — 80 6 — 36 7 *- 42 8 — 48 9 — 54 110 — 60 11 — 66 12 — 72 7 times 1 are 7 2 — 14 8 — 21 4 — 28 6 — 36 6 — 42 7 — 49 8 — 66 9 ^ 68 10 — 70 11 — 77 12 — 84 8 times 1 are 8 2 — 16 8 — 24 4 — 82 5 — 40 6 — 48 7 — 66 8 — 64 9 — 72 10 — 80 11 — 88 12 — 96 9 times 1 are 9 2—18 3—27 4—86 6 — 46 6—54 7—68 8—72 9—81 10 — 90 11 — 99 12 — 108 10 times 1 are 10 2 — 20 3 — 30 4 — 40 5 — 50 6 — 60 7 — 70 8 — 80 9 — 90 10 — 100 11 — 110 12 — 120 11 times li Eire 11 2 — 22 3 — - 33 4 — . 44 6 — 55 6 — . 66 7 __ 77 8 — 88 9 _ 99 10 — _ 110 11 _ 121 12 — • 182 12 times 1 are 12 2—24 3—86 4—48 5—60 6—72 7—84 8—96 9 — 108 10 — 120 11 — 182 12 — 144 It appears from this table, that the multiplication of the same two numbers, in whatever order taken, produces tho MULTIPLICATION TABLE. same result ; thus 5 times 6, and 6 times 5 are 30 : — the reason will be explained when we treat of multiplioation, There are, therefore, several repetitions, which, although many persons oonoeive them unnecessary, are not, perhaps, quite unprofitable. The following is free from such an objection : — m i.H'i lii.'li! liiiii Twice 2 are 4 „ 8 - 6 .. 4-8 „ 5-10 „ 6-12 » 7-14 „ 8-16 „ 9-18 5 times 7 are 85 „ 8-40 ,. 9—45 10 times 8 are 80 „ 9 — 90 „ 10 —100 „ 11 —110 6 times 6—86 » 7-42 „ 8-48 .. 9-54 11 times 2 — 22 „ 8 — 88 „ 4-44 „ 6 — 66 „ 6 — 66 „ 7-77 „ 8 - 88 „ 9 — 99 8 times 8 — 9 » 4-12 „ 6-15 „ 6-18 „ 7-21 „ 8-24 „ 9-27 7 times 7—49 „ 8—56 „ 9-68 8 times 8—64 .. 9-72 12 times 2 — 24 „ 8 — 86 „ 4-48 „ 6 60 „ 6-72 » 7 — 84 „ 8 — 96 „ 9 —108 „ 10 —120 „ 11 —182 „ 12 —144 4 times 4—16 „ 5-20 „ 6-24 .. 7-28 .. 8 - 82 „ 9—86 9 times 9 81 10 times 2 are 20 „ 8—80 „ 4—40 „ 6—50 „ 6—60 ,. 7—70 5 times 5 — 25* „ 6-80 " Ten," or " eleven times," in the above, scarcely requiiAA to be committed to memory; since we perceive, that to multiply a number by 10, we have merely to add a cypher to the right hand side of it : — thus, 10 times 8 are 80 ; and to multiply it by 11 we have only to set it down twice : — thus, 11 times 2 are 22. 4>unti i|rthin| ,4 or 48 rui I il.i! i. , ..■ i<>- "UV^jij^i >/ llii,' TABLE OF MONEY. are 30 .—the nultiplication, ioh, although I not, perhaps, Qrom such an ies 8 are 80 9 — 90 10 —100 11 -110 M 2 .i. 22 8 .. 8B 4 .» 44 6 — 65 6 — 66 7 _ 77 8 •-. 88 9 — 09 8 2 8 4 6 6 7 8 9 10 1 2 — 24 — 86 - 48 60 — 72 — 84 — 96 — 108 — 120 — 182 -144 >ly requiiAii ire, that to I cypher to 30 J and to ce :— thus, The following tables arc required for reduction, the impound rules, &o., and may be committed to memory convenience suggests. TABLE OF MONEY. ; A farthing is the smallest coin generally used in this Symbott, «|>untry, it is represented by . . . . | northing! make 1 half^senny, k 2 • halfpence 1 • 4 or • • ^48 24 or 480 504 pence 12 960 1,008 240 or 252 or • 1 penny. d. • • shillings 20 21 1 shilling, 1 pound, 1 guinea. a £ The symbols of pounds, shillings, and pence, are placed £ i. d jer the numbers which express them. Thus, 3 „ 14 „ 6, lounH, three pounds, fourteen shillings, and sixpence. Some- likmea only the symbol for pounds is used, and is placed f. d. before the whole quantity ; thus, £3 „ 14 „ 6. 3 9^ means Ibree shillings and ninepence halfpenny. 2s. 6}d. means two ■hillings and sixpence tiiree farthmgs, &c. When learning the above and following tables, the pupil i|hould be required, at first, to commit to memory only those liortions which are over the thick angular lines; thus, in the one just given : — 2 farthings make one halfpenny ; 2 half- pence one penny ; 12 pence one shilling ; 2U shillings one ipund ; ana 21 shillings one guinea. I, |, ^, really mean the quarter, half, and three quartera a penny, a. is used as a symbol, because it is tne first "jtter of " denarius," the Latin word signifying a penny; s. ras adopted for a similar reason — '^ solidus," meaning, in le same language, a shilling; and £ also—'' Libra," signify* a pound. s. d 2 6 make one half Crown. 6 one Crown. ' 13 4 one Mark. •M, WEIGHTS. il':;^ m 'I AVOTRDUPOISE WEIGHT. Its numo is clorivcjd tVoni French — find ultimately from Latin words signifying " to huv»' ' '»t." It is used in weighing heavy articles Drams SymhoU . muku 1 ounoe, uz. . . 1 pound, ^. . . 1 quarter, q. . 1 hundrcd,cwt. hundreds 20 . 1 ton, t. 14 lbs., and in some cases 16 lbs., make 1 stone. 20 stones ... 1 barrel. TROY WEIGHT. It is so called from Troyes, a city in France, where it was first employed ; it is used in philosophy, in weighing gold, &c. Symhoh. Ora'n" ...... grs. make 1 pennyweight, ^wt. pennyweights 20 . .1 ounce, . (fk. 256 or • ounces 10 . • • • * • 7,168 448 or 1,792 35,840 pounds 28 . . 28,072 112 or 2,240 quarters 4 678,440 80 or 24 480 or 5,760 ounces 240 or I 12 . 1 pound, . lb. A grain was originally the weight of a grain of com, taken from the middle of the ear ', a pennyweight, that of the silver penny formerly in use. APOTHECARIES WEIGHT. In mixing medicines, apothecaries use Troy weight, but subdivide it as follows : — Symbolt • make 1 scruple, 9 • . 1 dram, 3 1* ounce, 5 1 pound, lb. The " Carat," which is equal to four grains, is used in weighing diamonds. The term carat is also applied in estimating the fineness of gold ; the latter, when pei-faotiy Grains 20 . • scruples 3 • . drams 8 • 60 or • 480 24 or 288 • 5,760 96 or ounces 12 MEASURES. I ultimately '•t." It ia Symfiols ounce, 02. pound, tb. quarter, q. bundred,owt. ton, t. Htone. barrel. mce, where losophy, in Symhoh. ' . ?*"*• weight, ^wt. . lb. in of com, ht, that of )y weight, Symboli scruple, 9 dram, 3 ounce, f pound, ft). is used in ipplied in perfdotiy 4 pure, is said to be "24 carats fine/^ if there are 23 parts gold, and one part some other material, the mixture is said to be "23 carats fine ; " if 22 parts out of the 24 are gold, it is " 22 carats fine,'' &c. ; — the whole mass is, in all oases, supposed to be divided into 24 parts, of which the number , consisting of gold is specified. Our sold coin is 22 carats , fine ; pure gold being very soft would too soon wear out. ! llio degree of fineness of gold articles is marked upon them at the Goldsmith's Hall; thus we generally perceive " 18" on the cases of gold watches ; this indicates that they are "18 carats fine " — the lowest degree of purity which is stamped. crm. A Troy ounce contains . 480 An avoirdupoise ounce . 437| A Troy pound . . 5,760 An avoirdupoise pound . 7,000 A Troy pound is equal to 372*965 French grammes. 175 Troy pounds are equal to 144 avoirdupoise ; 175 Troy are equal to 192 avoirdupoise ounces. CLOTH MEASURE. make 1 nail. . 1 quarter. . 1 yard. . 1 Flemish ell . 1 English ell. 1 French ell Inehei 24 Bail! • 9 or 86 4 • 16 or 27 12 or 46 20 or 54 24 or quarters 4 8 5 6 Vi LONG MEASURE. (It is used to mesiure Length.) «^ • make 1 inch. 144 or 482 2,376 8,024 05,040 120,960 760,820 967,680 inehei. 12 86 or 198 26Z 7,920 10,080 8M,640 16ior 2:1 or 660 840 5,280 6,720 yw4« 5i 7 220 or 280 or 1,760 2,240 perches 40 40 820 or 820 oil 8 1 foot. 1 yard. 1 English perch 1 Irish perch. 1 English ftirlonig 1 Irish Airlong. fUrlon^s 8 1 English milsk llrishmik "^ (il'lll 'In! 'il l;ii^'! ;!i;il MEASURE 3 (i9r^J English miles make Three miles make one league. -. , , _ „ 60 nautical, or geographioal miles ; wlfiich are equal to one degree, or the three hundred and sixtieth part of the cir- cumference of the globe — as measured on the equator. 4 inches make 3 inches 3 palms 18 inches 6 feet 6 feet 120 fathoms 1 hand (used in measuring aorses). 1 palm. l^pan. 1 cubit 1 pace. 1 fathom. 1 cable's length. 100 links, 4 English perches (or poles), 22 yards, 66 feet, or 792 inches, make one chain. Each link, therefore, is equal to 7fA inches. 11 Irish are equal to 14 English miles. The Paris foot is equal to 12792 English inches; the Boman foot to 11*604 ; and the French metre to 39S83. MEASURE OF SURFACES. A surface is called a square when it has four equal sides and four equal angles. A square inch, therefore, is a surface one inch long and one inch wide ; a square foot, a surface one foot long and one foot wide, &c. neke 1 sq. foot. 1 square jrard. 1 sq. Enjr. perch. 1 sq. Irisn perciv Square inches 144 . • • a • 1,396 or square feet " • • • 89,304 63,504 373^ or 441 or 10.890 17,640 43,660 70,560 37,878,400 35,158,400 sq. yards 30| 49 1,668.160 9,540,160 1,310 or 1,960 or 4,840 7,840 3,097,600 5,017,600 oq. perchei 40 40 1. • • 6,373,640 10,160,640 160 or 160 or 103,400 103,400 •q. roodi. 4 4 4,014,489,600 6,603,809,600 3,560 or 3,660 or I sq. EiiL 1 sq. Iris^ rood, rood. 1 statute acre. 1 plantation acre. •q. aeni. 640 640 1 sq. En^ 1 sq. Iris mile, mile. The English, called also the statute acre, consists of 10 square chains, or 100,000 square links. The English acre being 4,840 square yards, and the Irish, or plantation acre, 7,840 ; 196 square English are equal to 121 square Irish acres. The English square mile being 3.097,600 square yards, and the Irish 5,017,600; 196 English square miles are equal tc 121 Irish : — we have seen, however, that 14 English are equal to 11 Irish linear .miles, . six 10 |cu |a Si isid( 8 I 32 iiiillJ MEASURES. sh miles make •e equal to one art of the cir- equator. 1,'-. liorses). yards, 66 feet, , therefore, is to 14 English iglish inches; Btre to 39S83'. ts four equal 5h, therefore, de ; a square ide, &c. 1 sq. foot. I square yard. 1 sq. Ene. perch. 1 sq. Irish perciv I sq. Eng. rood. 1 sq. Irish rood. I statute acre, plantation acre. sq. £n^. mile, sq. Irish mile. onsists of 10 ud the Irish, are equal to juare yards, e miles are 1 14 English MEASURE OF SOLIDS. The teacher will explain that a cube is a solid having %ix equal square surfaces ; and will illustrate this by lodels or examples — the more familiar the better. A Icubic inch is a solid, each of whose t^< sides or faces is la square inch ; a cubic foot a solidi . f a(^ of whose osr [sides is a 55'Wrtre/oo^, &c. [Cubic inches '1,728 ..... iaAk.a 1 cub^f* f4i>iV Icubic feet 27 . . . icubii 212 Aug. Sept. 212 243 181 163 122 92 31 365 31 334 304 273 243 i? 212 184 153 123 92 62 365 335 304 274 ■'A , J; Oct. Nov. 273 242 214 183 163 122 163 92 61 30 365 334 304 335 k 304 273 245 214 184 214 123 92 61 31 363 ^ % Dec. 334 303 276 241 183 133 122 91 61 80 3(59 w 10 TIME. '(i' i ')<< i I'M ■ r Mir" iiili!' T t : \\ To find by thiss table the distance between any two days in two different months : KuLE. — Look along that vertical row of figures at the head of which stands the first of the given months ; and also along the horizontal row which contains the second ; the number of days from any day in the one month to the same day in the other, will be found where these two rows intersect each other. If the given day in the latter month is earlier than that in the former, find by how much, and subtract the amount from the number obtained by the table. If, on the contrary, it is later, ascertain by how much, and add the amount. When February is included in the given time, and it is a leap year, add one day to the result. Example 1, — How many days are there between the fifteenth of March and the fourth of October? Looking down the vertical row of figures, at the head of which March is placed, and at the same time, along the horizontal row at the left hand side of which is October, we perceive in their intersection the number 214 : — so many days, therefore, in- tervene between the fifteenth of March and the fifteenth of October. But the fourth of October is eleven days earlier than the fifteenth ; we therefore subtract 11 from 214, and obtain 203, the number required. Example 2. — How many days are there between the third of January and the nineteenth of May 'f Looking as before in the table, we find that 120 days intervene between the third of January and the third of May ; but as the nine- teenth is sixteen days later than the third, we add 16 to 120 and obtain 136, the number required. Since February is in this case included, if it were a leap year, as that month would then contain 29 days, we should add one to the 136, and 137 would be the answer. During the lapse of time, the calendar became inaccu- rate : it was corrected by Pope Gregory. To understand how this became necessary, it must be borne in mind that the Julian Calendar, formerly in use, added one day every fourth year to the month of February ; but this being somewhat too much, the days of the months were thrown out of their proper places, and to such an extent, that <^3ch had become ten days too much in advance. Pope Q'regorji to remedy this, QXm^-\^ ^^'** ^^*^? according long I TIME 11 reen any two of figures at ven months ; contains the y in the one found where le given day the former, nt from the contrary, it 3 amount, u time, and between the r ? Looking vhich March iontal row at eive in their therefore, in- ! fifteenth of days earlier om 214, and (otween the Looking as 3ne between as the nine- id 16 to 120 were a leap we should me inaccu- mderstand mind that day every this being ?re thrown stent, that je. Pope according the Julian style, would have been the 5th of October )82, should be considered as the loth ; and to prevent le recurrence of such a mistake, he desired that, in jlace of the last year of every century ijeing, as hitherto, leap year, only the last year of every fourth century lould be deemed such. The " New Style," as it is called, was not introduced il^ito England until 1752, when the error had become |leven days. The Gregorian Calendar itself is slightly laccurate. . To find if any given year be a leap year. If not the last year of a century : ^ Rule. — Divide the number which represents the iiven year by 4, and if there be no remainder, it is a leap year. If there be a remainder, it expresses how Jong the given year is after the preceding leap year. Example 1. — 1840 was a leap year, because 1840 divided >y 4 leaves no remainder. Example 2. — 1722 was the second year after a leap year, )ecause 1722 divided by 4 leaves 2 as remainder. If the given year be the last of a century : Rule. — Divide the number expressing the centuries 3y 4, and if there be no remainder, the given one is a leap rear ; if there be a remainder, it indicates the number j)f centuries between the given and preceding last year |)f a century which was a leap year. I Example 1. — 1600 was a leap year, because 16, being i^jiivided by 4, leaves nothing. Example 2. — 1800 was two centuries after that last year lof a century which was a leap year, because, divided by 4, it leaves 2. DIVISION OF THE CIRCLE. Thirds 60 3600 or 210,000 77,760,000 seconds 60 3,600 or 1,296,000 Symbols. make 1 second " 1 minute ' 1 degree ® (degrees 360 1 circumference. minutes 60 Every circle is supposed to be divided into the same number of degrees, mmutes, &c. ; the greater oc less, theio- 12 DEFINITIONS. fore, the circle, the greater or less each of these will be. Tl following will exemplify the applications of the symbols s 60° 5' 4" 6'" ; which means sixty degrees, five minutes, fot aecondj, and six thirds. IbstI iii^'i III I DEFINITIONS. 1. Arithmetic may be considered either as a scieno or as an art. As a science, it teaches the properties of numbers ; as an art, it enables us to apply this kno^v ledge to practical purposes ; the former may be calles theoretical, the latter practical arithmetic. 2. A Unit, or as it is also called. Unity , is one of th? individuals under consideration, and may include manj units of another kind or denomination ; thus a unit of | ,the order called " tens" consists of ten simple units. Or it may consist of one or more parts of a unit of a higher denomination ; thus five units of the order of ** tens" are five parts of one of the denomination called " hundreds ;" three units of the denomination called " tenths" are three parts of a unit, which we shall presently term the . " unit of comparison." 3. Number is constituted of two or more units; strictly speaking, therefore, unity itself cannot be con- ^eidered as a number. 4. Abstract Numbers are those the proJ)erties of which are contemplated without reference to their appli- cation to any particular purpose — as five, seven, &c. ; abstraction being a process of the mind, by which it sepa- rately considers those qualities which cannot in reality exist by themselves ; thus, for example, when we attend only to the length of anything, we are said to abstract from its breadth, thickness, colour, &c., although these are necessarily found associated with it. There is nothing inaccurate in this abstraction, since, although length cannot exist without breadth, thickness, &c., it has pro- perties independent of them. In the samo way, five, seven, &c., cau be considered only by an abstraction of the mind, as not applied to indicate some particular things. 64 A^licate 'Numbers are exactly the reverse of ras ling Iwg DEFINITIONS. It 3r as a scienc more units )straot, being applied to indicate particular objects— five men, six houses. 6. The Unit of Comparison. In every number lere is some unit or individual which is used as a Standard : this we shall henceforward call the ^' unit )f comparison." It is by no means necessary that it thould always be the same ; for at one time we may (ipeak of four objects of one species, at another of four )bjects of another species, at a third, of four dozen, or four scores of objects ; in all these cases four is the lumber contemplated, though in each of them the idea jonveyed to the mind is different — ^this difference arising from the different standard of comparison, or unity issum'od. In the first case, the " unit of comparison" ras a single object ; in the second, it was also a single lobject, but not of the same kind ; in the third, it became la dozen ; and in the fourth, a score of objects. Increas- [ing the " unit of comparison" evidently increases the [quantity indicated by a given number ; while decreas- ling it has a contrary effect. It will be necessary to |bear all this carefully in mind. 7. Odd Numbers. One, and every succeeding alter- [nate number, are termed odd ; thus, three, five, seven, &c. 8. Even Nwinbers. Two, and every succeeding alter- [nate number, are said to be even ; thus, four, six, eight, j&c. It is scarcely necessary to remark, that after taking iway the odd numbers, all those which remain are even, [and after taking away the even, all those which remain are odd. We shall introduce many other definitions when treat- ing of those matters to which they relate. A clear idea of what is proposed for consideration is of the greatest importance ; this must be derived from the definition by which it is explained. Since nothing assists both the understanding and the memory more than accurately dividing the subject of inatruction, we shall take this opportunity of remarking to both teacher and pupil, that we attach much impor- tance to the divisions which in future shall actually be madf^, or shall be implied by the i)rder in which the different heads will be examined. '•*::;' ' - '•»' •"^f *'■ ' • b2 ' 14 SECTION I. ;;il,. Ji i:' 'M f-n \i::w ON NOTATION AND NUMERATION. ' 1. To avail ourselves of the properties of numbers, we must be able both to form an idea of them ourselves, and to convey this idea to others by spoken and by written language ; — that is, by the voice, and by characters. The expression of number by characters, is called notation^ the reading of these, nvmtration. Notation, therefore, and numeration, bear the same relation to each other as writing and readings and though often confounded, they are in reality perfectly distinct. 2. It is obvious that, for the purposes of Arithmetic, we require the power of designating all possible num- bers ; it is equally obvious that we cannot give a dif- ferent name or character to each, as their variety is boundless. We must, therefore, by some ' .eans or another, make a limited system of words and signs suffice to express an unlimited amount of numerical quantities : — ^with what beautiful simplicity and clear- ness this is effected, we shall better understand presently. 3. Two modes of attaining such an object present themselves ; the one, that of combining words or cha- racters already in use, to indicate new quantities; the other, that of representing a variety of different quan- tities by a single word or character, the danger of mistake at the same time being prevented. The Romans simplified their system of notation by adopting the prin- ciple of combination ; but the still greater perfection of ours is due also to the expression of many numbers by the same, character. i , «> 4. It will be useful, and not at all difficult, to explain to the pupU the mode by which, as we may suppose, an idea of considerable numbers was originally acquired, and of which, mdeed, although unconsciously, we still avail ourselves ; we shall see, at the same time, how methods of simplifying both numeration and notation were naturally suggested. .J' .: u-.\v'\a. NOTATION AND NbMERATlON. 15 Let ns suppose no system of numbers to be as jei con- Istructed, and that a heap, for example, of pebbles, is )laced before us that we may discover their amount. tf this is considerable to cannot ascertain it by look- ling at them all togetner, nor even by separately in- specting them; we must, therefore, have recourse to that contrivance which the mind always uses when it lesires to grasp what, taken as a whole, is too great for its powers. If we examine an extensive landscape, as the eye cannot take it all in at one view, we look sue- jcssively at its different portions, and form our judg- lent upon them in detail. We must act similarly with reference to large numbers ; since we cannot compre- 'lend them at a single glance, we must divide them mto sufficient number of parts, and, examining these in mccession, acquire an indirect, but accurate idea of the entire. This process becomes by habit so rapid, that it seems, if carelessly observed, but one act, though jit is made up of many : it is indispensable, whenever we lesire to have a dear idea of numbers — which is not, lowever, every time they are mentioned. 5. Had we, then, to form for ourselves a numerical system, we would naturally divide the individuals to be reckoned into equal groups, each group consisting of some number quite within the limit of our comprenen- i ; if the groups were few, our object would be attained dthout any further effort, since we should have acquired m accurate knowledge of the number of groups, and of the number of individuals in each group, and therefore |a satisfactory, although indirect estimate of the whole. We ought to remark, that different persons have Ivery different limits to their perfect comprehension of Inumber ; the intelligent can conceive with ease a com- Iparatively large one ; there are savages so rude as to be lincapable of forming an idea of one that is extremely IsmaU. 6. Let us call the ni mber of individuals that we choose Ito constitute a group, the ratio ; it is evident that the llarger the ratio, the smaller the number of groups, and Ithe smaller the ratio, the larger the number of groupi^-^ Ibttt the smaller the number of groups the better. m 16 NOTATION AND NUMERATION. it' 7. If the groups into which wo have divided the objects to be reckoned exceed in amount that number of which we have a perfect idea, we must continue the process, and considering the groups themselves as indi- viduals, must form with them new groups of a higher order. We must thus proceed until the number of our highest group is sufficiently small. 8. The ratio used for groups of the second and higher orders, would naturally, but not necessarily, be the same as that adopted for the lowest ; that is, if seven indi- viduals constitute a group of the first order, we would probably make seven groups of the first order constitute a group of the second also ; and so on. 9. It might, and very likely would happen, that we should not have so many objects as would exactly form a certain number of groups of the highest order-^ some of the next lower might be left. The same might occur in forroiag one or more of the other groups. We might, for example, in reckoning a heap of pebbles, have two groups of the fourth order, three of the third, none of the second, five of the first, and seven indi- viduals or "units of comparison." 10. If we had made each of the first order of groups consist of ten pebbles, and each of the second order consist of ten of the first, each group of the third of ten of tlie second, and so on with the rest, we had selected the decimal system, or that which is not only used at present, but which was adopted by the Hebrews, G-reeks, Ilomans, &o. It is remarkable that the language of every civilized nation gives names to the different groups of this, but not to those of any other numerical system; its very general diffusion, even among rude and barbarous people, has most probably arisen from the habit of counting on the fingers, which is not altogether abandoned, even by us. 11. It was not indispensable that we should have used the same ratio for the groups of all the different orders ; we might, for example, have made four pebbles form a group of the first order, twelve groups of the first order a group of the second, and twenty groups of the second a gtoup of the third order i^—m such » filii:': I n.MmH *ND NUMERATION. 17 ided the t number itinue the i as indi* a higher er of our ad higher the same ven indi- WQ would jonstitute , that we ctly form order — me might ips. We ' pebbles^ bhe third, len indi- >f groups ad order rd of ten selected used at f Greeks, ^age of di£ferent umerical [ig rude en from is not Id have dififerent pebbles of the groups such » case we had adopt* a «iy«t ' i exactly like thai ti, »e found in the taf>l 'f iiiun« (|>SAg'" '0? in w'hicl n^ur farthings make a -loiip ot tin- - Wr pe^ f, tw 'a piiiice a group of the order s/n/Un^ twenty i*hillin;^.i a group of the order pounds. Wliilt must ; admitted that the use of the same system i r nppli Ate, as for abstract numbers, would greatly simplify our arithmetical processes — as will be very evident hereafter, a glance at the tables given already, and those set down in treat- ing of exchange, will show that a great variety of systems have actually been constructed. 12. When we use the same ratio for the groups of all the orders, we term it a common ratio. There appears to have been no particular reason why ten should have been selected as a " common ratio" in the system of numbers ordinarily used, except that it was suggested, as already remarked, by the mode of counting on the fingers ; and that it is neither so low as unnecessarily to increase the number of orders of groups, nor so high as to exceed the conception of any one for whom the system was intended. 13. A system in which ten is the " common ratio" is called decimal, from ^^ dccem," which in Latin signifies ten : — ours is, therefore, a " decimal system" of numbers. If the common ratio were sixty, it would be a sexagesi- mal system ; such a one was formerly used, and is still retained — as will be perceived by the tables already given for the measurement of arcs and angles, and of time. A quinary system would have five for its " com- mon ratio ;" a duodecimal^ twelve ; a vigesim/ily twenty, &c. 14. A little reflection will show that it was useless to give different names and characters to any numbers except to those which are less than that which consti- tutes the lowest group, and to the different orders of groups ; because all possible numbers must consist of individuals, or of groups, or of both individuals and groups : — in neither case would it be required to specify more than the number of individuals, and the number of each species of group, none of which numbers — as is evident — can be greater than the common ratio. This !• NOTATION AND NUMERATION. ft is just what wo have done in our numerical nyfliem, except that we have formed the names of some of tho eroups by combination of those already used ; thus, " tons of thousands," the group next higher tlian thou- Bands, is designated by a combination of words already applied to express other groups — whicli tends yet further to simplification. 15. ARABIC SYSTEM OP NOTATION: Units of Comparlion, First eroup, or units of the second order, Second group, or units of tlie third order, Third group, or units of the fourth order, Fourth group, or units of the fifth order, Fifth group, or units of the sixth order, Sixth group, or units of the seventh order, Charaettrt. 1 3 8 4 6 6 7 8 9 10 100 1,000 10,000 Name$. One Two Throe Four Five Six . Seen FJiht Nw.e Ten Hundred Thousand Ten thousand Hundred thousand 100,000 Million . . 1,000,000 16. The characters which express the nine first num- bers are the only ones used ; they are called digits^ from the custom of counting them on the fingers, already noticed — " digitus" meaning in Latin a fincer ; they are also called significatU figures^ to distingui^ them from the cypher, or 0, which is used merely to give the dibits their proper position with reference to the decimal point. The pupil wiU distinctly remember that the place where the ** units of comparison" are to be found is that imme- diately to the left hand of this point, which, if not ex- pressed, is supposed to stand to the right hand side of all the digits — thus, in 468*76 the 8 expresses *^ units of comparison," being to the left of the decimal point ; in 49 the 9 expresses " units of comparison," the deci- mal point being understood to the right of it. 17. We find by the table just given, that after the nine first numbers, the same digit is constantly repeated, its position with reference to the decimal point being, however, changed : — that is, to indicate each succeeding group it is moved, by means of a cypher, one place fiixther to the left. Any of the digits may be used to u 300, 7 the cy figure, 5 four well ud 3 iu th 18. two va. valuo is units n means i six hun ord'jr of of comp 19. V ject, is ordinary quite sul a child, appropri stance, t times of [several c figures — changed jto scores acquaint! they will very defi p^upon to iperfcctly Iteacher i NOTATION AND NUMERATION. express its rospoctivo number of any of the groups :— thus 8 would bo eight *^ units of comparison ;'' 80, eight groups of the first order, or eight " tons" of siuiple unit« ; 800, eight groups of the second, or units of the third order ; and so on. We might use any of the digits with the different groups ; thus, for example, 5 for groups of the third order, 3 for those of the second, 7 for those of the first, and 8 for the " units of compari- son ;" then the whole set down in full would bo 5000, 300, 70, 8, or for brevity sake, 5378 — for we never u»e the cypher when we can supply its place by a significant fi<;ure, and it is evident that in 5378 the 378 keeps the tour places from the decimal point (understood), just as well as cyphers would have doae ; also the 78 koeps the 3 iu the third, and the 8 keeps tho 7 in the .econd place. 18. It is important to remember that each din^it has two values, an absolute and a relative; the absolute value is the number of units it expresses, whatever these uaits may be, and is unchangeable ; thus 6 always moans six, sometimes, indited, six tens, at other times six hundred, &c. The ndative value depend.-H on the ord'jr of units indicated, and on the nature of the ^^ unit of comparison." 19. What has been said on this very important sab- ject, is intended principally for the teacher, though an ordinary amount of industry and intelligence will be quite sufficient for the purpose of explaining it, even to a child, particularly if each point is illustrated by an appropriate example ; the pupil may be made, for in- stance, to arrange a number of pebbles in groups, some- times of one, sometimes of another, and sometimes of several orders, and then be desired to express them by figures — the " unit of comparison" being occasionally changed from individuals, suppose to tens, or hundreds, or to scores, or dozQaa^ &c. Indeed the pupils must be well acquainted withn-these introductory matters, otherwise 1 they will contract the habit of answering without any very definite ideas of many things they will be called r^upon to explain, and which they should be expected {perfectly to understand. Any trouble bestowed by the [teacher at this period will oe well repaid by the ease 20 NOTATION AND NUMERATION. and rapidity with which the scholar will afterwards advance , to be assured of this, he has only to recol- lect that most of his future reasonings will be derived from, and his explanations grounded on the very prin- ciples we have endeavoured to unfold. It may be taken as an important truth, that what a child learns without understanding, he will acquire with disgust, and will soon cease to remember ; for it is with children as with persons of morer advanced years, when we appeal suc- cessfully to their understanding, the pride and pleasure they feel in the attainment of knowledge, cause the labour and the weariness which it costs to be under- valued, or forgotten. 20. Pebbles will answer well for examples ; indeed, their use in computing has given rise to the term calcu' latiouy ^^ calculus" being, in Latin, a pebble : but while the teacher illustrates what he says by groups of par- ticular objects, he must take care to notice that his remarks would be equally true of any others. He must also point out the difference between a group and its equivalent unit, which, from their perfect equality, are generally confounded. Thus he may show, that a penny, while equal to, is not identical with four farthings. This seemingly unimportant remark will be better appre- ciated hereafter ; at the same time, without inaccuracy of result, we may, if we please, consider any group either as a unit of the order to which it belongs, or so many of the next lower as are equivalent. 21. Roman Notation. — Our ordinary numerical cha- racters have not been always, nor every where used to express numbers ; the letters of the alphabet naturally presented themselves for the purpose, as being already familiar, and, accordingly, were very generally adopted — for example, by the Hebrews, Greeks, Komans, &c., each, of course, using their own alphabet. The pupil should be acquainted with the Bottaiu notation on account of its beautiful simplicity, and its being still employed in inscriptions, &c. : it is found in the follow- ing taUe : — NOTATION AND NUMERATION. 21 tenrards to recol- I derived iry prin- he taken J without and will I as with peal suc- pleasure ause the »6 under- ; indeed, rm calcU' but while >8 of par- tbat his He must p and its lality, are ; a penny, igs. This 3r appre- aaccuracy oy group rical cha- •e used to naturally g already idopted — tans, &c., The pupil tation on )eing still ROMAN NOTATION. 1 " Chttraeter$. I. II. III. Anticipated change IIII. or IV. Change . V. VI. VII. VIII. Anticipated change IX. Change . . X. XI. XII. XIII. XIV. XV. XVI. XVII. XVIII XIX. XX. XXX., &c. Anticipated change XL. Change . . L. LX., &o. Anticipated change XC. Change . . C. CC, &c Anticipated change CD. Change . . D. or Iq. Anticipated change CM. Change . . M. orClQ. f. orlQO. Xumbert Expretud. One. Two. Three Four. Five. Six. Seven. Eight. Nine. Ten. Eleven. Twelve. . Thirteen. . Fourteen. Fifteen. . Sixteen. . Seventeen. Eighteen. Nineteen. Twenty. Thirty, &c. Forty. . Fifty. , Sixty, &o. , Ninety. , One hundred. . Two hundreds -&c . Four hundred. , Five hundred, &o . Nine hundred. . One thousand, &c. Five thousand, &o. X. or CCIoo . Ten thousand, &c. IDOO- • Fifty thousand, &o. CCCIooQ. . One hundred thousand, &a 22. Thus we find that the Romans used very few characters — ^fewer, indeed, than we do, although our system is still more simple and effective, from our apply- ing the principle of ^' position," unknown to them. They expressed all numbers by the following symbols, or combinations of them: I. Y. X. L. C. D. or I^. M., or CL{3. In constructing their system, they evidently had a quinary in view ; that is, as we have said, one in which five would be the common ratio ; for we find that they ohanged their character, not only at ten, ten times 22 WOTAT ^j^ Awl* fruMERATION. m 'h '.''•■■ * ten, &c., but also at five, ten times five, &o. : — a purely decimal system would suggest a change only at ten, tea times ten, &c. ; a purely quinary, only at five, five times five, &c. As far as notation was concerned, what they adopted was neither a decimal nor a quinary system, nor even a combination of both ; they appear to have supposed two primary groups, one of five, the other of ten " units of comparison ; " and to have formed all the other groups from these, by using ten as the common ratio of each resulting series. 23. They anticipated a change of character; one unit before it would naturally occur — that is, not one " unit of comparison," but one of the units under consi- deration. In this point of view, four is one unit before five; forty, one unit before fifty — tens being now the units under consideration ; four hundred, one unit before five hundred — hundreds having, become the units con- templated. 24. When a lower character is placed before a higher its value is to be subtracted from, when placed after it, to be added to the value of the higher ; thus, IV. means one less than five, or four ; VI., one more than five, or six. 25. To express a number by the Eoman method of notation : — Rule. — Find the highest number within the given one, that is expressed by a single character, or the " anticipation " of one [21] ; set down that character, or anticipation — as the case may be, and take its value from the given number. Find what highest number less than the remainder is expressed by a single charac- ter, or " anticipation ; " put that character or " anticipa- tion " to the right hand of what is already written, and take its value from the last remainder : proceed thus until nothing is left. Example. — Set down the present year, eighteen hundred and forty-four, in Roman characters. One thousand, ex- pressed by M., is the highest number within the given one, indicated by one character, or bvan anticipation; we put down and talu one thousand from the given number, which leaves I .5 ■i and leave num tion,' and IS, NOTATION AMD NUMERATION. 23 le more eight hundred and forty-four. Five hundr€ ' >v tho highest number within the last remainder (e i^fA Hundred and forty-four) expressed by one character, c/ an " antici- pation ;" we set down D to the right hand of M, MD, and take its valu9 from eight hundred and forty-four, which leaves three hundred and forty-four. In this the highest number expressed by a single character, or an " anticipa- tion," is one hundred, indicated by C \ which we set down \ and for the same reason two other Cs. MDCCC. This leaves only forty-four, the highest number within which, expressed by a single character, or an "anticipation,*' is forty, XL — an anticipation ; we set this down also, MDCCCXL. Four, expressed by IV., still remains; which, being als< added, the whole is as follows : — MDCCCXLIV. 26. Position. — The same character may have dif- ferent values, according to ^le place it holds with refer- ence to the decimal point, or, perhaps, more strictly, to the ^^ unit of comparison." This is the principle of position. 27. The places occupied by the units of the different orders, according to the Arabic, or ordinary notation [15] , may be described as follows : — units of comparison, one place to the left of the decimal point, exprcissed, or understood ; tens, two places ; hundreds, three places, &c. The pupil should be made so familiar with these, as to be able, at once, to name the " place" of any order of units, or the " units" of any place. 28. When, therefore, we are desired to write any number, we have merely to put down the digits expres- sing the amounts of the different units in their proper places, according to the order to which each belongs. If, in the given number, there is any order of which there are no units to be expressed, a cypher must be set down in the place belonging to it ; the object of which is, to keep the significant figures in their own posi" tions. A cypher produces no effect when it is not between significant figures and the decimal point ; thus 0536, 536 0, and 536 would mean the same thing — ^the .1 24 NOTATION ANB NUMERATION. second is, however, the correct form. 536 and 5360 are different ; in the latter case the cypher affects the value, because it alters the position of the digits. KxAMPLE. — Let it be required to set down six hundred and two. The six must be in the third, and the two in the first place ; fw this purpose we are to put a cypher between the and 2 — thus, o02 : without the cypher, the six would be in the second place — ^thus, 62; and would mean not six hundreds, but six tens. " ' • 29. In numerating, we begin with the digits of the highest order and proceed downwards, stating the num- ber which belongs to each order. To facilitate notation and numeration, it is usual to divide the places occupied by the different orders of units into periods ; for a certain distance the English and French methods of division agree ; the English billion is, however, a thousand times greater than the French. This discrepancy is not of much importance, since we are rarely obliged to use so high a number, — ^we shall prefer the French method. To give some idea of the amount of a billion, it is only necessary to remark, that according to the English method of notation, there has not been one billion of seconds since the birth of Christ. Indeed, to reckon even a million, counting on an average three per second for eight hours a day, would require nearly 12 days. The following are the two methods. . . , i ENGLISH METHOD. Trillions. Billions. Millions. Unite. 000-000 000*000 000-000 000-000 Billions. jHundredi. Tms. Units. FRENCH METHOD. Millions. Thousands. Units. Hund. Tens. Unlta. Huod. Tern. Uaits. BhimI. Tf nk UniU. 30. Z7m of Periods. ^Lei it be required to read off the following number, 576934. We put the first point to the left of the hundreds' place, and find that there are exactly two periods — 576,934 ; this does not always ooQur, as the highest period is often imperfeet, consisting only of oae or two digits. Dividing the number thus NOTATION AND NUMERATION. is into parts, shows at once that 5 is in the third place of the second period, and of course in the sixth place to the left hand of the decimal point (understood) ; and, therefore, that it expresses hundreds of thousands. The 7 being in the fifth place, indicates tens of thousands ; the 6 in the fourth, thousands ; the 9 in the third, hun- dreds ; the 3 in the second, tens ; and the 4 in the first, units (of ^* conaparison"). The whole, therefore, is five hundreds of thousands, seven tens of thousands, six thousands, nine hundreds, three tens, and four units,— or nore briefly, five hundred and seventy-six thousand, nine hundred and thirty -four. 31. To prevent the separating point, or that which divides into periods, from being mistaken for the decimal point, the former should be a comma (,) — the latter a full stop (•) Without this distinction, two numbers which are very different might be confounded : thus, 498-763 and 498,763, — one of which is a thousand times greater than the other. After a while, we may dispense with the separating point, though it is conve- nient to use it with considerable numbers, as they are then read with greater ease. 32. It will facilitate the reading of large numbers not separated into periods, if we begin with the uuits of comparison, and proceed onwards to the left, saying at the first digit " units," at the second *' tens," at the third " hundreds," &c., marking in our mind the deno- mination of the highest digit, or that at which we stop. We thfen commence with the highest, express its number and denomination, and proceed in the same way with each, until we come to the last to the right hand. Ex \MPLE. — Let it be required to read off 6402. Looking At the 2 (or pointing to it), we say "units;" at the 0, "tens;" At the 4, "hundrMs;" and at the 6, "thousands." The ktter. therefore, being six thousands, the next digit is four hundreds, &c. Consequently, six thousands, four hundreds, no tens, and two units ; or, briefly, six thousand four hun • dred and two, is the reading of the given number. 33. Periods may be used to facilitate notation. The pupil will first writ« down a number of periods of cyphers .1 i • n 26 NOTATION AND NUMERATION. m to represent the places to be occupied by the yarioua orders of uaits. He will then put the digits express- ing the different denominations of the given nuuiber, under, or instead of those cyphers which are in corres- ponding positions, with reference to the decimal point — • beginning with the highest. Example. — Write down three thousand six hundred and fifty-four. The highest denomination being thousands, will occupy the fourth place to the left of the decimal point, [t will be enough, therefore, to put down four cyphers, and under them the corresponding digits — ^that expressing the thousands under the fourth cypher, the hundreas under the third, the tens under the second, and the units under the first : thus 0,000 3,654 A cjrpher is to be placed under any denomination in which there is no significant figure. Example. — Set down five hundred and seven thousand, and sixty-three. '' 000,000 ^ ^ : ■■' ■'■ ' ■ '-■' 607,063 ■ After a little practice the periods of cyphers will become unnecessary, and the number may be rapidly put down at once. , % > 34. The units of comparison are, as we have said, always found in the first place to the left of the decimal point ; the digits to the left hand progressively increase in a tenfold degree — those occupying the first place to the left of the units of comparison being ten times greater than the units of comparison ; those occu- pying the second place, ten times greater than those which occupy the first, and one hundred times greater than the units of comparison themselves ; and so on. Moving a digit one place to the left multiplies it by ten, that is, makes it ten times greater ; moving it two places multiplies it by one hundred, or makes it one hundred times greater ; and sc of the rest. If all'the digits of a quantity be moved one, two, &c., places to the left, the whole is increased ten, one hundred, "^c, times — as the case may be. On the other hand, moving I :■:( NOTATION AND NUMERATION. 27 a digit, or a quantity one place to the right, divides it by ten, that is, makes it ten tim-^s smaller than before ; moving it two places, divides It by one hundred, or makes it one hundred times smaller, &c. 35. We possess this power of easily increasing, or diminishing any number in a tenfold, &c. degree, whether the digits are all at the right, or all at the left of the decimal point ; or partly at the right, and partly at the left. Though we have not hitherto considered quantities to the left of the decimal point, their relative value will be very easily understood from what we have already said. For the pupil is noW aware that in the decimal system the quantities increase in a tenfold 'degree to the left, and decrease in the same degree to the right ; but there is nothing to prevent this decrease to the right from proceedmg beyond the units of comparison, and the decimal pomt ; — on the contrary, from the very nature of notation, we ought to put quantities ten times less than units of comparison one place to the right of them, just as we put those which are ten times less than hundreds, &c., one place to the right of hundreds, &o We accordingly do this, and so continue the notation not only upwards, but downwards, calling quantities t(' the left of the decimal point iw^^er*, because none of them is less than a whole " unit of comparison ;" an^ those to the right of it decimals. When there are deci- mals in a given number, the decimal point is actuallj expressed, and is always found at t^o- right hand side of the units of comparison. 36. The quantities equally distant from the unit of comparison bear a very close relation to each other which, j^ indicated even by the similarity of their names ; those which are one place to the left of the units of com- parison are called " tens," being each identical with, or equivalent to ten units of comparison ; those which are one place to the right of the units of comparison are called " tenths," each being the tenth part of, that is, ten times as small as a unit of comparison ; quantities two places to the kft of the units of comparison are called " hundreds," being one hundred times greater ; and those two places to the right^ " hundredths," being one it < :tf ' ;ti S8 NOTATION AND NUMERATION hi !■ " iniil hundred times less than the units of comparison ; and so of all the others to the right and left. This will be more evident on inspecting the following table :— Aacrnding Scrlev. or Intogvn. One Unit . . 1 Ten . . '. , 10 Hundred . . 100 Thousand . . 1,000 Ten thousand . 10,000 Hundred thousand 100,000 -^ &o. Dencendintf Series, or Declnudi. 1» One Unit. •1 Tenth. ^••i.-,;,pn. •01 Hundredth. •001, Thousandth. •000,1 Ten-thousandth. '000,01 Hundred- thousandth. We have seen that when we divide integers into periods [29], the first separating point must be put to the right of the thousands ; in dividing decimals, the fiurst po^nt must be put to the right of the thousandths.- . . 37. Care must be taken not to confound what we now call ^* decimals," with what we shall hereafter desig- nate " decimal fractions ;'' for they express equal, but not identically the same quantities — the decimals being what shall be termed the " quotients" of the corres- ponding decimal fractions. This remark is made here to anticipate any inaccurate idea on the subject, in those who already know something of Arithmetic. 38. There is no reason for treating integers and deci- mals by different rules, and at different times, since they follow precisely the same laws, and constitute parts of the very same series of numbers. Besides, any quantity may, as far as the decimal point is concerned, be ex- pressed in different ways ; for this purpose we have merely to change the unit of comparison. Thus, let it be required to set down a number indicating five hun- dred and seventy-four men. If the *' unit of compari- son" be one maUy the quantity would stand as follows, 574. If a band of ten men, it would become 57*4 — ^for, as each man would then constitute only the tenth part of the *' unit of comparison," four men would be only four-tenths, or 0*4 ; and, since ten men would form but one unit, seventy men would be merely seven units of comparison, or 7 ; &c. Again, if it were a band of one humdred men, the number must be written 5'74 ; and y «tly, if a band of a tlumsand meny it would be 0574 m ■• n digits NOTATION AND NUMERATION. 29 Should the " unit " be a band of a dozen, or a score men, the change would be still more complicated ; as, not only the position of the decimal point, but the very digits also, would be altered. 39. It is not necessary to remark, that moving the decimal point so many places to the left, or the digits an equal number of plaoes to the right, amount to the same thing. Sometimes, in changing the decimal point, one or more cyphers are to be added ; thus, when we move 42*6 three places to the left, it becomes 42600 ; when we move 27 five places to the right, it is '00027, &c. 40. It follows, from what we have said, that a deci- mal, though less than what constitutes the unit of com- parison, may itself consist of not only one, but several mdividuals. Of course it will often be necessary to indi- cate the " unit of comparison," — as 3 scores, 5 dozen, 6 men, 7 companies, 8 regiments, &c. ; but its nature does not affect the abstract properties of numbers ; for it is true to say that seven and five, when added together, make twelve, whatever the unit of comparison may bv; : — provided, however, that the same standard be applied to both ; thus 7 men and 5 men are 12 men ; but 7 men and 5 horses are neither 12 men nor 12 horses ; 7 men and 5 dozen men are neither 12 men nor 12 dozen men. When, therefore, numbers are compared, &c., they must have the same unit of comparison, or — ^without altering their value — they must be reduced to those which have. Thus we ma^' consider 5 tens of men to become 50 individiial men — the unit of comparison being altered from ten men to one man, without the value of the quantity being changed. This principle must be kept in mind from the very commencement, but its utility will become more obvious hereafter. f EXAMPLES IN NUMERATION AND NOTATION. . , , Notation, ; ■ I 1. Put down one hundred and fbur 2. One thousand two hundred and forty 3. Twenty thousand, three hundred and forty-five c 104 1,240 20,345 30 NOTATION AND NUMERATION. 5. 6. 7. 4. Two hundred and thirty-four thousand, five •^"•. hundred and sixty-seven . 234,567 Three hundred and twenty-nine thousand, seven hundred and seventy-nine . 329,779 Seven hundred and nine thousand, eight hun- dred and twelve . . 709,812 Twelve hundred and forty-seven thousand, four hundred and fifty-seven . . 1,247,457 8. One million, three hundred and ninety-seven thousand, four hundred and seventy-five . 1,397,475 9. Put down fifty-four, seven-tenths . . 54-7 10. Ninety-one, five hundredths . . . 9105 11. Two, three-tenths, four thousandths, and four hundred-thousandths . . . 230404 12. Nine thousandths, and three hundred thou- sandths ..... 000903 13. Make 437 ten thousatid times greater . 4,370,000 14. Make 2-7 one hundred times greater . 270 15. Make 0056 ten times greater . ; « 0-56 16. Make 430 ten times less . • ..> • ^3 17. Make 2*75 one thousand times los9 . ' . 000275 Numeration 1. read 132 ' ,. 2. — 407 3.-— 2760 4. — 5060 5. __ 37654 6. -- 8700002 7. read 8540326 8. 9. 10. 11. 12. 5210007 6030405 560075 3000006 0-0040007 13. Sound travels at the rate of about 1142 feet in a second ; light moves about 195,000 miles in the same time. 14. The sun is estimated to be 886,149 miles in diameter; its size is 1,377,613 times greater than that of the earth. 15. The diameter of the planet mercury is 3,108 miles, and his distance from the sun 36,814,721 miles. 16. The diameter of Venus is 7,498 miles, and her dis- tance from the sun 68,791,752 miles. 17. The diameter of the earth is about 7,964 miles; it is 95,000,000 miles from the sun, and travels round the latter at the rate of upwards of 68,000 miles an hour. 18. The diameter of the moon is 2,144 miles, and her dis- tance from the earth 236,847 miles. 19. The diameter of Mars is 4,218 miles, and his distance from the sun 144,907,630 miles. 20. The diameter of Jupiter is 89,069 miles, and his dis- tance from the sun 494,499,108 miles. NOTATION AND NUMERATION. 31 234,5C7 329,779 709,812 1,247,457 1,397,475 64-7 9105 2-30404 000903 4,370,000 270 0-56 43 000275 6 : 7 5 > 07 'eet in a e time. iameter; rth. 8 miles, her dis- |es; it is le latter her dis- Idistance hisdis- 21. The diameter of Saturn is 78,730 miles, and his dis- tance fVom the sun 907,089,032 miles. 22. The length of a pendulum which would vihrate neconds at the e<]^uator, is 39-011,684 inches ; in the latitude of 45 degrees, it is 39- 116,820 inches; and in the latitude of 90 degrees, 39-221,956 inches. 23. It has heen calculated that the distance from the earth to the nearest fixed star is 40,000 times the diameter of the earth^s orbit, or annual path in the heavens ; that is, about 7,600,000,000,000 miles. Now suppose a cannon ball to fly from the earth to this star, with a uniform velocity equal to that with which it first leaves the mouth of the fun — say 2,500 feet in a second — it would take nearly ,000 years to reach its destination. 24. A piece of gold equal in bulk to an ounce of water, would weigh 19-258 ounces ; a piece of iron of exactly the same size, 7*788 ounces; of copper, 8-788 ounces; of leadj 11-352 ounces; and of silver, 10-474 ounces. c NoTx. — The examples in notation may be made to answer for numeration ; and the reverse. QUESTIONS IN NOTATION AND NUMERATION. [The references at the end of the questions show in what paragraphs of the preceding section the respective fkuswers are principally to be found.] 1. What is notation .? [1]. '' 2. What b numeration .? [1]. 3. How are we able to express an infinite variety of numbers by a few names and characters } [3] . 4. How may we suppose ideas of numbers to have been oridnally acquired .^ [4, &c.]. 5. What is meant by the common ratio of a system of numbers.^ [12]. 6. Is any particular number better adapted than another for the common ratio } [12] . 7. Are there systems of numbers without a comm&n ratio .> [11].^ 8. What is meant by quinary, decimal, duodecimal, vigesimal, and sexagesimal systems ? [J3j. 9. Explain the Arabic system of notation ? [15] . 10. What are digits.? [16]. 11. How are they made to express all numbers } [17] . \l 33 NOTATION AND NUMERATION. 12. What IB moant by tbeir absolute and relative ralues ? [18]. ,,j 13. Are a digit of a higher, and the equivalent num- ber of uoitB of a lower order preoisely the same thing ? l20]. 14. Have the characters we use, always and every where been employed to express numbers ? [21] . 1 5. Explain the Roman method of notation ? [22, &o.] . 16. What b the decimal point, and the position of che different orders of units with reference to it? [26 and 27] . 17. When and how do cyphers affset significant figures? [28]. 18. What is the difierence between the English and French methods of dividing numbers into periods ? [29] . 19. What is the difference between integers and decimals ? [35] . 20. What is meant by the* ascending and descending series of numbers ; and how are they related to each other? [36]. 21. Snow that in expressing the same quantity, we must place the decimal point differently, according to the unit of comparison we adopt ? [38]. 22. What effect is produced on a digit, or a quantity by removing it a number of places to the right, or left, or similarly renrioving the decimal point ? [34 and 39] |i ^i tion.' same. l./Ji' 'I - f 83 SECTION II. THE SIMPLE RULES. SIMPLE ADDITION. 1. If numbers are oha..^ed by any arithmetical pro- cess, they are either increased or diminbhed; if in- creased, the effect belongs to Addition; if diminished, to Subtraction. Hence all the rules of Arithmetic are ultimately resolvable into either of these, or combina- tions of both. 2. When an^ number of quantities, either different ^ or repetitions of the same, are united togcvher so as to form but one, we term the process, simply, '^ Addi- tion.'* When the quantities to be added are the same^ but we may have as many of them as we please, it is called ^* Multiplication ;'* when they are not only the safMy but their number is indicated by one of tkenij the process belongs to ^^ Involution.'' That is, addition re- stricts us neither as to the kind, nor the number of the quantities to be added ; miiltiplication restricts us as to the kind, but not the number; involution restricts us both as to the kind and number : — all, however, are really comprehended nnder the same rule — addition. 3. Simpk Addition is the addition of abstract num- bers ; or of appliccte numbers, containing but one deno- mination. The quantities to be added are called the addends ; the result of the addition is termed the svm, 4. The process of addition is expressed by +> called the plus, or positive sign ; thus 8+6, read 8 plus 6, means, that 6 is to be added to 8. When no sign is prefixed, the positive is understood. The equality of two quantities is indicated by =, thus 9+7=16, means that the sum of 9 and 7 is eooai to 16. ? 34 ADDITION. W ^ ^;'!i Quantities connected by the sign of addition, or that of equality, may be read in any order ; thus if 7+9=16, it is true, also, that 9+7=16, and that 16=7+9, or 9+7. 5. Sometimes a single horizontal line, called a vin' culu?Hy from the Latin word signifying a bond or tie, is placed over several numbers ; and shows that all the quantities under it are to be considered, and treated as but one ; thus in 4+7=11, 4+7 is supposed to form but a single term. However, a vinculum is of little consequence in addition, since putting it over, or remov- ing it from an additive quantity — ^that is, one which has the sign of addition prefixed, or understood— does not in any way alter its value. Sometimes a parenthesis () is used in place of the vinculum; thus 5+6 and (5+6) mean the same thing. 6. The pupil should be made perfectly familiar with these symbols, and others which we shall introduce as we proceed ; or, so far from being, as they ought, a great advantage, they will serve only to embarrass him. There can be no doubt that the expression of quafitities by characters, and not by words written in ftdl, tends to brevity and clearness ; the same is equally true of the processes which are to be performed — ^the more con- cisely they are indicated the better. 7. Arithmetical rules are, naturally, divided into two parts ; the one relates to the setting down of the quan- tities, the other to the operations to be described. We shall generally distinguish these by a line. To add Numbers. Rule. — ^I. Set down the addends under each other, so that digits of the same order may stand in the same vertical column — units, for inst^ince, under units, tens under tens, &c. II. Draw a line to separate the addends from their ium. - -V. -; : III. Add the units of the same denomination together, beginning at the right hand side. IV. If the sum of any column bo less than ten, set it down under that column j but if it bo greater, for every ten it down Mothin V. 8. E ADDITION. 3&. ten it contains, carry one to the next column, and put . down only what remains after deducting the tens ; if r Mothing remains, put down a cypher. • ^ V. Set down the sum of the last column in full. 8. Example.— Find the sum of 542+3754-984— 542) 375 f afldends. . .* ■'■ 984 \ 1901 sum. Wo have placed 2, 5, and 4, T^hich belong to the order J "units," in one column; 4, 7, and 8, which are "tens," in another ; and 5, 3, and 9, which are " hundreds," in another. 4 and 5 units are 9 units, and 2 are 1 1 units — equivalent to one ten and one unit ; we add, or as it is called, " carry" the ten to the other tens found in the next column, and set down the unit, in the units' place of the " sum/' The pupil, having learned notation, can easily find how many tens there are in a given number ; since all the digits that express it, except one to the right hand side, will indicate the number of "tens" it contains; thus in 14 there are 1 ten, and 4. units ; in 32, 3 tens, and 2 units ; in 143, 14 tens, and 3 units, &o. The ten obtained from the sum of the units, along with 8, 7, and 4 tens, makes 20 tens ; this, by the method just men- tioned, is found to consist of 2 tens (of tens), that is, two of the next denomination, or hundreds, to he carried, and no units (of tens) to be set down We "carry," 2 to the hun- dreds, and write down a cypher in the tens' place of the sum." The two hundreds to be " carried," added to 9, 3, and 5, J hundreds, make 19 hundreds ; which are equal to 1 ten (of hundreds) ; or one of the next denomination, and 9 units (of hundreds) ; the former we "carry" to the tens of hundreds, or thousands, and the latter we set down in the hundreds' place of the "sum." As there are no thounands in the next column — that is, nothing to which we can "carry" t^ } thousand obtained by adding the hundreds, we put it down in the thousands' place of the " i^uiii ,'" in other words, we set down the sum of the last column in full. 9. Reason of I. (the first part of the rule).— We put units of the same denomination in the same vertie'U columnt Hi I 36 ADDITION. that we may easily find those quantities which are to be added together ; and tliat the value of each digit may be more clear from its being of the same denomination as those which are under, and over it. Ukason or II. — We use the separating line to prevent the ■um from being mistaken for an addend. Keasoiv or III. — We obtain a correct result only by adding units of the same denomination together [Sec. I. 40} :— hun- dreds, for instance, added to tens, would give neither hundreds nor tens as their sum. We begin at the right hand side to avoid the necessity of more than one addition ; for, beginning at the left, the process would be as follows — 642 876 984 1,700 190 11 .1., l.COd 800 100 _JL 1,901 . . ^(f: The first column to the left prod ■.''■^~. '■ 47-85 82-57 66-46 146*88 EXEBCISES. •4785 •8257 •6646 1*4588 •004785 •008257 •006646 '0X4588 ■■:■■:■■■ > <.■:. t-«';;ni{;' :^ X 9-- (Add the following numlMNS.) ■ f . 6 8 8 7 Addition. f 4 r 2 9 7 6 (4) 6 6 6 6 Multiplication. '>iiii InTO^tion. (6, 4 4 4 (6) (7) C8 9 CO <8 9 (3 9 — 9 (8) 4 4 4 »o 4 5 5 6 6 'Vi- "iN (10) 6768 2341 6279 (11) 8T07 2465 6678 (12) 2367 $246 1239 (13) 6978 8767 1236 (14) mm 4679 1236 (IS) 7647 1239 8789 i ADDITION. « ►e, their position also by (W) (17) (18) (19) (20) (21) 6673 8767 8001 6147 84567 78456 1287 4567 2788 8745 47$91 45678 addends 2345 1234 4667 • 6789 41234 91234 « (22? (23) (24) (25) (26) (27) 76789 84667 78789 84676 73412 36707 46767 124/6 89123 01007 78767 70760 46770 46678 84657 46679 47076 86767 J adding ly plain WQ come (28) (29) (30) (31) (82) (m ^Sec. I. K)ints of 46697 76767 23466 45678 23745 87967 37676 46677 78912 91234 67891 82785 36767 76988 84667 66789 23466 64127 vertical hanged; 1 '• ! '-• • ■^:t\-]^)\ 1-1,. ; 004785 (84) (35) (36) (87) (38) (89) 008257 30071 46676 37645 47656 76767 45676 006546 45667 37412 67466 12346 12345 84667 12345 87373 12346 67891 37676 1^45 )1458S 47676 -TTT" 45674 67891 10707 71267 67891 V (40) (41) (42) (43) (44) (45) l%\ ■' 71234 19123 93466 45678 46679 76756 \ [5. 1 12498 67345 13767 34667 84667 34667 5 1 91379 67777 87124 12345 12345 12845 «P ' 1 92456 i 1, 88899 12466 99999 76767 67891 - 1 (46) (47) (48) (49) (50) (51) (15) 1 87676 78967 84667 47676 67678 67667 7647 I 12677 12345 12345 12345 12845 84567 1239 ■ 88991 78767 77766 67671 67912 23456 S 1789 ■ 23478 12671 67345 10070 46767 76799 40 i m ) ADDITION. (62) (68) (64) (66) (66) (67) 76769 67567 767846 478894 876767 676 12846 19807 476784 767867 128764 4689 76776 84076 467007 412345 845678 87 46G66 18707 128456 671284 912846 84028 (68) (69) (60) (61) (62) %) 74564 6676 76746 67674 42^87 0-87 7674 1567 71207 75670 56-84 fit-278 876 63 100 86 27^93 8-127 6 6767 66 77 62*41 >5-r (64) »"._ "■ (66) ; - (66) (67) 08*786 * 85*772 : :': -00007 6471-8 20-766 \ 6034*82 •06236 663-47 00*258 57*8563 •0572 21-502 10*004 712-52 •21 0-00007 (68) ;; • '• (69) <. (70) (71) ^ 81*0285 0*0007 8456-5 676-84 876*03 5000- -87 -- ■ f: 4000-005 4712*5 427* 8456*302 218 -ST 6*53712 87*12 •00 7 2758- 72. ^£7654 + ^250121 + iBlOO + ^£76767 + ^£675 =£135317. 73. £10 + £7676 + £97674 + £676 + £9017 ^=£115053 74. £971 +£4004-£97476+£30+£7000+£76734 =£182611. p; ,sv .-. ,: ^ 76. 10000 + 76567 + 10+76734 + 6763 + 6767+1 =176842. 76. 1 + 2 + 7676 + 100 + 9 + 7767 + 67=15622. /7. 76 + 9970 + 33 + 9977+100 +67647+676760 ar=764663. ADDITION. 41 (71) »76-84 K)0*006 !18-6r 58- .. , and to E. .£1900 ; what is the Ans. JS5168. 78. -75 4- -6 + -756 + •72544--345+-54--005+-07 — 3'7514. 79. •4+74-47+37007-f7505+747-077=934-004. 80. 5605+4-754--007+3614-h4'672=101-619. 81. •76-|-0076+76 + -5+5+-05.=82-3l76. 82. •5+-05+-005-f-5+50+500=555-555. : 83. :367+56'7-f762+97-6+471=1387-667. '.: 84. 1-|--1 + 10H--01 + 160+-001=17M11.. . 85. 3-76+44-3+4761-|-5-5=529-66. i » 86. 36-77-H4-42+11001 + -6=42-8901. 87. A merchant owes to A. JS1500 ; to B. J6408 $ to C. .£131 jD. dB50 sum of all his debts ? 8S. A merchant has received the following sums :^ £200, £315, £317, £10, £172, £513 and £9 ; what is the amount of all ? Ans, £1536. 89. A merchant bought 7 casks of merchandize. No. 1 weighed 310 ft) ; No. 2, 420 tt» ; No. 3, 338 ft» ; No. 4, 335 ft) ; No. 6, 400 ft) ; No. 6, 412 ft) ; and No. 7 429 ft) : what is the weight of the entire ? ■ '.,; . *" • iw' .• •-- ??■ .^. i . Aiu. 2644 ft. 90. What is the total weight of 9 casks of goods :— Nos. 1, 2, and 3, weighed each 350 lb ; Nos. 4 and 5, each 331 ft) ; No. 6, 310 ft) ; Nos. 7, 8, and 9, each 342 ft) ? Am. 3048 ft). 91. A merchant paid the following sums : — ^£5000, £2040, £1320, £1100, and £9070; how much was the amount of all the payments ? Aiu. .£18530. 92. A linen draper sold 10 pieces of cloth, the first contained 34 yards ; the second, third, fourth, and fifth, each 36 yards ; the sixth, seventh, and eighth, each 33 yards; and the ninth and tenth each 35 yards; how many yards were there in all > Ans. 347. 93. A cashier received six bags of money, the first held £1034 ; the second, £1025 ; the third, £2008 ; the fourth, £7013 ; the fifth, iS5075 ; and the sixth, M9 : how much was the whole sum ? Ans. £16244. 94. A vintner buys 6 pipes of brandy, containing as. follows :— 126, 118, 125, 121, 127, and 119 gallons; how many gallons in the whole ? Atis. 736 gals. 9^^ What is th« tot»l weight of 7 Vftakg^ Nq^ \^ 091H I Hi 42 ADDITION. ^ I:" ■'*<^ ^y i' I taining, 960 lb ; No. 2, 725 !b ; No. 3, 830 lb ; No. 4, 798 lb ; No. 5, 697 lb ; No. 6, 569 fib ; and No. 7, 987 ft) } ; V . • Ans. 5566 lb. 96. A liiercBant bought 3 tons of butter, at iS90 per ton ; and 7 tons of tallow, at i£40 per ton ; how much is the price of both butter and tallow ? Ans. £dbO. 97. If a ton of merchandize cost jS39, what will 20 tons come to ? Ans. j£780. 98. How much are five hundred and seventy-three ; eight hundred and ninety-seven ; five thousand six hun- dred and eighty-two ; two thousand seven hundred and twenty-one ; fifty-six thousand seven hundred and seventy- one ? Am. 66644. 99. Add eight hundred and fifty-six thousand, nine hundred and thirty-three ; one million nine hundred and seventy-six thousand, eight hundred and fifty-nine ; two hundred and three millions, eight hundred and ninety- five thousand, seven hundred and fifty- two. ^'^ Ij? nxfrv .. V. r' ,d?L;i . /; , = I ♦. ^7W. 206729544. 100. Add three millions and seventy-one thousand; four millions and eighty-six thousand ; two millions and fifty-6ne thousand ; one million ; twenty-five millions and six ; seventeen millions and one ; ten millions and two ; twelve millions and twenty-three ; four hundred and seventy-two thousand, nine hundred and twenty-three ; one hundred and forty-three thousand ; one hundred and forty-three millions: Ans. 217823955. 101. Add one hundred and thirty-three thousand; seven hundred and seventy thousand ; thirty-seven thou- sand ; eight hundred and forty-seven thousand ; thirty- three thousand ; eight hundred and seventy-six thousand ; four hundred and ninety-one thousand. Ans, 3187000. 102. Add together one hundred and sixty-seven thou- sand ; three hundred and sixty-^even thousand ; nine hun- dred and six thousand ; two hundred and forty-seven thousand ; ten thousand ; seven hundred thousand ; nine hundred and seventy-six thousand; one hundred and ninety-five thousand ; ninety-seven thousand. A76S, 3665000. 103. Add three ten-thousandths-; forty-four, five ^nths } 4^e hundredths ; six thousandths, eight ten-thou- tenti 1( dredl ten- ADDITIOn. 43 > ; No. 4, d No. 7, . 6566 ft». t iS90 per iow much ns. £bbO. kt will 20 ns. £1^0. ity-three ; I six hun- idred and Iseventy- s. 66644. and, nine idred and line ; two i ninety- 5729544. bousand ; lions and lions and tod two ; [red and y-three ; dred and 823955. lousand ; en thou- ; thirty- oueand ; 187000. an thou- i&e hun- y-seven ; nine ed and P65000. , five n-thou- Mndths ; four thoasand and forty-one ; twenty-two, one tenth ; one ten-thousandth. Ans. 4107*6573. 104. Add one thousand ; one ten-thous^andth ; fitei hun- dredths ; fourteen hunitred and forty ; two tenths, three ten-thousandths ; five, four tenths^ four thousandths. ; . M v t, .>i . I V ^■.. Alts. !a445'6544. 105.^ The circulatioh of promissory notes for the four weeks ending February 3, 1844, was as follows : — Bank of England, about ^22 1,22^,000 ; private banks of Eng- land and Wales, ^£4,980,000 ; Joint Stock Banks of England and Wales, ^£3,446,000 ; all the banks of Scot- land, ^2,791,000 ; Bank of Ireland, ^£3,581, 000 ; all the other banks of Ireland, j£2,429,000 : what was the total circulation? -v -Ati*. iE38,455,000. 106. Chronologers have stated that th0 creation of the world occurred 4004 years before Christ ; the deluge, 2348 ; the cjiU of Abraham, 1921 ; the departure of the Israelites, from Egypt, 1491 ; the foundation of Solomon ^s temple, 1012 ; the end of the captivity, 536. This being the year 1844, how long is it since each of these events } Ans. Frotn the creation, 5848 years ; from the deluge, 4192 ; from the call of Abraham, 37(35 ; from the de- parture of the Israelites, 3335 ; from the foundation of the temple, 2856 ; and from the end of the captivity, 2380 107. The deluge, according to this calculation^ occur- red 1656 years after the creation; the (^11 of Abraham 427 after the deluge ; the departure of the Israelites, 430 after the call of Abraham; the foundation of the temple, 479 after the departure of the Israelites ; and the end of the captivity, 476 after the foundation of the temple. , How many yelirs from the first to the last ? -Atw. 3468 years. 108. Adam lived 930 years ; Seth, 912 ; Enos, 905 Cainan, 910 ; Mahalaleel, 895 ; Jared, P62 ; Enoch, 365 Methuselah, 969 ; Lamcch, 777 ; Noali, 950 ; Shem, 600 Arphaxad, 438 ; Salah, 433 ; Heber, 464 ; Peleg, 239 Reu, 239 ; Serug, 230 ; Nahor, 148 ; Terah, 205 ; Abra- ham, 175 ; Isaac, 180 ; Jacob, 147. What is the sum of all theur ages ? Ans. 12073 years 13. The pupil should not be allowed to leave addition, iil 44 ▲DDITIOir. until he can, with great rapidity, continually add any of the nine digits to a given quantity ; thus, beginning with 9, to add 6, he should say :-— 9, 15, 21, 27, 33, &c., without hesitation, or further mention of the numbers. For instance, he should not be allowed to proceed thus : 9 and 6 are 15 ; 15 and 6 are 21 ; &c. ; nor even 9 and 6 are 15 ; and 6 are 21 ; &c. He should be able, ulti- mately, to add the following — i* iHk.-i '..ff •• 6688 V r. •:■'..-.■.-. • = -"■ -• •• 4766 ^ . ' ' ' .,■ ■•,■• , 0342 19786 in this manner : — 2, 8 ... 16 (the sum of the column ; of which 1 is to be carried, and 6 to be set down) ; 5, 10 ... 13; 4, 11 ... 17; 10, 14... 19. i* !;■ QUESTIONS TO BE ANSWERED BY THE PUPIL. <"; ' 1. To how many rules may all those of arithmetic ba reduced? [1]. *' s . , , 2. What is addition .? [3]. 3. What are the names of the quantities used in addi- tion? [3]. 4. What are the signs of addition, and equality ? [4]. 5. What is tne vinculum ; and what are its effects on additive quantities ? [5] . 6. What is the rule for addition ? [7]. ^ 7. What are the reasons for its different parts ? [9] . 8. Does this rule apply, at whatever side of the deci- mal point all, or any of the quantities to be added are found? rm. 9. How is addition proved ? [10]. >>. . . 10. What is the reason of thu proof? [10]. ^i.' <<. -U; .''. ■ U *' . .";*..' 4,.. <.«';■ i ' : ' ^ ^. 's y •* ■■ ' •UBTRACTIOir. 45 nj. ; ■vt .;i 1 ''-'fL I SIMPLE SUBTRACTION. 14. Simple subtraction is confined to abstract numbers, and applicate which consist of but one denomination. Subtraction enables us to take one number called the subtrahmdy from another called the minuend. If any- thing s left, it is called the excess ; in commer 'jial con- cerns, it is termed the remainder ; and in the mathema- tical sciences, the differeiue. 15. Subtraction is indicated by — , called the minus, or negative sisn. Thus 5 — ^==1) read five minus four equal to one, mdicates that if 4 is substracted from 5, unity 18 left. Quantities connected by the negative sign cannot be taken, indifferently, in any order ; because, for example, 5-^—4 is not the same as 4~-3* In the former case the positive quantity is the greater, and 1 (which means -f-l[4j) is left; in the latter, the negative quantity is the greater, and — 1, or one to be subtracted, stiu remains. To illustrate yet further the use and nature of the signs, let us suppose that we have five pounds, and owe four ; — the five pounds we have will be repre- sented by 5, and our debt by —-A ; taking the 4 from the 5, we shall have 1 pound ( + 1 ) remaining. Next let us suppose that vre' have only four pounds and owe five ; if we take the 5 from the 4 — that is, if we pay as far as we can — ^a debt of one pound, represented by — 1, will still remain; — consequently 5—4=1; but 4— 5=— 1. 16. A vinculum placed over a subtractive Quantity, or one having the negative sign prefixed, alters its value, unless we change all the signs but the first : — thus 5 — 3+2, and 5 — 3+2, are not the same thing; for 5—3+2=4 ; but 5 — 3+2 (3+2 being considered now as but one quantity) =0 ; for 3+2=5 ; — ^therefore — 3+2 is the same as 5 — 5, whkjk leaves nothing; or, in other words, it is equal to 0. K, however, we change all the signs, except the first, the valuo of the quantify is 4« #urrtiA€tfOir. i ^i/f=» not altered by the vinculum; — thus 6—3+2=4; and 5—3—2, also, is equal to 4. Again, 27- 4+7-8 -87. < Th« following exaoipb will show how the yinoulum affects numbers, according as we make it include an additive or a subtractivo quantity :— 48+7-3-8+7-2s*4d. "' ''^'*"""' " .' **'■" AUti — 5 — ffXT — o An . what la under the Tinenlnfli being .„ , *o-J-/— «J— O-i-/ — -iaa^J, additive, it is not neoeuarjr to >■"' " change any liltni. '** * ilfi I 7 '^i isu/D'V-..: ii .stmov? Vji\ii} n: y. After subtracting any denomination of i&e sub- 18. SUITRACriON. 47 trahcnd from the oorretpondioff part of the minuend, set down what is left, if any thing, in the place which belongs to the same dcaomination of the *' remaioder.*' VI. But if tlwro is nothing left, put down a cypher — provided any di^t of the *^ remaindeH' will be more dis- tant from the decimal point, and at the saipe side of it. 18. £xAWPi.B 1.— Subtract 427 from 792. --u t- '» ? T92 minuend. k'^i.j«fru; •>iin.--» ' %'[> : ■•» ■ 427 subtrahend. " -^ ' y^v. .^^ ::i^ i' tt 865 remainder, difference, or excess. « <^'f^>'' We cannot take 7 units from 2 units j but " borrowing," as it is called, one of the 9 tens in the minuend, and consider- ing it as tm units, we aUd it to the 2 units, aad t^en have 12 units ; tojkiog 7 from 12 units, 5 are left :-^we ^ut 6 in the units' place of th'' "remainder." We may ctraliend partM of ■ ? nd. 1 J. 3 the total remainder. They are to be set down under the deno- minations which produced them, since they belong to these denominations. Reason of VI. — Unless there is a significant figure at the same side of the decimal point, and more distant from it thar. the cypher, the latter — not being between the decimal point and a sigftaificant figure— will be useless [See. I. 28], and may therefore be omitted. 20. Proof of Subtraction. — Add together the re- mainder and subtrahend ; and the sum would be equal to the minuend. For, the remainder expresses by how much the subtrahend is smaller than the minuend ; addinff, therefore, the remainder to the subtrahend| should make it equal to the minuend ; thus ',* 8754 minuend. •'"***' ;■' ' ^ §??? subtrahend. 2915 difference Bum of difference and subtrahend, 8754sminuend. Or ; subtract the remainder from the minuend, and what is left should be equal to the subiarahend. For the remainder is the excess of the minuend above the subtrahend ; therefore, taking away this excess, should leave both equal ; thus 8684 minuend. "^ Pnoor : 8634 minuend. 7985 subtrahend. ^ - 649 remainder. 649 remainder. New remainder, 7985=:subtrahend. In practice, it is sufficient to set down the quantities once ; thus 8684 minuend. 7985 subtrahend. 649 remainder. Difference between remainder and minuend, 7985=8ubtrahend. 21. To Subtract y whin the quantities contain Deci- mals. — The rule just given is applicable, at whatever side of the decimal point all or any of the digits may be found; — ^this follows, as in addition [11], n-om the very nature of notation. It is necessary to put the decimal JKnt of the remainder under the decimal points of the minuend and subtrahend; otherwise the digits of the remainder will not, as they ought, have the same value as the digits from which they have been derived. !" i i ! I- '.i 5 III' .v;^i *' ^:!«;:! il '1 Bv. ?* .■ ■ iii'' ! W , .,,^,1 w :!';''i 1 .ll 'I "-i^ ^^k ,' .:^' l| ' fip SUBTRACTION. ExABiPLB.— Subtract 427 85 from 56304. - ^ - .. ... 663-04 427-85 136 -19 ^yU^rh $inoe the di^t to thie right of the deoknal point ia th,Q remainder, indLoatefl what is lefb after the oubtractioa of the tenths, it expresses so many tenths ; and sinqe the digit to the left of the decimal point indicates what remains after the subtraction of the units, it eiqpresses so man^ unitsj^ — all this is shown by the position of^ the decimal point. ■t<^.--> 22. It follows, frosu the principles of nolntloii [See. I. 402, ^1^^ however we may alter the decimal points of the minuend ivid sul^trahend, as long ^s they staad io the same yertic||.l column, the digits of the difference are not chfuaige4 ; ^us^ in the following examples, the 0ame digits are fouo^ in ^ the remainders : — k 4862 8547 486*2 864*7 48*62..y.r<-i 36-47 . ^4862 •8547 '0004862, •0003547 ■Tr- f "i/-^ »*■<.' " 816 7H.4-81.6 8-16.^r* 'OSIS •0000816 Wv f ,^. E3i;erpi8es in subtraction. *! ,„^- From Take (1) 1969 1408 7432 6711 (3) 9076 4567 18*t46 4377 (5) 8176 2907 (6) 76877 46761 prr- .L(t9'.iij i-'hi^ v; ;. ■' .tabm w J (»• (9) Profm 86167 67777 71284 TalM 61876 46699 48412 (10) (11) (12) 90007« 876704 745674 899934 ^7610 876780 -> '• -IT ■ ■ •!■■ .- a -J* (18) (14) Frop 6(7001 9788876 567674 Take 85690 4124767 476476 SfKf: ~~~~ ,b^ " i I III " i* I i ' >i(», (16) (W) (17) (18) 473676 63107^^ 876576 821799 8767016^ 240940 SUBTRACTION. 61 (19 (20) (21) (22) (23) (24) From 345676 234100 4367676 845673 70101076 67360000 Take 1799 990 266669 124799 37691734 81237777 (25) From 1970000 Take 1361111 (26) (27) (28) (29) 7010707 6734.'^001 1674561 14767674 3441216 47134777 1123640 7476909 (30) 4007070 8713916 (31) From 7045676 Take 3077097 (32) 37670070 26716645 (33) 70000000 9999999 (34) 70040500 56767767 (35) 50070007 41234016 (36) From 11000000 Take 9919919 (37) 8000001 2199077 (38) 8000800 377776 (39) 8000000 62358 (40) 4040053 220202 (41) From 86-73 Take 42-16 (42) 865-4 73-2 (43) 594-763 85-600 (44) 47-630 0-078 (45) 52-137 20-005 , - (46) From 0-00063 Take 0-00048 (47) 874-32 5-63705 (48) (49) 57-004 47632- 2-3 0-845003 (60) 400-327 0-0006 I 51. 745676-567456=178220. 62. 566789—75674=501115. 53. 941000—5007=936993. 64i 97001—50077=46924. 56. 76784—977=76767. 56. 56400—100=56300. 67. 700 Ans. £1269. 78. Paris is about 225 English miles distant from London ; Rome, 950 ; Madrid, 860 ; Vienna, 820 ; Copenhagen, 610 ; Geneva, 460 ; Moscow, 1660 ; Gib- raltar, 1160; and Constantinople, 1600. How much more distant is Constantinople than Paris ; Home than Madrid ; and Vienna than Copenhagen. And how much less distant is Geneva than Moscow ; and Paris than Madrid } An^. Constantinople is 1375 miles more dis- tant than Paris ; Home, 90 more than Madrid ; and Vienna, 210 more than Copenhagen. Geneva is 1200 mil«« less distant than Moscow; and Paris, 635 less than Madrid. 79. How much was the Jewish greater than the English mile ; allowing the former to have been 1*3817 miles English > Atis. 0*3817. 80. How much is the English greater than the Roman mile ; allowing the latter to have been 0*915719 of a mile English .^ Ans. 0*084281 81. What is the value of 6-34*15-4 .? Ans. 14 Ans. 33 Am. 52-94 82. Of 43+7-3— 14.? 83. Of 47*6-2+1-244-16— 34? 84. What is the difference b etween 16+ 13—6—81 + 62, and 15+13—6—81+62 ? Ans, 38. 23. Before leaving thif rule, the pupil should be abla MULTIPHCATION. 53 b 10901 IS. 1192, jntaining 680 gal- ining ? ) gallons, ig 16800 I tt) ; how • weight ? ; 8991 ft), ind, gives son ; how ns. 1276. 90; toC. e has hut ,s. iB1269. tant from ina, 820; 560 ; Gib- low much lome than how much 'aris than more dis- liid ; and a is 1200 635 less to take any of the nine digits continually from a given number, without stopping or hesitating. Thus, sub- tracting 7 from 94, he should say, 94, 87, 80, &c. ; and should proceed, for instance, with the following example 6376 4298 1078 in this manner : — 8, 16. . .8 (the difference, to be set down) ; 10, 17.. .7; 3, 3...0; 4, 5...1. QUESTIONS TO BE ANSWERED BY THE PUPIL. 1. What is subtraction .? [14], 2. What are the names of the terms used in subtrac- tion ? [14]. 3. What is the sign of subtraction ? [15]. 4. How is the vinculum used, with a subtractive quantity.? [16]. 5. What is the rule for subtraction } [17]. 6. What are the reasons of its different parts } [19], 7. Does it apply, when there are decimals } [21]. 8. How is subtraction proved, and why ? [20] . 9. Exemplify a brief mode of performing subtrac- tion ? [23]. SIMPLE MULTIPLICATION. 24. Simple multiplication is confined to abstract numbers, and applicate which contain but one denomi- nation. Multiplication enables us to add a quantity, called the multiplicand^ a number of times indicated by the mtUti- flier. The multiplicand, therefore, is the number mul- tiplied ; the multiplier is that hy which we multiply ; the result of the multiplication is called the product. It follows, that what, in addition, would be called an " addend," in multiplication, is termed the " multipli- cand ; " and what, in the former, would be called the " sum," in the latter, is designated the " product." The quantities which, when mvdtiplied together, give the D ■■rl I ^11 m m 54 MULTIPLICATION. product, are called also factors, and, when they are integers, submultiples. There may be more than two factors ; in that case, the multiplicand, multiplier, or both, will consist of more than one of them. Thus, if 5, 6, and 7, be the factors, either 5 times 6 may be con- sidered as the multiplicand, and 7 as the multiplier — or 5 as the multiplicand, and 6 times 7 as the multiplier. 25. Quantities not formed by the continued addition of any number, but unity — that is, which are not the products of any two numbers, unless unity is taken as one of them — are called prime numbers : all others are termed composite. Thus 3 and 5 are prime, but 9 and 14 are composite numbers ; because, only threes multiplied by orte, will produce " three," and only^re, multiplied by one, will produce " five," — but, three multiplied by three will produce " nine," and seven mul-i tiplied by two will produce " fourteen." 26. Any quantity contained in another, some number of times, expressed by an integer — or, in other words, that can be subtracted from it without leaving a re- mainder — is said to be a measure, or aliquot part of that other. Thus 5 is a measure of 15, because it is contained in it three times exactly — or can be sub- tracted from it a number of times, expressed by 3, an integer, without leaving a remainder ; but 5 is not a measure of 14, because, taking it as often as possible from 14, 4 will still be left;— thus, 15—5=10, 10—5= 5, 5 — 5=0, but 14 — 5=9, and 9 — 5=4. Measure, Bubmultiple, and aliquot part, are synonymous. 27. The common measure of two or more quantities is a number that will measure each of them : it is a measure common to them. Numbers which have no common measure but unity, are said to be prime to eacJi other ; all others are composite to each other. Thus 7 and 5 are prime to each other, for unity alone will measure both ; 9 and 12 are composite to each other, because 3 will measure either. It is evident that two priTM numbers must be prime to each other ; thus 3 and 7 ; for 3 cannot measure seven, nor 7 three, and — except unity — there is no other number that will mea- sure either of them. Tl yet are T each yet 2^ num MULTIPLICATION. 55 they are than two ] tiplier, or j ?hus, if 5, i y be con- I iplier — or 1 iltiplier. ; i addition 3 not the I taken as others are le, but 9 )nly threej I only five, j ■but, threi seven mul- (? le number her words, ring a re- t part of |j cause it is J be sub- I 1 by 3, an is not a s possible 10—5= Measure, quantities : it is a have no Irne to eadi Thus 7 ilone will |ch other, that two thus 3 56, and — Iwill mea- Two numbers may be composite to each other, and yet one of them may be a prime number ; thus 5 and 25 are both measured by 5, still the former is prime. Two numbers may be composite^ and yet prime to each other ; thus 9 and 14 are both composite numbers, yet they have no common measure but unity. 28. The greatest common measure of two or more numbers, is the greatest number which is their common measure ; thus 30 and 60 are measured by 5, 10, 15, and 30 ; therefore each of these is their common mea- sure ; — but 30 is their greatest common measure. When a product is formed by factors which are integers, it is measured by each of them. 29. One number is the multiple of another, if it contain the latter a number of times expressed by an integer. Thus 27 is a multiple of 9, because it con- tains it a number of times expressed by 3, an integer. Any quantity is the multiple of its measure, and the measure of its multiple. 30. The common multiple of two or more quantities, is a number that is the multiple of each, by an integer ; — thus 40 is the common multiple of 8 and 5 ; since it is a multiple of 8 by 5, an integer, and of 5 by 8, an integer. Tho least common multiple of two or more quantities, is the least number which is their common multiple ;— thus 30 is a cmnmon multiple of 3 and 5; but 15 is their least common multiple ; for no number smaller than 15 contains each of them exactly. 31. The equimultiples of two or more numbers, are their products, when multiplied by the same number ; — thus 27, 12. and 18, are equimultiples of 9, 4, and 6 ; because, multiplying 9 by three, gives 27, multiplying 4 by three, gives 12, and multiplying 6 by three, gives 18. 32. Multiplication greatly abbreviates the process of addition ; — ^for example, to add 68965 to itself 7000 timos by " addition," \^ould be a work of great labour, and con- sume much time ; but by " multiplication," as we shall find presently, it can be done with ease, in less than a minute. 33. At first it may seem inaccurate, to have stated [2] that multiplication is a species of aadition ; sine"; we can know the product of two quantities without having MULTIPLICATION. ill ■" = recourse to that rule, if they are found in the multipli- cation table. But it must not be forgotten that the mul- ttiplioation table is actually the result of additions, long since made ; without its assistance, to multiply so simple a number as 4 by so small a one as five, we should be obliged to proceed as follows, 20 perfoaTning the addition, as with any other addends. The multiplication table is due to Pythagoras, a cele- brated Greek philosopher, who was born 590 years before Christ. 34. We express multiplication by X ; thus 5X7= 35, means that 5 multiplied by 7 are equal to 35, or that the product of 5 and 7, or of 5 hy 7, is equal to 35. When a qiantity under the vinculum is to be multi- plied by any number, each of its parts must be multi- plied — for, to multiply the whole, wo must multiply each of its parts ;— -thus, 3X 7+8 — 3=3 X 7 + 3 X 8— 3X3; and 4+5X8+3—6, means that each of the terms under the latter vinculum, is to be multiplied by each of those under the former. 35. Quantities connected by the sign of multiplication may be read in any order; thus 5X6=6x5* This will be evident from the following illustration, by which it appears that the very same number may be considered either as 5X6, or 6x5, according to the view we take of it :— • 6 are 6X 3< shoi pro( a la] with! tlicnl Tak[ for and rapic if Quantities connected by the sign of multiplication! w MULTIPLICATION. 57 multipli- the mul- ons, long 30 simple hould be inds. bS, a cele- 90 years s 5X7= to 35, or lal to 35. be multi- 36 multi- multiply t-3X8— of the plied by fplioation This )y which )nsidered we take I IlicatioDi are multiplied if we multiply one of the facto:is ; thus 6X7X3 multiplied by 4=6X7 multiplied by 3X4. 36. To prepare him for multiplication; tSe pupil should be made, on seeing any two digits, to name their product, without uientioning the digits themselves. Thus, a huge number having been set down, he may begin with the product of the first and second digits ; and then proceed with that of the second and third, &o. Taking 587034925867 for an example, he should say :— 40 (the product of 5 and 8) ; 56 (the product of 8 and 7) ; 42 ; 18 ; &c., as rapidly as he could read 5, 8, 7, &c. To Multiply Numbers. 37. When neither multiplicand, nor multiplier ex- ceeds 12 — KuLE. — Find the product of the given numbers by the multiplication table, page 1. The pupil should be perfectly familiar with this table. Example. — What is the product of 5 and 7 1 The mul- tiplication table shows that 5x7=35, (5 times 7 are 35). 38. This rule is applicable, whatever may be the relative values of the multiplicand and multiplier ; that is [Sec. I. 18 and 40], whatever may be the kind of units expressed — provided their absolute values do not exceed 12. Thus, for instance, 1200X90, would come under it, as well as 12X9 ; also '0009 X 0*8, as well as 9X8. We shall reserve what is to be said of the man- agement of cyphers, and decimals for the next rule ; it will be equally true, however, in all cases of multiplica- tion. 39. When the multiplicand does, but the multiplier does not exceed 12 — Rule. — ^I. Place the multiplier under that denomi- nation of the multiplicand to which it belongs. II. Put a line under the multiplier, to separate it from the product. - " ■ •' • III. Multiply each denomination of the multiplicand by the multiplier — ^beginning at the right hand side. 68 MULTIPLICATIOIf. IV. If the product of the multiplier and any digit of the multiplicand is leus than ten, set it down under that digit ; but if it be greater, for every ten it contains carry one to the next product, and put down only what remains, after deducting the tens ; if nothing remains, put down a cypher. y. Set down the last product in full. 40. Example. 1.— What is the product of 897351x4? 897351 multiplicand. 4 multiplier. 3589404 product. 4 times one unit are 4 units ; since 4 is less than ten, it gives nothing to be " carried," we, therefore, set it down in the units' place of the product. 4 times 5 are twenty (tens) ; whicli are equal to 2 tens of tens, or hundreds to be carried, and no units of tens to be set down in the tens' place of the product — in which, therefore, we put a cypncr. 4 times 3 are 12 (hundreds), which, with the 2 hunareds to be carried from the tens, make 14 hundreds; these are equal to one thousand to be carried, and 4 to be set down in the thousands' place of the product. 4 times 7 are 28 (thou- sandH^, and 1 thousand to be carried, are 29 thousands ; or 2 to DC carried to the next product, and 9 to be set down 4 times 9 are 36, and 2 are 38 ; or 3 to be carrried, and 8 to be set down. 4 times 8 re 32, and 3 to be carried are 35 ; which is to be set down, since there is nothing in the next denomination of the multiplicand. Example 2.— Multiply 80073 by 2. 80073 2 160146 Twice 3 units are 6 units ; G being less than ten, gives nothing to be carried, hence we put it down in the units' place of the quotient. Twice 7 tens are 14 tens ; or 1 hundred to be carried, and 4 tens to be set down. As there are no hundreds in the multiplicand, we can have none in the pro- duct, except that which is derived from the multiplication of the tens ; we accordingly put the 1, to be carried, in the hundreds' place of the product. Since there are no thou- sands in the multiplicand, nor any to be carried, we put a cypher in that denomination of the product, to keep any significant figures that follow, in their proper places. del thl to MULTIPLICATION. 59 any dint )wn un£}r it contains only what : remains. 41. Reason or I. — The multiplier is to be placed under that denomination uf tlie multiplicand to which it belongs; since there ia then no doubt of its value. Sometimes it is necessary to add cyphers in putting down the maltiplior ; thus* Example 1.— 478 multiplied by 2 hundred— 478 multiplicand. 200 multiplier. 51x4? Example 2. — 6&9 multiplied by 8 ten-thousandths^ 680* multiplicand. 0-0003 multiplier. ihan ten, it it down in ity (tens) ; be carried, 9' place of ypner. 4 Ireds to be are equal wn in the 28 (thou- isands ; or set down . and 8 to )d are 35 ; \ the next ten, gives he unittj' . hundred re are no the pro- iplication }d, in the no thou- we put a ceep any s. Hkahon or II. — It is similar to that given for the separating line in subtraction [19]. Reahon or III. — When the multiplicancT exceeds a certain amount, the powers of the mind are too limited to allow us to multiply it at once ; we therefore multiply its parts, in suc- cession, and add the results as we proceed. It is clear that the sum of the products of the parts by the multiplier, is equal to the product of the sum of the parts by the same multi- plier :— thus, 587 X8 is evidently equal to 500x8-j-30x8-f 7 X8 For multiplying all the parts, is multiplying the whole ; since the whole is equal to the sum of all its parts. We begin at the right hand side to avoid the necessity of afterwards adding together the subordinate products. Thus, taking the example given above ; were we to begin at the left hand, the process would be — 897351 4 ■ 8200000:^800000X4 860000= 90000X4 28000= 7000X4 1200= 800X4 200= 50X4 4= 1X4 3589404=8um of products. Reason or IV. — It is the same as that of the fourth part of the rule for addition [9] ; the product of the multiplier and any denomination of the multiplicand, being equivalent to the uum of a column in addition. It is easy to change the given example to an exercise in addition ; for 897861 X4, is the same iliijjL^ as 897351 897351 897351 897351 8589404 I ■'- 60 MULTIPLICATION. h !., !^ i&\ REAtoir or V.—- It follows, that the last product is to be set down in full; for the tens it contains will not be increased i they may., hercforo. be set down at once. Thlfe ► .^0 includes all cases in whi^h the absolute value ot iic dibits in the multiplier does not exceed 12. Their relative value is not material : for it is as easy to multiply by 2 thousands as by 2 umts. 42. To prove multiplication, when the mnltiplier does not exceed 12. Multiply the multiplicand by the mul- tiplier, minus one ; and add the multiplicand to the pro- duct. The sum should be the same as the product of the multiplicand and multiplier. £xAMPLE. — MuUinly 6432 by 7, and prove the rwult. 6432 multiplicand. 6ass7 (the multiplier)— 1 6432 38592 multiplicand X 6. 7(=6+l)_6432 multiplicandxl. 45024 a 45024 multiplicand multiplied by 6^ ^ \z=:7. We have multiplied by 6, and bv 1, and added the v'esults ; but six times the multiplicand, plus once the multiplicand, is equal to seven times the multiplicand. What we -^^tain from the two processes should be the same, for we lu»ve merely used two methods of doing one thing. EXERCISES FOR TH& PUPIL. Multiply By (1) 76762 2 (6) 768452 6 (0) 866342 11 (2) 67466 2 (6) 466769 7 (8) 78976 6 (7) 854709 8 (4) 57346 5 Multiply By (8N 45678? f • Multiply (10) 738679 12 (11) 476387f 11 (12) 8J12976S 12 or th(| call not pre be MULTIPLICATION. 64 to be set icreoaed t absolute I exceed it ia 08 lier does ;he mul- the pro- »duct of ult. 1=7. •results ; )Ucand, B -^htain ve hikve |346 6 8f f 2 43. To MxUtiphj when the Quantities contain Cyphers^ or Decimals. — The rules already given are applicable: those which follow are consequences of them. When there are cyphers at the end of the multipli- cand (cyphers in the middle of it, have been already noticed [40])— lluLE. — Multiply as if there were none, and add to the product as many cyphers as have been neglected. For The greater the quantity multiplied, the greater ought to be the product. ExAMPLK.— Multiply 56000 by 4. 56000 4 224000 4 times 6 units in the fourth place from the decimal point, are evidently 24 units in the same place ; — that is, 2 in the fifth place, to be carried, and 4 in the fourth, to be set down. That we may leave no doubt of the 4 being in the fourth place of the product, we put three cyphers to the right Land. 4 times 5 are 20, and the 2 to be carried, make 22. 44. If the multiplier contains cyphers — » Rule. — Multiply as if there were none, and add to the product as many cyphers as have been neglected. The greater the multiplier, the greater the number of times the multiplicand is added to itself; and, therefore, the greater the product. Example.— Multiply 567 by 200. 567 200 113400 From what we have said [35], it follows that 200x7 is the same as 7x200 ; but 7 times 2 hundred are 14 hundred ; and, consequently, 200 times 7 are 14 hundred ; — that is, 1 in the fourth place, to be carried, and 4 in the third, to be set down. We add two cyphers, to show that the 4 is in the third place. 45. If both multiplicand and multiplier contain cyphers — ... Rule. — ^Multiply as if there were none in either, and add to the product as many cyphers as are found in both. d2 62 MULTIPLICATION. ii!,i Hi: l^-'l X ", 1 ■1 !i . i Each of the quantities to be multiplied adds cyphers to the j>roduct [48 and 44]. EXAMPLE.- -Multiply 46000 by 800. 46000 800 368000C0 8 times 6 thousand would be 48 thousand ; but 8 hundred times six thousand oueht to produce a number 100 times greater — or 48 hundred thousand ; — that is, 4 in the seventh place from the decimal point, to be carried, and 8 in the sixth place, to be set down. But, 5 cyphers are required, to keep the 8 in the sixth place. Auer ascertaining the position of the first digit in the product — from what the pupil already knows — there can be no difficulty with the other digits. 46. When there are decimal places in the multipli- cand — Rule. — -Multiply as if there were none, and remove the product (by means of the decimal point) so many places to the right as there have been decimals neglected. The smaller the quantity multiplied, the less the product. Example. — Multiply 5*67 by 4. 5-67 4 22-68 4 times 7 hundredths are 28 hundreths ; — or 2 tenths, to be carried, and 8 hundredths — or 8 in the second place, to the right of the decimal point, to be set down. 4 times 6 tenths are 24 tenths, which, with the 2 tenths to be carried, make 26 tenths ; — or 2 units to be carried, and 6 tenths to be set down. To show that the 6 represents tenths, we put the decinal point to the left of it. 4 times 5 units are 20 imits, which, with the 2 to be carried, make 22 units. 47. When there are decimals in the multiplier — E.ULE. — Multiply as if there were none, and remove the product so many places to the right as there are decimals in the multiplier. The smaller the quantity by which we multiply, the lesf must be the result. MULTIPLICATION. Example.— Multiply 563 by 07 563 007 63 59-41 3 multiplied by 7 hundredths, is the same [351 as 7 hun- dredths multiplied by 3 ; which is equal to 21 hundredths ; — or 2 tenths to be carried, and 1 hundredth— or 1 in the second place to the right of the decimal point, to be set down. Of course the 4, derived from the next product, must be one place from the decimal poiMi, 6ie. 48. When there are decimals in both multiplicand and multiplier — BuLE. — Multiply as if there were none, and move the product so many places to the right as there are decimals in both. In this case the product is diminished, by the smallness of both multiplicand and multiplier. Example 1.— Multiply 56-3 by -08. 56-3 •08 4-504 8 times 3 tenths are 2*4 [46] ; consequently^ a quantity one hundred times less than o— or -08, multiplied by three- tenths, will give a quantity one hundred times less than 24 — or -024 ; that is, 4 m the third place from the decimal point, to be set down, and 2 in the second place, to be carried. Example 2.— Multiply 5-63 by 000005. 5-63 000005 00002815 49. When there are decimals in the multiplicand, and cyphers in the multiplier ; or the contrary — Rule. — Multiply as if there were neither cyphers nor decimals ; then, if the decimals exceed the cyphers, move the product so many places to the right as will be equal to the excess ; but if the cyphers exceed the deci- mals, move it so many places to the Up as will be equal to the excess. The cyphers move the product to the left, the decimals to the right ; the effect of both together, therefore, will be equal to the difference of their separate effects. 64 MULTIPLICATION. K " Example 1.— Multiply 4600 by 06. 4600 006 2 cyphers and 2 decimals } excess = 0. 276 Example 2.— Multiply 47-63 by 300. 47-63 300 2 decimals and 2 cyphers ', excess =0. 14289 Example 3.— -Multiply 85-2 by 7000. 85-2 7000 1 decimal and 3 cyphers j excefls=2 cyphers. 596400 Example 4.— Multiply 578-36 by 20. 578-36 20 2 decimals and 1 cypher ] excesses 1 decimal. 116672 Multiply By exercises (13) 48960 6 (17) 7468 80 (21) 748660 800 FOR THE (14) 75460 9 (18) 770967 900 PUPIL (15) 678000 8 (16) 67800 6 Multiply By (19) 147005 4000 ^20) 66976748 80000 Multiply By (22) 634900 80000 (23) 60000 800 (27) *21375 6 (24) 86000 6000 Multiply By (26) 62786 2 (26) 8-7563 4 (28) 0-0007 8 1' • . Mull By Mul^ B> MULTIPLICATION. = 0. 0. uyphera. ecimal. (16) 57800 6 l20) 6976748 30000 (24) 86000 6000 (28) 0-0007 Multiply By Multiply By (29) 66341 0-0003 (33) 876-432 0-04 (30) 86637 0005 (84) 78000 0-3 (31) 72168 0-0007 (36) 61-721 6000* (32) 2176-38 06 (36) 82 0-00007 -00224 In the last example vre are obliged to add cyphers to the product, to make up the required number of decimal places. 50. When both multiplicand and multiplier exceed 12— Rule. — ^I. Place the digits of the multiplier under those denominations of the multiplicand to which they belong. II. Put a line under the multiplier, to separate it from the product. III. Multiply the multiplicand, and each part of the multiplier (by the preceding rule [39]), beginning with the digit at the right hand, and taking care to move the product of the multiplicand and each successive digit of tli3 multiplier, so i»i;iiy places more to the left, than the preceding product, as the digit of the multiplier fthich produces it is more to the left than the signifi- cant iiguie by which we have last multiplied. J\. Add together all the products ; and their sum will be the product of the multiplicand and multiplier. 51. EXAMPLE.- -Multiply 5634 by 8073. 5634 8073 16902=product by 3. 39438 —product by 70. 45072 =rproduct by 8000. 45483282=product by 8073. The product of the multiplicend by 3, requires no ot^/^^ -\ ; nation. 66 MULTIPLICATION. Mii 'ikil 7 tens times 4, or [35] 4 times 7 tens are 28 tens : — 2 hun- dreds, to be carried, and 8 tens (8 in the second place from the decimal point) to be set down, &c. 8000 times 4, or 4 times 8000, are 32 thousand : — or 3 tens of thousands to be carried, and 2 thousands (2 in the fourth place) to be set down, &o. It is unnecessary to add cyphers, to show the values of the first digits of the different products ; as they are sufficiently indicated by the digits above. The products by 3, by 70, and by 8000, are added together in the ordinary way. 62. Reasons of I. and II. — ^They are the same as those given for corresponding parts of the preceding rule [41]. Rgasojv of III. — We are obliged to multiply successively by the parts of the multiplier ; since we cannot multiply by the whole at once. Reasopt of IV. — The sum of the products of the multipli- cand by the parts of the multiplier, is evidently equal to the product of the multiplicand by the whole multiplier ; for, in the example just given, 5634 X 8073 = 5634 X 8000 -f- 70 -f 3= [34] 5634X8000-1-5634x70-1-5634x3. Besides [35], we may consider the multiplicand as multiplier, and the multiplier as multiplicand; then, observing the rule would be the same thing as multiplying the new multiplier into the different parts of the new multiplicand ; which, we have already seen [41], is the same as multiplying the whole multiplicand by the multiplier. The example, just given, would become 8073X5634. 8073 new multiplicand 5634 new multiplier. We are to multiply 3, the first digit of the multiplicand, by 6634, the multiplier; then to multiply 7 (tens), the second digit of the multiplicand, by the multiplier ; &c. When the multiplier was small, we could add the different products as we proceeded; but we now require a separate addition, — which, however, does not affect the nature, nor the rfeasons of the process. 53. To prove multiplication, when the multiplier ex- ceeds 12 — Rule. — Multiply the multiplier by the multiplicand; and the product ought to be the same as that of the multipl'cand by the multiplier [35] . It is evident, that we could not avail ourselves of this mode of proof, in the last rule [42] ; as it would have supposed the pupil to be then able to multiply by a quantity greater than 12 "cai R the ders, left inir t of'th Ex 3785. Takin The niultip have b or orai acted these i EXAl T T Taking The be con£ Exab : — 2 hun- lace from les 4, or 4 mds to bo to be set show the ; as they 3 products 3 ordinary ; as those 41]. iccessively mltiply by 5 multipli- [ual to the er ; for, ia F70HF3= i], we may altiplier as the same e different ready seen plicand by Id become icand, by le second When the oducts as which, of the tns plier ex- plicand ; ifc of the lent, that m the [pupil to than 12 MULTIPUCATION. 67 54. We may prove multiplication by what is called " casting out the nines." Rule. — Oast the nines from the sum of the digits of the multiplicand and multiplier ; multiply the remain- ders, and cast the nines from the product : — what is now left should be the same as what is obcained, by cast- ing the nines, out of the sum of the digits of the product ofthe multiplicand by the multiplier. Example 1.— Let the quantities multiplied be 9426 and 3785. Taking the nines from 9426, we get 3 as remainder. And from 3785, we get 5. 47130 75408 65982 28278 3x5=15, from which 9 being taken, 6 are ieffc. Taking the nines from 35677410, 6 are left. The remainders being equal, we are to presume the multiplication is correct. The same result, however, would have been obtained, even if we had misplaced digitd, added or omitted cyphers, or fallen into errors which had counter- acted each of her : — with ordinary care, however, none of these is likely to occur. Example 2. — Let the numbers be 76542 and 8436. Taking the nines from 76545:, the remainder is 6. Taking them from 8436, it is 3. 459252 229626 6x3=18, the 306 168 remainder from which is 0. 612336 Taking the nines from 645708312 also, the remainder is 0. The remainders being the same, the multiplication may be considered right. Example 3. — Let the numbers be 463 and 54. From 463, the remainder is 4. j From 54, it is 0. 1852 4x0=0 from which the renuuiider is 0. 2315 From 25002 the remainder is 0. 68 MULTIPLICATION. jll: The remainder being in each caee 0, wo are to suppoM that the multiplication is correctly performed. This proof applies whatever be the position of the decimal point in either of the given numbers. 55. To understand this rule, it must be known that ** a number, from which 9 is taken as often as possible, will leave the same remainder as will be obtained if 9 be taken as often as possible from the sum of its digits." Since the pupil is not supposed, as yet, to have learned division^ he cannot use that rule for the purpose of casting out the nines; — nevertheless, he can easily effect this object. Let the given number be 563. The sum of its digits is 5+6-1-3, while the number itself is 5004-60-f.3. First, to take 9 as often as possible from the sum of its digits. 5 and 6 are 11 ; from which, 9 being taken, 2 are left. 2 and 3 are 5, which, not containing 9, is to be set down as the remainder. Nest, to take 9 as often as possible from the number itself 5G3^=:500-|-60-f3=5xl00+6xl04-3=5x9^i-|-6x 94.14.3,= (if we remove the vinculum [34]), 6x99-|-5-f- 6x9-f-6-|-3. But any number of nines, will be found to be the product of the same number of ones by 9 : — thus 999s=: 111x9; 99=11x9; and 9=1x9. Hen ce 5x99 expresses a certain number of nines — being 5x11x9 ; it may, there- fore, be cast out ; and for a similar reason, 6x9; after which, there will then be left 5-f-6-f-3 — from which the nines are still to be rejected ; but, as this is the sum of the digits, we must, in casting the nines out of it, obtain the same remain- der as before. Conseq^uently "we get the same remainder whether we cast the nmes out of the number itself, or out of the sum of its digits." Neither the above, nor the following reasoning can offer any difficulty to the pupil who has made himself as familiar with the use of the signs as he ought : — they will both, on the contrary, serve to show how much pimplicity, is derived from the use of characters express- ing, not only quantities, but processes ; for, by means of such characters, a long series- of argumentation may be seen, as it were, at a single glance. 56.^*^ Casting the nines from the factors, multiplying th« resulting remainders, and casting the nines from this product, ^%^ MULTIPLICATION. 69 ) suppose n of the own that possible, ined if 9 s digits." e learned irpose of m easily I digits is mm of its cen, 2 are to be set mber itself ►T1+6X lX99-f^H- iind to be hus 999= expresses ay, there- ;er which, nines are digits, we \e reraain- 'emainder If, or out img can himself )ught : — )w much I express- means lion may lying th« product, will leate the same remainder, as if the nines were oast from the product of the factors,'^ — provided the multiplication has been rightly performed. To show this, set down the quantities, and take away th« nines, as before. Let the factors be 573x464. Casting the nines from 5-f-7-f-3 (which we have just seen is the same as castmg the nines from 573), we obtain 6 as remainder. Casting the nines from 4-|-6-f-4, we get 5 as remainder. Multiplying 6 and 5 we o btain 30 as product j which, being eaual to 3xl0=r3x~9+l=:3x9-|-3, will, when the nines are taken away, ^ive 3 as remainder. We can show that 3 will be the remainder, also, if we cast the nines from the product of the factors ; — which is effected by setting down this product ; and taking, in suc- cession, quantities that are equal to it- -as follows, 573x464 (the pro duc t of the factors)^ 5x100+7x10+3 X 4xl00-f.6xl0+4=s 5 x99-hl-f-7x9-f-l-f-3 X 4 x 99+1+6 x9+l-h4= a 5x99+5+7x9+7+3 X 4x99+4+6x9+6+4. 5x99, as we have seen [55], expresses a number of nines ; it will continue to do so, when multiplied by all the quan- tities under the second vinculum, and is, therefore, to be cast out; and, for the same reason, 7x9. 4x99 expresses a number of nines ; it will continue to do so when multiplied by the quantities under the first vinculum, and is, therefore, to be east out ; and, for the same reason, 6x9. There will then be left, only 5+7+3x4+6+4, — from which the nines aiire still to be cast out, the remainders to be multiplied together, and the nines to be cast from their product ; — but we have done all this already, and obtained 3, as the remainder. XXERGISSS FOR THE PUPIL. Multiply ^y Products Multiply By Produoti (87) (38) (89) (40) 766 782 997 767 766 456 846 847 (41) 667 789 (42) 466 791 (48) 767 789 (44) 745 741 70 MULTIPLICATION. 57. If there are cyphers, or decimals in the multipli- cand, multiplier, or both ; the same rules apply as when the multiplier does not exceed 12 [43, &c.] . II , fl 1 ■ 11 (1) 4600 67 262200 (2) 2784 620 EXAMPLES. (8) (4) 82-68 7856 26- 0-32 (6) (6) 87-96 482000 220- 0-37 1726080 849-68 2513-92 19351*2 178340 Contractions in Multiplication. 58. When it is not necessary to have as many deci- mal places in the product, as are in both multiplicand and multiplier — KuLE. — ^Reverse the multiplier, putting its units' 'placA under the jlact of that denomination in the multipli- cand, which is the lowest of the required product. Multiply by each digit of the multiplier, beginning with the denomination over it in the multiplicand ; but adding what would have been obtained, on multiplying the preceding digit of the multiplicand — unity, if the number obtained would be between 5 and 15 ; 2, if between 15 and 25 ; 3, if between 25 and 35 ; &o. Let the lowest denominations of the products, arising from the different digits of the multiplicand, stand in the same vertical column. Add up all the products for the total product ; from which cut off the required number of decimal places. 59. ExAMPLi 1.— Multiply 5-6784 by 9-7324, so as to have four decimals in the product. Short Method. 66784 42379 511056 39749 1703 113 22 55-2643 Ordinary Method. 5-6784 9-7324 22 113 1703 89748 511056 55-26446016 7136 568 52 8 MULTIPLICATION. n 9 in the multiplier, expresses units; it is therefore pu6 ander the/ourf A decimal place of the multiplioand — that being the place of the lowest decimal required m the product. In multiplying by each succeeding digit of the multiplier, we neglect an additional digit of the multiplicand ; because, as the multiplier decreases, the number multiplied must in- crease — to keep the lowest denomination of the different pro- ducts, the same as the loweab denomination required in the total product. In the example given, 7 (the second digit of the multiplier) multiplied by 8 (the second digit of the mul- tiplicand), will evidently produce the same denomination as 9 (one denomination higher than the 7), multiplied by 4 (one denomination lower than the 8). Were we to multiply the lowest denomination of the multiplicand by 7, we should get [46] a result in the Jifth place to the right of the decimal point ; which is a denomination supposed to be, in the present in- stance too inconsiderable for notice — since we are to have only Jo 'T decimals in the product. But we add unity for every ten that would arise, from the multipl cation of an addi- tional digit of the multiplicand ; since every such ten consti- tutes «ne, in the lowest denomination of the required product. When the multiplication of an additional digit of the multi- plicand would give more than 5, and less than 15 ; it is nearer to the truth, to suppose we have 10, than either (), or 20 ; and therefore it is more correct to add 1, than either 0, or 2. When it would give more than 15, and less than 2f , it is nearer to the truth to suppose we have 20, than either 10, or 80 ; and, therefore it is more correct to add 2, than 1, or 3 ; &c. We may consider 5 either as 0, or 10 ; 15 either as 10, or 20 ; &o. On inspecting the results obtained by the abridged, and ordinary methods, the difference is perceived to be inconsiderable. When greater accuracy is desired, we should proceed, as if we intended to have more decimals in the product, and afterwards reject those which are unnecessary. Example 2.— Multiply 8-76532 by -5764, so as to hav^ 3 decimal places. 8-76532 4675 4383 613 n 6-05} 72 MULTIPLICATION. There are no units in the multiplier: but, as the rule directs, we put its units^ place under the tnird decimal place of the multiplicand. In muUipWing by 4, since there is no digit over it in the multiplicand we merely set down what would have resulted from multiplying the preceding deno* mination of the multiplicand. Example 3.— Multiply -4737 by -6731 so as to have 6 ilecimtJ places in the pioductL •47370 1376 284220 33159 1421 47 •318847 tVe have put ^ he units' place of the multiplier under the siith decimal place of the multiplicand, adding a cypher, or su^iposing it to be added. Example 4.— Multiply 84-6732 by ^0056, sc as to have four decimal places. 84-6732 4234 508 •4742 Etjlmple 5.— Multiply ^23257 by 243, so as to have four decimal places. 23257 342 465 93 ^ 7 •0565 We are obliged to place a c^her in the product, to make up the required number of decimals. 60. To multiply by a Composite Nuniber — Rule. — Multiply, successively, by its factora. MULTIPLICATION. 73 Example.— Multiply 732 by 96. 96»«8yl2- therefore 732x96«732x8xl2. [35]. 732 8 5856, product by 8. 12 70272, product by 8 X 12, or 96. If we multiply bv 8 only, we multiply by ft quantity 12 times too small ; and, therefore, the product will be 12 titnea less than it should. We rectify this, by making the product 12 times greater — that is, we multiply it I ; 72. 61. When the multiplier is not fj^acti/ a Composite Number — Rule. — Multiply by the factors of the nearest com- r ite ; and add to, or subtract from the last product, 80 many times the multiplicand, as the assumed compo- site is less or greater than the given multiplier Example 1.— Multiply 927 by 87. 87=rr7xl2-f3; therefore 927 x 87=927 X 7x li+i— 927x7x12+927x3. [34]. 927 7 6489. 12 :927 X 7. 77868=«927x7xl2. 2781=927x3. 80649=927 X 7 x 12+927 X 3, or 927 x 87. If we multiply only by 84 (7 X 12), we take the number to be multiplied 3 times less than we ought ; this is rectified, by adding 3 times the multiplicand. Example 2.— Multiply 4 32 by 79. 79=81-2=9 x 9-2; therefore 432 X 79 =432 x 9x9-2=432 x 9 x9-432x 2. 432 I 9 3888: 9 34;»92= &64: e432x9. :432x9x9. c432x2. 34128»432 x 9 x 9-432 x 2, or 432 X 79. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 1.1 11.25 Li|28 |2^ ■JO '*^^ HMHi ■tt U& 12.2 Sf Ug 110 u i.4 il.6 6" Fhotographic Sdmoes Corporadon 4^ '^ V 23 WBT MAIN STMIT WnSTIR,N.Y. 14SM (n*)t7a-4SM \ I i\ 74 MULTIPLICATION. :!■• :l^ t; 'ill rii'l , 111 i;- li' ■n;':- In mnltiplying by 81, the composite number, we have taken the number to be multiplied twice too often ; but the inaccu- racy is rectified by subtracting twice the multiplicand from the product. 62. This method is particularly convenient, when the multiplier consists of nines. To Multiply by any Number of Nines, — KuLE. — ^Kemove the decimal point of the multipli- cand 80 many places to the right (by adding cyphers if necessary) as there are mines in the multiplier ; and subtract the multiplicand from the result. ExAMPLE.—Multiply 7347 by 999. 7347 X 999=7347000-7347=7339653. We, in such a case, merely multiply by the next higher conTcnient composite number, and subtract the multiplicand so many times as we have taken it too often; thus, in the example just given — 7847x999=7347x1000-1=7847000-^7347=7339658. 63. We may sometimes abridge multiplication by considering a part or parts of the multiplier as pro- duced by multiplication of one or more other parts. Example.— Multiply 57839268 by 62421648. The mul- tiplier may be divided as follows : — 6, 24, 216, and 48. 6=6 24=6x4 216=24x9 48=24x2 57839268, multiplicand 62421648 , multiplier. 347035608 : : : product by 6 (60000000). 1388142432 : : product by 24 (2400000). 12493281888 : product by 216 (21600). 2776284864 product by 48. 3610422427673664 product by 62421648. The product by 6 when multiplied by 4 will give the pro« duct by 24 ; the product by 24, multiplied by 9, will give the product by 216 — and, multiplied by 2, the product by 48. 64. There can be no difficulty in finding the places of the first digits of the different products. For when there are neither cyphers nor decimals in the multiplicand— ^d di^fii^g multiplication, we may suppose that there are ^eitl^er t/^^ {^p,] — ^the lowest denomination of each pro- duct mul^ by tens I will eachi the Wh( necei givei MULTIPLICATION. 76 ive taken B inaccu- ind from fhen the nultipli- phers if ir ; and t higher tiplioand i> in the 39653. /ion bj as pro- is. he mul* id 48. le pro- iye the oeB of there and— ire are pro- daot, will be the same as the lowest denomination of ths multiplier that produced it; — thus 12 units multiplied by 4 units will give 48 units ; 14 units multiplied by 4 tens will give 56 tens; 124 units multiplied by 35 units will be 4340 units, &o. ; and, therefore, the beginning of each product — if a significant figure — ^must stand under the lowest digit of the multiplier from* which it arises. When the process is finished, cyphers or decimals, if necessary, may be added, according to the rules already given. The vertical dotted lines show that the plaees of the lowest digits of the respective multipliers, or those parts into which the whole multiplier has been divided, and the lowiest digits of their resulting products are — as they ought to be— of the same denomination. 48 bein^ of the denomination units, when multiplied into 8 units, will produce units; the first digit, therefore, of the product bv 48 is in the units' {)lace. 216, being of the deno- mination hundreds when multiplied into units will give hun- dreds ; hence the first digit of the product by 216 will be in the hundreds* place, &c. The parts into which the multi- plier is dividea are, in reality, 60000000 2400000 21600 48 =:62421648, the whole multiplier. We shall sive other contractions in multiplication hereafter, at the proper time. EXERCI8K8. 45. 745x466tjs889720. 46. 476X767ai865002. 47. 845X579-B199755 48. 476X479«228004. 49. 897x979a878168. 60. 4-59X705=8235-96. 51. 767X407«812169. 62. •457X-606=-276942. 68. 700X810>b667000. 64. 670X910a600700. 66. 910X870=791700. 56. 6001*4X70=850098. 57. 64-001 X40s2560 04. 68. 91009x79—7189711. 69. 40170x80—8218600. 60. 707X604=427028. 61. 777 X •407=816-239. 62. 7407X4404=32620428. 63. 6767X1807=7687469. 64. 67 -74X -1706=11 -666444 65. 4567X2002«:^148134. 66. 7 -767X801 -2=2389 -4204 67. 9600X7100=68160000. 68. 7800X9100=70980000. 69. 6700X6700=44890000. 70. 6000X7600=38000000. 71. 70-814x901 07=68808-87098. 72. 97001X76706=7440568706. 78. 98400 X67407ad6295818800. 74. -56007X46070— 25242-8649& ■■■ -ll III •■;. :il^ 76 MULTIPLICATION. ii: : ii'i ■^'i :^^- 75. How many shillings in J&1395 ; a ponnd being 20 shUlings ? Ans, 27900. 76. In 2480 pence how many farthings; four far- things being a penny ? Atu. 9920. 77. If 17 oranges cost a shilling, how many can be had for 87 shiUings ? Ans. 1479. 78. How much will 245 tons of butter cost at £25 a ton? '5^ ■ ' ;: -4tw. 6125. 79. If a pound of any thing cost 4 pence, how much will 112 pounds cost ? Ans. 448 pence. 80. How many pence in 100 pieces of coin, each of which is worth 57 pence ? Ans. 5700 pence. 81. How many gallons in 264 hogsheads, each con- taining 63 gallons ? Ans. 16632. 82. If the interest of £1 be iS0'05, how much will be the interest of iS376 ? Ans. j&18'8. 83. If one article cost i&0'75, what will 973 such cost ? Ans. JE729-75. 84. It has been computed that the gold, silver, and brass expended in building the temple of Solomon at Jerusalem, amounted in value to £6904822500 of our money ; how many pence are there in this sum, one p jund containing 240 ? Ans. 1657157400000. 85. The following are the lengths of a degree of the meridian, in the following places : 60480*2 fathoms in Peru ; 604866 in India ; 60759*4 in France ; 60836*6 in England ; and 60952*4 in Lapland. 6 feet being a fathom, how many feet in each of the above ? Ans. 362881*2 in Peru; 362919*6 in India; 364556*4 in France ; 365019*6 in England ; and 365714*4 in Lapland. 86. The width of the Menai bridge between the points of suspension is 560 feet ; and the weight between these two points 489 tons. 12 inches being a foot, and 2240 pounds a ton, how many inches in the former, and pounds in the latter ? Ans. 6720 inches, and 1095360 pounds. 87. There are two minims to a semibreve ; two crotchets to a minim ; two quavers to a crotchet ; two semiquavers to a quaver : ana two demi-semiquavers to a semiquaver : how many demi-semiquavers are equal to seven semibreves ? Ans. 224 65| in w( He duct carric the MULTIPLICATION. 77 and being ns. 27900. four far- ins, 9920. ny can be [ns. 1479. at £25 a ins. 6125. low much 48 pence. t, each of OO pence, lach con- t. 16632. auch will r. jei8-8. >73 such 6729-75. Iver, and lomon at of our lum, one 400000. e of the horns in B0836-6 being a Ans, 56-4 in apland. en the >etween ot, and brmer, ounds. two two era to equal 224 88. 32,000 seeds have been counted in a ringle poppy ; how many would be found m 297 of these ? Ans. 9504000. 89. 9,344,000 eggs have been found in a single cod fish ; how many would there be in 35 such ? Ans. 327040000. 65. When the pupil is familiar with multiplication, in working, for instance, the following example, 897351, multiplicand. 4, multiplier. 3589404, product. He should sa> : — 4 Tthe product of 4 and, 1)? 20 (the pro- duct of 4 and 5), 14 (the product of 4 and 3 plus 2, to be carried), 29, 3o, 35 ; at the same time putting down the units, and carrying the tens of each. QUESTIONS TO BE ANSWERED BY THE PUPIL. 1. What is multiplication ? [24]. 2. What are the multiplicand, multiplier, and pro- duct ? [24]. 3. What are factors, and submultiples ? [24] . 4. What is the difference between prime and compo- site numbere [25] ; and between those which are prime and those which are composite to each other 7 [27] . 5. What is the measure, aliquot part, or submultiple of a quantity ? [26]. 6. What is a multiple ? [29]. 7. What is a common measure ? [27]. 8. What is meant by the greatest common measure ? [28]. 9. What is a commoTi multiple ? [30]. 10. What is meant by the least common multiple ? [30]. . 11. What are equimultiples ? [31]. 12. Does the use of the multiplication table prevent multiplication from being a species of addition ? [33] . 7 13. Who firot constructed this table ? [33]. :|7,14. What is the sign used for multiplication ? [34]. 15. How are quantities under the vinculum i^ected by the^ign of multiplication ? [34] . 16. show that quantities connected by the toga, of multiplication may be read in any order ? [35]. i-i % DIVISION. U- It. S .1 'li' 17. What in the rule for multiplication, trben neither multiplicand nor multiplier exceeds 12 ? [37]. 18. What iff the rule, when only the multiplicand exceeds 12? [39]. 19. What Is the rule when hoth multiplicand and multiplier exceed 12 .^ [50] . 20. What are the rules when the multiplicand, mul- tiplier, or both, contain cyphers, or decimals } [43, &c.] : and what are the reasons of these, and the preceding rules .? [41, 43, &c., 52]. 21; How is niultiplication proved ? [42 and 53]. 22. Explain the method of proving multijplication, by *' casting out the nines [54] ;" and show that we can cast the nines out of any number, without supposing a knowledge of division. [55] . 23. How do we multiply so as to have a required number of decimal places ? [58] . 24. How do we multiply by a composite number [60] ; or by one that is a little more, or less than a composite number? [61]. 25. How may we multiply by any number of nines ? [62]. : 26. How is multiplication very briefly performed ? [65]. , ^^. , SIMPLE DIVISION. - 66. Simple Division is the division of abstract num- bers, or of those which are applicate, but contain only one denomination. Division enables us to find out how often one number, called the divisor , is contained in^ or can be taken from another, termed tiie dividend ; — the number expressing how often is called the quotient. Division also enables us to tell, if a quantity be divided into a certain number of equal parts, what will be the amount of each. When the divisor is not contained in the dividend any number of times exactly, a quantity, called the remainder^ is left after the division. 67. It will help us to understand how greatly divi- sion abbreviates subtraction, if we consider how long a process would be required to discover — ^by actually sub* tractinf while, by dii 68. dividen(| th9 di^ means, 69. divided,! divisor of addil was to 1 3 3 we divi( Thel sionally fore, wh will son already , 27 but ^ in these the e£fiM is the 81 and th< what ifa yinculu Whe iion ar< will be 25-r5: 70. ^vid«] II 1 1 ; 'I DIVISION. ^ n neither Itiplicand >and and nd, mul- i3, &o.] : (receding 53]. iliicatidn, t we can posing a required Br [60J ; )mpo8ite ' nines? brmed r it num- in only umber, in from dressing enables lumber vidend sd the r divi- long a Tsub- tracting it^-bow often 7 is contained in 8563495724, while, as we shall find, the same thing can be effected by division, in less than a minute. 68. Division is expressed by -r , placed between the dividend and divisor ; or by putting the divisor under i thQ dividend, with a separating Ime bjBtween: — thus ^ 6 , 6-r3=2, or-^2 (read 6 divided by 3 is equal to 2) o means, that if 6 is divided by 3, the quotient will be 2. 69. When a quantity under the vinculum is to be divided, we must, on removing the vinculum, put the divisor under each of the terms connected by me sign of addition, or subtraction, otherwise the value of what was to be divided will be changed ; — ^thus 6 -f- 6— 7-5-3= 5—}- "5 — •^; for we do not divide the whole unless we divide aZ2 its parts. -^ ... - The fine placea between the dividend and divisor occa- sionally assumes the place of a vinculum ; and there- fore, when the quantity to be divided is subtractive, it will sometimes be necessary to change the signs — as 1 J J. 1. J r^^n xi. 6 . 13—3 6+13 — 3; ahready directed [1€]: — thus ~+ ' For when, as 2 ■ 2 ^ , 27 16—6+9 27—15+6—9 3 ^,;>-f' 3' .:*, 1-.. >•■• 3 in ihese cases, all ishe terms are put under the vinoulumy the efleet — as fiu^as the subtractive signs are coneemed— is the same fis if the vinculum were removed asltogether ; and then the signs should be chaiiged i<^ ^ain to what they must be considered to have been before the vinculum was affixed [16 j. '''^^* *' When quantities connected by tiie sign of multiplica- tion are to be divided, dividing any one Qf the factors, will be the same as dividing the pvodiiet ; ihus, 5X 10 X 25+5=s r-X 10X25 ; for each is equal to 250. ;.t 5 To Divide Qaa/ntines, a ' ' ■ ' 70. Whev the divisor does not ^ceed 12) nor t|ie dividend 12 times the divisor ? ^ , r : ^s& : usMftrt^ 'n 80 DIVISION. Vl V 1 'V 111' 'liii' Ill Rule. — ^I. Find by the multiplication table tbe greatest number which, multiplied by the^ divisor, will give a product that does not exceed the dividend : this will be the quotient reauired. n. Subtract from the dividend the product of this number and the divisor ; setting down the remainder, if any, with the divisor under it, and a line between them. EzAMPLK. — ^Find how often 6 is contained in 58; or, in other words, what is the quotient of 5K divided by 6. ^ We learn from the multiplication table that 10 times 6 are 60. But 60 is greater than 58 ; the latter, therefore, does not contain 6 10 times. We find, by the same table, that 9 times 6 are 54, which is less than 58 :— conseciuently 6 is con- tained 9, but not 10 times in 58 ; hence 9 is the quotient ; and 4 — ^the difference between 9 times 6 and the given num- ber — ^is the remainder. 4 4 58 4 The total quotient is9-J-g, or9g; that is, ■g-=9g. If we desire to carry the division farther, we can effect it by a method to be explained presently. 71. Reason or I. — Our object is to find the greatest num- ber of times the divisor can be taken from the dividend ; that is, the greatest multiple of 6 which will not exceed the num- ber to be divided. The multiplication table shows the pro- ducts of any two numbers, neither of which exceeds 12 ; and therefore it enables us to obtain the product we require ; this must not exceed the dividend, nor, being subtracted from it, leave a number equal to, or greater than, the divisor. It is hardly necessary to remark, that the divisor would not have been subtracted as often as possible fVom the dividend if a number equal to or greater than it were left ; nor would the quotient answer the question, haw often the divisor could be taken from the dividend. Reasoiv or II. — We subtract the product of the divisor and quotient from the dividend, to learn, if there be any remainder, what it is. When there is a remainder, we in reality suppose the dividend divided into two parts; one of these is equal to the product of the divisor and quotient — and this we actually divide ; the other is the difference between that product and the given dividend — this we express, by the notation already explained, as still to be divided. In the eaum- , . 68 64-M 64 , 4_^ . 4 pie given, ^«-^«^+jj=:®+^. 72. When the divisor does not exceed 12, bni Uie dividend exceeds 12 times the divisor— n. the dil III minati iv] duct from consic minatil V. table the 1 ^i8or, will g end: this 1 DIVISION. 81 Qt of this under, if , \en them. 58; or, in 6. times 6 )fore, does le, that 9 7 6 is con- quotient ; ven num- 4 1 effect it •est nam- pnd; that ^he num* the pro- 12; and ire; this from it, r. It is not hare end if a Ottid the sould be dirisor be anj I we in one of nt — and Mtween by the ee&4m- ni the KuLE. — ^I. Set down the dividend with a line under it to separate it from the future quotient : and put the divisor to the left hand side of the dividend, with a line between them. n. Divide the divisor into all the denominations of the dividend, beginning with the highest. III. Put the resulting quotients under those deno- minations of the dividend which produced them. IV . If there be a remainder, after subtracting the pro- duct of the divisor and any denomination of the quotient from the corresponding denomination of the dividend, consider it ten times as many of the next lower deno- mination, and add to it the next digit of the dividend. y. If any denomination of the dividend (the preced- ing remainder, when there is one, included) does not contain the divisor, consider it ten times as many of the next lower, and add to it the next digit of the dividend — putting a cypher in the quotient, under the digit of the dividend thus reduced to a lower denomi- nation, unless there are no significant figures in the quotient at the same side of, and farther removed from tne decimal point. YI. If there be a remainder, after dividing the " units of comparison," set it down — as already directed [70] — with the divisor under it, and a separating line between them ; or, writing the decimal point in th*. quotient, proceed with the division, and consider each remainder ten times as many of the next lower deno- mination; proceed thus until there is no remainder, or until it is so trifling that it may be neglected without inconvenience. 73. Example.— What is the quotient of 64456-^7 % Divisor 7)64456 dividend. 9208 quotient. ^ 6 tens of thousands do not contain 7, even once ten thou- sand times ; for ten thousand times 7 are 70 thousand, which is ereater than 60 thousand ; there is, therefore, no digit to be put in the ten-thousands' place of the auotient-^wa do not, however, put a cypher in that place, nnoe no digit DIVISION. W: of the quotient can be further removed from the decimal point than this cypher ; for it would, in such a case, produce no effect [Sec. I. ^]. Considering the 6 tens of thousands as 60 thousands, and adding to theoe the 4 thousands already in the dividend, we have 04 thousands. 7 will ^' go" into (that is, 7 can be taken from^ 64 thousand, 9 thousand times ; for 7 times 9 thousand are b3 thousand — which is less than 64 thousand, and therefore is not too larse ; it does not leave a remainder equal to the divisor — and therefore it is not too small : — 9 is to be set down in the thousands* place of the quotient ; and the 4 already in the dividend beiiig added to one thousand (the difference between 64 and 63 thousand) considered as ten times so many hundreds, we have 14 hun- dreds. 7 will go 2 hundred times into 14 hundreds, and leave no remainder ; for 7 times 2 hundreds are exactly 14 hun- dreds : — 2 is, therefore, to be put in the hundreds* place of the quotient, and there is nothing to be carried. 7 will not go into 5 tens, even once ten times ; since 10 times 7 are 7 tens, which is more than 5 tens. But considering the 5 tens as 50 units, and adding to them the other 6 units of the dividend, we have 56 units. 7 will go into 56, 8 times, leav- ing no remainder. As the 5 tens gave no digit in the tens* place of the quotient, and there are significant figures farther removed from the decimal point than this denon^ination of the dividend, we have been obliged to use a cypher. The division being finished, and no remainder left, the required quotient is found to be 9208 exactly ; that is, Y^"aa9208. 74. ExAMPLK 2.— What is the quotient of 73268, divided 6)73268 < m; . '■'■•'"". '^' 122111 '• ■■'^'^ "' •. ■ ''' '- We may set down the 2 units, which renwin after the units of the quotient are found, as represented ; or we may proceed with the division as follows— 6)73268 irf'n-'Va.H/fciU/. 12211-333, &c. Conndering the % units, left from the units of the divi- dend, as 20 tenths, we perceive thai 6 will go into them three tenths times, and leave 2 tenths— ^ippe 3 tenths times 6 (»6 times 3 tenths [35]) are 18 tenths :— we put 3 in t^e tenths* place of the quotient, and consider the % tenths re- nuuning, as 20 hundredths. For siofiil^ reasons, 6 wiU gp into 20 hondredtlML 3 hundredths times, and leaie 2 hiin- dredj they I as* re ring, denoil put d| tinal 75.1 DIVISION. 83 .U'.i.V e divi- them times |iit)ie re- hun- dredths. Considering these 2 hundredths as 20 thousandths, they will cive 3 thousimdths as quotient, and 2 thousandths a» remainder, &c. The same remainder, constantly recur- ring, will evidently produce the same digit in the succesoive denominations of the ((uotient ; we may, therefore, at onoo put down in the quotient as many threes as will leave the final remainder so small, that it may be neglected. 75. ExAMPLB 3.— Divide 47365 by 12. 12)47365 3947-08, &c. In this example, the one unit left (after obtaining the 7 in the quotient) even when considered as 10 tenths, does not contain 12 : — there is, therefore, nothing to be set down in the tenths' place of the quotient — except a cypher, to keep the following digits in their proper places. The 10 tenths are by consequence to be consiaered as 100 hundredths, 12 will go into 100 hundredths 8 hundredths times, &o. This may be applied to the last rule [70], when we desire to continue the division. Example. — Divide 8 by 5. 8-^6 = 12, or 1-37, &o. 76. When the pupil fully understands the real deno- minations of the dividend and quotient, he may proceed, for example, with the following 5 )46325 In this manner : — 5 will not go into 4. 5 into 46, 9 times and 1 over (the 46 being of the denomination to which 6 belongs [thousands], the first digit of the quotient is to be put under the 6 — ^that is, under the denomination which produced it). 5 into 13, twice and 3 over. 5 into 32, 6 times and 2 over. 5 into 25, 5 times and no remainder. When the divisor does not exceed 12, the process is called «Aor^ division. 77. Reason of I. — In this arrangement of the quantities^ which is merely a matter of convenience — the values of the digits of the quotient are ascertained, both by their position with reference to the digits of the dividend, and to their own decimal point. The separating lines prevent the dividend, divisor, or quotient from being m any way mistaken. Reason of II. — We divide the divisor successively into all the parts of the dividend, because we cannot divide it at onoe into the whole : — the sum of the numbers of times it can be subtracted from these parts is evidently equal to the number 84 DIVISWS. fJ of timei it can be subtractcfl from their sum. Thus, if 5 goes into 500, 1(N) times, into 50, 10 times, anil into 5, once; it will go into 500-H>0-}-6 (rsoOa), lOO-f-lO-j-l ( = 111) times Tiie pupil perceives by the exnmples given nbove, tlitit, in dividing the uiviBor successividy into the piirts of tlie dividend, each, or any of thc»e purtu does not necuasurily consist of ono or more digits of tlie divi«lend. Tlius, in hnding, for example, the quotient t)44o»J-^7, we are not obliged to consider tlto part4 as »)Oi)t)0, 4000, 4(X), 50, and : —on the contrary, to render tlio dividend suited to the process of division, we alter its form, -while, at the same time, we leave its value unchanged ; it be- comes Thousands. Hundreds. Tens. Units. 68 -f 14 4. 4- 68 (=64450). Each part being divided by 7, the different portions of th« dividend, with their respective quotients, will be, , 'j Thousands. Hundreds. Tens. Units. ' *'••'*■ ' -■ '• 71 63 14 66 =3 64466. , /,f r 9 2 8= 9208. We begin at the left hand side, because what remains of the higher denomination, may still give a quotient in a lower; and the question is, how often the divisor will go into the dividend — its ditferent denominations being taken in any oon- yenient way. We cannot know how many of the higher we shall have to add to the lower denominations, unless we begin with the higher. Rrason or III. — Each digit of the quotient is put under that denomination of the dividen, onuo; it tirncH e, tiiHt, in ) (liviiivnii, !ii»t of uno ' exDiuplc, ' tiie piiit.f pondor tlio • its form, ed; it be- U56). ns of the ■ o .■ 'i-';' : ■♦•; ins of the a lower; > into the any oon- liglier we we begin ut iin'icr \>, beoiiuso t number le divisor ridend : — of times lend; tba lies it can total re- )d as of !i quotient, er as of ent; and limes ex- El it some it is the n" only Thus, in otient, it or units • of huiv Bn(L A cypher must ho added [Sec. I. 28], when it is rcquir'.d, to give signltiuaiit figures tlieir proper vuluo — which is r .ver the case, except it comes between tlium and the deoim'>l point. IIkahow of VI. — We may continue the process of division, if we please, as long as it is posible to obtain quotients of an}/ denomination. Quotients will be produced altiiough there are no lunger any .signiticant figures in the dividend, to which wo can add the successive remainders. 78. The Hiiiullor the divisor the larger the quotient — for, the smaller the parts of a given quantity, the greater their number will be ; but is the least possible divi- sor, and therefore any quantity divided by will give the largest possible quotient— which is infinity. Hence, though any (juantity multiplied by is equal to 0, any number divided by is equal to an infinite number. 5 1 It appears strange, but yet it is true, that ;^=^ ; for each is equal to the greatest possible number, and one, therefore, cannot be greater than another— the appa- rent contradiction arises from our being unable to Ibrm a true conception of an infinite quantity. It is necessary to bear in mind also that 0, in this case, indicates a quantity infinitely small, rather than absolutely nothing. 79. To prove Division. — Multiply the quotient by the divasor ; the product should be equal to the divi- dend, minus the remainder, if there is one. For, the dividend, exclusive of the remainder, contains the divisor a number of times indicated by the quotient ; if, there- fore, the divisor, is taken that number of times, a quantity equal to the dividend, minus the remainder, will be produced. It follows, that adding the remainder to the product of the divisor and quotient should give the dividend. 6832 Example 1. — Prove that — ^=1708. ,: 4)6832 1708 Proof. 1708, quotient. 4, aivisor. 6832, product of divi- sor and quotient, equal to the dividend. 85643 5 Example 2.— Prove that —T=—= 12234 =. Proof. ..„_ , Proof. .^ , . j 12234 or ■' ' ^ 12234 7 . '■- .. ,. , -7 \ 85638=dlTidend minui 6, the remain 'ar. 85638-|-6=^iTideiir can be subtracted from the dividend. Hence, if 8 will go into 56 7 times, it will go into 5600 (a quantity 100 times greater than 66) 100 times more than 7 times— or 700 times. Example 1.— What is the quotient of 568000^4 % 568 ,,^ ,^ ^ 568000 ^,^^^ -7-=142 J therefore — j— = 142000. Example 2.— What is the quotient of 4060000^5 % '" ~=81-2; therefore —g— =812000 [Sec. I. 39.]. 81. When the divisor contains cyphers — Rule. — ^Divide as if there were none, and move the quotient so many places to the right as there are cyphers in the divisor. The greater the divisor, the smaller the number of times it ean be subtracted from the dividend. If, for exsmnle, 6 can be taken from a quantity any number of timeSt lOO times 6 ean be taken from it 100 times le&s often. 66 Example.— What i» the quotient of g;^ ? 66 . therefore ^^ goo 07. 82] Ri quoti^ betwc tities I the divisc DIVISION. 87 ) 284 067 ) 457 coniun 3 ap]ili. lem. remove re havo be the divisor go into greater I Hi 9.]. re the ^hers mes it 6 can imes 6 82. If both dividend and divisor contain C3rphers — KuLE. — Divide as if there were none, and move the quotient a number of places equal to the difference between the numbers of cyphers in the two given quan- tities : — if the cyphers in the dividend exceed those in the divisor, move to the left; if the cyphers in the divisor exceed those in the dividend, move to the right. We have seen that the effect of cyphers in the dividend is to move the quotient to the left and of cyphers in the divisor, to move it to the right; when, therefore, both causes act together, their effect must be equal to the difference between their separate effects. Examples. (1) (2) 7)68 7)6800 9 m 70)68 (4) (6) (6) 70 )6300 700 )680 700 )6800 90 900 0-9 90 0-9 9 In the sixth example, the difference between the numbers of cyphers being =0, the quotient is moved neither to the right nor the leu. 83. If there are decimals in the dividend — Rule. — ^Divide as if there were none, and move the quotient so many places to the right as there are deci- mals. The smaller the dividend, the less the quotient. Example. — ^What is the quotient of •048-4-8 ? 48 *048 ^=6, therefore -g-=:006. 84. If there are decimals in the divisor — Rule. — ^Divide as if there were none, and move the quotient so many places to the left as there are deci- mals. The smaller the divisor, the greater the quotient. Example. — ^What is the quotient of 54-f--006 % 54 54 -^■=9, therefore:^=9000. 85. J£ there are decimals in both dividend and di* visor — ' ^ Rule. — ^Divide as if there were none, and move the quotient a number of places equal to the difference 88 DIVISION. between the numbers of decimals in tlie two given quan- tities : — if the decimals in the dividend exceed those in the divisor, move to the right ; if the decimals in the divisor exceed those in the dividend, move to the left. We have seen that decimals in the dividend move the quotient to the right, and that decimals in the divisor move it to the left; Trhen, therefore, both causes act together, the eifect must be equal to the difference between their separate effects. 1 ; i 1 1 i : /J pi I' '^I 1 t , Examples. ; 1 1 * 1 ' '*! (1) 5)45 9 (2) 5) -45 •09 (3) (4) •05)45 '5) -045 900 -09 (5) •005) 450 90000 (6) •05) -45 9-00 86. If there are cyphers in the dividend, and deci- mals 4n the divisor — ■- — ^ -«:' ~ ' ■ lluLE. — Divide as if there were neither, and move the quotient a number of places to the left, equal to the number of both cyphers and decimals. Both the cyphers in the dividend, and the decimals in the divisor increase the quotient. - : . » i Example. — What is the quotient of 270-^-- 03 ; ^■' ?J=9, therefore, 270^-03 = 9000. ,. 87. If there are aecimals in the dividend, and cyphers in the divisor — Rule. — ^Divide as if there were neither, and move the quotient a number of places to the right eqiial to thenumber of both cyphers and decimals. Both the decimals in the dividend, and the eypners m the divisor diminish the quotient. Example. — What is the quotient of •18-J-20 *? ,, r 4r=^. therefore - = 009. , - 2 20 - The rules which relate to the management of cyphers and decimals, in multiplication and in division — though numerous — ^will be very easily remembered, if the pupil merely considers what ought to be the eilovl 4^t' either I quan- lose in in the left. )ve the ir move ler, the eparate (6) •05) -45 9-00 I deci- move [ual to in the t\ • ;•";:;!< /■phers move lal to in the phers lougb pupil « (18) 8)10000 DIVISION. * EXERCISES. (14) (15) (16) 11)16000 8)70170 6)68530 89 (17) 20)86528 (18) ^000)47865 (19) (20) 40)66020 80)75686 (21) 12)63-075 ' (22) 10) -08766 (23) (24) •07)64268 •00)67-368 (25) •0005)60300 (26) • 700) •03576 (27) » (28) " ' •008)57 •362 400)63700 ^'j-' ■ .. ...-, i .••■■ '• (29) 110)97-634 t* t 88. When the divisor exceeds 12 — . •* The process used is called lo'ng division; that is, we perform the multiplications, subtractions, fee, in full, and not, as before, merely in the mind. This will bo understood better, by applying the method of long divi- sion to an example in which — the divisor not being gr ater than 12 — it is unnecessary. Short Pi viKioQ : the same by Lon;^ Division. „i ,r., 8)5763172 , *. ., - 8)5763472(7204Si . . .V '• ' * • " • 56 720434 .»-i,!U.!." 16 16 rjv/ -.^c ,. 84 82' .1 . n* I: •7ii 'f^ In the second method, we multiply the divisor by the different ports of the quotient, and in each cose set down I . i n- 90 DIVISION. j.i,.'t J' If. the product, suhtrcut it from the corresponding portion of the aiyidend, write the remainder, and oring down the re- quii,-ed digits f£ the dividend. All this must be done when tne divisor becomes large, or the mem(»ry would be too heavily burdened. 89. Rule — ^I. Pnt the divisor to the left of the divi- dend, with a separating line. II. Mark off, by a separating line, a place for the quotient, to the right of the dividend. 90 III. Find the smallest number of digits at the left hand side of the dividend, which expresses a quantity not less than the divisor. IV. Put under these, and subtract from them, the greatest multiple of the divisor which they contain ; and set down, underneath, the remainder, if there is any. The digit by which we have multiplied the divisor is to be placed in the quotient. v. To the remainder just mentioned add, w, as it is said, '^ bring down" so many of the next digits (or cyphers, as the case may be) of the dividend, as are required to make a quantity net less than the divisor ; and for every digit or cypher of the dividend thus iHTought down, except one^ add a cypher after the digit last placed in the quotient. YI. Find out, and set down in the quotient, the numher of times the divisor is contained in this quan- tity ; and then subtract from the latter the product of the divisor and the digit of the quotient just set down. Proceed with the resulting remainder, and with all that succeed, as with the last. YII. If there Is a remainder, after the units of the dividend have been *^ brought down" and divided, either place it intiO the quotient with the divisor under it, and a separating line between them [70] ; or, putting the decimal point in the quotient — and adding to the re- mainder as many cyphers as will make it at least equal to the divisor, and to the quotient as many cyphers minus one as there have been cyphers added to the remainder — ^proceed with the division. DIVISION. 91 Drtion of I the re- me when 1 be too ;he divi- ] for the I the left |uantity im, the ontaiD ; here is divisor as it is ^ts (or as are ivisor; 1 thus Q digit it, the quan- act of down. Uthat >f the either g the e re- equal phers the 90. Example 1.— Divide 78325826 by 82. 82)78325826(955193 738 82 will not go into 7 ; nor into 78 ; but it will go 9 times into 783 : — 9 is to be put in the quotient. The values of the higher denominations in the quotient will be sufficiently marked by the digits which succeed them — it will, however, sometimes be proper to ascertain, if the pupil, as he proceeds, is acquainted with the orders of units to which they belong. 9 times 82 are 738, which, being put under 783, and sub- tracted from it, leaves 45 as remainder ; since this is less than the divisor, the digit put into the quotient is — as it ought to be [71] — the largest possible. 2, the next digit of the divi- dend, being brought aown, we have 452, into which 83 goes 5 times ; — 5 being put in the quotient, wo subtract 5 times the divisor from 452, which leaves 42 as remainder. 42, with 5, the next digit of the dividend, makes 425, into which 82 goes 5 times, leaving 15 as remainder ; — ^we put anotheir 5 in the quotient. The last remainder, 15, with 8 the next digit of the dividend, makes 158, into which 82 goes once, leaving 76 as remainder ; — 1 is to be put in the quotient. 2, the next digit of the dividend, along with 76, makes 762, into which the divisor goes 9 times, and leaves 24 as remain- der ', — ^9 is to be put in the quotient. The next digit being brought down, we have 246, into which 82 goes 3 times exactly ] — 3 is to be put in the quotient. This 3 indicates 3 units, as the last digit brought down expressed units* ^ ^ 78325826 ^^^,„ ' . -jt- Therefore — gs — =955193. 92 DIVISION. Example 2. ih'i B r i if'--' -Divide G421284 by G42. 642)6421284(10002 642 1284 1284 642 goes once into 642, and leaves no remainder. Bring- ing down the next digit of the dividend gives no digit in the quotient, in which, therefore, we put a cypher after the 1. 'i'he next digit of the dividend, in the same way, gives no digit in the quotient, *in which, consequently, we put another cypher ; and, for similar reasons, another in bringing down the next ; but the next digit makes the quantity brought down 1284, which contains the divisor twice, and gives no remainder : — we put 2 in the quotient. 91. When there is a remainder, we may continue the division, adding decimal places to the quotient, as follows — Example 3.— Divide 79G347 by 847. ..;, . . .. . , 847) 796347 (940- 19, &c. 7623 ]!■ " 3404 ' 3388 1670 847 8230 7623 92. The learner, after a little practice, will guess pretty accurately what, in each case, should be the next digit of the quotient. He has only to multiply in his mind the last digit of the divisor, adding to the product what he would probably have to carry from the multiplica- tion of the second last : — if this sum can be taken from the corresponding part of what is to be the minuend, leaving little, or nothing, the assumed number is likely to answer for thie next digit of the quotient. l*i -"i " 93. Reason of I. — This arrangement is merely a matter of convenience; some put the divisor to the right of the dividend, and immediately over the quotient — believing that it is more convenient to have two quantities which are to be multiplied' together as near to each other as possible. Thus, in dividing using DIVISION. 93 digit ill ifter the ^y, gives we put )ringing juantity ice, and lue the •Hows — guess e next 3 mind b what Iplica- i from luend, likelj • '• • 'i 1 » tterof idend, more tiplied' riding 6425/ 54 51. Ul87&^ ' : ' « 102 ■ '■ . ^ ■ " ; ■ ' , ■ 485 432 ^ . , 53, &c Reason of II. — This, also, is only a matter of convenience llKAsoN OF III. — A smaller part of the dividend would give no digit in the quotient, and a larger vrould give more than one. Rkasoiv of IV. — Since the numbers to be multiplied, and the products to be subtracted, are considerable, it is not so convenient as in short division, to perform the multiplications and subtractions mentally. The rule directs us to set down each multiplier in the quotient, because the latter is the sum of the multipliers. Reason of V.— One digit of the dividend brought down would make the quantity to be divided one denomination lower than the preceding, and the resulting digit of the quotient also one denomination lower. But if we are obliged to bring down two digits, the quantity to be divided is two denomi- nations lower, and consequently the resulting digit of the quo- tient is two denominations lower than the preceding — whiclij from the principles of notation [Sec. I. 28], is expressed by using a cypher. In the &ame way, bringing down three figures of the dividend reduces the denomination three places, and makes the new digit of the quotient three denominations lower, than the last — two cyphers must then be used. The same reasoning holds for any number of characters, whether significant or otherwise, brought down to any remainder. Reason of VI. — We subtract the products of the different parts of the quotient and the divisor (these different parts of the quotient being put down successively according as they are found), that we may discover what the remainder is from which we are to expect the next portion of the quotient. From what we have alresidy said [77], it is evident that, if there are no decimals in the divisor, the quotient figure will always be of the same denomination as the lowest in the quantity from which we subtract the product of it and the divisor. Reason of VII. ^-The reason of this is the same as what was given fo» the sixth part of the preceding rule [77]. It is proper to put a dot over each digit of the divi- dend, as we bring it down ; this will prevent our forget- ting any one, or bringing it down twice. 94. When there are cyphers, decimals, or both, tho rules already given [8), &c.] are applicable. It 94 DIVISION. El M m m .., 95. To prove the Division. — ^Multiply the quotient by the divisor ; the product should be equal to the divi- dend, minus the remainder, if there is any [79] . To prove it by the method of ** casting out the 266)71 inines" — Rule. — Cast the nines out of the divisor, and the quotient ; multiply the remainders, and cast the nines from their product: — that which is now loft ought to be the same as what is obtained by casting the nines out of the dividend minus the remainder obtained from the process of division. Example. — Prove that -vv— saallSl/j. Considered as a question in multiplication, this becomes 1181 X 54=63776-2=63774. To try if this be true, Casting the nines from 1181, the remainder is 2. ) n ^^ a a from 54, „ isO. j'^^^"^" Casting the nines from 63774, the reminder is . . The two remainders are equal, both being ; hence the multiplication is to be presumed right, and, consequently, the process of division which supposes it. The division inyolves an example of multiplication; since the product of the divisor and quotient ought to be equal to the diTidend minus the remainder [79]. Henoe, in proving the multiplication (supposed), as already explained [o4], we indirectly prove the division. . (30) 24)7654 ' . . , EXERCISES. ,.-..• (31) (32) (S3) 15)6783 ... 16)5674 17)467$ 318ff 452f», T 35410 275 (34) 18)7831 (35) (36) (37) 19)5977 21)6783 22)9767 435t»i 314H 323 443f^ (38) 23)767500 (39) " • '= (40) • 390)5807 1460)6767600 33369J^| M 14-8897 . 4635-3425 64-25) DIVISIOIf. 95 quotient the divi- t out the and the be nines »ught to be nines ed from becomes 'ue, !X0=:0 . mce the qaentlj, i; since equal to proyine m, we (33) 7)4675 275 [37) 9767 443S 5-3425 : (41) 26 6)77676700 808424*6094 (44) 5l'26 )128-70686 2-2808 (42) 67*1 ) -1842 •002 (48) '158 ) '829740 5-4232 (46) 14-85 )269-0625 18-75 (46) '0087 ) 666 150000 In example 40 — and some of those which follow— after obtaining as many decimal places in the quotient as are deemed necessary, it will be more accurate to consider the remainder as equal to the divisor (since it is more than one half of it), and add unity to the last digit of the quotient. CONTRACTIONS IN DIVISION. 96. We may abbreviate the process of division when there are many decimals, by cutting off a digit to the right hand of the divisor, at each new digit of the quotient ; remembering to carry what would have been obtained by the multiplication of the figure neglected — unity if this multiplication would have produced more than 5, or less than 15 ; 2 if more than 15, or less than 25, &c. [59]. Example.— Di^de 754-337385 by 61-347. Ordinary Method. Contracted Method. 61-347)754-33 7385(12-296 61- 347) 754- 337385 (12- 296 61347 14086,7 122694 1817 33 1226|94 590398 552123 38 36 2755 8082 46730 61347 14086 12269 lilJ 1227 "590 552 "si 87 ^ 96 1'1VI.-»I0X. According ns the denominations of the quotient become smiill, their products by the lower deuumination of the divisor become incunsidcniblc, and may bo neglected, and, conse- quently, the portions of the dividend from which they would have been subtracted. What should have been carried from the multiplicatiun of the digit neglected — since it belongs to a higher denomination than what is neglected, should still be retained [69]. 97. We may avail ourselves, in division, of contri- i vances very similar to those used in multiplication | [60]. To divide by a composite number — ' ' . Rule. — Divide successively by its factors. * Hi- Example. --Divido 98 by 49. 49=7 x 7. 7)98 7)14 , .. ■ . _ 2=98^7x7, or 49. ., Dividing only by 7 we divide by a quantity 7 times too small, for we are to divide by 7 times 7 ; the result is, therefore, 7 timei too great : — this is corrected if we divide again by 7 98. If the divisor is not a composite number, we cannot, as in multiplication, abbreviate the process, except it is a quantity which is but little less than a number expressed by unity and one or more cyphers When this is the case — Rule. — Divide by the nearest higher number, ex- pressed by unity and one or more cyphers ; add to re- mainder so many times the quotient as the assumed exceeds the given divisor, and divide the sum by the preceding divisor. Proceed thus, adding to the remain- der in each case so many times the foregoing quotient as the assumed exceeds the given divisor until the exact, or a sufficiently near approximation to the exact quotient is obtained — the last divisor must be the given, and not the assumed one. The last remainder will be the true one ; and the sum of all the quotients will be the true quotient ;:: - 1 -'^'' t become he divisor id, conae- ley would ried from longs to a id Btill be f contri- iplication s t'l ■ times too therefore, in by 7 iber, we process, ) than a cyphers iber, ex- d to re* assumed 1 by the I remain- quotient he exact, quotient and not the true the true DIVISION. 97 ExAMPY.E.-Divido 9870G3425 by 998. 987()03.,425=:987 (»0342r)-f.l00 0. ' ' 1975 ,751= 987003 x 2-f-4 2') h- 1000. 4 J01=l975x2-f-75 1-14000. 0- 7.,090=4 X 2+701 -f- 1000. 001a040=7 X24-9-1- 1000. 0000,420=1 01 x24^-f- 1000. 00004a0208=:01 X2-I--4-J-998 that h. the last quotient is 0'0004, and '0208 id the last renuiinder. f 987663 1975 4 0-7 001 00004 all the quotients are < The true quotient is 989642- 7 104, or the sum of the quotients. And the true remainder 0*0208, or the last remainder. Unless we add twice the preceding quotient to each succes- sive remainder, we shall have subtracted from the dividend, or the part of it just divided, 1000, and not 998 times the quotient — in which case tlie remainder would be too small to the amount of twice the quotient. — We have used (a) to sepa- rate the quotients from the remainders. There can be no difficulty when the learner, by this process, comes to the decimals of the quotient. Thus in the third line, 4701 gives, when divided by 1000, 4 units as quotient, and 701 units still to be divided — that is, 701 as remainder. 4'701 would express 4701 actually divided by 1000. A number occupying four places, all to the left of the decimal point, when divided by 1000, gives units as quotient ; but if, as in 709*0 (in the next line), one is a decimal place, the quotient must be of a lower denomination than before — that is, of the order tenths; and in 010*40 (next line), since two out of the four places ar^ decimals, the quotient must be hun- dredths, &c. ff," * . In adding the necessary quantities, we must carefully bear in mind to what denominations the quotient multi- plied, and the remainder to which the product is to bo added, belong \ 98 DIVISION. I /I i'i'' M4 >1. 7867674 -J-O712«8i0?,^. )2. 30707(J0-^467000=6-7198. EXCRCIIES. 47. 66789H-741=76*2f 48. 478(M)7+971=498j«f 49. 9.7076-f-47000=20f5i 50. 667897 -^842=:674 2] 61. 62. 63. 07651684-7894es867. 64. 67470-^.3900=17 -8. 65. 69000-i-47600=l •4496. 66. 767674-40700=1 '8862. 57. 0114692-1.704824=8. 68. 9676744.^-910076=10 -6829. 69. 740070000.^741000=098•7449. 60. 9410607111.^46678=206048-1182. 61. 454076000-1.400100=1 184 -9068. 62. 7876476767 .4-845670=21889 '640. 68. 47 •6782075.^26 •175=1 -8177. 47 •655.1.4 •6=10 •59. 756^98 J.76^78612=9-866. 75 • 8470-j-8829=:196 • 7798. 0•l-^7•6845=0•0000181. 64 65. 66. 67. 68. 5378.^0 • 00096=6602088 • 88, &0. 69. If iS7d00 were to be divided between 5 persons, how much ought each person to receive ? Ans. iS1500. 70. Divide 7560 acres of land between 15 persons. Ans. Each will have 504 acres. 71. Divide jBSSSO between 60 persons. Ans. Each will receive £4S, 72. What is the ninth of ^6972 ? Ans. iBlOS. 73. What is each man's part if JS972 be divided among 108 men? Am. £9, 74. Divide a legacy of jS8526 between 294 persons. Ans. Each will have £29, 75. Divide 340480 ounces of bread between 1792 persons. Ans. Each person's share will be 190 ounces. 76. There are said to be seven bells at Pekin, each of which weighs 120,000 pounds ; if they were melted up, how many such as great Tom of Lincoln, weighing 9894 pounds, or as the great bell of St. Paul's, in London, weighing 8400 pounds, could be made from them ? Ans. 84 like great Tom of Lincoln, with 8904 pounds left ; and 100 like the great bell of St. Paul's. 77. Mexico produced from the year 1790 to 1830 a quantit 6,178,1 farthinj 78. thread i thread I be req ^ 77026; 79. to 567 525948 thirds it trave About minute, 80. . year 1' 1827, ( the pro and of 2429-5' 81. J saved tl half of may ea with an 99. ] able to EZAN He w be brou Bubtrac 28 (9 t Temain( 0, or K be carr rowed; DIVISION. 99 quantity of gold which was worth i&6,436,443, or 6,1 78,985^280 farthings. How many dollars, at 207 farthings each, are in that sum ? Arts. 29850170 nearly. 78. A sinele pound of cotton has been spun into a thread 76 miles in length, and a pound of wool into a thread 95 miles long ; how many pounds of each would, be required for threads 5854 miles in length ? Ans, 77'0263 pounds of cotton, and 61*621 pounds of wool. 79. The earth travels round its orbit, a space equal io 567,019,740 miles, in about 365 days, 8765 hours, 525948 minutes, 31556925 seconds, and 1893415530 thirds ; supposing its motion uniform, how much would it travel per day, hour, minute, second, and third ? Ans. About 1553480 miles a day, 64691 an hour, 1078 a minute, 18 a second, and 0'3 a third. 80. All the iron produced in Great Britain in the year 1740 was 17,000 tons from 59 furnaces ; and in 1827, 690,000 from 284. What may be considered as the produce of each furnace in 1740, one with another ; and of each in 1827. Am. 2881356 in 1740; and 2429-5775 m 1827. 81. In 1834, 16,000 steam engines in Great Britain saved the labour of 450,000 horses, or 2 millions and a half of men ; to how many horses, and how many men, may each steam engine be supposed equivalent, one with another ? Ans. About 28 horses ; and 156 men. 99. Before the pupil leaves division, he should be able to carry on the process as follows : — Example.— Divide 84380848 by 87532. . ,' 87532)84380848(964 < 560204 . / 'v . 350128 He will say (at first aloud) 4 (the digit of the dividend to be brought aown). 18 (9 times 2) : (the remainder after Bubtractmg the right hand digit of io from 8 in the dividend). 28 (9 times 3 -f- the 1 to be carried from the 18) ; 2 (the remainder after subtracting the right hand digit of 28 from 0, or rather 10 in the dividend). 48 (9 times 5 -f- the 2 to be carried from 28, and 1 to compensate for what we bor- rowed when we considered in the dividend as 10) ; (th# 100 DIVISION. lit It*' -i I 'Hi remainder when we subtract the right hand digit of 48 from 8 in the dividend). 67 (9 times 7 + the 4 to be carried from the 48) ; 6 (f the remainder after subtracting the right hand digit of 67 from 3, or rather 13 in the dividend). 79 (9 times 8 -f. the 6 to be carried firom the 67 + the 1, for what we borrowed to make 3 in the dividend become 13) ; 5 (the reminder after subtracting 79 from 84 in the divi- dend). As the parts in the parentheses are merely explanatory, and not to be repeated, the whole process would be, First part, 4. 18: 0. 28 ; 2. 48 ; 0. 67 ; 6. 79 ; 5. Second part, 8. 12:2. 19:1. 32; 0. 45:5. 53 j 3. Third part, 8 ; 0. 12; 0. 21 ;0. 30; 0. 35; 0. The remainders in this case being cyphers, are omitted. All this will be very easy to the pupil who has prao« tised what has been recommended [13, 23, and 65]. The chief exercise of the memory wUl consist in recol- lecting to add to the products of the different parts of the divisor by the digit of the quotient under oonsideration, what is to be carried from the preceding product, and unity besides — when the preceding digit of th« dividend has been increased by 10 ; then to subtract the right hand digit oi this sum from the proper digit of the dividend (increased by 10 if necessary). Questions ton the ptjpit. 1. What is division ? [66]. 2. What are the dividend, divisor, quotient, and re- mainder ? [66] . 3. What is the B%n of division ? [68]. 4. How are quantities under the vinctdum, or united by the sign of multiplication, divided > [69]. 5. What is the rule when the divisor does not exceed 12, nor the dividend 12 times the divisor.? [70]. 6. Give the rule, and the reasons of its different parts, when the divisor does not exceed 12, but the dividend is more than 12 times the divisor } [72 and 77]. 7. How is division proved .? [79 and 95] . 8. What are the rules when the dividend, divisor, or both contain cyphers or decimals ? [80] . 9. What is the rule, and what are the reasons (^ its different parts, when the divisor exceeds 12 ? [89 and 93] . GREATEST COMMON MEASURE. 101 f 48 from e carried the right ind). 79 ;he 1, for ome 13) ; the divi- )lanatory, ' 79 ; 5. ; 0. mitted. las prao« %ud 65]. in recol- rts of the deration, luct, and dividend ;he right t of the and re- r united k exceed different but the p ind77]. nsor, or 18 of its iiid93]. 10. What is to be done with the remainder? [72 and 891. 11. llow is division proved by casting out the nines ? [95]. 12. How may division be abbreviated, when there are decimals ? [96] . 13. How is division performed, when the divisor is ft composite number ? [97]. 14. How is the division performed, when the divisor is but little less than a number which may be expressed by unity and cyphers ? [98] . 15. Exemplify a very brief mode of performing divi* sion. [99]. THE GREATEST COMMON MEASURE OF NUMBERS. 100. To find the greatest common measure of two quantities — Rule. — ^Divide the larger by the smaller ; then the divisor by the remainder ; next the preceding divisor by the new remainder : — continue this process until nothing remains, and the last divisor will be the greatest common measure. If this be unity, the given numbers are prime to eocA oiAer. r - Example. — ^Find the greatest common measure of 32^2 81144248. «CB1)4248(1 3252 "996)3252(3 2988 264)996(3 792 ■ ^^ ^ . • 204)264(1 204 "i0)204(3 180 : r •. . :\ "24)60^^2 . . m . '. ,■~^ .,....:;.; , .. • 12)24(2 ,: .: ' ■ , -. 24 102 GREATEST COMMON MEASURE. f^ I 996, the first remainder, becomes the second divisor *264, the second remainder, becomes the third divisor, &c. 12, the last divisor, is the required greatest common measure. 101. Reason of the Rule.— Before we prove the correct- ness of the rule, it will be necessary for the pupil to he satis- fied that " if any quantity measures another, it will measure any multiple of that other ; " thus if 6 go into 30, 6 times, it will evidently go into 9 times 30, 9 times 5 times. Also, that ** if a quantity measure two others, it will measure their sum, and their difference." First, it will measure their sum, for if 6 eo into 24, 4 times, and into 36, 6 times, it will evi- dently go into 24+36, 4-t-6 times :— that is, if-;r=4, and— = 24 , 86 ,j^ ^ ** 6,6'+6"^+®- Secondly, if 6 goes into 36 oftener than it goes into 24, it is because of the difference between 36 and 24 ; for as the differ- ence between the numbers of times it will ^o into them is due to this difference, 6 must be contained in it some number **. *!. * • • 36_^ ^24_.36 24/ 36— 24\ of times :— that is, since— =6, and -^==4» ■^— «• I or — « — ) s=6 — 4=ss2, a whole number [26]— or, the difference between the quantities is measured by 6, their measure. This reasoning would be found equally correct with any other similar numbers. 102. Next ; to prove the rule from the given example, it is necessary to prove that 12 is a common measure ; and that it is the greatest common measure. It is a common measure. Beginning at the end of the process, we find that 12 measures 24, its multiple ; and 48, because it is a multiple of 24 ; and their sum, 24-f^8 (because it measures each of them) or 60 ; and 180, because it is a multiple of 60 ; and 180-|-24 (we have also just seen that it measures each of these) or 204 ; and 204+60 or 264 ; and 792, because a multi- pie of 264 ; and 792+204 or 996 ; and 2988, a multiple of 996 ; and 2988+264 or 8262 (one of the given numbers) and 8252+ 996 or 4248 (the other given number). Therefore it measures each of the given numbers, and is their common measure. 103. It is also their greatest common measure. If not, let some other be greater ; then (beginning now at the top of the process) measuring 4248 and 3252 (this is the supposition), it measures their diffSerence, 996; and 2988, because a multiple of 996 ; and, because it measures 3252, and 2988, it measures their difference, 264 ; and 792, because a multiple of 264 ; and the diffsrence between 996 and 792 or 204; and the difference between 264 and 204 or 60 ; and 180 because a multiple of 60 ; and the difference between 204 and 180 or 24 ; and 48, because a multiple of 24 ; and the difference between 60 and 48 or 12. Bttt measuring 12, it cannot be greater than 12. i or GREATEST COMMON MEASURE. 103 >p *264, kc 12, asure. correct- 1)6 satis- measure times, it measure ire their will evi- 24, it is e differ- them is number B6— 24 x 6 / between rith any cample, easure; process, use it is leasures eof 60; each of I multi- of 996; 8252+ leasures re. not, let » of the tion), it nultiple leasures 34; and fference Bof 60; because B or 12. In the same way it could be shown, that any other common measure of the ffiven numbers must be less than 12 — and con- sequently that 12 is their greatest common measure. As the rule might be proted from any other example equally well, it is true in all cases. 104. We m&j here remark, that the measure of two or more quantities can sometimes be found by inspection * Any quantity, the digit of whose lowest denominatioo is an even number, is divisible by 2 at least. Any number ending in 5 is divisible by 5 at least. Any number ending in a cypher is divisible by 10 at least. Any number which leaves nothins when the threes are cast out of the sum of its digits, is divisible by 3 at least ; or leaves nothing when the nines are cast out of the sum of its digits, is divisible by 9 at least. EXERCISES. 1. What is the greatest common measure of 464320 and 18945 > Am, 5. 2. Of 638296 and 33888 ? Ans. 8. 3. Of 18996 and 29932 ? Ans. 4. 4. Of 260424 and 54423 > Ans. 9. 5. Of 143168 and 2064888 > Ans. 8. 6. Of 1141874 and 19823208 ? Ans. 2. 105. To find the greatest common measure of more than two numbers — BuLE. — Find the greatest common measure of two of them ; then of this common measure and a third ; next, of this last common measure and a fourth, &c. The last common measure found, will be the greatest common measure of all the given numbers. Example 1. — ^Find the greatest common measure of 679, 5901, and 6734. By the last rule we learn that 7 is the greatest common measure of 679 and 5901 ; and by the same rule, that it, the greatest common measure of T and 6734 Tthe remaining number), for 6734 -{- 7 = 962, with no remainaer. Therefore 7 is the required number. Example 2.— Find the greatest common measure of 936, 736, and 142. 104 LEAST COMMON MULTIPLE. 1 I I The greatest common measure of 936 and 736 is 8, and the common measure of 8 and 142 is 2 ; therefore 2 is the greatest common measure of the giyen numbers. 106. Reabon or the Rule. — ^It may be shown to be correct in the same way as the last ; except that in proring the num- ber found to be a eomman measure, we are to begin at the end of all the processes, and go through all of them in succession ; and in proving that it is the greatest common measure we are to begin at the commencement of the first process, or that used to find the common measure of the two first numbers* and proceed Buccessively through all. EXERCISES. '■'■ ^- 7. Find tlie greatest common measure of 29472, 176832, and 1074. Ans. 6. 8. Of 648485, 10810, 3672835, and 473580. Atu. 5. 9. Of 16264, 14816, 8600, 75288, and 8472. Ans 8. THE LEAST COMMON MULTIPLE OF NUMBERS. 107. To find the least common multiple of two quan- tities — KuLE. — ^Divide their product by their greatest com- mon measure. Or ; divide one of them by their greatest common measure, and multiply the quotient by the other — the result of either method will be the required least common multiple. £xAMPLE. — ^Find the least common multiple of 72 and 84. 12 is their greatest common measure. 72 1^=6, and 6 x 84=504, the number sought. 108. Reason or the Rule. — It is evident that if we mul« tiply the given numbers together, their product will be a multiple of each by the other [30]. It will be easy to find the smallest part of this product, which will still be their common multiple. — ^Thus, to learn if, for example, its nine- teenth part is such. From what we have already seen [69], each of the factors of any product divided by an^ number and multiplied by the {>roduct of the other factors, is equal to the product of all the i&ctors divided by the same number. Hence, 72 and 84 being the given numbers — and multi of 72 mum 1. Ans. 2. 3. 4. 5. 6. 1( mor< E thet of \ last mul E 3 9 27, MHMM LEAST COMMOir MULTIPLE. 105 (6 18 8, aud we 2 is the to be eonreet ig the num- I at the end succession ; aeasure, we ess, or that it numbers. )f 29472, . Ant. 5. !. Am 8. BERS. wo quan- test com- r greatest by the required 2 and 84. t. ' ire mul- ^ill be a y to find be their its nine« e factors by the r all the B4 being Z£2lM (the nineteenth part of their product)a.-X84, or 72x !f. Now if I?and^ be equiralent to integers. I? X 84 will bea multiple of 84, and ^X72, will be a multiple of 72 [29] ; ,72X84 72 g^ *" 19"* 19^ ' and ,84 72 X— will each be the common 19 multiple of 72 and 84 [301. But unless 19 is a common mei^ure 72 84 of 72 and 84, — and — cannot be both equivalent to integers. Therefore the quantity by which we divide the product of the given numbers, or one of them, before we multiply it by the other to obtain a new, and less multiple of them, must be the common measure of both. And the multiple we obt%in will, evidently, be the least, when the divisor we select is the greatest quantity we can use for the purpose — that is, the greatest common measure of the given numbers It follows, that the least common multiple of two .numbers, prime to each other, is their product. xxsncisKs. 1. Find the least common multiple of 78 and 93. Ans. 2418. ^ 2. Of 19 and 72. Ans. 1368. 3. Of 464320 and 18945. ^w*. 1759308'480. 4. Of 638296 and 33888. ilm. 2703821856. 5. Of 18996 and 29932. Ans. 142147068. 6. Of 260424 and 54423. -Atm. 1574783928. 109. To find the least common multiple of three or more numbers — Rule. — ^Find the least common multiple of two of them ; then of this common multiple, and a third ; next of this last common multiple and a fourth, &c. The last common multiple found, will be the least common multiple sought. Example. — Find the least o(»nmon multiple of 9, 3, and 27. 3 is the greatest common measure of 9 and 3 ; therefore . X 3, or 9 is the least common multiple of 9 and 3. o 9 is the greatest common measure of 9 and 27 ; therefore ^ X 9, or 27 is the required least oonunon multiple. 106 LEAST COMMON MULTIPLE. k : (• 'II' |K"'«' 110. Reason of the Rule.— By the last rale it is eyident that 27 is the least common multiple of 9 and 27. But since 9 is a multiple of 3, 27, which is a multiple of 9, must also be » multiple of 8 ; 27, therefore, is a multiple of each of the given numbers, or their common multiple. It is likewise their least common multiple, because none that is smaller can be common, also, to both 9 and 27, since thej were found to have 27 as their least common multiple. exercises. 7. Find the least oommoD multiple of 18, 17, and 43. Am. 13158. 8. Of 19, 78, 84, and 61. Ans. 1265628. 9. Of 51, 176832, 29472, and 5862. Ans. 2937002688. 10. Of 537842, 16819, 4367, and 2473. Ans. 8881156168989038. 11. Of 21636, 241816, 8669, 97528, and 1847. Ans. 1528835550537452616. QUESTIONS. 1. How is the greatest common measure of two quan- tities found .? [100]. 2. What principles are necessary to prove the correct- ness of the rule ; and how is it proved ? [101, &c.]. 3. How is the greatest common measure of three, or more quantities found f [105] . 4. How is the rule proved to be correct ? [106] . 5. How do we find the least common multiple of two numbers that are composite ? [107]. 6. Prove the rule to be correct [108]. 7. How do we find the least common multiple of two prime numbers ? [108.] ' 8. How is the least common multiple of three or more numbers found ? [109]. 9. Prove the 7uie to be correct [110]. In future it will be taken for grcmted that the ;)iipa is to be asked the reasons for each rule, &o.. the 1 18 eyident But since lust also be lach of the lause none d 27, since ultiple. r, and 43. ii 002688. 68989038. 452616. >wo quan- J correct- &c.]. three, or 06]. e of two e of two liree or le ;>upiA 107 SECTION III. REDUCTION AND THE COMPOUND RULES. The pupil should now be made familiar with most of the tables given at the commencement of this treatise. REDUCTION. 1. Redaction enables us to change quantities from one denomination to another without altering their value. Taken in its more extended sense, we have often practised it already: — thus we have changed units into tens, and tens into units, &c. ; but, considered as a separate rule, it is restricted to applicate numbers, and is not confined to a change from one denomination to the Tica:^ higher, or lower 2. Keduction is either descending^ or ascending. It is reduction descending when the quantities are changed from a higher to a lower denomination; and reduction ascending when from a lower to a higher. Reduction Descending. 3. Rule. — ^Multiply the highest given denomination by that quantity which expresses the number of the next lower contained in one of its units; and add to the product that number of the next lower denomina- tion which is found in the quantity to be reduced. Proceed in the same way with the r^ suit ; and continue the process until the required denomination is obtained. Example. — ^Reduce £6 I65. 0| J. to farthings. £ s. d. 6„ 16 m 11 »0i 136 shiUing8»X6 „ 16. 12 1632 pence sjC6 „ 16 „ 0. 4 6529 farthings=£6 „ 16 „ 0^ 108 REDUCTION. H" ^m If We multiply the pounds by 20, and at the same time add the shillings. Since multiplying by 2 tens (20^ can j^iye no units in the product, there can be no units of shillings in it except those derived from the 6 of the IQs. : — we at once, therefore, put down G in the shillings^ place. Twice (2 tens' times^ 6 are 12 (tens of shillings), and one ^ten shillings), to be added from the 16«., are 13 (tens of shillings) — which we put down. £6 16«. are, consequently, equal to 1365. ^ 12 times 6(2. are 72d. : — since there are no pence in the given quantity, there are none to be added to tne 72d. — we nut down 2 and carry 7. 12 times 3 are 36, and 7 are 43. i2 times 1 are 12, and 4 are 16. £0 16s. are, therefore, equal to 1632 pence. 4 times 2 are 8, and | (in the quantity to be reduced) to be carried are 9, to be set down. 4 times 3 are 12. 4 times 6 are 24, and 1 are 25. 4 times 1 are 4, and 2 are 6. Hence £6 16s. Old. are equal to 6529 farthings. 4. Reasons or the Rux.b. — One pound is equal to 20f. ; therefore any number of pounds is equal to 20 times as many shillings; and any number of pounds and shillings is equal to 20 times as many shillings as there are pounds, plus the shillings. It is easy to multiply by 20, and add the shillings at the tame time ; and it shortens the process. Shillings are equal to 12 times as many pence; pence to 4 times as many farthings; hundreds to 4 times as many quarters ; quarters to 28 times as many pounds, &c. EXERCISES. 1. How many farthings in 23328 pence ? Atu. 93312. 2. How many shillings in Jg348 ? Aiis. 6960. 3. How many pence in £38 10s. ? Ans. 9240. 4. How many pence in £5S 13s. ? Ans. 14076. 5. How many farthings in £58 135. ? Ans. 56304. 6. How many farthings in £59 13s, 6}^. ? Ans. 57291. 7. How many pen^e in £63 05. 9d. ? Ans. 15129. 8. How many pounds in 16 owt., 2 qrs., 16 fib. ? Ans. 1864. 9. How many pounds in 14 cwt., 3 qrs., 16 fib. ? Ans. 1668. 10. How many grains in 3 fib., 5 ex., 12 dwt., 16 grains.^ Ans. 19984. 1 graij 175!S i; 1- 11 1( 56 242( 1' REDUCTIOIf. 109 le time add )an jgive no hillings in we at once, ice (2 tens' illin^s), to -which we nee in the I 7a/.— we 7 are 43. therefore, Hluced) to 5. 4 times 6. Hence 1 to 20f . ; 1 as manj s is equal » plus the igs at the pence to as man J ? Arts, I to. 76. >6304. * Ans, 5129. 16 fc.? 16 ib.? 11. How many grains in 7 ib., 11 os., 15 dwt., 14 grains ? Ans. 45974. 12. How many hours in 20 (common) years ? Ant, 175200. 13. IIow many feet in 1 English mile ? Ant, 5280. 14. How many feet in 1 Irish mile ? Ant. 6720. 15. How many gallons in 65 tuns ? Ant. 16380. 16. How many minutes in 46 years, 21 days, 8 hours, 56 minutes (not taking leap years into account) ? Ant. 24208376. 17. How many square yards in 74 square English perches ? Ans. 2238*5 (2238 and one hiOf)* 18. How many square inches in 97 square Irish perch- es ? Ans. 6159888. 19. How many square yards in 46 English acres, 3 roods, 12 perches ? Ans. 226633. 20. How many square acres in 767 square English miles } Ans. 490880. 21. How many cubic inches in 767 cubic feet ? Ant. 1325376. . 22. How many quarts in 767 pecks ? Ant. 6136. 23. How many pottles in 797 pecks ? Ant. 3188. Reduction Ascending. 5. BuLE. — ^Divide the given quantity by that number of its units which is required to make one of the next higher denomination — the remainder, if any, will be of the denomination to be reduced. Proceed in the same manner until the highest required denomination is obtained. Example. — ^Reduce 856347 farthings to pounds, &c. 4 )856347 12 )214086^ ::j 2 0)17840 „ 6i 892 „ „ 6J«=856347 forthings. 4 divided into 856347 farthings, gives 214086 pence and 3 larthings. 12 divided into 214086 pence, gives 17840 shillings and 6 pence. 20 divided into 17840 shiUings, gives £892 and no uiillings ; there is, therefore, nothing in the shillings' place of the result. ^2 110 REDUCTION. U-' ^1 lill liyide by 20 if we divide by 10 and 2 [See. II. 97]. de by 10, we have merely to cut off the unitSt if We divide To divide any, [Seo. I.' 34], ii^hich will then' be the units of shillings in the result ; and the quotient will be tens of shillings : — dividing the latter by 2, gives the pounds as quotient, and the tens of shillings, if there are any in the required quan- tity, as remainder. 6. Rbasons or thx Rui.c. — ^It is evident that every 4 farthings are equivalent to one penny, and every 12 penoo to one shilling, &o. ; and that what is left after taking away 4 farthings as often as possible fVora the farthings, must be farthing^, what remains after taking away 12 pence as often as possible from the pence, must be pence, &o. 7. To prove Jteduelion. — ^Reduction ascending and descending prove each other. Example.— JC20 175. 2|(2.i==20025 farthings; and 20021 farthing8=:j£20 17^. 2ld. £ 8. d. Reduction 20 „ 17„2i 20 417 12 5006 4 farthings. 4)20025 Reduction •{ 12 )5006| . 20)417 „ 2 4 )20025 f 12)5006} Proof I 20]lTr„ 2 Proof *20 „ 17„ 2| 417 12 5006 4 ^20 „ 17 „ 2j 20025 farthings. I' « 5: EXERCISES. 24. How many pence in 93312 farthings? Ans, 23328. 25. How many pounds in 6960 shillings? Ans. £3^, 26. How many pounds, &c. in 976 halfpence ? Ans, £2 Os. Sd. 27. How many pounds, &c. in 7675 halfpence ? Ans, £15 195. 9^d. 28. How many onnceS) and Dounds in 4352 dramB ? Ans, 272 oz., or 17 tt>. REDUCTIOK. Ill JO. II. 971. e units, if r Bhillings lillings : — >tient, and ired quan- t every 4 2 penoo to g away 4 , must be >e as often ling and aid 2002i ,2 \ 17„ 2{ iet-rthinga. Am. r.ie348. Ans, ? Ans, drams? 29. How many owt., qrs., and pounds in 1864 pounds ? Ans. 16 owt., 2 qrs., 16 Ih. 30. How many hundreds, &o., in 1668 pounds. Ans, 16 ft). many pounds Troy in 115200 grains? 14 owt., 3 qrs 31. How Ans. 20. 32. How many pounds in 107520 oz. avoirdupoise ? Ans. 6720. 33. How many hogsheads in 20658 gallons ? Ans, )27 hogsheads, 57 saflons. 34. How many . 47. How many lines in the sum of 900 feet, the 112 R£DUCTIOIC. 6/ < |« length of the temple of the ron at Balbec, 450 feet ita breadth, 22 feet the circumference, and 72 feet the height of many of its columns ? Ans. 207936. 48. How many square feet in 760 English acres, the inclosure m which the porcelain pagoda, at Nan-King, in China, 414 feet high, stands ? Ans. 33105600. 49. The ^eat bell of Moscow, now lying in a pit. the beam which supported it having been burned, weighs 360000 ft), (some say much more) ; how many tons, &c., in this quantity ? Ans. 160 tons, 14 owt., 1 qr., 4 lb. QUESTIONS FOR THE PUPIL. 1. What is reduction ? [11. 2. What is the difference between reduction descend- ing and reduction ascending ? [2]. 3. What is the rule for reduction descending ? [3] 4. What is the rule for reduction ascending? [5]. 5. How is reduction proved ? [7], Questions faumded on the Table page 3, Sft, 6. How are pounds reduced to farthings, and farthings to pounds, &c. ? 7. How are tons reduced to drams, aitd dnams to tons, &c. ? 8. How are Tror pounds reduced to graims, and grains to Troy pounds, &c. ? 9. How are pounds reduced to gnuns (apothecaries ells, &c. ? 11. How lire yards reduced to ells, or ells to yai^s, &o.? 12. How are Irish or English miles reduced to lines, or lines to Irish or English miles, &o. ? 13. How are Irish or English square miles redueed to square inches, or square inches to Iridi or English fquare miles, &c. ? and COMPOUND RULES. 113 ) foet its feet the sres, the n-Kingy 00. 1 a pit , weighs ,4fi>. descend* [5]. irthings Urns to bi, and lecaries ^Is, re- f'rench yai*ds, lines, edueed Snglish 14. How are cubic feet reduced to cubic inches, or cubic inches to cubic feet, &o. ? 15. How are tuns reduced to naggins, or naggins to tuns, &c. ' , 16. How are butts reduced to gallons, or gallons to butts, &c. ? 17. How are lasts (dry measure) reduced to pints, and pints to lasts, Sic. ? 18. How are years reduced to thirds, or thirds to years, &c. ? 19. How are degrees (of the circle) reduced to thirds, or thirds to degrees, &o. ? THE COMPOUND RULES. 8. The Compound Bulq/i, are those which relate to applicate numbers of more than one denomination. if the tables of money, weights, and measures, were constructed according to the decimal system, only the rules for Simple Addition, &c., would be required. This would be a considerable advantage, and greatly tend to simplify mercantile transactions. — ^If 10 far- things were one penny, 10 pence one shilling, and 10 shillings one pound, the addition, for e.tample, of £\ 9s. s(d. to £6 Ss. 6id. (a point being used to separate a pound, then the *' unit of comparison," from its parts, and 0*005 to express | or 5 tenths of ft penny), would be as follows — £ 1-983 6-865 Sum, 8*848 The addition might be performed by the ordinary rules, and the sum read off as follows — " eight pounds, eight shillings, four pence, and eight farthings." But even with the present arrangement of money, weights, and measures, the rules alroad;^ given for adcUtion, sub* traction, &o , might easily have been made to include the ad(Ktion, suotraotion, &c., of appUeate numbers consisting of more than one denominatioii , since the m 114 COMPOUND ADDITION. principles of both simple and compound rules are p^'e- oisely the same — the only thing necessary to bear carefully in mind, being the number of any one de- nomination necessary to constitute a unit of the next higher. iit^t'.. k, 1 iMM''. < SH' I!' COMPOUND ADDITION. 9. Rule. — ^I. Set down the addends so that •quanti- ties of the same denomination may stand in the same vertical column — units of pence, for instance, under units of pence, tens of pence under tens of pence, units of shillings under units of shillings, &c. II. Draw a separating line under the addends. ^ m. Add those quantities which are of the same denomination together — ^farthings to farthings, pence to pence, &c., beginning with the lowest. IV. If the sum of any column be less than the num- ber of that denomination which makes one of the next higher, set it down under that column ; if not, for each time it contains that number of its own denomination which makes one of the next higher, carry one to the latter and set down the remainder, if any, under the column which produced it. If in any denomination there is no remainder, put a cypher under it in the sum. 10. Example.— Add together £52 I7s, 31(2., £47 5s. 6M., and £66 1^. 2^(2. £ 8. d. 52 17 3f 47 5 6} ^addends. \ 66 .("•4 ■('.:;.'^ 8. a. 17 3n 14 211 166 17 OJ uW- f Tf* I and I ir.ake 3 farthings, which, with |, make 6 far- things ; tnese are equivalent to one of the next denomina- tion, or that of pence, to be carried, and two of the present, or one half-penny, to be set down. 1 penny (to be carried) and 2 are 3, and § are 9, and 3 are 12 pence— equal to one COMPOliND ADDITION. 115 ^ of the next denomination, or that of shillings, to be carried, and no pence to bo set down; we therefore put a cypher in the pence' place of the sum. 1 shilling (to be carried) and 14 are 15, and 5 are 20, and 17 are 37 shillings — equal to one of the next denomination, or that of pounds, to be carried, and 17 of the present, or that of shillings, to be set down. 1 pound and C are 7, and 7 are 14, and 2 are 16 pounds — equal to 6 units of pounds, to be set down, and I ten of pounds to be carried ; 1 ten and 6 are 7 and 4 are II and 5 are 16 tens of pounds, to be set down. 11. This rule, and the reasons of it, are the same as those already given [Sec. II. 7 and 9] . It is evidently not so necessary to put a cypher where there is no remainder, as in Simple Addition. 12. When the addends are very numerous, we may divide them into parts by horizontal lines, and, adding each part separately, may afterwards find the amount of all the sums. . 6i€f., Exa.mple: r- £ s. d. 57 14 21 32 16 4 £ s. d. 19 17 6 ' =r 151 7 11 8 14 2 '' 32 5 9J >=:404 11 d, 10. 47 6 41 ' 32 17 2 ' ■ 56 27 3 4 9 2 = 253 3 11 4 52 4 4 37 8 2 f 13. Or, in adding each column, we may put down a dot as often as we come to a quantity which is at least equal to that number of the denomination added which is required to make one of the next— carrying forward what is above this number, if anything, and putting the last remainder, or — ^when there is nothing left at the end— -a cypher under the column : — ^we carry to the next column one for every dot. Using the same example — >;- ■ -v. r>- r^ : .♦ ,-.;4. ^;•• vnj'=. 116 COMFOUND HULES £ 3. d. 57 •14 2 32 16 4 19 •17 •6 8 •14 2 32 5 9 47 •6 4 32 17 2 56 •3 •9 87 4 2 52 4 4 87 8 2 404 11 lo ."& U' 2 pence and 4 are 6, and 2 are 8, and 9 are 17 pcnce-^ equal to 1 shilling and 5 pence : we put down a dot and carr^ 5. 5 and 2 are ^ and 4 are ll, and 9 are 20 pence — equal to 1 shilling and 8 pence ; we put down a dot and carry 8. 8 and 2 are 10 and 6 are 16 pence — equal to ] shilling and 4 pence ; we put down a dot and carry 4. 4 and 4 are o and 2 are 10 — which, being less than 1 shilling, we set down under the column of pence, to which it belongs, &c. We find, on adding them up, that there are three dots ; we therefore carry 3 to the column of shillings. 3 shillings and 8 are 1 1, and 4 are 15, and 4 are 19, and 3 are 22 shillings — equal to 1 pound and 2 shillings ; we put down a dot and carry 1. 1 and 17 are 18, &c. Care is necessary, lest the dots, not being distinctly marked, may be considered as either too few, or too many. This method, though now but little used, seems a convement one. 14. Or, lastly, set down the sums of the farthings, shillings, &o., under their respective columns ; divide the farthings by 4, put the quotient under the sum of the pence, and the remainder, if any, in a place set apart for it in the sum — ^under the column of farthings ; add together the quotient obtained from the farthings and the sum of the pence, and placing the amount under the pence, divide it by 12 ; put the quotient under the sum of the shillings, and the remainder, if any, in a place allotted to it in the sum — under the column of pence ; add the last quotient and the sum of the shil- lings, and putting under them their sum, divide the latter by 20, set down the quotient under the Bum of COMPOUND ADDITION. 117 pence— ind cari7 56 — equal , carry 8. Uing and &re o and set down We find, therefore 8 are 11, equal to carry 1. marked, f. This lent one. irthings, ; divide n of the it apart ^; add igs and i under der the y, in a amn of le shil- ide the sum of the pounds, and put the remainder, if any, in the sum — under the column of shillings; add the last quoti')nt and the sum of the pounds, and put the result under the pounds. Using the following example — £ 8. d. 47 9 2\ 862 4 lU 51 16 2\ 97 4 6 541 13 2? 475 6 4 6 11 m 72 19 n 1651 82 47 13 farthings. 4 4 3 86 50 1655 6 2| The sum of the farthings is 13, which, divided by 4, gives 3 as quotient (to be put down under the pence), and one farthing as remainder (to be put in the sum total — under the farthings). M. (the quotient from the farthines) and 47 (the sum of the pence) are 50 pence, which, bemg put down and divided by 12, gives 4 shillings (to be set down under the shillingsV and 2 pence Tto be set down in the sum total — under tne pence). 4«. (the quotient from the pence) and 82 (the sum of the shilhngs) are 86 shillings, which, being set down and divided by 20, gives 4 pounds (to be set down under the pounds), and 6 shillings ^to be set down, in the sum total — under the shillings). JC4 (the quotient from the shillings) and 1651 (the sum of the pounds) are 1655 pounds (to be set down in the sum total — under the poundsj. The sum of the addenda is, therefore, found to be ^1655 6s. 2\d, 15. In proYinff the compound rules, we can generally avail ourselvos of tb^ methods used witii the ttiuple ^uIm [See. U. 10, &c.] 118 m\' W '' COMPO»,ND ADDITION. EXERCISES rOR THE PUFIXi Money, (1) (2) (3) « (^> . £ ». d. £ s. d. £ 8. d. £ 8. d. 76 4 6 68 14 7 76 14 7 84 3 2 67 9 9 69 16 6 67 16 9 96 4 Oi 49 10 8 72 14 8 76 19 10 41 6 183 4 11 (5) (6) (7) « (^> £ 8. d. £ «. d. £ 8. d. £ 8. d. 674 14 7 767 16 6 667 14 7 327 8 6 466 17 8 472 14 6 476 16 6 601 2 111 676 19 8 667 16 7 647 17 6 864 6 627 4 2 423 8 10 627 14 3 121 9 8i (9) (10) (11) (12) £ g. d. £ a. a £ 8. d. £ 8. d 4667 14 6 76 14 7 3767 13 11 5674 17 6i 776 16 7 667 13 6 4678 14 10 4767 16 Hi 76 17 9 67 15 7 767 12 9 8466 17 lOj 61 10 6 4 2 10 11 6 6984 2 24 44 6 6 6 3 4 3 4 11 8762 9 9 (13) (14) (15) (16) £ s. d. £ 8. d. £ 8. d. £ 8. d. 9767 6i 6767 11 6i £764 17 61 634 7 lU 7649 11 2k 7676 16 94 7467 16 6 66 7 7 4767 16 101 6948 17 H 6743 18 Oi 7 12 lOi 164 1 1 6786 7 6 67 6 6i 5678 18 8 92 7 2i 6326 8 2i 432 6 9 439 « <1^> (18) (19) (20) £ ». d. £ 8. d. £ 8. d. £ 8, d. 14 71 6674 16 7i 6674 1 H 4767 14 74 677 1 4767 17 61 4767 11 10| 743 18 74 6767 2 6 1646 19 7i 78 18 lU 7674 14 64 8697 14 Ik 8246 17 6 19 lOi 7 18 84 6634 01 4766 10 61 6044 4 1 760 6 4 67 66 47 34 cwt. 76 37 14 (4) s. d. 8 2 4 Oi 6 COMPr>UND ADDITION. 119 (21) (22) (23) (24) £ 8. d. £ s. d. £ 8. d. £ 8. d. 674 11 lU 476 14 7 674 18 8i 674 17 6i 667 14 lOj 576 15 Qi 46 16 74 128 12 2 476 4 11 76 17 74 476 4 6i 667 7i 347 16 Oi 676 11 8 677 16 04 679 18 94 476 18 94 463 14 94 678 6 81 476 6 64 (25) (26) (27) (28) — ^"^p £ s. d. £> 8. d. £ 8. d. £ 8. d 676 4 7i 649 4 6i 876 8 219 6 7 7 6 7 19 9^ 6 82 11 81 732 19 04 16 64 56 11 11 Oi 667 9i 734 19 9i 128 5 24 127 8 2 764 2 64 666 14 44 12 29 6 6i Avoirdupoise Weight. (29) (30) (31) (32) cwt. qrs. lb cwt. qrs. lb cwt. qrs. lb cwt. qrs. !b 76 3 14 44 1 16 14 8 17 66 8 14 (38) (34) (35) (86) owt. qrs. Vb cwt. qrs. R> cwt. qrs. lb cwt. qrs. lb 76 1 19 88 2 17 470 8 16 667 2 19 66 8 18 69 2 20 764 1 7 4 1 20 47 2 17 3 6 8 14 67 8 2 81 2 14 67 1 15 1 18 767 1 11 (37) (38) (39) (40) cwt. qrs. n> cwt. qrs. n> cwt. qrs. lb cwt. qrs. lb 767 1 16 476 1 24i 447 1 7 14 12 12 44 1 17 766 8 214 676 1 6 8 4 7 667 8 18 767 1 16 467 1 7i 6 16 676 1 667 2 15 668 1 6 7 8 841 2 11 973 1 12 428 04 14 87 14 2 1 16 11 66 47 3 1 11 16 87 47 1 2 16 27 67 68 1 2 17 26 128 8 12 130 COMPOUND ADDITION. Pi 1 Troy Weight. 1 (41) (42) (*»? 1 n> oz. dwi L grs. lb oz. dwt. grs. lb oz. dwt. grs. 7 5 9 6 9 7 88 7 9 8, 9 8 6 5 6 6 7 6 7 80 9 5 6 8 8 7 6 4 8 7 6' 21 11 18 (44) (45) (46) lb oz. dwt . grs. lb oz. dwt. grs. lb oz. dwt. grs. 67 9 12 14 87 3 7 12 57 10 14 11 67 9 11 11 11 12 3 11 10 ' 66 8 10 5 16 14 46 9 9 8 74 6 6 8 44 12 10 13 22 8 7 5 12 3 6 4 67 8 9 10 11 10 13 14 Cloth Measure. (47) (48) (49) ^ (50) yds. qrs. nlR. yds. qrs. nls . yds. qrs . nls . yds. qrs. nls. 99 3 1 176 3 3 37 8 2 2 1 47 1 8 47 2 2 3 5 8 2 76 3 2 7 3 3 2 3 224 2 * (61) (52) (53) (54) yds. 567 qrs. nls , yds. qrs. nls . yds. qrs. nls. yds. qrs. nls. 3 2 147 3 8 157 2 1 166 1 1 476 1 173 1 143 3 2 176 8 1 72 3 8 148 2 1 1 2 54 1 6 2 1 92 3 2 54 3 673 2 3 Wine Measure. ■"■■ (66) (56) (W) is. hhds. gls. ts. hhds. gls. ts, hhds. gls. 99 8 9 89 8 8 76 8 4 80 39 7 8 4 67 8 44 98 8 46 76 1 66 1 66 87 2 27 44 2 7 5 8 4 41 1 26 54 2 17 6 02 27 407 3 21 yr«. 99 88 77 2^ COMPOUND ADDITION. 121 ) iwt. gn. 9 8 8 6 7 6 y a* gls. 4 u Time. (68) (69) (60) jra. ds. hrs. ins. yrs. ds. hrs. ihd. jrs. dik lars. ms. 1 1 wt. grs. 14 11 11 10 9 8 7 6 13 14 (50) qrs. nls. 2 1 8 2 3 (64) — irs. : qIs. 1 1 3 1 1 2 8 1 99 359 88 77 120 9 8 7 56 67 49 205 115 2 42 60 6 6 90 76 1 1 8 2 50 57 58 59 127 7 60 120 9 44 76 121 11 44 6 47 3 41 8 9 11 17 61. What is tbe sum of the following: — three hun- dred and itinety-siz pounds Toar shillings and two pence ; five hundred and seventy-three pounds and four pence halfpenny ; twenty-two pounds and three halfpence ; four thousand and five pounds six shillings and three farthings? Ans. ^£4996 lOs. 2id. 62. A owes to B je567 16*. 7|rf. ; to C ^£47 16*. ; and to D £56 Id. How much does he owe in all? Ans, je671 12*. S^d. 63. A man has owing to him the following sums :-— £3 10*. 7d. ; £46 7^d. ; and £52 14*. 6d. How much is the entire ? Ans. J&102 5*. 8^d. 64. A merchant sends off the following quantities of butter :— 47 cwt., 2 qrs., 7 lb ; 38 cwt., 3 qrs., 8 ft) ; and 16 cwt., 2 qrs., 20 lb. How much did he send off in all ? Am, 103 cwt., 7lb, 65. A merchant receives the following quantities of tallow, viz., 13 cwt., 1 qr., 6 lb ; 10 cwt., 3 qrs., 10 ft); and 9 cwt., 1 qr., 15 ft). How much has he received in all } Ans. 33 cwt., 2 qrs., 3 ft). 66. A silversmith has 7 ft), 8 oz., 16 dwt. ; 9 fc, 7 oz., 3 dwt. ; and 4 &, 1 dwt. What quantity has he ? Atis. 21 ft), 4 oz. 67. A merchant sells to A 76 yards, 3 quartiers, 2 nails ; to B, 90 yai'ds, 3 duartfers, Snails ; and to C, 190 yards, 1 nail. HdW miich has he sold in all ? Am. 357 yards, 3 quarters, 2 nails. 68. A wine tnerbhant receives from his correspondent 4 tuns, 2 hogsheads ; 5 tuns, 3 hogsheids ; and 7 tuns, 1 hogshead. How much is the entire ? Ans, 17 tuns, 2 122 COlfPOUFD ADDITION. I' -' 69. A man has three farir.<4, the first contains 120 acres, 2 roods, 7 perches ; tbe second, 150 acres, 3 roods, 20 perches ; and the third. 200 acres. How much land does he possess in all ? An^. 471 acres, 1 rood, 27 perches. 70. A servant has had three masters ; with the first he lived 2 years and 9 months ; with the second, 7 years and 6 months ; and with the third, 4 years and 3 months. What was the servant's age on leaving his last master, supposing he was 20 years old on going to the first, and that he went directly from one to the other ? Ans. 34 years and 6 months. 71. How many days from the 3rd of March to the 23rd of June ? Am. 112 days. 72. Add together 7 tons, the weight which a piece of fir 2 inches in diameter is capable of supporting ; 3 tons, what a piece of iron one-third of an inch in diameter will bear ; and 1000 lb, which will be sustained by a hempen rope of the same size. Ans. 10 tons, 8 cwt., 3 quarters, 20 lb. 73. Add together the following: — 2d.y about the value of the Koman sestertius ; 7^^., that of the dena- rius; 1^^., a Greek obolus ; 9d.y a drachma; £3 15s. a mina ; J8225, a talent ; Is. 7d., the Jewish shekel ; and iE342 3*. 9d.y the Jewish talent. Ans. ie571 2s. 74. Add together 2 dwt. 16 grains, the Greek drachma; 1 ib, 1 oz., 10 dwt., the mina ; 67 lb, 7 oz., 5 dwt., the talent. Ans. 68 lb, 8 oz., 17 dwt., 16 grains. QUESTIONS FOR THE PUPIL. 1. What is the difference between the simple and compound rules? [8]. 2. Might the simple rules have been constructed so as to answer also for applicate numbers of different denominations } [8] . 3. What is the rule for compound addition ? [9]. 4. How is compound addition proved .^ [15]. 5. How are we to act when the addends are numei^ ous ? [12, &c.] 16. under farthi COMPOUND SUBTRACTION. 123 Mns 120 acres, 3 aw much rood, 27 the first icond, 7 s and 3 ving his m going 3 to the i to the a piece •ting; 3 inch in istained tons, 8 >ut the e dena- J3 155. el ; and ■ achma; vt.y the e and ted so fferent COMPOUND SUBTRACTION. 16. Rule — ^I. Place the digits of the subtrahend under those of the same denomination in the minuend- farthings under farthings^ units of pence under units of pence, tens of pence under tens of p^**ce, &c. U. Draw a separating line. III. Subtract each denomination of the subtrahend from that which corresponds to it in the minuend — beginning with the lowest. IV. If any denomination of the minuend is less than that of the subtrahend, which is to be taken from it, add to it one of the next higher — considered as an equi- valent number of the denomination to be increased ; and, either suppose unity to be added to the next deno- mination of the subtrahend, or to be subtracted from the next of the minuend. V. If there is a remainder after subtracting any denomination of the subtrahend from the correspond- ing one of the minuend, put it under the column which produced it. YI. If in any denomination there is no remainder, put a cypher under it — ^unless nothing is left from any higher denomination. 17. Example.— Subtract £56 IZs. 4J(f., from £96 7s. 6id. £ s. d. 96 7 6|, minuend. 56 13 4|, subtrahend. umei^ 39 14 11, difference. We cannot take J from j, but — borrowing one of the pence, or 4 farthings, we add it to the |, and then say 3 far- things from 5, and 2 farthings, or one halfpenny, remains : we set down \ under the farthings. 4 pence from 5 (we have borrowed one of the 6 pence), and one penny re- mains : we set down 1 under the pence (l^rf. is read " three halfpence"). 13 shillings cannot oe taken' from 7, but (bor- rowing one from the pounds, or 20 shillings) 13 shillings from 27, and 14 remam : we set down 14 in the shillings' place of the remainder. 6 pounds cannot be taken from 5 (we have borrowed one of the 6 pounds in the minuend) ;i I It 124 COMPOUND SUBTRACTION. but 6 from 15, and 9 remain : we put 9 under the units of pounds. 5 tens of pounds from 8 tens (we have borrowed one of the 9), and 3 remain: we put 3 in the tens of pounds' place of the remainder. 18. This rule and the reasons of it are substantially the same as those already given for Simple Subtraction [Sec. II. 17, &o.] It is evidently not so necessary to put down ovphers where there is nothing in a d«>inomination of the remainder. 19. Compound may be proved in the same way as simple fubtraotion [Sec. II. 20]. ^' & JlvoirdupoUe Weight. (19) . (20) cwt. qrs. Id cwt. qrs. R) From 200 2 26 276 2 l6 Take 99 8 16 27 2 7 100 8 11 ■ ' (21) ^ cwt. qrs. ID 9664 2 25 9074 27 (22) cwt. qrs. 9^ 654 476 8 6 MP from fake (1) £ s. d. From 1098 12 6 Take 484 16 8 SZERCISEi. (2) (8) £ 9. d. £ 9. d. 767 14 8 76 15 6 486 18 9 14 6 £ 9. d. £ 9. d. 47 16 7 97 14 6 89 17 4 6 16 7 From Take 668 16 10 1 1 £ 9. d. From 98 14 2 Take 77 16 8 (7) (8) £ 9. d. £ 9. d. 47 14 6 97 16 6 88 19 9 88 17 7 (9) (10) £ 9. d. £ 9. d. 147 14 4 660 15 6 120 10 8 477 17 7 From Take 1 £ 9. d. From 99 18 8 Take 47 16 7 (12) (18) (14) £ 9. d. £ 9. d. £ 9. d. 767 14 bk 891 14 U 676 18 H 476 6 74 677 16 6| 467 14 9i 33 42 y iB15 tney iS6 1 (15) £ 9. d. From 567 11 51 Take 479 10 10^ (16) (17) (18) £ 9. d. £ 9. d. £ 9. d, 971 Oi 487 16 478 10 7 11 14 47 11 Ok 3^ 23 fl muc 27 1 COMPOUND SUBTRACTION. 1S3 units of )orrowed pounds' iaily the [Seo. II. ovphers nuer. B simple £ 9. d. 97 14 6 6 16 7 (10) £ ». d. ^60 15 6 77 17 7 (H) 8. d. 18 11 14 H (18) 8. d, 10 11 Oi [22) qrs. fk 3 5 (23) lb OS. dwt. gr. vVom 554 19 4 fake 97 16 15 Troy Weight. (24) ID OS. dwt. gr. 940 10 17 23 (24) lb OS. dwt. gr. 917 14 9 798 18 17 457 9 2 18 Wine Measure. (26) (27) (28) (29) is. hhds. gls. tfl. hhds. gls. is. bhds. gls. ts. hhds. gis. From 818 15 64 027 804 054560 1 Take 29 2 26 8 42 100 8 51 27 2 25 2 62 Time. (80) (81) (82) yrs. da. hs. ms. yrs. ds. ha. ms. yrs. da. ha. nis. From 767 181 6 80 476 14 14 16 567 126 14 12 Take 476 110 14 14 160 16 18 17 400 15 291 20 16 16 33. A shopkeeper bought a piece of cloth containing 42 yards for jS22 10;., of which he sells 27 yards for £15 155. ; how many yards has he left, and what have they cost him ? Ans, 15 yards ; and they cost him £6 155. 34. A merchant bought 234 tons, 17 cwt., 1 quarter, 23 ft), and sold 147 tons, 18 cwt., 2 quarters, 24 m ; how much remained unsold ? Ans, 86 tons, 18 cwt., 2 qrs. 27 1b. 35. If from a piece of cloth containing 496 yards, 3 quarters, and 3 nails, I cut 247 yards, 2 quarters, 2 nails, what is the length of the remainder ? Ans, 249 yards, 1 quarter, 1 naU. 36. A field contains 769 acres, 3 roods, and 20 perches, of which 576 acres, 2 roods, 23 perches are tilled ; how much remains untilled ? Ans, 193 acres, 37 perches. 37. I owed my friend a bill of iB76 165. 9^^., out of which I paid £59 175. lO^d. ; how much remained dae ? Ans, £16 185. lOid, 126 COMPOUND MULTIPLICATION. i '' 38. A merchant bought 600 salt ox hides, weighing 561 cwt.,2 ib; of which he sold 250 hides, weighing 239 cwt., 3 qrs., 25 lb. How many hides had he left, and what did they weigh ? Ans, 350 hides, weighing 321 cwt., 5 lb. 39. A merchant has 209 casks of butter, weighing 400 cwt., 2 qrs., 14 lb; and ships off 173 casks, weighing 213 cwt., 2 qrs., 27 lb. How many casks has he left ; and what is their weight ? Ans. 36 casks, weighing 186 cwt., 3 qrs., 15 lb. 40. What is the difference between 47 English miles, the length of the Claudia, a Roman aqueduct, and 1000 feet, the length of that across the Dee and Vale of Llangollen } Ans. 247160 feet, or 46 miles, 4280 feet. 41. What is the difference between 980 feet, the width of the single arch of a wooden bridge erected at St. Petersburg, and that over the Schuylkill, at Phila- delphia, 113 yardj3 and 1 foot in span ? Ans. 640 feet QUESTIONS FOR THE PUPIL. 1. What is the rule for compound subtraction ? [16]. 2. How is compound subtraction proved ? [19]. to whi^ out a 22. I W COMPOUND MULTIPLICATION. 20. Since we cannot multiply pounds, &o., by pounds, &c., the multiplier must, in compound multiplication, be an abstract number. 21. When the multiplier does not exceed 12 — Rule — ^I. Place the multiplier to the right hand side of the multiplicand, and beneath it. II. Put a separating line under both. III. Multiply each denomination of the multiplicand by the multiplier, beginning at the right hand side. IV. For every time the number required to make one of the next denomination is contained in any pro- duct of the multiplier and a denomination of the multi- plicand, carry one to the next product, and set down the remainder (if there is any, after subtracting the number equivalent to what is carried) under the denomination COMPOUND MULTIPLICATION. 127 weighing weighing 1 he left, weighing weighing '3 casks, ;asks has 16 casks, 9h miles, md 1000 Vale of 280 feet, feet, the ected at tt Phila- 140 feet ? [16]. pounds, ication. t hand plicand e. make Qy pro- multi- wn the lumber ination to which it belouffs ; but should there bo no remainder, out a cypher in that denomination of the product. 22. EXAMPL2.— Multiply £Q2 Us. lOd. by C. £ $. d. 62 17 10, multiplicand. 6, muUiplier. 377 7 0, product. Six times 10 pence are CO pence ; these are equal to 5 shillines (5 times 12 pence) to be carried, and no pence to be set down in the product — we therefore write a cypher in the pence place of the product. 6 times 7 are 42 shillings, and the 5 to be carried are 47 shillings— we put down 7 in the units^ place of shillings, and carry 4 tens of shillings. 6 times 1 (ten shillings^ are 6 (tens of shillings), and 4 (tens of shillings) to be earned, are 10 (tens of shillings), or 5 pounds (o times 2 tens of shillings) to be carried, and nothing, (no ten of shillings) to be set down. 6 times 2 pounds are 12. and 5 to be carried are 17 pounds— or 1 (ten pounds) to be carried, and 7 (units of pounds) to be set down. 6 times 6 j^tens of pounds) are 36, and 1 to be carried are 37 (tens or pounds). 23. The reasons of the rule will be very easily understood from what we have already said [Sec. II. 41]. But since, in compound multiplication, the value of the multiplier has no connexion with its position in reference to the multiplicand, where we set it down is a mere matter of convenience ; neither is it so necessary to put cyphers in the product in those deno- minations in which tnere are no significant figures, as it is in simple multiplication. 24. Compound multiplication may be proved by re- ducing the product to its lowest denomination, dividing by the multiplier, and then reducing the quotient Example.— Multiply £4 3<. 8<2. by 7. £, s. d. Proof : 4 3 8 29 5 8 7 20 29 5 8, product. 585 12 7 )7028, product reduced. 1 2)1004 26 p3' 8 quotient reduced 4 3 8s=multiplicand. 128 COMPOUND MULTIPLIGATIOIf. £29 6». Sd. axe 7 times the multiplioand ; ift therefbre, th« process has been rishtly performed, the seTeath part of this should be equal to the multiplicand. The quantities are to be ** reduced," before the diTition by 7, since the learner is not supposed to be able as yet to divide £29 6s. 6d. EXAl » 1 VZKllCISSS. I ' 9. d, 9 8. 7i. 2 6. £ », d. £ 1. 76 14 74X 2= 168 2. 97 13 6iX 3= 298 8. 77 10 74X 4= 810 4. 96 11 7iX 6=: 482 18 li. 6. 77 14 6|X 6^ 466 7 li 6. 147 18 8iX 73=1038 18 Oi. 7. 428 12 7iX 8=3429 1 0. 8. 672 16 6 X 9=6166 8 6. 9. 428 17 8 X 10=4288 12 6. 10. 672 14 4 Xll=t399 17 8. 11. 776 16 6 X12:=^321 6 0. 12. 7 lb at 59. 244. ^, will cost £1 16«. ^d. 13. 9 yards at 10«. lUd. W» will cost £4 iSa. Bid. 14. 11 gallons at 13«. 9d. 4^* will cost £7 11«. 3d. 16. 12 h at £1 3s. id. ^, will cost £14. 25. When the multiplier exceeds 12, and is a com- posite number — BuLE. — ^Multiply sucoessively by its factors Example 1.— Multiply £47 13s. 4d. by 56. £ 8. d. 47 13 4 7 56=7x8 £ s. d. S33 13 4»47 13 4x7. 8 The 26. numb< Rui EXAl The [Sec. I] the pai multip 2669 6 8=47 13 4x7x8, or 56. Example 2.— Multiply 14s. 2d. by 100. t. d. 14 2 10 100=10x10 : «. d. £7 1 8-44 2x10. 10 £7016 8»14 2x10x10, or 100. COMPOUND MULTIPLICATION. 129 e|bre, the rt of this Bion by 7, to divide Example ^. — Multiply £8 25. ^. by 700 £ s. d. 8 2 4 10 £ s. d. 81 3 4i Bs8 2 4x10. 10 «8 2 811 13 4: 4x10: 7 S8. 5id. iU. 8d. a corn- 6681 13 4 =8 2 4x10x10x7, or 700. The reason of this rule has been already given [Sec. II. 60]. 26. When the multiplier is the sum of composite numbers — KuLE. — Multiply by eacli, and add the results. ExAMPLE.^Multiply £Z 14s. 6d. by 430. £ 8. d. 3 14 6 10 " J — — — _ £ s. d, £ t. d. 37 5 x3=lll 15 0, or 3 14 6x30. 10 372 10 0x4=1490 0, or 3 14 6x400. 1601 15 0, or 3 14 6x430. The reason of the rule is the same as that already given [Sec. II. 62]. The sum of the products of the multiplicand by the parts of the multiplier, being equal to the product of the multiplicand by the whole multiplier. XXXRCIBE8. £ 8. d. £ 8. d. 16. 3 7 6 X 18= 60 15 0. 17. 4 16 7 X 20= 96 11 8. 18. 5 14 6kX 22=125 19 11. 19. 2 17 6 X 86=103 10 0. 20. 8 16 7 X 66=214 8 8. 21. 2 8 6 X 64;;?139 4 0. 22. 8 4 7 X 81=261 11 8. 28. 9 4 X 100a« 46 13 4. 24. 16 4 X1000=^16 13 4. 25. 100 yards at 9«. iid. ^, will cosib £46 17 6. 26. 700 gallons at 18«. 4^. ^, will cost 4iS6 13 4. 27. 240 gallons at 6«. 8d. ^, will cost 80 0. 28. 860 yards at 18«.4<<.4r>wiU coat 240 0. i^t I i 1 130 COMPOUND MULTIPLICATION. tr ! 27 If the multiplier is not a composite number— Rule. — Multiply successively by the factors of the nearest composite, and add to or subtract from the pro- duct so many times the multiplicand as the assumed composite number is less, or greater than the giv^'V multiplier. Example 1 —Multiply £62 12s. Qd. by 76. £ s. d. 62 12 6 8 76=8x9+4 501 9 £ s. d. 4509 0=62 12 6x8x9, or 72. 250 10 0=62 12 6x4. 4759 10 0=62 12 6x8x9+4, or 76. Example 2.— Multiply £42 3«. 4d. by 27. £ s. d. 42 Z 4 4 27=4x7-1 168 13 4 7 1180 13 4=42 42 3 4=42 $. 3 3 d. 4x4x7, or 28. 4x1. 1138 10 0=42 3 4x4x7-1, or 27. The reason of the rule is the same as that already eiven [Sec. II. 61]. "^ ^ EXERCISES. £ 8. d. £ 8. d. 29. 12 2 4 X 83= 1005 13 8. 30. 15 04X146= 2193 3 OJ. 31. 122 5 X102= 12469 10 0. 32. 963 01x999—962040 2 6i. SV8. When the multiplier is large, we may often con- veniently proceed as follows — lluLE.---Write once, ten times, &c., the multiplicand, and, multiplying these respectively by the units, tens &o., of the multiplier, add the results. COMPOUND MULTIPLICATION. 131 r— . I of the the pro- issumed 76. 7. iy given m con- Jicand, I, tens ISxAMPLE.—Multiply £47 16s. 2d. by 5783. 5783=5 X 1000+7 X lOO-f-8 x 10+3 X 1. £t s. d. £ s. d. Units of the multiplicand, 47 16 2x3= 143 8 6. • - •• ' ...■. 10 -.t: I Tens of the multiplicand^ 478 1 8x8= 10 Hundreds of the multiplicand, 4780 16 8x7: 10 3824 13 4. 33465 16 8. Thousands of the nmltiplicand, 47808 6 8x5 = 239041 13 4. Product of multiplicand and multiplier =276475 11 10. BZERCISX8. £ 8. 88. 76 14 84. 971 14 85. 780 17 86. 73 17 87. 42 7 ' 88. 76 gai 89. 92g.M.. £ 8. 92= 7067 18 76 = 74077 16 92 = 71889 14 9013 10 :^ w':i X X X 7^X122 1 7iXl62= 6866 11 lOi. ^8 at £0 13 4 ^. will cost £50 18 14 2 ^, will cost 65 8 d. 8. 8. 8. 8. ./ 1 fr a' ■m^ 4. 4. 40. What is l'^^ difference between the price of 743 ounces of gold at £3 lis. lO^d. per oz. Troy, and that of the same weight of silver at 62d. per oz. ? Ant. £2701 2t. S^d, ,;, , tfc ^^ ..^,;: ;^,,,, 41. In the time of King John (money bemg then more valuable than at present) the price, per day, of a cart with three horses was fixed at U. 2d. ; what would be the hire of such a cart for 272 days ? Am. ^15 175. 4d. 42. Veils have been made of the silk of caterpillars, a square yard of which would weigh about 4 grains ; what would be the weight of so many square yards of this texture ap would cover a square English mile ? Am. 2151 fc, I OB., 6 dwt., 16 grs., Troy. ^ , .. .. J';/ QUESTIONS TO BE ANSWERED BY THE PUPIL. 1. Can the multiplier be 9,Xi. applicate number ? [20J. 2. What is the rule fox compound miiltiplicatioii when the multiplier does not exceed 12 ? [21]. ' '" , ,' 3. What is the rule when it exceeds 12, and is a composite number ? [25]. ,,, ■**»1/P,ls.j!i'"" i >: i32 COMPOUND DIVISION. 4. When it is the sum of composite numbers ? [26]. 6. When it exceeds 12, and not a composite number ? [27]. 6. How is compound multiplication proved ? [24], COMPOUND DIVISION. 29. Compound Divii^on enables us, if we divide an applicate number into any number of equal parts, to ascertain what each of them will be ; or to find out how many times one applicate number is contained in another. If the divisor be an applicate, the quotient will be an abstract number — for the quotient, when multiplied by the divisor, must give the dividend [Sec. II. 79] ; but two applicate numbers cannot be multiplied together [20]. If the divisor be abstract, the quotient will be applicate — for, multiplied by the quotient, it must give ihe dividend — an applicate number. Therefore, either divisor or quotient must be abstract. 30. When the divisor is abstract, and does not ex- ceed 12 — Rule — I. Set down the dividend, divisor, and sepa- rating line — as directed in simple division [Sec. II. 72]. n. Divide the divisor, successively, into all the deno- minations of the dividend, beginning with the highest. III. Put the number expressing how often the divisor is contained in each denomination of the dividend under that denomination — and in the quotient. ' IV. If the divisor is not contained in a denomina- tion of the dividend, multiply that denomination by the number which expresses how many of the next lower denomination is contained in one of its units, and add the product to that next lower in the dividend. V. " Reduce" each succeeding remainder in the same way, and add the product to the next, lower denomi- nation in the dividend. ' V ' ' ' h"^- - ; VI. If any thing is left after the quotient from the lowest denomination of the divideivd is obtained, put it COMPOUND DIVISIOIf. 13t ' [26]. lumber ? 24]. ivide an arts, to fiud out ained in ill be an plied by 9] ; but together will be ust give J, either not ex- id sepa- n. 72]. le deno- ;hest. J divisor d under nomina- by the t lower md add le same lenomi- om the put it down, with the divisor under it, and a separating line between : — or omit it, and if it is not less than half the divisor, add unity to the lowest denomination of the quotient. 31. Example 1.— Divide ^72 6s. Ud. by 5. £ s. d. 5)72 14 9 4| 5 will go into 7 (tens of pounds) once (ten times), and leave 2 tens. 5 will go into 22 (units of pounds) 4 times, and leave two pounds or 40s. 40s. and 6s. are 46s., into which 5 will go times, and leave one shilling, or 12c^. 12^. and 9d, are 21K. — ^Divide successively by the factors. EXAMPUE.— Divid* £12 17«. 9:. 12.' 24 17 e-h 24= 1 81. - ' '• k '■ . ^ ■ ^:... ; 18.. 676 13 8-e- 36=16 4i. ; !#. 447 12 2-*- 48an 9 6 6, 16. 547 12 44- 66= 9. 16 7. 16. 9740 14 64-120=81 3 6i. •■■ n. 740 18 44-49=15 2 81. 85. When the fSvisor exceeds 12, and is not a com- ^site number— Rule. — ^Proceed hy itie method of long division; but in performing 4ihe mnltiplioation of the remainders by the numbers which make them respectitely a deno- mination lower, and adding to. the produots of ,that next lower denominaUon whatever is already in ih^ dividend, set down the multiplier^, &c. obtained. I^la^e the ,^ii»- tient as directed in long division [Sec. 11. 89]. • -^ > COMPOUND DIVISION. 135 ExAMPLB.— Divide £87 Ifis. 4d. by 62. £ s. d. £ s. d. 62)87 16 4 (1 8 4. 62 25 ^ ' • • • 20 multiplier. afaiUings 516 («=:25x 20+16) - "20 ' 12im:nip" pence 244(=is\^xl2-t-4) 186 "68 4 multiplier farthings 232 (:=:58x4) 186 "46 62 goes into £87 once (that is, it gives £1 in the quotient), and leaves £25. £25 are equal to 5005. (25x20), which, with I65. in the dividend, make 516s. 62 goes into 516.9. 8 times (that is, it gives 8s. in the quotient), and leaves 20s., or 240d. ('20x 12) as remainder. 62 eoes into 240, &o. Were we to put f in the quotient, the remainder would h« 46, which is more than half the divisor ; we consider the quotient, therefore, as 4 farthings, that is, we add one penny to f3) the pence supposed to t)e already in the quotient. £1 8s. 4d. is nearer to the true quotient than £1 8s. 3^(2.[32]. This is the same in principle as the nle giyen aboye [30]— but since the numbers are large, it is more convenient actually to set down the sums of the different denominations of the divi- dend and the preceding remainders (reduced), the products of the divisor and quotients, and the numbers by which we multi* ply for the necessary reductions: this prevents the memory from being too much burdenea [Sec. II. 93]. 36. When the divisor and dividend are both applicate numbers of one and the game denomination^ and no reduction is required — Rule. — Proceed us already directed [Seo. 11 70| 72, or 89]. :.;:::" h. ,:> : v' '• \--» 136 COMPOUND DIVISION. ExAMPUB. — ^Diyide JB45 by £5. ;> , ^ i . j f m . ' ' ; £ £5)45 That is £5 is the ninth part of £45. 37. When the divisor and dividend are applicate, but not of the same denomination ; or more than one de- nomination is found in either, or both — Rule. — Reduce both divisor and dividend to the low- est denomination contained in either [3], and then pro- ceed with the division. Example.— Divide £37 5s. 9Jd. by Zs. 6irf. 8. d. £ 8. d, 3 64 37 5 91 12 20 PmI If. ;i 42 4 170 farthings. 745 12 8949 4 170)35797(211 340 179 170 Therefore 3«. eid. is the 211th part of £37 5s. 9|d. AS OS 97 not being less than the half of 170 [32], we consider it equal to the divisor, and therefore add 1 to the obtained the last quotient. XZERCIBES. ■V i--'- i'ii £ 8. 18. 176 12 19. 134 17 20. 4786 14 21. 73 16 22. 147 14 23. 167 16 24. 68 15 25. 62 10 26; 8764 4 27. 4728 16 28. 8^4 29. 6286 2 80. 4698 4 d. 2 8 7 7 6 7 2 71 £ 8. d. 191= 18 6. 183= 14 9. 443=slO 13 104 271= 5 6i 973=0 487=0 761=0 3 6 1 6i^ 419= 2 Oi^ 468=18 14 317=14 18 Oi 6|. 6|. llj. 6i. 4i. 54-|. 261=81 10 Hi 875= 2 .f.98423 19 9 8i. 44. COMPOUND DIVISION. 137 i , t i sate, but one de- the low- aen pro- . 18 the J. 9\d. isider it >btained ;;. '•"♦T > J.v 31. A cubic foot of distilled water weighs 1000 ounces what will be the weight of one cubic inch ? Ans 253*1829 grains, nearly. 32. How many Sabbath days* journeys (each 1155 yards) in the Jewish days' journey, which was equal to 33 miles and 2 furlongs English ? Ans. 50*66, &c. 33. How many pounds of butter at llf^. per ib would purchase a cow, the price of which is JS14 15f. } Ans. 301*2766. ;, i .'A'...> rl it '-' •-••^it I'. ■ .1 \ • :> 1 < 1 I i, 1 h ■'< .;;v Hit • r •i. w. ' I T,/: '( ^I<;: ,'. - ' • J . '• V ." 1 j* . /" ■ - " •'< 'Si> / °» }♦ . , u . J- . : ( .--r » ' V- " ' ' ' L •iij • t '■ -t-; 1' I ' ' ' ,, :-■ . • ;» ,r-. i'l ' .^> Vf; i.'r • nh to :/■■ it ■ / i' * f ;.U -Si* - :;■. . L i . ! ' ■ • :-.;''~^ vr[y m'tv' ■•:•;';: ■ •• ► - . . .' f .n:;u .1: il ;n>r 4 *vi 138 >U.' nttr*. " . ''5 '^ '»-. V' ' It; i^iiiil SECTION IV. • ' ' FRACTIONS. , ." V \ 1. If one or more units are divided into equal parts, and one or more of these parts are taken, we have what is called a fradvm. Any example in division — before the process has been performed — may be considered as affording a fraction : — thus f (which means 5 to be divided by 6 [Sec. II. 68] ) is a fraction of 5 — its sixth part ; that is, 5 being divided into six equal parts, | will express one of them ; or (as we shaU see presently), if unity is divided into six equal parts, five of them will be represented by |. ''^ ' ■^'- 2. When the dividend and divisor constitute a frac- tion, they change their names — ^the former being then termed the numerator^ and the latter the denominator ; for while the denominator tells the dmonwnation or kind of parts into which the unit is supposed to be divided, the numerator numerates them, or indicates the number of them which is taken. Thus ^ (read three- sevenths) means that the parts are " sevenths,'* and that " three" of them are represented. The numerator and denominator are called the terms of the fractions. 3. The greater the numerator, the greater the value of the fraction — ^because the quotient obtained when we divide the numerator by the denominator is its real value ; and the greater the dividend the larger the quotient. On the contrary, the greater the denomina- tor the less the fraction — since the larger the divisor the smaller the quotient [Sec. II. 78] : — Whence | is greater than 4 — which is expressed thus, |]^4 ; but | is less than 4 — ^which is expressed by f 4. Since the fraction is equal to the quotient of its numerator divided by its denominator, as long as this quotient is unchanged, the value of the fraction is tho ^me, though its fDrm may be altered. Hence we can multiply or divide both terms of a fraction by the same ifiumber without affecting its Talue ; since this is equally to inoi which 5. and fivl The it eqna\ which subjoii Thel that h< the na 6. T as eitl identic .;.'. CJ In belouj five b La times FRACTIONS. 139 »• 1 J parts, kve what las been 5tion : — [I. 68]) ; divided ; or (as iz equal a frao- t^ then inator ; Hon or [ to be Eites the i three- nd that tor and e value iien we its real ;er the omina- divisor 3 4 is but I of its la this is the VQ can 3 same qually to increase or diminish both the dividend and divisor — whioh does not affect the quotient. '"' ' ' " 6. The following will represent unity, seten-scvcnths, and five-sevenths. I ? I i-J__iJi " I ^"'^y 1 UILD [HHu The very faint lines indicate what | wants to make it equal to unity, and identical with ^. In the diagrams which are to follow, we shall, in this manner, ^bnerally subjoin the difference between the fraction and unity. The teacher should impress on the mind of the pupil that he might have chosen any other unity to exemplify the nature of a fraction. 6. The following will show that f may be considered as either the ^ of 1, or the \ of 5, both — though not identical — being perfectly equal. i.,,j,_«.;,> . ''.\'\r.:, « , . :..-••■' -' *■ | of 5 UnitS. "1 ' V' •' ■i','» W Unity. I f of 1 un it Trmn "1 1 v^ti -»)»■ 1 1 > .>-■ i * * » ■■ ; en In the one case we may suppose that the five parts belong to but one unit ; in the other, that each of the five belongs to different units of the same kind. Lastly, I may be considered as the \ of one unit five times as large as the former ; thus — -.i I of 1 unit. ^. I of 5 units. .'''.' \\'i. r.,:.,fr "i«4 equal to ,r^i:^.- iU: ■ r VI -_-t> ' I ; I ->■ ! 140 rHACTIOIfS. I' t I h I I*' J: s ■ I ! 7. If its numerator is equal to, or greater than its denominator, the fraction is said to be improper; be- cause, although it has the fractional form, it b equal to, or greater than an integer. Thus } is an improper fraction, and means that each of its seven parts is equal to one of those obtained from a unit divided into five equal parts. When the numerator of a proper fraction is divided by its denominator, the quotient will be ex- pressed by decimals; but when the numerator of an improper fraction is divided by its denominator, part, at least, of the quotient will be an integer. It is not inaccurate to consider } as a fraction, since it consists of '* parts " of an integer. It would not, however, be true to call it part of an integer ; but this b not required by the definition of a fraction — ^which, as we have ^Ad, consbts of " part," or " parts " of a unit [1]. 8. A mixed number is one that contains an integer and a fraction; thus If — ^which b equivalent to, but not identical with the improper fraction }. The fol- lowing will exemplify the improper fraction, and its equivalent mixed number — ' r Unity. I -.»• ■;■. ,1-, l-i :',> y* » • ' '^j Unity -h ^ %•? <°v,-\- invi}' I If 9. To reduce an improper frtiction to a mixed number An improper fraction b reduced to a mixed number if we divide the numerator by the denominator, and, after the units in the quotient have been obtained, set down the remainder with tl^o divbor under it, for denominator ; thus } b evidently equal to If — ^as we have already noticed when we trearted of division [Sec. 11. 71]. 10. A simple fraction has reference to one or mora integers; thus ^ — ^whioh means, as we have seen [6], the ySvc-seyenihs of one unit, or the one-seventh of ^ti units. 11. refer (three^ the fol three-f four pf parts, .i.;« .t... 'lyi » M4*~i!.'-i VJUCTIOKA. 141 er than iti yroper; be- it is equal 1 improper rts 18 equal d into five er fraotioQ i^ill be ex- itor of an ator, part, tion, flince rould not, but this tt — ^which, rts " of a n integer ^ to, but The fol- and ltd •■».! number imber if id, after it down linator ; already i. >r mora mf6], otfiv 11. A eompownd fraction suppoaes one fraction to refer to another ; thus f of | — represented also by | X | (three-fourths multiplied by four-ninths), means not the four-ninths- of unity, but the four-ninths of the three-fourths of unity : — that is, unity beinff divided into four parts, three of these are to be divided into nine parts, and then four of these nine are to be taken j tbus — i nt« ■" ^ •s *l« „». '•. 12. A comjplez fraction has a fraction, or a mixed a number in its numerator, denominator, or both ; thus -, 4 which means that we are to take the fourth part, not of unity, but of the f of unity. This will be exem- plified by— u J f A^ {. • /^ -> \^ -j"t r^, are complex fractions, and will be better understood when we treat of the division of fmcioT^s. 13. Fractions are also distinguished by the nuiare of their denominators. When the denominator is unity^ followed by one or more cyphers, it is a de/imal frao- Hon — ^thus, /g-, j^^sj ^^- y ^U other ir&e.thuB are vulgar —thus, f , f , ^f T, &c. Arithmetical processes may often be performed with fractions, without aclttally dividing the numerators by the denominators. Since a fraction, like an inteser, may be increased or diminished, it is capable of audi- tion, subtraction, &c. :i*M{>W^i >4 y 142 FRACtlOMS 14. To reduce an integer to a fraction of any deno- mination. An integer may be considered as a fraction if we make nnity its Nominator : — ^tbus {- may be taken for 5 ; since *=5. ■ ■"'^' -:■ -■ >J-*'" -" '^^'^^ W^' may give an integer any denominator We please if we preyionsly multiply it by that denominator ; 25 30 35^ ^ 25 6X5 5 ^ thus, 5=y, or -g-,or y, &c., for — =j^=j=6; , 30 5X6 5 and — = =~=5. &c. i 6 1X6 1 'T { TJ XXERCI8E8. 1 1. Reduce 7 to a fraction, having 4 as denominator Am. V- 2. Beduce 13 to a fraction, having 16 as denomina- tor. Ans. W. 3. 4=V. I 4. 19=V. I 5. 42=«i.V. | 6.'71 = «|J«. 15. To reduce fractions to lower terms. ">' Before the addition, &c., of fractions, it wfll be often convenient to reduce their terms as much as possible. For this purpose — * BuLE.^I>ivide each term by the greatest common measure of bot^. 40 5 ^ 40 40-5-8 5 I We have already seen that we do not alter the quotient — which is the real value of the fraction [4]— if we multiply or divide the numerator and denominator by the same number. What has been siud, Sec. 11. 104, will be usefully remem- bered here. t minations. 25 26 27 17. [ other — BuLi minatioi of the i to be t remaiui Exam farthini quired i Beabi 960 fan il ofapoi 31. 32. 33. 34. 3 qua 35. 36, 37 Tioy FRACTIONS. .•.4 '/ 143 ' any deno- if wemaJce Een for 5; 16. To find the value c^ a fraction in terms of a ower denomination — Rule. — ^Reduce the numerator by the rule already irefl [See. III. 3] , and place tke denominator under it. Example. — ^What is the value, in shilMues, of f of a pound ? we please \£^ reduced to 9hiIHBgsx3i6(i^. ; therefore £| reduced to sMl- Tbe reason of the rule is the same as that ahready giyen [See. III.. 4}. The j of a pouiul becomes 20^iiii^8 as much if the " unit of coEDT>arison" is changed from a pound to a shilling. We may, if «fe please, obtain tho value of the result- ing fraction by actually ^forming i^ diyisjon [9] ; thr^r V- = 1S'- ' — ^bence j|^|r=l<5«. i> i,;;f) ,/., ■ or :5 5 } lominator denomina- '! = •««. I be often ^ possible. common I " ■ f [uotient— ultiplv or » number. y remem- "i- IF- ^e shall, it deno- 25. jei|f=14s. ed. 26. ^?=17«. 4d. 27. £lf=rl%. .■0'iiVlk "i X*^ ^ri- 2S. £l—t5l 30. £j.^ld. ■ < {'■• Tio" 17. To express One quantity as the fraction of an- other — Rule. — ^Reduce both quantities to the lowest deno- mination contained in either — if they are not already of the same denomination ; and then put that which is to be the fraction of the other as numerator^ ^pd the remaining quantity as denominator, - '^^'' * ' • Example.— What fraction of a pound is 2|d. t £jtaB960 farthings, and 2|h ii-j.ii , v:**.'.i.-.i < ii ii-- ^■.' -jk. ; 1, !■ 1 i- 144 VULGAR FRACTIONS. "'J -ii ,! few,. f. ''»0 »;rrf-,'>' .J.J i',.. QUESTIONS. ''Jf'-* ^"-"'^ ^''■' ■''!' 1. What is a fraction ? [1]. 2. When the divisor and dividend are made to oon- stitute a fraction, what do their names become ? [21. 3. What are the e£fects of increasing or diminisning the numerator, or denominator ? [3] . 4. Why may the numerator and denominator be mul- tiplied or divided by the same number without altering the value of tli4 fraction ? [4]. 6. What is an improper fraction ? [7]. ' ».' • ' 6. What is a mixed number ? [8]. 7. Show that a mixed number is not identical with the equivalent improper fraction ? [8] . 8. How is an improper fraction reduced to a s^ixed number? [9]. - -> 'I ^r 9. What is the difference between a simple, a com- pound, and a complex fraction ? [10, 11, and 12] ; 10. Between a vulgar and decimal fraction ? [13J. 11. How is an integer reduced to a fraction of any denomination ? [14] . 12. How is a fraction reduced to a lower term? [15]. 13. How is the value of a fraction, found in terms of a lower denomination ? [16]. ;";..' '-""' 1^ 14. Hpw do we express one quantity as the fraction of another ^ [17], ,^|,4.^ fe|:^ m in^it .roi/'^/,ni L ••::>, VULGAR FRACTIONS. 4f .^^1L A .^^^'.^ ADDITION. '*., y^;,.,ni^ xani jr: 18. If the fractions to be added have ii cpmmon denomhiiator-^^ , i.^--. ■ . -hc. .>.;*; i^*,^i *f .;■.-; Rule: — Add all the numerators, and plioe the com- mon denominator under their sum. • . ' J Example.— 4 4.|«ay. \,^i .,iv "., rroHnp/* umW r^r* Rbamit of the Rule.— If we add together 6 and '6 of any kind of individuals, their earn must be 11 of tht'iame kind of individuals—sinoe the process of addition has not changed the!r instanj Additf 1. 2. 3. 4. 5. 6. 7. 8. of ■•"ft oT .:'[ '-■) TULOAR FRACTIONS. 145 iade to oon* cumiiiisjiiQtf itor be mul- out altering > r '■ .' I ■-•(i mtical witt to a mixed lo, a com- 1 121; / m- apQ of anjr ver term ? terms of e fraction ■ "*'.'..-•» .'V ■ ml {.■.,<•. cpminoD he com<- 6 of any «n« kind changed thtfr nature. But tlie nniti to be added were, in tlie prMent instance, seTenthe; therefore their sum consists of sevenths. Addition may be illustrated as follows : — I Unity. ±1: ::::::] ? 4- 4 V u. VZXRCI8XS. 9. 10. 11. 12. 13. 14. TW— TBi— ^« 4 l.ll _, S8 g B I l» 5 2 O rtuznd ll — T3 33 5a_ ffiniTil—slI— i" /I* w 2. i+j+H- 4* T<^"4"Tf4"nr" 19. If the fractions to be added have not a common denominator, and all the denominators are prime to each other — , i '■ . . .;! Rule. — ^Multiply the numerator anj denominator of each fraction by the product of the denominators of all the others, and then add the resulting fractions — ^by the last rule. ; .■«..j Example. — What is the sum of f +l+f • 2 3 4 2x4x7 3x3x7 4x3x4 66 63 48 167 3+4+7=«3x4x7"*"4x3x7"'"7x3x4'*84"^g4"'"84"'84 Having found the denominator of one fraction, we may ai once put it as the common denominator ; since the same factors (the eiven denominators) must necessarily produce the same product. 20. RxAsoir or tmx Rulx. — To bring the fractions to a common denominator we have merely multiplied the nume- rator and denominator of each by the same number, which [4] does not alter the fraction. It is necessary to find a common denominator; for if we add the fractions with- out so doing, we cannot put the denominator of any one 24-34-4 of them as the denominator of their sum; — thus ^ . o for instance, would not be correet—eince it would suppose all the quantities to be thirds, while some of them are fourths and sevenths, which are le»t than thirds ; neither would 2-4-8-f-4 ^L^^ be correot-Hunoe it would suppose aU 9C ihtm, to be t ; Kit t- im 1 1 ii \f' 146 VULQAR FRACTIONS. seTenths, although some of them are thirds and fi>art]i9* which are greater than seyenthif* 21. In altering the denominators , we has^e onlpr changed the parts into which the unit is supposed to be divided, to an equivalent number of others which are smaller. It is neces- sary to diminish the size of these parts, or each fraction would not be exactly equal to some number of them. This will be moire evident if we take only two of the above fractions. Thus, to add } and j, 2 8 2X4 3xg 8 9 17 8+1—3^4+4^3 12"T'12'*12 These fractions, before and after they receive a common denominator, will be represented as follows : — I Unity. i _l mi r. •r- ' J - I equal to 1 • j-';s equal to h We have increased the number of the parts just as much as we have diminished their size ; if we had taken parts larger than twelfthis, we could not have found any numbers of them exactly eqaivalent, respectively* to both I and |. f':„- i. v'H' , ^• XXXRCI8XS. 21. |-H44-^?J< 22. il+ff 24. ll-NSW-ift-^fil?. 22. If the fractions to be added l^a^e not a oommon denominator, and all the denominators ar0 not prime to each other — Proceed as directed by ^e last nde ; or^— Rule.— -Find the least commoii miiltble of all the denominators [Sec. II. 107, &c.] , this will be the copnmon denonuiiatb]^ midtiply the nnmerator of each fraction into 1 tiplel ratoii Ex maltil 4-- 1^1 23. rator common 1 luulti ifl instrti we ra auiub ■ . <*• . ]■ diniia the ot Wh tions m'llti ^1 denon v<"f* :'•.:.-> ■ stanc< '-.b ■ woulc -.!] 1 40320 ^ '""' '' "^ ll iiooy t;ermf 25. = 20. ■ 27. 28. 2.). 31. 2 tioT 1 ha\ [1' 1 VULGAR FRACTIO^•^. /47 into'tiie quotient obtained on dividing the oommou mul* tiple by its doni^ininator — this vrill give the new nume- rators ; then add the numerators as already directed [18]. Example. — Add 3*7 -f- A + i\- 2S8 is the least common 6 4 3 288-fa2x6 multiple of 32, 48, and i 2 ; tUerefcwre jf» 4-45472^^ — $88 288-f-48x4 288^72X3 45 , 24 *^J^_8l + 2ri8 + 288 ^288+2b8'*'288~~288' 23. Reasopt or the Rule. — We have multiplied each na;n(^ rator and denoininator by the saiae number (the least-coiMMi 1 multiple of the denominators 1.4]; — tiioGe 6 X 288 -j- 32 " 288 "~ ■^'" instil nee ;s=B _ ' , . "' . — For we ohtaia the same quotient, wiiethrr 3J+i88" ^ we multiply the divisor or divide the dividend hy the siunc ouinber — as in both Ciises we to the very same ai«f>iiiir diiuiaish the number of times the one can betiubtracted from the other. ^ When the denominators are not prime to each olhcT tli-c frac- tions we obtain have lower terms if we make the least common multiple ef the denominators, rather than the prodiftt of the denominators, the common denominator. In the present in- stance, had we proceeded according to the I'tst rule [Itf], w« would have ^ . 6 8 3 172S0 18432 4'308 found — a.— -i. Ts — — — L. _L. ■ . ■- 82^48^72^1 iUo'J2^1 10592^1 lUd;*2" 40320 .. 4032am ^^3 In this case it is necessary, before performing the addition f 10 and 2Z]j to reduce the fractional pai'ta to a common aenominatoF. 27. Reason' of the Rule. — The addition of mixed num- bers is performed on the same principle as simple addition; but, in the first example, far instance, eig/it of one denomina> tion is equal to o7ie of the next — while in shnpie addition [Seo li. 8], ten of one denomination is equal to one of the next. KXERCTSES. 55. 3f-him-14||=29i3». 56. 403-}-38i4.4a3=110i. 57. 81?4-6;f4-ll=99A. 58. 92A+37t^-{.?M372||. 59. 173T^-f854.91j^;«273"' -8-i nl61 60. 454.32= 61. 8U4-2n:-xx^fj. 62. 10yVH-7i==26H. 53. 102,.fllAr:322^. 154. ll|T8|2l9t^ seen that aa any dcuomi- imber, there- ly reduced tn 1090 IT ■ 145 12 * 1^ 7 3 I', 16 • ts ; then, if to a mixed y in»tegers in fraction, sot jgery. VULGAR FRACTIONS. 149 QUESTIONS. 1. Wliat is tilt' rulo for adding fractions which have a conimon denominator? [IS]. 2. How are fractions brought to a common denomi- Qator r [19 and 22]. 3. What is the rule for addition Avhen the fractions have diflforent denominators, all prime to each other ? [19j. 4. What is the rule when the denominators are not the same, but are not all prime to each other ^ [22] . 5. How is a mixed number reduced to an improper fraction? [24]. 6. How are mixed numbers added ? [2<^] . as 8 eighths unit and 4 set down. I T the addition to a common SUBTRACTION. 28. To subtract fractions, when they have a common denominator — Rule. — Subtract the numerator of the subtrahend from that of the minuend, and place the common deno- minator under the ditforcnce. Example. — Subtract f from -ji . 7 4 7-4 3 9~9~* 9 ~9" 29. Reason of the Rule. — If we tiikc 4 individuals of any kind, from 7 of the same kind, three of them will remain. In the example, we take 4 (ninths) from 7 (nintlis), and 3 are left — which must be ninths, since the process of subtraction cannot have changed their nature. The following Avill exemplify the subtraction of fractions : — mixed num* iple addition; me denomina- addition [Seo the next. -99,V. .Z|^l37f||. Ti c O « « c^ {2 rt ^', u t-t M'^ .O -O ia Unity. , I 3 4 =27^1^. ,Sil-Hi^^if'J V ::^:r\ H ISO VULGAR FRACTIONS. EXKnCISEB. 1. 2. 3. 4. 5. 1 1 __3, 1 n_¥zf* 17 3 ifl 18 — If; 2l 7 7 2 J ■TT.; IT- 6. 7. 8. 9. 10. IB _ 7 TQ _ a .1 •2 1* 8 8 — i 7 4 a TT 1 f— -T1 1 4 _ a .' 51 51 — ff- 30. If the subtrahend and minuend have not a com- mon denominator — Rule. — Reduce them to a common denominatejff [19 and 22] ; then proceed as directed by the last rule. Example. — Subtract f from J. 7 ri_.H:» 40 ,33 ■ 1 ,j •-, ,j. CI. Rtason or THR Rule. — It is similnr ^ that nlre.idj^ given [20] for reducing fractions to a coimnou denominator, previoualy to adding them. KXT^RCISES. 11 12 13. 5 7 4ff* 15. 10. 17. 18. 1 1 (5 1 5 1 _., 7 R n Ttl~Ti5 — afful" 47 _ 5 •rio7 >.3 1(3 — I 1)2 8* 3i; 4 1 _5 J (1 lit 2 8" 7 5'.) 3-'U.^2.1 14. 14_13 _2 32. To subtract mixed numbers, or fractions from] mixed numbers. If the fractional parts have a common denominator — Rtle — I. Subtiact the fractional part of the subtra- hend from that of the minuend, and set down the differ- ence with the common denominator under it : then subtract the integral part of the subtrahend from the integral part of the minuend. II. If the fractional part of the minuend is less than that of the subtrahend, increase it bj adding the com- mon denominator to its numerator, and decrease the integral part of the minuend by unity. ^ .^ Example 1. — 4g from 9|. Of minuend. 4'i subtrahend. I ' 5 J difference. 3 eighths from 5 eighths and 2 eighths (=|) remain. 4 from 9 and 5 remain. VULGAU FRACTIONS. Example 2.— Subtract 12J from 18|. 181 minuend. 12^ subtrahend. 151 5^ difference. 3 fourths cannot be taken from 1 fourth ; but (borroAving one from the next denomination, considering it as 4 fourths, and adding it to the 1 fourth) 3 fourths from 5 fourths and 2 fourths (=^) remain. 12 from 17, and 5 remain. If the minuend is an integer, it may be considered as a mixed number, and brought under the rule. Example 3. — Subtract 3| from 17. 17 may be supposed equal to 17f; therefore 17— 3js=s 17o_34. But, by the rule, 17^-3| = 1G5-3| = 13^. 33. Reason of the Rule. — The principle of this rule is the same as that already given for simple subtraction [Sec II. 19] : — but in example 3, for instance, Jioe of one denomina- tion make one of the next, while in simple subtructiuu ten of one, make one of the next denomination. 34. If the fractional parts have not a common deno- minator — Rule. — Bring them to a common denominator, and then proceed as directed in the last rule. Example 1. — Subtract 42] from 56i. 56^=5Gy*7r, minuend. 42i=42f5, subtrahend. 14y^^, difference. 85. Reason of the Rule. — We are to subtract the dif- ferent denominations of the 8ui)trahend from those which cor- respond in the minuend [Sec. II. 19] — but we cannot subtract fractious unless they have a common denominator [30]. \ t fZERCISES. 19. 20. 21. 22 23 27f-3^=24^. 12f-12S=|. 147A '51' 24'. 82iii-7iH=74|f 25. 76{— 72t^=.3|^. 26. 67|— 34^?,=32^. 27. — "■■ " - 28. 29. 30. 31. 32. 971-3211=64^. 60^-41^^^=191. 92i-90J^=:2X. 1001— 9f=90|. 60-y\=59^. 12^- 102=11- a t •ft, 152 VULGAR FRACTIONS. ^'^! r.y- i QUESTIONS. 1 . What is tho rule for the subtraction of tVactlIo uumbor, cr the pro. 'lie integer «ca bj the 1 evidently »al extent tfocted by ier; tliu's we re luce ' ha vino- a ed Jn tlie I integer, able. be multi- tainecl in 38. The mtcgral quantity which is to form one of the factors may consist of more than one denomiuatioc. Example.— What is the ^ of X5 2s. 9d. ] . ,. ; ? 5 s. 2 d. Ox?: £ 5 s. 2 (1. 9x2 o O £ .3 s. 8 1. 2. 3. 4. |X2-1?. ^X8=0f. ixl2=0L 5. fffXSO ■Ji. .;.;x3G=34. if X 20=19. 11. 12. 13. 22x^-4J. 14. tVx17=.Ij)5. 15. 143 -'^=61?. Ajuf Am EXERCISES. G. 27 X 1=1 2. 7. Ax 18=3?. 8. kx8=:7i. 9. 21x2=^. 10. 15xi=3. 16. How much is -^^^ of 26 acres 2 roods ? 20 acres 3 roods. 17. How much is i*- of 24 hours 30 minutes? 7 hours. ' - ■ • • . : *v^.' 18. How much is -^^j^ of 19 cwt., 3 qrs., 7 ft> ? Ans 7 cwt., 3 qrs., 2 ft). 19. How much is f| of ^29 .' Avs. £\\t =£S 19# 39. To multiply one fraction by another — Rule. — Multiply the numerators together, and under thftir product place the product of the denominators. l)ljtAMPLE.— Multiply ^ by |. ' ' ■ \^ ■ 'L-^ v > 20 ' ■■ '-■ ••- 4 5^4x5. 9 e 9x6' 54* 40. Reason of thr Rule. — If, in the example given, we were to multiply ^ by 5, the product (V) would be 6 times too great — lince it was by the sixth part of 5 (f), we should have multiplied. — But the product will become what it ought to be (that is, G times smaller), if we multiply its denominator by 6, and thus cause the size of the parts to become 6 times less. Wo have already illustrated this subject when explain- ing the nature of a compound fraction [11]. 22. 5xt-X?=^25. 23. ixj^y 32. How much is the | of f .? 33, H jw much is the | of ^ ? EXERCISES. 24.i3x^!=!|l. 25.^X^Xf, "■ 26.|x*=».V. .145 28. 29. 30. 31. ,1 If) V20_ r V 1 » 1 -^ ^ A=*Tr Ans. ^. Ann. ■j-V p. p IM Vl/LUAR FRAvrtoyif. f. 41. When we multiply one proper fraction hv r-j 'tlii»i% we obtain a product snialler than either ot' the t ." • .^rs,— Nevertheless such multiplication is a species d uJ U- lion; for when we add a fraction once, (that is, when we take the whole of it,) we got the fraction itsolf a.^ result ; but when we add it less than once, (that is, take- ao much of it as is indicated by the fractional multiplier, ) we most i>ecessarily get a result which is Ktss tlun vvhcit M'e took the whUe of it. jiusides, the multiplication of a fraction by a fraction supposes nvultiplicatioii by one number — the nuineratw of the multiplier, and (whidr will b« seen presently) clivisfon by ano.tber — the denominator of the fnukiplier. Ileace, when the ber. Rule. — Reduce mixwl numbers to improper fractionaf r24j, and then pro^&eed aecordiufj to ih^e last rule. ExAMPLC 1. — Mukiplv Y by 4,f. <* 4|=»*-' ;. thewmtTO i«x4f;: ExAMPU^ 2.— Multiply 5" by 6|. 5HV' and G.5'=»V ; therefore 5^x0^=-*/ xV='IIt*. 43 Rkason or the I.',ui.f:. — We n>er«ly piat the raixedl numbers iuto a luuvo eonreuieut tomt, without ulttt-riug theii- To obtain the requiretr product, we might multiply e«ob j^nrt of tiie muUvpHcund by caeh f^vl e t 9n EXERCISES. 3:9. 3^x19^ xMOJT- " 40. (5?x"X*X*=2-\. 41. 12,xl3;xf>'=1097^-(; 42. 3-xl4^xl5=818^ 43. 14xl5rVx3i=^749yV5. ,3 w41 12-? I- , '■ 'K .. 34. 8-'x^=7^. 35. 5vx^=ilU. 36. 4ix7ix3=rl01|. ^ w' fnXo Xtt XT7f=s5!^i5r» 38. 5jxl6xlii, we can reduce the numerator of one fraction and iht.* denominator of another to lower terms, by dividin;^ both by the same number : — thus, to multiply 'j by :} . Dividing both 8 and 4, by 4, we get in their piact.s, 2 and 1 ; and the fractions then are ^ and ^, which, multiplied together, become ^}X| = y\- This is the same as dividing the nmiicrator and denomina- tor of the product by the same number ; for 3 4 3><4-f4__3x^l /_3 1\ 3 =ijx7-i-4~"2x7 \"~2^7/ ""14' 8X7= 50* the raixecil •riug tlieii- tiply e«ch er.— Thus.^ 2-V =ioa; av 818^ 4 QUESTIONS. 1 . How is a fraction multiplied by a whole number or the contrary ? [36] . 2. Is it necessary that the integer which constitutess one of the factors should consist of a single denomina- tion .? [38]. 3. What is the rule for multiplying one fraction by another? [39]. 4. Explain how it is that the product of two proper fractions is less than either .? [41]. 5. What is the rule for multiplying a fraction or a mixed number by a mixed number ? [42] . 6. How may fractions sometimes be reduced, before they are multiplied .? [44 and 45] . I of o t ofSfJ I5 '■ ! : i 158 VULGAR FRACTIONS, DIVISION. 46. To divide a vulgar fraction by a whole number— liuLE. — Multiply the denominator of the fraction by the whole number, and put the product under its nu- merator. •. -' v.; Ttf^J.? . . *. .• Example.— f ~ 4 = ^^ =r yV 47. Rkason of the Rule. — To divide a quantity by 8, for instance, is to make it 3 times smaller th^-ii before. But it is evident that if, while we leave the number of the parts the same, we make their ,size 3 times less, wc make the fniction itself 3 times less — hence to multiply the deuominator by 3, is to divide the fraction by the same number. A similar effect will be produced if we divide the numera- tor by 3 ; since the fraction is made 3 times smaller, if, while wc leave the size of the parts the same, we make their 8 g ♦ \ 2 number 3 times less; thus --r- 4=:-^^=-. But since the numerator is not always exactly divisible b}' the divisor, the method given in the ru/e is more generally applicable. ~ The division of a fraction by a whole number has been already illustrated, when we explained the nature of a complex fraction [12]. EXERCISES. 1 3 . O 4 1. 7r-T--ii — Q-. Q 10.1 o t O.-^-^lJ — 2(5"* 4. -^^9= . 1 ■ff5- 5. H-^3= 6. 2H-8=,V. .1 1 ■38"' O 1 s.^r. 3 10. |-r-ll=5fV- II - - 7 -i-42 — » 7 7 _• 1 4 1 12. ^-rl4=g»,. 48 It follows from what we have said of th.e multi- plication and division of a fraction by an integer, that, when we multiply or divide i'^ numerator and denomi- nator by the same number, we do not alter its value — since we then, at the same time, equrJly increase and decr'iase it. 49. To divide a fraction by a fraction — liULE. — Invert the divisor (or suppose it to be in- verted), and then proceed as if the fractions were to be multiplied. VULGAR FRACTIONS. 157 Example. — Divide | by J. 6 3 6 4 2 number—. fraction by der its nii- aiitity hy 8, before. But the parts the the frjiction inutor by 3, the numera- smaller, if, make their t since the divisor, the )le. iiniber has the nature ^5 3 ■11=/ ?^' the raulti- ;ger, that, i denomi- ;s value — rease and to be in- ere to be 7*4 7^3" 5x4_20 Reason of the Rule. —If, for instance, in the example just given, we divide f by 3 (the numerator of the divisor), we use a quantity 4 times too great, since it is not by 3, but the fourth part of 3 (?) we are to divide, and the quotient (;j^l^) is 4 times too small. — It is, however, made what it ought to be, if we multiply its numerator by 4 — when it becomes |y, which was the result obtained by the rule. 50. The division of one fraction by another may bo illustrated as follows — «5;t- 5.3, ■!■■«'& The quotient of 4"^f must be some quantity, which, taken three-fourth times (that is, multiplied by f ), will be equal to ^ of unity. For since the quotient multiplied by the divisor ought to be equal to the dividend [Sec. II. 79] , 4 is f of the quotient. Hence, if we divide the five-sevenths of unity into three equal parts, each of these will be one-fourth of the quotient — that is, precisely what the dividend wants to make it four-fourths of the quotient, or the quotient itself. 51. When we divide one proper fraction by another, the quotient is greater than the dividend. Nevertheless such division is a species of subtraction. For the quo- tient expresses how often the divisor can be taken from the dividend; but were the fraction to be divided by unity, the dividend itself would express how often the divisor could be taken from it ; when, therefore, the divisor is less than unity, the number of times it can be taken from the dividend must be expressed by a quantity greater than the dividend [Sec. II. 78] . Besides, divid- ing one fraction by another supposes the multiplication of the dividend by one number and the division of it by another — ^but when the multiplication ie by a greater '^2 i!''ii m:- W M ti P h »• ! 1 158 VULGAIl FRACTIONS. number than the division, the result is, in reality, that of multiplication, and the quantity said to be divided must be increased. EXERCISES. HI _. 2 1 ■ ■5~S — 2- 16. 17. 18. .3 3 [5* 5 ^^6 — 3,5' 2 . I 9 15 . 5 11 10 15_. 9 1 21 91 1 . J 2 52. To divide a whole number by a fraction — KuLE. — Multiply the whole number by the denomi- nator of the fraction, and make its numerator the deno- minator of the product. Example. — Divide 5 by 2. .^a- ._3_5x7_35 ' '7" 3 ~3' This rule is a consequence of the last ; for every whole num- ber nuiy be considered as a fraction having unity for deno- minator [14]; hence 5-^'^=fH-^=t Xi=V- It is not necessary that the whole number should consist of but one denomination [38]. Example. — Divide lis. 3J(/. by ^. EXERCISES. 22. 3-^*-=6?. 23. 11 -Hm. 24. 42H-ra=^C4. ZO. — j^ — Oy. =03. l')_:_il:- 28. 8-^l|=8t. 21). 14^ '^,=38. 30. 10-1.^=32. J£17 lis. £\0 Ss. £d 9s. Aid 2\d. 20. 27. 9^^ 31. Divide £1 \Qs. 2d. by |. Ans. 32. Divide MS IS*-. Ad. by f. Ans. 33. Divide .£5 0^. \d. by |i. Ans. 53. . To divide a mixed number by a whole number, or a fraction — . ' Rule. — Divide each part of the mixed number accord- ing to the rales already given [46 and 49], and add the quotients. Or reduce the mixed number to an improper fraction [24], and then divide, as already directed [40 and 49] . Example 1. — Divide 9^ by 3. 93^3=9^3+^^3=34.-^=3^. ^ " Example 2.— Divide 14/- by 7. '' ' -^' • l4fV=W; therefore Uj^-hHtt -^]== it X: lC?f ,1 251 ■ It VULGAR FRACTIONS. 159 64. Reason or the Rule. — In the first exnniple we have divided each part of the dividend by the diviaor and added the results — which [Sec II. 77] is the same as dividing tlie Airhole dividend by tlie divisor. In the second example wo lirve put tho m'xecl number into a more convenient form, without altering its value. ■ I EXERCISES 34. 8?^17=?,». 35. 51^^-3=:17>, 36. 187,',^/,«398.V3-. 37. i:);VV41=r'iJ. 00. I0,.iij-p-^ff— 1/ , 8Y^. •IQ 4.3 3 5 . 4 1.^^«i fl4 5P HI. io„ Tj-r-j, — •»■ -^ f) a ■; .i • 42. loa ^,-^A=l^'8-i{^. 43. l8J-^ll=^l. ho. To divide an integer by a mixed number — KuLE. — Reduce the mixed number to ;in improper fraction [24] ; and then proceed as already directed 1 52]. Example. — Divide 8 by 4 J. 4^==^', theref()re \^^^=R^'i=Sxis=nh Rkason of thr Rule. — It is evident that tlic improper frJiction which in equal to the divisor, is contained in the divi- dend the same number of times as the diviaor itself. EXERCISES. 46 14-^11=7,-^. 47 21-1-14 '*=i-''.^. Ans. £2 fi.^. W^d. r, by a 44. 5-^3^:==l'). I HO. lO-r-ll.jjj, — I3J1. [ 4S. Divide £7 16.?. 7d. by 3^ 49. Divide £3 3s. 3d. by 4f Ans. 14s-. U'^d. 56. To divide a fraction, or a mixed aumiy' mixed number — lluLf-:. — Reduce mixed number!? to imprnp r fracti )n [24] ; and then proceed a.s already direct ■•! [4.-J]. Example 1. — Divide J by 5^. Oj;=as fj , inoreiore j--.j^=j-;- ^ — ^ x vi — io/;- Example 2.— Divide Hj% by 7^. A 8,WHi and 7^=«Vi therefore S^V-r- 7^ = ^^ 47 Rkkhos nr tke Rule. — We (as in the htst rule) merely change the mixed numbers into others more cunvonieublv divided — without, however, altering their value 160 VULGAR FRACTIONS. IV i ' tl' ' 50. 51. 52. 53. 54. 58. 3 . '^ 5 2 t 3 15 . 14 2;i 3 2-^^S ff^- AT . K\ 1 7 EXERCISES. 57. ii^8j=«^H- 58. l^-7-2|+5^-j-Og — j-(P(, KQ OI . 3_l_5 1 9_ When the divisor, dividend, or both, arc com- pound, or complex fractions — Rule. — Reduce compound and complex to simple fractions — by performing the multiplication, in those which are compound, and the division, in those which are complex ; then proceed as already du:ected [49, &c.] Example 1. — Divide f of I by ^. 5 of §=|g [39], therefore ^X^-hf-I^H'sB-XHMs- 4 Example 2. — Divide -^ by f . J=^r5 [46]> therefore ^-t.|=5\4-|=4^X§=2¥B- EXERCISES. 60. 61. 62. 4 V ■' • S.— f* 4.1JL • 5 V 3 r\0*^ ^T2 ~T¥ ^ IT — ^^OT}"' 3 5 . J_ — 02 21 no 22.2 V''/ 1 t 7 97"^^ — fairs- 64. -• 3 =25. 00. -r-r-jaXar 66. Tff 4 ^ . 3v5 ^22 I 243^ ^ ^i* h li: i Jlj.! 1 QUESTIONS. 1 . How is a fraction dived by an integer .? [46] . 2. How is a fraction divided by a fraction ? [49]. 3. Explain how it occurs that the quotient of two fractions is sometimes greater than the dividend ? [51]. 4. How is a whole number divided by a fraction.^ [52]. 5. What is the rule for dividing a mixed number by an integer, or a fraction ? [53] . 6. What are the rules for dividing an integer, a frac- tion, or mixed number, by a mixed number } [55 and 56]. 7. What is the rule when the divisor, dividend, or both are compound, or complex fractions ? [58] . =3 J 58 • y rr* arc com- to simple , in those lose which [49, &c.j K-'^a— Us- 12 k =243|g 46] . [49]. at of two id? [51]. fraction ? umber by r, a frac- [55 and idend, or VUI.QAR rRACTIOr^d. 16) MISCELLANEOUS EXERCISES IN VULGAR FRACTIONS. 1. How much is -} of 186 acres, 3 roods? An^. 20 acres, 3 roods. 2. How much is f of 15 hours, 45 minutes ? Am. 7 hours. 3. How much is ^Vj^'j of 19 cwt., 3 qrs., 7 lb ? Aii9. 7 cwt., 3 qrs., 2 lb. 4. How much is ^\%\ of i^lOO ? Ans. £3Q 9s. a If one farm contains 20 acres, 3 roods, and another 26 acres, 2 roods, what fraction of the former is the latter ? Ans. -^^^g. 6. What is the simplest form of a fraction express- ing the comparative magnitude of two vessels — the one 4 containing tuns, 3 hhds., and the other 5 M- hhds 7. shilling ? Ans. I'Ss 8. What is tho sum o? p. and j^cL ? Ans. 7j\d. What is tlie suin of | lO^d. tuns, 2 of a pound, and | of a jU. A7l$ ^1 9. What is iho siuii of £1, f5., and 3s. m-d. 10. Suppose I havt^ !;• of a ship, and that I buy ^ more ; what is my outii*e hhare ? Ans. |^. 11. A boy divided his marbles in the following manner : he gave to A I- of them, to B ^V? ^^ ^ h ^^d to J) j-, keeping the rest to himself; how much did he givo away, and how much did he keep ? A'ns. He gave away of them, and kept j^j^. 12. What is the sum of | of a yard, | of a foot, and \ of an inch ? Ans. 7 inches. 13. What is the difference between the ^ of a pound, and d\d. ? Ans. lis. 6^d. 14. If an acre of potatoes yield about 82 barrels of 20 stone each, and an acre of wheat 4 quarters of 460 ft) — but the wheat gives three times as much nourish- ment as the potatoes ; what will express the subsistence given by each, in terms of the other ? Ans. The pota- toes will give 4^i times as much as the wheat ; and th(? wheat the //^ part of what is given by the potatoes. 15. In Fahrenheit's thermometer there aro ISO de- grees between th^ boiling and freezing pqiutfl j in tij.at I %> 'ill' 162 DEriMAL rHACTIONS. i! &• ii!!, It '1- h ' i^ iiii of Reaumar only 80 ; what fraction of a degree in the latter expresses a degree of the former ? Ans. |. 16. The average fall of rain in the United Kingdom is about 34 inches in depth during the year in the plains ; but in the hilly countrios about 50 inches ; what fraction of the latter expresses the former ? Ans. ^^. 17. Taking Chimborazo as 21,000 feet high, and Purgeool, in the Himalayas, as 22,480 ; what fraction of the height of Purgeool expresses that of Chimborazo ? Ans. Iff 18. Taking 4200 feet as the depth of a fissure or crevice at Cutaco, in the Andes, and 5000 feet as the depth of that at Chota, in the same range of mountains ; how will the depth of the former be expressed as a fraction of the latter } Ans. \\. DECIMAL FRACTIONS. 59. A decimal fraction, a« already remarked [13], has unity with one, or more cyphers to the right hand, for its denominator ; thus, j-^js-^ is a decimal fraction. Since the division of the numerator of a decimal fraction by its denominator — from the very nature of notation [Sec. I. 34] — is performed by moving the decimal point, the quotient of a decimal fraction — the equi- valent dedrnal — is obtained with the greatest facility. Thus y/g^^=-005 ; for to divide any quantity by a thousand, we have only to move the decimal point three places to the right. 60. It is as inaccurate to confound a decimal fraction with the corresponding decimal, as to confound a vulgar fraction with its qn.otiePt. — For if 75 is the quatinii of ^1°, or of VoV, uid is distinct from either ; so also is -75 the quotient of | or of ^^^ , and equally distinct from either. 61. A decimal is changed into its corresponding deci- mal fraction by putting unity with as many cyphers as it contains decimal places, under it, for denominator — having first taken away its decimal point. Thus '5646= ^«5S& ; '008= yAv, ^ 62. as VI couvt form byth 63. dednu Ri this V C'han; alrea< Fxi Exj Thi 1. 2. 3. 4. 13 €25. 14 15 yard. 16 ton. 64 Rt III. Ex DECIMAL FRACTIONJ. 103 Tee in the Kingdom the plains ; lat fraction high, and at fraction limborazo ? fissure or 'eet as the noun tains ; sssed as a [13], hand, ked ight h{ il fraction, lal fraction )f notation le decimal -the equi- st facility, itity by a )oint three al fraction d a vuliijar lie quofdnii ir ; so also Uy distinct ading deci- cypheis as )minator — 18 -5646= 62. Decimal fractions follow ezactt/ the same rales as vulgar fractions. — It is, however, generally more convenient to obtain their quotients '[5&j , and then per- form on them the requirea processes of addition, &c., by the methods already described [Sec. II. 11, &c.] 63. To reduce a viUgar fraction to a deamal, or to a decimal fractwn — ' . . . RuLE.-^Divide the numerator by the denominator — this will give the required decimal ; the latter may be changed into its corresponding decimal fraction — as alreavf; Jc«cribed [61]. FxAMPL^ 1. — Reduce ^ to a decimal fraction. Example 2. — ^What decimal of a pound is l\d. ! 7J IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 11.25 u 122 ,2.2 £ Hi HI lit u 20 6" Fhotograiiiic Sciences CorpQratiQn \ v\ 23 WIST MAIN STRHT WIUTn,N.Y. I49M (716)t72^S09 ^.^ ^ i> o^ lt» CIRCULATINU DECIMALS. !i|i fU' Example 1.— What yulgar fraction is equivaleiit to *3'1 Ans. f . Example 2. — ^What yulgar fraction is equivalent to •^7854"? Aru.m^, 75. Reason or I. — ^ will be found equal to *111» &o.— or •1'; therefore^ (=s8x^)s-883. &c.s(8Xlll* &o.) For if we multiply two equal <;^uantities by the same, or by equal quantities, the products will still be equal. In the same way it could be shown that any other digit divided by 9 would give that other digit as a repetend. — And, consequentljr, a repetend of any digit will be equal to a vulgar fraction having the same (ligit for numerator, and 9 for deno- minator. Reabow or II. — ]fV ^ill ^^^ '0101, &o.~-or '^01' as quotient. For before unity can be divided by 99, it must be considered as 100 hundredths; and the quotient [Sec. II. 77] will be one hundredth, or '01. One hundredth, the remainder, must be made 100 ten thousandths before it will contain 99 ; and the quotient will be one ten thousandth, or '0001. One ten thousandth, the remainder, must, in the same way, be considered as ten million' eths; and the next quotient will be one millioneth,or '000001 — and so on with the other quotients, which, taken together, will be Ol-f-OOOl+OOOOOl-f-Ac, or 010101, &c.~represented by '^01'. ^ (=:87XtV=87X.^01') will give '878787, &c.— or •>37' as quotient. Thus 010101, &c . ^■'- -'-^ ■'■ - 87 ■ •■'■■ '- ' ■/• •■ . .' 70707 80808 878787, &c.r=87X- 01'. In the same way it could be shown that any other two digits divided by 99 would give those other digits as the period of a circulate—And, consequently, a circulate having any two digits as a period, will be equal to a vulgar fraction having the same digits for numerator, and 2 nines for denominator. For similar reasons w^^ will give '001001, &c., or *^00r as quotient. But 001001, &c., X (for instance) 6d8s>'663o68, &c Thus 001001001, &o. 8008008008 6006006006 6006006005 m 668668668568, &c.a-668X''001 . In the taaa way it could be shown that any other three digitf 4itidi4 Igr 999 wovU ^Tt a eimUitiog itfiimal hatiag iheit Blent to '3'1 quivalent to '111, &o.— or &c.) For if , or by equal \y other digit etend. — And, Etl to a vulgar id 9 for deno- [' as quotient. I>e considered '] will be one must be made i the quotient Dusandth, the I ten million' or 000001— cen together, —represented .—or '^37' as er two digits 9 period of a ng any two a liaying the lator. , or '^OOr as '663568, &c I. ' three digit! Atiag thfM CtnCL LATINO DECIMAM. 169 digits as a period. — And, consequently, a circulating decimal having any three digits as period will be equal to a vulgar fraction having the same digits for numerator, and 3 nines for denominator. We might, in a similar way, show that any number of digits divided by an equal number of nines must give a circulate, each period of which would consist of those digits. — And, sonsequently, a circulate whose periods would consist of any ligits must be equal to a vulgar fraction having one of its periods for numerator, and a number of nines equal to the number of digits in the period, for den^ominator. 76. If the decimal is a mixed repetend or a mixed circulate — BuLE. — Subtract the finite part from the whole, and set down the difference for numerator; put for deno- minator so many cyphers as there are digits in the finite part, and to the left of the cyphers so many nines as there are digits in the infi^iite part. Example. — ^What is the vulgar fraction equivalent to •97^8734' ? There are 2 digits in 97, the finite part, and 4 in 8734, the infinite part. Therefore 978734-97 978637 . ^. . , , - ,. -999900 ^999900' " *^" "^"''"^ ^"^S'^' ^'^^'**'°- 77. Reason of the Rule. — ^If, for example, we multiply •97>8784' by 100, the product is 97 •8784=97-f--8734. This (by the last rule) is equal to 97+|^f j, which (as we multiplied by 100) is one hundred times greater than the original quantity — but if we divide it by 100 we obtain r^jr\~JS^}^7f* which is equal the original quantity. To perform the addition of ys^ and vli^^f 7f, we must [19 and 22] reduce them to a common denominjvtor — when they become 97x99itA00, 878400 _97X9999, 8734 ^^^ "•"oQoonnnft oooorin I oooonn ' ' (Since vyvV ^ 99990000 ^99990000 999900 ^999900 10000-1) 970000—97 97x10000-1 , 8734 97x10000-97 ' 9999SS f 8734 999900 ^999900 8734 978734-97 999900 +999900'^ 999900 result obtained by the rule. The same reasoning would hold with any other example. 999900 978637 .. . , ,- ,, 999900 * ^'^^•^ *• e^caotly the BSBRCUIBS. 7. •574'-^J. •83^25 " •147^658 8. 9. •147^658'r,l|: 10. •432>0075'-.| 11. 875-49'66'»875| 12. 301-82'756'--301 !{W^ ^Tl'r-r*!*****'**-:-*^— ■I t 170 OIRCCLAtlNO DECIUALS. » : 78. Except where great acctiracy is required, it is not necessary to reduce circulating decimals to their equi- valent vulgar fractions, and we may add, and subtract them, &e., like other decimals — ^merely taking care to put down so many of them as will secure sufficient accuracy. 79. it may be here remarked, that no vulgar fraction will give a finite decimal if, when reduced to its lowest terms, the denominator contains any prime factors (fac- tors that are prime numbers — ^and all the factors, can be reduced to such) except twos or fives. For neither 10, 100, 1000, &c., nor any multiples of these — ^as 30, 400, 5000, &c., nor the sum of any of their multi* pies— as 6420 (5000+400+20), &c., will exactlv con- tain any prime numbers, but 2 or 5. Thus | (consi- dered as r — J toill give an exact quotient ; so also will } ^ considered as ^ )• But | toiW wo^ give ., , 10 tenths 100 hundredths, one ; for } (considered as = » or = &c.) does not contain 7 exactly. For a similar reason 4 '"^i^^ '"^^ gi^e ^n exact quo . . _ ,40 tenths, 400 hundredths, tient ; smce 4 (considered as = or = &o.) does not exactly contain 7. 80. A finite decimal must haire so many decimal places as will be equal to the greatest number of twos, or fives, contained as factors in the denominator of the original vulgar fraction, reduced to its lowest terms. Thus ^ will give one decimal place; for 2 (found ortce in its denominator) is contained in 10 (5x2) ; and 10 tenths ....,,. ,. .. ,, , therefore 5 (=i) will give some digit (m the tenths' place [Sec. 11. 77]), that is, one decimal as quotient. ^ (3 \ ■= g ) will give two decimal places ; because 2 being found twice as a factor in its denominator, 6 wOi' not bd enough to consider the numerator m bo red, it is not their equi- md subiract king care to ire sufficient Igar fraction to its lowest factors (fae- factors, can For neither >f these— —as their multi- exactly con- ns I (consi- ient ; so also will not give hundredths, exact quo hundredths, ""7 ~ iimal places OS, or fires, the original r 2 (found 'X2) ; and igit (in the decimal as i; because nominator, mtor M so CIRCULATING DECIMALS. 171 many tenths ; for 30 tenths (=i) cannot give an exact quotient — 30 being equal to 3X2X5, which contains 2, but not 2X2. It will, however, be sufficient to reduce . , , , , 300 hundredths the numerator to hundredths ; because ;; loill give an exact quotient — ^for 300 is equal to 3X2X 2X5X5, and consequently contains 2X2. But 300 hwndredths divided by an integer will give hundredths — or two decimals as quotient. Hence, when there are two twos found as factors in the denominator of the vulgar fraction, there are also two decimal places in the quotient. /ff ^== 2^2x2x5 ^ contains 2 repeated three times as a factor, in its denominator, and will give three decimal places. For though 10 tenths — ^and therefore tJXlO tenths — contains 5, one of the factors of 40, it does not contain 2X2X2, the others; consequent!) it will not give an exact quotient. — Nor, for the same reason, will 6X100 hundredths. 6X1000 thousandths 6 X 1000 thousandths, wiZ/give one — that is, Tq (=i^o) ^"^ leave no remainder ; ftr 6X1000 (=6X2X2X2X5X 5X5) contains 2X2X2X5. But 6 X 1000 thousa/ndths divided by an integer will give thousandths — or three decimals as quotient. Hence, when there are three twos found as factors in the denominator of the vulgar frac- tion, there are also three decimal places in the quotient. 81. Were the fives to constitute the larger number of factors — as, for instance, in jV zizy ^^-^ ^^^ same reason log would show that the number of decimal places would be equal to the number of Jives, It might also be proved, in the same way, that were the greatest number of twos or fives, in the denominator of the vulgar fraction, any other than one of those num- bers given above, there would be an equal number of decimal places in the quotient. 82. A pure circulate will have so many digits in its period as will be equal to the least number of nines, which ffonld represent a quantity measured bf the denosuna* i^ 172 CIRCULATING DECIMALS. m ill tor of the original vulgar fraction, reduced to its lowest terms. For we have seen [74] that such a circulate will be equal to a fraction having some period for its nume- rator, and some number of nines for its denominator — that is, it will be equal to some fraction, the numerator of which (the period of the circulate) will be as many times the numerator of the given vulgar fraction, as the quantity represented bj the nines is of its denominator. For if a fraction having a given denominator is equal to another which has a larger, it is because the numerator of the latter is to the same amount larger than that of the former — in which case the increased size of the nu- merator counteracts the effect of the increased size of the denominator. Thus i=f f ; because, if the numerator of f f is 5 times greater than that of f , the denominator of 11^, also, is five times greater than that of f . Let the given fraction be -j^^. Since ■JL=*^3846ir)' ; und •^384615'=f f ^f i| ; J^, also, is equal to f f ffif ; - and, therefore, whatever multiple 384615 is of 5, 999099 is the same of 13. — ^But 999999 is the least multiple of 13, consisting of nines. If not, let some other be less. Then take for numerator, such a multiple of 5, as that lesser number of nines is of 13 — and put that lesser number of nines for its denominator. The numerator of this new fraction will [75] form the period of a circulate equal to the original fraction. But as this new period is different from, 384615 (the former one), the circulate of which it is an element, is also different from the former circulate ; there are, therefore, two different circulates equal to ^^ — that is two different values, or quotients for the same fraction — ^which is impossible. Hence it is absurd to suppose that any less number of nines is a multiple of 13. *t, > ' 83. The periodical obtained does not contain a finite part, when neither 2 nor 5 is found in the denominator of the vulgar fraction — ^rednoed to its lowest terms. For [761 a finite part would add cyphers to the right hand of the nines in the denominator of the vulgar fraction^ obtained from the circulate. But cyphers would suppose the denominator of the original firaitioii to ooatiaa twoi, or fiye»—flinot no other prime fttttoip CIRCULATING DECIMALS. IW could give cyphers in their multiple — the denominator of the vulgar fraction obtained from the circulate. 84. If there is a finite part in the decimal, it will contain as many digits as there are units in the greatest number of twos or fives found in the denominator of the original vulgar fraction, reduced to its lowest terms. Let the original fraction be ^. Since 56=;2X2X 2X7, the equivalent fraction must have as many nines as will just contain the 7 (cyphers would not cause a number of nines to be a multiple of 7), multiplied by as many tens as form a product which will just contain the twos as factors. But we have seen [80] that one ten (which adds one cypher to the nines) contains one two, or Jive ; that the product of two tens (which add two cyphers to the nines), contains the product of two twos or fives ; that the product of three tens (which add three cyphers to the nines) , contains the product of three twos or fives^ &c. That is, there will be so many cyphers in the denomi- nator as will be equal to the greatest number of twos or fives, found among the factors in the denominator of the original vulgar fraction. But as the digits of the finite part of the decimal add an equal number of cyphers to the denominator of the new vulgar fraction [76] , the cyphers in the denominator, on the other hand, evidently suppose an equal number of places in the finite part of a circulate : — there will there- ^re be in the finite part of a circulate so many digits as will be equal to the greatest number of twos or fives found among the factors in the denominator of a vidgar fraction containing, also, other factors than 2 or 5. 85. It follows from what has been said, that there is no number which is not exadhi contained in some quantity expressed by one or more nines, or by one or more nines followed by cyphers, or by unity followed by cyphers. Contractions in Multiplication and Division • (derived from the properties of fractions.) 86. To multiply any number by 6— RuLB. — ^Remore it one place to tho left hand, tad diTido tiie remit \^ d m P ,{,;. W'^' • ^ !ii; kiM; i m CONTRACTIONS. ■ 1 ■ .3680. ExAMPLB.— 736x5=^:.'^y«=3680. Rbason.— 5=sV ; therefore 736x5a«736xy« 87. To multiply by 25— Rule. — Remove the quantity two places to the left, and divide by 4. Example.— 6732x25«=«'»V»»«168300. . Reason.— 25=»»J» ; therefore 6732x25— 6732x » J*- «8. To multiply by 125— Rule. — Remove the quantity three places to the loft, and divide the result by 8. Example.— 7865xl25="«|««»«€83125. Reason.— 125==»V"; therefore 7865xl25»:7865=r»V*- 89. To multiply by 75— Rule Remove the quantity two places to the left, then multiply the result by 3, and divide tne product by 4. Example 685x75=s"*J»-21i=saoc5*o«.y375 -75 30«. 100 X i ; therefore 685 X 75 =. 685 X Reason. lOOx^ 90. To multiply by 35— Rule. — To the multiplicand removed two places to the left and divided by 4, add the multiplicand removed one place to the left. Example 1.-67896x35 » "••«•• + 678960 « 1697400 ^:678960=2376360. Reason.- 35s=>f»4-10j therefore 67896x35= 67896 x »r+io. Many similar abbreviations will easily suggest themselves to both pupil and teacher. 91. To divide by any one of the multipliers — Rule. — Multiply by the equivalent fraction, inverted. Example.— Divide 847 by 5. 847-i-5«:847-f-V=847x t\-169-4. Reason. — We divide by any number when we divide by the fraction equivalent to it ; but we divide by a fraction when we invert it, and then consider it as a multiplier [49]. 92. Sometimes what is convenient as a multiplier will not be equally so as a divisor ; thus 35. For it is not so easy to divide, as to multiply by *|^ + 10, its equivalent mixed number. DICIMAL8. 17l> tbe left, and the loft, and r866=»V*- he left, then by 4. 15. ■ . ^^•^■ <75=.685x ilaoes to the removed one )« 1697400 5 » 67896 X ihemselyes to averted. ■f.¥=W7x livide by the ion when we -« itiplier will it ig not 80 I equivalent QUESTIONS FOR THE PUPIL. 1 Show that a decimal fraction, and the oorrespond- hig decimal are not identical [59] . 2. How is a decimal changed into a decimal frao lion? [61]. 3. Ai-e the methods of adding, &o., vulgar and deci- mal fractions di£ferent ? [62]. 4. How is a vulgar reduced to a decimal fraction ? [63]. 5. How is a decimal reduced to a lower denomina- tion .? [64]. 6. How are pounds, shillings, and pence changed, at once, into the corresponding decimal of a pound ? [66, 67, and 68]. 7. How is the decimal of a pound changed, at onee^ into shillings, pence, &c. ? [701. 8. What are ttrminatt and circulating decimals ? [71]. 9. What are a repetend and a periodical, a pure and a mixed circulate ? [72] . 10. Why cannot the number of digits in a period be equal to the number of units contained in the divisor r [73]. 11. How is a pure circulate or pure repetend changed into an equivalent vulgar fraction ? [74] . 12. How is a mixed repetend or mixed circulate reduced to an equivalent vulgar fraction ? [76] . / 13. What kind of vulvar fraction can tproduce no equivalent finite decimal ? [79]. 14. What number of decimal places must neceet arily be found in a finite decimal ?■ [80] . 15. How maby dieits must be found in the periods of a pure circulate ? [82] . 16. When is no fiiite part found in a repetend, or circulate ? [83]. 17. How many digits must be found in the finite part of a mixed circulate ? [84]. 18. On what principal can we use the properties of fractions as a means of abbreviating the processes of multipUcaUon and division ? [86, &o!j !,.,•»-»■ m ti» ■Xi. 17^ 11 !l lit;'.''' •I,! Ih- ■i ■»?.■ '■% m SECTION V. I PROPORTION. 1. The rule of Proportion is called also the golden m^, from its extensive utility ; in some cases it is termed the rule of three — because, by means of it, when three numbers are given, a fourth, which is unknown, may be found. 2. The rule of proportion is divided into the simpie^ and the compound. Sometimes also it is divided into the direct^ and inverse — ^which is not accurate, as was shown by Hatton, in his arithmetic published nearly one hundred years ago. 3. The pupil, to have accurate ideas of the rule of proportion, must be acquainted with a few simple but important principles, connected with the nature of ratios, and the dodrine of proportion. The following truths are self-evident : — If the same, or equal quantities are added to equal quantities, the sums are equal. Thus, if we add the tanu. quantity, 4 for instance, to 5X6 and 3x10, which are equal, we shall have 5X6+4=3X 10+4. ' Or if we add equal quantities to those which are equal, the dums will be equal. Thus, since ^ ., ;. 6x6=a8XlO, and 2+a«4 , / ^ .? .J 6x6+2X2=8x10+4. 4. If the same, or equal quantities are suJttraeied from others which are equal, the remainders will be equal. Thus, if we aubtu- zt 3 from each of the equal quantities 7, and 5+2, we shall have ,,,,_ 7-«s=:6+2-«. . . And smce 8=6+2, and 4=3+1. ; . ^ ^^; r - «-4=«+2-8+r. 5. If equal quantities are muUipHed by the aame, or by equal quantities, the produott will be equal. Tliiif FROFORTION. 177 if we multiply the equals 6 1-6, and 10+1 by 3, we shall have 5+6x8— iopxs. And since 4+9=13, and 3X6=18. • 4+9X8x6al8Xl8. 6. If equal quantities are divided by the same, or hj equal quantities, the quotients will be equal. Thus if we divide the equab 8 and 4+4 by 2, we shall have 8 4+4 2"" 2 And since 20=17+3, and 10=2X6. 2017+8 10"" 2X6* 7. Ratio is the relation which exists between two quantities, and is expressed by two dots ( : ) placed be- tween them — ^thus 5 : 7 (read, 5 is to 7) ; which means that 5 has a certain relation to 7. The former quantity is called the antecedent^ and the latter the c&Mequemt, 8. If we invert the terms of a ratio, we shall have their inverse ratio ; thus 7 : 5 is the inverse of 5 : 7. 9. The relation between two quantities may consist in one being greater or less than the other — then the ratio is termed arithmetical ; or in one being some mvl' tiple or part of the other — ^and then it is geometrical. If two quantities are equal, the ratio between them is said to be that of equality ; if they are unequal it is a ratio of greater inequality when the antecedent is greater than the consequent, and of le»ser inequality when it is less. 10. As the arithmetical ratio between two Quantities is measured by their differenu, so Ions as this ai£ference is not altered, the ratio is unohanffed. Thus ^e ratio of 7 : 5 is equal to that 15 : 13 — fox 2 is, in eaoh case, the difference between the antecedent and consequent. Hence we may add the same quantity to both the antecedent and consequent of an arithmetical ratio, or may subtract it from them, without changing the ratio. Thus 7 : 5, 7+3 : 6+3, and 7—2 : 5—2, are equal aridimetioal ratios. But we cannot muUijUy or divide ihe termf <3i aa aikli- fi'S ■■ \l J 178 PROPORTION. 1 ■, ■^1 ■ . I'lll '1 1 i , ;ii 1^ m! ' "'\ 1 i! I ( 1 il ■j r il uotical ratio by the same number. Thus 12x2 : 10X2, 12-r2 : 10-r2) and 12 : 10 are not equal arithmetical ratios; for 12X2—10X2=4, 12-f-2— 10-i-2=l, and 12-10=2. 11. A geometrical ratio b measured by the quotient obtained if we divide its antecedent by its con8e<|uent ; — therefore, so Ions as this quotient is unaltered the ratio is not changed. Hence ratios expressed by equal fractions are equal ; thus 10 : 5=12 : 6, tor y=y . — Hence, also, we may multiply or divide both terms of a geometrical ratio by the same number without altering the ratio; thus 7X2 : 14X2=7 : 14— because 22^=2 . - 14X2 14 But we cannot add the same quantity to both terms of a geometrical ratio, nor subirad it from them, with- out altering the ratio. 12. When the pupil [Sec. IV. 17] was taught how to express one quantity as the fraction of another, he in reality learned how to discover the geometrical ratio between the two quantities. Thus, to repeat the ques- tion formerly given, ** What fraction of a pound is .2\d. P^ — which in reality means, *' What relation is there between 2\d. and a pound ;" or '* What must we consider 2^d.j if we consider a pound as unity ;*' *^ or,*' in fine, ** What is the value of 2^ : 1" — We have seen [Sec. I. 40] that the relation between quantities cannot be ascertained, unless they are made to have the same *^ unit of comparison :'* but a farthing is the only (init of comparison which can be applied to both 2\d. and £1 ; we must therefore reduce them to farthinss — ^when the ratio of one to the other will be- come that of 9 : 960. But we have also seen that a geometrical ratio is not altered, if we divide both its terms by the same number ; therefore 9 : 960 is the same ratio as ^f j : f|f , or ^f ^ : 1. — That is, the ratio between 2\d. and^£l may be expressed by 2\d, : £1, or 9 : 960, or j^j : 1 ; or, the pound being considered as unity, the farthing will be represented by ^j. 13. The geometrical ratio between two numbers is the •ame as that which exists between the quotient of the frMtion which represents their ratio, and unity. T^us, PBOPORTION. 179 m the last example 9 : 960 aud vf v • ^ ^^^ c^l^^l ratios. It i6 not necessary that we should bo able to express by iutegers, nor even by a finite decimal, what part or muf- tiplo one of the terms is of the other ; for a geometrical ratio may be considered to exist between any two quan- tities. Thus, if the ratio is 10 : 2, 5 ( y ) is the quantity by which we must multiply one term to make it equal to the other ; if 1 : 2, it is 05 ^|), a JinUt decimal ; but if 3 : 7, it is M28571' (f ), an tnfinUe decimal — in which case we obtain only an approximation to the value of the ratio. But though tlie measure of the ratio is ex- pressed bv an infinite decimal, when there is no quantity which will exactly serve as the multiplier, or divisor of one quantity so as to make it equal to the other — since we may obtain as near an approximation as we please — there is no inconvenience in supposing that any one number is some part or multiple of any other ; that is, that any number may be expressed in terms of another— or may form one term of a geometrical ratio, unity being the other. '"'' ' 14. Proportion^ or analogy^ consists in the equality of ratios, and is indicated by putting ==, or : :, between the equal ratios ; thus 5 : 7==9 :ll,or5:7::9:ll (read, 5 is to 7 ab 9 : 11), means that the two ratios 5 : 7 and 9:11 are equal ; or that 5 bears the same relation to 7 that 9 does to 11. Sometimes we express the equality of more than two ratios ; thus 4 : 8 : : 6 : 12 : : 18 : 36, (read, 4 is to 8, as 6 is to 12, as 18 is to 36), means there is the sarnie relation between 4 and 8, as between 6 and 12 ; and between 18 and 36, as between either 4 and 8, or 6 and 12 — it follows that 4 : 8 : : 18 : 36 — for two ratios which are equal to the same, are equal t^ each other. When the equal ratios are arithmetical, the constitute an arithmetical proportion ; when geometri cal, a geometrical proportion 15. The quantities which form the proportion are called proportionals ; and a quantity that, along with three others, constitutes a proportion, is called a fourth proportional to those others. In a proportion, the two outside terms are called the extremes^ and the two middle terms the meams ; thus in 5 : 6 : :7 : 8, 5 and 8 are the p.' i! ^Jr r ■• 180 PKOFORTIOfr. '. f » f ( If: I'.. I a it' extremes, 6 and 7 the means. When the same qnantiij IS found in both means, it is ealled the mean of thie extremes ; thus, since 5:6: : 6 : 7. 6 is /Ae metm of 5 and 7. When the proportion is arithmetical, the mean of two quantities is called their ariikmeikal mean ; when the proportion is geometrical, it is termed their geome* trioM mean. Thus 7 is the arithmetical mean of 4 and 10 ; for, since 7— 4:=10— 7, 4 : 7 : : 7 : 10. And 8 is the geometrical mean of 2 and 32 ; for, since 1=^9 2 : 8: :8 : 32. 16. In an arithmeiical proportion, ** the turn of the means is equal to the sum of the extremes." Thus, since 11 : 9: : 17 : 15 is an arithmetical proportion, 11— 9=s 17— 15 ; hut, adding 9 to hoth the equal quantities, we have 11-9+9=17-15+9 [3] ; and, adding 15 to these, we have 11-9+9+15=17-15+9+15; but 11—9+9+15 is equal to 11 + 15 — since 9 to he sub- tracted and 9 to be added =0 ; and 17— 15+9+15ss 17+9 — since 15 tabe subtracted and 15 to be added ^sO : therefore 11 + 15 (the sum of the extremes) =17+9 (the sum of the means). — The same thing might be proved from ofif other arithmetical proportion ; and, therefore, it is true in every case. 17. This equation (as it is called) , or the equality which exists between the sum of the means and the sum of the extremes, is the test of an arithmetical proportion : — that is, it shows us whether, or not, four given quantities constitute an arithmetical proportion. It also enables us to find a fourth arithmetical proportional to three given numbers — since any mean is evidently the difference between the sum of the extremes and the other mean ; and any extreme, the difference between the sum of the means and the other extreme — For if 4 : 7: :8 : 11 be the arithmetical proportion, 4+11=7+8 [16] ; and, subtractmg 4 from the equals, we have 1 1 (one of the extremes) =s7+8— 4 (the sum of the means, minus the other extreme) ; and, subtracting 7, we have 4+11—7 ^the sum cf the extremes minus one of the means) =s8 (the other mean). We might in the same way find the remainins extreme, or the remaining Anjf other arithmetioal proportion would hav* PROPORTllUf. 18i 06 quantitj ean of tM scm of 5 and he mean of ean ; when heir geome- mean of 4 9. And 8 if ace t=A» «iMn of the Thus, since n, 11-9= antities, we ding 15 to + 15; but to be Bub- +9 + 15=: added ^sO: i) =17+9 J might be >tion ; and, lality which sum of the tion : — that 1 quantities > enables us three given B difference ther mean; sum of the proportion, the equals, (the sum of btracting 7, I minus one night in the e remaining pfould ban answered Just as well — Whence what we hare said is tmo in all oases. 18. EzAMPLK. — ^Find a fourth proportional to 7, 8, 5. Makinc the required number one of the extremes, and putting the note or interrogation in the place of it, we have 7 : 8 : : 5 : ? ; then 7 : 8 : : 5 : 8+5—7 (the sum of the means minus the given extreme, b6) ; and the proportion com- pleted will be 7: 8:: 5:6. ^^* Making the reqmred number one of the means, we shall have 7 :9 : : V: 5, then 7:8:: 7+5-8 (the sum of the extremes minus the given mean, SB 4) : 5; ana the proportion completed will be 7 : 8 : : 4 : 5. As the sum of the means will be found equal to the sum of the extremes, we have, in each case, completed the pro- portion. ; r !: r frift 19. The arithmetical mea/n of two quantities is half the sum of the extremes. For the sum of the means is equal to the sum of the extremes ; or — since the means are equal — ^twice one of the means is equal to the sum of the extremes ; consequently, half the sum of the means— or one of them, will be equal to half the sum of the extremes. Thus the arithmetical mean of 19 and 19+27 27 is -—^ — (=^) > "^^ ^® proportion completed is 19 : 23 :: 23 : 27, for 19+27=r23+23. 20. If with any four quantities the sum of the means is equal to the sum of the extremes, these quantities are in arithmetical proportion. Let the quantities be 8 6 7 6. As the sum of the means is equal to the sum of the extremes 8+5—6+7. Subtracting 6 from each of the equal quantities, we have 8+1— 6&=6+7— 6 ; and subtracting 5 from each of these, we have 8+6—6—5=6+7—6—6. But 8+5—6—6 is equal to 8—6, since 5 to be added and 6 to be subtracted are =0 ; and +6+7— 6— os 7—6, smoe 6 to be added and 6 to >e subtracted sO ; i2 i MM 182 PROPORTION. iJ^ V if; . MM' i.1 Ill t *i [ , j ii ri ii . iherefore 8+5 — 6 — 5=6+7 — 6 — 5 is the same as 8—6=7^ — 5 ; but if 8 — 6=7 — 5, 8 : 6 and 7 : 5, are two equal arithmetical ratios ; and if they are two ajual arithmetical ratios, they constitute an arithmetical pro- portion. It might in the same way be proved that Any othet ibur quantities are in arithmetical proportion, 4f the suia of the means is equal to the sum ot the iztremes. 21. In a geof/utrical proportion, " the product of . he means is equal to the product of the extremes." f huSy since 14 : 7 : : 16 : iS is a geometrical proportion, y =;y [li] ; but, multiplying each of the equal quanti- fles by 7, we have (yx7)=yx7; and multiplying jach of these by 8, we have 14 X 8=16 X7(VX7X8):— out 14X8 is the product of the extremes; and 16x7 e the product of the means. The same reasoning would aold with any other geometrical proportion, and there- ^re it is true in all cases. Mtx«,MB \-v';\^'«'aL^ii>^ ajS^T ^f f < 22. This equation (as it is called), or the equality of the product of the means and the product of the extremes, (S the test of a geometrical proportion : that is, it shows us whether or not four given quantities constitute a geometrical proportion. It also enables us to find a fourth geometrical proportional to three given quanti- ties — ^which is the object of the rule of thru ; since any mean is, evidently, the quotient of the product of the extremes divided by the other mean ; and any extreme, is the quotient of the product of the means divided by the other extreme. For if 7 : 14 : : 11 : 22 be the geometrical proportion, 7X22=14X 11 ; and, dividing the equals by 7, we have 14X11 32 (one of the extremes) = — r^ — . (the product of the means divided by the other extreme) ; and, dividing thbse 7X22 by 11, we have — j-;— (the product of t^e extremes di- vided by one mean)=sl4 (the oth^r mean). We might iq'the j^ame way find the remaining meati or the remain- ing extreme. Awif other proportion would have answered just as well— and therefore what we have said is true bi fiivery case. «JM '.' ^^vnAirJ ^y« PROPORTION. yM same as 7 : 6, are two etjtuil itical pro- oVed that roportion, im ot the rodud of xtremes." roportion, lal quanti- lultiplying 7X8):— rnd 16X7 ling would md there- at .-^I (quality of extremes, , it shows nstitute a to find a en quanti- since any net of the y extreme, (iivided by proportion, 7, we have iuot of the iding thbae bremes di- We might he remain- 3 answered ud 18 true 23. EzAMPLK. — ^Find a fourth proportional to 8, 10, and 14. Making the required quantity one of the extremes, we shall 10x14 have 8 : 10 : : 14 : ?; and 8 : 10 : : 14 : 8 -(the product of the means divided l»y the given exif eme, ^17*5) . \ And the proportion completed will be . jino<|ifitj 8 : 10 :: 14:17-5. ^"^ ""'^ Making the required number one of the means, we shall 8x14 have 8 : 10 : : ? : 14^ and 3 : 10 : : (the product of the extremes divided by the gwen niean, a=ll'2) : 14. And the proportion completed will be . ^ii^Mr^s 8: 10:: 11-2: 14. - ^ - -^-^^ '>^^'«. >^f «.«t* » EXXRCISE8. Find fourth proportionals 1. To 8, 6, and 12 . Ant. 24. 2. „ 6, 8 3 4. 8. , 8, 6 8 . . 16. 4. , 6, 12 4 . 8. 5. , „ 10, 150 M 68 . 1020. 6. , „ 1020, 68 „ 150 . . 10. // 7. , „ 150« 10 „ 1020 68. 8. , ,, 68, 1020 » 10 . . 150. Vi 24. If with any four quantities the product of the means is equal to the product of the extremes, these quantities are in geometrical proportion, ^Let tjhe Qiwntities be ^ , ^^7^^^ Vw/C. V^ As the product of the means is equal to the prodi.r;t of the extremes, 5x24=20x6. *'* 5X24 20r t\ 184 PROPORTION. t i i 91 1 11:1 ■' ' 4 '■ ■ > '< M : ■ i 1 ' , i i ■ i ■;, f, : ,; ,; f, ■J !'■■ ' .tj'i ^1 ■ ' ' . ■>i \ ' ,' ''■J' i'' i : tion. It might, in the same way, be proved that immt other four quaDtities are ia geometrical proportion, if the product of the meami is equal to the product of the extremes. 25. When the first term is unity, to find a fourth proportional— BuLE. — ^Find the product of the second and third. Example.— >What is the fourth proportional to 1} 12, and 271 5 1 : 12 : : 27 : 12x27=324 - We are to divide the product of the means by the ^iren extreme ; but we may neelect the divisor when it is unity — since dividing a number by unity does not alter it. XXSRCI8XS. Find fourth proportionals 9. To 1, 17, and 8 . Ans, 186. 10. „ 1, 28 „ 20 . 460. 11. o 1, 100 „ 78 . . 7800. 12. „ 1, 63 „ 110 . . 6880. 18. „ 1, 16 ., 1284 . . 18510. 26. When either the second, or third term is unity— Rule. — ^Divide that one of them which is not unity, by the first. ExABiPLE.— Find a fourth proportional to 8, 1, and 5. 8 : 1 : : 5 ; |. We are to divide the product of the means by the given extreme ; but one of the means may be considered as the product of both, when the other is unity. For, since multi- plication by unity produces no eflTect, it may be omitted. Bzsncucs. «« fJij 1 Find fourth proportionals. 14. To 6, 20, and 1 -l" Ant, 4 16. » 6, 1 ., 20 • • 4. 16. >» 7, 21 „ 1 . . 8. 17. t> 8, 24 „ 1 8. 18. t> 6, 1 „ 60 . . 8i 19. »» 17, 1 „ 68 • . 4. 20. «» 200, 1000 „ 1 . » 6. 21. •t 200, 1 ,,1000 • . 6. 27. When the means are equal, each is said to be ikt geometriealmeaD of the extremes ; and the priDduet W HOLE OP PROPORTIOIf. 186 I that 4Mjr portion, if luct of the 1 a fourth 1 third.' ol, 12,aiid rthe ^iren J is unity — it. is unity— not unity^ ad 5. the given red as the nee multi- nitted. S^!'/ 1' > y-- • ~ id to be i produet of the extremes is equal to tke mean multiplied by itself. Hence, to discover the geometrical mean of two quan- tities, we have only to find some number which, multi- plied by itself, mil be equal to their product — that is, to find, what we shall term hereaftez, the square root of their product. Thus 6 is the geometrical mean of 3 and 12 ; for 6X6=3X 12. And 3 : 6 : : 6 : 12. 28. It will be useful to make the pupil acquainted with the following properties of a geometrical proportion — We may consider the same quantity either as a mean, or an extreme. Thus, if 5 : 10 : : 15 : 30 be a geometrical proportion, so also will 10 : 5 : : 30 : 15 ; for we obtain the same equal products in both cases — ^in the former, 5x 30=10 X 15 ; and in the latter, 10 X 15=5X30— which are the same thing. This change in the proportion is called inversion. 29. The product of the means will continue equal to the product of the extremes — or, in other words, the proportion will remain unchanged — If we alternate the terms ; that is, if we say, '* the first is to the third, as the second is to the fourth" — If we ** multiply y or divide the first and second, oi the first and third terms, by the same quantity" — If we " read the proportion backwards** — If we say *' the first term plus the second is to the second, as the third plus the fourth is to the fourth" — If we say *^ the first term plus the second is to the first, as the third plus the fourth is to the third"«-&c. RULE OF SIMPLE PBOPORTION. if^h 30. This rule, as we have said, enables us, when three quantities are given, to find a fourth proportional. The only difficulty consists in stating the question; when this is done, the required term is easily found. In the rule of simple proportion, two ratios are given, the one perfect, and the other imperfect. 31. Rule — ^I. Put that given quantity which belongs to the imperfect ratio in the wird place. II. If it appears from the nature of the question that the required quantitj must be greater than the other. /.' J86 KULE OF PROPORTION. m' . k ;4 ■| ".' . Mi;'?. 1. ■ Vli -i.' or ^yen term of the same ratio, put the larger term of the perftri ratio in the eecond, and the smaller in the first place. But if it appears that the required quantity must be less, put the larger term of the perfect ratio in the first, and the smaller in the second place. III. Multiply the second and third terms together, and divide the product by the first. — The answer will be of the same kind as the third term. 32. Example 1. — ^If 5 men build 10 yards of a waH in one day, how many yards would 21 men build in the same time i it will facilitate the stating, if the pupil puts down the question briefly, as follows — using a note oC interrogation to represent the required quantity — ' J' ' ; men. " > 10 yards. .^. 21 men. '• ' '-'-^ • •• - ■'"*' ■ ? yards. ^-'- "''''-^'' ' ' ■ ' 10 yards is the given term of the imperfect ratio — ^it must, therefore, be put in the third place. 5 men, ana 21 men are the quantities which form the perfect ratio ; and, as 21 will build a greater number of yards than 5 men, the required number (x yards will be greater than the given number — whence, in this case, we put the larger term of the perfect ratio in the second, and the smaller in the first place — ,„ . » 5 : 21 :: 10 : ? " ; "^ And, completing the proportion, ,|^ ,^^^ ,i '^ ' 6 : 21 : : 10 : ?Lji2=42, the required number. o Therefore, if 5 men build 10 yards in a day, 21 men will build 42 yards in the same time. 33. Example 2. — ^If a certain quantity oS bread is sufficient to last 3 men for 2 days ) for how long a time ought it to last 5 men ? ^ This is set down briefly as follows : , ,,, , «k .? ■ • ■fi\-:.iu-^ v. i. 3 men. • i.i .f'V>-^ '■ ::■■• ;. ; i f ,f, • t.-, -, ? days. 2 days is the given term of the imperfect ratio — ^it must^ tiierefore, be put in the third place. The larger the number of men, the shorter the time a given Quantity ^ bread will last them ; but this shirttr time is the th larger term ) smaller in he required ■ the perf&i id place. ks together, answer will \ waQ in one )same time' ts down the irrogation to io — it mast, ?h form tlio iber of yards be greater it the larger ) smaller in lumber. <1 men will is sufficient )ught it to >— it must^ me a given bime is the RULE OF PROPORTIOIf. 187 required quantity — hence, in this case, the ereater term of the perfect ratio is to be put in the first, and the smaller in the 8eo r^i -j dutu'^ *ai *♦] 5 : 3 : : jS : -r— =sl|, the required term. * 34. Example 3.— If 25 tons of coal cost £21, what will be the price of 1 ton ? ' • • ■' ' • ' ^'^"^ 1x21 21 25 : 1 : : 21 : -^ pounds je^g^l^*- ^i^- " ' ' It is necessary in this case to reduce the pounds to lower denominations, in order to divide them by 25 ; this causes the answer, also, to be of different denominations. 85. Reason of I. — It is conyenient to make the required quantity the fourth term of the proportion — that is, one of the extremes. It could, however, be found equally well, if consi- dered as a mean [28]. Reason of II. — ^It is also convenient to- make quantities of the same kind the terms of the same ratio ; because, for in- stance, we can compare men with men, and days with dayto— -> but we cannot compare men with days. Still there is nothing inaccurate in comparing the number of one, with i\k% number of the other ; nor in comparing the number of men with the quan' tity of work they perform, or with the number of loaoe$ they eat ; for these things are proportioned to each other. Hence we shall obtain the same result whether we state example 2, thus or thus 6 3 2 2 8 j£'i When diminishing the kind of quantity which is in the per- fect ratio increases that kind which is in the imperfect — or the reverse — the question is sometimes said to belong to the inverse rule of three ; and different methods are given ^r the solution of the two species of questions. But Hatton, in his Arith- metic, (third edition, London, 1758,) suggests the above gene- ral mode of solution* It is not accurate to say ** the inverse rule of three" or •* inverse rule of proportion ;" since, although there is an inverse ratiot there is no inverse pr&portion. Reason of Ill.-^We multiply the second and third terms, an^. divide their product by the first, for reasons already given [22]. The answer is of the same kind as the third term, since neither the multiplication, nor the division of this term has changed its nature ; — 20$. the payment of 5 days divided by 6 188 RULE or PROPORTION. 20f 20f. 11 n mA iil^ Ir.. ^^1 p . ■' ^T«t-r- M the payment of one day; and "-r-, the payment of one day multiplied by 9 giyes =^ X 9 aa the payment of 9 6 days. If the fourth term were not of the same kind ai the third, it would not eomplete the imper/eet ratio, and therefore it would not be the required /otirtAjiroporltona/. 36. It will often be conyenient to divide the first and second, or first and third terms, bj their greatest com- mon measure, when these terms are composite to each other [29]. ...:„' ,,.. , , : r: r .o EzAMPLK.— If 36 owt. cost £24, what will 27 cwt. cost 1 36: 27:: 24:1 Diyiding the first and second by 9 we have '''^' 4: 3:: 24:? And, dividing the first and third by 4, l:3::6:3x6»X18. - ? BZXnCISKB TOR THK PUPIL. Find a fourth proportional to 1. 5 pieces of cloth : 50 pieces : : iS27. Ans, £270 2. 1 cwt. : 215 cwt. : : 50«. Ans, 10750i. 3. 10 ft : 150 ft : : 5s, Ans, 75s, 4. 6 yards : 1 yard : : 27^. Aiis, As, 6d, 5. 9 yards : 36 yards : : I8s, Ans, 72s, 6. 5 ft : 1 ft :: 15«. Ans, 3s. 7. 4 yards : 18 yards : : Is, Ans, 4s, 6i. 8. What will 17 tons of tallow come to at £25 per ton? Ans, £425.,, 9 ' If one piece of cloth cost iS27, how much will 50 pieces cost? iln«. ) 10. If a certain ^quantity of provisions would last 40 men for 10 months,'^ how long would they suffice for 32 ? Ans, \2\ months. 11. What will 215 cwt. of madder cost at 50f. per ewt. ? Ans. 107501. 12. I desire to have 30 yards of cloth 2 yards wide, with baise 3 yards in breadth to line it, how much of tbe Utter 9haU I require ? Ans, 20 yards. -.',U th« paymeat wjmuki of 9 M the third, therefore it he first and eatest oom- lite to each owt. ooet? : (' if 'I'n. ■' ..; of .'" i-j!- ins. £270 i £25 per oh will 50 Id last 40 Be for 32 ? t 50s, per ftrdfl wide, much of • -^'■■J'im:V.t AULE or PROPORTlOir. 180 13. At lOf. per barrel, what will be the priee of 130 barrels of vsrley ? Arts. £65. 14. At 5s. per &, what will be the price of 150 lb of tea? Ans. 750s. 15. A merohant agreed with a carrier to bring 12 cwt. of goods 70 miles for 13 crowns, but his waggon being heavily laden, he was obliged to unload 2 cwt. ; how far should he carry the remainder for the same money ? Ans. 84 miles. 16. What will 150 owt. of butter cost at ^£3 per owt. ? Ans. je450. f ' 17. If I lend a person iS400 for 7 months, how much ought he to lend me for 12 ? Ans. £233 6s. 6d. 18. How much will a person walk in 70 days at the rate of 30 miles per day ? Atis. 2100. 19. If I spend £4 in one week, how much will I spend in 32 ? Ans. .£208. 20. There are provisions in a town sufficient to sup- port 4000 soldiers for 3 months, how many must be sent away to make them last 8 months ? Am. 2500. 21. What is the rent of 167 acres at £2 per acre ? Ans. ie334. 22. If a person travelling 13 hours per day would finish a journey in 8 days, in what time will be accomplish it at the rate of 15 hours per day ? Ans. 6|} days. 23. What is the cost of 256 gallons of brandy at 12f. pergaUon? Ans. 3072s. - . (^ 24. What will 156 yards of doth come to, at £2 per yard ? Am. iS312. 25. If one pound of sugar cost &2., what will 112 pounds come to ? Am. 896<2. 2b. If 136 masons can build a fort in 28 days, how many men would be required to finish it in 8 days? Am. 476. 27. K one yard of calico cost Od.^ what will 56 yards come to ? Am. 336d. 28. What will be the price of 256 yards of tape at 2d. per yard ? Am. 5\2d. 29. If iSlOO produces me £% interest in 365 days, what would bring the same amount in 30 days ? An$ £1216 \3s. U, J..^; ■i.-'.>< ' L -|» ij \90 RULE or PROPORTION. ■■•: \ fi: ' i V • 30. What shall I receive for 157 pair of gloreB, at lOd, per pair ? Ans. I570d. 31. What would 29 pair of shoes come to, at 95. per pair? Ans. 26ls. : ' • —^ 32. If a farmer lend his neighbour a cart Korse which draws 15 cwt. for 30 days, how long should he have a horse in return which draws 20 cwt. ? Ant. 22^ days. 33. What sum put to interest at iS6 per cent, would give £6 in one month ? Ans. jgl200. 34. If I lend J^OO for 12 months, how long ought £150 be lent to me, to return the kindness ? Am. 32 months. 35. Provisions in a garrison are found sufficient to last 10,000 soldiers for 6 months, but it is resolved to add ad many men as would cause them 'to be consumed in 2 months ; what number of men must be sent in ? Ans. 20,000. 36. If 8 horses subsist on a certain quantity of hay for 2 months, how long will it last 12 norses ? Ans. 1 J months. '"• ^ ' ^ ' ;• • - • -- 37. A shopkeeper is so dishonest as to use a weight of 14 for one of 16 oz. ; how many pounds of just will be equal to 120 of unjust weight ? Am. 105 lb. 38. A meadow was to be mowed by 40 men in 10 days ; in how many would it be finished by 30 men r Am. 13^ days. 37. When the first and second terms of the proportion are not of the same denomination ; or one, or both of them contain different denominations — RuLE.^-^&educe both to the lowest denomination con- tained in either, and then divide the product of the second and third by the first term, ^amiii i^ 4:i tf d ' 'i EzAMPLB 1. — If three ounoet of tea cosi ISd. what will 87 pounds cost? ..The lowest denomination contained in either is ounces* d. ii sq^.i 'i^h {■'■*'<)• lb 87 16 d. 15: 1392x15 d. =6960==£29. W mi% Y! X t... ; av+^.^2_ounce8.^^.j^,^ ^.^ ,^jj ,-. ^.^^.^ There is evidently the same ratio between 3 oz. and 87 lb as between 8 oz. and 1892 oz. (the equal of 87 tb). RULE or PROPORTION. >f gloTOB, at to, at 9j. per ; horse which d he have a It. 22| days. ' cent, would ; ought iSl 50 \ 32 months, sufficient to s resolved to be consumed he sent in ? mtiiy of hay •rses ? Am. use a weight of just will )5ib. men in 10 by 30 men r // . le proportion , or both of aination con- duct of the what will 87 r is ounoei. >z. and 87 lb ExAMPLC 2.~If 3 Yards of any thing cost 4f . 9|d., i ean be bought for £2 ? ' ' "» • '«<*[*-" -> i^£29 Is. 5(1 19 qrs. 28 _>• '* 549 lbs. rrz 'd ?j:*' 'i-'^f* r!*«irii* 30 28 qrs. »'j"'i ^ ""I <.»•'. -. 58. If 12 !b, 6 OS., 4 dwt., cost £150, what will 3 lb, 1 01., 11 dwt., cost ? Ant, £37 10«. 59. If 10 yards cost 17«., what will 3 yards, 2 qrs. , cost? Ant, bt. \l\d, 60. If 12 cwt. 22 ib cost £19, what will 2 cwt. 3 qrs. cost ? Ant, £4 bt, 8\d, 61. If 15 01., 12 dwt., 16 grs., cost 19«., what will 13 OS. 14 grs. cost? Ant, \bt, \Qd, 38. If the tlurd term consists of more than one deno- mination-^ "' *"" Rule. — ^Reduce it to the lowest denomhiatlon which it contams, then multiply it by the second, and divide the product by the first term. — ^The answer will be of that denomination to which the third has been reduced, and may sometimes be changed to a higher [Sec. RULI OF PROPORTIOIf. 193 , at £\S p«r wiH be the ir, et £5 per I will bo the ne to, at 3s. oome to, at », at 7d, per lOf. 6d. per hat will no 138 7«. 2\d. w much will r. 10^. £6ibl5t.} 760 12f., at lat will 3 fb, ^arde, 2 qrs. ill 2 cwt. 3 ., what will n one dono- ation which , and divide | ' will be of 6n reduced, ;her [Sec. ■uMPLi 1.— If 8 yardf cost 9«. 2kl., what will 827 yards at The lowejt denomination in tbe third term it fkrthingt. ydi. ydi. «. d. 827^441 £ '• c^* 8:^27:: 9 2^ : — x— ^fitfthinge-SO 1 6|. 110 penoe. 441 farthings. ExABfPLE 2.— If 2 yards 3 qrs. cost lljit., what will 27 yards, 2 qrs., 2 nails, cost ? The lowest denomination in the first and second is nails, snd in the third farthings. yl"- V* ?3?y-5* A\ 442x45 2 3 : 27 2 2:: 11 qr. 110 qr. 4 4 44 nails. 442 nails. 11| •• — ^j — iarthings—9*. 6d. 45 farthings. Redaoing the third tsnn generally easbles ns to perform the required multiplication snd diTision with more fseUity.— It is iometimesi howeTer* unneoeassry. ExAMPLK.— If 3 lb cost j&3 11«. 4|<2., what will 96 lb cost? lb Xb £ s. d. ^ '• fj Qi. ^ '• <*• ^ '• ^' 8:96::3 11 4}: 3^^ »8 11 41x82-114 4 8 EXERCISES. Find fiMirth proportionsls to '■■'■ 62. 3 tons : 14 tons : : ^£28 lOf. Ant, 199 10«. 63. 1 ewt. : 120 cwt. : : 18#. 6d, Ans, £U\, 64. 6 barrels : 100 barrels : : 6«. 7d, Am. £6lU.ed 65. 112 lb ! 1 lb :: £3 IOj. Ans. Hd. 66. 4 lb : 112 lb : : bjrd. Ans. \2s. 3d. 67. 7 ewt., 3 qrs., 11 lb : 172 cwt., 2 qrs., 18 lb : : JB3 f«. 4f£. Ans. £»1 bs. 4d. mm 194 RULE OF PROPORTION. 68. 172 cwt., 2 qrs., 18 ft) : 7 cwt., 3 qrs., 11 lb : : £87 65. 3d. Ans. £3 19s. A\d. 69. 17 cwt., 2 qrs., 14 lb : 2 cwt., 3 qrs., 21 lb : : £73 Ans. iS12 3s. 4d. 70. .£87 6*. 3d, : £3 I9s. 4^d. : : 172 cwt., 2 qrs., 18 ft). Ans. 7 cwt., 3 qrs., lift*. 71. £3 195. 4^d. : ie87 65. 3d,i:7 cwt., 3 qrs., 11 Bt>. Ans. 172 cwt., 2 qrs., 18 tt>. v>:>*T c 'f be |if:il r i. fl f^r ';"I. 72. At I85. 6d. per cwt., what will 120 Cwt. cost? Ans. £in. 73. At 3\d. per pound, what will 112 ib come to? Ans. £1 105. 4d. " • ' "''■' ■'' • 74. What will 120 acres of land come to, at 145. 6d. per acre ? Ans. £S1. 75. How much would 324 pieces come to, at 25. 8Jrf. per piece ? An>s. £43 lis. 6d. 76. What is the price of 132 yards of iploth, at I65. 4d. ver yard ? Ans. £107 16*. 77. If 1 ounce of spice costs 35. 4 78. If 1 ib costs 65. Sd.y what will 2 cwt. 3 qrs. come to ? ins. £102 135 4d. 79. If £1 25. be the rent of 1 rood, what wUl be the rent ( i 156 acres 3 roods ? Ans. £689 145. 80. At IO5. 6d. per qr., what will 56 cwt. 2 qrs. be worth? Ans. £118 135. 81. At 155. 6d. per yard, what will 76 yards 3 qrs. come 60 ? Ans. £59 95. 7^d. 82 What will 76 cwt. 8 tt» come to, at 25. 6d. per ft>? -4715. £1065. 83 At 14?. 4£?, per cwt., what will be the cost of 12 cwt. li qrs. ? Ans. £8 195. 2d. 84. How much will 17 cwt. 2 qrs. come to, at 195. lOd. ^er cwt. Ans. £17 75. Irf. 85 If 1 cwt. of butter costs £6 65., what will 17 cwt , 2 qrs , 7 lb, come, to ? Ans. £102 125. lO^d, ' 86 If 1 qr. 14 ft) cost. £2 155. 9rf., what will be the cost of 50 cwt., 3 qti.y 24 lb ? Ans. £378 l6s.S\d. I ,1:, RULE OF PROPORTION. 195 llft>::£87 lft»::£73 i.,2qr8., 18 \ qrs., 11 ft». I dwt. eost? h oome to ? , at 14f. 6d. y at 2s. S^d. iloth, at 165. kt will 18 & 3 qrs. come will be the rt. 2 qrs. be yards 3 qrs. 2s, 6d. per e cost of 12 to, at ids. nil 17 cwt , d. will be the [et.8\d. 87. If the shilling loaf weigh 3 lb 6 oz., when floor sells at £1 13s. 6d. per cwt., what should be its weight when flour sells at j£l 7s. 6d ? Am. 4 ib 1|| oz. 88. If 100 lb of anything cost £25 6s. 3d.y what will be the price of 625 ft) ^ Am. iB158 4*. OJrf. 89. If 1 lb of spice )OSt 105. 8d.y what is half an oz. worth ? Ans. Ad, 90. Bought 3 hhds. of brandy containing, respectively, 61 gals., 62 gals., and 62 gals. 2 qts., at 65. Sd. pef gallon ; what is their cost } Ans. jS61 I65. 8<2. 39. If fractions, or mixed numbers are found in ono or more of the terms — Rule. — Haying reduced them to improper fractions, if they are complex fractions, compound fractions, or mixed numbers — ^multiply the second and third terms together, and divide the product by the first — ^according to the rules already given [Sec. lY. 36, &c., and 46^ &c.] for the management of fractions. Example. — ^If 12 men build 3f yards of wall in | of a week, how long will they require to build 47 yards % 3f yardsss^" yards, therefore 3 K^ 47 i^ V : 47 : : I : ^fe — ^ g ^®®^ nearly. ^ 40. — ^If all the terms are fractions — Rule. — ^Invert the first, and then multiply all the terms together. Example. — ^If | of a regiment consume || of 40 tons of flour in f of a year, how long will | of the same regiment take to consume if? I : ? :: I : ?Xf^|=?X|Xf«^^^262-8 days. This rule follows from that which was given for the division of one fraotion by another [Sec. IV. 49]. 41. If the first and second, or the first and third terms, are fractions — Rule. — ^Reduce them to a common denominator (should they not have one already), and then omit the denominators m 196 RULE OF PROPORTION. W-i |s|« i\wm^. ExAMPLS.— If j of )l cwt. of rice costs £2, what will -fg of a cwt. cost 7 ; ■a If Reducing the firactionis to a common denominator, we have And omitting the denominator, ' 20 : 27 : : 2 : ?^=X2-7=je2 14». This is to^relj multiplying the first and second, or the ^rst and third terms by the common denominator — ^which [30] does not alter the proportion. . ■ ' • . t ■ ■ -■•■■■; I , : XZERCISES. 91. What will | of a yard cost, if 1 yard costs 135 6^. ? Am. 10«. l^d. 92. If 1 ib of spioe costs }«., what will 1 fi> 14 oz. Gost.^ Ans. Is. 4f^. 93. If 1 oz. of silyer costs d|5., what ^1 } oz. cost ? Ans. 4s. 3d. 94. How much will \ yard come to if | cost |5. ? Atis. -^jS. 95. If 2\ yards of flaoQel cost 3J-5., What is the price of 4f yards } Ans. 65. Ad. 96. "What will 3f oz. of silver cost at Q^s, per oz. ? Ans. £\ la. A\d. 97. If tV of a ship ooats dg273^, what is ^ of her worth I Ans. J^27 I2s. Id. 98. If 1 lb of silk costs 16|5., how many pounds can I have for 37^5. } Ans. 2} ib. 99. What is t^e |)rice of 49,^ yards of cloth, if 7f cost iB7 18*. 4d. > Ans. £51 3*. l|^frf. 100. If JglOO of stock is worth £98^, what will ^2362 8*. 7^d. be worth .> Ans. ^358 7s. Id. 101. If 945. is paid for 4| yards, how much can he bought for i&2|\ ? Ans. 24 yards, nearly. MISCELLANBOUS EXERCISES IN SIMPLE PROPORTION. 102. Sold 4 hhcls. of tobacco at lOid. per ib : No. 1 weighed 5 cwt., ij (jrs. ; No. 2, 6 cwt., 1 qr., 14 lb : No. 3, 5 cwt., 7 ft» ; and No. 4, 5 cwt., 1 qr., 21 fc. What was their price ? Ans £^04 lAs. 9d. RULE OF PROPORTIOIf. 197 it will A of or, we liave r. or the Srst ch [30] does [ costs 135 1 & 14 oz. } oz. cost ? } cost |5. ? is tlie price s. per oz. } ■ff of her pounds can cloth, if 7f t wiH £362 etch can be »0RTI0N. ft) : No. 1 14 lb : No. ft». What 103. Suppose that a bale of merchandise weighs 300 Tb, and costs Js 15 4s. 9d. ; that the duty is 2d. per pound ; that the freight is 255. ; and that die porterage home is Is. 6d. : \how much does 1 lb stand me in ? £ s. d, 15 4 9 cost. 2 10 duty. 15 freight. 16 porterage. lb 300 ib 1 : 19 20 881 la 300)4575 3 entire^ost. 15|i2. Answer. 104. Beceived 4 pipes of oil containing 480 gallons whidi cost 5s. 5^d. per gallon ; paid for freight 45. per pipe; for duty, 6d. per gallon.; for porterage, l5. per pipe. What did the whole cost ; and what does it stand me in per gallon ? Am. It cost iS144, or 65. per gallon 105. Bought three sorts of brandy, and an equal quantity of each sort : one sort at 55. ; another at 65. ; and the third at 75. What is tl^e cost of the whole — one gallon with another ? Ans. 65. 1^. Bought three kinds of vinegar, and ^n equal quantity of each kind: one at 3f^. ; another at 4^.; and another at 414' per quart. Having mi^ed them I wish to know what the mixture cost me per quart ? Ans. 4d. 107. I^ought 4 kmds of salt, 100 barrels of each ; and the prices were 145., I65., 175., and 195. per barrel. If I mix them together, what wiU the miztuire have cost me per barrel f Aiss. t^s. ^d. 108. How many reams of paper at <95. 9(2., and 125. Sd. per ream jshaU I have, if I buy £55 worth of both, bat an equ^ (^uini^ty of eadh ? Am, 50 re^ms ofeacti. 109. A vintner paid JS171 for three kinds of itine : one kmd was £S 10s. ; another £9 5s. ; and tiie third ":^ f^ mm l'%yk ■ VI »:i ■ 198 RULE OP PROPORTION. j£10 155. per hhd. He had of each an equal quantity, the amount of which is required. £ s. 8 10 9 5 10 15 28 10, £ 5. £ 28 10 :] L71 10, the price of three hogsheads of each. £171x3 £28 10 =18 hhds. 110. Bought three kinds of salt, and of each an equal quantity; one was 14s., another 165., and the third 195. the barrel ; and the whole price was £490. How many barrels had I of each ? Ans. 200. 111. A merchant bought certain goods for jS1450, with an agreement to deduct £1 per cent for prompt payment. What has he to pay ? Ans. iS1435 105. 1 12. A captain of a ship is provided with 24000 fb of bread for 200 men, of which each man gets 4 S> per week. How long will it last ? Ans. 30 weeks. 113. How long would 3150 ib of beef last 25 men, if they get 12 oz. each three times per week ? Ans. 56 weeks. 114. A fortress containhig 700 men who consume each 10 fi> per week, is provided with 184000 lb of provisions. How long will they last ? Ans. 26 weeks and 2 days. 115. In the copy of a work containing 327 pages, a remarkable passage commences at the end of the I56th page. At what page may it be expected to begin in a copy containing 400 pages ? Ans. In the 191st page. 116. Suppose 100 cwt., 2 qrs., 14 lb of beef for ship's use were to be cut up in pieces of 4 fib, 3 ib, 2 fib, 1 fib, and | fib — there being an equal number of each. How many pieces would there be in all > Ans. 1073 ; and 3| fib left. 117. Suppose that a greyhound makes 27 springs while a hare makes 25, ana that their springs are of equal length. In how many springs will the hare be overtaken, if she is 50 springs before the hound ? RULE OF PROPORTION. 199 quantity, Is of each. h an equal third 195. How many or £1450, for prompt J5 10*. 1 24000 ft» ts 4 ft) per IS. 25 men, if Afis. 66 consume iOOO lb of . 26 weeks 7 pages, a the 156th begin in' a 1st page. f beef for 3 ft), 2 fib, r of each. ins. 1073 ; 27 springs ngs are of 6 hare be id? ^ Th^ time taken by the greyhound for one spring is to that ri^uired by the hare, as 25 : 27, as 1 : ||, or as 1 : lA^y2]. The greyhound, therefore, gains ^y of a spring durmg «very spring .of the hare. Therefore ^ : 50 : : 1 spring : 50-f.^2,=675, the number of springs the hare will make, before it is overtaken. 118. If a ton of tallow costs £35, and is sold at the rate of 10 per cent, profit, what is the selling price ? Am. JS38 105. 119. If a ton of tallow costs £317 105., at what rati must it be sold to gain by 15 tons the price of 1 ton ? Am. £A0. 120. Bought 45 barrels of beef at 2l5. per barrel ; among them are 16 barrels, 4 of which would be worth only 3 of the rest. How much must I pay? Am. jB43 l5. 121. If 840 eggs are bought at the rate of 10 for a penny, and 240 more at 8 for a penny, do I lose or gain if I sell all at 18 for 2d. ? ^715. I gain ^d. 122. Suppose that 4 men do as much work as 5 women, and that 27 men reap a quantity of corn in 13 days. In how many days would 21 women do it ? Am. The work of 4 men=sthat of 5 women. Therefore (dividing each of the equal quantities by 4, they will remain equal), 4 men's work , , ^ the work of 5 women „ T (one man's work)= -. . Con- sequently \\ times the work of one woman=s-l man's work : — that is, the work of one man, in terms of a woman's work, is \\ ; or a woman's work is to a man's work : : 1 : 1|. Hence 27 me-ixo work = 27 x 1| women's workj then, in place of saying — 21 women : 27 men : : 13 days : ? say the work of 21 women : the work of 27x11 (=33?) 333 y 13 women :: 13 : — ^r — =:20|f days. J^ 123. The ratio of the diameter of a circle to its circumference being that of 1 : 3*14159, what is the circumference of a circle whose diameter is 47*36 feet? Am. 148-78618 feet. 124. If a pound (Troy weight) of silver is worth 665., 200 RULE OF PROPORTIOlf. .it) • v. ■ ! %A h ,i It »' ■■:l|:i I . lit I 1 wliat is the value of a pound avoirdupoise ? Am. £4 Ot. 2\d. 125. A merchant failing, owes iB4088t87l Ito his creditors ; and has property ta the amount of ill2677dl7 105. 1 \d. How much per cent, ean he pay ? Ans. iS30 lbs. Sid. 126. K the digging of an English mile of canal costs £1347 Is, 6d.y what will he the eost of an Irish mile ? Ans. iS1714 165. 9|e2. 127. If the rent of 46 acres, 3 roods, and 14 perches, is £100, what will be the rent of 35 acres, 2 roods, and 10 perches? Am. £75 185. 6^4. 128. When A has tr&velled 68 days at the rate of 12 miles a de^, B, who had travelled 48 da3'9, overtook him. How many miles a day did B travel, allowing both to have started from the same plaee ? Ans. 17. 129. If the value of a pound avoirdupoise weight be £4 05. 2|ill be finished in 1|- days. In how many days would C be able to do it by himself? ^915. 2^}f days. ^ ^ RITLB OF PROPORTION. 201 Am. £4 ft to his 12677517 Ans. iS30 anal costs rish mile ? I perches, ■oods, and be rate of overtook allowiikg ns. 17. weight be t for one is tenant ; 38. 4id. oes to j64 a? Ans. days; B PTOuld all it wholie— > in a day. i« whole— in a day. le whole— > in a day. » in a day. work :: 1 lie work) : fit. ^y=* ! = B TTi t^ill Id C s I A, B, and C*s work in one day=£| of the wholesj^}} ^5^ V* «.- whole^^^ • • • mi 1 whole of the work : : 1 Sttbtraet iag ) B'8 work in 1 day^j. G'8 work in one day remiaintf «qual to Then JJJ^ (C's work in one day) day : 2 ^||, the time required. 134. A ton of coals yield about dOOO cubic feet of gas ; a street lamp consumes about 5, and an argand burner (one in which the air passes through the centrd of the ftame) 4 cubic feet in an hour. How many tons of coal would be required to keep 17493 street lamps, and 192724 argand burners in shops, ^c, lighted for 1000 hours.? Ans. 95373^. 135. The gas consumed in London reauires about 50,000 tons of coal per annum. For how long a time would the gas this quantity may be supposed to pro- duce (at the rate of 9000 cubic feet per ton), keep one argand light (consuming 4 cubic feet per hour) con- stantly burning .? Ans. 12842 years and 170 days. 136. It requires about 14,000 millions of silk worms to produce the silk consumed in the United Kingdom annually. Supposing that every pound requires 3500 worms, and that one-fifth is wasted in throwing, how many pounds of manufactured silk may these worms be supposed to produce .? Ans. 1488 tons, 1 cwt., 3 qrs., 17 ft). 137. If one fibre of silk will sustain 50 grains, how many would be required to support 97 ib ? Ans 13580. 138. One fibre of silk a mile long weighs but 12 grains ; how many miles would 4 mmions of poundSy annually consumed in England, reach ? ' ' Ans. 2333333333^ miles. 139. A leaden shot of 4| inches in diameter weighs 17 ft» ; but the size of a shot 4 inches in diatheter, is to that of one 4^ inches in diameter, as 64000 : 91125 : what is the weight of a leaden ball 4 inches iiS diameter ? Ans. 11-9396. 140. The sloth does not advance more than lOQ yards in a day. How long would it take to cra^ from Dublin to Cork, allowing the distance to be 160 Engliab miles ? Ant. 2816 days; or 8 years, nearly. ^^ : I;- 202 COMPOUND PROPORTIOn. I Iv '1 ^ 141. English race horses have been known to go at the rate of 58 miles an hour. In what time, at this Telocity, might the distance from Dublin to Cork be travelled over ? Ans. 2 hours, 45' 31" 2" 142. An acre of coals 2 feet thick yields 3000 tons ; and one 5 feet thick 8000. How many acres of 5 feet thick would give the same quantity as 48 of 2 feet thick? Ans. 18. 143. The hair-spring of a watch weighs about the tenth of a grain ; and is sold, it is said, for about ten shillings. How much would be the price of a pound of crude von, costing one halfpenny, made into steel, and then into hair-sprines — supposing that, after deducting waste, there are (Stained from the iron about 7000 grams of steel ? Ans. iS35000. COMPOUND PROPORTION. 42. Compound proportion enables us, although two or more proportions are contained in the question, to obtain the required answer by a single stating. In compound proportion there are three or more ratios, one of them imperfect, and the rest perfect. 43. Rule — ^I. Place the quantity belonging to the imperfect ratio as the third term of the proportion. ll. Put down the terms of each of the other ratios in the first and second places — ^in such a way that the antecedents may form one column, and the consequents another. In setting down each ratio, consider what effect it has upon the answer — ^if to increase it, set down the larger term as consequent, and the smaller as ante- opdent ; if to diminish it, set down the smaller term as consequent, and the larger as antecedent. jn. Mi^tiply the quantity in the third term by th9 product of all th. quantities in the second, and divide the result by the product of all those in the first. 44. Example 1. — ^If 5 men build 16 yards of a wall in 20 days, in how many days would 17 men build 37 yards 1 * ' The question briefly put down £32], will be as fi^Uowf ! [■>'. f -iS I COMPOUND PROPORTIOir. 203 Q to go at e, at this } Cork be )000 tons ; I of 5 feet of 2 feet about the about ten . pound of stee], and deducting bout 7000 lOugh two lestion, to tting. In atios, one ig to the ion. ler ratios that the nsequents der what set down ' as ante- r term as n by th^ nd divide It. wall in 20 irds? l[>ll0Wj|! 16 Yards i ®®°<^^t>®'*'' which give 20 days. , 20 days imperfect ratio. ? days, the number sought, ny J I conditions which give the required number of dayi. The imperfect ratio consists of days — therefore we are to put 20, the given number of days, in the third place. Two ratios remain to be set down — that of numbers of men, and that of numbers of yards. Taking the former first, we ask ourselves how it affects the answer, and find that the more men there are, the smaller the reauired number will be — since the greater the number of men, tne shorter the time required to do the work. We, therefore, set down 17 as antecedent, and 5 as consequent. Next, considering the ratio consisting of yards, we find that the larger the number of ^ards, the longer the time, before they are built — therefore increasing their number increases the quantity required. Hence we put 37 as consequent, and IC as antecedent ; and the whole will be as follows : — 17 : 5 : : 20 : I i^ 16 : 37 ;^;. And 17 : 5 : : 20 : — ttTTTS"— 13-6 days, nearly. 16:37 ITXB^ ^"^ ^ 45. The result obtained by the rule is the same as would be found by taking, in succession, the two proportions supposed by the question. Thus If 5 men would build 16 yards in 20 days, in how many days would they build 37 yards ' 16 : 87 : : 20 : T^ — number of days which 6 men would require, to build 37 yards. If 6 men would build 87 yards in — ^ — days, in how many 16 , days would 17 men build them ? - 17 K 20x37 . 20X37^K . i7«. 20x5x87 .. ^ «„«i^. 17 : 6 : : -^ : — ^X6^17= ^^^^^ , the number of days found by the rule. 46. Example 2. — ^Tf 3 men in 4 days of 12 working hours each build 37 perches, in how many days of 8 working hours ought 22 men to build 970 perches % d04 COMPOUND PROPORTION. 'Vl i ir • 1 • ■<-',i:l. 22 : 8 :: 4 8 : 12 87 : 970 3 men. 4 days. '< ' 12 houri. 37 perches. T days. 8 hours. 22 men. 970 perches. 8X12X970X4 22X8X87 * ^'^ '••'*/• The number of days is the quantity sought ; therefore 4 days constitutes the imperfect ratio,^and is put in the third place. The more men the fewer the days necessary to per- form the work ; therefore, 22 is put first, and 3 second. The smaller the number of working nours in the day, the larger the number of days; hence o is put first, and 12 second. The greater the number of perches, the greater the nvimber of days required to build them ; consequently 17 is to be put first, and 970 second. 47. The process may often be abbreviated, by divid- ing one term in the first, and one in the second place ; or one in the first, and one in the third place, by the same number. Example 1. — ^If the carriage of 32 cwt for 5 miles costs hs.y how much ixdll the carriage of 180 cwt. 20 miles cost ? 32 5 160 20 8 160x20x8 32x5 =160 Dividing 32 and 160 by 32 we have 1 and 5 as quotients. IXviding 5 and 20 by 5 we have 1 and 4 ; and the propor- tion wiU be — 1 1 5 4 8 : 5x4x8»160 48. We arc to continue this kind- of division as long as possible — ^that is, so long as any one number will measure a quantity in the first, and another in the second place ; or one in the first and another in the third place This will in some instances change most of the quantitiei into unity — which of course may be omitted. COMPOUND PROPOBTIOlf. •0. Mirly. therefore 4 in the third tssarj to per- seoond. The r, the larger 12 second. r the riMmber 17 is to be 1, by divid- iond place ; loe, by the ' miles coats miles cost ? ExAMPLK 2— If 28 loads of stone of 15 cwt. each, build a wall 20 feet Ions and 7 feet high, how many loads of 19 cwt. will build one 323 feet long and 9 fbeti high f 19 : 15 : : 2» : 15x323x9x28 ^^^ . ' {^ V 20 : 323 19x20x7 7 : 9 Dividing 7 and 28 by 7, we obtain 1 and 4^8absiitttting these, weliaye 19 : 16 :: 4 : 1 20 : 323 1:9 Dividing 20 and 15 by 5, the quotients are 4 ani 3 : 19 : 3 : : 4 : ? . 'I 4 : 323 1:9 ' Dividing 4 and 4 by 4, the quotients are 1 and 1 : 19 : 3 :: 1 : 1 1 : 323 1:9 Dividing 19 and 323 by 19, the quotients are 1 and 17 : 1 : 3 :: 1 : 3xl7x9»459. In this process we merely divide the first and second, or first and third terms, by the same number — which [29] does not alter the prt^rtlon. Or vire divide the numerator and denominator of tno fraction, found as the fourth term, by the same number — ^which [Sec. IV. 15]. does not alter the quo- tient. - quotients, the propor- ion as long umber will the second ;hird place B quantitiei EX£BCI8E8 IN COMPOUND PROPORTION. 1. If iS240 in 16 months gains iS64, how much will £60 gain in 6 months ? Ans. £9. 2. With how many pounds sterling conld I mim £5 per annum, if with £450 I gain £30 in 16 monthd ? Am. JBIOO. 3. A merchant agrees with a carrier to brihg 15 cwt. of goods 40 miles for 10 crowns. How niuch ought he to pay, in proportion, to have 6 cwt. carried 32 milcB ? Ans, 16«. r i- .!■! 206 COMPOUND PROPORTION. 4. If 20 owt. are carried the distance of 50 miles for £5y how much will 40 owt. cost, if carried 100 miles ? Atu. i£20. 5. If 200 ft of merchandise are carried 40 miles for 3s. y how many pounds might he carried 60 miles for je22 I4s. 6d. Arts. 20200 &>. 6. If 286 fb of merchandise are carried 20 miles for 3s. y how many miles might 4 cwt. 3 qrs. be carried for JB32 6*. Sd. ? Ans. 2317-627. 7. If a wall of 28 feet high were built in 15 days b^ 68 men, how many men would build a wall 32 feet high in 8 days ? Ans. 146 nearly. 8. If 1 n> of thread make 3 yards of linen of 1|- yards wide, how many pounds of thread would be required to make a piece of linen of 45 yards long and 1 yard wide ? Ans. 12 ib. 9. If 3 lb of worsted make 10 Yards of stuff of 1^ yards broad, how many pounds would make a piece 100 yards long and 1| broad ? Ans. 25 tt>. 10. 80000 cwt. of ammunition are to be removed from a fortress in 9 days ; and it is found that in 6 days 18 horses have carried away 4500 cwt. How many horses would be required to carry away the remainder in 3 days? Ans. 604. 11.3 masters who have each 8 apprentices com i£36 in 5 weeks— each consisting of 6 working days. How much would 5 masters, each having 10 apprentices, earn in 8 weeks, working 5^ days per week — the wages being in both cases the same ? Ans, jSllO. 12. If 6 shoemakers, in 4 weeks, make 36 pair of men's, and 24 pair of women's shoes, how many pair of each kind would 18 shoemakers make in 5 weeks ? Ans. 135 pair of men's, and 90 pair of women's shoes. 13. A wall is to be built of the height of 27 feet ; and 9 feet high of it are built by 12 men in 6 days. How many men must be employed to finish the remain- der in 4 days ? Ans. 36. 14. If 12 horses in 5 days draw 44 tons of stones, how many horses would draw 132 tons the same dis- tance in 18 days ? Ans. 10 horses. 15. If 27s. are the wages of 4 men for 7 days, I'i COMPOUND PROPORTIOir. 207 50 miles for 100 miles ? [ 40 miles id 60 miles 20 miles be carried in 15 days fall 32 feet inen of 1^ be required EUid 1 yard stuff of 1| i piece 100 e removed it in 6 days aany horses inder in 3 ' > • . ■ * . 3 cam £36 lays. How pprentices, -the wages 36 pair of my pair of 5 weeks ? l's shoes, f 27 feet ; in 6 days, he remain- of stones, same dis- r 7 days, what will be the wages of 14 men for 10 days ? Ans. £6 15#. I. . . .. .. 16. If 120 bushels of com last 14 horses 56 days, how many days will 90 bushels last 6 horses ? Ant, 98 days. 17. If a footman travels 130 miles in 3 days when the days are 14 hours long, in how many days of 7 hours each will he travel 390 mues ? Ans. 18. 18. If the price of 10 oz. of bread, when the corn is 45. 2d. per bushel, be 5^., what must be paid for 3 lb 12 oz., when the corn is 5i. dd. per bushel ? Atis. 3s, 3d. 19. 5 compositors in 16 days of 14 hours long can compose 20 sheets of 24 pages in each sheet, 50 lines in a page, and 40 letters in a line. In how many davs of 7 hours long may 10 compositors compose a volume to be printed in the same letter, containing 40 sheets, 16 pages in a sheet, 60 lines in a page, and 50 letters in a line ? Ans. 32 days. 20. It has been calculated that a square degree (about 69X69 square miles) of water gives off by evapora- tion 33 millions of tons of water per day. How much may be supposed to rise from a square mile in a week ? Ans. 48519-2187 tons. 21. When the mercury in the barometer stands at a height of 30 inches, the pressure of the air on every square inch of surface is 15 ib. What v^ill be the pres- sure on the human body — supposins its whole surface to be 14 square feet ; and that the barometer stands at 31 inches? A.ns. 13 touB 19 awt, ,^ ,^^,, j; QUESTIONS IN RATIOS AND PROPORTION. 1. What is the rule of proportion ; and is it ever called by any other name ? L^] • .. 2. What is the difference between simple and com- pound proportion ? [30 and 42] . 3. What is a ratio .? [7]. 4. What are the antecedent and consequent ? [7] . 5. What is an inverse ratio ? [8] . 6. What is the difference between an arithmetical and a geometrical ratio ^ [9] . sii: 208 COMPOUND PROPORTION. I'' ■I .t;t 7. How can we know whether or not an arithmetical or geometrical ratio, is altered in value ? [10 and 11]. 8. How is one quantity expressed in terms of an other? [12]. 9. What is a proportion, or analogy ? [14] . ' 10. What are means, and extremes ? [151. 11. What is the arithmetical, or geometrical mean of two quantities ? [19 and 27]. 12. How is it known that four quantities are in arith- metical proportion ? [16]. 13. How is it known that four qnatftitses turfe in geo- metrical proportion ? [21] . 14. How is a fourth proportional to three quantities found .^ [17 and 22]. 15. Mention the principal changes which may be made in a geometrical proportion, without destroying it.? [29]. ^ ' 16. How is a question in the simple rule of three to be stated, and solved ? [31]. 17. Is it necessary, or even correct, to divide the rule of three into the direct, and inverse ? [36] . 18. How is the question solved, when itbe first oi second terms are not of the same denomination ; or one, or both of them contain different denominations ? [37] 19. How is a question in the rule of proportion solved, if I9ie third term consists of more Hhan one denomina- tion ? [38]. 20. How is it solved, if fnuHaons f three to » divide the ib]. ioe fifst Of ion ; or one, ions? [37] rtion solved, e den(Hnina- ed ni;ini%eni tand Mrdj ipovnd pro- in 't^e rulo ' ali^ether \i.i': il n ■■ ■■i'.f: '/ <1 *' • '• .. *'• ., ARITHMETIC. PART II. SECTION VI. PRACTICE. 1. Praciloe is so called from its being the method of calculation practised by mercantile m6n : it is an abridged mode of performing processes dependent on the rule of three — ^particularly when one of the terms is unity. The statement of a question in practice, in general terras^ would be, " one quantity of goods is to another, as the price of th« former is to the price of the latter." The simplification of the rule of three by means of practice^ is principally effected, either by dividing the given quantity into ** parts," and finding the sum of the prices of these parts ; or by dividing the price into ** parts," and finding the sum of the prices at each of these parts : in either case, as is evident, we obtain the required price. 2. Farts are of two kmds, " i^iquot" and " aliquant." The aliquant parts of a nuniber, are those which do not measure it — that is, whicfh eannot be multiplied by any integer so as to produee it : the aliquot parts are, as we have seen [Sec. IE. 26]^ tliose which measure it. 3 To find the aliquot parts of any number — Rule. — ^Divide it by its least ^i^or, and the result- ing quotient by its leait divisor :— proceed thus until the last quotient is unity. AU the liivisors are the prime aliquot parts ; and the pMdudt of every two, every Uiree, ko.j of them, are the eomp&umd m^fixii parts of the given number. m 210 PRACTICE. are the compound aliquot parts. 4. Example. — What are the prime, and compound aliquot parts of 84? 2)84 2)42 . . - : 3)21 ' i - , 7)7 "I The prime aliquot parts are 2, 3, and 7 ) and 2x2s= 4l 2x^-= 6 2x7=14 3x7=21 2x2x3=12 2x2x7=28 2x3x7=42 J All the aliquot parts, placed in order, are 2, 3, 4, 6, 7, 12) 14, 21, 28, and 42. 5. We may apply this rule to applicate numbers. — ^Let it be required to find the aliquot parts of a pound, in shillings and pence. 240(i.=j£l. >? 2)240 r 2)120 ■ ' ^ / -' ' 2)60 - ' 2)30 ;,.. 3)J[5 . , 6)^ 1 The prime aliquot parts of a pound are, therefore, 2d.y 3(2., and 5(2. : and the compound, d. ^ ' ■ ■. " .2x2= 4 ■ \ ■"•■ ' 0-'^-. ■;-;»* ■• 2x5= 10 Mfv, f ; 2x2x2= 8 ». d, > - a u^ vW, 2x2x3=12=1 2x2x5= 20= 18 2x3x5= 30= 2 6 2x2x2x2= 16= 1 4 2x2x2x3= 24= 2 2x2x2x5= 40= 3 4 2x2x3x5= 60= 5 2x2x2x2x3= 48= 4 2x2x2x2x5= 80= 6 8 2x2x2x3x5=120=10 ') ■ 'y.i''''i ».■.!• Li I • -.i ■•:■ : 'V'i' ■.i\r. ' • t ! PRACTICE. 211 ind aliquot uot parts. 4, 6, 7, 12k rs. — ^Let it 1 shillings sfore, 2d.y \ • i 'hi «>V-v And placed in order — vn — 1 T« — I Tff — 1 _ TS — ^=12=1 d, 2 3 4 5 6 8 ■tv 16= 20= 24= 30= 40= 48= 60= 80= 1 vz i: I J— 1=120=10 1 1 2 2 3 4 5 6 d, 4 8 G 4 8 Aliquot parts of a shilling, obtained in the same way— s. d. ^,*Z8 T5— C d. JL-sl '^-11 =2 I. Aliquot parts of avoirdupoise weight — s. d, 1=6* Aliquot parts of a ton. ton cwt. qr. 1 1 ¥Z »TZ ■R — 1 5— I T— 1 ss i 2 4 11=5 2 = 8 2!=10 4=16 5 =20 -J=10 =40 Aliquot parts i^ a cwt. cwt. lb r 4 7 8 1 1 . 1=14 4=16 i=28 i=56 Aliquot parts of a quarter qr. lb U '. U m Aliquot parts may, in the same manner, be easily obtained by the pupil from the other tables of weights and measures, page 3, &c. i *: ; . 6. To find the price of a quantity of one. denomina- tion — the price of a " higher" being given. Rule. — Divide the price by that number which ex- presses how maoy times we must take the lower to make the amount equal to one of the higher denomina- tion. Example, — What is the price of 14 ft) of butter at 72s. per cwt. 1 We must take 14 lb, or 1 stone 8 times, to make 1 cwt. Therefore the price of 1 cwt. divided by 8, or 72s.-i-8=95., is the price of 14 lb. The table of aliquot parts of avoirdupoise weight sho'vi^ that 14 lb is the i of a cwt. Therefore its price is the { of the pric« of 1 cwt. 21S Iki :. I . ?' F1CACTICE. EXERCISES. What is the price of 1. I cwt., at 295. 6d. per cwt. } Ans. Is. A{d. 2. ^ a yard of cloth, at 85. Qd. per yard } Ans. As. 3d 3. 14 ft) of sugar, at 45«. Qd. per cwt. } Ans. bs. ^\d 4. What 19 the price of | cwt., at 50«. per cwt. ? fi s. d. ';, 50s.=2 10 The £ 5. , ■ li, tfn. cwt. price of 2=4 is 15 0==2 lO-i-2 „ of 1=1-5-2 is 12 0=1 6^2 Therefore the price of 24-1 qr8.(=J cwt.) is 1 17 6 J cwt., or 3 qrB.=2-f-l qrs. Bat 2 qr8.=r| cwt. ; and it8 ice 18 half that of a cwt. 1 qr.=s^ cwt.-^>2 ; and its price price 18 half the price ei 2 qrs. Therefore the price of f cwt. h half the price of 1 cwt. plus the half of naif the price of one cwt. What is the price of 5. I oz. of cloves, at 9*. 4d. per ft) ? Atis. S^d. 6. 1 nail of lace, at ]5«. 4d. per yard? Ans. ll^d. 7. I ft), at 235. 4d. per cwt. ? Ans. l\d. S. f lb, at 185. Sd. per cwt. ? Ans. l^d. 7. When the price of more than ont "lowjr" deno- jnination is required — KuLE. — Find the price of each denomination by the last rule ; and the sum of the prices obtained wUl be the required quantity. Example. — ^What is the price of 2 qrs. 14 lb of sugar, at 455. per owt. ? 9. d. 45 price of 1 cwt. — [or ^ of 1 cwi cwt. And 22 6, or 45?. h-2, is the price of 2 qrs., 2 qr8.=4 14 «)=.{, or I of 2 qrs. 5 71, or 45«. -f.8=r229. 6rf. -j.4, is the price of 14 lb, the I of 1 cwt., or the I of 2 qrs. And 28 1^ is the price of 2 qrs. 14 lb. 2 qn.asf of 1 cwt. Therefore 45*. (the price of 1 cwt.) -^2, et 26$. 6(2., i» the price of 2qr». 7 6 PRACTICE. 213 Ans. 4$. 3d Ans. 5«. S\d er cwt. ? . d. } £ s. 5 0=2 lO-i-2 I 2 G=l 5^2 i ^ cwt. ; and its , and its price e of f cwt. iu ' the price of | ns. 3^d. Ans. ll^d. i. owjr" deno- lation by tbe .ined wUl be fb of sugar, or I of 1 cwi >riceof2qr8., Qd.-^4, is the 10 ^ of 1 cwt., 8. qrs. 14 lb. )flcwt.)-f-2, 14 lb is the I of 1 cwt., or the | of 2 qrs. Therefore 45«.-j-8, or 22«. 6d.-f.4=5s. 7\d., is the price of 14 lb. And 22s. Qd.-^s. 7|rf., or the price of 2 qrs. plus the price of 14 lb, is the price of 2 qrs. 14 Vb. EXERCISES. What is the price of 9. 1 qr., 14 ib at 46*. 6d. per cwt. ? Ans. 17s. 5\d. 10. 3 qrs. 2 nails, at 175. 6d. per yard? Ans. 15s. 3^d. 11.5 roods 14 perehfes; it 35. lOrf. ^er acre ? Ans. 6s. l^d. 12. 16 dwt. 14 grs., at £4 4s, 9d. per oz. ? Ans, £3 10s, 3id. 13. 14 lb 5 oz., at 255. 4d. per cwt. } Ans. 3s. 2^d. 8. When the price of one '^ higher" denomination is required — liuLE. — ^Find what number of times the lower deno- mination must be taken, to make a quantity equal to one of the given denomination ; and multiply the price by that number. (This is the reverse of the rule given above [6]). Example. — What is the price of 2 tons of sugar, at 505. per cwt. ? 1 cwt. is the ^jf of 2 tons ; hence the price of 2 tons will be 40 times the price of 1 cwt.— or 50s. x40=£100. 50s. the price of 1 cwt. multiplied by 40 the number of hundreds m 2 tons, gives SOOOs . ^ j or JCIOO as the price of 40 cwt., or 2 tons. E^tkRCIS^S. What is the price of 14. 47 cwt., at l5. Sd. per lb ? Ans. ^2438 135. 4d 15. 36 yards, at 4d. per nail .? Ans. iE9 125. 16. 14 acres, at 55. per perch .? Ans. £^60, ^ 17. 12 ft), at l^d. per grain ? -4ii5. iB504. 18. 19 hhds., at 3d. per gallon ? Ans. iB14 195. 3d, 9. When the price of more than one "higher" deno- minatio*! is required — m^ m ^^ 214 PRACTICE "i \m -| Rule. — Find the price of each by the last, and add the results together. (This is the reverse of the rule given above [7]). Ex^PLE.—What is the price of 2 cwt. 1 qr. of flour, at 2$. per stone 1 1 stone is the ^ o{2 cwt. Therefore 2s., the price of one stone, multiplied by 16, the number of stones in 2 cwt., gives 32s., the price of 16 stones, or 2 cwt. There are 2 stones in 1 qr, ; therefore 2s. (the price of 1 11 stone) x2=4s. is the price of 1 qr. And 32s.-f-4s.=36s.=s |1 £1 16s., is the price of 2 cwt. 1 qr. . , ,• j EXERCISES. .'. - . ; {. What is the price of ' ^ 19. 5 y:.rds, 3 qrs., 4 nails, at 4d. per nail ? Ans. £1 I2s. 20. 8 cwt. 14 ft), at 3d. per ft) ? Ans. £8 Us. 6d. 21. 3 ft) 5 oz., at 2\d. per oz. ? Ans. 9s. ll\d. 22. 9 oz., 3 dwt., 14 gre., at ^d. per g". ? Ans, £\3 155. 4^d. 23. 3 acres, 2 roods, 3 perches, at 5^. per perch } Ans. iB140 16*. ?- • 10. When the price of one denomination is given, to find the price of any number' of another — Rule.— Find the price of one of that other denomi- nation, and multiply it by the given number of the latter. ' im Example. — What is the price of 13 stones at 255. per cwt.? ^ 1 8tone=s=| cwt. Therefore i .'.I r ' 8)25s., the price of 1 cwt. divided by 8, gives 3 1 1, the price of 1 stone, or I of 1 cwt. Multiplying this by 13," the number of stones, V^' we obtain £2 7^ as the price of 13 stones. 1 stone is the J of 1 cwt. Hence 25s.-f-8=3s. IW., is the price of one stone ', and 35. 1^. x 13, the price of 13 stones. 2 2 2 2 2 2 3 aliq ber B tion give reqi E 46 wha arti< or 2 Hi 3 3 £6i 1 aliq oft 1 giv qui last, and add B of the rule 1 qr. of flour, cwt., •r 2 cwt. the price of 1 s.-|-4s.=36s.=5 nail ? Ans, . £S Us. 6d, )s. n^d. g-r. ? Ans, per perch? n is given, to ther denomi- imber of the 38 at 255. per Ivided by 8, I, or I of 1 cwt. aes, stones. \s. Ihi.., is the of 13 stones. PRACTICE. EXERCISES. 215 What is the price of 24. 19 ft), at 2d. per oz. .? Ans. £2 10*. Sd. 25. 13 oz., at 1*. 4d. per ft) ? Ans. Is. Id. 26. 14 ft), at 2s. 6d. per dwt. .? Ans. ^£420. 27. 15 acres, at 18*. per perch } Ans. i£2160. 28. 8 yards, at 4d. per nail ? Ans. £2 2s. Sd. 29. 12 hhds., at dd. per pint ? Ans. iei26. 30. 3 quarts, at 9ls. per hhd. ? Am. Is. Id. 11. When the price of a given denomination is the aliquot part of a shilling, to £nd the price of any num- ber of that denomination — Rule. — Divide the amount of the given denomina- tion by the number expressing what aliquot part the given price is of a shilling, and the quotient will be the required price in shillings, &c. Example. — What is the price of 831 articles at 4d. per '^ 3)831 217s.=£lZ 175., is the required price. 4d. is the.^ of a shilling. Hence the price at 4d. is | of what it would be at Is. per article. But the price at Is. per article would be 831s.:— therefore the price at 4d. is 831s.-t-2^ or277». . ^ EXERCISES. . .' What is the price of 31. 379 ft) of sugar, at 6d. per ft) ? Ans, £9 9s. 6d. 32. 5014 yards cf calico, at Sd. per yard? Ans. £62 13*. 6d. 33. 258 yards of tape, at 2d. per yard ? Ans. £2 3s. 12. When the price of a ^ven denomination is the aliquot part of a pound, to find the price of any number of that denomination — KuLE. — Divide the quantity whose price is sought by that number which expresses what aliq|uot part the given prioe is of a pound. The quotient will be the re- quired price in pounds, &o. *..,>. m m IB' W'-^ 216 PRACTICE. Example. — ^What is the price of 1732 lb of tea, at 5i pertt>? 5s. is the | of £1 ; therefore the price of 1732 lb is the I of what it wo^ld be at £1 per ft). But at Jpl wr lb it would be J&1732; th^efore i^t 5<. p^f^ lb it is 43,733^4n £433. EXERCISES. , I What is the price of 34. 47 cwt., at 6s. Sd, per cwt. ? Jns. £15 I3s. Ad. 35. 13 oz., at As. per oz. ? Ans. £2 I2s. 36. 19 stones, at 2s. 6d. per stone ? Am. £2 7s. 6d. 37. 83 ft), at 1*. Ad. per ft) ? ^^. .£5 10*. 8d. 38. 115 qrs., at 8d. per qr. ? Ans. £3 16s. 8rf. 39. 976 ft), at 10*. per ft) .? 4^. ^88. 40. 112 ft), at 5^. per ft) .^ ^7». j|S2 65. 8^. 41. 563 yards, at lOd. per yard ? Ans, £23 9 J. 2d. 42. 112 ft), at^ 5*. per ft) } Ans. £2% 43. 795 ft), at 1*. 8rf. per ft) ? il«5. 466 5«. 44. 1000 ft), at 35. Ad. per ft) ? ^9U. £\m \3s. Ad, " - i 13. The cqmplm^ of the price is wliat it want^ of a pound or a slbilluig» When th|B complement of the price is the aliquot part or parts of a pound or shilling, but the price is not — . BuLE. — Find the price at £1, or l5. — as the case may be — and deduct the price of the quantity calculated at the complement. .^^ 7 Example. — ^What is the price of 1470 yards, at 135. Ad. per yard % 6s. Sd. (the complement of 135. 4^.) is ^ of £1. From X1470, the price at £1 per yard, «^VftW<( i99i, % pnc^ a^t %, 84, (tl^e cp^pkm^njt) and thediffbrence, 980, wiH be the price at 13s. ^^ j^r ya^ 1470 yasda at 13». 4d., plus 1470 at 65. 8(2., are equal to 1470 ait I3j(. Ad.4^6s. Sd, oe at £1 per yard. Hentie the pricp of 1470 at IS*. 4^.=the pric« of 147Q at £1, vaJsm the price of 1470 at 65. od. per yajrd. ... , j, ^ . . Ifi^'. PRACTICE. 217 r tea, at 5i 32 lb is thf. f 1 p?r lb it EXERCISES. 16 135. 4d. £2 7s. 6d. 9. 8d. I*. 8d, d. ^23 9s, 2d, 5s, )6 13*. 4d, Wft»tfif of a liquot part ;* not — B the case calculated at 13s. 4d. ippfem^ujt) e equal to HenOa the ^1, ^nfflyu« i'!.i. What is the price of 45. 51 ft), at 17«. eef. per ib? ^7». £44 12s. 6« 46. 39 oz., at 7d. per oz. ? Ans. £1 2t. 9d. ' 47. 91 tt>, at 10<;. per i^ } Ans. £3 \5is. ibd, 48. 432 cwt., at IQs, per cwt ? Am, £345 I2s, 14. When neither the price nor Its complenient is the aliquot part or parts of a pound or shilling — Rule l.--l-Diyide the price into pounds (if there are any), and aliquot parts of a pounid or shilling ; then find the price at each of these (i>y preceding rules) :— <- the sum of the prices will be what is' required. * Example.— What is the price of 822 lb, at £5 I9s, ZU, per lb 1 £5 I9s. 3|d.=X5-f.l9«, Z}d, But 19^. ZUr 8. 10 6 2 d. 8 6 £ [hr ^^ is ^^ **^« ^*«* 'vtvi <^ i <>f ^^ la^ Hence the price at £5 19s. 3f<2. is equal to £ 822x5 8 38 • la sfa >! '£ 8. d. trr411Q 0, the price at = 411 I = 274 » 102 15 = 52 9 = 17 ^ £ s. d. 5 per lb. £1 or la jelorO 6 8 £\ or Q 2 6 „ £jU or U •_L «• A of 'ffmr orO 11 It n n n n And £4903, 14 10^ is the pricj9 at £b 19 3^ The price at the whole, is evidently equa^ to the sum of the prices at each' of the 'tofa;' -■' •- ' * '^ If the price wex^ ^5 19f. 3|(^. per ^, we should sub- tract, and not add the price at \d. per lb ; and we then would have £49S>Z, Os. %^d. aa the answer, 15. Rule 2. — ^Find the price at a pound, a shillinff, a penny and a forthing ; then multiply each 'hf their I' 2l8 PRACTICC^ i-'' V PI respectiye numbers, in the given price ; and add the products. Using the sama example — £ a. d, £ 9. 20)822 (the price at £1)X 5=^4110 12)41 2 (theprioeatl«.)Xl9= 780 18 8 6 (the price at li/.)X 3= 10 5 17 1 l(the price at iriee at the oe. A the giTen £ 14932 ». 28 6 U 1903 14 10} xt higher it the price iven price erft? Sd, per lb. re 7s. 9d, .2d. Ans.£55 ins, jSlO shillings. Rule. — Multiply the nunihor of articles by half the number of shillings ; and conuidcr the tens of the pro- duct as pounds, and the units doubled^ as Hhillings. Example.— What i( the price of 640 lb, at IQs. per lb 1 646 8 610 8 2_ £516 109. 2?. being the tenth of a pound, there are, in the price, half as many tenths as shillings. Therefore half the number of shillings, multiplied by the number of articles, will express the number of tenths of a pound in the price of the entire. The tens of these tenths will be the number of pounds ; and the units (bei^^ tenths of a pound) will be halfthe required number of shillings — or, multiplied by 2 — the required num- ber of shillings. In the example, 16s., or £-8, is the price of each article. Therefore, since there are 046 articles, 046 xX-8=ii5 10-8 is the price of them. But 8 tenths o^ a pound (the units in the product obtained), are twice a.s many shillings ; and hence we are to multiply the units in the product by 2. EXERCISES. 'r"' What is the price of 54. 3215 ells, at 6s. per ell ? Ans. j£964 lOs. 55. 7563 lb, at 8*. per ft) } Ans. je3025 45. ' 56. 269 cwt., at 16*. per cwt. ? Ans, iB215 4*. ' 57. 27 oz., at 4^. per oz. ? Ans. £5 8s. 58. 84 gallons, at 14^. per gallon ? Ans. £5S I6s. 19. When the price is an odd number of shillings, and less than 20 — Ru]t.E.^~Find the amount at the next lower even number of shillings ; and add the price at one shilling. Example. — ^What is the price of 275 lb, at 175. per lb ? 275 •> ; , 8 ^^:,lj(. tw •ill vf The price at 16s. (by the last rule) is 220 Tlie price at Is. is 275s.=s . ^ IS 15 Hence the price at 16s.-|-ls^ or 17s., is £>2&Z 15s. •J- * If •!f-, I I ■fcJr 220 PRACTICE. The price at lis. ia equal to the price at I65., plus price at one Bhilling. the EXERCISES. 59. 86 01., at 5s. per oz. ? Ans. £f2\ IO5. . 60. 62 cwt., at 19j. per cwt. ? Ans. £d8 18s, 61. 14 yards, at I7s. per yard ? Ans. £\l 18s. 62. 439 tons, at lis. per ton ? Ans. £241 9s. 63. 96 gallons, at 7s. per gallon ? Ans. £33 12s. 20. When the quantity is represented by a mixed number — Rule. — ^Find the price of the integral part. Then multiply the given price by the numerator of the frac- tion^ and divide the product by its denominator — the quotient will be the price of the fractional part. The sum of these prices will be the price of the wbole quan- tity. Example.— What is the price of 8J lb of tea, at 5s. per £ s. d. ThepriceofSIbisSxSs.ss 2 The price of I ft is ^4^'— 3 9 An^ tbe price of 8| ft is 2 3 9 The pric^ of } of a pound, is evidj^ntly f of, the price of a pound. . ^ EXERCISES. What is the price of 64. 51 dozen, at 3s. 3d. per dozen I Ans. 17s. 104^. 65. 273i flb, at 2s. 6d. per ft? Ans. £3.4 3s. IJrf. 66. 530J ft, at i4s. per ft ? -Ans. 371 10s. Bd. ' 67. 1781 oWt., at 17s. per cwt. ? Ans. £151 12s 68. 762} cwt., at £1 12s. 6^. per cwt. ? Ans. £1239 4s. 6rf. - 69. 817f\^> cwt., at £3 7s. 4d, per ewt.? Ans, /^751 IU.6U. mil abb| Enc II mul lipii PRACTICE. 221 )«•, plus the 18«. |l 185. 9*. 33 12*. ' a mixed •t. Then ' the frac- ator — the irt. The ole quan- at 55. per i 1 I price of a *. Urf. ed. 151 125 . £1239 rr ? ^915. The rules for finding th( >rice of several deno- minations, that of one being given [7 and 9], may be abbreviated by those which follow — Avoirdupoise Weight. — Given the price per cwt., to find the price of hundreds, quarters, &c. — IluLE. — Having brought the tons, if any, to cwt., multiply 1 by the number of hundreds, and consider the product as pounds sterling ; 5 by the number cf quar- ters, and consider the product as shilling; 24, the number of pounds, and consider the product as pence : — the sum of all the products will be the price at £\ per cwt. From this find the price, at the given number of pounds, shillings, &o. Example. — What is the price of 472 owt., 3 qrs., 16 lb, at JC5 95. 6(2. per cwt. ? £, s. d. Multipliers 472 5 3 10 2^ 472 17 10| is the price at £1 per cwt. 2364 9 3^ the price, at £5 per cwt. 212 16 Of the price, at 95. (£>i^y.^.^ U 16 5\ the price, at U. (X^v-^2.) 2589 1 ^\ the price, at £^ 95. W. At £1 per cwt., there will be £>\ for every cwt. We mul- tiply the qrs. by 5, for shillings ) because, if one cwt, co§t8 £1, the fourth of 1 cwt., or one quarter, will cost the fourth of a pound, or 55. — and there will be as many times 55. as there are quarters. The pounds are multiplied by 2\ ; becf^use if the quarter costs 55., the 28th part of a quarter, or 1 lb, must cost the 28th part of 5«., or 21d. — and there will be as many times 2^d, as there aire pounoS' EXERCISES. What is the price of 70. 499 cwt., 3 qrs., 25 &, at 255^ 11<^. per cwt. ? Ans. £647 175. 7^^. ^^ - - "^ 1 .?»* .|..^e. , 71. 106 cwt., 3 qns., 14 &, at I85. 9d. per owt. ? Ans, £100 35^ lOi^. ' -, % ■ u 222 PRACTICE. /.,; .\ 72. 2061 cwt., 2 qrs., 7 ft), at 165. Qd.^ per cwt. ? Ans. £1700 155. 9i^. 73. 106 cwt., 3 qrs., 14 ft), at 95. Ad, per cwt. ! Ans. je49 175. 6^. 74. 2Q cwt., 3 qrs., 7 ft), at 155. Qd. per cwt. ? ^tw. iB21 25. S\d. 75. 432 cwt., 2 qrs., 22 ft), at 185. 6i. per cwt. ? -4w5. £400 45. 10^(i:. 76. 109 cwt., qrs., 15 ft), at 195. 9^. per cwt. ? Ans. £107 155. 4|rf. 77. 753 cwt., 1 qr., 25 ft), at 155. 2d, per cwt. } Ans, £571 75. Sd, 78. 19 tons, 19 cwt., 3 qrs., 27|ft), at £19 195. 11 J^. per ton } Ans, £399 195. Qd, 22. To find the price of cwt., qrs., &c., the price of a pound being given — Rule. — Having reduced the tons, if any, to cwt., multiply 95. Ad. by the number of pence contained in the price of one pound : — this will be the price of one cwt. Divide the price of one cwt. by 4, and the quotient will be the price of one quarter, &c. Multiply the price of 1 cwt. by the number of cwt. ; the price of a quarter by the number of quarters ; the price of a pound by the number of pounds ; and the sum of the products will be the price of the given quantity. Example.— What is the price of 4 cwt., 3 qra., 7 ft), at 8(i. per ft>. ? S. d, ■-.-■•. -.^ r". s- 9 4- -^s -. > . .-■ .; 5. d. 4)74 8 the price of 1 cwt. X4, will give 298 8 the price of 4 cwt. 28)18 8 the price of 1 qr. X3, will give 66 the price of 3 qrs. 8 the price of 1 ft) X 7, will gi ve 4 8 the price of 7 ft). 20)359 4 And the price of the whole will be £17 19 4 At \d, per ft) the price of 1 cwt. would be 112d. or 95. Ad. : — therefore the price per cwt. will be as many times 95. Ad. as there are pence in the price of a pound. The price of a q^uarter is | the price of 1 cwt. ; and there will be as many limes the price of a quarter, as there are quarters, &o. :i ,■ m PRACTICE. 223 per cwt. ? wt. r Ans. 7t. ? Atu, per cwt. ? wt, ? Ans, per cwt. ? 19*. Hid. e price of f to cwt., itained in ice of one e quotient of cwt. ; ters; the 1 the sum quantity. ., 7 lb, at ■■■h EXERCISES. '•e ;i ■.•■ eof4owt. le of 3 qrs. :eof7 lb. 9».4rf.:— 9«. 4d. as )rice of a i as many What is the price of I 79. 1 cwt., at ed. per ib ? Atis. £2 16s. 80. 3 cwt., 2 qrs., 5 lb, at 4d. per fib ? Ans. £6 12s. Ad, 81. 51 cwt., 3 qrs., 21 lb, at M. per fi> ? Ans. JS218 25. M. 82. 42 cwt., qrs., 5 lb, at 25e2. per S^ r Ans. jS490 10*. 6rf. ' . • I 83. 10 cwt., 3 qrs., 27 lb, at bid, per ft) ? Ans. £26\ Us. 9d. ■ 23. Given the price of a pound, to find that of a ton — Rule. — Multiply £9 6s. 8d. by the number of pence contained in the price of a pound, v^-. „ y~,^.^^'r: Example. — What is the price of a ton, at Id, pei' lb ? ' '" • '''.-. •"• ' £ 8. d, < •' '■■■ -.- '-- ■■"■-' i 9 6 8 .0 . 7 65 6 8 is the price of 1 ton. If one pound cost IJ., a ton will cost 2240^., or £9 6s. Sd. Hence there will be as many times £9 6s. Sd. in the prica of a ton, as there are pence in the price of a pound. t exercises. What is the price of 84. 1 ton, at 3d. per lb ? Ans. iB28. 85. 1 ton, at 9rf. per lb ? Ans. £84. 86. 1 ton, at lOd. per lb .? Ans. je93 6s. Sd, 87. 1 ton, at 4d, per lb ? Ans. £37 6s. Sd, ' The price of any number of tons will be found, if we mul- tiply the price of 1 ton by that number. 24. Troy Weight. — Given the price of an ounce — to find that of ounces, pennyweights, &c. — * . ^ Rule. — Having reduced the pounds, if any, to ounces, set down the ounces as pounds sterling ; the dwt. as shillings ; and the grs. as nalfpence : — this will give the price at jSl per ounce. Take the same part, or parts, &c., of this, as the price per ounce is of a pound. i- i: 224 PRACTICE. m ' :i * ' ■ i I f :i EzAMPLK 1 . — What 18 the price of 538 oz., 18 dwt., 14 grs., at lis. 6d. per oz. ? lU. 6d. =|l+§+^^2. : //'• £ s. d. 2 )538 18 7 is the pHce, at iCl per birnce. 10)269 9 3A is the price, at 10». per ounce. 2) 26 18 111 is the price, at Is. per oUkice. 13 9 5| is the price, at 6d. per ounce. And 309 17 iB| is the price, at ll». 6J. per ounce. 14 halfpence are set down as 7 pence. If ohe biihce, or 20 dWt. cost £1, 1 ^wt. or the 20th part of an ounce will cost the 20th part of £1 — Oris.; and the 24th ^rt of 1 dwt., or 1 gr. Will cost the 24th part of 1«. — or ^. Example 2. — What is the price of 8 oz. 20 grs., at £Z 2s. 6d. per oz. ? £ s. d. 8 10 is the price, at £1 per ounce. 3 A- i 24 2 6 is ihe pride, at £3 per ounce. Price at £l-f-10= 16 1 is the price, at 2s. per ounce. Price at 2s.-t- 4=r 4 0| is the price, at Qd. per ounce. And £25 2 7\ is the price, at £Z 2s. 6d. per oz. EXERCISES. What is the price of S8. 147 oz., 14 dwt., 14 grs., at 7s. 6d, per oz. } Ans. £56 7s, U^d. 89. 194 oz., 13 dwt., 16 grs., at lis, 6d. per oz. ? Ans. ieill 18s. l6}d. 90. 214 oz., 14 dwt., 16 grs., at 12«. 6<2. per oz. ? Ans. iei34 4s. 2d, 91. 11 ft), 10 oz., 10 dwt., iio gfs.,'iit lbs. per oz. } Ans. iE71 6s. 5rf. , 92. 19 ft), 4 oz., 3 grs., at £2 5s, 2d, })er 6z. ? Ans, JS523 18s. ll^^;. 93. 3 oz., 5 dwt., 12 grs., at £1 6s. 6d, per os. ? Ans. £4 7s. Sid, 18 dwt., 14 ice. nee. loe. ce. r ounce. le 20th part 9.; and the Lth part of grs., at £Z er ounce. > per ounce, per ounce, per ounce. s. 6d. per oz. . per oz. } . per oz. ? .per oz. .' per oz. ? ? Ans, iZ. . pet 6<. ? PRACTICE. 225 25. Cloth Measure. — Given the price per yard — to find the price of yarda, quarters, &c. — Rule. — Multiply j£l by the number of yards; bs. by the number of quarters ; Is. 3d. by the number of nails ; and add these together for the price of the quantity at £\ per yard } Take the dame part, or parts, &c., of this, as the price is of ^1. Example 1. — What is the price of 97 yards, 3 qrs., 3 nails, at 8s. per yard 'i £1 5s. Is. Zd. Multipliers 97 3 2 2 )97 17 6 is the price, at £1 per yard. 5)48 18 9 18 the price, at 10s. per yard. From this subtract 9 15 9 the price, at 2s. per yard. And the remainder 39 3 is the price, at 8s. (10s.— 2s.) If a yard costs jCI, a quarter of a yard must cost 5s. ; and a nail, 0/ the 4th of a yard, will cost the 4th part of 5s. or Is. Zd. Example 2. — ^What is the price of 17 yards, 3 qrs., 2 nails, at £2 i.- 9d. per yard ? £1 5s. Is. 3d. Mu ..v5ers 17 3 2 17 17 6 is the price, at £1 per yard. 2 35 15 is the price, at £2 per yard. The price at £1-^- 4=s4 9 4^ is the price, at 5s. The price at 5s.h-10s0 8 111 is the price, at Qd. The price at 6(2.-^ 2a=0 4 5} is the price, at 3(2. And £40 17 9| is the price, at £2 5s. 9d. EXERCISES. What is the price of 94. 176 yards, 2 qrs., 2 nails, a Ids, per yard ? Ans. iei32 95. 4ld. 95. 37 yards, 3 qrs., at £1 5s. per yard ? Ans, £47 3«. 9d, 96. 49 yards, 3 qrs., 2 nails, at £1 10s, per yard? Ans. £1A 16s. 3d. 97. 98 prds, 3 qrs., 1 nail, at £1 lbs, per yard? Ans. £\12 18*. b\d. ' '^ 226 PRACTICE. i' ill' ■'!■ iHfi?^ ^ » 98. 3 yards, 1 qr., at 175, 6^^. per yard? Ans £2 16s. lO^d. 99. 4 yarda, 2 qrs., 3 nails, at £1 2s. Ad. per yard ? Ans. £5 4s. S^d. ,. 26, Land Measure. — Kule. — Multiply £1 by the number of acres ; 55. by the number of roods ; and 1 ^d. by the number of perches : — the sum of the prduucts will be the price at £1 per acre. From this find the price, at the given sum. Example. — What is the rent of 7 acres, 3 roods, 16 perches, at £>Z Ss. per acre ? . V . £> s. d. 1 5 U Multipliers 7 3 16 ^,,>'j Sum of the products 7 17 0, or the price at £1 per acre. 3 23 11 the price at £3 per acre. 3 18 6 the price at IO5. per acre. 27 9 6 the price at £3 IO5. per acre. Subtract 15 8^ the price at 2s. per acre. And 26 13' S^ is the price at jC3 85. If one acre costs £1, a quarter of an acre, or one rood, must cost bs. ; and the 40th part of a quarter, or one perch, must cost the 40th part of 5«. — or \kd, ., 'I ' > i. EXERCISES* "What is the rent of 100. 176 acres, 2 roods, 17 perches, at £5 6s. per acre ? Ans. £936 Os. Sd. 101. 256 acres, 3 roods, 16 perches, at £6 6s. 6d, per acre.? Ans. ^21624 Us. 6}rf. 102. 144 acres, 1 rord, 14 perches, at £5 6s. Sd. per acre .? Ans. £769 I65 103. 344 acres, 3 rtiods, 15 perches, at £4 Is. Id. per acre ? Ans. i£1398 I5. Id. 27. Whie Measure. — To find the price of a hogs- head, when the price of a quart is given — * ^' Rule. — For each hogshead, reckon as many pounds, and shillings ag there are :|^nce per quart. PRACTIOE. 227 Ans £2 per yard ? 21 by the ; and ^\d. rduucts will I the price, 3 roods, 16 CI per acre. per acre, per acre. iOs. per acre, per acre. £3 8«. ae rood, must perch, must JB5 6s. per £6 65. 6d, ) 6s, Sd. per £4 Is. Id. of a hogs-* any pounds, hx.KKfhx. — What M the price of a hogshea4rat 9(2. per quart* Atis.£d9s. ^ --' ♦. ]. >^n i;; . ; ;.., .| One hogshead at Id. per quart would be dSx4, since there are 4 quarts in one caUoa, and 68 gallons in (^e hhd. But 68X4^. ■ What is the price 4f 104. 1 hhd. at I8d, per quart ? Aiv^. dS]i8 18«.n kvi 105. 1 hbd. at 19 Ans. £2fi 4s. 108. 1 hhd. at 2s. 6d. pet quart ? Ans. £Z\ 10s. When the price of a pint Ss given, of course we know that of a quart. - ' ■ 28. (j^iven the price of a quart, to find that of a tun-— BuLE. — Take 4 times as many pounds, and 4 times as many shillings as there are pence per quart. Example. — ^What is the price of a tun at 11(2. per quart '\ & 9. ' ' '' .fcl'j' >J ' r%f 11 11 4 ,;».)] ntnw .'•• 'J' •I 46 4 is the price of a tun. t J. V -lUj- •J Since a tun contains 4 hogsheads, its price must be 4 times the price of a hhd. : that is, 4 times as many poUnds and shil- £ngs, as pence per quart [27]. EXERCISES. What is the price of 109. 1 tua, «t 19<;. per quart .^ Ans. 0fd 16s. 110. 1 tun, at 20rf. per quart .> Ans. £S4. * 111.1 tun, at 2s. per quart ? Ans, iSlOO 16s. 112. 1 tun, at 2s. Gd. per quart .> Ans. £\26. 113. 1 tun, at 2s 8(2. per quart } Sns. JS134 8s. 29. A number ef Ariides.'-^ireik the prltie of 1 article in penoe, to find that of any number-— BiH.E.'— vDivide the aumber hiy 12, ^ liul^gn i&d y . '.-^ ! ■ii">7 5; ii^' I: ■ .,■ 228 PITACTICE. It • ■« , I penoo; and' multiply the quotient by the number of pence in the pri^e* Example.— What is the price of 438 articles, at 7d eaoV^ 36«. 6(7., the price at Id, each. 20 )255 6 £12 15 6 the price at 7d. each. 438 articles Md. «aeh will cost 438(f.s:86«. 6(2. At 7(f. «ach, they will cost 7 times as muojti— or 7X36«. Qd.m^bs. Qd.-r^ £1216$. Qd, ^X .7.rk ;.?•,■ ,fv, 'i^vt .1/^1, .h; ,?>f|{f I .•.'. ^ EXERCISES. f . .:^0? , : .. What is the price of ,,-ur ■ di mdli 114. 176 lb, at Zd. per ft) .? Ans, £2 4s. U5. 146 yards^ at 9(2. per yard ? 4ns, £5 ds, 64 116. 180 yards, at 10^(2. per yard ? :Ans. £7 17*. 64 117. 192 yards, at 7^d. per yard ? Ans. £6. 118. 240 yards, at 8^(2. per yard ? Ans. £S 10s 30. Wages. — Having ihe wages per day, to find their amount per year — BuLE. — Take so many pounds, half pounds, and 5 pennies sterling, as there are pence per day. ExAMPLE.^-'What are the yearly Wages, at 5d* per day ? 1 10 5 -^-"^ ..:,^,.,,_,.„ .. 5 the number of pence per day. 7 12 1 the wages per year. One penny per day is equal to 865(f.s=240(f.+120(/.4-5(2.=3 iCl-|-10«.4'54. Therefore any number of pence per day, most be equal to £,1 10«. .6(2. multiplied by thac number What is the amount per year, at ' l * ' 119. 2d. per day.? Ans. £4 lis. Sd. ' ' 120. 7d. per day .? Arts. iBlO 12*. 1 Id. 121. 9d. per day? Ans. jS13 13i. 9d. 122. 14«E. per day? Ans. jS21 5«. lOd. 123. 21. 3:.*^:--J4. IW'^ lumber of .t7d eaoV^ ch. It--.' ,\-i At7<2. «ao1i, ji '. . .v)! £7 17*. 6/i iB6. £8 10s ly, to find ads, and 5 per day 1 iay. I20rf.-f-ftrf.=» ST day, mast J, (ii ■ .^,i';!j: PRACTICE. 229 Xi-m ... u ' BILLS OF PARCELS. Mr. John Day 2>u6/tn, 16tA ^prt?, 1844. [ Bought of Richard Joties. $. d. 15 yards of fine broadcloth, at 13 6 per yard 24 yards of superfine ditto, at 18 9 27 yards of yard wide ditto, at 8 4 16 yards of drugget, at . . 3 12 yards of ser^e, at . .13 32 yards of shsuloon, at . .18 n £f s. d. 10 2 6 22 10 11 5 5 4 i-± 2 13 "J «' Ans. £53 4 10 ■ai<>ii Mr. James Paul, Dublin, Qth May, 1844, j Bought of Thomas Norton. 9 pair of worsted stockings, at 4 6 per pair 6 pair of silk ditto, at . . 15 9 17 pair of thread ditto, at .54 23 pair of cotton ditto, at . 4 10 14 pair of yarn ditto, at .24 18 pair of women's silk gloves, at 4 2 19 yards of flannel, at . .:V'r dtl 4-* It Mr. James Gorman, 40 ells of dowlas, at 34 ells of diaper, at 31 ells of Holland, at 29 yf^rds of Irish cloth, at 174 yards of muslin, at 13| yards of cambric, at I 7| per yard Ans. £23 15 Dublin, nth May, 1844. Bought of John Walsh & Co 1 6 per ell 1 H „ 5 8 „ 2 4 per yard 7 2^ „ 10 6 ; * ^ ,A $4 yar4i of printed calico, at 1 2J An*. £54 5 10| l2 230 PRACTICE ^^^1 '■ »i 1 f 1 «.( ml •ill- Dublin, 20th May, 1844. Lady Denny, Bought of Richard Mercer 9^ yards of silk, at . .12 9 per yard 13 yards of flowered do., at 15 6 „ 1 If yards of lustring, at . 6 10 „ 11 3 „ • ,• ! I .-'F 10 8 „ 18 „ --4 ^-— —■ ' — -— — — o? — - 14 yards of brocade, at 12| yardfi of satin, at ll| yards of velvet, at ( t Mr. Jonas Darling, Ans. £44 15 10 Dublin, 2lst May, 1844. Bought of William Koper. s. d. 4 per lb 51 6 151 lb of currants, at 17| lb of Malaga raisins, at 19^ lb of raisins of the sun, at 17 lb of rice, at 8^ Yb of pepper, at . 3 loaveo of sugar, weight 32^ tt>, at 81 13 oz. of cloves, at . . .09 per oz. 31 1 6 » 7» ^''1 'i'>''ii.', Ans. £3 13 DuUin, 27th June, 1844. Mr. Thomas Wright, Bought of Stephen Brown k Co. 252 gallons of prime whiskey, at 6 4 per gallon 252 gallons of old malt, at .68 „ 252 gallons of old malt, at .80 ,| 0| Ans. £2C4 12 MISCELLANEOUS EXERCISES. What is the price of 1. 4715 yards of tape, at irf. per yard? Ans. £4 ISs. 2^d. 2. 354 ft), at l^d. per !b ? Aru. £1 16*. lO^d. 3. 4756 ft) of sugar, at I2\d. per ft) > Ans. iE242 15*. Id. ,. ; j 4. 425 pair of silk stockings, at 6*. per palt ? Ans. £\21 IQs. ,a PRACTICE. 231 ay, 1844. trd Mercer fay, 1844. liam Roper. ., - , 1 ^ T / £3 13 0| ttnc, 1844. )wxi & Co. £264 12 ,rd ? Ans Ans. £242 >air ? Ans. 6. 3754 pair of gloves, at 2s. Qd. ? Ans. £469 5« 6. 3520 pair of gloves, at Slf. 6d. ? Ans. £616. 7. 7341 cwt., at £2 6s. per cwt. ? Ans. £16884 6s, 8. 435 owt. at £2 75. per cwt. ? Ans. £1022 os. 9. 4514 cwt., at £2 17<. 7^d. per owt. ? Ans, £13005 19*. 3-H. QUESTIONS IN PRACTICE. ^.l '♦; 1. What is practice ? [11. ■'■' lad <. I -r'M ':i 2. Why is it so called ? [1]. ■* ''^* • -^- '^ *' i 'r.?v^ 3. What is the difference between aliquot, and aliquant parts? [2]. • 'V »rv.;; :; . ;:•. v, v < -i ij^/iv^ 4. How are the aliquot parts of abstract, and of applicate numbers found } [3] . 5. What ia the diflference between prime, and com- pound aliquot parts ? [3] . ^ 6. How is the price of any denomination found, that of another being given ? [6 and 8] . 7. How is ^e price of two pr more denominations found, that of one being given } [7 and 9]. ,'.iy-m 8. The price of one denomination being given, how do we find that of any number of another ? [10]. 232 PRACTICE M \ 1*" 9. When the price of any denomination is the aliquot part of a shilling, how is the price of any number of that denomination found ? [11]. '^ V-i h t/,' tr,*:: ? 10. When the price of any denomination is the aliquot part of a pound, how is the price of any num- ber of that denomination found ? [12]. 11. What is meant by the complement of the price r [13]. 12. When the complement of the price of any deno- mination is the aliquot part of a pound or shilling, but the price is not so, how is the price of any number of that denomination found ? [13]. VJ . 13. When neither the price of a given denomination, nor its complement, is the aliquot part of a pound or shilling, how do we find the price of any number of that denomination.^ [14, 15, 16,. and 17]. . 14. How do we find the price of any number of articles, when the price of each is an even or odd num- ber of shillings, and less than 20 ? [18 and 19] . ; 15. How IS the price of a quantity, represented by a mixed number, found ? [20] . 16. How do we find the price of cwt., qrs., and lb, when the price of 1 cwt. is given ? [21]. 17. How do we find the price of cwt., qrs., and lb, when the price of 1 lb is given } [22] . 18. How is the price of a ton found, when the price of 1 ft) is given ? [23] . 19. How do we find the price of oz., dwt., and grs., when the price of an ounce is given r [24] . 20. How do we find the price of yards, qrs., and nails, when the price of a yard is given } [25] . 21. How do we find the price of acres, roods, and perches? [26]. 22. How may the price of a hhd. or a tun* be found, when the price of a quart is given ? [27 and 28] . j 23. How may the price of any number of articles be found, the price of each in pence being given ? [29]. , 24. How are wages per year fo^ind, uose per day being given? [30] ^Qu t^. ■■\<'. A> - L-'- i '.•V TARl AMD TRET. 233 the aliquot ber of that on is the any num- the price r any deno- )r shilling, ,ny number • .:»■ • ' lomination, i pound or number of dumber of ' odd num- lented by a s., and lb, s., and fby I the price y and grs.. , aiad nails, roods, and r be found, IS]. articles be : day being ; '-vj *>»■' i , 'I If TARE AND TRET. ,. ' ,;. f .=i 3>. Tko fi^oss weight is the weight both of the^ goodb, and of the bag, Hic, in which they are. ..,,...., ,',.. u Tare is an allowance for the bog, &c., which contains the article. Sutile is the weight which remains, after deducting the tare. Tret is, usually, an allowance of 4 R> in every 104 ib, or ^ of the weight of goods liable to waste, after the tare has been deducted. Cloff" is an allowance of 2 ft) in every 3 cwt., after both tare and tret have been deducted. What remains after making all deductions is called the 9^, or neat weight. DiiFerent allowances are made in different places, and for different goods ; but the mode of proceeding is in all cases very simple, and may be understood from the following — Wl-"'. u'.i ^>i A.. EXERCISES. i 1. Bought 100 carcasses of beef at IBs. 6d. per cwt.; gross weisht 450 cwt., 2 qrs., 23 lb ; tret 8 ft> per car- cass. What is to be paid for them ? ' ; ' ; cwt. qrs. tt). ,,,. ;, 'j^ . 100 carcasses. ' Gross 450 2 23 ' ; 8 tt> per carcass Tret 7 16 "" cwt. qrs. Ib. Tret, on the entire, 800lbsB7 16 443 2 7 at 18s. 6(2. per cwt.»£410 55. lOJii. 2. What is the price of 400 raw hides, at 19«. lOd. per cwt. ; the gross weight being 306 cwt., 3 qrs., 15 lb ; and the tret 4 fib per hide ? Ans. iS290 3s. 2|i. 3. If 1 cwt. of butter cost £3, what will be the price of 250 firkins ; gross weight 127 cwt., 2 qrs., 21 & ; tare 11 lb per firkin.^ Ans. £309 Ss. O^d, ,- 4. What is the price of 8 cwt., 3 qrs., 11 ft>, at 15*. 6d. per cwt^, allowing the usual tret? Ans. £6 11#. \i^ ii .1 K'i H 734 TARE AND TRET. 5. What 18 the price of d cwt 21 lb, at 185. 4^^. per owt., allowing the usual tret ? Ans, £7 4«. S^d. 6. Bought 2 hhds. of tallow; No. 1 weighing 10 cwt., 1 qr., 1 1 ft), tare 3 qrs., 20 lb ; and No. 2, 1 1 cwt.. qr., 17 ft), tare 3 qrs., 14 ft); tret 1 lb per cwt. What do they come to, at 30«. per cwt. ? ^' '' ' *"^' '- owt qrs. ID. Gross weight of No. 1, 10 111 Gross weight of No. 2, 11 17 owt. qrs. lb. • V wt. qr TareO 3 TareO 3 20 14 21 1 2 3 6 , 19 2 22 19^ ft 13 6 •T-^f/. Gross weight, . Tare, Suttle, Tret 1 lb per cwt. Net weight, 19 2 241. The price, at 30«. pet cwt., is £29 5s. 7fl}d. It is evident chat the tret may be found by the following proportion — owt, owt. qrs. lb. lb. lb. • •') n .* .0 * 1 : 19 2 22 : : 1 : 19||. ' ^ >f^ k ^ 7. What is the price of 4 hhds. of copperas ; No. 1, weighing gross 10 cwt., 2 qrs., 4 lb, tare 3 qrs. 4 lb ; No. 2, 11 cwt., qr., 10 ft), tare 3 qrs. 10 ft) ; No. 3, 12 cwt., 1 qr., tare 3 qrs. 14 lb; No. 4, 11 owt., 2 qrs., 14 ft), tare 3 qrs. 18 ft) ; the tret being 1 lb per cwt. ; and the price IOj. per cwt. } Atu, jS20 175. 1«K 8. What will 2 bags of merchandise come to ; No. 1, weighing gross 2 cwt., 3 qrs., 10 ft); No. 2, 3 cwt., 3 qrs., 10 ft) ; tare, 16 ft) per bag ; tret 1 ft) per cwt. ; and at Is. Sd. per ft) } Am. £59 2s. 8{d. 9. A merchant has sold 3 bags of pepper; No. 1, weighing gross 3 cwt. 2 qrs. ; No. 2, 4 cwt., 1 qr., 7 ft) ; ^0. 3, 3 cwt., 3 qrs., 21 lb ; tare 40 fb per bag ; tret \ ft) per owt. ; and the price being 15d. per ft). What 4q they come to } Ans. £74 Is. 7|| t 18*. 4\d. 4s. S^d. ing lOowt., cwt.,Oqr., What do \ 8. lb. 20 • at 3Qf. pet be following as; No. 1, qrs. 4 lb ; lb ; No. 3, 11 owt., 2 ag 1 S> per £20 17*. to; No. 1, 2, 3 cwt., per cwt. ; )r; No. 1, qr., 7 ft); bag; tret fb. What 1,3 cwt., 3, 3 cwt., for every it do they TARK AND TRET. 236 No. 1, No. 2, No. 3, GroM, lure, owt. qrs. lb. 3 1 12 3 3 7 3 2 15 i n !• .'. i.i' 10 3 3 6 ii 1-t J ' ,M. lb. Tare 30 Tare 30 Tare 30 ,7 i I 90k>3 qrs. 6 lb. Sifttle, 10 St. St. 20 : 70 St. 70 Suttle, Tret, 0=s70 stones. lb. lb. :: 8 : 28 lb. fn>.- I Bt. 1 .rv/ lb. 12 \'i 1 12 IV»': r. Net weight, 68 4, at 10^. Qd. per stonestje35 16s. 7}({. ] 1 . Sold 4 packs of wool at 95. 9d. per stone ; weigh- ing, No. 1, 3 cwt., 3 qrs., 27 lb. ; No. 2, 3 cwt., 2 qrs., 16 tb. ; No. 3, 4 cwt., 1 qr., 10 lb. ; No. 4, 4 cwt., qr., 6 lb : tare 30 lb per pack, and tret 8 lb for every 20 stone. What is the price } Am. £49 Ids. 2^^jd. 12. Bought 5 packs of wool ; weighing. No. 1, 4 cwt., 2 qrs., 15 tt) ; No. 2, 4 cwt., 2 qrs. ; No. 3, 3 cwt., 3 qrs., 21 lb ; No. 4, 3 cwt., 3 qrs., 14 Bb ; No. 5, 4 cwt., qr., 14 lb : tare 28 tt» per pack ; tret 8 ft) for every 20 stone ; and at 11«. ^d, per stone. What is the price ? Ans, £11 \bs. 8|frf. 13. Sold 3 packs of wool ; weighing gross, No. 1, 3 cwt., 1 qr., 27 tb ; No. 2, 3 cwt., 2 qrs., 16 lb ; No. 3, 4 cwt., qr., 21 lb : tare 29 ft) per pack ; tret 8 ft> for every 20 stone ; and at \\s. Id. per stone. What is the price } Am. £4\ 13*. l%\\d. 14. Bought 50 casks of butter, weighing ffross, 202 cwt., 3 qrs., 14 lb ; tare 20 ft) per cwt. W. .at is the not weight ? cwt. qrs. ft). Gross weight, 202 3 14 Tare, . . 36 25i cwt. qrs. ft). 202 3 14 20 qrs. owt. f V 4040 lb. 10 1 14^ Net weight, 166 2 16^ 5 ss 1 of the last, > sthe tare on 3 qr. 14 ft). 2i==Jofthela8t,i ,, ,^, ,. ^^,,, Tare, 4067llb»36 cwt., Oqr., 25^Ib. i.-^A T l-i 236 TARE AND TRET. 15. The grosfl weight of ten hhds. pf tallow is 104 cwt., 2 qrs., 25 fib ; and the tare 14 ib per cwt. What is the net weight } Ans. 91 cwt., 2 qrs., 14} ib. 16. The gross weight of six butts of currants is 58 cwt., 1 qr., 18 lb ; and the tare 16 lb per owt. What is the net weight ? Ans. 50 cwt., qr., 7h} ft>. 17. What is the net weight of 39 cwt,, 3 qrs., 21 9b ; the tare being 18 lb per cwt. ; the tret 4 tt> for 104 ib ; and the clo£f 2. fb for every 3 cwt. ? 07 cwt. qrs. lb. 39 3 21 18: lb. cwt. 16=J 2«i-*-8 5 2 2 23 24 ir cwt. qrs. lb. Gross weight, 39 3 21 Tare, 6 1 13 Suttle, . Tret=^th, or ir Tare, 6 1 13 2 ft) in 3 cwt. is the ,-ivth part of 3 cwt. Hence the cloff of 32 cwt. 26 lb is its y^f th part, or \ 33 2 2 1 1 4 32 26 22 "A A 4 Net weight, 32 '18. What' is the net weight of 45 hhds. of tobacco ; weighing gross, 224 cwt., 3 qrs., 20 lb ; tare 25 cwt. 3 qrs. ; tret 4 ib per 104 ; cloflF2 flb for every 3 cwt. ? Ans. 190 cwt., 1 qr., 14/y lb. 19. What is the net weight of 7 hhds. of sugar, weighing gross, 47 cwt., 2 qrs., 4 ib ; tare in the whole, 10 cwt., 2 qrs., 14 ib ; and tret 4 tt» per 104 & ? Ans, 35 cwt., 1 qr.. 27 ib. 20. in 17 cwt., qr.. 17 lb, gross weight of galls, how much net ; allowing 18 lb per cwt. tare ; 4 lb per 104 ib tret ; and 2 ib per 3 cwt. cloff? Aw. 13 owt., 3 qrs., 1 ib nearly. t^^ ^'■■^./rw'i ; I QUESTIONS. il fe^V.. s.. 1. What is the gross weight ? [31]. ^^^ : 2. What is tare ? [31]. ^.^,_.J'l 3. What is suttle } [31]. M -- > "4. What is tret.? [31]. '' 5. What is cloff? [31]. '' ^ ^ 6. What is the net weight ? [31]. 7. Are the allowanoes made, always ilie lame? [31]. ix-^ How is 104 ^t. What rib. 'ants is 58 L What is rs., 21 fi> ; or 104 ft) : wt. qrs. R). 39 3 21 6 1 13 33 2 1 1 2 4 32 26 22 32 4 tobacco ; p 25 cwt. 3 cwt.? of sugar, le whole, ► ? Ans, of galls, 4 lb per 13 cwt., • ''^''i' [31]. 237 'i m^ A^U-.'^AfiftL 'i-U liifi Ui4'. I'j o■-^ ir)~~.rji)ui z.^ j4-» i* i.' fc' 3i«.4 ; *-n^ i 'j ;-hi^f.( ivji'>in' j'-^f-lj' 'I'l iff'. SECTION VII. i!'?'!!'/ INTEREST, &c. .i'ff ii:-: /*- ~.j.*ii^j 1. Interest is the price which is allowed for. the use of money ; itxl^^pends on the plenty or scarcity of the latter, and the risk vrhich is run in lending it. Interest is either si'aple or compound. It is simple when the interest due is not added to the sum lent, so as to bear interest. ■-'■■'■*' '^\< ' ^^ - It is compound when, after certain periods, it is made to bear interest — ^being added to the sum, and considered as a part of it. The money lent is called the principal. The sum allowed for each hundred pounds " per annum" (for a year) is called the " rate per cent." — (per iSlOO.) The amount is the sum of the principal and the interest due. SIMPLE INTEREST. • -->.■ 1 C\ r^ f» ••.#'.' '. .m 2. To find the interest, at any rate per cient., on any sum, for one year — ^^ t,, - -s-?^*,: • ;; » ^ >? Rule I. — Multiply the sum by the rate per cent., and divide the product by 100. ^ -, ,. Example.— What is the interest of £672 14«. Zd. for one year, at 6 per cent. (£6 for every £100.) , , . . ^ £ s. d. /,^. .-. t •• ,■_■■- - ■'>>■■• h-i^ , crii 6 ■'} .■■■'fcX 40-36 5 6 20 '!> . J-/»M 7*25 The quotient, £40 7s. 3ii; ;•"■•■ ', ent^'K': • '; '. • ent. ^ • I) per cent. r one year, ir, at 5 per ae year, at le year, at cent, con- denomina- are lower. the given id., tv one oent.^'''' INTEREST. ^ 239 At 5 per cent, the interest is the ^V ^ ^he principal ; at 10s. per cent, it is the i^ of what it is at 5 per cant. There- fore, at £5 10s. per cent., it is the sum of both. ri> j T-M'! '■ "S 1: EXERCISE3. 5. What is the interest of ^2371 195. 7^d. for one year, at £3 15*. per cent. > Am. ^£13 18s. U^d. 6. What is the interest of £84 11«. 10^. -for one year, at £4 5». per cent. ? Am. £3 1 1*. lO^d. 7. What is the interest of £91 Os. S^d. for one year. At £6 I2s. 9d. per cent. .? Am. £6 Os. I0\d. 8. What is the interest of £968 5*. for one year, at £5 14s. 6d. per cent. ? Am. £55 8*. 8d. 5. To find the interest of any sum, for several years — Rule. — ^Multiply the interest of one year by the num- ber of years. :•' '*/' ' '-- -^"'i' Example.— What is the interest of £32 I4s. 2d. for 7 years, at 5 per cent. ? !• ,' ' v) 20 )32 14 2 ' ^ .:: \l'\[ ISjJ '."* 1 12 8^ is the interest for one year, at 5 per cent. 7 And 11 8 11^ is the interest for 7 years, at 5 per cent. This role requires no explanation. ... , ^ ,; '. ' exercises. ' '• '-'■ 9. What is the interest of £14 25. for 3 years, at 6 per cent. ? Am. £2 105. 9d. 10. What is the interest of £72 for 13 years, at £6 105. per cent. ? Am. £60 165. O^d. * 11. What is the interest of £853 05. e^d. for 11 years, at £4 125. per cent. ? Am. £431 125. 7f rf. 6. To find the interest of a given sum for years, months, &c. — •<• • ' " Rule. — Having found the interest for the years, as already directed f2, &c.], take parts of the interest of one year, for that of the months, &c. ; and then add the results. -^ -♦ r .,, . .. ! ! :■ ■ f w If \i ■1 ■ If f •1 1 1 ^' li ' . 4 240 INTEREST. Example.— What is the interest of £86 8^. 4d. for 7 yean and 5 months, at 5 per cent. I «' y, o<1t^ ax .?ii.i>t; -rhf >i't 20 )86 8 4 4 6 5 is the interest for 1 year, at 5 per cent. £ s. d. 30 4 11 is the interest for 7 years. 4 6 5 -^3=: 1 8 9f is the interest for 4 months. ■ 1 8 9? ^4= 7 21 is the interest for 1 month. ^ And 32 11^ is the required interest. : i' » \'Ji F.XERCISES. ■ lings, and pence are of a pound, the interest of any other sum« for the same time anJ rate, mast be the same part or parts of that other sum — since the interest of any sum is proportional to tiie intereS'^ ui £1. Example 2.— What is the interest of £14 2s. 2il. for 6 years and 8 months, at 6 per cent. 1 6s. Sd. is the J ofu pound. ;••• jjt -! [ i £> s. d. 3)14 2 2 i(j r 5)4 14 0} is the interest, at 5 per cent. . 18 9'f is the interest, at 1 per cent. 5 12 101 is the interest, at 6 (5+1) per cent. ;k EXERCISES. ! I 20. Find the interest of iBlOQO 17*. 6d. for 1 year and 8 months, at 5 per cent. } Ans. i290 18s. l^d. 21. Find the interest of £976 14s. 7d. for 2 years and 6 months, at 5 per cent. ? Atu. iS122 Is. 9}d. 22. Find the interest of £780 17s. 6rf. for 3 years and 4 months, at 6 per cent. .? Ans. £156 3s. 6^;?. 23. What is the interest of £197 lis. for 2 years and 6 months, at 5 per cent. } Am. £24 13s. lO^d. 24. What is the interest of £279 Us. for 7^ months, at 4 per cent, r Ans. £6 19s. 9^d. 25. What is the interest of £790 16s. for 6 year and 8 months, at 5 per cent. ? Ans. £263 12f . 242 INTEREST. ii 1 26. What is the interest of iBia4 2s. 9d. for 3 years and 3 months, at 5 per cent. ? Ans. ^20 3*. 5^4. • *>, ^• : 27. What is the interest of iei837 4^. 2d. for 3 years and 10 months, at 8 per cent. .? Am. iB563 8*. Sd. 9. When the rate, or. number of years, or both of them, are expressed by a mixed number — Rule. — Find the interest for 1 year,- %t 1 per cent., and multiply this by the number of pounds and the frac- tion of a pound (if there is one) per cent. ; the sum of these products, or one of them, if there is but one, will give the interest for one year. Multiply this by the number of years, and by the fraction of a year (if ther& is one) ; and the sum of these products, or one of them, if there is but one, will be the required interest. Example 1. — ^Find the interest of £21 2s. 6d, for 3| years at 5 per cent./? _ • u. iC21 2s. 6d.^l00=r4». 2JJ. Therefore >^-'*'^ '^5^'* "^^ / 4 2^ is the interest for 1 year, at 1 per cent. . /'! 1 1 1| is the interest for 1 year, at 5 per cent. M 3 3 5|^istheinterestfor3 years, atdo. 15 10| is the interest for | of a year (£1 Is. l|rf. x|), at do. 3 19 3^ is the interest for 3| years, at do. Example 2. — ^What is the interest of £300 for 5| years, at 3| per cent. ? j^^OO ^1()0=b3 is the interest for 1 year, at 1 per cent ':vJj>; f:»rit imrl Aii; .|i|^- 9- is the interest for 1 year, at 3 per cent. a y^': .; r. 2 6 is th« interest for 1 year, at £ J ( £3 X J) '["; 11 , 5 is the interest for 1 year, at 3| per cent. ^^m r 3| years f^ •rnl'ii'i Hid., t. t. Xf),atdo. 5^ years, per cent, per cent. i (i^sxf ) 2 per cent. ^} per cent 11 5s.-^2) 1. 6Jd.-i-2) at 3} do. ^ EXERCISES. 28. What is the interest of ^£379 2s. 6d. for 4 J years, at 5f per cent. .? Am. £91 5s. 5d. 29. What is the interest of iE640 105. 6d. for 2^ years, at 4^ per cent. ? Ans. £72 \s. 2j\d. 30. What is the interest of £600 10*. 6d, for 3^ years, at 5f per cent. } Ans. £115 2s. O^^d. 31. What is the interest of £212 8a\ \^d. for 6| years, at 5 J per cent. ? Ans. £81 8*. 6^d. 10. To find the interest for days, at 5 per cent. — Rule. — Multiply the principal by the number of days, and divide the product by 7300. Example. — What is the interst of £26 4s. 2d. for 8 days ? £ s. d. 26- 4 2 8 .. 209 13 4 20 4193 . 12^ 7300)5032a?G^-fM. 43800 -since the remainder 6520 The required interest is 6||f, or Id. is greater than half the divisor. The interest of £1 for 1 year is JC^V' ^^^ ^^^ ^ ^^^ 3^-^365= SQvTgg^ =7300 ; that is, the 7300th part of the principal. Therefore the interest of any other sum for one day, is the 7300th part of that sum ; and for any number of days, it is that number, multiplied by the 7300th part of the principal — or, which is the same thing, the principal multiplied by the number of days, and divided by 7300. exercises. 32. Find the interest of £140 10^. for 76 days, at 5 per cent. Ans. £1 9^. ^f^-^d. 33. Find the interest of £300 for 91 days, at 5 per cent. Anr,. £3 14.?. 9^d. 34. What is the interest of £800 for 61 days, at 5 '• ! ■ I i per cent. } Ans. £6 13^. ^d. -\,-X!\ 244 INTEREST. ' 4 I ! 1 1 ^ 1 ,1,. , 11. To find the interest for days, at any other rate — KuLE. — Find the interest at 5 per cent., and take parts of this for the remainder. Example. — What is the interest of JC3324 6s. 2d. for 11 iays, at £Q 10s. per cent. I X3324 6s. 2d. x ll-4-7300=£5 Os. 2\d. Therefore £ ,s. d. 5)5 2| is the interest for 11 days, at 5 per cent. 2)1 Oi is the interest for 11 days, at 1 per cent. 10 ia the interest for 11 days, at 10s. per cent. And 6 10 21 is the interest for 11 days, at £6 10s. (£5-|- Xl+lOs.) This rule requires no explanation. EXERCISES. 35. What is the interest of ^£200 from ine 7th May to the 26th September, at 8 per cent. } Ans. £,6 4s. b^d. 36. What is the interest of £150 15j. 6d. for 53 days, at 7 per cent. .? Ans. £1 lOs. l^d. 37. What is the interest of £371 for 1 year and 213 days, at 6 per cent. .'' Ans. £35 5*. Od. 38. What is the interest of £240 for 1 year and 135 days, at 7 per cent. .? Ans. £23 Os. ^\d. Sometimes the number of days is the aliquot part of a year ; in which case the process is rendered more easy. Example. — What is the interest of £175 for 1 year and 73 days, at 8 per cent. 1 1 year and 73 day8=l^ year. Hence the required interest ia the interest for 1 year-f-its fifth part. But the interest of £175 for 1 year, at the given rate is £14. Therefore its interest for the given time is £14-f-£^5=£14+£2 16s.= £16 16s. .. .. ;• 12. To find the interest for months .^ at 6 per cent — Rule. — If the number expressing the months is even^ multiply the principal by half the number of months and divide by 100. But if it is odd^ multiply by the half of one less than the number of months ; divide the result by 100 ; and add to the quotient v^hat will be obtained if we divide it by one less than the number of months. k.-.-^.- ..:-\i -vj. --va ■ -Ji.!/;- INTEREST 245 her rate — , and take I. 2d, for 11 jfore per cent, per cent. )s. per cent. 3 10s. (£5+ ne 7tli May ins. £6 4s. 6d. for 53 ar and 213 2ar and 135 uot part of I more easy, r 1 year and ired interest le interest of Therefore its [+£2 16s.=: )r cent — nths is everij of months tiply by the ; divide the rhat will be i number of Example 1.— -What is the interest of £72 6s. 4c?. for 8 months, at G per cent ? £> s. d. [ 72 6 4 ! £2-89 20 5 4 17-85s. Tho required interest is £2 175. lOld, 12 l0-2id. 4 Q'9Q=:\d. nearly. Solving the question by the rule of three, we shall have— £100 : £72 6s. M. : : £6 : £72 6s. 4r/.x8x6 12 : 8 100xT2 == (divid- ing both numerator and denominator by 6 [Sec. IV. 41). £72 6s. 4d .x 8x6-h6 £72 6s. 4f/.x8 ,^. .^. ^ ^ 100x124^^^ .^ 100x2 = (dividmg both A A '\ r. ox ^72 6s. 4d.xS^2^ numerator ana denommator by 2) — I00v2^» 2 — £72 6s. 4(/.x4 "^ loo >t — that is, the required interest is equal to the given sum, multiplied by half the number which expresses the months, and divided by 100. Example 2. — What is the interest of £84 6s. 2d. for 11 months, at 6 per cent. ? 11=10+1 10-i-2=:5. £ s. d. 84 6 2 5 One less than the given number of months=10. £4-21 10 10 20 £ s. d. 4- 30s. 10)4 4 3| is the interest for 10 months, at 6 per cent. 12 8 5^ is the interest for 1 month, at same rate. 3-70d. And 4 12 9 is the interest for 11 (10-f-l) months, at 6 «* 4 2'80f =?d. nearly. The interest for 11 months is eviden tly the interest tvT 11 — 1 month, plus the interest of 11 — 1 month -hll — 1 . M ) I :< 246 INTEREST. . i . '■ EXERCISES. 39. What is the interest of ^2250 17*. ed. for 8 months, at 6 per cent. ? Ans. £\0 Os. 8|rf. 40. What is the interest of £d7\ 15*. for 8 months, at 6 per cent. ? Ans. £>22 \ls. 4^d. 41. What is the interest of je840 for 6 months, at 6 per cent. ? Ans. ,£25 4*. 42. What is the interest of £3790 for 4 months, at 6 per cent. ? Ans, £75 16*. 43. What is the interest of £900 for 10 months, at 6 per cent. .? Ans. £45. 44. What is the interest of £43 2*. 2d. for 9 months, at 6 per cent. ? Ans. £1 18*. 9^d. 13. To find the interest of money, left after one or more payments — Rule. — If tlio interest is paid by daySy multiply the sura by the number of days which have elapsed before any payment was made. Subtract the first payment, and multiply the remainder by the number of djiys which passed between the first and second payments. Subtract the second payment, and multiply this remain- der by the number of days which passed between the second aad third payments. Subtract the third pay- ment, &c. Add all the products together, and find the interest of their sum, for 1 day. If the interest is to be paid by the w^ or month^ substitute weeks or months for days^ in the above rule. Example. — A person borrows £117 for 94 days, at 8 per cent., promising the principal in parts at his convenience, and interest corresponding to the money left unpaid, up to the diflferent periods. In 6 days he pays £17 ; in 7 days more £20 ; in 15 more £32 ; and at the end of the 04 days, all the money then due. What does the interest come to "? £ days. £ day. 117X 6== 702x11 lOOx 7= 700x1 If r77rt 80x15=1200x1 f— *'^^'"- 48x66=3168x1. The interest on 5770 for 1 day, at 5 per cent., is 15*. ^Id. Therefore 6d. for 8 8 months, lonths, at 6 months, at months, at r 9 months, ifter one or nultiply the psed before it payment, )er of days payments, his remain- etween the third pay- md find the or month, ove rule. lys, at 8 per onvenience, unpaid, up £17 ; in 7 lid of the 91 he interest INTEREST. 247 is 15s. 9^^. £ a. (I. ■ <■ I " 6)0 15 9} is the interest, at 5 pei cent. ^^ ^ ^ ** ^he interest, at 1 per cent. 3)0 18 11^ is the interest, at 6 per cent. 6 4 is the interest, at 2 per cent. And 1 5 3^ is the interest, at 8 per cent., for th^ given sums and times. If the entire sum were 6 days unpaid, the interest would be the same as that of 6 times as much, for 1 day. Next, £100 due for 7 days, should produce as much as .£700, for 1 day, &c. And all the sums due for the different periods should produce as much as the sum of their equivalents) in I day. . EXERCISES. 45. A merchant borrows iS250 at 8 per cent, for 2 years, with condition to pay before that time as much of the principal as he pleases. At the expiration of 9 months he pays i280, and 6 months after iK70 — leaving the remainder for the entire term of 2 years. How much interest and principal has he to pay, at the end of that time ? Ans. £\27 16s. 46. I borrow i£300 at 6 per cent, for 18 months, with condition to pay as much of the principal before thf time as I please. In 3 mouths I pay £.QQ ; 4 months after ,£100 ; and 5 months after that Mlb. How much principal and interest am I to pay, at the end of 18 months ? Ans, £79 15s. 47. A gives to B at interest on the 1st November, 1804, JB6000, at 4^ per cent. B is to repay him with interest, at the expiration of 2 years — having liberty to pay before that time as much of the principal as he pleases. Now B pays . . . , , '• . , £ The 16th December, 1804, . . 900 The nth March, 1805, . . 1260 The 30th March, ... 600 The 17th August, . . . 800 The 12th February, 1800, . . 10^ , How much principal and interest is he to pay on th# l8t November, 1806 .? Ans. £1642 9s. 2|ff |rf. 48. Lent at interest £600 the 13th May, 1833, for A * • 24ft INTEREST. i 1 year, at 5 per cent. — with condition that the may discharge as much of the principal hcfore as he pleases. Now he pays the 9th July £2i)^ anu the 17th September jC150. How much principal and interest is he to pay at the expiration of the year? Ans. iE266 13*. 6,,V- 14. It is hoped that the pnpil, from what he has learned of the properties of proportion, will easily un- derstand the modes in which the following rules are proved to be correct. Of the principal, amount, time, and rate — given any three, to find the fourth. Given the amount, rate of interest, and time ; to find the principal — RuLE.—Say as jSlOO, plus the interest of it, for the given time, and at the given rate, is to iSlOO ; so is the given amount to the principal sought. Example. — What will produce £802 in 8 years, at 5 per cent. *? ,, . •.,.1,' .• . , . , f '•!,' ■ MO (=:£5x8) 18 the interest for £100 in 8 years at the given rate. Therefore £140 : £100 : : £862 :^^^^^191==£Q151is.^^. 140 When the time and rate are given — • £100 : any other sum : : interest of £100 : interest of that other sum. . By alteration [Sec. V. 29], this becomes— £100 : interest of £100 : : any other sum : interest of that sum. And, s^ing " the first + the second : the second," &c. [Sec. V. 2y] we have — £100 -f- its interest : £100 : : any other sum + its in- terest : that sum — which is exactly the rule. EXERCISES. ,'•!! 49. What principal put to interest for* 5 years will amount to £402 10^., at 3 per cent, per annum ? Ans. £350. 50. What principal put to interest for 9 years, at 4 per cent., will amount to £734 Ss. ? Aiis. £540. I INTEREST. 249 )re ii.. limn JE20L auu 'incipal and ■ the year? rhat he has II easily uu- g rules are —given any lue ; to find ' it, for the } ; so is the ars, at 5 per years at the : intercfit of ,1 interest of jecond," &c. im -)- its in- years will im ? Ans. i i-1 7/ it years, at 4 51. The amount of a certain principal, bearing inter- est for 7 years, at 5 per cent., is i^334 16*. What is the principal ? Ans. i^248. l"). Given the time, rate of interest, and principal — to liiid the amount — UuLK. — Say, as iillOO is to iBlOO plus its interest for the i^iven time, and at the given rate, so is the given sum to the amount required. Example. — What will X272 come to, in 5 years, at 5 per cent. ? £125 (=£100-f-£5x5) is the principal and interest of JCIOO for 5 years J then — 272x12') cClOO : £125 : : £272 : —^^^ =£340, the required amount. We found by the last rule that £100-|-its interest : £100 : : any other sum-f-its interest : that sum. Inversicm [Sec. V. 29] changes this into, £100 : £100-f-its interest : : any other sum : that other sum-f-its interest — which is the piwsent rule. EXERCISES. 52. What will i£350 amount to, in 5 years, at 3 per cent, per annum .'* A7is. £402 10*. 53. What will J£540 amount to, in 9 years, at 4 per cent, per annum ? Ans. iB734 8s. 54. What will .£248 amount to, in 7 years, at 5 per cent, per annum ? Ans. £334 \6s. 55. What will £973 4.?. 2d. amount to, in 4 years and 8 months, at 6 per cent. .? Ans. £1245 I4s. l^d. 56. What will £42 3.?. O^d. amount to, in 5 years and 3 months, at 7 per cent. ? A'tis. £57 13*. lO^d. 16; Given the amount, principal, and rate — to find the time — Rule. — Say, as the interest of the given sum for 1 year is to the given interest, so is 1 year to the re- ][uired time. 250 INTEREST. '■t Example. — Wheu would ii281 13s. 4J. become £338, ak i) per cent. ? £14 Is. 8d. (the interest of £281 13s. 4d. for 1 year [2]) : £56 6s. Sd. ^ , ^56 6s. m. (the given interest) : : 1 : £T4"T7~87 ='*> ^^^ lequlred number of years. 17. Ileiice briefly, to find the time — ^Divide the interest of the given principal for 1 year, into the entire interest, aiid the quotient will be the time. It is evident; the principal, and rate being given, the interest is prtpon'ionai to the time } the longer the time, the more the interest, vvnd the reverse. That is — The interest for one time : the interest for another : : the former time : the latter. Hence, the interc^fc of the giver sum for one year (the interest for one time) : the given interest (the interest of the same sum for apMher time) : : 1 year (the time which produced the former) : the time sought (that which pro- duced the latter) — which is the rule. EXERCISES. 57. In what time would ^£300 amount to £372, at 6 per cent. } Ans. 4 years. 58. In what time would £211 5s. amount to £230 5s. 3«?., at 6 per cent. ? Ans. In 1 year and 6 months. 59. When would £561 15s. become £719 Of. 93^., at 6 per cent. } Ans. In 4 years and 8 months. 60. When would £500 become £599 3s. 4^., at 7 per cent. } Ans. In 2 years and 10 months. 61. When will £4.'-i6 9s. Ad. become £571 8s. \\d.y at 7 per cent. } Aas. In 4 years and 5 months. 18. Given the amount, principal, and time — to find the rate — Rule. — Say, as the principal is to £100, so is the given interest, to the interest of £100- -which will give the interest of £100, at the same rate, and for the same time. Divide this by the time, and the quotient will be the rate. INTEREST. 251 >me Je338, at 1 year [2]) : "8^=4, the ■Divide the the entire ; given, the he time, the another ; : e year (the interest of time which which pro- £372, at 6 t toie230 (ar and 6 ^ Os. 93^., bhs. ?.,at 7 per hs. e — to find so is the will give the same int will be Example. — At what rate will £350 amount to £402 10s in 5 years ? £350 : £100 : : £52 10s. : ^^^^*><12^=£15, thein 350 terest of £100 for the same time, and at the same rate Then ^=3, is the required number of years. We have seen [14] that the time and rate being the same, £100 : any other sum : : the interest of £100 : interest of the other sum. This becomes, by inversion [Sec. V. 29] — Any sum : £100 : : interest of the former : interest of 100 (for same number of years). But the interest of £100 divided by the number of years which produced it, gives the interest of £100 for 1 year — or, in other words, the rate. EXERCISES. 62. At what rate will i£300 amount in 4 years to i£372 ? Ans. 6 per cent. 63. At what rate will i£248 amount in 7 years to £334 16s. ? Ans. 5 per cent. 64. At what rate will i6976 14*. 7d. amount in 2 years and 6 months to £1098 16s. 4^d. ? Ans. 5 per cent. Deducting the 5th part of the interest, will give the in- terest of £ KuLE I. — Find the interest due at the first time of payment, and add it to the principal. Find the interest !?i 252 INTEREST. ir 4 kSK^-t, of that sum, considered as a new principal, and add it to wliat it would produce at the next payment. Con- sider that new sum as a principal, and proceed as before. Continue this process through all the times of payruent. Example. — What is the compound interest of JC97, for 4 years, at 4 per cent, half-yearly ? £ s. d. 97 3 17 7| is the interest, at the end of Ist half-year. 100 17 7 J is the amount, at end of Ist half-year. 4 8^ is the interest, at the end of 1st year. 104 18 3J is the amount, at the end of 1st year. 4 3 11| is the interest, at the end of 3rd half-year. 109 2 3 is the amount, at the end of 3rd half-year. 4 7 3i is the interest, at the end of 2nd year. 113 9 6^^ is the amount, at the end of 2nd year. 4 10 9 1 is the interest, at the end of 5 th half-year. 118 4 is the amount, at the end of 5th half-year. 4 14 5 is the interest, at the end of 3rd year. 122 14 9 is the amount, at the end of 3rd year. 4 18 2 j is the interest, at the end of 7th half-year. 127 12 11 [ is the amount, at the end of 7th half-year. 5 2 li is the interest, at the end of 4th year. 132 15 OJ is the amount, at the end of 4th year. 97 is the principal. ,: And 35 15 OJ is the compound interest of £97, in 4 years. 20. This is a tedious mode of proceeding, particularly when the times of payment are numerous ; it is, there- fore, better to use the following rules, which will be found to produce the same result — lluLE II. — Find the interest of £1 for one of the payments at the given rate. Find the product of so many factors (each of them iSl-fits interest for one payment) as there are times of payment ; multiply this product by the given principal ; and the result will be tb^-s principal, plus its compound interest for the given we INTEREST. 253 ad add it at. Con- roceed as J times of £97, for 4 If-year. ar. ir. ir. alf-year. If-year. par. ar. If-year. If-year. ar. ar. If-year. If-year. ar. I 4 years. rticiilarly is, tliere- \i will be 3 of the net of so ; for one tiply this t will be the given time. From this subtract the principal, and the remain* der will be its compound interest. Example 1. — What is the compound interest of £237 for 3 years, at 6 per cent. ? £0G is the interest of £1 for 1 year, at the given rate ; and there are 3 payments. Therefore £1-00 (£l-|-£0-6) is to be taken 3 times to form a product. Hence lOGxlOOx 106x£237 is the amount at the end of three years; and lOOxl 0Gxl06x£237— £237 is the compound interest. The following is the process in full — 1-06 the amount of £1, in one year. 106 the multiplier. 11236 the amount of £1, in two years 1-06 the multiplier. 1-191016 the amount of £1, in three years. Multiplying by 237, the principal, £ s. d. we find that 282- 270792=282 5 5 is the amount • and subtracting 237 0, the principal, I we obtain 45 5 5 as the compound interest. Example 2. — What are the amount and compound inte- rest of £79 for 6 years, at 5 per cent. 1 The amount of £1 for 1 year, at this rate would be £105. Therefore £105xl05xl05xl05xl 05x105x79 is the amount. &c. And the process in full will be — £ 105 . : 1-05 :.r • ' : - 11025 the amount of £1, in two yeacs. 11025 . i ■ 1 21551 the amount of £1, in four years. 11025 ..• ^ 1-34010 the amount of £1, in six years. 79 ^ ^ .. £ .s. d. £105 86790=105 17 4i is the required amount. 79 And 26 17 4^ is the required interest M 2 254 INTEREST t Example 3. — What are the amount, and compound interest of £27, for 4 years, at £2 10s. per cent, half-yearly. The amount of £1 for one payment is £1025. Therefore £1 025 X 1 025 X 1025 X 1025 X 1025 X 1025 X 1025 x 1025 x27 is the amount, &c. And the process in full will be £ -' ■ ' 1-025 ' •• ' 1025 105063 the amount of £1, in one year. 105063 1- 10382 the amount of £1, in two years. 1- 10382 1-21842 the amount of £1, in four years. 27 £ s. d. £32-89734=32 17 lU is the required amount. 27 And 5 17 llj is the required interest. 21. EuLE III. — Find by the interest table (at the end of the treatise) the amount of £l at the given rate, and for the given number of payments ; multiply this by the given principal, and the product will be the required amount. From this product subtract the principal, and the remainder will be the required compound interest. Example. — What is the amount and compound interest of £47 lOi'. for years, at 3 per cent., half-yearly % £47 105.=£47-5. We find by the table that £1-42576 is the amount of £1, for the given time and rate. 47- 5 is the multiplier. £ 8. d. 67-7236=67 14 5? is the required amount. 47 10 And 20 4 5 J is the required interest. 22. Rule i requires no explanation. Reason or Rule II. — VVhen the time and rate are the same, two principals are proportional to their co«*re8ponding amounts. Therefore £\ (one principal) : £106 (its corresponding amount) :: £106 (another principal) : £106 X 106 (its corresponding amount). 'I o fo £ m md interest ply. Therefore i X 1025 X I full will be ir. ars. iars. imount. interest. (at the end ta rate, and this by the required ncipal, and interest. nd interest le and rate. te are the responding [imount) : : rebpoudiug \ I INTEREST. 255 Hence the amount of £1 for two years, is £1 OCX 1*06— or the product of two factors, each of them the amount of £1 for one year. Again, for similar reasons, £1 : £1-06 :: £1-06x1 '06 : £1 '06x1 '06x1 '06. Hence the amount of £1 for three years, is £1-06x1 06x1 '06— or the product of three factors, each of them the amount of £1 for one year. The same reasoning would answer for any number of pay- ments. The amount of any principal will be as much greater than the amount of £1, at the same rate, and for the same time, as the principal itself is greater than £1. Hence we multiply the amount of £1, by the given principal. Rule III. requires no explanation. 23. When the decimals become numerous, we may proceed as already directed [Sec. II. 68] . We may also shorten the process, in many cases, if we remember that the product of two of the factors multiplied by itself, is equal to the product of four of them ; that the product of four multiplied by the pro- duct of two is equal to the product of six ; and that the product of four multiplied by the product of four, is equal to the product of eight, &c. Thus, in example 2, 11025 (=105x105) Xl-1025=105xl05xl05xl05. EXERCISES. 1. What are the amount and compound interest of £91 for 7 years, at 5 per cent, per annum ? Ans. £128 Os. lid. is the amount; and £37 0^. lld.j the com- pound interest. 2. What are the amount and compound interest of £142 for 8 years, at 3 per cent, half-yearly? Ans. £227 I7s. A\d. is the amount ; and £85 17^. 4i<^., the compound interest. 3. What are the amount and compound interest of £63 bs for 9 years, at 4 per cent, per annum ? Ans, £90 05 compound interest 4. What are the amount and compound interest of £44 55. Qd. for 1 1 years, at 6 per cent, per annum } 5^d. is the amount ; and £26 Ids. 5frf., tnti r- 266 INTEREST. m Ans. i^$4 l5. 5d. is the amount; and jS39 155. Bd.y the compound interest. 5. What are the amount and compound interest of je32 4*. ^d. for 3 years, at £2 10*. per cent, half- yearly ? A71S. iE37 7*. 8^6?. is the amount ; and £b 2s. lOJid.y the compound interest. 6. What are the amount and compound interest of iB971 05. 2\d. for 13 years, at 4 per cent, per annum } Ans. iE1616 155. U^d. is the amount; and ^£645 155 9^d.y the compound interest. 24. Given the amount, time, and rate — to find the principal ; that is, to find the present worth of any sum to be due hereafter — a certain rate of intcict being allowed for the money now paid. Hole. — Find the product of as many factors as there f»,re times of payment — each of the factors being the amount of £1 for a single payment ; and divide this pioduct into the given amount. Example. — ^What sum would produce £834 in 5 years, a^: .*> per cent, compound interest ? The amount of £1 for 1 year at the given rate is £105 ; and the product of this taken 6 times as a factor 105 x 105 X 105 X 105 X 105. which (according to the table) is 1-27628. Then je834-M-27G28=£653 ds. 2^d., the required principal. 25. Reason of the Rui.e. — We have seen [21] that the amount of any sum is equal to tlie amount of <£1 (for the same time, and at the same rate) multiplied by the principal ; that is, The amount of the given principal=the given principal X the amount of £1. If we divide each of these equal quantities by the same number [Sec. V. 6], the quotients will bo equal. Therefore — The amount of the given principal -f- the amount of £l=the given principal X the amount o*' £l-T-the amount of £1. That is, the amount of the given principal (the given amount) divided by the amoun- of ci1, is equal to the principal, or quantity required — wlucii ia the rule. EXERCISES. 7. What ready money ought to be paid for a debt of j£629 175. Hj^., to be due 3 years hence, allowing 8 per cent, compound interest } Ans. i£500. m •w INTEREST. 257 155. 8d., aterest of ent. half- ; and £d nterest of • annum } £645 155 ) find the F any sum ic«t being PS as there being the iivide this in 5 years, 8 is £105 : Rtor 105 X e table) ia rincipal. 1] that the br the same pal ; that is, principal X )y the same Therefore — of £l=the f £1. That en amount) )rincipal, or a debt of Qy allowing 8. What principal, put to interest for 6 years, would amount to £2QS Os. A\d.y at 5 per cent, per annum } Am. JS200. 9. What sum would produce £1^2 195. \\\d.'m 14 years, at 6 per cent, per annum } Ans. iE328 125. Id. 10. What is ^£495 195. ll^^^., to be due in 18 years, at 3 per cent, half-yearly, worth at present. Ans. £>\n 2s. S^d. 26. Given the principal, rate, and amount — to find the time— Rule I. — Divide the amount by the principal ; and into the quotient divide the amount of iBl for one pay- ment (at the given rate) as often as possible — the number of times the amount of i£l has been used as a divisor, will be the requued number of payments. Example. — In what time will £92 amount to £106 135. 0|rf., at 3 per cent, half-yearly 1 £106 IZs. 0^(^-^£92=^•15927. The amount of £1 for one payment is £103. But 115927 -r- 103 = 11255; 11255 -f- 1-03 = 109272 ; 109272 -^ 103 = 10609 ; and 10C09-j-103-^103 ; 103-^103=1. We have used 103 as a divisor 5 times ; therefore the time is 5 payment§^ or 2^ years. Sometimes there will be a remainder after divid- ing by 103, &c., as often as possible. In explaining the method of finding the powers and roots of a given q;iiintity, we shall, hereaftfa, notice a shorter method of ascertaining how often the amount of one pound can be used as a divisor. 27. Rule II. — Divide the given principal by the given amount, and ascertain by the interest table in how many payments £1 would be equal to a quantity nearest to the quotient— considered as pounds : this will be the required time. Example. — In what time will £50 become £100, at 6 per cent, per annum compound interest '? £100.^50=2. We find by the tables that in 11 years £1 will become £1-8983, which is less ; and in 12 years that it will become £20122, which is greater than 2. The answer nearest t6 the truth, therefore, is 12 years. ■^M I'M EP? ftvQ INTEREST. 28. Reason of Rule I. — The given amount is [20] equal to the given principal, multiplied by a product which contains ns many factors as there arc times of payment— each factor being the amount of <£1, for one payment. Hence it is evi- dent, that if we divide the given amount by the given prin- cipal, we must have the product of these factors ; and that, if we divide this product, and the successive quotients by one of the factors, we shall ascertain their number. Reason of Rule II. — We can find the required number of factors (each the amount of £1), by ascertaining how often the amount of £1 may be considered as a factor, without forming a product much greater or less than the quotient obtained when we divide the given amount by the given principal. Instead, however, of calculating for ourselves, wo may have recourse to tables constructed by those who have already made the necessary multiplications — which saves much trouble. 29. When the quotient [27 J is greater than any amount of i£l, at the given rate, in the table, divide it by the greatest found in the table ; and, if necessary, divide the resulting quotient in the same way. Continue the process until the quotient obtained is not greater than the largest amount in the table. Ascertain what number of payments corresponds to the last quotient, and add to it so many times the largest number of pay- ments in the table, as the largest am^urU in the table has been used for a divisor Example.— When would £22 become £535 12s. O^J., at 3 per cent, per annum 1 £535 12s. 0^(;.-^22==24-34560, which is greater than any amount of £1, at the given rate, contained in the table. 24-345G0-f-4-3839 (the greatest amount of £1, at 3 per cent., found in the table) =5 55339 ; but this latter, also, is greater than any amount of £1 at the given rate in the tables. 5-55339^4-383y=12G677, which is found to be the amount of £1, at 3 per cent, per payment, in 8 payments. We have divided by the highest amount for £1 in the tables, or that corresponding to fifty payments, twice. Therefore, the required time, is 50-f-504-8 payments, or 108 years. EXERCISES. 11. When would ^£14 6s. Sd. amount to £\S 2s. S^d. at 4 per cent, per annum, compound interest ? Am. In e years. > INTEREST. 259 [20] equal ch contains )uch factor ) it is evi- iven prin- id that, if tuts by one }d number ; how often )r, without e quotient the given rse/vea, wo who have saves much than any , divide it ttecessary, Continue )t greater tain what quotient, r of pay- the table 12$. O^d., r than any the table. ) per cent., is greater he tables, lie amount ents. We tables, or refore, the rs. 2s. S^d. t ? A'NJi. 12. When would £54 2s. 8d. amount to £76 3*. 5^., at 5 per cent, per annum, compound interest ? Ans, In 7 years. 13. In what time would £793 0*. 2ld. become £1034 13.5. 1 0^^., at 3 per cent, half-yearly, compound interest ? Ans. In 4^ years. 14. When would £100 become £1639 7s, 9rf., at 6 per c Ans. In 24 fears. QUESTIONS. 1. What is interest } [1]. 2. What is the diiforence between simple and com- pound interest } [1]. 3. What are the principal, rate, and amount } [ ]• 4. How is the simple interest of any sum, for 1 }ear, found.? [2 &c.]. 5. How is the simple interest of any sum, for several years, found } [5] . 6. How is the interest found, when the rate consists of more than one denomination } [4] . 7. How is the simple interest of any sum, for year8| months, &c., found .? [6]. 8. How is the simple interest of any sum, for any time, at 5 or 6, &c. per cent, found } [7] . 9. How is the simple interest found, when the rate, number of years, or both are expressed by a mixed number } [9] . 10. How is the simple interest for days, at 5 per cent., found.? [10]. 11. How is the simple interest for days, at any other rate, found .? [11]. 12. How is the simple interest of any sum, for months at 6 per cent., found } [12]. 13. How is the interest of money, left after one or more payments, found } [13]. 14. IIow is the principal found, when the amount, rate, and time are given } [14]. 15. How is the amount found, when the time, rato, and principal are given .^ [15]. 'M 200 DISCO L.VT. IR. How is tho timo fouri'l, when the amoutij priu cipnl, und rato are given ? [1<^]. 17. How is the rate t'ountl, when the amount, princi pal, and time arc given ? [1^]. 18. How are the amount, and compound interest found, when the principal, rate, and time are given ? [19]. 19. How is the present worth of any sum, al com- pound interest for any time, at any rate, found .' [2 IJ . 20. How is tlic time found, when the principal, rate of compound interest, and amount are given } [20] . I r ih DISCOUNT. '50. Discount is money allowed for a sum paid before it is due, and shoui I be such as would be produced by what iS paid, were it put to interest from the time the payment isy until the time it ou,ght to be made. Tlie jpri'Mnt worth of any sum, is tliat whicli would, at the rate allowed as discount, produce it, if put to interest until thj sum becomes due. 31. A bill is not payable until three days after the timo mentioned in it , these are called days of grace.. Thus, if the tune expires on tht 11th of the month, the bill will not be payable until the 14th — except the latter falls on a Sunday, in which case it becomes payable on the preceding Saturday. A bill at 91 days will not be due until the 94th day after date. 32. When goods are purchased, a certain discount is oft^n allowed for prompt (immediate) payment. The discount generally taken is larger than is sup- posed. Thus, let what is allowed for paying money one year before it is due be 5 per cent. ; in ordinary circumstances £95 would be the payment for £100. Bat £95 would not in one year, at 5 per cent., produce more than £99 15.9., which is less than £100 ; the error, however, is inconsiderable when the time or sum is 3raall Hence to jSnd the discount and present worth at any rate, we may generally use the following — ■ • DISCOUNT. 261 riouui, prlu )unt, princi itercst found, ? [19]. iim, ill com- 11(1? [21J. •incipal, rato ? [2GJ. I paid before produced by tho time the le. that which roducc it, if lya after the ys of grace. e month, the ept the latter s payable on 3 will not be n discount ia ent. than is sup- aying money in ordinary it for ieioo. jnt., produce ; the error, sum ia small vorth at any I ( 33. Rule. — Find tho interest for the sum to be paid, at the discount allowed ; consider this as discount, and deduct it from what is due ; tho remainder will be the required present worth. Example. — £02 will be duo in 3 months ; what should bo allowed on immediate payment, the discount being at the rate of G per cent, per annum } The interest on XG2 for 1 year at 6 per cent, per annum is £3 14.9. 4^d. ; and for 3 months it is lh>\ 7jt/. Therefore £02 minus I85. 7fd.=£,(Ji is. 4^ci., iu the required present worth. 34. To find the present worth accurately — KuLE. — Say, as iSlOO plus its interest fnr the given time, is to iilOO, so is the given sum to the required present worth. Example. — What would, at present, pay a debt of jC142 to be due in months, 5 per cent, per annum discount being allowed I £> £ s. £0 £, i()() y 140 £ .<■ d. 102-5 (100+2 10) : 100 : : 142 : — i0f5~=^^^ ^^ ^ This is merely a question in a rule already given [14]. ■ ■ ■ ' . ,, EXERCISES. 1. What is the present worth of i6850 15*., payable in one year, at 6 per cent, discount } Ans. £,S02 1 \s. \0}d 2. What is the present worth of iS240 10«., payable in one year, at 4 per cent, discount ? Ans. i£23 1 5«. 3. What is the present worth of j£550 IO5., payable in 5 years and 9 months, at 6 per cent, per an. discount ? Ans. £409 55. lOi^. 4. A debt of £1090 will be due in 1 year and 5 months, what is its present worth, allowing 6 per cent, per an. discount ? Ans. £1004 \2s. 2d. 5. What sum will discharge a debt of £250 17*. 6^., to be due in 8 months, allowing 6 per cent, per an. discount ? Ans. £241 4s. e^d. 6. What sum will discharge a debt of £840, to be due in 6 months, allowing 6 per cent, per an. discount ? Ans. £815 IO5. SJrf. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 11.25 u> 12.0 lU HiolQgraphic Sdmces Corporalion 23 WIST MAIN STRIIT WHSTIR,N.V. I4SM (71*)I73-4S03 6 <\ 262 DISCOUNT. 11; ^ 4 m ii : r hi 7. What ready money now will pay a debt of iJ200, to be due 127 days hence, discounting at 6 per cent, per an. ? Ans. £l9o I8s. 2\d. 8. What ready money now will pay for J21000, to be due in 130 days, allowing 6 per cent, per an. discount ? Ms. ie979 1*. Id. ■ '-• . '-^ ^ " • 9. A bill of iS150 105. will become due in 70 days, what ready money will now pay it, allowing 5 per cent, per an. discount } Ans. ^149 \s. bd. 10. A bill of £>\A0 lOs. will be due in 76 days, what ready money will now pay it, allowing 5 per cent, per an. discount } Ans. £139 Is. Q^d. 11. A bill of jeSOO will be due in 91 days, what will now pay it, allowing 5 per cent, per an. discount ? Ans. £296 65. \\d. 12. A bill of £39 55. will become due on the first of September, what ready money will pay it on the preceding 3rd of July, allowing 6 per cent, per an. ? Atis. £38 185. l\d. 13. A bill of £218 35. ^d. is drawn of the 14th August at 4 months, and discounted on the 3rd of Oct. ; what is then its worth, allowing 4 per cent, per an. discount } Ans. £216 85. \^d. 14. A bill of £486 185. Sd. is drawn of the 25th March at 10 months, and discounted on the 19th June, what then is its worth, allowing 5 per cent^ per an. discount.? Ans. MA12 95^ ll^d. 15. What is the present worth of £700, to be due in 9 months, discount being 5 per cent, per an. ? .^915. £674 135. 11^^. •; ; f i^;?/ jjj ;k > ' 16. What is the present worth of £315 12*. 4\d.^ payable in 4 years, at 6 per cent, per an. discount ? Ans. £254 10s» l\d. 17. What is the present worth and discount of £550 105. for 9 months, at 5 per cent, per an, ? Ans, £530 125. 0|i. is the present worth; and £19 lis, \l\d. 8 the discount. ^^iumnih- >fh.>,:! , \ .* 18. Bought goods to the value of £35 135. 8i. to be 9aid in 294 days; what ready money are they now worth, 6 per oent^ per an. discount being allowed ? Ans, £^4. 05. ^d. ..^, ...,. „, , ,.,,, fcl.'^-t- lebt of iesOO, 6 per cent. £1000, to be m. discount ? in 70 days, I 5 per cent. ^6 days, what per cent, per ys, what will count ? Alls. on the first ly it on the nt. per an. ? of the 14th 3rd of Oct. ; ent. per an. \,-.r. ?' of the 25th ) 19th June, entk per an. to be due in an. > 5 12s. 4\d., n. discount? »untof JS550 Aru. iS530 • 17*. U\d. U.Sd. to be e they now ig allowed? ■- X U,i.. COMMISSION. 263 19. If a legacy of igeOO is left to me on the 3rd of May, to be paid on Christmas day following, what must I receive as present payment, allowing 5 per cent, per an. discount ? Am. ii^581 4; . 2\d. 20. What is the discount of i^756, the one half pay- able in 6, and the remainder in 12 months, 7 per cent, per an. being allowed ? Ans. i237 145. 2\d. 21. A merchant owes iSllO, payable in 20 months, and £224y payable in 24 months ; the first he pays in 5 months, and the second in one month after that. What did he pay, allowing 8 per cent, per an. ? Ans. iS300. QUESTIONS FOR THE PUPIL. 1. What is discount ? [30]. 2. What is the present worth of any sum ? [30] . 3. Whsit are days of grace? [31]. 4. How is discount ordinarily calculated ? [33] 5. How is it accurately calculated ? [34] . COMMISSION, &c 35. Commission is an allowance per cent, made to a person called an agent, who is employed to sell goods. Insurance is so much per cent, paid to a person who undertakes that if certain goods are injured or destroyed, he will give a stated sum of money to the owner. Brokerage is a small allowance, made to a kind of agent called a broker, for assisting in the disposal of goods, negotiating bills, &c. 36. To compute commission, &c. — KuLE. — Say, as iSlOO is to the rate of commission, so is the given sum to the corresponding commission. Example. — ^W^hat will be the commission on goods worth £437 5s. 2d., at 4 per cent. ? £100 : £4 : : £437 5s. 2d. : ^^"^^i^"' ^' = ^17 95. ^d., the required commission. 37. To find what insurance must be paid so that, if the goods are lost, both their value and the insurauco paid may be recovered—- T! 264 COMMISSIOl^ \':^& t I KuLE. — Say, as £100 minus the rate per cent, is to jSlOO, 80 is the value of the goods insured, to the required insurance. •'(ij , : Example. — What sum must I insure that if goods worth J&40U are lost, I may receive both their value and the in- surance paid, the latter being at the rate of 5 per cent. ? £95 : £100 : : £400 : ^^^^ X 400 ^ ^^^ ^^ ^3^ If £100 were insured, only £95 would be actually received, since £5 was paid for the £100. In the example, £421 Is. Okd. are received ; but deducting £21 Is. Oid.t the insurance, £400 remains. EXERCISES. 1. What premium must be paid for insuring goods to the amount of iS900 15^., at 2^ per cent. ? Ans. £22 \^s.A\d. -'■ ■ f • '^'^ 2. What premium must be paid for insuring goods to the amount of £7000, at 5 per cent. } Ans. £350. 3. What is the brokerage on £976 17*. Gd.^ at 5s, per cent. .? Ans. £2 8*. I0\d. 4. What is the premium of insurance on goods worth £2000, at 7| per cent. .? Ans. £150. 5. What is the commission on £767 145. 7d.y at 2^ per cent. > Ans. £19 3*. lOf rf. - 6. How much is the commission on goods worth £971 145. 7d., at 55. per cent. .? Ans. £2 85. 7-g\d, 7. What is the brokerage on £3000, at 25. 6d. per cent. ? Ans. £3 155. 8 How much is to be insured at 5 per cent, on goods worth £900, so that, in case of loss, not only the value of the goods, but the premium of insurance also, may be repaid .? Ans. £947 75. 4j%d. 9. Shipped off for Trinidad goods worth £2000, how much must be insured on them at 10 per cent., that in case of loss the premium of insurance, as well as their recovered } Ans. £2222 45. 5^d. may QUESTIONS FOR THE PUPIL. 1. What is commission ? [35]. 2. What is insurance ? [35] . 3. What is brokerage } [35] PURCHASE OF STOCK. 265 per cent, is to isured, to the if goods worth ue and the in- S per cent. ? 421 Is. Qld. stually received, ie, £421 Is. OH. insurance, £400 insuring goods cent. ? Ans. insuring goods Ans. ie350. 7s. 6d.y at 5f. m goods worth 14*. 7d.y at 2 J goods worth 2 8s. 7s%d. at 2s. 6d. per cent, on goods )nly the value i alsO) may be ti ^22000, how ' cent., that in I well as their s. 5^d. '. 4. How are commission, insurance, &c., calculated.^ [36]. ; ;- * ( i , J, ; i*i. ^,,: . i t , ; 5. How is insurance calculated, so that both the in- surance and value of the goods may be received, if the latter are lost ? [37] . PURCHASE OP STOCK. 38. Stock is money borrowed by Government from individuals, or contributed by merchants, &c., for the purpose of trade, and bearing interest at a fixed, or variable rate. It is transferable either entirely, or in part, according to the pleasure of the owner. If the price per cent, is more than .£100, the stock in question is said to be above^ if less than jKlOO, below " par." Sometimes the shares of trading companies are only gradually paid up ; and in many cases the whole price of the share is not demanded at all — they may be £bO, £100, &c., shares, while only £d, £10, &c., may have been paid on each. One person may have many shares When the intesest per cent, on the money paid is con- siderable, stock often sells for more than what it origi- nally cost ; on the other hand, when money becomes more valuable, or the trade for waich the stock was contributed is not prosperous, it sells for less. 39. To find the value of any amount of stock, at any rate per cent. — Rule. — Multiply the amount by the value per cent., and divide the product by 100. Example. — ^\Vhen £69 1 will purchase £100 of stock, what will purchase £642 '? £642x69| 100 =£443 155. 7^d. It is evident that £100 of stock is to any other amount of it, as the price of the former is to that of the latter. Thus £100 : £642 : : £69^ £642X69^ 100 EXERCISES. 1. What must be given for £750 16*. in the 3 per cent, annuities, when jS64| will purchase £100 ? Ans. £481 95. O^i. : 866 EQUATION OF PAYMENTS. I'^l '?;.j ''>';r 4 II': f 3 ■J ■, 'IS' 5 ,lii W'VMi IrK^I i^^i m. 2. What must be given for i&1756 7s. 6d. India stock, when i&196| will purchase iSlOO r Ans. iS3446 17i. 8f ri. 3. What is the purchase of i£9757 bank stock, at jei25f per cent. ? ^^. £,\22bl As. l\d. QUESTIONS. 1. What is Stock? [38]. ' 2. When is it ahove^ and when hdow " par" ? [38] . 3. How is the value of any amount of stock, at anj rate per cent., found ? [39] . EQUATION OF PAYMENTS. 40. This is a process by which we discover a time, when several debts to be due at different periods may be paid, at once^ without loss either to debtor or creditor. KuLE. — Multiply each payment by the time which should elapse before it would become due ; then, add the products together, and divide their sum by the sum of the debts. Example 1. — A person owes another £20, payable in 6 months ; £50, payable in 8 months ', and £90, payable in 12 months. At what time may all be paid together, without loss or gain to either party ? £ £ 20x 6=r 120 ' 60 X 8= 400 _90x 12=1080 160 160) 1600(10 the required number of months. 160 Example 2.— A debt of £450 is to be paid thus : £100 immediately, £300 in four, and the rest in six months. When should it be paid altogether? 100 X 0= 300 X 4=rl200 60 X 6 = 300 450 450)1500(34 months. 1350 m m =^i EQUATION OF PAYMENTS. 267 8P of months. 41. We have (according to a principle formerly used [13]) reduced each debt to a sum which would bring the same interest, in one month. For 6 times jG20, to be due in 1 month, should evidently produce the same as X20, to be due in 6 months— and so of the other debts. And the interest of £1600 for the smaller time, will just be equal to the interest of the smaller sum for the larger time. EXERCISES. 1. A owes B ieeOO, of which ^2200 is payable in 3 months, j£150 in 4 months, and the rest in 6 months ; but it is agreed that the whole sum shall be paid at once. When should the payment be made } Ans. In 4^ months. — 2. A debt is to be discharged in the following man- ner : I at present, and \ every three months after until all is paid. What is the equated time ? Ans. 4^ months. 3. A debt of £120 will be due as follows : ^£50 in 2 months, i^40 in 5, and the rest in 7 months. When may the whole be paid together ? Ans. In 4\ months. 4. A owes B iSllO, of which £60 is to be paid at the end of 2 years, j£40 at the end of 3^, and i^20 at the end of 4^ years. When should B receiye all at once } Ans. In 3 years. 5. A debt is to be discharged by paying | in 3 months, ^ in 5 months, and the rest in 6 months. What is the equated time for the whole } Ans. 4^ months. QUESTIONS. 1. What is meant by the equation of payments ? 2. What is the rule for discovering when money, to be due at different times, may be paid at once ? [40]. -i,!; ■ar.M^-;-.. 1 ir' Bptei. ii|l 1^ MhI ||1|| fm HiL 1 Pplll 1 B.g^»tt^>^ HK'^' Ib^pk lr«v MK , " ! &i ^' ! d^ ■! 1 ■<'l '.1 I'J ;.. M^, .■I i SECTION VIII. ' EXCHANGE, &c. ' 1. Exchange enables us to find what amount of the money of one country is equal to a given amount of the money of another. Money is of two kinds, real— or coin, and imaginary — or money of exchange, for which there is no coin ; as, for example " one pound sterling." The par of exchange is that amount of the money of one country actually equal to a given sum of the money of another ; taking into account the value of the metals they contain. The course of exchange is that sum which, in point of fact, would be allowed for it. 2. When the course of exchange with any place is above " par," the balance of trade is against that place. Thus if Hamburgh receives merchandise from London to the amount of ill 00,000, and ships off, in return, goods to the amount of but ^50,000, it can pay only half what it owes by bills of exchange, and for the remainder must obtain bil^s of exchange from some place else, giving for them a premium — ^which is so much lo«t. But the exchange cannot be much above par, since, if the pre- mium to be paid for bills of exchange is high, the merchant will export goods at less profit ; or he will pay the expense of transmitting ai^d insuring coin, or bullion. 3. The nominal value of commodities in these countries was from four to fourteen times less formerly than at present ; that is, the same ap^ount of money would then buy much more than now. We may estimate the value of money, at any particular period, from the amount of corn it would purchase at that time. The value of money fluctuates from the nature of the crops, the state of trade, &c. i 1^ ti nount of the [uount of the imaginary — no coin ; as, f the money sum of the the value of exchange is be allowed any place is st that place, rom London return, goods ily half what aainder must else, giving :t. But the , if the pre- is high, the or he will ing coin, or ese countries 2rly than at would then ite the value 3 amount of be value of ps, the state i * EXCHANOE. 26d In exchange, a variable is given for a fixed sum ; thus London receives di£fereut values for £1 from different countries. ■ " Agio is the difference which there is in some places batween the current or cash money, and the exchange or banJc money — which is finer. The following tables of foreign coins are to be made familiar to the pupil. , , FOREIGN MONEY. MONEY OF AMSTERDAM. Flemish Mirney. . . make 1 grote or penny. • • .1 stiver. Tcnninga 8 10 or 320 800 1920 grotes 2 40 or 100 240 stiver! 20 50 or 120 or guilders 2i 6 1 florin or guilder 1 rixdollar. ' 1 pound. ■ Pfennings 6 72 or 1440 grotes 12 skillings 240 or I 20 rfenniugs Pence 12 or 2 192 • 884 676 32 or 64 96 MONEY OP HAMBURGH. Flemish Money. make 1 grote or penny 1 skilling. 1 pound. Hamburgh Money. make 1 schilling, equal to 1 dtiver schillings 16 . 1 mark, marks 2 1 dollar of exchange. .-.v- 3 • 1 rixdollar. 32 or 48 or We find that 6 schillings=l skilling. Hamburgh money is distinguished by the word *• Hambro.** " Lub," from Lubec, where it was coined, was formerly used for this purpose ; thus, *' one mark Lub." Wo exchange with Holland and Flanders by the pound sterling. * M r i'' : ■I .( I.-! ,1 I »70 EXCHAIfQB. Ii '!< : 1 I I' I i I' I 1 < 100 or I 10 81 livres=80 francs. Rees 400 PORTUGUESE MONEY. Accounts are kept in milrees and rees. . . make 1 crusado. crusados 1000 or I 2k 4800 I 12 1 milree. 1 moidere. SPANISH MONEY. Spanish money is of two kinds, plate and vellon ; the latter being to the former as 32 is t-o 17. Plate is used in exchange with us. Accounts are kept in piastres, and maravedi. make 1 real. . 1 piastre or piece of eight Maravedies 34 . . . 272 or reals 8 1088 875 1 piastres 82 or 1 4 . • • • 1 pistole of exchange. 1 ducat. AMERICAN MONEY. In some parts of the United States accounts are kept in dollars, dimes, and cents. Cents 10 , . , make 1 dimeii I dimes 10 . . . . . 1 dollar. In other parts accounts are kept in pounds, shillings, and pence. These are called eurreneyy but they are of much less Talue than with us, paper money being used. ke. f < f ' ' ■ i > » I < .1 BOtt. livre. ecu or crown imes. deoime. frano. orusado. milree. moidere. \; the latter in oxohango hvedi. ieoe of eight Lchange. are kept in B.ke 1 dimoi 1 dollar, lillingg, and if much less Pfenning! 12 EXCHANQS DANISH MONEY. 192 or •killing! 16 . mark! 6 1152 |90 or 6 DanishcsiS Hamburgh marks. 271 make 1 skilling. 1 mark. . 1 rizdollar ,»W.'"i*> i VENETIAN MONEY, Dienari (the plural of denaro) 12 . . . . make Isolde. Iioldi 20 ... . llira. lire !oldi 6 4 . . 1 ducat current. 8 . . 1 ducat effective 1488 1920 /!. » AUSTRIAN MONEY. Pfenning! 4 240 or 860 Grains 10 creutzeri 60 . I florin! "; r •ti i make 1 creutzer 1 florin. » '1 rixdoUar. NEAPOLITAN MONEY. . '^' . . make 1 carlin. carlin! ^ 100 or I 10 . . ' . . 1 ducat rewA9 MONEY OP GENOA. Lire soldi 4 and 12 make tsoudo di cambio, or crown of exchange. 10 and 14 1 scudo d'oro, or gol I crown. OF GENOA AND LEGHORN. Denari di pezza 12 .... make 1 soldo di pezza. 240 or I aO soldi di pezza * Denari di lira 12 240 or 1880 soldi di lira 20 . 116 or I 61 . .\\i\; ! 1 pezza of 8 reals, make 1 soldo di lira. llira. ^ 1 pezza of 8 reals Penning!, or oen 12 676 or I 48 I !kiUings SWEDISH MONEY. •..>.; make 1 skilling. 1 rizdollsi 272 EXCHANOE. RUSSIAN MONET. i' . ♦ ; i f:N' ;( !!l: Copeoi 100 make 1 itible. make 1 rupee. EAST INDIAN MONEY. Cowries 2660 Huueei 100,000 . . . llao. •' • 10,000,000 . . 1 crore. The cowrie is a small 'shell found at the MaldlTea* and near Angola : in Africa about 6000 of them pass for a pound. Tlie rupee has different values: at Calcutta it is 1«. Hid. the Sicca rupee is 2s. Oid. ; and the current rupee 2$. — if we divide any number of tliese by 10, we chanse them to pounds of our money ; the Bombay rupee is 2s. Su. , &c. A sum of Indian money is expressed as follows; 5*88220, which means 6 lacs and 88220 rupees. 4. To reduce bank to current money — RrLE. — Say, as jElOO is to iElOO + the agio, so is the given amount of bank to the required amount of current money. Example. — How many guilders, current money, are equal to 463 guilders, 3 stivers, and 1^^^ pcnnings banco, agio being 4S ? 100 : 104f :: 463 g. 3st. 13|fjp. : ? " 7 7 20 700 65 733 9263 stivers. 16 45500 148221 pennings. -' Multiplying by 65, and adding 64 to the will give 9634429 P'^^^*' Multip lying by 733 ; ' and dividing by 45500 )7062036^ " p™'- will give 15520^ pennings. , 16 )155209 2 0)9700 9 ; . r -^ - And 485 g. st. 9f ||4^ p. is the amount sought. 6. We multiply the first and second terms by 7, and add the numerator of the fraction to one of the products. This is the same thing as reducing these terms to fractions having 7 for their denominator, and then multiplying them by 7 [Sec. V. 29]. For the same reason, and in the same way, we multiply the first and third terms by 65, to banish the fraction, < without destroying the proportion. EXCHANGE. 273 The remninder of the process is according to the rule of proportion [8eo. V. 81]. We roduoe the anttwer to pennings, Btiversi and guilders. EXERCISES. 1. Reduce 374 guilders, 12 stivers, bank money, to current money, agio being 4^ per cent. ? Ans. 392 g., 5 St., 3^y p. 2. Keduce 4378 guilders, ^ stivers, bank money, to current money, agio being 4| per cent. ? Atis, 4577 g., 17 St., 3,% p. 3. Reduce 873 guilders, 1 1 stivers, bank money, to current money, agio being 4} per cent. } Ans. 91(> g., Sst., lUtp. 4. Reduce 1642 guilders, bank money, to current money, agio being 4|| per cent. ? Am. 1722 g., 14st., 6. To reduce current to bank money — Rule. — Say, as iE 100 + the agio is to jBlOO, so is the given amount of current to the required amount of bank money. Example. — How much bank money is there in 485 guil- ders and 9|^^^ J pennings, agio being 4| ? 100 :: 4f5 o' 9||f»^ : 1 7 20 104f 733 45500 700 1 ■lU, v)1. 9700 16 M I- ,»■;•'*■ 'f^' 33351500 ' ■ ' ' 155209 Multiplying by 45500 the denominator, . .;,... . 7062009500 I and adding 25957 the numerator, we get 7062035457 TOO 33351500)l 9i342481990 • : Quotient 148221f$ 1 6)14822111 20)9263 ' ' ■'.li i .- •mr^zr 463 3 13|| is the amount sought. *,'.; 274 EXCHANGE. •»;n EXERCISES. . .*T m U' ■', 5. Reduce 58734 guilders, 9 stivers, 11 pennings, current money, to bank money, agio being 4| per cent. ? Ans. 56026 g., 10 st., 1 IfB p. 6. Keduce 4326 guild«!rs, 15 ponnings, current money, to bank money, agio being 4^ per cent. ? Ans. 4125 g., 13 St., 2i|^ p. 7. Reduce 1186 guilders, 4 stivers, 8 pennings, cur- rent, to bank money, agio being 4f per cent. ? Ans 1136 g., 10 St., 0|f^ p. 8. Reduce 8560 guilders, 8 stivers, 10 pennings^ current, to bank money, agio being 4 J per cent.. Ans. 8183 g., 19 st., 5||f p. 7. To reduce foreign money to British, &c. — Rule. — Put the amount of British money considered in the rate of exchange as third term of the proportion, its value in foreign money as first, and the foreign money to be reduced as second term. Example 1. — Flemish Money. — How much British money is equal to 1054 guilders, 7 stivers, the exchange being 33«. 4d. Flemish to £1 British ? ' •■ ' 33s. 4x1. : 1054 g. 7 st. : : £1 : ? 12 20 ^ ^ 400 pence. 21087 stivers. 2 400)42174 Flemish pence. £105-435 =r £105 8«. 8id. £1, the amount of British money considered m the rate, is put in the third term ; 33s. 4d., its value in forei^ money, in the first; and 10^4 g, 7 st., the money to be reduced, in the second. 9. How many pounds sterling in 1680 guilders, at 33*. Zd. Flemish per pound sterling } Ans. iB168 8s. 10. Reduce 6048 guilders, to British money, at 33s. 11^. Flemish per pound British? Aiis. jS594 7^. 11. Reduce 2048 guilders, 15 stivers, to British money, at 34s. bd. Flemish per pound sterling ? Ans iB198 8j. 6HK pcnnragfl, per cent. ? ent money, IS. 4125 g., inings, cur- at. ? Ans pennings^ per cent. . ). — considered proportion, the foreign itish money ;e being 33s. m the rate, ■ei^ money, be reduced, guilders, at J. iB168 8s. ley, at 33s. ^£594 7s. to British ing? Ans I EXCHANGE. 275 IJt.;, How many pounds sterling in 1000 guilders, 10 stivers, exchange being at 33s. 4d. per pound sterling ? Ans. iSlOO Is. Example 2. — Hamburgh Money. — How much British money is equivalent to 476 marks, 9 skillings, the exchange being 33s. 6d. Flemish per pound British ? d. m. 8. s. 33 12 6 476 32 £1 402 grotes. 15232-f-194=15251i grotes. 402 )15251^ £37-9386=£37 18s. 9d. Multiplying the schillings by 2, and the marks by 32, reduces both to pence. 13. How much British money is equivalent to 3083 marks, 12| schillings Hambro', at 32s. 4d. Flemish per pound sterling ? An^. £2bA 6s. Sd. 14. How much English money is equal to 5127 marks, 5 schillings, Hambro' exchange, at 36s. 2d. Flemish per pound sterling .'* Ans. £378 Is. 15. How many pounds sterling in 2443 marks, 9^ schillings, Hambro', at 32s. 6^. Flemish per pound ster- ling } Am. je200 10s. 16. Reduce 7854 marks, 7 schillings Hambro', to British money, exchange at 34s. Wd. Flemish per pound sterling, and agio at 21 per cent. } Ans. £495 15s. O^d. Example 3. — French Money. — Reduce 8654 francs, 42 centimes, to British money, the exchange being 23f., 50c., per £1 British. f. 23 c. 50 f. 8654 c. 42 8654-42 — so7ka=£368 5s. 5^d. 42 centimes are 42 of a franc, since 100 centimes make 1 franc. 17. Reduce 17969 francs, 85 centimes, to British money, at 23 francs, 49 centimes per pound sterling } Ans. £765. 18. Reduce 7672 francs, 50 centimes, to British money, at 23 francs, 25 centimes per pound sterling ? Ans. £330. . . T- u- 276 EXCHANGE. n f^ 19. Reduce 15647 francs, 36 centimes, to British money, at 23 francs, 15 centimes per pou^d sterling? Ans. £61 o 18s. 2^d. 20. Reduce 450 francs, 58} centimes, to British money, at 25 francs, 5 centimes per pound sterling ? Ans. ^176 Us. KxAMPLE 4. — Poiiugtiese Money. — How much British monpy ik equal to 540 milrees, 420 rees, exchange beiug at 5v. h'.i per milree 1 m. m. r. s. d. 1 : 540-420 : : 5 G : 540'420x5s. 6(/.=£148 12«. 3Jc/. Ill this case the British money is the variable quantity. avA o.s. Or/. 18 that amount of it which is considered in the v'rtte. ■ ; 'i ihe rees are changed into the decimal of a milree by putting them to the right hand side of the decimal point, since one ree is the thousandth of a milree. 21. In 850 milrees, 500 rees, how much British money, at 5*. 4d. per milree ? Ans. .£226 16*. 22. Reduce 2060 milrees, 380 roes, to English money, at 5.-?. 6-^-d. per milree ? Ans. i2573 0*. \0}d. 23. In 1785 milrees, 581 rees, how many pounds sterling, exchange at 64^ per milree } Ans. i£479 lis. 6d. 24. In 2000 milrees, at 5*. 8^d. jpcr milree, how many pounds sterling.? Ans. iS570 16*. Sd. Example 5. — Spanish Money. — Reduce 84 piastres, 6 roal8, 19 maravedij to British money, the exchange being 40r/. tho piastre. p. p. r. m. d. 1 : 84 6 19 : : 49 : 1 « 8 8 34 678 reals. 34 272 23052 maravedi 49 272)1129548 4152-7, &c.=£17 %s. 02(/. I '^ i EXCHANGE. 277 to British id sterling? ^ to British id sterling ? luch British nge being at 8 12s. Zld. ible quantity; kidered in the ' a milree by locinial point, nuch British I 165. iglish money, many pounds ^715. M7Q milree, how d. astres, 6 reals, being 40f/. th« d. EXERCISES. 25. Reduce 2448 piastres to British money, exchange at 50d. sterling per piastre ? Ans. £510. 26. Reduce 30000 piastres to British money, at 40d. per piastre ^ Ans. iE5000. 27. Reduce 1025 piastres, 6 reals, 22J-f ^ maravedi, to British money, at ^d^d. per piastre } Ans. ^£167 155. 4d. Example 6. — American Money. — Reduce 3765 dollars to British money, at 4s. per dollar. 4s.=£\ : therefore 5 )3765 dol. del. s. £ 753 is the required sum. Or 1 : 3765 : : 4 : 753 28. Reduce £2^2 3«. 2^d. American, to British money, at 66 per cent. ? Ans. £176, 29. Reduce 5611 dollars, 42 cents., to British money, at 45. b^d. per dollar .? Ans. iE1250 175. 7d. 30. Reduce 274r» dollars, 30 cents., to British money, at 45. 3^d. per dollar ? Ans. ^£589 65. 2^d. From these examples the pupil will very easily under- stand how any other kind of foreign, may be changed to British money. 8. To reduce British to foreign money — Rule. — Put that amount of foreign money which is considered in the rate of exchange as the tiiird term, its value in British money as the first, and the British money to be reduced as the second term. Example 1. — Flemis'i Money. — How many guilders, &c., in i^236 14$. 2d. Britislr, the exchange being 345. 2d. Flemish to £1 British 1 £ £ s. d. 5. d. 1 20 io h 240 236 U 20 5. 34 12 4734 12 56810d. 410 410 pence. 240)23292100 12 )97050-4, &c. 20)8087 6 £404 7 6^ Fleriieh. N 2 278 EXCHANGE. iiliiil iiiiii^ if J We might take parts for the 34s. 2d. — 34s. 2d.^£l 4- 10s.+4s.+2ti. £ jei = 1 105.= i 4s.== I _ f £ 236 118 47 s. 14 7 6 2d.=ji^ii,of\)ll9 d. 2 1 10 H £404 7 EXERCISES. 6^ Flemish. 31. In iElOO Is.y how much Flemish money, exchange at 33s. 4d. per pound sterling? Ans. 1000 guilders, 10 stivers. 32. Reduce £168 8s. 5yJ^rf. British into Flemish, exchange being 33s. 3d. Flemish per pound sterling ? Ans. 1680 guilders. 33. In iJ199 lis. lO^^^d. British, how much Flemish money, exchange 34s. 9d. per pound sterling .? Ans. 2080 guilders, 15 stivers. 34. Reduce £198 8s. G^d. British to Flemish money, exchange being 34s. dd. Flemish per pound sterling .? Ans. 2048 guilders, 15 stivers. Example 2. — Hamburgh Money. — How many marks, &c., in £24 6s. British, exchange being 33s. 2d. per £1 British I £1 : £24 6s. :: 33s. 2d. : 'i 20 20 12 20 486 398 398 grotes. t .■■ 35. 20 )193428 2)9671 8 pence. 16)4835 schillings, 1 penny. 302 marks, 3 schillings, 1 penny. Reduce £254 6s. 8d. English to Hamburgh money, at 32s. 4d. per pound sterling.? A7is. 3083 marks, 12| stivers. 36. Reduce £378 Is. to Hamburg money, at 36s 2d. Flemish per pound sterling } Ans. 5127 marks^ 5 schillings. 37. Reduce £536 to Hamburgh money, at 36s. 4d, per pound sterling > Ans. 7303 marks. EXCHANGE. 279 mish. ey, exchange )00 guilders, nto Flemish, ind sterling ? nuch Flemish rling ? Ans. to Flemish h per pound ly marks, &c., T £1 British I )tes. enny. ings, 1 penny. Hamburgh A71S. 30B3 mey, at 36s 5127 marks^ r, at 36s. 4d. 38. Reduce £495 lbs. O^d. to Hamburg currency, at 34*. lid. per pound sterling ; agio at 21 per cent. .? Am. 7854 marks 7 schillings. Example 8. — French Money. — How much French monoy is equal in value to £83 2s. 2d., exchange being 23 francs 25 centimes per £1 British ? £ £ s. d. f. 1 : 83 2 2 : : 23-25 : 1 20 20 20 1G62 12 12 240 19946 23-25 240 )463744-50 19322-7, or 19322f. 70c. is the required sum. 39. Reduce j2274 5s. 9d. British to francs, &c., ex- change at 23 francs 57 centimes per pound sterling } Ans. 6464 francs 96 centimes. 40. In JE765, how many francs, &c., at 23 franca 49 centimes per pound sterling .? Ans. 17969 franca 85 centimes. 41. Reduce £330 to francs, &c., at 23 francs 25 cen- times per pound sterling t Ans. 7672 francs 50 cents. 42. Reduce £734 45. to French money, at 24 francs 1 centime per pound sterling ? Ans. 1769 francs 42^ centimes. - ,; , , k > Example 4. — Portuguese Money. — How many milrees and rees in £32 65. British, exchange being 5*. 9d. British pt milree ? . , , s. d. £> s. 5 9 : 32 6 :: 1000 : ? . .12 20 69 646 12 7752 1000 69)7752000 ,«i{uired sum. 112348 rees=112 milrees 348 rees, is tii« If ; J 1 if I I ) if f*-.!; •/ If m ■i;;!! ill! m (If, 'I- I m-m 3 ll 280 EXCHANGE. 43. Eeduoe i£226 165. to milrees, &c., at5«. 4<;?. pet milree ? Am. 850 milrees 500 rees. 44. Reduce iE479 175. 6d. to milrees, &c., at 64^d. por milree .? Ans. 1785 milrees 581 rees. 45. Reduce i£570 16s. 8d. to milrees, &c.,at55. S^d. per milree ? Ans. 2000 milrees. 46. Reduce iE715 to milrees, &c., at 5*. Sd. per mil- ree .? Ans. 2523 milrees 529y\ rees. Example 5. — Spanish Money. — How many piastres, &c., in £62 British, exchange being 50(/. per piastre I d. £ 1 : ? 60 62 20 ^• 1240 12 50 )14880 297-6 piastres. , 3 48 reals. 34 r. m. i97 32Af , is the required sum. 60 )1632 ^ 32i| maravedis. ^7. How many piastres, &c., shall I receive for i£510 sterling, exchange at 50d. sterling per piastre .'* Ans. 2448 piastres. 48. Reduce i25000 to piastres, at 40d. per piastre ? Ans. 30000 piastres. 49. Reduce iei67 15*. Ad. to piastres, &c., at 39i^. per piastre } An^. 1025 piastres, 6 reals, 22}|^ mara- vedis. 50. Reduce iE809 95 Sd. to piastres, &c., at 40f ^. per piastre .? Ans. 4767 piastres, 4 reals, 2y«^^ maravedis. Example 6. — American Money. — ^Reduce £176 British to American currency, at 66 per cent. £f £ £ 100 : 176 : : 166 : : 166 100 )29216 £292 35. 2^d.^ is the required sum. i 58. Ad. pet :c., at 64|rf. !.,at 55. 9>\d. Sd. per mil- piastres, &c., [uired sum. ;ive for £510 istre } Ans. per piastre } fcc, at ^^\d. 22}|^ mara- , at 40f i. per ij maravedis. 76 British to EXCHANGE. EXERCISES. 281 Ans. 51. Reducd £753 to dollars, at 4s. per dollar ? 3765 dollars. 52. Reduce £532 4s. Sd. British to Americau money, at 64 per cent. } Ans. £872 lis. 3d. 53. Reduce £1250 17*. 7d. sterling to dollars, a . 4s. b^d. per dollar ? Ans. 5611 dollars 42 cents. 54!" Reduce £589 Qs. 2^^d. to dollars, at 4s. 3d. per dollar ? Ans. 2746 dollars 30 cents. 55. Reduce £437 British to American money, i 78 per cent. } Ans. £777 17s. 2\d. 9. To reduce florins, &c., to pounds, &c., F) -mish — Rule. — Divide the florins by 6 for pound , and — adding the remainder (reduced to stivers) to ^ le stivers — divide the sum by 6, for skillings, and ouble the remainder, for grotes. Example. — How many pounds, skillings, ai d grotes, in 165 florins 19 stivers ? f. St. 6)1 65 19 £,21 13s. 2df., the reqi "red sum. 6 will go into 165, 27 times — leaving 3 florin , or 60 stivers, which, with 19, make 79 stivers ; 6 will go intc 79, 13 times — leaving 1 ; twice 1 are 2. 10. Rkason of the Rule. — There are 6 \ mes as many florins as pounds ; for we find by the table tl at 240 grotes make <£l, and that 40 (^f") grotes make I flor n. There are 6 times as many stivers as skillings ; since 96 ] ennings make 1 skilling, and 16 ("^ ) pfennings make one stive . Also, since 2 grotes make one stiver, tlie remaining stiver 1 are equal to twice as many grotes. Multiplying by 20 and 2 would reduce the floiins to grotes ; nnd dividing the grotes by 12 and 20 would I'educe them to pounds. Thus, using the same example — f. St. 165 19 20 ... red sum. 3319 2 12 )6():J8 20 )553 2 £27 13s. 2d.i as before, is the result. 1382 EXCHANOt. JSf r .»■ > I ; i CV>^«| i EXERCISES. 56. In 142 florins 17 stivers, how many pounds, &e., Am. ie23 165. 2d. 57. In 72 florins 14 stivers, how many pounds, &c., Ans. £12 2s. 4d. 58. In 180 florins, how many pounds, &o. ? Ans. £30 11. To reduce pounds, &o., to florins, &c. — Rule. — Multiply the stivers by 6 ; add to the product half the number of grotes, then for every 20 contained in the sum carry 1, and set down what remains above the twenties as stivers. Multiply the pounds by 6, and, adding to the product what is to be carried from the stivers, consider the sum as florins. Example. — ^How many florins and stivers in 27 pounds, 13 skillings, and 2 grotes ? £ s. d. 27 13 2 165fl. 19st., the required sum. 6 times 13 are 78, which, with half the number (f ) of grotes, make 79 stivers — or 3 florins and 19 stivers (3 twenties, and 19) ; putting down 19 we carry 3. 6 times 27 are 162, and the 3 to be carried are 165 florins. This rule is merely the converse of the last. It is evident that multiplying by 20 and 12, and dividing the product by 2 and 20, would give the same result. Thus £ s. d. 27 13 2 20 663 12 ' 2 )6638 20)3319 166fl. 19st, the same result as bcfere. exercises. 59. How many florins and stivers in 30 pounds, 12 skillings, and 1 grote } Ans. 183 fl., 12 st., 1 g. 60. How many florins, &o., in 129 pounds, 7 skil- lings .? Ans. 776 fl. 2 st. 61. In 97 pounds, 8 skillings, 2 grotes, how many florins, &o. ? Am. 584 fl. 9 st. )Ounds, &c., [>ounds, &c., ? Ans. ieao • > the product 30 contained mains above is by 6, and, ,ed from the in 27 pounds, sum. imber (f) of rs {Z twenties, es 27 are 162, It is evident e product by 2 lit as b«fdre. pounds, 12 mds, 7 skil- , how many ARBITRATION OF EXCHANGES. QUESTIONS. 283 1. "What is exchange ? [1]. 2. What is the difference between real and imagin- ary money? [1]. 3. What are the 'par and course of exchange ? [1]. 4. Whatis«^o? [3]. 5. What is the difference between current or cash money and exchange or bank money ? [3] . 6. How is bank reduced to current money ? [4]. 7. How is current reduced to bank money ? [6l. 8. How is foreign reduced to British money ? [7]. 9. How is British reduced to foreign money ? [8]. 10. How are florms, &c., reduced to pounds Flemish, fcc? [9]. 11. How are pounds Flemish, &c., reduced to florins, fcc? [11]. ARBITRATION OF EXCHANGES. 12. In the rule of exdmnge only two places are con- cerned ; it may sometimes, however, be more beneficial to the merchant to draw through one or more other places. The mode of estimating the value of the money of any place, not drawn directly, but through one or more other places, is called the arbitration of exchanges, and is either simple or compound. It is " simple " I when there is only one intermediate place, " compound " when there are 7nore than one. All questions in this rule may be solved by one or more proportions. 13. Simple Arbitration of Exchanges. — Given the course of exchange between each of two places and a third, to find the par of exchange between the former. Rule. — Make the given sums of money belonging to the third place the first and second terms of the propor- tion ; and put, as third term, the equivalent of what is in the first. The fourth proportional will be the value of what is in the second term, in the kind of money contained in the third term. 284 ARBCfRATION OK EXCHANGES. I I iir I I Example. — If London exchanges with Paris at lOJ. per franc, and with Anisterdam at3-ls. 8^/. per £.1 sterling, what onght to bo the course of oxciiange, between Pans and Amsterdam, that a merchant may without loss remit from London to Amsterdam through Paris I £1 : \0d. : : 34s. 8^/. (the equivalent, in Flemish nione^, of jGl) : 1 the equivalent of lOt/. British (or of a franc) in Flemish money. . Or 240 : 10 :: 34*. M. : 24Q^ =17^., the re- quired value of 10^/. British, or of a franc, in Flemish money. £1 and \0d. are the two given sums of English money, or that \yhich belongs to the third place ; and 34s. 8J. is the given equivalent of XL It is evident that, 17ic/. (Flemish) being the value of lOt/., the equivalent in British money of a franc, when more than 17iQ lis. lyW^- i^ it makes payments by way of Holland. 5000 ruble8=n=£1041 13s. Ad. British, or £1875 Flemish; but £1875 Flemish=£1032 2s. 2^*^. British. \ -.1 I ARBITRATION OF £X(;lfANaC8. 285 is at lOd. per sterlinjj, what len Pari8 and )S8 remit from Lemish monejr, of a franc) in 17^., the ro- lomish money. ;li8h money, or 345. 8(2. is the > value of lOd.t vhcn more than ant will gain if idirectly receive e than its equi- >se of exchange er hand, if less he will lose by a franc for lOd Q franc : — while, . to Amsterdam id Amsterdam exchange be- 3, what is the Paris ? Ans. 5000 rubles ; and London tersburgh and and Holland ound sterling, ithod for Lon- direct .? Ans. ikes payments ^1875 Flemish; h. :■! 14. Compound ArhU ration of Kxchanges — To find what should bo the courso of exchange between two places, through two or more others, that it may be on a par with the course of exchange between the same two places, directly — Hulk. — Having reduced monies of the same kind to the same denomination, consider each course of exchange as a ratio ; set down the diflorent ratios in a vertical column, so that the nnteoedcnt of the second shall be of the same kind as the consequent of the first, and the antecedent of the third, of the same kind as the conse- quent of the second — putting down a note of interroga- tion for the unknown term of the imperfect ratio. Then divide the product of the consequents by the product of the antecedents, and the quotient will be the value of the given sum if remitted through the intermediate places. Compare with this its value as remitted by the direct exchange. 15. Example. — £824 Flemish beins; due to me at Am- sterdam, it is remitted to France at IGa. Flemish per franc ; from France to Venice at 300 francs per 60 ducats ; from Venice to Hamburgh at \00d. per ducat ; from Hamburgh to Lisbon at 50i/. per 400 rees ; and from Lisbon to England at 5s. 8(/. sterling per milree. Shall I gain or lose, and how much, the exchange between England and Amsterdam being 34.S'. 4r/. per £1 sterling % IM. : 1 franc. 300 francs : 60 ducats. 1 ducat : 100 pence Flemish. 50 pence Flemish : 400 rees. • 1000 rees : 68 pence British. ? : £824 Flemish. 1x60x100x400x68x824 ... , ... — - — — - — , en -ifM^r. — ==(" "^6 rcducc thc tcrms 16x300x1x50x1000 "^ 17x824 [Sec. V. 47]) 25 £560 6s. 4ld. But the exchange between England and Amsterdam fcl £824 £824 Flemish is £480 sterling Since 34s. 4d. : £824 : : £1 =£480. 34s\ Ad: I gain therefore by the circular exchange £560 6«. 4|a. minus £480=:£80 6s. 4^. 286 ARBITRATION OF EXCHANQCS. If commisaion is charged in any of the places, it must bo deducted from the value of the sum which can be obtained in that place. Tho process given for the compound arbitration of ex- change may bo proved to bo correct, by putting dovrn tho different proportions, and solving them in succession. Thus, if UM. are equal to 1 franc, what vfrill 300 francs (=00 ducats) be worth. If the quantity last found is the value of 60 ducats, what will be that of one ducat (alOOd.), &c. 1 U'-'f EXERCISES. 3. If London would remit jfilOOO sterling to Spain, the direct exchange being 42^d. per piastre of 272 maravcdis ; it is asked whether it will be more profit- able to remit directly, or to remit first to Holland at 35.?. per pound ; thence to France at IQ^d. per franc ; thence to Venice at 300 francs per 60 ducats ; and thence to Spain at 360 maravedis per ducat ? Ans. The circular exchange is more advantageous by 103 piastres, 3 reals, 19|^ maravedis. 4. A merchant at London has credit for 680 piastres at Leghorn, for which he can draw directly at bOd. per piastre ; but choosing to try tho circular way, they are by his orders remitted first to Venice at 94 piastres per 100 ducats; thence to Cadiz at 320 maravedis per ducat ; thence to Lisbon at 630 rees per piastre of 272 maravedis ; thence to Amsterdam at 50d, per crusade of 400 rees ; thence to Paris at I8^d. per franc ; and thence to London at 10^^. pe^* franc ; how much is the circular remittance better than the direct draft, reckon- ing ^ per cent, for commission ? Atis. £14 12*. 7\d 16. To estimate the gain or loss per cent. — Rule. — Say, as the par of exchange is to the course of exchange, so is iSlOO to a fourth proportional. From this subtract ieiOO. : Example. — ^The par of exchange is found to be IS^d. Flemish, but the course of exchange is 19d. per franc ; what is the gain per cent. ? lS\d. : 19f?. : : £100 ; ^ii| jM PROFIT AND LOSS. 17. This rule enables us to discover how much we gain or lose in mercantile transactions, when we sell at certain prices. Given the prime cost and selling price, to find the gain or loss in a certain quantity. Rule. — Find the price of the goods at prime cost and at the selling price ; the difierence will be the gain or loss on a given quantity. . . Example. — What do I gain, if I buy 460 lb of butter at 6(/., and sell it at Id. per ib ? The lotal prime cost is 460d. y 6==2760<;. The total selling price is 460ti.x7=3220rf. The total gain is 3220c2. minus 27&0d.^4Q0d.—£l 18s. 4d. 288 PROFIT AND LOSS. \fir f ' EXERCISES. 1. Bought 140 ft) of butter, at lOd. per ft), and sold it at 14d. per ib ; what was gained .? Ans. £2 6s. 8d. 2. Bought 5 cwt., 3 qrs., 14 ft) of cheese, at £2 12s. per cwt., and sold it for £2 18s. per cwt. What was the gain upon the whole } Ans. £1 Ids, 3a. 3. Bought 5 cwt., 3 qrs., 14 ft) of bacon, at 34*. per cwt. . and sold it at 36^. 4d. per cwt. What was the gain on the whole ? Ans. 13s. 8^d. 4. If a chest of tea, containing 144 ft) is bought for Qs. Sd. per lb, what is the gain, the price received for the whole being ^£57 105. .? Ans. £9 10s. 18. To find the gain or loss per cent. — Rule. — Say, as the cost is to the selling price, so is j£100 to the required sum. The fourth proportional minus i^lOO will be the gain per cent. FrcAMPLE 1. — What do I gain per cent, if I buy 1460 lb of beef at 3 ice. T)rice. eilD 135. 4J. 4d.) is the gain price plus the pi'ice plus the as any multiple an equimultiple :40, and loses 9 cent. What is as £40 (what to what it cost. it the horse cost. Ser cent, or \ d have been 3 19 lA is the diflference between cost and selling price. * I-** 9| is what he should have received above cost. 8 15 b 12 14 llj is his total loss. EXERCISES. 5. Bought beef at 6d. per ft), and sold it at 8d. What what was the gain per cent. } Ans. 33|. 6. Bought tea for 55. per ft), and sold it for 35. What was the loss per cent. ? Ans. 40. 7. If a pound of tea is bought for 6s. 6, are bought for £9 85., and sold for £\\ 185. llc^.,how much is gained per cent. ? Ans. 21 j^-^. 9. When wine is bought at 175. Qd. per gallon, and sold for 27.V. 6f/., what is the gain per cent. } An^. 57|. 10. Bought a quantity of goods for j£60, and sold tliem for £lb ; what was- the gain per cent. } Ans. 25. 11. Bought a tun of wine for .£50, ready money, and sold it for i:^54 105., payable in 8 months. How much per cent, per annum is gained by that rate } Ans. 13|. 12. Having sold 2 yards of cloth for ll5. 6^?., I gained at the rate of 15 per cent. What would I have gained if I had sold it for 125. } Ans. 20 per cent. 13. If when I sell cloth at 75. per yard, I gain 10 per cent. ; what will I gain per cent, when it is sold for 65. ^d. } Am. iB33 ll5. b\d. Is. : 85. dt:. •: £110 : £133 lis. ^d. And £133 II5. b\d. — £100=£33 11.?. 5;c?., is the required gain. 19. Given the cost price and gain, to find the selling price — Rule. — Say, as .£100 is to iBlOO plus the gain per cent., so is the cost price to the required selling price. Example. — At what price per yard must I sell 427 yards of cloth which I bought for 19s. per yard, so that I may gaiu 8 per cent. ] 108x195. 100 : 108 : : 19s. : — j-^-^j— =£1 O5. ^d. This result may be proved by the last rule. 290 PROFIT AND LOSS. [t'4^ 5 ',, r '8 1 \H f f 1 J fSI* "1 EXERCISES. 14. Bought velvet at 4s. 8d. per yard ; at what price must I sell it, so as to gain 12^ per cent. ? Ans. At 6*. 3d. 15. Bought muslin at 55. per yard ; how must it bo sold, that I may lose 10 per cent. .? Ans. At 4*. Qd. 16. If a tun of brandy costs iS40, how must it be sold, to gain 6|^ per cent. .? Ans. For jE42 10*. 17. Bought hops at £4 16s. per cwt. ; at what rate must they be sold, to lose 15 per cent. } Ans. For £4 Is. l\d. 18. A merchant receives 180 casks of raisins, which stand him in 16^. each, and trucks them against other merchandize at 28*. per cwt., by which he finds he has gained 25 per cent. ; for what, on an average, did he sell each cask } Ans. 80 lb, nearly. 20. Given the gain, or loss per cent., and the selling price, to find the cost price — Rule. — Say, as jSlOO plus the gain (or as £100 minus the loss) is to £100, so is the selling to the cost price. Example 1. — If I sell 72 lb of tea at Cs. per lb, and gain 9 per cent., what did it cost per lb 1 .€100x6 109 : 100 : : 6 : — ^^^p^Ss. M. What produces £109 cost £100 ; therefore what pro- duces 6s. must, at the same rate, cost 5s. 6rf. . , Example 2. — A merchant buys 97 casks of butter at 30s. each, and selling the butter at £4 per cwt., makes 20 per cent. ] for how much did he buy it per cwt. 1 30.:" '?t:' 292 FELLOWSHIP. 22. Single Felloicship. — Rule. — Say, as the whole Btock is to the whole gain or loss, so is each person's contribution to the gain or loss which belongs to him. Example. — A put £720 into trade, B £340, and C £960 ; and they gained £47 by the traffic. What ia B'& share of it ? £ 720 ■'■■:■. . 340 9G0 2020 : £47 :: £340 £47x340 2'.120 ' =£7 185. 2\d. Each person's gain or loss must evidently be proportional to his contribution. EXERCISES. 1. B and C buy certain merchandizes, amounting to ^£80, of which B pays ^^30, and £bO ; and they gain £20. How is it to be divided .? Ans. B £7 10s., and £12 105. 2. B and C gain by trade £182 ; B put in £300, and £400. What is the gain of each } Aiis. B £78, and C £104. 3. 2 persons are to share £100 in the proportions of 2 to B and 1 to C. What is the share of each } Ans. B£66|, C £33|. ,, 4. A merchant failing, oWef to B £500, and to C £900; but has only £1100 to meet these demands. How much should each creditor receive } Ans. B £392^, and C £707f 5. Three merchants load a ship with butter ; B gives 200 casks, C 300, and D 400 ; but when they are at sea it is found necessary to throw 180 casks over- board. How much of this loss should fall to the share of each merchant ? Ans. B should lose 40 casks, C 00, and D 80.' 6. Three persons are to pay a tax of £100 accord- ing to their estates. B's yearly property is £800, C^s j £600, and D's £400. How much is each person's share .' Ans. B's is £44^, C's £33^, and D's £22|. 7. Divide 120 into three such parts as shall be to each other as 1,2, and 3 } Ans. 20, 40, and 60. FELLOWSHIP. 293 IS the whole 3ach person's igs to him. £340, and C What is Bfe -=£7 18s. 2ld. be proportional jes, amounting ^50; and they .ns. B £7 105., 1 put in ^£300, ? Ans. B ie78, the proportions ihare of each } ;500, and to C ihese demands. Ans. B £3924, ith batter; B , when they are 80 casks over- 'all to the share se 40 casks, C f £100 accord- ty is £800, C's I person's share .' 2'2|. as shall be to \i D, and 60. ^. A ship worth £900 is entirely lost ; } of it be- Itwiged to B, 1 to C, and the rest to D. What should be the loss of each, £540 being received as insurance .'' Ans. B £45, C £90, and D £225. 9. Three persons have gained £1320 ; if B were to take £6, ought to take £4, and D £2. What is each person's share ? Ans. B's £660, C's £440, and D's £220. 10. B and C have gained £600 ; of this B is to have 10 per cent, more than C. How much will each receive ? Ans. B £3i4f , and C £285f 1 1 . Three merchants form a company ; B puts in £150, and C £260 ; D's share of £62, which they gained, comes to £16. How much of the gain belongs to B, and how much to C ; and what is D's share of the stock ? Ans. B's profit is £16 16*. 7j\d., C's £29 3s. 4^^d. ; and D put in £142 125. 2^d. 12. Three persons join ; B and C put in a certain stock, and D puts in £1090 ; they gain £llO, of which B takes £35, and £29. How much did B and G put in ; and what is D's share of the gain } An^. B put in £8^9 65. ll^^d.y C £687 35. 5^id. ; and D's part of the profit is £46. 13. Three farmers hold a farm in common ; one pays £97 for his portion, another £79, and the third £100. The county cess on the farm amounts to £34 ; what is each person's share of it .? Ans. £11 185. ll^frf. ; £9 145. 7^fd. ; and £12 6s. A\%d. 23. Compound Fellowship. — ^Rule. — Multiply each persoiii's stock by the time during which it has been in trade j and say, as the sum of the products is to the whole gain or loss, so is each person's product to his share of the gain or loss. Example. — ^A contributes £30 for 6 months, B £84 for 11 months, and C £96 for 8 months ; and they lose £14: What is C'« share of this loss ? 30 X 6=180 for one month. ) 84x11=^924 for ofie month. > =£1872 for one month. 96 X 8=768 for one month. \ 1872 : £14 : : £768 : ^1^^^38£6 Is. 4^^., C's bhare n' ' 294 FELLOWSHIP. 24. Reason or the Rule. — It is clear that £30 contributed for 6 months are, as far as the gain or the loss to be derived from it is concerned, the same as 6 times £30— or £180 con- tributed for 1 month. Hence A's contribution may be taken as £180 for 1 month ; and, for the same reason, B's as £924 for the same time; and C's as £768 also for the same time This reduces the question to one in simple fellowship [22]. EXERCISES. ■<*■ i ■"!■ 1 w ii i 1: 14. Three merchants enter into partnership ; B puts in ^89 5s. for 5 months, C £92 15^. for 7 months, ana D iess 10*. for 11 months; and ihey gain £86 16s. What should be each person's share of it ? Ans. B's £25 105., C's £37 2s., and D's £24 4s. 16. B, C, and D pay £40 as the year's rent of a farm. B puts 40 cows on it for 6 months, C 30 for 5 months, and D 50 for the rest of the time. How much of the rent should each person pay } Ans. B £21/^, C £13y\, andD£4yV 16. Three dealers. A, B, and C, enter into partnership, and in a certain time make £291 13s. Ad. A's stock, £150, was in trade 6 months ; B's, £200, 3 months ; and C's, £125, 16 months. What is each person's sliare of the gain ? Ans. A's is £75, B's £50, and C* £166 135. Ad. 17. Three persons have received £665 interest ; B had put in £4000 for 12 months, C £3000 for 15 months, and D £5000 for 8 months ; how much is each person's part of the interest } Ans. B's £240, C's £225, and P's £200. 18. X, y, and Z form a company. X's stock is in trade 3 moiiths, and he claims -^^ of the gain ; Y's stock is 9 months in trade ; and Z advanced £^56 for 4 months, and claims half the profit. How much did X and Y contribute ? Ans. X £168, and Y £280. ' It follows that Y's gain wjjs -j^. Then ^ : -^ : : iC756x4 : 604=X's product, which, being divided oy his number of months, will give £168, as his contribution. Y's share of t: stock may be found in the same way. | 9. Three troops of horse rent a field, for which they \ pay £80 ; the first se&t into it 56 horses for 12 days, the i :30 contributed 1 to be derived -or £180 con- i may be taken ,n, B'8 as £924 the same time wship [22]. FELLOWSHIP. 295 I srsbip ; B puts 7 months, ana gain ^£86 165. it ? Ans. B's I rent of a farm. I for 5 months, 3W much of the nto partnership, 4d. A's stock, ), 3 months ; and >erson's share of , and C'« iei66 565 interest; B for 15 months, is each person's C's i2225, and X's stock is in , the gain; Y's Ivanced ^£756 for How much did ^nd Y ie280. :JL:: £756x4: rhis number of Ion. Y'B share of , for which they for 12 days, the second 64 for 15 days, and the third 80 for 18 days. What must each pay .? Ans. The first must pay i217 10*., the second ^£25, and the third iB37 10*. 20. Three merchants are concerned in a steam vessel ; the first, A, puts in ^£240 for 6 months ; the second, B, a sum unknown for 12 months ; and the third, C, ^£160, for a time not known when the accounts were settled. A received jgSOO for his stock and profit, B £600 for his, and C £260 for his ; what was B's stock, and C's time } Am. B's stock was £400 ; and C's time was 15 months. If £300 arise from £240 in 6 months, £600 (B's stock and profit) will be found to arise from £400 (B's stock) in 12 months. Then £400 : £160 :: £200 (the profit on £400 in 12 months) : £80 (the profit on £160 in 12 months). And £160+ 80 (£160 with its profit for 12 months) : £260 (£160 with its profit for some other time) : : 12 (the number of months 260x12 in the one case) : -i/^q iqq (the number of months in the other case)=13, the number of months required to produce the difference between £160, C's stock, and the £260, which he received. 21. In the foregoing question A's gain was £60 during 6 months, B's £200 during 12 months, and C'a £100 during 13 months ; and the sum of the products of their stocks and times is 8320. What were their stocks > Ans. A's was £240, B's £400, and C's £160. 22. In the same question the sum of the stocks is £800 ; A's stock was in trade 6 months, B's 12 months, and C's 15 months; and at the settling of accounts, A is paid £60 of the gain, B £200, and (J £100. What was* each person's stock.? Ans. A's was £240, B's £400, and C's £160. QUESTIONS. 1. What is fellowship .? [21]. 2. What is the difference between single and double fellowship ; and are these ever called by any other names.? [21]. 3. What are the rules for single, and double fellow- ship ? [22 and 23] . f ■♦' I «■■■■' ', m ml II t'- 296 BARTER. BARTER. 25. Barter enables tlie merchant to exchange one commodity for another, without either loss or gain. KuLE. — ^Find the price of the given quantity of one kind of merchandise to be bartered ; and then ascertain how much of the other kind this price ought to pur- chase. Example 1. — How much tea, at 8s. per ft, ousht to be given for 3 cwt. of tallow, at £1 16s. 8a. per cwt. 1 £ s. d, . 1 16 8 -a if ir m 6 10 is the price of 3 cwt. of tallow. And £5 10s.-f-8s.=13^, is the number of pounds of tea which £5 10s., the price of the tallow, would purchase. There must be so many pounds of tea, as will be equal to the number of times that 8s. is contained in the price of the tallow. Example 2. — I desire to barter 96 lb of sugar, which cost me Sd. per lb, but which I sell at 13(Z., giving 9 months' credit, for calico which another merchant sells for lid. per yard, giving 6 months' credit. How much calico ought I to receive ? I first find at what price I could sell my sugar, were I to give the same credit as he does — If 9 months give me 5d. profit, what ought 6 months to give 9 : 6 ^J?=.3i^. 9 9 "* Hence, were I to give 6 months' credit, I should charge lljfi. per lb. Next — As my selling price is tO my buying price, so ought his selling to be to his buying price, both giving the same credit, 17 :5> uld purchase. IB will he equal to in the price of the ■ > of sugar, which at 13d., giving 9 merchant sells for ; How much calico ij sugar, were I to i mght 6 months to t, I should charge )rice, so ought his ng the same credit 96x8d., or 768(i.i of yards. EXERCISES. 1 . A merchant has 1200 stones of tallow, at 2s. 3\d, the stone ; B has 110 tanned hides, weight 3994 tb, at 5f rZ. the lb ; and they barter at these rates. How much money is A to receive of B, along with the hides } Ans. £40 lis. 2\d. 2. A has silk at I4s. per !b ; B has cloth at 12^. 6^. which cost only 10*. the yard. How much must A charge for his silk, to make his profit equal to that of B ? Ans. 17 s. 6d. 3. A has coffee which he barters at lOd. the lb moro than it cost him, against tea which stands B in lO^-., but which he rates at 125. Gd. per ft). How much did the coffee cost at first .? Ans. 3s. Ad. 4. K and L barter. K has cloth worth 8*. the yard, which he barters at 9*. 3d. with L, for linen cloth at S.s*. per yard, which is worth only 2s. Id. Who has the advantage ; and how much linen does L give to K, for 70 yards of his cloth } Ans. L ^ives K 215f yards ; and L has the advantage. 5. B has five tons of butter, at i£25 IO5. per ton, and IQi tons of tallow, at i£33 lbs. per ton, which he barters with C ; agreeing to receive ^150 Is. 6d. in ready money, and the rest in beef, at 21*. per barrel. Hoi|f many barrels is he to receive .'' Ans. 316. 6. I have cloth at Sd. the yard, and in barter charge for it at I3d.j and give 9 months' time for payment ; another merchani. has goods which cost him 12^. per lb, and with which he gives 6 months' time for payment. How high must he charge his goods to make an equal barter ? Ans. At I7d. 7. I barter goods which cost 8d. per ft), but for which I charge 13^., giving 9 months' time, for goods which are charged at 17 d., and with which 6 months' time are given. Required the cost of what I receive ? Ans. I2d. 8. Two persons barter ; A has sugar at 8d. per lb, charges it at 13^., and gives 9 months time ; B has at I2d. per lb, and charges it at I7d, per ft). How time must B give, to make the barter equal ? 6 months. El;? 298 ALLIGATION. QUESTIONS. U' : I ' 1 i' I i$ m^.W^ 1. What is barter ? [25]. 2. What is the rule for barter ? [25]. ALLIGATION. 26. This rule enables us to find what mixture will be produced by the union of certain ingredients — and then it is called alligation medial; or what ingredients will be required to produce a certain mixture — when it is termed alligation alternaie ; further division of the subject is unnecessary : — it is evident that any change in the amount of one ingredient of a given mixture must produce a proportional change in the amounts of the others, and of the entire quantity. 27. Alligation Medial. — Given the rates or kinds and quantities of certain ingredients, to find the mixture they will produce — Rule. — Multiply the rate or kind of each ingredient by its amount ; divide the sum of the products by the tfumber of the lowest denomination contained m the whole quantity, and the quotient will be the rate or kind of that denomination of the mixture. From this may be found the rate or kind of any other denomination. Example 1. — ^What ought to be the price per ft, of a mixture containing 98 lb of sugar at 9a. per lb, 87 lb at 5f/., and 34 lb at 6e d. d. 9x98 = 882 6x87 — 435 6x34 = 204 219 219)1521 Ans. Id. per lb, nearly. Thp price of each 8iigar» is th^ number of pence per pound inultiplied' by the number of pounds ; and the price of the whole is t)ie siim of the pric98. ^ut if 219 lb of sugar have cost 1521c?., one !b, or the 219th part of this^ iQU9t cost the 219th part of 1521d., or ^^^d. a=7d., nearly. :; !t' ALLlOATIOrr. 399 nixture will be mtfl — and then Qgredients will re — when it is ivision of the lat any change given mixture I the amounts rates or kinds nd the mixture jach ingredient roducts bj the itained in the )e the rate or •e. From this : denomination. oe per ft), of a )er ft), 87 ft) at learly. )ence per pound ;he price of the b of sugar have mii^t cost the M E:iFAMPLE 2. — What will be the price per ft) of a mixture oontaininc 9 tt> 6 oz. of tea at 5s. 6d. per ft), 18 ft) at 6&. per ft), and 46 ft) 3 oz. at 9s. 4^d. per ft) f ft) oz. s. 9 6 at 5 6 3 9 18 46 d. £ 8. 6 per ft)= 2 11 per lb= 5 8 4J per ft)=21 13 d. I' 73 16 9 63 1177 )29 12 Ans. Qd. per oz. nearly. 1177 ounces. And 6rf.xl6=8s., is the price per pound. In this case, the lowest denomination being ounces, we reduce the whole to ounces ; and having found the price of an ounce, we muitiply it by 16, to find that of a pound. Example 3. — A goldsmith has 3 ft) of gold 22 carats fine, and 2 lb 21 carats fine. What will be tne fineness of the mixture *? In this case the value of each kind of ingredient is repre- sented by a number of carats — tt)8 3x22 = G6 2X21 = 42 i 5 )108 The mixture is 21} carats fine. EXERCISES. 1. A vintner mixed 2 gallons of 'jdne, at 145. per gallon, with 1 gallon at 125. , 2 gallons at 95., and 4 gallons at 85. What is one gallon of the mixture worth ? Am. 105. 2. 17 gallons of ale, at 9^. per gallon, 14 at 7^d., 5 at 9^d., and 21 at 4^^., are mixed together. How much per gallon is the mixture worth } Ans. 7-^d. 3. Having melted together 7 oz. of gold 22 carats fine, 12^ oz. 21 carats fine, and 17 oz. 19 carats fine, I wish to know the fineness of each ounce of the mixture ? Ans. 20|f carats. 28. Alligation Alternate.— Given the nature of the mixture, and of the ingredients, to find the relative amounts of the latter — Rule. — ^Put down the quantities greater than the given mean (each of them connected with the difference 300 ▲LUQATION. 1 !■■ l-ii 1 ^^ II.' f ' 'I* i' i illva: between it and the mean, by the sign — ) in one column ; put the differences between the remaining quantities and the mean (connected with the quantities to which they belong, by the sign + ) in a column to the right hand of the former. Unite, by a lino, each plus with some minus difference , and then each difference will express how much of the quantity, with whose difference it is connected, should be taken to form the required mixture. If any difference is connected with more than one other difference, it is to be considered as repeated for each of the differences with which it is connected ; and the sum of the differences with which it is connected is to be taken as the required amount of the quantity whose difference it is. Example 1.— How many pounds of tea, at 5?. and 85. per lb, would form a mixture worth 7s. per !b ? Prico. DifTerences. Price. S. S. The mean=8— 1- s. s. 2-f-5=the mean. 1 is connected with 2s., the difference between the mean and 5s. ; hence there must be 1 lb at 5,9. 2 is connected with 1, the difference between Ss. and the mean ; hence there must be 2 lb at 8s. Then 1 lb of tea at 5s. and 2 lb at 8s. per lb, will form a mixture worth 7s. per lb — as may be proved by the last rule. It is evident tb^t any equimultiples of these quantities would answer equally well ; hence a great number of answers may be given to such a question. Example 2. — How much sugar at 9d., Id., 5d., and lOd , will produce sugar at 8^. per lb ? Frlffes. DilTerences. Prices. The mean=s d. d. d. d. 9_1 14.7) 10-2 34.5 J =*^® "®*^' 1 is connected with 1, the difference between Id. and the mean ; hence there is to be 1 lb of sugar at Id. per lb. 2 is connected with 3, the difference between bd. and the mean ; hewie there is to be 2 lb at 5rf. 1 is connected ^vith 1, the difference between 9rf. and the mean ; hence there is to be 1 lb at 9(/. And 3 is connected with 2, the difference between lOd. and the mean; hence there are to be 3 lb at 10c2. per K). I h ALLIGATION. 801 in one column ; ning quantities itities to whioh m to the right 1 plus with some ace will express difference it is quired mixture. L more than one as repeated for 3onnected ; and is connected is [>f the quantity Eit 5$. and Ss. per e mean. atween the mean 2 is connoctcd Ban ; hence there and 2 lb at 8^. ft) — as may be these quantities imber of answers d., 5c2., and lOd , le mean. reen Id. and the Id. per lb. 2 is '. and the mean ; Btcd ''vith 1, the e there is to be Serence between be 3 lb at lOd. Consequently we are to take 1 lb at Id., and 2 lb at 5J.. 1 tb at 9d.j and 3 n> at 10c/. If we examine what mixture these will give [27], we shall find it to be the given mean. Example 3. — What quantities of tea at 45., 6«., 8i., and 9s. per lb, will produce a mixture worth 55. ? Pricei. Difterenoei. Prices. The mean= :the mean. 3, 1, and 4 are connected with Is., the differeoe between 4s-. and the moan ; thorefore we are to take 3 lb -f* 1 lb + 4 ]h of tea, at 4s. per lb. 1 is connected with Ss., Is., and 45., tho differences between 85., 6s., and 95., and the mean; tlutrefore we are to take 1 lb of tea at 85., 1 lb of tea at G5., and 1 lb of tea at 95. per lb. Wo find in this example that 85., 65., and 95. are all con- nected with tlie same 1 ; this shows that 1 lb of each will be required. 45. is connected with 3, 1, and 4; there must be, therefore, 3-|-l-|-4 lb of tea at 45. Example 4. — How much of anything, at 35., 45., 55., 75., 8.$., 95., II5., and 125. per lb, would form a mixture worth 0,?. per lb 1 • Prices. Diflerences. Pricei. 5. 5. 5. 5. 1 lb at 35., 2 lb at 45., 3 lb at 75., 2 lb at 85., 34-5-1-6 (14) !b at 55., 1 lb at 95., 1 lb at lis., and 1 lb at 125. per tt>, will form the required mixture. 29. Reason of the Rule. — The excess of one ingredient above the mean is made to counterbalance what the other wants of being equal to the mean. Thus in example 1, 1 lb at 55. per lb gives a deficiency of 25. : but this is corrected by 25. excess in the 2 lb at 85. per lb. In example 2, 1 lb at 7d. gives a deficiency of Id, 11^ at 9d. gives an excess of Id'. ; but the excess of Id. and ihe deficiency of Id. exactly neutralize each other. Again, it Is evident that 2 lb at 5^. and 8 lb at lOd. are worth ju6t as much a3 6 lb at 8d. — that is, Sd. will be the average price if we mix 2 lb at 6d. with 3 lb at lOd. O 2 302 ALLIGATION. ■;!)• i ^^ EXERCISES. 4. How much wine at 8*. 6d. and 9*. per gallon will make a mixture worth 8*. lOd. per gallon } Ans. 2 gallons at Ss. 6d.y and 4 gallons at 95. per gallon. 5. How much tea at 6s. and at 35. 8d. per lb, will make a mixture worth 45. 4d. per lb ? Ans. 8 ib at 6s. J and 20 ft) at 35. 8d. per lb. 6. A merchant has sugar at dd., lOd., and 12d. per ft). How much of each kind, mixed together, will be worth 8^. per ft) } Ans. 6 tt> at 5^., 3 lb at lOd.y and 3 ft) at I2d. 7. A merchant has sugar at 5dJ., 10^., I2d.y and 16^:? per ft). How many ft) of each will form a mixture worth lid. per ft) ? Ans. 5 ft) at 5d.y 1 ft) at lOd.^ 1 ft) at 12d., and 6 ft) at IG^?. 8. A grocer has sugar at dd.^ 7d.y 12d.y and 13d. per ft). How much of each kind will form a mixture worth lOd. per lb ? Atis. 3 ft) at 5i., 2 ib at7rf., 3 ft) at 12^?., and 5 ft) at 13d. 30. When a given amount of the mixture is required, to find the corresponding amounts of the ingredients — Rule. — Find the amount of each ingredient by the last rule. Then add the amounts together, and say, as their sum is to the amount of any one of them, so is the required quantity of the mixture to the corresponding amount of that one. Example 1. — What must be the amount of tea at 45. per tt>, in 736 lb of a mixture worth 55. per ft>, and contaimng tea at 6s., 8s., and 95. per tt) ? To produce a mixture worth 5s. per ft), we require 8 ft) at 4s., 1 at 8s., 1 at 6s., and 1 at 9s. per lb. [281. But all of these, added together, will make 11 lb, in which there are 8 lb at 45. Therefore ft) 11 ft) :8 ft) 736 ft) 8x736 11 '' ft) oz. =526 4|^, the required quantity of tea at 45. That is, in 736 ft) of the mixture there will be 536 ft> 4^ oz. at 45. per ft). The amount of each of the other ingre- die its may be found in the same way. i ALLIGATION. 303 er gallon will on? Ans. 2 gallon. I. per lb, will Ans. 8 S) at and I2d. per jther, will be • at 10^., and 2d.y and 16d nixtnre worth lOd.j 1 fi) at 2d.^ and 13^. rm a mixture lbat7d., 3 1b re is required, ingredients— redient by the r, and say, as hem, so is the corresponding ■ tea at 4s. per tnd containing e require 8 lb 281. But all wnich there lired quantity be 536lb4y«^ e other ingre- Example 2. — Hiero, king of Syracuse, gave a certain quantity of gold to form a crown ; but when he received it, suspecting that the goldsmith had taken some of the gold, and supplied its place by a baser metal, he commissioned Archimedes, the celebrated mathematician of Syracuse, to ascertain if his suspicion was well founded, and to what extent. Archimedes was for some time unsuccessful in his researches, until one day, going inito a bath, he remarked that he displaced a quantity of water equal to his own bulk. Seeing at once that the same weight of different bodies would, if immersed in water, displace very different quan- tities of the fluid, he exclaimed with delight that he had found the desired solution of the problem. Taking a mass of gold equal in weight to what was given to the goldsmith, he found that it displaced less water than the crown ; which, therefore, was made of a lighter, because a more bulky metal — and, consequently, was an allay of gold. Now supposing copper to have been the substance with which the crown was adulterated, to find its amount — Let the gold given by Hiero have weighed 1 lb, this would displace about -052 lb of water ; 1 lb of copper would displace about -1124 lb of water j but let the crown have displaced only -072 lb. Then Gold differs from 072, the mean, by — 020. Copper differs from it by . . 4-*0404. , Copper. Differences. 0»ld. Hence, the meansK=- 1124 --0404 •020-f-052=«the mean. Therefore '020 lb of copper and -0404 lb of gold would produce the alloy in the crown. But the crown was supposed to weigh 1 lb j therefore •0604 lb (020+0404) : -0404 Mb : : 1 lb : :0404-f 1 ft t=-669 lb, the quantity of gold. And the quantity of copper. •0604 669=-331 lb is EXERCISES. 9. A druggist is desirous of producing, from medicine at 5*., 6s., 8*., and 9s. per ft), 1^ cwt. of a mixture worth 75. per lb. How much of each kind must he use for the purpose } Ans. 28 lb at 5s., 56 ft) at 6*., 56 ft) at 8s., and 28 ft) at 9s. per ft). 10. 27 ft) of a mixture worth 4*. 4d. per ft) are re- qiured. It is to contain tea at 5^. and at 35. Qd, per 304 ALLIGATION. il lb. How much of each must be used ? Am. 15 ft) at bs.y and 12 ft) at 35. 6d. 11. How much sugar, at 4d.y 6d., and Sd. per ft), must there be in 1 cwt. of a mixture worth 7d. per ft) } A71S. 18| ft) at 4d.y 18f ft) at 6d., and 74| ft) at 8^. per ft). 12. How much brandy at 12*., 13*., 14*., and 14*. 6d. per gallon, must there be in one hogshead of a mix- ture wortji I3s. Qd. per gaiiun ? Ans. 18 gals, at 12*., 9 gals, at 135., 9 gals, at 14s., and 27 gals, at 14*. 6^. per gallon. 31. When the amount of one ingredient is given, to find that of any other — Rule. — Say, as the amount of one ingredient (found by the rule) is to the given amount of the same ingredient, so is the amount of any other ingredient (found by the rule) to the required quantity of that other. Example 1. — 29 lb of tea at 4s. per lb is to be mixed with teas at 6s., 8s., and 9s. per lb, so as to produce what will be "Yorth 5s. per lb. What quantities must be used ? 8 2b of tea at 4s., and 1 lb at 6s., 1 lb at 8s., and 1 lb at 9s., will make a mixture worth 5s. per lb [271. Therefore 8 lb (the quantity of tea at 4s. per lb, as found "by the rule) . 29 lb (the given quantity of the same tea) : : 1 lb (tne quantity of tea at 6s. per ib, as found by the rule) : 1x29 lb 8 (the quantity of tea at 6s., which corresponds with 29 lb at 4s. per lb)=3| lb. We may in the same manner find what quantities of tea at 8s. and 9b. per lb correspond with 29 fib— or the given amount of tea at 4s. per lb. Example 2.— A refiner has 10 oz. of gold 20 carats fine and melts it with 16 oz. 18 carats fine. What must b« added to make the mixture 22 carats fine ? 10 oz. of 20 carats fine==10x20 = 200 carats. 16 oz. of 18 carats fine=16 x 18 = 288 26 : 1 ::488 : 18}§ carats, the fineness of the mixture. 24 — 22=:2 carats baser metal, in a mixture 22 carats fine. 24 — 18j§s=:5j\ carats baser metal, in a mixture 18} J carats fine. Then 2 carats : 22 carats : : 5^^ : 57^^ carats of pure ■3 Ans. 15 ft) at ad 8d. per lb, th 7d. per ib ? 74| ft) at 8^. 14*., and 14*. head of a mix- 8 gals, at 125., Is. at 14*. 6d. nt is given, to redient (found ime ingredient, (found by the er. 3 be mixed with Lce what will be used! Ss., and 1 lb at 271. Therefore ad Dy the rule) . : : 1 lb (the . 1x29 lb rule) : 8 ds with 29 R) at antities of tea at he given amount I 20 carats fine What must b< 'ats. 16|| carats, the re 22 carats fine. I mixture 18-|^ f carats of pure ALLIGATION. 305 gold— required to change 5A carats baser metal, into a mixture 22 carats fine. But there are already in the mixture 18 [^ carats gold; therefore 57yV-18i^=38i§ carats gold are to be added to every ounce. There are 20 oz. ; therefore 26x38} 5=1008 carats of gold are wanting. There are 24 carats (page 5) in every oz. ; therefore ^f y" c{irats=42 oz. of gold must be added. There will then" be a mixture containing oz. car. car. 10x20 = 200 16x18 = 288 42x24 =r 1008 68 : 1 oz. :: 1496 : 22 carats, the required 'Oneness. EXERCISES. 13. How much tea at 6*. per ft) must be mixed with 12 ft) at 35. Sd. per ft), so that the mixture may be worth 45. 4d. per fb .^ Aiis. 4f ft). 14. How much brass, at I4d. per ft), and pewter, at lO^d. per ft), must I melt with 50 ft) of copper, at \6d, per ft), so as to make the mixture worth l5. per ft) ? Ans. 50 ft) of brass, and 200 ft) of pewter. 15. How much gold of 21 and 23 carats fine ^ust be mixed with 30 oz. of 20 carats fine, so that the mix- ture may be 22 carats fine r Ans. 30 of 21, and 90 of 23. 16. How much wine at 75. 5d., at 55. 2^., and at 45. 2d. per gallon, must be mixed with 20 gallons at 65. Sd. per gallon, to make the mixture worth 65. per gallon } Am. 44 gallons at 75. 5<^., 16 gallons at 65. 2d., and 34 gallons at 45. 2d. QUESTIONS. 1. What is alligation medial .? [26]. 2. What is the rule for alligation medial ^ [27] . 3. What is alligation alternate .'' [26]. 4. What is the rule for alligation alternate ? [28] . 5i What is the rule, when a certain amount of the mixture is required } [30] . 6. What is the rule, when the amount of one or moro of the ingredients is given ? [31]. 306 SECTION IX. INVOt^UTION AND EVOLUTION, &c. 1. Involution. — A quantity which is the product of two or more factors, each of them the same number, is termed a power of that number ; and the number, mul- tiplied by itself, is said to be involved. Thus 5X6X5 (=125) is a " power of 5 ;" and 125, is 5 " involved." A power obtains its denomination from the number of times the root (or quantity involved) is taken as a factor. Thus 25 (=5X5) is the second power of 5. — The second power of any number is also called its square ; because a square surface, one of whose sides is expressed by the given number, will have its area indicated by the second power of that number ; thus a square, 5 inches every way, will contain 25 (the square of 5) square inches ; a square 5 feet every way, will contain 25 square feet, &c. 216 (6x6X6) is the third power of 6. — The third power of any number is also termed its cube ; because a cube, the length of one of whose sides is expressed by the given number, will have its solid contents indicated by the third power of that number. Thuei a cube 5 inches every way, will contain 125 (the cube of 5) cubic, or solid inches ; a cube 5 feet every way, will contain 125 cubic feet, &c. 2. In place of setting down all the factors, we put down only one of them, and mark how often they are supposed to be set down by a small figure, which, since it points out the number of the factors, is called the index^ or exponent. Thus 5^ is the abbreviation for 6X5 : — and 2 is the index. 5* means 6X6X5X6X6, or 5 in the fifth power. 3* means 3X3X3X3, or 3 in the fourth power. 8'' means 8X8X8X8X8X8X8, or 8 in the seventh power, &c. 3. Sometimes the vinculum [Sec. II. 5] is used in con- junction with the index ; thus 5+8^ means that the sum of 6 and 8 is to be raised to the second power — ^this :! i i INTOLUTION. 307 &c. le product of e number, is Qumber, mul- ^hus 5X5X5 " involved." le number of jn as a factor. of 5.— The id its square; js is expressed licated by the aare, 5 inches of 5) square 11 contain 25 hird power of Iso termed its Df whose sides have its solid that number, ttain 125 (the J 5 feet every ictors, we put often they are J, which, since is called the )breviation for <5X5X5X5, <3X3, or 3 in <8X8X8X8, is used in cfon- is that the sum d power — ^tihis is very diiferent from 524-8^, which means the sum of the squares of 6 and 8 : d-j-S^* being 1G9 ; while 5^ +8^ is only 89. 4. In multiplication the multiplier may be considered as a species of index. Thus in 187X5, 5 points out how often 187 should be set down as an addend ; and 187X5 is merely an abbreviation for 187+187+187+ 187+187 [Sec. II. 41]. In 187«, 5 points out how often 187 should be set down as a factor ; and 187* is an abbreviation for 187X 187X 187X 187X 187 r—that is, the " multiplier" tells the number of the addends^ and the " index" or " exponent," the number of ike factors. 5. To raise a number to any power — Rule. — Find the product of so many factors as the index of the proposed power contains units — each of the factors being the number which is to be involved. Example 1. — What is the 5th power of 7 ? , . 7» =7x7x7x7x7=16807. " ,t Example. 2. — ^What is the amount of £1 at compound interest, for 6 years, allowing 6 per cent, per annum % The amount of £>\ for 6 years, at 6 per cent, is — 106 X 106 X 106 X 106 X 106x106 [Sec. VII. 20], or ro6"=i-4i852. " ^v '■■ -: '"^ '■"''■' . ■'■■ -• ; We, as already mentioned [Sec. VII. 23], may abridge kU tt, V EXERCISES. 1 8=243 - '1 2*. 20'«=:l6240000000000. ■" ■ ' 3. 3'=2187. - • ' 4. 105«=1340095640626. '"^ ■• ' ; OD 6. 106«=1 -340095640625. :> ih " .U. - 6. To raise a fraction to any power — . \-<: n Rule. — ^Raise both numerator and denominator to 11 that power. ,y:,,iU,, >^j>. ^ > . •* ,. ■ . .;. *.- Example.— (|)3=fx|Xf=J?. To involve a fraction is to multiply it by itself. But to multiply it by itself any number of times, we must multiply its numerator by itself, and also its denominator by itself* that number of times [Seo. IV. 39]. .ttji > r ti;i*<'i^>''fc M 308 EVOLUTION. P • 1.' I '»' EXERCISEB 7 ( .i\i ai87 /';<\7 2 187 U; — fffSBT" / 5 \ 5 :« 1 1! 5 7. To raise a mixed number to any power — Rule. — Reduce it to an improper fraction [Sec. IV 21 ! ; and then proceed as directed by the last rule. •^.1 > ExAMPLE.~(21)*=(f)*=Vlf'- 10 11. 12. 13 EXERCINES. /'112\3 185193 K^^n) — T2;5r • /'Q2S5 04:?r;34 3 \^l) 1^8 If • " " " .22 I t 43fil 129 ^smr ^ • .4 2018442977 B257I? il 8. JEvolution is a process exactly opposite to involution ; since, by means of it, '/e find what number, raised to a given power, would produce a given quantity — the num- ber so found is termed a root. Thus we " evolve " 25 when we tt ke, for instance, its square root ; that is, when we find what number, multiplied by itself, will produce 26. Roots, also, are expressed by ezponenis — but as these exponents are fractions, the roots are called ^^ fractional powers." Thus 4^ means the square root of 4; 4^ the cube root of 4 ; and 4^' the seventh root of the fifth power of 4. Roots are also expressed by ^y called the radical sign. When used alone, it means the square root — thus ^3, is the square root of 3 ; but other roots are indicated by a small figure placed within it — thus ^5 ; which means the cube root of 5. ^^7' (7^), is the cube root of the square of 7. 9. The fractional exponent, and radical sign are some- times used in conjunction with the vinculum. Thus 4—3% is the square root of th« difference between 4 and 3 ; y^5+7, or 5+7*, is the cube root of the sum of 5 and 7. '<■' ';^'---^ -^ i^i'" ' ^••'..'"-. ' ■ ■•" 10. To find the square root of any number — Rule — ^I. Point off the digits in pairs, by dots ; put- ting one dot over the units' place, and then another dot over every second digit both to the right and left of the units' place — ^if there are digits at both sides of the decimal point. i EVOLUTION. 309 ir- on [Sec. IV last rule. *i to involution ; ir, raised to a ity — the num- " evolve" 25 ; thatis, wben ■, will produce I — ^but as these id ^^ fractional t of 4 ; 4* the I the fifth power led the radical ,are root — thus ts are indicated IS ^5 ; which s the cube root sign are some- culum. Thus nee between 4 oot of the sum iiber — by dots ; put- len another dot rht and left of *oth sides of the II. Find tho highest number the square of which will not exceed the amount of the highest period, or that which is at the extreme left — this number will be the first digit in the required square root. Subtract its square from the highest period, and to the remainder, considered as hundreds, add the next period. III. Find the highest digit, which being multiplied into twice the part of the root already found (consi- dered as so many tens), and into itself, the sum of the products will not exceed the sum of the last remainder and the period added to it. Put this digit in the root after the one last found, and subtract the former sum from the latter. * IV. To the remainder, last obtained, bring down another period, and proceed as before. Continue this process until the exact square root, or a sufficiently near approximation to it is obtained. 11. Example.— What is the square root of 22420225 } 22420225(4735, is the required root. 16 ;. 87)642 609 943)3302 2829 9465)47325 47325 22 is the highest period ; and 4^ is the highest square which does not exceed it — we put 4 in the root, and subtract 4^, or 16 from 22. This leaves 6, which, along with 42, the next period, makes 642. We subtract 87 (twice 4 tens-|-7, the highest digit which we can now put in the root) X 7 from 642. This leaves 33, which, along with 02, the next period, makes 3302. We subtract 943 (twice 47 tens -f^? the next digit of the root) x3 from 3302. This leaves 473, which, along with 25, the only remaining period, makes 47325. We subtract 9465 (twice 473 tens -(-5, the next digit of the root) x5. This leaves no remainder. The given number, therefore, is exactly a square ; and its square root is 4735. ^.7 , / , vi^. 12. Reason of I.— We point oS the digits of the given square in pairs, and consider the number of dots as indicating k' :): f 1.1- 310 EVOLUTrOW Mlt t* • the number of digits in the root, since neither one nor tvro digits in the square can give more or less than one in the root ; neither three nor four digits in the square can give more or less than two in the root, &c. — which the pupil may easily ascertain by experiment. Thus 1, the smallest single digit, will give one digit as its square root; and 90, the largest pair of digits, can give only one — since 81, or the square of 0, is the greatest square which does not exceed 99. Pointing off the digits in pairs shows how many should be brought down successively, to obtain the successive digits of the root — since it will be necessary to bring down one period for each new digit ; but more than one will not be required. Reason of II. — We subtract from the highest period of the given number the highest square which does not exceed it, and consider the root of this square as the first or highest 4Jgit of the required root ; because, if we separate any number into the parts indicated by its digits (568, for instance, into 500, 60, and 3), its square will be found to contain the square of each of its parts. Reason of III. — We divide twice the quantity already in the root (considered as expressing tens of the next denomina- tion) into what is left after the preceding subtraction, &c., to obtain a new digit of the root; because the square of any quantity contains (besides the square of each of its parts) twice the product of each part multiplied by each of the other parts. Thus if 14 is divided into 1 ten and 4 units, its square will contain not only 10^ and 4^, but also twice the product of 10 and 4. — We subtract the square of the digit last put in the root, at the same time that we subtract twice the product obtained on multiplying it by the part of the root which pre- cedes it. Thus in the example whic^i illustrates the rule, when we subtract 87X7, we really subtract 2x40x7+72. It will be easily to show, that the square of any quantity contains the squares of the parts, along with twice the pro- duct of every two parts. Thus 22420225=4736^—4000+700+30+6^. 4000 "=16000000 6420225 " )■. <■■ 2X4000X700+700^= 6090000 ,. . ' " ' 330225 ; 2X4000X30+2X700X30+30'= 282900 47325 j:. ^m. j', ,'i r - :>, 2X4000X6+2X700X6+2X30X5+6«=47325 Reason of IV. — Dividing twice the quantity already in the root (considered as expressing tens of the next denomi- nation) into the remainder of the given number, &c., gives the next digit; because the square contains the sum of twice the products (or, what is the same thing, the product k ^-1 EVOLUTION. 311 er ono nor tvro )ne in the root ; in give more or ipil may easily !8t single digit, the largest pair square of U, is many should be jessive digits of lown one period be required. 58t period of the s not exceed it, first or highest rate any number or instance, into Qtain the square mtity already in 3 next denomina- atraction, &c., to e square of any tch of its parts) each of the other units, its square ;wice the product ) digit last put in twice the product 5 root which pre- strates the rule, X40X7+72. I of any quantity th twice the pro- 1000+700+30+6 a Eintity already in the next denomi- umber, &c., gives lins the sum of thing, the product of twice the sum) of the parts of the root already found, multiplied by the new digit. Thus 22420225, the square of 4735, contains 4000=*-|-700"-f-30^+53 ; and a/«o twice 4000X 700 4- twice 4000 X 30 + twice 4000x5 ; plus twice 700x30+ twice 700x5; plus twice 30X5: — that is, tlie square- of etich of its parts, with the sum of twice the product of every two of them (which is the same as each of them nnO*\..ied by twice tlie sum of all the rest). This would, on examination, be found the case with the square of any other number. If we examine the example given, we shall find that it will not be necessary to bring down more than one period at a time, nor to add cyphers to the quantities subtracted. 13. When the given square contains decimals — If any of the periods consist of decimals, the digits in the root obtained on bringing down these periods to the remainders will also be decimals. Thus, taking the example just given, but altering the decimal point, we sh all have ^2 24202- 25 =47 3- 5; ^224 2-0225=47-35; ^22-420225 = 4735 ; ^ 22420225 = 4735 ; and y^0022420225= -04735, &c. : this is obvious. If there is an odd number of decimal places in the power, it must be made ex'en by the addition of a cypher. Using the same figures, ^2242022-5=1497-338, &c. 2242022 -50 (1497-388, &o r 1 24)124 96 289)2820 2601 2987)21922 20909 i <^ 29943)101350 89829 299463)1152100 898389 2994668)25371100 23957344 1418756 In this case the highest period consists but of a single digit- ami the given number is not a perfect square. There must be an even number of decimal places ; since m numbw of decimals in the_root will produce an odd numbex in th» aquaxe [See. II. 48]— as may be proved by experiment 312 EVOLUTION. 11 'if, EXERCISES. 5 if ,, i rf ' f 1 •• 1 14. y 195364=442 16. ^328829=578 16. y - 0676= -26 17. ^87- 65=9 -3622 18. y 86 1=2 9 -3428 19. y 984064=992 20. y5=2- 23607 21. ^•6=-707106 22. ^91 '9681=9-59 23. ^238144==-188 24. ^32 -^62=5 -69 25. y* 881776=: -676 14. To extract the square root of a fraction — Rule. — Having reduced the fraction to its lowest terms, make the square root of its numerator the nume- rator, and the square root of its denominator the deno- minator of the required root. Example. — ^J=f. 16. Reason of the Rule. — The square root of any quan< tiiy must be such a number as, multiplied by itself, will pro- duce that quantity. Therefore ^ is the square root of 4 ; for g X ^=^- The same might be shown by any other example. Besides, to square a fraction, we must multiply its numera- tor by itself, and its denominator by itself [6] ; therefore, to take its square root — that is, to bring back both numerator and denominator to what they were before — we miuit take the square root of each. 16. Or, when the numerator and denominator are not squares — Rule. — Multiply the numerator and denominator together ; then mak^ the square root of the product the numerator of the require! root, and the given denomi- nator its denominator ; or make the square root of the product the denominator of the required root, and the given numerator its numerator. Example. — ^What is the square root of | ? (|)i s=« 6 6X4 =4-472136-5-5=s-894427. 17. We, in this case, only multiply the numerator and denominator by the same number, and then extract the square 4 4x6 4x4 /4\i root of each product. For ^=jr;;j^, or ^;^. Therefore r? J 6~6X6' 5X4' 4 >5X6/ \5X4/ VSxT :a;' j ;1 1 ii . ^J f. r >ti EVOLUTION. 313 lommator are ! 18. Or, lastly— Ki'LE. — lloduco the given fraction to a docimal 'Sue IV. 03J, and extract its square root [13j EXERCIHEH. 29. 30. 74535a 80002-) 1 31. /5 \i 845154J 20 / 22 \i_28 -5300852 (,37/ ~ 37 27 /14\i 14 \I0/ "14'9'0002'J5 28. /3\^_ 6 -2 ^998 \i3/ "" 13 19. To extract the square root of a mixed number — llui.K. — Reduce it to an improper fraction, and tliou proceed as already directed [14, &c.] Example.— y2j=y'^=Hli- {tY- EXERCISES. 32. v'^ilJ^Tl 33 r»i 35. yl7|=:4 1GS3 36. y 0^=2 5208 37. yi3'=3-6332 34. v'lc'.=l-01858 20. To find the cube root of any number — Rule — I. Point off the digits in threes, by dots — putting the first dot over the units' place.y and then proceeding both to the right and left hand, if there are digits at both sides of the decimal point. II. Find the hiojhest digit whose cube will not ex- ceed the highest period, or that which is to the left hand sido — this will be the highest digit of the required root ; subtiact its cube, and bring down the next period to the remainder. III. Find the highest digit, which, being multiplied by 300 times the square of that part of the root, already found — being squared and then multiplied by 30 times the part of the root already found — and being multiplied by its own square — the su7n of all the pro- ducts will not exceed the sum of the last remainder and the period brought down to it. — Put this digit in the root after what is already there, and subtract the former ium from the latter. IV. To what now remains, bring down the next mm 314 EVOLUTION. J. ■6'ilU ^> period, and proceed as V)';forc. CoDtinue this process uutil the exact cube root, or a sufficiently near approxi- mation to it, is obtained. Example.— What is the cube root of 179597069288 ? 179597069288(5042, the required root. 125 300x5'x6 30x5x6» O'XO 300x50*x4 30x56x4' 4^X4 300x564«x2 30x564x2» 2'x2 54597 60016 3981069 3790144 190925288 190925288 We find (by trial) that 5 is the first, 6 the second, 4 the third, and 2 the ]ast digit of the root. And the given number is exactly a cube. 21. Reason or I. — We ^aint off the digits in threes, for a reason similar to that which caused us to point them off in twos, when extracting the square root [12]. Reasox or II. — Each cube will be found to contain the cube of each part of its cube root. Reason of III. — ^The cube of a number divided into any two parts, will be found to contain, besides the sum of the cubes of its parts, the sum of 3 times the product of each part by the other part, and 3 times the product of each vart by the square of the other part. This will appear fro»> the following : — - ■''< ' . 179597069288 5000'=125000000000 „ J » 54697069288 8 X6P00'x000+3 X5000x600»-f600'= 60616000000 ■ff.i. : 3x560(yx40-f 3x6600x40»-i-40»=s: ' : 8x5640'X2-|-3x5640x2'4-2'== 3981069288 3790144000 190925288 190925288 Hence, to find the second digit of the root, we must find by trial some number which — being multiplied by 3 times the square of the part of the reot already found — ^ita square being EVOLUTION. 315 3 this process near approxi- )7069288 1 required root. e second, 4 the And the given in threes, for a oint them off in to contain the [ivided into any the sum of the product of «ach lot of each part Etppear fron. the 697069288 000000000 697069288 616000000 981069288 790144000 190925288 190925288 we must find by by 3 times the its square being multipUod by 3 times the part of the root already found — and being multiplied by the squiire of itself — the sum of the pro- ducts will not oxceod what remains of the given number. Instead of considering the part of tlie root already lound as so many tens [12] of tlie denomination next following (iis it really is), which would add one cypher to it, and two cypliei's to its square, we consider it as so many units, and multiply it, not by 3, but by BO, and its square, not by 3, but by 300. For 300 X 5* X 6 4-80 X X (V^-fO'XO is the same thing as 8X^>0'XH-|-3X50X6*+0'X0; since we only change the post- Hon of the factors 100 and 10, which does not alter the product [Sect, II. 35]. It is evidently unnecessary to bring down more than one period at a time ; or to add cyphers to the subtrahends. Urahon of IV. — The portion of the root already found may be treated as if it were a single digit. Since into whatever two parts we divide any number, its cube root will contain the cube of each part, with 8 times the square of each multi- plied into the other. 22. When there are decimals in the given cube — If any of the periods consist of decimals, it is evident that the digits found on bringing down these periods must be decimals. Thus ^179597-069288=56-42, &o. When the decimals do not form complete poriods, the periods are to be completed by the addition of cyphers. Example. — What is the cube root of '3 ? 0-300(-669, &o. 216 800X6»X6 80X6X6* 6X6' 800X66'X9 80X66X9' 9X9' •669, &c. And 84000 =71496 12504000 =11922309 581691, &e. ^•3aBc'669, &c. And -8 is not exactly a cube. It is necessary, in this case, to add cyphers ; since one decimal in the root will give 3 decimal places in the cube ; two decimal ^laces in the root will give six in the cube, &c. [Sec. II. 48.] EXERCISES. 38. y38«=3- 207534 39. y 39=3 •391211 40. y 212=5 -96 2731 41. ^123505992=498 42. yi90109376=67d 43. y 458314011= 771 44. ^ 483- 736 625=7 -85 45. y;63605a=-86 46. 3^ 999=9 •996 666 47. ^•979146667=-998 W':'n' ff'l III m If- 'I*' ^i l';V- 1 316 EVOLUTION. Example. 23. To extract tlie cube root of a fraction — Rule. — Having reduced the given fraction to ita lowest terms, make the cube root of its numerator the numerator of the required fraction, and the cube root of its denominator, the denominator. 24. Reason of the Rule. — The cube root of any number must be such as that, taken three times as a factor, it will produce that number. Therefore f is the cube root of yfj ; /or f X f X s = tIt- — ^^® same thing might be shown by any other example. Besides, to cube a fraction, we must cube both numerator and denominator ; therefore, to take its cube root — that is to reduce it to what it was before — we must take the cube root of both. 25. Or, when the numerator and denominator are not cubes — Rule. — Multiply the numerator by the square of the denominator ; and then divide the cube root of the pro- duct by the given denominator ; or divide the given numerator by the cube root of the product of the given denominator multiplied by the square of the given numerator. Example. — ^What is the cube root of ^ ? ^ 5-277632-1.7 (3)«=^ixr „, „ ,, . . -753947. ^7x3« This rule depends on a principle already explained [16]. 26. Or, lastly— Rule. — Reduce the given fraction to a decimal [Sec. IV. 63] , and extract its cube root [22] . 48. 49. 60. 8-653497 \dJ 9 \11/ S-O EXERCISES. 61. ay^^ 604079 651725 8 ay- 62. / 3 \ * ay- 941036 560907 63. /2\i C^)' •472163 27. To find the cube root of a mixed number — Rule. — Reduce it to an improper fraction ; and then proceed as already directed [23, &c.] Example.— ^3^=^"^=l-54. ; - ''t ■ '1 EVOLUTION. 317 ion — Taction to ita numerator tbe the cube root EXERCISES. t of any number a factor, it will ube root of y ^j? ' be shown by any til numerator and -that is to reduce be root of both. enominator are le square of tbe root of the pro- ivide tbe given act of tbe given 3 of the given 2 ^7 = -753947. plained [16]. 1 to a decimal [22]. =•941036 =:=• 560907 ==•472163 54. (28nJ=3-0G35 55. (7})i=l-93098 50. (9^)i=20928 57. (713)i=41553 58. (32/j-)i=31987 59. (54)1^1-7592 2S. To extract any root whatever — KuLE. — When the index of the root is some power of 2, extract the square root, when it is some power of 3, extract the cube root of the given number so many times, successively, as that power of 2, or 3 contains unity. / Example 1. —The 8th root of 65536=-s/ V\/65536=4. Since 8 is the third power of 2, we are to extract the square root three times, successively. ^ number — iction ; and then Example 2.— 13421 7728 fi=yVl342rr728=8. Since 9 is the second power of 3, we are to extract the cube root twice, successively. 29. In other cases we may use the following (Hutton Matliemat. Diet. vol. i. p. 135). Rule. — Find, by trial, some number which, raised to the power indicated by tbe index of the given root, will not be far from the given number. Then say, as one less than the index of the root, multiplied by the given number — plus one more than the index of the root, multiplied by the assumed number raised to the power expressed by the index of the root : one more than the index of the root, multiplied by the given number — plus one less than the index of the root, multiplied by the assumed number raised to the power indicated b the index of the root, : : the assumed root : a sti nearer approximation. Treat the fourth proportional thus obtained in the same way as the assumed number was treated, and a still nearer approximation will be found. Proceed thus until an approximation as near as desirable is discovered. ' ^ Example.— What is the 13th root of 923 1 ^ Let 2 be the assumed root, and the proportion will be 12x9234-14x2^3 : 14x923-j-12x2'^ :: 2 : a nearer I approximation. Substituting this nearer approximation for 2. in the above proportion, we get another approximation, «rhich we may treat in the same way. I f I •: ^ i; gfv:^a. 318 EVOLUTION ¥A :i^ PJJ :' ?l EXERCISKS. 60. (96698)1=6-7749 61. (66457)TT=2-7442 62. (2365)t==31-585 63. (87426)4=5084-29 64. (8-965) 1=1-368 13 65. (075426) f4=-(;46988 30. To find the squares and cubes, the square and cube roots of numbers, by means of the table at the end of the treatise — Tl is table contains the squares and cubes, the square imd cub6 roots of all numbers which do not exceed 1000 but it will be found of considerable utility even when very high numbers are concerned — provided the pupil bears in mind that [12] the square of any number is equal to the sum of the squares of its parts (which may be found by the table) plus twice the product of each part by the sum of all the others ; and that [21] the cube of a number divided into any two parts is equal to the sum of the cubes of its parts (which may be found by the table) plus three times the product of each part multi- plied by the square (found by means of the table) of the other. One or two illustrations will render this I 3ufficiently clear. Example 1. — Find the square of S73456. 873456 may be divided into two parts, 873 (thousand) and I 456 (units) . But we find by the table that 873'=762129 and | 456'=207936. Therefore 762129000000=873000' 796176000=873000 xtwice 456 207936=456* i 1 And 762925383936=873456 Example 2.— Find the cube of 864379. Dividing this intc 864 (thousand) and 379 (units), we find S6?bi644972544 B64 =746496, 379^=54439939, and 379'=143641 Therefore 644972544000000000=864000* : 848765952000000 =3x864000^X379 372317472000=3x864000x379* 54439939=379" WR» And 645821682323911939=^864.Tf9' M LOGARITHMS. 319 6)4=508429 ))Ul-368 126)t2=C46988 the square and table at the end i ibes, the square ,ot exceed 1000 f even when very the pupil bears mber is equal to oh may be found each part by the ] the cube of a jqual to the sum be found by the each part multi- of the table) of will render this 573 (thousand) and 873'=762129 and ce 456 Dividing this int« I gg7«644972544 =143641 i4000x379' 5R« 31 In finding the square and cube roots of larger numbers, we obtain their three highest digits at once, if we look in the table for the highest cube or square, the highest period of which (the required cyphers being added) does not exceed the highest period of the given number. The remainder of the process, also, may often be greatly abbreviated by means of the table. r QUESTIONS. 1. What are involution and evolution .'' [1]. 2. What are a power, index, and exponent } [1 & 2] . 3. What is the meaning of square and cube, of the square and cube roots .? [1 and 8J. 4. What is the difference between an integral and a fractional ladex .? [2 and 8] . 5. How is a number raised to any power ? [5] . 6. What is the rule for finding the square root ? [10]. 7. What is the rule for finding the cube root ? [20] . 8. How is the square or cube root of a fraction or of a mixed number f^imd > [14, &c., 19, 23, &c., 27]. 9. How is any rcoi -ind ? [28 and 29]. 10. How are the e . t s and cubes, the square roots and cube roots, of numoers found, by the table ? [30] . ;'(/.• LOOARTIHMS. •r: W 32. Logarithms are a set of artificial numbers, which represent the ordinary or natural numbers. Taken along with what is called the base of the system to which they belong, they are the eqtmls of the corres- ponding natural numbers, but without it, they are merely their representatives. Since the base is un- changeable, it is not written along with the logarithm. The logarithm of any number is that power of the base which is equal to it. Thus 10^ is equal to 100 ; 10 is the base J 2 (the index) is the logarithm, and 100 is the corresponding natural number. — ^Logarithms, therefore, are merely the indices which designate certain powers of some base. 33. Logarithms afford peculiar facilities for calcu- lation. For, as we shall see presently, the multiplica- tion of numbers is performed by the addition of their p w f ! I- .1 It . , ,' ■- If >l r I 1 hi ^''?l :* 320 LOGARITHMS. logarithms ; one number is divided by another if we subtract the logarithm of the divisor from that of the dividend ; numbers are involved if we multiply their logarithms by the index of the proposed power ; and evolved if we divide their logarithms by the index of the proposed root. — But it is evident that addition and subtraction are much easier than multiplication and division ; and that multiplication and division (particu- larly when the multipliers and divisors are very small) are much easier than involution and evolution. 34. To use the properties of logarithms, they must be exponents of the same base — that is, the quantities raised to those powers which they indicate must be the same. Thus 10* X 123 is neither 10'' nor 12% the former being too small, the latter too great. If, therefore, we desire to multiply 10* and 12^ by means of indices^ we must find some power of 10 which will be equal to 12 3, or some power of 12 which will be equal to 10*, or finally, two powers of some other number which will be equal respectively to 10* and 12^, and then, adding these powers of the sarne number, we shall have that power of it which will represent the product of 10* and 123. This explains the necessity for a table of logarithms — we are obliged to find the powers of some one base which will be either equal to all possible numbers, or so nearly equal that the inaccuracy is not deserving of notice. The base of the ordinary system is 10 ; but it is clear that there may be as many different systems of logarithms as there are different bases, that is, as there are different numbers. 35. In the ordinary system— which has been calcu- lated with great care, and with enormous labour, 1 is the logarithm of 10 ; 2 that of 100 ; 3 that of 1000, &c. And, since to divide numbers by means of these .loga- rithms (as we shall find presently), we are to subtract the logarithm of the divisor from that of the dividend, is the logarithm of 1, for 1=L°=10»-»=10« ; -1 is the logarithm of -1, for •l=i=i£'=3l0»-»=10-» ; and 10 io» for the same reason, —2 is the logarithm of '01 ; —3 that of -001, &c. LOGARITHMS. 321 inotHer if we na that of the multiply their i power ; and the index of it addition and iiplication and dsion (particu- re very small) tion. J, they must be uantities raised 3t be the same, he former being efore, we desire ndices, we must ^ual to 123, or 10* , or finally, li will be equal I, adding these ave that power f 10* and 123. jf logarithms — 5 one base which ers, or so nearly of notice. The it is clear that s of logarithms ere are different las been calcu- us labour, 1 is lat of 1000, &c. of these -loga- are to subtract )f the dividend, »=10« ; — 1 ifl -»s--10-* ; and im of -01 ; -3 36. The logarithms of numbers between 1 and 10, must be more than and less than 1 ; that is, must bo \ some decimal. The logarithms of numbers between 10 I and 100 must be more than 1, and less than 2 ; that 1 is, unity with some decimal, &c. ; and the logarithms of I numbers between "1 and '01 must be —1 and some deci- l mal ; betw-sen '01 and '001, —2 and some decimal, &c. j The decimal part of a logarithm is always positive. , 37. As the integral part or characteristic of a posi- tive logarithm is so easily found — being [35] one less than the number of integers in its corresponding num- ber, and of a negative logarithm one more than the number of cyphers prefixed in its natural number, it is not set down in the tables. Thus the logarithm corresponding to the digits 9872 (that is, its decimal ■ part) is 99440-^ ; hence, the logarithm of 9872 is 3 •994405 ; that of 987-2 is 2*994405 ; that of 9-872 is 0-994405 ; that of -9872 is- 1*994405 (since there is no integer, nor prefixed cypher) ; of -009872— 3*994405, &c. : — The same digits, whatever may be their value, \ liave the same decimals in their logarithms ; since it is the integral part, only, which changes. Thus the i logarithm of 57864000 is 7-762408; that of 57864, is 14-762408; and that of '0000057864, is— 6-762408. 3 38. To find the logarithm of a given number, by the % table — I The integral part, or characteristic, of the logarithm I may be found at once, from what has been just said [37] — When the number is not greater than 100, it will be found in the column at the top of which is N, and the decimal part of its logarithm immediately opposite to it in the next column to the right hand. If the number is greater than 100, and less than 1000, it will also be found in the column marked N, and the decimal part of its logarithm opposite to it, in the column at the top of which is 0. If the number contains 4 digits, the first three of them will be found in the column under N, and the fourth at the top of the page ; and then its logarithm in vhe same horizontal line as the three first digits of jthe given number, and in the same column as its fouitb 4 322 LOGARITHMS. I', I at- 'r4 If the number contains more than 4 digits, find the logarithm of its first four, and also the difference be- tween that and the logarithm of the next higher num- ber, in the table ; multiply this difference by the remain- ing digits, and cutting off from the product so many digits as were in the multiplier (but at the same time adding unity if the highest cut off is not less than 5), add it to the logarithm corresponding to the four first digits. Example 1. — The logarithm of 59 is 1770852 (the charac- teristic being positive, and one less than the number of integers) . Example 3.— The logarithm of 338 is 2-528917. Example 3.— The logarithm of 0004587 is —4 001529 (the characteristic being negative, and one more than the number of prefixed cyphers) . Example 4.— The logarithm of 28434 is 4-453838. For, the difference between 453777 the logarithm of 2843, the four first digits of the given number, and 453930 the logarithm of 2844, the next number, is 153 ; which, multi- plied by 4, the remaining digit of the given number, pro- duces 012; then cutting off one digit from this (since we have multiplied by only one digit) it becomes 01, which being added to 453777 (the logaritlim of 2844) makes 453838, and, with the characteristic, 4453838, tLe required logarithm. Example 5.— The logarithm of 873457 is 5 941242. For, the difference between the logarithms of 8734 and 8735 is 50, which, being multiplied by 57, the remaining digits of the given number, makes 2850 ; from this we cut off two digits to the right (since we have multiplied by tuo digits), when it becomes 28; but as the highest digit cut off is 5, we add unity, which makes 29. Then 5941213 (the logarithm of 8734) -f 29=5-941242, is the required logarithm. 39. Except when the logarithms increase very ra pidly — that is, at the commencement of the table — the differences may be taken from the right hand column (and opposite the three first digits of the given number) where the mean differences will be found. Instead of multiplying the mean difference by the remaining digits (the fifth, &c., to the right) of the given number, and cutting off so many places from the product as are equal to the number of digits in the multiplier to obtain the proportional part'—ot what is to bo idded ^^ LOQARITIIMS. 323 digits, find the diifercnce bc- tt higher num- by the remain- ■oduct so many t the same time ess than 5), add four first digits. )852 (the charac- imher oiintegers). 528917. gj is -4 601529 ne more than the J 4-453838. logarithm of 2843, r, and 453930 the 53 ; which, multi- iven number, pro- om this (since wc les 61, which being nakes 453838, and, uired logarithm. is. 5941242. Lthms of 8734 and 57, the remaining • from this we cut I multiplied by two 5 highest digit cut ^hen 5-941213 (the required logarithm. to the logarithm of the first four digits, we may take the proportional part corresponding to each of the re- I maining digits from that part of the column at the left I hand side of the page, which is in the same horizontal I division as that in which the first three digits of the \ given number have been found. \ Example.— What is the logarithm of 839785 1 i The (decimal part of the) logarithm of 839700 is 924124. ; Opposite to 8, in the same horizontal division of the page, we find 42, or rather, (since it is 80) 420, and opoosite to ; 5, 20. Hence the required logarithm is 9"^ 24-|-4204.20= ■ 924570 J and, with the characteristi )-9z-+ '0. I 40. The method given for finding the proportional part — or 1 what is to be added to the next lower logarithm, in the table— i arises from the ditFerence of numbers being proportional to the I difference of their logarithms. Hence, using the last example, 1 100 : 85 : : 52 (924170, the logarithm of 839800—924124, 62X85 (the logarithm of 839700) : ■ ,qq - , or the difference (the mean ^difference may generally be used)Xhy the remaining digits of ■the given number -r 100 (the division being performed by cut- ting off two digits to the right). It is evident that the number of digits to be cut off depends on the number of digits in the multiplier. The logarithm found is not exactly correct, be- [cause numbers are not exactly proportional to the differences f their logarithms. The proportional parts set down in the left hand column, ave been calculated by making the necessary multiplica- ions and divisions. 41. To find the logarithim of a fraction — Rule. — Find the logarithms of both numerator and ienominator, and then subtract the former from the increase very ra of the table— the ight hand column nd. difference by the right) of the given s from the producl in the multiplierJ hat i» to b« addel atter ; this will give the logarithm of the quotient. Example.— Log. ^} is 1-672098 - 1-748187== - 1-923910. Ve find that 2 is to be subtracted from 1 (the character- stic of the numerator) ; but 2 from 1 leaves 1 still to be gj . lubtracted, or [Sect. II. 15] — 1, the characteristic of the he given numberjr [uotient. We shall find presently that to divide one quantity by nother, we have merely to subtract the logarithm of the latter rora that of the former. 42. To find the logarithm of a mixed number — Rule. — ^Reduce it to an improper fraction, and pro- ftd as direoted by the last rule. 324 LOGARITHMS. I 43. To find the number which corresponds to a given logarithm — If the logarithm itself is found in the table — Rule. — Take from the table the number which cor- responds to it, and place the decimal point so that there may be the requisite number of integral, or decimal places — according to the characteristic [37], Example. — ^What number corresponds to the logarithm 4?14314l We find 21 opposite the natural number 163 ; and look- ing along the horizontal line, we find the rest of the logarithm under the figure 8 at the top of the page ; therefore the digits of the required number are 1638. But as the characteristic is 4, there must in it be 5 places of integers. Hence the required number is 16380. 44. If the given logarithm is not found in the table — Rule, — Find that logarithm in the table which i& ne^t lower than the given one, and its digits will be the highest digits of the required number ; find the difference between this logarithm and the given one, annex to it a cypher, and then divide it by that differ- ence in the table, which corresponds to the four highest digits of the required number — the quotient will be the next digit ; add another cypher, divide again by the tabular diiFerence, and the quotient will be the next digit. Continue this process as long as necessary. Example. — What number corresponds to the logarithm 5-654329 ? 654273, which corresponds with the natural number 4511, is the logarithm next less than the given one ; therefore the first four digits of the required number are 4511. Adding a cypher to 56, the difierence between 654273 and the given logarithm, it becomes 560, which, being divided by 96, the tahiJar difference corresponding with 4511, gives 5 as quo- tient, and 80 as remainder, Therefore, the first ^ ye digits (.f the required number are 45115. Adding a cypher to 80, it becomes 800 ; and, dividing this by 96, we obtain 8 as the next digit of the required number, and 32 as remainder. The integers of the required number (one more than 5, the characteristic) are, therefore, 451158. We may obtain the decimals, by continuing the addition of cyphers to the re* mainders, and the division by 96. LOGARITHMS. 325 nds to a given able — )er which cor- ^t so that there al, or decimal n I the logarithm • 163 ; and look- of the logarithm jrefore the digits lie characteristic ;ors. Hence the d in the table- table which i& ts digits will be mber ; find the I the given one, t by that diffcr- the four highest tient will be the e again by the rill be the next necessary, to the logarithm iral number 4511, ne ; therefore the re 4511. Adding 273 and the given ivided by 96, the , gives 5 as quo- le first jive digits ig a cypher to 80, 6, we obtain 8 as I 32 as remainder, more than 5, the iTe may obtain the yphers to the re- 45. "Wo arrive at the same lesult, by subtracting from the difference between the given logarithm and the next less in the table, the highest (which does not exceed it) of those proportional parts found at the right hand side of the page and in the same horizontal divi- sion with the first three digits of the given number— r continuing the process by the addition of cyphers, until nothing, or almost nothing, remains. Example. — Using the last, 4511 is the natural number corresponding to the logarithm 654273, which differs from the given logarithm by 56. The proportional p.^rts, in the same horizontal division as 4511, are 10, 19, 29, 38, 48, 58, 67, 77, and 86. The highest of these, contained in 56, is 48, which we find opposite to, and therefore corresponding with, the natural number 5; hence 5 is the next of the required digits. 48 subtracted from 56, leaves 8 j this, when a cypher is added, becomes 80, which contains 77 Tcorres- pondin^' to the natural 'number 8) ; therefore 8 is tne next of the required digits. 77, subtracted from 80, leaves 3j this, when a cypher is added, becomes 30, &c. The inte- gers, therefore, of the required number, are found to be 451158, the same as those obtained by the other method. The rules for finding the numbers corresponding to given logarithms are merely the converse of those used for finding the logarithms of given numbers. Use of Logarithms in Arithmetic. 46. To multiply numbers, by means of their loga- rithms — Rule. — Add the logarithms of the factors ; and the natural number corresponding to the result will be the required product. ExAMPLE.-87x24==l-939519 (the log. of 87) 4.1-380211 (the log. of 24)=3-319730; which is found to correspond with the natural number, 2088. Therefore 87x24=2088. Reason of the Rule. — This mode of multiplication arises from the very nature of indices. Thus 6*X5''=5X5X5X5 multiplied 6X5X5X5X6X5X5X5; and the abbreviation for this [2] is 6'*. But 12 is equal to the sum of the indices (logarithms). The rule might, in the same way, be proved oorrect by any other example. V 2 326 LOQARITIIMS 47. When the characteristics of the logarithms to be added are both positivo, it is evident that their sum will be p( ^itive. When they are both negative, their sum (diminished by what is to be carried from the sum of the positive [36j decimal parts) will be negative. When one is negative, and the other positive, subtract the less from the greater, and prefix to the diflference the sign belonging to the greater — bearing in mind what has been already said [Sec. II. 15] with reference to the subtraction of a greater from a less quantity. 48. To divide numbers, by means of their logarithms — Rule. — Subtract the logarithm of the divisor from that of the dividend ; and the natural number, corres- ponding to the result, will be the required quotient. Example.— 1134 -1.42 = 3054613 (the log. of 1134) — 1-623249 (the log. of 42) = 1431364, which is found to correspond with the natural number, 27. Therefore 1134-r 42=27. Reason of the Rule. — This mode of division arises from the nature of indices. Thus 4*-j-4'=[2] 4X4X4X4X4-^4X ^ ^ 4X4X4X4X4 ^ ^ 4X4X4 , , ,^ ^^ . ,. 4X4= 4X4X4 — =4X4X43^4^=4X4, the abbreviation for which is 4^. But 2 ia equal to the index (logarithm) of the dividend minus that of the divisor. The rule might, in the same way, be proved correct by any other example. 49. In subtracting the logarithm of the divisor, if it is negative, change the sign of its characteristic or inte- gral part, and then proceed as if this were to be added to the characteristic of the dividend ; but before making the characteristic of the divisor positive, subtract what was borrowed (if any thing), in subtracting its decimal part. For, since the decimal part of a logarithm is positive, what is borrowed^ in order to make it possible to subtract the decimal part of the logarithm of the divisor from that of the dividend, must be bo much taken away from what is positive, or added to what is negative in the remainder. We change the sign of the negative characteristic, and then add it; for, adding a positive, is the same as takingl away a negative quantity. LOGARITHMS. 327 garitlims to be their sum will ive, tlicir sum m the sum of gative. When ibtract the less rence the sign nind what has jference to the tity. nr logarithms — le divisor from number, corres- id quotient. log. of 1134)- hich is found to Therefore 1134-r vision arises from <4X4X4X4-^-4X I, the abbreviation lex (logarithm) of :he rule might, in er example. the divisor, if it •acteristic or inte- were to be added )ut before making ve, subtract what acting its decimal »f a logarithm is make it possible logarithm of the nust be so much added to what is characteristic, and! the same as ta3dng| 60. To raise a quantity to any power, by means of it^i logarithm — Rule. — Multiply the logarithm of the quanity by the index of the power ; and the natural number oor- rebponding to the result will be the required power. Example. — Raise 5 to the 5th power. The logarithm of 5 is 0C9897, which, multiplied by 6, gives 3-49485, the logarithm of 3125. Therefore, the 5th power of 5' is 3125. Heabon or THE Rule. — This rule aT!»o follows from the natare of indices. 6" raised to the 5th power is 6x5 multiplied by 5x5 multiplied by 6x5 multiplied by 6X6 multiplied by 5X5, or 5x5x5x6x5x6x6x5x5x5, the abbreviation for which is [2] 5"". But 10 is equal to 2, the index (logarithm) of the quantity, multiplied by 6, that of the power. The rnle might, in the same way, be proved correct by any other example. 51. It follows from what has been said [47] that when a negative characteristic is to be multiplied, the product io negative ; and that what is to be carried from tae multiplication of the decimal part (always positive) is to be subtracted from this negative result. 52. To evolve any quantity, by means of its loga- rithm — Rule. — ^Divide the logarithm of the given quantity by that number which expresses the root to be taken ; and the natural number corresponding to the result will be the required root. Example.— What is the 4th root of 2401. The logarithm of 2401 is 3380392, which, divided by 4, the number expressing the root, gives -845098, the logarithm of 7. Therefore, the fourth root of 2401 is 7. Reason of the Rule. — This rule follows, likewise, from the nature of indices. Thus the 6th root of 16"* is such a number as, raised to the 6th power — that is, taken 6 times as a factor— would produce 16". But 16 z » taken 5 times as a factor, would produce 16". The rule might be proved correct, equally well, by any other example. 53. When a negative characteristic is to be divided — Rule I. — If the characteristic is e.rac% divisible by the divisor, divide in the ordinary way, but make the characteristic of the quotient negative. )"t, t^': "if ' ■ 4\ 328 LOGARITHMS. II. — ^If the negative characteristio is not exactly divisible, add what will make it so, both to it and to the decimal part of the logarithm. Then proceed with the division. Example. — Divide the logarithm — 4'837564 by 5. 4 wa nts 1 of being divisible by 5 ; then -4-837564^5=- — 5-f.l^375G4^5=:l-367513, the required logarithm. Reason or I. — The quotient multiplied by the divisor must give the dividend ; but [51 J a negative quotient multiplied by a positive divisor will give a negative dividend. Reason or II. — In example 2, we have merely added -f-1 and — 1 to the same quantity — which, of course, does not alter it. If- J ■!.' QUESTIONS. 1 . What are logarithms ? [32] . 2. How do they facilitate calculation } [33] . 3. Why is a table of logarithms necessary ? [34] . 4. What is the characteristic of a logarithm ; an(^ how is it found ? [37] . 5. How is the logarithm of a number found, by tho table.? [38]. 6. How are the " differences," given in the table used.? [39]. 7. What is the use of '* proportional parts .?" [39] . 8. How is the logarithm of a fraction found ? [41]. 9. How do we find the logarithm of a mixed num- ber ? [42]. 10. How is the number corresponding to a given logarithm found } [43] . 11. How is a number found when its corresponding logarithm is not in the table } [44] . 12. How are multiplication, division, involution and evolution effected, by means of logarithms ? [46, 48, 50, and 52] . 13. When negative characteristics are added, what is the sign of their sum } [47] . 14. What is the process for division, when the cha- racteristic of the divisor is negative ? [49] . 15. How is a negative characteristic multiplied } [51], f 16. How is a negative characteristic divided ? [53] ;i r from the greater al to the common number of terms. than the number IB- EXERCISES. 5. The extremes of an arithmetical series are 21 and 497, and the number of terms is 41. What is the common difference .^ Ans. 11-9. 6. The extremes of an arithmetical series are 127||- and 9|, and the number of terms is 26. What is the common difference ? Ans. 4^. 7. The extremes of an arithmetical series are 77|f and f, and the nu iber of terms is 84. What is the common difference ? Ans. |f 9. T^ find any number of arithmetical means between two given numbers — Rule. — Find the common difference [7] ; and, ac- cording as it is an ascending or a descending series, add it to, or subtract it from the first, to form the second terra ; add it to, or subtract it from the second, to form the third. Proceed in the same way with the remain- ing terms. We must remember that one less than the number of terms is one more than the number of means. Example 1. — Find 4 arithmetical means between 6 and 15 21. 21—6 = 15. 7q^=3, the common difference. And the series is — 6 . 64-3 . 6+2x3 . 6-1-3x3 . 6-j-4x3 . 6-h5x3. Or6 . 9 . 12 . 15 . 18 . 21. Example 2. — Find 4 arithmetical means between 30 and 20 10. 30—10=20. 2rrT=4, the common difference. And the rieries is — 30 . 26 . 22 . 18 . 14 . 10 4+1- 26 . 22 . 18 . 14 This rule is evident. exercises. 8. Find 11 arithmetical means between 2 and 26 Ans. 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, and 24. 9. Find 7 arithmetical means between 8 and 32 Ans. 11, 14, 17,20, 23,26, 29. 10 Find 5 arithmetical means between 4^, and 13|. Ans. 6, 7^, 9, 10^, 12. 332 PROGRESSION. ■ w l-l 'i. 3 ' I ' 10. Given the extremes, and the number of terms— to find any term of an arithmetical progression — Rule. — Find the common difference by the last rule, and if it is an ascending series, the required term will be the lesser extreme plus — if a descending series, the greater extreme minus the common difference multiplied by one less than the number of the term. Example 1. — What is the 5th term of a series containing 9 terms, the first being 4, and the last 28 ? 28-4 „ — Q — =3, is the common difference. And 4-1-3x5 — l=a 16, is the required term. Example 2. — What is the 7th term of a series of 10 terms, the extremes being 20 and 2 ? 20-2 ^ — — Q~"=2, is the common difference. 20 — 2x7—1=8, is the required term. 11. Reason or the Rule. — In an ascending series th« required term is greater than the given lesser extreme to the amount of all the differences found in it. But the number of differences it contains is equal only to the number of terms which precede it — since the common difference is not found in iXxB first term. In a descending series the required term is less than the given greater extreme, to the amount of the differences sub- tracted from the greater extreme — but one has been subtracted from it, for each of the terms which precede the required term. EXERCISES. ■' - 11. In an arithmetical progression the extremes are 14 and 86, and the number of terms is 19. What is the 11th term .'' Ans. 54. 12. In an arithmetical series the extremes are 22 and 4, and the number of terms is 7. What is the 4th term } Ans, 13. 13. In an arithmetical series 49 and £ are the ex- tremes, and 106 is the number of terms. What is the 94th term } Ans. 6*2643. 12. Given the extremes, and common difference— to find the number of terms — Rule. — Divide the difference between the given ex- tremes by the common difference, and the quotient plus unity will be the number of terms. iber of terms— ession — >y the last rule, aired term will ding series, the ence multiplied • series containing nd 4+3x5-1= eries of 10 terms, 10 — 2x7^1=8, nding series th« er extreme to the at the number of number of terms le is not found in is less than the differences sub- irS been subtracted ;he required term. le extremes are s 19. What is imes are 22 and hat is the 4th I J are the ex- What is the 1 diflference— to n the given ex- le quotient plus PROGRESSION. 333 Example. — How many terms in an arithnfetical series of which the extremes are 5 and 26, and the common differ- ence 3 ? 26-5 ^ — 2 — =7. And 74-1=8, is the number of terms. 13. Reason of the Rule. — The greater differs from the lesser extreme to the amount of the differences found in all the terms. But the common difference is found in all the terms except the lesser extreme. Therefore the difference between the extremes contains the common difference once less than will be expressed by the number of terms. EXERCISES. 14. In an arithmetical series, the extremes are 96 and 12, and the common difference is 6. What is the number of terms .'' Ans. 15. 15. In an arithmetical series, the extremes are 14 and 32, and the common difference is 3. What is the number of terms ? Ans. 7. _ .. 16. In an arithmetical series the common difference is ^, and the extremes are 14| and a1. What is the number of terms .'' Ans. 8. 14. Given the sum of the series, the number of terms, and one extreme — to find the other — Rule. — Divide twice the sura by the number of terms, and take the given extreme from the quotient The difference will be the required extreme. Example. — ^One extreme of an arithmetical series is 10 the number of terms is 6, and the sum of the series is 42 I What is the other extreme ? — g — —10 = 4, is the required extreme. 15. Reason of the Rule. — We have seen [5] that 2 X the sum = sum of the extremes X the number of terms. But if we divide each of these equal quantities by the number of terms, we shall have 2 X the sum sum of extremes X the number of terms the number of terms 2 X the sum Or the number of terms =: sum of the extremes. And sub- the number of terms tracting the same extreme from each of these equals, we shall have . r-i': .jj .,.-.. 834 FROaRESSlOfr. 2 X tbft sum ^ftnttflTt.rflTnft=tlift sumof the extremes the number of terms — the same extreme. twice the sum Or the number of terms minus one extreme^ the other ex- treme. EXERCISES. 17. One extreme is 4, the ntimber of terms is 17, and the sum of the series is 884. What is the other extreme ? Ans. 100. 18. One extreme is 3, the number of terms is 63, and the sum of the series is 252. What is the other extreme } Ans. 5. 19. One extreme is 27, the number of terms is 26, and the sum of the series is 1924. What is the other extreme.? Ans. 121. 16. Geometrical Progression. — Given the extremes and common ratio — to find the sum of the series — Rule. — Subtract the lesser extreme from the product of the greater and the common ratio ; and divide the difference by one less than the con\mon ratio. Example. — In a geometrical progression, 4 and 312 are the extremes, and the common ratio is 2. What is the sum of the series. 312x2-4 — 2 1 = 620, the required nmrber. 17. RfiAsoN OF THE RuLE. — ^Tho rulo may be proved by setting down the series, and placing over it (but in a reverse order) the product of each of the terms and the common ratio. Then Sum X common ratf o = 8 -f- 16 -(- 32, &c. . + 812 + 624 Sum= 4 + 8 + 16 4-32, &c. . +312 . And, subtracting the lower from the upper line, we shall have Sum X common ratio — Sum = 624 — 4. Or Common ratio — 1 X Sum = 624 — 4. And, dividing each of the equal quantities by the common ratio minus 1 642 (last term X common ratio) *-4 (the first term) 2 Ji I ai tl 9; A gum =3 common ratio — > 1 Which is the rule. b; is bi IF IT tl K A IS .P. Iti b is a e m of the extremes Bssthe other ex* of terms is 17, lat is the other of terms is 63, hat is the other of terms is 26, bat is the other i the extremes the series — from the product ; and divide the ratio. in, 4 and 312 are What is the sum number. Qay be proved by it (but in a reverse the common ratio. c. . + 812 4- 624 ;c. . 4-312 . line, we shall have -4. Or -4. ies by the common ^i (the first term) I 1-1 PROGRESSION. EXERCISES. 335 20. The extremes of a geometrical series are 512 and 2, and the common ratio is 4. What is the sum : Ans. 682. 21. Th-e extremes of a geometrical series are 12 and 175692, and the common ratio is 11. What is the sum ? Ans. 193260. ,. 22. The extremes of an infinite geometrical series are yV ^^^ ^) ^^^ I'o i^ the common ratio. What is the sum } Am. |. [Sec. IV. 74.] Since the series is infinite, the lesser extreme=0, 23. The extremes of a geometrical series are '3 and 937-5, and the common ratio is 5. What is the sum ? Ans. 1171-875. 18. Given the extremes, and number of terms in a geometrical series — to find the common ratio — RuLE.-T-Divide the greater of the given extremes by the lesser ; and take that root of the quotient which is indicated by the number of terms minus 1. This will be the required number. Example.— 5 and 80 are the extremes of a geometrical progression, in which there are 5 terms. What is the com- mon ratio 1 80 v-=16. And ^16=2, the required common ratio. 19. Reason ot the Rule. — The greater extreme is equal I to the lesser multiplied by a product which has for its factors i the common ratio taken once less than the number of terms- [ since the common ratio is not found in the first term. That I is, the greater extreme contains the common ratio raised to a power indicated by 1 less than the number of terms, and mul- , tiplied by the lesser extreme. Consequently if, after dividing i by the lesser extreme, we take that root of the quotient, which is indicated by one less than the number of terms, we shall I obtain the common ratio itself. S ' ' • ' •' EXERCISES 24. The extremes of a geometrical series are 49152 and 3, and the number of terms is 8. What is th*) eommon ratio ^ At^' 4. 35. The extremes of a geometrical series are 1 and 336 PROGRESSION. U':J;$ r, J.'- *v I ■■*■ W 15625, and the number of terms is 7. What is the common ratio ? Ans. 5. 26. The extremes of a geometrical series are 201768035 and 5, and the number of termd is 10 What is the common ratio ? Ans. 7. 20. To find any number of geometrical means be twoen two quantities — Rule. — Find the common ratio (by the last rule)^ and — according as the series is ascending, or descend- ing — multiply or divide it into the first term to obtain the second ; multiply or divide it into the second ta 'obtain the third ; and so on with the remaining terms. We must remember that one less than the number of terms is one more than the number of means. Example 1. — ^Find 3 geometrical means between 1 and 81. 4/-r-=3, the common ratio. And 3, 9, 27, are the re« quired means. Example 2. — Find 3 geometrical means between 1250 and 2. 1250 , , 1250 1250 1250 «,^^^ ,^ 4/-2--5. And -5- 5^ 5X55<5' «' 250, 50, 10, are the required means. This rule requires no explanation. EXERCISES. 27. Find 7 geometrical means between 3 and 19683 r Ans. 9, 27, 81, 243, 729, 2187, 6561. 28. Find 8 geometrical means between 4096 and 8 ? Ans. 2048, 1024, 512, 256, 128, 64, 32, and 16. 29. Find 7 geometrical means between 14 and 23514624? Ans. 84, 504, 3024, 18144, 108864, 653184, and 3919104. 21 . Given the first and last term, and the number of terms — to find any term of a geometrical series — Rule. — ^If it be an ascending series, multiply, if a descending series, divide the first term hy that power of the common ratio which is indicated by the number of the term minus 1. PROGRESSION. 337 What is the ical means be , 27, are the re* QS between 1250 m3and 19683? Example 1.— Find the 3rd term of. a geometrical series, of which the first term is 6, the last 1458, and the number of terms 6. The common ratio is ^— ^-=3. Therefore the required term is 6x3'==54. Example 2. — Find the 5th term of a series, of which the extremes are 524288 and 2, and the number of terms is 10. 524288 524288 The common ratia^ — « — =^* ^'^d — 74— s* 2048, is the required term. 22. Reason of the Rule. — In an ascending series, any term is the product of the first and the common . .tic taken as a factor so many times as there are preceding terms — since it is not found in the^r«/ term. In a descending scries, any term is equal to the first term, divided by a product containing the common ratio as a factor so many times as there are preceding terms — since every term but that which is required adds it once to the factors which constitute the divisor. EXERCISES. 30. What is the 6th term of a series having 3 and 5859375 as extremes, and containing 10 terms ? Am. 9375. 31. Given 39366 and 2 as the extremes of a series having 10 terms. What is the 8th term ? Ans. 18. 32. Given 1959552 and 7 as the extremes of a series having 8 terms. What is the 6th term ? Afis. 252. 23. Given the extremes and common ratio— to find the number of terms — Rule. — Divide the greater by the lesser extreme, and one more than the number expressing what power of common ratio is equal to the quotient, will be the required quantity. Example. — How many terms in a series of which the extremes are 2 and 256, and the common ratio is 2 ? 256 -s-=128. But 2^:^128 . There are, therefore, 8 terms. The common ratio is found as a factor (in the quotient of the greater divided by the lesser extreme) once less than tht number of terms. 1 i: it* 338 M PROGRESSIOIf. EXERCISES. 33. How many terms in a series of whicli the first is 78732 and the last 12, and the common ratio is 9 ? Ans. 5. 34. How many terms in a series of which the ex- tremes and common ratio are 4, 470596, and 7 ? Ans. 7. 35. How many terms in a series of which the ex- tremes and common ratio, are 196608, 6, and 8 ? Ans. 6. 24. Given the common ratio, number of terms, and one extreme — to find the other — Rule. — If the lesser extreme is given, multiply, if the greater, divide it by the common ratio raised to a power indicated by one less than the number of tci-is. Example 1. — In a geometrical series, the lesser extreme is 8, the number of terms is 5, and the common ratio is 6; what is the other extreme ? Ans. 8x6»-'=10368. Example 2. — In a geometrical series, the greater extreme is 6561, the number of terms is 7, and the common ratio is 3 J what is the other extreme '? Ans. 6561-s-3'~'=9. This rule does not require any explanation. - EXERCISES. 36. The common ration is 3, the number of terms is 7, and one extreme is 9 ; what is the other ? Ans. 6561. 37. The common ratio is 4, the number of terms is 6, and one extreme is 1000 ; what is the other ? Ans. 1024000. 38. The common ratio is 8, the number of terms U 10, and one extreme is 402653184 ; what is the other ? An^s. 3. In progression, as in m^ny other rules, Uie appUcatioin of algebra to the reasoning would greatly simplify it. MISCELLANEOUS EXERCISES IN PROGRESSION. 1. The clocks in Venice, and some other places strike the 24 hours, not beginning again, as ours do, after 12. How many strokes do they give in a day ^ Ans. 300. 2 A butcher bought 100 sheep ; for the first he gave Is.y and for the last ^9 195. What did he pay for ft,?] PROGRESSION. 339 L tbe first ia ratio is 9 ? licli the ex- \ 7 ? Ans. 7. lich tbe ex- d 8 ? Ans. 6. )f terms, and , multiply, if ,0 raised to a ►er of tei-is. lesser extreme men ratio is 6; .0368. greater extreme jommon ratio is all, supposi ig their prices to form an arithmetical series ? Ans. ii5U0. 3. A person bought 17 yards of cloth ; for the first yard he gave 2s., and for the last 10*. What was the price of all ? Ans. £^ 2s. 4. A person travelling into the country went 3 miles the first day, 8 miles the second, 13 the third, and so on, until he went 58 miles in one day. How many days did he travel ? Ans. 12. 5. A man being asked how many sons he had, said tliat the youngest was 4 years old, and the eldest 32, and that he had added one to his family every fourth year. How many had he ? Ans. 8. 6. Find the sum of an infinite series, ^, ^, ^, &c. Ans. |. Ans. A jr of terms is 7, ? Ans. 6561. ber of terms is , other ? Ans, oer of terms is it is the other ? le appUcaUonof lify it- [gression. ^er places strike trs do, after 12. J ? Ans. 300. for the first he it did he pay for 7. Of what value is the decimal '463' .'* 8. What debt can be discharged in a year by monthly payments in geometrical progression, the first term king d£l, and the last ^2048; and what will be the common ratio ? An^. The debt will be i£4095 ; and I the ratio 2. 9. What will be the price of a horse sold for 1 far- tliing for the fii'st nail in his shoes, 2 farthings for the second, 4 for the third, &c., allowing 8 nails in each I shoe } Ans. £4473924 bs. 3|^. 10. A nobleman dying left 11 sons, to whom he be- [queathed his property as follows ; to the youngest he Igave £1024; to the next, as much and a half; to the [next, 1^ of the preceding son's share ; and so on. What has the eldest son's fortune ; and what was the amount jof the nobleman's property } An^. The eldest son re- Eeived £59049, and the father was worth £175099. QUESTIONS. 1. What is meant by ascending and descending ^eries? [1]. , r ,- , ., 2. What is meant by an arithmetical and geome- trical progression ; and are they designated by any other lames ? [2 and 3] . 3. What are the common difierence and common Utio.? [2 apdS]. I: l) J ■ I P ■' 540 ANNUITIES 4. Show that a continued proportion may be formed from a .scries of either kind ; [2 and 3]. 5. What are means and extremes r [4]. t). How is the sum of an arithmetical or a geome« trical series found ? [5 and 16]. 7. How is the common difference or common ratio found r [7 and 18]. 8. How is any number of arithmetical or geometrical means ioiind ? [9 and 20] . 9. How is any particular arithmetical or geometrical mean found ? [ 1 n nd 2 1 ] . 10. How is the number of terms in an arithmetical or geometiicnl series found ? [12 and 23]. 1 1 . How is one ext/fline of an arithmetical or geome- trical series found t [14 and 24]. '■■? ANNUITIES. 2"). An annuity is an income to be paid at stated tini'^s, yealy, half-yearly, &c. It is either in jwssession^ liiat is. entered upoi. already, or to be entered upon iiiim'idiatoly ; or it is in reversion, that is, not to com- mence until after some period, or after something has occurred. An annuity is certain when its commence- ment and termination are assigned to definite periods, contingent when its beginning, or end, or both are uncertain ; is in arrears when one, or more payments are retained after they have become due. The amounll of an annuity is the sum of the payments forborne (in| arrears),' and the interest due upon them. When an annuity is paid off at once, the price given] foi* it is called its present worth, or vahie- — which ought to be such as would — if left at compound interest untill the annuity ceases — produce a sum equal to what wouldl be due from the annuity left unpaid until that timej This value is said to be so many years'* jpurchase ; tljatl is, so many annual payments of the income as would bfi| just equivalent to it. 26. To find the amount of a certain number of pay| ments in arrears, and the interest due on them — ANNUITIES 341 lay be formed • 1 or a geome* common ratio or geometrical or geometrical an avlibmetical 3tical or geome- e paid at stated [her in jwssessioiiy he entered upon ■ is, not to com- ;r something has n its commence- definite periods,! ,d, or both are I r more payments ae. The amowm lents forborne (in m. . . 3, the price given /itc— which ouglitl und interest unti I aal to what wouldl i until that time.l rs' purchase ; tliatl icoine as would bel in number of paJi Ion them — Rule. — Find the interest due on each payment ; then the sum of the payments and interest due on them, will be the required amount. Example 1. — What will be the amount of £1 per annum, unpaid for G years, 6 per cent, simple interest being allowed? The last, and preceding payments, with the interest due on them, form the arithmetical series £l-f-j£05x5. £l-|-£05x 4 . . £lx£ 05 £1. And its sum is £ l+^l-j-X 05x5x f=£2-f£ 25x3=£6'75=£6 15s., the required amount. Example 2. — If the rent of a farm worth £60 per ann m is unpaid for 19 years, how much does it amount to, at 5 per cent, per an. compound interest ? Tn this case the series is geometrical / and the last payxnenfc with its interest is the amount of £1 for 18 (19 — 1) y*»Rr« multiplied by the given annuity, the preceding payment with its interest is the amount of £1 for 17 years multiplied ly the given annuity, &c. The amount of £1 (as we find by the table at the end of the treatise) for 18 years is £2-40662. Then the sum of the series is — £2-40662xl05x60-60, ^ . , T^^Zn [16j=sl832-4, the required amount. The an: unt of £1 for 18 years multiplied by 105 is the same as the amount of £1 for 19, or the given number of years, which is found to be £2 527. And 105 — 1, the divi- sor, is equal to the amount of £1 for one payment minus £1 ; that is, to the interest of £1 for one payment. Hence •A MIT. ^2527x60-6d the required sum will be :Qg =£ioo2-4. It would evidently be the same thing to consider the annuity as £1, and then multiply the result by 60. Thua Z^^ — t X 60 = £1832-4. For an annuity of £60 ought -Oo to be 60 times as productive as one of only £1. Hence, briefly, to find the amount of any number of payments in arrears, and the compound interest due on them — Subtract £1 from the amount of iEl for the given number of payments, and divide the difference by the interest of i£l for one payment ; then multiply the quo- tient by the given sum. 'U \\ 49 ANNUITIES. Ix: I». :| H. V t £ 27. Reahon or thk Rulk. — Eiich payment, with its inN^- real, evidently conatitute a separate aniuutit ; an I the Huni (U • must be the sum of these amounts — which Ibrni a derreanin^ scM'ics, because of the decreasing interest, arising from the decreasing number of times of payment. When simple interest is allowed, it is evident that what Is ilue will be the sum of an arithmetical series, one extreme of wliich is the first payment plus the interest due upon it at tho time of tlie last, the other the last payment ; and its common dilference tlie interest on one payment due at the next. But when compound interest is allowod, what is due will be tlie sum of a geometrical series, one extreme of which is the first payment plus the interest due on it at the last, the other tlie last payment; and its common ratio £\ plus its intcre>t for the interval between two payments. And in each case tiie interest due on the first payment at the tjime of the last will be tlie interest due for one leas than the number of payments, since interest is not due on the first until the time of the second payment. « EX£RCIS£S. 1. What is tho amount of iB37 per annum unpaid for 1 1 years, at 5 per cent, per an. simple interest } Ans. ieaOS 155. 2. What is the amount of an annuity of jSlOO, to continue 5 years at 6 per cent, per an. compound inte- rest .? Ans. £563 14s. 2\d. 3. What is the amount of an annuity of i£356, to continue 9 years, at 6 per cent, per an. simple interest ? Ans. ^£3972 19*. 2\d. 4. What is the amount of iS49 per annum unpaid for 7 years, 6 per cent, compound interest being allowed } Ans. £.A\\ bs. n\d. 28. To find the present value of an annuity — Rule. — Find (by the last rule) the amount of the given annuity if not paid up to the time it will cease. Then ascertain how often this sum contains the amount of £\ up to the same time, at the interest allow.».d. Example. — What is the present worth of an annuity of £12 per annum, to be paid for 18 years, 5 per cent, com- pound interest being allowed % An annuity of £12 unpaid for 18 years would amount to £28-13238 X 12 = £337-688£G. "1;* ANNUITIES. 343 with its int>^- 1 1 the 8UIU \ would yroduco in perpetuity into £1, and the quotient will be tne sum required to produce an annuity of £>\ per anunm in perpetuity. Multiply the qu jtijut by the number of pounds in the given annuity, and the product *vill be the required present worth. Example. — What is the value of an income of £17 for ever \ Let us suppose that £100 would produce £5 per cont. per an. for ever : — then £1 would produce £-05. Thvivefore, to produce £1, we require as many pounds as will be ciual to the number of times £05 is contained in £1. ButTpTss' £20, therefore £20 would produce an annuity of £1 for ever. And 17 times as much, or £20x17=340, which would produce an annuity of £17 for ever, is the required present value. i. , '. ' , EXERCISES. 8. A small estate brings £25 per annum ; whi <; is its present worth, allowing 4 per cent, per annum inte- rest .? Ans. £625. 9. What is the present worth of an income of £347 K 'II ir j !!■• •:i 344 ANNUITIES. in perpetuity, allowing 6 per cent, interest? Ans £5783 6s. Sd. 10. What is the value of a perpetual annuity of j£46, allowing 5 per cent, interest ? Ans. £.920. 30. To find the present value of an annuity in rever- sion — liuLE. — Find the amount of the annuity as if it were forborne until it should cease. Then find what sum, put to interest now, would at that time produce the same amount. Example. — What is the value of an annuity of £10 per annum, to continue for 6, but not to commence for 12 years, 5 per cent, compound interest being allowed ? An annuity of £10 for 6 years if left unpaid, would be worth £680191 ; and £1 would, in 18 years, be worth £11-68959. Therefore £680191 w=£28 5s. 3d., is the required present worth. 11G89^9' EXERCISES 11. what is the present worth of £75 per annum, which is not to commence for 10 years, but will con- tinue 7 years after, at 6 per cent, compound interest } Ans. iei55 9s. l^d. 12. The reversion of an annuity of £175 per annum, to continue 1 1 yearjj, and commence 9 years hence, is to be sold ; what is its present worth, allowing 6 per cent, per annum compound interest } Ans. £430 7*. \d. 13. What is the present worth of a rent of £45 per annum, to commence in 8; and last for 12 years, 6 per cent, compound interest, payable half-yearly, being allowed } Ans. £117 2s. S^d", 31 When the annuity is contingent, its value depend? on the probability of the contingent circumstance, or circumstances. A life annuity is equal to its amount multiplied by the value of an annuity of £1 (found by tables) for the given ago. The tables used for the purpose are calcu- lated on principles derived from the doctrine of chances, observations on the duration of life in different cu'cum- i^tances, the rates of compound interest, &c. POSITION. 345 QUESTIONS. 1. What is an annuity ? [25]. 2. What is an annuity in possession — in reversion — cortiiin — contingent — or in arrears ? [25] . 3. What is meant by the pre^ient worth of an an- nuity r [25] . 4. How is the amount of any number of payments in arrears found, the interest allowed being simple or compound r [26] . 5. How is the present value of an annuity in posses- sion found ? [28]. 6. How is the present value of an annuity in per- petuity found ? [29] . 7. How is the present value of an annuity in rever- sion found f [30] . resent worth. POSITION. 32. Position, called also the " rule of false," is a rule which, by the use of one or more assumed, but false numbers, enables us to find the true one. By means of it we can obtain the answers to certain questions, which wc could not resolve by the ordinary direct rules. When the results are really propo)tional to the sup- position — as, for instance, when the number sought is to be tmdtipfied or divided by some proposed number ; or is to be increased or diminished by itself^ or by some given mullipk or part of itself — and when the question contains ox\\y one prtposUion^ we use what is called sing'le position, assuming only ofie number ; and the quantity found is exactly that wliich is lequired. Other- wise — as, for instance, when the number sought is to be increased or diminished by some absolute number, which is not a known multiple, or part of it — or when tico p'opositions, neither of which can be banished., are coii- tiiined in the pioblem, we use double position, assuming two numbers. If the number sought is, during the process indicated by the question, to be involved or evolved, we obtain only an approximation to the quan- tity required. IJ ii ■ 346 POSITION. 33. Single Position. — Rule. Assume a number, and perform with it the operations described in the question ; then say^, as the result obtained is to the number itsedy so is the true or given result to the number required. Example. — What number is that which, being multiplied by 5, by 7, and by 9, the sum of the results shall be 231 1 Let us assume 4 as the quantity sought. 4x5-f-4x7-f- 4x9=84. And 84 : 4 :: 231 ; Iri|i^=:ll, the required number. 34. Reason of the Rule. — It is evident that two num- bers, multiplied or divided by the same, should produce pro- portionate results. — It is otherwise, however, when the same quantity is added to, or subtracted from them. Thus let the given question be changed into the following. What number is that which being multiplied by 5, by 7, and by 9, the sum of the products, plus 8, shall be equal to 239 ? Assuming 4, the result will be 92. Then we cannot say 92 (84-f8) : 4 : :• 239 (231-f 8) : 11. For though 84 : 4 : : 231 : 11, it does not follow that 844-8 : 4 : : 231-J-8 : 11. Since, while [Sec. V. 29] we may multiply or divide the first and third terms of a geometrical proportion by the same number, we cannot, without destroy- mg the proportion, add the same number to, or subtract it from them. The question in this latter form belongs to the rule of double position. ] . • EXERCISES. 1. A teacher being asked how many pupils he had, replied, if you add j, i, and } of the number together, the sum will be 18 ; what was their number .'' Ans. 24. 2. What number is it, which, being increased by ^, 1, and i of itself, shall be 125 .? Ans. 60. 3. A gentleman distributed 78 pence among a num- ber of poor persons, consisting of men, women, and chil- dren ; to each man he gave 6Q 135. Ad. ; for the horse, £13 6*. ^d. ; and for the chaise, i£40. 5. A's age is double that of B's ; B's is treble that of C's ; and the sum of all their ages is 140. What is the age of each } Ans. A's is 84, B's 42, and C's 14. 6. After paying away \ of my money, and then j of the remainder, I had 72 guineas left. What had I at first } Ans. 120 guineas. 7. A can do a piece of work in 7 days ; B can do the same in 5 days ; and C in 6 days. In what time will all of them execute it .'' Ans. in ly^^ days. 8. A and B can do a piece of work in 10 days ; A by hhnself can do it in 15 days. In what time will B do it .'' Ans. In 30 days. 9. A cistern has three cocks ; when the first is opened all the water* runs out in one hour ; when the second is opened, it runs out in two hours ; and when the third is opened, in three hours. In what time will it run out, if all the cocks are kept open together : Ans. In y\ hours. 10. What is that number whose i, ^, and ^ parts, taken together, make 27 } Ans. 42. 11. There are 5 mills; the first grinds 7 bushels of corn in 1 hour, the second 5 in the same time, the third 4, the fourth 3, and the fifth 1. In what time will the five grind 500 bushels, if they work together .'' Auo. In 25 hours. 12. There is a cistern which can be filled by a cosvk in 12 hours ; it has another cock in the bottom, by which it can be emptied in 18 hours. In what tine will it bo filled, if both are left open } Ans. In 36 Ldlivs. 35. Double Position. — Rule I. Assume two con- venient numbers, and perform upon them the processes supposed by the question, marking the error derived from each with + or — , according as it is an error of txcess, 9r of defect. Multiply each assumed number into the error which belongs to the other ; and, if the errors are both plus, or both minus, divido the dijference of the products by the difference of the errors. But, if one is a plus, and the other is a minus error, divide the sum of 348 POSITION. ir. the products by the sum of the errors. In either case the result will be the number sought, or. an approxi- mation to it. Example 1. — If to 4 times the price of my horse £10 is added, the sum will be £100. What did it cost '? Assuming numbers which give two errors of excess- First,, let 28 be one of them, Multiply by 4 112 Add 10 From 122, the result obtained, subtract 100, the result required, and the remainder, +22, is an error of excess. Multiply by 31, the other assumed number and G82 will be the product. Next, let the assumed number be 31 Multiply by 4 124 Add 10 From 134, the result obtained, subtract 100, the result required, and the remainder, -f-34, is an error of excess. Multiply by 28, the other assumed uum. and 952 will be the product From this subtract 682, the product found ;ibove, divide by 12)270 and the required quantity is 22-5=£22 10s. Difference of errors=34— 22=12, the number b^ which we have divided. 36. Ueason of the Rule. — When in example 1, we mul- tiply 28 and 31 by 4, we multiply the error belongim. to each by 4. Hence 122 and 134 are, respectively, equal to the true result, plus 4 times one of the errors. Subtracting lOO, the true result, from each of them, we obtain 22 (4 times i \e error in 28) and 34 (4 times the error in 31). But, as numbers are proportional to their equim Itiples, the error in 28 : the error in 31 : : 22 (a multiple of t lie for- mer) : 34 (an equimultiple of the latter). And from the nature of proportion [Sec. V. 21]— POSITION. 349 In either case or. an approxi- my horse £10 is b cost 1 8 of excess — ned, ired, rcesii. iied number iuct. suit obtained, suit required, error of excens. her assumed aum. 3 the produc t. oduct found ibove, 2 10s. lumber h\ which sample 1, we mul- belongin) to each , equal to the true ibtracting lOO, the 1 (4 times i le error lelr equim Itiples, nultiple of t iie for- r. 211- The err or in 28 x34=th e error in 31x22. But 682= the error in Bl^-the required number X 22. And 952=tlie error in 28-f-the required number X 34. Or, since to multiply quantities under the vinculum [Sec II. 34], we are to multiply each of them — 682=22 times the error in 31-|-22 times the required number. 952=84 times the grror in 28-J-34 times the required number. Subtracting the upper from the lower line, we shall have 952 — 082=34 times the error in 28 — 22 times the error in 81-J-34 times the required number — 22 times the required number. But, as we have seen above, 34 times the error in 28=22 times the error in 31. Therefore, 34 times the error in 28 — 22 times the error in 31=0; that is, the two quantities cancel each other, and may be omitted. We shall then have 952 — 682=:34 times the required number — 22 times the re- quired number; or 270:=34— 22 (=12) times the required uuuiber. And, [Sec. V. 6] dividing both the equal quanti- ties by 12, 270 34-22 -^ (22 '5) = — r^ — times (once) the required number. 37. Example 2. — Using the same example, and assuming numbers which give two errors of defect. Let them be 14, and 16 — 14 16 4 4 56 10 06, the result obtained, 100, the result required, — 34, an error of defect. 16 64 10 74, the result obtained, 100, the result required, 544 — 26, an error of defect. 1-1 364 Difference of errors = 34 — 26 = 8. 8)186 22-5 =£22 10s., is the required quantity. In this example 34=four times the error (of defect) in 14; ami 2b = four times the error (of defect) in 16. And, since numbers are proportional to their equimultiples, The error in 14 : the error in 16 : : 34 : 26. Tl srefore The er ror in 14X26=th e error in 16 X34. But 644^the required number — the error in 1<>X34 And o64=the required number — the error in 14x26 Q 2 .: r I . t V i 1 POSITION. If we subtract the lower from the upper line, we shall have 644 — 364=(rcinoving the vinculum, and changing the sign [Sec. II. 16]) 84 times the required number— 20 times the required number — 34 times the error in 16+26 times the error in 14. But we found above that 84 times the error in 16s=:26 times the error in 14. Therefore — 84 times the error in 16, and-j--6 times the error in 14:=0, and may be omitted. We will then have 644 — 364=34 times the required number — 26 times the required number ; or 180=8 times the required number ; and, dividing both these equal quantities by 8, 180 8 -g- (22*5) ==sg times (once) the required number. 38. Example 3. — Using still the same example, and as- eimung numbers which will give an error of excess, and pa error of defect. I it them be 15, and 23 — 1.5 4 GO iO 70, the result obtained. 100, the result required. — 30, an error of defect, 23 690 30 23 4 92 10 102, the result obtained. 100, the result required. 4-2, an error of excei>s. 15 30 Sum of errors = 30 -f 2 = 32. 32)720 22" 5 = J£22 10s., the required quantity. In this example 30 is 4 times the error (of defect) in 15 ; and 2, 4 times tiie error (of excesfe) in 23. And, since numbers are proportioned to the eqii {multiples, The error in 23 : the eri or iii 15 : : 2 : 30. Therefore The error in 23 X30=tho error in 15 X^- But 61)0=the required number-f-the error in 23x30. And 30=the required number — the error in 15x2. If we add these two linos together, we shall have 690-|-30= (^removing the vinculum) 30 times the required number-f twice the required number -f- 30 times the error in 23 — twice the error in 16. But we found above that 30 X the error in 28=2 X the error in 15. Therefore 30 X the error in 23 — 2xtho error in 15=0, POSITION. 351 ed numbpf. n error of excei>s. 30. Therefore Tor in 23X30. and may be omitted. We shall then have 690-f 30-=the re- quired number X 30 -f- the required number X 2 ; or 720=32 times the required number. And dividing each of these equal quantities by 32. 720 32 -^2 (22'5)=^ times (once) the required number. The given questions might be changed into one belonging to single position, thus — Four times the price of my horse is equal to £100 — £10; or four times the price of my horse is equal to £90. What did it cost ? This change, however, supposes an effort of the mind not required when the question is sol ved by double position. 39. Example 4. — What is that number which is equal to 4 times its square root +21 1 "*_ Assume 64 anf^ y64=8 4 32 21 81— ^^81= 9 4 36 21 53, result obtained. 64, result required. -11 81 8ui 57, result obtained 81, result required 6t 1536 891 13)645 The first approximation is 49 61 5 4 It is evident that 11 and 24 are not the errors in the assumed numbers multiplied or divided by the same quantity, and } therefore, as the reason upon which the rui~ is founded, does jnot apply, we obtain only an approximation. Substituting this, however, for one of the assumed numbers, we obtain a j still nearer approximation. 40. Rule — II. Find the errors by the last rule ; then I divide their di£ference (if they are both of the same kind), or their sum (if they are of different kinds), into the product of the difference of the numbers and one of the errors. The quotient will be the correction of that j error which has been used as multiplier. 352 POSITION. |V ;■ Example. — Taking the same as in the last rule, and m Burning 19 and 25 as the required number. 19 4 76 1& 86 the result obtained. ' 100 the result required. —14, is error of defect. 110 the result obtained. 100 the result required. -|-10, is error of excess. The errors are of different kinds j and their sum is 14-j. f0=s24; and the differenoa of the assumed numbers is 25— 19=6. Therefore 14 one of the errors, is multiplied by 6, by the difference of the numbers. Then divide by 24)84 and 35 is the correction for 19, the number ■which gave an error of 14. 19-|-(the error being one of defect^ the oorreotion is to be added) 8-5=s22 5=£22 10s. is the required quantity. 41. Reason of the Rule. — The diflFerence of the resultu arising from the use of the different assumed numbers (tho diflFerence of the errors) : the diflFerence between the result ob- tained by using one of tho assumed numbers and that obtained by using the true number (one of the errors) : : the diflFerence between the numbers in the former case (the diflFerence between the assumed numbers) : the diflFerence between the numbers in the latter case (the diflFerence between the true number, and that assumed number which produced the error placed in the third term — that is the correction required by that assumed number). It is clear that the diflFerence between the numbers used I produces a proportional diflFerence in the results. For the results are diflFerent, only because the diflFerence between the assumed numbers has been multiplied, or divided, or both— in accordance wifh the conditions of the question. Thus, in the present instance, 25 produces a greater result than 19, because 6, the diflFerence between 19 and 25, has been multi' plied by 4. For 25x4=19x4-f-6x4. And it is this 6X4 which makes up 24, the real diflFerence of the errors. — The diflFerence between a negative and positive result being the I sum of the diflFerences between each of them and no result. | Thus, if I gam 10«., I am richer to the amount of 2is. than if | I lose 14.V. , -. ^. . ., ^.w. POSITION. 353 last rule, and at* numbers. Then r 19, the number ceived the following answer. Your age is now | mine, but 5 years ago it was only \. What arc ages ? EXERCISES. 13. What number is it which, being multiplied by 3, the product being increased by 4, and the sum divided by 8, the quotient will be 32 ? Ans. 84. 14. A son asked his father how old he was, and re- of their Ans.' SO and 20 15. A workman was hired for 30 days at 2s. 6d. for every day he worked, but with this condition, that for every day he did not work, he should forfeit a shilling. At the end of the time he received £2 14*., how many days did he work ? Ans. 24. 16. Required what number it is from which, if 34 be taken, 3 times the remainder will exceed it by \ of itself .> Ans. 58f 17. A and B go out of a town by the same road. A goes 8 miles each day ; B goes 1 mile the first day, 2 the second, 3 the third, &c. When will B over- take A } A. B. Suppose 5 1 8 2 ••A 8 40 4 15 5 5)25 15 -5 7 ii 20 Suppose A. 7 8 56 28 7)^ -4 5 B. 1 2 3 4 5 6 7 28 20 1)15 5-4=1 We divide the entire error by the number of days iu each ! case, which gives the error in one day. 18. A gentleman hires two labourers; to the one he I gives Qd. each day ; to the other, on the first day, 2d.j on the second day, 4d., on the third day, 6d., &c. In I how many days will they earn an equal sum } Ans. In 8. 19. What are those numbers which, when added, 854 FOSITION". ■ ! 1^ ^S p.,. muko 23 ; hut wlicn one is halved and tlio other doubled, give LMjual results r Ans. 20 aud 5. 20. Two contractors, A and 13, are each io build a wall of equal dimensions ; A employs as many men as fl.wsli 22i pt>rehes in a day ; B eniploys the first day as many as finish 6 perches, the second as many as fiijish 9, the third as many as finish 12, &c. In what time will they have built an equal number of perches r Ans. In 12 days. 21. What is tint number whose ^, |, and f, miPM- plied together, make 24 ? Suppose 12 rroduct=iH 1=^ Suppose 4 J =2 J— 1 Product=2 3 — li 81 result u!>+ained. 24 rcsijlt required. +57 64, the cube of 4. 3 result obtained. 24 result required, —21 1728, the cube of 12. 3648, product. 57+2l=:78 57-21=78. 3G288 To this product 3G48 hi added. 78 )39036 is the sum. And 512 the quotient. ^512=8, is the required number. We multiply the alternate error by the cube of the supposed number, because the errors belong to the g\th part of the cube of the assumed numbers, and not to the numbers themselves ; foi', in refjlity, it is the cube of some number that is required — since, 8 being assumed, according to the question we have ?X?X«-|?=24; orl^x8'=24. 22. What number is it whose i, |, |, and ^, multi- plied together, will produce 6998| ? Ans. 36. 23. A said to B, give me one of your shillings, and I shall have twice as many as you will have left. 3 answered, if you give me 1^., I shall have as many as you. How many had each .'' Ans. A 7, and B 5. POSITION. 355 other doubled, ich io build a many men as he first day as nany as fiiiisi; In what timo • of perches r and ^, multi- Lit ol)tained. lit required. cube of 12. lis product tided. le sum. quotient. er. of the supposed part of the cube bers themselves ; that is required uestion we have , and ^, multi- f. 36. shillings, and have left. B ave as many as and B 5. 21. There are two numbers which, when added to* ^.ther, make 3u ; but the ^, i, and ^, of the greater re equal to i, ^^ and \^ of the lesser. What are they } :{ns. \2 and "is. 2'). A gentleman has 2 horses and a saddle worth £.yj. The saddle, if set on the back of the first horse, s\\\ make his value double that of the second ; but if ict on the back of he second horse, it will make his alno treble that of the first. What is the value of acli horse } Ans. £3D and ^240. 26, A gentleman finding several beggars at his door, ;iivc to each 4cl. and had 6^/. left, but if he had given /. to each, he would have had I2d. too little, llo lany beggars were there .'* Am. 9. It is so likely that those who .are desirous of studying liis subject further will be acquainted with the method f treating algebraic equations — which in many case? iffords a so much simpler and easier mode of solvin^^ ui stions belonging to position — that we do not deem necessary to enter further into it. • QUESTIONS. 1. What is the difference between single and double ttsition? [32]. 2. In what cases may we expect an exact answer by |iese rules .'' [32] . 3. What is the rule for single position ? [33] . 4. What are the rules for double position } [35 and )]. MISCELLANEOUS EXERCISES. 1. A father being asked by his son how old he was; [plied, your age is now | of mine ; but 4 years ago was only | of what mine is now ; what is the age of kli? Ans. 70 and 14. j 2. Find two numbers, the difference of which is 30, |id the relation between them as 7| is to 3^ 't Ans^ and 28. j3. Find two numbers whose sum and product are lual, neither of them being 3 ^ Am. 10 and 1 J, ^^-v^. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 1.1 lii|21 125 ■tt lii 122 ■if |«£ 12.0 u ■Iwu M L25 iU iM — A" Photographic Sdmces Carporation 23 WIST MAIN STRHT WIISTM,N.Y. 14SM (71*) ■72-4903 it ;\ 356 EXERCISES. 4. A person being asked the hour of the day, answered, It is between 5 and 6, and both the hour and minute hands are together. Hequired what it was? Am, 27y\ minutes past 5. 5. What is the sum of the series |, }, |, &c. ? Ans. 1. 6. What is the sum of the series |, tV) t'jj t's!?* &o- • Ans. l\. v.'f r 7. A person had a salary of £75 a year, and let it remain unpaid for 17 years. How much had he to, receive at the end of that time, %llowing 6 per cent, per annum compound interest, payable half-yearly ,' I An9,^£204 I7s. lO^d. 8. Divide 20 into two such parts as that, when the I greater is divided by the less, and the less by the greater, and the greater quotient is multiplied by 4, and the lessj by 64, the products shall be equal.? Ans. 4 and 16. 9. Divide 21 into two such parts, as that when the I less is divided by the greater, and the greater by the lo'«*, and the greater quotient is multiplied by 5, and tl»- less by 125, the products shall be equal ? A'iu.\ 3^ and 171. ]•' A, B, and C, can finish a piece of work in lOj days; B and C will do it in 16 days. In what time wii A do it by himself r Ans. 26| days. 1. A can trench a garden in 10 days, B in 12, an Cm 14, In what time will it be done by the three if| they work together } Am. In 3iVt ^^y^- 12. What number is it which, divided by 16, wil| lettve 3 ; but which, divided by 9, will leave 4 } Aw\ 6- i3. What number is it which, divided by 7, wil leave 4; but divided by 4, will leave 2 ? Ans. 18. 14. If iSlOO, put to interest at a certain rate, will at the end of 3 years, be augmented to iB115'762' (compound interest being allowed), what principal an interest will be due at the end of the first year ? M jei05. . ,. , 15. An elderly person in trade, desirous of a littlj respite, proposes to admit a sober, and industrious youi^ person to a share in the business ; and to encoura him, he offers, that if his circumstances allow him EXERCISES. 357 he day, answered, hour and minute t it was? Am, ., \. &c. ? Ans. 1. a year, and let it much had he to )wing 6 per cent, rable half-yearly ! as that, when the less by the greater, [ by 4, and the less Ans. 4 and 16. , as that when the bhe greater by the iiltiplied by 5, and be equal? Ans. • iece of work in 10 In what time wil days, B in 12, ui ne by the three if days, livided by 16, wa ill leave 4 ? A4 divided by 7, vil 2 ? Ans. 18. , certain rate, will Qted to £115762] what principal an e first year? A<)i desirous of a littl| ad industrious youn • and to enooura )ances allow him advance iSlOO, his salary shall be iB40 a year ; that if he is able to advance ^3200, he shall have £55 ; but that if he can advance i£300, he shall receive iE70 annually. In this proposal, what was allowed for his attendance simply? Ans. £25 a year. '■■' 16. If 6 apples and 7 pears cost 33 pence, and 10 apples and 8 pears 44 pence, what is the price of one apple and one pear ? Ans. 2d. is the price of an apple, and 3^. of a pear. 17. Find three such numbers as that the first and \ the sum of the other two, the second and ^ the sum of the other two, the third and \ the sum of the other two will make 34 ? Aits. 10, 22, 26. 18. Find a number, to which, if you add 1, the sum will be divisible by 3 ; but if you add 3, the sum will be divisible by 4 ? Ans. 17. 19. A market woman bought a certain number of eggs, at two a penny, and as many more at 3 a penny ; and having sold them all at the rate of five for 2/f., she found she had lost fourpence. How many eggs did she buy ? Atis. 240. 20. A person was desirous of giving 3rf. a piece to some beggars, but found he had 8af. too little ; he there- fore gave each of them 2d.^ and had then Zd. remain- ing. Required the number of beggars? Ans. 11. 21 . A servant agreed to live with his master for iE8 a year, and a suit of clothes. But being turned out ttt the end cf 7 moi^ths, he reeeived only £2 13». Ad. and the suit of clothes ; what was its value ? Ans, £4 16*. 22. There is a number, consisting of two places of figures, which is equal to four times the sum of its digits, and if 18 be added to it, its digits will be in- verted. What is the number? Ans. 24. tJ< « 23. Divide the numbcJl* 10 into three such parts, that if the first is multiplied by 2, the second by 8, and the third by 4, the thre« products wiU be equal? Ans. 24. Divide the number 90 mto four such parts that, if the first is increased by 2, the second dimmished by 2) the third midtipUed by 2, and the fourth divided by 858 EXERCISES. %tJ 2, the sum, difference, product, and quotient will bo equal; ^tw. 18, 22, 10, 40. ^.r.,.. 25. What fraction is that, to the numerator of which, if 1 is added, its value will be ^ ; but if 1 be added to the denominator, its value will be | ? Aiis. j*j. 26. 21 gallons were drawn out of a cask of wine, which had leaked away a third part, and the cask being then guaged, was found to be half full. How much did it hold ? Ans. 126 gallons. .. r. ' , •. ■• 27. There is a number, i of which, being divided by 6, ^ of it by 4, and |^ of it by 3, each quotient will be 9 .? Am. 108. 28. Having counted my books, I found that when I multiplied together i, a, and f of their number, the product was 162000. How many had I .'' Atis. 120. 29. Find the sum of the series 1 4-^+^ + 1, &c. ? Ans. 2. f;M :. ■• •." -■':■ . :■ ■■,.■-■."■■. , '■, ... : 30. A can build a wall in 12 days, by getting 2 days' assistance from B ; and B can build it in 8 days, by getting 4 days' assistance from A. In what time will both together build it } Ans. In 6^ days. 31. A and B can perform a piece of work in 8 days, when the days are 12 hours long ; A, by himself, can do it in 12 days, of 16 hours each- In how many days of 14 hours long will B do it } Ans. I34. }«'■■• 32. In a mixture of spirits and water, ^ of the whole plus 25 gallons was spirits, but ^ of the whole minus 5 gallons was water. How many gallons were there of each ? Ans. 85 of spirits, and 35 of water. 33. A person passed } of his age in childhood, y'j of it in youth, | of it +5 years in matrimony ; he had then a son whom he survived 4 years, and who reached only I the age of his father. At what age did this per- son die ? Ans. At the age of 84. .;(; r;r ir 34. What number is that whose ^ exceeds its } by 72.? Am. 540. 35. A vintner has a vessel of wine containing 500 gallons ; drawing 50 gallons, he then filb up the cask with water. After doing this five times, how much wine and how much water are in the cask.'* Am. 295^^7 gallons of wine, and 204 /^|,^ gallons of water. Qt will bo • of which, le added to 4_ k of wine, I the cask full. How divided by aotient will hat when I lumber, the Afis. 120. i + i, &c.? bting 2 days' 18 days, by at time will t in 8 days, limself, can If many days of the whole lole minus 5 Bre there of • ahood, yV ^^ )ny ; he had who reached did this per- eds its \ by ntaining 500 up the cask how much I cask? Ans. Ins of water. EXERCISES. 359 * <). A mother and two daughters working together m»4 spin 3 ft) of flax in one day ; the mother, by herself, )an do it in 2^ days ; and the eldest daughter in 2^ days. In what time can the youngest do it ? Ans, In 6^ days. 37. A merchant loads two vessels, A and B ; into A he puts 150 hogsheads of wine, and into B 240 hogs- heads. The ships, having to pay toll, A gives 1 hogs- head, and receives I2s. ; B gives 1 hogshead und 36s. besides. At how much was each hogshead valued } Am. £4 12*. 38. Three merchants traflSc in company, and their stock is iS400 ; the money of A continued in trade 5 months, that of B six months, and that of C nine months ; and they gained i2375, which they divided equally. What stock did each put in.? Ans. AiB167|f, B iei39af , and C £93^^. 39. A fountain has 4 cocks. A, B, C, and D, and under it stands a cistern, which can be filled by A in 6, by B in 8, by C in 10, and by D in 12 hours ; the cistern has 4 cocks, E, F, G, and H ; and can be emptied by E in 6, by F in 5, by G- in 4, and by H in 3 hours. Suppose the cistern is full of water, and that the 8 cocks are all open, in what time will it be emptied? Ans. In 2f^ hours. 40. What is the value of '2^97' > Ans. i|. 41. What is the value of '5416' ? At^. |f . 42. What is the value of •0^76923' } Ans. yV- 43. There are three fishermen, A, B, and C, who have each caught a certain number of fish ; when A'a fish and B's are put together, they make 110; when B's and C's are put together, they make 130 ; and when A's and C's are put together, they make 120. If the fish is divided equally among them, what will be each man's share ; and how many fish did each of them catch ? Ans. Each man had 60 for his share ; A caught 50, B 60, and 70. 44. There is a golden cup valued at 70 crowns, and two heaps of crowns. The cup and first heap, are worth 4 times the value of the second heap ; but the cup and second heap, are worth double the value of the first 360 EXERCISES. u. it }i w ill ' h W'' m heap. How many crowns are there in ench heap } Am 50 in one, and 30 in another. 45. A certain number of horse and foot soldiers are to be ferried over a river ; and they agree to pay 2\d. for two horse, and 3\d. for seven foot soldiers ; seven foot always followed two horse soldiers ; and when they were all over, the ferryman received £>2b. How many horse and foot soldiers were there ? Am. UQOO horse, and 7000 foot. 46. The hour and minute hands of a w^tch are to- gether at 12 ; when will they be together i^ain ? Ans, at 5fy minutes past 1 o^clock. 47. A and B are at opposite sides of a wood 135 fathoms in compass. They begin to go round it, in the same direction, and at the same time ; A goes at the rate of 11 fathoms in 2 minutes, and B at that of 17 in 3 minutes. How many rounds will each make, before one overtakes the other .? Ans. A will go 17, and B 161. 48. A, B, and 0, start at the same time, from the same point, and in the same direction, round an island 73 miles in circumference ; A goes at the rate of 6, B at the rate of 10, and at the rate of 16 miles per day. In what time will they be all together again T Ans. in 36^ days ( . ' ' • r .It V. if ... -f ♦ ■>; i • ■ ■ ■ '' ' t ■•. ,-::r// r I 1 ..(. i'!- ji'iii :■ MIS V - r-i"[-'. >! u\ ,.' i heap ? Ant t soldiers are i to pay 2^d. )ldiers ; seven ijid when they How many ;. 2000 horse, W9'teh are to- a^ain ? Ans. ' a wood 135 ound it, in the A goes at the at that of 17 h make, hefore go 17, and B time, from the ound an island the rate of 6, f 16 miles per ogether again T '[ .;:.!■■// ,i: / t .MC7 362 LOGARITHMS. ffl , ^ I' ii pp N. too 1 S 3 4 5 « 7 8 • D. 43'2 000000 000434 000868 001301 001734 002166 002598 003029 003461 003891 41 1 4321 4751 5181 5609 6038 6466 6894 7321 7748 8174 428 88 2 8000 9026 9451 9876 010300 010724 011147 011670 011993 012415 124 134 3 012837 013359 013680 014100 4531 4940 5360 6779 6197 6616 420 166 4 7083 7451 7868 8284 8700 9116 9532 9947 020361 020775 416 207 5 031189 031603 032016 032428 032841 023252 023664 034076 4486 4896 112 348 6 5306 5715 6125 6533 6942 7350 7757 8164 8571 8978 408 390 7 9384 0789 030196 080600 031004 031408 031812 032216 032619 033021 404 331 8 033424 J83J26 4327 4628 6029 5430 6830 6330 6629 7028 100 373 9 7426 7835 8323 8620 9017 9414 9811 040207 040602 040998 397 110 041393 041787 042182 042576 042969 043362 043755 044148 044540 044932 393 38 1 5323 5714 6105 6495 6885 7275 7664 8053 8442 8830 390 76 3 9218 9606 9993 060380 050766 051153 051538 051924 052309 052694 386 113 3 053078 053463 053846 4230 4613 4996 5378 5760 6142 6624 383 161 4 6905 7386 7666 8046 8426 8305 9185 9563 9942 060320 379 189 5 060698 061075 061452 061829 063206 062582 062958 063333 063709 4083 376 237 6 4458 4832 5206 5580 5953 6326 6699 7071 7443 7816 373 365 7 8186 8557 8928 9298 9668 070038 070407 070776 071145 071614 370 303 8 071882 072250 072617 072985 073352 3718 4085 4451 4816 6182 366 340 9 5547 5912 6276 6640 7004 7363 7731 8094 8457 8819 163 360 130 079181 079543 079904 080266 080626 080987 081347 081707 082067 082436 35 1 082785 088144 083503 3861 4219 4576 4934 5291 6647 6004 357 70 3 6360 6716 7071 7426 7781 8136 8490 8845 9198 9552 355 104 3 9905 090258 090611 090963 091315 091667 093018 092370 092721 093071 352 139 4 093422 3773 4122 4471 4320 5169 5618 6866 6215 6562 349 174 6 6910 7257 7604 7951 8298 8644 8990 9335 0681 100026 346 309 6 100371 100716 101059 101403 101747 102091 102434 102777 103119 3462 343 344 7 3804 4146 4487 4828 6169 5510 6851 6191 6531 6871 341 378 8 7210 7649 7888 8227 8565 8903 9241 9579 9916 110253 338 313 9 110590 110926 111263 111599 111934 112270 113605 112940 113275 3609 335 333 130 113943 114277 114611 114944 115278 115611 115943 116276 116608 116940 83 I 7271 7603 7934 8265 8595 8926 9256 9586 9915 12024i; 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3d 3 1709 1835 1943 3058 3174 3391 3407 3633 3639 9765 116 46 4 3873 3988 3104 3230 8336 8463 3668 8684 8800 8915 116 68 6 4031 4147 4363 4379 4494 4610 4736 4841 4967 6073 116 70 6 6188 6303 5419 6534 6630 6765 6880 6996 6111 6336 115 81 7 6341 6457 6573 6687 6803 6917 7083 7147 7363 7377 115 93 8 7493 7607 7733 7836 7961 8066 8181 8396 8410 8535 115| 104 9 380 8639 8754 8868 8983 9097 9313 9836 9441 9666 0669 114| 579784 879898 580013 680136 580341 680356 680469 680683 680697 680811 114 11 1 680936 581039 1163 1367 1381 1493 1608 1733 1836 1950 114 33 3 3063 3177 3391 3404 3618 3631 3746 3868 3973 8083 114 34 3 3199 8313 3436 3539 3633 8766 3879 3993 4106 4318 113 46 4 4331 4444 4567 4670 4783 4896 6009 6133 6336 6348 113 67 6 6461 6574 6686 6799 6913 6034 6137 6360 6363 6475 113 68 6 6687 6700 6813 6935 7037 7149 7363 7374 7486 7699 Ui 79 7 7711 7833 7935 8047 8160 8373 8384 8496 8608 8730 Ui 90 6 8833 8944 9056 9167 9379 9391 9603 9615 9736 9838 m 103 9 9930 590061 590173 590384 390396 690507 690619 690730 690843 690953 113 890 591066 501176 691387 691399 691610 691631 691733 691843 691966 693066 111 11 1 3177 3388 3399 3610 3631 3733 3843 3964 8064 8173 111 33 2 3386 » 8397 3508 3618 3739 8840 8950 4061 4171 4383 HI 83 a 4393 4503 4614 4734 4834 4945 6033 6165 6376 6386 no 44 4 5496 6606 6717 6837 6937 6047 6137 6367 6877 6487 110 66 6 6697 6707 6817 6937 7037 7146 7356 7366 7476 7686 110 66 6 7695 7805 7914 8034 8134 8343 8358 8463 8673 8681 no 77 7 8791 8900 9009 9110 9338 9387 9446 9656 9666 9774 109 88 8 9883 9993 600101 600310 600319 600438 600587 600646 600766 000864 109 09 9 600978 601083 1191 1399 1408 1617 1635 1734 1843 1961 109 LOGARITHMS. 367 |N.| I »4a 639873 1618 364A 1787 1068 r81A 1674 mo 080 1807 1671 Sl'i 040 389 613 780 963 181 898 810 »0 4914 6180 7441 6600 0964| Ml 306 34631 683600 8773 6041 0306 7667 8836 640070 1880 3676 8830 644986 •46060 6173 6396 7406 7699 8686 8768 9861 0964 661084 661906 8C90I 61671 64331 76031 8061 I O40304I 14441 37011 89441 9808 8619 4781 6940 )36 138 138 134 117 106 93 176 66 33 667146 8840 0648 660743 1036 3136 4811 6404 6673 7849 9496 8640 4869 6061 6419 7663 8881 6601 U6 133M 3647 8763 4073 6183 667967 0667 660863 9066 8944 4430 6613 6701 7067 660093 670103 1869 9633 3684 4841 6006 7147 8906 0441 9 680683 1799 3868 3009 6199 6360 7374 8406 0615 600730 669140 670300 1476 9630 8800 4067 6111 7969 8410 0656 667887 868» 9787 660083 9174 8863 4648 6730 6000 80»4 601843 9964 4061 6165 6367 7366 8463 8556 S00646 1734 680607 1836 9879 4106 6335 6363 7486 8608 0736 500849 660357 670436 1603 975A 8015 6073 6336 7377 8535 0660 m m m III! 1191 119 1191 lie ii» 680811 1000 8085 4318 6348 6475 7699 8730 0838 ii; 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7022 4i 9 2 7666 7711 7766 7800 7646 7890 7934 7979 8024 8068 4S 14 3 8113 8167 8202 8247 8291 8336 8381 8435 8470 8514 4S 18 4 8569 8604 8648 8693 8737 8782 8826 8871 8916 8900 4> 33 5 9005 0049 9094 9138 9183 9227 9272 9316 9361 9405 4i 37 6 9450 9494 9539 9583 9628 9072 9717 9761 9606 9850 4( 8-2 7 9895 9939 9983 990028 990072 990117 990161 990206 990260 990294 44 36 8 990339 990383 990428 0472 0516 0561 0606 0650 0694 0738 4t 41 9 0783 0827 0871 0916 0960 1004 1049 1093 1137 1183 41 980 991226 991270 991315 991859 991403 991448 991493 991536 991560 99163fi 41 4 1 1669 1713 1768 1802 1846 1890 1936 1979 2023 3067 41 9 2 2111 2156 2200 3244 3288 2333 2377 3421 2465 350fi 41 13 3 2554 2598 3642 3686 3730 2774 3819 2863 3907 3961 44 18 4 2996 3039 3083 8127 3172 3216 3360 3304 3346 3393 4 22 6 3436 8480 8624 8568 8618 8667 3701 874S 8789 3639 4 26 6 3877 3921 3965 4009 4053 4097 4141 4186 4229 4279 4 31 7 4317 4361 4405 4449 4493 4637 4581 4626 4669 4719 35 8 4767 4801 4845 4889 4933 4977 6031 6065 6108 6163 40 9 6196 6240 6284 6328 6372 6416 6460 6504 6647 6691 990 995635 995679 995723 996767 995611 995854 995S99 995942 995986 996030 i 4 1 6074 6117 6161 6205 6349 6293 6337 638C 6424 6464i 9 2 6512 6565 6699 6643 6687 6731 6774 6818 6663 6906 13 3 6949 6993 7037 7080 7134 7168 7213 7265 729fl 7343 IH 4 7886 7430 7474 7517 7561 7606 7648 7693 7736 777S 2a 7823 7867 7910 7954 7998 8041 8085 812S 817u 8216 26 6 8259 8303 8347 8390 8434 8477 8521 8564 860G 1 8652 31 7 8695 3739 8782 8826 8369 8913 8956 900(1 1 9049 1 90S7 35 8 9131 9174 9218 9261 9305 9348 9392 9438 947fi 1 9523 40 9 956S 960S 9662 9696 9738 9783 9826 9870 1 9912J 9957 9734A1 3913 4874 4834 6394 5t53 6213 «671 7129 7586 8 D. 973497 8959 4430 4880 5340 5709 6358 6717 7175 7633 073548146 40051 4C 44661 46 49361 46 63861 46 68451 46 978043 978089 8500 8956 9413 9867 980333 0776 1339 1683 3135 6804 6763 7330 7678 8546 9003 9457 9913 980367 0831 1375 1738 3181 A TABLE OF 8QUAHG8, cUUES, AND ROOTS. 377 983588 8040 8491 8943 4393 4843 6393 6741 6189 6637 978135 8591 9047 9603 9958 980413 0867 1330 1778 3336 40 88 34 81 136 73 17 61 (05 987085 987130 983633 3085 3536 3987 4437 4887 6337 6786 6334 6683 983678 a 3130 8581 4083 4483 4933 6883 6830 6379 4i 6727 7533 7979 8435 8871 9316 9761 7677 8034 8470 8916 9361 9806 987176 7632 8068 8514 990206 990350 0650 149 1093 192 »3d 177 tl9 i60 roi 41 >81 )31 160 4il 0694 1187 8960 4i| 9405 9850 990394 0738 1183 991536 1979 3431 3863 8304 3746 4186 4626 6066 6504 991580 3033 3465 2907 3348 8789 4339 4669 6108 6647 991635 3067 2509 2951 8392 3833 4373 4713 6153 6591 B98 J37 774 213 S48 1)85 521 956 393 826 995943 995986 6380 6434 6818 6863 7256 7399 7693 7736 8139 817-i 8564 8608 9000 9043 9436 9479 9870 9913 996030 6468 6906 7343 7779 8316 8653 90S7 9532 9957 No. 84|Uftr«> 1 3 3 4 5 7 8 9 10 11 1-i 13 14 Id 16 17 16 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 36 37 38 39 40 41 42 43 44 46 46 47 4ti 49 60 51 52 83 54 55 66 57 58 69 60 61 62 63 Cube. 1 4 9 16 35 36 49 64 81 100 131 144 169 196 326 256 289 324 361 400 441 484 629 576 635 676 729 784 841 9U0 961 1034 1089 1156 1235 1396 1369 1444 1521 1600 1681 1764 1849 1936 3U36 2116 2309! 3304; 3401; 3500: 2601 2704! 3809; 3916 3026! 3136| 33491 3364 3481 3600 3731 3844 3969 Sq. Root. ,Cub« Root 1 8 27 64 135 216 343 512 739 1000 1331 1738 2197 2744 8375 4096 4913 6833 6859 8000 9361 10648 12167 13834 15635 17576 19683 21952 24389 27000 39791 33708 35937 39304 43875 46656 60653 64873 69319 64000 68931 74088 79507 65184 91135 97336 108833 110593 117649 135000 133651 140608 148377 157464 166376 175616 185193 195113 205379 216000 336981 338338 S50047 00000001- 4143186 1- 7330508 1 ■ 0000000 r 3360680 1- 4494897 1- 6457513 1 83843713 3-0000000 3 3-1622777|3 3-3166348 3 3-4641016,3 3-6055513,3 74165743 87-298333 00000003 1331056:3 34-20407 3 35889893 4731360:3 583575713 6U04158|3 79583153 89897953 00000003 0990195S 19615343-000000 000000 359931 442350 587401 709976 817131 913931 000000 080084 154435 323980 289428 351335 410142 4663U2 619843 571282 620741 669403 '714418 ■768934 '80-2039 '843867 ■8844U9 -934018 -963496 No. 39150-263 38516183 4773-256 5677644 6568543 7445636 8309519 9160798 0000000 6-0837635 6*1644140 6-2449980 3245553 4031243 4807407 6674385 6333496 7082039 6-7833300 6-8566646 6-9-28-2033 0000000 0710678 1414384 3111036 3801099 3484693 4161986 4833148 6498344 6167731 6811457 7459667 8103497 '8740079 •937C.»W 036589 072317 107233 141381 174802 3 307534 3-339613 371066 3019-27 332222 361976 3-391211 3-419953 448217 476037 603398 630348 656893 683048 608836 634-241 6593061 684031 3-7084301114 64 65 66 67 68 69 70 71 73 73 74 76 76 77 78 79 80 81 83 83 84 85 86 87 88 89 90 91 93 93 94 95 96 97 98 99 100 101 103 103 104 105 lOG 107 im 109 110 HI 113 113 Square 732511 756286 779763 802953 825862 848501 870877 89-2996 91486 936497 3-957893 3-979057 115 116 117 118 119 120 131 1*32 133 124 135 136 Cube. 4096 4335 4356 4489 46-34 4761 4900 6041 5184 6339 6476 66*26 5776 5929 6084 6341 6400 6561 6724 6889 7056 7-235 7306 7569 7744 7931 8100 8381 8464 8649 8836 9035 9316 9409 '-."■;' i 98C1 10000 1G.01 10404 10609 10816 11035 11336 11449 11664 11881 13100 13331 12544 13769 12996 13-236 13456 13689 13924 1^161 14400 14641 14884 15129 15376 1562a 15876 Sq. Root. 263144 274635 287496 300763 314433 338509 343000 357911 873348 389017 405334 421875 438976 456533 474553 493039 612000 631441 651 368 671787 692704 614125 636056 658503 681472 704969 729000 763571 778688 804357 830584 857375 884736 913673 941193 970399 1000000 1030301 1061208 109-2727 1124864 1167625 1191016 1226043 1269712 1295029 1331000 1367631 1404928 1443897 1481544 1530875 1.560896 1601613 1643033 1685169 1728000 1771561 1815848 1860867 1906624 1953125 3000376 9- 9- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 10- 11- 11- 11- 11- 11- 11- Cube Root 0000000 4 -000( «0 0632.>77 4-0-207* 1 1-240.184 4-04l24« 1853,ri8 -246-2113 3066230 3666003 8-4-26M98 8-485-2814 5440037 6033353 6603540 7177979 7740644 8317609 8881044 944-2719 0000000 0553851 11043.36 1651514 3195445 3736185 327.1791 3808315 4339811 48683.10 5393930 6916630 6436508 6953597 7467943 9-7979590 9-8488578 8994949 9498744 0000000 0498756 0995049 1488916 1980390 10-2469508 10-2956301 3440804 3923048 4403065 4880885 5356538 6830052 6301458 6770783 7-238053 7703396 8166538 4*890973 8637805 4-904868 0087121 4*918685 9544512 4*932434 00000004-946088 0453610 4-959675 090536514-973190 1355387;4- 986631 1803399 6-000000 2249722 6-013298 06154^ 081656 101566 131-285 140318 160168 179339 198336 317163 335834 3.54331 372659 4-290841 308870 326740 344481 36-2071 379519 396830 414005 431047 447960 464745 481405 497941 514367 530656 646836 66-2903 878857 594701 610436 626065 641689 657010 672339 687648 702669 717694 73-2624 4*747469 762203 776856 791420 805896 830384 834588 848808 863944 4*876999 378 SQUARES, CUBES, AND ROOTS. I* ■ Ns. 127 8<|imrt. Cub*. Sq. Root. Cube Iloot No. Square. Ciib«. Sq. Root. Cubt Rii.it 1H129 2048383 11-2694277 5 -0-26626 190 36100 6859000 13-7840488 5-74899: 138 16.181 2097152 11 •31.17085 5 •0.19684 191 36181 6967871 13-8-20-2750 5 -7.58!)():, 1.29 lUb41l 2146089 11 -.3578167 6-0.T2774 192 36864 7077888 13-8564065 3 -768»;ls 130 ItiUOJ 2197000 11-4017543 5 06.5797 193 37249 7189057 l3-8924410'5-778f»!)(i 131 17i(U 2248091 ll-44d.V231 5 0787.53 194 37638 7301384 13^9-28.18S3 3-78S!t»iO 13i 17424 2299968 11-4891263 3-091643 195 38026 7414875 13-964-2400|3-79881tO 183 17689 2352637 11-332.5626 5-104469 196 38416 7529330 14-0000000 5 -8087 r*6 134 179A6 2406104 11-3758369 5- J 17230 197 38809 764.5373 14-0356688 5- 81 8H4S 18a 18W.) 2460376 11-6189300 3-1-299-28 198 39204 7762392 14-0712473 6-8-28»:6 136 184U6 26164iV6 11-6619038 6-14-2563 199 39601 7880599 14-1067380 3-838-2:j 137 1SJ769 2571353 11-7016909 6-165137 200 40000 8000000 14-1421.156 5-84803.; 138 19U44 2628072 11-747.3144 5-167649 201 40401 8r20601 14-1774409 5-8577tiii 139 19321 2685610 11-7898261 5-180101 202 40804 8242408 14 21-26704 6 •8674WI 140 19600 2744000 11-8321506 5-19-2494 203 41-209 8.165427 14 2478068 3-8771.1II 141 19881 2803221 11 8743421 6-2048281 204 41616 8489064 14 •28-28,569 6-88676) 142 20164 2863288 11-9163763 3-217103 205 42025 86151-25 14-3178211 3-896.36^ 148 2U449 2924207 11-9882607 5-2-29321 206 42436 8741816 14-3527001 3-90.5941 144 20736 2985984 12-0000000 5-241483 207 4^2849 8:369743 14-3874946 6-9154S3 145 21026 3048625 12-0415946 5-253688 208 43264 8998912 14-4-22-2051 5 -924911;) 146 21316 8112136 12-0830460 6-26.5637 209 43681 91233-29 14- 4568323 6-93447.1 147 21609 3176523 12-1243567 3-277632 210 44100 9261000 14-4913767 5-9439'21 148 21004 8241792 12-1635'i5l 5-289572 211 44521 9.193931 14- 6-258390 5-9.-..134I 149 22201 3307949 12-2065566 3-301459 212 44944 95-28128 14-5602198 5-9627.31 150 . 22300 3376000 12-2474487 5 -31 3-293 213 45369 9663.597 14-5945195 5-972091 151 22801 3442951 12-2882056 5-325074 214 45796 9800344 14-6287388 5 -981 426 153 23104 3511808 12-3288280 6-3.16903 215 46225 9938375 14-66-28783 5 •99072; 153 23409 8581577 12-3693169 6-318481 216 46656 10077696 14-6969385 6-OO00li« 154 23716 8652264 12-4098736 5-360108 217 47089 10218313 14-7309199 6-009244 155 24025 3723875 12-4498996 6-371685 218 47524 10360232 14-7648-231 6-0184rJ3 156 24336 3796416 12-4899960 6-333213 219 47961 10.50.1459 14-7986486 6-0276)0 157 24649 3869893 12-5299641 6-394691 2-20 48400 10649000 14-8323970 6-036811 158 24964 3944312 12 -.5698061 6-406120 -221 48841 1079.1861 14-8660687 6-04594.3 159 25281 4019679 12-6095-202 5-417501 •222 49284 10941048 14-8996644 6-05504! 160 25600 4096000 12-6491106 6-428835 223 49729 11089567 14-9331845 6-064126 161 25921 4173281 12-6885775 6 -4401-22 224 60176 11-239424 14-9a66-295 6-073i:ii 162 26244 4251528 12-7-279221 5-451362 •226 60626 11390625 15-0000000 6-O8-2-20I 163 26569 4330747 12-7671463 5-46-25.56 2-26 61076 11543176 15-033-2964 6-091199 164 26896 4410944 12-8062485 5 • 473704 227 51529 11697083 15-0665192 6-ioono 165 27226 4492125 12-8452326 5^484S06 228 61984 1185-2352 16-0996689 6-1091U 166 27656 4574296 12-8840987 5-495865 •229 62441 12008939 16-1.1-27460 6-118033 167 27889 4657463 12-9228480 6-506879 230 62900 12167000 16-1657509 6-126925 168 28224 4741632 12-9614814 5-517848 •231 63361 12326391 16-1986342 6-135792 169 28561 4826809 13-0000000 6-628775 232 63824 12487168 15-2315462 6-144634 170 28900 4913000 130384048 6-639668 233 64289 12619337 16-2643376 6-153449 171 29241 6000211 13-0766968 5-650499 234 64756 1281-2904 16-2970.585 6-162-239 172 29584 6088448 13-1148770 6-661298 235 66225 12977875 16-3297097 6-171005 173 29929 6177717 13- 1629464 5-67-20,56 236 65696 13144256 16-3622915 6-1797!: 174 30276 6268024 13-1909060 5-682770 237 66169 13312053 16-3948043 6-1884(13 176 30625 6359376 13-2287666 5-593446 238 66644 13481272 15-4272486 6-1971.54 176 30976 6451776 13-2664992 6-604079 239 67121 13651919 16-4596-248 6 •205*21 177 31329 6345233 13-3041347 5-614673 240 67600 138-24000 15-4919334 6-2144t)4 178 31684 6639752 13-3416641 6 -626-226 241 68081 13997621 15-5-241747 6-2-23084 179 32041 5735339 13-3790882 6-635741 242 68564 14172488 15-5563492 6-231679 180 32400 5832000 13-4164079 6-646216 243 69049 14348907 15-6884573 6-2402»l 181 32761 6929741 13-4636-240 5-656651 244 69536 14526789 15 -6204994 6 -243800 182 33124 6028368 13-4907376 5-667061 •245 60025 147061-25 15-65-24758 6-2,57321 183 33'4S9 6128487 13-6277493 6-677411 246 60616 14886936 15-6843871 6-265820 184 33856 6229504 13-6646600 6-687734 247 61009 150692-23 15-7162336 6-274300 186 34225 6331625 13-6014706 5-698019 248 61504 15252992 16-7480157 6-2B->7tiO 186 34596 6434856 13-6381817 5-708267 249 62001 16438249 15-7797338 6-291194 187 34969 6539203 13-6747943 5-718479 250 62500 15625000 15-8113883 6-299604 188 35344 6644672 13-7113092 6-7-28654 261 6300L 15813-251 16-84-29795 6-307993 189 85721 6761269 13-7477271 5-738794 1 252 63504 16003008 16-8746079 6-316369 T8. SQUARES, CI;BES, AND ROOTS. 379 lib*. 8q. Root. Cubt RooiHNo 1» ilSOOOO [)67871 [)778H8 1890r»7 3013:34 414875 rm536 645373 76.239i 88051)9 000000 1.20601 •24-2408 3H:)4J7 4H9a04 «K>1'25 741816 869743 95)891-2 II 233-29 (•261000 1393931 15-28 1-28 1663597 1890344 1938375 )077696 )-218313 136023-2 1503439 I6IB000 793861 1941048 089567 '2394*24 3906-25 643176 697083 352352 !008999 167000 326391 •487168 ,649337 812904 977876 144256 312053 481272 651919 824000 997621 172488 348907 526789 7061-26 886936 069223 252992 438249 625000 813251 003008 13 13 13 13 13 14 14 14 14 14 14 14 14 14- 14- 14- 14- 14- 14- 14- 14- 14' 14' 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15 15- 15- 15- 15- 16- 16- 15- 16- 16- 16- 15- 16' 15 15 15 16 15 15 15 15 15 16 15 78404886 ■ 8-20-2750 5 8)64065 5 89244405 9283883 5 9642400 15 0U000005 0356688 5 07124735 1067360 5 •1421356 5 •177446915 21-26704 6 2478068 28-28569 3178211 3507001 3874946 4222051 4568323 4913767 •6258390 •5602198 •5915195 ■6287388 -6628783 - 6969385 16 •7309199 6 7648231 7986486 8323970 8660687 8996644 9331845 9066-295 0000000 0332964 •0665192 •0996689 •13-27460 •1657509 -1986842 -2315462 -2643375 •2970585 •3297097 •3622915 -3948043 •4-272486 ■4596248 •4919334 •5-241747 6563402 6884573 6204994 65-24758 6843871 •7162336 •7480157 •7797338 •8113883 ■84*29795 •8746079 74889' 758i)(i. 768!)> 778ft!w| 788!»tio 7988110 8087 !i6 818rt4s| '828i:6 '838-2TI '84803:>i ■8577t;i)| •8674HI •877131) •8867(ii •89636' •90.'.!)41 •915483 •924911;) •93447;) •9439'21 •953341 •96-2731 ■972091 •93\m •990T}: 0000(H) 009244 018463 ■O27fi.)0 036811 045943 055049 0641261 O8-2-201 091199 100170 109119 118033 1269-25 135792 144634 1534411 162-23!) 171005 '1797-!7 ' 188463 ■ 197154 •2058-21 •214464 •223084 ■23167!)! ■2402M •-2488O0 •2573il •2658-2t)i ■274300 •28-2760 ■'291 IW •299604 ■307993 •316369 .Square \m ■iVl [2)) mi 2)7 1 2)!' iiiU 201 mi ;ri") JjiJdl 1267 \:m ■2d!) 270 L'71 |!7-2 m t\ pi) B78 in 7!) p> hi m hi 64009 61516 6502) 65.-)36 66049 66">64 670H1 676U0 68121 68644 6i)l6:) 6<)6<>li 7U.'2.J 70756 71-23!) 718)4 723.'il 72900 73141 73984 74)29 75076 75625 76176 7t)7-29 77-231 77841 78400 78it61 79524 80089 80656 812-2;-) 81796 82369 82944 83521 84100 84681 85-264 85849 86136 87025 87616 83209 88804 89401 90000 90601 91-204 91809 92416 93025 93G36 94249 94864 95481 93100 96721 97344 97969 98596 99226 Cub*. Sq. Root. I t 16194-277 15 16387061115 16581375115 I67772l«!lfl 16974593; 16 17173512:16 17373979116 176760001 1 6 17779581 16 17f)847-28 18191447 18399744 18609625 18321096 19034163 10243832 19465109 10683000 19902511 •J0123648 20346417 20570824 20796875 21024576 21253933 21484952 21717639 21952000 22188041 22425768 2-2665187 '22906304 231491-25 3.33936.56 23639903 23887872 24137.569 24389000 24642171 24897088 25153757 25412184 26672375 25934336 26198073 26163592 26730899 27000000 27270901 27643608 27818127 28094164 28.372625 2865-2616 28934443 29218112 29503629 29791000 30080231 3037 132« 30664297 30969144 31265876 Cube Rooi 90597.17 9.17.3775 9687194 0000000 0312195 06-23784 0934769 1-245155 1554!)44 I86U41 217-2747 2480768 2788206 3095064 3401346 .3707055 4012195 4316767 46-20776 4924225 6-227116 56-29454 5831240 61.32477 04.33170 6733320 70.3-2931 7332005 7630546 7928556 8226038 8522995 88194.30 9115345 9410743 9705627 0000000 0293864 0587221 0380075 11724-28 1464282 1755640 2046505 '2336879 2626762 •2916165 3205081 3493516 3781472 4068952 4355958 4642492 4928567 621415516 649923316 5783958 6068169 6351921 6635217 6918060 7-200461 7482393 ■324704 ■341.3-26 349604 357861 ■.366095 .174311 ■.182504 390676 ■398828 40695H 41.506s 423 15H 431-2-2H 439277 447305 455315 463301 471274 479224 487154 495065 o029,>6 510830 518681 5-26519 .534335 5421.33 549912 55767-2 565415 573131) 680344 588532 696202 603354 611439 619106 6-26705 634287 641852 649399 656930 664444 671940 6794-20 686382 694329 701759 709173 716570 723951 731316 ■738865 ■745997 ■753313 ■760614 ■767899 •775169 •7824-23 •789661 •796834 ■804092 .Vo. Sqtisre. Cube. 316 99856 317 100489 318 1011-24 319 101 761 ! 320 102400] 121; 10.104 1 322 1036841 ;123 1043-29 324 104976 1-25 105625 ;!26 106276 J-27! 1069-29 ;l-23' 1 07584 3 29 '1082 41 :130 I08:)00 331 109561 :i32}l 10-224 333 110389 3341111556 335! 112-2-25 :i36 11-2896 337 j 11.3569 138|ll4244 .l:Vj; 114921 310lll5600 311^1 16-281 312 1161)64 3431 17649 344:118336 545 1190-25 316;119716 .147,1 '20409 343' 121 104 3191121801 350 1 122500 351 i 1-2.1-201 352' 1-23904 353.124609 354J 1-25316 355 1 1260-25 8q. Root. Ciib* Root 356 357 358 12673(1 127449 1-28164 359 128881 360 1 129600 361 1 1.10321 362!l31044 .3631131769 364 13-2496 365133-228 .166 133956 367 134639 368 13.54-24 369'l36161 370 '136900 371 j 137641 372! 138384 373 1391-29 3741.19876 375;i40e-25 376 141376 3771142129 378il42384. 315.5419617 3185.501 31 7 321574.12 17 3-2461759 17 .12768000.17 33076161 17 33.136-248 17 33698-267 17 3401-2-2-24! 18 341281-25 34645976 34965783 3.52875.52 .15611-289 35937000 .16-264691 36594368 .169*26037118 37259704 37595375 37933056 38*27-27.53 38614472 38958219 39304000 39651.S21 40001683 40353607 40707584 41063625 41421736 41781923 42144192 42508549 42875000 43243551 43614-208 43986977 44361864 44738875 45118016 45499293 45382712 46268-279 46656000 47045881 47437928 47832147 48228544 48627125 49027896 49430863 49836032 50243409 50653000 61064811 51478348 61895117 6-2313624 52734375 63157376 53682633 64010152 7763333 6 804493«j6 83265456 860.571116 8885433:6 91647*29,8 944.158 4!6 972-2008 6 0000000 0277564 0564701 0831413 1107703 1.183571 1659021 1934054 2-208672 248-2876 *27666G9 3030052 330302816 35755986 38477636 41195-26 6 4390389 16 4661853 4932420 5-202592 5472370 .5741756 6010752 6279360 6547581 6815417 7032869 7349940 7616630 788-2942 8143877 8414437 8679623 8944436 9208879 9472953 9736660 0000000 0262976 052.5589 0787810 1049732 1311265 1672441 1833261 2093727 2353841 2613603 287301; 313*2079 3390796 3649167 3907194 4164878 4422221 ■811-284 -818162 •825621 -83*2771 •8.19901 -847021 ■8.54 1*24 -861212 •868285 •87.5344 ■ 83-2.138 •889419 -896435 -903436 ■9104-28 -91739G -924.155 -931.101 ■938232 ■945149 •952053 •9.58943 ■935819 -972683 -979532 -986368 -993191 •000000 •006796 •013579 ■0-20349 ■0*27106 > 033850 •040581 ■047298 •0.54004 •060696 ■067376 •074044 ■080699 •037341 ■093971 ■ 100688 •107194 ■11.1786 •120367 '126938 •133492 •140037 •148569 •1.53090 •159699 •166098 •172530 •179054 •185516 •191966 ■198405 •204832 ■21124S ■217662 •224045 •230427 i I S80 SQUARES, CUBES, AND ROOTS. i i I ii N*. Square. 879 U3G41 880144400 8811145161 38a|14d«'J4 888 '146689 384I1474A6 885|l48i2a5 3S6j 148996 387 149769 388 160A44 389 390 391 39a 393 304 39&I 396 397 398 399 400 401 Mi 403 404 405 406 407 408 409 410 411 4ia 413 414 41& 416 41 418 419 4*iO 4'il 4'i-2 423 4-^4 425 426 427 428 429 430 431 432 433 484 435 436 437 438 139 Culit. 8q. Root. CuIm Root 151321 162100 152^31 153664 154449 155236 156025 156816 157C09 158404 159201 160000 160801 161604 162409 163216 164025 164836 165649 166464 167281 168100 168921 169744 170569 171396 172225 173056 173889 174724 175561 176400 177241 1780S4 178929 179776 180625 181476 182329 183184 184041 184900 185761 186624 187489 188356 189225 190096 190969 191844 192721 440 193600 441 194481 19- 19- 19- 19- 10- 19 10 10 19 10 19 10 19 10 19 19 19 54439939 54872000 A5306341 65742968 50181887 66623104 67066625 67512466 67960603 68411072 58863869 60319000 59776471 60236288 60698457 61162984 61629875 62090136 62570773 63044792 63521199 6400000U 64481201 64964808 65450827 65939264 66430125 66923416 67419143 67917312 68417929 68921000 69426531 69934528 70444997 70957944 71473375 71991296 72611713 73034632 73560059 74088000 74618461 75151448 75686967 76225024 76765625 77308776 77854463 78402752 78953589 79507000 80062991 80621568 81182737 81746504 82312875120 82881856 20 83453453 20 84027672 84604519 85184000 86766121 4670223 4985887 6192213 6448203 6703868 6959179 6214169 6468827 6723156 6977156 7230829 7484177 7737199 7989899 8242276 '8494332 8746069 19-8997487 19-9248588 9499373 9749844 0000000 0249844 0499377 ■20-0748509 20-0997512 1-2461 IB 1494417 1742410 1990099 2237484 2484567 3731349 2977831 3224014 3469899 3715488 3960781 4206779 4450483 4694895 4039015 5182845 5426386 6669638 5912603 6155281 6397674 '6639783 '6881609 •7J-23152 ■7364414 '7605395 20-7846097 30-8086520 8326667 8566536 8306130 9045450 9284495 9523268 9761770 0000000 336797 443 1968641 243166 349504 366841 362167 368482 281079 20- •20- -20- -20- 20- 30' 30 20- •20 30 30 20 30 20 20 20 -20 30 30 30 30 30 30 -20 20 30 30 30- 7' 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- 7- No. Square Cuba. 443 106-249 444 197186 446:108036 446 198916 447 199809 374786 448 300704 449 301001 387362 450,30-2500 293633 299894 306143 812383 318611 324829 331037 337234 3434-iO 349597 355762 361918 368063 374198 '380322 '386437 ■392642 '398636 ■4U4720 ■410795 •416859 -422914 '438959 434994 441019 447034 45304U 459036 465022 470999 476966 482934 488872 494811 600741 506661 7-51'2571 7-618473 7-624365 630348 636131 541986 547842 7*653688 7-659526 7-565355 451203401 452|-204304 453f205-209 454I3O6II6 455|207036 456 ■207986 457 458 459 460 461 462 463 208849 209764 210681 211600 2^2S21 213444 214369 Sq. Root. •20- •20- 30- 21- 164 465 46i) 467 168 469 470 471 472 473 474 476 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 -671174 497 7-576985 7-682786 7-588579 7-594363 7-600138 7*605905 503 7 '611663 498 499 500 501 502 2 15-296 2162*25 217166 218089 219024 319961 220000 ■221841 2-22784 223739 224676 225625 226576 ■227529 228484 229441 230400 -231361 232324 233289 234256 235-235 -236196 237169 238144 239131 '240100 241081 242064 -243049 -244036 245025 246016 247009 248004 249001 350000 351001 25-2004 353009 604 364016 86860888 86938307 87638884 88131136 88716680 89814638 89916893 00518849 91135000 91733851 02345408 92059677 93576664 94196376 94818816 95443993 96071912 9670*2679 97336000 97972181 986111-28 99262847 99897344 100544625 10I194G96 101847563 102503232 103161709 103823000 104487111 106164048 105833817 106496424 107171876 107860176 108531333 109216353 109902339 110593000 111384641 111980168 112678687 113379904 1140841-25 114791'2d6 116501303 116214373 116930169 117649000 118370771 119095488 1198-23167 1-20553784 121-287376 122033936 1*22763473 133506992 134351499 126000000 125751501 126506008 137363637 1380-24064 31 21 31 31 31- 31- 21- 31- 31- 31- 31- 21- 31- 21- 21- 21- 21- 21- 21- 21- 21- 21- 21- 21- 21 21 21 21 21 21 21 21 21 31 31 21 21 21 21 21 21 21 23 Cuba Kxot 0387960 7 ■6174l{ 7 ■ 7- 7^ 7- 0475653 071.1075 0050381 1187121 1433746 1660105 1806301 318-2034 •2867606 360*2916 •2837967 3072768 3307390 3541665 '3776583 ■4009346 ■424'2863 ■4476106 ■4709106 ■4941853 ■5174348 -6406692 -5638587 -6870331 -61018-28 -6333077 -6564078 '6794834 '70-25344 -7'266610 •7486633 •7716411 '7944947 '8174-242 '8403-297 '8633111 '886068617 •9089033 7 9317133 9644984 977'2610 0000000 '23'0-227155 7 33-0454077 23-0680765 •23*0907320 113344417 1359436 7 16851987 18107307 32 32 33 33 33 -23 22 23 23 33 33 •22 22 32 3036033 7 •2'261108i7 34859557 3710676|7 3934968,7 31691367 3383079:7 3606798 7 3830*293:7 4053565 7 32*43766167 *2a -44994437 623U-i 6'2HHh4 7 -63100: 640;ij| 64tHl'); 657114 6A»U»| 7-6«rt7ti« 7-6744*1 7^680(i!i(i 7^6857M 7^69l3;j Tim 7I944J 72M}! 730611 73 85681 mit '86761 '87'JM '87831 '88371 '8H1I0I ■89441 ■80971 ■9051 ■0104 'HWli ■9-21II - 9-2641 -93171 -937« -94-^ •om 96811 7' 17- 7- 7' 17' 17' 7' 7- 7- 7- 7- rs. SQUARES, CUBES, AND ROOTS. 381 100888 938307 538384 fill'iOl'il 716680 31 314633 31 0387900 7 61741) ■047Afl53|7-tl*.23U'i •0713076 7 • 0960381 17 •||87I3»|7 016393 1618849 136000 738861 1346408 31 31 31 31 31 1069077 31 1670064 1196376 1818816 6443993 IM)71913 1438746 1660106 1896301 3183084 3867606 '360-2916 •3837967 •307-i768|7 •8307390 7 •3641666 7 •3776683 7 •4009346 7 6703679 31 7336000 7973181 8611138 19363847 19897344 10544636 ) 1 194006 ) 1847 603 )3603333 )3I61709 l)3S^23l»C0 )4487111 •il 31 31 31 31 31 31 4343863 4476106 4709106 4941868 6174348 ■640669317 •6038687 7 •6870831 •61018-28 •6333077 •6664078 •0794834 6'2HhmI 64U:iil 64tki'j: 7-66niil 667 lul 6A3U»|I tfrtH7ti(| 67 444 '6i:iUiiJ •6847m| •ttou; ■0U7wl •70-2fr«| •7( 71! 7l944l| 73(H '7:M)«id '73til!ii| •74l7i .lu. Ik|iwr«. Cuba. 7-76Mi •763911 •7694 •7749 70-36344 7 •780491 31- 31 31 31 31 31 31 31 31 31 )6164048l31 )&833817 )6496434 )7171876 17860176 M631333 )93163d3 )9003339 10693000 1384641 11980168 13678687 13379904 140841-3& 14791-266 16601303 16314373 16930169 17649000 18370771 19096488 19833167 •30663784 31-387376 •23033936 -2-2763473 33606993 34361499 >7 (7- 7- 7' 7' 7 7-266610 7486633 7716411 •7944947 •8174-243 •8408-207 •8633111 ■88606S6i7 •9089033!7 •9317133 7 •9644984 7 977-26107 Q000000;7 7 31 -23 •23-0-2-27166 •23-0464077 7- •23 -0680766 7 ■ •23-09073-20 7' 33-1138444 7 33-1369436 7 33-1686198 7 33-1810730:7 -23-3086033|7 •23-2-26110817 -23-3485966 7 33 -371067617 -ttlfiTi 33-2934968,7 -9-Jlll 33-8159186 7-9: „ „_ 33 -33880797 -931 36000000-23 • 3606798 7 ' 937i 36761601 33- 3830-293;7 -94! 36506008 33*4053665 7-947 •27-268637 23-4376616|7 •9M 380-24064 -23-4499448.7 96811 78ft 7914 7< 807914 81 8ii '8-i4J ■8-2971 •83&H •844)j •84601 •8dUi •8i •86-2-; 867611 ■87'i( ■87( -8H371 •88! •mi •899(1 -905li Oil MI& JJ6036|l-2S7876-26 .)0(lr26OO3a 1-20664316 407 367049,1.103 23843 Iq. Root. Cuba Root I.11096.')l 2 -23 •269081 13187-23-29 26U100 13i(iiI000 'dii3i i:)ai338:ii 2(3-2144! 134317728 ,13f20:U61> ,-)l4'264i96 }l;,|'26J3-26 266366 207389 ■264334 269361 270400 371441 •27-2484 273039 274676 276636 276676 •2777-29 278784 279841 ■280900 2H1U61 283034 284089 286166 286-236 '287396 388369 380444 290631 291600 •29-2681 ■293764 ■294849 296936 297036 208110 299309 300304 301401 303600 30.1li01 304704 306809 306916 308035 309136 310349 311364 31-3481 313600 314731 31/>844 551 n )il ft9 Si; 02; iti3i31696g 64;318096 31 93-25 330866 S7 331489 l3iMM)6697 136796744 186590876 187388096 188188413 138991833 139708369 140608000 1414-20761 14-2336648 lia066667 143877834 1447031-26 146631676 146363183 147197963 1480.16889 148877000 149731-201 160668768 161419437 16-2373304 163130376 153900656 164854153 1667-20873 166690810 167404000 168340431 1693-20088 160103007 160989184 161878026 16-2771336 168667333 164666693 166469149 166376000 167384161 168196608 16J1 1-2377 170031464 170963876 171879616 173803693 173741113 174676879 176616000 170668481 1776043-28 178463647133 179406144!33 93-473-20517-963874 •23-4944438 7-9686-27 33-5166605:7-078878 638856317 0791 13 5610-2887-984844 A83I796|7 -989570 006809117-994788 di74l70l8000000 6495088,1^-006-206 671668l|8Ol010:i 00301 1418016696 7 16633418 030779 7376340 8 •0'25957 7596134,8 781671518 8036086 18 8-264344 8 84731938 8691933 8 33 -23 2-2 33 -23 23 23 33 -23 33 33 33 -23 'i'2 33 •2i 33 33 33 33 No. Squart. 8910463 91-28786 9316899 9561806 0783500 -23-0000000 '23 -03 17-289 0434373 0661363 0867938 1064400 1300670 1616788 178-2606 1948-270 3163736 180363136 1813314g«i 183384363 23-2379001 33-2694067 33-3808936 3033604 3338076 3463361 36664-29 3880811 4093998 4307490 46-20788 4733893 4946803 51695-20 5373046 5684380 6796533 6008474 6330336 6431808 6643191 6864386 7066393 7i!76-210 '. 4868'13 7697386 7907546 8117618 •23- ■23- 23- 33- ■23- •23- -23- -23- 33- 33- •23- -23- 23- -23 • 33 • 33 • •23- -23- 33- 33- 081139 036393 041461 046603 061748 066886 063018 067143 073363 077374 083480 087679 09-207 -2 097769 10383!» 107918 11-2980 118041 ' 133096 ■13814 ■133187 ■1382-23 ■143363 -143-276 -153-294 - 168306 -163310 - 163309 -17330-2 • 178-289 - 183369 -188-244 79;336^241 180 336400 68i:33766i ">82'338734 683! 339889 )81'.ill066 J86|a*3336 686 343396 687i.M4669 688 346744 6891346921 6901348100 691 1349381 Cuba. 8q. RooU Cvbaltool 33- 38- 33- 8- 8- 6' 3- 8- 8- 8- s- 8 ••253371 8-267-i63 8- -263149 8-367039 3-271904 8-286773 693:360464 393 361649 391363836 696 364036 696'3563I0 366409 367604 368801 360000 361301 363404 303609 364816 306036 367336 368449 369664 370881 373100 373331 374644 375769 37699(t 3782-26 379466 380689 3819-24 383161 384400 385641 386884 338139 697 698 699 600 001 d03 603 004 006 606 607 608 ti09 610 611 013 613 614 616 616 617 018 619 630 631 633 633 347474 634 026 020 037 638 6-29 630 339376 343970634 188250482 184-2^20009 186193000 18616i»411 187149^248 188132517 180110-234 190109875 191103976 193100033 193100562 194104539 196113000 1961-2-2941 197137368 198166387 199176704 -200-201636 ■2013.10066 •203363003 -203307473 304336469 306370000 306135071 •207474089 308637867 -209684684 3106448?6 311708736 31-2776178 jl 3847 193 314931799 316000000 317031801 318167-203 319366337 •230348864 ■2314461-26 -233646016 333648543 334755712 -235866639 336981000 338099131 -239330938 330346397 '231475544 333608375 333744396 •234335113 336039033 •237176659 388338000 339483061 340641843 341804367 ■24 •24 '24 -24 -24 ■24 34 34 34 34 34 -24 34 -24- •24- •24- '24- '24- '24- '24- -24- -24- •24- -24- -24- '24- 34- 390036 391870 393139 394384 396641 396900 344140636 •246314376 246491333 347673152 243858139 8827506 8587-200 87467-28 8056068 9165315 9374184 958397 -23-9701676 -24-0000000 -24- 0-208243 0416.306 06-24188 0831393 1089416 1346762 1468939 1660019 1867733 3074369 3330O-29 3487113 '2693233 -2899166 -24-3104916 24-3310601 3615913 3731153 89-26318 4131113 4335834 4540885 4744765 4948974 5158013 6356883 5560683 5764115 5967478 6170678 34-6378700 -24-6576660 34-6779'254 34-6981781 34-7184143 34-7386338 •24-7588868 -24-7790334 24-7991936 •24 -81 93473 24-8394347 24-8596053 -24-3797106 '24-8997992 34-9198716 939927 1; 9599679 97999-20 0000000 0199930 0399681 '25-0599283 35-0798724 -24- •24- 24- 25- 35- 35- 350047000 125-0998008 218MA 286498 •291844 296190 301080 805865 810094 815517 830885 835147 3-29954 334755 339551 344341 349136 353905 858678 863446 368309 37-2967 377719 38-2465 ■387206 ■391943 ■396673 -401398 8-406118 8-410333 415542 420246 8-4-24945 3-439683 8 •434327 8 • 439010 8 -448688 8-448360 8-463028 8-457691 8-463848 8-467000 3-471647 476239 480926 485558 490185 494806 499428 504086 508642 513348 517840 5-22432 527019 531601 536178 540760 545317 549879 65M37 65S990 563538 8-568081 8-672619 m 38-i SQUARES, CUBES, AND ROOTS. No. Square Cube. G3I 398161 tiH-2 39I)4-24 6.33 400689 634 401956 i3;> 403-225 >i36 404496 i37 403769 i33i407044 i39 408321 640 409600 64li41088l 642;41'2164 64341341f.' 644414736 645:416025 646417316 Sq. Root. 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 66-2 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 686 686 887 688 690 691 692 693 418609 410904 421201 422500 423801 425104 426409 427716 429025 430336 431649 432964 434281 435600 436921 438244 439569 440896 442225 443556 444889 446224 447 .'>61 448900 450241 451584 452929 454276 455625 451)976 45S329 459634 461041 462400 46376i 465124 466489 467856 469225 470596 471969 473344 474721 476100 477481 478864 480249 251239591 25' 252435968 20' 2536.361.3?;2o 25484010425' •J5G047875i2d' 2572.59456 25 •258474853>25 259694072:25 2609171 19;25 262144000-25 263374721 !25 264609288:25 265847707 25 267089984125 >683.36125'25 269586136125 •>70340023|25 272097792125 27.3359449i25 27462500025 275894451 125 Cube Root 277167808 278445077 279726264 281011375 25 282.30041625 283593393125 284890312:25 286191179125 287496000'25 288804781 J25 290117528:25 291 434247 125 292754994 25 294079625 295408296 296740963|25 298077632125 299418309|25 300763000i25 .30211171X125 .303464448 25 304821217 306182024 .307546875 303015776 310288733 311665752 313046839 314432000 315821241 317214568 318611987 320013.504 321419125 322828356 324242703 325660672|26 327082769 26 328509000 329939371 331373888 3.32812557 1197134 1396102 1.094913 1793566 1992063 2190404 2388589 2586619 2784493 2982213 3179778 3.377189 3574447 3771.551 3968502 41G5301 4361947 4558441 4764784 4950976 5147016 5342907 5538647 5734237 5929078 6124969 6.320112 6515107 6709953 6904652 7099203 7203607 7487864 7681975 737,5939 8069758 8263431 8456960 8650343 8843582 9036677 9-229628 94224.35 9615100 •9307621 0000000 0192237 0384331 •0576284 0768096 0959767 ■1151297tS •1342687 ■1533937 ■1725047 ■1916017 ■2106848 •2297641 •2488095 ■2678511 ■2863789 ■3058929 ■3248932 •677152 ■681681 •686205 •590724 •695-238 •699747 •604252 •603763 •613248 •617739 •622226 •626706 •631183 •635665 •640123 •644.585 •64904'1 •653497 •657946 •662391 •666331 •671266 •675697 •680124 •634546 •633963 •693376 •697784 •702188 •70658 •710983 •71.5373 •7197.59 •724141 •7^28518 •732892 •737260 •741624 •745986 •760340 •754691 •759038 •763381 •767719 •772053 •776333 •780708 •785029 •789346 •793659 •797968 •802272 •806572 •810868 •815160 •819447 •823731 •828009 •832285 •836556 •840823 •845085 •849344 No. 'Square, 694 481636 695 48.30'26 696 434416 6971485809 693:487204 6991488601 700!490000 701 1491401 702 '492804 703'494209 704:495616 705497025 706'493436 707J499849 7085012()4 709.502681 710504100 711 '505521 712J506944 713;508369 714'509796 715i5ir225 716 512656 717 718 719 514089 6 16524 516961 720518400 72l|519841 722 1 52 1284 7231522729 724|524176 7iJ5 1525625 726527076 727.528529 723'529984 7-29I531441 730 532900 Cube. Sq. Root. 334266384 26 33570237626 .3.37153536 26 338608873'26 340068392 26 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 743 749 750 751 752 753 754 766 341632099 343000000 344472101 346948408 347428927 348913664 350402626 351895816 353393243 354894912 356400829 357911000 359425431 360944128 362467097 363994344 365525876 367061696 368601813 370146-232 ,371694959 373248000 374805361 376367048 377933067 379503424 381 0781 125 382667176 384240583 385828352 387420489 389017000 390617891 392223168 d37289j393832837 538766 396446904 39706,5375 398688266 400315553 401947-272 403583419 534361 535824 540225 541696 643169 544644 546121 547600 ,549081 650564 552049 563536 555026 556516 415160936 653009 559504 561001 405224000 406869021 408518488 41017-2407 411830784 41.3493626 41683-2723 418508992 4-20189749 5625001421375000 564001|423564751 565504'425259008 .567009:426957777 668516 4-28661064 d70026'430368875 756|671636 432081216 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 26 ■27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 27 -27 27 27 27 27 27 27 27 27 27 •3438797 •3628627 •3818119 •4007676 •4196'396 •4386081 •4675131 •4764046 •4962826 •6141472 •6329983 ■6618361 •6706606 •6894716 •6082694 •6270639 fa •6468262 ■6646833 •6833281 •7020598 •7207784 •7394839 •7681763 •7768667 •7956-220 •8141764 •83-28167 •8614432 •8700577 •8886693 •9072481 •9268240 ■9443672 •9629376 •9814761 •0000000 •0186122 •0370117 •0564986 •0739727 •0924344 •1108834 ■1-293199 •1477439 •1661654 •1845644 -2029410 •2213162 •2396769 •2580263 •2763634 •2946881 •3130006 •3313007 •3496887 •3678644 •3861279 •4043792 •4226184 •4408455 •4690004 •4772633 •4964452 Cube Root :s. SQUARES, CUBES, AND ROOTS. 383 ube> Sq. Root. Cube Uoot !55384 •26-3438797 '02375 26-3628527 153536'26-3818119 8-8535<)g 8-857849 8-«6209.i ?qii«re. 5Tl.j7304467097 126 - 7020598 8 - 93366iS 3994344 26 • 7207784 8 - 937813 5525876 -26-7394839 i7061 696 26-7581763 18601813 26-7768557 0146-232-26-7955-220 »393243 1894912 3400829 7911000 9425431 0944128 26- 26- 8-94-2014 8-9461811 8-950344 8-654503 ri694969 26 - 81 41764 3 - 968658 26-83-28157 8-962809 r3248000 ?4805361 6367048 ■933067 r9503424 26-8514432 26-8700577 •26-8886593 26-9072481 n078125l26-9258240 J2657 176 •26-9443872 342405H3 26-9629375 35828352 26-9814751 966957 971101 8-975240! 8-979376 8-983505' 3-987037 S-99176;1 8-995t'83 37420489 39017000 90617891 a^-223168 93832837 95446004 97065375 98688256 003155531-27 01947272 '27 •27-0000000 9-000000 •27' 27' 27 27 27 27 27 03583419 05224000 06869021 108518483 10172407 0185122 0370117 0654985 0739727 ■09-24344 •1108834 -1-293199 •1477439 •1661554 1845544 •20-29410 2213152 2396769 2580263 ,11830784 27-2763634 13493625-27-2946881 15160936 27-3130006 1683'27-23 -27 •3313007 9 18508992'r27- 3495887 •20189749 27-3678644 a 1875000 -27 •3861279 123564751-27-4043792 L25259008 27 -422618419 12695777' 1-28661064 E30368875 t3'2081216 27-4408455 27-4590604 27-4772633 •27-4954452 O04U3I oosd 01232!! 016431 O206i9| 0246M 028715' 'O3'280-l| •036» •040963 •045941 -04911^ -0531 -057'2 9-061311 0653 0694 07347 •0775 •08161 •08561 •0896; •09367 •097701 •1017 •1057 •1097 |.)S|'>74.)64 [.i!)jj760Sl |oi)'j77600 611579121 di'")f>i)644 bio82169 |64iJ33(j96 .•)3.V225 ut);J86756 ::.s3-289 ,j|589824 i9i.)91361 [()j.392900 l|.j94441 :'.)95984 ,)i)97529 4i'<99076 [o:i>02176 7|i)03729 |S.i05284 |!io06S41 ;t5084OO :ii09961 1)11524 1)13089 til 4656 'til6-2-25 !til7796 1|)19369 lii-20944 6-22521 |i>2410U 625GS1 1)27264 628849 630436 632025 633616 635209 336804 638401 640000 641601 643^204 1644809 146416 1648025 W9636 151249 652864 154481 656100 157721 159344 160969 i«2d9G 164225 165856 17489 19124 0761 El" Cube. Sq. Root. 4.35519512 437245479 438976000 440711081 442450728 444194947 445943744 4476971-25 449455096 451217663 452984832 454756609 456533000 45 'MOll 460u. '648 461889917 463684824 465484376 467^288576 469097433 470910952 4727-29139 474552000 476379541 473211763 480043687 48 1 8903'" 4 4837.366 > 4855876 i 4874434 .1 489303872 491169069 493039000 494913671 496793038 198677257 500566184 502459876 504358336 506261573 503169592 510032399 612000000 513922401 515849603 517781627 519713464 52166012' 523606616 525557943 627514112 529475129 631441000 633411731 535387328 537367797 539353144 541343375 64333«4B8 546338513 547343432 549353269 27- •27- •27- •27- 27- 27- •27- 27- •27- 27' •27 ■ •27' •27' 27' •27' 27 27' •27 27 27 •27 •27 •27 27 •27 •27 •27 •23 28 28 23 28 28 28 28 28 28 23 28 28 28 28 28 28 28 28 23 28 28 23 23 23 28 28 23 28 28 23 23 29 28 28 28 Cube Root 5136330 9- 5317998 9- 549954619 • 568097 jjO- 586228419 • 6043475|9■ 6224546 9" 6405499 6586334 6767050 0947648 71281-29 7303492 7483739 9' 7663368 9' 784388019' 802377519' 8-208555|9' 8388218 9 320 321 8567766 8747197 8926514 9105715 9284801 946,3772 9642629 9821372 0000000 0178515 035691 0535203 071.3377 08914.18 1069380 • 1247^2-22 • 1424946 1002557 '1780056 •1957444 •21347-20 •2311334 2488938 2665881 2842712 •3019434 •3196045 •3372546 •3543938 •3725219 -.3901391 •4077454 4253408 44-29253 4604989 4730617 4956137 6131549 5306352 6482048 6657137 8332119 6006993 6181760 No. 113781 117793 121801 1 '25805 129806 133803 137797 141788 145774 1497.57 153737 157714 161636 165656 1696-22 173585 177544 181500 1854.53 189402 19.3347 197289 •201229 205164 209090 213025 216950 2-20373 224791 •228707 232619 •237528 240433 '244335 248234 252130 2560-22 •259911 263797 267680 271659 27543.i •279308 ■283178 •287044 •290907 •294767 •293624 •302477 •3063i3 •810175 -314019 •317860 ■321697 •325532 ■329363 ■333192 ■337017 ■340338 •344667 •343478 •352236 .Square. 672400 674041 822 675684 323 824 325 826 827 328 329 830 331 332 833 834 835 836 837 833 839 340 J41 8421708964 843 844 315 846 847 348 349 350 351 852 853 354 355 356 357 858 859 360 861 862 863 864 865 866 367 868 369 870 371 372 873 874 375 876 877 378 879 830 331 710649 712336 7140-25 Cube. 551368000 553337661 555412243 557441767 I Sq. Ilo-Jt. 677329 678976 6iW25 561515625 68-2276 563559976 683929 565609283 635584 56760355^J 637241 5697^22789 638.900 571737000 690561 573356191 69-22-24 575930368 693389 578009537 695556 5800.'»3704 697225 582182875 693896 5842770.56 700569 .586376253 70^2244 5884'^0472 703921 590589719 705600 592704000 707231 594323321 596947638 599077107 601211584 60.3351125 715716|605495736 717409 6076454-23 '28' •23' •28' ^M096 832 719104 7-20801 7-22500 7-24201 725904 7-27609 7-29316 7310-25 732736 734449 736164 737881 739600 741321 743044 744769 746496 748225 749956 751689 763424 755161 766000 768641 760884 762129 763376 7656'25 767376 769129 770884 772641 774400 776161 777924 559476224|28' 28' ■2S' ■iS •28' 28 ■28 •28 ■iS •28 -28 -23 -28 28 •28' 23 •28 •29 29 •29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 609800192 611960049 614125000 616295051 613470-208 620650477 622335864 6-250-26375 6272-2-2016 629422793 6316-287 1-2129 633839779 636056000 633*277381 640503928 642735647 644972544 647214625 649461896 651714363 653972032 656234909 653503000 660776311 663054843 6653.33617 667627624 669921875 67'2221376 674526133 676336162 6791&1439 681472000 6337973^1 686128963 6356421 6530976 6705424 6379766 7054002 72-28132 74021.57 7576077 7749891 79-23601 8097206 8-270700 8444102 8617394 3790582 8963666 9236646 9309523 9432297 9654967 9827535 0000000 017-2363 0344623 0516731 0688337 0360791 10.12044 1-204.396 1376046 1547595 1719043 •1890.390 •20616.37|9 •223-2784 9 Cube Root 2403330 2574777 2745623 2916370 3037013 3257566 34'28015 3593365 3768616 3938769 4103823 4278779 4448637 4618397 4788059 4957624 •8127091 5296461 5465734 8634910 5803989 -697'297«!9 614186819 6310648;9 64793-2519 6647939i9 6dl6442|9 6984848 9 I 359902 363705 367505 371302 375096 378837 382675 386460 390242 394020 .397796 4015t>9 405339 409105 412869 4166.30 4-20387 424142 427894 431642 435383 439131 44-2870 446607 450341 454072 457800 461525 465247 •468966 •472632 •476395 •480106 •483313 •487618 •491-2-20 •494919 •498615 •502303 •605993 ■509635 ■513370 ■517051 ■5-20730 ■524406 •528079 •531749 •5.35417 •639082 •54-2744 -646403 -650059 -663712 -557363 -561011 -564656 -568'298 •671938 -576574 •679208 •682840 •686468 •690094 oa4t SQ'SVfcRES, CUBES, AND ROOTS. mi 'S'S' ■> f HPli No. 883 884 886 686 887 888 889 890 891 89a 893 894 896 Squart. 897 Cube. 899 900 901 90-2 903 904 905 906 907 908 909 910 911 913 913 914 916 916 917 918 919 930 931 933 933 934 936 936 937 938 939 930 931 933 933 934 935 936 937 938 1^939 P40 41 779689 781456 783335 784996 786769 788644 790331 793100 793881 795664 797449 799336 801035 803816 804609 8064Q4 808301 810000 811801 813604 815409 817316 819035 830836 833649 834464 836381 838100 839931 831744 833669 835396 837335 839056 840889 843734 844661 846400 848341 850084 861939 863776 855636 867476 859339 861184 863041 864900 866761 868634 870489 873356 874336 876096 877969 879844 881731 883600 886481 688463387 690807104 693154136 695606466 697864103 700337073 703695369 704969000 707347971 709733388 713131967 7145I69S4 716917376 719333136 731734373 734160793 736573699 739000000 731431701 733870808 736314337 738763364 74131763d 743677416 746143643 748613313 761089439 763571000 756058031 768550338 761048497 763531944 766060876 768576396 771095313 773630633 776131569 778638009 781339961 783777448 786330467 788889034 791433136 794033776 796397983 799178753 801763089 S04357000 806934491 809557668 813166337 814780504 817400375 830036856 833656953 835303673 837936019 830584000 833337631 Sq. Root. Cube Root 39- 39- 39- 39- 39- 39- 39- 39- 39- 39- 39- 39- 39- 30- 30- 30- 30- 30- 30- 30- 30- 30- 30- 30- SO- SO- SO- SO- SO- SO- SO- SO- SO- 30- SO- SO - 30- 80- 30- 30- 30- 30- SO- 30- 30- 30- 30- 39-7153169 39-7331375 39-7489496 39-7667531 7836433 7993389 8161030 83-38678 8496-231 8663690 8831056 89983-28 9165606 9333591 9499383 9666481 9833387 0000000 11166630 0333148 0499584 0665938 0833179 0998339 1164407 1330383 1496369 1663063 1837766 1993377 3138899 33-24339 3489669 30-3634919 30-3830079 3983148 3150138 3315018 3479818 3644539 3809151 3973683 4138137 4303481 4466747 4630934 30-4795013 30-4959014 6133936 5386730 6450487 5614136 6777697 6941171 6104567 30-6367857 30-6431069 30-6594194 No. Square 593716 597337 9-600955 9-604570 9-608183 9*611791 9-615898 9-619003 9-6-23603 9-626201 9-6*29797 9-633390 9-636981 9-640369 9-644154 9-647737 9-631317 9-654894 9-658468 9-662040 9-663609 9-669176 9-67-2740 9-676S03 9-679860 9-683416 9-686970 9-690331 9-694069 9-697616 701138 704699 708-237 711773 9-715305 9-718835 9-733363 9-738888 9-739411 9-733931 9-736448 9-739963 9-743476 9-746986 750493 753998 767500 761000 9-764497 9-767993 771484 774974 778463 783946 785439 788909 9-7933861 998 9 -796861 1 999 943 943 944 945 946 947 948 949 960 931 953 933 934 935 956 937 938 939 960 961 963 963 964 965 966 967 968 969 970 971 973 973 974 975 976 977 978 979 980 931 983 933 984 985 986 987 983 989 990 991 993 993 994 995 996 997 Cube. 30-6757333 9-79933411000 887364 889349 891136 893035 894916 896809 898704 900601 903500 904401 906304 908*209 910116 91-2033 913936 913849 917764 919681 931600 933331 935444 937369 939396 931323 933136 933089 937034 938961 940900 942841 944784 946739 948676 930635 953576 954539 956484 958441 960400 963361 964334 966389 968356 9703*25 973196 974169 976144 978131 980100 93-2081 984064 986049 988036 990035 99-2016 994009 996604 998001 1000000 835896888 a38561807 84133-2384 843908635 846590336 849378133 851971393 854670349 867376000 860085351 86-2801408 865533177 863350664 870983875 87373*2816 876467493 879317913 881974079 884736000 837503681 890377138 893056347 896841344 898633133 9014-28696 904331063 907039333 909853ti>'-9 913673c 9154986 918330043 93116731 9*240104-24 936839376 9-29714176 933574333 935441353 938313739 941193000 944076141 946966168 949863087 933763904 9556716*25 958586356 961504803 964430-273 967361669 970399000 973343371 976191488 979146657 983107784 983074875 938047936 991036973 994011993 997003999 1000000000 Sq. Root. Cube Bom 30 14 30 30 SO 31 31 31 31 31 31 31 31- 31- 31- 31- 31- 31- 31- 31- 31- 31* 31* 31- 31- 31- 30-693018!^ 30-7033051 30-7345830 30-7408533 30-7571130 30-7733651 7896086 8068436 8330700 8333879 8544973 8706981 30-8868904 30-9030743 "93497 9354166 9615751 9677351 9833668 0000000 0161348 0333413 0483494 0644491 0805405 0966336 ? 1369841 133764eJ 1443330 •1608739 1769145 •19-29479 •3089731 3349900 3409987 3669993 3739915 3889757 3049617 3309196 31 3368793 31-3538308 3687743 3847097 4006369 4166561 4334673 4483704 4643651 4801535 4960316 5119035 5377665 6436306 '6694677 6753068 •6911380 •6069613 31 31- 31- 31- 31- 31- 31- 31* 31- sr 31- 31- 31' 81 31 31 81 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9 9 9 9< 9 9 9 9 9 9 9 9 9 9 9 S- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9- 9 9- 9- 8O280 80627 8097J 81319 8I66g 8-2011 833i; 83703 8304; 83399 83731 8442J 8476 8511; 854S 857« 86U 8648 86S'I 8716 8751 8780 8819 88ol 888(1 89-21 893j 6337766ilO rs. TABLES. 385 396888 561807 !232384 9086i5 590536 278123 TABLE OF THE AMOUNTS OF £1 AT COMPOUND INTEREST. 30-692018i^ 30-7083051 30-7246830 30-7408523 30-7671130 30-7738651 971392 30-7896086 1670349 '376000 )085351 2801408 5823177 3250664 30-8068436 30-8220700 30-8382879 30-8644972 30-8706981 ^.„ 30-8868904 0983875 30-9030743 372-2816 30-" 92497 6467493bt t>3&4166 921791-2 30-9616761 11974079 30-9677251 14736000 30-9838668 J7 603681 31-0000000 )0277128 31 -0161248 )3056347 31-03-22413 95841344 31 -0483494| 9863-2125131 -0644491 014-28696 04231063 07039232 |09853'>'-9 12673* -0 31-0805405 31-09662361 31-! 126984 31 -128764a 31-14482301 115498b 131-16087-29 118330048 31-1769145 121167317 31-19-29479 h24010424 >268d9375 )-29714176 31-2089731 31-2249900 31-2409987 J3-2574833 31 -2669992 9-92i )3d441352 )383137S9 »41 192000 944076141 946966168 31*2729915 31-2889767 31-3049517 31-3209195 31 3368792 949862087 31-3628308 952763904 3r 3687743 956671626 31-3847097 958588266 31-4006369 961604803 31-4166561 964430272 31 •43-24673 967361669 31-4483704 970299000 31 -464-2654 97324-2271 31-4801526 976191488 31-4960316 979146667 31-6119026 982107784 31 •6277855 985074876 31-6436206 938047986 31 -6694677 991026973 81- 6788068 994011992 81 •6911380 997002999 31 ^OeOClS 9-8(1 9- 9-80973J 9- 9-8U 9- 9-82351J 9-8-27« 9- 9-83 9- 9- 9-844-2J 9-847e 9- 9-854 9-85791 9-861t 9-864 9- 9 -8718 9- 9-878 9- 9-88i 9-8S8j| 9-89-]l| 9- S-»a!)5 9-90'a 9-9051 9- 9091 9-911 9-9ia 9-9191 9- 9-9-261 9-9-23 9- 9-9J 9-9» 9-914 9-941 9-94! 9-9S 9-9S( 9-931 9- 9- 9- 9- 9- 9-911 9- 9- 9 9- 9- ; 9 per cent 4 per cent 6 per cent 6 per cent Vo. oi Piiy- inente 3 per cent 4 per cent 6 per rent 6 per ceat 1- 03000 1-04000 1-06000 1 -06000 Su 2-15659 2-77-247 3-68567 4 64938 1- 06090 1-08160 1-10250 1 - 12360 27 2-22129 2-88337 3-73346 4-8-2-236 1-09273 1 - 1-2486 1-1676-2 1-19102 28 2-28793 2-99870 3-92013 6-11169 1- 1-255] 1 • 16986 1-21551 1-26-248 29 2-35657 3-11865 4-11614 6-41839 1-159-27 1-21666 1-276-28 1-338-23 30 2-4-27-26 3-24340 4-32194 6-74349 J -19405 1-26532 1-31010 1-41852 31 2-60008 3-37313 4-83804 6-08810 1-2-2987 1-31593 1-40710 1-50363 32 2-57608 3-50806 4-76494 6-45339 1-26677 1-36857 l-4774.> 1-59385 33 2-65-233 3-64838 6-00319 6-84059 1-30477 1-4-2331 1-86133 1-68948 34 2-73190 3-79432 5-25335 7-26102 1-34392 1-48024 1-62889 1-79088 35 2-81386 3-94609 5-61601 7-68609 l-3S4'i3 1-53945 1-71034 1-89830 36 2-89828 4-10393 5-79182 8-147-25 1-42676 1-60103 1-79586 2-01-2-20 37 2-98523 4-26809 6-03141 8-63609 1-46853 1-66507 1-88565 2-13-293 38 3-07478 4-43881 6-38548 9-164-25 1-61259 1-73168 1-97993 2-2609C 39 3-16703 4-61637 6-70476 9-70361 1-55797 1-80094 2-07893 2-39656 40 3-26204 4-80102 7-03999 10-28672 1-60471 1-87-298 2 18-287 2-54035 41 3-35990 4-99306 7-39199 10-90288 1-65-285 1-94790 2-29-202 2-69-277 42 3-46070 6-19278 7-76169 11-65703 1-70243 2-0-2582 2-40662 2-85434 43 3-66452 6-40049 8-14967 12-26046 1-76351 2-10685 2-5-2695 3-02560 44 3-6714^ 8-61651 8-66716 12-93648 1-80611 2-19112 2-65330 3-20713 45 3-78160 6-84118 8-98501 13-76461 1- 86029 2-27377 2-78596 3-39956 46 3-89504 6-07482 9-434-26 14-69049 1-91610 2-36992 2-925-26 3-60354 47 401190 iS- 31782 9-90597 16*46592 1- 97359 2-46172 2-07152 3-81975 48 4-13225 6-67053 10-40127 16-39387 •2-03J79 2-56330 3-2-2510 4-04893 49 4-266-22 6-83335 10-9-2183 17-37750 •2-09378 2-66584 3-38635 4-29187 60 4-38391 7-10668 11-46740 18-42018 lOOOOOOOOO 8l-fi397766ilO- J'ABLE OF THE AMOUNTS OF AN ANNUITY OF £1. 3 |wr cent 4 per cent 1-00000 I 2-03000 I 3-09090 4-18363 5-30913 6-46841 7-66246 I 8-89-234 110-15911 111-46388 ll-2- 80779 It4- 19203 115-61779 117-08632 118-59891 120-15688 121-76159 123-41443 ]i5- 11687 p6- 87037 B8- 67648 BO -53678 |32- 45288 34-4-2647 ^6-46926 1- 2- 3- 4- 8' 6' 7- 9 10 12 13 15 16 18 20 21 23 25 27 •29 31 34 36 39 41 00000 04000 12160 24646 41632 63297 898-29 21423 58279 00611 48635 02580 6-2684 '29191 0-2359 8-2453 •69751 64541 •671-23 •77808133 •969-20 33 6 per cent 1- 2- 3- 4- 4- 6- 8' 9 11- 12 14 15 17 19 -21 23 25 28 30 24797 61789 08260 84691 00000 05000 15250 31012 62563 80191 14201 64911 02656 67789 20679 91713 71298 69863 87886 68749 84037 • 13238 •63900 -06695 ■71926 -60621 -43047 -60200 •72710 6 per cent No. of Pay. inentii 1-00000 2-06000 3-18360 4-37462 6-63709 6-97632 8-39384 9-89747 11-49131 13-18079 14-97164 16-86994 18-88214 21-01606 '23-27697 •26-67253 28-21-288 90565 76999 78569 99273 392-29 99683 60-6161 64-86461 30 33 36 39 43 48 >88| i6l| 26 27 28 29 30 31 32 83 34 35 36 37 38 39 40 41 42 43 44 46 46 47 48 49 60 3 per cent 38-65304 40-70963 42-93092 46-21888 47-87641 60 -00-268 62-60276 65-07784 57-73018 60 -46-208 63-27694 66-174-22 69- 16945 72-23423 76-40126 78-66330 82-023-20 85-48389 89-04841 71986 50146 100-39650 104-40839 108-64066 112-79687 4 per cent 92- 96- 44- 47- 49- 62- 56 69 62 66 69 73 77 81 85 90 95 99 104 110 118 121 126 132 139 148 163 31174 08421 96788 96629 08494 32833 70147 20953 85791 65222 69831 70-226 97034 40915 02561 82654 81960 01238 41288 02939 87057 •94539 •26321 ■83373 ■86708 6 per cent 61 64 88 62- 66- 70- 78- 80- 85- 90' 95' 101 107 114 120 1-27 135 142 151 180 168 178 188 198 209 11346 86913 40258 3-2271 43885 76079 29829 06377 06696 32031 83632 62814 70984 09802 79977 83976 •23178 99334 14300 70015 68516 11942 •02639 ■42666 6 per cent 69- 83- 88- 73- 79- 84- 90- 97- 104' 111- 119 127' 185 146 164 186 176 187 199 213 -226 -241 268 373 34799 390 16838 70678 53811 63980 06819 80188 88978 34318 18878 48476 12087 38812 904-20 08846 78198 04788 96064 60768 76803 74361 •60812 •09861 ■66458 ■95840 ■88690 386. ' TABLES. ' TABLE OF THK PRESENT VALUES OF AN ANNUITY OF £1. i It I ( .Vo. '.f Hiiy. moiiU a 6 7 8 9 10 11 12 13 14 15 16 17 IS 19 20 21 22 23 24 25 3 (ter cent 0- 1- 2- 3- 4- 6- 6- 7- 7- 8- 9- 9- lo- ll- 11- 97087 91347 82861 71710 57971 41719 2302t^ 01969 78011 53020 25262 95400 63496 29607 93794 12-56110 13-16612 75351 32380 87748 41502 93692 44381 93554 41315 4 per cent 1 2 3 4' d 6 6 7' 8' 8' 9' 9 10 11 11- 12- 12- IS- IS- 14- 14- 14- 15- 15- 96154 88619 77519 6^999 45182 24214 00205 73274 43533 11089 76058 38507 98565 56312 11849 65239 16567 65940 13394 59032 02916 45111 85684 24696 62208 6 percent 1 2 3 4 5 6 6 7 7 8 8 9 9 10 10 11' 11 12- 12- 12- IS- IS- 13- 14- 95238 85941 75325 54595 32948 07569 78637 46321 10782 72173 30641 86325 39357 89864 37965 83777 27406 68958 085S2 46221 8-2115 16300 48857 79864 09394 U per cetM 1 2 3 4 4 5 6 6 7' 7 8 8 9 9 10 10 10 11 11- 11- 12 12- 12- 12- 94340 83339 67301 46510 21-236 91732 68238 •20979 80169 36009 88687 38^84 85268 29498 71-225 10589 47726 82760 15811 46992 76407 04158 30338 55036 78335 S'n. of I'ny- rneiili 26 27 38 29 SO 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 43 49 50 3 per cent 17 18 18 19 19 20 20- 20 21- 21 21 -22 •22 22 23 23 -23 23- 24- 24 24- 25" 25- 25" •25- 87684 32703 76411 18846 60044 00043 38877 76579 13184 48722 83-225 16724 49246 80822 11477 41-240 70136 98190 25428 61871 77545 02471 26671 50166 72977 4 per cent 15 16 16 16 17 17 17 18 18 18 18 19 19 19 19 19 20 20 20 20 20 21 21 21 21 •98-277 '3-2958 66306 98371 29203 68849 87355 14764 41119 66461 90328 14-258 36786 •58448 79277 99305 18562 37079 64884 72004 88465 04293 19513 34147 48218 6 per cent 14-37618 14-64303 14-89812 16-14107 16-37-245 16-69-281 16-80267 16 00-255 19290 374l£ 16 64685 16-71128 16-86789 17-01704 17-15908 17-29436 17-42320 17-64591 66-277 77407 88006 98101 18-07716 18-16872 18-25592 16- 16- 17" 17- 17" 17" 6pcrce 13-0031 13-210) 13-40UI 13- 590; 13-764; 13^9iM 14-08* 14-23(); 14-36iil 14-498! 14"6-J0i 14-731): 14-846 14-P49 15-04S 15- 13.^ 15 -2j; 16-30« 16-383 16-4.5J 16-5-il 18-68!> i6-7o: 15-761 IRISH CONVERTED INTO STATUTE ACRES. Iriflh. Statute. Iriih. Statuta. Iriah. Statute. R. P. A. R. p. T, A. A. R. P. Y. A. A. R. p. I 1 1 '?! 1 1 3 19 H 20 33 1 23 1 '2 3 3 3 38 10 30 48 3 16 3 4 26 3 4 3 17 15 40 64 3 6 3 4 6 14^ 4 6 1 36 21 60 80 3 38 i 6 8 3 6 8 15 26| 100 161 3 37 1 10 16 6 6 9 3 36 1 300 323 3 34 i -20 32 12 7 11 1 14 6 300 486 3 32 1 1 34 24 8 12 3 33 11 400 647 3 29 1 i 8 9 17| 9 14 3 13 17 600 809 3 36 i 3 1 34 ll| 10 16 31 22| 1000 1619 3 13 1 VALUE OF FOREIGN MONEY IN BRITISH, Silver being 5a. per ounce. 1 Florin is worth 16 Scliillings (Hamburg) I Mark (Frankfort) . 1 Franc 1 Milree (Lisbon) 8 Reals ». d. 1 8 1 1 6 M 1 7 1 9 1 4 8 1 8 If 1 Dollar (New York) . Skillings (Copenhagen) Lira (Venice) Lira (Oenoa) Lira (Leghorn) . . Ruble . . . I.I 4 ? U 01 I I ANNUITY OF £1. 4 per cent 15 16 16 16 17 17 17 18 18 18 18 19 19 19 19 19 20 20 •20 20 20 21 21 21 21 98277 32968 66306 98371 29203 68849 87355 14764 41119 66461 90828 14258 36786 58448 79277 99305 •18562 37079 ■54884 ■72004 •88465 ■04293 •19513 •34147 ■48218 6 per cent 16- 16- 17- 17- 17 17 17 6pcrc( 14-37518 14-64303 14-89812 15-14107 16 -37-245 18-69-281 16-80267 16- 00255 19290 374l£ 16-546851 16-71128 16-86789 17-01704 15908 29436 17-42320 17-64591 66-277 77407 88006 17-98101 18-07716 18-16872 18-35692 13-0031 13-210) 13-4()(il 13-590; 13-764!l 13-9iM 14-08J