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MONTREAL: PRINTED AND PUBLISHED BY JOHN LOVELL, AND FOR SALE AT ALL BOOKSTORES. 1874, i '\ .r I EnteJred, according to the Act of the Provincial Parliament, in the year one thousand eight hundred and fifty-nine, by John Lovell, in the Office of the Rccristrar yf the Province of Canada- of the ( ▼)00ii til iOii('hc:\s and Pia oonnnen iniiu'ovci ' The Irish X;i to adiipt soriu'wlia Tuaii v alt titat the <'Xi,'('[)ti()i fi'oiji the iu .'1.11 est it WW th( on :.ho su he coiisidi other exc( I'*y far 1;^ h( pod I will tend tisefi.d thill soi'i<'3 of the s-ectioi of the bo( ^vork, as ^vhat has Jii'fives ufc con/bunde( measure w( i^inco t ibers is tlui J of tens and |why4he ru] |»n the Iblh PREFACE. i\ proparincj tli(> followiiii:; work (iinclortakcn at tho su^^crostion of the Oliicf SuiH>fiiircii(lciit of HdiK-iitioii for rppci- (-iiiiJiiIii), it. hiu T)ooii the coriHtiUit aim of tlio Author to present it to ('anadiati icaclicrs and studcntrf as a thoroui^dily rt'hal)l(' Trc.itlsc on tho Theory and rractic'O of NuniborH, and as an Arithmutic, in some di'gi'ce, i'o/nim'nsuiate with the hii^hor (pialifications of teachers anil the improved methods of instruction now <^enei'ally found in our schools. The Arithmetic now oHired to tho pul)ru' is ))ased u[)()n the Irisli Xational Treatise; — in fact, it was at iirst intended merely to adapt that work to tho decimal currency, and to al>breviatc tho somewhat tedious reasons here ;^iven for tlie varioiis rules. Ho man v altei'ations and impro. '"nents suijLjcsted tiiemselvcii, however, that the original desii;!! wa. speedily al)audoned, and, witii the exception of the ./st ten or tii'teen pa;j,es, which are taken entire from the v/ork in Vpicstion, the "realise, as at present issued, is, in :i.ll essential respects, an entirv.. '^ new book. Xt'vertheless, a-i it WIS the sole ol)ject of the Author to pre])ari! a couijdrfe text-book on !.he subject of Arithmetic, he has not hesitated to a(lo[)t whatever he consid(!red i^^ood, eitlier in the Irish National or in the numerou;! other excellent works on the subject. Fy far the ,interostin,i':; and usefvil than it has hitherto been. It will bo observed, that a thorouj;)! sorii'3 of review examples has been given at the close of each of the .sections np to the seventh, and a very extensive set at the end of the book. This is deemed an important feature in the present work, as in some degree insisting upon that careful revision oi' what has been learned from time to time, withotit which, tho pup'l arrives at the end of tho book with all the rules and ])rinci[)les f^o I cuniounded with one another, as to render his knowledge in a gre;" I measure worthless, \ Since the ouly difference between simple and denominate num- , bers is that the one increase and decrease according to the scale ot tons and the other according to dilferent scales, there is no reascei • why4he rules relating to them should be separatecl ; and therefore in the followmg pages no distinction is made between simple ami -/ /•' '/. PKEFACK. compound rulcH. A Honiewbat cxtcndcil experience has convinced the Author that, except to the niereiit hc^inneis, tiie 8cienc(! of Arithmetic is more successfully jjicscnted by this than by tiu' ordi- nary method of makinji; the pupil learn one set of rules for sim[»lo numbers and a o letely different set for compound numbers. It will bo observed tiiat towai'ds the end of the Treatise the rules are mainly deduced algebraically. Some teachers may not, at fitst, be disposed to re_i;ard this as an i.nprovenu'iit, but it viis not adopted until after cai-eful deliberation and consultation with many of the most successful teachers of Arithmetic in the I'rovince. It is ni-;KACE. convinced si-ioncH! of ,'i()v sinipio is:(> the rules lot, at iiiHt, not a(loi>tod nany <>t ^^'^ . It is gen- [110 sort, the u rropovtion >pted by the he deduction pleat possibk'. the rule m a ided in all the lay be furth'M- St' abbreviated lie. Once tho !r independerl uleshe is able, „v himself th« l>rest, Discount, , Uie pupil ia j^ obvious that to hin\ in th« vaova])hs. l^ct •„v c unuitting to coinniit tbo )\\"\\ conviction lU^rf. In short, ,c the pupil rc- rcracnd)ev all uiics; and tlu t ,;s of lil'e, these imded with one After many •uccessi'ul mode ,upii to deduce ,y the resultmg Icted tc the He- is thought, the imcration ar** so concinely worded that they may be advfintageously committed to memory by the pupil. The examination questions thi'oughout the work have lieen care- fully pre))ar(>d, and are d('siu;ned both to enable the sell-taught stu- dent to test, at each section, the extent and thoroughness of liis knowledge of the principles therein conlained, and also to guide the pujiil as to what piiiiciples ami dellnitions !ir»' of siu-h im- portance that they reiiuire to be connnitted to memory. Tiiis latttr object is further secured by the ai'rangeuM'nt of type — all the defi- nitions ami leading principles being piiiited in large tyj)(>, the cvpla- natlons, reas(ms, and remarks, in small type, ami the problems in a size inte-mediate to the two. Great pains have been taken to render the wording of the ndes as perfect as possible ; and it will be ob.served that, in order to catch the eye when glancing over the page, tliey are invariably printed in Italics. It is bclievc'.l that the sections on PropovtioTi, Fi-actions, Interest, &e., contain a larger amount of information, and a better selection of examples, than are commonly given ; and that the section on the Properties of Numbers and the tiin'erent scales of Notation will tend very materially tiO enlarge the pupil's ac(iuaintance with the general principles jf the science of Arithmetic. Although the Preface is not the proper place for discus.^-ing methods of teaching Arithmetic, the Author cammt refrahi from urging upon his fellow-teachers the following points : 1st. The pupil should be thorougidy drilled upon the use of the signs and symbols of Arithmetic, because these constitute the lan- guage proper to the subject. 2d. He should be re(|uired to commit to memory all the essential >, definitions, and also the tables of money, weights, and measures. ' The teacher would do well to examine his pupils on these taldes once a month or oftener, since if the pupil has to tm-n l»ack to his book for each table as it is required, it is not to be expt'cted that his progress will be very rapid or thoiough. It maybe fairly (jues- tioned, whether more tliau half the dilliculty and obscurity that cling to the subject of Arith.nietic does not arise fi-om the I'aet that the pupil is not familiar with the signs, the tables, and th.o {iruJcipUis of notation. I': 3d. The teacher .should give his class, from time to tinn^, (pu^s- ,|ions of his own construction, either to solve at home or as ordinary school-room work, and the pujjil should be encouraged and recpured to write questions themselves under ejich rul(.>. This is an ini))ortant ((jxercise, and no teacher who once adopts it will ever throw it iside. 4th. In all operations in which there are both multiplication and / S T JS \ i'UEi'Acii:. *vj ilivision, tlio pupil sliouM ho t,;iii,t:;ht to first Indicate tlio processes by uieii iippiopriate sif^iis, uiui then cancel U8 tUr as pOHsiMe. nth. Tli(» te.icher is respectfully reinlndcd, that without fieipient and Ihorouj^h reviews there can be no ri'al pi'o;4ress. Exju'rii nee has sliowu that from one-third to one-half of the time devoted to Arithmetic can be profital)ly devoted to revision and recapitulation. t>th. The teacher should re(|inre from his pupil the absolutely correct answer to eaeh tiuestion. "" J^'var enoujf/i '^ is productive of j^reat mischief to the pupil, as it encourages a habit of such careless- .■;e«s in his oi)erations, that no conlidence can be placed on his results. iu Is not enoui^li tliat the pupil undiislands the jirinciples — jilfhou^'h this of course is important. It is p(»ssible so to train the pu|)il that his operations in Arithmetic shall be at once rapid and accurate, and this shoidd be the aim of the teacher. Toronto, Dia/ubct', l8yy. li processes by e. liout frtMiuont Kxitrrinicc 110 (It'Vdtcd to capitulMtion. tho absolutely pnHluetivo ol" such fiut'lcss- 1 oti liis ri'sults. |)U'S — altlioujib the pu])!! that, 1 ai'curate, and ^1 PREFACE TO THE SECOND EDITION. Tfie Author onibraces tho opi)ortunity affonlod by tho issuo of a Socoiul Edition, both to tiiauk his fellow- toachors in Canada for the kind and flattering reception thoy have given his work, and to otlbr a few words of explanation on what, as far as he can learn, is the only ieaturo that does not meet with very general approval. He refers to the union of the Comj)ound with tho Simple Rules. It hjis been objected to the arrangement adopted in the National Anth- metic, that a pupil must know the Simple Rides before he can work problems in Reduction or in the Compound Rules. Now this is audoubtedly true, and would be a fatal objection to any such ar- rangement in an Elementary or Primary Arithmetic. The National «, however, an advanced or second book on Arithmetic, and tho pupil is assumed to have progressed through an elementary text- Oook before he enters it. If the National Arithmetic were designed for beginners, where would be the necessity for a First or Elementary book on Arithmetic V The objections have arisen altogether from a misconception of tho design of tho book. Tho pupil is supposed to have worked tlirough some elementary text-book on arithmetic, and to have ac(iHirod a certain amount of practical skill in arithmetical operations. lie then commences tho National, and, in progressing through it, not only meets with additional and more advanced practical exercises, but also learns the reasons and the mutual rela- tions of tho several rules. In tho Elementary he is taught how to multiply an abstract by an abstraqt number, or an appVcato by an abstract number. In the National he is shown that these operations, though differing in detail, are essentially the same in principle ; and b«^ inj thus enabled to generalize and classify. s PUKFA(E. Another objection urged is, tliat if the National Arithmetic be dcsif^iicd for a second book on the; science, tfie siiiii)lt' probh'ins jrivoM lit the cornmcncfiaerit oC each nde, and indeed the euilier rides tlu'inselvcs, should not l)e inserted. This is also a mistake. The object has been to exhibit a gradual progression from the simple to tljc more difficult — to show that the most simple and the most complicated problems depend essentially upon the same principles. Indeed, were the National Arithmetic intended merely as a second practical work on arithmetic, three-fourths of it might have been omitted, and nothi^ig given but the few rulea omitted in the Elc mentary. »s^. |i CONTENTS. SECTION I. PAOB Definitions 17 Notiition and Numeration 1^ Arabic Notation '-^ 1 IJoman Notation 23 Exercises in Notation J53 Exercitjes in Numeration 34 Denomination of Numbers 34 Tables of Money, Weights, and Measures 35 Reduction Descending .*. 50 Reduction Ascending 81 Recapitulation to Section T fi3 Miscellaneous Exercises on Section I C8 Examination Questions on Section I (50 SECTION IT. Fundamental Rules 62 Addition 03 Proof of Addition 07 A])plication 71 Recapitulation 74 Examination Questions on Addition 75 Subtraction 70 Proof of Subtraction 79 Application , 82 Recapitulation , 83 Examination Questions on Subtraction 84 Multiplication 85 Multiplication Table 87 ! I |i 10 coNTErra PAQB To Multiply by a Composite Number 90 To Multiply when the Multiplier contains Decimals 94 Proof of Multiplication 95 Contractions in Multiplication 97 Exercises in Multiplication lOU Examination Questions on Multipllcition 101 Division 102 General Rule for Division 105 General Principles 108 Proof of Division 109 General Principles Ill To Divide by u Composite Number 112 To Dinde when both Divisor and Dividend are Denominate Numbers lir> To Divide when the Divisor or Dividend or both contain Decimals 114 Contractions in Division 115 Exorcises in Division.*. 110 Examination Questions on Section II 117 Miscellaneous Exercises on Sections I and II ., 118 SECTION III. Properties of lumbers 120 Table of Prime Numbers 125 To Resolve a Number into its Prime Factors 120 To Find all the Divisors of a Number 127 Number of Divisors 128 To Find a Common Divisor of two or more Numbers 129 To Find the Greatest Common Mea. e of two Numbers 129 To Find the Greatest Common Measure of more than two Numbers 130 Second Method of finding the Greatest Common Measure 131 Least Common Multiple 132 Scales of Notation 136 To Reduce a Number from One Scale to Another 137 To Reduce a Number from Any Scale into the Decimal 139 Fundamental Rules in Different Scales 141 Duodecimal MuJjtiDlieation 143 ■» 5;:? Ml PAGE ■ yo 94 5)5 97 100 101 102 10.5 108 lO'J Ill 112 )enoniinaie 11:; til contain 114 115 110 117 118 120 126 120 127 12S 129 121) lan two 130 131 132 » 130 137 13!) 141 143 CONTKNTri. .. H PAOK Examination Questions on Section III * 147 Miscellaneous Exercisos on Sections I-I II 149 SECTION IV. Vui;;aran(l Decimal fractions.. ,^ 150 General Principles 151 Delinitions of Fractions 152 Reduction of Fractions 154 To Reduce a Mixed Number to a Fraction 155 To Reduce an Improper Fraction to a Mixed Number 156 To Reduce a Fraction to its Lowest Terms 156 To Reduce several Fractions to a Common Denominator... 157 To Reduce several Fractions to their Least Common Denomina- tor 158 To Reduce a Compound Fraction to a Simple One 159 Cancellation 160 To Reduce a Complex Fraction to a Simple One 161 Reduction of Denominate Fractions 102 To Reduce one Denominate Number to the Fraction of Ano- ther 164 Addition of Fractions 166 Addition of Denominate Fractions 168 Subtraction of Fractions 109 Multiplication of Fractions 171 To Multiply a Denominate Number by a Fraction 173 Division of Fractions 174 To Divide a Denominate Number by a Fraction 176 Multiplication and Division of Complex Fractions 178 Examination Questions on Vulgar Fractions 179 Miscellaneous Exercises on Vulgar Fractions 180 Decimals and Decimal Fractions 182 To Reduce a Decimal Fraction to its Corresponding Decimal 182 To Reduce a Decimal to a Decimal Fraction 182 To Reduce a Vulgar Fraction to a Decimal 183 To Reduce a Denominate Number of Several Denominations to au Equivalent Decimal of a given Denomination 183 To Find the Value of a Decimal of a Denominate Number 184 ,M I :v ill Mi ul i ■I , ■i 1 12 00NTENT3. pagh Circulating or Repeating Decimals 186 To Determine the Number of Places in the Decimal correspond- ing to a Given Vulgar Fraction 189 To Reduce a Pure Repetend to a Vulgar Fraction 190 To Reduce a Mixed Repetend to a Vulgar Fraction 191 Addition of Circulating Decimals 194 Subtraction of Circulating Decimals 194 Multiplication of Circulating Decimals 19£ Division of Circulating Decimals 196 Miscellaneous Exercises in Decimals 19G Examination Questions on Section IV 197 Miscellaneous Exercises on Sections I-IV 198 SECTION V. Ratio 200 Proportion 20G Simple Proportion 208 Compound Proportion 213 Conjoined Proportion 218 Examination Qu,A-nswers I'AOK 288 y or to 288 !ncy to 289 2W) 200 200 204 206 297 209 300 • •••■•••• 0*'f) 30H • •••••••• Oi'> 314 • •••••• 1 1 X. *f l]U\ ;U8 Loga- 3'''2 • ■••••« *} ^"X 324 lims... 825 lis 326 326 329 331 333 337 340 PAOB ro!=ition 345 Single rosition 34() Double Position 34S Coniponiul Interest 354 Annuities 357 Annuities at Simple Interest 358 Annuities at Compound Interest 300 Table showing the Amount of an Annuity of ^1 or £1 for any number of payments 3()2 Table showing the Present Value of an Annuity of $1 or £1 363 Examination Questions on Section XI 366 Examination Problems 3(>7 Arithmetical Recreations 378 Tables of Logarithms 381 Tables of Squares, Cubes, and Roots 307 Answers to Miscellaneous Exercises 405 Answers to Examination Problems 410 SIGNS USED IN THIS TREATISE. -j- the sign of additior ; as o-{-7, or 5 to be added t(^ 7. — the sign of subtraction ; as 4 — 3, or 3 to be sub- tracted from 4. X the sign of multiplication ; as 8 X 0, or 8 to bo multiplied by 9. -:- the sign of division ; as 18 -v- 6, or 18 to be divided by 6. ( ) which is used to show that all the quantities uiMted by it are to be considered as but one. Thus (4-|-3 — 7) X6 means 4 to be added to 3, 7 to be taken from the sum, and to be multiplied into the remainder. The latte?" is equivalent to the ivhole quantity within the brackets. = the ^ign of equality; aso+G — ll, or 5 added to 6 is equal to 11. f >^, and |<|, mean that J is greater than i, aad that I is less than ^. ^ : is the sign of ratio or relation ; thus, 5 : 6, means the ratio of 5 to 6, ai?d is read 5 is to 6. :: indicates the equality of ratios ; thus 5 : 10:: 7 : 14, means that there is the same relation between 5 and 10 q,s between 7 and 14 ; and is read 5 is to 10 as 7 is to 14. \/ the radical sign. By itself, it is the sign of the square root ; as j/S, which is the same as 5^, the square root of 5. |/3, is the cube root of 3, or 3^. ^4-. is the 7th root of 4, or 4^, &c. , Example. [ ^{(8— 3+7) X4-^6! -f 31]x I'd-^IO'^X 5^=: 556-25, (fee, may be read thus : take 3 from 8, add 7 to the difference, multiply the result 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 55G-25, &c. These signs are fuUi/ explained in their nroper places. m Til -. V " TISE be added to 7. 3 to be bub- [), or 8 to bo I to be d>ided antities ur^ted (44-3— 7) XG from the swm, The latteA" is brackets. r 5 added to 6 ;r than ^, and s, 5 : 6, means : 10::7 : 14, en 5 and 10 us IS 7 is to 14. le sign of the 52, the square |/4 is the X |''9-^102x 8 from 8, add 4, divide the Liotient and to Libe root of 9, I, multiply the will be equal )roper places. ARITHMETIC. i 1* SECTION I. DEI'IXITIONS. 1. Science is a collection of the general principles or leading truths relatiug to auy branch of knowledge, ar- ranged in systematic o'-der yu as to be readily remembered, referred to, and applied. 2. Art is a collection of rules serving to facilitate the performance of certain operations. The rules of Art are based upon the principles of Science. 3. Aritlnnetic is both a Science and an Art. 4. As a Science, Arithmetic treats of the natnre and piopertics of numbers ; as an Art, it teaches the mode of ap'ilying this knovvler as writ'nuj and rfiid'nnj, and, though often confound(!v., Jiey are in reality pcrtectly distinct 10. It is obvious tlwi.. for tlie pnri)Oses of Arithmetic, vc i rc(|ui'e the power of designatiuj^- .11 possible numbers; it is eijually obvious; that we cannot give a dift'crent name, or character to each, as their variety is boundless. We must, therefore, by some i»ieans or anolhcr, make a limited system of words and signs sufiice to ex^^ress <.n un- limited amount of numerical quantities. With what boantiiul t^im- plicity and clearness this is efl'ected, we shall better undei stand pies- eutly. 11. Two modes of attaining such an object {ircscnt themselves; the one, that of corribinmg words or characteis aliCfidy in use, to in- dicate new quantities ; the other, that of representing a vaiiety of different quantities by a sincfle word or <.' aiacter, the danger of mis- take at the same time being prevented. The Komans simi)litied their system of notation by adopting the piiiiciple of t'owiiv?(pjal groups, each / oup consisting of some numbei- qiiite within the limit of our comprehension ; if th(* groups were few, our object would be attained without any fiuthor ellbit, since wo should have aecpiircd an accurate kimwiedge of the number of groups, and of the number of individuals in each group, and thereibrc a satisfactory, although indii'ect estimate of the whole. W<} (»uj;l)t to rt'inark that difTort'nt pcr.'^ons have very difftront IhiiitR to tlieir i)i.'rfc(!t c wripri'hensi.) . of niunlxT. The iiiU'lIi!,"'nt can crncoivc willi oase a coinparutivi'ly Irirr ono; thore an; saviim-p so nide as to he incaitahlc of foimuig' au idea of oiiu that is extremely stnalh 14. Let us call the number of individuals that we choose to con- stitute a group, the ratio ; it is evident that the larger the ratio, the smaller the number of groups; and the smaller the ratio, the larger the number of groups. 15. Tf the groups into which we have divided the objects to be '\ reckoned, exceed in amount that number of which we have a perfect id^-a, we must continue the process, and, considering the groups thcm- ' j-^ selves as individuals, must form with them new groups of a higlier f order. We must thus proceed until the number of our highest group .• is sufficiently small. 16. The ratio used for groups of the second and higher orders, ^ woidd natural y, but not necessarily, be tiie same as that adoptetl tor I the lowest; that is, if seven individuals cnnstitute a group of 'Me fir^t " o.der, we should probai)ly make seven groups of tiie liist o.d-r jou- siitute a group of the second also ; and so on. r*'. It I'iiglvt, and very likely would happen, that we should not hi.v.( s) many objects as would exactU) form a certain number of g -ri-j oi" t'le higjiest order — some of the next lower migiit be left. T.;j x\ ,1' might occur in forming one or more of the other groups. Wo mighty for example, ia rediooiog a heap of pebbles, hav« twt? 20 NOTATION [Skct. 1. I* iil!:! '! I I; It groups of the fourth ordci-, three of the third, none of the second, five of tlie first, and seven individuals or tjiaiplo unit^i. 18. If wo h id nuulo faeh o. the first order of groups consi.-^t of ton pebbles, each of the second order consist of ten of tlio liisf, each fj;i()Mp of Miiiil ol t; ii of the ■•.ccoiid, and .so on with the u-sl. wc l)ad tclcrtcu .. (If'crrmf system, or that which is not only iistd at present, but -viiich was adopted Ity tlie Hebrews, (Jreeks, llonians, &e. It is roniarkal)!e that the hin;;ua;^o of eveiy civilized nation gives names to tlie (lilieifut of flsis, but not to those of any other numerical system, its very j^eiicial Uin'u.-ion, even ann-np: rude and barbarous people, has most probably arisen Irom the habit of counting ou the liii;j,ers, which is njt ullo^elher abandoned, even by us. 19. It was not indispensable that we slaudd have used Mie same ratio lor the Lrroups of all ilie dilierent ordeis. We niij;ht, for cx- uniple, have made iour pebl.les form a ^r(>up of the first oi'der, twelve fiioujis of the tirst oi(lr\' a f^renp of iIk' second, nnd twenty groups of tlii" second a j^roup ol tl.r iliiid oidrr. In such a ctise v.c had adopted a system (!>:actly like that to lie fotmd in the t;ibU' of 8terlin<;- ujoncy, in which four fai things make a ^Moup of th».' (M(!cr of petic(\ twelve pciuic a ji;roup of the oidi'r ul' n/iJ//iii(/.^, twentv shillinj:S . a group of the order of poitudu \\'liile it must be admiite*! that tlie use of the same system for applicate, as I'or a))stract nunibcs, would greatly simplify our arirhiiKlical processes — as will be evident heie- nltei — a glp'.icf^ at the tables ,<:;iven further on, and (hose set lown in treating ol" exchange, will show that a great variety of systems have actually been constructed. 20. When we use the same uUio for the gi^oujis of all tli-^ orders, we term it a c-niuDnni. ratio. '^I'liere appears to be no particular rcaiun why tai should have been selected as a "common latio" in the sys- tem of numbers ordinarily used, except that it was sujrgcst'- d, as already remarked, by the mode of counting on the fingei?; and that it is neither so low as unnecessarily to ineiea:-e the number (tf OJ'deia of groups, nor so high as to exceed the conception of any one lot whom the system was intended. (See Section 111.) 21. A system of numbers is called binary^ ternary^ quaternary^ quiiiory, senary^ septenary^ octcuary^ notiory^ di'itary, taalcuary or ihioik'iHfi'y^ according as iivo^ fhrrc^ four, Jive, si.v, semi, ciylit, -ithtc^ 1(71, e'eeex, or twelve, is the coimnon. ratio. The dciiary and di/.o- denary s\ stems are jnorc cemmonly know)) as ilic decimal and duo- decimal systems. Ouis is thcrefoio a dccinud or acnary system of nund)ors. If the common ratio were sixty, it would be a nexaycnmal system, Such a one was formerly used, and is still, to some extent, retained — tie will be perceived by the tublea hereafter given for tlie meaeure* [Skct. L lie sccuikI, fl\0 ups consi.-f of I'll of tlic liist, 1 with the u'sU ot oiiiy iisctl ill ici'ks, H()iu:;iis, 'Ivi!i/.('d nation to those of any en amoii}; vuilc n the ii:\bit ct' jaiidouCLl, even ! usod the same nnpht, for ox- thc iii>t ordc'i', lud, Hiid tucnty such a citse Vic in tl'V t;>M<' less tlian tiiat whieh constitutes the lowest p;roup, and to the di[/'(^rent .ir.7/'>'.s' of ftionps; beeaun? all po.ViMe nundnMs nui.-t vonsi.st of indi- vidu.il^, or of pronps, or of both indivi(iiials and ;,M0ups. In neither ease would it l)e refiuired to spceity more than the innnbisr of indi- viduals, and t]i(> nunil)er of each species of group, none of which nu'iil.eis — as is evident — ean bo irrealer than the eonunon ratio. 'Ihi'^ is piveisv'ly what we liave dune in our nunierieal .system, exeei)t t')at we have formed the name of .somo of the groups i)y combinin;^ those already used. Thus, ''tens of thousands," the group next lii'^her than ' thousands, is dosi^naled by a eotnbination of words iilieady applied to express other gioups — which tends still further to piinpliiication. 23. Arabic ayaicm of Xctaiion : — K'iniP'i. CharaeUfB. rOne 1 Two 9 'i'lirco 8 Fnar 4 ' Fivo 6 Six 6 S"v'(>n 7 Eicrlit 8 inio 9 Ten 10 lliniilrcfl 100 Tli(.:v-:.ii'l 1.0(M> Ten Thoa ~fitul 10.000 liiin.lroilThou -.I'nd 10(t,onO Million . 1,000,000 Units of cntr.pari.son, or siniplo units, First £rioiip. or units of tho sopoikI ord -r, Hi'i'Dmi ^rniip. or units of tlic lliir«l ordtT, J'iiir;! ^roiip. or units of tiic Jou-.tli orikr, FoMrlli proiip, or utiits of tin- Hl'tli ordiT, Fii'tii irroap, or units of rlu' t-lwh (ir(l,>r. Sixth ;_'roup, or units of the seventh order, 24. The ehanetors which express the first nine numbers are tho only ones used. They are called dir/its, from the custom of counting them on the fingers, already noticed, — "digitus" meaning in Latin a finger; and they Imve also been called K/r/nificant Jli/nre.t, to dis- tinguish them from the cipher, or 0, whieh has no value when stand- ing alone, and which is- used merely to give the digits then' proper position with reference ",o the dechnal point. 25. The drx'mal point is a point or dot used to indicate the posi- tion of the .simple units. The pupil will disi^inetly remember that the place where tho "simple units" are to be found is that immediately to the left-hand of this pomt, which, if not expressed, 5s Huppoacd to stand at the right-hand side of all the digits. Thus, in 4C8-'76 the ^ expresses 22 KoTATtON t9«or I I •il "simplo unitfl/' being to the loft of the decimal point ; in 4'J the U exj)resse9 " simple units," the Uccimal point being understood at the right of it. 26. Wo find by the tabic just given, that, after the first nine num- bers, the same digits are constantly repeated, tlieir jjositions v\ith reference to the decimal point being, however, changed ; that is, to indicate succeeding groups, the digit is moved, by means of a cipher, on(^ place faither to the left. Any one of the digits may bo used to express its icHpective number of any of the groups: — thus 8 would be eight "simple units"; 80, eiglit groups of the first order, or eiglit " tens" of simple units; 800, eight groiips of the second, or luiits of the third order ; and so on. We might use any of tlve di^rits with ditlerent groups; thus, for example, 5 for groups ot tiie third (udor, 3 for those of the sec(md, 7 for those of the first, and 8 lor the "eim- pie units," then the whole set er of pebbles in <;roiips. sometimes of one. sonii tiiiieh of another, ancl soineliiries (f several orders, and then be de- pired to express them by charaeters — the *• uidt of c be under- vaiiu-d or forsrotten. Pebbles will answer well for examples— Indeed, their use In computing [Sect. L nt ; in 49 the 9 nderatood at the le first nine nnm- V positions v\ilh iigod ; that is, to cans of a ci[)hor, may be tisod to -thus 8 wouM be t order, or ciglit cond, or units of )f tlve (lij/its with 1 thf thiid order, 1 H for tlic "Kim- 5000, ;iOO, 70, S, cipher, whin the and it is evident Lhe decimal point one; also the 7S le seeond place. It las two values, is the number of 1 is unchangeable; at other tinu'd le order of units IS 111 .., iiitciKlod jirind- .ly niul int.-Uict-nco •M-n to acli!l<1. jifir- .lo; tlic imi.il niiiv )i!ps. soiuctimos CI ■s. aii'l tlu'n be de- III ■'' beiiii! occasion- CIS, or to scores, or th these iiitroduC' .eririL' wilLoiit any )on to explain, ami y trouble bestowed and raj) (^ity w.th J this, he has only d fronn, and hisc.\- )iirt'd to unfold. It understandinff, he for it is with ehil- ;)eal snece.s'ifiilly to the attal mom of ;osts to be under- uiie lu computiug of I' A.Kt» It-m . AND NUMEUATIOtt. ROMAN SYSTEM OF NOTATION. sd 2 <. Oiu- ordinary numerical characters have not been always, or everyN\herc, used to express numl)ers; the letters of the ali)habet niiui.^lly presented themselves for the purpose, as being already fan.il ar, and, accurdinKly, were very generally adopted— for cx- a;npl'' by the IIelMew««, (Jreeks, Romans, &c., each, of cour.'^e, usinc; 'theii- own alphabet. Tlie i)upil shoidd be acquainted witu the Itoman notation on accoimt of its beautiful simplicity, and its being vtill employed in inscriptions, &o. : it is found in the lullowing table; '- Characters. Numbers expressed. I, . . . One. II, . . . Two. III, . . . Three. Antiupatedcluingellll, or IV, Fotir. *^iiaii^c . . . V, . . Five. V[, . . Six. VII, . Seven. VIII, . Eight. A.ntlt.ipatefl change IX, . . Nine. ChaiK^e . . . X, . . Ten. XI, . . Eleven. xn, . Twelve. XIII, . Tiiirteen. XIV, . Fourteen. XV,. . Fifteen. XVI, . Sixteen. XVII, . Seventeen. XVIII » • . Eighteen. XIX, . Nineteen. XX, . Twenty. XXX, &c ., Thirty, &c his £,'1 en rise to the term cnlcuhition; ^' ci[\cn\n»''' bcinti, in Latin, a pebble; b'.it while tile teacher illustrates what he says by croups of particular objects, he must take care to notice that his rouuirka would bo equally true oi" any others, llo Hiu>t also j)oint out tlio ditferenee between a L'roup and its equivn.- lent unit, wliich, from their perfeet equality, are ireriernlly confonndod. Thus, he ?!iay show tiiat a penny, while eqwtl to, is not hlentichl with four fa"thi nrs. This set'miuirl}' unimportant roinarK will be better appreriated iievinfrer; at the same time, without, inaennracy of result, we may, if we pleas-'. e; symbols, or com- er Vli). Ill con- inrv in view ; ihiit lie coviwov rath ; ot only at ton, ten ('. A purely deci- eii times ten, &c. ; As fur as nota- r n decimal nor a 0} appoiv to liave ol' t ii " units of I'oujis from these, series. ono unit before )lc unit," but one view, four is one s beiiip; now the before five hun- ted. 31. From the table (28) it will be seen that as often as any letter is repeated, so many times is its value re- peated. Tims I, standing alone, denotes one, II denotes two, / a single character, or " anticipation'''' ; Mpnt that character or ''''anticipation'''' to the right hand of vjhat is mtdreadg written, and take its value from the last remainder ; proceed \th'iis until nothing is left. fExAMTLR.— Set <1oAvn tho number cish'oon hnndrpd nnd forty-fou'", in Iloriiiiii charactiM's. One thousand expressed by M is tlie hicrlicst. number with- #|n the given one, indicated by one character or by an •' anticipation " ; we put idown I ™' fBTid take one thousand from the given number, whicli loaves eight hundred nnd torty-fonr. Five liundred, D, is tlie biu'-hcst number witiiin tlie last re- •Jinainder (eight hundred and forty-four) expre.-sed bj' one character, or an "an- iticipation '';' we set down D to tho right hand of M, ■I MD. nnd tflke its value from eight hundred nnd forty-four, which leaves thro'^ hnn- ;^drcd ai d forty-fonr. In this the highest number exprosred by a sinude charac- I'ter, or an "anticipation," is one hundred, indicated by C; which we set down, C^and for the same reason two other C's. MDCCC. This leaves only forty-four, the hiarhest number within v/hirh, expressed by a bingle character or an '" anticipation," is forty, XL,— an *' anticii)ation ;" we set this down also, MDCCCXL. Four, expressed by IV, still remains; which, being also added, the whole is as follows:— MDCCCXLIV, ! ! t 26 Exercise 1. tSlOT. 1, 33. Express the following numbers in the Roman notation :-» 1. Twenty-five. 2. Forty-three. 8. Sixty-seven. 4. Eighty-nine. 6. Ninety-eight. 6. One hundred and thirty-seven. ^. Three hundred and seventy-one. 8. Four hundred and two. 9. Six hundred and seventeen. 10. Nine hundred and ninety-nine. 11. One thousand four hundred and forty-six. 12. Three thousand eight hundred and five. 13. Eight thousand six hundred and seventy. 14. Twelve thousand one hundred and sixty-nine. 16. Four hundred and ninety-seven thousand, six hundred and eighty-two. 1. XXV. 4. LXXXIX. r. cccLxxi. 10. CMXCIX. 18. VMMMDCLXX. Answers. 2. XLIII. 5. XCVIII. 8. CDII. 11. MCDXLVI. 14. XMMCLXIX. 8. LXVII. 6. CXXXVII. 9. DCXVII. 12. MMMDOCCV. 16. CDXCVMMDCLXXXII. Exercise 2. 34. Read the following expressions: — 1. XCVII. 2. CCLXXII. 8. DCLXVIII. 4. CMIX. 5. XV. 6. VMMMXXXIII 7. XVDCCCLXXXVin. 8. DCXLVMCMIV. 9. XXVXXV. 1. Ninety-seven. 2. Two hundred and seventy-two. 3. Six hundred and sixty-eight. 4. Nine hundred and nine. 6. Fifteen thousand. 6. Eight thousand and thirty-three. 7. Fifteen thousand eight hundred and eighty-eight. 8. Six hundred and forty-six thousand nine hundred and four. 9. Twenty-five thousand and twenty-five. ^ To fi paces DC certain d tke Engl %encli. ilrely ob liethod. I tSaoT. ; l^tt. 33-38.] ANi) NUMEfeAtlOSf. ii >man uotation :•— ARABIC SYSTEM OF NOTATION. le. six hundred and 8. LXVII. 6. CXXXVII. 9. DCXVII. 12. MMMDCCCV. 8. DCLXVIII. 6. VMMMXXXIII 9. XXVXXV. ight. idred and four. 35. In the Common or Arabic system of Notation, the ^amo diaractcr may have diflcrent values, according to the place it holds Ifith reference to the decimal point (25), or perhaps more strictly to 4ie simple units. This is the principle of position. 36. The places occupied by the units of the diflerent orders (2?), Iftay be described as follows : — simple units, one place to the left of tihe decimal point, expressed, or understood ; tens, two places ; hua- di-eds, three places, &c. 37. When, therefore, we are desired to write any number, we liB'-e merely to put down the digits expieshing tiie amounts of the different units in tlieir proper placos, according to the order to which each belongs. If, in the given number, there is any *' place" in which there is no digit, a cipher must be set down in tliat place, when (iequired to keep another digit in its own poulion. — But a cipher Sroduces no effect, when it is not between one or more digits and the ecimal point; tlms, 0536, r>36*0, and 530 would mean the same thing — the first 13, liowcver, incorrect. B3G and 53(')() are different ; in the latter case the cipher affects the value, because it alters the position of the digits. ExAMPLB. — Let it bo required to sc'. down six hundred and two. The six must be in tile tidnl, and the two in tiie tlT-st pluce; lor this purimso wo aro to i>ut a cipiuif between the 6 and 2— thus, G02. Without a cipher, the six Woiihl be in the second phice— thus, 62; and would mean, not six liundreds, but 1^ tens. 38. In numerating, we begin with the digits of the highest order, ilid proceed downwards, stating the number which belongs to each Itt-der. To fiiollitate notation and numeration, it is usual to divide the ijlaces occupied by the different orders of units into periods. For a certain distance, the English and French methods of division agree ; t" e English billion is, however, a thousand times greater than the ench. This discrepancy is not of much importance, since we arc rely obliged to use so high a number ; — we shall prefer the Fi encli ethod. To give some idea of the amount of a billion, it is only cessary to remark, that, according to the English method of not • n, there has not been one billion of seconds since the birth oi ^ rist. Indeed, to reckon even a million, counting on an average tliree per second for eight hours a day, would require nearly 12 daya. e following are the two methods; 'r "■! r is NOTATION [Bect. } c e 125 W o W K O 2 a o o .2 • "^ .2 — ;s n: «; ^ '■" 9 c B bJ ?=,: 2s r^- ^§ -^9 s .2 .2 g^' t5§ £-1 ^1 •=! g^ T2- 5^ 5« i| 0;= Ui= C/j;^ '^'"•^ '^''^ ^.2 '^ 5 <. § t- 2 .22 c t' JC-n '?.?; '^ •J''r'- •«,->« ''^i;!; . '-ica '■'■'5 ■'" TI^ -- '■" 'c ^^ c ^ '-'-''-'■■' '■'^ - "3 ^^ •"• 'c ^ ;r: '^ t~- irj - ^ . '^ "^ • r: >■ £ '^ ^ «r c r' ^ •— i' t— .— i- V :r: w V. ;::; i* 5*, p ;^ •*-. v c, vj ?■ v '». 5; cj -- ■■" S ~ '-^ 'c; ' •« ,_> " '" c "H •/. "^ ;:^ '/:;::; ^ '/^ ,5 "S '■'^' •— "H ^ - "i '/' Z *" '_2 X3 ''^ '"^ ^». ""^ — • 7I> •"* ■■* .— _i ■"■ c C* *"* '""' C" •--■ ■" c"" ^*" •"" ** *™ r" '^^ P H c i:::^ H :>^ :5 H '/; ;5 H o':5 h 6^ p [- ti h S :i^ f-' ^' t h T :^ c- :^ 3 ;3 3 8 3 :J :? 3 8 8 8 ;i :3 8 ;j 3 3 » ;} ;3 ^; ;; r; ;. s 3 8 ;! 8 rfj ; ■ I ; ; ; ■ p :::::;::::;:::::::::: 2 . . . . v) . (T; .:'....;.. . :=:'/)••••£ m c nr c •;: ;.2 ::::: 5 ;;:::.£ ;;:;;:;:; ; 1§ ::: :|H g :;: :5 '/::: :S £::::::•:: : O'-s ,• : ;^S i ••%.= ••• •«^:f: •• • ;;;q'.= § ; ^-cf"* J • • •-« : • '. '^>' ; = ." :^ : : : : : '£--r5S ■ : =;-« ;5 2 : : S'^- --f. .;:: c-- ^: «■:;:£:;:: : «:-^±S :S-,= .s : :5::§| : : ^ : .2 1 i :|j : : : : ni^^-i :FIt^l iP^^f^i :Ppc:.^i ifSi M : i _ " '- - .'- '^ '- — .'- - " 'pis •'^ c '.'. '.', ^ %- i o <— E 5 ; .' ^ - "i - ^^ • • c ~ r c c £ c -5 39. r^'sr o/" Periods. — For the purpose f'f roijfVnp: or writing r • bera, we divide thorn by separating points, into j»eriods — tlie ; seinunting point being the decimal point, cxpiesi^^d or undereti and the other separating points being phiced after every thiid ili. or place, to the riglit and I'.'ft of the decimal point, .^nch period 1 three places — of which one or more may be occupied \v digits. 1 lowest place in every period — or that to the ?v;/'-^ hand, is "imits"' place of that period: and the highest, fr.t' " hnndrc place. And this is true, whether the period is to tlu 'i^A't or to : right of the decimal point. 40. The pe'.'Iod to the left of the decimal point oon:.i *rs the ; pie units. The first period to tlie left of th(> mat.s'' period, con! the tJiovwnd^ ; and the tirst period to the right of it, the thoiisait'. The second period to the left of the imits' period, contain.^ the : lin)ix ; and the second to tlie I'ight of it, the inillio'ntjix. The ♦! period to the left of the units' pei'iod, contaiiis the bit Hon a .■ and : third to the right of it, the hillionths. The fonrt'» period to the 1 of the units' pei'iod, contains the trillions; and the fourth t'^ riglrt of it, Uio trillionths. The fifth period to the left of the ei. perl "!, t^te- - limi h ■■■ tlu> The i limi < : pcri'd t' liia.li \:>\ of till' I 'I l-i^iL of TllL' p^ ttnfl ■ I' tion. i>> ii:| oagui to pefi"! ail two. tlii-> • any niinili i'lif \s *tt£UliVL,' ( I o '; r. 'L''-i ■■ > .^ — ' r— 1 t 3 11 It 3. 1 1 ./ -a T.XANir We i'lit a [/ utt is, it «"«x presses tlM -amo pi M&b lliUU;: '>i m [Bkct. 09, -iO.] AND NUMEIIATION. 29 9t P o c r: !< r. ■- a el S -n *? o ^ -J, ^ ^ ■/. ^j ^ :il h ■ ^- 1: E- h ::^ F^ t:- « « ;) 8 ;; 8 3 f^ ;{ 8 -L-t- -i - -/i • • ~ r f ;: £ c •- *^. ?— r" t~* >-^ c^ ^ p;)d, contains the -\adriUtona ; and the fifth to the riglit of it, the ldriUid, Cv)ntaius the scjifU- s ; and tlii' eiglith to the right of it, t'ne sc/jfil/ioNd'is. The ninili iud to tlie left of the units' period, contains tlie octUVions ; and tl-e |ui to the right of it, the oci'dlionfliK. The tcntli i>eriod to thi» left ithe iniits' period, contuiiis the tionUllouti ; and the icntli to the it of it, the douiiiio/it/is, % The pupil should be in-idc perfectly furiiiliar w'th the nunios of tlie periods JbI ui'the pliti'cs in c:',e!i p'.'ii.ij— :>(> ;h to be able, ^vil!iout the i>liu'htcst hesila- Uw. to name the period iUid [/iaee.l » whleh any (h':iit belon^'s, or into whieii it "it to bi! lint. Wlieii lie can ii';i'l or write any owi.; di'^it, bclonu'ini; to any lOil and i)l!u:e, he slionld h" lanulit to read and nrite a nninber eonsistini; of I. fliree, j'our, iVe.^ diuiis, whether they arc close togetlier, or separated by ■ iMiinher of 'dubers. ■^ 'i'hc w hoK^ of wiiat has been said above wi!) be-'onic move evident from aa #||eulivc consideration of the fvllowiny table; 5 Cl-I C tn O o 2 Cm o a u> O H <4-. c a a o m ^ o « a o 'a M 3 a o a '5 vid'tig- or writinp; r\: into }>oriods — tlio I ie>v-^d or undeic-ti after ovory tliiid I'i. ,'inl. .Fi.nch period i •upicd \v digits. 1 ■ rir>''- h;\nd, is • d, the "hiindrc' is to tht iii'i or to t oint C01KXV9. the ;' '/? .-^ ^' .i* ^•-< ^ J ," ,2 .'-' .^ ^ ,','■" h^i,^ ."^ .-< -" ."? .1^ ■- ,'■ J3 ^ ,'^ »ij ^ ,"" H ~^ I— I r'l ^- -H :-! i— ' I— I H >-^ r--" b^ 1-^ >-H H -- >-- H >— >--< H 1— nil. H — ^— 1-* — ' f-' L-^ "-J ►^ c~ »- ' :J 1 7 4 5 G 2 S o 7 '1 o 5 1 'i o (i -1 7 8 b 'J 'J 5 4 7 1 ^5 U 7 S U ;l 1 7 4, 5 2, S 3 7. 4 6 3, 5 1 2, 8 6 4, 7 3 S, 2 9 5, 4 7 1, 3 2, 7 8 9 T^ ■o t;; 'O 'a -3 r^ r2 -« f— t 1-3 o b ;-l ;-< ^ i^ ^ u* ^ (4 *> ' «J ^ V Ol Ol V \^ Ch Ch u^ P-. l-H e- Ph Ch Ch Ch Ph -a t-l t^ n '3 T3 rd rfl ^ « ^ CO o» v-t r-( f Llie ',», and lind that there are CiCtirf/// two periods ui;;. i"Mti,!Ki4; this does not always occur, as the liJLrhest or lowert jieriod i.s |n imperfect, consistinij only of one or two d;::its. Dividing the nnnibev into parts, shows at once that 5 is in the thirc' place of the second period jat is, in the llnn' hundreds of iii lis. tliive tens of units, and fi'ur units"; or more t(rielly, '■ fi .reqtilp' hundred and seventy-six tliousund nine hundred and tbiriy-four." iatiuis w 4i. To prevent the separatinc point or that whieh divides into peri'>(!; Bn.\:\i from beinii nnstaUeii for tlie decimal point, the former slionld be a comma (."-sevtM ni the latter a fiill .^tcp (.) Witiioiit this di^t netion, two munbers wldcli are vcr. ar(| <'"'> (iilVeient uii-xlit l)c confounded: tlms, 4!)8.7r.;.?. and 40S,T0;J, one of wiiicli is iini^li'S t tliousund times irrealer tiian the other. After a while we may dif^jiense wi: clpjurs 1 the seiKiratinir point, tlioui;rii it is conveniout to retain it with large numbuL-expr as tlicy are then ro.ul wilh' greater ease. ii. T 42. To write down any integral or whole number^ it is W'O*^' fipgt per ncc'js.vari/ to rcvic.ynber the order of the periodf<^ avd that every pcric.liffgj^!\ yir contains three places, each of which niust be Jillcd, either by a diyit oifjrgtpl.a cipher. The one, tivo, or three diyita, belonglny to the hiyhcst /jn^lh© yimpl are Jirst icrittcn in thctr appropriate places ; then the next lower peril than tho in filed with the difjits, oi ciphers belonyitig to it; afterwards i/'ithan 'he next ; and so on, ii.il the lohole number is set down. on^ \)h\.w Example.— Let it be required to write tho number seventy-three trillior^ , ! two huixlred and nine billions eii:hteen thousand and mx. The hitrhe.vt peri- '^"■*^'-'^ '[' here n\ehtioned is that of trillion,-, which \\\: know to be the tifth to the lei; i digits oi' the decimal [)oi!it (4u) We tlierel"ore set down the digits 73, beurii tc in Uii (,yliol,> is that wo are to [)iU in lour complete periods, or twelve places between the 3 in . ' ,., the decimal point The next perin!)cr of c/.j'hcrs. Example. 1. — Let it l.<^ reqiur>-d to write the number, seven hundred ;r sixteen tliousand a d eighty-nine billiouths. Now we know (40) that billh : ' ncciipy the 0th place to the riirht of the decimal jxii t. Were we to ]i\nfr : decimal point i7nnt((l!<>toJ)f bi'fore the disits themselves, thus, •71(!o^t>. il. Aoiild express not so many billio ths but so many millionths : .since nuliio )ceupy the ()th and billi ' hs the 9th place. It is obvious, then, that to t' the (lieita their i)roper value, we must insert three eiplievs between ti em & tb*:< d««imal puinti oud tb« number i» tbeu eorruetly written ■0UU,71O,U)3i9( sad'' dogi 1)8. V/e on«. place times less Quiintitio!; none of tl right of it the (lecin hanii ^-ide 46. T thfi .Tvfe/ii to- tal .ast"! of , and six tln•llsalld^ ; tii; . i''; or inure briefly, '• fi ■ thiriy-four." Iiich divides into pcrim;. r should bo fi oomma (.>- (» niunbers wiiicli are wr Ex \.Mi'LE 2.— Writo tlie number six thousand two hundred and onehun- dr'l ■•< 1 tiiiiionibH. From { .0) wo know that hundrpdtlia of trilHonths occu y tho lHli place. Tiie given digit.s{f)201) bcinp only four in nuinbor, renti I" tii ' ^ "l Vo Have, ile we may diF|iei)sc' wi; oMurs botwoen iho trivon digits and the docima u il with large numbtr; e3qiri!:->io.i, •0iX),G72,7()l,2, repreBCuts the given ni 41. Tho simple imits are, as we have ?aicl, always found in the : number, it is werf/fipg^ i.crl.fl to ihc left of tho docimul ])oint. The digits to the left and that sini])le units, being ten times greater than g to tlw fii(//icst jjcvKiht simple units ; those occupyii\g the second place, ten times greater 7n the next lover pcri( thsai those which occupy the finst, and one hundred times greater to it; afterwards i/ithati tlie units of comparison themselves; and so on. Moving a digit (jwn. OU€f place to the left, multiplies it ))y ten — that is, makes it ten times tor; moving it two places, multi()lics it by one hundred — that is, )er seventy-throe trillionP.^ " '. i'' i i i.- j. i r ii ^ ir u .i i six. The lii"be.vt i)erji '^W^'^ It ouc hundred times greater; and so of the rest It all the < be the fifth to the leU < dlsts of il. quantity be moved one, two, &c., places to the left, the ligits 78, boarii g in n,i (;^rij0i,. is increased ten, one hundred, &c., times — as the case may ; of'^nionV^wM ^" '^^*' ^^^^cr hand moving a digit, or a (luantity one place to [•lidd, that of n illions, lifthi^ riy/'ii'fly "■t tljc ielt. And the pupil snust remember that the (lu-ui- decimal the last di'' l-*^^ increase in a tenfold degree to the left, and decrease in the " sa|iie degree to the right wherever the decimal point may happen to b^l V.'e therefore put quantities ten times less than simple units otfj^ [)!ace to the right of them, just as we ptit those which are ten tiilies less than hundreds, &c., one place to the right of hundreds, &c. antiri(>s to the left of the decimal point are called intef/cr/i because e of them is less than a ichole simple " uint"; and those to the t of it, c^w'ma/.s. When there are decimals in a given number, decimal [)()int is alwai/n expressed, and is found at the rigiit- d side of the simple units. ; 46. The periods to the left of the decimal point may bo called I aseendln(f, and those to the right of \t the deseendinef series: — iC'i together, however, they constitute but one series, which is an '. .g or a descending series, according as it is read from right lo li-oni left to right. Periods that are equally distant frotn ti <^ < of »^timpariaon bear a very closo relation to each ther» whlo.U i4 y )er. 1 hus, -(38 is ru, iand and seventy-eiul ce first ascertain In iit to the ricjht of il ill vot bring the vu. ',S6' digits and the u'c ber, seven hundred ;r ki!ow(.10)th!U biilh ! ■ Were we to ])lii(f ! Ives, thus, ^Kid^t*. il. ioiiths : since niiliio ; lions, tlieii. that to t' leis bet,^^•eell ti eni iU" •iUen •000i7ia,0«9t flT' 52 i'iOTAHON [Slot. T. II ^1 I n f indicated even by the similarity of tiieir names ; the only difterenoo l)oin<^ in the terminations (-lo). We have seen aldo, that when wo divide integers into periods (40), the first separating point must be put to the right of the tliousands. In dividing decimals into period^, the first point must be put to the right of the thousandths also. 47. Care must l)e taken not to confound wh:it we now call " dc cirnals," with what we .shall hereafter designate "decimal fractions"; f'oi- they express ecpial, but not identically the same quantities — tlic decimals being wliat sliall be termed the " quotients" of the coi- respondiiig decimal iVactions. This remark is made here to anticipate any iuaecuiate idea on the subject, in those who already know some- thing of arithmetic. 48. There is no reason foi' treating integers and decimals by dif- ferent rules, and at dillercnt times, ,^ince they follow precisely the same laws, and constitute parts of the very same series of numbers (4(5). Besides, any quantity may, as far as the decimal point is con- ceincd, be expres-sed in different ways ; for this purpose we huvi; merely to change the unit of comparison. Thus, let it be requiied tu set down a number indicating five hundred and seventy-four men. tf the unit be one man, the ([uantity would stand as follows, 574. If a band of ten men, it would become 57 "4 — for as each man wouM then constitute only the tenth part of the " unit of conjparison," iouv men would be only four tenths, or 0'4 ; and since ten men woudi form but one unit, seventy men would be merely seven simple units, or 7, &c. Again if it were a band of one hioidrcd men , the number must be written 5*74 ; and lastly, if a band of a ihou.'ovd men, it would be 0*574. Should the " unit " be a band of a dozen, or a seoro of men, the change would be still more complicated ; as, not only the position of a decimal point, but the very digits also, would bo altered. 49. It is not necessary to remark that m.oving the decimal point so many plaoes to the left, or the digits an equal number of places to the ri(/ht, amounts to the same thing. Sometimes in changing the decimal point, one or more ciphers are to be added ; thus, when we move 42*6 three places to the left, jt becomes 42GuO ; when we move 27 live places to the right it id .0-00027, &c. 50. It follows fi'om what we have said, that a decimal, though 'ess than what constitutes tlic unit of eomparison, n.ay itself Of coiuse It. nature of the "■ sinijtlt; consist of not will often bo r units " ; as 3 F &c. But i' ., .e docs .^' aifect tiie abstract propoities of ;.1 numbers ; ... true to :-y that seven and five, whcu added 0"ly one, but several individuals, ^ary to iin'.ic.ie t' , o dozeri, G men, companies, 8 reguueiits. [Slot, t the only diflerouco aldo, that when wo ting point must be [jinKils into periods, isandths also. t we now call "dc ilet'iiJial fractions"; inic quantities — the tients" of the coi- e here to anticijiaic iheady know bomc- nd decimals by dif- ollow precisely llie series oi' nunibei - cimal point is con- i puipose we havd et it be required to seventy-four men. s follows, ri74. i\ s each man would • i" coniparison," I'ou, .'e ten men would ven simple units, Arto. 47-^IJ AND l^UMEUATIOIS. 33 / inert, the number t/ioii.s(ind 'incn^ it I dozen, or a scoro ted ; as, not only its also, would be the decimal point inber of places tu or more ciphers )laces to the left, to the right it la \ decimal, thouyij isoii, Uiay itself :«. Of course it. of the " Hini[)l'j ies, 8 j'cgimeiits. :t propcitles of Ivc, wLcii added together, make twelve, whatever the unit of comparison may ho : — provided, however, that the same standard be applied to both ; thus V men and 6 men are 12 men; but 7 men and 5 horses are neither 12 men nor 12 horses; 7 men and 5 dozen men arc neith- er 12 men nor 12 dozen men. When, therefore, nuuil)ers ard to be 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 may consider 5 tem of men to become 50 individual men — the unit 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. Exercise 3. 51. Write down the following Numbers: — 1. One hundred and ninety-four. 2. One thousand and seventy-six. 3. Twenty thousand five hundred and eight. 4. Two hundred and one thousand and tln-ee, 6. Eighty millions four thousand and thirty-three. 6. Sixteen quadrillions five hundred and ninety-seven trillions threo billions forty-four millions and ninety-one. 7. Ninety-seven hundredths. 8. Six hmidred and forty-three thousandths. 9. One hundred and twenty -two thousand and eighty-nine millionths. 10. Thirty-nine tenths of millionths, ^^ 11. Sixty-three hundredths of trillionths. ^ 1 2. Seventeen billions four thousand and one, and nine hundred and sixty-seven billionths. 13. Seven trillions eight hundred and two billions twenty-three thou- sand and eleven, and nine thousand nine hundred and ninety- nine billionths. '/4. One quadrillion one trillion one billion one million one thousand one hundred and one, and one trillionth. 15. Eight hundred and ninety-six trillions and two, and nine hundred and four hundredths of millionths. Aiiswers. 1. 194. 2. 1076, 3. 20508. 4, 201003. 5. 80004033. 6. 16597003044000091. 7. •97. 8. -643. 9. -122089. 10. •0000089. 11. •00000000000063. 12. 1 7000004001 -000000967. 13. 7S0200 023' >1 1 -000009999, 14. iooiooi('Oi>!oiioroooooooooooi, 15. 896000000000002-00000904, Hi ■ f :■ ri fif i 84 • DENOMINATION OF NUMBERS. \Stxn. L l Artt ' ' < Exercise 4. m I 1 ' 1 62. Read tbe followin;; numbers : — fl aKi .^^1 the i 1. 904. 7. 604-03. 9 ni'ili 1 1 2. 7060. 8. 90767M)04003. S V'* -I ' unit ■J M t 3 90004, 9. 9001-00070306. fl 4. 40300201. 10. 1237-9134071342918. W to 0.' 6 7060504030. 11. -OOIOOIOOIOOIOI. M niim 6. 70003000000400. 12. 100-2003004005006007. fl 'I I -I Answers. 1. Nine hundred and four. 2. Seven thousand and sixty. 3. Ninety thousand and four. 4. Forty millions three hundred thousand two hundred and one. 5. Seven billions sixty millions hve hundred and four thousand ami thirty. 6. Seventy trillions three billions and four hundred. 7. Six hundred and lour, and three hundredths, %, Ninety thousand seven hundred and sixty-seven, and four thcic sand and three niillionths, 9. Nine thousand and one, and seventy thousand three hundred and six hundredths of raillionths. 10. One thousand two hundred and thirty-seven, an^ nine trillion, one hundred and thirty-four billion six hundred and peventv- one million three hundred and forty-two thousand nine huu dred and thirteen tenths of trillionths. A. One hundred billion one hundied million one hundred thousand one hundred and one hundredths of trillionths. x2. One hundred, and two quadrillion three trillion four billion five million six thousand and seven tenths of quadrillion ths. ON THE DENOMINATION OF NUMBERS. 53. When two numbers have the same unit they are Baid to be of the same denomination ; when the units are not the same, they are said to be of different denomina- tions. For example, 16 shillings and 28 shillings are two numbers of the same denomination ; but 23 shillings and three farthings are not of the same denomination, the unit *of 23 shillings being one shilling, .and of three farthings, one farthing. The kind of unit always expresses the de- nomination* iiuat lot' ll i. ][flEcr. L Arti. M-5».] MONEY. WFJGHT?, AND MI'ASURES. 35 03. ,) 306. 71342918. J' 00101. 'i 4005006007. k 'i ndred and one. four thousand ami ed. t'en, and four thtic three hundred aurl , an'^ nine trillion, ndred and peventV' houBUud nine hun hundred thousand ths. jn four billion five adriUionths. IBERS. e unit they are tlie units are Irent denomina- lillings are two shillings and lation, the unit |;hree farthings, v^resaes the do- Evwn in a1)stia'jt or stinfjlo namlxTs, difFercnt names ai<: L;ivci« to th'! units as we [jrojfe- 1 to the ri^ht or left .if the (lueiuial point, viz , siiU'ile units, or units of the first nnlcr; tens, or units of the second ordi* r . hundreds, or unit -; of clie third order, ttc. ( ''>nsidered in this relation to ea"li other, these units raay he re^r irded aa are cents, tive-cent pieces, ten-cent pieces, and Lwcnty-cont pieces. -^ '' - - ' 100 cents (c) make I dollar, marked $ AVOIRDLTOIS WEIGHT 58, Is used in weij^liing lioavy articles. Its name is llerived from French — and ultimately from Latin words Bi'>nirvinjr "to have weiti^ht." Us (h.'uominations are tons, hundredweights, quarters, pounds, ounces, and drams. TABLE. 16 drams make 1 ounce, marked oz. 16 ounces (( 1 pound. (( lb. 25 pounds (( 1 quarter, (( qr. 4 quarters (( 1 hundredweight, '* cwt. 20 cwt. ({ 1 ton, u t. d. oz. 16 = 1 lb. 256 = 16 =: 1 qr. 6400 = 400 = 25 = . cwt. 25G()0 = lOUO Z^ 100 = 4 = 1 t. 612000 = 32000 = 2000 = 80 = 20 = 1. It was formerly the custom to allow 2S lbs. to the quarter, 112 lbs. to the iundredwi'i'.'ht, and 2240 to the ton. This has now fallen into disuse: and •luotiir niercliants in Canada the qr.. cwt., and ton are uuivers.illy considered Its respectively equal to 25 lbs.. iOO lbs., and 2(m)0 lbs. Tiic Custom Hou;;e9 eontiniic to rciiard the cwt. as equal to 112 lbs., and some few articles are Rtill ^ei^hed by the old cwt. by farmers and others. The English cwt. is 112 Iba. TROY WEIGHT. 59. The denominations of Troy Weight are pounds, ounces, pennyweights, and grains. TABLE. 2i grains (grs.) make 1 pennyweight, marked dwt. 20 penny^-cights '• 1 ounre, '* oz. 12 ounces " 1 pcuii^ " lb. \\ i Rl '^1'^ 28 MONEY, WEIGHTS, [SeOT. I. 4-ET8. I I ;i grg. dwt. 24 = 1 oz. 480 = 20 = 1 lb. 67tiO = 240 = 12 = 1. This wolp;htw.is introduced into Europe from Cairo, in E;?ypt, and was flvFt adopted in Troyes, a city of France— whence its narno. It is used in philo«o- piiy, in wciRhing gold, precious stones, &c. Note.— The oridn of all weiirhts used in Enfrland. was a jjrnin of wheat tfiki'n from the middle, of the ear and well dried. A weight ecuiid to 32 ofthesi' fraiins was ral'ed a pevn-i/ir. i(jht, beini; equal to tho weiirht of a silver penr.\ tl;eii in use; -JO of these pennyweights constituted an ounce, which was tli- IJlli part ot'a pound (Lat. '• uucia," a I2th [lart— compare "■ivcJi,"' the two!i;ii part of a foot). In later times tho pennyweisiht caniii to be divided into 'li eq.ial parts iiK^tead of o2, but these ttiil retain tlie name of g.'nins. The " Carat,"' which is eq^'al to about four irrni!is(s(>iiiev,ii)it lef,;! than Tr.M,- grains), is used in w^.i/liinf; dianiDUds. The term cuat is also ;'ppli(>d in ( sti- niaiiuir the fineness ;>f Enid : the lalter, whi n jierfectly pure, is s;iid to be '•2\ c-)itii.s' Hall; tl^M-, ir's . !hi^ indirato? tlmt iVliicJj is stampi'd. grs. 4S0 43Vi 5,7flO 7,9J- miles (nearly) make u (( a li u TABLE. 1 inch, marked in. 1 foot, « ft. 1 yard " yd. 1 rod, pole, or perch, rd. or p. 1 furlong " fur. 1 mile, " m. 1 league, " lea. 1 degree or 360th part of the earth's circumference. m. 12 36 198 7920 63360 ft. = 1 = 8 = 16i = 660 = 5280 yd. 1 5.^ = 220 = 1760 = rd. 1 40 820 fur. 1 8 = m. 1. Each link 100 links, 4 rods, or 22 yards, make 1 Gunter'a chain, therefore is equal to 7-hu) inches. Eleven Irish are equal to 14 English miles. The Paris foot is equal to 12 792 English inches, the Roman foot to 11*604 English Aiches, and the French metre to 39'383 English inches. 4 inches make 1 hand (used in measuring horses). 3 inches " 1 palm. 18 inches *' 1 cubit. 3 feet " a common pace. 5 feet a Roman pace. 6 feet " a fathom. 20 fathoms " a cable's length. SQUARE MEASURE. 62. This measure is used for estimating artificers' work, such as flooring, plastering, painting, paving, &c., and, in gliort, any kind of work where surface alone is concerned. It is always employed in measuring land,- and hence it is frequently called Land Measure. (!■;' III 11 40 MONEY^ WEIGHTS, [;;-oT. L 1 f oot = 12 inches. xn — — __ — ic yard — each edge |eing 3 feet in length. The top, thich is equal to the base, contains X 3 or 9 square feet ; hence, if it "rere only one foot in height it would )ntain 9 cubic feet ; but it is 3 feet height, and must therefore contain X 3 or 27 cubic feet. A cubic yard fen contains 3 x 3 x 3 or 27 cubic et. 3 feet. 4) ^ Similarly it may be shown that a cubic foot contains 12X12X12 or 1728 cubic inches. The denominations of Cubic Measure are cords, tons, ubic feet, and cubic inches. [728 cubic inches 27 cubic feet *40 c. ft. of round timber, or 50 c. ft. of sq. or hewn timber TABLE. make 1 c. ft. marked c. ft. a 1 nM^^n TTrl «< ■(■■ 1 cubic yd. 1 ton. c. yd. *' ton. 128 cubic feet make 1 cord of firewood, marked c. c. in. c. ft. 1'728 = 1 c. yd. 46G56 = 27 = 1. A pile of cord-wood 4 feet high, 4 feet wide, and 8 feet long, con- lins 128 cubic feet or one cord. One foot in length of such a pile called a cord-foot. It is equal to 1() solid feet, and is consequently Equivalent to the eighth part of a cord. CLOTH MEASURE. 65. The denominations of Cloth Measure are French ^lls, English ells, Flemish ells, quarters, nails and inches. * A ton of round timber is that quantity of timber wbicb, when hewn, will lake 40 cubic feet. V III r 1' I i 42 MONEY, WEIGHTS, [Shot. 1 '■ i Kl>i. <'» TABLE. 4, 2 J inches (in.) make 1 nail, marked na. / 4 nails '* 1 quarter, '* qr. -:i^ e'i 8 quarters " 1 Flemish ell, ^* . Fl. e. '% 4 quarters '* 1 yard, <* yd. ''H Th 5 quarters '' 1 English ell, " E. e. W^"^ 6 quarters *' 1 French ell, " F. e. I i in. na. 2i = 1 9 = 4 27 = 30 = 45 = 64 = 12 lo 20 24 qr. 1 3 4 5 6 Fl. e. 1 n = n = 2 = yd. 1 Eng. e. H = 1 Fr. e. H = H = 1. Note, — The Scotch ell contains 4 quarters H inch. DRY MEASURE. 66. By this are measured all dry wares, as grain, beans, coal, oysters, &c. The denominations of Dry Measure are chaldrons, bushels, pecks, gallons, quarts, and pints. TABLE. 2 pints (pt.) 4 quarts 2 gallons 4 pecks make 1 quart, ** 1 gallon, " 1 peck, " 1 bushel. marked qt. " gal. " pk. " bu. 3G bushels " 1 chaldron, " ch. pt. qt. 2 = 1 8 = 4 IG = 8 gal. = 1 pk. = 2=1 bu. 64 = 32 = 8=4 = 1 ch. 2304 = 1152 = 288 = 144 = 36 = 1. Onr Standard of Dry Measure is tlie Winclicstcr bushel. This is an upright cylinder whose intornardiameter is IS^ inches and depth S inches. It contiii'is '2]."l)4r cubic inches or 77'()'27 lbs. Avoirdupois of pure distilled water itt G2' Fahr. and "0 in. barometer. The standard unit of Dry Measure in the United States is also th.e Winchester bu.shel, so called because the sfanlavd measure was formerly kept at Winchester, England. The staiidard unit ol' Dry Measure in Great Britain is the Impt'rial bushel, which is an upriirht cylinder whose in- tern.al diameter is ]8TSI» inches and depth 8 inches inche barom It continn.'^< 2--lS-li<2 cnblR atur at C2° Fiihr. and 80 hi. Grain is often bought and sold by weight, allowing for a bushel. CO lbs. of wheat, 56 lbs. of rye, 50 lbs. of Indian corn, 48 Ibi*. of barley, 84 lbs. of oats. 60 lbs. of peas, 5Q lbs. of beans, 40 lbs. of buckwheat, 60 lbs. of timothy or red clover seed. l 4 ■ 8 I 32 jo 1 6 to; '.-2 VKA m i [Seot. 1 STS. 66-68.] AND MEASURES. 48 na. qr. Fl. e. yd. E. e. F. e. Fr. e. 1. nch. ^ares, as graiD> are chaldrons, qt. gal. pk. bu. ch. ch. 1. Tiiis is an riprfcrht nches. It contni'ia tillod watt'v at 02' asure in tlie Uniti'il standnni nioasurc mit ol' Dry Measure cvlirdci wliose in ^inr. 2'.?lS-li<2 fuhin '■ Fiilir. aud 80 i-j, a bushel, 60 Ihs. of F, 84 lbs. of oats. 60 of tiniotiiy or red LIQUID MEASURE. 67, Liquid Measure is used for measuring all liquids. The denominations of Liquid Measure aio tuns, pipes, lea ;s, oarrej • TABLE. • J. yuia. 4 gills (g .) make 1 pint, marked pt. 2 pints C( 1 quart, (( qt. 4 quarts ' il 1 gallon, « gal. 31^ gallons u 1 barrel. (( bar. 2 barrels (( 1 hogshead J hhd. 2 hoo-slieads " 1 pipe, (( pi". 2 pipes (( 1 tun; u tun. g- pt. 4. = 1 qt. 8 = 2 = 1 gal. 32 = 8 = 4 = 1 bar. 308 = 2o2 = 126 = 31i --= 1 hhd. ')U) = 504 = 252 -. 6;; — 2 z = 1 pi. r.v2 = 1008 = 504 - 126 = 4 : = 2 = 1 tun Ji'A = 2016 = 1008 = 252 - 8 : = 4 = 2 = 1 The Eufflish Imperial pallon contains 2T7'274 cubic inches or 10 lbs. avolr- |ipois of [Hire distilled v.ater, weiiihed at a tomperature of 62° Fahr. and under |l>;iro;noti-ic pressuro of 30 inches. In the Unitwl States the wine sjallo'i contains 231 cubic Inches, and the persiajio 1 232 cube inches. The irallon of Grout Britain is therefore about |aal to 1-2 gallo.s United States Wine Measure. By an Act of the Imperial Parliinie't, 182(5, the Imperial pallon of 277-274 ibic inches, was adopted as the only gallon, and is therefore the standard for ith liquid and dry mea -ure. Beer 's usually sold by the jrallon ; sometimes, however, in casks of 5 pals., ?als., 20 gals., &o. The beer barrel contains o6 gallons, and the ho.Hihead 54 ^Uons. TIME MEASURE. 68. Time is naturally divided into days and years — le former measured by the revoluti(m of the earth on its S:is, and the latter by the revolution of the earth round le sun. ' The denominations of Time Measure are years, months, feeks, days, hours, minutes, and seconds. . M l»^ ^ »—' m ■ u MONEY, WEIGHTf^, [Sect. I TABI miTiUte, hour, day, week, lunar month, marked rain, h. u 60 seconds (sec.) make 60 minutes *' 24 hours ** 7 days « 4 weeks " 13 hmar months or 12 calendar months or [• make 1 civil year, marked yr. 365 1 days (nearly) a i( u d. wk. mo. Bee. 60 3G00 86100 604800 31 557600 mm. 1 60 = 1440 = 10080 = h. 1 24 168 5'i5960 = 8766 = da. 1 1 3 Co A wk. 1 yr. = 1. The twelve colenrlar months, into which the civil or legal year is divided, and the number of days in each, are as follows . First mo'ith. Ja' uary, Socond " FobrnaVy Third " Fourth " Fifth Sixth Sevontb " Eiichth " Ninth " Tenth •' Eleventh" Twelfth " March April, May, June, July, Auj2;ust. September, October. November, December, has 81 days. " 28 " " 80 " " 81 " 30 " 81 " 81 80 81 80 81 4i li «< i( (4 44 44 44 The number of days in the respective months may be recalled by recollect- ing the following well-knowu lines : Thirty days hath September, April, June, a' id November : February lias twcuty-eisht alone, A:k1 all the rest have thirty-one ; But leap-year comino- once in four, February then has oiie day more. The number of days .in each month may also be recollected by counting th> months on thf fan >• finsjors and three intervening spaces. Thus, January n the lir.st finger; February in space between fir,--t and second fingers: Man'). (• second fln^i-r: Ajiril in second spnce ; May on third fli ger; June in tlii ■, space: July on f.>iirth finger: August on first finger (since there are no m ;v spaces); September in first space, itc. Now. when counted thus, nil l!. months having 31 days cotne on the fingers, and all having 30 only fall into ii spaces. The solar year is the time elapslncr from the passage of the sun from eitln: solstice back to the same again, a! id is equal to 3C.5d. 51i. 4Sm. 4Ssec. The sidereal year is the time betv.-een two successive coijunctioiis. of i-r sun with some stiir, and is equal to 8fi5d. fh. 9m. 14*sec. The civil or losal vear is that in common use amoou difTorerit natior.G.r.ruij equal to 305 days for three years in succession an-' to G06 days for the fourtlj p« T| >;ir i! K>;i ry wi illMl (/'■I, .-[ ia'^il^| tar. Til blar 4 Im. i| e>icl llend:! |aiui'!i| iY, thj Inies < iia. 68, 69.1 AND MEASURE. 45 year, marked yr. u be recalled by recollect- This afMitio; n\ day Is given to m-ory fourth yonr. In or/Aol'the Idu s of Miirch in the Uonian calendar- coiTosiiondfnfi with tlio '24th of Fobru- V with us. Tho day was callciterca/ Yi'itr. bccaiiso k'\psaday more than a CiMumon year. This c a-rection was made by .Julius isasar emperor of Home, and hence the civil year is often called the Juliau ar. Tho addition of one day every four years would be strictly cornet, if the 1 ir year contained !3(j5d. 6h. ; but it only contains o(j5d. 51i. 4sm. 48s., or 1 i-'s less than :3Cr)d. Gh. Adding 1 day every 4 years, irives us then an error at e.^ci'ss of 44m. 43s., or about JJ days for every 400 years. Thus the Juliau Xlfiidar was behind the solar time, sixe the Julian year was lon-rer than tho iliiral vear. This error, at the time of Pope Grejrory XIII., ainounteci to 10 vs which ho corrected in 15S2 by snppressinf,' lU days in the month of Octo- V. the (hu after the 4th being called the 15th. Hence this calendar is some- nies called the Grtgoriaii calenda7\ Tliis correction was not adopted In England till 1752, when the error iou:-ted to 11 days. By Actof i^arliament, 11 days after the 2nd of September rcre therefore omitted. The civil year, by tlie same act, was made to com- jeiice on the 1st of January, instead of the 25th of March, as it had done pro- Jiously. Dates reckoned by the old motliod or Julian calendar, are called Old Style; |nd those reckoned by the neto mtthod are called New Style. To change any date from Old to New Style, we must add 11 days to it; id if the given date in Old Style is between the 1st of January and the 25th of larch, we must add 1 to the year in New Style. Russia still reckons dates according to Old Style. The diflference now jouuts to 12 days. 69. To ascertain whether a year is Leap Year. Divide the given year by 4, andif there is no remainder it is leap Year. 27ie remainder, if any., shows how many years have lapsed since a Leap Year occurred, Tiius, dividing the year 1847 by 4, the remainder is 3 ; hence it 3 years since the last Leap Year, and the ensuing year will be jap Year. To this rule there is an exception ; for we have seen that a snlar year is |m. 123. less than a Julian year, whicli is 3G5}- days. This error, in 400 years, mounts to about 3 days; consequently if a day is added every ,/ottr •-• ) 0'f± 304 273 243 212 181 151 120 90 90 59 31 305 335 804 274 243 212 182 151 121 120 89 Oil 30 365 j oo4 304 273 242 212 181 161 Jlllli July' 1 151 I8li 12( 150| 92 122 01 91 81 01 365 30 335 305 304 334, 273 303! 243 273! 212 2421 lti2 2l2i 2121243 181j212 153 184 1221153 92123 01 31 92 62 3 65 1 81 334,365 304,335 273 304 243 274 273 304 242 273 214 245 183214 153 184 122 153^ 92 123 61 i 92 80 61 - 1 ]: J- 1-. 305 00 81' ( 4805 304 335 T/ie moniliii counted from any day of^ are arranged in the Ly hand vertical, column ; those counted to the sat7ie day of^ are in li' upper horizontal line ; the days between these periods are found in ii aaalc of intersecfion., in the same vmy as in a common table of mi- tiplicalion. If the end of February be included between the t\i- jmints of time, a day must be added in leap years. Ex VMPLK 1. — lIoNv many dims .ire there from tlio 15th of March to the '■. (tf October? Lookinjr down' the'vfrtical rov/ of numbcr.s at, the hoiid of \vi i. Octol)t'r is placed, and nt the same timo along the h^)rizontid row at the I: hand side of whi^h is March, wo. pt'rceive in tiieir intersection the number 'Zl\ fo many days, therefore, intervene between the 15th of M;;ioh to the IHtli' October. But the 4th of October is 11 da.ys earlier than the 10th: we thenlnh subtract 11 from 214, and obtain 208, tl;e number required. Example 2. — How many days are there between the 3rd of January and ti. 19th of May? Looking as before in the table, we lind that 120 days i'nterv i- between the 8rd of January and the Srd of May ; but as the 19th is 16 days lai'. than the 3rd, wo add 16 to 120, and obtain 136, "the iiumber required. Since February is in thl^ case included, if it were a leap year, as that men;; would theu contain 29 days, we should add 1 to the 186, and 187 would be tL answer. EXAMPLES. 1. How many days from May 3rd to the 4th of next July ? Ans. 62 days. 2. How many days from July 4th to the 25th of next Decemlicr! ■ A71S. 174 d;!};'! 8. How many days from Marab 21st to the 23rd of the next S^ ;■* tembcry Ans. 180 ua\: [Bkot, j^Tf, 70-72] AND MEA&UllKS. 47 iOM ANY DAY OF O.N I IN THE SAME YEAK, day of y Aug/Sopt. Oct. Nov.' Dc V2VI i)jl81 2 lf;;j 122; 92 61 31 CG5 3 334 3 304 248 273 304 212 242 273 184:214,245 ins: 183 214 123;153 184 92 122 153 62! 92 123 31j or 02 3651 80' 335 '365 3'': 2';: 24-1 2ir 1 :■: Vs., 2l273'304'334 61! 3li 865: 2|243 274'304 335 3i e arranged in the Irji ante day of, are hi "tk ^ periods arc found in fl common table of mi-- luded betweeji the h: irs. 15th of >rarch to the 4i: k'ns iir, the hotid of \\\A(:-. )iiz()nt;il row at the l;:; I'beelion the miiiibi-v '2!1 of Mi!!ch to the mtb, II tho 15th: we ibertfui, rod. le 8rd of January and th • liat 120 days interv.i.. the 19th is 16 days laltn*i ber required. eap year, as that mcnt; 6, and 187 v/ouid be tt: of next July ? Ans^. 62 days, of next Decern) icr; Am. 174 d;;;.r I3rd of the next ^vy Ans. 180 duvi 4. Ilow many days from September 23rd to the 21st of the next r-ciiy ■ Ans. 179 days. 5. How many days from June 21st to the 22nd of tiie next De- liiilx'rV Arts. 181 day.s. 6. How many days from December 22nd to the 2 1st of tho noxt M y Ans. 181 days. 7. iiov/ m;iny days from Murch 21st to the 21st of the next Juno ? Ans. 92 duys. 8. How many days from January 13th, 1818, to Soptoniber I7th rai; same your ? Ans. 248 dayd. 71. The vnit ofUme is tho basis of that of Length, Muss, and Pressure: |e t'oiinecLior.s being as follows : — , A pound nrejifture means that amonnt of pressure which is exerted towards 18 car ponna preiif' divided iiito.'idOi'qiial piirt-s I'lillud (lefin'os, as in tlie .siil)join- ed flgiiro. biiic} a de^xrce is ^iini)ly *''<* r.,'(, pfirt dl' ilio circiinifiTuofti of till' circle, it in obvious that its U'n;rth iiiiiHtd(;|ieiid ii|M>ii the size of tliocirtdu. If the fire urn ft- re ice ho ."(JO miles in length, tlien a de(,M 3e of that circle will he one mi'ft loiiir; if the circle ho JJGO iiiche.-j in eirciinil'erence, then a degree will he one inch. Sec. The div sioii of the circumference of tlio circle into 156 ) equal parts took its orl!,'iii iVotii tlie Itiipth of the year, which, in roinul nuinhers, was siip- posi'd to contain ;36o da\s or 12 montlis of 30 dnya each. The 12 nigna corre- spond to the 12 months. The term yninnte is from the Latin »w.inMic feet? 19. How many (juarta in 707 jjcck.^v 20. How many pints in 797 pecks? REDUCTION ASCENDING. Ans. 400880. A7i.. Then wo iniilti- is equal t(» 12 pence, sot down the result. t'(ni.<(e cat'li penny is ^rtbing given In the we deduce the at quantity which one of its units; \wer denomination \ continue the pro- Ans. 93312. Ans. 6900. Ans. 9240. J| Ans. 14076, Ans. 56304. Ans. 67291. Ans. 16129. Ans. 1666, Ans. 1491, I grains? Ans, 19984, 79. ExAMPLn. — Reduce 850347 fiirthings to poimds, &c. 4)85!) 347 12)21 4O80|d . 2 0)r7840s. Ojd. £892 Os. Ofd. = 856347 farthings. E:??T,A.VATioN'.— W divide the fiirthines by 4, ^<'e(7?/. sliillinirs are equal to one pound. By this process we have luced 856347 farthings to £892 Os. (Sid. From the above example and solution we deduce the flowing — RULE. Divide the given number by that number lohich it takes of the ven denomination to make one of the next higher. Set down the lainder, if any, and proceed in the sa?ne manner with each sue- isive denomdjiation till you come to the one required. The last loticnt^ with the several remainders annexed^ will be the answer re- tired. Exercise 6. 1. Reduce 32756 farthings to pounds, shillings, and pence. Ans. £34 2s. 6d. 2. Reduce 23547 troy grains to pounds, &c. Ans. 4 lb. 1 oz. 1 dwt. 8 grs. I 1 •ft r )» I?! i 'M i i 5^ HEDUCTIOl^. (Bk&t. t. 8. Reduce 39'7024 yards to miles, furlongs, &c. Ans. 225 m. 4 fur. 26 r. 1 yd. 4. How many hours are there in 28635 seconds ? Ans. 7 h. 57 min. 15 sec. 6. How many cwt., qrs., and pounds in 1666 pounds? Ans. 16 cwt., 2 qrs. 16 lb. 6. How many cwt, &c. in 1491 pounds? Ans. 14 cwt. 3 qrs. 16 lb. 7. How many pounds troy in 115200 grains? A7is. 20. • • How many pounds in 107520 oz. avoirdupois? Ans. 6720. 9. How many cubic feet, &c. in 1674674 cubic inches? Ans. 969 feet, 242 inches. 10. How many yards in 767 Flemish ells? Ans. 575 yards, 1 quarter. 11. How many leagues in 183810 feet? Ans. 11 lea, 1 m. 6 fur. 20 rd. 12. How many cubic yards in 138297 cubic inches? Ans. 2 c. yds. 26 ft. 57 in. i3. How many cords of wood are there in 67893 cubic feet? Ans. 530 cords, 53 cub. ft. 14. In 3561829 seconds, how many weeks? Ans. 5 wks. 6 dys. 6 h. 23 min. 49 sec, 15. In 1597 quarts, how many bushels? Ans. 49 bushels, 3 pks. 1 gal. 1 qt 16. In 1000 cord-feet of wood, how many cords? Ans. 125 cords 17. In 10,000" how many degrees? Ans. 2° 46' 4o' 18. In 70,000 square links, how many square chains? Ans. 7 square chains. 19. In 11521 grains apothecaries' weight, how many pounds? Ans. 2 lbs. 0| 3 03 1 gr. 20. In 26025 square feet, how many roods? Ans. 2 r. 15 sq. p. 17 sq. yds. 8 sq. ft. 36 sq. in. I W REDUCTION OF THE OLD CANADIAN CURRENCY TO THE NEW OR DECIMAL CURRENCY. I If II 80. Example. — Reduce £76 143. lOfd. to cents. £76x400 = 14s. X 20 = 10Jd.=48far.x5-i-12 = 80400 cents. 280 " 80697H cts. Explanation. — "We multiply £76 by 400, heca^ise each pound is equal to 4 dollars or 400 cents; next we multiply 14, the number of shil- lings, by 20, because each shilling i3 equal to 20 cents; lastly we mul- £76 14s. 10}d. tlply the number of farthiuiprs in the pence and farthings by 6 and divl'de the re Bult by 12, because each farthing is equal to t\ of a cent. That «aoh farthing is equal to j^ of a cent is evident from the fact that Arts. 79, 80.] RECAPITULATION. 53 fur. 26 r. 1 yd. ards, 1 quarter. 23 min. 49 sec. pks. 1 gal. 1 qt q. ft. 36 sq. in. | [CY TO TEE 4S farthings (or one ehilliDg)aro equal to 20 cents; or 12 farthings equal 6 cents, or one farthiug equal y'j of a cent. From the above example and solution we deduce the following — RULE- Multiply the pounds by 400, the shillings by 20, and take five- twelfths of the number expressing how many farthings there are in the given pence and farthings. Add the three results together and their sum will be the number of cents required. Consider the last two figures as cents, and the result will be dollars and cents. Note.— "We take five-twelfths of the farthings by multiplying them by five and dividing the result by twelve. Exercise 7. cents are there in £3 Vs. l^d,? Ans. 1842-,^^ cents. dollars are there in £29 188. S^d. ? Ans. 11965^ cents, or $119-65^ cents, centfe are there in ll^d. ? Ans. 18 J cents, dollars and cents are there in £69 15s. 6d. ? Ans. 27910 cents, or $279-10. dollars and cents in 18s. 8|d.? Ans. $3'74^. dollars and cents in £17 ICs. 6|d. ? Ans, |7l-29iV dollars and cents in £87? Ayis. $34800. dollars and cents in 15s. llfd. ? Atts. $3'19,V. dollars and cents in £16 6s. 2d.? Ans. $65*23^. 9s. lid. to dollars and cents. Ans. $9'98J-. 1. Ilowmany 2. How many 3. 4. 6. 6. How many How many How many How many 7. How many 8. How many 9. How many 10. Reduce £2 ^m the fact that RECAPITULATION. I. Science is a collection of the general principles or leading truths of any branch of knowledge systematically arranged. II. Art is a collection of rides serving to facilitate the performance of certain operations. III. The rules of art are based upon the principles of science. IV. Arithmetic is both a science and an art. V. The science of arithmetic discusses the properties of numbers and the principles upon which the elementary operations of arithmetic are founded. VI. The science of arithmetic is called Theoretical Arithmetic. VIL The art of arithmetic is called Practical Arithmetic, ^ ■ u liMr- 54 RECAPITULATION. [Skct. I. VIII. Practical Arithmetic is the application of i^iihs based upon the science of numbers^ to practical purposes, as the solution of problems, &c. IX. Numbers are expressions for one or more thinorg of the same kind. X. Unity ^ or the unit of a number, is one of the equal things which the number expresses. XI. Numbers are divided into two classes, viz. : simple or abstract numbers ; and applicate, concrete, or denomi- nate numbers. XII. An applicate, concrete, or denominate number is a number whose unit indicates some particular object or thing. XIII. A simple or abstract number is a number whose unit indicates no particular object or thing. XIV. Numbers may be expressed either by words or by characters. XV. The expression of numbers by characters is called Notation. XVI. The reading of numbers, expressed by characters, is called Numeration. XVII. The characters we use to express numbers are either letters or figures. XVIII. The expression of numbers by letters is called Eoman Notation. XIX. The expression of numbers hy figures is called Arabic Notation. XX. In the Roman Notation only seven numeral letters are used, viz. : I, V, X, L, C, D, M. XXI. When these letters stand alone, I denotes one, V fiuc^ X teny 1j fifty, C one hundred, D five hundred, M one thousand. XXII. All other numbers are expressed by repetitions and combinations of these letters. XXIII. In combinations of these numerical letters, every time a letter is repeated its value is repeated ; also when a letter of a lower vnlue stands before one of a higher, its value is to be subtracted ; but when a letter of a lower comes directly after one of a higher value, its value is to be added. Skct. I.] KECAPITULATION. 55 ore things of ! of the equal I 3 number is a XXIV. A bar or dash written over a letter or combina- tion of letters, multiplies the value by one thousand. As we have already a character for one thousand, viz., M, and can, by repeating it, express two or tlwee thousand, we do not dash the I, or combinations into which it enters. XXV. Anciently, IV was written IIII ; IX was writ- ten Villi ; XL was written XXXX, .) 90. How lire pints rcducrd to clialdroiH? ((50 ftiid 79 ) Answer tiftt-r inodol. 91. What lire tho denoiiiinaiio- of Liqiiul nioasuro? Give tho table. (C7.) 92. How .iro tnii.s reduced to u'' (OV and 7S.) Answer after model. 93. Vv'bat arc tlie denomination.'^ Time ineaKureV Give tbe table. (08.) 94. How aro second.s rcfhiced t'> , „rs'' (^08 and 79.) Answer after model. 9.*). Name tho montb.s and tbe niimber ot days in eacli. (OS.) 90. What 13 tho Solar year and it^^ lenytb ?— tho Sidereal year and its length ?— tho (Mvlj year and its length? (OS.) 97. How can wo ascertain whether any frivcn year be Leap year? (09.) 98. Show that the unit of time ia tho ba.si3 of tho units of length, mass or capacitj', and weight. (71.) 99. What are the denominations of Circular measure ? Givo tho table. (72.) 100. Upon what does tho lenprth of a degreo depend? (72.) How are degrees reduced to seconds ? (72 and 78.) ■ t ! ■ |?:i h m » til a* ;1 SECTION II. FUNDAMENTAL RULES. 1. Aritbmetic may be divided into four parts : — 1st. The Aritlimetic of Whole Numbers, or that which treats of the properties of entire units. 2nd. The Arithmetic of Fractions, or that which treats of the parts of units. 3rd. The Arithmetic of Katios, which treats of the re- lations of numbers, whether integral or fractional, to each other and to the unit 1. 4th. The Application of Arithmetic to practical and useful purposes. 2. The Arithmetic of Whole numbers includes Addi- tion, Subtraction, Multiplication, Division, Involution, Evolution, &c. 3. The Arithmetic of Fractions may be divided into two parts : — 1st. Vulgar or Common Fractions, in which the unit is divided into any number of equal parts. 2nd. Decimal Fractions in which the unit is divided according to the scale of ten. 4. The Arithmetic of Eatios relates to the comparison of numbers with respect to their quotients, and embraces Proportion and Progression. 6. Addition, Subtraction, Multiplication, Division, are called the fundemental rules, or ground rules of Arith- ...J IKTS. 1-10.] ADDITION. 6d ml Us length?— which treats actical and ivided into T' tic, bi^cmsc all the other operations of Arithmetic are h.rl')rinud bv means of them. 6. Whatever operations we may perform upon a num- )er, we can only either increase it or c/iminish it. It* we increase it, the process belongs to addition ; if we diminish It, to subtraction. All the rules of Arithmetic are there- ore resolvalde into these two. Multi[)lication is only a short method of performin'jf a peculiar kind of addition, in [which the addends are all the same ; and division is merely |an abridged method of performing a particular kind of subtraction, in which the same quantity is to be taken |away from a given number as often as possible. When any number of quantities, either (Htfcrcnt, or repctitlona of the same, are united together so as to ibrm but one, wo term the )rocoss, simply, " Addition." When the quantities to be added are the snme^ but we may have an many of them us we please, it is called Multiplication ; " when they are not only the same, but their num- Iber is indicated by one of them ^ the process belouf^s to " Invohition." That is, addition restricts us neither as to the kind, nor the number jof the quantities to be added ; multiplication restricts us as to the [kind, but not the number ; involution restricts us botli as to the kind land number. All, however, are really comprehended under the same Irule — addition. ? of Arith- ADDITION. 7. The sura of two or more numbers is a number which contains as many units, and no more, as are found in all Ithe given numbers. 8. Addition is the process of finding the sum of two or Imore numbers. 9. The quantities to be added together arc called ad- Idends, and the result of the addition is called the sum of Ithe addends. 10. Only those quantities can be added which have Ithe same unit, or, in other words., which are of the same Idenomina 'ion. Thus it is evident that 6 days and V miles cannot be added, since Ithe result would neither be 13 days nor 13 miles ; nor can 5 shillings land 3 pence be added, as the result would neither be shillings nor Ipence. Similarly, we cannot add units and tens, or tenths and hun- jdredths, or units and sevenths, &c. I r I .-f ill ). 'II i\ if u ADDITION. [SlOT. II. 11. Hence, in writing down the addends preparatory to adding, we must be careful to set units of the same de- nomination in the same vertical column, i. e. units under units, tens under tens, hundreds under hundreds, l' till! iilxivo operations wo have coinnioncod contlmialiy at tho liiifiicst or U't'L-liarid order. Tim liundrcils addcil *i,iliiiid mid live liiindrtMJ, whicli wi- .set down; Ihc ti-iis nddcil mako KJ toi s, ciinal to 1 iiiuidied and (i ten-, and tho iinili addod, lllal^o Ul units, t«(]ual to 1 toil and G units, uU of wliioU wo sot down in tliiir appropriate coliiniiiH. Next con.siderin'.; tlu^ partial sums I'HiO. 1(»0, and Ifl, n.s so many now aiMond.s, wi' |»ro(.'ood .iiinilarly with them and , and (>. Uut, from tho [.rinciplcs of noUition (Soc. 1.), llicso luM niimbcr.s (/. .'. |iiO(t, COO, 70, and <») may ho writion in oiio line, tlius, 1078, whicli ti'crcforo is tlio sum of tlio addonds !)^7 and (»-0. In (, '), (HI), {l\'), O')- tlio saiuo rosiilt b oLlainod by a fligUtly dillVrcnt process. In (fl) we have commenced at tho ienp, .^n(l In (III), (IV), and (V), at tho units or lowest order. (IV) is simply (III) with tho nnneeosMiry O's omitted. (V')is(IV) soiiiowh.'Lt modiiliMl as follows: — ;» uii/(.s and 7 unit.s make 10 ttniii, iqiial to iiuiln, whitrii wo set down, and one t^/t wideh we carry to tb>) next ooluiiui or eolnmn of tens ; 1 ten and s iciis tiiak(! !> tons, and S tons make 17 tens, eijnal to 7 tons, wliicli wo sot down, and 1 hnndre(l, wliioh wo carry to thi' eoliiiiui of luindfeds; 1 hui'drod and ('» hnndri^ds make 7 hiindroi's, and !> hun"ide theif srnn by so many an make one of the denotnination 'next hiyhev {Arts. 13 and It). Set down the remainder and carry tha quotient to the n^xi higher denomination {Art. 14). H. M\ 66 ADDITION. [Skct. II. last. Proceed in the same manner through all the denominations to the i ^ m 16. We commence at the lowest order or tenths of thousar,d(hp. Tliore being nothing to add to tho 9 tentlis of thonsundih.'s, mo sftnpiy set down the 9 in its afipropriate cohimn. Nest wh add the thousandths, thus: — 2 thousandths and G ihousandtLs are S thousandihs and 4 thousimdths are 12 thoiit>aii.'ll-.-, which are equal to 2 thousandths and 1 hundredth. The '2 thousandths we write down in its own coUunn and carry the hundredth to tlie column of hundredths. Ne.\t we a'!(l the column of hundredthp, thus: — 1 hundredth (carried) and 6 hundredths make 7 hundredths and 9 hundredths make 1G hundredths, and 6 hundredths make 22 hundredths «nd C hundredths make 28 hundredths, which are equal to 8 hun- dredths and two tenths. Wc set down ilie 8 liuucnvdtlirf nii- addends. 66 14 2i) £166 17 Oi I and i make three farthings, which, with f, make 6 farthings; these are ♦equivalent to one of the ne.\t denomination, or that of pence, to he c.irried, and two of the present, or one halfpenny, to he set down. 1 penny (carried) and 2 are .S, and (> are 9, and 3 are 12 pence— equal to one of the next denomination, or that of shillings, to be carried, and no pence to he set down ; we therefore pui, a cipher in the pence place of the sum. 1 shilling (carried) and 14 are 15, and 5 are 20, and 17 are C7 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 dowu. 1 pound and 6 ari^ 7, and 7 are 14, and 2 are 16 i)ounds— eqn.al to 6 units of pounds, to be set down, and 1 ten of pounds to be carried ; 1 ten and 6 are 7 and 4 are 11 and 5 are 16 tens of pounds, to be set dowu. When the addends are very numerous, we may divide them Into two of more parts by horizontal lines, and. Oidding each part separately, may aftef wards And the amount of all the suma. [Sbct. II. itions to the r.dths. TliPro ousiindtlis, ^\o mil. NoNt wt' , 6 Ihoiisiindtbd I tliOu^ali•'!t.!i.■^, Iredth. Tho 2 linn and cany Next we add h (carried) and rcdths make IG idredtha ftnd G equal to 8 hun- )i(l cany tlin lie loiitlib, \vi! >t down, iHud nils, equal to saddrdmakp ^'iththp three lousand, both ivest denornina' its and 42 cents nts. But«'V«'iy ?fore equal to 7 in their propwi ollars. 58. e^d., and ABm 16-19.] [ngs ; these are |l)e carried, and [carried) and 2 fdenoDiination, w*; therefore and 14 are 15, loininnlion, or khillintrs. to be lis— equal to 6 jl ; 1 ten and 6 into two cil ly, may after" ADDITION. EXAMPLE £ s. d. bi 14 21 82 16 4 jE 8. d. 19 17 6 • = 151 7 W 8 14 2 82 5 9. £ 8. d. 47 6 ~41 .=404 11 10 82 17 2 56 27 8 4 9 2 • =253 3 11, 52 4 4 87 8 2 67 Or, in addins; each column, we may put down an a.sterisk, thus*, as often as we come to a quantity which is at Iwist equal to that number of the denomina- tion added which is required to make one of tin- next— carryin;.' forward wliat is above this nuniher, if anythins, and putiinir the last reniaiiidor, or— when there is nothing left at the end— a cypher uiidi-r the column ; — we curry to the next column ono for every asterisk. ' Usiny the same exum])le. £ s. d. 57 *I4 2 82 16 4 19 ♦17 *6 8 *14 2 32 5 *9 47 *6 4 82 17 2 56 *3 *9 27 4 2 52 4 4 87 8 2 404 li 10 2 pence and 4 are B, and 2 are 8, and 9 a.e 17 pence— equal to 1 shillinz and 5 pence ; we put down u dot or an asterisk a. id carry .5. 5 and 2 are 7. and 4 are 11 and 9 are 20 pence— equal to 1 siiilliiiii and 8 pence ; we put down a dot or an asterisk and carry 8. 8 and 2 are 10 and 6 are 10 pence— equal to 1 sliillini? and 4 pence ; wo put do^"n a dot and cirry 4 4 and 4 are >5 Jiiid 2 are 1<> — which boina; less than 1 shillinir. we set down under column of pence to which it be- hiiiirs, vtc. We find on addins them np, tiiat there .ire tliree dots; we there- fore carry 8 to the column of shillinss. '-i sliillinsxs and 8 .are 11, and 4 are 15, and 4 are 19, and 3 are 22 shillinjrs-eqiial to I pound and 2 shillings: we put down a dot and carry 2. 2 and 17 are 19, \,c. Care is necessary, lest the dots, not lie) n? distinctly marked, may bo con- sidered as either too few or too many. This method, though now but little used, seouas a convenient one. PROOF OF ADDITION. 19. First Method. — Go tlwoujh the process again, beginning at the top and adding downwdvds. This method of proof is merely doing the same work twice, in a slightly different manner. Second Method. — Separate the addends into two parts. Add each, part separafeh/, in the usual way, and then add their sums. If the Last sum is the satne as that found by the first addition, the loork luay be presumed to be correct. TwL- ni.:il.od of pr of is f :u idcd on lac iixiou that "the 'vhole is cq-.al 'o tlic sum of all its jarts." ^< ■pf- a r' m 68 ADDITIOJr. [Sbct. IL Akt. 19. |i 1 Example.— Find the sum of 509267, 235809, 72910, and 83925. OPERATION. 509267 235809 PROOF BY SECOND METHOD. 509267 72910 . 235809 83925 72910 83925 Sum 901911 Partial sums 745076 156835 First partial sum.. ..745076 Second partial sum 15683i,' Proof..... . 901911 EXERCISL ; 0. ,. (2) (3) (4) (5) (6) Dollars. Bushels. Days. Acres. Dollars. Pounds 16 76 765 yy2 5832 98764 26 48 881 446 8907 8753 18 69 872 872 4671 76 61 81 315 969 6789 9889 120 264 2333 2679 26199 117482 (7—30) The sum of the numbers in each row of the following table, whether taken vertically or horizontally, or from corner to corner, is 24156. Let the pupil be required to make these 24 distinct additions. • TABLE. 2016 252 2448 684 2880 1116 3312 1548 1 4212 1656 4248 3852!l296'3492 i 936 3132 972 3564 1404 3600 1836 4032 2268 103 2700 54o 576 3108 1008 3204 1440 3636 1872 4068 2304 144 2736 2772 612 2808 216 2412 648 2052 288 2484 720 2916 1152 1692 388813323528 ! 1 208814284 1 1728'3924!l368 1 1 324 2520 758 2952 2124 360 2556 792 2988 828 3420 1260 4320'l764 3960 1800 3996 2232 72 2664 604 3096 1044 2844 2160 396 2592 4356 2196 36 3240 1476 3672 1080 3276 1512 3348 1188 432 2628 1908 3708 1 13744 i L ! 1980| i 4176 1584 3780 1620^ 1 3384 1224 3816 3024 8C.4 3466 468 3060 900 4104 2340 180 1944 4140 2376 (81) 74564 7674 376 6 82620 ( 3 20 10 34' 81 576 4712 6 5376 (45) £ 8. 4567 14 776 15 76 17 51 44 5 5516 14 (49) cwt. qrs. lb 76 3 1^ »7 2 U U 1 11 128 8 16 «?• This table is formed by raultip^lfMliU»numUr«ip the roagic square ofll b^-*^' lie, whether .. is 24156. )ns.- Akt. 19.] ADDITION. 69 (.31) (82) (83) (84) (35) (86) 745G4 5676 76746 67674 42-37 0-87 1614: 1667 71207 75670 56-84 r'i73 376 63 100 36 27-92 "rW 6 6767 56 77 62-41 5s '^^ 82620 m (83) (89) (40) 3-785 85-742 0-00007 6471-1^ 20-706 6034-82 0-06236 663-47 0-253 57-8563 0-572 21-502 10-004 712-52 0-21 0-0007 31 808 (41) (42) (43) (44) 81-0235 0-0007 8456-5 576-34 576-03 5000-0 0-37 4000-006 4712-5 427-0 8456-302 213-5 6-53712 37-12 007 2753-0 5376-09062 MOx\EY. - (45) (4G) (47) (48) £ 8. d. £ 8. d. £ a. d. £ 8. d. 4567 14 6^ 76 14 7 3767 13 .1 5674 17 6i 776 15 7i 667 13 6 4678 14 10 4767 16 Hi 76 17 9f 67 15 7 767 12 9 3466 17 lOf 51 10| 5 4 2 10 11 5 5084 2 2:^ 44 5 6 3 4 3 4 11 8762 9 9 5516 14 3f ::i-^» ' . ? ¥. 'm 1 'l ™, H I 11- 1 AVOIRDUPOIS WEIGHT. (49) (50) cwt. qrs. lb. cwt. qrs. lb. 76 3 14 476 1 2U 37 2 15 756 3 '^H 14 1 11 767 1 16 567 973 2 1 15 12 128 8 16 (51) (52) cwt. qrs. lb. cwt. qrs. lb. 447 1 7 14 2 12 676 1 6 8 3 7 467 1 n 2 16 563 1 6 7 8 428 oj 14 1 ■ m 70 ADDITION. [Sbot. It Abt. id.j TROY WEIGHT. (68) lb. oz. dwt. Rra 7 5 9 6 6 6 7 9 5 6 8 21 11 18 (56) yrs. ds. bra. ma. 99 359 9 56 88 8 67 77 120 7 49 265 115 2 42 (59) yds. qrs. nls. 5G7 3 2 476 1 72 3 8 5 2 1 1122 2 2 (54) lb. oz. dwt. KTS. 57 9 12 14 67 9 11 11 66 8 10 5 74 6 5 3 12 3 5 4 TIME. (5T) yrs. ds. hrs. ms. 60 90 50 6 76 1 57 8 58 6 12 CLOTH MEASURE. (60) yds. qi,, uls. 147 3 3 173 1 148 2 1 92 3 2 67. 0- 68. 56 69. 0- (55) 70. 0- lb. oz. dwt. grs. 71. 0- 87 3 7 12 72. Ac 11 12 3 thirty-thve 16 14 eiaht imnc 44 12 10 18 Hundred a 67 8 9 10 Lm^\ A 73. Ac and eight} million; ti ten millior dred and s (58) hundred ar yrs. ds. hrs. ms. lions. 60 127 7 50 74. Ad 120 9 44 and seveni 76 121 11 44 fbrty-seveu 6 47 3 41 seveuty-six 8 9 11 17 75. A ' hundred ar two handle thousand ; ninety-five (61) yds. qrs. nls. 167 2 1 113 3 2 1 2 64 3 (68) (64) $978-63 $09-42 492-29 189-87 83-43 674-29 729-47 86-43 9-00 982-78 CANADIAN MONEY. (65) $7l9-'i3 912-99 68-68 50-00 9-73 (62) yds. qrs. nl.s 156 1 1 176 3 1 54 1 573 2 3 (66) $9863-47 986-10 91-S9 7-45 •98 |2292-82 1. How the Gulf oi milos long ; Erie, 260 m and the Riv 2. The ton, 25000 Montreal, 'J these seven 3. In th $^165000; ^^lOOOOODO products, $] ous other pi value of Can 4. A wh airo'.mt of 1 iBT. 19.1 ADDITION. V 55) . dwt. prs. 1 12 12 3 16 U 5 10 18 J 9 10 (58) 8. hrs. ms. i1 1 50 20 9 44 21 11 44 47 3 41 9 11 17 67. 0-4 + 74 •474-37-007+'76-05+747-077 = 934-004. 68. 56-05+4-75+0-007 + 36-14+4-672 = 101-619. 69. O-76-t-0-OO76+76-hO*5 + 5-|-0-05-.= 82-3176. 70. 0-5+0005+5-H50 + 500=:555-505. 71. 0-367 + 56-7+762 + 97-0+471 = 1387-667. 72. Add eight hundred and fifty -six thousand, nine hundred and thirty-thvee ; 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. Am. 206729544. 73. Add three millions, and seventy-otie thousand ; four millions, and eighty-six thousand ; two millions, and fifty-one thousand ; one million; twenty-five millions, and six; seventeen millions, and one; ten millions, and two • twelve millions, and twenty-three ; four hun (Ired and seventy-two thousand, nine hundred and twenty-three; one hundred and forty-three thousand ; one hundred and forty-three mil- lions. Ans. 217823955. 74. Add one hundred and tbhty-three tliousand ; seven hundred and seventy th(jusa;id ; thirty-seven thousand ; eight hundred and forty-seven thousand; thirty- three thousand; eight hundred and seventy -six thousand ; four hundred and ninety-one thousand. Ans. 3187000. 75. Add together one hundred and sixty-seven thousand ; three hundred and sixty-seven thousand ; nine hundred and six thousand ; two hundred and forty-seven thousand ; ten thousand ; seven hundred thousand ; nine hundred and seventy-six thousand ; one hundred and uiuety-five thousar 1 ; ninety-seven thousand. Ans. 3665000. ii il! Mil V APPLICATIONS. 1. How many miles is it from the lower end of Lake Huron to the Gulf of St. Lawrence, passing through the River St. Clair, 25 milos long ; Lake St. Clair, 20 miles ; River Detroit, 23 miles ; Lake Erie, 250 miles; Niagara River, 34 miles; Lake Ontario, 180 miles; and the River St. Lawrence, 750 miles long? Ans. 1282 miles. 2. The city of Toronto lias a population of about 50000 ; Hamil- ton, 25000; Kingston, 15000; London, 10000; Ottawa, 10000; What is the population of A71S. 230000. Montreal, 75000 ; and Quebec, 45000. these seven cities taken together ? 3. In the year 1856 Canada exported : — Produce of the mine, 6;! 65000; produce of the sea, $500000; produce of the forest, |U00000i)0 ; animals and their produce, .$2500000 ; agricultural products, $1500001)0; manufactures and ships, flUOOOOO ; and vari- ous othor products to the amount of $2235000. What was the total value of Canadian exports for that year ? Ans $32000000. 4. A wholesale merchant sells, during the year, goods to the uiro-mt of $11080 in Toronto; $9427 in Gait; $1798 in Berlin: 1-: n..i ,t I ■ ri|[;| !•l^■l,■:f| ADDlTIOK. tSECT. ll Abt8. 19-2( I ¥ \\'-i\ $164'>S in IlaRiilton ; $7496 in Guelph ; $6429 in Woodstock ; $5297 in Ctiatliam ; and $8426 in Goderich. Required the amount of tlic year's sales. Ans. $06376. 5. The Grand Trunlc Railway is 962 miles long, and cost $60000000 ; the Great Western is 229 miles long, and cost $14000000 ; the Ontario, Simcoe, and Huron is 95 miles long, and cost $3300000 ; the Toronto and Hamilton is 38 miles long, and cost $2000000. What is the aggregr.te length and cost of these four roadw ? An.s. Length, 1824 miles, and cost $79300(*00. 6- The circulation of promissory notes for the four weeks endin;^ February 3, 1844, was as follows: — Bank of England, about £21228000; private banks of England and Wales, £4980000 ; Joint Stock Banks of Englnnd dnd Wales. £3446000 ; all the banks of Scotland, £2791000; Bank of Ireland, £3581000; all the othei banks of Ireland, £2129000 ; what was the total circulation ? Ans. £38455000. 7. Chronologers have stated that the creation of the world occurred 4004 years before Christ ; the deluge, 2S48 ; the cull of Abraham, 1921 ; the departure of the Israelites from Egypt, 1491 ; the foundation of Solomon's temple, 1012 ; the end of tee captivity, 536. This behig the year 1859, how long is it since each of these events? Ans. From the creation, 5863 years ; from the deluge, 4207 ; from the call of Abraham, 3780; from the departure of the Israelites, 3350; from the foundation of the temple, 2871 ; and from the end of the captivity, 2395. 8. Add together the following: — 2d., about the value of the Roman sestertius; 1h\., that of the denarius; Hd., a Greek obolus ; 9d., a drachma; £3 15s., a mina ; £225, a talent; Is. 7d., the Jew- ish shekel ; and £342 3s. 9d., the Jewish talent. Ans. £5*^1 2s. 9. Add together 2 dwt. 16 grains, the Greek drachma ; 1 lb. 1 oz- 1 dwt., the mina : 67 lb. 7 oz. 5 dwt., the talent. Ans. 68 lb. 8oz. 8dwt. 16 grains. 10. What was the population of the British provinces in North America in 1834, the population of Lower Canada being stated at 649005, of Upper Canada, 336461 ; of New Brunswick, 152156; of Nova Scotia and Cape Breton, 142548 ; of Prince Edward's Islauvl, 32292 ; of Newfoundland, 76,000 ? Ans. 12S1 4 &^. 11. A owes to B £567 16s. 7id. ; to C £47 16s.; and to D £56 Os. Id. How much does he owe in all? Ans. £671 I'is. 8|d 12. A man has owing to him the following sums : — £3 10s. 7d. ; £46 Os. 7^d. ; and £52 14s. 6d. How much is the entire ? Ans. £102 5s. 8^.1 13. A merchant sends off the following quantities of butter :— 47 cwt. 2 qrs. 7 lb. ; 38 cwt. 3 qrs. 8 lb. ; and 16 cwt. 2 qrs. 20 lb. How much did he«eend oif in all? Ans. 103 cwt. 10 lb. 14. A merchant receives the following quantities of tallow, viz. :— Arts. 19-20.} ADDITION. 73 e. PSVl ; and 13c\yt. 1 qr. lb. ; lOcwt. 3qr3. 101b.; and Ocnvt. 1 qr. 161b. How much has he rceeivod in all ? Ans. 33 cwt. 2 qrs. 6 lb. 15. A silversmith has 71b. 8oz. 16d\vts. ; 91b. 7 oz. 3 dwts. ; and 4 lb. 1 dwt. What quantity has he ? Ans. 21 lb. 4 oz. 16. A merchant s;'lls to A, 7<» yards 3 quarters 2 nails; to B, 90 yards 3 quarters 3 nails; and to C, 190 yards 1 nail. How much has he .«old in all ? Ans. 357 yards 3 (piarters 2 nails. 17. A merchant in Toronto sells goods to the followin;:^ amoimts during the week, viz. : — Monday, i;i^ 129-38; Tuesday, §711-43 ; Wed- nesday, $119-87; Thursday, !5?lb80-42; Friday, .^lyoi-G.'i ; Saturday, S2498-91. Required the whole amount of the week's sales. Ans. $6444.66 18. Looking over ray last montli's expenditure, I lind that I have paid the following sums, viz.: — Baker's bill, !!^r)-73 ; Butcher's bill, ^20-91; Groceries, $12-75; Fruit, $3-29; Kent, $10 25; Servants' wai.'e^!, $10; Tailor's account, $17"87 ; Shoemaker's bill, $11*03 ; and sundries, $9*47. Requited how much I paid in all. Ans. $107*90. 19. Add together $007-19; $298-97; $789-87; $1723-10; and ,S123-nO. Ana. $3542-13. 20. A farmer sells seven loads of wheat, the first containing 176^' lbs., the second 1827 lbs., the third 1329 lbs., the fourth 1901 lbs., the fifth 1666 lbs., the sixth 1879 lbs., and the seventh 1185 lbs. What was the aggregate weight of the seven loads, and how many bushels did they contain? Ans. 11550 lbs. or 192.^ bushels. Note. — The biislivls are found V.y dividing the aggregate weiglit by 00 lbs., the weight of one bushel. 21. Having eftected an insurance on my housohold furnit'arc, &c., I am required to make a detailed statenjcnt of its value. I find this to bo as follows :— Carpets, $250-00, table and bed linen, $90-88, beds and bedding, $173-00, furniture, $791-23, pictures and engravings, §207-18, books, $1649-19, plate and plated ware, $307-18. Keciuired the total value of my household furniture. Ans. $34t;9-26. 22. Toronto has a population of 45000, ITanillton, 20000, Erock- ville, 4000, Prescott, 2500, Kingston, 15000, Ottawa Citv, 10000, Chatham, 4000, Goderich, 2000, London, lOOOO, Port Hope, 4000, Cobourg, 5000, Montreal, 70000, and Quebec, 50000. Wliat is the eutire population of these 13 cities and towns? ^In*-. 241600. I'M, '-', 'M. 'i t b; 20. The pupil should not be allowed to leave addition until he can read up the column without hesitation. For instance, in the following questions, which are in.serted for the sake of pi-aetice in rapid addition, he shoukl not be permitted to sjull the columns thus, 6 and 4 are 10, and 4 are 14, and 4 arc 18, and 5 are 23, &c., but should be required to read them, i. ^. 76 SUBTRACTION. [Sect. II 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. S). Why do we draw a lino under the addends f (V.) Why do wc begin to add at the lowest denomination ? (VI.) Way do we divide the nuni of the unitH of any one denomination by as many as make one of tlie next higlier '( (VII.) How do we prove addition ? (19) Upon what a.viom is llieUiid method of proof founded ? (19) So far as tiie result is concerned, does it mak any diflerence where w« commence to add '/ (12) Exhibit the work when we commence addinpat the left-hand side, or highest denomination. (12) Wheti tlie addends are very numerous, what plans may wp adopt ? (18i Upon wliat i)rincij)i(' docs the lormer oftlic Ke plans proceed f (19) What (liflerent rubs were formerly made in ail — 3-t-9,) the wliole cxfueajsion is to be considered as one (luaiitity. The iio<;iitive si;.:;ii b»'i'ore such an expre.'^t^iou indicates that the vdlnc of the whole expression within the biack(!tfl, is to be subtracted, or, what anioinittj to the same thiii<;, that the nunibets haviii,:i; the bij^'n + befbre them are to bi? HubtnicliMl, iiiid those having the sign — , added. Hence a minus sign before a Ijrackft, has the effect of changing the signs of all the — ininnoi d Wo (tet over thin difllciilty hy 181 or li>0 + 8n + l CO .hidoriiiK tlio inii iiond to bo, not 8(iO+-4(» + 2, but 7<>0 + 14ii +2, or in otlior word.>*. wo hoi-row one of the order of hiindrtMLs niid ridiico it t<» tons Now wo hnvo 1 unit from 2 uidtR and I unit roinuins; <1(» unifrt or (i to'H from 140 units (»r 1 1 tons, and 80 units or 8 toi s remain ; (J'lO unitH or 6 hundrods, from 700 units or 7 hundreds, and 100 unitB or 1 hundn-cj roniiiin. EXA.MPLK 4.- from 9 cwt. 1 qr Exi'LANATIOK. OPKRATION cwt. qrs. lb. cwt. qrs. lb. 9 8 I 2 8 = T = 8 8 6 2 8 7 Let it be required to subtract 3 cwt. 2 qrs. 7 lbs. 8 lbs. -As we can'ot subtract 2 qra. from 1 qr. we I'orrow 1 cwt and reduce it to quarters. Tlic 9 cwi. 1 qr. 8 lb. W(< then consider as 8 cwt, 5 (|rs. 8 il). and Irorn It subtriict tlie 8 cwt. 2 qrs. 7 lb. Thus, 7 lbs. from 8 lbs. and 1 lb. reniuii ^; 2 (jr.s. from .'> qrs. and g — — qrs. rernuin: and 8 cwt. from 8 cwt. and 6 cwL 6 8 1 ft 8 1 remain. 28. Hence, to find the difference between two numbers, we deduce the following : — ^ RULE. Write the subtrahend under the minuend, so that units of tfic same denomination ma)/ be in the same vertical column (24). iJraw a Urn under the subtrahend to separate it from the remainder. Subtract each digit in the subtrahend from the one over it in t/ie minuend, k- ginyiing at the lowest denomination. When the units of anyone denomination of the vxinuendfall sfioH of those of the same denomination in the stibtrahcnd, borrow one of the next /tigher denom,inatioii in the minuend, redme it to its equiva- lent units of the required denomination., add them to the units of that denomination given in the minuend, and from their sum. subtract the units of that denomination given in the subtrahend. 29. The following is the complete work of a question in Subtraction: Example 6.— From 6400 lbs. oz. dwt. 7-0006 grs. take 987 lbs. 3 oz. 17 dwt. 22-6349 grs. OPERATION. 19 24" 9 9 9 dwt. /T-O 6 grs. Minuend. 17 22-6 3 4 9 Subtrahend. (10) 9 9 11 5 3 ;p ;p n ^^00 lbs. oz. 9 8 7 3 5 4 12 8-3 6 5 7 Kemainder. inr^v Aki». 28-31.] SUBTRACTION. 79 Exi'LAN TioN. — Here. ii» wc innnoi tnk" tenths of thouhandths of a grain fioiii ('. i.nJlis of thijii:'.ainUli9 of a gniln, wc borrow ono gralu, thore being no I ■ iths, liiinlri ii!< lit. '.o ii'ii of ilic orik'r of tontiis of jjjralns. Borrow ono tenth and ther« ri'iniiii ;• t 'ii'iJ.^ ami the oiio ti-nth we borriwod Is i qiial to 10 hiindretlths. iJf.rrow 1 lii:ii !,L';llli. thtre rcin.ilii 9 hundrodtha. and the (»ne hundredth we iiu.Ku.cd iHeo.tuI to HI thotiMiiKllhs. Borrow 1 thousandth, there reuiuiM 0, and ill"- I tli"ii.;tM<'th Is en lai to It) of the ord'ir of tenlhi of thousandths— the order for vsiiicli it wu.-i iieces^iary to borrow. 1(» of the order of tenths of thcmsandtlis iif u.'iin* a!id »'» of the order of tcntlus of thousandths of grains, make 10, Iron wiii li ;;i\" •> of till- ordei- of tt-nllis uf thousandths tenths and '■) tenths remain. Au'ain, as we c.innot take '^2 urains from G grains, wo borrow from the next iiviii!;ihle hiiriier nnier, wliieh, in this case 18 hundreds of pounds. 1 of the or li-r of hull. Ire Is of pounds reduced, as above, to its equivalent lower donomi- n.'i'ioti, i- equiil to :t I'-ns ofllis.. '.» units of Ihs 11 oz. 19 dwt. 2i gr.s. 21 grains, lul le i fo (>. make ;iO irrains, and 2-' trains from 'M grains, leave S gral la ; 17 dwt. from I'l ilwt. leave 2 dwt.; :< oz. from 11 oz. leave 8 oz ; 7 units of lbs. from 9 II i's (jf ihs. leave 2 units of lbs. ; S lens of lbs. from 9 tens of Iba. leave 1 ten of li><. We 'annot take 9 hundreds of Ibi from 8 hundreds of lbs, so wo are com- pilled to borrow 1 of the order of thousands of lbs., which Is equal to 10 liun- ,lie(i- of Ib^.. and 3 hundreds of iVi.s., make I'i liundretls of lbs. ; 9 hundreds of lbs. from |:} hundreds of lbs. aivd 4 hundreds of lbs. remain; thousands of Iba. tVi'iii 5 thousands of lbs. and 6 thousands of lbs. remain. 30. If !iny diixit of the minuend be smaller tha i the corresponding digit of the subtrahend, practically, we can proceed in either of two way.s. First, we may increa.^e that denomination of the minuend which is too small, by borrow- iiii; one from the next higher (considered a.s so many of the lower denomination, ,»r that which is to be increased), and adding it to those of th'> lowi-r, already \\ the minuend. In this cavse we alter the form, but not the value of the minuend; which, in the example given below, would become — hundreds, tens, units. 7 8 12 = 792, the minuend. 4 2 7 = 427 , the subtrahend. 8 6 5 = 805, the difference. Or, secondly, we may add equal quantities to both minuend and subtrahend, which will not alter the difference; then we would have — buiidi'ods. tens, units. 7 9 2 + 10 = 792 + 10, the minuend +10 4 2 + 17 = 427 + 10 , the subtrahend + 10. 8 6 5= 365 + 0, the same difference. In this mode of proceeding we do not use the giv&n minuend and subtrahend, but others which jiroduce the same remainder. lit i '•^ ■ i sail ilemainder. PROOF OF SUBTRACTION. 31 . First ^'ethod. — Add together the remainder and subtrahend; ike ^iim xhould be equal to the minuend. For ti ■' remainder expres.ses by how much the subtrahend is smaller than the m nueii 1 : addins, therefore, the remainder to the subtrahend, should makf it equal to iti' minuend; thus, 8754 minuend. 5a39 subtrahend. } 6am of difference andsu' 2915 difference. ^ 8754 =: miuuend. J ft ■ "Si i] ^H 80 SUBTRACTION. [Sect. II. Second Mltjiod. — Subtract the remainder from the minuend, and what is left should be equal to the subtrahend. For tho remainder is tlie exci-ss of the mlnncnd over the subtruhend; thcrclbro, tal^iiij; away tbis excess should leave both equal ; thus, 80^4 Tiiiniicnd. 7'JS.j sui^trahond. Pkoof : SO.'U niinuend. 404 v)? 153 220202 (10) 52-1^7 20-005 (15) 40C-3270 )3 006 4 = 9'7'7'7n. ,=: r)99()!». ", zr To-iol. ') = 6 -91!. 1 =r 1-G7r.. )t= 9'?-M01. ) I 2-274. ~~* 1-70. ~~ 4-97909. ■ = 176-09??. =r 7*170, kstBTR ACTION. MONEY. 81 (88) From $9876-43 Take 987-49 $8888-94 (42) From $1234-50 Take 999-96 $234-54 (89) $427-63 197-21 $230-42 (43) $671-98 99-67 $572-31 (40) $721-73 91-00 (44) $286-29 611-89 (46) £ 8. d. From 1098 12 6 Take 434 15 8 (47) £ s. d. 767 14 8 486 13 9 £663 16 10 (51) £ 3. d. From 98 14 2 Take 77 15 3 (62) £ 8. d. 47 14 6 88 19 9 (48) £ 8. d. 76 15 6 14 5 (53) £ 8. d. 97 16 6 88 17 7 (49) £ 8. d. 47 16 7 39 17 4 (54) £ 8. d. 147 14 4 120 10 8 AVOIRDUPOIS WEIGHT. (66) cwt. qrs. lb. From :100 2 24 Take 99 3 15 (57) cwt. qrs. lb. 175 2 15 27 2 7 (58) cwt. qra. lb. 9664 2 23 907a 24 100 3 (60) lb. oz. dwt. grs. From 554 9 19 4 Take 97 16 15 TROY WEIGHT. (61) lb. oz. dwt. grs. 946 10 17 23 (41) $16-25 9-75 (45) $7.19 1-86 46V 9 2 18 L. (50) £ B. d. 97 14 6 6 15 7 (55) £ 8. d. 560 15 6 477 17 7 (59) cwt. qrs. lb. 554 476 3 5 (62) lb. oz. dwt. grs. 917 14 9 798 18 17 I. I, I I . , !■ ."m il ■ t i ' ^ y I i 1 ^'* ii^j ■^ top* /I nl "■ 1 I. in * 11 ? tie ri M / & 82 BUBTKACTION. OECX. II, TIME. I' "i]' i a^Jii (88) (64) (65) ?67 ds. hrs. ms. yrs. (Is. hrs. nis. yrs. (h. lirs. ni8. From 131 6 30 475 14 13 10 507 12G 14 12 Take 476 110 14 13 IGO 16 13 17 400 15 291 20 16 17 APPLICATIONS. 1. A shopkeeper bought a piece of cloth containing 42 yards for £22 lOs., of which he sells 27 yards for £15 15s. ; how many yards has he left, and what have they cost him V Ans. 15 yards; and they cost him £6 15s. 2. A merchant bought 234 tons, 17 cwt., 1 quarter, 23 lb., and Bold 147 tons, 18 cwt., 2 quarters, 241b.; liow much romaiued un Bold? Aus. 80 tons, 18 cwt. 2 qrs. 24 lb. 3. In 1856, the revenue of Canada was as follows: — customs, $4500000; public works, §500000; crown lands, ^,500000; and casual, $320000. For the same year the expenditure was as follows : — interest on public debt, &c., >?lfMH)000; civil government, ^225000; legislation, $450o00 ; administration of justice, $450000 ; education. $380000; collection of revenue, $940000; pubhc works, &c., $1755 000. How much did the total revenue of that year exceed the tolal expenditure? ' Ans. $620000. 4. The census of 1852 gives the population of Upper Canada an 962004, and that of Lower CJauada as 890201. By how much did tho population of the former exceed that of the latter? A7is. Vri43. 6. Upper Canada contains 147832 square miles; Lower Canada. 209990 square miles; Nova Scotia and (^a}>e Breton, 18746 sciuaro miles ; New Brunswick, 27620 square miles ; Piince Edward's I.-iland, 2173 square miles; Newfoundland, 30000 square miles; and Hudson's Bay Territory, 243C0O0 square miles. By ho\y much does the aggre- gate extent of these Bridsh North American Provinces fall short of the total area of the United States — the latter being 2930116 square miles? Ans. 57756 .square miles. 6. A merchant has 209 casks of butter, weighing 400 cwt. 2 qrs. 14 lb. ; and ships oil' 1 73 casks, weighing 213 cwt. 2 qrs. 24 lb. How many casks has he left ; and what is their weight ? Am. 36 casks, weighing 186 cwt 3 qrs. 15 lb. 7. If from a piece of cloth containing 496 yards, 3 quarters, and 3 nails, I cut 247 yards, 2 qrs., 2 nails, wliat is the length of the re- mainder. Ans. 249 yards, 1 (juarter, 1 nail. 8. A jBeld contains 769 acres, 3 roods, and 20 perches, of wh.ici* 576 acres, 2 roods, 23 perches, are tilled ; how much remains ud- tUle(/ Aps. 193 acres, 87 perches. fl.. Arts. 81, S2. 1 RECAI'lXrLATION. 83 SS '; i! «. hrs. ni!>. 10 14 12 15 2 yards for many yards lim £6 158. 28 lb., and maiued un : qrs. 24 lb. : — customs, 0000; and i follow.s: — t, $226000; educatiou, &c., I'.llory ed the tolnl $620000. Canada as uch did tho ns. 71.48. r Canada. 146 S(iviaro d's I.^laiid, Hudson's the aggrc- 1 short oi" 6 S([uare are niilos. wt. 2 qis. lb. How qrs. 1 5 lb, irters, and of the re- er, 1 nail. of wldcii mains nn- 9. I owed my friend a bill of £7^ lOs. 9id., out of which I paid £69 17s. 10|. ; how much remaiiicd due? Aiis. £16 18s. lO^d. 10. The population of London is 23(Un41, and th; * of Paris is 10.58202. How much docs tho population of London exceed that of Paris ' A)is. 180987^. Aufi. 14. Ann. 88. Arts. 52-94. 7 perches. 11. The yiopulation of Liveipool is 384205, and tlmt of New York 515547. How much does the population of New York exceed tliat of Liverpool? ^^'•'"■- l^l;'''-^- 12. Laive Huron contains 20000 square miles: by how much docs it exceed the area of La!v(\s Eri<' anil Ontario— tho former containing 11000 square miles, and the laltcr 70i»t) ,7-n8 on his books against his cus- tomers: his debts amount to ,^19478-25. How much is he worth after paving what he owes? Ann. $15918-29. 14. What is the value of 6-3 + 15-4? 15. Of48+(7-3~14)? 16. Of 47-0-(2-f 1-24 + 16-0-84? 17. What is the dilfercnco between 15 + 13 — 6 — 81 and 15 + 18 — («l._81 + 62)? ^^"«- l***^- 32. Before the pupil leaves subtraction he should be able to take any of the nine digits, continually, from a given nuMiber, without stopping or hesitatmg. thus, in subtracting 7 continually from 94, he should 'say, 94, 87, 80, 73, 60, 59, &c. In the following examples, which are inserted fo,- practice, he should not be allowed to spell the subtraction, thus, G from 9 and 3 remain, 4 from 2, we can't, but 4 fiom 12 and 8 remain, &c. ; but should be re((uirod to read as fol' lows:— 6, 9.. 8; 4, 12.. 8; 9, l.*^.. 4; 10, 11..1; 10, 18..8, &c. (18) 9800046048019181097800041081829 191847813191081478199910 1 99846 (19) '74321913047128098706540456007139 1 3423450789 1 2;i4567891 28 1 5^7891 2 RECAPITULATION. I. Subtraction is the process of finding the diiference be* tween two numbcT-s. IL The greaver of the two numbers is called the mi- nuend. Si I' 04 QUESTIONB. [flacT. n ! I!J III. The smaller of the two nnmbery is called the au^ trahend. IV. What is left after making the '^ubiraction is called the reynainder or ditt'eience. V. Only quantities of the same denomination can be subtracted. VI. Subtraction is indicated by the sign — , which is called minus, or the negative sign. VII. When several numbers are inclosed in brackets, they are to be considered as constituting only one quantity. VIII. When a negative sign precedes the first bracket it indicates that all tbc quantities within the brackets are to have their signs changed when the lirackets are removed. IX. When quantities are removed into brackets, pre- ceded by the negative sign, all their signs must be changed. X. We begin subtraction at tlie lowest denomination, be- cause it is sometimes necessary to borrow from the higher denominations and reduce. XL Instead of tlms borrowing and r<;ducing, we may consider any denomination in the minuend increased by as many units of that denomination as make one of the next higher, and then add one to the next higher denomination in the subtrahend. This is merely ad(Ung the same quan- tity under different forms to both minuend and subtrahend, and consequently cannot affect the value of the remainder. (30.) QUESTIONS E ANSWFRLiJ BY THE PUPIL. Note.— JVwOT&ers in Roman numerals, thus (V), refer to the Recapittila- tion; those in Arabic numerah, thus (25), refer to the articles of the iSection 1. What is Subtrfiction ? (I.) 2. What is the ininuond ? (II.) 3. Wiiat i.s tlic (liTivation of tlie word minuend t (22) 4. Wlial is the subtrahend? (III.) 5. What i.s the doriviition of the word suhtrahendt (22) <5. What is th(^ remainder? (IV.) 7. Wiritliind of quantities can bo subtracted' (V.) 8. How is subtraction indicated? (VI.) 9. Wlien several numbers are inclosed togeth'" in brackets, how are tlicy to be tnlven ? (VII and 2(i.) 0. What effect has a neerailve si-ni preceding brackets? (VIII and 20) Jl, When quantities arc removed into brackets, luecvdt^l by \\\v^ higy ~, wi)ftl must be done with them? {IX and 26) .r ■• -b » 12, What is the ruU fur eubtraciiou i (28) Arts. 82 : 13. W U. W 15. W 16. \'\ 17. W 13. He 19. r 20. II 33. her as u nnilti[)li 34. mulf'rpli 35. plicand we mult 36. as man\ ft/ the prot addition The of the factor^ 37. divided 33. integral composi 39. taking 1 in the n 1st. he eqna 2nd. will be . multipli 3d. m [Sect. U ,d the i'v.'/ 1 is called m can be , winch is brackets, 3 quantity, bracket it :ets are to removed, kets, prc- 3 changed, nation, be- he higiier we may xsed by as ' the next omination line quan- btraliend, emainder. tL. Re.capitnhu ' the /Section 7 are they to id 20) Arw. 82 39.1 MULTIPLICATION. 85 13. TVhy must wo put units of the same denomination ;u the same vertical ooiuuin? ('i4) 14. "Wlicii a didt in the subtraheiul \s crre^t ir aan tlie •orresj'ondlufe digl* in tlie iiiinui id, wlut it. done? (27 L.va'nplc 'A, ov 2in 15. Wimtofl'or plat, may l«ft adopted ? (>30) 16. Upon ^vh!lt pri cipk' does tiiis i)lan jinxeed? (XI.) 17. Why do wo bcyin to .subtract iit the riglit-liand sido? (X.) 13. How do wo prove subtraction? (-'n) 19. L'pon what principles arc these n.ctliods of i)roof founded? (31) 20. Illustrate the difference between upeiliiii/ and readiny in subtraction. MULTIPLICATION. 33. iVIultiplicitLlon i.s a short process of taking one num- ber as n.any times as there are units in another. Hence Tiuilti[)lication is a sliort method of performing addition. 34. The number to be taken or mulliiilied is called the m>f/fiplican(/, and in addition would l)e called an addend. 35. The number denoting how many times the multi- plicand is to be taken, or, in other words, that by which we multiply, is called the multiplier. 36. The number arising from taking the multiplicand as many times as thei^ are units in the multiplier, is called the product, and corresponds to the sum of ike addends in addition. The multiplicand and multiplier are called the factors of the product because they make or produce it, (Lat. factor^ "a maker, agent, or producer.") 37. A prime numher is one which cannot be exactV divided by any lohole number, except tlic unit one and itsei/. 38. A composite number is the product of two or vr^'Vi integral factors, neither of which is unity. Thus IG io a composite number, and its factors are 8 and 2, or 4 and 4. 39. Since the product is the result wliich arises from taking the multiplicand as many times as there are units in the multiplier, it foUows : 1st. If the multiplier be equal to unity, the product will be equal to the multiplicand. 2nd. If the multiplier be greater than unity, the product will be as many times greater than the multiplicand as the multiplier is greater than unity. 3d. If the multiplier be less than unity, that is, if it be i 86 MULTIPLICATION. [Sect. U, Aets. 40-4f !{ ir I a proper fraction, the product will be as many times less than the multiplicand as the multiplier is less than unity. 40. Let it be re([uired to multiply any two numbers together, say 7 and G. If \v(3 iimko in n horizontal lino as many stars as 7 thcro aro nnits in the inultiiilicuud, and make as iiiuny , ' , Hucli linos of stars as there arc units in the mulLij)lior, It \i manifest that thi^ entire niimbor of stars will re[>resent the number of units which result from takinu tho multiplicand as many times as there uro units in the multiplier. But it is evident that we may consider the 42 stars i.j the uhovo fi<:uro, either as 7 star-s taken 6 times,' or ns C stars taken 7 times, that is, x 7=4'2=7 x 6. Hence either of the factors may be used as muUipUer without altering the product. 41. Let it he required to multiply tho number 8 by tho composite luirri' bcr 6 of which the factors are ii and 2. 8 I* ♦ * « l|! * # * ' o 1 r z\* * ♦ * * I" x> *,-2 8 X 3=24 f, ^ 24 X 2=48 ^ ' 1 1* * f* * * * * * >i< « 4< 8x2=16 16x8=48 ♦ * * * 8 X 6=48. * If T.'e write S star^ in a iiorizontal lino and mnko fi s'leli lines, we shall evidently have in nil y x (5=48, tlie number of units in all th(! lines. But wo may eonaider the G lines as 2 .vets of 8 lines each, and in eaeh set of 3 lines there are 8 x;j=24 units. Therefore in tho 2 sets there are 24 < '2=4S iinit.s. /..cain wo may (.Mnsfidcr tho Hues as i» sets of 2 lines e.-ich, and in eaeh hct of 2 lines there' aro Sx2=lC units. Therefore in 3 such sets there are 16 X. 3=48 unit-. Hence 8 x G=4S 8 X .3=24 and 24 x 2=4S=S x 6 8 X 2=16 and 10 x 8=48=8 x And as the { ame inay he shewn for any other composite number as well as for 6, wo may conclude that. When the multiplier is a composite number we msty multiply by each of the factors in succession, and the last p'l luct will be the entire product sought. , or tons of oents, iniil- tiplii'd hv 11. iivike 41 tontlia of dolhirs. and '■\ tenths uo ciinird, niaku i7"tont.ljs, cqiul to 7 tt-nlhs, and 4 units to carry; (i units, null iplicd l»v 11, nialcti ttl units, and 4 units wo oarrird, luako 70 units, equal to t> un ts to .sot down, and 7 tons to oarry ; !) t"ns. multiplied by 11, make 99 tens, and 7 t.-ns, make lifti tons, equal to (i tons and lu liuiulriMls; s imu- drcd.s, multipliod by 11. m:»ko SS hundreds, a;id It), make 9S Imndreds. eciunl to 8 hundreds and 9 thousands; 7 thou.^ands. iiiultipli d l)y 11, make 77 tliousands, and 9, make SO thou.sands, equal to li ihousandd and S tens of tliousands. Example 2.— Multiply 3 cwt. 2 qrs. 11 lbs. 7 oz. G drs. by 7. KxPLAVATiov.— 7 times 6 drams nro 42 drama, 0(iual to lu drams to set down, and "2 oz. to carry ; 7 times 7 oz. are 49 oz.. and 2 oz.. make '>[ oz.. equal to ^ oz. to sot down, anil '^ li>«. to carry; 7 times 11 Ibd. are 77 Ib.s., and 8 11)S., make 80 lbs., eq lal to 5 il)s. to Hot down, and :? qr.s. 25 1 5 3 10 to carry: 7 times 2 qi - are 14 qrs., and 8 qrs., make 17 qrs., equal to 1 q^". to .set down, and 4 cwt. to carry ; 7 Alo'aa 8 cwt. are 21 cwt., and 4 cwt., make 25 cwt. OPEnATIOK. cwt. qrs. lbs. oz. dr, 8 2 11 7 C 7 IP'i IV m !■' ■•'ii ight-hand or V intefi'ial EXEKCISE 12. Multiply By 0) 48960 6 (2) 75460 9 (3) 678000 8 (4) 57800 6 244800 Multiply By (8) 5-2736 2 (0) 8-7563 4 (7) 0-21375 6 (8) 0-0067 8 10-5472 Multiply By (9) $767-62 2 (10) $672-56 2 (11) $789-76 6 (12) $573-46 6 $1535-24 Multiply By (13) 866342 11 (14) 738579 12 (15) 4716375 11 (10) 8429763 12 17. Multiply .€32 8.^. 18. MultiDlv£4:j \u 6id. by 5. . 4W. by a.. A ns. Ans. £162 2s. £348 lis 8^d .2d \i li i i I i 't 90 MULTIFi^rCATION. [Sect. II. 19. Multiply £126 ISa. O^d. by 12. Ans. £1807 16a. 3d. 20. Multiply 10 cwt. 3 qrs. B lbs. by 8. Ans. 32 cwt. 1 (p-. 15 lbs. 21. Multiply 7 yda, 8 qra. 1 ua. by 7. Ans. 54 yds. 2 qra. 3 na. 22. Multiply 11 oz. 10 dwt. 19 gra. by 12. A71S. 11 lbs. 6 oz. 9 dwt. 12 gr. 48. When the multiplier is a composite number, and can be resolved into two or more factors, neither of which is greater than 12, wo deduce from (41) the folio .ving:— RULE. Multiply by each of the factors in snccension and the last product toill be the entire produce sought. Example 1. — Multiply 3 hiu 7 min. 14 sec. by 04. Explanation.— Multiplying 8 lirs. 7 mIn. 14 sec. by 8, wt' obtain 1 day brs. f)7 min. 52 stc, which wc npnln inultinly by 8, and obt;.in 8 days 7 hrs. 42 min. TiC pec, wliich is the product of 8 trs. 7 uilu. 14 B«c., by 8 tloies 8 or 64. OPKUATION. hrs. min. sec. x64=8x8 8 7 14 8 1 67 53 8 7 43 66 Ana. Example 2.— Multiply 796-437 by 132. Explanation.— We first multiply tho OPERATION. 796-487 X 132=11X18 11 8760-807: 12 =11 times multlpUv'and. given number by eleven, or, in other words, take it 11 times, and tlien take this re.siill 12 times, which Is evidently equivalent to taking the given number 12 times 11 or 18'J times. 106129684=:12 times 11 Umes multiplicand. Example 3.--Multiply 16 cwt. 8 qrs. 11 lbs. by 270. OPBRATION. cwt. qrs. lb. 16 8 11X270 8 50 2 8 9 455 23 10 Explanation.— 270=10 times 27 or 10x8x9. If, therefore, we take the given mnltiplicanil 3 times, and then this product 9 times, and then this second pro- duct 10 times, it is evident we shall have, in eflect taken the given multiplicand 3x9x10 or 270 limes. ' 4552 20 Exercise 13. 1. Multiply 1169-78 by 36. 2. Multiply $796342-3 by 121. 3. Multiply $33460 by 144. 4. Multiply 735 by 648. 6. Muftiply £3 7s. 6d. by 18. Ana. $011208. Ans. 96857418-::. Ans. $4818240. A71S. 476280. Ans. £60 15s. Od. ike itravaiirti-j multiply tho other words, :e this result 'quivftlcnt to Ilea 11 ur 13'J Arts. 48,49.J MULTIPLICATION. 01 6. Multiply £6 Ms 6Ad. by 22. Ans. £125 I9fl. lid. 7. Multiply £;i 49. 7d."by810. Ans. £2616 Vli i. 6d. 8. Multiply 11 "Wt. 3 qrs. 14 lb. 7 oz. by 64. A71S. 642 cwt. 1 qr. 4 lbs. ] 10 oz. 9. Multiply 2G bush. 3 pks. 1 gal. 1 qt. 1 pt. by 4U. Ans. I;il9bu3h. pks. 1 gal. 1 qt. Ipt. 10. Multiply 2 yds. 1 2 qrs. 2 iia. 2 in. by 6'S. Am. 168 yds. 3 qrs. 2 na. Oin. 11. Multiply 5 days 171U-S. 33niin. Usee, by 28ti . A71S. 1650 days, 15 hrs. 16 mill. 48 sec. 1 49. When the mulfipl'K'aiid is a denominate number und the multiplier is greater tlian 12, but noi, a eomposite i.cunber, we proceed according to the following : — RULE. Take the nearest composite Humber to the given tmiltipliei', mxil- infill nucctsnivelji by ts factors^ and add to or subtract from the jiioduct so mani/ times the multiplicand as the assumed composite number is less or greater than the given multiplier. Example 1.— Multiply £62 12s. 6d. by 76. Explanation.— We take '16=9 x / 8 + 4, and thus we pet 72 times tho inultipllc.ind, luul to it adding 4 times tho inultiplicftud, obtain tho desired product, viz., TU times the multipU- cund. 72 times multiplicand. 4 times multiplicand. 76 times multiplicand. Instead of multiplying as above, we might have multiplied by 7 and \^ and increased tho result oy times the multiplicand, or wo misrht have multiplied by » and 11, and decreased tho result by once the multiplicand, Ac. Example 2.— Multiply 17 lbs. 3 oz. 7 dr. 2 .scr. 16grs. by 789. OPERATION. lbs. or. dr. scr. grs. 17 3 7 2 6x9= 9 times multiplicand. 10 OPKKATIOM. £ ». d. C2 12 6 8 W\ 9 4.*^9 10 £1759 10 178 8 1783 X 8 = SO times multiplicand. JO 7 12132 10 1 1 = 700 times multiplicand. 1386 7 2 2 = 80 times multiplicand, 155 11 7 1 4 = 9 times multiplicand. 18675 5 8 1 4 = 789 times luultiplicand. -:'l' ]\ IMAGE EVALUATION TEST TARGET (MT-3) .• V^ M? ^ A 1.0 :f I.I I'll 1.25 2.5 •^ I4£ ill 2.0 JL4 IIIIII.6 V] m. "c^l / Photographic Sciences Corporation s. iiF ;{V #f^^ \ \ %^ ^ 6^ k ^9)'- 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 r^^ o ■it; jWj ™ ^0 "^SslC d2 MtTLTIPLICATIO^r. [SacT. n. ExPLANATifvV. -Wc divifle the given multiplier into 700 + S0 + 9, and obtain the 8 partial products, which we add to ..her, for tho entire product. Example 3.— Multiply 3 wks. . vs 17hrs. 21 inin. 12 sec. In- 4*736. OPERATIC. wks. ds. h. mil sec. wkij. ds. h, uiiP "(ec. 3 6 17 21 12x6= 23 5 8 . '^ = C timrs mnltiplioiui' 10 118 5 16 80 0= 80 times multiplicfiiK' 0x7= 2772 2 8 20 0= 700 times mtiUiplicnnr. 2" 20 0x4 = ir)841_5 5 20 = 4000 times multiplican.l Ans. 18756~4 9~23~^12 = 4736 times niiiltiplicaiKi. Example 4.^3Iuitiply £47 IGs. 2d. by 5783. 5783 = 5x1000 + 7x100 + 8x10 + 3. 89 4 5 3? 0x3 10 396 7 20 0x7 10 .€ s. d. 47 10 2x8 = 10 OPERATIOX. £ s. d. 14;3 8 0= product by units of the naultlpllcand. 47S 1 8x8= 3S2t 13 4 = product by tens of the multiplicnnd. 10 4780 10 8x7= 33465 10 8 = product by hundreds of the multiplicand. 10 47808 6 8x5 = 239041J3 4 = product by thousands of the multiplicand, Ans. 270475 11 10 = product by entire multiplier. Exercise 14. Avfi. £1005 13?. fiil Ans. 902040 2s. 54 i. 1. Multiply £12 2s. 4(1. by 83. 2. Multiply £903 Os. Ofd. by 909. 3. Multiply £3 6s. 5|a. by 3178. Ans. £10556 18s. 4^d. 4. Multiply 16 bush. 3 pks. 1 gal. by 078. Ans. 11441 bush. 1 pk. Ogal. B. Multiply 23 m. 6 fur. 33 rds. 4 yds. by 247. Ans. 5892 m. 2 fur. 10 rds. Sil^ yds. 6. Multiply 3S. 16° 30' 45" by 721. Ans. 2559 S. 25° 30'45". 50. It may be proper here to caution the pupil ajrniiist the absurd attempt to multiply one dciiomiuate number by anotlier. Multiplication Is merely a t)articuhir kind of aidition. and when we are required to multiply a quanUty )y any nnmber, we are simply required to repeat it as many tinics as it".'r( are units in tiie multiplier. It is evident, then, that to talk of muUifiU im: £19 19s. lljd.. by £19 lOs. llfd., or. iu other M-on!s, of adding or niK a'tict; £19 19s. llfd. £19 19s, lljd. times, is simply ridieulous. Nevertheless, p-icnr pains have bee, 1 taken to show that 2s. Od. maybe muiiiplied by '2s Od.. and that the product will be either 3i-il. or Os. 3d. !! Und(nible(lly. 2s (id. ean be ♦taken 2J titnes, ami the result will bo C-*. 8d. ; or it can be taken one-eit.iith of a time, and the result will be 3Jd. ; but this is a very dilfererit tliinrrfruin taking it 2s Cd. times. la fact it is quite a» Donsenslcal to talk of taking/ Ann. 49-51 2.S. Od. 2.S, 6< times; or. Lion, which tie uriltiplii fully sliOA'u 51. Li OPEKATION. 729 478 5832 5103 2910 348462 hundreds, i.' cohnnn. aii( ducts tjjget! Ilene the mult number, Multi f rately, beg separate li /ij/ure by products t( EXAMF OPERATION. 7423 6709 66S07 519610 44538 49800907 Mul By 6. Mult Y. Mult 8. Mult 0. Multi [Sbct. II. f9, and obtain uct. . 12 sec. Ijv imulliplioiui' irmiltiplicfiiic! ;iniiUiiilic:in(". ! riiulti!)lifiinil 1 nmltiidicaiKi tlpllcand. iplicnnd. multiplicand, multiplicand. 005 ISp. 8(1. 1040 '2s. 5|i. »56 18s. 4^.,1. 1 pk. Ogal. ) nls. 3| yds. 25° 30' 45". bsnrd attempt pti is merely a ply a quiinlity ini( s as tl".n nf niulli[il\ ii!i: ; or ri'p(iilii:t; rtlicU'ss. jri'';i' by 2s, 0(1., iiii'l 2> M. ciui 1)1' ion onc-c'ii.lith lit tliinp: IVoin talk of takii);/ AhXb. 49-51.] MULTIPLICATION. 93 2s. fid. 2.S. 6d. times as it would bo to t;ilk of taking 6 lbs. of bot'f (> lbs. of beef times; or. 7 bars of music 7 b:ir-. of music times, e rociuired to multiply 729 by 478. OPEKATION. 729 4T8 5832 5103 2916 348462 ExPLANAVioN. — From the preceding examples it is evident that when units are iiuiltifilied into a'.iy order whatever, th'- pvn- duct will always be of that order. Ih re, then, we first multii)ly by the 8 units, a.s iii (47). Ne.\t wo multiply by the 7 teii.s. thus: — 9 units, multiplied by 7 tens, jjive 0-] tens, equal to -i ten.s, which wo set down in the column of ten.<, and <; liundred.s, whicli we carry ; 2 tens, multiplied by 7 tens, ^dve 14 hundreds, and 6 hundreds which we carried, make twenty hundreds, iqual to hundreds to set down and 'i thousands to carry, &g. Ne.xt wo multiply by the 4 hundre(ls as follows: — 9 units multiplied by 4 hundreds, i.'ive -Jl! hundreds, equal to si.\ hundreds to set down in tlie hundreds cohnnn, and 8 thousands to c.irry, etc. Lastly, we add the several partial pro- ducts t^i^ether. Iletiee, when the multiplicand is an abstract number, the multiplier being j^reater than 12 and not a composite number, we have the following: — RULE, Multiply the multiplicand by inch figure of the multif Aer sepa- rately^ begiuning with the lowest, and ivrite the partial products in separate lines, placing the first figurr of each line directly nnder the figure by which you multiply, and, lastly, ad/l the several partial products together. Example.— Multiply 7423 by 6709. OPBBATION. 74-23 6709 66807 519610 44538 49800907 Explanation. — Here, as there are no tens in the multif/^ltr» we may either proceed directly to the hundreds after multiplyinaf by the units, or we may set down a under the tesis, and then write the pnfduct by the hundreds in tho same line, always re- membering to place the first digit of the partial product under the figure by which we are multiplying in order that all the digits of the same order may come in the same vertical column. EXEUCI [SE 15. 0) (2) (8) (4) c6) Multiply 325 765 732 997 Mi By 95 765 456 345 347 6. Multiply 7071 by 556. 7. Multiply 15607 by 3094. 8. Multiply 39948123 by 6007. 0. Multiply 27*78588 by 0867. Ans. 3931476 Ans. 48288058 Ans. 239968374861 ^•'j^- 27416327796. V'' 1 u 94 MULTIPLICATION. i.8bO|. y1. 62. Let it be required to multiply 63-5 by 97. orERATioN. Explanation.— Since (51) any order, m'-.itiplied by units, will 08-5 give tl:at order — fonths, multiplied by units, will ^ive tentlis. •97 Hence It is obvious tliat tenths, iuiiltii)iied by tenths will pivc tlu> next lower order, or hundredths; and also that tenths, uiulti])!!* li 4 44ft by Jiiindrcdtlis, will ^,nve the next lower order afrain. or 11. mu- fttlo san.itlis. In the above exanii-Ie, therefore, we proceed thus.— Ti tenths, multiplied by 7 liundredths, frive 3f) Ihout^andths, equal to Gl-595 5 thousandths to set down and 8 hundredths to carry; 3 unit multiplied by 7 hundredths, give 21 hundredth.s, and three hiin drcdths wc carried, make 24 hundredths, equal to 4 hundredths to set down and 2 tenths to curry; 6 tens, multiplied by 7 liundredths, give 42 tenths, and '.' tenths we carried, make 44 tentlis, equal to 4 tenths and 4 units. Again, .'. teiitlis, multiplied by 9 tenth.*, give 45 hundredths, equal to 5 LundredtLs to set down and 4 tenths to carry, &c. 63. Strictly speaking, all examples in multiplieation of decimals should be worked according to the aLovo method. An attentive consideration of the reasonings in (52) will, however, show that the lowest digit of the pro- duct of any two numbers containing decimals, must al- ways be a number of places to the right of the decimal point, equal to the sum of the decimal places, in both multiplicand and multiplier. Hence, when the multiplicand or multiplier, or both, contain decimals, we deduce the following — RULE. Multiply as though there were no decimals, and then remove the decimal point in the product as many places to the left as there are decimals in both the multiplicand and the multiplier. Example 1.— Multiply 5-63 by 0-00005. OPEBATioN. Explanation. — We multiply 563 by 5, and remove the dec!- 563 mal point seven places to the left, since there are ^ve decimal 5 places in the multipner and tuo in the multiplicand, that is, we nave taken a number a hundred times too great a hundred 2815 thousand times too often, and the product 2815 is therefore ten An6. •0002815 million times too great, and to make it what it should be, we divide it by ten millions ; or, in other words, remove the deci- mal point seven places to the left. Example 2.— Multiply ''•073 by 5-12. OPERATION. 2-073 6-12 4146 2078 10865 10-6' 87e Explanation. — We multiply as though both were whole nunr.« bers, and cut ofF^tJ« decimals, since there are three in the multi- plicand and two in tt 3 multiplier. Ai'.rs. 52-v'>4.1 TakJn" ♦i^p nl AiiTS. 52-54.1 MULTIPLICATION. 95 Exercise 16. Multiply By •003296 6-182 Prndiu-t •0190574'72 4. Multiply 8-2517 by -023. 5. Mulriplv GrOOl by 340. r,. Multiply 482000- bv -37. 7. Multiply 8782-4 by -00917. 8. Multiiav 87-96 by 220. (2) 41-78 •0629 2-627962 («) 86-1284 2-0006 Ans. -0747891. Alls. 2i76o-;;4. Ans. 178340. Ans. 34-684608. Ans. 19c51-a PROOF OF MULTIPLICATION. 54. If the mnUiplier is not greater than 12, multiply the multi- pUciinrl bi/ the multiplier., minus one., and add the multiplicand to the jn'odad. The siivi should, he the same as th^ product of the mul- (ijiUcand hy the whole multiplier. If the ninltiplier be greater than 12 and the multipli- caii I an abstract number : — f'lRST Method. — Multiply the multiplier hy the multiplicand^ find if the product thus obtained agree with the other, the work may hi' considered correct. This mcthtxl of proof depends upon the principle (40) that the product ot two nuinbera is the same whichever is taken as multiplier. Second Miothod. — Divide the product by one of the factors^ and if the qiiotient thus obtained is equal to the other factor, the work is "orrcct. Til is is simply reversing the operation, i. e., breaking up the product into its fiK.'tors. Third MethOi). — Divide the sum of the digits of the multiplicand h>/ 9 ajid sti, ..Mvn the remainder ; divide also the sum of the digits of (h^ multiplier by 9 and set doion the remainder ; multiply these two rcinninders together, divide the sum of the digits in their product by 9, Old 'f the remainder thus obtained is equal to the remainder obtained bi/ di. iding the sum of the digits in the product of the multiplicand and the multiplier by 9, the work is generally correct : if these two last remainders are different, it must be xorong. ExAMPL , 1. — Let the quantities multiplied be 9426 and 3786. Taking the ninos from 9426, we get 3 as remainder. And from 3785, we get 6. 47130 75408 659S2 28273 8 X 6 =: IS), from which 9 b«log tak*")* A are left- Takin" ♦>^o nines ft-om 85677410, 6 8r«l«lt i' ^i I U I 06 MULriPLICATlON. [Sect. 11. The remaindors hein;-' equal, wo are to prespme the multiplication Is ror- roct. The .same result, liowivcT, woulil have been obtained even if we liu'l .'.iM»lace occur. Example 2. — Lot the numbers be 76542 and 8436. Takini: tlie nines from 70542, the remainder is 6. Takinj,' them from 8106, it is 3. 4r)!i252 229626 0x3 = 18, the remainder from which is 0. 300168 612336 Takiny the nines from 645708312 also, the remainder is 0. The remainders being the same, the multiplication may be considerc/^ correct. Note. — This proof applies, whatever may be the position of the decimal point in either of the given numbers. Example 3. — Let the numbers be 4*63 and 5-4. From 4'03, the remainder is 4. From 5*4, it Is 0. 1S52 4x0 = 0, from which the remainder is 0. 2315 From 25 002 the remainder is 0. 55. The principle on which this process depends is, that if anj number is divided by 9, and the sum of its digits also be divided b) 9, the remainders are, in both cases, the same. Thus taking the number 7S25, wo have : l£2.i _ 7 000+800 + 20 + 5 _ Ifl Oil _|_ &ilfl ^ ^ _j_ i = 7 X -wo-a ■+ 8 X i^s + 2 X J^ + I = 7 X (111 + 1) + 8 X (11 + J) + 2 X (1 + -J) + I = 777 + 5+88+1 + 2+1 + 1 = 777 + 88 + 2 + -S- + I + f + f = 777 + 88 + 2 + ■' + »-^^ + » 9 Hence the remainder arising from the division of 7825 by 9 is evidently the same as that arising from dividing 7+8+2+5 or 22, Pi'hich is the sum of its digits, by 9. 56. Casting the nines from the factors, multiplying the resulting rem;iindei8, and casting the nines from the product, will leave tlie Bame remainder as if the nines were cast from the product of the factors — provided the multiplication has been correctly performed. Thus, let the factors be 573 and 484. Casting the nines from 5 + 7+3 (which we have just seen is the same as CRSting flie nines from 573), we obtain 6 us feviainder. Casting the nines from 4+6 + 4, we get 5 as remnhufer. Multiplying 6 and 5 we obtain 80 as product, which, whcu the nines iivc taKen 8Wft/, Will giv*? 3 fts j^ remainder. Aiits. 65-6T. We can the product • taking, in su 573 X 401 =(5x U) ri 5x =(5 X 09 II <09 5 X 99 es plied by all cast out ; an .lines; it wi first brackc^t There will t to be cast oi cast from th remainder. 67. ] AJJix a Reaso Tote AJix a Reaso III. Toi Affix t Reaso IV. To 1 Affix t Reaso V. To in Affix fourth of Reasc VI. To : Affix t Reasc VII. To Affix i Reasf VIII. T of [Sect. U. lication Is cot- k'en if we hail rs which had ,f are likely u> Q which is 0. be considcrci** of the decimal is a 5, that if anv e divided b) - -I) + « 7825 by 9 is t-2+5 or 22, the resulting vill leave the ;'oduct of the )erformed. I is the same as the nines t'voin 80 as product, sr. Aute. 65-67.] MULTIPLICATIOlf. 97 We can show that 3 will bo the remainder, also, if we cast the nines from the product of the factors:— whicli is effected by setting down this product, ind takinjf, in succession, quantities that are equal to it — as follows:— 573 x4Gt=(tho product of the factors). =(5x 100 + 7 X 10 + 3) X (4x100 + 0x10-^4) = j 5x (99 + l) + 7x(9 + l) + 8 {• X |4x(99 + l) + 6x(9xl) + 4 j. =(5 xO0 + 5 + 7x 9 + 7 + 3) X (4x99 + 4 + 6x9 + 6 + 4.) 5x99 expresses a number of nines it will continue to do so when multi- plied I'y all tlie quantities withiu the second braclvets, and is, therefore, to be cast out; and. for a similar reason, 7x9. Aanin 4x99 expresses a number of .lines; it will continue to do so when inultipied bv the quantities wiihin the first brackets, and is. therefore, to be cast (uit; and for a .similar reason, 6x9. Tliore will then be left only (5 + 7 +3) x (4 + 6 + 4)— from which the nines are '".11 to 1)0 cast out, the remaiiuiern to be multiplied tosether, and the nines to oe cast from their product;— but we have done all this already, and obtained 3 as remainder. CONTRACTIONS IN MULTIPLICATION. 67. I. To multiply by 5 : AJfiJi: a to the multiplicand and divide the remit hy 2. Reason 6=.^ II. To multiply by 15 : Affix a to the multiplicand and to the result add half of itself. Reason _5=:104-Y- III. To multiply by 25 : Affix two Os to the multiplicand and divide the result by 4. Reason 25 = ^10. IV. To multiply by 125: Affix three Os to the multiplicand and divide the remit by 8. Reason 125=:J-V-*^. V. To multiply by 75 : Affix two Os to the multiplicand and from the result take one- fourth of itself. Reason 75 = 100-J-^-o. VI. To multiply by 175 : Affix two Os — multiply the result by 7 and divide by 4. Reason VJ^ = ^K VII. To multiply by 275 : Affix two Os — multiply the result by 11 and divide by 4. Reason 275 = ij_o_a. VIII. To multiply by 13, 14, 15, &c.; or by 1 with either of the other digits affixed to it : &. ^1 'I'l I I' fl i 98 MULTIPLICATION. ibBCT. 11 Example. 21)25 X 13 C975 Midt'tp'}/ b>i the links' fnnrc. o-f flic mvltiplkr^ and wi'Ue each Jifjure of ihc. ]n riiol firoduct one placp. to the rii/fit of that from uhith it arhe/f ; — frunlhi, add the jiurtial pindoct to the mvllipli- Ani^. ;5<»i!'J5 cand, and the result uill be (he anmcr rcf/ttlrtd. 1IEA80N— ■ '^ is tht> sntnn in cffi ct as If we uotiiiilly iiinltiiilied by tlio coniinuM tnctliou. Wo uureiy uiaJco the multiiilicand serve for the secoiid purliiil product. IX. To multiply by 21, 31, 41, etc., or by 1 with either of the other signiticant figures prefixed to it : Example. Mvltiply b>/ the tens' f (jure of the mvltiplicr, 3tt5 X 21 cold write the first fumre of the partial product in ^SO the tens'* place ; fnullii, add this partial product to " the imdfiplicaiid, and the result will be the answer Ans. 7G05 required. Rs' SOX.— The ronson (if this method of eontrnction is eubstantially tho eatne as that of tlie pri^ccdiiiu, X. To multiply by 10 , 102, 103, 104, &c., or by 10 with either of the other dig.'s affixed to it : Multiply by the nnits\fif}urc of the midtipLer and urite the partial product^ thus obtained, two places to the rir/ht of the multiplica^id : finally, add the partial product to the nvaliiplicand. Reason.— Substantially the same as No. 8. XI. To multiply by any number of nines: Remove the decimal point of the midtirdicand, so many places to the right {by affixing (I's if necessary) as there are nines in the muU tiplier ; and subtract the multiplicand from the result. Example 1.— Multiply 7347 by 999. 7847 X 999=73t70no-7S47=7.^396o3. We. in such a case, inertdy multiply hy tho next liigher convenient com- fiosite number, and subtract the muUii)licand as many times aa ■«'€ have taken t too often ; thu-s'in tho example just given — 7347 X 999=7:547 x (10'oO-l)=7847000-7847=783965a Example 2.— Multiply 678943 by 999999. 678943 X 1 onooon=:ti7S9430ornoo 67894:3x1=; 678943 67S943 X 999999=:67&94232105T Example 3.— Multiply 78-9645 by 99993. 78-9fi45 X 100000=78964.50 78-9645 X 7 = 552-7515 78-9645 X 99993 =7895897-2485 XIL When it is not necessary to have as many decimal places in the product, as are in both multiplicand and mul- tiplier— A.RT. 6T.j Rever of that d required j Multi noviinafi been obtal — utnfy, tiiten 15 Let different column. Add the requi Exam mal3 iu tl t 9 in th decimal pli rcjuireil it In rnn dfional di number m f.'iont pro( duct. In h5 8 (the s ihuniniitioi denuminat tion of the i.jClitof th eiit instani mals in tl the multi] constitute; multi plica 5, and less Oor20; a 'it would £ pdse we 1 aiM 2 thai or 20 ; &c On ini the differ* Bired, we duct, and Art. 5it.l MULTIPLICATION. 99 Rp.verae the multiplier^ jtuttinrj its units* place under the. place of that denomination in the multiplicand, which is the lowest of ths required product, Midtiply btj each diffil of the multiplier begin ninfj with the de- nominntiini over it in the multiplicand; but addincf ichat would have been obtai)ied, on muitipli/ing the preceding digit of the mulliplicand — unit I/, if the number obtained would be between 5 and 15 ; 2, ?/ be- tiuen 15 and 25 ; 3, if betu-een 25 and 3u, dc. Let the lowest denominations of the products, arising from the different digits of the multiplicand^ Mind in the same vertical column. Add up all the products for the total product ; from which cut off the required number of decimal places. Example 1. — Multiply 5G784 by 9*7324, so as to have four deci- mals iu the product. Bbort method. Ordinary tnethod 567S4 66784 42379 97324 611056 22 7136 897-19 113 568 1703 1708 52 113 89T48 8 22 611056 65-2643 65-2644 6016 9 In the multiplier expn?saes units; it ia therefore put under the fourth decimal place of the mulliplicand— that being the place of the lowest decimal rojuirefi in tlie product. In rnultiplyiiig by each succeodins; digit of the multiplier we neslect an ad- di\ional diirit of the multiplicand; because, as the multiplier 3) a result in iha fifth place to tlie i.jCht of the decimal point; which is a denomination supposed to be, in the pres- ent instance, too inconsiderable for notice — since we are to have only./bttr deci- mals in the product. But we nud unity for every ten that would arise, from the multiplication of an additional diaitof the niultiplicand ; since every ^f'Ti constitutes one iu the lowest denomination of the requireii product. When the multiplication of an additional digit of the multiplicand would give wore than 5, anli(M-? (;{.j) 4. Wliat is tlio |,i-()(lucty (8(5) 5. Wiiy lire tlie miiUiplior a.^d mult plica -d callnd tlio lactors of tbo product? (■•ii"') f^. Wliat is a primf number? (:37) 7. What is !v (^<)iii|)(»sito nmiil)i'r? (-33) 8. If tlie iiiiilii|>lit'r 1)« gicater ibaii u ity, hov7 will tho product compare with the multi|>licrii. ly (:V.)) 9. If the multiplier b ■ equal to unity, how wil the product compare with the m iltiplicand? (-'50) 10. If tlie multi|ilier be less than unity, how will tho product compare with the multiplicaul? (IJ!)) 11. Slioiv tliat either of the factors may be used as multiplier without altering the value of the product. .40) I'i. Show that wiien tho m'iitii)lier is a composite number we may obtain the entire product by multiplying by eaeii of the fa(!t(»rs i i succession. (41) 13. Bv whom was the TniiltiplicaLiou table calculated? (42) 14. How was it calculated ? (42) 1 ■. What is the sign of multiplication? (43) 16. How do we multiply a quantity consisting of several factors connected by tho siirn «)f f.iultiplieation? (44) 17. How do we multiply a quantity consisting of several terms, connected by the siifns + aud — enclosed within a bracket? (45) 15. "Whatismeant by (7 + :i-2 + 5) x (0 + 8-7):' (45) 19. Why do we begin multii)lyin'^ a number at the riirht-hand side? (-10) 20. What is tho rule for multiplication when tho multiplier is not greater than 12? (47) 21. What is the rule when the multiplier is a composite number, none of its fiieiors beinsx greater than 12? (4S) 22. Wliat s the rule when the multiitlicand is a denominate number, and the muliiplier greater thau 12, hut not a composite number? (49) 23. Show tile absurdity of attempting to multiply one douominate number by another. (50) greater 24. 25. When the multiplieand or multiplier, or both, contain decimals, what is the than 12, but not a composito number, what is the rule '' (51) rule? (.^i:3) 26. Give the reason of this rule (52 and 53) ^7, Uow do wti ^rove multij;jlicaUuu wheu the multiplier is less than 12? (54) mi nn lOif DIVISION. tStCT. II. ARTS. 5^-''J 29, »() 81 28. How (lowo prove mnUlpllpfttlon A^hcn the miiltlpUcnntl is an nb>*tract num- ber iin'l tfio iiiulli|ilit'r l.s >friiiU-r thiiii l-'.- (.Vj) ITjioii what (Iocs till' proof by cuatlng out tlio nines rtppend? (.V)) I'rovf tills prliiclpK'. (.tr*) I'rovp I lull fiisii tr ilif iiliii's IVoui tlic f'attnrti. iimltiiilylns: tlio rcsnltinc vo- niaiinliTs, mill ciLstinc llu' niiu'.-t tioin ihc |)mili,tt, w'nl lf;i\(' tlu' s;ii:ic ro- nniiwiliT as if till- iiiiu-.s wiio cu.sl Imiii ilu- piodiicl ol' the (actur.s. (Ijij) 82. Wliut .siioit nicllKuls liave wo J'or niiiltip yliiu hv T), ti.') and l2o1 (57) 83. Wliat hlioit nu'liuMls of miiiliplyliitr by i:> aiui VtY (57) 84. llow may wo iiiultli)lv i)y lTr>y How by 27.')? (f>7) 85. How may \\v, muliii.ly by 13, 14, If), &c. ? How by 101, 102, 108, &c.J (57) 80. How may wo multiply l)y 21,81,41, ^Vc.? (<)1) 87. liow may wo muiliply by any minibcr of nines? (57) 83. ilow may wo contract th\) woik wlien wo require only a limited number of dccimula? (57) DIVISION. 68. Division is the process of finding how many times one number is contained in another. 58. Tlie number by which we divide is called the divisor. 60. The number to be divided is called the dividend. 61. The number obtained by division, that is, the number which shows //ow many times the divisor is con- tained in the dividend is called the quotient (Lat. quotics, "how many times.") 62. If the divisor be less than the dividend, the quo- tient will be greater than unity. If the divisor be equal to the dividend, the quotient will be equal to unity. If the divisor be grc ater than the dividend, the quotient will be less than unity. 63. It is sometimes found that tha dividend does not con- tain the divisor an exact number ot times ; in such cases the quantity left after the division is (ailed the remainder. The remainder, being a part of the dividend, is, of course, of the same denomination. The remainder must 1 e less than the divisor — otherwise the divisor would be contained once more in the dividend. 64. Division is merely a short mdhodof perfbrniinga particular kind of subtraction (Art. 6, Sec. II.) The divi- dend correspond.^ to the minuend, the divisor to tliO subtralul qv'tfif'nf] siiiec it Bubtra'.:tj It will I If wo oonsii giibtriiotin- tlio siiino tl| 65. dcnd col quotient j divisor \ maludei equal td 66. one y times led the ndend. is, the is con- le qiio- uotient tiotieut lotcon- ses the is, of 3rM ise id t rid. iing a divi- -> the AKT8. 5S-69.J DIVISION. 103 subtrahend, and the remainder to the difToronoe. Tlio q}f')(irni has no corresponding^ quantity in subtraction — siuce it siinnlv t(dl.s how many times the divisor can bo subtr.vjte I tVo.n tho dividend. It will holp IIS to uiulorstaiiil lutw crontly division iiM)rcvi;itos siibtmotion, If Wi! consider liow loiiu' u pi'ot't'.-s woiilil hii nquirt'd to (JiMovcr— by iiotiiullvr Biilirr.ictiii.' it— liow ot'tfii 7 is cotitaiiicil in H.^tl -l'.).")V_M, \\h\li. us wo fihuU flmi, tlic sjiiuo thiiiy; can be cIlectiMl by ifiriaion lit less tliuii u luiimte. 65. Since the quotient shows how many times the divi- dend contains tlie divisor, it f'oHows that the divisor and quotient aio the /(triors of the dividend. Hence if tho divisor and quotient be multipbed toi.;ether, and the re- mainder, if any, added to tlie product, the result will bo equal to tlie dividend. 66. We have three ways of expressing the division of one quantity by another : — 1st. By the sign ■—■ written between them; thus, 15 -f- 3=5. 2nd. By the sign : written between them ; thus, 15 : 3=5. 3rd. By writing the dividend above and the divisor be- low a horizontal line; thus, ■i.J'- = 5. Two quiintitles written thus ^'r conitituto what is called a fraction, and tho expression is ru ul .sLi'-eJeuerit/ifi. it is usual and proper to write the remainder obtained in division, in the form of a fraction; thus 17+''> irives 5 as .i rioUeiit and 2 as a remainder. Now the remainder, 2. is written above the line, and divisor .'} below tlie line; the whole (i;i(»iient beinir expressed thus .'ij (read live and two-thi^d^); tho mt.aniiig of svhieh is, that 3 is contained in 17, 5 tunes and i of a time. 67. When a quantity consisting of several terms con- nected by the sign of multiplication is to be divided, divi- ding any one of the factors will be the same as dividing the pi'oduct; thus 5 X 10X^5 -f-5 = |X 10X25, for each is equal to 250. 68. When a quantity consisting of several terms con- nected by the signs + and — , contained within brnckets, is to be divided, it is necessary, on removing the bi ackets, to put the divisor under each of the terms of the quantity; 6+3-7+9 3 7 9 thus (6 + 3-7H-9)-^3, or =- + 4-; for 3 3 8 3 we do not divide the wh(de unless we divide a/l its parts. 69- It will be seen from (68J that the horizontal line ' $1 ■11 .i.il 'M ... ii ! i ^m 104 DIVISION. ISSOT. II. ABT8. 70-7i which separates the dividend from the divisor assumes the place of a pair of brackets when the dividend consists of several terms ; and, therefore, when the quantity to he divided is subtractive, it will sometimes be necessary to change the signs, as already directed (26) ; thus : 6 13—3 6 + 13—3 2V 15—6 + 9 27—15 + 6—9 - + = ; but = 2 2 2 3 3 8 Example 1. Let it be required to divide 798 by 3. OPEBATION. 8)T98 Explanation.— Place the divisor a little to the left of the divl" (lend and separate them by a short curve line. Also draw fi fitruis^ht line beneath tne dividend. 793 700 + 90 + 8 600 + 190 + 8 Now — = 8 600 + 180 + 18 600 180 18 = — + — +— = -200 8 8 8 3 8 8 + 60 + 6r=266(See68). Iii&tead of going through this long operation it is evident that, we may f»roceed as follows ;"8 units ii to 7 hundreds will go 2 (hundreds) times anil eave a remainder 1, which being of the order of hui dreds, is equal to 10 tens; 1<^ tens and 9 tens make 19 tens, and 3 into 19 goes 6 (tins) times and K-ax e^ a remain.ler 1, which, being of the order of teis is equal to 10 units: lit nnits and 8 units make 18 uiiits, and 8 units into 18 units goes 6 (units i times. Example 2. Let it be required to divide 917 lb. 13 oz. 12 tii* by 4. OFBBATION. EXPLANATION. — Placiig the dividend aid divisor as before, se proceed thus : 4 in 9, 2 (hundreds) timi-s and 1 over; 1 hiri- lb. oz. dr. dred, equal to 10 tens, and 1 ten muke 11 tei s , 4 in 11, 2 (teiif?) 4)917 13 12 times and 8 over; 3 L<^i!f, equal to 80 units, and 7 units make 87 units; 4 in 87, 9 times aid 1 over, which is 1 lb. because the &17 229 7 7 are pounds (63); lib., equal to 16oz. and 13oz. make 29 oz., 4 in 29, 7 times and 1 over, which is 1 oz.. since the 29 are oz. ; 1 oz, is equal to 16 drams and 12 drams make 28 drams ; 4 in 28, 7 times. Observe that any order divided by units gives that order in the quotient. Example 3. Let it be required to divide 9789 by 26. Explanation. — i .icing the dividerd ard divisor as be- fore, we say 2ii in 9 (thousands) no times; 26 in 97 (hun- dreds), 3 (hundreds) times. "We place tho 3 (hundreds) to the right of the dividend and mnltiidying tlie divisor 26 by it, get 78 hundred, which we subtract from the 97 hundred, and obtain a remainder 19 hundreds. 19 Ini: dreds are eqiuil to 190 tons, and 8 tens, make 198 tens ; 26 in 198. 7 (tens) times. Multiplying tho 20 by the 7 tens, we cet 182 te; s, which, tub- traded fr^ ml08 tens', leaves a remainder of 16 tc s. 10 (ens are equal to 160 ui its and 9 units make 169 units: 26 ii IGO, goes 6 times, and leaves a remainder 13. Tins VA should lie divided by 26, but since 13 does not eontiiin 26, the division cannot be effected, a' d we can only i dicnte it, which we do ty placing the 26 under the 13. as is explained in (Ait. 66). The complete quotient is therefore 876^| read 376 and thirteen-twenty-sixtlir or 376 and 13 divided bv 26. OPERATION. 26)9789(876 78 198 182 169 iri6 13 rem. Ana. 376ja 71. deduce, Bcrjiv to fhc Ion will conta of the qu lowest use Multi the produ order in the next dividend Froce heeyi divii Exam OPERATION 7)'JS765 14109? becomes 14 Thus 2 to 60 luuu thou.sandth thousandth Exam: OPERATION 12)124789 10399 r\ or 12)124789 10399-0 Exam OPF-KATK 9) jeiOaO 1 i;220 1 in 18, 2,1.^ 72. ] of the di £1986 1- that £22 bU siniik Noiw ABT8. TO-72.] DIVISION. 105 71. From tbo preceding illustration and example we deduce, for the division of numbers, tlie following general RULE. Bcfiinnimj with thf highest order of units in the dividend^ pass on to the lower orders v d the fewest number of figures be found that will contain the divisor ; divide th.ese figures by it, for the first figure of the quotient ; this figure idll be of the same order as that of the lowest used in the partial dividend. Multiply the divisor by the quotient figure sofoundj and subtract the product from the dividend, bnng careful to place units of the same order in the same vertical column. Reduce the remainder to units of the next lower order., and add in the units of that order found in the dividend: this will furnish a new dividend. Proceed in a similar manner until units of every order shall have been divided. Example 1.— Divide 98766 by 7. OPERATION. Explanation —Here we say 7 in 9, 1 and 2 over ; in 28, 4 7/Js7(i5 and over, In 7, 1 and over; in 6, times and 6 over ; in 65, 9 and 2 over Beneath this 2 we write the divisor 71 to indicate 14109? its division Wc may, however, carry on the division by con- sidering the 2 units reduced to tenths, «fec., and tlie quotient becomes 14109'2S57. Thus 2 units, equal to 20 tenths, 7 in 20, 2 and 6 over; 6 tenths are equal to 60 liundredths, 7 in GO, 8 times and 4 over; 4 hundredths are equal to 40 thousandths, 7 in 40, 5 and 5 over; 5 thousandths are equal to 50 tenths of tliousandlhs, «fec. Example 2.— Divide 124789 by 12. OPERATION. 12)124789 10399 r's or 12)124789 Explanation. — Here again we may either stop at the units and write the remainder 1 over the divisor 12, or we may redttoe the 1 unit to tenths, &c., as iu the second operation. 10399-083 + ExAMPLE 3.— Divide £1986 14s. 7id. by 9. OPEKATION. 9)i;i9aG 14 7i Explanation.-- 9 in 19, 2 and 1 over; 9 in 18, 2 and over; 9 in 6, and 6 over; jEG are cqnal to 120a. and 148. make 134s. ; 9 in 184, 14 and 8 over ; 8s. are equal to 96d. and 7d. malic 103d.; 9 in 103, 11 times and 4 over; 4d. are equal to 10 farthings and 2 farthines malie 13 farthings ; 9 i;220 14 Hi in IS, 2, i. e., one ninth of 18 farthings is 2 farthings, written thus id. 72. In example 3, we are, in reality, required to find one-ninth of the dividend. The obvious meaning is, not that 9 is contained in £198(3 14s. 7id. £220 14s. U^d. times., which would be nonsense, but that £220 14s. ll^d. is the ninth part of £1986 14s. 7^d. : so also in all similar questions. ^'oL\vithstaading this, all such examples are reducible to a spegiegf C^-ii J .i..''-_V I II 106 DIVISION. I.3ECT. l-k of subtraction. Thus, in the above example, we, for the momerit, consider the divisor 9 tn be of the same denomination as the dividend, and ascertain how many times £9 will go into (i. c., can be subtract! c" from) £198t>. Wc get, as a result, 220 times, and a remainder ot£G. Then we argue, from the principles already established, that since £9 is contained in £1980 220 times, with a remainder of £6; £220 is contained in £1986 9 times, with a remainder of £6 ; that is, that the ninth part of £1980 is £220, with a remainder of £0. Next reducing this £0 to shillings, and adding in the 14s., we obtain a total of UMs., and we find that 9s. is contained in 134s. 14 times, with a remainder of 8s., whence we conclude that 14s. is contained in 134s. 9 times, with a remainder of 8s., that is, that the ninth part of 134s. is 14s., with a remainder of 8s., or that the ninth part of £1080 14s. is £220 14s., with 8s. still undivided, &c. Example 4.— Divide 978904 by 3429. Explanation.— 8429 into 97S9 (the stnallost nunN bcr of fifriiros that will contain tlic «livisor) froos 2 times, we tiuTc'lbie put 2 in tlic quotient. ^MuJtiplyinfr 8'i!;l9 hy 2. we get (i'^.'JS, whieh we siiblruct iVoin li7h9; in d obtain as n'nui.iuicr 2931, whieh we reduee to tlie nesf, lower order (ten.-) and add in the 6 tens, oA'l[) into 2!(o (i goes 8 times. We therefore j)lace 8 in llie qiiotieit. Multiplying 3429 by 8 we m^t 274o2, vliieh we snbtrait from 29310, and obtain 1884 ns a remainder. lUdneii;); tu units fti d addintr in the 4, or what amounts to ti ») pame thinir, briri^ring down the 4 and writintr it after t>'o 1884 we ^et 18844 ^ and 3429 into 18844 goes 5 timi::., with a remainder 1699, under which we write the divisor 3429. 73. When the dividend is an absttact number, it is evident that bringing down the next ligure and writing it to the right of the ve- maindcr, is the same in elfect as reducing the remainder to the next lower denomination and adding in the units of that order found in the dividend. Thus, in the last example, bringing down the 6 and writing it directly to the right of the hist remainder, 2931, makes the next partial dividend 29316, which is the same as reducing the 2931 to the next lower ordei foimd in the dividend. OPERATION. 8429)978964(2S5-i«S CS58 29316 27432 1SS44 17145 1699 to the next lower order and adding to the result the 6 of that order ExAMPL'i:. 5.— Divide 6121284 by 642. OPERATION. 642)! •421284(^10002 G-J2 1284 l:;i;4 Explanation.— 642 goes once into 642, and leaves no remainder. Briniring down the next di;:it ot iho liividoiid gives no ciigit in tlic quotient, iti wliieli, thrre- fore. we piit a cipher alter the 1. Tiie next digit of ti:0 dividend, in tlie M'.me way. gives no digi; in the qi.o- tienr, in wliieli, oont-eqiienliy, we ))i:t another cipher, luul, for is'milar reaMHis, another in bringing down the tux! ; but the next digit mal} cipher J must be put in the Quotient, AKT3. 73-75 74. W decimal plac EXAMI' 75. W determinin figure of t the first tu' right ligur< from the p the first int quotient fit dividend, 1 diminished divisor, th< ,>-U-«j [Sect. /X he moment, he dividona, e subtrat'tt c iliider of £<). hat since 10 £0; £220 is t is, that tho L'Xt roduoiiig )talof 134s., a remainder 14s. 9 times, [34s. is 14s., 1086 14s. is :mallost mini- ) froc's 2 times, iltiplviiif.' 8i!;',9 .)in 1)789; iiiid CO to llio iicsf, i4-Jl) into 2!(y (I the quotic'it. ii we .subtiiii t er. II('d!icii;(; TOlintS to VtM lit: it after ion goes 5 tiuK;.'^ evident that it of tlie Te- ■ to the next er found in the 6 aud , makes the ng the 2931 i' that order 2, and leaves 9 = 12U :0Z. 719 645 74 16= = drams in oz. 444 74 1184 1116 =dram8. lb. oz. dr, 16 15 99.1 Explanation.— 129 in 279, 1. e., tho 129th p. qr. and we therefore place a in tho quarters' place of the quotient. We next reduce qrs. to lbs. by multiplying by 25 and adding;- in the 14 lbs. of the dividend. We thus obtain 2189 lbs., of which tho 129th part is 16 lb , with aii undivided remainder of 125 lbs. lieducing 125 lbs. to oz.. and ad- ding in the 9 oz., we obtain 2()0',) oz., of which the 129th jtart is 15 oz., with an undivided remaindir of 74 oz. Reducing the 7t oz. to drams, we obtain 1184 drams, of which the r29tb part is J' (h-aiiis. With an undivided remainder of 23 drams, under which we phice the divisor 129 to iwdicate its di- vision. Thus we find the total Juotient to be 2 cwt. qr. IG lb. 5 oz. 9x^8 drs. 23 remainder. 76. The general principles on which the operations in division depend are : — 1st. The quotient arising from the division of the whole dividend by the divisor, is equal to the sum of th.'. quo- tients arising from the division of the several parts of tLy dividend by the divisor. (68) 2nd. The divisor and quotient are the factors of the di- vidend. (65) 3r.l. The product of the divisor, by the enfie qnotien', is equal to the sum of the products of the divisor bv tii© several parts of the quotient, (45) several p;. (j^uotient, ^45) [Sect. II. A ma. Td-:8.) Division. 109 129. -129 in 279, i.e., f 279 cwt., is 2 indcr (if '21 cwt. reduce to qmir- nj? by 4 and lul- rs. The 12!)tli eqiinl to qr. place a in the f the quotient. qrs. to lbs. by ) and addinfi- in dividend. Wc Ib.s., of wliich 16 lb , with iui ider of 125 lb.-. to oz.. and ad- we obtain 2()ii!) 129th part is 15 ided remainder ng the l'^ oz. to 1184 drams, of :)art is 9 drams, 1 remainder of k'hich we place iiidicate its di- fliid the total wt. qr. 10 lb. erations in • the whole thi', quo- arts of ti.y s of the ('i- e qnofien'-, Bor b^ th© We ask how many timps the divisor la contained In a part of the dividend, and thus a part of the quotient is found; the product of the divisor by this pari i-i lak.-n fmm the dividend, showing how much of the latter remains un- divided ; then a part of the remaining dividend is taken and another part of the q.iotient i.s I'onnd, and the product of the divisor, by it, Is taken away from wliat before remained; and thus the operation proceeds till the icAo^e of the dividend is divided, or till the remainder is less than the dixnaor. 11, We begin at the left-hand side, because what re- mains of the higher denomination may still give a quo- tient in a lower ; and the question is, how often the divisor will go into the dividend — its different denominations be- ing taken in any convenient way. We cannot know how many of the higher we shall have to add to the lower de- nominations, unless we begin with the higher. PROOF OP DIVISION. 78. First Method. — Multiply the quotient by the divisor^ and to thf product add the remainder, if any ; the result should be equal to til '. dividend. (65) Example 8.— Divide £5681 13s. 4d. by TOO. £ 8. 700)56S1 18 5600 81 20 1633 1400 233 12 2300 2800 d. jE 4 (8 d. 4 PEOOP. £ s. d. 8 2 4 10 81 8 4 10 811 18 4 T 6681 18 4=je8 2s. 4d. x 700=dividend. Second Method. — Subtract the remainder, if any, from the divi- dend; divide f.he dividend, thus diminished, by the quotient; and if the result is equal to the given divisor, the work is right. This is merely doing the same work by a different method. Third Method. — Cast the nines out of the divisor nnd quotient, and multiply the remainders together ; add to their product the re- mainder, if any, after division, and cast the nines out of this sum ; the remainder thus obtained should he equal to the remainder obtained by casting the nines out of the dividend. Since the divisor and quotient answer to the multiplier and multiplicand, and the dividend to the product, it is evident that the principle of casting out the 9s. will apply to the proof of division as well as to that of multiplicatioo. ■is ^^' ' 5: til 110 DIVISION. [Sect. II FonRTii Method. — Add the remainder and the respective products of the dii'isor into each tjuotient fjure together ; and if the sum is equal to the dividend, the -work is right. This mode of proof depeiuls upon the principle that the whole of a QiMn- tity is equal to the sum of all ilt parts. Example 9.— Divide U1856 by 97. 97)147856(1524 97* 608 485* 285 194* 416 888* 28» 147856 Note.— The asterisks show the lines to be added. Exercise 18. (1) 12)876967 73080fif (5) $ Ct8. 9)6789-60 (2) 7)891023 12/289 (6) $ Ct3. 11)4298-76 $754-40 $390-79 1\ (3) 9)763457 84828f (1) £ 8. d. 4)19 6 4 4 16 7 (4) 8)65432-978 8179-12225 (8) wks. ds. hra. min. 9)69 4 19 30 7 6 4 50 9. 10. 11. 12. 13. 14. 16. 16. 17. 18. Divide Di-vide Divide Divide Divide Divide Divide Divide Divide Divide 19. Divide 20. Divide 21. Divide 798965 by 6423. £176 14s. 6d. by 12. 56789 by 741. 6785158 by 7894. £4728 16s. 2d. by 317. $07896-64 by 429. 970763 by 6. 71234 by 9. 977076 by 47600. 7289 lbs. 6 oz. 4 drs. 2 scr. Ans. 14 lbs. £157 16s. 7d. bv 487. 7867674 by 9712. 422 m. 3 fur. 38 rds. by 87. Ans. 124im- Ans. £14 14s. Hd. Ans. 76'fJt. Ans. 859Hif- £14 18s. 4-//7-d. .4ns. $228-19j|f. Ans. 161793-8333+. Ans. 7914f. A71S. 20ffeSt. 13 grs. by 498. 7 oz. 6 dr. scr. Ans. 6s, Ans. ■^^498 gr- ,. -^^- 411. 487. SlO-g^^VV 5jd. Ans. 11 m. 3 fur. 14 rds. [Sbct. IL ive products the sum is leaf a quan- '978 12226 (8) s. hrs. min. 19 30 4 60 14s. 6id. .ns. 76|lt. I8s. 4^,%-d. l228-19Hf. 93-8383+. ns. 7914f ioia.7. or 5|d. -AV SlO-gWs-- ■ur. 14 rds. 4ttl6. IjVh^.j' 79, If DIVISION. GEN'ERAL PRINCIPLES. Ill di' (1 a given dividend a certain nuTiber of vi.ne.-?, the same divisor will be contained mdonble that eliN i(ieti(i tii,i.tfi drt .rany fimes ; in three times that dividend thrice as jiiixny iimcs, &t. Kenijd, When the ui visor remains the same, multiplying the dividend by any nnmh<^r has the effect of multiplying the quotient by the same number. Thus 9-r-3-3 ; 9 X 2 or lS-r-:i=:0-0 x 2, 9 x 5 or 45-^3=15=3 x 5, &c. 80. if a given divisor is contnined in a given dividend a certain nuiB'ter of times, the s>ame divisor will be contained in half that dividend half as many ui.iea ; in one-third of that dividend one-third as many times, &e. IIoixCO, When the divisor remains the same, dividing the divi- dend by any number has the effect of dividing the quo- tient by the same number. Thus 43-T-3=16; -V-r^ or 24-j-3=S— V-; -V-J-3 or 6-5-3=2=^-, &o. 81. If a given divisor is contained in a given dividend a certain number of times, half that divisor will be contained in the same dividend t^oice as many times, one-third of that divisor thrice as many limes, &c. Hence, Wlien the dividend remains the same, dividing the divisor by any number has the effect of multiplying the ijaotient by that number. Thus 4S-f.6=^: ; 48-; •« or i8-r-3=16=8 x 2 ; 48-^-5 or 48-f-2=24=8 x 8, &c. 82. If a given divisor is contained in a given dividend a certain lEimiiber of times, iwice that divisor will be contained in the same dividend oply lialf as many times, three times that divisor only 07ve- ,third as many times, &c. Hence, When the dividend remains the same, multiplying the divisor by any number has the effect of dividing the quo- tient by the same number. Thus 48-5-2=24; 4S-f-twice 2 or 48-i-4=12=hfilf of 24. 4S-reight times 2 or 4S-i-lC=3=one-eighth of 24, &c. 83. If a given divisor is contained in a given dividend a certain number of times, twice that divisor is contained in twice that dividend the same number of times ; thrice that divisor in thrice that dividend the same number of times, &c. Hence, When the divisor and dividend are both multiplied by the same number, the quotient will remain unchanged. Thus 12-r-4=3 ; 24 or twice 12-^8 or twffco 4=3 ; 72 or thrice 24t-24 or thrice 8=3, &c. 84t If a given divisor is contained in a given dividend a certain <«i ■ iM« l' ri S. s'h3 nrti 11^ DiVlSIOlt. [Slot. It. nnmber of tiineB, half that divisov \z contained in half that dividend the same number of times ; one-third that divisor in one-third that divic end the same number of times, &c. Hence, When the divisor and dividend are both divided by the same number, the quotient will remain unchanged. Thus 48-r24=2 ; 24 or half of 48^J2 or half of 24=2, Ac. TO DIVIDE BY } COMPOSITE NUMBER. RULE. ^b*— Divide the dividend by one of the factors of the divisor ; (hen the resulting quotient by another factor ; and. so on till all the factors are used. The last quotient will be the answer. Multiply each remainder by all the preceding divisors and add their products to the first remainder, if any, for the true remainder. When the divisor is separated into oiily two factors, the rule for finding the true remainder maybe thus expressed— Multiply the last remainder by the first divisor, and to their product add the first remainder, if any ; the result will be the true remainder. Example.— Divide 118 lbs. by 12. 8)7i3 OPEEATION. lat remainder = 1 lb. 4)289-1 2Dd remaiDder=8x8 = 9 lb. «)39-8 8rd remainder=5 X 4 X 8=60 ib. "5-6 tnie remainder 70 lb. 70 lb. Ans. m That dividing by the factors of a number will gire the sanae quotient as dividing by the number itself, follows directly from Art. 84. In tho last example, dividing by 8 distributes the 718 lbs. into 229 parcels of 8 lbs. each, and leaves a remainder of 1 " ■. ; dividing next by 4 distributes the 289 parcels into 59 still larger parcels, each containing 4 of the smaller or 3 lb. parcels, and leaves a remainder 8, which is not 8 lbs. but 8 parcels, each of 8 lb. ; lastly, dividing the 59 by 6 distributes it into 9 large parcels of 72 lbs. each, and leaves a remainder 5, which is, of course, 5 of the 12 lb. parcels. Hence the reason of the rule for finding thS5797(210iVb tlmee. 840 179 170 87. In the above and all similar (questions we are required to find wliat fraction the divisor is of the dividend ; or, in other words, how often the divisor is contained in, or can be subtracted from, the divi- dend, and the quotient must necessarily be an abstract number. E^iAMPLE 2. — Divide 729 cwt. 3 qrs. 16 lb. by 3 qrs. 9 lb. % oz. qrs. lbs. 3 9 25 oz. 7 cwt. qrs. lbs. 729 3 16 4 84 16 2919 25 511 84 14611 5888 L861 oz. 72991 16 487946 72991 1361)1167856 oz. (Smm times. 10808 8705 8106 r)996 t>404 "692 III : 1 114 DIVISION, EXEKCISE 20. Divide £'8'.»r.s 18s. 7 Ad. by £401 12s. O^d. Am. Divide l(»-27 ni. 1 fur. i'mh. by 17 m. 5 fur. 27 rds. Divide t*17l Is. l(4d. by £57 On. 7id. Divide lb. [) o'l. 8 dwts. 12 ^rs. by 5 dwts. 9 grs. 5. Di\ *i!5('0 ucTCs 8 roods 80 ids. by Ul acres G rda. 2. 8. 4 LBeot. H. 1 SHI US?. ^lr».v. 58. A71S. 8. ^w«. 48(5. Am. 26. 88. ^\iic'n tlic dividend alone contains decimal places, tlie prec'cdiiig rides are sufficient ; but when the divisor contains dcMuiiiaLs, it becomes necessary to pre})are the (Quantities for the di\'ision according to the Ibllowiu^' — RULE. Remove the decimal point ax tnauji places to the rujht in loth the dividend and the divisor, as there are decimals in the divisor, and then proceed as in Art. 7 1 . This is simply luidtiplyiiig both dividend and divisor by the same number, and therefore (Art. 88) does not affect tlie (juotient. Thus removing the decimal pohit one place to the right, in both dividend and divisor, is equivalent to multij)lying each by 10 ; two places, the Bame as multiplying each by 100 ; three places, by lOOO, &c^ Example 1.— Divide 87-6 by -0009. MultiplyiiiR each by lOOUO, or, in other words, removing the decimal point four places to the rif,']it, tn oacli, (since there arc four decimals in the divisor,) gives 118 STiiJOO-r-^iiAUd tills (Ait. 88) must yivo the same quotient as 87*C-r'00U9, therefore 87-6-i-'0009=876000-r9=97388-33, &o. ExA-MPLE 2.— Divide -06 by 8-984. •06-5-8'934= 60-4- 8984. 8934)60-000(0'OOCT, &c. 58-604 6-8960 62538 1422 Removing the decimal point three places to the right, In each, we get 60-^8984, and we then proceed thus : 8984 into 60 (units), (unitn) times; st»t down with the decimal point after it ; 8934 into 600 (tenths), times ; into 6000 (hundredths) times , into 60000 (thousandths), 6 (thousandths) times, «&c Example 3.—- Prepare 93-004-7- -0000069 -for division. Ans. 93-004-=-*0000009=930040000-r-69. Exercise 21. 1. 43^-0006947-430000000^-6947. 2. 9378-92-^9-7891 = 93789200-*-97891. 8. 4-96723-f-23-934=:4967*22s-2o9S4. 4. •793-^ •49=79-3-7-49. 6. 6. 7. 8. 9. 10. 11. Artb. SS-94.] DIVISION. Hi 5. •OOl-rr)74-037 = l-4-674937. 6. lUvido 47-i 1000. Kkason.— This contraction is multiplying both the d'- ■ ' .i. u. ''visor by8. For tlu) irwe reirminder, therefore, we tout "ivide the r ui k jr, .* any, by 8. 93. To divide by 75, 175, 225, or 27o. Multiply tlie dividend by 4, and divide the product by 300, 700, 900, oj' 1100, as the case may be, Keason.— 75=2^e, 175=^1^, 6lc. For the true remainder, divide the re- mainder, if any thus found, by 4. 94. When there are many decimals in the dividend and but few are required in the quotient, we may abbre- viate the division by the following — RULE. Proceed as in Art. 71 till the decimal point is placed in the qtw- tient, and then cut (ff a digit to the right hand of the divisor^ at each neio digit of the quotient ; remembering to carry what would have been obtained by the multiplication of the digit neglected — unity if this multiplication would have produced more than 5 and less than 16 ; % if mor^ than 1 5 and less than 25, &c. I ; 116 DIVIBION. [I*WT. IL ExAMi'LE.— Divide 754-337386 by 61-347. Ordinary Method. «1847)7f)48)U-886(l'2 296 dlU47 14086 122rt» 1817 1226 690 562 88 86 Contrfictod Mothod. eia47)76in87'd»6(12-296 6184T 7- 4- ;{-8 9-4 89B 128 2-768 8-082 140S67- 122694- 18178- 12269- 4-6780 6904- 6621- 896- 368- Acoordinp as the denominations of the quotient become emal?, Iheir pro- ducts by the lower dcnonii nation of tlio divisor become inconsiderable, and may be nct;lected, and conHequontly, the portions of the dividend from Mlilch they would have been subtracted. What should have been carried from f lio mullipiicaUon of tlie dijjit nepiectcd— since it belongB to a higher denomlnatloa than whut ia uegleclcd— must still be rotuined. Exercise 22. 1. The Ontario, Simcoe, and Huron Railway is 98 miles in length, and cost $3300000, What was the cost per niUc ? 2. The Ilideau Canal ia 126 miles in length, and cost $3860000. What was the average cost per mile V 3. Thr- (M'Jtance of the earth from the sun is 95270400 inilca. How loii /(;u: • i^ take a cannon ball, going at the rate of 28800 miles per uu,^ , ach the sun ? 4. The national debt of France is 1145012096 dollars, and the number of inhabitants is 35781628. What is the amount of indebted- uess of each individual ? 6. The national debt of Great Britain is 8764112127 dollars, and the number of inhabitants is 27476271. What is the amount of in- debtedness of each individual ? 6. What is the ninth part of $972 ? 7. What is each man's part, if $972 be divided equally among 108 men? 8. Divide a legacy of $8526 equally between 294 persons. 9. Divide 340480 ounces of bread equally between 792 persons. 10. A cubic foot of distilled water weighs 1000 ounces. What will be the weight of one cubic inch ? 1 1 . How many Sabbath days' journeys (each 1155 yards) in the Jew- ish day's journey, which was equal to 33 miles and 2 furlongs English? 12. How many pounds of butter, 19 cents per lb., would purchase a cow, the price of which is $47'50 ? 13. Divide 978*634 by 96-34.762. imrt. It.] DIVISION. 117 14. Dirlflo 7'>9 bush. 1 pk. 1 gal. 1 qt. 1 pt. I)y 297. 10. UiviUe 179 cwt. 8 qr. 4 lb. hi oz. by 9 lb. 7 oz. 8 dm. IH. Tho circumference of the earth Ls about 26000 miles ; if a TOH80I fluils 9'.i m. 4 fur. 7 rd». a day, how lotig will it require to Muil round tho earth ? QinCSTIONS TO BE ANSWERED BY THE PUPIL. NoTH. — T^*' niimhfira after the questions refer to the article ofths aection. 1. What li division? (69) «. What Is tho divisor ? (59) 8, What la tho dividend ? («0) 4. WlmtUthoquotlent? What la tho derivation of tho word "iiiiotlont?" (01) C. Explain when tho quotient will bo C(iiiai to unity, and wtion creator or loss than unity. (B2) 6. Under what clrcuinstancoa does a remainder arise In division ? (G8) 7. What b tho donondnation of tho remainder? (08) 8k Why can It never bo aa ^roat as tho divisor ? (oJi) 9. What Is tho correspondence between tho minuend and tho nubtrabond In subtraction and the divisor and tho dividend In division? (M) 10. What may wo consider as tho factor."* of tho dividend? (6.'5) 11. How many ways have wo of expressing tho division of one quantity by another? What arc they? (60) 12. When a quantity consistinj? of several terms, connected by tho sijjn x, Is to be dlvldeion be proved ? (7S) 31. The divisor remaining nnchangod, what etfect has multiplying the divi- dend by any tmmber ? (79) 22. Tho divisor remaining unchanged, what effect has dividing the dividend by any number ? (SO) 23. Tho dividend remaining unchanged, what effect has dividing tho divisor by any number? (81) 24. Tho dividend remaining unchanged, what effect has multiplying the divisor by anjr number? (S2) ; 25. What is tho ettect upon the quotient when the divisor and the dividend , are both multiplieu by the same number? (88) ] 36. What is tho effect upon the quotient when the divisor and the dividend 1 are both divided by tho same number ? (84) ' 27. How do we divide by a composite number? (85) 28. When wo divide by tho divisors of a composite divisor, how do wo ob- tain tho correct remainder ? (86) 39. When tho divisor is separated into only two factors, how may the rulo for obtaining the correct remainder bo worded ? (85) 80. When the divisor and the dividend are both denominate numbers, what is the rule ? (80) SI. When one denominate number is nivided by another, what kind of a number must the quotient always be ? (87) 1,1 i! 118 MISCELLANEOUS EXERCISE. [Sect. IL 83. In the qnostion " Divide 87 lb. 2 oz. 15 dr. by 1 lb. 9 OE. 11 dr.," what are wo in roalitv required to do? (87) 88. When the divisor contains decimals, how do we proceed? (88) Upon what principle do wo do tliis ? (b8) 34. How do wo divide by 1, followed by any number of Os ? (S9) 86. How do we contract the work when dividing by 25'/ How by 15. 85. 46, or 55? (IK), 91) 8f). How do we divide by 125 ? How bv 75, 175, 225, or 275? (92, 93) 87. How do we abbreviate the work when there are manv decimals in th© dividend and but few are required in the quotient? (94) SflCT. 1 plantci oats Jlow 1 li). ^7764 Exercise 23. MISCELLANEOUS EXERCISE. {On preceding mles.) 1. Multiply 789643 by 999998. 2. Read the following numbers: 67818420-021030046, 72000OOO-O0O0OqO72, 1001000100-0010000010000001. 3. Express 709, 4376, 9999, 86004, and 3947596 iu Roman nu- merals. 4. Multiply 749 lb. 10 oz. avoirdupois by 72. ' 5. What is the price of 17 pairs of gloves at 4s. 7fd. per pair ? 6. The planet Neptune is 2850 millions of miles from the sun. How long would it take a locomotive to travel from the sun to Nep- tune, at the rate of 30 miles an hour ? 7. Reduce £729 17s. 6^d. to dollars and cents. 8. From $10000 subtract $9876-23. 9. Write down five hundred and twenty billions, six millions, two thousand and forty-three, and five thousand and sixteen trillionths. 10. Reduce 7964327 inches to acres, roods, &c. 11. Add together the following quantities: $729-43, $16-70, $976-81, $9987-17, $429*00, $129-19. 12. Multiply 6 weeks 4 days 3 hours 17 minutes by 429. 13. Take the number 741, and, by removing the decimal point : (1) multiply it by 1000000 ; (2) divide it by 100000 ; (3) make it mil- lions ; (4) make it billionths ; (5) make it trillionths ; (6) make it hun- dredths of thousandths ; (7) make it tenths. 14. Multiply 78-96 by -00042. 15. How many hogsheads of sugar, each containing 13 cwt. 2 qrs. 14 lbs., may be put on board a ship of 324 tons burden ? 16. A farmer's yearly income was 9287 dollars. He paid for re- pairing his house 186 dollars, for hired help on his farm 4 times as much lacking 95 dollars, and for other expenses 1902 dollars. How much does he save yearly ? 17. How many suits of clothes can be made from a piece of cloth containmg 39 yrds. 2 qrs. 3 nls. ; each suit requiring 3 yrds. 1 qr. 2 nls. V 18. There is a farm consisting of 732 acres ; 25 acres of which is ..aisnr?.'' i-Kae 9BCT. 11.] MISCELLANEOUS EXEPwClSE. 119 1 6 grs. of silver to planted with corn and potatoes; 197 acrea sowti with rye; 156 with outs ; 97 with wheat ; 199 In pastured ; and the remainder ia meadow. How many acres of meadow? 19. Bought 9G acres 3 roods 17 perches of land, for which I pay 1^7764 ; what did I jiay for it per perch ? 20. A lady, having 812 dollars, paid for a bonnet 20 dollars, for a shawl 7t' dollars, for a silk dress 97 dollars, and for some delaines 83 dollars ; how much had she remaining V 21. A silversmith received 30lb. 8 oz. 14 dwt make 12 tankards: what would the weight of each tankard beV 22. I bought four fields ; in the first there were t> acres 3 rds. 12 perches; in the second, 7 acres 2 roods; in the third, 9 acres and 13 perches; in the fourth, 5 acres 2 roods 36 porches. How much in all. 23. A merchant expended 294 dollars for broadcloth, consisting of three different kinds ; the first at 5 dollars a yard ; the second at 7 dollars; and the third at 9 dollars a yard. He had ns many yards of one kind as of another — how many yards of each kind did he buy ? 24. A silversmith made three dozen spoons, weighing 6 lb. 9 oz. 8 dwt. ; a tea-pot weighing 3 lb. 2 oz. 16 dwt. 16 grs. ; two pair of silver candlesticks, weighing 4 lb. 6 oz. 17 dwt. ; a dozen silver forks, weighing 1 lb. 8 oz. 19 dwt. 22 grs. ; what was the weight of all the articles ? 25. Reduce £972 lis. lljd. to dollars and cents. 26. Reduce 179 lbs. 3 oz. 3 dr. 1 scr. 14 grs. to graln.s. 27. There is a house 56 feet long, and each of the two sides of the roof is 25 feet wide ; how many shingles will it take to cover it, if it require 6 shingles to cover a Sfjuare foot ? 28. A merchant bought 4 bales of cotton ; the first contained 6 cwt. 2 qr. 11 lb. ; the second, 5 cwt. 3 qr. 16 lb. ; the third, 8 cwt. qr. 7 lb. ; the fourth, 3 cwt. 1 qr. 1 7 lb. He sold the whole at 1 5 cents a pound ; what did it amount to ? 29. A merchant has 29 bales of cotton cloth, each bale containing 57 yards; what is the value of the whole at 15 cents a yard? 30. A man willed an estate of $370129 to his two children and wife, as follows : to his son, ^139468 ; to his daughter, $98579 ; and to his wife the remainder. How much did he will to his wife ? 31. Divide £1694 16s. OHd. by £9 19s. llfd. 32. Reduce £19 19s. llfd. to dollars and cents. 33. A merchant having purchased 12 cwt. of sugar, sold at one time 3 cwt. 2 qrs. 11 lb., and at another time he sold 4 cwt. 1 qr. 15 lb. ; what is the remainder worth, at 15 cents per pound? 34. Bought 4 chests of hyson tea ; the weight of the first was 2 cwt. qr. 17 lb. ; the second, 3 cwt, 2 qrs. 15 lb. ; the third, 2 cwt, \ (\r, 20 lb. ; the fourth, 5 cwt. 3 qr. 17 lb. ; what is the value of the wliole at 37^ cents a pound ? »5. Express 100200300709 in Roman numerals. 'it 5? ^ i 120 PROPEETIES OP miMBEtlS, ETC. Sect. III. 86. Divide 43-2 by '76-843T. 37. Divide 123-4 by -000000066. 38. From $2789-27 take 17 times |63-29. 39. Add together $278*43, $417-16, $11-27, $2110*40, $723-15, and £29 6s. llfd. and divide the sum by 173. 40. In 1857 the total number of volumes in the Common School and other Public Libraries of .Canada West was estimated at 491644 and the number of libraries at 2076. How many volumes were there upon an average to each library ? SECTION in. Properties of Numbers, Prime Numbers, Measures, Greatest Common Measure, Least Common Multiple, Scales of Nota- tion, AND Application of the Fundamental Rules to Differ- ent Scales. Duodecimals. 1. A divisor, or measure of a number, is a number wbich will divide it exactly ; that is leaving no remainder. 2. A multiple of a number is a number of which the given number is a divisor. 3. An integer, or integral number, is a whole number. 4. Integers are either prime or composite, odd or even. 5. An Even Number is that of which 2 is a divisor. 6. An odd number is that of which 2 is not a divisor. 7. A Prime Number is one which has no integral divi- sor except unity and itself, thus 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, &c., are primes. 8. A Composite Number is a number which is not prime ; or is a number which has other integral divisors besides unity and itself, thus 4, 6, 9, 10, 12, 14, 15, IG, 21, &c., are composite numbers. 9. The Factors of a number are those numbers which, when multiplied together, produce or make it. 10. Factors are sometimes callefl measures, submulti- ples, or aliquot parts. 11. A Common Measure of two or more numbers, is a number which will divide each of them without a remain- der ; thus 7 is a common measure of 14, 35, and 63. 12. Two or more numbers are prime to one another when they have no common divisor except unity ; thus, 9 and 14 are "prime to each other." .1 Arts. 1-19.] PE0PERTIE9 OP NUMBERS, ETC. 121 Hence all prime numbers are prime to each other ; bot composite numbers may or may not be prime to one another. 13. Commensurable Numbers are those which have some common divisor. Thus 55 and 33 are commensurable, the common divisor being 11. 14. Incommensurable Numbers are those which arc prime to one another. Thus 55 and 34 are incommensurable. 15. A Square Number is one which Is composed of two equal factors. Thus 25=5 X 5 is a square number ; so also 64=3 x 8, &c. 16. A Cube Number is one which is composed of three equal factors. Thus 343=7 x 7 x 7 is a cube number ; so also 27=3 x 8 x ,3, &c. 17. A perfect Number is one which is exactly equal to the sum of all its divisors. Thus (i=l + 2 + 3 is a perfect number; so also 28=1 + 2 +4 + 7 + 14 is a perfect number. All the numbers known to which this property really bclonss, are the eiffht following : 6; 28; 496; 8128; 813550336; 8589S69056; 137438G91S28 ; and 230584800S139952128. NoTK.— All perfect numbers terminate with 6, or 28. 18. Amicable Numbers are such pairs of integers that each of them is exactly equal to the sum of all the divisors of the other. Thus 220 and 284 are amicable; for, 220=1 +2 + 4 + 71 + 142, which are all the divisors of 28 1, and 284=1 +2 + 5 + 11 + 4+10+22+20 + 44+55 + 1 10, which aro all divisors of 220. Other amicable numbers arc 1729G and 18416; also 9363583 and 9437056, 19. By the term properties of number s, is meant those qualities or elements which are inseparable from them. Some of the most important properties of numbers are the following— I. The sum of two or more even numbers is an even number. II. The difference of two even numbers is an even number. III. The sura or difference of two odd num^ors is an even number. IV. The sum of three, five, seven, &c., odd numbers, is an odd number. . w m \ 122 PROrERTIES OF NUMSEES, ETC. [Sect. III. V. The snm of two, four, six, eight, &c., odd numbers, is an even number. VI. The sum or difference of an even and an odd num- ber, is an odd number. VII. The product of two even numbers, or of an even and an odd number, is an even number. VIII. If an even number be divisible by an odd num- ber, the quotient will be an even number. IX. The product of any number of factors will be even if one of the factors be even. X. An odd number is not divisible by any even number. XI. The product of any number of factors is odd if they are all odd. XII. If an odd number divide an even number, it will also divide half of it. XIII. Any number that measures two others must like- wise measure their sum, their difference^ and their iwodnct. Thus, if 6 goes into 24 four times, and into 18 three times, it will go into 24 + 18 or 42, three phis four, or seven times. Also, If 6 goes into 24 four times, and Into 42 seven times, it will go into 42—24 or 18, seven minus four, or three times. Lastly, if C goes into 24 four times, and into 12 twice, it will evidently go into 12 tiaaes 24, twelve times 4 times, or 48 times. XI 7. If one number measure another, it must like- wise measure any multiple of that other. Thus, if 7 measures 21, it must evidcntlj' measure 6 times 21, or 11 times 21, or 17 times 21, &c. XV. Any number, expressed by the decimal notation, divided by 9, will leave the same remainder as the sum of its digits divided by 9. (See Art. 55, Sec. II.) This property of the numljer 9 affords an ingenious method of proving each of the fundamental rules. The same property belongs to the number 8 ; for 3 is a measure of 9, and will therefore be contained an exact number of times in any number of 9s. But it belongs to no other digit. The preceding is not a necessary but an incidental property of the num- ber 9. It arises from the law of increase in the decimal notation. If the ra'dix of the system were 8, it would" belong to 7 ; if the radix were 12, it would be- long to 11 ; and, universall}', it belongs to th- number that is one less than the radix of the system of notation. XVI. If the number 9 be multiplied by any single digit, the sum of the figures composing the product will make 9. Thus 9 X 4=36, and 8 + 0=9 ; so also 8 x 9=72 and 7 + 2=9. XVII. If we take any two numbers whatever ; then one of them, or their sum, or their difference^ is divisible by 3. 'lAW/i'K AW. 19.] PR0PERTIK3 OF NUMBERS, ETa 123 : n times 21, Thns, take 11 ancl iT; thoutrh nolthcr the Tiumbers themselves, nor their 8nm, is divisible by 8, yet their dilForence is, for it is 0. XVIII. Any number divided by 11, will leave the same remainder as the sum of its alternate digits in the even places, reckoning from the right, taken from the sura of its alternate digits in the odd places, increased by 11, if necessary. • • • • Take any number as 3840.5603, nnd mark the alternate flffurea. Now che mm of those marlied, viz : 8 + + ()4-8=17. The sum of the others, viz : 8 + 4 + 6+0=12. And 17—12=5, tho remainder sougiit That is, 88405603 divided by 11, will leove 5 remainder. Again, take 5847362, tho sum of tho marked flffures is 14 ; tho sum of those not marked is 21. Now 21 taken from 25, (i. e. 14 increased by 11) leaves 4, the remainder sought=remainder obtained by dividing 5847862 by 11. XIX. Any number ending in 0, or an even number, 13 divisible by 2. XX. Any number ending in 5 or is divisible by 5. XXI. Any number ending in is divisible by 10. XXIL When two right-hand figures are divisible by 4, the whole is divieible by 4. XXIII. When the three vight-hand figures are divisible by 8, the whole number is divisible by 8. XXIV. When the sum of the digits of a number is di- visible by 9, the number itself is divisible by 9. XXV. When the sum of the digits of a number is divi- eible by 3, the number itself is divisible by 3. XXVI. When the sum of the digits, standing in the even places, is equal to the sum of the digits standing in the odd places, the number is divisible by 11. Thus to illustrate the last five properties. The numoer 7416 is divisible oy 4, because 16, the last two digits. Is divisible by 4. — -^ is divisible by 8, because 416, its last three digits, is divisible by 8. is divisible by D, because tho sura of its digits, 7 + 4 + 1 + 6—18, is divisible by 9. is divisible by 8, because the sum of its digits, 7 + 4 + 1 + 6:=18, is divisible by 8. So also the number 4667821 is divisible by 11, sinoo the sum of the digits in the odd places, 1+8 + 6+4=14=2 + 7 + 6, the sum of tho digits in the even places. XXVII. Every composite number may be resolved into prime factors* For since a composite number la producfed by multiplying two or more fac- tors together, it may t^vidootly be resolved into those factors ; ami if these factors themselves are componit", they also may be resolved into other factors, *nd thus the analysis may be continued until ail tho ftictors are jtr'ime uuu.bors. r. ; •! .1 124 PROPERTIES OF NUMBERS, ETC. taEOT. in. 4!! XXVIII. Tho least divisor of any number is a prime number. For every whole number Is either prime or composite (Art. 4) : but & com- posite number can be resolved into factors ^XXVII): consequently, itmhatA divisor of any number must be a prime numoer. XXIX. Every prime number, except 2, if increased or diminished by 1 is divisible by 4. (See table of prime numbers on next page.) XXX. Every prime number except 2, is odd ; and therefore terminates in an odd digit. Note.— It must not be inferred from this that all odd numbers are prime. XXXI. All prime numbers, except 2 and 5, must ter- minate with 1, 3, 7, or 9. Every number that ends in any other dig't than 1, 3, 7, or 9, is a composite number. For all prime numbers, except 2, must end in an odd digit (XXX), and aU numbers ending in 5 are divisible by 5. XXXII. Every prime number, except 2 and 3, if in- creased or diminished by 1, is divisible by 6. 20. To find the prime numbers between any given limits. RULE. Write down all the odd numbers^ 1, 3, 5, 7, 9, &c. Over evert/ third from 3 vn'ite 3 ; over every fifth from 5 write 5 ; over every seventh f^om 7 write 7; over every eleventh from 11 write 11 ; and so on. Then all the numbers which are thus marked are composite ; and the others., together with 2, are prime. Also the figures tMis placed over^ are the factors of the numbers over which they stand. EXAMPLE. Find all the prime numbers 1 ess than 100. 8 8-5 1 3 5 1 9 11 13 15 n 3T 5 8 8-11 5-T 19 21 813 23 25 27 85 29 31 7 33 817 35 87 39 41 43 45 47 49 51 53 5-11 819 8-7 518 8-23 55 57 69 61 03 65 67 09 71 8-5 711 8 6-17 8-29 IB 75 77 79 • 81 83 85 87 89 M8 8,31 5.19 811 91 93 95 97 99 1 173 2 179 8 •181 5 191 7 198 11 197 18 199 17 211 19 223 23 227 29 229 81 23;? 87 239 41 241 43 251 47 257 53 263 59 269 61 271 67 277 71 281 73 283 79 293 83 307 89 811 97 313 10! 817 io;3 831 107 337 109 347 118 349 127 358 131 3.')9 137 367 139 378 149 879 151 38S 157 SSL im 391 167 401 ABT. 20.] PEOrEliTlJiB OF NUMBEUS, ETC. 125 ere are prim)', , must ter- at ends in e number. XXX), and ali 3-5 15 n Ml 5-1 33 35 •IT 61 53 •23 G9 11 •29 87 89 Hence, rejecting all the numbers which have superiors, the primes Ies3thanl00arel,3, 6,7, 11, 13, 17, 19, 23, '29, 31,37,41,43,47,53, 69, 61, 67, 71, 73, 79, 83, 89, 97, together with thp number 2. This proceaa may bo extended Indefinitely, and '* the method by which ])rlni 3 are found even by modern coniputtitors. I'- ^as invented by Eratos- thenes, a learneu librarian at Alexandria (iiorn •< C. 275). He inserlbed the series of odd numbers upon parchment, then cuttmg out such numbers iw he found to be composite, nis parchment with its holes somewhat resembled a sieve; hence, tbia method is called '•' Eratosthenea' Sieve.^'' TABLE OF PRIME NUMBERS FROM 1 TO 3407. ^» 1 173 409 659 941 1223 1511 1811 2129 2423 2741 8079 2 179 419 661 947 1229 1523 1823 2131 2437 2749 3088 3 •181 421 678 958 1231 1531 1831 2137 2441 275:3 3089 6 191 431 677 967 1237 1543 1847 2141 2447 2767 3109 7 198 433 683 971 1249 1549 1861 2143 2459 2777 3119 11 197 439 691 977 1259 1558 1867 2153 2467 2789 3121 18 199 443 701 983 1277 1559 1871 2161 2473 2791 3137 17 211 449 709 991 1279 1567 1878 2179 2477 2797 8168 19 223 457 719 997 12*3 1571 1877 2203 2503 2801 8167 28 227 461 727 1009 1289 1579 1879 2207 2521 2803 8169 29 229 463 783 1013 1291 1533 1889 2213 2531 2819 3181 31 23;3 467 739 1019 1297 1597 1901 2221 2589 2833 3187 87 239 479 748 1021 1301 1601 1907 2237 2543 2887 3191 41 241 487 751 1081 1803 1607 1913 2-«9 2549 2843 3203 43 251 491 757 1083 1307 1809 1931 2243 2551 2851 3209 47 257 499 761 1039 1819 1618 1933 2251 2557 2857 8217 53 263 508 789 1049 1321 1619 1949 2267 2579 2861 8221 59 269 509 773 1051 1327 1621 1951 2269 2591 2879 32-29 61 271 521 787 1061 1861 1627 1978 2278 2593 2887 8-251 67 277 523 797 1063 1367 1637 1979 2281 2609 2897 8253 71 281 541 809 1069 1873 1657 19S7 2287 2617 2908 3257 73 283 547 811 1087 1881 166;3 1993 2293 2621 2909 8259 79 293 557 821 1091 1399 1667 1997 2-297 26^33 2917 3271 83 807. 563 828 1098 1409 1669 1999 2309 2647 2927 8299 89 811 569 827 1097 1423 1693 2003 2311 2657 2939 8801 97 318 571 829 1103 1427 1697 2011 2sm 9-659 2953 8807 10? 817 577 839 1109 1429 1699 2017 2Sii9 266:5 2957 8313 io;3 831 687 853 1117 1433 1709 2027 2841 2671 2968 3319 107 837 598 857 1123 1439 1721 2029 2347 2677 2969 8823 109 347 599 859 1129 1447 1723 2039 2351 2683 2971 3329 113 349 601 868 1151 1451 1733 2053 2357 2687 2999 3331 127 353 607 877 1153 14f)3 1741 2063 2371 2689 3001 8343 131 859 818 881 1163 1459 1747 2069 2377 2693 3011 3847 137 867 617 883 1171 1471 1753 2081 2381 2699 3019 3359 139 878 619 887 1181 1481 1759 2083 2;i33 2707 3028 3;3C1 149 879 631 907 11S7 1438 1777 2087 2389 2711 3087 8371 151 3S3 ftll 911 1193 1487 1783 20S9 2393 2713 8041 3373 157 889 643 919 1201 1439 1787 2099 2899 2719 8049 ass9 16;3 397 647 929 1218 1493 1789 2111 2411 2729 3061 3391 167 401 653 937 1217 1499 leoi 2118 2417 2781 8067 3407 When it is required to determine whether a given number is a prime, we first notice tlie terminating figure ; if it is different from 1, 3, 7, or 9, the num- ber is composite; but if it terminate with one of the above digits, we must endeavour to divide it with some one of the primes, as found in the table, commencing with 8. There is no necessity for trying 2, for 2 will divide only the even numbers. If we proceed to try all the (successive primes of the table until we reach a prime which is not less than the square-root of the number, without finding a divisor, we may conclude with certainty tU^t ttie ttvuabcr is a^'rtnte, ifirji m 126 rUOrEKTlEb OF NUMBKUS, ETC. [bKOT. HI. The reason wby wo noed not try any primes groati-r tlmn tlio square-root of the number, isdniwii fn»m tho followlnjf conHidcrutioii: If ucoinpoaitts num- ber is resolved into two lUctors, one or wliich is lows tlian the squiire-rool of tiie number, the other must be greater tlmn the square-root. The square of the last nriine niven in our lable is ll(i0764l>; hence, tliis table is HUllicieutly extended to enable us to determine whether any niiuiliir not exceodini,' 11C07()49 is a prime. It is obvious that numbers may be' proiioMd which would require by tliis method very great labor to determine whetlitT tiuy arc primes, still this is the only sure uud general method a^ yet discovered. , 21. To Resolve a Composite Number into its Prime Factoids. RULE. Divide the given number by tlic smallest number which mil divide it without a remainder; then divide the quotient in the same wv///, and thus continue the operation till a quotient is obtained which con be divided by no number greater than 1. 21ic several divisors witli Ihc last quotient, will be tJie prime factors required. i^Vd-XXVlI.) Reason.— Every divimon of a number, It is plain, resolves It into two/iW^ torSy viz. : the divisor and the quotient. But according to the rule, the divisors, In every case, are the nma/lent numbers that* •will divide tlic given number ot the successive quoiients without uremaluder; consequently they are all pHme numbers. (19-XXVlIl.) And since the division is continued till a quotient i.i obtained, which cannot be divided by any number but unity or itself, it follows that the Za*^ quotient must also be a prime number; for, a prime number is one which cannot be exactly divided by any whole number except %mity and itself. (Art. 7.) NoTK. — Since the lea it divisor of every number is a prime number, it is evident that a composite number may be resolved into its prime factors by dividing it continually by any prime number that will divide the given niiui- ber and the successive quotients without a remainder. Hence, A composite number ean be divided by any of its pt'i/me factors without a remainder, and by the product of any two or more of them, but by no other nmmber. Thus the prime factors of 42 are 2, 8, and 7. Now 42 can be divided by 2,.^, and 7 ; also by 2 x 3, 2 x 7, 3 x 7, and 2x8x7; but it can be divided by no other number. Example 1. — Resolve 210 into its prime factors. OPEEATION. 2)210 8)106 5)35 Wc first divide the given number by 2, which Is the least number that will divide it vithout a remainder, ant! which is also a prime number. "We next divide by 8, then by 5. The several divisors and the la£t quotient are tlio prime factors reciuired. Am, 2, 8, 5, and 7. 1 PROOF. 2x8x5x7=210 Example 2. — Resolve '728 into its prime factors. OPERATION. 2)728 2)864 2)182 7)91 ii Therefore, 2 x 2 x 2 x 7 x 13. or 2» X 7 X 18, are the prime factors of 788. AKrt. 21-22.1 rKlME FACTORS, ETC. 127 RIME Factor.^ Exercise 24. 1. Resolve 11368 into its prime fuetora. 2. What are the prime factors of 2934 ? ',]. What are the prime factors of 1011? 4, What are the piime factors of 1000? 5. What are the prime factors of 1024? (j. Wliat are tlie prime factors of 32320? 7. What are the prime factors of 707 ? 8. What arc the prime factors of 1118? Ans* 2' X 7' X 29. Ans. 2x3'xl«8. Ans. 3 X 337. Ans. 2^ X 5'. A7tS. 2'". Ans. 2"x5x 101. Ans. 7 X 101. Ans. 2 X 13 X 43. DIVISORS. 22. From Art. 21, Note, fur finding all the divisors of any number, we deduce wlie following — RULE. Resolve the number into its prime factors ; form as many seriet of terms as there are pritne factors, by 7naking 1 tlio first term of each si'i'/es, the first power of one of the prime factors for the second term^ ilie second power of this factor for the third term, and so on until we reach the highest that oceurred in the decomposition. Then multiply these scries together^ and the partial products thus obtained will be the divisors sought. Example 1. — What are the divisors of 48? Here we find 48=2* x8. Tlioroforo our series of terms will bo 1 "IC and 1 "'S; uiulti2)lviiifr thos(i uwtbur. i"2"4"8--lG 1-3 .2--4--8 1-2--4-S-16-8-G-12-24-48 Tlierofore the divisors of 4^ are 1, 2, 8, 4. 0, 8, 12, 16, 24, and 48. "We l)egiu each series with 1, bocuu.se, were we not to do so, the different powers of the prime factors would not themselves a[)peur timong the partial 'iroducts. Example 2. — What are the divisors of 360. The prime factors of 360 are 2'''X8»x6 and therefore the series are 1 ' :l"8"9andl-5. 2"4 1-2-4-8 1-3-9 OPERATION. l"2"4"8'3--6"12"24"9-18'*36"72=products of Ist and 2nd series 1"5 l"2-4"S"3-6--12"24"9-18"36-72-5-10"20"40-16-80"60"120"45"90"180-860. Therefore the divisors of 860 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, ISO, 860. '•''''»'» > > . . «'. ^ » * The small figures written to the right of the factors and above the lin«j are called exponents, and show how often the digit is taken as factor. • •- .-.% 128 UUEATEHT COMMON MEA81IUK. [ftltCT. Ill, 1. What arc the divisors of 100? Ans. 1, 2, 4, 5, 10, 20, 2fi, 50, 100. 2. What arc the divisors of 810? J ( 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 46, 54, 81, 90, 135, 162, ^"*- "j 270, 405, 810. 3. What are the divisors of 920 ? Ann. 1, 2, 4, 5, 8, 10, 20, 23, 40, 46, 92, 115, 184, 280, 460, 920, 4. What arc the divisors of 25000? J ] 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 62r), ^^**' \ 1000, 1250, 26U0, 3125, 5000, 6260, 12500, 25000. NUMBER OF DIVISORS. 23. Since the series of terms which we multiplied to- gether, by the last rule, to obtain the divisors of any num- ber commenced with 1, it follows that the number of tiTius in each series will be one more than the units in the expo- nent of the factors used. Hence, to find the number oi divisors of any number without actually setting them down, we have the following-^ RULE. Resolve the number into its prime factors and express them as in examples 3, 4, and 6, in Art. 21. Increase each exponent by unity and multiply the resulting numbers together. The product will be the number of divisors. Example. — How many divisors has 4320? 4820=2' X8'x5. Iloro the exponents are 5, 3, and 1: each of which being Increased by one, we obtain 6, 4, and 2, the continued product of which is 6x4X2=48=tho number of divisors sought. 1. How 2. How 8. How 4. How 6. How 6. How 7. How b. How Exercise 26. many divisors has 88200? many divisors has 3500 ? many divisors has 6336 ? many divisors has 824? many divisors has 49000? many divisors has 81000? many divisors has 75600 ? many divisors has 25600 ? Ans. 108. Ans. 24. Ans. 42. Am. 8. Ans. 48. Ans. 80. Ans. 120. Ans. '6'6. GREATEST COMMON MEASURE. 24. The greatest common measure, or greatest com- mon divisor of two or more numbers, is the greatest num- ber that will divide each of them without a remainder. [ftKCT. Ill 20, 25, 50, 10(\ 90, 135, IGL', 280, 460, 920. 250, 500, 625, 25000. lultiplied to- of any mini" iber of tiTins in the expo- any number e following- rcss them as in \onent by unity )duct will be the h of which boing iluct of which is Ans. 108. Ann. 24. Ann. 42. Alls. 8. Ans. 48. Ans. 80. Ans. 120. Ans. Ji3. reatest com- reatest num« mainder. AfiTS. 23-20,] OKIuATErjT COMJJON MEASUUE. 120 25. To find a comraon divisor or corainou measure of two or more nujnbers — RULC, Resolve the gioen numbers into their |H4hm factors., then if any factor be common to all, it would be a comm*->H measure. If the fjivon numbors havo not a mmmon fitHor they cannot Imvo a common measure greater than un; y, and couscfjuently arc cither prime ntnnbers or are prime to eacii other. (Arts. 7 and 12.) Example. — Find a common divisor of 14, 35, and 63. 14=2X7; 35=.^x7, and 63=8x3x7. The factor 7 Is common to all tho jrlven num'oors, and Is thurofore a common measure of them. Exercise 27. 1. Find a common divisor of 21, 18, 27 and 36. 2. Find a common divisor of 21, 77, 42 and 35. 3. Find a common divisor of 26, 52, 91 and 143. 4. Find a common divisor of 82, 118 and 146. An6. 3. Ans. 7. Ans. 13. Ans, 2. 25. To find the greatest common measure of two quantities — RUr.E. Divide the larrjer by the smaller ; then the divisor by the rema'n- der ; next the preceding divisor by the new remainder: — continue this process nntil nothing remains, and the last divisor leill be the greater,t common measure. If this be unity, the given members arc prime to each other. Example. — Find the greatest comraon measure of 3252 and 4248. 8252)4248(1 • 8252 996)3252(3 2983 264)996(3 792 204)264(1 204 60)204(3 180 24)60(2 43 12)24(2 24 996, tho first remainder, becomes the second divisor; 264, the second re" ■ 1 i i i 1 i i ^^H II w i 1 ! il m\ 130 OREATKST COMMON MEASURE. [Sect. Ill, ninlndor, bocotnos tho third divisor, &o. 12, the l»»Ht divisor, Is the required greatest common tnciisiire. 1'kook. — In order tiM'stfthli.sh the truth of thh nilo.'ft b ncccs.siiry to re- moiribor (19-XIII. nn t'i, It iii.^o imiisiiros 21, ii iiiiiIti|)1o nf 12; bi'wmsi) 12 incdMurc 24, It mcaaiircs 4S. u iniiltiplo of 24 ; liccaii.-ic 12 im as. iires 12 and iil>o 4s, It nit asiiros 60, wliich Is tliclr aiitn ; Ik'cuiim' 12 nn'tisincs Tru, It meuHiiros IM), u niiiKlj)!!* of 6((; b<'cftiiso 12 nuMWuri-s IhO.aiid also 21, it iiit as- uroH their sum, wliich is 2(l4, bocaiisc 12 minstircs 201. and lilu wIm' (io. it inc as- iirt'S tlu'ir «um, 2*)l; bt'caiiso 12 m»'asur»'8 204. it iiiciwiin'S 792. a miiltiplc or 204; and bi'Caii.so 12 riu'asiireH 7U2, and also 204, It iiicat^nrcs tliclr sum, wl.ii h Is 1)96; because 12 mc.isiir.'s i)S<6, it moasiircs 2US^t. a mnltlpli' of OUO; anil Ic causo 12 measures 2yss. and also 204, ll measures their sum, o2r)2; and bccau.sr 12 measures 82.')2, and also O'.IO, it measures their sum, wliicli Is 424R, 12, tlieje- foro, measures eucli of the u'ivou numbers, arid i.s a common meusuro; next It h\ their (jreures their diller- •nee, which is 204; becauFc 1)3 mensuics 204. It also measures 792 a multiple (f 2(}4; and becau.so 18 measures 792, and also 91)0, it lueusuros their dilT'ircnce, ivhich Is 204; iiccause 18 measures 2W. and also 204. it tiutisures tlu-lr (iilVir ence, whicli is 00, because 18 measures 00. il jucasures l.vO, a iiiulliple of (0 ; and becaus') 18 measures IbO. and also 204, it me.nwurcs their l'21 ; jumI 1 ausc 1;! measures 60, and also 48, it measures their diU'erenec, wliicli is 12. 'I'liai. i*^^, |;> measures or divides 12 — a preater number measures a less, wliieli is impos.'-ilili. Therefore 18 is not a common measure! of 82.V2 and ^24S; and in a simii ir manner it may bo shown that no number greater than 12 is a conimon ua'asure. Therefore 12 is the greatesl common measure. As the rule might be proved for any other example equally well, it Is Irun in all ca6Qs. Exercise 28. 1. What is the greatest common mea.surc of 296 and 407 '? Jrts. 37. 2. What is the greatest common measure of 5(i6 and 808? A71S. 2L'. 8. What is the greatest common measure of 74 and 84? Ans. 2. 4. Wliat ia tlie greatest common measure of 182{) and 2555? A71S. 3G5. 6. What is the greatest common measure of 550 and G72 ? Ans. 4. 27. To find the greatest commori measure of more than two numbers — RULE. Find the greatest common measure of two of them ; ihen^ of this common measure and a third ; next of this last common meastire and a fourth, &c. Tfie last common measure found will be the greatest comm,on measure of all the giveti nuynhers. Example 1.— Find the greatest common measure of 679, 5901, and 6734. Arts 2T-2S 1 (IKKATI.ST COMMON MLASl'UE. lai Is the requlrpfl well, ;t is liijo By tho ta^t ruli> wp flnrl that 7 Is the trroafost common measure of rtll) nntl B901 ; and hy tho Mim<( ruio tliiit It U iho irr ntvut. commi> ■ iiioa.Huro of 7 iirul niM (the remalnin'X iiiunhcr), tor 61\'l »-7=Ufl"2, with no roniuladiT. Therol'oro 7 is tbo reqiili'od Miiintx-r. ExAMFLC 2. — Fiud the greatest cutuinon measure of 036, 736, and 112. The aroitpst, rornmon moasnro of O'^rt nml TW \i ^, a d tlip (rroafost common monsiae "f S and ll'i I.h 'i; tlicclorc 2 is liio tficutist common mcanuro of ilio givon 11 um bora Thi« mil! muv l)o shown to tt« corroot In tho samo waj' m tho last; except that In proving; tiio numhcr ('o:inil to lii« a coirirnov m('a-*iin', wf nro to licplii at tho ciul of (ifJ till! pro I'sscs, tutil -jn tlni»(;.'li nil of tlu-iii in hiicccsslon ; iiiiil In proving that If is Hw (/r<'Ofefif coJiniori mi' i -.in'i'. wo iin- to bccrln ut thu cotn- mciicomciit of the Hrst jiros'i'ss. or ih it nsfd to llii'l ilic coiuttion nicuburo of tho two Urst uuiubcrs, and procuotl suoci'rtsivcly tluoiigli u/l, EXKRCISK 4t). 1. What id the greatest eonimun measure of 110, 1 10, and (180? A IIS. 10. 2. What is the greatest common meaauro of 1326, n0'.)4, and 1120? Alls. 112. 3. Whiit is the greatest common measure of 'ir.8, 922, and 37r» ? Ann, TIh'V have none. 4. What is the greatest common measure of 201, li'.)0, 1115, and 200(5? Ans. 17. SECOND METrK^D. 28. It' is manifest that tho greatest common measure or greatest common divisor of two or more numbers, must be their greatest common factor, and tliat this greatest common factor must be thir product of all the prime factors that ai'C cominon to all the gi^'en nurn])erH. Hence to find tlie greatest common jncasure of two or more numbers, we liave the foUowing — RULK. Resolve each of the given numbers into its prbne factors ; and the product of those factorSy which are common to all^ wili be the greatest common measure. Example 1. — What is the greatest common measure of 1365 and 1905? 8)1095 m 3)1065 6) !55 7)91 13 Hence, 3, 5, 7, and 13 arc the prime fiicturs. 5)665 7)183 10 nence, 3, 5, 7, and 19 are the prime factors. rl\ I - hhi ! 1.1 liil 132 LEAST COMMON MULTIPLE. [Sect. III. .,11 Ml And the factors thiit are common to botH are 3, ^, 7. Ilenci (6x7=105 =grcate8t coir.mon nie;i.sure. Ex/MPLE 2. — What is the greatest common measure c ^8, 126, and 1G2? 10.?— 22x33, 12G=2x8'»x7. niul 1^2-2^^*. Ilt'iicc. tlie f!i( !<>rs tliat are couimon ure 2 and 3", and the greater ' mmbn iiit'abuie=2 Xo'— 18. EXKRCISE 80. 1. Work hji fJih r.'iclhod all the ^'vcccdhu/ examples. 2. Wliut l-^ t!ie greatest cor.)nionii'e:ir,urcoi'5r), 84, 140, 108? ,^7?s. 28. 3. What iH the greatest cuiamou iu<;asure of 241920, 380160, (lUTiO, 108680? Am, 84560. 4. What is the greatest conimon measure of 10800, 28040, and 2100? • Ana. 40. LEAST COMMON MULTIPLE. 29. One number is a coiiimon iiuilliple of two or more others wLen it can be (li\ idcd bv eacL of them without a remainder. 30. One nnml)er is the least ccAiinion muhiple (1. c. m.) of two or more others when it is the least number tliat can be divided bv each of them witiiout a remainder. 31. It is evident tliat a dividend will contain a divisor an exact number of times, when it contains, as factois, cvm/ factor of that divisor; and hence, the question of finding the least common multiple of seveial numbers is reduced to finding a number which shall contain all the prime factors of each number and none oiliers. If the numbers have no common prime factor, their product Avill be their least common multiple. Siippof-e wo wish to sco what is tho least romtoon iniiltii.ie of 0. 12, 10, 20, and 35. Kcsolvin:^ tiiose into thoir prime liictoii, wl- olitain 1)=32, 12--2-X3, ]C=2«, 20=2-y5, and 85=L7xr). Now it is plain tliut 2* must tnter into tho least comjnon innltiple as a fac^tor, and, since 2-* is a niulUpU>. of 2^. we do not rcnsider 2^ also a factor of tlie least common multiiiie. So also 3^ must he a factor of the least common miiUiplo; and since it contains 8, wo do not again uiu'liidy b 3. I.astl.y, T) and 7 must enter inio the lea&t common multiple. The factors of the lo:ist common multiple are then 2'*, 3-, 5 and 7; and those, mulMplfed togfther, give 2'*x32x5x7— 5040=least common multiple. Hence, to find the least common multiple of two or raore numbers, wc have the following — RULE. lieaolve the mcmhers into their prime factor;^ (Art. 21), select all the dij'ere7it factors ivhich cccw\ observing when the same factor has fev,.. catct ■ mmon Ante. 29-S2.1 LEAST COiMMON MULTIPLE. 1S3 dijfVrmt powers, to fake the h'u/hcst power, llie co)i tinned product of the. factors thus selected vjill be the least common multiple. ExEiunsE 31. 1. What i.^ tl.e least common multiple of 8, 9, 10, 12, 25, 82, 75, and SO? Hare 8==2\ 9 = 8^ 10 = 2x5, 12=:2''x3, 25-5', 32=:2*, ''o^S' x 3, 80=2* X 5. Therororo the least common multiple=2* x 3' x 5* = 70200. 'i. What is the least common muUiplc of 6, 1, 42, 9, 10, and 630? Ans. 2x3^x5x7 = 630. ;]. What is the least common multiple of the nine digils? Ans. 2^ x3-x 5x7 = 2520. 4. What is the least common multiple of 6, 9, 12, 15, 18, 21, and 30? Am. 1260. 5. What is the least common multiple of 670, 100, 335, and 25? Ans. 6700. 6. What is the least common multiple of 8, 10, 18, 27, 36, 44, and 396? Ans. 11880. SECOND METHOD. 32. We ruay also ihi!l tlie least common mulliple of two or more unmhurs by the following' — RULE. . TT^'fi^*!' the f/iven v}tmh<:r^ in a lliw, wi/h ttoo polittii between them. Divide hfi the LiiAST mc-tioer wlilrii null dUide (unj ttro or more of them without a rei)ialud';r, luid set the quotients and the undivided numbers in a line below. Divide this li?ie and set down the results as before ; thus continue the operation till there are no two numbers which can be divided by an-/ number greater than 1. Tlie continued product of the divisors and the numbers in the last line will be the least common, muliip'e so'uiht. I t' Example 1. — What is the least common multiple of 16, 48, and 103? 2)1(5. .48. . 108 2)3. .24 • • 54 2)4. .12.. 27 2)2. . 6- 2T 8)1. .3.. 27 1 . . 1 . . 9 Ans 2x2 )<2x2 x3 ;<9=432=leii.8.. 9.. 9 8)1 . , 8 . . 8 1.. 6.. 1 12x8x6=:216 1.. 1.. 1 2x2x3x3=86=1. c. 2)12 . III. IS. .36 2)6. 9. . 18 9)3. 9. . 9 3. 2x2x9 . 1. x3= . 1 103. Exercise 32. 1. Find the least comtnon multiple of 12, 2Cv, phoI 24. Ans. 120. 2. Find the least common multiple of 14, 21, 'o, 2, and 68. Ans. 126. 3. Find the least common multiple of 18, 12, 39, 216, and 234. Ans. 2808. 4. Find the least common multiple of 8, 18, 15, 20, and TO. Ans. 2520. 6, Find the least common multiple of 24, 16, 18, and 20. Ans. 720. 6. Find the least common multiple of 60, 50, 144, {55, and 18. Ans. 25200. 7. Find the least common multiple of 27, fA, 8i, 14, and 63. Ana. 1134. iSsct. Hi. 7, X 2, will be tlie th line ; the It ast imiltiple of 4, l-i, iple 6.. 9. •3 . . 9 . . 18 . 9 3. . 1.. 1 X 9 X 8=108. Ans. 120. 63. Ans. 126. and 234. Ans. 2808. iVO. A71S. 2520. 1 20. Ans. 720. md 18. Ans. 25200. ad 63. Ans. 1184. THIRD METHOD. 33. The least common multiple of several numbers is most expeditiously found by the following — RULE. Write the given numbers in a line ; take any one of them ar, did/- sor, and strike out of each of th^ given numbers all the factors that are common to it and the assumed number. Arrange the imcancelled factors of the given numbers^ and the ?m- cancelled numbers in a line ; take the least other number tohich K'xactty contains otie or more of thein^ and strike out all the factors of the num- bers in the second line which are common to any of them and the sec- ond assumed number. Proceed thus until the assumed numbers cancel all the factors of the given numbers. Multiply all the assumed numbers together for the least common multiple of the given numbers. Example 1. — What is the least common multiple of 16, 27, 45, 60, 88, 96, 100 ? Assume 100 ;^..27..${3..0P..88..p'^..;P0 Assume 24 4.. 127.. 9.. ^..n^-U Assume 99 ^.. 3.. n 100 X 24 X 99=287600=1. c. m. ExPLANATTOK.— 4, ;v factor of 100, reduces Id to 4, 3S to 22, and 06 id 24; 6, another factor of 100, reduces 45 to 9; and 20, another factor of 100, reduces CO to 8. The numbers in the second line then are 4, 27. 9, 3, 22, and 24. Wo assume 24, of which a factor, 4, cancels 4; another factor, 2, reduces 22 to 11: ami another factor, 8, reduces 27 to 9 and 9 to 8. The numbers in the third lino then are 9, 3, and 11. For this lino wo assumed 99, of t\'hich a factor, 3, can- cels 8 ; another factor, 9, cancels 9; and a third, 11, cancels 11. Now since the least couimon multiple of a Bevies of sumbora Is a nnmber which still contains all the prime factor.? of each number, and none others, it is manifest that the least common multiple of the given numbers will be the ?amo as the least common multiple of 100, and 4, 27, 9, 3. 22, ami 24, because only those factors which were commc^n to the given numbers and 100 were struck out. Similarly, the least common multiple ot 100, 24, and 9, 8, and 11, will be the same as the least common multiple of lOO, and the numbers in the second line, since only those factors which were common to 24 and the numbers of tho sec- ond line are struck out. Finally the least common multiple of 100, 24, and 99, is equal to the least common multiple of the given numbers. Example 2. — What is the least common multiple of 120, 40, 39, 65, 88, and 16? Assume 120 ;?p. .^0. .29'. .«fJ. .88. .16 Assume 13 J3.J3..11.. 2 Assume 22 JLl.. % 120 X 13 X 22=84320=1. c. m. Explanation.— We first a.ssame 120. Now this cancels 120 and 4ft. Also 8, a factor of 120, reduces 89 to 18, and 5, another factor, rcdncs «>.') to 18. Also 8, another factor, reduces 88 to 11 and 16 to 2. Next atisume 18 ; this can- cols 13 and 18. Next assume 22, of which 11, one factor, cancel* the 11, and another ftictor, 2, cancels 2. '.J^ %xi .; i ^^^H : , 1 l\ 13G SCALES OF NOTATION. (Sect. 111. ExA^f^LK 3.— Find the least common multiple of 12, 16, 20, 24, 80, 48, 66, and 64. Assume 96 Assume 10 n. j^. .%^. .%^, .^(^. .^^. M' M 96x70=0'720=l. c. m. Exercise 33. 1. What is the least common multiple of 300, 200, 150, 60, 60, 75, and 125 ? Ans. 3000. 2. What is the least common multiple of 20, 60, 16, 165, 210, 63, and 27 ? Ans. 41580. 3. What is the least common multiple of 12, 132, 144, 60, 96, and 1728 ? Ans. 95040. Work also hy this method all the preceding questions in least com- mon multiple. / DIFFERENT SCALES OF NOTATION. 34. The radix or base of a scale of notation is its com- mon ratio. Thus in our system the radix is 10 ; in the duodecimal system the radix is 12,'2/=8166323 duodenary. 12)TT8602-8 12)64875-2 12)5406-8 12)450-6 12)8T-6 1-1 Exercise 84. 1. Change 692885 from the decimal to the duodenary scale. Am. 2470fe. 2. Express the common number 8700 m the quinary scale. Ans. 104800. 8. Express 10000 in the undenary scale. Ans. 7B71. 4. Express a million in the senary scale. Ans. 33233344. 5. Express 10000 in he octenary scale. Ans. 23420. 6. Transform 12345664321 into the duodenary scale. Ans. 248664^^69. *I. Express 10000 in the nonary scale. Ans. 14041. 8. Transform 300 from the common to the binary scale. Ans. 100101100. Example L— Transform 2313042 from the quinary to the octe- nary scale. OPERATION. V. 8)2818042 8)131810-7 8)10100-5 8)811-2 8)20-1 1-a Explanation.— We divide hero as before, bear- Inp in mind, however, that the ratio is no longer ten, but Jive. We proceed thus.— 8 in 2, lo times ; twice five (the radix) is ten, and 8 nmlie tliirteen; 8 in 13. 1 and 5 over; 5 times 6 are 25. and 1 make 26 ; 8 in 26, 8 time s and 2 over ; twic(; 5 are 10, and 8 make 18, S in 18, once and 5 over, &c. Therefore 2818042 gwf«ary=l21257 ocU lary. Note. — ^The Roman Numeral written over the number indicates (he radix of the scale. AbT9. 89-40.J tRANSFORMATiON OP SCAliid. m ry to the octc- fixAMPLS 2.— Transform 878<18 from the undenary to the duode- nary scale. Observe the first two figures here are not tbirty- Bevon, but 8x11 + 7=40. We sav 12 into 40, 8 tlmea and 4 over ; next, 12 into 4x11 + 8=or 62, Ac OPBRATION. XI. 12)878^8 scale. 12)84456-8 12)3132-4 12)294-9 878^13 tmrf«»ary=24W48, duodenary. Ana. 12)26-9 12)2-4 ExAMPLS 3, — Transform M23/ from the duodenary to the nonary OPKRATION. xir. 9)11971-1 9)1649—4 9)200-3 9)28- « "s-s Observe, here we say 9 into t ten, 1 and 1 over; 9 tnt.) 16, (1 X 12 + 4) 1 and 7 over ; 9 into 86. (7 x 13 + 2) 9 and 5 over; 9 into 68, (5x12 + 3)7; 9 into t, 1 and 1 over. Ami we proceed in the other lines in the sama manner. <423{)0 detntiy. ^bW) denary. \hOO denary. \bOO denary. SirK^e tlie results all agree when reduced to the deuary scale, we conclude the work id correct. 6. Transform 182713 nonarif, into the ternmy, duodenary ^ and octenari/ scales, and prove the rehults by reducing all four numbers to the denary scale. 7. Tran.sform <2/290 duodenary^ into the nonary, senary, guater- nary, and binary scales, and prove the result by reducing all five numbers to the decimal scale. FUNDAMENTAL RULES. 41. The fundamental rules of arithmetic are carried on in the different scales as with numbers in the ordinary or de- cimal scale ; observing that, when we wish to find what to carry in addition, subtraction, multiplication, &c., we divide, not by ten, but by the radix of the particular scale used. Example 1.— Add together 34120, 3121, 13102, 81410, 12314, 112243 and 444444 in the senary scale. OPERATION. Observe the sum of the first line is 14, which, divided by 6, the VI. radix of the scale, gives us 2 to set down and 2 to carry; the sum 84120 of the. seco d line is 16, which, divided by the radix, 6, gives us 4 8121 to set down and 2 to carry, &c. 13102 81410 12314 112243 414444 1144042 Am. Example 2. — From 43^6 take 9^09, in the undenary scale. OPERATION. Observe, here we say 9 from 6, we cannot, but 9 from 17 (1 bor* XI. Towed=ll and 6) and 8 remains, &c. 43^76 9^09 • ("-i 85068 142 TRANSF' >UMAT10N OF SCALES. [Pb'T. Ill r V ;■■■ a . 1 i-XAMPLE 8. — Multiply 8426 by 667, in the octenary ncale. RATION. OkEBATION VJIL 8i:!6 6(17 262(t4 21 66 Ohserve, we Bay 7 tlmrs 6 aro 42, 8 (th* rndix) Into 42 6 to carry aud 2 to set down ; 7 tiinoa 2 are 14 and 6 tuuko l'^, oqiiiil tu 8 to set down and 2 to carry, &o. 2460172 Ana. Example 4. Divide 671384 by 7876, in the nonary scale. OPIBATION. IX. IX. 7876)6 ri»84(75n? J An8. 61786, 62424 43828 7501 Here 7876 will oo Into 67188 7 tlnie.« (obsPrvo It would go 8 timeii la the decimal scale); and 7876 niultiplled by 7 gives 61 780. this being Pubtrncted, gives n remnindor, 6242, to which we bri. g down tbi- next digit, 4, and proceed as in voujinun d ri- fiion. Note. — After the units' figure is brought down, we may eitlier write the remainder in the form of a fraction, as in example 29, or we may place a point, and annexing Os, continue the division as in the following example. Observe, this point is called the decimal or denary point only in the decimal system. In every other scale of notation it takes its name from the system — thus, in the duodenary or duodecimal system it is called the duodenary or duodecimal point, in the senary system, the senary point, &c. Example 5. — Divide <134567 by e473, in the duodenary scdl^ OPERATION. XII. XII. «473)n34567CW- Ic, Sm, 95/06 768e6 67829 97807 95«!06 K91-0 e47-8 e45-90 te2 79 Exercise '61. 1. Multiply 262 by 252, in the senarij scale. Ans. 122024. 2. Divide 82e75721 by 62lv 'AMI by 6666, in the duodfinnri/ scale, ulns. I<36e296. 1 I. Dlvi.i;' iDlolOoboi by 100101, in tho binari/ scale. Ans. 10010 ruWtJT. 43 All the methods of proof ^iven in Sec. II., for the fiiiriaui'ti'al rules in the common Real e, apply to the 'arious oilier sciles ; but it must be remembered that, in using the priiK-'ip'C of the proof by nines for multiplication and divi- sion, we use, not nine^ but a number one less than the radix ot* t!^e scale. T)ms, In i»i.plyinct this prindplo to tho proof In Example 4, fievenn cast out of .'»7-'64, i:ive a ri'iriniiulor ij ; sevons cast out of 675, uivo a remainder 4, 4 x 8, Mil! .^^rw^cnsi out, give rt remuiuder 5; eovena cast out of 51117.344, give a ro- it llie radix be 12, we cast out the lis; if the radix be 6, we cast out the 43. Numbers containing digits to the right of the sep- ar.itijig point, ave dealt with according to the rules given ill Arts. 53 and 88, Sec. II. Example. — Multiply 37*14/3 by 6'le<, in the duodenary scale. Wfl place the separating point in the product ao aa to have sevt^n digits to the r pht of it, because there are four to the risbt of the point in the multiplicand and three la the multi« piier, and 4 + 3=7. (Art. 68, Sect. II.) ornit \TioN. XII ;3TI4<3 frht 833^549 3714!!:^ 19uS516 DUODECIMAL MULTIPLICATION. 44. The term hiodeciinal is commonly applied to a set of denominate fractions having 1 foot (linear^ square^ or cubic measure) for their unit. The foot is supposed to be divided into 12 equal parts, called jormes ; < achof which is divided into 12 equal parts, called seconds., throu<.'h/nnd /, parallel to aA\ they will di- vide the wnolo surface into tlio sniull figures, AhsK bsrc, &c. !/ And, since Afj=^l foot, and A/=l foot, AMj is a square ■• ■ foot.m likewise is each of tlio other flpurcs, o»rc, crrd, Ac H k e mO Now it Is evident that there aro as many vertical rows of those square feet as there are linear feet in AE, and as many squares in each row as there aro linear feet in All, that is in this case the number of eguare feet in the surface =4x3=12. As the same method of proof would apply in any elmilar case, it appears that — 77ie area of a rectangular mirface is found in square feet, and fractions of a square foot, by imdtiplying the number expressing how many linear feet, MH Hb wM !» ■ IKHS ^l^^Kt' '^1 • m ' ii 1 I lie DUODECIMAL MULTIPLICATION. lbi:cT. IIL Example 2.— Multiply 3 ft. 2' 1" 4'" by 1' 8" 1' OPERATION. 8 ft. 2' T ■ 4" 18 7 8 2 7 4" 9 7 10 1 10 6 8 4 4- 2" !"■ 8'- 8 4 -dTW. Exercise 38. 1. Multiply 4 ft. r 6" 10'" by 9 ft. T 11" 11'". ^ws. 44 sq. ft. 9' 1" 8'" 0"" 6'"" 2""". 2. Multiply 19 ft. 10' 3" by 11 ft. 2' 7". Ans. 222 sq. ft. 8' 0" 5'" 9"". 8. Multiply 9" r" 4'" by 7'" 3"" 11"'". Ans. 5"" 10'"" 4""" 11'""" 8"""" 8""'"". 4. How many square inches, &c., are there in a sheet of paper 9S inches and 5 inches 7" 4'" wide? Ans* 4' 6" 8'" 6'"' or 51^ sq. inches. 6. What is the superficial contents of a sheet of glass whose length is 7 ft. 4' 11" and breadth 3 ft. 2' 2" ? A7is. 23 sq. ft. 6' 9" 7'" 10". 61. The solid contents are found by multiplying to- gether the length, breadth, and thickness. Example. — How many cords of wood are there in a pile 79 ft. 8 inches long, 4 ft. 2 inches wide, and 7 ft. 11 inches high? OPSRATION. iST UBTHOD. 8EC0KD METHOD f7-8 79 8' 4-2 4 2 11^ 2268 18 8- 4" 818 8' 287-e4 7-e 881 11' 4" 7 11' 214848 141774 804 8' 4" 8" 2323 7' 4" No. of it in oord=;^)1626'^8(18 64469 duodeuary t% = — 20HHf com. scale. 76tf 714 bit 64-0 2627 10' 8" 8"+128. (number of ft In cord) =^HHf cords. 4n«. •A+ r*. ^c, of a aquare/oot. Arts. 61, 52.] QUESTIONS. 147 iiitriin Qiiiniti' Exercise 39. 1. Multiply together 15 ft., 1 ft., 1 ft. 2', and 8'. Aw^. 11 cubic ft. 8' = 11 cubic ft.. 1152 cubic in. 2. Multiply together 53 ft. 6 in., 10 ft. 3 in., and 2 ft. Alls. 1098 cubic ft. 9'. 8. How many cords of wood in a pile 10 ft. long, 5 ft. high, and 7 ft. wide? A71H. 2 cords 9t cubic ft. 4. How manv co:d3 of wood are there in a pile 4 ft. wide, 5 ft. 3 iii. high, and 70 ft. long? Ans. l\\\\. 5. What are the exact cubic contents of a block of mai'ble 4 ft. 7' 8" long by 9 ft. 6' wide and 2 ft. 11' thick? Am. 128 cubic ft. 6' 5" 2'". 6. How many bricks, 8 inches long, 4 inches wide, and 2 inches tliick, will it require to make a wall 23 ft. long, 20 ft. high, and 2 ft. 6 inches thick? Aiis. 33750 bricks. ^■'i 'M.iif 14 I m 7D MBTHOD. 52. It is sometimes askerl h'lw we cirn multiply f ct, innhcs, fee, by foct, inches, vtc, while we caiiiiot m iltiply pnimils, shilliagd a -d peace by pound.', Bhillinsrs mrl ponce. The answer is very simple. 1st. When we say that feet m iltipli-'fl by feet give square feet, wo morely use, as we have seen, (Art. 46), an abbreviattul form of e.vpr. ssion for the follo\v- Ing, viz: that "the number of sq laro fc^'t eoniainud in any rcctancrular s-nfaco, is equal to the prorlnct of two niunberd, one of which repnsenta the nnmb^' of liiiinir feet in one side; and the other the number of linear feet in tlie adjucent side." 2nd. When we are m-iltiplvins: tostethep primes, seconds, &c.. we .iro merely muUiplyinec together a sot of factors hnving 12 or powers of 12 for de- nominators; and when we say t.ha.t seconda multiplied hyfourfJiH, trive Nuthn: primes, multiplied by seoiuiti, give thirds; Ac, we simf>ly nie.nn that the pro- du t of any two of t'icse fractions is a fraction havinyrfor its denominator a i)0.v- er of 12, which power is in^licated by tlie sum of the imiic^'s of thi' factors. It is hence obvious that duolocimal miltiplioation affords i.o support what» ever to the idea that money may be multiplied by money. QUESTIONS TO BE ANSWERED BY THE PUPIL. Note. — 77c3 numbers after the qup.Htiona refer to the avtu lee of the Section, 1. What is the r.ieasiire of a number? (I) 2. What is the mitldpl-^ of a number? (2) 8. What is an integr f (3) 4. Of how many kinds are integers? (4) 5. What is an even number ? (5) 6. What is an add number ? (6) 7. What is a prime immber ? (7) 8. What is a composite number ? (8) What are the factors of a number? (9) Hy wliat other" names are factors known? (lO) 9 10. 11. 12. I'i. 14. 10. What is a common mea.^nrf of two or morn numbers? (11) When are two or more nwmXn'.ra prime to each other? (12) .\re idl prime numbers prim* to each other? (12) Are all compo.sito numbers prime to each other? (12) What are commensurable nurabcraV (13) 148 QUESTIONS. [Sect. III. ■ ' -i How do all perfect numbers terminate ? (18) 16. What are {ncommsvsiiraovfi nnmuers^ yt.-*) 17. What is a agunre number^ (15) 18. What is a cube niimbfr ? (16) 19. What, i.-i a perfect nutnhcr:' (17) 20. Mention suD'e rjfrlect numbers. (17) 21. What arw aniicohle numbers? Mention some amicable numbers, 22. What is meant ly the properties of'numher'H f (19) 28. Whiit is the sum of two or nmre evi-n numbers? (19-1.) 24. Wluit is the (iifl'erence of two eveii numbers ? (19-11.) 25. Whiit is the sum of 3. 5, 7, Ac., odd numbers? (ly-IV ) 2G. Whiit is the Si.m of 2, 4, 6. 8, &c . odd numbers? (18-V.) 27 Whnt is the si.m or difference of an odd and an even number? (19-VI.) 28. Vrheu is the pioduct of any number of factors c .i-n ■■ (19-(X.) 29. When is the product of any iiuiiibir of factors o^ld ? (19-XI.) 80. When will a number measure the sum, dijherente and product of two num- bers? (19-XIlI.) 81. If the number & be multiplied by any si gle digit to what is the sum of the diorits in the product t qual? (19-XVI.) 82. By wl.at is any number endins in divisible ' 19-XIX. «frc.) 33. By what is any i umber endi g in 5 divisible ? (l^-^^XJ 84. By what is any number endinfr in 2 divisible ? (19-XIa.) 85 When is a number divisible by 4 ? (19-XXlI.) 86. When is a number divisible by 8? (19-XXIII.) 87. When is a riuinbtr divisible bv 9 ? (19-XXlV ) 88. When is a number divisible by 8? (19-XXV.) 89. When is a number divisible by 11 V (19-XXVI.) 40. Show that every composite number may bo resolved into prime factors. (19- XXVII.) 41. Show that the /eos^ divisor of any number is a prime number. (19 XXVIIT) 42. With what digits must all inime numbers except 2 and 5 terminate ? (19- XXXI.) 48. TTow do y.u find the prime numbers b^^tween any limits? (20) 44. What is this process called and why? (20) 46. When it is required to ascertain whether a given number is prime or not, what is the first thin:; we do? (20) 46. When we try the primes of the table as divisors, which Is the highest wa need use ? (20) 47. Why is it unnecessary tc try any divisor greater than the square root of tLn number? (20) 48. How do we resolve a composite number into its prime factors? (21) 49. By what numbers can a composite number be divided ? (21 -Note ) 60. What is the rule for finding all the divisors of a number? (22) 61. How do we find simply how many divi^-ors a number hns? (28) 62. What is the srreatest common measure of two or more numbers? (24) 63. How do we find a common meawire of two or more numbers? (25) 64. How do we find the greatest common measure of two numbers? (26) 55. Prove the rule in Art. 26. 56. How do we find the G. C. M. of three or more numbers? (27) 67. What is the seco-d method of fl ding the G. C. M. ? (28) 68. Upon what principle does this method rest? (28) 59. What is a common multiple of two or more numbers? (29) 6(1. What is the least ommon multiple of two or more numbers? (30) (il. Give the first rule for findinsr the 1. c. m. of two or more numbers. (81) 62. Give tlio second rule. (82) What is the reason of this rule? (82) 63. Give tlie most convenient and expeditious rule for finding the 1. c. m. of several numbers. (03) 64. What is meant by the radix or hase < f n system of notation ? ( M) 65. How do we read numbers in difierent scides? ("5) 66. E.xpress the number 284213 quinary as in Art. 86. $7. What does the e.\pression 10 always represeni? (37) fiKOT. til.] MISCELLANEO. S EXEllClSfi. 149 fi8. What \a the highest diirit used in a y scale? (38) b',*. llovv do we rediicH a number tVoin one scale to another? (39) 70. Wliat is the rule tor lruii.-loim»ng a number from any scale into the deci- mal? (40) 71. How are tiie fundamental ofier.itions carried on 1 " the diflFerent scales? (41) 7'2. Ho Ti. llow are operation.s in tiie ditl'erent scales proved ? (42) 74. What are diiodeoiuials? (44) 75. Give the table of duodecimals. (44) 76. What is a prime? (45) 77. How is the area of a rectangular surface found? (46) 78. What is the rule for duodecimal multiplication? (48) 79. Uow may the rule for finding the denoujii.ation of the product be concisely worded? (48) 80 How are solid contents found ? (51) 81. Show that duodecimal multiplic aior> affords no support to the idea that money may be multiplied by money, &c. (52) Exercise 40. MISCELLANEOUS EXERCISE. On preceding rules. 1. Add together $'729'18, fYlo-SO, $166'78, £93 148. V^d., £276 19a. lOid., $497-81, and £275 4s. llfd. 2. Multiply 47 miles 6 fur 17 per. 4 yds. 2 ft. 7 in. by 576. 3. How many divisors has the number 243uOO? 4. From 713427 octenary take 4234434 quinary and give the answer in both scales. 5. Divide 79-342 by -00006378. 6. Express 79423 and 231567 in Roman numerals. 7. What is the 1. e. m. of 5, 7, 9, 11, 15, 18, 20, 21, 22, 24, 28, 80, 33, 35, 36, 40, 42, 44, 45, 48, and 50? 8. Give all the readings of 376342. 9. Multiply 64276-3427 by 9999993000. 10. Transform 78263 nonary into the quinary and undenary scales and prove the results by reducing all the numbers to the sep- tenary scale. 11. Form a table of all the prime numbers less than 200. 12. Reduce £rt72 7s. 7d. to dollars and cents. 13. What is the G. C. M. of 243000, 891, 37800 and 35100? 14. Give all the readings of 6 yards 3 qrs. 3 nails 2 inches. 16. Write down as one number, seven hundred and forty-two quln- tillions, nine hundred and five billions, seventy-eight thousand and fourteen, and eighty-seven million, two hundred thousand and eleven tenths of trillionths. 16. Read the following numbers — 71300100200401-000000070402 134900101000100100-000200020002 47000uOOOi.i020007-0000000000027S ^ I'Ji 150 VULGAE FRACTIONS. [Sect. IV 17. Add together £178 Ifis. 4|d., £97 15p. lUd., £693 19s. llfd., £210 lis. 9id., £678 14s. 7id., £197 138. llfd., £117 6s. 5d., and £91 Is. Ifd. 18. What are the prime factors of 276000? 19. Multiply 6 ft. 2' 7" 9" 10'" by lo ft. 11' 11" 11'" 7"". 20. Divide 7 1 22. Since fractions, like integers, are capable of being increased or diminished, they may be added, subtracted, ^ ; 6, f ; 29, \»-, Ac. REDUCTION OF FRACTIONS. 24. Since (Art. 8) multiplying both numerator and denominator by the same number does not alter the value of the fraction, we may reduce an integer to a fraction having any proposed denominator, by the following: — Arts. 20-25.] REDUCTION OF FRACTIONS. 155 ex fractions. f unity, but of tb» RULS. Write the integral number in the form of a fraction having I for its denominator. (Art. 23.) And multiply both numerator and denominator of the resulting expression by the proposed denominator. (Art. 8.) Example 1. — Reduce 16 to a fractioD having 11 for its deuomi- nator. Ekamplf. 2. — Reduce 173 to a fraction having 31 for its denomi- nator. Exercise 41. 1. Reduce 29 to a fraction having 12 for its denominator. Ann. *,*«*• 2. Reduce 243 to a fraction having 3 for its denominator. Atis. ^^*. 8. Reduce 7, 23, and 101 to fractions having 13 for denominator. Ans. f^, Vrf^ ^^\^' 4. Reduce 4, 37, 126, 73, and 1007 to fractions having 101 for de- nominator. 6. Reduce 204, 7011, and 1999 to fractions having 207 for danomi- nator. li' 26. Let it be required to reduce the mixed number 8/^ to an Improper fraction. 8pV 13 equal to the whole number 8, and the fraction /y, and by (Art 24.) 8=ff , therefore 8rV=f f + Ti=?f • Hence, to reduce a mixed nnmber to an improper frac- tion, we deduce the following — RULE. Multiplying the whole member by the denominator of the fraction^ to the product add the given numerator and place the turn over the givx... denominator. Example 1. — Retluce 73^ to an improper fraction. OPKRATioN. Explanation.— We multiply the whole number, 78, by 9 and 78| add in the numerator, 4. This gives us 661, which we write over 9 the given denominator, 9, and the resulting fractiou, 2|i, is the — improper fraction sought. 4|i Am. Example 2. — Reduce 276^5 to ^^ improper fraction. *• ., «' 156 bEt)UCTION OJ* PRA0TION8. Exercise 42. ta«0T. IV 1. Reduce the mixed numbers, 78-,^-, 18-,V, and 128^?, to improper tr0c*".;n8. Ans. ^'^\ ^.'^jii, and ^ri\ 2. Reduce i mixed numbers 884|, 673 ,»j, 4792/5, and ^m^^ to impro -r fractions. Ans. ^%^^, ^^U^, ^U^^, and ^V/L 26. Since every fraction indicates the division of the numerator by the denominator — to reduce an impn^per fraction to a mixed number, we have the following — RULE. Divide the numerator by the denominator and the quotient mil ht tlie required mixed number. Example 1. — Reduce ^^^ to a mixed number. a?4=204+7=29M"». Example 2. — Reduce 2^21)^11 to a mixed number. 20047 -»- 11=1 822 iV-^n*. Exercise 43. 1. Reduce the improper fractions ^^^^ ^^W and ^-y^-P- to mixed numbers. Ans. 81-,^, 474**;,, and 1675^' 2. Reduce the improper fractions ^^H^, ^^oS and ^1^^^ to niixod numbers. Ans. 88^^, 158^^, and 78, 27. To reduce a fraction to its lowest terms — RULE. Divide both terms by their greatest common measure. This is simply dividing both terms by the same number— which does not afifect the value of the fraction. (Art. 8.) The greatest common measure may be found by (Art. 26, Sec. III.) or, very frequently, by inspection. Example 1. — Reduce f g to its lowest terms. Greatest common measure=25. Dividing both terms by 26; f|=l Ans. Example 2. — Reduce ^|f to its lowest terms. Greatest common measure of 126 and 162=18. Dividing both terms by 18 we get Tsa=S Ana. Exercise 44. 1. Reduce ^f |o to its lowest terms. 2. Reduce |tM8 to its lowest te ms. -4ns. Tin. 28. tneasure (luce the us the tliefiac NOTI Dumber! ._ a a a « » — \ :i < i if a7H 1 _ 30U — fffiS 5= 551 ■ 1. 2. S. 4. 5. ;'■ -i ! tSECT. IV I ^j„ 26-29.] le quotient mil ht REDUCTION OF FRACTIONS. 157 8. Reduce HtH *"^ tiJ *o t^^''" lowest terras. Aru. ^ and |. 4, Reduce Ji^g, iWi and ^/JJ^f to their lowest terms. Ans. H, iV, and^UJg-. 28. Instead of dividlnrj both terms bi; their greatest common tnea!f any common meaxure. We /tun rt- dnce the fraction to lower termx^ atid^ continuing the diviuou as long «,< the terms have a common measure^ we shall finally have reduced tlie fraction to its lowest terms. Note. — It is advisable to commit to memory tlie properties of numbers given in Art. 19, Sec. Ill, from XVIII to XXIV. ExAMPLK 1. — Reduce ^HfioJ to its lowest terms. ?J""-" dividing by 10. (XXI. of Art 19, St>c. Ill ) z=^l\m i\UU\u\g by 8. (XXIII. of Art. 10, 8ec. I 'I.) = im (lividln- by 9. (XXIV. of Avt. 10, S..c. Ill ) = JHs dividing by 3. (XXV. of Art. 19, Sec. Hi.) Example 2. — Reduce ||?f to its lowest terms. m\ dividing by 5. (XX. in Art. 19, Sen. III.) -= IjI dividing by 9. (XXIV. in Art. 19 Sec. III.) .-= I j- dividing by 3. (XXV. In Art. 19, See. III.) :s ll Alls. Exercise 45. J. Reduce ^J to its lowest terms. 2. Reduce Tlb^^u to its lowest terms. 3. Reduce H^IH^& to its lowest terms, i. Reduce iViVo" to its lowest terras. 5. Reduce -^g, ^^iVj and if JS^ to their lowest terms. Ans. -j^i-, ^l^:, and ^U- Ans. H- Ans. -jH,V(j. Ans. ^. Ans. ^Vtf- 29. To reduce fractions of different denominators to equivalent fractions having the same denominator — - ^i 1 i'''i i '1 M^l ; I'm )y 25 ; f f =J Ans. RULE. Multiply each numerator by all the denominators except its oion for a ne%o numerator, and all the denominators together for a new denominator. This is merely multiplying both numerator and denominator of each fyoo- tion by the same quantity, viz: the product of all the other denominators, and conseqi ently (Art. 8.) it does not alter the value of the fraction. Example 1. — Reduce f, -,^,- and f to a common denominator, 8 X 11 X 9=:297=l8t numerator. 7x 4x 9=252=2nd numerator. 6x 4 xll=220=8rd numerator. ' 4 X 11 X 9=396— common denominator. Tb9r«fore the e<|uivaleiit fracUons are 2 g 2, i||, and ||}, 1 , '■* 3. 158 REDUCTION OF FKACTI0N8. [Sbot. IV. Example 2. — Reduce i, |, ^, and ^^ to equivalent fractions hav- ing it cornojon denominator. 1 V 6x7x11 =886=1 «t numprator. 8 * 2 X 7 X ll=462=2n(l numerator. 4 - 2 x6 X 21=440- 8n1 numerator. « X 2 X ft X 7=680=4tli nurnerntor. 'J vfty 7 y ll=r:770=coninion denominator. And the equivalent fractious are j'JSi *?8. tJil »D H^^ft, jW.^o. Reduce -i**!-, ]i^ and V!* to fractions having a common denominator. Am. im,Hthi\hh. Reduce ?, -n-, -j^j, ^ and ^ to fractions having a common denouii- nator. .4n.,. \n\h -.^V,«4, AW, ^'^^^^. and -.^HJ'.V. Reduce -j^, f , and -^"j to a common denominator. J„o .6 4 4. ^Tl- «n/1 6iC. ^»iS. 100 l» TOOI» «"" 1 00 1* Reduce ^, ^, ^, and j^,- to a common denominator. ^ns. Kn, iMH,H!S,and5V,V Reduce i, |, f , and ^ to a common denominator. 6. 30. To reduce fractions to equivalent fractions having their least common denominator — RULE. Find the least common multiple of all the denominators. (Art. 88, Sec III.) Multiply both terms of each fraction by the quotient obtained by dividing this least common multiple by the denominator of that fraction. This is merely maltiplyiag both terms by the same qaantity, as in Art. 29. Example 1. — Reduce i, i^, f, and ^\ to their least common de- nominator. The least common multiple of 4, 12, 8, and 16, is 48. Multiplying both terms of the 1st fraction by 12 (i. e. -V-) 1* becomes ff. " " " 2rHl " by 4 (i. e. fj) it becomes JJ. •* " ♦• 8rd " by 16 (1. e -V) it becomes Vi- *• M u 4tb .. ty 3 (1. e. ff) it bncomrs |J. The equivalent fractions bavliig their least common denuroiaatur, are therefore if |f, ||, and |J. ii^;- tg Ai.t:<. 0.1 C .J KliDLCTlON or FRACTIONS. 159 \-fi mAl^%.u^^ «, H&, and ifi.V Fx.v.MPLn 2. — Redueo J, ,*'(, Jg, J^, J2i a^^ I *^ ^^^e'r ^caat com- mon (Icnomiaator. Tho loftst common multiple of 5, 11, 20, 44, fiS, nnd 4, Is 220. Tho iiiiilti,illi'r for bulh turma of the first t'niution U 22*^=44, f<»r second, .',',•--'20; f<.r the thirrl, Vo° = ll; 'or the fourth, V4*=5; for the flltb, \\'=*\ M(l for iho sixth, ^'^-^.W. Mul iiilyin:z by these numboro, we obtain J{||, JJg, }J8, iJJ, jVa, and iJJ for the rcqulrod fractions. Exercise 47. 1. Reduce jj, J, ^, J, and / mh HU> HU, and ^UJ. 8. Reduce \^^ |, f, H, /j-, ^jj, f, and |^ to their least common de- nominator. Ans. mh ini W4^o^ hh, mh %m. im. and im- m I linators. {Art. 31. Let it be required to reduce ff of y^ to a simple fraction. II uf rV means t2 times tV of ^. We get ^, of xTi '• e divide t"t by 17, when we multiply the denominatoT 11 by 17 (Art. 8). Therefore yV of iV=Tr"-TTi and to multiply this result by 12, we multiply the numerator, 6, by 12, (Art 8.) Therefore \^ of i\=^-;^=/^. 11x17 Hence to reduce a compound fraction to a simple one we deduce the following — RULE. Multiply all the numerators together for a new numerator ^ and all the denominators together for q new denominator. hi, ; ^1 3.11 i Bii'' 160 REDUCTION OF FRACTIONS. [Sect. IV. Example. — Reduce § of ^ of to a simple fraction. Note.— Id all cases the answer must be reduced to its lowest terms. Ans. ■^\: Ans. -,\. Ans. -^Q. Exercise 48. 1. Reduce f of | of -,^,- of f f to a simple fraction. 2. Reduce f of | of ^ of -, yfo of f^ to a simple fraction. 8. Iicduc' ^^ of 1^1 of 3^ to a simple fraction. 4. Reduce | of f of i^,- of |? to a simple fraction. Ans. /sW 32. Since the several numerators of the compound fraction form the factors of the numerator of the simple fraction, and also the several denominators of the com- pound fraction, ohe factors of the denominator of the sim- ple fraction, it follows (Art. 8.) that, — Before applying the rule in (Art. SI) we may east out or cancel all the factors that are common to a numerator and a denominator of the compound fraction. Example 1. — Reduce -^ of ^ of f of |^ of \% to a simple frac- tion. STATEMENT. OANCEtLED. 2 2 3 6 4 8 22 85 6x4x8x22x85 ^ x ^ x 3 x ^2)2 x 3^ 1 ;;rx7x^x;2/x;^"3 ^''*- — of-of-of— of — = 11 7 6 27 16 11x7x6x27x16 8 Here 6 and 27 contain a common ftctor, 8, which Is cast our, and these numbors thus reduced to 2 and 9. Next thJM 2 reduces 16 to 8, and the 9 is re- duced to 8 by the third numerator, which Is thus cancelled. A$;ain, 11 cancels n (the first denominator) and reduces 22 to 2, and this 2 reduces the 8, befopi obtained from the 16, to 4. Next, this 4 is cancelled by the 4 in the numerator. Ajraii, 7 cancels the 7 in the deiiominator and reduces the 86, in the numerator, to 5, and this 5 cancels the 5 in the denominator. All the numerators are n'>w reduced to unity, as also all the denominators but the fourth, which id 3. The 1x1x1x1x1, resultiug fraction is therefore 1x1x1x8x1 but this is simply ). Example 2. — Reduce -ff of ^ of f of ^^ to a simple fraction. STATEMENT. 7 4 8 65 — of-of-of — = 11 6 5 20 11x6x6x20 7x4x8x55 _ 7x^x3x^^ 2 6 OANCBLLEP. 2x5"l0^'"' Note.— If any of the ti-rms of the compound fraction are whole or mixed numbers, they miist be reduced to fractions (Arts. 28 mnd 26.) The process of cancelling excippUQed ftl^QV^ §l)ould always be adooted when possible, [Sect. IV. tion. lowest terms. Ans, -,^,-. ion. Ans. -^\. Ans. Jq. the compound ' of the simple rs of the com- -tor of the s!m. ^ast out or cancel a denominator oj to a simple frac- I ^ 8 i cast our, and these 8, and the 9 is re- • Afraln, 11 cancels duces the 8, befop) t In the numerator. ), tn the numerator, umerators are n')W 1, which id 8. The pie fraction. CtLEI>. 2x5 10 ^'"* re whole or mixed I Iwoys be adooted Abts. 32, 83.] EEDUCTION OF FRACTIONS. Exercise 49. 161 1. Reduce ^ of f of § of -^^ to a simple fraction. Ans. -■^. 2. Reduce % of ^ of -^^^ of /j- of j^ of \'i to a simple fraetirm. 3. Reduce f of -,*,- of 5^ to a simple fraction. Ans. j. 4. Reduce ^ of -^^ of ^^l of -/'^H,- of |f of 2^ to a simple fraction. Ans. -/i". 6. Reduce ,\ of ^ of ,\ of |^ of f 2 of 6^ to a sinople fraction. 6. Reduce ^ of -|\ of 154 to a simple fraction. Ans. ■^. Ans. 24. y 33. Let it be required to reduce the complex fraction -, to a simple fraction. '* Since (Art. 8) we may multiply both numerator aid denomi ator of a frac- tion by the samo lujnil^er, wiriiout alteriiisr its value — we may multiply both terms of ttu' jiiven fcotion by S, i. e., by the denominator with its terms in- verted, without alteiiny its value. Therefore I = J-^-^ = ^J =.,.=. J 6 y 4 x"3' Hence to reduce a complex fraction to a simple one, we deduce the following : — RULE. Reduce the expression {Arts. 23 and 25) to the form of '- ~- . frution ' i, c, reduce both numerator and denominator to simple fractions. TJten nniHipl>i the extremes or outside numbers tofjvther for a new numerator^ and the means or intermediate numbers tvyetloer for a new denominator. Example 1. — Reduce _ to a simple fraction. 4^ ft 7 TT 9 X 11 99 14 = 7 A Ans. Ti NoTF.— Factors that are common to one of the extremes and one of the means, are to bo struck out or cancelled. i,Art. 32). Example 2. — Reduce -^^ to a simple fraction. 9 E n_ ^ 7x_9 ^ 63 ^ Q.a. ^^^ ^0 10 10 10 I ■ , ■!: Illlii: 162 EBDUCTION OF FEACTIONS. Exercise 60. [SaoT. IV. H 1. Reduce nr to a simple fraction. Ans. ^. H 2. Reduce hTI to a simple fraction. Am. ^. 15| 8. Reduce -=t to a simple fraction. Am. 2. 11| 3i ^f 4. Reduce j^k' "^ '"<^ | to simple fractions. -4ns. ^5f , il, and H- 6. Reduce — ~ » -5- and -^ to simple fractions. Am. gVi 3H, and -j^. 6. Reduce J4|^ tI' 1^1 ' ?^ a^^ minute=-i-Y"* minute. Hence, to reduce a denominate fraction from a higher to a lower denomination, we have the following — RULE. Take the number expressing how many of the next lower denomi- nation make one of the given denomination ; also, the number, ex- pressing how many of the next lower again make one of this aenomi- nation, and so on till the required denomination be reached. Write the fractions formed bji these ivimhers as namerafors, vrith 1 as denominator, as the gir)cn fraction in the form of a compound fraction^ which reduce to a simple fraction. {Art. '61 ) Example 1. — Reduce ?| of a £ to the fraction of a penny. % of Y of ■L,^ = i§'i=lGO pence. Example 2. — Reduce ^ of f of \\ of a furlong to the fraction of A foot. % of $ of H of Y of V- of f =300 ft. An9, mil M i {'■ 164 EEDUCTION OF FRACTIONS. Exercise 52. [Sect. IV. 1. Reduce 4i5^ of a bushel to tlie fraction of a quart. Ans. ^9^ Qt. 2. Keduce ^ of a gallon to the traction of ^ of ^ of a gill. Ans. ^^K 8. Reduce § of 2 pecks to the fraction of ^ of | of a pint. Ans. ^K 4. Reduce ^} of a pound to the fraction of a scruple. ^4;?,"?. ^W^ scr. 5. Reduce t^bHitt of ^ of J of -^j; of ^.p- of a lb. avoiidupois to the frac- tion of a dram. Ans. ^V?^ di'« 38. To find the value of a denominate fraction in terms of a lower denomination — RULE. Divide the numerator by the denominator according to the rule given in Art. 71, Sec. II. This is only actually performing the work which the fraction indicatefi. (Art, 8.) Example. — What is the value of j^ of a mile? 11 miles -»- 13 18)11 miles (6 fur. 80 per. A^^ yds. An9. 8=fur. in a mile. 88=number of furlongs. T8 10 40— perches in furlong. 400= perches. 890 6j=yards in a perch. 55=number of yards. 62 11 ExEuciSE 63. 1. What is the value of -,\ of a bushel and also of ^ of a lb. avoirdu- pois? Ans. 1 pk. gal, qt. Ifr pt. and 13 oz. llf^ drams. 2. What is the value of -,^a of a yard of cloth ? Ans. 2 qrs. na, 1 ,Aj inches. 8. What is the value of | of a lb. troy ; and altfo of -^^^--^ ^q, mile ? Ans. 10 oz. 13 dwt. 8 grs. ; and G2 acres, 1 rood, 8 sq. per. 4 sq. yds. 2 ft. 79x1^ in. [Sect. IV. ARts. 88,30.1 REDUCTION OF FRACi'IONS. 1C5 Ans. V/ qt. gii 1. A)is. H\ int. Ans. HK Anti . aiifl scr. ois to thf t htc- Ans. ^^ dr. fracti ion in Iff to ike ruii iction indicaten. IS. 4. What is the value of 9 of a furlong ; and of f of a £? A71S. 35 rdd. 3 yds. ft. 2 iu, ; and lis. 5}d. 39. Let it be required to reduce 28. "jd. to the fraction of £7 18s. 2s. 7}d 127 f irthinsrs. 127 — — Therefore 2s. TJd.= of £7 18. jE7 ISs. 75S4 furthiiigs. " 7584 Hence, to reduce one denominate nunaber to the fraction of anoth- er, we deduce the following — RULT!. Reduce both quaniilies to the lowest denomination contained in eitlier. Thcyi place that quantity/ which is to be the fraction of the other as numerator f^'d the remaining quantiti^ as denominator. Example 1. — Reduce 3 days 4 hours to the fraction of a week. 8 days 4 hours = 76 hours. 1 week=16S hours. And the required fraction is j'u'a = l? ■^'**' Example 2. — What fraction is 3 lb. 4 oz. 2 dr. 2 scr. 7 grs. of 63 lb. 4 oz. 7 dr. Apothecaries' weight ? 8 lb, 4 oz. 2 dr. 2 scr. 7 grs. = 19367 grs. 68 lb 4 oz. 7 dr. =365220 grs. And the fraction is sWiVs ■^^**- I'M t lb. avoirdu- • llf drams. 1 Rf inches, •q. iniie? 8 sq. per. 4 Exercise 54. 1. What fraction is 6 bush. 1 pk. 1 gal. 1 qt. 1 pt. of 50 bush. ? Ans. ^^^}is. 2. What fraction is 35 per. 9 ft. 2 in. of a furlong? Ans. f. 3. What fraction is 7 h. 12 ra. of a day? Ans. ■^. 4. W^hiit fraction is 2 sq. yds. 2 ft. 120 in. of 3 sq. per. 13^ yds. 1 ft. 72 in.? Ans.:^. 5. What fraction is 7 oz. 7 dr. 2 scr. 14 grs. of 21 lbs. Apoth.? Ans. -ii\jS' 6. Reduce 9 min. 48 sec. to the fraction of a day. Ans. 7200* 7. Reduce 16 bush. 1 pk. 1 pt. to the fraction of 69 bush. Ans. -,^4SV' 8. Reduce 3 qrs. 3^ na. to the fraction of an ell Eng. Avs. ||. 9. Wliat part of a lb. Troy is 13 dwt. 7 grs. ? Ans. oVeV 10. What part of 54 cords of wood is 4800 cubic feet? Ans. f|. 1G6 ADDITION OP rULOilE FtlACTlONS. ADDITION OF VULGAR FRACTIOXS. (Sect. IV 40. Addition of fractions is the process of finding- a single fraction which shall express the value of all tho fractions added. Addition may be illustrated as follows : — i ^ I ■ nil Ml I II H ■!■! DM] 'f m 41. In order that fractions may be added they must have a common denominator. Thus l + l makw neither ^ nor |; but if we reduce them to equivalent fiac. tlons having a common denominator, as xi ^^^ t5» we are enabled to add them and thus obtain for their sum }l. These fractions, before and after they receive a common denomi^ nator, will be represented as follows : — Unity. § 1 ^ I equal to equal to "We have increased the number of the parts just as much as we have diminished their size. 42. For the addition of fractions we have therefore the following : — RULE. Reduce compouvd and complex fractions to simple owes, and all to a comtnon denomi7iator. {Arts. 29, 31, and 33=) Add all the numerators together, and beneath their stem place the common denominator. Reduce the resulting fraction^ when it is an improper fraction, to a mixed number. (Art. 26.) Note. — If mixed numbers occur among the addends, the integral portions arc to be added separately and their sum added to the sum, of the fractions [Sect. rV xVS. of finding- a ue of all tho -d they inus( > oqiiivaloiit fl-ac, Wed to add thern mmon denomi^ much as we lerefore the neSf and all 'M place (he fraction, to the i7itegral to the s^um AbTS. 40-42.] ADDITION OF VULGAR FRACTIONS. J[g7 Example 1. — Add together -,*,-, -^,-, -^^ f^f and \\. Here, since the fractions have already a comnmn denominator, we have gin'uly to add the nuuierators and place 11, the commoa denominator, beneath their sum. Thus TV+A-HT\-^T\^-H= '*''^''^\''^''^^ =rr=2TV ^n«. Example 2. — Add together f, ^, f , f and \\. These fractions reduced to their least common denominator by Art. 80, be- come f », u, u, n> \h And ?a+g4 + ?5 + J.! + J|=?5i^i^— ^^^=Va'^=H=8T'* An,. Example 3,— -Add together ^ ^, -A- and ^ of ^f A of tl of 5^. i of * of W of 1? of 5J is equal to J (i4r#. 81). The fractions to bo addi-d iire' therefore ? + J + fs + J. Theso reiiuced to a common denominator {Arz. 29), bttcome J a 'JO . 2 40 4 I asao 1 aesi — huum — OiiBSB J/na o5o~ SoaS ~ 3?)«o ' SobU — 3obO — ^'Soui ■""<'• Example 4.— Add together 9^, 11 J, 16 J, 43 J, and rr Heve the last frnetion is a complex fraction and is equal to j|. 9i + llJ + 16i+43? + i=:9 + ll + 16 + 43 + a + | + -J + S + *). And 9 + 11 + 10 + 43=79. Also ,f + 4-1- a + if -^H — n«o + 3iTo + 3frr)+ SifTj + Sno — Ti-fcT— "jcny* Therefore the sum of the given quantities is 79 + 8^^=82/^. Example 5. — Add together f , f , aud 5|. Here addin^ the three fractions together we obtain YnV' ^^^ their sum, to ^hlch we add the integral number 5 and thus obtain the entire sum 6|^. Exercise 65. 1. Add together ii -ff and yY Ans. ^—2^. 2. Add together ^V, A. tV» iS^ \h and ^'^, Ans. ^. ■ 4 — "4* 3. Add together 4^, II4, 16f, 215 and l^. Ans. 71-i-V-=73f 4. Add together I6|i, llii, \S^\, 17if, and 112||. Ans, 177^f. 5. Add together 4}, 1^ and ^j. 6. Add together i, %, 5., 1, ^^ ^, 1 and 7. Add together ^, f , and |. 8. Add together |, f , f , f and -j^p 9. Add together ^, -•-, J^, 1, ^ and |. ^725. 2^3-. ^ns. 3^fii. Ans. 1/jV 10. Add together 16^3^, 47a-, 2IJ4, y'V, and 19, Ans, 104^|. 11. Add together 17^, 43f, IGS-^-, 207^^ and 506|af. ^ws. 943JJ. m 1C8 ADDITION OF V ULG AE Fli ACTIONS. [Skot. IV. 12. Add together 6a, II4, JL, 16/^, h ^t and 17|i 4ns. 53 J? 3 A71S. GOJ-ei 13. Add together 1, ?^, i and 681. 14. Add together 173,?^, 84 and 91 ff 15. Add together lif, 2||, 3|i and 4|a. Ans. 13a 3 M6' l«S. 273 2 9 A.. JL. J 16. Add together 17. Add together 7, ili,"l8",'2(l| and"79yV 8) 12» 48» 24' 16 -^^, f , A, and 4. .^ws. 3/ 4« .^ws. 142 M 18. Add together |, 7/y, and f of ^ of 10^. Ans. 1 1 y U 19. Add together ^ i of 3^ of j% of 2^, and 20i ".4 5' i\ • 1 I ;i 20. Add together 3f, 111 and 14 ■r 'r\t A .1 J ^ xl 1 _i? o a _ 1» a Ins. 15 4 8' J ^_ 14 0' Ans. 292 a 1. Add together 1 of 5% of f, |"of f, | of l/^ and Vi of I of i of i of 1 of 1 Ans. 112 6 1 22. Add together 4U, 105§, 300?, 2413 and 472i J 6 a u' 2) 9> '4» ■S' 23. Add together 92y\, 37 yV and 7f ^ws. 11 61? 5 !.»" ^/i5. 137-2.!^ Ty 8 103 24. Add together 21^, 35^, — ^ and | of f. ^ns. 61f 22. 25. Add together 23 of 3|, JJ^S 2 J of 4i of If, and 4 0fy3j0f2i0fl^ ^^5. 34 H? a 14 4 0- 43. In order to add denominate fractions they mnst not ovly have a common denominator^ but they must be fractions of the same unity i. e., must be of the sam • denomination. Thus £f, fs. and Jd. cnnnot be added together, as the result would be neither f of a pound, f of a shilling, nor f of a penny. But if we reduce them all to the frnction of a pound, or nil to the fraction of a sbillinp, or all to the fraction of a penny, it is obs ious that we may then add the resulting fractions, Laving first reduced them to a common denomina- tor. Hence, for the addition of denominate fractions, we have the following — RULE. « Reduce all the fractions to the same denomination {Arts. 86 and 37). Reduce the resulting fractions to a common denominator {Arts. 29 aoid 30). A dd (as in Art. 42) and find the value of the resulting fraction {Art. 88). ^18.48,44 EXAMI r^h.+ EXAMI of a penn] 152A-+4| Note.' of each fri Exam) a gallon. 1. Wha 2. Add 3. Add 4. Whj 2 2 5. Whi 6. Ad( 7. Wb £ A4 findin We denomi ulso of tractioj [Sect. ly. AUTO. 48, 44.] SUBTRACTION OF VULGAR FRACTIONS. 169 1 1711. ns. 53j»3 ^'5. 691 «!. *. 2733? 5 , « •* '5 • ■ /25. 13^-' •» » s. 142X''-. 20| "^ I*. 15 J, ^-. ns. 29 j' 3. 7o and 4i *• lH^'-" ^21.'" 1375 'i 7 y 8' Iws. 61 f. , and 4| 341133 J 44 0- would be le fraction "iiiy then denomina- )ns, we ■ 86 and r {Arts, esultinf) Example 1. — Add together f^ of a day and ^ of an hour. S of a day ^ 2 of V= V =-'3' "f an hour. JL4h.-f-fh.= Vf + 2"r = ¥r'-=5^?li.=5h. 35m. 42f sec. Example 2. — Add together -^^ of a pound, f of a shilling, and ^ of a penny. ,V of a £ -,V '>f -V- of -V- = 'tt" of a penny=152A pone 2 of a shilling = J of u = "A of a penny = ^ pei.ce. 280 + 308 + 165 152A+4R^=156+ =157M! pence=133. l^fd. 385 Note. — In place of proceeding as above, we may find the value of each fraction separately (Art. 38) and add the results. Example 3. — Add together f of a bushel, ^ of a peck, and -j\ of a gallon. j of a bushel = 3 pks, cal. 1 qt. 11 pts. I of a pork = gal. qts, y\ of a gal^_j^ li\ P^8- Sum = 1 bush. pk" g^.., \ qt. O^g pta. An9. Exercise 56. 1. What is the sura of y'r '^^* Apothecaries' weight, ^ oz. y*y dr. and | scr. ? . 1.9. 4 oz. 6 drs. 2 scrs. IBi^f- grs. 2. Add together ^ yd. | ell Eng. and -f qr. ylw5. 3 qrs. 3 na. Iffa in. 3. Add together 4 of a yard, ^ of a foot, and | of an in. Ans. 7 inches. 4. What is the sum of ^j of a mile, ^ of a furlong, and 2^ of a yard ? Ans. 5 fur. 16 rds. yds. ft. 3/^^^ in. 5. What is the sum of 1 wk. i day, 1 h.? Ans. 2 days 2 h. 12 m. 6. Add together £1, ?^s., and y^d. Ans. 3s. Ig-^d. 7. What is the sum of f of 21s."| of 5s. s. of £3 12s. 6d. £-i7_ and ||d.? ^«s. £3 12s. 4i|d. %^ SUBTRACTION OF VULGAR FRACTIONS. 44. Subtraction of vulgar fractions is the process of finding the difference between two fractions. We have seen that before fractions can be atldcd thoy must have a common denominator and tli it when denominate fraction.s a; o to be added they must ho also of the same denominatioi/, aud this is manifestly the case a'so in the sub- traction of fractions. I m ,if 170 SUBTRACTION OF VULGAR FRACTIONS. [Skct. IV. Hence, for the subtraction of fractions, we have the fol- low ing : — RULE. Reduce compound and complex fractions to simple ones and all to the same denomination, if not already such. Reduce both of the resulting fractions to a common denominator. Subtract the numerator of the subtrahend from the numerator of the minuend^ and beneath the difference write the comtnon denominator. Note. — In the case of mixed numbers it frequently happens that the fractional part of the subirahond is greater than the fractional part of the minuend. When this occurs, instead of reducing both quantities to iniproper fractions and then applying the rule, it is much better to borrow unity from the integral part of the minuend and coH' Bidering it as a fraction, having the common denominator, add it tci the fractional part of the minuend. (See 3rd, 4th and 6th Example,') below.) Example 1. — From f take iV Here reducing | and iV to ji common denominator they become ^ and x>rV 8^ Example 2.— From f of f of ^^ of 49 take -^ of i of i Here J of ^ of 5Vi of 49= J. And||ofJon=i- And f-J=l!-3''o=/o- ^w«- Example 3.— From 192-,\ take 16{-^. ^ and f ,^ reduced to a common denominator become yVV an^^ Iff* 192,2^—16^ = 192tS%-16H| = 19H-lj^^-16m = 191^1- Here, since we cannot subtract |ff from ^Vb ^« ^'^^^ *" borrow 1 from tbe Integral part of the minuend, and considering it as \}f, add it to I'^V We tliiia reduce 192rVff to 191ff | and then make the subtraction. Example 4. — From 29^\ take 16f. 29A - 16^ = 29H - 16H = 28 + l|f - 16H = 28^ - 16^ = 12H. Ans. Example 5. — From 11 Yt^^ take 67 ^^. 117^ - 67H = ii7iH-67w = iiQ+HH-Qim = ^^^n - Example 6. — What is the difference between ^ of f of f of 2^ days and f of f of 5^ hours? I of i of f of 23 days=f of a day=; of V of an hour=if9 hours=17| hours; and f of J of 5| hours=S| hours =lsV hour. And 17| h.-lA h.=17/j-lA = ia,. 17. MuUi[ MuUij Find t Find 1 Multii of 1 18. Findt 19. Multip 20 Findt 100 121 46. fractioDj Mulii and divid Note. having 1 f Exam EXA)J ^ofi :0N9. (SiccT. IV nno nurnrroior tnonnnatur. n\ numerator to 8, ABT9. 4S, id.] MULTIPLICATION OF VULQAE FRACTIONS. 173 Exercise 68. What is the product of ^^ x g ? What is tilt' product of ^ x ^ ? Wliat Ls the [)ro(iuct of ,V x A ? Multiply together 5, '^ and xh ^ Multiply toj^'other 14, 15, -^-tj and 8j. 1. (I ««• '6. 4. 5. U. 7. 8. 9. Ketiuiitd the pi-oduct of |j, ^, /, , ,V and 209. 10. Find the value of 0^ x 11^ x 1(5,^ x -^^ x „^,r U. i'uid tiie value of f of -,\ of y'g of 77 x f of ^»a Multi[)ly together -,''o> ^f, iv a"J li- Ketiuired tlie product of J, -,''|-, -,\, ^f,|| and §•. Ueciuired the product of Jf, V") A> '^1> )^ ""^ 5. Ans. ^. 74 '4. of y^. of 91 12. 13. 14. 15. 1(). 17. 18. 19. 20. 1. a 7^ 4} Multiply together -?-, i-, — , — , -A, and li. fj b 8' 9^' f '7iV Multiply i: of 8 by ^ of 19. Multiply y'o of 7 by H of 87-,V ' Find tlie value of tif x ^ x ^ x ^. Find the value of 3| x 4g x 15. Multiply i-of 81 of i% of 9i by 8/,- x H of 6^ _a 27 81^ ^^ i ,, 81^^ of 1-a- 2i X . 128 Find the value of x - " x m 98i Multiply $8 iV by I of f of H. ,.. , t. , r75i ^oVsi X A- of 28 Fii)d the value of ~ x ^ — — ? ^7- x 6 A 1 r ot Of X -jV of 24 Ans. IH^. Ana. 9 J. ^ HH. i\. X 6JJ. -4n«. 1127f Ans. j^j. Ana. 10^. Ana. 403^. -4«s. 2,^1,. ^7i.s. 2()8i. of^of^oflSi ^ws. 47295^J. 200 121 j4^ 9 Ans. — X t X 14» X 15 ^ ^ ^Mfi. 17|Jg. 46. To multiply an integral denominate number by a fraction, we have the following : — RULE. Multiply the denominate number hy the numerator of the fraction and divide the result by the denominator. Note. — This is merely conslderincr the denominate number as a fraction having 1 for its denominator (Art. 23), and applying the preceding rule. Example 1.— How much is f of $129-63. 9 $518 52 „^,, ^ ■JL- =57.61 J. j^ng^ Example 2. — How much is ^^j- of | of 10 lb. 6 oz. 4 dr. Avoir. ? fr of J of 10 lb. 6 oz. 4 dr. =-. /, of 10 lb. 6 oz. 4 dr. = — ' "^- ^^^' "" '^ 8 lbs. 4 oz. 14tt drams. Ana. %% 174 DIVISION OF VULGAR FRACTIONS. [Sect. IV. Exercise 59. 1. How much is 1^^^ of 4 days 5 h. ? Ans. 6 days 38 m. 20 sec. 2. How much is ^| of £29? Ans. £8 19s. 6fd. 8. How much is ^- of 186 acres 3 roods? Ans. 145 acres i rood. 4. Howmuchis^f of f of ^^-of 23^ times24h. 30 m. ? ^/is. lh.38in. 6. How much is f of J of U of § of 33 bush. 2 pk. 1 gal. ? Ans. 2 bush. 2 pk. gal. 3 qt. l^f pt. 47. From the principles already established, it is evi- dent that — 1st. When the multiplier is less than unity, the prod- uct is less than the multiplicand. 2ud. To multiply a fraction by a whole number, we may either multiply the numerator of the fraction or divide the denominator by that number. (Art. 8.) 3rd. To multiply a whole number by any fraction hav- ing unity for its numerator, we simply divide the whol(« number by the denominator. Thus, to multiply by i, i, J, |, ^ti *c-. we divide by 2, 8, 4, 7, 11, Ac. 4th. When multiplying by a mixed number of whicli the fractional part has unity for its numerator, it is better to multiply by the integral part of the multiplier first and then by the fractional part, afterwards adding the two partial products together. DIVISION OF VULGAR FRACTIONS. 48. Let it be required to divide | by W- Here we are required to divide f by j\, that Is, by yV of 8. Now if we divide f by 5, we use a divisor 11 times too great, and the qno« tient is 11 times less than the required quotient. Therefore, to obtain the correct quotient of f-J-rr) after dividing f by 5, ^^e shbll have to multiply the result by 11. But (Art. 8) we divide the fraction 1^ by 6, when we multiply the denoirji- nator 7 by 6, and we multiply the result by 11 when we multiply the numera- tor 8 by 11. 8x11 Therefore f -f-Vr-r 7x5 :f X — V ^dividend x divisor with its terms inverted. Hence for the division of fractions we have the follow- ing:— [Sect. IV. s 38 m. 20 sec. J. £8 19s. 6fd. 5 acres 1 rood. Ans. lh.38m, gal.? il. 3 qt. m pt. led, it is evi- iy, the prod- number, we ion or divide fraction hav- e the whol(ii ,7,11, Ac. )er of which , it is better ier first and ng the two \t, and the quo. Iding f by 6, \tq )lytbo denotTji- )ly the numcra- terms inverted. ; the follow- AST3 47,49] DIVISION OF VULGAR FRA0TI0N8. 175 RULE. P.'lace compound and complex fractions to simple ones ; whole and mixed uuiiio:r.s' to improper fractions. fiK'eri the terms of the divisor and proceed as in multiplication. In iidJitiou to the foregoing analysis, the following may be given as a pi-ooi' of tiic truth of this rule. I j_ j«j. = — because the dividend of any question In division may be made Tr the numerator and the divisor the denominator of a fraction. Now since we may multiply both terms of the fraction -j- by any number we may multiply them by -V-, *• e., the denominator with its terms inverted. f f X V- |x ■«" (because xV X .'^ = 1) = f X Y • whence Therefore -- = , .. the truth of the rule. Example 1. — Divide -j^ by -i\-. Example 2.— Divide | of t^^- by -^t of 8|. I of -h - A of \^ - U -^ M = H X H = A ^^^^ Example 3.— Divide 8^ by S^\. 8}-3A- = V-^f = ¥ X H = ^ X ¥ =H = 2H Ans, ga gl Example 4.— Divide -j^ of ^V of -»- x 8f by iV of -rr x 4|. TT °z STATEMENT. TERMS OF DIVISOR INTERTED. ^xAxWxV-f-AxMlxV=A-x-A-xWx^xVxHfx/tf 3 i — .— y ___ X ;;7 w cancelled. XX % 85 n"" i'' i"" m i 2 u 8 ? ^ Exercise 60. 86 86 2x3 6 zz\ Ans. 1. Divide i of I by f of 8j. 2 Divide || by \\ and divide the result by -^^ 3. Divide 82,\ by 264V 4. Divide 2t^ by | + i 6. Divide If by f of 2| of 16 of Sf of ^. 6. Divide 2^ by (§ -r- -3^ of 9.) 7. Divide 48^ by f + f of 6. 8. Divide Oi by f of 1^ 4- ■^. Ans. yfy. Ans. ^. Ans. 3-i\y^. Ans. 1-,V Ans. ^i<£. Ans. 1-}c. Ans. 19i|. f ■ .'»! I ^li m V\-^^l m f m»: m^-^ 176 DIVISION OF VULGAE FRACTIONS. 9. Divide 4^ of 3^ by 2^ of 6^. 10. Divide ^by-i. 11. Divide ^ of 7-,;^,- l)y -^^ of IVf 12. Divide IH ot iii of i of U by g of A of | of 5. 13. Divide ^^ by ^i 14. Divide i% by ^. 15. Divide 14J of ^ by ^ of S-jAj of -^-. 16. Divide 15i of 1- of -- of '^'- by H of A of 1- of . ■I f 3 -^ 7 4f 8i ^ [Sect. IV, Ans. l^K. Ana. 6-,V,-. Anr. ^JD_ -'4 J * Ans. '6'^l. Ans. I, Ans. i. Ans. li%^,\. n 49. To divide an integral denominate number by a fraction ; — RULE. Multiply it hji the denominator and divide the result by the nume- rator of the fraction. Note.— This is. in effect, merely considering the denomintite nnmher ns fraction having 1 for its dLMiojiiinator (Art. 2:1) iind applying the foregoing rule. Example.— Divide 6 days 17 hours 11 minutes by ■^. 11 edaysHh. llm. x U 6 days 17h. 11m, j- j'^ = 6 days 17h. Urn. x g- = ^ = 14 days ISb. 8Cm. 12 sec. Ans. Exercise 61. 1 ^- 1. Divide £8 1 4s. 6fd. by-^y. Ans. £8 8s. S^d. 2. Divide Im. 5 fur. 91 yds. 2 feet by 2 J of IjV Ans. 2 fur. 124 yds. 2 ft. 3. Divide 3 acres, 3 roods and 3 perehes by ^. Ans. 6 acres 1 rood 5 per. 4. Divide £7 16s. 2d. by f. Ans. £17 lis. 4id. 50. To reduce a fraction havinp^ a complex fraction in its numerator or de'iominator or both to a simple fraction we have simply to apply as often as necessary the ritle given in Art. 33, Note. — Pai'ticnlar attention must be paid to the relative lennth and hfiavw£sn of the separating lines, as they determine th^^ varioua numerators and denominators. [Sbct, IV, Ans. 1 1^. Jns. 6/;,. Jnr. f,^o. Ans. 'S'^l. Ans. I. Ans. k A71S. l^%% of 2} A71S. 28iV.:^-- umber bv a U by the nunu- late number ns foregoing iuIl. m. X 11 ifi. £8 8s. 5id. . 124 yds. 2 ft. s 1 rood 5 per. £17 lis. ^d. x fracti(m in iple fraction iry the rule roliitive lenafh 10 th^^ varioua Akt9. 40, 50] DITIilON OF VULGAR FRACTIONS. 177 Example 1.— Simplify 7* \ ii OPERATION. n =• ¥_ i 4 V 65x2x198 13x83 ^} ) 3 5 15x35 4xlRx35 85 ■=12-3^ Example 2.— Simplify 6 6 4 }| ¥ 2x198 5 6 it 6 i OPERATION. 13 1 20 24 13 18x13 20x24 3. ¥_ 15 2 2__ 17 15 13x18 20 X 24 7x6 17 13xl3x 17 2873 "20 X 24 X 7 X 5 16800 •Ana. I: li^ t^ff' ?• i| ! HI ,1 Br — I 178 DIVISIOi'? OF VULGAK FBACTIONS. [^-;vr. IV, m 1 1. Multiply 8i 9 f i i 5 • 4i 4 7 2. Divide 6^ 9^ S ^ m R DivirlA H 8| by 5 of 32 8i Ana. 2-i\\. I 6y -4n«. y|8^. by 5i 6 IT 8f 16| T -4n5. 8||. 61. From what has already been said, the truth of the following principles is evident. 1st. When the dividend is equal to the divisor, the quotient will be 1. 2nd. When the dividend is greater than the divisor, the quotient will bo q-reater than 1. 3rd. When the uividend is less than the divisor, the quotient will be leF*^ an 1. 4th. The quotient wni be as many times greater or less than 1 as the dividend is greater or less than the di' Jsir. uSiaj^' :i-;vx. IV, Ans. 2^,V Ans. tI^^. Ans. S{1 ;e truth of the 3 divisor, the 1 the divisor, e divisor, the greater or less 1 the c1-ivi33r. ART. 61.] QUESTIONS. 179 5th. To divide a fract'iji by a wli-jle Luraher, we miiy either divide tu6 numera'or or multiply the denomiuatur by that number, eth. To divide a whole number by a fraction having I for its numerator, we simply multiply the whole uuuj.bei l)y the denominator of the fraction. Thus, to divide by ^, J, J, ^ &c., we multiply by 2, 3, 5, 7, JLo. QUESTIONS TO BE ANSWERED BY THE PUPIL. NoT)-.— -77^6 numerala after the Qiuf-iions lejcr to the numbered articles of the Section. 1. What is a fraotion ? (1 and 8) 2. Wliut d "(.'.i o'-ery tVarlioi inlicit''? (3} 8. What is ilie dcnoiiiinator of a rriU'tiuii, and why I's it ('.".llod so? (4) 4 Wliat is the muncrator of a fru'tiou, and why is it so culled ? (4) 6. What arc the terms of a fraction? (■">) fi. How is the val;i« of a fraction obtal od? (<») 7. Whi'H is the fraction ('(piai to 1, and when !.'rcatc.r or less than 1' (7) 8. What t'ff -ct has multipiyiug tho numerator of a fraction by any niiin. ber ? (S) 9. IIow does multiplying: the denominator of a fraction bj' any number affect tiic value of the fraction? (8) 10. How does multiplying both terms of a fraction by tho same number affect its value ? (S) 11. How does dividing tho numerator by any number atfect tho value of tho fraction ? (8) 12. How does dividing: tho denominator by any number affect the value of th«> fraction? (S) 13. How docs dividina: both numerator and denominator by the s; e num! cr affect tiie value? (6) \\ Into wiiat classes are fractions divided^ (!)) 15. What is the distiactiou between vtilgar and decimal :' ;.uons? (10 and 11) IG. W at is the meanincr of the word •' vulerar " a^^ applied to fractie,; 5? (11) 17. I'.niiiiu'iate the six (lilT''rci)t kinds of vulgar fractions. (12) 1^. What is :i proper fi'aeti.)n? (1:3) I;'. What is an improper fraction? (15) 't\ What is a mixed number? (10) 21. To what must an inipn.per fraction always be equid? (17) 22, Wliat is a simple fraction? (lb) 2;J, What is a coiupound fraction? (H?) 24. What is a complex fraction ? (21) 2.1. How may we convert an intoirer into a frr.cM'in ? (2") How may wo reduce a whole number to a fraction having a given deuoml* nator ? (24) How is a rnixeil number reduced to an iin[)roper fractio ' ? (2.'i) How is an iiiipr<(|,t'r fraction re(iuce./ How are fractions reduced to a common denominator? (2i>) flow are fra.-tions reduced to tlielr h a.'^t. '-oiiUMon '. ! ;nni:iutor? (30) How ia a compound iiactio.i reduced to k tiuiple one? (SI) M ^ii 19 J^0 180 MISC^LLANKOUS EXimclSE. [Sect. IV. 88. 84. 85. 86. 87. 8a 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50 61. 62. R3. 64. 66. 66. 67. 68. What is mennt by cancelUne? (32) Upon what principle may we cancel factors common to numerator and de- nominator? (32 and 8) How do wc rudneu complex frnctions to simple ones? (88) What is a denominate fraction t (34) III what does reduction of denominate fractions consist? (85) How do we reduce a deooiuioate fraoiion from a lower to a higher denomt- nation? (30) How do we n liuce a denominate fraction from a higher to a lower denomj. nation? (37) How do we fi: d the value of a denominate fraction ? (89) How do wo reduce one dei.oniinate number to tlie fraction of another? (39) Wiiat is addition of fmctlons ? (40) What kind of fiiiction^ only can bo a'Med* C41) What is the rule for addition of frnctions? (42) When mi.xed numbers are to be add<(l how do wc proceed T (42, note) Wliat is the rule for the addition of donominute fractious? (43) What Is the rule for the subtraction of frnctioi s? (t4) What is the rule for multiplication of frnctions? (45) Give a proof ot the truth of this rule. (45) How do we multiply an integral denoniinnte number by a fraction? (46) How may we multiply n fraction by a whole nunihor? (47) How do we multiply a who'e number by a fraction having 1 for numera- tor? (47) How do wo multiply a whole number by a mixed number, the fractional part of which has 1 for numerntor? (47) What is the rule for division of fractions? (48) Give a proof of the truth of this rule. (48) How do we divide an inliKrul dcnomiiuito number by a fraction ? (49) How do wo divide a frnction by a whole inimbcr? (51) How do we divide a m hole number by a fraction having 1 for its numera- tor > (6J) XERCISE 63. MISCELLANEOUS E. .A^.'ISE ON VULGAR FRACTIONS. 1. The Ottawa River is 801 rii-^slong; the Gatineau 420 miles, tlie Chaucli6ie 100 miles, tL3 l.n'helieu 160 miles, and the Niagara 3u miles. The entire k'^i.^i'^ of the St. Lawrence, from the upper end of Lake Superi^ ( the Sea is 2000 miles. How will the lengths of these different rivers be expressed as fractions o) that of the St. Lawrence V 2. The population of Goderich is f of that of Peterborough, the popu- lation of Peterborough is 1^ of that of Brockville, the popula- tion of Brockville is 1^ of that of Prescott, the population of Pr<'f-cott is ^ of that of O^x./i City, the population of Ottawa C ity is 2^ of that, ji a ort Hone , and the population of Fort li(;pe is 4^5 of thi of '''"":,aio. What fraction is the population of Goderich of tlu . oi Toronto ? 8. What will 6| poimds of tea cost, nt 65j cpnts per lb. ? 4. Suppose I have ^ of a ship, and that I buy -,^^- more ; what is Uiv vntirc share ? »:■ f^y- [Sect IV. 8scT. IV.1 MISCELLANEOUS EXERCISE. 181 numerator aDd de- «) (85) to a higher denomt- to a lower denoml- ) •action of another? ed f (42, note) s? (43) a fraction ? (46) 47) .ing 1 for numerft. Qber, the fractional fraction? (49) X 1 for its nuinera- FRACTIONS. :u 420 miles, the and the Niagara 'rence, from the miles. How wil) d as fractions o) rough, the popu- nlle, the popula- e population of lation of Ottawa )ulation of Port is the population lb.? 01 e ; what is my 6. A boy divided his marbles in the following manner : he gave to A ^ of them, to B -,^„, to C ^, and to D ^, keeping the rest to himself; how many did he give away, and how many did he keep ? 6. Find the value of ^tz?^ of li±^^ of ii±.li 7. What cost 1670 -fj pounds of coffee at 12j cents per pound? 8. A tree whose length was 136 feet, was broken into two pieces by falling ; | of the length of the longer piece equalled J of the length of the shorter. What was the length of the two pieces respectively ? 9. A farmer bought at ono time 97^ acres of land, for 1000 dollars; at another, 127o acres for lS75i dollars ; at another, 500^ acres for 6831 dollars; and at another, 333-^ acres for 4013, \ dollars. What was the whole quantity of laud that he purchased, and the sum that he paid for it ? 10. Find the value of (12|;-8J-1t1o+A) x 4^ x C^A-Ci), and also of(|-l-5)-(f-^3,^). 11. What is the value of 19} barrels of fiour at $6f a barrel? 12. What is the value of 376|^ acros of land, at $75| per acre? 13. Bought at one time 147| bushels of coal, and at another time 320^ bushels. Having consumed 166^ bushels, I desire to know what quantity of the coal pui chased is still on hand. 14. Divide liliof f) by 71 ; and find the value of f 4-^ 1 4 2i + l_ 1 15. If 11} bushels of wheat sow 7* acres, how many bushels will it re- quire to sow one acre V 16. Multiply the sum of 3|, 4^^, and 4^, by the difference of 7^ and 5^ ; and divide the product by the sum of 94^ and 93^. 17. Divide 2 by the sum of 2|, |, and 4 ; add li — I to the quotient; and multiply the result bv the difference of 5^ and 4^-. 18. Find the value of (Ki) x (H+2a) x(2A-- U) x {Z^^-~})- and also of (1|^2^) + (5^ ^3^). 19. A person dies worth $40000, and leaves ^ of his property to his wifo, ^ to his son, and the rest to his daughter. The wife at her • death leaves f of her legacy to the son, and the rest to her daughter; but the son adds his *'■ rtuno to his sister's and gives her ^ of the whole. How much will the sister gain by this, and what fraction will her gain be of the whole ? ^: ii i 182 DECIMAL FRACTIONS. ^ [Skct. IV I AnT9. M -fi DECIMALS AND DECIMAL FRACTIONS. 52. A decimal fraction is a fraction having unity with one or more Os to the right of it for denominator : Thus y/rtff, -jl.ff, |3„, j-BUsny &c., are decimal fractions. 53. A decimal fraction is reduced to its corresponding decimal by dividing tlie numerator by the denominator; but since (Art. 52) this denominator is unity followed hv one or more Os, we divide the numerator by the denomina- tor when we move the decimal point as many places to thr left in the numerator as there are Os in the denominator. Example 1. Reduce j^^s 'o ^ decimal. Ans. -743. 2. Reduce jffj^jji^iliyTy to adecimal. ^^'- OOOK^LVe. Exercise 64. 1. Reduce ^^^'^t, j^^l-f,-^^ and j\ to decimals. Ans. *6G7, -00008 and -7. 2. Reduce jTsmms and rcijVi/^TCo *« decimals. Ans. -0000023 and -000: '176, S. Reduce jijl^^-ii^ 0-17(5 t» a decimal. Atis. -00027>JC4&. 64. It is as inaccurate to confound a docimnl fraction with its correspondin«^ decimal na to confound a vulgar traction wiih its quotient : Thus the value of ^ is -VT). so also the vnlue nf -j^'o" is -75, but -75 and iVo are no more identical than are j and -75. 55. To reduce a decimal to its corresponding decimal fraction : — RULE. CovMfJer the siff7iiflcant part of the decimal as numerator c^A beneath it write for denominator 1 followed by as many Os as thm are places in the decimal. Example 1. Reduce '043 to a decimal fraction. Ans. jf j;;. 2. Reduce -00000576 to a decimal fraction. Ans. x^jyc^^soj. Exercise C6. 1. Reduce '73, -002 and -0003 to decimal fractions. Ans, f'siT, rij-oC") and j-^^^-^i' 2. Reduce -137 and -000006943 to decimal fractions. An9 -•'*'' nn<\ - ^'^^^ — 8. Reduce -13578967 and -023004003 to decimal fractions. 55. I f-acLioMd, quotients tion, &c., Tor fraction Divf'} the rcrpii iiuj deciii This i cates. Exam tion. 2. Ri Ji.na. TB^Q-Q^-Q^^', ana y^< 1. Reduc 'L Reduc 3. Reduc( 4. Reduci 5. Reduci 57. I pound. f(l=-75fl h( as tl GJi-6-7.)d Next if we Ts. f)}fl:=7-J Tiieicibre Hen( ♦Tho a reuialcid( , [Sect IV I Anrs. M^I] DECIMAL FRACTIONS. 18.) [ONS. nng unity with nator: fractions. ! correspond in ff i denominator; ity followetl hv the denomiria- ly places to tb( denominator. Ans. •743. An». -00092076. B7, -00098 and -7, )23 and -OOOMHe, Ans. •00027^'C4S, fraction with its Igar fraction ^viih also the viiluo of identical than are )nding decimal IS numerator avd many 0« as thcrt Am. Y^^o- ins. xi7(Tao7So5' ictions. uul _2.'^no4003. 55. Decimal fractions follow exactly the paine mips as vulgar f-actiona. It i.s, however, generally more convenient to obtain their quotient.^, ami then perform on them the required processes of addi tion, &c., by the mcthodd already described (Sect. II). To reduce a vulgar fraction to a decimal or to a decimal fraction — RULE. DiviiJc- the nnynerntor bjf the detiominator and the guotu>iit tcitl b the required '^ decimal''^ ; the latter may be changed to its correspond- ing decimal fraction by (Art. 56). Tliis i3 mori'ly actually performing the division which the fraction indl- catts. Example 1. — Ueduce J to a decimal and also to a decj'nal frac- tion. 8)7^ •875 Jw."?. =-,B(fiAj Ana. 2. Reduce VV to a decimal. 16)9^ '6626 Am. Exercise G6. Ann. *6 and 5=<1ecimal of a Tnlnute=8 mJii. 80 sen. 24)1205S3=(Iociinal of an hour=12 h. 8 m. 80 sec. 7)3;6n24.S(>fi=(lccimul of a day=8 days 12 h. 8 m. 80 sec, '^n«.~^50O.317'2=(lecimal of a weck=8 days 12 h. 8 in, 80 sec. Example 2. — Reduce 187 lb. 13 oz. 11 drams to the decimal of a ton. OPERATION. 10)11' drams. 16)i:i-«^75 ounces, 2000)1 87 -sSmOSTj lbs, •093927734375 ton. Ana. Hero we divide the 11 drams by 16 nnd thus obtain •6875 to wliich we prefi.x tlie given 18 oz. Next we divide tiiis by ^ and obtain •8,')546875 to wliicii we blii: down the 1S7 lb. and divide the result 'jy 20U0, the number of lbs. in a ton. NoTB.— To divide by 2000 remove the decimal point three places to thi; left and divide by 2: .«>tmilaily to divide by GO, 20, «fec., remove the decini.il point one place to the left and divide by 6, 2, &c. Exercise 67. 1. Reduce 3 yds. 2 ft. 1 in. to the decimal of a furlong. Ans. -01679+, 2. Reduce 3 dwt. 17 grs. Troy, to the decimal of a pound. Ans. -01545138+. 3. Reduce 2 scr. 7 grs. to the decimal of a pound, Apoth. Ans. -0081 597+. 4. Reduce 5 fur. 35 per. 2 yd. 2 ft. 9 in. to the decimal of a mile. Ans. -73603+. 5. Reduce 3 qr. 2 na. to the decimal of a yard. Ans. -875. 6. Reduce 5s. to the decimal of 13s. 4d. Ans. -375.* * Reduce Ks. first to the fraction of I83. 4d. and then reduce the resulting ftwjtion to a decimal. Thuf fi» '•eduped to the fi-actlon of 188. 4d.=y«iaj=2=-87^ (Sect, iv given denonii- )at member which higher denomina ) the one required, tcs, 80 seconds, to 3ec. 80 sec. ) the decimal of a 11 drnms by 10 and (vliich we prefix tlie l^e divide tliis bv 1 to wliicli we brii; divide the result 'iy )s. iu a ton. three places to thn •einove the decinidl )ng. Ans. -01679+, lound. ns. -01545138+, poth. Ins. -0081 597+. nal of a mile. Ans. -73603+. Ans. '81i). Am. -375.* duce the resulting K%U. AT, 68.] DECIMALS. 18d 7. Reduce 12 h. 65 min. 21 sec. to the decimal of a day. Atis. •6384876. 8. Reduce ^ of ^ of 6fd. to the decimal of £f Ana. •0120634-. 9. Reduce f of | of a mile to the decimal of 3^ inches. Ann. 3620-671428+-. 10. Reduce i^ of f of 3^ lb. Avoir, to the decimal of f of an oz. Ans. 9-2444+. 11, Reduce 3 pk. 1 gal. 1 qt. 1 pt. to the decimal of a bushel. Ana. -921876. OPERATION. •7826 8 2-8475 12 4-1700 12 2 0400 Ana. 2 ft 4 in. 204 lines. 68. Let it be required to find the value in terms of a lower de- nomination of -7825 of a yard. Explanation. —91nc« fhero are 8 feet fn a yard, it is evident that any docinial of a yiird in three times as great a decimal of a foot. Hence to reduce the decimal of a yar 1 to a decimal of a foot we inultinly it by y. This gives us two feet and •3175 of a foot. Slmil: ly multiplying the decimal of a foot by 12 r*.jclucrs It to an oqnivaloiit decimal of an Inch. We thus find -3475 of a foot equal to 4 Inches and -17 of an inch. Again, multiplying this lo'^t by 12 redu- ces It to the decimal of a line, an IMAGE EVALUATION TEST TARGET (MT-3) £>< Q.r i/i 1.0 !f I.I 1.25 12,8 i5 I4S """ — " b 140 ■H IM 1.8 U 11 1.6 V]

ri? ^'h ri7 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 186 CIRCULATING DECIMALS. L^-ECT. IV. Example 2.— :Find the value of -7863625 of a pound Apothecaries weight. OPERATION. •7868626 12 ©•4868500 oz. 8 8-4908000 -hy aS, 5fV> sV. 6^-, h aiid sV- 77. To reduce a pure repetend to an equivalent vulgar fraction : — RULE. Pnt the period for numerator, and as many nines as there are 'j,taces in the period for denominator. ■ = " third " 90=:1st Dcnoininaror. since the repetend contains one place in the finite, and one pluce in tht' Infinite part. 9900-^2n(l Denominator, since the repetend contains two places in the finite part and twr. in the infinite part. 99v)9000=:3rd Denominator, since the infinite part of the decimiil contains /oii) places and the finite part thi cc places. ? 1 V' Hi n m 192 OIECHLATINO DECIMALS. [Sktt. it Hence, •73=5S=-H, 123^=^1^=4^^^ and 'ln2092=UUUl Reason. — Let it be required to reduce '978734 to an equivalent v" -"KT fraction. Let* =-978734 (I) Then 100 a: =97-8734 (II) And 1000000 x =978734-8734 (III) ; subtracting (II) from (III) gives 999900 a: =978734—97. 978734—97 Whence x = = Whole repeicnd minus the finite part for a 999900 numerator ; and as many 9s as there are places in infinite part, followed by as i any Os as there are places in finite part for de- nominator. The rule may also be expl iincd as follows : — Taking the same example •978T34 and multiplying it by 100, we got •978784 X 100=97"8734=97 f -8734=97 + SJSS (Art. 77.) Now, since we multiplied by 100 this result is 100 times too great There- • « fore ■978734=V5ij + «m^ ^^^ to add these fractions we must reduce them to a common denomiuator when they become : 97 X 9990 8T34 + =(8ince 9999=10000-1) 07 X (10000-1) 99990U 8734 999900 " 999900 97 X 10000-97 999900 8T34 999900 970000-97 999900 8734 999900 999900 978784-97 = z=.Whole repeiend minus Jlnite part for nv/merator; and cut 999900 many 9« c/« there are places in the finite part, followed by as many 0« as there are places in the finite part for denominator. Whence the truth for the rule it maiifest Exercise 72. ' 1. Reduce -8325, -147658, and -4320075 to their equivalent vulgar fractions. Ans. mi=HU, UUhh, and MMU%=MM^H. 2. Reduce 875-4965 and 30182756 to their equivalent mixed num- bers. Ans. 81 o^^^ and 301-}-ffi. 8. Reduce -083, -0714285, and -123456 to their equivalent vulgar fractions. Ans. iV, tx^ and -aViVifo-- 4. Reduce -7034, -96432, -00207, and -14327*1 tj their equivalent vulgar fractions. Ans, f§f|f, fff, if^> and Uholt Aets. t8, 7 79. wliidi it Ist. if we mi Thus, 2nd. duced to Thug 4, C, 8, IC For exi 3rd. of {dace! same nu: Take /; repc'oid, c tend to the Thus, 1 having the Here th of 1,2 and 3 Therefo: 4th. 1 having a many pla or more \ Thus, -4 7-6 5th. 1 the last preeedint \.ays be i 6th. I gather th places : si [8iOT. IV. equivalent (III) gives part for a finite plj4io83rjolo. l)ivi 2. \V/ at i.-< fh( I'lilClio)! ■; J, t f.,t m^- ""^^' '■■' i» <'' Am. r7'iq|4. How isa V 5. llmv woiili Ans. 8-8235294117G4706i||. What are c 6. Wliuii are r P'ls (:r lion cau 7. Find the difference between 17*428571 sq. ft. and 100'8 sq. in. 8. What is the value of '91789772 of two acres? ||7. Wlion cau i S3-5. Siiow iliat 1 (fl'J. How L'an \v corri'snoi 0. ir the lU'oi what is t 21. Sliow that 2,^. What are p 2;j. How is a pi 2j. H(i\v is a in 25. Show the ti '^6. Show that ; [8«CT. ivl 4^. fc2.J EXA^iI^■ATION QUESTIONS. 197 Ana. 0:^19. Reduce 11-287 and 1'0128571 to vulgar fractiona. n«. •8ir>rj4io880olo. IHvidc 47';M5 by 1'70. Ans. 'MS.!!, i'loin Ma-.Vi tak(> in-7nr.'j. J. Wliiit is I ho difloit'ucu between '7^4 of a lb. and 198 of an oz. uvoirtlupoi.s? .;, Ilow iiiaiiy yards of carpet 2 ft. 6^ in. wide will be required to IMAL!=i. sr or by a finiu^ ar fractiona^ dividi ndiuy decimal. i = 062. cover a floor 27-3 ft. long and 20-16 ft. wide? i. Multiply 3-145 by 4-2')7. 5. How many finite pi;. cos are there in the decimals coiTespouding to A, ./4, -i\, iW, lA", ttiitl /aV4-? 6. Add together 81?^, 6M2G, 328^J, and 5-624. 7. RcJuco I qu.mUty. 4-4—283 i-()-i-2Gjy 6-8ofn\ 2-Hof2-27 of J + : to a simple 2-25 / 1-136 M ^ !!' Ans. "0 Am. 246 Ans. VT^i ;235294117G470oi 1 2. 3. 4. ECIMALS. rnal. II. 18. h. lo. 1. IS4 J4, and from the: ecimal of a mile. nd 100-8 sq. in. QTT-'TIOXS TO BE ANSWERED BY THE PUPIL. y.OTv..— J he II ambfTS after the quentiona rcferto the artides of t'le Section. V'l,:ir is (I (Icoinitil frik'tlon ? (H2) ^V'l at. is the di^iiiiction between a decimal and its correspondlntj decimal I'laeiiuii? (."^1^ and Art. 47, Sec. I.) How i.-i ii (U'ciiuitl reduced to its corre.*pondl g decimal fraction? (W) lIow is a viilji.ar fraction reduced to a (leolmal ? (r)6) llmv svoiiln you reduce 4 oz. 17 dwr. 1() ;i\s. to the decimal of a lb. ? (57) lloA- woull you find Uio value of •7i;it5 of a French ellV (58) W'iiut, is meant i>y rcpeatini; or ciieulatinc: docimula? (6 >) Wiiiit is a rei)elend, period, or t.'ircio''' (Gl) V.Mial is a sinL-lo repoteiid. and how is it e.vprossod? (fii and CH) Wliat iti a cireuhiting decimal or compound veiK'U-ud, aud how is it express ed ? (ti4 a d G5) - * * Wlial is a pure repetend ? (66) Wli.tt isami.xed repetend? (67) m V Wiiat are simple repetends? Give an examnlo. (68) What are (li^simiiar repeiends ? Give examides. (69) V/hat are coterminous re|)orends? Give examples. (70) 6. Wlien ixTP repetends said to be both similar and coterminous? Give exam pes (71) Whon can a vnlauie num- ber «if i)lace8, and pive the rule. (79) 29. Show thflt any pure leiiuteud uiay bo transformed into a mixed repetend? (79) 80. Show that two or more repetends may be made similar and coterminous, (79) 81. How are circulatlnp decimals added? (80) 82. How are circiilalinp decimalh subtiaeted ? (81) 88. ilow do wo multiply ohculatinp decimals together? (82) 84. Ilow do we diviue oue circulating decimal by another? (88) Exercise V8. MISCELLANEOUS EXERCISE. (On preceding Rules.) 1. Transform 4812181 quinary^ into the nonart/^ ternary, and oete- nary scale.', and prove the results by reducing all four numbera to the decimal scale. 2. Write down seven hundred and two trillions seven millions thirty thousand and seventeen, and four millions and seventy-six tenths of quadrillionths. 8. Divide 976*432 by -OuOOOOQe, (2|-}--5625-l-5 + -i^)-*-y- 4. What is the value of ™" '" "^ ' ;";- '^"^'^^ 19 6 6. 6. 7. Divide 97 lb. 8 oz. 4 dr. 1 scr. 17 grs. by 9 lb. 7 oz. 7 dr. 2 scr. A wall is to be built 15 yards long, 7 feet high, and 13 in. thick, with a doorway 6 ft. high and 4 ft. wide ; how many bricks will it require, the solid contents of each being 108 cubic inches? Multiply ' ft. 6' 4" 7'" by 11 ft. 7' 9" ll'" 4^-f^ - ■h Also change 8. Find the value of , .^-^ i ot fa+i of f. 9. Reduce 782436 pints to bushels, &c. 10. Find the least common multiple of 77, 42, 27, 21, 83, 14, 7, 11, 63, and 30. 11. Divide 36<879i2 by 28e4 in the duodecimal scale. 3762814 from the nonary to the decimal scale. 12. How many divisors has the number 150528? 13. Find the value of -1234625 of 2 weeks and 2 days. 14. Multiply 27 lb. 4 oz. 3 dr., avoirdupois, by 728|^. 15. Add together $98-17, $42-29, £16 38. 8^d., $97-10, $127.87^, and from their sum subtract £67 1 7s. 7id. 16. Reduce '8, fractious, '^> -9123, and -003327 to their nlent vulgar 4' 6£0T. lY.; v.. Take it ma it 18 . Redi 19. Divid givi wor 20. Add 21. Writ( 22. Find 23. Find secc 24. Howi to I whe 25. What tern galh NOTE.- 26. Reduc 27. From 28. Find t 29. Transf bina bers 80. What 81. Reduc 32. Findt 33. What and 34. Divide '.'nra- bers to the decimal scale. What are the prime factors of 7680 ? Reduce 72 miles, 3 fur., 7 per., 2 yds., 1 ft., 7 in. to lines. Find the price of 97 pairs of gloves at 47 cents per pair. What is the worth of a pile of cord wood 73 feet long, 4 feet wide and 11 feet high, at $3-62^ per cord ? Divide 93-723 by 29-4173. How many bushels of oats are there in 73429 lbs. ? What is the worth of 719630 lbs. of wheat at $1-80 per bushel ? Add together $72'14 and $93-76 ; multiply the sum by 9*47 and divide the product equally among 1 1 persons. Find the G. C. M. of 21389 and 180781. * These qiu«t ions thonah apparently di thou It are not so in reality — they are (iletiigaed for exeruKO in caQcelling, and do Qot require much work. 200 EATIO, [Sect. V. 39. Reduce -,V, I, ?, 3^^, H. u., and ^ to equivalent fractions, havii a common denominator. 40. Purchased 1*7 yards of cotton at 11 cents per yard, 19 yards of ribbon at 3U cents a yard, 14|- yards of silk at $2'17 a yard, a parasol $4-'75, a bonnet $11-50, 0*7 yaidsof slieetiii^-; at 27 centd a yard, 15 yards of French merino at $1-374 a yard, and trim- mings $7"93. Re(j[uired the amount of my bill. SECTION Y. ar RATIO AND PROPORTION. 1. Two numbers having the same unit may be com- pared with one another in two ways, 1st. By considering how much greater or less one is than the other ; and 2nd. By considering how many times one contains the other. 2. Eatio is the relation which one number bears to another with respect to magnitude, when the numbers are compared by considering, not how much greater or less one is than the other, but how many times or parts of a time one contains the other. Hence : The ratio of two numbers is the quotient arising from the division of one by the other. Thus the ratio of 18 to C is 8, since 18^-6 = 8, the ratio of 7 to 21 is i, slnco • ^" *1 — 'iT — S- 3. The ratio of ono number to another, when measured with respect to their difference, i.i soinotimcs cnlled urUhwoticjtl ratio, to distinguish it from the ratio considered as in (Art. 2), wliieh is called geometrical ratio. In the followinji pajres, whenever the term ratio is used, treometrical ratio is meant; we sliall use the term difference in place of arithmetical ratio. 4. Since ratio simply expresses the quotient arising from the division of one number by another, -and since (Art. QQ, Sect. II.) we have three ways of indicating divi- sion, it follows that we have three ways of expressing the ratio of one number to another. Thus the ratio of 9 to 4 is expressed either by 9 -s- 4, or by -J, or by 9 :4. The ratio of 7 to 13 is indicated either by 7 -i- 13, or by./g, or by 7: 13. B. Eatio can exist only between numbers of the same kind. to 3. [Sect. V. ctions, having 1, 19 yards of 217 a yard, a If.-: at 27 ceiitd lid, aud trim- ay be com- one is than on tains the er bears to lumbers are • or less one 's of a time .rising from to 21 is i, slnc'j »'1th respect to nguish it from tio. (imetrical ratio ;ul ratio. 3nt arising- -and since ;ating divi- re.ssing tliu or bv 9 :4. r by 7 : 13. f the same i-ir-J RATIO. 201 Thus it is obvious that no comparison with respect to maErnitnde can be mule bi'tween 6 hours at)il 11 poundft. or between ID (fai/a and 16 niilefi, &o., i. ■•.. tiii'se iiuuibi-rsj aro not of liie sauie kind, and therefore no ratio can exist l.'clwien tliiiin. 6. Numbeis are of the same kind when they are of the same denomination, or when they have the same unit, or when one can be multiplied so as to exceed the other. 7. The two given numbers which constitute the ratio are called the ierms of (he ratio ; when spoken of together they are called a couplet. 8. The first term of a couplet is called the antecedent ; the last term, the consequent. When the ratio is expressed in the form of a fraction, the nu- mciator is the antecedent aud the denominator the consequent. 9. Ratio is either direct or inversey simple or compound. iO. A Direct ratio is that which arises from the divi- sion of the antecedent by the consequent. 11. All Inverse or Inverted Ratio is that which arises from the division of the consequent by the antecedent. Tlins tlio inverse ratio of 15 to 3 is 3 : 16 or y'g, or 3-r-15, or \. 12. An Inverse Ratio is sometimes called a reciprocal ratio. Thui the reciprocal ratio of 16 to 3 Is 3 : 15 or A = J=inver8e ratio of 16 to 3. 13. The reciprocal of a quantity is unity divided by ihaf. quantity. Tlius the reciprocal of 8 is f; of 11, y^; of ?.!; vi* ¥; "f i 9 '. «'f tV V%*c. 14. When the direct ratio of two numbers is expressed by points^ the Inverse or reciprocal ratio is expressed by inverting the order of the terms; tuhen by a fraction, by inverting the fraction. 15. A Simple Ratio is one that has but or^e antecedent and one consequent. Tlius 9 : 8, 7 : 11, 18 • 2, &c., are simple ratios. 16. A Compound Ratio is a ratio produced by com- pounding or multiplying together the corresponding terms ol' two or more simple ratios. Thus, the simple ratio of 9 : 8 is 8. the simple ratio of 24 : 2 is 12. The ratio c<»mpounded of tliese is 216 : 6 = b6. 17. It must bo distinctly remembered that a compound ratio is of the same nature as any other ratio, and. lilie a biuiplo ratio, consists of one antecedent and one conseqnent. The term compound ratio is used merely to iudicuto t-U^ origin ol tac ratio in particular cases. ^1 ii - A '"■ t :,.'|i i i » I I (N ,1 i t ih. 41 202 EATIO. [SSOT. V. 18. Ratios are compounded hy multiplying togetJier all the ante- cedents for a new antecedent^ and all the consequents for a new conse- quent. Thus, the ratios compounded of 2 : 7, 2 : 8, o : 11, and 4:3 is 2x2x6x4:7 y8xllx3urS0:96d. Exercise 79. 1. What is the ratio of 27 to 3 ? 2. What is the ratio of 7 to 11 ? 3. What is the ratio of 9 to 27 ? 4. What is the ratio of 42 to 5 ? 5. What is the ratio of 72 to 6? Am i Ans. ^^, Ans. ^ Ans. 8f. Ans. 12. Required the 6. 5 to 25. 7. 49 to 7. 8. 83 to 7. 9. 187 to 11. 10. 19 to 162. 11. 23 to 299. 12. 147 to 21. Required the ratio of the following numbers: — Ans. \. ]ld. $17 to $8-50. Ans. 7. Am. llf. Ans. 17. Ans. 2. Ans. 3. Ans. 28. Am. 2(;§. 20. 21. 22. 28. 24. 25. 26. 84. 85. 36. 87. 38. 44. 45. 46, 47. 48. 7 to 21. 12 to 2. 27 to 6. 9 to 36. 19 to 57. 81 to 9. 187 to 17. 14. $93 to $31. 15. 14 bus. to 2 pks, 16. 40 m. to 12 fur. 17. 24 lb. to 12 oz. 18. 17 shillings to £51. 19. 16 acres to 30 sq. per. inverse ratio of the following numbers : Ans. 3. I 27. 6 days co 4 weeks. Ans. 4|. Ans. ^. 28. 11 min. to 30 sec. Ans. ^^. Ans. f. i 29. 4 lbs. to 12 oz. Av. Ans. -^\. Ans. 4. 30. 3 qts. to 43 gals. Ans. dl^. 31. 70 per. to 2 miles. 32. 7 Flem. ells to 9 Eng. ells. 33. 11 oz. to 68 scruples. 72 to 18. 512 to 32. itoi. ^tof. Ans. 5. Ans. {. Ans. 1^. Required the reciprocal ratio of the following numbers : — 7 to 42. Am. \ : -4^=6. 39. \ to i. Ans. 8 : 2=4. 40. 42 to 28. Ans. 4V : h=i- 41. 17 to 68. 42. 19 to 17. 43. Required the ratios compounded of the following ratios: — 2 to 3, 5 to 7 and 1 to 7. Ans. 10 to 147. 8 to 6 and 17 to 3. Ans. 136 to 18. 9 to 8, 7 to 6, 6 to 6, 4 to 3 and 2 to 1. Atis. 2520 ; 864. 1 to 7, 1 to 3, 3 to 1 and 5 to 1. Ans. 15 : 21. 2 to 5, 8 to 7, 4 to 5, 21 to 2 and 1 to 9. Ans. 504 : 3150, 18. Since the antecedent of a couplet is a dividend, the consequent a divisor, and the ratio the quotient, it follows from the priucipli'S established in Arts. 79-84, Sect. II., that;— ^ tl ASTB. 1&-20.] EATia 203 Ist. Multiplying the antecedent of a couplet or dividint* the consequent by any number multiplies the ratio by that number. Thus the ratio of 28 to 112 = f The ratio of 28 x 8 to 112 = f = i x 3 = three times the ratio of 28 to 112. 2nd. Dividing the antecedent of a couplet or multiply- ing the consequent by any number divides the ratio by that number. Thus the ratio of 64 to 16 = 4. The ratio of 64 -i- 2 to 16 = 82 ; 16 = 2 — 4 -f- 2 = half the ratio of 64 to 16. 3rd. Multiplying or dividing both antecedent and con- sequent of a couplet by the same number does not alter the value of the ratio. Thus the ratio of 18 to 6 is 8. The ratio of 18 X 7 : 6x7 = 126 ; 42 = 8 = ratio of 18 -j- 2 : 6-i-2=:9:8. 20. Since any number of ratios to be compounded to- g:ether may be expressed as fractions and then compound- ed by the rule for multiplication of fractions (Art. 45, Sect. IV.) jt follows that : — When several ratios are to be compounded together, we may, before multiplying the corresponding terms together, cancel any factor that is common to an antecedent and a consequent. Example 1. — Compouud together 4 : 17, 34 : 55, 11 : 2, 13 : 7, and 21 : 65. 4 : lit OPERATION. H XX X^ 8 ;Z }-=*x3: 6x6 /f I or 12 : 25 Ans. %X '. ^^ Example 2. — Compound the following ratios : — OPERATION. Explanation.— 17 cancels 17 and re- duces 34 to 2 and this 2 cancels 2, the third consequent; 11 reduces 55 to 5 ; 18 reduces 65 to 5 and 7 reduces 21 to 8. The only antecedents now left are 4 and 8 which multiplied together make 13, and the only remaining consequents are 5 and 5 which multiplied together make 25. The ratio 12 to 25 is therefore the ratio compounded of all the given ratios. Example 3. — Find the ratio co.npounded of the following ratios : — u 9 W 2 %% X^ XX 13 5.=9x2 : 13 or 18: 13 Ans. OPERATION. 1 X& XX m Xi m %9 =1 4 Am. i \ 1 !■ I 1^ jji^.J 204 BATiO. [Sbot. v. Exercise 80. 1. Find the ratio compounded of 9 : 16, :^5 : 81, 841 : 18 and 48: 100. Ans. 55 8 : 8. 2. Find tlie ratio compounded of 18 : 25, 7 : 9, 11 : 12, and 91 : 49. Ans. 143 : 150. 3. Find the ratio compounded of 1 : 2, 2 : 3, 8 : 4, 4 : 5, 5 : 6 and 7 : 11. ' Ans. 1 : 66. 4. Find the ratio compounded of 2 : 6, 8 : 11, 14 : 17 ai,d 187 : 112. Ans. 2 : u. 6. Find the ratio compounded of 3 : 5, 7 : ;i», 11 : 18, 15 : 17 and 19 : 21. Ans. 209 : 663. 21. If the antecedent of a couplet be eqval to the con- sequpnt, the ratio is equal to 1 and is called a ratio of equality. If the {intecedent be greater than the consequent the ratio is greater than 1 and is called a ratio of (jr eater ine- qua lilt/. If the antecedent be less than the consequent the ratio is less than i, ■''.nd is ca,lled a ratio of less inequality. Thus the ratio of 7 : 7 = 1 is a ratio of equality. The ratio of 7 : 2 =■-. 2^ is a ratio of grcatei- InGquality. The ratio of 7 : 14 = i is a ratio of less inequality. Exercise 81. In examples 1-43 of Exercise 79 point out which are ratios o^ greater and which ratios of less inequality. 22. Ratios are compared with ont another by expressing them in the form X:-- 20d PEOPOETION. PROPORTION. LBlwt. V. 27. Proportion is an equality of ratios. Thus, the ratios 15 ' 8 aud 25 ' 5 constitute a proportion, since 16 : 8 = 6 = 23:6. 28. The terras of the two couplets are called propor- tionals. 29. Proportion may be expressed in two ways, Ist. By placing =, the sign of equality, between the ratios. 2Dd. By placing four points, thus : : , between the two ration. Thus, we may express the proportion existing between 15, 8, 26, and 6 by 16:8 = 25 6, or l>y 15 8 -25.5. We read either of them by saying the ratio of 15 to 8 equals the ratio of 28 to 6 ; or simpi v 15 is to 8 as 25 is to 5. NoTK.— The sign : ; is supposed to be derived f^om =r, tha sign of equality, the tour pointe bGiag merely the eeetrerniUea of the lines . 30. In every proportion there must be four terms^ since there must be two couplets, and each couplet consists of two terms. 31. When three numbers constitute a proportion, one of them is repeated so as to form two terms. Thus, if 18, 6, and 2 are proportionals. 18 : 6 : : 6 : 2. In this case the 6, 1. e., the term repeated, is called the middle term or a mean proportional between tho other two nuiribers. The 2 is called the third term >r a third proportional to the other two numbers. 32. It is important to remember the distinction between ratio and proportion. A ratio consista of two termsy an antecedent and a consequent. A proportion consists of two couplets or four t«rms. One ratio may be grtater or less than another One proportion cannot be greater or less than another, since equality does not admit of degrees. 33. The outer terma of a proportion are called the eX' trem.es, and the two intermediate ones, the means. Thus in the proportion 8 : 17 : ; 21 : 11». 8 and 119 are the extremes. 17 and 21 are the means. 34. If four quantities be proportionals, the product of the extremes is equal to the product of the means. 6:li::18:88. Xi^en 6x88 = llxia Aufs. 27^6.] PROPORTION. 207 I This may ho establishod In the following mnnner :— 6 11 = /t and 18 88 = 15, ftiK since 11 la b3, TT = 4S (Art. 27). Now, sirnio multiplyinj? equals by liie same uuuiber doea not destroy their equality, if wo multiply these fractions by 11 wo get 6 = — — — ; and multiplying each of these by 83, wo have 6 x 88 = 18 X 11; but 6 and 33 are the extremes, and 18 and 11 are the means; there- fore in any geotnetrlc.d proportion the product of the extremes equals the pro- duct of the means. The same fact may be established more generally as follows : — Let a, 6, c and d be any four proportionals irhatever, Then ab.e.d ct But a :b = ^ and c : d = ^ b d a Therefore ;- = ^ 6 d a ti d = b •< c. Therefore, «tc. 35. This principle then may be considered the tent of a Kcometrlcal pro- portion. If the product of the extremes equals the product of the means, the four quantities are proportional; If the products are not equal, the numbers are not proportional. 36. It follows from Art. 34 that : — \st. If the product of the means be divided by one extreme^ the quotient will be the other extreme. 2nd. If the product of the extremes be divided by one mean, the quotient will be the other mean. and hence, Zrd. If any three terms of a proportion be given^ the fourth may be found thus : 2nd term x Zrd term Multiplying each of these equals by & x rf, we have But a and d are the extremes and b and o are the means, x«t ttrrwt Ath term. 2nd term =: 1st term x Ath term Srd term. Zrd term = Ut term x Ath term 2nd term. 2nd term x Zrd term l.o^ '?r7n. Example 1. — What is the fourJh p >'» jortional to 7, 11 and 35? 4th term = 2nd term x 3rd U i 11 X 35 = 65 Ana. 1st term. 7 Example 2. — The first, second and fourth terms of a proportion ore 9, 16 and 128. Required the third term. ^ 8rd term s Ist X 4th 9 X 128 2ud 16 = 72 Ana. ' it*' ( 1 In -^«■.*. *;.)**t*/,U ,. 208 SIMPLE PROPORTION. [Sect. V. EXKRCISE 83. V The second, third and fourth terras of a proportion arc 17, )1 and 93 J. What is the first term ? Ans. 2. ''he first, third and fourth terms of u proportion are 21, 63 and 39. Requirod the second term. Ans. 18. 8. The first three terms of a proportion are 2, 3 and 7. What is the fourth term ? Ans. 10^. 4. The last three terms of a proportion are 91, 88 and 104. Re- quired the first term. Ans. 11. Find the fourth proportional to 6. 4 yds. 18 yds. and $96. Ans. $432 6. 6 lb. 2 lb. and $3-76. Ans. ^l' 60. 1. 1 cwt. 215 cwt. and |;7 50. Ans. $1612-50. 8. 6 miles, 1 mile and 27 shillings. Ans. 4s. Cd. 9. 10 lb. 150 lb. and £6 3s. 9d. Ans. £92 163. 3d. 10. 4 days, 27 days and $100. - Ans. $675. 37. It will be useful to remember the following properties of a Geometrical proportion. As the proofs are given in every common work on Algebra, it has not been thought advisable to inscit thtm here ; a, 6, c and d stand for any four proportionals whatever. It ri -.b-.'.cd Alternately a:o::b:d Inversely h -.a: :d:c By Composition a + b :b::c + d:d By Division a — b -.b : -.c — d- d By Conversion a:a + b ■.•.c:o + d Or a -.a— b '.•.c:c — d Or if 15:6: 10. 4 16.10::6:4 6:16::4:10 16 + 6:6: :10 + 4:4, or21 •6:14 :4 16-6-6: .10-4:4, or 9:6: :6- 4 16 15 + 6: :10 10 + 4, or 16- 21 •:10:14 15 : 15 - 6 : • 10 . 16 - 4, or 15 : 9 : : 10 : 6 38. Proportion in Arithmetic is usually divided into simple, compound and conjoined. SIMPLE PROPORTION. 39. Simple proportion is frequently called the Eule of Three, because when three terms are given, by means of them a fourth may be found. It is also sometimes called the Golden Rule from its extensive utility. 40. Example.— If 16 barrels of flour cost $112, what will 1^9 barrels cost ? In this and every other question in Simple Proportion there are two ratios, one of which is perfect {i e. has both terms piven) and the otiier imperfect, and fVom the nature of proportion wo know that these two ratios must be both of the same kind, that is, they must be both ratios of greater inequality or both ratios of less inequality. Now in the above example, the ratio of |112 to the arswer is a ratio of less inequality since it is evident that, if 16 barrels cost $112, 129 barrels will cost more. Therefore the other ratio is also a ratio of less inequality and must be written 16 129. *;'. A»Ts 8T-41.1 SIMPLE PROPORTION. 200 And sinco tbo ratios are equal barrels, dollars. 16: 129: : 112: Ant. 1 1 <> V 1 '20 Also (Art. 80) ^n«. = - ,„— =$903. 10 Pkoof.— Set 908 in the fourth place, thus: 16: 1-29: 112: 908 and SCO if the product of extrorin's = product of moans (Art. 85.) 10x908 = 14448=129x112. From the preceding ill ii.st rations and principles wc de- duce for Simple Proportion the following general RULE. Set the given term of the imperfect ratio in the third place, and the letter a;, to represent the answer, in the fourth. jlJien, if by the nature of the (question, the ratio of the third term to the anawer is a ratio of greater inequaliti/, make the remaining ratio a ratio of greater inequalitg also ; but if the ratio of the third term to the answer be a ratio of less inequality, make the other ratio a ratio of less inequality also. Lastly, {Art, 30,) multiply the second and third terms together, divide the product by the first term, and the quotient will be the an- siver in the same denomination as the third term. Pkoof. — Multiply the first term and the answer together, and, if the product is equal to the product of the second and third terms, tne work is correct. {Art. 35.) Example 1. — If a man can walk 155 miles in 12 days, how many miles can he walk in ^ . days^ Here the imperfect /atio Is IfiS miles to «, and, in order to ascertain wheth- er it U a ratio of greater or less ineqimlity. we have merely to ask the following simple question : If a man can walk 155 miles in 12 days, can he walk more or less in 60 days? Evideiitly more. Therefore the ratio of 155 cr is a ratio of les^ inequality, or, in other words, the antecedent must be the leaat of the two numbers, and the st itement is Whence the answer = davs. miles. 12:'60:: 155 : or. 60x155 12 ■=775 miles. 41. Sinet>the second and third terms multiplied together, corsti- tuifi a dividend, and the first term is a divisor, it is manifest, from the principles of division (Arts. 79-84, Sect. II.), that we may cancel any fact jr that is common to the first term and either of the other term.s. Thus in the last example we have 12 : 60 : : 155 : u and, dividing the first and second by 12, we get 1:5:: 165 : x and 155 x 5=775 Ans. Example 2.—If 96 bushels of wheat cost $128, what will 15 bushels cost? As the answer to the question must be in dollars, the imperfect ratio Is $128 : (P, and from the nature of the question, we know that 15 bushels wlU cost O 4 ' ,(| ' 1 a ■ m 210 SIMPLE PROPORTION. (fi«ot. V. TTpro n2 rp«1ucos Ofl to ft and 128 to 4, and 8 ouaoula tt ttud I'vilucea 15 lo 5. 1m8 than 00 buslu-ls , wh thereforo pliicp t"). tho smnU.-r of fho rirnnlnlnR terms, In tliu Htconit pl(n-e, mid tbo otln't- teriu, (>U, iu tLe Jtrat place. Hence th« eiuteinent Is 96 10 bu&Ueb . ^Ii6 : 0. OPEHATION. mh. $ 3 6x4= %20 Am. The teaclier would do well to insist upon his pupils performing all qu stions in Proportion by analysis. ThuH, to solve the lust niiostion, we hetrin as follows: If 96 bushels cost ft28, 1 biKshtl will cost A>,, of ll-JH, or !|il-:{:}||. Then if 1 busboi cost 91-33), 16 buabt'la will cost IT) tinu's us much, which i.s I'JO. Example 8. — I!' 27 men can mow GO acros of grass in a day, how many acres can 1)3 men mow ? OPERATION. men. ff 81 8 81x20 acres. 20 ■=206H acrcH Ans. 8 Ilore the Imperfect ratio fs 60 : o? acres, and since 93 men will evidently mow more than 27 men, we muke 93 the necorui term and 'J7 the JirMt. Hence the stafemenl is 27 : 98 : : «0 : x. Then 8 reduces 27 lo 9 and 98 to 81, and 3 iiKidit reduces 9 to 8 imd 6(i to 20, and the answer ia equal to 81 multiplied by 20, ai.d divided by 3. This question may be performed thug by analysis : If 27 men mow 6i) acres a day, 1 man will mow ,\ of CO acres, or 2] acres ; 98 tuen will tberei'uro mow 93 tiuies 2^ acre8=206S Ans. Exercise 84. 1. If 11 baskets of peaches cost $13'42, what will 87 baskets coat? Ans. $106'14. 2. If 28 cords of wood cost $266, what will 25 cords cost ? Ans. $237-50. 8. If a man receives $2',)*20 for 16 days' work, for how many days should he work for $83-60? ' ^n.9. 45f i^ dayn. 4. If 16 bags of potatoes are sold for $12-80, what will 156 bags bring? Am. $124-8(i. 6. If a stick 7 feet long cast a shadow of 6 feet, what will be the height of a tree which casts a shadow of 1 1 2 feet long ? Ans. 156| feet. 6. If a stack of hay will feed 27 cows for 99 days, how long will it feed 55 cows? Ans. 48f days. 7. If 9 bushels of peas sow 6 acres, how many bushels will be re- quired to sow 48 acres ? Ans. 8 6| bushels. 8. If 3 men put up 73 perch'^s of fencing in 2 days, how long will they take to put up 803 perches? Ans. 22 days. 9. If 176 pails of maple sap make 100 lbs. of sugar, how much sugar will 1128 pails make? Ans. 640^ Ihs. XO. If it cost $20-88 to weave 108 yards of cloth, what will it cost to weave 466 yards? -4w«. $89*90. ^ i\ ifiiot. V. mnlnlnj? terms, :e. iii'iice the 128 tu 4, and 8 • his pupils ^sis. 96 buslu'ls cost cost |l-a3|, 15 in a (lay, how : 05 acres, and w more than 27 erm and 27 the 27 : 98 : : 60 : (r. 1 81, and 3 iijraiii 1 tlie answer ia d divided by S. as, or 2) acres ; »asket9 coat ? Ins. 1106-14. )St? ns. $237-50- w many days 45^^55 days, nil 166 bags ws, $124 -SI*. will be the ong? 156| feet. V long will it IS. 48f days. will be re- 86| bushels. low long will ns. 22 days. much sugar 640H Ihs. vill it cost to ins. $89-90. k f Allf«.4l,48.] SIMPLE PROPORTION. 211 11. If f Irt pay for the carringc of 72 barrels of flour, for the carriage of how many hurrels will $1278 pay? An.^. 67B1 barrelrt. 12. If 11 mon ploiij^h 165 acres in a week, how many acres would 3 men [jlough In the same time ? Ann. i6 acres. 13. If 4 Ij:inels of flour make 250 four-pound loavca of bread, how many such loaves will 67 bands make? Ans. 4187^ loaves. 14. If lUO biislicls of apples make 16 barrels of cider, how many barrels of cider will 38 bushels of apples make V Anx. 3^ barrels. 15. If 90 men can build a wall in 12 days, how many men could build it in 15 days? Aria. 72 men. 16. If 17 days' work pay for two barrels of flour, for how many bar- rels will 279 days' work pay? Ans. 32ff barrels. 17. If a tniin travel 27 miles per hour, how far will it travel in 24 hours? Am. 648 miles. 18 If 7 cows make SO lbs. of butter a week, how much may be ex- pected from 23 cows? Ans. 98f lbs. 42. If any of the teiins contain fractions or mixed numberSf apply the rules in Section IV. Example 1. — If f of a basket of pejiches cost ^ of a dollar, how much will /,- of a basket of peaches cost ? OPKBATION. f : ^^f w^'.x. Therefoie answer = ^ x -,\ -4- f = $^ x -,\ x f = 19^f cents. Example 2. — If ^^ of a bushel co?^t ,*,- of a pound, what will -fi- ef a bushel cost? OPEllTAION. )''g \\\:\ £ ,*, : X. Therefore answer = -j*, x ^^^ -f- -'^^ = A x j^ x -^ = k^Y- - lis. 10|d. NoTR. — If the first term be a fraction, invert it and connect it to Ae others i>y the sign uf multiplication. ExEP.cisE 85. 1. If -,\ of a ship cost Sr f 0-0669. )t of the tions — either, and lat will 76 his i-educed 1 pk. 1 gal. sq. yds. 2 IU-246. jer oz. ? . $192-50. cost? l«.s. $1-19. gis. cost. ? IS. $3-167. : dwt. cost dns. $150. n. 30 sec, h. 14 min. Flem. ells (is. 10|^d. >s. cost ? 13 9s. Ofrl )le passage )age might 68 y 91st page. ^ 10. If the rent of 46 acres, 3 roods, and 14 perches be £100, what will e the rent of 35 acres, 2 roods, and 10 perches? Ans. £75 18s. 6Hf ?d. 11. When A had travelled 68 days at the rate of 12 miles a day, B, wlio had travelled 48 days, overtook him. How many miles a day did B travel, allowing botli to have started from the same place? Ans. 17. 12. If 21^^ shillings pay for 16| lbs. of prunes, how many pounds can be bouglit for 32t shillings? Ans, 24/^(iV lbs. 13. A ton ot coiil yields about y.iUO cubic feet of gas; a street lamp coii.sumt'S about 5, and an aigand burner (one in which the air pas.>es through the centre of the flame) 4 cubic feet in an hour. Ho'.v many tons of coal would be required to keep 17493 street lamps, and 192724 argand burners in shops, &c., lighted for 1000 hou s? Ans. 95373|. 14. Tlic gi.s consumed in London requires about 50000 tons of coal per annim. For how long a time would the gas this quantity may be supposed to produce (at the rate of 9000 cubic feet per tt»u), koi'p one argand light, (consuming 4 cubic feet per hour) coiisraiuly burning ? Ans. 12842 years and 170 days. 15. Suppose 11270 lbs. of beef for a ship's use were to be cut up in pice s of 4 lb., 3 lb , 2 lb., 1 lb., and ^ lb. — there being an equal number of each. How many pieces would there be of each ? Ans. 1073 ; and 3^ lb. left. 16. The sloth docs not advance more than 100 yards in a day. How long wouid if take to crawl from Toronto to Kingston, allowing the distance to be ISO miles ? Am. 3168 days, or about 8f years. 17. Suppose that a greyhound makes 27 springs while a hare makes 25, and that their springs are of equal length. How many springs must the hound make to overtake the hare, if the latter has a start of 50 springs ? Ans. 675, COMPOUND PROPORTION. 44. Compound Proportion is an equality between a compound ratio and a simple ratio. Thus 7 : 11 conipounded with 22 . 21 : :34 51, is a compounrl rntlo. Or 7 X 22 ; 11 X 21 : : 34 ■ 51, and applying Art. 40 wo have 7 x 22 x 61 == 11 X 21 X 34. 45. Compound Proportion is also called the Double Rido of Three. It enables us to obtain the answer by a single statement, although two or more questions are con- tuiued \ii the questioa. -f 1 ::l m*':- 214 COMPOUND PROPOIITION. [Sect. V. 46. In Compound Proportion there are three or more ratios, one of which is imperfect and all the others perfect. 47. Let is be required to solve the following question : If 18 men dig a trench 30 yards long, in 24 days, by working 8 hours a i]ay, how many men will dig a trench 60 yards long, in 64 days, working 6 hours a day ? Let ns suppose the tfme to be the same in both cases, and this questio.; be- comes the same as tht^ ftillowing If 18 men dig 30 yards of trench, how many men will dig 60 yards ? Here it is evident the answer will be the same fraction of lb that 60 yards is of 80 vards; or, in other words, the required number of men = |„ of 18 men. Next let us take into account the numb* r of days ; but suppose they work the same number of Lours per day in botli oases. The question then becomes : If |2 of 18 men require 24 days to dig a trench, how many men will disr it in 64 days? In this case it is plain that th<'"answer will bo the same fraction of |^ of 18 men that 24 days is of 04 days ; that i(», the required number of men = §1 of |g of 18 men. Lastly, let us take into co. sideration the time worked each day. The question then beconn's If ?J of ",' of 18 men dig a trench in a certain nnmbor of days, working 8 hours per day, how many men will dig it working 6 hours per : : 18 : Answer. 6 SS The answer is equal to the continued product of the third terra, and all the second terms, divided by the conlinued product of all the first terms. From the preceding principles and illustrations, we de- duce the following general RULE FOR COMPOUND PROPORTION. Place that number which is of the same kind as the answer in the . ird U.rm^ and the letter x to represent the answer in the fourth teitn. TJien take the other numbers in pairs, or two of a kind, and ar- range them as in simple proportion. linally midtipUj together all the second terms and the third term, divide the ra^vlt by the product of the first term, and the quotient will be the fourth term or answer required. Note — Since the third term and second terms multiplied together constitute a dividend, and the first terms multiplied together a divisor, we may (Arts. 79-84, Sect. II) cancel any factors that are common to ((n^* of tlie first terms and to the third term or au^' of the tiecond terms. lupre Abt8. 46, 47.] COMPOUND PROPORTION. 2li Example 1. — If 6 compositors, in 16 days, 11 hours long, can compose 25 sheets of 24 pages ia each sheet, 44 lines in each page, and 40 letters in a line; in how many days, each 10 hours long, may 9 compositors compose a volume, to be printed in the same letter, consisting of 36 sheets, 16 pages to a sheet, 50 lines to a page, aiV 45 letters to a line? STATEMENT. 9 comp. 10 hours : 5 comp. : 11 hours. 25 sheets : 36 sheets. days. 24 pages 44 lines : 16 pages. : 50 lines. ":: 16: 40 letters : 45 letters. TIHE CANCELLED. K 'X9 3f 'ii^ tU u %mH 7P, i^P 1 4 Ans. 8x4=12 days. Explanation.— The fmporfpct rntlo Is that of 16 days to an nnknown num- ber of days. We place tliis ratio to the right-hand side, us in Simple Propor- tion. Now we compaie each pair of toria» with thia ratio, in order to decide whether they constitute a ratio of greater or leas inequality. Thus, if 5 com- fjositors re{^aire 16 days, will 9 compositors require more or less? Evidently ess ; theretoro it is a ratio of greater inequality, and we must write it 9 ; 6. Next, if n hours to the day require 16 days, •will 10 hours to the day require ha ore or less;'— more; therefore we raust write H) : 11. Next, if 25 sheets re« quire 16 days, will 86 she<*t3 require more or less? — move; therefore we writo 25 : 86. Next, if 44 lines to a page require 16 days, will 50 lines to a page re- quire more or less? — move; tlierefore we write 44 : 50. Lastly, if 40 letters to a line require 16 days, will 45 letters t-o a line requke more or lebs t— more; therefore we write 40 : 45. The statement is now complete, and we cancel as follows : 5 cancels 6, the first consequent, and reduces 25, tii j third antecedent, to 5, and 6 cancels this 5, and reduces 50, the fifth consequent, to 10. and 10 cancels this 10 and 10. the second antecedent. Again, cuncels the first antecedent and reduces 86, the third consequent^ to 4, and 4 cancels this 4 and reduces 44, the fifth antecedent, to 11, and li cancels this 11 and 11. the secoml consequent. Again, 8 reduces 24 to 3 and 16 to 2, 3 cancels this 3 and reduces 45 to 15. 2 cancids the 2 re- sulting from the 16 and reduces 40 to 20, and 5 reduces this 20 to 4 and the 16 resulting from 45 to 3. Lastly, 4 cancels this 4 and reduces 16, the third torm, to 4. There remain but 3 and 4 which multiplied together make 12. Ana. Example 2. — If 24 men can saw 90 cords of wood in 6 days when the days are 9 hours long, how many cords can 8 men saw in 36 days, when they are 12 hours long? STATEMENT. 24 men 6 days ^ hours 8 men 36 days 12 hours, :. \ ' cords. 90: X. SAME CANCELLED. ,2 ) 10 ^1 '.'.9^'.X. ^-.ii^ C Ans. 10 X 2 X 12 = 9 : 12 ; 240 cords. 74 : r Here the imperfect ratio is 90 : Ans. If 24 nr, • saw 90 cords, will 8 men saw more or loss? — leos; therefore it is a ratio of greater inequality, and we write 24 : 8. Next, if 6 days saw 90 cords of wood, will 86 da^s saw more or less?— more; therefore it is a ratio of less inequality, and we write 6 : 86, Last- ly, if 9 hours per day saw 90 curds, will 12 hours per day saw more or lost.?-* luore ; therefore it is a ratio of less inequality, and we writeil ; 12. 216 COMPOUND PROPORTION. [Sbct. V. Example 8.— If 248 men, in 6^ day.s, of 11 hour' each, dig a trench of 7 degrees of hardness, 232^ yards long, 3| wide, said 2^ deep ; in how many days, of 9 hours long, will 24 men dig a trench of 4 degrees of hardness, 337^ yards long, 5| wide, and 3^ deep? STATEMENT. 24 : 248 men. 9 : 11 hours. 7 : 4 degrees. 282J : 837.'< yds. long. 8i : 5i,' vds. wide. 2| : 8i yda deep. . : : Si days -.Ana. or, • r v-^tn ?:¥ ?-: \ 465 ■ 075 " 5 ■ • 5 ~ -V- : -?- V-:» The answer will be (H^ x V" x 1" x ^t^ x ¥ x 5 x ¥')-5-(¥ x ? x ^ ^i|ayijLjw many men must be em- ploye J to linish the remainder in 4 days? Ans. 36. 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 miles? Ans. 18. If the price of 10 oz. of bread, when the flour is Is. lU^d. per stone, is Id., what must be paid for 3lb. 12 oz. when the Hour is 2s. 6d. per stone? Ajis. 8d. If 5 compositors in 16 days of 14 hours long, can compose 20 sheets of 24 pages in each sheet, 60 lines in a page, and 40 let- ters in a line, in how many days of 7 hours long may 10 com- positors compose a volume to be printed in the same letter, con- taining 40 sheets, 16 pages in a sheet, 6U lines in a page, and 50 letters in a line ? Ann. 32 days. If 338 men, in 5 days of ten hours each, dig a trench of 5 de- grees of hardness, 70 yards iong, 3 wide, and 2 deep, what length of trench of 6 degrees of hardncF yards wide, and 3 deep, may be dug by 240 men in 9 days ot 12 hours each ? Ans. 36 yards. If a pasture of 18 acres will feed 6 horses, for 4 months, how many acres will feed 1 2 horses for 9 months ? Ans. 72 acres. If 25 persons consume 300 bushels of corn in. one year, how much will 139 persons consume in 7 years at the same rate? Ans. 11U76 bushels. If 32 men build a wall 36 feet long, 8 feet high, and 4 feev wide, in 4 days, in what time will 48 men build a wall 864 feet long, 5 feet high, and 3 feet wide ? Ans. SO days. If a regiment of 679 soldiers consume 702 bushels of wheat in 336 days, how many bushels will an army of 22407 soldiers consume in 112 days? Ans. 7722 bushels. If 12 tailors in 27 days can finish 13 suits of clothes, how many tailors in 19 days of the same length can finish the clothes of a regiment of soldiers consisting of 494 men? Ans. 648 tailors. If 17 head of cattle consume 5 acres 2 roods 10 perches of pas- ture in 30 days, how many acres would be consumed by 40 head in 51 days? Ans. 22 acres 1 rood. If 180 bricks, 8 inches long, and 2 wide, are required for a walk 20 feet long, and 6 feet wide, how many bricks will be required for a v/alk 100 feet long and 4 feet wide ? Ans, 600 bricks. 218 CO^^JOI^'ED PKOPOETION. CONJOINED PROPORTION. [Sect. V. 48. Conjoined Proportion is a kind of Compound Pro- portion, in which the ratio of one of the terms to its corre- sponding term is made to depend on equivalencies among the intermediate terms of the proporti m. 49. Conjoined Proportion is sometimes called th 3 Chain Rule from the peculiar manner in which the differ- ent pairs of terms are linked, as it were, together. It re- lates principally to exchanges between diflferent countries, in respect to specie, weights, and measures, but is ai)p]ica- ble to jommon business transactions. 60. Example ] . — Suppose 1 yards of velvet in Toronto cost as much as 9 in Montreal, and 16 in Montreal as much as 24 in Puiis, how noany yards in Toronto will cost as much as 64 in Paris ? Explanation. — This question may be stated as a problem in Compound Proportion as follows : The imperfect ratio is 7 yard- Toronto to an unknown 9 : 16 J . . 7 . jB nnmber of yards Toronto, Then, if 9 yards Montreal pay for 24: 54 f ■ 7 yards Toronto will 1(5 yards puy tor more or It'ss? — mor ; therefore we write 9 : 16. Next, if 24 yards Paris pay for a certain number ( —^ — j yards Toronto, will 54 yards Paris pny for more or ^ y -^ less? -more; therefore we write the ratio 24:54. Now (Art. 47) the answer 16x54x7 = ; and it is evident that we may consider all the factors of the nu- 9x24 merator as antecedents, and nil the factors of the denominator as consequents, and then make the statement thus : STATEMENT. T yds. Toronto = 9 yds. Montreal. 16 *' Montreal = 24 " Paris. 64 " Paris = a; " Toronto. Since th^ left-hand numbers coristitute a dividend and the riRht-hand num- bers a divisor, we may eancol factors that are common. Merely writing the numbers and doing this Ave have— SAME CANCELLED. ^ — 9 J! 4 J^ = '^4* ii^^ = x = 4:x1=z28 yds. Ajts. From the preceding principles and illustrations we de- duce the following ; S AsTS. 4^50. Write ,ngn of equ be on oppoi Multipi dend and a required te) EXAMPI goats as mi many horse STA1 26 sheep = S3 goats = 38 cowa = X horses^: Here, sin term, 60 shee Note.—! equal in iinn EXAMPL in JEIamiltoi 30 lbs. in Q ton as mucl 57 lbs. in H lbs. iu Guel] stat: 19 Guelph 1 Hamiltoi 30 Quebec 8^ Boston 10 London X Hong Kc J. If 17 coi of tea 1 work, a of peac many c If 6 lbs. ( for 1 bi to 4 to> Abts. 46 50] CONJOINED PROPORTION. RULE FOR CONJOINED PROPORTION. 219 Write the equivalent terms, as they oceitr^ right and left of the sign of equality^ taking care that terms of the same name shall always be on opposite sides. Multiply all the terms on the same Hde as the odd term for a divi dend and all on the other aide for a divisor. The quo^ eiU mil be tfie required term. Example 2. — If 25 sheep eat as much hay as 19 goats, and 88 goats as much as 10 cows, nnd 38 cows as much us 22 horses, how many horses will eat as much as 60 sheep ? STATEMENT. 26 sheep =19 goats 83 goats =10 cows 88 cows =22 horses X horse8=60 sheep Or writing the f^ numbers merely, j 3 SAME CANCELLED. ^^ = ;?o ' cancelling and ap- 1 ^ «o Z ooXX 4 plying the rule. [ Z-i^ Ans. 4x2=8 horses. Here, since the term 25 sheep Is on the left-hand side, we put the odd term, 60 sheep, on the rli^ht-hand side. Note. — The sign =: in such questions, merely means equal in value, or equal in time, or equal in ^ect, &c Example 8. — If 19 lbs. of tea in Guelph cost as much as 20 lbs. in Hanailton, and T in Hamilton as much as 9^ lbs. in Quebec, and 30 lbs. in Quebec as much as 29f lbs. in Boston, and 8j- lbs. in Bos- ton as much as 6^ lbs. in London, and 10 lbs. in London as much as 67 lbs. in Hong Kong ; how many lbs. in Hong Kong are worth 100 lbs. iu Guelph? STATEMENT. 19 Guelph =20 Hamilton 1 Hamilton =9^ Quebec 80 Quebec =29f Boston 8^ Boston = 6^^ London 10 London =57 Hong Kong X Hong Kong=100 Guelph SAME CANCELLED. 10 X9=^0 g /=9^ ;p=^/ 19 x=W^ Ans. 10 X 9^ X 5i = 5065 lbs. Exercise 88. 1. If 17 cords of wood are equivalent to 116 lbs. of tea, and 87 lbs. of tea to 28 barrels of flour, and 19 barrels of flour to 84 days' work, and 92 days' work to 57 baskets of peache?, and 81 baaicets of peaches to 24 dollars, and 12 dollars to 2 tons of coal; how many cords of wood may be purchased for 85 tons of coal ? Ans. 135f. 2. If 6 lbs. of tea are worth 29 lbs. of sugar, and 17 lbs. of sugar pay for 1 bushel of wheat, and 27 bushels of wheat are equivalent to 4 tons of coal, and H tons of coal purchase 15 cows, and 29 1 220 EXAMINATION QUESTIONS. ISect. V. u cows «ost $1160 ; how many pounds of tea can be ptirchused for $20? -l?ts. iiiJi;;?,. 3. If 11 bushels of barley pay for 21 bushels of potatoes, i.iid 11) bushels of potatoes for 29 bushels of oats, and 115 bushels of outs fur 4i buohels of wheat, and 1-1^ bushels of wheat for ZS bushels of poas, and 60 bushels of peas for 55 bushels of rye, and 75 bushels of rye for 11^ bushels of clover seed; for Low many bushels of barley will 36 bushels of clover seed pay ? Ahs. 81 ^. 4. If 16 baskets of pears pay for 29 turkeys, and 17 turkeys for 7 days' work, and 7^- days' work for 187 loaves of bread, and 3^- loaves of bread cost as much as 4 lbs. of veal, and veal is 1 1 cents per pound, and $7"92 pay for 63 lbs. of sugar ; how many pounds of sugar will 21 baskets of pears purchase? A7(s. 40-l|. 6. Suppose A can do as much work in 7 days as B can in 1 1 days, and B as much in 5 days as C can in 8 days, and C as much in 15 days as D can in 21 days, and D as much in 11 days as E can in 5 diys ; in how many days would A do as much woik ns E can do in 42 days? Avs. 26^. 6. If 7 barrels of flour pay for 23 cords of wood, and 6 cords of wood pay for 11 cwt of beef, and 46 cwt. of beef cosL £28, and £77 pay for 9 sheep, and 5 sheep are worth as much as 8 tons of coal ; how many barrels of flour may be purchased for 9 tons of coal? Ans. 13^. 7. If 153. in N. England be the same in value as 20s. in N. York, and 24s. in N. York the same as 22s. 6d. in N. Jersey, and 309. in N. Jersey t!ie same as 20s. in Canada ; how many pounds in N. England are the same in value as £240 7s. 6d. in Canada ? Ans. £288 Oa. l.\ 19. •JO. 21. 22. 23. 24. 1'}. 2T. 28. 29. .SO. [',[ .'5J. '■■> t) '. 84. 85. PA h:. 39. 40. 41. QUESTIONS TO BE ANSWERED BY THE PUPIL. Note. — T7ie numbers following the questions refer to the numbered arti- cles of Che sficiioii. 1. In how many ways may one number bo compared with another with re- spect to ir.",cnitude ? (I) ■ What is ratio y (2) \J:\v is the difference between the Geometrical and the Arithmetical ratio Oi ..umbers? (8) How many ways have we of expressing the ratio of one number to anoth- er? (4) Between what kind of quantities only can ratio exist? (5) Wlioii are qiuuititits said to be of the same kind? (6) 7. What is a couplet? (7) 8. What is the antecedent?— the consequent? (8) 9. How many kinds of ratio are there ? (9) 10. What is a direct ratio? (10) 11. What is an inverse ratio ? (11) 12. What is the reci[)rocal of a quantity ; iS) 13. What is a reciprocal ratio? (12) }4, Uow is the reciprocal ratio of two numbers expressed? (14) 8. 5. 42 4:l 47. 48. 52, 63. 54. 55. ISkct. V. Skct. v.] EXAMINATION QUESTIONS. ^21 rtliiised for \r>. US. 'iU^ii?,. cs, iii.d ■]•) 16. bushtlsi of 17. 18. leat for 28 lels of rye, 19. \ ; for how ■-'0. pay? 21. Aus. SVH- rkeys for 7 ^ 22. ■ad, and 8^ 23. veal is 11 how manv 24. Avs. 404f 2.'). 20 in 11 days, 27. as much in j 28. ys aa E can 29. woik as Vj Avs. 26^. .SO. •ds of wood 111. "1 8, and £77 ;;■"..' 3 8 tons of 84. )r 9 tons of i 30. Ans. \H' \ r.«. . York, and t h:. I 30s. in N. 3-i. unds iu N. 39. lada? 40. !. £288 9a. 41. XL. 1 42 1 4:1 nbei'ed artU fl 44. 1 ^ 43. her with re- 4G aietical ratio 47. 48 )er tu anotb- 49 50 ': 51 52. 63. 54. 56. Show that "ropfprncnl rntio" and "inverse ratio" are Intorchangeable terms. (12) Wliiit Is a siinplo ratio? (15) Wliiit Is u CMinpoimd ratio? (ICi) SliKH' a cnnipounil ratio does not differ in nature from a oimple ratio, why is the term us.s.i? (17) IIow are ratios cf>m|)oiindod together? (IS) jIow does in iltiplyiii:; the antecodent or dividing the consequent of a coup- let by anv nciiiher. afto t the ratio? (19) How does dividing the antecedent or multiplvlnff the ronst-quent of a coup- let by any nuinb.T, affect th • ratio? Why? (19) How does ri nltiplyinix or dlvidini; both antecedent and consequent of a coiipiot by any number, affi'ct the ratio? Why? (19) How does It happen tiiat we in;iv cau'el any fa.Uors common to an antece* dent and a coiiscque t. before compoiindiuir ratios togettier ? (20) When la a ratio called a . tio of equality f (21) When is a ratio c died a ratio of greater inequnlityt (21) When is a ratio called a ratio (flexn inequaliti/f (21) How are ratios compared with one another? (22) When equal rata»8 aro added together, what ia the nature of the resnlting ratio? (2:5) What effect has adding the same number to both terms of a ratio? (25 and 2(5) Wt-.at i9 Proportion? (27) What aro the terins of the two equal ratios called ? (2S) H )W many ways are there (»f e.xpressing [Proportion? (29) What is the supposed derivation of the siijn : .? (29 — Note) How many terms must there be in every projiortion ? ("0) When tliree numbers constitute a proportion, what is the repeaied terra called ? What is the last term culled ? (81) Point out the distinctions between ratio and proportion. (32) What are ''■ en'tr ernes'" and "■means'''' f (S'.i) Prove that if four quantities are proportional, the product of the extremes is equal to the product of the means. (34) What Is the test of treometrical ratio? (3.5) Deduce from this principle a rule for finding anyone of the terms when the other three are given. (So) If r : w : : X : y, what does the proportion become ? 1st, by compn 1J22 illSClCLLANEOUS EXEKCl9fi. [SlOT V. B6. What Is Cunjotned Proport'on? (48) 67. Why is It Eonvflmt's culled the Chain Rule ? (49) 68. Give the rule f»»r C'onioined Proportion. (50) 69. In what sensu is the sign = taken in these statements? (50) 1 Exercise 89. MISCELLANEOUS EXERCISE. {On preceding Rules.) What is tlie ratio compounded of the ratios 7 : 8, 17 : 1 1, 28 : 29, 819 : 119, and 16 : 69? 2. Reduce £119 16s. 6-^d. to dollnrs and cents. 8. How many days are there liom I'ith March to the 17th of the following February ? 4. Compare together the following ratios, and point out which is greatest and which least, 9 : 13, 21 : 27, 7 : 10, and 11 : 16. 6. From 76-23478 take 19-134229i. 6. Multiply 71324< undenary by 23421 qumary and divide the re- sult by <4e7 duodenary. Give the answer in each scale. 7. If 5'63 cubic inches of water weigh 3'254 ounces avoirdupois, what will be the weight of 79 cubic inches of nitric acid having a specific gravity of 1*220? 8. Divide 63 yds. 8 qrs. 2 na. 1 in. of ribbon equally among 17 persons. 9. What is the value of -913625 of an acre at 67 cents persq. yard? 10. Multiply ^ of f of I of 20 bushels by -5 x -6 x i 11. Of the ratios 6 : 7, 17 : 8, 23 : 11, and 88 : 176, point out (1) which is the greatest, (2) which is the least, (3) which are ratios of greater inequality, (4) "which are ratios of less inequality, (6) what is the ratio compounded of these ratios. 12. The population in Canada in 1861 was 1842265, and in 1857 it was estimated at 26^71437. What was the rate per cent, of increase ? 18. From one-half of two-thirds of eighteen twenty-ninths subtract one-eighth of two-thirds of five-sevenths. 14. Deduct 7 per cent, from 1 1 feet. 16. What is the value of 79 lbs. of tea at £-00168 per ounce? 16. If 8 men in 2^ days, working 12 hours a day, can cradle a field of wheat containing 20 acres, in how many days can 4 men, working 10 hours a day, cradle a field of wheat containing So acres ? . . 17. Find the value of (J of ^ x -02 x •456)-4-(|f of \ of \ of 51). 18. A certain number is divided by 5, the result is divided by |, this result by ^^ and this last result by \. The last quotient is 2 ; what was the original number ? .1^ [6 tor V, Ssct. v.] MISCliLi.ASKOLS tXEiiClSE. 223 7: 11,28:29, le 17th of the i out which is md 11 : 16. divide the re- i scale. I avoirdupois, ic acid having Uy among 17 per sq. yard ? point out (1) lich are ratios nequality, (5) id in 1857 it per cent, of Qths subtract mce? 3radle a field can 4 men, oulaining 85 iofSl). 2d by f , this uotient is 2 ; 19. If 50 barrclsi of flour in T'.ironto are worth 125 yards of cloth in New York, and 8o yanlfr of cloth in New Y'ork tt bales of cotton in Charleston, and 18 baJea of cotton in CharlcMon 8^ hogs- heads of sugar in New Uil'^ani* ; how many liogsheads of sugar in Now Orleans are worth 100«> barrolsj of flour in Toronto? 20. Multiply 73-47 by '00H3, and divide :(ie result by 17'2346. 21. Keduce 2 roods 7 per. 4 yds. 3 ft. 1J7 in. to the decimal of 7 acres. 22. Deduct -73 of 11 furlongs fnxm ^ of J of ^ of 70 miles. 23. From 274*812 nonary take 1,1<4<)U,0U) biliary, and multiply the result by 5555 septenari/. Give the answer in all three scales. 24. Find the 1. c. ra. of 44, 275, 18, I'JO, 209, and ii25. 25. If 60 men in 6 weeks of 5 working days, of 10 hours each, build an embankment 800 yards in length, 18 feet in mean breadth and 11 ft. in mean height, how many men will make an em- bankment 8742 feet long, 2<) feet wide and 8 ft. liigh, in 10 weeks, of 6 days each, and eleven working hours to each day ? 26. How many divisors has the number 172000? 27. Multiply 42-7 by 9*7i23. 28. Deduct 27 per cent, from $73-42. 29. What are all the divisors of 6300 ? 30. If f of f of 3^ lbs. of cofiee cost ^ of J of |^ of i of a dollar, what will f of -7 of -6 of H of 90 lbs. cost ? 31. If |2739'18 be divided among 7 men, 2 women, and 11 children, so that each child shall have f of a woman's share, and each woman ,\ of a man's share, what will be the amount received by each ? 32. What is the reciprocal ratio of ?^ : ^ ; the direct ratio of 98 and the inverse ratio of f of I ? 33. Add together f of 6^ yards, f of f of 8J ft., and ^ «f tV of inches. G4. What is the ratio compounded of 28 : 7, 4 : 11, 6 : 5, 13 and 38i : 3 ? S5. A pint contains 9000 grains of barley, and each grain is one third of an inch long. How far would the grains in 23 bush. 2 pUs. 1 gal. 1 qt. 1 pt. reach if placed one after another ? 36. Reduce -^^^g to its lowest terms. 37. Add together ^, |, ^, and f in the octenary 'iale, 08. If 17 sheep eat as much grass as 6 cows, and 26 cows require 27A acres, and 12 acres supply 13 horses, and 11 horses eat a3 much as 28 goats, how many goats will eat as much as 68 sheep ? 89. Suppose that 50 men, by working 5 hours each day, can dig, in 54 days, 24 cellars, which are each 86 feet long, 21 feet wide, and 10 feet deep, how many men would be required to dig, in 27 days, 18 cellars, which are ecch 48 feet long, 28 feet wide, and 9 feet deep, provided they work only 8 boars each day ? / 17, 7i'tf Hi, '1 'S^' ■ "1 V- 1 1 .1 i. /• . J 1 m I y^ j^- t i f 1 JhH t; V^H^^^^^H ^B 1 1 1 I^H ! ! i i \ ii^4 PRACTlClt. SECTION VI. [SiCT. vi. PRACTICE. 1. Practice is so called from its being the method of calculation practiced by raercantile men ; it is an abridged mode of performing processes dependent on the Kule of Three — particularly when one of the terras is unity. Tb« statement of a question In practice, In general termn, would bo— One quantity of goods : another quantity of goods: '.price of former : price of latter, 2. The simplification of the Rule of Three by means of pruclice, 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 of each of these parts ; in either case, as is evident, we obtain the required price. ' 3. An Aliquot Part is an exact or even part. ThuA, 2 shillings i!« an aliquot part of a pound ; 12^ cents is an aliquot part of a dollar; 6 months, 4 months, 8 months, 2 months, 1| months are aliquot parts of a year, Ac. TABLE OF ALIQUOT PARTS. Parts of $1. Parts of a year. "^mo^nth/ P-t««^^^- Parts of Is. ParU of ft cwt.* of W'l Ibi. 50cts — 6m'th8= i lfiday8= ilOs = i 6d= i 56 lb = i 88^ zz ■ 4 = ilO = 4 6s8d = *!4d= i 8 = i n =i68 =i8d=i 28 lb = i 25 ss ; ■ 161b = 1 :;o . ; 1 ■A = j: 6 = J 4s =1 2d= t 141b = i I16» - "^ H = i B = i 88 4d = A 1*= \ 8 1b = ,^, m := 1 =tV 8 = ^\ 2s 6d = i ld=A 7 1b = ,^g 10 ^ T^S 2 = tV 28 = Vo parts of a qr. % — tV 1 = ,'5 Is 8d = j-V of 28 lbs. IS 4rt = rt 14 lb =4 6 = A Is 8d = tV 7 1b = J 4 = *> 18 =,V 8i lb = ^ 3 — " ^ If lb = ^s * Although we allow but 100 lbs. to the cwt. In Canada, it is often neces- sary to make calculations with the old cwt. of 112 lbs. This arises from the ItTB. l-«.] PRACTICE. 225 «rti of A cwt.* •V of 11'^ Ibi. a lb = i 1 lb = i ' 1^ = ^ \ lb = i lb = ^, ' lb = i'. : jarts of a qr. of 28 lbs. lb = i lb = 1 lb = A , lb = ^, i a often neces- 1 ses from the 1 Example 1. — Find the price of 2783 yards of silk at $3*87^ per yard. OPRHATTON. 20 0. i 2788 The cost uf 2Tba yurds ut iifS'STi— cost at |8-»> cost at 87| 8 cei ts. 27fi3 ydH. at !?8 comes to 8 times as mnrh ai> at fl ; 1. «»., 8349 to 8 tim»'H S'ZTSJ, or *H:tt9. 874 <'"• •qiiiil" "i^ els. + 12J cents, 12i 0. \ mhV> hence, 2788 yds. ut 8TJ ccnts^iirlcc at 25 rent8 + [irlce at 124 847 07} cunts. Since 2783 yards at \\ come to $27S3. and 25 cents=i of Ans. $939262i a dollar; 27S;i yards at 25 cents conic to \ f)f ♦278J», 1. e., to $ri95"7fi. Atrain, be'uiisc 2783 ynnts at. 2."* cc ts come to $695-75 and 12^ cents oqiml.s \ of 25 cents, 2788 yards ut 12| cents would come to \ of $695 75; 1. e., to .<8f7-87f Then 2783 yurd.'« nf $3 :57J=priceatf8 Hprice at 25 cents+nrlce at 12* centi =48349 + 1^095 75+ j847 87i=.-98926Jf Example 2. — Wluit is the cost of 972 oz. of geld dust at £8 148. 8|d. per oz. ? orEltATIOI*. x= coft at £8 =^ cost at 10 = cost at 8 4 = COM at 10 = cost at 6 = cost at li lOa. \ 972 8 £2916 8b. 4d. ■ 486 lOd. 162 6d. i 40 10s. lid. i 20 6 6 1 8d day £3629 16 8 =c()6t at £3 14 8i ExAMPLL 8. — Find the price of 729 days' work at £] 7a. l^d. per 5d. , i Is. 8d. j i 6d. id. OPERATION, £729 0= price at £1 182 6 = price ot 6 60 15 = price at 1 8 16 8 9 = price at 6 16 2i= price at 0^ i;987 18 lli=priceat £1 7 li Example 4. — What ia the cost of 624 bush. 1 pk. 1 gal. 8 qt. of wheat at $2-87^ per bushel ? OPERATION. 50 cts. 25 cts. 12i cts. 624 2 $1248 = price of 624 bu8h. at f2'00 812 = price " " at 50 156 = price " " at 25 78 = price " " at 12* 11794 = price of 624 bneh. at $2-87i fact that the latter is still in common use in Great Britain, several of the States of the American Union, &c. The aliquot parts of the dom cwt. of 100 Ibe, are the same as the aliquot parts of $1. J h ■ ^ly *- 226 PEACTICB. [Sbci VI. I- > 1 pk. i $?-87j = price of 1 bush. •71 1 = price of 1 pk. •SSfJ = price of 1 pill. Igal. 2qt. 1 qt. j -171* = price of 2 qt. I -08^1 = price of Iqt. Then $1794 1 84*9 ai-34J» = price of 1 pk. 1 gal. 8 qt. = prto«? of 624 bushels at $2871 per bushel. price of 1 pk. 1 gal 8 qt. at |2-67i ptT bush. $n05Q-i*gl = price of 624 bush. 1 pk. 1 gal. 8 qt at $287i per biisb. Example 5.— What is the price of 96 acres 1 rood 14^ per. a1 £1 lis. 6|d. per acre? 10s. Is. 8d. lid. Jd. t 96 7 je672 0- price of 96 acres at £7 48 = " " " at 10 6 0=" " " at 1 8 12= « " " at H 6= " " " at 0} £726 18 = price of 96 acres at £7 11 5i 1 rood iOper. 4 per. iper. * I To £7 11 5i 1 17 lOi + i 8 9i+|S Si + tfS = price of 1 rood. = price of 10 perches. = price of 4 perches. = price of i perch. £2 11 7 +,J5 f. = price of 1 rd. 14* per. at jE7 lis. S^d. per ac £726 18 = price of 96 acres. Ans. £129 98. 7d. + jj^ f. = price of 96 acres 1 rood 14t per. Example 6. — What is the cost of 964f| square yards of plaster* ing at 22^ cents per square yard ? 20cte. 2ictfi. 964 $192-80 = cost of 964 yds. at 20 cts. 2410 = cost of 964 yds. at 2* cts. $31 6-90 = cost of 964 yds. at 22* cts. •16* = cost of U ot a yd. at 22* cts. ^.!^=16*cent». IC An9. 121706* = cost of 964}* yds. at 22* cts. per yd, AuT. 8.] PRACTICE. 227 at $2-87i per busb. ood 14^ per. at 1 8 H Of 'ards of plaster* Exercise 90. 1. Required the value of 92647 lbs. of tea at 35 cents per lb. Ans. $32426-46. 2. What is the cost of 94937 puilsat Is. 5d. each? Ans. .£m''23 14s. Id. 8. What is the worth of 95972 boxes at 7^ cents ? Ans. |7197-9(). 4. Wliat is ti\o cost of 62 acres at $28-80 per acre V Ans. $1785-60. 5. Find the p. ice oi'-iolO lbs. at 32. V cents per lb. Ans. ^$75(»-75. 6. Find tho pt ici- of 21 1 7 bags at o7^ cents each. Ans. $793-87^. 7. Find the p;ice of 7506 piiir of shoes at Is. 9fd. a pair. A71S. £680 4s. 7^d. 8. What is the value of 1217 lbs. of coffee at 17.V cents per lb? A71S. $212-97^. 9. Find the price of 2103 cords of wood at §3 07 A per cord. Ans. ^6466-72^ 10. What is the cost of 2006 oz. of gold di'st at £3 ISs. lO^d. per oz. ? Ans. £82;)6 2s Od. 11. Required the value of 6 oz. 18 dwt. 20 grs. of silver at $1-55 per oz. Alls. 10-75j^. 12. What la the cost of 98 yds. 3 qrs. 1 ua of cloth at £1 15s. per yard? ^u.s. £172 18s. 5^d. 13. What is the rent of 344 acres 3 roods 15 per. at £4 Is. Id. per acre? Aiis. £1398 Is. O'^^d. 14. What is the price of 5 oz. 6 dwt. 17 grs. of mercury at 5s. lod. peroz. ? Ans. £1 lis. l^'^d. 15. Find the price of 4 yards 2 qrs. 3 nails of satin at £1 2s. 4d. per yard. Ans. £5 4s, 8^d. 1C>. Find the price of 32 acres 1 rood 14 perches at £1 16s. per acre. A71S. £58 4s. Ud. 17. Find the price of 3 gals, 5 pts. of spirits of wine at 7s. 6d. per gallon. A71S. £1 7s. 2^d. 18. How much will 724 bushels of apples come to at $1-67 J per bushel? ^n,s-. $1212-70. 19. What is the cost of 721 bush of wheat at $1-93 J per bush. ? A71S. |1396-93|. 20. What is the cost of 4514 rods of fencing at £2 17s. I^d per rod ? Ans. £13005 19s, 3d. 21. Wiiat is the price of 3749^ acres at £3 153. 6d. per acre ? A71S. £14153 173. OJd. Allowing 112 lbs to the cwt., find the value of — 22. 17 cwt. 1 qr. 17 lbs. ut £1 4.s. 9d percwt. Ans. £21 10s. 8,Vfd. 23. 78 cwt. 3 qrs. 12 lbs, at $11-55 per cwt. A7is. $910 80. 24. 20 tons 19 cwt. 3 qrs 27^ lbs. at £10 10s. per ton. Ans. £220 9s. IHd. nearly. 20. 219 tOOS 16 QWt. 3 qrs. at $45-50 per ton. Ans. $10002-60|. I ) 228 BILLS OF PARCELS. [Sect. VL Exercise 91. BILLS OF PARCELS. (No. L) QuKBEC, I6th April, 1859. Mr. John Day, Bought of Richard Jones. 8. •d. £ s. d. 16 yards of fine broadcloth, at 13 6 per yard, 10 2 6 24 yards of superfine ditto, at 18 9 " 22 10 27 yards of yard wide ditto, at 8 4" 11 5 16 yards of drugget, at 6 8" 600 12 yards of serge, at 2 10 " 1 14 82 yards of shalloon, at 1 8 " 2 13 4 Ans. £53 4 10 (No. 2.) Mr. James Paul, Montreal, 24^/i June, 1859. Bought of TnaMAS Norton. s. d. 9 pair of worsted stockings, at 4 6pairof silk ditto, "at 15 1*7 pair of thread ditto, at 5 23 pair of cotton ditto, at 4 14 pair of yarn ditto, at 2 18 pair of women's silk gloves, at 4 19 yards of flannel, at 1 6 9 4 per pair. u 4 2 per yard, Ans. £23 15 4^ (No. 3.) Mr. William Filbert, Toronto, 10th July, 1859. Bought of George Price. 75J lbs. of sugar, at ^f cents per lb., 63 lbs. of tea, at 93 126 lbs. of butter, at 13 35^ lbs. of raisins, at 18 J 17 lbs. of sago, at 16 23 lbs. of rice, at 9 68^ lbs. of starch, at 22 Ans. |105-02f. ipril, 18R9. D Jones. £ s. j-ard, 10 2 22 10 11 5 5 1 14 2 13 d. 6 4 .ns. £53 4 10 June, 1859. Norton. f pair, u (( Abt. 8.1 BILLS OF PARCELS. 229 (No. 4.) Hamilton, I2th Augmt^ 1869. Mr. John James, Bought of James Thomas. $ cts. 198 Sangster's National Arithmetic, at O'oO 19Y Robertson's Philosophy of Grammar, at OoO 83 Hodgins' Geography, at 1-00 57 Sangster's Algebraic Formula, at 0'12^ 217 Strachan's Canadian Penmanship, at 0*37^ 143 Hodgins' Geography of British Provinces, at 0'45 227 Sangster's Elementary Arithmetic, at 0*30 Ana. $521-26 (No. 6.) ^ Niagara, l^th September^ 1859. Mr. Alex. Leith, Bought of Lawrence Merger. s. d. 9^ yards of silk, at 12 9 per yard, 13 yards of flowered ditto, at 15 6 '* 11 J yards of lustring, at (5 10 " 14 yards of brocade, at 11 3 *' 12^ yards of satin, at 10 8 " 11 J yards of velvet, at 18 " • Am. £44 16 10 ^'l V! ■ t^ ri I m' ns. £23 16 4^ Ins. |105-02f. (No. 6.) Kingston, llth July., 1869. Dr. Alex. Hamilton, Bought of Timothy Pestle. 14 oz. ipecacuanha, at $0*67 23 '* laudanum, at 0-89 17 " emetic tartar, at 1*25 25 " cantharides, at 217 27 " gum mastic, at 0-61 66 " gum camphor, at 0'27 Ans. $136-94 .tH.| I; : i 230 TARE AND TRET. [Sbot. VI. (No. 7.) London, C.W., Ut May^ 1859. Mr. Jas. Grey, Bought of Michael Lewis. s. d. 15^ lbs. of currants, at 4 per lb,, 17^ lbs. of Malaga raisins, at 5^ '* 19| lbs. of sun raisins, at 6 " 17 lbs. of rice, at 3^ '• 8i lbs. of pepper, at 1 6 " 8 loaves of sugar, weight 32^ lbs., at.... 8J- " 13 oz. of cloves, at 9 per oz. Ans.£Z 13 6i TARE AND TRET. 4, Tare and Tret is the name given to a rule by means of which merchants calculate the amount of certain allow- ances which were foimeily made in buying and selHn;^ goods by weight in large quantities. They were as fol- lows : 1. Tret, an allowance for waste in weighing*. 2. Tare, an allowance for the actual or supposorl weight of the box^ bay^ barrel^ &c., containing the ^ool.;-. And 3. Cloff, an allowance of 2 lbs. in every 336 for the turn of the scale in retailing goods. ^ Of these the only one known in Canada is Tare ; and as this is always set down in full in the invoice, Tare and Tret, as a rule, has no existence in Canadian mercantile transactions, and has therefore been altogether omitted. tion QUESTIONS TO BE ANSWERED BY THE PUPIL. NoTK — The numbers after the questions refer to the articles of the sec- 1. What is Practice ?(1) 2 Why is it HO called ?(1) 8. Of what niU) 's rractice merely a modification ■ (1> 4. What would hi* the peneial stiitemi'iit of a qufsiion in Prnctico? (1) 6. How is the process of finding the price of a number of articles 6iini)lifled by Practice ? (2) 6. What is an aliquot part ? (3) 7. Wbat are the aliquot parts of a dollar ? (8) An. 4) MISCELLANEOUS EXEECI8E. 231 8. What are the aliquot parts of a year ? (8) 9. Wliut are the liliquot parts of a m<)nth? (8) 10. What are the aliquot parts of a £ V (a) 11. What are the aliquot par^s of a shilliu)?? (8) 12. What ar« the aliquot parts of a cwt. (11? ">3.)f (8) i ■« Exercise 92. 4ns. £3 13 6i ry 336 for ti.c- MISCELLANEOUS EXERCISE. {On preceding liules.) 1. Take the number 70204, and by removing the decimal point (1) multiply it by 100000; (2) divide it by 10000; (3J make it thousandths ; (4) mak^ :t tenths of hillionths ; (5) make it tenths ; and (6) make it hundredths of billionth^. 2. Divide 427-1 by -0000637. 8. What will 19 tons 19 cwt. 3 qr3. 27^ Iba. of hops cost, at £19 19s. llfd. per ton? 4. Add together 73-723, 11-842, 16-713, 19 034, 713-218437, and 12-345678. 6. Of the ratios 5 : 7, 9 : 13, 12 : 17, and 7 : 10, point out (1) which is greatest, (2) which is least, (3) what is the ratio compounded of these"? 6. If 1 ac; f land cost $80-50, what will 25 acres, 2 roods, 85 rods cost ? 7. What is the G. C. M. of 144, 485, and 63. 8. What is the price of 7439 cords of wood at |3-68| a cord? 9. Reduce HH'r^, iMI^B, i^BS^, and m^ *<> t^eir lowest terms. 10. If 34^ bushels of turnips are worth 17 bushels of potatoes, and 9 bushels of potatoes 59^ lbs. of tea, and 6 lbs. of tea 11^ stone of flour, and 13 stone of flour $3-60, and 38 cents pay for 12 lbs. of bread; how many bushels of turnips are worth 119 lbs. of bread ? 11. If 27 men in 7 days, working 8 hours a day, paint 42 floors, each 20 feet long and 16 feet wide, with three coats of paint to each; in how many day, of 11 hours each, will 64 men paint 77 floors, each 24 feet long and 22 feet wide, giving each 5 coats of paint ? 12. Take the number 7449164 and by removing the decimal point, make it f 1) One hundred thousand times greater. (2^ One million times less. (3) Hundredths of quadrillionth& 4) Thousandths. 5) Tenths of billiontha. 6)Tenthfl|i '1; % /v*' m / !|! 282 PESCENTAGli. [Skot. Vil, 15, Reduce 72342 nonary t. j^uivalent expressions in the duodena- ry^ senary^ and ternary scales, and prove the results by reducing all four numbers to the decimal scale. 14. Express in the decimal scale the greatest and least numbers that can be formed with six digits in the binary^ quaternary^ senary, octenary, and duodenary scales. 16. Write down all the divisors of 1728. 16. What is the 1. c. m. of the first fifteen even numbers, 2, 4, 6, 8, &c.? 17. From 97-91342 take 18-1234567. 18. What would be the cost of painting a ceiling 20 ft. 7 in. long and 19 ft. 6 in. 7" wide, at $287^ per square yard? 19. Divide 916 acres, 3 roods, 17 per., 7 yards, by 43 acres, 1 rood, 2 per., 17 yds. SECTION VIL PERCENTAGE, CujIMISSION, BROKERAGE, STOCKS, INSU- RANCE, CUSTOM-HOUSE BUSINESS, ASSESSMENT. 1. The term Per Cent, is derived from the Latin word per, "by " or "for" and centum, "a hundred," and means "for a hundred." The term is usually employed to indi- cate the allowance paid for the use of money, but may also be used to express so much the hundred units of any other quantity. Thus, the term 6 per cent, on so many dollars, gallons, miles, days. &o., ■Ignifles $6 on every $100, or 5 gallons on every 100 gallons, or 5 miles on every 100 miles, or 5 days on every 100 days, &c. 2. When the rate per cent, is known, the rate per unit is easily obtained by dividing the rate per cent, by 100. Thus, 1 per cent, is eqnal 2 per cent, is equal 7 per cent, is equal 9 per cent, is equal 10 per cent, is equal 18 per cent, i^ equal 89 ptT cent, is equal 95 per cent, is equal 125 per cent, is equal 378 per cent, is equal to to to to to to to to m to r^% a . Too Too o_ I oo \S> Too 1_R_ Too Si Too OS Too or '01 p»r unit or '02 per unit, or -07 per unit or "09 per unit. or "10 per unit or -18 per unit or '39 per unit or '95 per unit or 1*25 por unit or 8'7d per unit ABM 1-S.] Pi. 2CKNTAGK. 2S3 imbers, 2, 4, 6, 8, ' ft. 1 in. long and 1? 43 acres, 1 rood, i per cent Is equal t ^ 4. How much Is dYi per cent, of $1624'60f Ans. |1421-43?6. 6. What is 12^ per cent on $904-70? Ans. |1 24-8876. 6. What is 8f per cent, on 1777-60 ? Am. 168-03^. 7. What is 2f per cent, of |7186-80? Ans. |160-6566. 8. A merchant iroporte 2740 boxes of oranges, and finds, upon re* ceiving them, that 20 per cent, of the whole quantity are de- cayed To how many boxes was his loss equivalent ? Ans. 648 boxes. 9. A gentleman purchases a farm for $7490, agreeing to pay 10 per cent, down, 17 per cent, at the end of the first year, 27 per cent, at the end of the second year, and 46 per cent, at the end ^ the third yttf. What is the amount of each payment ? Ans. $749 down. $1278-80 at the end of Ist year. $2022-80 at the end of 2nd year. $3445-40 at the end of 8rd year. 10. What is the difference between 4^ per cent, of $740 and 2} per cent, of $lft80? ; Am. $8-70. If I purchase 729 gallons of brandy and lose 11 per cent, by leakage, &c., h[||[^jnuoh have I remaining ? ■ ■ Ana. 648^^^- gallons. Add together 26 per cent, of $763*22, 16 per cent, of $847 16, and 6i per cent, of $123417. Ans, $403-486226. A person dying leaves an estate worth $17429-40 to be divided among his three sons. The eldest is to receive 43 per cent, of tiie whole, the second 37 per cent, of the whole, and the yourg- est son the remainder ; what is the share of each ? Am. The eldest receives $749464^, the second $6448-87^, and the youngest $3486*88. 1^. A merchant purchases vinegar to the amount of 68978 gallons, ' and finds, upon receiving it, that 86 per cent, had leaked away. What was his loss ? Ans. 24882 08 gallons. A brick kiln contains 29800 bricks, and it is found after burning that 17 per cent, of the entire quantity are worthless ; how many good bricks were there in the kilnf Ans. 24734. 11. 12. 13. 16. COMMISSION. 4. Commission is tbe percentage charged by agents, or commission merchants^ for their services in purchasing or selling goods, collecting billsi &c. Tbe venon who buys or sells goods for another iji called an Agent, a nUMrioa Merchant, a Factor, or • Oorr4lKpd>&d«ttt- Com- (8isrt. VXl ins. $1421'43'76. Am. 1124-3876. Am. |68-03i. Am. 1160-6566. id finds, upon re- quantity are de- ^alent ? Am. 648 boxes. ing to pay 10 per irst year, 27 per ' cent, at the end ih payment? end of Ist year, end of 2nd year, end of 3rd year, 1740 and 2} per An». $8-70. 11 per cent, by 648iVu gallons, ent. of $847 16, IS, 1403-486225. 10 to be divided 5 43 per cent, of , and the yourg- $6448-871, and 68978 gallons, >ad leaked away. t882 08 gaUons. id after burning iless ; how many Am. 24734. Arts 4-7.] bsokebaob. 235 B. To find the commission of any sum at a given rate per cent, is simply to find the percentage on that sum, and the rule employed is the same as that in Art, 3, viz : Multiply the given amount by the rate per unit expressed deci- mally. Example 1. — What \s the commission on $790*80 at 8 per cent.? Am. $790-80 x -03 = |23-724. Example 2. — A commission merchant sells goods to the amount of $7982-75 ; what is his commission at 2j per cent. ? Am. $7982-76 x -0275 = 219-626626. Exercise 96. 1. What is the commission on $1000 at 4^ per cent. ? Ans. $46. 2. What is the commission on $1678 SO at 2^ per cent. ? Ans, $37-76176. 8. What is the commission on $7531-19 at 8 J per cent. ? Am. $282-419626. 4. Find the commission on $50860 at \\ per cent. Ans. $6*8676. 6. Find the commission on $7862-50 at \\ per cent. Am. $137*61126. 6. An agent collects debts to the amount of $878.30; what is his commission at 2^ per cent.? Ans. f 21*9576. 7. A correspondent purchases teas for me to the amount of $7193-16 ; what have I to pay him for commission at 3^ per cent. ? $224-78625. 8. A commission merchant sells goods to the amount of $6734-10; what is his commission at 17 per cent.? Am. $1144797. 9. An agent sells 718 barrels of flour at ^7-13 a barrel ; what is his commission at 4^ per cent.? Ans. $217-57195. 10. A commission merchant disposes of 8243 bushels of wheat at $1-85 per bushel; what is the amount of his commission at 5f per cent. Am. $867.7871876. liii 'A, d by agents, a purchasing tn Agent, • Com- BROKERAGE. 6. Brokerage is the percentage charged by money dealers, called Brokers, for negotiating notes., mortgages^ bills of exchange, &c., or for buying or selling stocks, UBOKfiRAGE. isuoT. va Exercise 96. 1. What is ihe brokerage on |7893-87 at 2 per cent. ? Am. $157.8774. 2. What is the brokerage on $8000 at | per cent. ? Ana. $70. 8. Wiiat ia the brokerage on $8643-22 iit 1^ per cent. ? Ans. $108-04025. 4. What is the brokerage on $78963'80 at I per cent.? Anfi. $690-93325. 8. What is the brokerage on $1987*27 at ^ per cent. V Ana. $74-522626. 8. Coramission and Brokerage should both be com- puted on the amount of money collected or invested. For example: If I receive $10000 to invest, and charge 5 pe; cent., my brokerage would be $500 if I invested the whole $10000 , but if, as is usually the case, I am requested to deduct, from tht amount sent, my brokerage or commission, and invest the remainder, it would obviously be unjust to charge commission on the whole amount — i. e., on the sum invested and also on the sum I retain iot commission. Hence, in all cases, the sum actually expended is the proper basis upon which to compute the commission, bioke-.age, &c. 8. To compute coramission or brokerage when it is to be deducted in advance from a given amount, and the balance invested : — RULE. 1. Divide the given amount hy $1, plus the commission on $1, and the result will be the sum to be invested. 2. Subtract the part to be invested from the given amount^ and the remainder will be the commission or brokerage. Example. — A correspondent receives $16782, with instructions to deduct his commission at 3^ per cent., and invest the balance in sugar at 9^ cents per pound. How much sugar does he ship to his employer, and what is his commission ? OPERATION. $16782 -f- 1-035 = !5!l 6214-492 75 = sum to be invested. $16782 - $1621449275 = $567-50725 = conuiilssion. $16214-40275 -:- H cents = 170678-871 lbs. Ans. Explanation.— The commission on $1, at the rate of 81^ per cent, is $0035. Hence, for every time he receives Sl-035, he keeps $0-0.S5 lor commission, nml invests $1. It is plain, then, that if we divide the given amonnt, $16782, by $1-035, or. In other words, find how often the latter sum is containt;d in the for- mer, we shall find bow often be invests %1\ i. e., bow m^ny dollars be invests. I [StOT. Vli Ans. $lti1.8l1i. Ans. $70. It.? Ans. $108-04025. It.? Am. $690-93325. It.? Ans. $74-522625. both be coin- inveated, ind charge 5 pe; le whole § 10000 , deduct, from the jst the remainder, )n on the whole sum I retain fur ^ expended is the n, brokerage, &c. e when it is to ount, and the ission on $1, and ven amount, and th instructions to t the balance in S3 he ship to his sted. 1. ppr cpnt. Is $0035. )r commission, nml montit, $1(5782, by )ntaimid in the for- doUara he invests. ll^ftf^^ Abtb. »-18.] STOCK. 237 T/te work may he proved bij fnding the commission on the. sum in- « vested (Art. 5), and comparing it with the commission as founl 6// deducting the sum invested from the whole sum sent. If these are equal, t/ie work is correct. E.TERCISE 97. 1. An apjent receives $4000, with instructions to purchase Great Western Railway Stock. After deducting his brc eragc at 1^ per cent., how much money had he to invest, and what was hia brokerage ? A ns. Invested $3950-6 1 728. Commission $49-38271. 2. A merchant sends his agent $7500, with instructions to deduct his commission at 4^ per cent., and purchase laces with the re- minder. What is the commission, and what sum was expended in aces? Ans. Commission $322-96651. Invested $7177-03349. .S. A commission merchant receives $S470, with instructions to pur- chase the best brand of Canadian superfine flour at $6-40 per barrel. He is to receive out of this sum 5 per cent, on the amount ha invests. How many barrels of flour does he pur- chase? -4ns. 1260i^ barrels. A broker receives $11000, with instructions to invest it in Bank stock — deducting his brokerage at J per cent. What sum had he to invest ? Ans. $10904-584882. 5. If I remit to my agent $13000, instructing him to purchase broad cloth at $3.63 per yard, and he keeps 4^ per cent, on the sura invested, for commission , how much cloth does he send me, and what is his commission? Ans. 3427-0499 yards of cloth. $559 8086 commission. 4 STOCK ■ ^ 10. Stock is a term used to denote the Capital of moneyed institithions, as Banks, Railroad Companies, Gas Companies, Insurance Companies, Manufactories, &c. 11. Stock is usually divided into portions of $100 or £100 each, called shares^ and the different individuals owning thest are called shareholders or stockholders. 12. The Association of Shareholders is called a Com- pany or Corporatio:i ; and the Act of Parliament specify- ing their corporate powers, rights, and privileges is called a charter. 13. The nominal or par value of a share is its original cost of valuation. i\ > I 'i 'T : H n 238 BTOCK. [Sect. VI. I 14. The market or real value of a Hhare is the sum for which it can be sold. IB. The rise and fall in the value of Stock is reckoned at a certain per cent, on its nominal or par value. 16. When stocks sell for their original cost or valua- tion, they are said to be at 2^^'>' i when they sell for more than their original valuation, they are said to be at a pr(' mium or advance^ or above par ; when they do not bring their original cost or valuation, they are said to be at a dis- count^ or below par. NoTB. — Par Is a Latin word, and means ^qwjl or a titnts of (quaUf}/. Stock Is at par when a hundred-dollar ehjiro edls for $100; it is ahore por whHu it brlugs taore than iftlOO, and leloxo par when it will nut brinL' ati tnuch 88 $100. 17. Persons who deal in stocks are called stock-brokers or stock-jobbers. 18. To find how much stock either above or below par a given sum will purchase: — RULE. • Divide thefjiven amount by the worth of%l stock j and the resvlt will he the stock required. Example 1. — How much stock at 10 per cent, below par can be purchased for $25000? Am. $25(»00 -h 0-90 = $27111-111. Explanation. — When stocTc Is 10 per cent, bflow par, each share of $100 sells for only $90, i. e. $90 money will purchase $100 stock, ther.f.ire -tOOO Tooney will purchase $1 stock, and the given sum will purchase $1 stock as often as it (the given sum) contains $090. ExAMPLK 2. — How much stock at 15 per cent, premium may be purchased for $1000 ? Ans. $7000 -i- M5 = $6086-9565. Explanation. — When stock is 15 per cent. rt&or«j9ar, it remilres $115 money to purchase *100 stock, or $1'16 money to purchase $1 stock. Hence if we divide the whole sum to be invested by the value of $1 stuck, It is evident we must get the amount of stock produced. Example 3. — I own $16400 stock of the Bank of Montreal, and Bell out at 18 per cent, premium. What do I receive? Ans. $1 6400 x MS = $18532. Explanation. — Each $100 stock brings me $118 monev. or $1 stock brings $M8 moneys therefore $16400 stock must bring $16400x1 1^ money. Exercise 98. 1. A person has $9000 which he wishes to invest in Grand Trunk Railway shares, then selling at 17 per cent, discount, what amount of stock can he purchase ? Ans. $10843*373. 2. If I invest $8500 in Upper Canada Bank stock, which is selling 11 per cent above par, what amount of stock do I receive ? Ans. $7607'6e76. [Sect. VI. is the sum for 'k is reckoned /^alue. 3ost or valua- sell for more he at a pr(- (lo not bring to be at a dis- ntats of eqvaHfy. • ; it is (thore pur nut bring n^ much . stock-brokers I or below par id the result will slow par can be I = $27111-111. each share of $100 k, then f. re .tODO cliube $1 stuck as remium may be = $6086-9565. quires $115 money Hence if we divide ident we must get f Montreal, and y 1-13 = S18632. or $1 stock brings loney. Grand Trunk discount, what ns. $10843-373. ich is selling 1 1 receive ? '■f. ART*. 14-88.] lySUBAKOK. 289 6. If I rc.uit to my agent $17600, with inftructioni to deduct his biokcragu at 1^ per cent., and invest the remainder in (ireat Western Railroad stook, then 63lling at 7 per cent. pr(>niium, what amount of stock do I receive? . I16163-22. If I receive $2OQ0O, with instructions to deduce > oromission at 1} per cent., and invest the balance in stock, M is then sell* ing at 3 per cent, discount, what amount of st. ck do I remit to my employer? ^n«. $20263937. Mr. A. owns 200 shares in the Canada Life Assurance Company. The par value is $100 a share, the stock at a premium of 5^ per cent. ; if 1 purchase it through a broker who charges me | per cent, for the transaction, how much do my 200 shares cost me ? Ans. $21284-626. INSURANCE. 19. Insurance is a written agreement by "which an in- dividual or an iiicorp6rAted company becomes bound, in consideration of a certain sum paid in advance, to exempt the oNners of certain kinds of property, as houses, house- hold furniture, merchandise, ships, &c., from loss by fire, shipwreck, or other calamity. 20. The Written Instrument^ or contract between the parties, is called a Policy of Iniuraneei 21. The sum paid for the insurance is called the Premium, and is usually a certain per cent, on the sum for which the property is insured. 22. Houses, merchandise, furniture, cordinc; to the charactt^r and position of the tenement; vessels are ioeurvd fuir the voyage or the year. 23. To compute the premium for insurance for 1 year, or a specified time, we use the same rule as for Commis- sion or Brokerage. Example. — If I insiire ttiy house ind fmniturel for $7389^ at the- rate of If per cent, per annum, what premium miist I pay yearly? Ans. $7389 X •012ftc=$92«626. Exi»£AHATio».— 1| per cent, i. k fl*^ per $100, is equal to $0*0125 per dol- lar. The prenaium therefore will be as many times |0'013& as tbe lum UlSlU'Sd contains %l \ i. e. the premium will b« 00125 * 7889i 240 H 1 ^ 1 I^HnH^ I IN8UEANCK. Exercise 99. [6bct. VIL 1. What is the premium for insurance on $7600, at IJ per cent.? Ann. $131-25. 2. What is the premium for insurance on $8375, at f per cent.? Am. $62-8125. 3. What is the premium for insurance on $6000, at 1^ per cent. ? Ans. $112-50. 4. What is the premium for insurance on $5000 at $1*17 per cent. (i. e. per $100)? Ans. $58-50. 6. What is the premium for insurance on $6400, at $0*90 per cent. ? Ans. $57-60. 6. What is the premium for insurance on $4500, at $0*35 per cent.? Ans. $15-75. 7. What premium must I pay for insuring a cargo of flour worth $36000, from Quebec to Liverpool, at $3 per cent. ? Ans. $1080. 8. A firm, owaing four steamers running on Lake Ontario, effect an insurance with a company in Toronto to the amount of $27000 on each, paying $4-82 per cent. (i. e. 4-y^^^ per cent.) What is the total premium on the four steamers ? Ans. $5205-60. 9. What is the annual premium on an insurance for $39000, at 2^ per cent. ? Ans. $858. 10. A farmer insures his barns and their contents to the amount of $17800. What premium does he pay at ^ per cent ? Ans. $89. 11. A vessel running between Hamilton and Oswego is insured for $12350, at the rate of If per cent, per month. To what does the premium of insurance amount for 7 months, beginning with the 10th of April and ending with the 10th of November? Ans. $1236. 24. To find what sum must be insured on property so that, if destroyed, its value and the premium may both be recovered — RULE. Divide the value of the property 6^ $1, minus the premium on $1 at the given rate per cent. Example 1. — A ship-owner wishes to insure a vessel valued at $17450, so that if it be wrecked he may recover both the value , of the vessel and the premium. In order to do so, for what sum must he insure, at $4-60 per cent. ? Ans. $l7450-^•964=$l 8291 -40461. iBTS. 24-2S.] CUSTOM HOUSE BUSINESS. 241 ExPLANATioK - if I lni=uro poods to tl*(i> 'luo of |10r>, at 4-B per cent., and they are destroyed, I receive only .'-gS itO lowardd ray loss, since I paid $400 for insurance; timt is, for every $1 of my loss 1 roci-ive $0-954. Si co, tiicn, the recovery of $0"954 requires $1 to be fiisured the recovery of :*17150 will require as many dollars to be insuiod as ,'t0-9.')4 is contained times in .$17450. Proof,— $18291-40461 x •04<>=|841-4o461=tho premium, tind $18291-40461- $341-40461=$17460=value of the vessel. Example 2. — What sum must be insured on a house valued at $6000, at 3 per cent, so that in case of fire the value of both premium and property may be secured ? Ans. $6000^'97=^G185-5G7. Explanation — For every dollar I lose (taking premium into account) I receive 97 cents; that is, in order to receive 97 cents, I must insure for $1, and 111 order to receive $G000, without any loss, I must insure for $6000-!-97= $6185-567. t*'l Exercise 100. 1. For what sum must T insure a cargo valued at §17000, so that in case the whole is lost I may recover both the value of the property and the premium of 3^ per cent. ? Ans. $17616'58. For what sum must I insure on $22750 in order to cover both the premium of 6 per cent, and the value of the property insured ? Ans. $24202-127. What sum must be insured at 2J per cent, on property worth $15000 so that the owner may be secured against all loss? Am. $15345-2685. i. A steamer worth $33000 is insured at 5f per cent, for such a sum, that in case of its becoming a total wreck, the owners may re- cover both the worth of the vessel, and the premium of insu- rance. For what sum is it insured? Ans. $35013'2625. 8 4 i I, he premium on $1 4=|1 8291 -40461. CUSTOM HOUSE BUSINESS. 25. All goods coming into Canada from "Foreign coun- tries are required by law to be landed at certain places or ports called Ports of JE a try. 26. At every Port of Entry in Canada, the Government has an establishment called a Custom House, with one ot more officers attached to it, called Ci'stom-House Officers. 27. A certain charge called a I>i(i!/, fixed by Act of Parliament, is made upon nearly all goods entering Can* ada from Foreign countries. 28. It is the business of the Custom-House Officers to inspect the cargoes of all vessels entering at any of these Q I f \ 242 CUSTOM HOUBE BUSINESS. [StOT. VII. ports, to examine the invoice of goods, collect the duties, (fee, &c. 29. Besides the duties on merchandise, all vessels en- gaged in commerce are required to pay certain charges for the privilege of entering the port, &c. ; these charges are called harbor dues. 30. The duties levied by law on goods imported into Canada are of two kinds : Ist. Specific duties. 2nd. Ad Valorem duties. 31. A specific duty is a certain sum levied on the ton, cwt., lb., gallon, square yard, &c., of a particular kind of merchandise, as so much per square yard on woollens, flannels or cloths, so much per lb. on tea, so much per gallon on brandy, wine, elow part (16) 19. Wliat is the meaning of tlie term part (IG, note) 20. What are persons who deal in Stocks called ? (17) 21. When Stock is either above or below pur, how do we find how much of It a given sum will purchase? (18) 22. What, is Insurance? (19) 23. What is a Policy of Insurance? (2r>) 24. What Is meant by the Premium of Insurance ? (21) 25. For what lensth of time is property it-sitwA'^/ insured? (22) 26. How do we compute the premium of insurance ou any amount of goods, property, &c. ? (23) 27. How do we compute the amount for which we mnst insure In order to cover both the value of the property and the premium paid? (24) 23. How may the truth of this rule be proved ? (24) 29. What are Ports of Entry ? (25> 80. What is the duty ot Custom-House Offtcers ? (28) 81. What are duties ? (27) 82. Wliat are harbor dues? (29) 83. What different kinds of duties are levied on goods in Canada? (80) 84. What are specijic dutien f (31) S5. What is nn ac? valorem duty t (32) 86. What is the meaning of the term ad valorem t (82) 37. What is an invoice ? (82) 88. What is the rule for computing specific duties ? (84) 89. W* at is the rule for calculating ad valorem duties? (85) 40. What is a tax ? (86) 41. How are taxes imposed ? (9 and 88) I 'i '] ^ 'i f i't'y '■<> J :* i 246 INTEREST. [Skct. nil. SECTION VIII. INTEREST, DISCOUNT, EQUATION OF PAYMENTS, AND PARTNERSHIP. 1. Interest is the sum allowed for the use of money, and is usually reckoned at a certain rate per cent, per an- num ; that is, so many pounds for the use of £100 for one year, so many dollars for the use of 1100 for one year, &c. Note.— The term per cent, means per hundred ; per annum meaos per year. 2. Interest differs from Commission, Brokerage, &c., in that the latter are computed at a certain per cent, with- out regard to time, while interest is calculated at a certain rate per cent, for one year, and consequently for longer and shorter periods in like proportion. 3. The Principal is the sum lent. 4. The Rate per cent, is the sum paid for the use of each hundred dollars, pounds, &c. 6. The Rate per unit is the sum paid for the use of each dollar, pound, &c. 6. The Interest is the whole sum received for the use of the principal. 7. The Amount is the sum obtained by adding together the principal and the interest. Thus, if I lend $200 for a year, on the agreement that I am to receive inter- est at the rate of 7 por cent, (.jyer annum, understood), at the end of the year 1 receive back the $200, and in addition $14 for interest. Here, 8200-00 la the principal. 7 00 is the rate per cent. 007 is the rate per unit. 1400 is the interest. 314'00 is the anioun*-=principal +Intere8t. 8. Interest is either Simple or Compound. 9. Money is lent at Simple Interest when the Interest is not added to the principal so as so bear interest. Thus, If $100 be lent at simple interest at 5 per cent, the principal re>' mains unchanged, beiug always ^i^lOO, and the interent for each successive year Ant9 1-11.1 SIMPLE INTEEE81'. U7 10. Money is lent at Compound Interest when the in- terest, as it fal^s due from time to time, is added to the principal ; the bjm thus obtained constituting a new prin- cipal for the ensuing year, half year, quarter, &c., as the case may be. Thus, ff $100 be lent at 5 per cent, per annum compound interest, the prin- cipal chaiifres at tho end of each year; being $100 for the first ye.ir, $105 (L e. former principal + its interest) for the second, $110 25 for the third, «fec. Tho interest i.s consequently $5 for the, first year. $5-2J) for the second, |5 'SI 26 for the third, &.c. SIMPLE INTEREST. 11. Questions in Interest are dependent on Proportion, and may all readily be solved by one or more statements in the Rule of Three ; but in order to deduce special rules, we shall represent the different quantities by their initial letters, and thus obtain a series of algebraic formulae, which, translated, become the common arithmetical rules lor interest. It is to be presumed that the pupil has made sufficient progress in Algebra before he arrives at this point, to readily understand what follows. The opera- tions involved are of the simplest liind, and may without difficulty be compre- hended, even by those who] 'y ignorant in Algebra, The only part, however, absolutely necessary for working any problem in interest, is the interpretation of tho formula, i e. the arithmetical rule,aniX this we iiave always appended. A glance at the forraulsB and tho coriesponding rules will show how much less labor is necessary to remember the former than tho latter ; ai d indeed the pu- j>il should be required to deduce from time to time any formula^ he may fii.d It necessary to use. Note. — When two or more letters are written together thus, prt. the meaning is that the values of these letters are to be multiplied together. Thus, Prt means that the value of P is to be multiplied by the value of r, and that by the value of t. A-P When letters are written in the form of a fraction, thus the meaning is the same as in common arithmetical fractions ; i, e., that the part constituting the numerator is to be divided by the part constituting the denominator. A-P Thus, means that the value of P is to be subtracted from ' Pr the value of A^ and this difference is to be divided by the value of P multiplied by the value of r. J ff/tr^ I 248 8iMJ>LE INTERESt. [Skct Vllf .»(j?i \/:(i J! 12, time (i. I=Prt{I.) P = --{in r — t = Let V— Principal, l=Interest, A=Amount, r=Tate per unit, and t= e., number of years). Then because r=intere8t of $1 for 1 year, and f = number of yearti, r<=intere8t of $1 for the piven time, and /'7Y=lntere8t of given principal for piven time and at given rate. Therefore 7= P;^ and divi- ding each of these equals, Ist by it, 2nd by Pt, and 8rd by Fr, we get formulas (11.) (III.) and (IV.) in the margin. Again, because r<=liiterest of $1 at given rate and for given time, l + r^=the amonnt of ffil at given rate and time, and P times 1 + >Y, that is, P {\ + rt)=: amount of given principal at the given rate and time. There fore A = 2' (l+rt), which is formula (V.)in the maigin, and dividing each of these equals by 1 + 7't, we set ftirmula (VI.) in the margip. Taking (V.) (///.) Pt I_ Fr A = F {l+rt) (V.) A {IV.) r = t = 1+rt A-P Ft A-P Pr M— 1 {VI.) ( vn.) {VIII) {IX.) n — 1 r = -(X) n=tr+\ {XL) and iictufllly multiplying as indicated, the part with- in the brackets by i", we get A=P+Prt ; and sub- tracting P from each of these, we pet A—P=Prt. Dividing these equals, 1.st by Pt and 2nd by Pr, we get lorniulits (VII.) and (VIII.) in the margin. Lastly, if we are required to find in what time any Bum of money will amount to any given number of times itself at a given rate per cent, or, in other woids, in what time any principal will amount to n times that principal where n simply stands for the required nvmher of times, we have In formula (VIII ) m the margin, A—P nP—P t=— r; — = — IT — , because the amount is to be nP; Pr Pr and dividing both numerator and denominator of this fraction by P, we get formula (IX ) in the mar- pin, multiplying (IX ) by r we get tr=n—l ; and divi- ding these equals by t, we get formula (X.) ; and, again, adding 1 to each of the&e sanift eguula, we get formula (XI ) APPLICATIONS. 13. When the principal, rate per cent., anci time arc given, to find the interest — Rule / = Prt (i.) Interpretation. — TJie interest is found by multiplying the princi- pal by the rate per unit, and the resulting product by the time. ExAMPLU. — What is the interest on |342-20 for 7 years at 8 per cent. ? OPERATION. Here P = $343-20, r = -08, and t = l. Then I- Prt = |342-20 x -08 x 7 = $191 682. And. 14. When the interest, rate per cent., and time are given to find the principal — Rule. P= -(ii.) Interpretation. — The principal is found by dividing the interest by the product of the rate per U7iit and the tim/> -.«>>K • V A«w.ia-iti dlWPLE INTEBBdT. 249 per unit, and t= aunt is to be nP ; md time ar6 J years at 8 per ,nd time are ling the interest ExAWPLK.— What principal will give $207-60 interest m t^ years at 4| per cent. ? opaRATioir. Here /= $207-50, t=6-6, and r=-0475. / _$207jO 1207-50 Then P=^ - g.g ^ _^^^- .gQg^g =1672064. Aru. IB. When the interest, principal, and time are given, tojind the rate per cent. — Rule. r = -~ (iii.) Pt ^ ' Interpretation. — The rate per unit is found by dividing the in- terest by the product of the principal and time, arm the rate per cent, is found from the rate per unit by multiplying the latter by 100. £xAMPLE.~At what rpte per cent, will $729 18 give $10911 in- tefjst in 9 years ? OPFKATION. pore /'=$729 18, /=f 109-11, and ^=9. ^, / 10911 109-11 Then r = = 0-01662 = rate per unit Pt 729-18 X 9 6662 62 Therefore the rate per cent = 001 662 x 100 = 1-662 = 1! nearly. Ans. 16. When the interest, principal, and rate per cent, are given, to find the time — Rule. ' = Pr ("•) Interpretation. — The time is found by dividing the interest by the product of tlie principal and rate per unit. Example. — In what time will $850 give $89'76 interest, at 18 per cent. ? Here P = $850, /= $89 75, and r = -la _,. , J 89.75 89-76 897-6 = 0-8i2217 years =■ 9 moalhs 110-6 1106 22 days. 17. When the principal, rate per cent., and time are given, to find the amount--' Rule. A — F {\-\-rt){\,) Interpretation. — The amount is found by multiplying the prin- cipal by the amount of $1 for the given rate and time. Example. — To what sum will $789 80 amount in 11 years, at 8 per cent. ? OPBRATTON. Here P = $789-80, r = -03. and <= 11. Then A=:P{\-¥rt)= $78980 x 1-38 = 1060-484. Ana. NoTB.— (1 + rt) in this question = 1+ 8x11 = l.t-'38 = l-88L r 250 SIMPLE INTEREST. tflKOT. Vlli :is . I^H '■f ^^BM ',-■ j ^^^^■^H V ,'■' ^^^fflHI' 18. When the amount, rate per cent., and time are given^ toji: the principal — Rule. I^ (vi.) 1+rt ^ ' Interpretation. — 7%e principal is found by dividing the given amount by the amotmt of$l for the given time at the given rate. Example. — What principal put to interest at 74 percent, will amount to $2000 in 8 jears V DPI RATION. Here A - |2000, r = -075 and t = 8. mu r, '^ 2000 20000 ^,„,„ , Then F = ^-^--^ = -^:^ = 16-= ♦^^^ ^'^' 19. When the amount, principal, and time are given, to find the rate per cent.--' (vu.) Rule. r = - Pt Interpretation. — The rate per unit is found by subtracting the principal from the amount^ and dividing the difference by the princi- pal multiplied by the tim^. 21ie rate per cent, is found by multiply- ing the rate per unit by 100. Example.— -At what rate per cent, will $780 amount to $2783-8^1 in 23 years ? OPEBATION. Hero A =-• $2788-80, P = $780 and < = 28. -,. A-P $27S8-80-$780 $205880 ,.„„ Thenr= -^ = -$730738- = "$16790 = -1223 = rate per unit Honce rate per cent. = 12*28 = 12i nearly. 20. When the amount, principal, and rate per cent. are given, to find the time-— jl P Rdl^. t ■= —— — (viii ) Fr ^ ' Interpretation. — The time i& found by subtracting the principal from the amount, and dividing the difference by the principal multi- plied by the rate per unit. Example.— In what time will $666-33 amount to $983-73 at 12 per cent. ? operation. Here A = $988-78, P = $666 8;i and r = 12. Th - —"^ - ^88-78-666-38 _ 817-40 ®" ~ Pr ~ "666-33 X -12 ~ 79 9596 8174000 799696 8 years 11 months 19 days. Ana = 3 9695 years = Arts. 18-«8.] SIMPLE INTEREST. 251 le are given, mt to $2783-80 = 3 9696 years = 21. To find the time in which any sum will amount to any given number of times itself at a given rate per cent. — 1 RULK. t = n Interpretation. — To find the time in which a (jiven num will amount to n times itself at a given rate per cent.^ subtract 1 from n, and divide the remainder by the rate per unit. Example 1. — In what time will any sum of money amount to eleven times itself at 8 per cent. ? OPERATION Here n = 1 1 and r = 08. n-Ji _ 11-1 • ~ -03 Then t = 10 1000 ,„^ 08 = T = ^^^ ^''"- ^"*- Example 2. — In what time will §67-83 quadruple itself at 4t per cent. ? operation. Here n = 4, elnce the money is to qvatfrvple lts(lf, and r = '0476. _,^ ^ tt-l 4-1 8 80000 „„,„ Then t = = - -^^ = -- - = — _- = 68167 years. Atis. r '0476 -0475 476 '' 22. To find the rate per cent, at which any sum will amount to a given number of times itself in a given time— n- 1 r = -^ - (X.) Rule. Interpretat'on. — 77*« rate per unit is found by subtracting 1 from n, the number of times itself to which the given principal is to amount, and dividing the remainder by the given number of years. Example. — At what rate per cent, will a given sum amount to 26 times itself in 72 years? Here » = 25, < = 72. » - 1 25 Then r = t 72 opekation. - 1 24 = — = i = -884 = rate per unit. 72 Hence rate per cent. = 83^. An9, 23. To find to how many times itself & given sum will amount in a given time at a given rate per cent. — Rule, n = ir -^ I. (xi.) Interpretation. — The number of times, or n, is found by mul- tiplying the time by the rate per unit, and adding 1 to the product. Example. — To how many times itself will /owr cents amount in 20 years at 17 per cent. ? '.I ■-■!!>; i---. ( U2 UtUPtK INT£lt£flT. [8«oT. VlU. OPIRATIOK. Here t = 20 and r s= IT. Tbi-u n = «r > 1 = 90 K -ir •»■ 1 s= 8-4 4- 1 = 4 4 c 41 tlmea itaelf. An*. Exercise 104. 1. What ia the interest on |l72319 for 732 years, at 67 per cent. .♦ Ans. $!854-6Rl 303ft. 2 To what sura will $867"19 anaount in 6^ veara, at 6A per cent. V Ans. |1219-35277f'. 8. To how many tiinea itself will £2 lOs. 9^(1. amount in 11 years, at 72^ per cent. V Am. 8*975, or nearly 9 tinic."> 4. In what time will ^064-82 give $234*j6 interest, at 7 per cent. ? Ans. 5*12112, or 6 years 1 m. 13 days. 5. At what rate per cent, will ij^TuO amount to $1200 in 5 years? Ans. 14^ per cent. 0. In what time will any sum of money quadruple itself, at 23 per cent.? Ans. 13 years 16 duy.s. 7. Find the time in which $270 will give $87 interest, at 7 per cent. Ans. 4 years 7^^,- montii.s. 8. To what sum will $680 amount in 11^ years, at 11 per cent. ? Am. $1540*20. 9. What principal will amount to $2000 in 20 years, at 8 per cent. ? Am. $769-28 ,V 10. At what rate per cent, will any sum of money amount to 21 tiniin itself in 24 years? Ans. ^83^ per cent. 11. In what time will a given sum of money amount to 23 times itself, at 16 per cent. ? Ans. 187f years 12. Find the interest on $679*18 at 7| per cent., for 11*73 years. Ans. $617*4255. 18. At what rate per cent, will $950 amount to $1768*42 in 10 years? Ans. 8'502 per cent., or rather over 8^ per cent, 14. In what time will $666 amount to $1347*50, at 6 per cent, ? Ans. 17054 4- years, or 17 years 19 days. 16. In what time will $273 give $100 interest, at 9 per cent. ? Ans. 4 yeai's 25 days, 16. At what rate per cent, will $476*30 amount to $500 in 2 year.'-'? Ans. 2^^ per cent, 17. At what rate per cent, will $749*49 give $257 interest in 7 years? Ans. 4*898 per cent, 18. What principal will amount to $1111*11 in 11 years, at 11 per| cent ? Ans. $502*7C4T 19. Find the interest on £167*47, at 11 per cent, for 9 years. Am. £165 16s. lOf^^A SPECIAL RULES. 24. The Interest of $100 at 6 per cent, for one year, is $6; hrnce the Id- | terest on $1 at 6 per cent., for one year, is $0*06, aqd for iioo months it is ^o( $0-06; i. e., Iceni. t8«0T. VUI. I ABm24,2a.] SIMPLE INTEREST. 253 Itues itoelf. AM. Hence, to find the interest of $1, at per cent, per an- num for any number of months, we deduce the following — RULE. Divide the number of months by 2, and call the quotient cent$. Example 1. — What is the interest of |1 at 6 per cent, for 1 years and 9 months? OPRRATION. 7 years and 9 niontha=:9'S months, and 9d-f-2=46) o«nts=|0-46-\ Antt. Example 2. — Find the interest on |72-93 for 7 years and 8 months at 6 per cent. OP B RATIO w. 7 years 8 mo, =92 months, half of 92=46 oents= interest of |1 f«r gives rate and time. Then vO-46 x 72 98 =$88 5478. An«. Exercise 106. 1. Find the interest of $1 for 11 months at 6 per cent. An«. 6^ cent«. 2. Find the interest on $1 for 16 months at 6 per cent. Ans. $0-08, or 8 cents. .3. Find the interest on $1 for 9 years 8 months at 6 per cent. Ans. $0-58. 4. What is the interest on |1 for 16 years 3 months at 6 per cent. ? Ans. |-97i. 6. What is the interest on $1 for 1 1 years 7 months at 6 per cent. ? Ans. #0-695. 6. What is the interest on |1 for 12 years 6 months at 6 per cent. ? Ans. $0-746. 7. Find the interest on $279 40 for 3 years 2 months at 6 per cent. Ans. $53-086. 8. Find the interest on $189-70 for 6 years 7 months at 6 per cent. Ans. $74-9315. 9. Find the interest on $1463 for 3 years 11 months at 6 per cenh Ans. $343-805. 10. Find the interest on $28967'60 for 11 years 1 month at 6 per cent. Ans. $10263-3876. 26. Since In computing interpst the month Is taken as 80 days, two months will contain 60 days, and, by Art. 24, the interest on $1 at 6 per cent, for 8 months or 60 dai/'i is one cent, the interest on $1 at 6 per cent p^r annum, for 6 cfc/y«, will therefore be t^ of one cent ; i. e. one mill or ve^s of $1. Hence, to find the interest on |1 at 6 per cent, per an- num for days, we have the following — t; I' 3 w 3. .!■ 1 m 3!:- "i* » ''. 254 SIMPLE INTEEEST. [Sect VIIL Aivi3. 26, 2 ! i> RULE.* Call one-sixth of the number of days mills or t/tousandihs of a dollar. Example. — What is the interest on $1 at 6 per cent, for 16 days? OPBKATIOIC. 164-6= 2| uiill8=$0-002e. Ans. Exercise 106. 1. What is the interest on $1 for 2 days at 6 per cent, f Ans. $0 0003. 2. What is tlie interest on $1 for 7 days at 6 per cent. ? Ans. 10-001^. 8. What is the interest on $1 for 11 days at 6 per cent. ? Ans. $0-001 g. 4. What is the interest on $1 for 27 days at 6 per cent. ? Ans. $0-004^ 6. What is the interest on $1 for 47 days at 6 per cent. ? Ans. $0-007i 6. Required the interest on $1 for 8 months 12 days at 6 per cent. Ans. $0-042. 7. Required the interest on $1 for 66 days at 6 per cent. Ans. $0-011. 8. Required the interest on $1 for 2 years 2 months 19 days at 6 per cent. Ans. $0-183^. 9. Find the interest on $1 for 7 years 8 months 9 days at 6 per cent. Ans. $0-40 H. 10. What is the interest on $1 for 17 years 11 months 23 days at 6 percent.? Ans. H'Q1%1. 11. Required the interest on $1 for 12 years 7 months 17 days at 6 per cent. Ans. 0-757f. 26. To find the interest on any sum of money at 6 per cent, per annum for any time — RULE. Find th£ interest on $1 for the given time^ by Arts. 24 and 25, and multiply this by the given principal. Example. — What is the interest on $763-20 at 6 per cent for 6 years 7 months and 26 days ? • This is the method in common use for computtDs; interest for days; but, since it considers the year as containing only 360 days instead of 865, the result is too larjro by j';,, or ^s of itself Flence, when perfect accaracy is desired, tlif interest for the days when obtained by tbo rule most be diminished by ^^ part vf itself. Tbcr^'fo-- per < 10. Find tl I 11. To whi I per c li i2. To whi 6 pei 27. 1 per cent. Find th per cent, by Then a<. "f itself as num. Th^ am A together. 1 * In ord< M ^P retainv>d i make the iot ;1, AivT3. 26, 27.] bIMPLE INTEREST. 265 OFERATIOW. Interest on $1 for 6 years 7 months = $0895 Interest on $1 for 26 days = 4^ Tbci'i'fo'-e interest on |1 for 6 yrs. 7 months 26 days = $0 809J Then* $0*399^ x 768-20 =$804-7712. Ana. Exercise lOY. 1. Find the interest on $9n'30 for 1 months 17 days at 6 per cent. Ans. $:}4-704516. 2. Find the interest on $842-50 for 3 months 13 days at 6 per cent. Ans. $14-462916. 3. Required the interest on $573'83 at 6 per cent, for 2 years 11 months 10 days. Ans. $101-3766. 4. Required the interest on $642*30 at 6 per cent, for 6 years 9 months 19 days. Ans. $262-16545. 5. Required the interest on $1427'87^ at 6 per cent, for 6 years 6 months 7 days. Ans. 465*7252. 6. Find the interest on $709*63 for 4 years 7 months 16 days at 6 per cent. Ans. $197-040596. 7. Find the amount of $2463*20 at 6 per cent, for 7 years 7 months 22 days. Ans. $3592-9877. 8. What is tlie interest on $999-99 at 6 per cent, for 9 years 9 months 9 days? . ^ns. $586-494135. 9. What is the interest on $68*70 for 3 years 4 months 27 days at 6 per cent. ? Ans. $14-04915. 10. Find the interest on $742-63 at 6 per cent, for 3 years 28 days. Ans. $137-139. 11. To what sum will $200 amount in 7 years 4 months 11 days at 6 percent.? Ans. 288-366. i2. To what sum will $743-63 amount in 9 years 3 months 9 days at 6 per cent. ? Ans. $1157-460095. >' ■ ;i'i 27. To find the interest on any sum at any other rate per cent, for any given time — RULE. Find the interest on the given principal for the given time at 6 per cent, hy Art, 26. Then add to or subtract from this interest such a fractional part of itself as the given rate exceeds or falls short of 6 per cent, per a7i- iium. Ih^ amount is obtained by adding the interest and the principal together. * In order to obtain the cnrrect answer, this fraction when it occurs muit be retainv>(l in the form of a vulgar fraction; and in that case it is bett«r to make the intevest uf |1 for the given time the multiplier. 'WT 256 PARTIAL PAYMENTS. [Sect. VIII, „ % ii Example. — What is the interest on |450 for 8 years 6 months 11 days at 8 per cent. ? OPERATION. Interest on i*l at 6 per cent, for given time=|0-211l. Interest on $450 at 6 per cent, for given tirao=|0-2ll| x 450=$95-825. Hence interest on W50 at 8 per cent. $95-326 = $127-10. Ans. fior given time= $96-826 + (Wi« third of Note.— Siice 8-6 + 2=6+^ <>f 6 we find the interest at 6 per cent., and increase it by one third, of itself for the interest at 8 per cent. So for interest at ' f.r cent, we ehould find the interest at 6 per cent, and increase it by on e-hfiij . it' 'If. for 7 per cent, increase the interest at 6 per ci-nt. by one'sixth; at 14 per < nt., double the Interest at 6 per cent, and in- crease it by i of the interest at 6 per cent ; at 5 per cent., find the interest at 6 per cent and deduct (nie-nixth; at 4^ per cent, And the interest at 6 per cent, and deduct onc'/ourth, &c., dsc Exercise 108. 1. Required the interest on $1234*66 for 8 years 9 mouths 10 days at 1 per cent. Ans. ITSS'SeSS. 2. Required the interest on $9876'54 for 2 years 1 month 11 days at 3 per cent. Ans, $626-33'7245. 8. Required the interest on f'TlS'SO for 8 years 7 months 10 days at 8 per cent. Ans. $206 -6422. 4. To what sum will $555 'SS amount in 2 years 4 months 8 days at 12 percent.? Ans. $712-58546. 6. To what sum will $7766-55 amount in 100 days at 5 per cent. ? Ans. $7874-41875. 6. To what sum will $500 amount in 8 years 8 months 8 days at 16 percent.? -^n.'r. $1195-111. 7. What is the interest on $576 for 8 years 6 months 7 days at 5 per cent. ? Ans. $98-96. 8. What is the interest on $2478-91 for 2 years 6 months 11 days at 4i per cent.? ^ws. $282 285- Wlmt is the interest on $780 from May 9, to December 11, at 6 percent.? Ans. $28-08. What is the interest on a note of $1830-68 from August 16, 1851, to June 19, 1852, at 7 per cent.? Ans. $109-63489. Whnt is the amount of a note of $6200 from Sept. 8, 1858, to January 9, 1859, at 6 per cent. ? Ans. $6332-266. 9. 10 11. PARTIAL PAYMENTS. 28. To compute the interest, on notes or bonds, when partial payments have been made — RULE. If the interest he paid by days : Multiply the sutn by the number of days which have elapned before any payment was maae. Subtract the first payment, and multiply Abt. 28.1 the remt and sect this rem and thir Add for one ( Ifiki or month EXA^ following For VI the sum c at 6 per c Thefo II Fr AEt. 28.1 PARTIAL PAYMENTS. 257 t., and increase the remainder hy th^. number of days wMoh pas^se ' between the fmt and second paymetUi. Subtract the second pay, • , and multiply this remainder hy the number of days which passet Jween tfie second and third payments. Subtract the third payment ^ d'c Add all the products together, and fnd the interest of their sum for one day. If ike interwtt is to be paid by the week or month, substitute weeks 0T months for days, in the above rule. ExAMPLC. — How miK^ principal and interest have I to pay on the following note on the 10th November, 1859 ? Toronto, \9>th October, IS.'iS. For value received, I promise to pay Timothy Thomas, or order, the sum of six Imndred and twenty dollars, on demand, with interest at 6 per cent. Thomas Williams. The following endorsements were made on this note : — * 1868. — November 25th, there was endorsed $ 47-50 " December 28th, " *' " 108-93 1859.~February IJth, " " " 216-18 June 6th, ' " " " 60-10 September 2nd, " " " 133-26 u OPERATION. From 18th October to 25th November there are 88 days. " 26th Nov. to 28th December " 88 " " 28f,h Dec. to 11th February " 46 " " 11th February to 6th J ane " 115 " " 6th June to 2nd September " 88 " " 2nd September to 10th Nov. " 69 " Whole sum $620-00 for 88 days = $23560-00 for 1 day. First endora What ready money ought to be paid for a debt of $7111-11, to be due 7 years heuce, allowing G per cent, compound interest ? Ans. $4729-295. 6. What principal, put to interest for 6 years, would amount to £208 Os. 44d. at 5 per cent, per annum? Ans. £200. II I I ■f. DISCOUNT. 34. Disoount is an allowance made for payment of a debt before it is due. 35. The present worth of a debt payable at some future time, vvithout interest, is that sum of money which, bein,i( put out at legal interest, will amount to the debt by the time it becomes due. Thus, if I owe a man $100 and give him a note for that amount, payable one year hence without interest, i\\Q present value of my nolo is less than $100, since $100 being put out at interest lor 1 year at 6 per cent, will amount to $106. 36. Froin Art. 18 it ia evidont that to find tlie present worth of a note, piiyablo at some futin-e timt^, without interest, is !nt., payable 9 months hence. Ans. $79-59036. 8. What is the present worth of a note for $4360, payable one year 5 months hence, at 6 per cent. ? Ans. $4018-4331'7. 9. What is the present worth of a note for $1647. due 11 months hence, at 6 per cent. ? Ans. $1661 •13744, U '% 264 BANK DISCOUNT. [SBOi . VIU. n, 10. Required the present worth of a note for $2000 due 3 years 7 months hence, at 6 per cent. Ans. $lfi46 09O53, 11. What is the discount on a note for $2070-90, payable 1 year 7 months hence, at 5 per cent. V Ans. $15rOl!). 12. What is the present worth of a note of $970*63, payable in 11 mouths, at 8 per cent. ? A)is. $904-3 IH. Note. — When the pftytnonts are fo bo mflde at different timee, find the present value of the suina seoitratcly ; their sum will be the present vulne of the note, and, as before, this subtracted from the whole amount will give the dis« count. 18. What is the discount on $3024, the one half payable in 6 and th« remainder in 12 mouths, 7 per cent, per annum being allowed ? Ans. $1500464, 14. A merchant owes $440, payable in 20 months, and $896, 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 annum? ulna. $1200. BANK DISCOUNT. 37. Bank Discount is a charge made by a hank for the payment of money on a note before the note is due, and differs materially from discount as commonly calculated. 38. Banks consider the discount to be the same as the interest on the whole amount of the note, from the time it is discounted until the time it becomes due. Bank Dis- count is therefore greater than the true discount by the in- terest on the discount. 39. The three days of grace, which by mercantile usage, are allowed to elapse after a note falls due, before it is pay- able, are always included by banks in the time for which they calculate the discount. 40. Two kinds of notes are discounted at banks: Ist. Business notes or hmdnens paper. These are notes actually grl^en by one individual to another for property sold or value received. 2nd. Accommodation notos, called also accommodation paper. These are notes made for the purpose of borrowing money from the banks. 41. To find the bank discount on a note : — RULE. Add 3 day8 to the time which the note has to run before it becomes due, and calculate the interest for this time at the given rate per cent Example. — What is the bank discount on a note of $700, payabl<> in 69 days, allowing discount at 6 per cent. ? -Ml .. AET8. 8T-42.] BANK DISCOUNT. 265 OPERATION. Htife the time the note hna to run is 72 days = 2 months 12 dayt. Inti^rcst of :«1 at 6 per cent, for 2 months 12 dnya, in tO-012. Int(>rc»t of v'TOU at 6 pur cout. fur 2 munths 12 days=|0 012 x 700=|8 40. Ant. Exercise* 114. 1. What is the bank discount on a note for $986, having 2 years and 3 months to run, allowing discount at 7 per cenl. ? Ans. S166-8701. 2. If I have a note for $640, payable in 100 days, and get it dis- counted at the rate of 8 per cent, per annum, what discount am I charged? Ana. $14-6488. 3. I sell a horse and carriage for $563-80, and receive a note for that sum, payable, without interest, 91 days hence. Now if I get thia discounted at the rate of 6 per cent, per annum, what sum do I receive ? Ans. $554-967. :H: 42. It is often necessary to make a note of which the present value shall be a certain sum. Thus, suppose I require to receive from I'le bank $1000, and wish fo give my note, payable in 7 months, at 6 per cent., what amount must I put on the face of the note ? Now the Interest on $1 at 6 per cent, for 7 months and 8 days (i. e. days of c;raco) is 100355, and this will be the bank discount on |1 for 7 months at o per cpnt. To get the present value of |1, we subtract 100855 from |1. which gives us I0 964.'). Hence for every $0-9646 I receive, I must put $1 on the face of the note ; and therefore to receive $1000, 1 must put i. e. f 1036'806 on the face of the note. Paoor.— Face of note $1036-806 Bank discount on $1086-806 at 6 per cent per an. for 7 mon.. 86-806 Present value $100000 » ') Flence to find the face of a note, due at some future time and discounted at a given rate per cent, per annum, that shall have a known present value, we have the I'oUow- iiig- * These examples are worked by the rule priven in Arts. 26 and 27. If the absolutely correct answer is required, it must be obtained by deducting from thcic results ', of the interest mr the day>i used, as before explained, In ej(- amjjle 2, it wljl be ubserved, thia makes a (iiflFerence of 20 cent&, li i h t ■ n i 266 EQUATION OF TAYMENTS. [9«0T. VIII. n U L E . Mnd the present value of $1 for the sntne time {addhifj the thrre dni/H of (/race) and at thu name rate. ; diridc the re(/uired j)r<.sent value of the note bi/ this, and the tjuotieiit ivill he the face of the note. ExAv.rLK. — For what suin iriust a note be driiwn iit 8 luoiiths 18 days, so tliat discounted iinmi'uiately nt G per ceut. it sli^l. produce OPEItATION, Intorost on f 1 for fi months 21 (lny^^ ftt (J per cent. = $00485, and this tnkcn turn |1 t,'ivL'3 us $O!)")05 -prcs^'nt woriU of$l. Exercise* 115. 1. What sum must I put on the face of a note payable in 90 (hiya .^o that I may obtain $3750 when discounted at a bank at 7 p? Ans. $717-2471. EQUATION OF PAYMENTS. 43. Equation of payruents is the process of finding Ihe equaled or average time wlien two or more payments, cluu at (litferent times, muv be made at once without loss tu either party. 44. The averac^e time for the payment of several sums due at different limes is called tl.'.e mean time or equated time. 45. To find the equated time for any number of pay- ments : — RULE.f Mrxt muWph/ each debt by ihe time before it becomes due ; tli< :i d'wde the 6'«.'/? of thr frroducfs thus ohfairud by the sian of the pay- metis, and ihe quotieut zti'l be ihe equated time required. * Work l>y Arts. 2i] and 27. t Tliis r:i!o i.^ ll!l^< (i upon tlio snpposition thnt what is pt'inod liy kcr • in*? certain imyin^nls utter thoy become due Is tqiial lo wluit is U/St by payinT ether p-iynu'iits l>i foro fVoy be come dno. Tlr.s, 1 owtvor, is not (jxuotlv true ; lor the galu ia the interest, while the logo is equul only to tho AiiTU. 13 4.1 Note.— \1 (huH'l for s| one I time but i| with 1 ExAMil be paid i| nionth.s. what tiniel operI HlKKt X 11- 700O) month. II' In' cqn.'il ti Huni o> iho 'i'hat is, EXAMT £50, paya lime may EXAMl diately, $i paid idtog illsconnt. V is so triflin With lows :— Lei And Mnee Tlie iuterc Also inter< Uunce i>r Ar.d X jiuuiber ol I.'- A UTS. ^-46. J EQUATION OF PAYMENTS. 207 i thla taken fiom Note. — When thorc are both days and months, thoy must all be re- (hic'fd to flic HiiMu' tmit ; i. e., the payments mnst all be reckoned lor so many tiiiys, or so mniiy niuiitlis or parts ot a month. If one of the payments is due on the day from whieh the etpiated thiie is reckoned, the correspondin;:^ product will be notljing; but in lindint; the Knm of the debts, this payment must be added with the others. (See Example 3 below.) ExAMiM.E 1. — A mcrehant purehasos a vessel for !?7000, $2000 to be j)aid in 3 months, Jit^iiOOO in 5 months, and the balance in 11 months. Now if he wishes to make the whole in one payment for what time must liitj note bo drawn V Explanation.— TJip Intcrost of »2a00 for tlircL" iiiKiiths is oqnul to tint Intort'st of .TrOljiM) for one nioiiMi. Hiiiiilmly, (he iiitcri'st of tlio si'coml i)iiynient i.s *'(]ual to the iiiturcst of >-10ltil(l lur one nioiith, uiid till' iiitcrot of tlK< tliini i)!iyiiicnt is cqiiiil to tho iiitorcst of >f;l:{0(K) for one month. Hence, tho int(" est of the sevoial jiiiyiiniits, at the trivoii times, will ill- eqiiiil to ih.'it of jfl'.Hiiii) for on.' iiioiilii ; and if we divide tins .t49000 by the MUii ot Iho paymLius, $T00i), wo oljtain 7 luoutlhs for the eqiiutcd time. 'I'hat is, 17000 . $19000 : : 1 month : yl/i«.=*.-^^|'-^=7 nioutha. Example 2.— A person owes another £20, payable in 6 months; £.■)(), payable in 8 months ; and £'.)<), payal)le in 12 months. At what lime may all be paid to^'cther, without loss or gain to either partv V OPKUATinN. fjfMlOx ;{-$ Cllitdxl •2lHhl / U~ lit i^ir /: 1 ,siil)<)x 11= 33000x1 7000) $10000(7 months. Ans. OPERATION. 20 X 6= 120 ftOx 8= 400 90 X 12^=1080 160 160)ir.00(]0 months. Jn«. 1000 Example 3. — A debt of $450 is to be paid thus: $100 imme- diately, $3i)0 in four, and the rest in 6 months. When should it be paid altogether V discount, which (Art. 83) is always less than the interest : but the discrepancy is so triflinir as net to make any inuterial difference in tlie result. With tins cxeeiition, the rule is true, and may he demoi'Strated as fol- lows; — Let ^) = lirst payment., and ^:=tho time " bct'oro it becomes due; pizrother payment, and r=:tlie time bifoiv it becomes due; »!=: equated time, and /'=:tlie rale of interest per unit. And since 03, the equated time, lies between t and t' the time between t and x \f,z=zx—t, and tliat between t and x is = f — tr, Tlie interest of /> for the time x—t is (frt)m Art. \Z) pr {x—t). Al.so interest of />' for the time t—x is i)'r {t z-ic). Hence pr (ps—t = pr (t'—jr). Ar.d 00 ~ ^ — , which is the rule, and may bo similarb oved for any liuruber of payments, 1 1I i.ir 1 I. i ! ■ i -ll - — ■'■ Mi. '■"■II- ■.r' 2G8 PAETNEESHIP OE FELLOWSHIP. OPERATJON. $100x0= 800x4=1200 &0x6= 300 ESkci. VIIL 450 460)1500(81 months. Ang. 1860 150 460 (=' Exercise 116. 1. A owes B $600, of which $200 is payable in 3 months, $150 in 4 months, and the rest in 6 aionths ; but it is agreed that the whole sum shall be paid at one payment. When should the paymom be made? Ana. In 4| months. 2. A debt is to be discharged in the following manner : ^ at present, and i 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, $40 in 5, and the rest in 7 mouths. When may the whole be paid togeth er? A71S. In 4^ niontli? 4. I owe $1000 to be paid down, $1500 in one month, $600 in ;i months, $700 in 5 months, and $1400 in 7 months. For what time must my note be drawn so that the whole may be paid iu one payment? Ans. 3^^ nioiith.e, 6. Bought of Messrs. Hendrie & Robarts, goods to the following amounts, on the credit of six months : 15th of January, a bill of $3750. 10th of February, a bill of 8000. 6th of March, a bill of 2400. 8th of June, a bill of 2250. I wish on Ist of July to give my note for the amount ; at what time must it be made payable ? A7is. 3 Ist August. PAETNERSHIP OR FELLOWSHIP. 46. Partnership or Fellowship is the joining together of two or more persons for the transaction of business, agreeing to share the profits and losses in proportion to the amount of md Double rartncr.ship, or Partnorship TVith Time. Frtr oxamplo ; suppose A puts in $2(10 for 8 years, and B $.'5W» for 4 yonr?, find thny make 11 certain giUn or loss. This would give a case of CompoutKl Purtncivsljip. In yuoli cftsos it is plain that each man's share of the profit depends upon two circnnjutances : 1st. Tlie amount of his stock ; and 2nd. The period for which it in continued in the business. Also that when the times are equal, the shares ( f the gain or loss are .a.*! tlio stocks ; when tbc stocks are equal, tlie shares arc ns the times ; and when neither the times n»r the stocks are equaJ, the shares are as their products. Hence, for Coin pound Partnership we have the follow- ing : — nVLE. Multiply each man'n ftock by the time he continues it in trade ; then say, a."? the sum of , the pr-oducts is to each particular product, so is the whole gain or loss to each tnan's share of tite gai7i or loss. f J km. 62.] COMPOUND PARTNEE3HIP. 271 Example — A contributes $120 for 6 months, B $336 for 11 rooiUh.->, and C |?884 for 8 raontha ; and they lose $56. What is C's share of the losa y OPERATION. $120 X 6=$7'20 for one month 1 886 X ll=:;>t;',iG for one month V =$7488 for one month. 884 X 8=3072 for one month ) 40A70 V "tfK $7188 : $3072 : : $56 : C's share ; or' -"-^^~ - = $22-974. profit depends upon lave the follow- Expi.AiTATTON'.— It is clear that 1120 contributed for C month-* are, as far as the train or loss is concerneM*^0 In 12 u..jnth8; that I«, A's Btook and profit at the md of a months ". c old be $960 + $480=11440. then $1440 : 2400 : : $960 : B's Bt<\;lc ; or-*^T^-^|J'-^=$1600 B's stock. Again, B's stock : C's stock : : B's profit : C's profit for same time, via. : 12 640 x800 months. That is $1600 : $640 : : |800 : — riT^^r- = *820 = C's profit for 12 months. i^O Lastly. C's profit for 12 months : C's given profit : ; 12 months ; C's time ; 40(1 X 12 that Is, $820 : $400 : : 12 months : „-. =15 mouths, C's time. *l. In the foregoing question A'a gain was |240 during 6 months, B's $800 during 12 months, and C's $400 during 15 months; and the sum of the products of their stocks and times is 34560. What were their stocks? Am. A's was $ 960, B'a " 1600, C»8 '• 640. S. In the same questior the sum of the stocks is $3200; A's stock was in trade 6 mont'as, B's 12 months, and C's 15 months; and at the settling of accounts, A is paid $240 of the gain, B $800, and C $400. What was each person s stock ? Am. A's w(is $960, B's #1600, and C's $640. QUESTIONS TO BE ANSWERED BY THE PUPIL. NoTB.— 2%« rmmbwi /ollotoinff the que%tion$ refer to the arUelee of the Section. 1. What Is Interest? (1) What Is the meaning of the terms v)!'r cent, and per cmnwn t (1) In what respect does interest differ iroixx Commission and Brokerage? (2) What Is the principal ? (8) What is i:ii<»ant by the rate per c&ntf (4) Wha^ Jr >«^nt by the rate per unit f Xo) What t?3: Interest? (8) What amount ? (7) Of how many kinds is interest? (8^ Explain the distinction between Simple and Compound Interest. (9 and 10) In.uslnc formulas for interest, what is the meaning of the letters P. A. I, 8. 4. 6. 6. 7. 8. 9. 10. 11. 12. 18. 14. IS- IS. 17. la 19. t,anSrf (12) Deduce algebraically a fbll set of rules for Simple Interest. (12) How is the interest found when the prindpac, rate per cent., and time are flven ? (13) Note.— Answer this and succeeding similar questions by giving the form- ula. Interpret this formula. (18) When the interest, rate per cent, and time are given, what Is the rale for fl^nding the principal ? (14\ Interpret this rormula. (14) How is the rate per cent, found when the interest, principal, and time are given? (16) Interpret this formula. (16) When the interest, principal, and rate are glv«n, how Is the tims found ? (16) BjtOT. V?t!.3 QUESTIONS. 273 )0 B's stock. ;00, and C'b $640. )y giving the form- 20. 21. 22. 23. 24. 26. 27. 2S. 2i>. 80. ;!1. 82. U. 84. 85. 87. 88. SO. 40. 41. 4?:. 41. 4i.. 4'j. 45. 5(. 61. f-2. M. fit. 65. tu. 63. 59. fiO. 61. 62. In^orpret tTils formula. (Ifl) When the principal^ ■>ote, and fin"! arc gl' id, how Is tL j usnoriut fooort? (17) IntcM-pu t this *")rmnla. (17) \> hc.T the amuunt, rate, and time are given, Ii >w do we find the priiicips". * (18) Interpret this formula. (18) When tho amount, principal, and Umt are given, how do we find the v-to f (19) Interpret this formula. (19) When the amount, principal, and rate are given, how do wo find the time ? (vO) I nteriiret this formula. (20) llow do we find the time in which any sum of money will aroeant to any g.ven • .iiTiber of times itaelf at a given rate? (21) Interpret this furmula. (21) How do we find tiie rate at which any sum will amount to a given Buraber of times itself in a given time ? (22) Interpret tl'is formula. (22) When the time and r.ite are given, how do we find to how many times itself a given sum will amount? (23) Interpret this formula. (23) llow do we find the interest nn $1 at 6 per cent, per annam for any nuro- her of months? (24) How do we find the interest on $1 b.i 6 per cent, for any number of days? (25) How do we find the interest of any sum for any given time at 84»«r cant. ? (26) IIo'.v may we find the interest at any other r.ate than 6 per cent. V (27) llow do wc compute interest on notes, &c., when partial prtynienta Ktm nia we find the equtitod time of pay iieiit? (46) Whiit is Partnership or Fellowship? (46) What are the persons associated together in partnership called? (47) Wiiat is the monev employed in the business called ? (4vS) What is meant by' the dividend ? (49) Wliat is the distinction between Simple and Compound Fellowship? (60 and 52) By what i)ther name is Simple Partnership known? (50) What is the rule for Simple Partnership? (51) What is the rule for Compound Partnership? (62) 8 is the tim« found ? Il —".Of 274 PROFIT AND LOSS. SECTIOX IX. ISbot IX. PROFIT AND LOSS, BARTER, ALLIGATION, CURRENCIES, EXCHANGE, &c. PROFIT AND LOSS. 1. Profit and loss is a rule by which we are enabled to ascertain what we gain or lose in mercantile transac- tions. It also instructs us how much we must increase or diminish the price of our goods in order that our gain or loss may be so much per cent. • CASE L 2. To find the total gain or loss on a certain quantity of goods when the prime cost and selling price are given : FIRST RULE. Find the price of the cjoods at prime cost and also at the selling price, l^hc difference will be the whole gain or loss. Example 1. — What do I gain if I buy 201 corda of wood at ^S-^S per cord and sell it at $4-25 ? OPERATION. 207 cords @ $4-25 = 3>879-75 = wliole sum for which gooda were sold. 207 cords @ ^-73 = $782-46 = whole cost. m Difference = $97-29 = whole gain = Ans. Example 2. — If I purchase 900 bushels of wheat at $1*47 p(;r bushel and sell it at $1-25, what do I lose upon the whole transa(!- tion? OPERATION. 900 bushels (?^f 1-47 = $1.^23 = whole cost. 900 bushels @ |l-25 = $1125 = whole sum received for wheat $198 = whole loss = Ans. vum second rule. Find the difference between the buying and selling price of a htuhel, Ib.^ yard, dtc. Multiply the gain or loss per bushel, lb., yard, c&c, by the number of bushels f lbs,, or yards, and the result will be the whole gain or loss, * [Seot IX, \, CURRENCIES, we are enabled •cantile transac- nust increase or liat our gain or certain quantity price are given : also at the selling s. s of wood at p-16 I goods were sold. dieat at 81-47 p(:r the whole transai;- ,V' ed for wheat. selling price of a (£t., by the number whole gain or loss, Aets. 1-S.J PROFIT AND L083. 275 Example, — Bouj^ht 211 yards of flannel at 37^ cents per yard, and sold it at 45 cents. Required my total gain? OPEUATION. $0-375 = bnyiiiK piioo. $/J-45 = soiling prico. $0075 = gain per yard SO-075 x 211 = $15-325. Ans. Note. — This second rule affords the shorter method of finding the gain or loss. Exercise 119. - 1. Bought 317 lbs. of butter at 9 cents per lb., and sold it at 12^ cents. What wa3 my gain on the whole? Ans. $11 '095. 2. Bought 2138 bushels of potatoes at 87^ cents per bushel, and sold them at $r2o. What was my gain on the whole ? Ans. $694-86. 8. Bought 13 barrels of sugar, each weighing 317 lbs. net at 15 cents per lb., and so'.i the wliolo for ^735. How much did I gain or lose on the transaotion ? Ans. Gained $116-85. 4. Bought 17 kogs of wine, each containing 22 gallons, at $3-15 per gallon, ami puid in addition $26 33 for carriage, &c., and an ad valorem duty of 37^ per cent I sold the whole for $1625. What was my gain or loss? Ans. Loss $21-2176. CASE II. 3. Let it be requiied to find for what sura I must sell a house whica cost $2900 so that I may gain 15 per cent. TTi?r« for every $!00 the house cost me I am to receive $116, or for every $1 cost I ain to receive sfiriS. The s( iling price must evidently bo as many times $1'15 as the buying price contai .s.Jl; i. e., $1-15 x 2000 = ?.38%-00. Ann. Agftln ; If a person buys a horse for $'280, and afterwards sells it so A3 to lose 11 per ce t. ; how iniicli does lie, receive for it? Iloro for every $1 he paid for the horse he receives only $0 S9 (^ince he loses 11 per cent., i. e. 11 centn on the ^1.) Then, the selling prico iviU obviously be $0-89 x 230 = $204-70. Ana. Hence, to find at what price an article must be sold so as to gain or lose a specified per centage, the cost price being given : — RULR. Find (Art. 2, Sect. VIT.) how much must be received for each dol- lar of the: buffing price, and multiph ^his by the whole buying price. J7te result will be the selling price. Example 1. — Bought a quantity of oatmeal for $1793-80. For what must I sell it so as to gain 8 per cent. ? OPERATION. Here for every $1 1 expend I desire to receive $1-08 ; hence, the selling prlo« yt\\\ be $1-08 X 1T98-80 = $1937-804. Am. ■ \ ■i-m \' ,' ', t 11 t : '' 1 1' si i .ft t.i ^ \k 276 Example 2.- lose 8 per cent. I^EOFIT AND L088. tSKOT. IX. -Bought a lot of sheep for $7000, and am willing to For what sum must I sell ? OPERATION. Hore for every %\ 1 pxpend I nm willing to receive $0*97, and hence selllDg price will bo $0-97 k 7000 = |6790. Ane. Exercise 120. 1. Botjght cordwood at $8-25 per cord. At what rate per cord musf 1 sell it in order to gain 30 per cent? Arts. $4'22i 2. Bought a stock of goods for $13420. For how much must it h\ sold in order to gain B per cent? Ans. $14091. 5. Bought a quantity of wool at 11 cents n lb., and wish to sell so as to gain 15 per cent. At what rate per lb. must I sell it ? Ans. 12^§ cents. 4. Bought axes at $15'25 a doz., and desire to sell them so as to gain 23 per cent. At what rate per doz. must I sell? Am. $18-75J. 5. Bought a farm for $7890, and am willing to lose 1 1 per cent. At what price must I sell ? Ans. $7022-10. CASE III. 4. Let it be required to find what per cent, of profit a merchant makes by buying tea at 43 cents per lb. and selling it at 67 cents. Here the gain on each lb. is 24 cents. That h every 43 cents Invested gives a gain of 24 cents. Therefore every cent invested gains -it of 24 cents — \\ cents. And hence, the gain per cent = |J x 100 = aija = 55-8 per cent. Hence to find the rate per cent, of profit or loss when the prime cost and selling price are given, we have the following : — RULE, JFlnd the difference between the buying and selling price, and hence the gain or loss per unit. Multiply this by 100, and the result will be the gain or loss per cent. Example."— A speculator invests $44400 in stocks, and sells out for $50000. What per cent, does he make by the operation ? OPERATION. Here the whole gain is $50000 - $44400 = $5600. That is $44400 gain $5600, and therefore $1 gains J//^ = 7*A of a doUar. Hence gain per cent = ,?^ x 100 = VrS^ = 12-6. Ans. Note.— The above and all similar questions may be solved by Proportion. Thus this question is, if $44400 gain $5600, what will $100 gain ? fifiOU X 100 And th« «tat«ment ii ^44400 : flOO : : $5800 : Aru. = ^^^qq = 1*«« tSEOT. IX. md am willing to 7, and hence selllDg ite per cord musj Ans. $4-22i much must it b( Ans. $14091. wish to sell so aa 1 1 sell it ? Ans. 12^^ cents. ;hem so as to gain 1? Am. $18-75f. 1 1 per cent. At Ans. $7022-10. jrofit a merchant it at 67 cents. cents, por cent. It or loss when 1, we Lave the 7 pricCy and hence J gain or loss per cki5, and sells out )peration ? i^^^ofadoUar. veil by Proportion. In? MAOO ^ AuTS. 4, 0.] PROFIT AND LOSS. ExERCISK 121. 277 1. Bought tea at 60 cents a lb., and sold it at 87^ a lb. ; how much did I gain per cent. ? Ans. 46^. 2. Bought coffee at 13 cents and sold it at 11 cents a pound ; what wairi my loss per cent. V Ans. IS/j. 8. Bought flour at $6 20 a barrel, and*sold it at $7*80 ; what was the per cent, of profit? Ans. 25} per cent. 4. Bought cloth at $2*75 per yard, and sold it at $3-10; what was my gain per cent. ? Ans. 12i\ per cent. 5. Bought oats at $0-47 per bushel, and sold them at $0*56 ; what was my gain per cent. V Ans. 19^f per cent. 6. Bought meat at 12 cents per lb., and sold it at 10^ cents a pound ; what was my loss per cent. ? Ans. 12^ per cent. 7. Bought a horse for $93, and sold it for $127; what per cent, of profit did I make? Ans. S6?,^. 8. A man bought a farm for $6742-50, and sold it for $6000 ; what was his loss per cent? Ans. HkV?) V^^ cent. 9. If I purchase a house for $5700, a horse for $276, and pay $1987 32 for household furniture and a carriage, and then sell the whole for $8750, what is my gain or loss per cent. ? Ans. Gain 9'89 or nearly 10 per cent. 10. I purchase 723 yards of black silk velvet in Paris and pay $425 a yard ; I further pay 7 per cent, for insurance, $23*70 for car- riage, $2 70 for harbor dues, $3" 16 for wharfage and storage, and an ad valorein duty of 22 per cent., and then sell the whole for $6270; what is my gain or loss per cent.? Ans. Gain 3 1 '96749 or nearly 32 per cent. CASE rv. 5. Let it be required to find the prime cost of cloth which I sold for $4 and gained 10 per cent, thereby. Here the gain on $1 was 10 cents, or what T sold for fl'lOcost mo only $1. Therefore the cost price will contain $1 aa many times as the selling price contains $1-10. That is the cost prico=i-.fg=$3'C86. Ana. Hence, to find ih& cost price., the selling price and the gain or loss per cent, being given, we have the following : — RULE. Mnd the gain or loss per unit, and add it to unity if it be gain^ but subtract it from unity if it be loss. Divide the selling price by the quantity thus obtained^ and the re- sult will be ike cost price. Or say as ^100+gain per cent, (or as $100 — loss per cent.) is to 1 1 00 so is the selling price to the cost price, (1 iiV?' J Ji«^ BARTER. [Sect. IX. in Example. — Sold a quantity of coal for $719, ami lost 7 per cent. by the transaction ; what was the prime cost ? OHEKATIOX. 1st Rule.— Losb on $1 la 7 cents, or lor every $1 ptiid I receive |0 98. Hence co3t = iJl-'.i'j = !|;77;Jll9. 2nd KULE.-I98: 1100 :: 1719: ^««. = —„''— =1773 118. vo EXKRCISE 122. 1. For what did I buy a quanUty of sugar wu',ch I sold for )B24'60, losing 4 per cent. ? Ans. $26 C25. 2. A gentleman sold his library for $2360, which was 10 per cent, less than cost; what did ho give for it? Ans. $202222. A farmer sold his farm for $7400, gaining 11 per cent, on tho prime cost; what did he give for it? Atm. $6066-06^. 4. A merchant sold a quantity of silk velvet for $378940, gaining 17 per cent, by the transaction; required tho buying price? Ans. $3238-803. 6. Sold a lot of cattle for $2740, losing 13 per cent, by the transac- tion; what did I give for them? Ana. $3149-425. 8. V I '!. r BARTER. 6, Barter signifies an exchange of goods or articles o/ commerce at prices agreed upon so that neither party in the transaction may sustain loss. 7. 77ie principle of solution depends nponjindincf the value of tlui commodity whose price and quantity are given, and thence the equira- lent quantity of a second commodity of a given price, or Hie equiva- lent price of a given quantity of a second commodity. Example 1. — How much tea at $1'10 per lb. ought to be given for 712 lb. of sugar at 13 cents per lb. ? OPEUATION. 712 lbs. of snpftf at IS cents per lb.=$9'2-56, and $92-5G-«-$M0=84-1454 Iba. =84 lbs. 2J oz. Ans. Example 2. — I desire to barter 96 lbs. of sugar, which cost me 8 cents per lb., but which I sell at 13 cents, giving 9 months' credit, for calico which another merchant sells for 17 cents per yard, giving 6 months' credit. How much calico ought I to receive ? OPERATION. 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 mo 5 cents profit, what ought 6 months to give ? 6x5 80 „ 9:6::5: -^y- __ ^ =3^ cents. Hence, wero I to give 6 months' credit, I should charge 8 + 3^=11 J cents per lb. Next— [Sect. IX. lost 7 per cent. receive |0 93. 8. old for !?(24'60, Aiis. $25 025. ms 10 per cent. A7ia. $2G22-2:i». ^er cent, on tho Uiff. $6066-06 i5. ;78940, gainitij? ying price? ins. $8238-803. by the trunsac- i7is. $3149-425. or articles o/ ther party in the value of Ovc p.nce the equira- or the equivn- ;ht to be given ^1 -1 0=84-1454 Iba. ^hich cost me 8 nths' credit, for yard, giving 6 were I to give to give? 8 + 3i=nj cent9 AKT8. 0-10.] ALLIGATION. 279 \s my sf>llin(; prico la to my buyintr price, so ought bis selllog to bo to blf buying price, both ijiving the satiiu credit Hi: 8:: 17 8j<17 ThP price of my sugar, thorcforo, Is 06 x 8 cenf I'i Ci'iils por yiini. henco .„ = (W, Ib the required iiuuibor of yarJu. = 12 cents. or (7'GS ; and of bis calico, 1. 2. 4. 0. ExKKCiSK 123! A Fins cofTce which he barters at 10 cents the lb. more than it cost him. agiiin.st tea which slumls B in $_', but which he rates at $2-50 per lb. llow much did the coflee cost at first? An». 40 cents. A has silk which cost $2-80 per lb. ; R has cloth at $250, which cos 9> r<^^ :« t ! tl h • ' ■ \i "' ' ■ ■!■ 280 ALLIGATION MEDIAL. [Sxox. lA ingredients, of which it is composed and their prices art' known. 11. Alligation Alternate enables us to find what pr^ portion must be taken of several ingredients, whose pricey are known, in order to form a compound of a given price. ALLIGATION MEDIAL. 12. Let it be required to find the price per lb. of a mixture con- taining 47 lbs. of sugar at 11 cents per lb., 29 lbs. at 13 cents, and 24 lbs. at 17 cents. OPSRATION. 47 lbs. at 11 cents=517 cents. 29 lbs. at 18 cent8=377 cents. 24 lbs. at 17 cents =403 cents. Then 100 lbs. cost 1802 conts and 1 lb. ■will cost Vo'V' - 18j\, cents. Hence for Alligation Medial we deduce the following—* RULE. Divide the entire cost of the whole mixture by the sum of the in gredients, and the quotient will be the price per unit of the mixture. Example 1. — What will be the price per lb. of a mixture of tea containing 7 lbs. at $0-50 per lb., 11 lbs. at $0-80, 19 at $1-06, and 3 lbs. at $1-28? OPEBATION. 7 lbs. 11 " 19 " 8 " f $0-50 = $3-50 $0 80 = $8-80 $106 = $20-14 $1-23 = $3-69 40 lb8.=:8nm of Ingredients. $86 18=»Total cost. 40)$36-13($0-90i§. Am. 860 •13 ExAMPLR 2. — A goldsmith has 3 lbs. of gold 22 carats fine, and 2 lbs. 21 carats fine. What will be the fineness of the mixture? In this case the value of each kiij. at UrJ. per lb (Jonseq'.iently we are to take 1 lb. at 7d. and 2 Ib.s. at 5d., 1 lb. at 9d. nn ) 8 lbs. at lod. If we e\;unine the price of the mixture these will give (Art. Iv } •we shall find it to be tho given mean. ExAMPLK 3, — What quantities of tea at 4s., 63., 8s., and 93. pec lb., will produce a mixture worth os. ? Prices. OPERATION. Dilil' ranees. Prices. 1+4 = 5 3, 1, and i ftre connected with Is., tho difference between 49. and tho mean; therefore we are to take ;3 Ibi?. + 1 lb. + 4 lbs. of tea, at 4s. per lb. 1 is connect, l with os., Is., and 4,s., tlie ditlvrenees between 8-., Gs., and 9s., and tho mean ; therefore we are to take I lb. of tea at 8s., 1 lb. of tea at 6s,, and 1 lb. at 9s. per lb. Example 4. — How much of any thing at 3s., 4g., 5s., 7s., Rs., 9s., lis., and 12s. per lb., would form a m-xture worth 6s. per lb. V Prices. OPE!!.VTTON. ni(Teioiic3f. Prices. AB^8. 3. 4. :J„ 1 lb. at ."^a., 3 lbs, at 4s., i^ lbs. at 7s., 2 lbs. at Ss., 3 + 5 + 6 ; i. c., 14 lbs. at fis,, I lb. ut 9s,, 1 lb, at lis,, and 1 lb. at 12^. per lb. will form tho required mixture. [SacT. XI. 7. the required price, .vitii 1, thi- (litfiTftico .1 '>j li)s. !<.t S.S. TIh'U re worih Ts per lb.— would answer equnlly i> sucli a question. 5d., and lOd., will d tho moan, S : Ikmk'.' wirh o, tlio ditlVn iic'^ iit oil. 1 is odniH'Ctid tluTC 1^ to ho 1 11'. :lt n lOd. aud the moan; t 5d., 1 lb. at 9d. an 1 jso will give (Art. \.\ ) 53., 83., and 93. pec '■oen 43. and tho moan ; )er lb. 1 is connocfcd nd 9s., and tho mean; at 6s., and 1 lb. at 9j. s., 5s., Ys., 8s., 9s., 1 6s. per lb. ? ■6 ; 1. c, 141be. at 5s., bo required mixturo. AB13. 14, 15.] ALLIGATION ALTEilNATE. 283 NoTB.— Tho principle upon which this rule proceeds Is that the cxces-s of one insrrediont abovo tiio mean is mudo to counlcrbalaneo what tlio otiior wanis of bein;; equal to tho moan. Tliu:' in examplo 7. 1 lb. at Hs. per lb. civos a c/t- fidency of 2s.; hut tliis is corroctod bv 'Js. crcexn in the 2 lbs. at Ks. per lb. In example 8, 1 lb. at 7d. t;ivf.< a (fi'Jiviein-if of !d.. 1 Ih. at 9d trivoa an ev- ciSH of Id. ; but tho excess of Id. and tho defloiency of Id. exactly neutralizo each other. Attain, it is evident that 2 lbs. .-it ~i(\. and 3 lbs. at tOd. aio worth just aa much as 5 lbs. at Sd. — that is, 8d. will bo tiiu average price if we mix 'J lbs. at 5d. with 3 lbs. ut lOd. Exercise 125. 1. How much wheat at $1*60, $1-40, $1-10, and 81 per bushel nm.st be mixed <-ogetiier in order to Ibiin a mixture, worth §1'25 per bushel ? Give at least two sets of answers. Ans. 85 bushels at i^liO. 15 at .-tl-CO, l.'- nt ?l-00. and 25 at,41-40. 85 bushels at $100, 15 at §1-40, 15 at $1-10, and 25 at .;-l-00. 2. How much wine at 60 cents, 50 cents, 42 ccnt.-^!, 38 cents, and 30 cents per quart, will muke a mixture worth 45 cents a quart? An..<. per lb. (Art. 14.) But all of these added together, will make 11 lbs. in whlrh there are 8 lbs. at 43. Therefor© lbs. lbs. lbs. lbs. lbs. oz. 11:8 : 736: 11 =535 4y\, the required quantity of tea at 4a. That i3, li, ,;}() lbs. of the mixture there will be MS lbs. 4^*7 oz. at 4s. per lb. The amount of each of tho other Ingredients may be found in the same way. Exercise 127. 1. A druT!?ist is desirous of producing, from medicine at fl*00, $1 20, $1-00, and $1-80 per lb,, 168 lbs. of a mixture worth $l*4t) per lb. ; how much of each kind must he use for the pur- pose? Ans. 281bs. at $l-uO, 56lbs. at $1-20, 56lbs. at $1-60, and 281b.-. at $1-80 per lb. 27 lbs. of a mixture worth 48. 4d. per lb. are required. It is to contain tea at 5y. and at 3s. 6d. per lb. ; how much of each must be used? Ans. lulbs. at Ss., and 12lbs. at 3.s. Cd. i. How much brandy at $2-40, $2-6(», $2-80, and |2-90, per gallon, must there be in one hogshead of a mixture worth $2 70 per gal- lon ? Ans. 18 ga»s. at $2-40, 9 gallons at $2'60, $9 gals, at $2-80, and 27 gals, at $2-90 per gallon. 2 EXCHANGE OF CURRENCIES. 17. Exchange of Currencies is the process of changing a sum of money expressed in the denomination of one country to an equivalent sum expressed in the denomina- tions of another country. 18. By the currency of a country is meant the coinSy or money, or circulating medium of trade of that country. 19. The intrinsic value of a coin is determined by the kind, purity, and quantity, of metal it contains. 20. The relative value or commercial value of a coin is its market value, and is fixed by law and commercial usage. • r" !il ><„ :?■■ / 280 EXCHANGE or criiKENClES. tSECT. IX. ;ll V S^'l :i \m FOREIGN MONEYS OF ACCOUNT, WITH THE PAR VA'"'s OF THE UNIT, AS FIXED BY COMMERCIAL USAGE, t.-.. ICEloSUD IN DOLLARS AND CENTS. Austria.— GO krcutzers — 1 llurin (.silvpr)= |0'4S5 IjELOium. — 100 coiits— 1 guilder or florin ; 1 guilder (silver)= -40 Brazil— 1000 ret'S=l U!ilreo= -823 IJUE.MKN.— 5 scli\vuro8=l g ; 72 gn>te8=l rix-dolliir (hilvcr)= -"ST BiiiTisii I.sinA.— 12 j)icc — 1 m ; 10 nnnas=l C'ornpuiiy''h* rupee = ^-IS Bi'ENos Ayuks.— y rials-1 iir currency (viirlablf), ineau vahio=: •90 Canton.— 10 cash t-1 caiK ..^cs; 10 CHnd. = l Hiace; 10 uiac;(f=i tael= l-lb Cape of Good Hope.— (J sti . ,=1 schiliug ; G sc'diings— 1 rix-dollar=.. olG Ckylon.— 4 i)ice=;l lanaiu; _ _.inaiiis=l rix-aollar = 40 Cuba, Colombia and Chili rials. =1 doi; r. = 1 00 Denmark. — 12 i)fciuiiug=] shilling; 16 skil; 'J{s=l uiarc; G njarcs= 1 rix-dollur. = ^2 England.— 4 farthings=: 1 penny ; 12 peiice= 1 shilling; 20 E,ail. = £l= 4-807 FiiANct:. — 10 centimes = 1 deciuiu ; 10 deciuies^l franc = l&C Grekce.~100 lepta=il drnchnie ; 1 draclinie (silver) = 'lIC Holland.— 100 cents = I florin or guilder ; 1 florin (silver) — -40 Hamburgh. — 12 pfenning =: 1 schiling; 1(! schil. =: 1 marc ; 3 marcs =1 rix-dollur = 84 Malta.— 20 grains = 1 taro ; 12 turi ~ 1 scudo ; 2| scudi = 1 pezza = . . . 100 Milan.— 12 denari = 1 .soldo ; 20 soldi = 1 lira = -20 Mexico.— 8 rials = 1 dollar = 100 MoxTK Video.— 100 ceritesimos = 1 rial ; 8 rials ^ 1 dollars -SSS Naples.— 10 grani =1 carlino; 10 oarlini = l ducat (silver) = "80 Noeway.— 120 skillings = 1 rix-dollar specie (silver) = I'd Papal States. — 10 bajocclii = 1 paolo ; 10 paoli = 1 scudo or crov.n = . . 1 00 Peru.— 8 rials = 1 dollar (silver) = 1 00 Portugal.— 400 rees = 1 cruzado ; 1000 ree3= 1 milree or crown =. . . . 112 Pr.ussiA.— 12 pfenning.s=l grosch (silver); 80 groschca = l thaler or dollar -69 Russia.- 100 copeoks — ■ 1 ruble (silver) = -78 Sardinia.— 100 centesimi=:l lira = "186 Sweden.— 48 skilliiigs = l rix-dolLir specie =: 1"06 fiiciLY.— 20 gi an\ = 1 taro ; 30 tari = 1 oncia (gold) = 240 Bpain. — 34 maravedis = 1 real of old plate %=. ; '10 8 renls= 1 j)'astre ; 4 piastres = 1 pistole of exchange. 20 reals vellon — 1 Spanish dollar = 100 * The current silver rupee of Bombay, Madras, and Bengal, is worth $0'444. In India also "^hey use cowries for coin. These ore stiuill shells found in the Maldives and elsewhere ; 2500 cowries make a rupee, uad 100000 rnpeei make a lac. t The cash, made of copper and lead, is said to be the only money coined in China. :f The old plate real is not a coin, but is the denomination in which ex* Qhanges are usually made, tSECl. IX. UNT, BY COMMERCIAL CENTS. I0-485 ,'er)= -40 •623 ;hilvor)= '"ST ^^'b^rupcezr -AAb leau vahio = ■91i uiucp=i tot;l= r4S =:lrix-dollar=.. '818 -40 100 arc; 6 njarcs= 1 •r>2 ,';20£ail. = £l= 4-807 1C=: ISC -ICC or) — -40 arc ; 3 marcs =1 84 1 = lpezza=... 100 -20 100 lur = -538 i'r)= -80 1'3 do or crov/n = . . 1 00 100 or crown =:.... 112 u = t thaler or -09 -78 -186 1-06 240 ; -10 hatige. 100 nd Benfral, is worth ore small shells found jee, v..\d 100000 rttpeec le only money coined ui nation in which ex- Ant. 21. J EXCHANGE or CT'RRENOIES. 287 Pt, PoMiN'GO.—inO centimes =1 dollar = |0-338 TusovNY. — 12 d<'niiri dl pi-zzii — 1 soldi dl ja'Zza; 2 soldi dl pozza = l pi'zza of S riuU; 1 pizza (-liver) = '90 'lURKRY. — 3 aspers = 1 para ; 40 paras =- 1 piastre (varialde) nbout "OOft Vbnicr.— 100 c.-nti'simt = 1 lira= -188 Tnitki) Statks of Ami:iiica.— 10 mills = 1 cent; 10 cents = 1 dime; 10 dimes = 1 dollar = 1-00 2il. The following table exhibits the commercial value of the Foreign coins moat frcquentlg met loith. OUINKA $510 BovKUEiON of Great Britain 4'867 CuowN of England 1"218 IIalf-Crown of i]iigUuul '008 SuiLi-iNO of Englaixl '244 TJoLLARof the United States 100 FuANvj of France 'ISI Fivf.-Franc I'lKCE of Fvance "93 LiVRE TouRxois of France '18^ FoaTY-FRANC I'lECii ' f Franco 7"C6 Orown of France 1"0S Louis-D'Or of France 4-58 Flotun of the Netherlands. "40 Guilder of the NothtM-lands '40 Fi-oitiN of Soiithorn Germany "40 TiiALKR or Rix-DoLLAR of Prubsla and Northern Germany "69 I'lx-DoLLAu of Bremen '1S\ Florin of Prussia '2,-1^ Marc-Banco of llamhiirgh "85 Florin of Austria and city of Augsburg. •48i' Flokix of Saxony, Bohemia, and Trieste •43 Florin of Nuremburg, Frankfort, and C're veld '40 Rix-DoLLAR of 3enmark I'OO 8?Ek.iE-DoLLAii of Denmark 1-05 ; Collar of Sweden and Norway 1 -OS MiLREE of Portugal 1'12 MiLUEr, of Madeira 1-00 MiLREE of Azores , i&\ Eeal-Vellon of Spain , -05 Eeal-Platk of Spain -10 Pistole of Spain 397 Eial of Spain '12 Pistareen -18 C:t0S3 Pistareen , "16 ?cUblr (silver) of Russia -75 Imperial of Russia ; T'SS \l » »' If ft" r ir*T^ m 2S8 EXCHANGE OF CUUEENClM. ISSOT. IX DounT.ooir of Mexico |1S-C0 IIalf-Joji of Portugal 6b'\ Lira of Tuscany and Lombardy '10 LiBA of Sardinia i8j OUNOB of Sicily 2.40 l»iroAT of Naples •{•() CiiowN of Tuscany 1"08 Florence Livbb. -ir) Genoa " 181 Geneva " 21 Leghorn Dollar 'flO Swiss Li VRE •27 ScuDO of Malta -40 Turkish Piasxkb 'OU Paqoda of India 184 RCPKE of India ^ 44^ Taei of China 1-48 22. In Canada all accounts were kept In pounds, shillings, pence, and farthings, previous to the adoption of the decimal coinage by Act of Provincial Parliament In 1858. In the United States also accounts were similarly kept Srlor to the adoption of Federal Money In 1786. In the States, at the time 'ederal Money was adopted, the Colomal currency or hills of credit had be- come more or less depreciated in value, i. e., a colonial shilling was worth less than a shilling sterling, \c., and the depreciation in value being greater in the currencies of some colonies than in others gave rise to the different valu*« of the present old currencies of the different States. or£i. or£|. or £^%. or£|. or £ Jo". TABLE OF CURRENCIES IN CANADA AND THE UNITED STATES. In Canada, Nova Scotia, New Brunswick, &c., $1 = 6s. In N. y., N. C, Ohio, and Mich., %\ = 8». In N. Eng., Va., Ky., Ten., la.. 111., Miss., and Missouri, $1 = 6s. In Penn., New Jer., Del., and Md., $1 = 7s. 6d. In Georgia and S. C, $1 = 4s. 8d. Note. — The remaining States use the Federal money exclusively. 23. To reduce dollars and cents to old Canadian Cur- rency, or to any State Currency : — RULE. Multiply the given sum by the value of $1 in the required cur- rency expressed as a fraction of a pound. The product vAll ht pounds and decimals oj a pound. Reduce {Art. 68, Sect, IV.) decimals to shillings^ pence, and farthings. Aew. 22 24.] EXCHANGE OF CURRENCIES. m Example \. — Reduce $493-72 to Old Canadian Currency. OPERATION. 4e3'72 y i -- £12318 -- £128 8s. Tjd. Ana. FiXAMri.tt 2. — Reduce $749'80 to New England Currency. OPERATION. T40S0 X ^5 = £224-94 = £224 1«^8. OJ.I. Anit. Example 3. — Reduce $1111'11 to New York Currency. OPF.RATION. null X i = £444-414 = £444 88. lO'JJ. Ans. Exercise 128. 1. Roduoc $1974-80 to Now .Jcr.iey Currency. Ans. £740 lis. 2. Koduce $765-43 to Michirran ('urroney. Ans. £30C 38. Cj^^d. 3. Reduce $817219 to Old Canadian Currency. Ans. £2043 Os. 11 §d. ling9, pence, and 24. To Refluce Old Candian Currency or any State CuiTcncy to dollars and cents : — RULE. Express the r/iven sum decimally and divide it h>/ the rnhie of a dollar expressed as a fraction of a pound ; the quotient will be dol' Inrs, cents, i'n!jive to importers to pay for the insurance and transportation of bullion and coin to meet their payments than to tranbmit bills of exchange. 44. By an old act of Provincial Parliament it was enacted that jElOO ster- lings or 100 sovereigns should be livalent to jElllJ Canadian money, i. o. to $444-444 or XI sterling = $4 -l-W. It was found however that this was very much bolow the real or intrinsic value of the sterling pound, accordingly, wiiile its legal value was only !*44'14,- the market or commercial value varied from 14 83 to $4' 86. By an act recently passed by the Provincial Parliament llic value of the pound sterling was fixed at ,*4866. Now the new par is equal to the old par phia nine and a-half per cent, of tlio old par, that is, .?4-444 + 9i per cent, of $4-4 14, which is, •422, make $4 860 = the new par. Consequently the ruto of exchange between Canada and Great Britain must reach the nominal premium or OJ per cent, before it is at par, uo- cordiiig lo the new standard. 45. Rates of exchange between Canada and Great 1j itain are commonly reckoned, at a certain per cent, on the old par of exchange, instead of on the new par. Example 1. — A merchant in Hamilton wishes to remit to London £749 OS. 0(1. sterling ; exchange being at 10 per cent, premium. How much must he pay for the bill of exchange ? OPERATION, Old commercial par of £1 sterling = $4 '444 To which add 10 per cent, of itself = -444 Gives price of £1 = 4-888 Then i;749 Ss. 6d. = £14Q-l'l& x 48S8 = $3C62-63i. Am. Example 2. — A merchant in Toronto wishes to remit 144479 francs to Paris, exchange being at a premium of 2 per cent. What will be the cost of his bill in dollars and cents V OPERATION. Commercial value of the franc = 18'6 cente. Add 2 per cent. = -373 " Gives value for remitting = 18"972 " Then 18-972 x 144479 = $27410-55.588. Am. Example 3. — What sum in dollars and cents will purchase a bill of exchange on Hamburg for 14607 marcs banco, exchange being ajb l| per cent discount ? OPERATTON. Comnr.erclal value of tbo marc banco = 88 cents. Deduct 4 per cent. = •625 " Gives value for remitting = 84-475 •* Tbcp 84-475 cents x U667 = |«056448. Ant 1 f. ( ■i '*ii I I I, •( 1 1 ~ I ym m f ' 294 AEBITRATION OF EXC^A^'GE. Exercise 182. [Sect. IX 1, II I wisli to remit ^leYSS^S to Paris, for how many francs and centimes can I obtain a bill — exchange beiii;:^ 5 iVancs 4 cen- times to the dollar ? Ans. 84597 francs 00 centimes. 2. What is the cost of a bill of exchange for 4000 marcs Itanco at one per cent, above par ? Ans. $1414. 8. How mnch must I give for a draft on New York for ISOOTS at 2^- per cent, prenuum V Ans. $o6480-YiJo! 4. What will a bill of exchange on St. Petersburg for 2500 ru!)le>- cost in dollars and cents, at 2 per cent, discount, the per being 75 cents per ruble? Atis. $1J; U-60. 5. What will be the cost of a bill of exchange on Great Britain for £800 sterling, at 8 per cent, premium V Ans. $384000 ARBITRATION OF EXCHANGE. 46. ArbitTation of exchange is the process of changing a given amount of the money of one country into an equiva- l lent sum of the money of another, through the medium of one or more intervening currencies with which the first and last are compared. Note. — Arbitration enables a person to ascertain whether it is more ad- vantagfous to draw or r<'rnit :i bill of exchange direct from ohe country to an- other or indirectly through other places. 47. When there is but one intervening country, the operation is termed simple arbitration ; when there are two or Qiiore intervening countries, compound arbitration. 48. All questions in arbitration of exchange may be solved by one or more statements in simple proportion ; it is more convenient, howevei', to consider tliem as problems in Conjoined Proportion, and work them by the rule given in Art. 50, Sec. V. Note. — Care must he taken to reduce all the moyien of Jie same country to the same dcnomiuaiiou before linking iliem as directed in il»e rule. Example 1. — A merchant in Toronto wishes to remit 2000 marcs uanco to Hamburg, and the exchange between Toronto and Ilanibiuf^- is 35 cents for one marc banco. He finds, however, that the ox- fhange between Toronto and Lisbon is $r08 for 1 mihee, that be- tween Lisbon and Paris is 6 miliees for 38 frnncs, and that iK'tucor. Paris and Hamburg is 19 francs for 10 marcs banco. How mudi wil/ Ue gain by thf» circuitous exchange V Abts. 40-48.1 ARBITKxiTION OF EXCHANGE. 295 OPERATION. STATEMENT. 108 cents 6 milreea 19 francs 1 milree. 38 francs. 10 marcs banco. SAME CANCELLED. „108 = 1 „ — sts/^ " = 3^' 200 ^^ = ^^ 2000 marcs banco = x ^^'";2pp^ = x. X — 200x3xl08=:$f)48. 2000 X 35=:$7O0'0O— what he lias to pay by chrect exchange. G48'00=:what he lias to pay by circuitous exchange. Diircrence=$ r)2"00=:what he gains by the latter mode. Example 2. — £824 Flemish being due to me at Amsterdam, it is remitted to France at 16d. Flemish i)er franc; i'rom France to Venice at 300 francs per GO ducats ; from Venice to Hamburg at lOOd. per ducat; from Hamburg to Lisbon at 50d. per 400 rees ; from Lisbon to England at 5s. 8d. sterling per milree ; and from England to Cana- da at $4'8GY per £1 sterling. Shall I gain or lose, and how much, tlie exchange between Canada and Amsterdam being la. Id. Flemish per dollar? statement. OPERATION. SAME CANCELLED. CB= 16d. Flemish = 1 franc. 800 franca = 60 ducats. 1 ducat — lOOd. Flemish. 50d. Flemish = 400 rees. 1000 rees = G8d. British. 240d. British = $4-8G'7. X = lyzVGOd. Flemish. IT X4-S67 X 3296 2 X hO~ 60. ;a= 1 ;p 1 = xm, 0P = ^00 !^ ^ U^ = 4-8G7 3296, ■= $2727*072 = amount remitted. Then since exchange between Canada and Amsterdam la Tb. Id. Flemish per dollar wo havo 8r>d Flomiah=100 centos. X " =:197V60d. Flemish. 197760x100 ^„„„^ ,„ X L .,^ . J ^ , .. V Here x = = $2820 \5S= sum I should have received had it been tranpmitted direct from Amsterdam to Canada. Hence by the circuitous exchange I gain the difference between $2727"07i and $232Go8 that is $400-492. Exercise 133. I. If London would remit £1000 sterling to Spain, the direct ex- change being 42id. per piastre of 272 maravcdis; it is nAcd whether it will be more profitable to remit directly, or to remit first to Holland at 35s. per pound; thence to France at 19^d. per franc; thence to Venice at 300 liaucs per GO ducata ; and ^henco to Spain at 360 maravcdis per ducat V 4^?w>'. The circular exchange is more advantageous by 103 piastres, 3 reals, 20 maravedi^. .i... i^^ i1 : i 296 EXAMINATION QUESTIONS. [Sbot. X li 2. A merchant wishes to remit $4888*40 from Montreal to London, and the exchange is 10 per cent. He finds that, lie can remit to Paris at 5 francs 15 centimes to the dollar, and to Hamburg at 35 cents per marc banco. Now, the exchange between Paris and London is 25 francs 80 centimes for £1 sterling, and between Hamburg and London 13| marcs banco for £1 sterling. How had ho better remit? Alts. If he remits direct to London he will obtain a bill for £1000. If he remits through Paiis he will obtain a bill for only £976 15s. 8:id. If he remits throdgh Hamburg he will obtain a bill for £1015 1 5s. 5d. Hence the best way to remit is through Ham- burg, and the next best way is direct to London. i. A merchant in Quebec wishes to remit 1200 marcs banco to Ham- burg, and the exchange of Quebec on Hamburg is o5 cents for 1 marc. He finds the exchange of Quebec on Paris is 18 cents for 1 franc; that of Paris on London, is 25 francs for £1 sterling; that of London on Lisbon, is 180 pence for 3 milrec.'^; that of Lisbon on Haiaburg, is 5 milrees for 18 marcs banco. How much will he gain by the circuitous exchange ? Alls. Direct exchange $420; circuitous exchange $375; gain $45. QUESTIONS TO BE ANSWERED BY THE PUPIL. Note. — The mimbers after the qnestions refer to the numbered articlea of the sfciion. 1. What is profit and loss? (1) 2. How do \vc find tiio toUil gain or loss on a quantity of goods when the cost pi ice and selling price are given? (2) 8. How do wo find at what price an article must be sold so as to gain or lose a specified percentngc, the cost i)ricc being given? (8) 4. llow do we fitid the rate per cent, of profit or loss? (4) 6. IIow do we find tiic cost price wheu tho selling price and the gain or losi 6. 7. 8. 9. 10. per cent, are given ? (6) Hi " What is barter? (6) What is alligation ? (S) Into what rules is alligation subdivided? (9) What is alligation medial? (10) Wliat is alligation alternate? (11) 11. How is allia.ition alternate proved? CIS) 12. Give the dilTerent rules for alligation ? (12. 14-1 fi) 13. What is niea-t by the exchange of currencii^sl' (17) • 14. What is meant by the currei cy of a country ? ( i vi 15. How is the intrinsic value of a coin determined r (19) Ifi. What fixes the conmiercial value of a coin ? (f'l) 17. How do you account for the fact that the $ it of difi'erent values in th» American States? (22> 18. Give the value of tho pound currency in Canada, acd in tlie dinFer jut States. (22) 10. How do we reduce dollars ajiA ccntJ to old Caoadian Mrency or to any state ouTannv ? (23) [Sect. X ^BTS 1-4.3 INVOLUTION. 207 mtreal to London, ai he can remit fo (I to Hamburg at between Paris and ling, and between 11 sterling. How n he will obtain a ti will obtain a bill irg lie will obtaia is through Ham- 3 direct to London. cs Iianco to Ham- ', is 00 cents for 1 iris is 18 cents for s for £1 sterling; niilrcos ; that of cs banco. How mitoua exchange : PUPIL. numbered articles )ods when the cost as to gain or lose a Qd the gain or losi 20. erit values In th» icdiifer Jilt States. M-renoy or to any 27. 2\ 29. ■30. 31. 32. Si. 84. .6. IIow do we refliice Oid Canadian Currency or any state currency to dollars iind C'.'nts? (24) H<»'.v do we reduce dollars and cents to sterling money ? (25) llmv do we roduee sterling money to dollars and cents? (20) Whiit is ft bill of E.\chanKe? ('28) E.vplain the terms drawer, d/*aiv6€,accept(/r, payee, holder, endorser, atid eiulornie. (29-^35) How i.s !i Idil acci'pted * (84) What is the diirenmce between a blank endorsomoiit and a special endorso' niont? O^.'S) What is r.i(\int by proteHting a bill ? (!^6, 37) Kxplain what is meant by the First, Second, and Third of Exchange. (39) What is the par of Exchange? (40) Explain the difference between the intrinsic par and the commercial par of Exchanire. (41, 42) What is the com'sc of Exchange? (43) Explain wh-it is meant by saying the par of Exchange between Canada and Britain is 9^ per cent. (44) Upon what is the rate of Exchange between Canada and Britain reckon- ed ? (45) What is arbitration of Exchange? (46) What is the difforeiioe between simple and compound arbitration ? (47) By what rule are qnesUons in arbitration of Exchange worked? (4S) SECTION X. INVOLUTION, EVOLUTION, LOGARITHMS, AND LOGARITHMIC ARITHMETIC. 1. A power of any number is the product obtained by multiplying that number by itself one or more times. Thus 26 = 5 X 5 is a power of 5; 81 = 8x8x3x3 is a power of 3, &c. 2. The number which, being multiplied once or oftener by itself, produces the power, is called the root of that power. Thus 5 is the root of 25, since 6 x 5 = 25; 8 is the root of 81, slnco 3 x 8 x 3 X 8 - 81. 3. The powers of a number are called i\iQjirst, secondy tldrd, fourth^ fifth^ <^t., according as the root is taken tiiice^ twice^ thrice, four times, five times, = 2 X 2 X 2 X 2 X 2 = 32 = Fifth power of 2. So also B'' means the eeventli power of 6; i. e., a number produced by takiny S beven timob as factor, &,c. 7. (5 + 8)'' means that the sum of 5 and S is to be squared as one number, and is a very diffcront tiling from 5" + 8^, which means the sum of tlie squares of D and 8. Tlius (5 + 8)2 = 182 _iG9, while 6' + 8" - 25 + 64 = 89. Therefore, (5 + 8)* = 25 + 80 + 64 = \s'c part squared, ^;JMfl twice product qf Aat part hy 2nd part, plus 2nd part squared. 8. The process of finding- a power of a given number by multiplying it into itself is called Involution. 9. To involve a number to any required power : — RULE. Take the yiven number as factor as mariy times as there are nrdts in the index of the reiiuired power aiid find the continued product of these factors. NoTK. — Fractions are involved hy nndtiplying both numerators and denominators as above, and mixed numbers should be reduced to fractions before applying the rule. Example 1. — What is the fifth power of V ? OPn RATION. Here tho index of the required power is 5, and henco the given number 7 must bo taken 6 times a.s factor. 7x7x7x7x7 = 16807 Ans. Example 2. — V/hat is the third power of f ? 27 Am. (J)3 — i-x-i-x-i— „. Ana. 64 Exercise 134. 1. Find the fifth power of 3. 2. Required the tenth power of 20. 3. Required tho si.ith power of \'0^. 4. Find the seveijth power of f. 6. Find the fifth power of f. ^, Required the third power of llf . Ans. 248. Ans. 10240000000000. Ans. 1-840095640625. Ans. fi^^^h- Avs.^^^ii^^xm.-^ [Sect. X. I AtmC-M.] •fivaLUirow. 209 > called its cube ; he expressed by cube will be ex- •. (See Art. 04, s a small figure I the root has to given power. •of 2. CM- of 2. rof K. er of 2. of 2. Dumber produced by iared as one niiinher, ic sum of the i^qiiiucs 9. ', plus twice product ' given number ition. power : — as there are vnifs inued product of both numerators mid be reduced to the given number 7 lu- J ; * ' luired to find the procfuct of 4' by 4^. 10000000000. :0095()40625. 7-- = 1481 ^^%% 4»=4x4x4 ., =4' = -4 X 4. Therefore 4» x 4="=(4 x 4 x 4) x (4 x i)=4 x4x4x4xi Hence ^o or more powers of the same number are multiplied .-^gether uy adding their indices or exponents. X O-J X 5» X 5' =5' + » + =» + » =6»», &c., &0. 11. Let it be required to divide 3" by 3^. 8" =8 X 3 X 3 X 3 X 3 and 3» =:3 X 3. 3» __ 3 X 3 X 3 X 3 X 3 ';r»~ 8T3 Therefore, 3»-r^»=- =' 3x8x3=8»-3'-». Hence, to divide one power of a number by another power of the same number, we subtract the index of the divisor from the index of the dividend. Thus, 7''+7»=7»-«=7» 8»^-T"8'^-3'^-*=3% &c., Ac. 12. Let it be required to find the third power of 7*. (:'')»=7^ X 7" X P=7 X 7 X 7 X 7 X 7 X 7=7''=7» X », Hence to '^nd any required power of a given power, we multiply the index of the given power by the index of the required power. Thus, (2')'=2* X 0=2=°; (3»)^=3'> x "=3", &c., Ac. Exercise 135. 1. Multiply together 4^, 4*, 4", and 4'. A?is. 4'». 2. Divide 13' » by 132. . ^ns. 139. 3. Find the fifth power of 3'\ Ans. 3' «. 4. Find the value of {{1* x 1'^)-h{1^ x1^)\^ Ans. 1^^. 6. Find the value of {(5^ x 6* x 6" x 5-*)-i-(5^ x 5^ x 5^ x 5")^=*. Ans. 5^*. EVOLUTION. 13. Evolution is the process of finding any required root of a given power. Note.— Evolution is the reverse of involution ; the latter teaclies how to find a power of u number by multiplyine: it into itself; t'uo former, how to And the root of 11 power by resolving it. into e(juiil fuctom. It follows tlmt powers and roots nre corrolatiVo terms. If one number is a power of another the latter is a rcot of the former. 14. A root of a number may be indicated by either of two methods. ;« t! ;i it*' 800 8QUAEE ROOT. [Bkot. X 1st. By using |/, called the radical sign (Lat. radix, a root). 2nd. r>y using a fractional index having unity fur i ,-< numerator, and the number expressing the degree of the root for denominator. Tbiis, The Bqimre root of 7 is expressed eitlior by ^/ 7 or by 7^. Tho wibc root of C i8 " " V 6 or by CJ. Tbo seventh root of 2 is " " V 2 or by 2». Note.— Tho fitruro placod in tho radical sign, or asdunoininator of the fmc tlona! index denotes tho root. A fractional index with numerator greater than one in sotnotiniesnscil; in Bncli easi-b tho dcnoruinato' denotes the root^ and the nuuieraior the power to be tui'' or 101^^ 15. Let it be required to oxtract the fifth root of 3**. The l^fth root of 3'* is e.-^pressed eif uer by Vs^S or by Sy. Taking the latter mode, we have 8"*«*=8»=3»»-i.» Hence, to extract any root of a given power of a number, we divide the index of the power by the index uf the root. Thus, Tho seventh root of 2»* is 2"-r-7=2» The fourth root of 2*" is 2' "—4=2% «tc., &c. EXTRACTION OF THE SQUARE ROOT, 16. To extract the square root of a number, is to find a number which, being multiplied once by itself, will p.o- [luce the given number, fUT'- •'«■ ,inmng\ Jf. qiuitient III. hand p{ /r.' V. dend, ca tained VI. ih root hi'lnq d< Vii triat. d until cil NOTl cannot I) quired a[ nti .1«.J fiOtTARE Roof. 801 'er of a nuTiiber, ]cx ui the r(j(jt , &C. :ooT. aber, is to find itself, will p/o- RULE. /. Point off (he given number into periods of two Jigtires each c /nning at tJh. aecimal point, II. Find the higheHt square contained in the le/t-hund pf''n> and li'ace its root to the right of the number^ in the place occii ■■.; Oy the {jiiotient in division. III. Subtract the square of the digit put in the root^j '-n the left- hand period, and to the retnainder bring down the next period to the iht, for a new dividend. I\ . Double the part of the root alreadg fo^md for a trial divisor. V. Find hoio many times the trial divisor is contained in the divi' dcnd, exclusive of the right-hand digit, and place the figure thus ob- tained both in the root ond at, > to the right of the trial divisor. VI. Multiply the divisor thus completed by the digit last put in 'he root ; subtract the product from th^e dividnid, and to the remainder bring down the next period for a new dividend. VII. Again, double the part of the root already found for a new :riat. divisor; proceed as in V. and VI. y and continue the process until all the periods are brought doivn. NoTK.— If the given numbor in not n perfect Fqnnro. its pxnct sqnaro root cannot be "o.md; but by annexing periods of ciphois, we can obtain any ic* quired approximation to it. Example 1. — What is the square root of 22420225 ? .... Explanation— TTei 22420225(4735, Is tho required root. 16 87)642 609 948)38i>2 2S29 0465H7325 47325 Explanation —Here 9.2 is the loft hand period, and tiie highest sqnm. in 22 is 16, of whicli the square roct Is 4. We place 4 in the root ari aubtract 16 tVoiii 22. This leaves a remainder6, towliich wc bri, trdir/n the next period, 42, and thusob din 642 for the new dividend. Onr /;xt 61 ep is to find the triril <1ii (toi\ which we obtain by doiiblir tlie part of the root ali-eadj' found Tin's feives us 8, (= 4 doubled) and ?e ask how many tim-'H 8 will po nto 64 (the dividend exclusive of the riffht hanfl dii^it). Bearini: in mind thr wo are to put the digit tl)U3 obtainef' both in the root and in the divisor, and Chat the completed divisor will be ove» 80, we find tliattho required diuit s 7, ^hlch wo accordingly place b<»th in the root and in the divisor. The complete divisor is 87, whicli multiplied by 7, gives 609, and this subtracted from 642, gives a re- mninder 33, to wiiich wo bring down the next period, 02, and thus get 3302 for t!\e next dividend. Ajraiii, doubling the part of the root already found, we obtain 94 ( = 47 liiilik'il) for a trial divisor, and as this will go into 330 (tlie dividend exclusive of 'lie right ha d digit) 3 times, wo place 3 both in the root and in the divisor. Multiplying the 943 thus obtained by 3 , subtracting and bringing down the ni'xt period, we get 47325 for the next (lividend. The next trial divisor is 946 ( - 4"'3 doubled) which will go into 47:^2 (the dividend exclusive of the right hand figure) 5 times; and wo therefore place 6 both in the root and in the divi- sor, Multi|»lying and subtracting, we find no remaiuder ; 4W6 is therefore tb« square root of 22420225. FBOOf .--4T85 X 4785 = 2242022ft i i 80^ «QITARE Root. titer. X J BXPT.ANATTON AND nEAflON. 17. '^<> miiy cnnsidor t-vi-ry miiiilnT as conbistinu <»f It. f«'«*, plus Its 7 scau-d by ihe letter a and the UDitti by the Ictttr b. Nuiiibv T + ft;' Numbc. rod = (a + ft)' =: a' + 2a6 + 6'. Hence the square of »i number is equal to the square of the tens, phis twice the product of the tens by the units, plus the square of the units. Thus, fi9 ■--- 60 + 9 And (OU)''' = (60 + 9)' = (fiO)" + 2 + 60x9+9' = 8C0O + 1080 + 81 = 4701. 18. Let it now bo required to extract the pquarc root of 4701. I. It is evident thiit tlu; s(ninre of ft niiinbLT consistinc of n sinfrlo dijrit can never coiitiiin more than two dii'ltsor less ihitn one; converholy the square rodt of a number of one or two difiits niu.st bo a number of one digit. Again tlio square of a number eoiihisliiip of two dii^its can never cy thn miitfl. plus the nqiiaro of tho unlt.s; and sinco wo havo sub- trii';t<'(l from it M, (or if tlio ciphtTH be annoxod JJfiOO) tho square of tho tens, tiio iciniiinili r, 11<51, must pontain twice tho nrodiict of the tens by the unitH. |ilii.s tliii sqtiiuc of tiie units; that Is, twien o tens x by a certain number or ii;iit^4, plus tlio bquiiru of th:it nninhor of units ; and because wo do not know [i-i yet what the unlth' figure of the root is, wu use twice the tens for a trial divisor. IV. Since wo arc now ^coking tho units' dljrlt of tho root, and since tons nuiltiplied by units can tclvo no digit of a lower order than tens, tho right hand (i'^rit of tho dividend can form no part of twice tho product of tho tens by tho units, and we ha^e whiiply to ascertain how often 12 tcus (=twico 6 tens) will j;i» in Mt»tens. Evidently 9 times. V. Lastly, we place tho digit thus found In the root, because it is a flguro of tho r'>ot, and in the illvi-'or, because tho dlvldoner of units. I. Wo are first then to look for tho digit in the place of hundreds, and since hnndreils squared can give no digit of a lower order than tenn of thousands or of a liiglior order than hundreds of thousands, we see that th.' square of the hiindreils can be found only in the left baud period. The liigliest square con- tfiiiied in tlie left hand period is 9, tho square root of which is the left hand digit of the entire root. II. After subtiactintr, we bring down the next period only, because we aro now looking f(»r tho digit in the place of tens in tho root. And since tens squared can give no digit of a lower order tlian hnndreda, tlie lowest period cannot enter into any part of the square of ten.s, much less can it enter into any part of twice the product of tho hundreds by the tons, and therefore when look- ing for the tens of the root, wo pay no attention to tlie right hand period of the square. III. The remainder of the process Is similar, and the reason for the various steps tho same as in example 1. ' 19. To extract the square root of a decinial — RULE. /. Annex one cipher', if necessary, in order that the number of deci?nal places may be even. 11. Foint off" into periods of two figures each., beyirining at th6 decimal point, and extract ths sgiMV^ root as in whole numbers^ re- V U to, l|^: % $ U I ^ I i"'t I El I 304 fcQtARfe ftOOT. tSrot. JC mfmhering that the number of decimal placet in the root will be equd to the number of pcrioda in tne square. Exercise 137. 1. Kxtract the square root of 195304. Am. 442. 2. Extract the square root of -0676. A ns. 'ir.. 3. Extract the square root of 984064. A71S. 992. 4. Extract the square root of 5, true to five decimal places. Ans. 2-28GO0. 5. Extract the square root of *6, true to six decimal places. Ans. -707100. 6. Extract the square root of 00-487120. Ans. I'Til. 1. Extract the square root of 797922G(i'297Cl 2001. Ans. 282475219. 8. Extract the square root of 0-0000012321. Ans. 0*001 1 1. 20. To extract the square root of a fraction — • RULB. /. Reduce mixed numbers to improper fractions, and compnnvd and complex fractions to simple ones, and the resulting fraction to its lowest terms. II. Extract the square root of both mtmerafor and denominator separately, if they have exact roots ; but if they have not both exact roots, reduce the fraction to its corresponding decimal, by Art. 56> Sec. IV., and then extract the root as in Art. 19. Example 1. — Extract the square root of 2^, OPERATION. V9 Ans. 2i=Jand|/2= =i=U. Example 2. — Extract the square root of 3^. OPKRATION. 8|=V = 3-42857142 and V3-42857142=1-8616. Exercise 188. 4. Find the square root of '1. 2. Find the square root of 7^. 3. Find the square root of 5^ . 4. Find the square root of ^\. 6. Find the square root of 13|-. Ans. ^. An9. T^-. Ans. 2-2^^.786. Ans. -63509. Ans. 8'63318. 21. Let it be required to extract the square root of 63518'42S Hpt$nary. ^» — tSrot. Ji 'oot will be equri Aktb. 20-22.) APPLICATION OF 8QUAEE UOOT. 305 Ans. 442. Ans. -20. Ans. 992. places. Ans. )laccfl. Ans. 2-28000. •7OT1O0. Ans. 7777. ns. 282475249, Ans. O'OOIII. ion — ' and compnimA g fraction to its nd denominator ! not both exact al, by Art. 56, 616. ins, \. [ns, T^-. ns. 2-2G.786. ills. -63509. \ns. 8'63318. of 68613-429 OPBIIATION. f)30i:3 4230(230 155 + 43)'2:l5 4G0)i:'.lB 4161 5051)1'JV{4J S()51 60525;41«180 «4j<.'>6-t_ 505336)V2'2:«00 8^t{]6;i44 •453C2S ExPLAMATioK. — Wo point off Into periods of two plucos «>ucb, aa in the deciuial or coininoi) HCiilo. Then tbo hi;rlic.st miiiurt^ In 6, tliu flrnt period Is 4. of wLlcli tlio Mxt douhlinfr tho 2 we ubtuiii 4, and wc And that this will \lo Into 23, thfl dividend e.xcliLsivu of tho ri^ht hand ficnro, 8 times. Flacin^r this 3 In hoth root and (liviMur, multipiyiudr (heurinK in mind that 7 is tho common ratio of tho systcn)) and BubtrHCtinir, wo obtain a remainder of 43, to whit b we brinp down the next period, 18, aud thus i{U 4313 for tlid iioxt dlvidoudf «bc. Example. — Extract the square root of 4731392 undenary true to ;wo places to the right of the separating point. OPKRATIOM. 473i392(218299. 4 41)78 41_ 428)32 IS 80/9 An$, 4352)11592 86/'4 4354-9)3999-00 8.594'<4 ^ 4855'79)404-07()«> 8^^-^744 66-5«G7 Exercise 139. 1. Extract the square root of 11333311 septenary. ^^. Extract the square root of 33233344 senary. 8. Extract the square root of 4234-10123 quinary. 4. Extract the square root of 888888-888 nonary. 5. Extract the square root of 248G64e/G9 duodenary. APPLICATION OF SQUARE ROOT. 22. A triangle is a figure having three sides, and con. sequently three angles. When one of the angles is a right angle, like the corner of a square, the triangle is called a right angled triangle. Ans. 2626. I )p>' Ans. 4344. Ans. 43-412. .' Ans. 888-88. { '^ ... Ans. 54373. ^■,ii 1 .1 (1 ! ■ : ■) ' M - a.iu.i.-: n /■..;■ 80(^ APPLICATION OF SQUAPvE ROOT. [Skot X. if ;i,' 23. In a right angled triangle, the side opposite the rujlt angle is called the kypothenuse^ and the sides containing the right angle, are called the base and the perpendicular, tl^. It is shown by elementary geometry that the square described on the hypothenuse of a right angled triangle is equal to the sum of the squares described on the other tv¥o sides. Or If A be tho hypothenuse, h the base, and p the perpei:dlcular; then h^—b^ + jp", a nd hence A = V ft" + p ^ b =z 'sj h^ - p 9 , 77iat is — tojind the hypothenuse of a right angled triangle when th* other sides are gioen we add the square of the base to the square of the perpendicular and extract the square root of the sum. To fin'' the length of the base we subtract the square of the per- pendicidar from the square of the hypothenuse and extract the squats root of the remainder. lofind the length of the perpendicular we subtract the square cf the base from the square of the hypothenuse and extract the squa'-e root of the remainder. 25. The following principles are also established I7 geometry :— Circles are to each other as the squares of their diameters. If the diameter of a circle be multiplied by 3 "141 6, the produ (he sqtiare of half the diameter. Example 1. — If the hypothenuse of a right angled triangle ■ wire an inch in diameter? Ans. 45000 lbs. 12. What length of rope must be tied to a horse's neck, in order that he may feed over an acre? Ans. 7*136-|-perches. 13. Four men, A, B, C, D„ bought a grindstone, the diameter of which was 4 feet ; they agreed that A should grind ofif his share first, and that each man should have it alternately until he had worn off his share ; how much did each man grind off? Note.— In this question we disregard the thickness of the grindstone. Af- ter the first has ground off hla portion, thert will remain f of the ston«. ¥? tft" \0Q CUBE ROOT. ^<3«o». X Thett the whole stone : part remaining • • square c. u;,._oter of whole Atone : square of diumeter of part remaining. (Art. 25.) That l8, 1 : t : : 4» 05 ■" and hence a;=4 x Vt=4'< V'75=-866x4=3-4W^ dlametur of stone after tlio first has ground off his portion. Similarly, after the second has ground oif his portion there will remain J of tho btoue, and after the third has taken his portion, i of the stone. Hence 1 : ^ : : 4» ; »% whence fl?=4'\/i=^2-828 ft.=diameter after 2nd had taken his portion. 1 : i : : 4» : »•, whence a5=4x '\/j"2 ft.=diameter after 8rd has taken off bis portion. Hence A takes ofT 4-8464=-58fi ft. = 6-432 Inches. B " 8464-2-828=-68Gft.=7 6:32 inches. C •» 2-828-2 ='828 ft. =99«6 inches. D " remaining 2 ft.= 24 inches. CUBE ROOT. 26. To extract the cube root of a number is to iind a number which taken three limes as factor will produce the given number — BULK. /. Poi7it off the number into periods of three figures each begin- ning at the decimal point. II. Find the highest cube contained in the left hand period and place its root to the right of the number^ in the place occupied by the quotient in division. III. Subtract the cube of the digit put in the root from the left hand period^ and to the remainder bring down the next period to the right for a new dividend. IV. Multiply the square of the part of the root already found by BOO foi' a TRIAL DIVISOR. K. Find how many times the trial divisor is contained in the divi- dend and put the figure thus obtained in the root. VI. Complete the trial divisor by adding to it : 1st. Tfie part of the root previously found x the last digit put in the root x 30 and • 2w(/. 2ne square of the last digit put in the root. VII. Multiply the divisor thus cornpleted by the digit last put in the root ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. YIII. Again multiply the square of the part of the root already found by 300 for a new trial divisor, find what digit to place next in the root as m V, complete the divisor by making the two additiont to the third divisor described in VI, multiply, subtract and bring down as directed in, VII, and continue the process until all the periods art^ brought down. Arts. 25-28.] CUBE ROOT. 309 rd has taken off Example.— What is the cube root of 4291V2932007 ? OPEBATIOX. Ist trial divisor = 7» x 300 1st. increment = 7 x 5 x 80 2 lid " = 6» ist complete divisor \!iid*rial divisor = 75' x 300 1st increment = 75 x 4 x SO 2nd " = 4» 2nd complete divisor = 14700 = 1050 = 25 = 15775 429172932007 848 86172 = l8t dividend. |7548 Ana. 1GS7.VJ0 90f'i0 IG 1696010 78S75 = product of comp. dlv by 5. 7297932 = 2nd dividend. C78C0C4 = product of comp. dir. by 4. 3rd trial divisor = 754' x 300 = 17050 tSOO 1 St increment = 754 x 8 x 80 = 67860 2nd " = 8'= 9 511868007 = 8rd dividend. <3id complete divisor = 170622669 611868007 = product of comp. div. by 3. Explanation.— After pointinpr off we find that the hiirhest cube number conti'ino ( in the lift hand period is 81-J, of which the cube root is 7. Wo there- fow iiliice 7 in Ihe root and subtract '64-i from the first period. This gives us a rcniaiiuler of t)6, to which we bring down the next period 172, and tlius obtain 80172 for H new dividend. Next we tiilce 7, the part of the root already found, square it and multiply the 49 tlius obtained by 800, this gives the first trial divisor 14700 which we find will go into the dividend 861^2 ^malsing due allowance for the increase of the divisor) 5 times. Next we complete the divisor by adding to It 1st, 7 X 5 X 30 = 1050, and 2nd, 5' = 25 which gives us 15775 for n complete divisor. This wo multiply by 5, the digit last put in the root, subtract the product 78875 from the Ist dividend, and to the remainder 7297 bring down the next period 932, &c., Ac. 27. Explanation and Reason.— "We have seen (Art. 17) that wo may consider every number as consisting of its tfMs, plus its units^ or if a = tens and h=. units, then Numbcr=flr + A,' and Number cubed = (u + 6)» = a» + Sa'S + odb"^ + &'\ Hence the cube of a number is equal to the cube of the tens, plus threo times the product of the tens squared multiplied by the units, plus three times the product of t!ie ti ns multiplied by the square of the units, plus the cube of th».> units. Thus 69 = (e)0 + 9); and 69" = (60 + 9)» = 6o» + 3 X 609 X 9 + 3 X 60 X 99 + 9» = 21600 + 97200 + 14580 + 729 fc 828509. 29. Let it now be recjuired to extract the cube root of 328809, 310 CUBE BOOT. [Sect. X, I. It is manifest that the cnbe of a single dipit can never contain more thnn three digits or less than one digit, and hence the cnbe root of a laimlHT (i. c,, perfect cube) of one, two or three digits niiist be i\ number of one digit. A^Miii the cube of a number consisting of two digits can never contain more than ., the digit in the units' place will frive (ha'' + 3rt6 + 6") ft =:3a»& + 3rti' + 7>» = the dividend. Now (6 tens) x8=(60)x 3=6 X 10x3=6x30, i. e., the product of any number jf teiis multiplied by 3 is equal ^o the product of that same number of units multiplied by 30. Hence we obtain tho constant multiplier 30. The additions we make then are 6x30x9=1620, and 9'=81, and thus we obtain the complete divisor 125Ul=(60)» x3 + 60x3x9 + 9',and multiplying thia by 9, we get ] (60)' X 3 + 60x3x9 + 9'} 9=60' x3 x 9 + 60 x3x9» + 9»=three times the square of the tens multiplied by the units, plus three times tho tens multiplied by the square of the units, plus" the cube of the units. Note. — When there are more than iwo periods, the reasons are analosrous, since we never have to do with more than tens and latiti of the root at one time ; i. e., when we arc seeking the second digit of the root, we call the first digit tens and the second, units; when we are seeking the third digit of the root, we consider the first two as so maiiy tens, and tiie third as unit.>». &c. Tht' reason for bringing down only one period at a time is similar to the reason for the same step in tho extraction of the square root (for which see Art, 18, E.xample 2). 29. To extract the cube root of a decimal — RULE. /. Annex two ciphers, if necessary, in order to make the last period cotnplete. II. Point off into periods of three places each, beginning nt the decimal point, and extract the cube root as in whole numbers, r^-mem- bering that the number of decimal places in the root will be equal to the number of periods in the cube. 1. What is the cube root 2. Extract the cube root 3. Extract the cube root 4. What is the cube root 5. What is the cube root 6. Find the cube root of v. Find the cube root of 8. Find the cube root of Exercise 141. of 62712728317? of 1953125. of 1076890625. of -697864103? of 102503-232? 179597-069288. 483-736625. •636056. Am. 3973. Ans. 125. Ans. 1025. Ans. -887. Ans, 46-8. Ans. 56-42. Ans. 7-85. Ans. -86. 30. To extract the cube root of a mixed number or a Vulgar fraction — RULK. Redme mixed numbers to improper fractions, and compound or 312 CUBE EOOT [8iCT. X m I complex fractions to simple ones^ and the resulting fraction to its low- est terms. II. Extract the cube root of both numerator and denominator separately, if they have exact roots ; but if they have not both exact roots, reduce the fraction to its corresponding decimal by Art. 5(1, Sect. IV, and then extract the root as in Art. 29. Example 1. — What is the cube root of 3| ? OPERATION. V8:- = V-y- = F^=5=ii- ^«*- Example 2. — Extract the cube root of IT^. OPBRATION. 17^=17-125, and Vn-125=2-C77, nearly. Exercise 142. 1. Extract the cube root of ^. 2. Extract the cube root of iV 5. Extract the cube root of \ of 2\, 4. Extract the cube root of 28f . 6. Extract the cube root of 32-i\. 31. In extracting the cube root of a number in any scale, other than the decimal, we proceed in the same man- ner, pointing off into periods of three figures each, finding a trial divisor and afterwards completing it as in the pre- ceding examples. Note.— In all scales having a radix higher than 3, the constant multipliers are 800 and 30; but as in the hinary and ternary scale we cannot use a digit so high as 8, these multipliers become respectively 1100 and 110 for the binary •cale, and 1000 and 100 for the ternary scale. Example 3. — Extract the cube root of 613412*132 septenary, OPBKATION. 613412-182(65-04 426 Ans. •4721. Ans. •5609. Ans. ■941. Ans. 3-0G3. Ans. 3-198. 6» =51x800=21300 6x80=240x5= 1560 6»= 84 23224 65''=6804x300 = 2521500 650!" =680400 ^ 300 =252150(;00 650 X 80 = 26100 X 4= 143400 4«= 22 $U$2223422 154412 152456 1623-132 1628-182000 1402-630321 220201846 IRT8 3l-fl2.J CUBE ROOT. 313 action to its low- fj'lCERCISE 143. I. Express one million in the senary scale and then extract its cube root. Ans. 244. Extract the cube root of61312'71 odenary. Ana. 166-32. .3. Extract the cube root of 10221012-102 ternary. Ana. 112-012. 4. Extract tlie cube root of teteet in the duodenary scale true to two places to the right of the separating point. Ans. e1-t2. 5. Extract the cube root of 421030-4412 quinary true to two places to the right of the separating point. Ans. 44'004. 32. Slrsce many tencliers prefer Horner's method of exttftctinp tho cnbe root to the common method, we shall give it here. Upon closely examining it the stu(l(*nt will And that the reasons for the several steps of the process are iiltMitical with those friven in Arts. 27 and 28. The constant multipliers SOU and ciU aiu still used, but in a disguised form. istant multipliers RULE. /. Point off as in the common method. II. Find the greatest cube in the Jlrst period on the left hand ; •olac". its root, on the right of the number for the first figure of the root, and also in col. I. on the left of the number. Tlien midtiplying this figure into itself set the product for the first term in col. II. ; and multiplying thii^ term by the sam^ fgure again, subtract this product from the period^ and to the remainder bring down the next period for a dividend. III. Adding the figure placed in the root to the first term in col. L, multiply the sum by the sam^ fg^^i^c^ (M the product to the first term in col. II., and to this sum annex two ciphers, for a divisor ; also add the figure of the root to the second term of col. I. IV. Find how many times the divisor is contained in the dividend^ and place the result in the root, and also on the right of the third term of col. I. Next multiply the third term, thus increased by the figure last placed in the root, and add the product to the divisor ; then multiply this sum by the same figure, and subtract the product from the dividend. To the remainder bring down tJie next period for a new dividoid. V. Find a new divisor in the same manner that the last divisor vms found, then divide, d'c, as before ; thus continue the operation till the root of all the periods is found. Example,— What is the cube root of 7S314'6, true to two decimal places, 3U APPLICATION OF THE CUBE ROOT. [Bbot. X. OPKRATION. i I Col. I. Ist term 4 2nd 3rd 4\h '• 8 " 122 " 124 6th " 1267 6th " 1274 OoL II. 16x4 = 4800, iBt divisor) 6044x2 = 78814-600(42 78 + . 64 14814 10088 629200, 2nd divisor) 688069x7 = 7th " 12818 64698700, 8d divisor) 64801244x8 = 4226600 8766488 460117000 438409952 Explanation.— Tho cubo root of the groatrst cube In 78 is 4, which is placed ill the root and also in column I, then multiplyinp this 4 b.v itself jrivos us 16 whioh iH the 1st term in column II, and a^ain niuliinlyinpthis 16 hy 4 gives us 64, the niimbor which we are to subtract from the nrst period 78. Subtracting and brii.ging down the next period 814 we get 14814 for tho next dividend. Now addintr 4, tho figure placed in tho root, to 4 tho Ist term in col. I. pivos U3 8, tho 2nd term in col. I, multii)lyinir this 8 by the 4. i. e., the figure iv t!io root, gives us 32 which vvo add to the 1st term of col. II, and aflfix two cipliers. We thus obtain 48i)() the second term of col. 11, which is our trial divisor. We then find that 4800 poos 2 times in the dividend. This 2 we place in tho root and also to the right of the sum of the 1st and 2nd terms of col. I. The Ist and 2nd terms of col. I, added together make 12 and the 2 of the root allixtd makes 122, the third term of col. I. Then we multiply this 122 by 2, the last digit put in the root, this gives us 244 which we add to -ISOO, the second term of col. II. and thus obtain 5044, the 8rd term. Lastly this third term multiplied by 2, gives us the number to subtract. Note —For examples in this method work any of the preceding questions. v"* APPLICATION OF THE CUBE ROOT. 33. Principles Assumed. — /. Spheres are to one another as th cubes of their diameters. II. Cubes and all other regular solids are to one another as the cubes of their like dimcndons, EXEUCISE 144. 1. If a cannon ball 3 inches in diameter weighs 8 lbs., what will be the weight of a ball of the same n^etal 4 inches in diameter? 8^ : 4=^ : : 8 lbs. : Ans. = \^^ Iba. 2. If a ball 3 inches in diameter weighs 4 lbs., what will be the weight of a ball that is G inches in diameter? Ans. 82 lbs. 8. If a globe of gold one inch in diameter be wor*i> $120, what is the value of a globe 3^ inches in diiiraeter? Ans. $614;'). 4. If the weight of a well proportioned man, 5 fei . 10 inches in height be 180 pounds, what must have been the weight of Goliath of Gatb, who was 10 feet 4^ iuches in h^U^ An^ J0161 lbs. Arts. 33, 34.] ROOTS OF niOHER ORDERS. 315 f). A person has a cube of clay whoso sides are 973 ft. long ; ho wishes to take out of the same 5 cubes whose aides are 45 feet, 62 feet, ;)0 feet, 80 feet, and 20 feet. He requires to know the length of tiie side of the cube that can bo formed out of the re- maining clay. Ans. 972'()9 ft. 0. Wliaf is the side of a cube which will contain as much as a chest 8 feet 3 inches long, 3 feet wide, and 2 feet 7 inches deep ? Ans. 47-9843 inches. 7. Four ladies purchased a ball of exceeding fine thread, 3 in. in diameter. Wliat portion of the diameter must each wind off so as to share off the thread equally ? Ans. 1st lady mu^ wind off '27432 inches. 2nd " " '34458 " 3rd " " -49122 '* *. 4th " " 1-88988 " Note.— This question is solved by a method similar to that adopted in Example 13, Exercise 140. IP EXTRACTION OF THE ROOTS OF HIGHER ORDERS. 34. When the index of the root is a power of 2 or 3, or a multiple of any power of 2 by any power of 3 — iceding questions. }., what will be RULE. Resolve the given index into its prime factors. Extract the root denoted by one of these factors^ then of this root, extract the root denoted by another factor ^ and so on till all the prime factors be used. Thus, for the 4th root extract the square root of tho square root. for the 6th root extract the cube root of the square root "" for the 8th root extract the square root of the square root of the square root, for the 12th root extract the cube root of the square root of tlid square root. for the 16th root extract the square root four times. for the 18th root extract the cube root of tho cube root of the square root, &c., «bo. EXKRCISK 146. 1. What is the fourth root of 19987173370 ? 2. What is the sixth root of 308915776 ? 3. Extract the ninth root of 40353607. 4. Extract the eighteenth root of 387420489. 5. Extract the twenty-seventh root of 134217728. Ans. 376. Ans. 26. Ans. 7. Ans. 8. Ans, 2, n^ 1016-1 lbs. 316 LOGARITHMS. [8lCT, X LOGARITHMS. 35. The Logaritlira of a number is the index of ilie power to which it is necessary to raise a given root or base, in order to produce the given nuraber. 36. The Base of a system of logarithms is the f.rrd mtmher to which all the logarithms of that system belong as indices. Tluis 10» = 1000; horo 3 ts callod tho loffarlthm of 1000, to tho hnso 10. So also 2* =32; hero 5 Is called the logarithm of 3'2, to the biiso 2, Ac, ,lf. 37. A System of Logarithms is a collection of die logarithms of a series of numbers corresponding to the same base. « Any number wlifttover may bo. taken as the hrx'P of the pyptcm ; but it is obvious Uiat somo numbers are'much more convenient than otiiers. 38. Two systems of logarithms have been constructed and tables calculated with great care. They are — 1st. The Common System or Briggean System, whose base is 10. 2nd. Napierian System, whoso base is 2*71 828. The Napierian System was invent?d by Baron Napier, and the pecii'ii.r hn^e 2'71828, was adopted chiefly beeause the lojriirithms having that base are iiioio simply expressed and more' easily calculated than any other. It has hence hmi called tho Natxi-dl System of LojraritLms. These 'Logarithms were also tm- nierly called Hyperbolic logarithm.s, from certain nilations found to exist tM> twee'n them and the asymptotic spaces of thy hyperbola, and which wen* "erroneously believed to be peculiar to them. The CoiTimon System was sliortly afterwards Invented by Bripps nr.d adopted by Baron Napier, and is tho system now ualvcffcally employed for thu purposes of calculation. 39. The Characteristic of a logarithm is the part whicb stands to thq left of the decimal point. 40. The Mantissa (liandfid) is that part of the log.v rithm which stands to the right of the decimal point. 41. Since 10 is the base of the common system of logarithms and at the same time the radix of our systora of notation, we have — 00000 = 10'; whence log. 100000 = 5 10000 ^ 10^ whence log. 1(000 SIS 4 1000 = lO"; whence log. 1000 = 8 100 = 10»; whence log. 100 ^ 2 10 r= lOS whence log. 10 = 1 1 ;s 10"; whence log. 1 = •1 r: 10-»; whence log. •1 r= -1 •01 = 10-"; whence log. •01 rr -2 •001 z^ 10-»; wli.nce log. •001 r= -8 •Opo; 5= 10-*; whi'i cc log. •0001 =: -i index of tlie ,i. 8,^-U.j L00xVniTII.\t8. S17 42. From thin It nppoarH that the lotfarifbm of any number between ! and 1 ) ivili [>~' iiioi''* lli.iii <) 1111*1 i'.'s.'^ til III 1 , i. o., will bu n frii tiofi or a (lecitji'ti ; »o u:-it tho |.i;iiir,ihin of any nimilnT hetwooii 10 and 100 will be tfreHtor tbMn 1 Hii I i>"«i4 t lun 2; 1. »., will bu 1 ami a fraction, ur a (iocimal; so »)«<> tjtbe lugA* ri'.iiiii ut'atiy number between lOU and lOiU will bu 2 and a duciuial, A«. Hence, the charaeteristic of any number containing ili/ltH to the left of the decimal point is posit ve and nu- nierieally one Ics^ than the number of such digits. Thus th(> nharactprisfin of 7842 la 3; of 973'2C It id 2; of $313426789 It id 8 'A ot'GOKtiJ it U t>; of 2075J l2GTStf it Is 4, Ac. ' 43. It iilso «|»ii<'ars, from Art. 41, that the logarithm of every number bo- twi'i'ii I and 'l will be less than and ik'reater than —1 ; that 18, it will bu equal tn - 1, /iluf) some decimal ; the logarithm of uvery number between '1 and "Oi will be less thiin —1 and urealer than —2; or. In other words, will be —2 plitt' !ii):iie decimal; so al^o tiie loirarithm of overy uumbur butweou '01 and '001 will be —8 plus .so>iie decimal, &c., «bc. Hence, the characteristic of the logarithm of a decimal is negative and numerically one greater than the number of Os which come between the decimal point and the first bi'y'uiticant figure. Thiia, the characteristic of the logarithm of '000001 is G; the characterintio of the lon:arithm of 'O0OU000U002J47 Lsll; the characteristic of the logarithm of •UJ027i92ti345 is 4, Ac, Ac. NoTK. — TTie negative sign affects onli/ the characteristic — the mantissa or decimal portion of a logarithm is ahvays positive. To indicate this it is customary to write the negative sign over the charac- krintic^ as in the above examples^ and not before it. Exercise 146. What aro the characteristics of the logarithms of the following numbers : 1. V23, 912C-4, 81234'567, 912678-96124567, 23-912342. Ans. 2, 3, 4, 6, and 1. 2. -027, -002134, -000000098, -8120714, -00000000021 34. _ _ Ans. 2, 3, 7, 1, and 10. 8. 1-1111111, 111111-11, 1000000000, 000000002162, 7^12-78. Ans. 0, 5, 9, 9, 0, and 1. 44. Since (Art. 11), to divide one power of a number by another power of the same we subtract the index of the divisor from the index of the dividend, fctil since common logarithms are indices 'o the base 10, let us take the number 4I2SO and successively dividing it by 10, examine the results. Numbers. Logarithms. 47230 = 4'674677 4728 = 8-674677 472-8 = 2-674677 47-28 = 1674677 4 728 = 0-674677 •4728 =1-674677 •04728 = 2-674677 •004728 .•••••iiMtiii = 8-674677 •, f: M: Ut di8 tOOA^ITHMd. tdicT. X. Hero we have simply nerformeii the samfi operation by ♦wo different tnotl\. odd, lat, dividing the iinmoerH by 10, iiiul 'iiid, from the logat'UUma corrfsiiond- Ing to too uuinbois, subtracting 1, thu loguritlim of 10. From this illustration it is evident that, — Ist. The characteristic of 'he logarithn\ of a number is dependent wholly upon the position of the decimal point in that number, and is not at all aflfected by the sequence of the digits that compose that number; and 2nd. The Mantissa or decimal part of the logarithm cf a number is dependent wholly upon the sequence of the digits that compose that number, and it is not at all affect- ed by the position of the decimal point. Note. — It is only common lof^aritbros (i, e., tbose bavln^c 10 for their base) that possess the important property of havinft the same mantissa for tbo same flgnre, wliiitber integral or decimal, or both, and it was this property tlmt iu- duced Briirgs to adont that base in preference to the Napierian base, 2"71828. 46. Sincj the onaractorisilc of the logarithm of any number does not de- pend upon the value of the digits composing that number, and is so easily found Dy attention to the rules fouml in Arts. 42, 48, it is customary to omit it alto- gether ill logarithmic tables, and merely give the mantissa. The annexed tables contain the logarithms of all numbers from 1 to 10000 calculated to 6 decimal places. When greater accuracy is required, tables cal- culated to a greater number of places I're used. By means of the proportional parts and difference given in the tables, the logarithm corresponding to ull nurabors whatever, may be found with sufficient accuracy for all practical pur- poses. 46. To find the logarithm of any number not greater than 100 — RULB. Find on the first page of the table of logarithms , the given number in the column marked No., and directly opposite to it, — in the column marked log., will be found the logarithm. Example l.—What is the logarithm of 47? Ans. 1*672098. NoTB.— By saying that 1'6720','8 is the logarithm of 47, we simply mean that the base 10, raised to the power 1'672098, is equal to 47, or briefly Example 2.--What is the logarithm of 93? Ans. 1-968483. 47. To find the logarithm of any number consisting of not more than four digits— RULK. Find, in the column marked N^ the first three digits of the given nnmber. Then the mantissa will he found in the intersection of the hori- zontal line containing these three digits and the vertical column at the head of which stands the fourth digit. To this mantissa attach the characteristic as found by the rules in Arts. 42, 48. ♦^i' Aii-m. 4&-43.) LoOaritumi did II Ans. 1-672098. lAns. 1-968483. ler consisting KxAMPr.E 1.— What is the logarithm of 7988? L>.>)klii« ill the 0(>l;iiiin niarkod N, wo flixl tho first thrco dIuit.sTOS, on piifW S93 in tliu fuiirth liorizuutal ilivlsion, uoiintiii); troin the top of the pa;;^ iind in the lu-^l lino but uiio uf that divlHiu i. Curi'yiii;^ thu oyc uloni; this horizontal lino till wo cutnu to thu verticul column, at thu head of which ntamh thu ru« iiiuinlng diiiit, 8, we obtain for tho mantissa of thu requirod loicarithin •'JO'iUJtt, lo which we protix the characteristic 3 (sluco there are four dibits to tho left of the dociuial point in thu givon number), and thus obtain thu required logarithm SUO.'l (50. Example 2.— What is the logarithm of -00000012^4? Tlio first three digits, viz: 123, are found in tho fourth lino of tho third liorizontal division on page 382, and at thu tntorsec-tion of this line with tho column lieadod 4, la found •091315. To this we attach tho cliaractcristie 7, (since there aro aioe Os, between thu decimal ])oint and thu first significant fig« ure) and thus obtain tho required logarithm, 7*001315. Exercise 147. 1. What are the logarithms of 6794, 67'94, 6794000, and -0005794? Ans. 3-7G2978, 1 -762978, 0-762978, and 4-762978. 2. What are the logarithms of 1-169, 11690, and TuHfoV-'jTj ? Ans. 0-067815, 4-067816, and 3-067816. 3. What are the logarithms of -734, 7340000000. and -0000000O734? Ans. 1-865694, 9-866696, and 9-865696. 4. What are the logarithms of 978-4, 9-784, 978400, and -9784? Am. 2-990516, 0-990616, 6-990616, and 1-990516. 48. To find the logarithm of a numher containing more than four dibits — RULE. First Method. — l^nd the maydissa corresponding to the loga- rithm of the first four digits by the last rule. Subtract this mantissa from the next following mantissa in the tables. Multiply the differ- ence thus obtained by the remaining digits of the given number, and cut off from the product as many digits as there were in the multiplier {but at the same time adding iinity if the highest cut off be not less than 6). Add the number thus obtained to the mantissa of the logarithm corresponding to the first four digits^ arid the result will be the man- tissa of the given number. Lastly attach the characteristic to this mantissa. Example 1.— What is the logarithm of 53803-2? OPERATION. The mantissa of the logarithm of 5380 (the first four digits) is •7307S2 ond the next following mantissa is -TSOSeS. Then from -780863 Bubtract -730782 ^t. 11 "niffereDce 81; and 81>(82 ^rtmalning digits of given number) 620 LOGAKI'TUMa tSEOT. X. e=2502, ii jva which we cut off two digits, efnco we multiplied by a numb'T of two digits, and eiace the liighest digit cut off is not lesa than 5, wo add unity u the part retained, whiuli g.v^es ua 26. Then mantissa of logarithm of first four digits '780782 Add 26 Mantissa of logarithm of given number -730809 To which attach the characteristic 4 and required logarithm =4-730308. KoTE.— Except at the beginning of the tables, where the mantissas increase '•i;ii(lly in magnitude, the difference maybe taken from the right hand column- (iiiiided D) and opposite the first three digits of the given number, where tin; UieuD difierenco of the mautiasas In that line will be found. Example 2.— What is the logarithm of 832-17242? CPERATIOK. Mantissa of logarithm of 8821 •92017* Dififercnce from column D=62 : and 62x7242=876684 from which we cut off four digits and add 39 •9202 M To which we attach the characteristic 2 and required logaTlthra=2-920214 49. The difference given In the column headed D in the tables, is that du» to an increiuent of one unit in the fourth figure of natural number, thus Logari th m of 6788 8-758761 Logarithm of 6789 8 - 758836 Difference of natural numbers=l; difference of logaTithms=75 • And since it Is shown In common works on Algebra that, with email incre- ments in the natural numbers the lorarithnis corresponding to them increase in arithmetical progression, in order to find the logarithm of any number between those given above, we consider that the increment of the logarithm to be adtled to 8'76d761, bears the same proportion to 76 (the increment for 1), that the in- crement of the natural number does to 1. For example. — Let it be required to find the logarithm of 5738*47. Hero the increment of the given number being -47, we form the proportion 1 : ^47 : : 75 : •47x 75=85^25, the increment to be added to 3758761, and this ad- dition having been made, we get 8*758796 for the logarithm of 5788-47, Similarly, if the increment of the natural number had been -047 or •0047, tlife corresponding increment of the log. would have been 8-525 or -8525. These illustrations sufficiently explain the reasons of the last rule. 60. Taking the same number as In the last article and dividing the differ- ence 75 by 10, we obtain 7-5 the difference corresponding to an increase of one unit in the fljfth place of the natural number ; the double of this, or 15 for two units, the treble or 22-5 for the three units, and so on ; and each of the num- bers thus obtained will be the increment of the logarithm corresponding to an increase of that nninber of units in the Jlfth place of the natural number. The increments thus obtained, and corresponding to each of the nine digits, are in- lerted in the left hand column of the tables, headed P. P. (Proportional Parts ) 61. The numbers In the column headed P. P., as already explained, ar^s the Increments in the logarithm for an Increase in the^fth place of the naturnl numbers. They express also the increments for the digits in the aiaeth, seventh, eighth; ninth, Ac, places of the natural number, when they are divided by 10, 1Q<), 1000, &C., as the case may be. 62. Hence to find the logarithm of any numher con- taining more than four digitsp— A.KTS. 49-53.] tSEOT. %, lied by a numbor ol m 5, wo add unity to I -780782 26 r •730809 ,rithm=4-730308. he mntitissas increase e right hand column- ft number, where tin/ i. 242? rom which we .•92017» 8? •92021J !d logarithm =2-920214 the tables, is that du« d number, thus 8-758761 , ......8-758836 f logarithms =75 that, with email incre. ng to them increase in f any number between ogarithm to bo added nt lor 1), that the iu- m of 5738-47. e form the proportion ) 3-758761, and this ad- ni of 5738-47. been -047 or -0047, tiie 25 or -8525. the last rule. Qd dividing the differ. to an increase of one of this, or 15 for two _nd each of the num. n corresponding to an natural number. The the nine digits, are in- (Proportional Parts ) already explained, ftr»J !A place of the natural in the siastfu, seventh, ey are divided by 10, my number con* LOGARITHMS. RULR. 321 SpcoNP Method. — Pind fit" nnnti'im of the logarithm correspond- ing to the Jirst four digits of the (,iven number. Firidiit the same horizo.dal ditisiou as that in which the 7nant^-< i \s foimd^ the proportional part in the column headed P. P., co.n- xponding to the digit in the fifth place of the given numbci', and set. :t lowu benedth the part of the mantissa alrcadg found, so that their \'ight hand digits mag be in the same vertical line. Find the P. P. corresponding to the digit in the sixth place of the given number, and set it down so that its right hand ^figure m.ag be one place to the right of the last. Find the P. P. corresponding to the digit in the seieath place of the given number and set it down one place to the right of the last, and so on till all the digits of the given number be used. Add the part of the mantissa already found, and the P. Ps. as written, together, and reject from the result all but the frt six digits to the left, adding 07ie to the last retained, if the highest of the rejected digits be not less than 5 — the result teill be the mantissa of tin: loga- rithm of given number. Lastly, attach the proper characteristic to this mantissa, and the result will be the required logarithm. Example l.--Whut is the logarithm of SSYi-ies ? OPEEATION. Mantissa of logarithm of 8372 =-922829 P. P. corresponding to -4 = 21 P. P. " to -06 = 31 P. P. " to -008= 42 Sum = ■9-22853152 Therefore required mantissa= -922854 and required log.=3922S54. Example 2. — What is the logarithm of 403567 ? OPERATION. Mantissa of logari tli m oi 403500 -- COS:-^ 14 P. P. correspdudiug to 60= 64 P. P. " to 7~ 75 Sum =0059155 Therefore required logarithm is 5 605910. EXERCIf.E 148. FIND THE L'^'GARITHMS OP TFIE FOLLOWING NUMBERS BY THE FIRST METHOD OBTAINiyO THK DIFFERENCES BY SUBTRACTION. 1. What are the logarithms corrcHponding to 8198217, 73'0245, and •843742? Ans. yia-^^S, l-8;'S73y, and T-926210. 2. Fiud the lo-javitjims eorrespondiiig to •00')2345(34 and -001007013. Arvs, 4-S702()l and 3-003035, % 3^2 LOGARITHMS. [Bkct. X. USING THE TABULAR DIFFERENCES. 8. fiiii the logarithms correspoading to 52'376 and 129*476. Ans. 1-719133 and 2*112189. USING THE PROPORTIONAL PARTS, 4. Find the logarithms corresponding to -000471398 and 9136712, Ans, 4-673387 and 6-960790. 5. Find the logarithms corresponding to 4-23429 and 763-12987. Ans. 0-626780 and 2-882598, 53. To find the logarithm of a vulgar fraction — RULE. Subtract the logarithm of the denominator from the logarithm of tlie numerator. 64. To find the logarithvn of a mixed number — RULE. Either reduce the mixed number to a fraction and proceed as in Art. 53, or reduce the fractional part to a decimal^ attach it to tltf whole number and proceed as in Arts. 48-52. 55. To find the natural number corresponding to anj, given logarithm — ^ RULE. First Method. — Find the logarithm in the table which is next lower than the given one, and the four digits corresponding to it wiU he the first four digits of the required number. II. Subtract this logarithm from the given logarithm, to the re- mainder annex one cipher and divide by the tabular difference corres- ponding to the four digits already obtaii\ed^ the quotient will be the fifth digit. ,, !/ ///. To the remainder attach another cipher and again divide by the tabular difference, the quotient will be the sixth digit, and thus proceed till a sufficient number of digits has been obtained. IV. Tlie characteristic of the logarithm shows where to place the decimal point. Note. — ^The number cannot he cr.rried with accuracy to more places than the logarithm has decimal places. (See Art. 56.) Example 1. — Find the number corresponding to the logarithm 4-923267. OPBBATION. , , Given lop. -928267 Next lower in tables, •928244=log. of 8880. Difference= 28 Tabular differenct=Ef52. Tbea 880004-02 i;iT«a Ui for digiU in 5th, 6tb, and 7th pUc«i. [Bbct. X, AwTS. 53-55.] LOGARITHMS. 323 •476. nd 2-112189. i 9186712. nd 6-960790. 3-12987. ^nd 2-882598, ion — logarithm oj ber — proceed as in Itach it to thf/ ding to anjy which is next ding to it wi!l hm^ to the re- ference corres- nt will be the yoin divide by igit^ and thus led. e to place the ore places thun the logarithm Hence the digits of the natural number are 8-380442 ; and since the charac- teristic is 4, i. 6. one lesn than the number of digits to the left of the decimal point, the required number is 88804 42. Second Mkthod. — Find the first four digits of the required mem- ber and also the difference between the given logarithm and the next lower in the table as in the last rule. II. Fi7fd in the same horizontal division of the table the highest P. P. that does not exceed this difference. Opposite to it in the column headed N. will be found the digit of the fifth place. III. Subtract this P. P. from the difference^ to the remainder annex one cipher and find the highest P. P. not exceeding the number thus farmed. Opposite to it in column N. mil be found the sixth digit. IV. Continue this process by the addition of ciphers^ till the re- quired number of digits be found. Example 2. — Find the natural number corresponding to the logarithm 3-563259. OPERATION. Given log. -553259 Next lower in the table -553155 = log. of 3674 DiflForence = Highest P. P. not greater than 104 = 104 98 corresponds to 8 for fifth place. 60 Highest P. P. not greater than 60= 49 corresponds to 4 in «ifl5^^ place. [place. Highest P. P. not greater than 110= 110 corresponds to 9 in seventh 110 Therefore digits of required number are 3574849 ; and since the characteristic is 3, there must bo four digits to the left of the decimal point. Hence requireu number is 3574'849. Exercise 149. by first method. 1. Find the natural numbers corresponding to the logarithms 4-137139, 0-718134 and 4-635421. -4ns. 13713-227, 5-225578 and -0004319376. 2. Of what numbers are 2-921686 and 1-922165 the logarithms? vlws. 835and -8359211. BY SECOND METHOD. 3. Of what numbers are 5-407968, 7-408386 and 3*416369 the logarithms? Ans. 255839-4, 25608588 and -0026083. 4. What are the natural numbers corresponding to the logarithms 4-877777 and 0-555556? Ans, 75470-,5168 and 3-6938. .ii,.' t fl ■^ 1 is 324 LOGARITHMIC ARITHMETIC. [9bot. X 66. In order to ascertain how many flgnres of these results may be relied upon as correct, let us take from the tables any logarithm, as 4'285635. Nov/ the real value of this logarithm if carried to a greater number of placos might be anything between 4-2856336 and 4-2856845, and might therefore dilTer from the given logarithm by very nearly -0000006, which is therefore the ex- treme limit of the error attached to tables of six places; 1. e. any difference less than -0<(i)0005 might occur without producing any change in the logarithm as given in tWe table. Now it is demonstrated in works treating of the theory of logarithms that the ditt'erence between the logarithms of numbers, which differ only by unity, is less than the modulus of the system divided by the smaller nuriiber. The modulus of the common systenj of logarithms is -4342945, and if we let n repre- sent the smaller number, the dififereuce between the logarithms of n and of n+1 is less th^n -434'2945-t-«. Now we have shown that the difference between the true logarithm and that given in tl-e table to six places, may be nearly equal to -0000005, \^rhich -4342946 -4342945 is therefore less than •4342945-r-n, or n is less than 7n7;?,nKJ^ But -tkt,,..- •0000005 -oooooos = 868589, That is, unless the number whose logarithm is givefi be less than 868589 its value cannot be found accurately beyond the Qrst fve digits, but if it bo less than 868589, then the flrst six figures found from the table will be correct. If tables of seven or eight places are used, the result can be depended on to seven or eight places, if the number be less than 868589 or if the mantissa bo less than -9378 ; but if greater, then the result can be relied on only to one less number of figures than the decimals of the logarithm. LOGARITHMIC ARITHMETIC. 57. The Arithmetical Complement of a logarithm is tlio remainder obtained by subtracting the logarithm from 10 Thus the arithmetical complement of 2-713426 is 10— 2-713426 = 7-286574. Exercise 150. 1. Find the arithmetical complements of 5*681642 and 0-714000. Ans. 4-368358 and 9-286000. 2. Find the arithmetical complements of 3-123456 and 7-213i49. Ans. 12-876544 and 16-786851. 3. Find the arithmetical complements of 6-124357 and 2-000837. A71S. 3-875643 and 11-900163. 58. To multiply two or more numbers together by means of logarithms : — RULE. I. Add their logarithms and the sum will be the logarithm of their product. II. Mnd the natural number correspotiding to this logarithm. Note 1. — For reason see Art 10. Note 2.— The following exorcises are all worked by the difference, and not by the proportional parts : [Shot. X. ay be relied «5. oer of pianos rcfore dit'er ifore the ex- ly (liffereno le logarithm arithms that ily by unity, uiiber. The 5 let n repre- of n ami of igarithm and 100005, •\Vhich •4342945 ^^^ ¥000005 be lesa thun digits, but if ) table will be depended on ' the mantispa )n only to one ithm is tlio from 10 = 7 •286574. •714000. 9-286000. •213149. 16-786851. •000837. 11-990163. )getlier by ogarithn of Xogarilhm. irenco, and not Arts. 56-60.] LOGARITHMIC ARITHMETIC. 825 Example.— Multiply 5631 by 47. Logarithm of 5681 =8-75058B " " 47=1-672098 5-422684 5-422590=: logarithm of 264600 94= Exercise 151. 5T Am. 264657 1. Multiply 61, 22, and 6r together. Am. 87230. 2. Multiply 52, 734, and 6 together. Ans. 229008. 3. Multiply together 35-86, 2-1046, -8372 and -00294. Ans. -185761. 4. Multiply -0C008764 by -86359. Ans. -000076686. 59. To divide numbers by means of their logarithms — RULE. I. Subtract the logarithm of the divisor from the logarithm of the dividend : the result will be the logarithvi of the required quotient. II. Find the natural number corresponding to this. NoTE.~for reason see Art. 11. Example 1.— Divide 6732-7 uy 478. OPEBATIOM. Losfirithra of 6732 7=3-S2S189 Logarithm of 478 =2-679428 Difference=l-148T61 1-14860.3 =logarithm of 14-0800 15i= 51 Ana. 140851 Example 2.— Divide -036584 by -00078593. OPERATION. Logarithm of -036584=2-563291 Logarithm of •00078593=4-895384 Difference =1-667907 l-667826=logarithm of 46-5400 87 81 = Ans. 46-5487 60. Instead of subtracting the logarithm of the divisor^ we may add its arithmetical complement — the result., with 10 subtracted from, the characteristic.^ will be the logarithm of the quotient. ii I ! i 326 LOGARITHMIC ARITHMETIC. t8«0T. X. Thus, in the last example the arithmetical complAraent of 4895884 is 18-104616, and this added to 2-568291 gives 11-667907, and subtracting 10 from this characteristic, gives us 1*667907, the same as obtained by the other method. Note.— This method of using the arithmetical complement is very con- venient when we have to divide one number by the product of several others. * Exercise 152. 1. Divide '6734 by -0009278. Ans. 725*8033. 2. Divide 437-89 by 62-735. Ans. 6-98. 3. Divide 93-217 bv -0007132. Ans. 130702-4. 4. Divide 9835267 *by the product of 23, 189 and 2-748. Ans. 823-839. 61. To raise a quantity to any power by means of logarithms — RULE. /. Multiply the logarithm of the given number by the index of the required power, the result will be the logarithm of the required power. 11. Find the natural number corresponding to this logarithm. Note.— For reason see Art. 12. Example 1. — Find the 10th power of 2. OPERATION. Logarithm of 2=0-301030. 0-801080 xl0==8-010800=logarithm of 1024. Ana. Example 2. — Find the 7th power of 2*71. OPEKATION. Logarithm of 'i-Tl = 0-4;?2969. Then 0-432969 x 7 =3-080783 = logarithm of 1078-45. Ans. Note.— In order to obtain the correct result when the characteristic hap. peus to be negatixe, it must bo recollected that the mantissa is always posi' tive. Exercise 153. fiXERCISE 1. What is the Bth power of 5 ? 2. What is the 6th power of 1-073? 3. What is the 4th power of -0279? 4. What is the 11th power of I'lll ? Ans. 8125. Ans. 1-5261. Ans. -00000060592. Ans. 3-1831. 62. To extract any root of a given number by meanu of logarithms — rule. /. Find the logarithm of the given number and divide it by the index of the required root, the result mil be the logarithm of the root. of 4-895884 is AflTS. 61-64.1 LOGARITHMIC ARlTHME'i.C. 327 11. Find the natural number c&t'responding to this logarithm. Note. — For reason see Art. 16. Example. — What is the cube root of 12345 ? OPEBATION. Lngarlthra of 12845=4-091491. Then 4091491-T-3=l-86as30-logaritlini of 23-11159. Ans. 63. To extract any root when the characteristic of the logarithm of the given number is negative : — RULE. /. If the characteristic is exactly divisible by the divisor, divide in the ordinary tvay, but make the characteristic of the gtiotient negative. JI. If the negative characteristic 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 22.— Extract the fourth root of •00'76542. _ OP-EBATION. Logarithm of -0076542=8-883899. Now since S is not exactly divisible by 4 we add— 1 to the characteristic and + 1 to the mantissa which gives us 4 + 1-883899 and this is evidently = 3^888399. Then 4 + l-883S99-r-4=T-4709747=logarithm of -2957S4. Ant. Exercise 154. 1. Extract the 7th r-oot of 913426000. 2. Extract the 11th root of 1-61342. 3. Extract the 5th root of '000007189. 4. Extract the 7th root of -002147. Ans. 19-0588, Ans. 1-04444. Ans. -0934817. Ans. -41575. 64. When the logarithms of two or more prime num- bers are given, the logarithm of any multiples of these factors by each other can be easily obtained by attention to the foregoing rules. Thus if the logarithm of 2 and 3 be given ;— 1st. We can obtain the logarithm of any i)ower of 2 or 3 by Art. 61, and any root of 2 or 3 by Art. 62. 2nd. We know the logarithm of 10 to be 1, and hence we can obtain the logarithm of 6, since 10^2=5 and also of 3-8 since 10^=3-3, hence we can also obtain the logarithm of any power or root of 5 or 8-3. 8rd. By Arts. 58, 59, we can obtain the logarithm of any power or root of 2, 8, 5 and 3-8 multiplied by any power or root of 2, 3, 5 or 8-3. Example. — Given the logarithm of 2 = 0-301030 and the loga- rithm of 3 = 0-477121. Find the logarithms of 500, 24, 54, 120, 75000, 16t, ^, and 13-5. M '." f ?*♦" '•-I 1 I j 528 LOGARITHMIC AEITILMETIC. [Sect. X. OPERATION. Since 5=10-r-2 the loprarithm of r>=\og. 10-log. 2=l-0-301000=:0-69S970. Then logarithm of 500=2'69'^970. 24=8x3-^3 x3.-. log. 24=(loc. 2)x3 + (lo-,'. 8.) log. 2. =0 301030 x8-0*903000 log. 3= -477121 Sam = 1-380211= log. 24. 54=27 X 2=8» X 2 .-. log. 54=(log. 8) x 3 + O-og. 2.) log. 3=0-477121 x8=l-431.163 log. 2= 0-301030 Sum =1-732393 = log. 54. 120=4x3x10=25x3x10.-. log. 120= (log. 2)x2 + (log. 8) + (log. 10.) log. 2=0-801080 x2=0'6l)20()O log. 8= 0-477121 log. 10= 1 Sum =2-079181 =log. 120. 75000 = 25 X 8 X 1000 = 53 x 3 x 1000 . • . log. 75000 = (log. 5) x 2 + (log. 3) «r + (log. lOOO.) log. 5=0-698970x2=1-397040 log. 8= 0-477121 log. 1000= 8 Sum=4-875061=log. 75000. 16| = 3-3 X 5 . • . logarithm of 16,5 = (log. 8-*3) + (log. 5.) Since 10-r3 = 3-3, log. 3-3=log. 10- loc. 8 = 1-0-477121 = 0-522S79 ■ ^ 0-69S970 Sum = ■'.221849 = log. 16|. logarithm 5 = i = -5 . - . bv changing only the characteristic = 1-698970 = logarithm i. 13-5= -5x27 = -5x3=».-. logarithm 13-5 = (log. 3)x3 + (log. -5) logarithm 3 = 0-4* 7121 x 3= 1-431363 logarithm -5 = 1-698970 Sum =1-180333=: log. 18-5. Exercise 155. 1. Given logarithm 2 = 0-301030 and log. Y = 845098, find the logarithms of 14000, 4'9, '00196, 1160, 1428'5'71428. •00000112 and 3-0625. Ans. Log. 14000 = 4-146128. ■ Log. 4-9 = 0-690196. Log. -00196 = 3'292256. Log. 1750 = 3-243038. Log. 1428-571428 = S-154902, Log. -00000112 = 6-049218. Log. 3-0625 = O-^B^'^IQ. NoTic.~142S-5'^'*e^> X 10000, also 3-0625=49—16. 1030=0-695970. Skot. X.] EXAMINATION QUESTIONS. 320 Example 2.— Given logarithm ^=: 1-098970 logarithm 3=0'477121 logarithm 11 = 1-041393 Find the logarithms of 49^ 363, 4-09, 2-4, 392-72, 293333^ and 19-965. A71S. Logarithm of Logarithm of Logarithm of Logarithm of 49^=1-694605. 363=2-559907. 4-09=0-611819. 2-4=0-388181. Logarithm of 392-72=2-594090. Logaritlim of 293333^ = 5-46736-2. Logarithm of 19-965 = 1-300270. ,,) >f 221849 = log. 16| 333=: log. 13-5. QUESTIONS TO BE ANSWERED BY THE PUPIL. Note. — The numbers after thei quentions rfiferto the mtmbered artiolm of the section. 1. What is the power of a nninher? (1) 2. What is the root of a number? (2) 3. Why is the second pctwer of a number called its square? C4) 4. Why is the third power of a number called its cube? (5) 5. What is the index or exponent of a power? (,G) 6. What is involution? (S;> 7. How do we multiply two or more ditferent powers of the same number to- gether? (10) 8. How do we divide any power of a number by another power of the same number? (11) 0. How do we find any required power of a given power? (12) 10. What is evolution? (18) 11. By what methods do we indicate a root of a number? (14) 12. H'^w do we extract any root of a givon power of a number? (!.')) 13. What is meant by extracting thi» square root of a number? (16) 14. What is the first step in extracting the square root of a number? (16) 15. Why do wo point olf into periods of two figures each ? (IS-I) 16. What is the second steii in tlie proci^s of extracting the square root? (16) 17. How do we know that the square root of the highest square in the left hand period is the highest disit of the root? (18-II> 18. What is the third step in the process of extracting the square root? (16) 19. Why do we bring down only the next period to the risjrlit? (IS-II in Ex. 2) 20. What is the fourth part of the process for extractini; the square root? (16) 21. Whv do we double the part of the root already found for a trial divisor? (18-111) 22. What is the next stop in extracting the .«-4t:!l ' ;i'l UiM. 10-14.] ARITUMETICVL PROGRESSION. 335 netical series 1, 3, )5, and the number oj OPEEATION Here we have given the first term 2, the last term 50 and the number of t»rnis 17 to tind 8, the sum of the series. Then« = («+0 "" = (2 + 50)x^^ = 52x~ =26x17 = 442. Ans. a o a 13. To find the common difference when the last term, the number of terms, and the sum of the series are given : RULE. d- 2n;— 2s (xii.) n{n — 1) Interpretation. — Take twice the product of the number of terma hy the last term, and from it subtract twice the sum of the series. Divide the resulting difference by the product of the number of terms by 1 less than the number of terms and the quotient will be the com- mon difference. Example. — In an arithmetical series the iist term is 80, the num- ber of terms 11, and the sum of the series 746, required the common ditt'erence. OPERATIOIf. Here we have given I, w, and s to find d and since i=80, n=H, and «=746 we bav« : «i = 2nl-2ii (2 X 11 X 80)-(2 x 746) (»-l) 11 X (11-1) 1760-1492 _ 26S _ ~Tl X To~ ~" ilo ~" "' 14. To find the number of terms of an arithmetical series when the first term, the common difierence, and the sum of the series are given : — RULE. d—la /2s '2a— d 2d )' (xiv.) Interpretation.—/. Subtract the common difference from twice the first term^ divide the remainder by twice the common difference^ square the quotient^ add the result to the qtwtient obtained by dividing twice the sum of the series by the common difference and extract the square root of this sum. 11. Next, from the common difference subtract twice the first term, divide the remainder by twice the common difference, and to the quo' tie.nt add the square root obtained in I. Tfie sum will be the number of terms. Example. —The first term of an arithmetical progression is Y, the common difference ^, and the sum of all the terms 142. What is the cumber of terms ? 836 AEITIIMETICAX PllOGKESSION. [Sect. XI. OPERATION. :r Here we have given a, d, and s, to find n, and since a = 7, d = J, and $ = 142, we have d~2a i/2« / i-14 2d~y 1/2S4 01^*)' = ~ 2xi i "*" ^ 2xi'V ~ I ,:5i Vrri6 ' "(27i)2 ^ _ 2U + Vll36 + T5Ci = - 27i + VlS92i = — 27* + 43 i = 16. Atis. Exercise. 156. 1. In an arithmetical series the first term is 4, the number of terms 17, and the sum of the series 884. What is the last term ? Alls. 100. 2. The extremes of an arithmetical series are 21 and 497, and the number of terms is 41. What is the common diflerence ? Alls. 11-,'',;. 3. In an arithmetical series the first term is 12, the last term 96, and the common diflerence is 6. Kequired the number of terms? Ans. 15. 4. In an arithmetical series the last term is 14, the common differ- ence 1, and the sum of the series 105. Required the number of terms? A7is. 15. 5. The first term of an arithmetical series is §, the common difler- ence I, and the sum of tiic series 1180. What is the last term ? Ans. 39i 6. If the extremes of aa arithmetical series are 8 and 170, and the sum of the series 4895, what is the common difference ? Ans. 8. 7. If the extremes of an arithmetical series are 5 and 27^, and the common difference 2^, what is the number of terms? Ans. 11. 8. If the fir.st term of a series is 2, the last term 478 and the num- ber of terms 86, what is the sum of the series ? Atis. 20640. 9. In an arithmetical series the last terra is 998, the first term 2 aud the common difference 6. What is the sum of the series? A71S. 88500. 10. In an arithmetical series the first term is 5, the nuniber of terms 11 and the common dilferonce 2-^. What is the last term? Ans. 271% 11. In an arithmetical series the last term is 199, the common differ- ence is 11 and the number of terms 19. Required the sum of the series? A.ns. li^OO. 12. The sum of an arithmetical series is JJ9840, and the extremes are 2 aud 478. What is the number of terms? Ans. 1C«). 18. The sura of an arithmetical series is 83500, and the extremes aie 998 and 2. Recjuired the common difference? Ans. 0, [Sect. XL Awn. 15, 16.] GEOMETRICAL PE0GRES3I0N. 837 a = 7, d = i, and ) =-27i + )2i = — i>7i + 43 i number of terms le last term ? Ans. 100. ind 41)7, and tie diflerence ? he last term 96, li the number of Ans. 15. e common differ- lired the number Ans. 15. e common difi'cr- t is the last term ? Ans. S9i. and 170, and the ifference ? Ans. 3. and 27i, and the terms? Ans. 11. |478 and the nuiii- ? Am. 20640. Ic first term 2 uud if the series? Ans. 88500. nunber of terms Ihe last term ? Ans. 2Vi. e common diflor- equired the sinn Ans. 1900. the extremes are Ans. lO). the extremes a'O 14. A snail crawls up a flag staff 130 feet high and upon reaching the top begins to descend. In what time will he again reach * the ground if he goes 2 feet the first day, 4 feet the second, 6 feet the third, and so on ? Ans. 15 days, 15 hours, 10 min. 27*264 sec. 15. The sum of an arithmetical series is 83500, the first term is 2 and the common difference 6, what is the last term ? Ans. 998. 16. A person wishes to discharge a debt of $1125 in 18 annual pay- ments which shall increase in arithmetical progression. Hovi much must his first payment be in order that the last may be |120? Ans. $5. 17. In an arithmetical series the extremes are 5 and 27^ and the num- ber of terms is 11. What is the common difference? Ans. 2^. 220 stones are placed in a straight line exactly 2^ yards apart, the first being 2|- yards from a basket, how far will a person go vhilst picking up the stones, returning with one at a time and depositing it in the basket? Ans. 69^^^ miles. The sum of an arithmetical series is 39840, the number of terms is 166 and the last term is 478. What is the first term ? Ans. 2. A person travelled from Toronto to Kingston, in 12 days, walk- ing 4 miles the firpl day, 6 miles the second, 8 miles the third, and so on. How far is Toronto from Kingston ? Ans. 180 miles. 21. The clocks of Venice strike from 1 to 24. How many strokes does one of these clocks make in the day f Ana. 800. 18 19, 20 GEOMETRICAL PROGRESSION. 15. Quantities are said to be in Geometrical Progres- sion when they increase or decrease by a common multi- plier. Thus 8, 12, 48, 192, &c., are in geometrical progression, the common ratio or common multiplier being 4. 100, 20, 4, J, 5*j, «Sic., are in geometrical progression, the common ratio being J. 16. In geometrical progression there are fi^'"e things to be considered : 1. fhe flrnt term. 2. Thelustterm. 8. The common ratio. 4 The number of terms. 0. Th6«umo/th«9«riH. W -nws^^ J3S GEOMKTRICAL PROGEESftlOJif. [SfiOT. kl ll> I if, I As ill aritlimrtical pronrressinn, these five quantities are so related that a'ly three oftlii'm b.'iri:; liivi-n the < ther two can be louiid, and hon'je there are 20 tliitlnct cases arising from their combihatioas. • 17. Kepresenting these five quantities by letters, thus, a = th" first term. I ■= the lant term. r = the common ratio. n =: the mtmber of terms. s = the sum of the series. the general expression for a geometrical series : ecomes a+ar+ nr^ + ar^ + ar* + 1*/-^ + , &c., •where the index of r is always one less than the number of the term. Thus in the third term the index of r is 2, which is one less than the num- ber of the term : in the fifth term the index of /' is 4, which is one less than the number of the term, Ac. Hence I = arn—^ ; that is, the last term is equal to the first term multiplied by the common ratio raised to that i)ower whicu is indiciited by one less thiiii tliie number of terms. 18. Since the sum of the series is equal to the sum of all the terms. « = a + ar+ar'^ + ar^+ arn^ + am ^+ar* \ multiplying by r we get «r =s ar + ar^ + ar^ + * ^ + ar» ^+arn—i + arn. Hence sr—s = art'^a ; or «(/'—!) = a(?'»— l), and therefore 8 — — - — . That is, the stmt of the series is found hy finding that power of the com- mon ratio which is e.vpressed by the number of terms — subtracting I fro7r, this, dividing the remainder by one less than the common ratio and multi- plying the quotient by the first term. Note. — The second of the above series is found from the first by multiply- ins; both sides of the equation by /■, and in subtracting we take the tertms of the upper series from the corresponding terms of the lower. Only the first threo or four and the last three or fo ir terms are written and between ar^ and ar"-^ there may be any number of intermediate terms. The urn a in the lower Beries is obtained by multiplying the term before a/*" 8 in the upper series, which is ar>* *, by r. 19. From the formula obtained in Art. 17 we get by \ransposing the terms, &c. it- I =r a = r s= f»SB I fii -1 lo g. I — log, a iog,r + 1 Arts. 17-2<).] GEOMETRICAL PEOGRESblOlf. 83d 80 relntpfl that a-^y bonce there are iJi) An.\ sabstltnting these values of ?, n, t\ n In tho formula obtained In Art. 18 we find rl—a § = r-1 Z(r»j-1)_ * "" (r-l)r" I e = e less than the num- h is owe less thun the first term multiplied tited by one less than al to +.he sum of iplying by r we get at power of the com- -Hnhtmcting 1 .froin Oil ratio and rnulti- he first by mulliplv- take the teriiiri oi tlm Only the first thrco itween ar^ and a/"*-^ ar" 3 in tho lower 8 in the upper series, t. 17 we get by In— I — a^—l i~ "T" pt—X — an — 1 *dd these together with tlio two formulas obtained in Arts. 17 and 18, ofrn— 1) r — I are the fun'L'imontal formulas of {jeometrical progression from which the other fifteen are derived by reduction. Thus, « = , gives formulaa for «, r, /, and a = 4 r-1 l{rn-\) $ = « = (r-l)r"-i n m In— I m—l — a ln-~l Q,H — 1 cf(r"— 1) 7—1 I = rtr»»— 1 " «, /, n, and a = 4 " 8, r, a, ow(f « = 4 " ?, a, r, OTUf 71 = 4 7b'-i = 7 x SiJ-i = 7 x 3» = 7 x CoCl = 45927. Ana. 2i2i. Given the first term, the common ratio, and the lasl term, to fiiid the sum of the series : — RULE. rl—a ' S = (VI.) r — 1. Interpretation. — Subtract the first term from the product of the common ratio by the last term and divide the remainder by one less than the common ratio. Example. — The first term of a geometrical series is 5, the com- raon ratio 4, and the last term lOdOOOO. What is the sum of all the terms ? OPERATION, Here o = 5, r = 4, and i = 1000000. rl-a 4x1000000-5 Then« = :^-^f-^= 13333313. ^n.. r— 1 4 23. Given the first term, the common ratio, and the number of terms, to Jind the sum of the series : — RULE. /r"-l\ \r-l.) (V.) Interpretation. — Mnd that power of the common ratio v)hich in indicated by the number of tenns, subtract one from it, a7id divide the remainder by one less than the common ratio. Lastli,, hiultiply the quotient thus obtained by the first term of the series, and the result will be the su7n of all the terms. Example. — The first terra of a geometrical series is 3, the com- mon ratio is 4, and the number of terms 9. Required the sum of the series, : (1 r i: g42 GEOMETRICAL PROGRESSION. [Skct. XL Heio rt = 3, r = 4, and n = 9. /r"-K „ 40— 1 „ 262144 — 1 „,„,,., , Then «=a( ) =8x - -- =8x ^ = 262148. Atia. \ r~\ J 4 — 1 8 24. To find the coinmon ratio when the first term, the last term, and the sum of the terms are given : — RUL8. s — l (XV.) Interpretation. — Divide the difference between the first term and the sum by tJie difference between the last term and the sum : the quo- tient will be the common ratio. Example. — The first term of a geometrical series is 1, the last term lOGSo, and the sum of all the terms, 29524. What is the com- mon ratio ? ^ Here a = 1,/: ^. 8- a 29524 Then r = = — g.-l 29524-19683 OPEnATIO* 19683, and s= 29524. 1__ __ 29528 _ '^ 9841 " 8. AfiM. Exercise 157. 1. A nobleman dying left 11 sons, to whom he bequeathed his prop- erty as follows: to the youngest he gave £1024 ; to the next, as much and a half: to the next 1^ of the preceding son's share; and so on. What was the eldest son's fortune ; and what was the amount of the nobleman's property ? Ans. The eldest son received £59049, and the father was worth £175099. 2. The first term of a geometrical progression is V, the last term is 1240029, and the sum of all the terms is 18C0040. What is the ratio? Ans. o. 3. What debt can be discharged in a year by monthly payments in geometrical progression, the first term being £1, and the last £2048 ; and what will be the common ratio? Ans. The debt will be £4095 ; and the ratio 2. 4. The ratio of the terms of a geometrical progression is ^, the num- ber of terms is 8, and the last term is lOG^^If. What is the suui of all the terms ? Ans. 307;, i '7 1 i J. In a geometrical progression the first term is 1, the number of t^nas 7, and the cgmmon ratin 3, what is the sum of the series y Am, 1093. ART. 24. J GEOMETRICAL PROGRESSION. 343 ; and what was the father was 6. The first terra of a geometrical progression is 1, the last term h 1007701)6, and the number of terms is 10. What is the sum of all the terms ? Ans. 12093235. 7 The (irst term of a geometrical progression is 6, the last term ia 3072, and the sum of all the terms is G138. What is the ratio ? Ans. 2. 8 The ratio of the terms of a geometrical progression is 2, the number of terms is 11, and tlie sum of all the terms is 20470. What is the last term ? Ans. 10240. 9. A gentleman married his daughter on New Year's day, and gave her husband 1 shilling towards her portion, and was to double it on the first day of every mouth during the year. What was her portion? Ans. £204 15s. K What will be the price of a horse sold for 1 farthing for the first nail in his shoes, 2 farthings for the second, 4 lor the third, &o., allowing 8 nails in each shoe V Tlu A71S. £4473924 58. 3|d. 12 first term of a geometrical progression is 4, the last term is 78732, and the number of terms ia 10. What id the rotio ? Ans. 5. A person travelling goes 5 miles the first day, 10 miles the sec- ond day, 20 miles the third day, and so on increasing in geo- metrical pr()gression. If he continue to travel in this way for 7 d;iys, how far will he go the last day ? Ans. 320 miles. 13. The first term of a geometricid progression is 5, the last term is 327680, and the ratio is 4. What is the sum of all the terms ? Ans. 436905. 14. A king in India, named Sheran, wished (nccording to the Arabic author Asephad) that Sessa, the inventor of chess, should him- self choose a reward. He requested the king to give him 1 grain of wheat for the first square, 2 grains for the second square, 4 giains for the third square, and so on ; reckoning for each of the 64 squares of the board twice as many grains as for the preceding. Sheran was angry at a demand apparently so insignificant ; but when it was calculated, to his astonishment it was found to be an enormous quantity. What was the number of grains of wheat, and what was its worth at $r50 per bushel, reckoning 7680 grains to a pint? Ans. 18446744073709551615 grains. 37529996894754 bushels. $56294995342131. II. The ratio of the terms of a geometrical progression is 3, the nutnber of terms is 10, and the sum of all the terms is 29.")240. What is the last term ? A7is. 196830. IC. The first term of a geometrical progression is 1, the last term is 2048, and the number of terms is 12. What is the sum of all the terms ? Ans, 4096, •■- V I ■ ■' 1 -.' &44 GEOMETRICAL PROGRESSION. [SSOT. XI. 17. The first term of a geometrical progression is 6, the ratio is 4, and the number of terms 9. Whiit is the last term ? Ana. 327660. 26. When tho common rut'o .- a poometrlcal series Is a i)roper frno- tlon, 1 c, It'ss than 1, the scries is a desccn^lin^ one, and wlion ihe number of terms becomes very larj^e r" becomes very small. In nn infinite descend- ing series r" becomes infinitely email, i. e., its value becomes = 0, and there- fore ara may be neglected, and tlie formula for finding the sum becomes ar>* — a —a a « = r— 1 r — 1 1 — r \vheD r is lets than 1 :— • = (Ml.) 1-r Hence for finding the sum of any Ivjiniti scries RULE. Interpretation. — Tlie &um of an infinite series is found by di- viding the first term by unity minus the common ratio. Example 1. — What is the sum of the infinite series l + i+a'o 4- OPERATION. Here a = 1 and r = 1 a 1 1 Then s = = = — = f = IJ. An«. 1-r l-l I • • Example 2. — What is the sum of the infinite series '784 ? OPERATION. Here a = ^^^*q and r = t^. Then s = rooo TBoS 1- = = {IJ. Ana. ToTo TooS Exercise 158. 1. What is the sum of the infinite series f, -3^^, ^%-, &c. ? Ans. ^. 2. What is the sum of the infinite series 4, 2, 1, ^, i, &c. ? Ans. 8. 8. What is the sum of the infinite series '79 ? Ans. 5 9. 4. What is the sum of the infinite series -1234 ? Ans. l^H, 26. To insert any number of means bet^^een two given extremes : Arts 25-3T.] POSITION, B45 ,ny injinite series RULS. If the Kcrku in an arithmetical one, find the eomtnon difference by formula IX. Art. 8. Than add thin common difference to the first tenn and the remit loi/l be the .second term ; add the common differ' owe to the .second and the rcKult will he the third term., dec. If the .series is a geometrical one, find the common ratio by for- mula XIII. Art. 20. Then multiply the first term, by the common ratio and the product will be the .second term ; multiply tlie second term by the common ratio and the result will be the third, d'c. Example 1. — Insert 7 arithmetical means between 3 and 61. OPERATION. Bincd there are 7 means and 2 extremes the number of terms Is 9. ^^ , l-a 81-3 48 „ Then d= = -^ — - = ~ = 6. n—l 9—1 8 1st term =3; 2nd=:8 + 6=9, 8rd=9 + 6=15, 4th = 15+6=2l; 5th-21 + 6=3T; 6th=27 f 6=33, and so on. And series is 3, 9, 15, 21, 27, 33, 89, 45, 51. Example 2.— Insert 6 geometrical means between 1 and 128. OPBRATIOy. Since there are 6 means and 2 extremes the number of terms is S. Then »•=( ^J„-^x = (~)8^=(128) ♦-2. Hence 2nd term = l x2=2; Brd term=2x2=4' 4th=4x2-8, &o. And series is 1, 2, 4, 8, 16, 82, 64, 128. Exercise 159. 1, Insert 9 arithmetical means between 2 and 92. Ans. 2, 11, 20, 29, 38, 47, 66, 66, 74, 88, 92. 2. Insert 4 arithmetical means between 7 and 60. Ans. 7, 15|, 24i, 32J, 41^, 60. 8. Find 8 geometrical means between 4096 and 8. Ans. 2048, 1024, 512, 266, 128, 64, 82, and 16. 4. Find 7 geometrical means between 14 and 23514624. Ans. 84, 504, 3024, 18144, 108864, 653184, and 3919104. S:ir it , r: ; 1 ',•■ I ' POSITION. 27. Position is a rule which enables us to solve, by means of assumed numbers, a class of problems which we could not otherwise solve without the aid of algebra. NoTB.-' Positidn Is also called the Bule of False, or the Rule of Tri^ «d4 Error. 34G 6IN0LE POSITION. [Beot. XI. 28. Position h diviled into : — Ist. Single Position — when only one aasuinerl num- ber is use .jiiirnl. Or, 43 : C3 . : 40 . —''.- =60. Ana. I'aoop— 00+ > (.fOO-i of 00=08. ExAMi'LE 2. — A teacher being askod liow many piipils he had, re- plied, if you add ^, ^, and \ of the niunber together, the sutii will bo iS; what was their uumber y OPEKATtON. Assutno GO to t)0 tho nnmbor of piipild. Then oiu'-third of (!0^'2i) oiii-foiirlii oftJO^lS one-sixth ol 00 = 10 Sum --45, but it shoiiM, by qiu>»(li»n, equal 18. V) Thou 45 : IS PEOoF.~iof24 + iof24 + Jof 24=13. Exercise* 160. ivbence 6': 6 : : « = 10, and one which we can tak*' 1. A gentleman distributed 78 pence among a number of poor per- sona, consisting of men, women, and children ; to each man he gave 6d,, to each woman 4d., to each child 'Jd. ; there were twice us many women as men, and three times as many children as women. How many were there of each ? An^. 3 men, 6 women, and 18 children. 2. A person bought a chaise, horse, and harness, for £G0 ; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness. What did he give for each? An8. He gave for the harness, £6 13s. 4d.; for the horse, £13 Os. 8d. ; and for the chaise, £40. 3. A's nge 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? Am. A's is 84, B's 42, and C's 14. 4. After paying away \ of my money ; and then \ of the remainder, I had 72 guineas left. What had I at lirst? Ans. 120 guineas. * Ail qup.sdons in position inay be solved by simplo analysis, and vory fre- quently tliis is the better method, and indeed the ti-ucher should insist iijxin the pupil thus jolving ciich problem. The following will servo as examples of the mi»de of solution. Kx.^Mi'T.i!. 6. —Since 140 is equal to A'.s age. + B's affe, + C's nee. nnd B's ace is equal to th'ce Umes C's, and A's to 6 tunes C's. it follows ihat 140 is equal to 1 +3 + 0=10 tmios C\s age, and heuco C's age is I'g of 140=14; B'i>=li k3=4. n. diiy he will do \ of the whole work, similarly in one day B will do J. ui d thu whole work. Therofore working together they will do * + \ + ii = V-iJ '• whole work, and they will require as many days to do the whole wors aa _ roMtaincd timeb in \, I. e„ l-7-i?J=lt§y days. 4w«, Af.T3. 83, 3i.i bOUBLE POSITION. ^49 34. Lot it be requiicd to find a value for x such aa to satisfy the equation, ax -^ b=. c. Ill such 11 case as-siime any two known numbers n and n' and perform on t!;e>■ ..hi- ■■*sm» ' \ , I * i; .. I lit I: ■If p :! i* 854 COMPOtmD tlJTEHESt. COMPOUND INTEREST. tBwt. XL 36. Let P = tho principal, / = the Interest, A = the amoant, i := the number of payments, jvnd r = the rate per unit for one payment. Then since r is the interest of $1 for one payment, the amonntof $1 for on* payment is 1 + r, and since the principal is always proportional to the amount: 1 : 1 + r : : P : P (1 +/•) = Amount of P at end of Ist period. 1 : 1 + r : : P (1 +r) : P (l + r)' = Amount of P at end of 2nd period. 1 : 1 + r : : P (l + r)9 : P (1 + r;' = Amount of P at end of 8rd i)erind. 1 : 1+r : : P (l + r)3 : P (l + r)« = Amount of P at end of 4th period. And so on ; hence at the end of the /<* period A = P (1 + r)(, which is I formula (I) in the margin. A = P(l+r)<(I) P = (l+ry /A (11) r = V^ ~1(III) Is log. A »- log^ log. (1+r) (IV) log, n log. (1+r) (V) Dividinjr eacli side of (1) by (l + r)« we get for- mula (II) in the margin. DJvifllng each side of (I) by P we get (1 + r) = p ; extracting the <<* root, and transposing the 1, we get formula (III). Obtaining as before (1+r)* = g- and applying the principle of logarithms we get log. (1 + r) x < = log. A — log, P, and dividing each side by log. (1+r) we get t = '"f ^ ~ ^^ - , which is (IV) ^ * log. (1+r) ^ ' of the margin. Lastly, to find the time in which any sum of money will amount to n times itself at a given rate per cent, compound interest, we substitute mP for A in fornmla (I), which gives us nP s P (1 +r)< and dividink each of these by P we get n = (1 +r)« whence log. n = log. (l+r)K *; or < s: r-~. — r, whlch is fOHnuIa (V). log. (1+r)' APPLICATIONS. Wlien the principal, rate per cent., and time are given tojind the amount ;— RXTLS. -4 = JO (1 + rj w log. A = log. P 4- log. (1 + r) x t. (1) Interpretation. — Multiply the logarithm of the amount of $1 for one payment by the number of payments, and to the product add the logarithm ^f the principal ; the result will be the logarithm of the amount. 11. Find the natural numker corre^otiding to this logarithm and the result mil be the answer. Example. — To what sum will |760 amount in 8 years, at 2 per cent., quarterly compound interest? CPRBATIOir. Here P s 760, r = *02, and t — 12, since there are 12 qnarters In 8 yean. Then A a P (1 + r)« or log. A = log. P + log. (1 + r) x « = 2-870061 + «<008i00 M IS ae 2^8261 e: log. of AuwM. Honoo amount « $96117. AKTB. I J.; h [i^tct. XI imonnt, t ss the at. mntofSl for on* to the amount : of Ist period, of 2nd period, of 8rd i>erind. of 4th period. 1 + r)*, which U + r)* we get for- » wo get (1 + r) and trani^poslng and applying the log. (1+r) X < = each side by log. ^i*, which 13 (IV) hloh any sum of » itself at a given est, we substitute hich gives us nP of these by P we = log. (l"»-r)K«; mula (V). me are given r) X t. (1) Tiount of $1 for troduet add the )garithm of the logarithm and years, at 2 per rters In 8 yean. X t = 2-870061 ♦ AHTB. 86-81) COMPOUND INTERfiSl'. 355 36. When the amount, rate, and time are given to find, the principal ;— RULK. A P — jz rj ; or log. P = log. A — log. (1 + r) x t. (II.) (1 + r; Interpretation. — TaJce the number expressing the amotmt of^l for one payment^ and raise it to the power indicated by the number of payments. II. Divide the given amount by the number thus obtained and the quotient will be the required principal. • BY LOGARITHMS. Take the logarithm of the amount of $1 for one payment^ and multiply it by the number of payments. Subtract the logarithm thus obtained from the logarithm of the given amount ; the remainder will be the logarithm of the required principal. Example. — What principal put out at compound interest, at the rate of 3^ per cent, half-yearly, will amount to |8'764'00 in 11 years? Here A = 8764, r Then P = ^ OPERATION. •085 and t = 22. or log. P = log. A — log. (1 + r) X t (1 +r> log. P = 3-942702 - 0014940 x 22 = 8942702 - 0-328680 = 8-614022. IlencGP =$4111-70. Ana. 37. When the amount, principal, and time are given to find the rate per cent. : — RULB. r =r /fA\ . , , log. A — log. P = «j/[ pj-1; orlog.{l + r)=-^ -j—^ (HI.) t Interpretation. — Divide the amount by the principal, and ex- tract that root of the quotient which is indicated by the number of payments. II. Subtract 1 from the root this obtained and the remainder will be the rate per unitf multiply this by 100, and the result will be the rate per cent. BY logarithms. Subtract the logarithm of the principal from the logarithm of the given amount^ and divide the difference by the number of pay- ments ; the result will be the logarithm of the amount of $1 for one payment. Find the natural number corresponding to this, and from it sub- tract 1, the result will be the rate per unit^ and this multiplied by 100 gives the rate per cent. I'd ( i ill w I 8dd OOilPOUND INTERK8T. idtOT. Xi. ExAMPLi. — At what rate per cent, compound interest, payable half-yearly, will $278 amount to $6742 in 27 yeura ? OPKRATIOK. Her« A = 6T42, P = 2T9 and t - 64. log. A — log. P _ 8-828t89 - 2'444045 _ 1-884744 t 64 ~ ~ 64 '02664S4. Henoe 1 + r = 1'06, r = 08, and rate per cent. = 6. Jns. Then log. (1 + r) = 38. When the amount, principal, and rate are given to find the time : — RULE. loa. A — log. P i - y- -—. J (IV.) hj. (1 + r) ^ ' i : -"^r : 4noN. — Subtract the logarithm of the principal from the k ritis,n of the given amount, and divide the remainder by the logarithm of i\ amount of $l/or one payment ; the quotient will be the number of t/ie payments. Example.— In what time will $729 amount to $7143 at 2^ pei cent, compound interest, quarterly ? OPERATION. Here A = 7148, P = 729 and r = -026. _. . log. A— log. P 8-858881 -2-862728 inen t = — : ;: = - log. (1 + r) 0-991168 ^ .- — — — -- = 92-42 payments 0010724 *^ ' 0-010724 S3 88-lOS years = 28 years 1 month 7'8 days. Ane. 39. To find in what time any sum of money will amount to n times itself at any given rate per cent, com- pound interest : — RULE. ^-log.{l+r)^^'> Interpretation. — Find the logarithm of the number expressing to how many times itself the given sum is to amount ^ and divide it by the logarithm of the amount of $1 f 44. An Annuity contingent or a life annvHy is that is to continue to be paid only so long as one or nioio individuals shall live. 45. A Per'£)Ciu'Uy is an annuity that is to continue for ever. 46. An Annuity is in arrears when one or more yi) ■ ments are retained after they have become due. 47. The amount of an annuiiy is the sum of the pay- ments forborne (i. e. in arrears) and the whole interest due upon them. 48. The present worth of an anmitty is that sum which, being put out at interest until the annuity ceases, would produce a sum equal to what would have been accumulated had the annuity been left unpaid until that time. 49. Annuities are calculated at both simple and com- pound interest. ANNUITIES AT SIMPLE INTEREST. 60. L<'t (1=- ft fil plo paymoiit of the annuity, <=nnmber of pnymenta, r= rate por unit for oni' period, and A— ainomit of the annuity. Then whon the annuity is forborne any i iiml>cr of payments, tlio last pay- ment boinf; niuilo at the time it falls due, is equal to it ; last paynn nt but one =« + inton'st on « for one \>nnnlft boa yet been dtscovored for the sutntnatlnn of » aeriM for flndint; the prenent valn« of an annuity at aimplo interest Tho rule gene- rally adopted for flading the prvsont value of an anuulty at simple Intereet i* the following:— Find the present worth of each payment by itself^ discounting from the time it falls due — the sum of the present worth of all tice payinentt will be the present worth of the annuity. Note. — The absolute absurdity of purchasini; annuities by simple inter- est is evident fVom the fitct that tfio intereat of the sum required to purchase an annuity, discounlinz at 5 per cent simple interest., actually exceeds the an- nuity; i. e., to piirohastf an annuity to continue only a limited number of years, requires a eum which will yinld a larger yearly interest for ever. Heuco the various rules Kiveii for finding tho preitunt valtTo of aunoltiea i\t simplo laterMt arti, in ullctit, valuoltisa. APPLICATIONS. 51. Wlicn the annuity, number of payments forborne, and the rate per cent, of interest are given, to find the amount : — RULK. (<-l)r = a< I (1 + (I.) Intbrpretation. — Multiply the rate per unit by one less than the number of payments and to half the result add 1. Multiply the number thus obtained by the product of the annuity by the number of payments^ and the result will be the recjuired amount. Example —If a pension of $600 per annum be ferborne 5 years, to what sum will it amount at 4 per cent, simple interest ? OPEBATION. Here a = 60O, < = 5, r = -04. Then u4 = a< 1 1 + ^^'' (I 4. -08) = 8000 X 1-08 = $8240. Ans. I = 600 X BJ 1 (B~l) X -04 2 f = 8000 K 52. When the amount of the annuity forborne, the number of payments forborne, and the rate per cent, of interest allowed, are given, to find the annuity : — a = BULX. 2A (II.) t{2 + {t- l)rf IWTERPBETATIOK. — Multiply the rate per unit by one le^s than the number nf paymcjits, and to the product add 2. Multiply this sum by the number of payments, and divide twice the given amount of the annuity by the product thus obtained ; the result will be the annuity required* u^:^] ^<^pp-' IT Iff I i 860 AlllfUITlIB. [eioT. iL Example. — What annuity, payable quarterly, will amount to |3226'25 iu 7 years, at 4^ per cent, per annum, simple intercut ? OPBRATIOH. Here since the rate in 4| per cent, per annum, or HHS per uuit per annum, the rate per quiirtor = •04ft -t- 4 = "01 126. Then < = 28, -4 = $8225-25 and r = -Ollga. 9A 822525 X 8 g450-50 *~ <|2 + (« - l)r| = 28j2 + (28-1) x 01125} "" 28 x (2 + -8087^ 6460-60 6450-60 ,,«. , , * , u = 2r-x 2^80376 = 64-606 = ♦^^ = ^"""^"^^ P'^"'°^ "'^ ^*°*'* "°"'^ annuity = $400. Ant. 53. The application and interpretation of the remain- ing formulae will be readily understood from the foregoing examples. Exercise 168. 1. In what time will an annuity of $1000 per annum, payable half. yearly, amount to $8365, allowing simple interest, at the rate of 6 per cent, per annum ? Ans. 14 payments, or 1 years. Note.— In this question we use formula IV, r being equal to *08 and a = 600. 2. If a rent of $460 per annum, payable quarterly, be forborne for 11 years, to what does it amount, allowing 6 per cent, per annum, simple interest ? jlrt^. $6646'«37|. NoTE.-Take a = $112-50, r = -015 and t = 44. 8. At what rate per cent, per annum, simple interest, will an annuity of $300, payable yearly, amount to $1680 in 5 years ? Ans. 6 per cent. 4. The rent of a farm is forborne for 8 years, and then amounts to $2080. Now assuming the rent to be paid half-yearly, and '^mple interest at the rate of 8 per cent, per annum allowed, lehat was the rent of the farm ? Ans. $200, a = r = ANNUITIES AT COMPOUND INTEREST. 64. Let A, a,ryt = mma quantities as In last articles, and idso let v sa present value of the annuity. Then, as before, the last payment of a forborne annuity bfir/(i paid wiu-n doe, = a; last payment but one, = a + interest of a for one piiynient —it + ar = a n + r) ; so also last payment but two, = a (1 +/•)"; last but thr = a (1 + r)\ &c., and first payment ~ u {I + r)«-i. Hence A, the amount of the annuity = a + a (1 + r) + a (1 + ry» -» " (l+r)» + &c + a a + r)*-S vhicb is a fioonaetrical series an == - a %.(H-r) (IV) l(yp. a—loff.(a—iJr) %. (1 + r) (VII) ■" ri(l+r)'"*(Ur)«+'P^"^^ *(IX) vr (X) ^=-!(XI) t; ::= a =z V =z ril+ry (XII) to ^ll!!LJLf , whioh Is formate I r of margin. Fonnulaa II, III, and IV are obtained frum formula I by tran&poaiUun, Ac Since the present value of an annuitr at conittound interest is that prinof- pal which put out at compound in> tereat fur the given time, would produce the amount of the annuity we have from Art. 86, formula I, whence by dividing by (l + r)<, we get formula V in the margin. Formulas VI and VII are derived from V. To find the present value of an annuity which is to commence after t years and then continue for « yeure, we hava ft-om formula V, t> for « + < years, = alone, v = - \ ,^ ~ > Therefore for t years to commenoo after « years, v = 1 i (1 + rV-K -- 1 (1 ■<■ r)« - 1 I r\ (l+r)»+« " (1 + ry J _ a I __1 1 i °''"" r I (1 + *•)* (1 + ry-Kf which it formula VIII in the mar- gin. When an annuity lasts fbr ever, as in the case of landed property, (1 + rV in formula V l>ecomes Infinitely great, and therefore ■-: — r- = — ■ = 0, and the formal* (1 + r)t » ' for finding the present valne of a perpetuity is reduced to the form given in IX. Formulas X and XI aro derived from IX. The present ■^. ue of a ft-eehold estate to a person to whom It will revert after a yei -a and then continue for ever, ia found ftom formula VIII and is representt^ by formula XII in the margin. 65. To facilitate the calcnlatlon of annuities the followinic tables are given, tfao first Bhowiii'.; the amount of an annuity of $1 at compound interest^ ana tbb second, the preaent value of an annuity of $1 at compound interest ul 862 ▲NNurrisft. (Sect. XL I, 1 Ik , TABLE OP THE AMOUNTS OF AN ANNUITY OF $1 OR £1. I No. of Payments. 8 p^r cent 4 per oflnt 5 per cent 6 per cent. 1 1-00000 100000 1-00000 1-00000 2 203000 2-04000 2-0.''»000 2-06000 8 8-09090 8-12160 8-15-250 8-18860 4 4-lS;3C3 4-24646 4-81012 4-37462 6 6-80918 6-41032 6-52563 6-63706 8 6-46S41 6-68297 6-80191 6-97532 7 7-66346 7-89829 8-14201 8-39S«4 8 8-89-234 9-21428 9-64911 9-89747 9 10-15911 10-58279 11-02666 11-49131 10 11-4C3SS 12-00611 12-57789 18-18079 11 12-80779 18-48635 14-20679 14-97164 12 14-1 9-203 15026S0 16-91718 16-86994 18 15-61779 16-62684 17-71298 18-88214 14 17-0863-3 18-29191 19-50868 21-01506 15 1869891 20-02359 21-67856 2.3-27593 18 2015533 21-82458 28-6.5749 25-67253 17, 21-70159 23-69751 26-84037 28-21288 18 2;i-41443 25-64641 28-13-288 80-90565 19 2511637 27 67123 80-53900 83-75999 20...... 26 S7037 29-77808 a3-06595 36-78559 21 28 07648 81-96920 86-71925 89-99273 22 80-03678 84-24797 88-60521 48-39229 28 82 45238 8661789 41-43047 46-99583 24 a4-42&47 89-03260 44-60200 60-81568 25 80-45926 41-64591 47-72710 64-86461 26 83-53804 44-81174 6111845 59-15639 27 40-70968 4708431 64-66981 63-70676 23 42-93092 49-96758 68-40258 68-52811 29 46-21885 62-96629 62-32271 73 63980 80 47-57541 66 08494 66-43885 79-05819 81 60-00238 69-82833 70-76079 84-80168 82 62 50276 62 70147 75 29829 90-88978 33 6507784 66-20953 80-06877 97-34316 84 67-730 l^J 69 '88791 85-06696 104-18876 85 60 40ir08 73-66222 90-32081 111-43478 86 6ii-27r)94 77-59831 95-83628 119-12087 87 66-174-22 81 •702-25 101-02814 127-26812 38 6915945 85-97034 10770954 135-90420 39 72-2;]4'i3 90-40!) 15 11409502 146-05846 40 75-401;*6 9502551 120-79977 154-76196 41 78-60330 99-82664 127 88976 166-04768 42 82-'y-2820 104-81960 135-28176 175-95054 43 8.'r 48389 110-01238 142-i)93a4 187-50758 44 69-04841 115-41288 15M43>1I> 199-75808 45 92-71938 121-02939 159-70015 212 74351 46 96-rj0416 126 87957 168-68,516 220-50813 47 100-39660 182-94539 139-26321 178-11P24 241-09861 48 104-40S;}9 188-02589 256-50-163 40 108 54065 145-a3;^7i 198-42666 272-95840 60 112-79637 152-6C70& 209-8479& 290-33590 (8«CT. XL Aet. 55.] ANNUITIEa 868 r OF |1 OR £1. TABLE OF PRESENT VALUES OF AN ANNUITY OF (1 OR £1. 71 6 per cent. 16 m 66 I 1-00000 2-06000 8-18860 4-37462 5-68706 607.')82 8-393''4 9-89747 11-49181 18-18079 14-97164 16-86994 18-88214 21-01506 2327593 25-67258 28-21288 80-90565 83-75999 86-78559 89-99273 43-39229 46-99583 60-81568 64-86451 69-15639 68-70676 68-52811 78 63980 79-05819 84-80168 90-88978 97 -348 16 104-18876 111-48478 119-12087 12726812 185-90420 145-05846 164-76196 165-04768 175-95054 187-50758 199-75S08 21274351 226-50813 a41-09&61 256-50-153 272-95840 290 -3:3590 s 1 No. of Payments. 8 per cent 4 per cent 6 per cent 6 per cent 1 1 0-97097 0-96154 0-95288 0-94340 1 2 1-91847 1 -8861 9 1-86941 1-88339 1 8 2 -82861 2-7T519 2-87519 2 07301 4 8-71710 3C-.:999 8-M595 8-46510 6 4-57971. 4-451S2 4 82948 4-21 '286 6 6-417 la 5-24214 5-(»7569 4-91732 7 6-23023 6-00205 5-78687 6-5S288 8 7 019C-9 073274 6-4G321 0-20979 9 7-78G11 7-485:^ 7-10782 6-80169 ■ 10 8-53920 8-11 0&9 7-72173 7-86009 I 11 9-25202 6-76058 8-80641 7-S668T 1 12 9-95400 9 -.-'.8507 8 86.326 8-88384 1 18 10-68496 99S5fi5 9-31)857 8-65268 1 14 11-29607 I0f,6812 9-S9S64 9-29498 1 15 1193794 11-11849 10879f^ 9-71226 1 16 12-56110 11 -65-239 10-88777 10-1 ons9 1 17 1316612 12-1 6.567 11-27406 10-47726 1 18 13-75351 12-65940 11-68958 10 82760 1 19 14-32380 1318394 12-08532 11-15811 1 20 14-87748 13-59032 12-46221 11-46992 21 15-41502 1402916 12-82115 11-76407 22 15-93693 14-46111 13-16800 1204168 23 16--44.361 14-85648 13-4SS67 12-30888 24 16-93654 15-24696 13-79S64 12-5r,<'86 25 17-41315 15 62208 1409894 12-78385 26 17-87684 15-98277 14-87518 1309316 27 18-32703 16-8-2958 14-64303 18-21058 28 18-76411 16-66306 14-89812 18-4(.616 29 1918846 16-98871 15-] 4107 18-69072 80 19-60044 17"29'208 15-87245 18-76488 81 20-00048 17-58849 15-59281 13-92908 82... .. 20 88877 r ■87355 15-SC267 14-08404 33 20-76579 1811764 1 6-00266 14-2o(i28 34 2118184 1841119 16-19290 14-86814 85 a 1 •48722 18-66461 16 -.8741 9 14-49824 86 fl-8.S926 18-90828 10-646>>-6 14 -62099 37 22-16724 19-14258 16-71123 14-73678 88 , 22-49246 19 86786 10-86789 14-84602 39 89-80822 19-68448 1701704 14-94907 40 im-11477 19-79-277 17-1.''.908 15-94630 1 41 23-41240 19-99305 17-2i»436 r'->-13S01 42 23-70186 20-lS;-.62 17-42320 .j-2'2464 48 23 -981 90 20 87079 17 54591 16-80617 1 44 24-25428 20-54S44 17-(;6277 15-3!i318 1 46 24-51571 20-7L'0O4 17-77407 15-4^588 1 46 24-77646 20-8S465 17-88006 16-52487 1 47 25-02471 9104298 179R101 16 58903 B 48 26-266n 21 1951 8 18 07714 15-65003 1 1 49 25-59166 21-50166 18-10872 15-71.757 » 1 ■* 26-72977 21 -72977 18-25.592 15-76186 \mf^^ iMfa iFKiiiiaaa I: l» 364 ANNUITIES. APPLICATIONS. ISect. XI 56. To find the amount of an annuity forborne for any number of years at compound interest : BULK. A=^^^l±rl=^ (..) Interpretation. — From the amount raised to the power inJ'ca, I bi/ the nurtiber of ■payments subtract 1 and multiply the remainder i.'i the annuity. Lastly : divide the sum thus obtained by the rate jicr unit and the quotient will be the required amount. By the Table, — Find from the t(d)le the amount of $1 for fl- given number of pcyments and at the given rate ; multiply it by lA given annuity and the quotient will be the amount. Example. — If a yearly rent of $400 be forborne for 23 year?, to what sum will it amount at G per cent, compound interest? :l if;- . \ *=!■ Then A OPBEATION. Here a =400, <=2.3, r= 05. ~ g|0-^^) ^ — U ^0 1 (1 -05)28-1 } _ 400 y 2-0714T6 828M0 5= $16671-80. Ans. By tub Table.— Amount of |1 at the given rate and time — |4r4S04T. Then $41-43047x400 = $16572-188. NoTS. — These two methods fiive results slightly different. This ari.sos from the fact that the table 8h<>\v8 only an lipproxiiuation to the corroct auionni of the annuity for $1 ; all the flgurea except the first live of its decimal btir.g re- jected. 67. To find the present value of an annuity at com. pound interest :— ''="]'-(Ti7y.}(^-> BULK. (I+r/ Interpretation. — Divide one by thai power of the amount of ?1 xrhicli is indicated by the ^number of payments and subtract ihe resmi fi\.7n L Mtdtiply the remainder by the quotient arimig from the dhif^inn of the given annuity by the rate per unit and the result will be ih required present value. By the Table, — Find the present value of an annuity of %\ fot the given number of paymenti and at the given rate^ and multiply tJiu bi/ the given annuity. [Sect. XI forborne for any Ak'trt U-b'i.] ANNCltlEB, the power ind'ca',1 [ply the remaiiitlcr '' v lined by the rate //( r i. mount of $1 /"'• '' •i ; multiply it b^ »■' nt. oovne for 23 ycaiv, to lid interest? 400 X 2-0714T5 828f)i10 "•o5~ - " -OS ndtime = t41-4C04T. ffereiit. This avisos from Ito tlif correct, (inionn' "l of its decimal btii.t' r»-'- In annuity at com- 365 EXAMPLB. — ^What i3 tlic present value of an annuity of $40, to continue 5 years, allowing 5 per cent, compound interest ? OP^RATIOH. Here a = 40, < = 6, and r = -05. <* \ , 1 J 40 ( , 1 I 4000 , _-„_. Then. = --j 1 -^^ = ,-^ X J 1-^^^ f = _^-- . 1-.T885) = 800 X -216.5 = til 3-20. Ane. Ob by tub Tarlb — Pre«ont value of an annaity of %l for given vito and tlniH = $i32y43 fttid $4-32948 x 40 = |173-179. Anff. 68. To find the present worth of a perpetuity : — RDLl. (rx.) a r Interpretation. — Divide the annuity by the rate per unit and ihe quotient will he the value of the perpetuity. Example. — What is the present value of a freehold estate of $75 —allowing the purchaser 6 per cent, compound interest for hia money ? OPERATION. Here a ~ T5, and r = -O* a 75 ^ = liafiO. Ant. 69. To find the present worth of a perpetuity in re- version : — RUUL a V= - Kl + r)' (XII.) of the amount of '*! land subtract ihe rf.N'.-.'''| Ihinfj from the dhlsir:n\ the \'emlt wiU be tht\ I an annuity of ^\ fon Irate^ and multiply thi"] Interpretation. — Find that power of the amount of $1 for one payment that is ^^^dicated by the number of payments that have to elapue before the annuity reverts, multiply this by the rate per unit and divide the given annv'ty by the product — the result will be the present value. Example. — What is the present value of the reversion of a per- [petuity of $'79-20 per annum, to commence 7 years hence — allowing jtlie buyer 4^ per cent for his n^oney? OPKEATIOK. Here a = 7920, t— 1, and r_~ '04IIS. Then r- a 79-20 79-20 |li98-8tf7. Am. r(l + ry -045 x CI •# ••itfi/ -046 x 1 -860862 _jr9-50_ ■061288rit >r f 866 ANNUITIES. liifft. tl [3*jaiii r h 60. With due attention to the forc^c'^j^ ii tcrpretations imd examples, the pupil will not expeii'^nc'^ iuy Jiilicuity m applying the remaining formiihe. Exercise 164. 1. What is the annual rental of a freehold e.state, purchased for $3oOO when the rate of interest is at 4 per cent. ? y'.ns. $120. 2. If a perpetuity of $503 can be purchased for $11260 ready money, what is the rate of interest allowed ? Ans. 6 per cent. 8. A freehold estate producing $75 per annum is mortgaged for the period of 14 years ; what is its present value, reckoning com- pound interest at 5 per cent, per annum ? Ans. $757-608. 4. Required the present value of a deferred annuity of $90, to be entered upon at the expiration of 12 years, and then to be con- tinued for 7 years at 4 per cent, compound interest. Ans. $387-39. 6. What is the present value of an estate whose rental is $1500, allowing 5 per cent, compound interest? A71S. $30000, or 2i) years' purchase. 6. For how many years may an annuity of £22 be purchased for £308 123. lOd., allowing compound interest at 4 per cent. ? Ans. 21 years. 7. What is the present value of an annuity of $154 for 19 years at 5 per cent, compound interest? Ans. $186ri3. 8. What annuity, accumulating at 3| per cent, compound interest, will amount to £600 in 40 years? Ans. £6 18s. lid. 9. In how many years ^iil pn annuity of $8 per annum amount to $187 yi 5625 at .i ^i^er cent, compound interest? Ans. 18 years. 10. What will an annuity of $74 amount to in 30 years at 4 per cent, compound interest ? Ans. $4160-28. QUESTIONS TO BE ANSWERED BY THE PUPIL. NoTR. — The nwnibera after tJie questions refer to the numbered artieU* of the tection. 1. When are quantities said to be In arithmetical progression ? (1) 2. What ftftt the extroaies? What the means V (2) 8. What five quantities are to be ccjnsldered in arlthmatical progression ? (8) 4. How are these related tc each olhury (8) 6. Uow many cams xrise from tb?se oomblnations f (8) t^jJOT. Xl. AKT. 00. "J F.X\.v11NaT10N peoblems. 367 I* 0- i i tcrpretations v ice any viiuicuity state, purchased for cent. ? j.ns. $120. 111260 ready money, Ans. 6 per cent, is mortgaged for the ilue, reckoning com- Ans. $757-608. nnuity of $90, to be aud then to be con- I interest. Ans. $337-39. lose rental is $1500, r 20 years' purchase. 2^ be purchased iov t at 4 per cent. ? Ans. 21 years. 54 for 19 years at 6 Ans. $1861-13. compound interest, Ans. £6 18s. lid. ;r annum amount to •est? Ans. 18 yeari?. ) years at 4 per cent. ^ Ans. $4150-28. 'HE PUPIL. the numbered artieU* ession ? (1) tical progression f (8) rv 1. a 9. 10. 11. 12. 18. 14. 15. 16. 17. la 19. 20. 21. 22. 28. 24. 25. 26. •27. 28 29. 30. 81. 82. ^>o(1iico thi- fundaJnontal fopmnliB for arithmetical propressici. (4-T) When iire qu;iinnio>i suM Lo b(> in geometrical projrressi.m ? ('S) What Ave inM.iou8? (16) Deduce the fundamental f»)rmul8e for geomotrical progression. (i7-l9) What rule do you use when dnding the sum of any inhnite series when the ratio is less than 1 ? (25) Prove this vuie. (2.5) How do we insert any number of arithmetical means between two given ex- tremes? (26) How do we insert any number of geometrical means between two ex- tremes? QiQ) What is position ? (27) Into what rules is porition divided ? (28) When is a single positior used ? (29) What class of questions require the use of double position? Give and prove the common rule for slnfflo position. (32) Give and prove the common rule for double position. (34) Deduce algebraically a complete Ret of rules for compound interest What is an annuity ? (40) When is an annuity said to be in posses-^ion ? (41) What ia a deferred annuity or an annuity in reversion ' (45^ What is a contingent annuity? (44) What is a perpetuity ? (45) When is an annuity said to be in arrears ? What is the amount of an annuity ? (47) What is the present worth of an annuity ? Deduce a set of rules for computing annuities at simple interest. Illustrate the absurdity and injustice of computing tno present value of an- nuities at simple interest. (pO) Deduce a sut of rules for aunmties at compound interest (54) (80) (85) (46) (4S) EXJCRCISI 16& EXAMINATION PROBLEMS. FIRST SERIES. 1. Write down as one number sereD trillions and ninety millions, and nineteen and four million two hundred thousand and six hun- dredths of trillionths. 2. Deduct 19 per cent, from $7580 and divide the remainder among A, B, C, and D, so that A may have $111*11 more than B ; B $90-90 more than C, and D one third as much as A, B and C together. 8. 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 ? 4. Reduce £179 14s. 8f d. to dollars and cents, and divide the resvlt t)y -00000048. ft. What ia the 1. c. m. of 44, 18, SO, 77, 66 and 27 ? ^'^p^ y ',' ^ n 868 KXAMmATIOK PfiOBUM^ IS] I '« I II 6. In what timo will any sum of money amount to 20 times itself at 6^ per cent, simple interest ? 7. Divide 7842163 octenary by 61861 nonary^ and give the answer in the duodenary scale true to two places to the right of the separating point. 8. Multiply 43 lbs. 3 oz. 17 dwt. 11 cts. by 783f 9. Find the sum of the series 1 +i+i+i, ad infinitunu 8 10. Divide ( of | of 192 by — ^ i 11. Extract the 17th root of 129140168. 12. 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 inverted. What is the number f SECOND SXRIX8. 18. Divide $897*48 among A, B, and G, so that B may have |98'40 less than A, and $69*18 more than C. 14. If 7 lbs. of wheat contain as much nutritive matter as 9 lbs. of rye, and 6 lbs. of ryo as much as 8 lbs. of oats, and 18 lbs. of oats as much as 21 lbs. of buckwheat, and 27 lbs. of buck- wheat as much as 20 lbs. of barley, and 24 Iba of barley as much as 25 lbs. of peas, and 11 lbs. of peas as much as 85 lbs. of potatoes ; how many pounds of potatoes contain as much nourishment as 16 lbs. of wheat ? 9 IB. Reduce | of 4^ of 7| of — of f of 8 oz. 4 drs. 2 scr. 8 grains 19i to the decimal of 1^1- of -63' of 2^ of ^ of 6^ times 7 lbs. 8 01., Apothecaries' Weight. J6. From 623*^2793 take 98*4267192 ; mark diitinctly the resulting tapetend. I'h If ii own a vei«el valued at $7498 and wish to insure it at a pre- mix.m of A\ per cent, so as to recover, in case of the destruo- tio I of the vessel, both the premium paid and the value of tb« vessel, for what sum must I insure ? 18. If 18 men in 20 weeks of 6 working d*yi each, working 11 hours a day, dig 11 cellar^ eaoh ao f««t long, 16 fott wfff r-.> EXAMINATION PKOBLEMB. 869 ) 20 timea itself at B may have |9S-40 and 5 feet deep ; how many men will be required to dig 24 cellars, each 22 feet square and 4 feet deep, in 86 weeks of 6 days each, working 9 hours per day? 19. A certain number is divided by 9 and the quotient multiplied by 17; the product is then divided by 300 and 38 is added to the quotient ; the result is next divided by 3, and from this quo- tient 31 is subtracted, and the reulting dlftorence divided by 12^. Now ^ of ^ of f of this lust quotient is 2^^^. Required the original number. 20. What is the 1. c. m. of 480, 768, 348, and 1176? 21. What is the G. C. M. of 17598, 46090, and 171347? 22. In a certain adventure A put in $12000 for 4 months, then add- ing |8i)00, he continued the Avhole two months longer; B put in 5j250r)O, and after three months took out $10000, and con- tinued the rest for 3 months longer; C put in $35000 for 2 months, then witlidiawing f- of his stock, continued the remain- der for 4 months longer; they gained $15000; what was the share of each ? 23. Three merchants traffic in company, and their stock is £400 ; the money of A continued iu trade 5 months, that of B 6 months, and that of G 9 months ; and they gained £375, which they divide equally. What stock did each put in? 24. A fountain has 4 pipes. A, B, C, and D, and under it stands a cis- tern, 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 pipes, 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 8 pipes are all open, in what time will it be t jiptied ? ) i drs. 2 SCI*. 6 grains I 1347'90 for carriage, $479 40 for in.su- rance, $169 83 for storage, wharfage, and harbour dues, and an cui valorem duty at 2} per cent, on the invoice of goods. Re- quired how many yards of cloth my agent ships to me and what J gain or lose per cent, on the whole transaction if I sell tlie goods for $25000. Transpose 1S4234 quinory into the ternory^ odcnarr,^ and duode- nary scales, and prove the results by reducing all four numbers to the denary scale. 9f What is the difference between ^ of 4^ of j-^ of -^ of I of £4.'i J fo 18s. ll^d., and 3^ of -- of -56 of 1-75 of 6^ times $97-18 ? 34. Given the logarithm of 2=0-3010oO 3 = 0-477121 13 = 1-113943 Find the logarithms of -jij, 19*5, 1125, 28-16, 05000, -0005, 152-1, and 8112. 86. Extract the cube root of 871^e/'72 duodenary true to two places to the right of the separating point. 86. A person passed ^ of his age in childhood, -^ of it in youth, \ of it 4-5 years in matrimony ; he had then a son whom he survived 4 years, and who reached only \ the age of his father. Av what age did this person die ? FOURTH SERIES. I 87. Divide 63 miles 3 fur. 7 per. 3 yds. 2 ft. 7 in. by 7 fur. 23 per. 3J yds. 38. Divide 6-3 by 000000274. 89. Tf ^ yards of cloth cost $|f , how much will 6-,^, yards cost ? itO. Find the interest on $4237-71 at 6^ per cent, for 167 vaars. 41. In what time will $67430 amount to $1000 at 8i per cent. ? 42, What are the amount and compound interest of $813-71 for 7 years at 4 per cent, half-yearly ? paid as follows, viz. : $300 down, $700 af tho end of 4 months. ''5750 at the end of 7 months, §850 at the end of 9 months, ^ ■ wu at the end of 13 months, and the balance at the end of 19 m'"ntU.j. Recjuired the equated time for the whole debt. « .:i EXAMINATION rilOBLEMB. 371 jars, allowing sinrf d instruct him to uvest the balance •eceive the goods I , $479-40 for insu- rbour dues, and an ice of goods. Ko- ips to me and what action if I sell the "ten an;, and duodc- ig all four uumbers of ^ of i of m 6i times ^OI-IS? 8-lG, C5000, -0006, w true to two places of it in youth, | of ,n whom he survived of his father. A\ in. by 7 fur. 23 per. •^ , , yards cost? for 1(57 years. it 8i per cent. ? St of ^813-71 for 7 $300 down, $7f^0 sit months, $850 at the iths, and the balance quated time for the 44. Deduct 23 per cent, from $t2r>0 and divide the remainder be- tween A, li, C, D, and E, so that A may have $1710 more than B, C §19-23 k's.*i than B, I) $42- 11 less than C, and E half as much as A, B, C, and D together. 45. What pinicipal put out at simple interest at 16 per cent, will amount to $3780-80 in 11 years? 46. I'ind the value of \ (=^ ^-2.\)-) >^ -40-^^ of •i42S.-)7 [ -f-8^ times (A4-| + j^— a^Vt?) j773x-l23i5--^?^,;)-f-?+>'^ + l''n [ -r27.4<>i2077 47. Add together 312312302 aud 2312132 (piafernar;/ ; multiply the sum by twenty-three thousand and {?h>von times 4284 qninarii ; from the product subtract 555 -h 414 + 333 4- 222 + 111 Henarij ; divide the reiuaiiiJer by 6642 septenary, and give the answer in the octcnary scale. 48. What is the square of '1 and also of 1 ? FIFTH BEKIES. 49. 50 61, 62 Read the following numbers : 100();5O0:5( 10(500 00070080009. 7000290034007 000000007 400209. ♦ Find tiie 1. c. m. of 2, 9, 16, 27, 48, and 81. In what time will any sum of money an-.ouut to 7 times itself at 6 per cent, per annum compound interest? How often will a coach vvlieel turn in going from Toronto to Brampton, ft-distance of 20 miles ; the wheel being 14 ft. Ivi in. in circnniCorcnce ? 63. How many divisors has the nutnber 1749000? 96 , i of 7 64. Divide % of -r by -.— 6^. A can do a piece of work in 12 days, and A and B together can do it in 5 days ; in what time can B alone do it ? What principal will amount to $>;899-77 in 11 years at 6 per cent, half ytarly, compound interest ? Divide the number 10 into three such parts, that if the first be multiplied by 2, the second by 3, and the third by 4, the three products will be equal. 58. There are three fishermen. A, B, and C, who have each caught a certain nunjber of fisli ; wlien A's 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 be divided eciually among them, what will be each man's sljare ; and iiow many li4i did each of them catch ? 66 67 872 EXAMINATION PBOBLEMB. 69. What is tho forty-ficventh term and also the sum of the first 98 terms of the series 7, 11, 16, 19, &e. ? 60. In what time will any sum of money amount to 21 times itself at 7 per cent, uompound interest f |i li SIXTH 8KRIES. 61. Divide $3700 among three persons, A, B, and C, bo that B may have ^387 less than A and $iyt)'87 more than C. 62. What are all the divisors of 6710? 68. What is the value of ] (ITtV— lO^O— (-4+^ -f--9— ^) } -!-(-8l}7C-+-i of 81) •6322032 xi of 4-f-(^ of 4^ of j^,- of 85^^-1-101) 64. Divide $7200 among 3 men, 4 women, and 17 children, giving eacl) man twice as much as a woman, and eacli woman three times as much as a child. Wliat is the share of each? 66. How many divisors has the number 25100? 9^ 66. What is the difference between 5 of 4^ of YY of ^ of £3 168. lUd. and A of 4| of -^^of i^jfy of H of -85 of ^jgl^*^ ^1*^83? » 67. Compare together the ratios 7 : 13, 9: 16, 8 : 15 and 10 : 19 and point out which is the greatest, which the least, and what the ratio compounded of these given ratios. 68. Divide 67-432 by 7-9036. 69. Reduce 9 per 9 yds. 7 ft. 1 20 in. to the decimal of A of | of ^ of 35 acres 2 roods. 70. Add together 170342, 2700357, 98-123456, 829-642.% 986-1284298, 9-870.342, and 813-9864234507. 71. In the ruins of Persepolis there are two columns left standing upright. The one is 04 feet above the plain and the other 50. In a straight line between these stands a small statue, the head of which is 97 feet from the top of the higher column and bO feet from the top of the lower, the base of which is 70 feet from the base of the statue. Rc(iuired the distance between the tops of the columns. '72. 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 ? EXAMINATION PROBLEMS. 878 n of the first 93 1 times itself at ', BO that B may C. (■8: nG-4-^of 31) S-of 85^5^-1-101) chiUlrcn, giving ch woman three [)f each ? f ^ of £3 168. of -^-^ of $1783? 5 and 10: 19 and St, and what the I of i of I of ^ of 3456, 829-6423, imns left standing II id the other 50. statue, the head column and bO which is 76 feet distance between e plus 25 gallons lUons was water. 8IVENTH SERIES. 73, Extract the square root of 401241-8424 in the quinary scale. 74. A father being asked by his son how old he was, replied, your ago is now ^ of mine ; but 4 years ago it was only \ of what mine i.s now : what is the age of each ? 76. Divide •7'284^ by -0032. 76. Extract the 11th root of 97294764-372. 77. Find two numbers, the dittercncc of which is 30, and the relation between them as 7i is to 3J^. 78. What is the 1. c. ra. of 35, 16, 18, 28, 62, 63 and 40? 79. Sum the series l-+-7 + 13-t-l'.>-f-&c., to 101 terms. 80. What is the ratio comnounded of 19 : 7, 11 : 66, 36 : 121, Uf : 29, 8:4jand4f:3V 81. Find two nimibers whose sura and product are equal, neither of them being 2, NoTR.— In thi- question take any numbor for the first of the two, as for ex- Ample 7. Then T + some otht-r nuiul)or = 7 x that other number. Assume fur this second number any other, iis 3. Then 7 + 3-10-7x8, elves an error of— 11. Assume some (tther for the second, ua .'3. Then 7 + 5 = 12—7 x 5, gives an error of-23. 82. 84. Then 2;)y8=61) 11x5=55 14 Whence second nuiubcr = — = U. 12 Find the value of ({(y^+4H4-3i— 16a^)x Mf -j-l^)x 36 times -142867. { -97 X -24378 x (1^^ x 4-4^^)} x (4A-2-,V)- 83. The hour and minute hands of a watch are together at 12 ; when will thoy be together again ? Given the logaritlnn of 2=0-301030 logarithm of 7 = 0-845098 logarithm of 11 = 1-041393 Find the logarithms of 3850000, 3181 -si, -0000154, yV, 1-671428 and 93-17. ■IGHTII SERIES. 86. Find the difference between die simple and compound interest of $700 in 3 years at 4^ per cent, per annum. 86. X, Y, and Z, form a company, X's stock is in trade 3 months, and he claims -,^ of the gain ; Y's stock is 9 months in trade ; and Z advanced $3024 for 4 months, and claims half the profit. How much did X and Y contribute ? ■I A: %. # ^%. s> ^ k5» ^ # o^ ?v or /A Photographic Sciences Corporation \ s ,v c^-^ .^ \\ ^ ■^ V <5 o .v^ Aj V ^•b"- 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 Va %3 i ^' ^^ .v« 374 EXAMINATION PROBLEMS. 87. There is a fraction which multiplied by the cube of 1^ and divided by the square root of IJ, r)rodiice3 f ; find it. 88. Find the cube root of 80677568161. 89. How much sugar, at 4d., 6d., and 8d. per lb. must there be in 112 lbs. of a mixture worth 7d. per lb. 90. Find three such numbers as that the first and J 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. Note. — Assume 40 as the sum of the three numbers. Then l8t + 2nd + Brd=40 aud lst + i(2n(l +3ril)=84. •.i(2nd +8rd)=6 and 2nd+8rd = 12. 2nd+Hl8t + 3rd)=84.-.|(lst + 3rd)=6nnd lst+3rd=9. 8rd +i(lst + 2rid) = 34 . •. J(,l>t + 2nd)=fi and lst + 2nd=8. Then addini: these together, twice (l3t + 2nd +8r(l)=29. • . lst + 2nd + 3rd~ 14i=:sum. But should equal 40— therefore error— —'25J. Similarly as.sume some other number and ajiply the rule, and the true sum 68 will be found, from which the numbers may be easily obtained. 91. Insert 4 arithmetical means between 1 and 40. 92. The sum of all the terms of a geometrical progression is 1860040, the lust term is 1 240029, and the ratio is '6. Wiiat is the first term? 93. 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 ? 28i 94. Multiply i of f of § of by | of ^ of J. 96. From a sum of money, $60 more than the half of it is first taken ajkay ; from the remainder, $3() more than its fifth part ; and again from the second remainder, $20 more than its fout Lh part. At last there remained only $10. What was the original sum? 96. A gentleman hires a servatit, and promises him, for the first year, only $60 in wages, but for each following year $4 more than the preceding. How much will the jcrvant receive for the 17th year of his engagement, and how much for all 17 years together ? NINTH SERIES. H 1 97. Write down as one number eleven trillions and eleven, and eleven tenths of hillionths. 98. Reduce £749 168. 5fd. sterling to dollars and cents. 99. What are the prime factors of 177408 ? lOo. At what rate per cent, per annum will $704 amount to $1 1 1 1 1 "U in 11 years at compound interest ? EXAMINATION PEOBLEMS. 375 ; of li and divided ,t. must there be in i i the sum of the the other two, the lake 34. •.i(2nd+8rd)=6and + 3rd=9. lst + 2n(\=8. 29.-.lst + 2nd + 3ra~ ule, and tbe true sum >R6ily obtained. 0. srresplon is 1860040, i' Willi t is the first apples and 8 pears ,1 ^,.f. t>oor 9 nd one pear \e half of it is first e than its fifth part •, more than its foui lli ^'hat was the original es him, for the first iwing year $4 more ■rvant receive for the luch for all 17 years ons and eleven, and nd cents. amount to $11 lU-li 101. How many scholars are there in a school to which if 9 be added the number will be augmented by one-thirteenth ? 102. Three diilerent kinds of wine were mixed togetlier in such a way that for every 3 gallons of one kind there were 4 of another, and 7 of a third : what quantity of each kind was there in a mixture of 292 gallons? 103. Divide £500 among four persons, so that when A has £i, B sha'.; have £^, C ^, and D ^. 104. What is the present worth of an annuity of $100 to continue 23 years, at 6 per cent, compound interest ? 105. Twenfy-five workmen have agreed to labor 12 hours a day for 24 days, to pay an advance made to them of $900 ; but hav- ing each lost an hour per day, five of them engage to fulfil the agreement by working 12 days: how many hours per day must these labor ? 106. A man has several sons, whose ages are in arithmetical progres- sion ; the age of the youngest is 5 years, the common differ- ence of their ages is 6 years, and the sum of all their ages ia 161. What is the age of the eldest V 107. If a man dig a small square cellar, which will measure 6 feet each way, in one day, how long will it take him to dig a simi- lar one that shall measure 10 feet each way ? 108. A servant agreed to live with his master for £8 a year, and a suit of clothes. But being turned out at the end of 7 months, he received only £2 133. 4d. and the suit of clotlies : what was its value ? TENTH SERIES. 109. 110. 111. What number is that of which ^, J^, and J added together, will make 48 V If an ox, whose girth is 6 feet, weighs 600 lbs., what is the weight of an ox whose girth is 8 feet ? Four women own a ball of butter, 5 inches in diameter. It is agreed that each shall take her share separately from the sur- face of the ball. How many inches of its diameter shall each take ? 112. Divide 71214-43 by 12'342 in the nonary scale and extract the square root of the quotient true to three places to the right of. the separating point. Five merchants were in partnersliip for four years ; the first put in $60, then, 5 months after, $800, and at length $1500, four months before the end of the partnership ; the second put in at first $600, and six months after $1800; the third put in $400, and every six months after he added $500 ; the fourth 113. 376 EXAMINATION P110BLEM6. dill not contribute till 8 months after the commen cement of the partnership ; he then put in $900, and repeated this sum every six months ; the fifth put in no capital, but kept the acv counts, for which the others agreed to pay him $1*25 a day. What is each one's share of the gain, which was $20000 ? 214. In what time will any sum of money amount to 1 6 times itself at five per cent, per auniun. Ist. at simple interest ? 2nd. at compound interest? 116. Three persons purchased a house for $9202; the first gave a certain sum ; the second three times as much ; and the third one and a half times as much as the two others together : wha< did each pay ? j16. a piece of land of 165 acres was cleared by two companies of workmen; the first numbered 25 men and the second 22 ; how many acres did each company clear, and what did the clearing cost per acre, knowing that the first company received $86 more than the second ? 117. The greater of two numbers is 15 and the sum of their squares is 346 : what are the two numbers? 118. To what sum will $1200 amount in 10 years at 9| per cent. 8im« pie interest? 119. If 496 men, in 6| days of 11 hours each, dig a trench of 1 de- grees of hardness, 465 feet long, 3^ wide, 2^ deep, in how many days of 9 hours long will 24 men dig a trench of 4 de- grees of hardness, 387^ feet long, 5| wide, and 3^ deep? 120. Four men, A, B, C, and D, took a prize of $6213, which fhey are to divide in proportion to the following fractions : {/pos- sible^ A, B, and C, are to have |5 ; B, C, and D, §^; A, C, nnd D, T^tf ; and A, B, and D, f of the prize. What does eacL re- ceive ? ELEVENTH SERIES. 121. Reduce -7, -83, -727, -91325 and 8-671347 to their equivalent vulgar fractions. 122. Reduce 713|H undenary^ and 12123-iyoWiJ quaternary to equivalent expressions in the denary s'^ale. 123. Add together 3f of 2^ of 7H of a £, 9f of 3f of a shilling, and 8| of 4^ of a penny, and divide the sum by \^ of 5-i\ of ^ of 3^d. 124. If 24 pioneers, in 2^ days of 12|- hours long, can dig a trench 139 75 yds. long, 4^^ yds. wide, and 2^^ yds. deep, what len>(th of trench will 90 pioneers dig in 4^ days of 9^ hours long, Uw trench being 4| yds. wide, and 8^ yds. deep? _ EXAMINATION PR0BLEM8. 877 commencement ot 1 repeated this sum il, but kept the acv V him $1-25 a day. i was S20000 ? t to 1 6 times itself I interest ? 2nd. at 2 ; the first gave a ich; and the third lers together : wha( Y two companies of the second 22 ; how hat did the clearing apany received $86 im of their squares at 9^ per cent, sim' ig a trench of *l de- 3, 2^ deep, in how ig a trench of 4 de« , and Z\ deep? $6213, which fhey ig fractions : ifpos- ad D, t^; A,C,rind What does eacL re- to their equival'^tit -^§xi quaternary to pif of a shilling, and 7 H of 5-A- of i of ig, can dig a trench I. deep, what length f 9^ hours long, Uw 125. A person, by disposing of goods for |182, loses at the rate of 9 per cent. : what ought they to have been sold for to realize a profit of 1 per cent. ? 126. In what time will any sum of money amount to 11^ times itself at 6 per cent, per annum ? 1** At simple interest ? 2°* At compound interest? /27. It is desired to cut off an acre of land from a field 16^ perches in breadth ; what length must be taken ? Express a degree {Q9-^i miles) in metres, when 82 metres are equal to 35 yards y Find 7 geometrical means between 3 and 19683, Sum the infinite series 7 + If + tV, &c. Four men bought a grindstone of 60 inches diameter. Now, how much of the diameter must be ground off by each man, one grinding his part first, then another, and so on, that each may have an equal share of the stone, no allowance being made for the axle ? Divide 100 guineas into an equal number of guineas, half- guineas, crowns, half-crowns, shillings, and sixpences, ttid reduce the remainder to a fraction of a pound. 128. 129, 130 i31, 62. TWELFTH SERIES. 138. The owner of tV o^ a s^P ao^tl -^[ of f of hia share for $12^^^ ; 2i- what would rj" of I of ^^^ ship cost at the same rate ? 184. At what rate per cent, per annum will $700-90 amount to $1679*40 in 5 years, compound interest being allowed? |86. A person paid a tax of iv per cent, on his income ; what must his income iiave been, when, after he had paid the tax, there was $1250 remaining V 136. The sum of £3 138. 6d. is to be divided among 21 men, 21 women, and 21 t lildren, so that a woman may have as much as two children, and a man aa much as a woman and a child ; what will each man, woman, and child receive ? .^7. Distribute $200 among \., B, C, and D, so that B may receive as much as A ; C as much as A and B together, and D as much as A, B, and C together. /88. Find the difference between Vf and V|. /39. Reduce ^WoS", 17^ + h + U^\, 2i| _ ^i, f of f x tV of H of f i, and 6347 -*- 2f, to their simplest forms. Z40. Find the cube root of 884786, and the fourth root of 96951^. [ I lit 878 ARITHMETICAL RECREATIONS u 141. A general levied a contribution of $520 on four villages, con- taining 250, 300, 400, and 500 inhabitants respective!}' : what must they each pay ? 142. A person had a salary of $520 a year, and let it remain unpaid for 17 years. How much had he to receive at the end of that time, allowing 6 per cent, per annum compound interest, pay. able half-yearly ? 143. Insert four arithmetic^ means between 2 and 79 ; also find thp. 9th term and the sum of the first 207 terms of the series 3, 7, 11, 15, &c. 144. A, B, and C, start at the same time, from the same point, and in the slime direction, round an island 73 miles in circum- ference ; A goes at the rate of 6, B at the rate of 10, and C at the rate of 16 miles per day. In what time will they be all together again ? \ ^ ARITHMETICAL RECREATIONS. lIKifr the third of 6 be 8 what must the fourth of 20 be ? 2. If the half of 5 be 7 what part of 9 will be 11? 8. Place fdor nines so that their sum shall be 100. 4. What par. 6t'6 pence is tlie third of two pence? 6. If a herring and a half cost IJd., how much will 11 herrings cost? 6. If 12 apples are worth 21 pears, and 8 pears cost a cent, what will be the price of 100 apples? 7. Find a number such that 5 shall bo the three-sevenths of it. 8. A hundred hurdles are so placed as to inclose 200 sheep, and with two hurdles more the field may be made to hold 400; how is this to be done? a 9. A gentleman who owned four hundred acres of land in the form of a square, desired t« keep 100 acres, alt-o in the form of a square, in one corner, and divide the re- mainder, abed ef, equally among his four sons, so that each son sliouid have his lot of the same shape as bis brother's. How may this be done ? d / 10. Place four tht'ees So as to make 34. 11. Write down IS in such a way that rubbing half of it out 8 shall remain. 12. Two thirsty persons cast away on a desert island, find an 8 gallon cask of water. They wish to divide it equally between them, but have no other measures than the 3 gallon cask, a five gallon cask and a three gallon cask. How cau they divide it? 18. How must a board 16 inches long and 9 inches wide be cut into two such parts, that when they are joined together they may form a square ? 14 Place the 9 digits in the accompanying figure, one digit to each division, in such a way that when added vertically, horizon- tally or diagoually, the suid shall always be the same, U. bur villages, coii- spectively .- what it remain unpaid at the end of that und interest, pay- 79 ; also find thft )f the series 3, 7, e same point, and miles in circum- e of 10, and C at } will thev be all S. rings cost? jnt, what will be the of it. sheep, and with two w is this to bo done T a in in •e- 80 aa d a / it 8 shall remain. d an 8 gallon cask of m, but liave no other i. and a three gallon be cut into two such form a square t t to each horizon- le. 18 19. 20. 21. 22. AEITHMETICAL R "ATI0N9. 879 15. Three persons bought n quantitv of .sugar we. — j 81 lbs., and wish to part it ociually between them. They h.ive no weighta but a 4 lb. weight and 1 7 lb. Wfii,'ht. How can they divide il ? 16. Suppose 2(5 hurdles can be placed in a rectangular form so as to inclose 40 .>*(jijare yards of ground; how can they be placed when two of them are taken away, so as to inclose 120 square yards t 17. A person has a fox, a goose, and a peck of oats to carry over a river, but on account of the smalTness of the boat ho can onlv carry over one at u time. How can this be done so ad not to leave the fox with the goose, nor tho goose with the oatay In a distant villr •^ of Canada, there was stationed a small detachment of tronp.s coiisistiti"; of a sergeant aiitl 24 men. Having coustructel tem- porary barracks, the sergeant divided them into 9 com|)artments, allotting tlia ' entre one to him elf, and the rest to his men. One evening ihe ser- geant, wishing to ascertain if all were in, visited each compartment, and finding 3 men in each, making 9 in each row, retired. Four men, how- over, went out, and the sergeant feelintf shortly afterwards uneasy, rd- turned to count his men, but still finding 9 in each row, retired again ; thti 4 men thm ciitne hack, bringing each anothir man with him, and the Bcrgt'ant upon noing his round once more, counted as before, and retired perfectly sail -fled. 'After he left, four more men were introduced, and cnce more the sergeant, entertaining a suspicion that all was not right, counted, but fliulirig ttie number still the sanje in each row, he left. No sooner had he Ictt, than four more men came in, making 12 stransers,; and once more the sergeant inspected the compartments to his !*atisfao« tion. Finally the 12 strangers left, taking with them 6 of the soldiow, and the sergea it counting once more retired to rest, persuaded that no one had gone out or come in, and that his suspicions were unfounded. How was this possible ? Write down 12 so that by rubbing out one half 7 shall remaiu. Place the first 25 numbers, 1, 2, 3, 4, 5, Ac, in the divisions of the accompanying Hgure, so that the columns added in any order, i. e., upwards, horizontally, or diagonally, may amount to the same sum. What is the difference between half-a-dozen dozen and six dozen dozen? If a cross be made of 13 counters as in the margin, nine may be reckoned in three ways, i. e., by counting from the bottom ©oOoo up to the top of the perpendicular line; from the bottom up to tho cross and then to the right : or from the bottom up to the cross and then to the left. Now take away two of the counters and with the others form a cross which sh.all possess the same property of counting 7„ine when thus reckoned. Seven out of 21 bottles being full of wine, 7 half full and 7 empty— it Is re- quired to distribute them among 8 persons, so that each may have the same quantity of wine and the same number of bottles. Two traveller'*, one of whom had with him 5 bottles of wine and the other 8, were joined by a third person, who, after the wine was drunk, left 8 shillings for his just share of it ; how la this to be divided between the other two? person having by accident broken a basket of eggs, offered to pay for them on the spot if the owner could tell how many ho had; to which he replied that he only knew there were between 60 and 100, and that when he counted them Hy 2's and 3's at a time none remained ; but when h( counted them by 5 at a time there W9re 8 remaining; how many egg? i^ad he ? -■ ^ I I I I U. ^. A ^-- 111 mA,... 380 ARITHMETICAL it£CREATIONS. 28. It Is required to And 4 such weights that thej weigh any number cf ponnda from 1 to 40. 27. In the accompanying flffurc It Is required lo fill seven out of the elttht points with counters ' *he fullowin;; majiner, I. e., the co .t- er has to start from an uuoccu- pied point, puss along the line and be depositeil at the other ex- tremity. Thus, in coninienciiig, the counter may start from any point, since all arc unoccupied, Btartins; from 1 the counter may be curried either to 6 or to 4 and there deposited, suppose It to bo dep<»sited at C, then the next counter may start from any point except C, and so on. 98, A braztn lion, placed In the middle of a reservoir, tarows out watrr from Its mouth, its eyes and its ri{:ht foot. Wl-on the water llow.'i rotn it, mouth alone, it fills the reservoir in C hours , from the right eye it lills it In 2 days; from the lei't eye in .3 days, and from the foot in 4 days. In what time will the basin bo filled by the water flowli g from all tliete apertures at once ? 29. Desire a person to think of any three numbers, each less than 10, and tltn tell him the numbers thought of. 80. Three men, Jones, Browji, and Smith, with their sons Harry, Tom, and Ned, had each a piece of land in the form of a square. Jones' piece \va,s 28 rods longer on each side than Tom's, and Biown's piece was 11 nidi longer on each side than Harry's. Each man possessed 68 square rods rehand the answer to a question in addition of a given number of lines, you writing the /leiwnd, fourth, eixtfif iic, addends, and some other terson the intermediate ones. *w..o^»- 1*^-- m'rvVr* ' ly number of pounds I , J rows out vraf-.r from le wfttcr Hows roin \U II the right eye it nib It ♦he fool in 4 days. In ■ flowli g from all thote h less than 10, and then • sons Hnrry, Tom, and uiiro. Jones' piece wius ,„wn's piece vas U rc-l, isesBed 63 sqinire rods (if ,D,6 were father and sou f of whom were bill "ks. beir provisions bcOiec*ITHMS OP NUMBERS PROM 1 TO 10,000, WITH DIPPEREN0E8 AND PROPORTIONAL PARTS. Numberfl fh>m 1 to 100. 1 N*. JLog> No. liOg. Nc. Log. No. liOC 1 No. IjOC 1 0-000000 21 1-922219 41 1-612784 61 1-785330 81 1-908485 1 0-30103U 2J 1-342423 42 1-623249 62 1-792392 82 1-913814 8 0-477121 23 1-361728 43 1-633468 63 1-799341 83 1-919078 4 0-602060 24 1-380211 44 1-6434.53 64 1-806180 84 1-924279 5 « 0-698970 25 26 1-397940 45 46 1-653213 65 1-812913 86 86 1-929419 0-778161 1-414973 l-6627f& 66 1-819544 1-934498 7 0846098 27 1-431364 47 1-672098 67 1-826075 87 1-939519 8 0-903090 28 1-447168 48 1-681241 68 1-832509 88 1-944483 9 0-964243 29 1-462398 49 1-690196 69 1-838(^9 89 1-949990 10 11 1-000000 90 1-477121 50 61 1-698970 70 1-846098 90 91 1-954243 1-04139C 31 l-4rfI362 1-707570 71 1-851258 1-959041 12 1-05 »181 32 1-506160 62 1-716003 72 1-867332 92 l-96r88 13 l-lft943 S3 1-618514 63 1-724276 73 1-863323 93 l-9684f3 14 1-146128 34 1-631479 64 1-732394 74 1-869232 94 1-973128 15 16 1-176091 35 96 1-544068 C5 66 1-740363 76 76 1-876061 95 96 1-977724 1-204120 1-656303 1-748188 1-880814 1-982271 17 1-230449 37 1-668202 57 1-755875 77 1-886491 97 1-986772 18 1-255273 98 1-679784 68 1-763428 78 1-892095 98 1-991226 19 1278764 39 1*591065 69 1-7:0852 79 1-897627 99 1-995635 20 1-901.030 40 l-fi02060 eo 1-776161 80 1-903090 iOO 2H)00000 I 8821 B82 L0OARITHM3. 1 j - pp N. 1 tl .1 4 9 6 r ^ , fl n. 100 (KXHKK)' 000434 1 000868 001.301 0017.34 (m21(>6 002,598 oo.'«mi \^.440 60.3S 6lt")<) (WlM T.iV' ''HI ((174 i.'m' R3 2 HWH) m2a\ 9451 9876, OKLliH) 010724, 011147 011.570101 1993012415 I2i 12 J 3 012K{7 013259 Ol.VJSO 014100; 4.521 4940 5.360 677J tl.>7 6<)l(i| IJI ICC. i 7/a 4486 4S9533 6942] 7.3.50 77.57 fll64 8.571 897« tits 2!)l) 7 9:«4 9789 0»)195 0.30ii(H) 031IHI4 0314O.S 031812 (/3«1« 0.32619 0,'i3O21 Ml .H.S1 8 033424 03;is20 4227 4628 6029 64,30 Rs;w ITM (ki29 7t)2.S |IMI 1 37;J 9 110 7426 7825 8223 042182 86201 9017 9414 9811 dvmi >V148 04(KJ02 ( 40998 1 ,«'7 041393 041787 042.576' 042969 04,3362 04.37.V 044540 ' 4^1932 38 I 6323 6714 6105 6495 6885 7275 766'., y).5.3 8442 WCIO :\w 70 2 9218 9606 9093 050380 050766 0511.53 051.^>.*r'/1924 0.52:i09 ( 52t);»4 :h, 113 3 oKiim 06.346.3 053816 42.30 4613 4996 6,37 ^ 6760 6142 6,524 :^3 151 4 6iHI5 7286 76«>6 8046 8426 8805 9UA 9.5()3 9942 0(0321) 379 18!» 6 06(1698 061075 0614.52 061829 062206 0625S2 0629 yi UA:m 06.3709 40.S3 376 227 6 44,')8 4BS2 £206 5.580 59.5.3 6.326 6C97 7071 7443 781.5 ;:^ 7 8186 RW7 8928 9298 9(i(W 07(JO;i8 070 «.7 070776 071145 ()•, 1514 .370 1 302 8 071882 0722.10 072617 0729.S5 07;il52 3718 ♦WW 4451 4816 51S2 :«;(•. I 3-lU 9 120 6547 5912 6276 6640 7004 7368 7731 8094 &157 8819 3ii3 ■m 1 2 079181 0795 J3 079904 080266 080626 0809,87 Oil J 17 081707 082Ot)7!08242(i 35 1 082785 03.3144 08;i503 3H(il 4219 4.576 4! '2 J .5291 6647 6001 Xu 70 2 631)0 6716 7071 7426 7781 31.36 8i90 &S45 9198 9.5,52 :\:,5 d 10-1 3 9905 0902.58 09061 1 090963 091315 091667 f/^vm 092370 092721 09.3071 ;iJ2 ■i 139 4 093422 3772 4122 4471 4820 6169 M18 5.S(i6 6215 6,562 .U'J 7 174 6 6910 72.'J7 7604 7951 8298 8641 8990 9,3:i.5 9681 100026 :ii(i fl 209 6 100371 1(HI7I5 101059 101403 101747 102091 ; MM 102777 103119 3462 ;{i:i 11 244 7 3804 4146 4487 4828 6169 6.510 5851 5191 6.531 6M71 ;iii 1) 278 8 72l(. 7549 7888 8227 85<)5 BWt 9241 9,579 9916 U02,5;i .T{>' 111 313 9 130 110590 110926 111263 111599 111934 1122:C li2605 112940 113276 3609 ;j,35 m IS 21 113943 114277 114611 114944 116278 ii5rn 11.5943 116276 116608 116940 32 1 7271 7603 79.34 8265 8595 »/?«, 92.56 9.5.86 9915 120245 Xf) r>4 2 120574 120903 121231 121.5(K) 121888 123 iUi 122544 122.S71 123198 .3525 328 2^ 97 3 SS52 4178 4504 48;J0 61.56 /^81 5806 6131 64.56 6781 ;i2,'5 4,- 129 4 7105 7429 77.53 miG 8399 8''«2 9015 9.368 96i)0,130.39 3858 4177 4496 4814 M3.3 6451 6769 6086 6403 .{18 11: 225 7 6721 7037 73,54 7671 7987 8303 8618 8934 9249 9.5<54 m 131 258 8 9879 140194 140.508 140822 1411.3/. UJ450 141763 142076 142;J89 142702 m 15(j 2i>0 9 140 143015 3327 3639: 3951 i 426:< 4574 1 4885 6196 6507 6818 31 1 m 178 201 146128 146438 146748 '147058 147.367 1 tA.7676 147985 148294 148603 148911 SO 1 9219 9527 98;i5 150142 15044^ J 3*^56 151063 151370 151(576 1619,82 ;W7 60 2 152288 152594 152900 3205 35U ' 3816 4120 44::4 4728 6032 :w 21 90 3 6Xi6 6640 5943 6246 6.54f 68,52 71.54 74,57 77.59 8061 m 42 120 4 a362 8664 8965 9266 95ff- 9868i 160168 160469 160769,161068 M 64 150 5 1613(58 161667 161967 162266 162.56-1 162863 31611 346<' 3758 405,5 209 85 180 6 4353 4650 4947 5244 654.. 6a38 61,34 6430 6726 7022 2!t7 106 210 7 7317 7613 7908 8203 849" 8792 9086 93.S0 1 96J4 9968 •2% 127 240 8 170262 1705.55 170848 171141 171 43i 171726 i 172019! 172311 172(503 172,S95 293 148 270 9 150 3186 3478 3769 4060 1769.59 4361 4641 1 4932 '177825 6222 6512 6802 178689 291 2S9 170 191 176091 176,381 176670 177248 177.536 178113 178401 28 1 8977' 9ri64 9.552 as39 180126, 180413 180699 180986 181272 181.5.58 2,S7 56 2 181844 18212!) 182415 182700 2985 3270 3.5.55 1 38,39 4123 4407 2,S5 20 841 3 4691 J 975 52.59 5542 5825 6108 6391 ! 6674 69.56 72:^9 2^3 40 112 4 7521 . 7803 8084 8366 8«U7 8928 9209 1 9490 9771 1 190051 i^\ I ^' 140 6 I90;«2 190612 190802 191171 191451 U17.'W 1920101192289 192.567 2A16 279 168 6 312i 34tJ3 3681 3959 42;i7 4614 4792 6069 5.346 662,3 278 ■ 101 1 121 196 7 5900 6176 6453 6729 7005 7281 7.556 78:« 8107 8;iS2 276 224 8 8657 8932 9206 9481 9755 i 200029 ^3031200.577 S«0,S50 201124 274 1 I^' 2521 9 201397 201670 1201943 1 202216 202488 mi 1 3033 3306 3577 3848 272 ■ t io« ■ 18:! LOQARITHMS. 88» HV1U JU.V/O' 67711 9047' DMO/5 MM ••H' )il71 f/ll'>93'0124l.'. ti:»7i Wilt' 1 44Htii 4«)t; JW7ll Hit7.S ; (Miii) 7i)2.s 04(HJ02 ('40'J«3 >V148: 'A19.J4' {•7t>0, 7071 070776 4451 8094 044540 I 4^<9;VJ 8442 HKJO 052»K» ( ^2t)'.l4 G142 (5.V24 !)!»42 0(l»32il ()f);i7l>9 4083 ' 7443; 781.^ 07U45,0; 1514 i:'.2 j'ji uo I 41') m lot :«i7 ;{:);« :;'.H) 3;y :u^ 4816 8-I57 6 IS-;. 8819 081707 08201)7 08242(i 6291 6047' 600 » &S46I 91981 9.Vi2 092.J70 1 09272 1,09,1071 68()6i 6215: aWi 93:« 9681 i 10002f) 102777 103119' 3462 3191 6531 1 6871 9579 9916 110253 H2940 1132751 3609 3il3 116276 9586 1228711 6131 1 93t)8i 1325801 57ti9 8934 142076 5196 1166081 9915' 123198 6456 9690 132900 6086 9249 i 142:189 I 5507 116940 120245 3525 1 6781 13WJ12 3219 6403 1 9564 142702 1 6818 ■A:6 ■.<,:>2 ;ii9 lilt") .m 311 ■xa 3W 328 1 32S :il8 ;ii6 m 3;i 1482941 1513701 44;',4i 7457' 1604691 346<> 64301 9380 i 1723111 5222' 148603 148911 151676 151982 4728 5032 7759 8061 160769,161068 3758 405.5 6726; 7022 96W1 9968 172603^72895 i 6512 5802 11 5 1781131 9 180986' 3839 6674 9490 Oi 192289 6069 6 78:« 31200577 3305 173401 178689 1812721181558 4123 4407 6956 72;i9 9771 1 190051 192567 28^t6 6;M6 6623 8107 a'182 2:00850 201124 j 35771 3»4* 3i19 ;w7 M m 2!)9 291 •21t5 293 291 — •( 2S9 2.S7 2,s5 i^\ 279 278 276 1 274 273 IT \. 1 'i i 4 S 6 7 N 9 D. 100 JIM 120 204391 201663 2«)49,34 20.5204 '205475 205746 206016 '2(H52.*<6 206,5.56 ■271 26 1 (;s26 7(»'.'6 7.3t).5 7(W4 7!>04 8173 8441 8710 8Jt79 9247 269 53 2 (/515 97.K3 21(KI5I 210319 210.VS6 2l08.Vi 21 1121 211:588 211654 211921 267 7!t 3 2l:;iM8 2121.54' 2720! 2986 i 3252 3518 i 378:5 4049 ' 4314 4.57!t 2(56 1(15 1 l,s»4, 6li>9; 5373 5('>;« ! 5902 6im i fti:50 6(594 69.57 ' 7221 264 IL' 5 74M4 7747 i 8010 8273 ' 8.5:« 87!»8 ' 90»V) 932,3 9.V«;5 ' 981(i 2(52 158 6 i'JOKW 220.370,2206.31 220S92 2211.5,3 221414 221675 2219,36 222196 2224,56 261 184 7 2716 2976; .•J2.36 3196 .37.55 4015 4274 4,\5:5 4792 5051 259 210 8 .5;i09: 5.568 6826! 6.56 4882 6107 5:5:52 226 67 3 .5.557 57.S2 6007 62.32 64561 6681 6905 71:50 7:5.54 757,s 225 89 4 7,S02 8026 8249 8473 8()96i 8920 9143 9:566 9,589 9,S12 2-23 112 6 2!HM»;{.5 2902.57 290480 2907021290925,291147 2913(59 291,591 291813 2920:54 222 134 6 2i56 2478 2699 2920 3141 a'563 3.584 3804 4025 424(5 221 1.56 7 4466 4687 4907 6127 6.347 5567 6787 6007 6226 6-14C> •2-20 I7S a 6665 6NS4 7104 7.323 7542 7761 7979 8198 8416 8aV) 219 201 9 8853 9071 9289 9507 9725 9943 300161 302:531 300378 302,547 300595 300813 302980 218 217 3010;}0 3012^7 301464 3016,81 301898 302114 ;i027o4 21 1 3196 3412 3628 3844 4069 4275 4491 4706 4921 61 .-5(5 216 42 2 5351 5.'66i 5781 59<)6 62111 6425 6(5:59 68.54 7068 7282 215 M 3 7496 7/101 7924 81.371 8351 8.564 8778 8991 9204 9417 213 85 4 96:50 C843:3100.56 310268 319481:310693 310906' 311118 311:530 311542 212 III!) 5 Jl 17.54 31.9661 2177 2;i89 2600 2812 3023, 32.34 34.15 36.56 211 I'.'7 6 3S67 10781 4289 4499 4710, 4920 61.301 5340 6551 6760 210 H8 7 5970 61801 63901 &599 m\9' 7018 7227' 74:56 7646 7.854 209 170 8 80(5.3 8272, 8481: 8689 8898 9106 9314 9522 97:50 99:5M 208 191 9 210 320146 '*<«)3.54 320562 3226:53 320769 .320977 '321 184 321391 321598 321805 322012 324077 20' 206 322219 «2426 3228.39 .323046 323252 3234.5Si 323665 32.3871 20 I 42.S2 4488 4694 4,899 5105 5.310 5516 5721 6926 6131 •205 40 ?. 6.3.36 6541 6745 6950 71.55 7:J;")9 7.563' 7767 7972 8176 •204 61 3 8.iS0 R.5,S.3: 87,87 8991' 9194 9398 9601 9805 3:50008 i.^'502 11 203 81 4 ^mt 3;i06l7! 0)819 331022 .3:51225 331427 331(5:50 3318321 2034, 22:5(i 202 101 5 24;i8 2640 2.842 3044! 3246 3447 3649 38.501 4051: 42,5.3 202 1?1 6 +4M 4()55 48,56 5057 i 62.57 6458 5658 68.59 1 60.59, 626(1 201 141 7 64'K) 6660 6860 70(vOi 7260 7459 1 7659 7858 8058, 8257 200 \ 10. 8 8456 8656 MK 9054 9-253 9,|51 9650 9849 340047 340246 199 18:! 9 340444 340642j34')P4l 341039 341:^^1 341435 .1 I 3416:12 34183U 2028 2226 198 1.4 884 LOixARIrBMB. pp N. 1 9 3 4 5 • T 8 i» ]d. 22U 342423 342620 1 342817 '34301 4 34.3212 34.3(09 343606 1 1 '343802 34.3999 3441!>6 197 19 1 4392 4589 4785, 4981 0178 6374 6570 6766 6962 6157 196 «9 2 6:«3 6.M9 6744 1 69;W 71.35 7.1'W 7528 n20 7916 8110 195 »« 3 KV)!i' RVWl 8694' HHH9 9<)83 9278 9472 96;i6 360S29 35102.3 351216 351410 35 1()03 .361796 19M9 193 97 6 2IK3 2379 2568, 2761 29.54 3147 33.39 15.321 3724 3916 193 llA 6 4108 4301 4493 46.S5 4876 6068 5260 64.52 6643 6H,34 192 139 7 6026 6217 6408! 6A!)9 6790 6981 7172 7.36.3! 7.'i54 7744 i91 154 8 7935 8125 8.'il6' .«()6 8696 8886 9<»76 9266 94.56 9646 190 174 9 230 9835 360029 360215 360404 360593 360783 360972 361161 seKtno 361639 .36.3424 189 189 361728 361917 362105 .362294 .362482 362671 .3628.59 .V>.3048 36.32.36 19 1 3(>I2 3H00I 3P88i 4176 4.3I0 67'.»6; 698.3 7169 187 M 3 7.1% 7.^42 7729 1 7915 8101 8287 8473 8659' 8845 90.30 186 74 4 9216 9401 95871 9772 99.58 370143 S7032S 370513 37(W!98 37OHS.3 181 93 6 .■J71l»68 S712.'i3 37 1437 '37 1622 371806 1991 2175 2360, 2514 272^ 184 111 6 2912 3096 3280 3HV4 3647 3s;n 4015 4198! 4;«2 4.5(i.5 184 lao 7 4748 49.t2 5115 5298 648! 6664 6846 6029! 6212 6.394 183 148 8 6577 67.W 6942 7124 7;«)6 7488 7670 7852 8034 8216 182 167 9 8398 8580 8761 8943 9124 9306 9487 9668 9849 380030 181 181 240 .380211 S8<3.392 38(W73 .38()7.'>4 m^M .381115 381296 .381476 3816.56 3818.37 18 1 2017 2197 2377 2557 2737 2917 sum 3277 34.56 3CX.W 180 35 2 3815 Sim 4174 4.153 4533 4712 4891 6070 5249 .542H 179 53 S 5606 5785 5964 6142 6321 6499 6677 68561 7034 7212 178 71 4 7.S90 7568 7746 702.3 8101 8279 n.v, ami 8811 8989 178 89 6 9166 9343 9520 96t.8 9875 390051 390228 .390405 390.582 3907.59! 177 106 6 390935 391112 391288 39 1464 miyil 1817 1993 2I69i 2.315 2.521 176 124 7 2697 2873 3048 3221 34(K) a575 3751 ,3926 4101 4277 176 142 8 4452 4627 4802 4977 5152 6326 5.501 5676 6850 602.5 175 159 9 250 6199 .397940 6374 6.548 6722 6896 7071 7243 398981 7419 7692 7766 174 173 398114 .398287 398461 .35)86.34 398803 ,399154! 399328 399501 17 1 9674 9847 400020 400192 400,305 400.5.38 400711 40aS,S3 4(/105fl 401228! 173 S4 2 401401 401573 1745 1917 208S> 2261 24.33 2005 2777 2949 172 61 S 3121 3292 3^164 36.35 3807 3978 4149 4.'«0 44.8 6048 5207 6367 5.526 5685 6,S44| cm\ 1.59 47 a 6163 6322 6481 6040 6799 6957 7116 7275 r\?a} 7592 1,59 63 4 7751 7909 8067 £226 8.384 «>42 8701 88.59 9017! 9176 1.58 79 6 9333 9491 96-18 9806 9964 440122 440279 'i4<)43T 440594 4407.52 1.58 95 6 440909 441066 441224 441381 ♦41.5,^ 1695 1857 2009 £>66. txl•^ 1.57 111 7 2480 2637 X793 2950 i;»06 8363 3410 85/0 3732' 3S89 15V , 126 8 4045 4201 4357 4513 46<$!) 4'«25 4fc*.' nr 629.3 6449 \f>'A 142 9 5604 0760 5915 6071 6226 A38S «f// »fii 6843 7003 153 | rOOAllITHMB. 38^ (PP\ N. jljilWO !M4\'.)6 197 6062, 6C.76 r4i9 6850 7592 42rM 176 60251 175 77661 174 >154 399323 W8;} 4(/l05fi i(i()6 2777' i320 44 I 5 (> 7 8 9 UO 1 2 ■1 o 4 5 () V M 9 1 o 3 4 5 6 7 8 9 4I7I5A 'um ;«i8 4HI5 7>W2 93'.r2 1(52308 3S<»3 6;K1 (5.S6S M»7 5t;s22 471292 27.16 4473)3 45()t<):< l0'.O 3171 401 >7 (5518 95 « 461018 462518 4012 65;J2 7016 8-105 09(50 .'rii.w 29(»1 4216 4;'f>2 6071 1816 3 4 9 tt t 8 9 l>. I 1^,15 4 50." 57 2O03 3624 61.''.0 6(570 81, H4 IH504 461108' 17712' A 13 W74 l.'i(H) 5721 71.'W 8551 OO.'W 462(5<»7 4191 6<5M() 71fi'» K<;43 4''91I6 1585 3049 4508 5962 1913(52 27<5S\ 480 ''I, 48020 4 158(5: 1720 3016 4442 5S(53 72.SO 8(592 «900!n) 400239 491502 2000 4294 5(^i3 70. 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S4 184041 1,S49(HI 1.S.5761 186624 187489 ias.-t.56 189225 19(H)96 190969 191844 19272: 19.36(H) 194481 514.39939 54S72(MII> 5.5:ioi;:M i 5574296.S 66181887 5()623I04 67066625 575124.56 67960603 68411072 68863869 593I9(K)0 59776471 602.36288 6069,84.57 61162984 61629875 620991.36 62570773 6.'J044792 6a521199 64()000(H) 64481201 64964808 654.50827 6.59.39264 66430125 66923416 67419143 67917312 68-117929 6St'21000 6942(i.531 6993452S 70H49i»7 709.57944 71473375 71991296 72511713 7;-J034632 7.3560059 7108.S(HM) 74618461 75151448 756.86967 76225024 767(>5625 77.308776 778514,8.3 78402752 7S9r);j589 79.5O7(K)0 8(M)62991 80(>2 1.568 81182737 81746.504 82312,S75 82,><8ia56 834534.53 84027672 84604519 a5I 84000 85766121 ! 19-4079223 I9•49.^5887 19-5192213 1 19-5148203 1 19-.570.3H.58 ! I9-.59.59179 1 19-6211169 '19-646S827 19-67231.56 19-69771.56- 19-72.30829 19-74H4I77 19-77.37199 19-79S989& 19-8242276 19-8494a32 19-874(;069 19-8997487 19-9248.5.S8 19-9499;i73 19-9749844 20-0000000 20-0249.S44 20-0499377 20-074.8599 20'""7512 20- .^61 18 20-l!944)7 20-1742410 20-19900! 19 20-22.37481 20-24,84507 20-2731349 20-29778:31 20.3224014 20-3169899 20-3715188 20-39(iO781 20-4205779 20-44.504,s;3 20-4694S95 20-49;«H»15 20-51.S2845 20-5426;M) 20-.5C.<396,3S 20-.59 12603 20-61.55281 20-6;597674 20-66.-{97.8,3 20-6,S8I609 20-71231.52 20-7364414 20-7097 20-.S08tW20 20-.S.326667 20-8566.536 20-88061:50 20-9045450 20-9284495 20-S)52,3268 20-9761770 21-0000000 7-236797 7-243l5<-. 7-249.501 7-2.55,841 7-262167 7-26,S4.S2 7-2747.H6 7-281079 7-2.S7.3<;2 7-293r):53 7-299894 7-.306l4;( 7-312:^8.3 7-3^8011 7-324,s:>;i 7-.3.31(«7 7-33723 1 7-343l2(t 7-349.5'.) 7-431994 741101!) 7-417034 7-4.5.3010 7-4590;J6 7-4(5.5022 7-47IK>9i) 7-476!«)6 7-18-2924 7-4,S,'S,S72 7-491811 7-.50^t741 7-.5()6(>*'p1 7-512571 7-51,S473 7-.5243(55 7-.5:i024.s 7-.5:!(5121 7-54198() 7-517842 7-5.53t5,S,s 7-5,59526 7-5(5,5:5.55 7-57 1174 7-576!)85 7-58278(5 7-5.8,^,579 7-594.363 7-6(ioias 7 60.5905 7-611662 No. .Square. Cube. Sq. Root. Cubeltooi 142 443 444 145 146 147 448 449 1.50 451 4,52 4.-3 454 15,5 4.56 4,57 158 4.59 460 461 462 463 4(54 465 466 4(57 ■WH 4(59 470 171 472 173 474 175 176 477 478 179 1,S0 181 482 l,s;{ 484 (,S5 1,S«5 4,S7 4,><8 I.S9 490 4! II 492 193 194 495 496 m 498 499 .500 .501 502 .503 504 19.5364 196249 1971.36 198025 19,K916 199,^09 2(K)7(»4 201(501 202.500 203401 204:504 20.5209 206116 207025 2079:56 208,H49 2097(54 210681 211(500 212.521 21.3444 214369 21.5296 21(5225 2171.56 21,Sfl,S9 219024 219901 220900 221841 2227,84 22:5729 224(576 22.5625 22(5576 227529 22.S484 229411 2:50400 2,31.3(51 2.32324 2.-5.3289 234256 2.35225 2:5619(5 2:571 6!t 2.38144 2,39121 240100 241081 24'20(54 213049 244036 24,5025 246016 247009 24,S(H)4 24!»0(!1 2.5O(M)0 25 UK 11 252004 2,5.3'H)9 254016 86.^5088.8 1 21 8(5!(:i8:507l21 87.52,S3S4i21 88121125 21 887 1(5.5:5(31 21 89.314(5^3! 21 8991.5.392,21 90518S49|21 911'2.5(KI0:21 917:5:5851 92.345408 92t 1.59(577 9.a576664 9419(5.375 21 94818816 9514,3993 96071912 96702579 97^56000 97972181 98611128 992.52847121 99897344 21 100544(525121 101194096 21 101847.563 102.503232 103161709 103S2.3(H)0 104187111 1051.54048 105^2:5817 10(5496424 107171875 107.S.5O176 10,S,53 1:5:53 109215:5.52 1(I9!N)22.39 110.592000 111284641 111980108 11-.'C.78.5.S7 li:5:579!»04 1140,^4125 11 47! » 1-2.56 11.5.')Oi:503 11(5214272 11 6! 1:50 1(59 117619000 118:570771 1190951,88 1198231.57 1205.53784 121287.375 1220239:5(5 1 •22763173 12:i50.5992 124-251499 125000000 125751.501 126506008 12726.^527 128024064 -02,57960 -0475652 -071.3076 -09.50231 -1187121 -142.3745 •l()()0I05 ■189(5201 •21.32031 -2,-567606 •2602!II6 ■2,S,379(57 •.30727.58 .3-507290 .'5.541.5(55 ■377.5;58.3 4009346 4242.8.5,3 447(5106 4709106 4941,8.53 -5171348 ■610(5.592 •5(5:5.S,587 •5870:531 -6101828 ■6;5;5:5077 ■65(54078 -67948:54 -702.5:544 ■72.5.5610 -74.^5(532 -7715111 -7944947 -8174242 -840.321>7 ■8(5.32111 -8S60686 -i»O.S902:5 -9317122 -9544984 ■9772610 •IKKXKXM) (t2-27 1.5.5 0454077 •0(5807(55 •0'.Mt7220 Ii:!.-54I4 ■1:3.594.3(5 ■1.5,S51!« •18107:50 •20:56(ja3 •2261108 •2485955 •2710675 ■29349(5,8 •:i 1.59 136 ■3:K5()79 •3(KI(5798 •:i8:50293 •40.5:5565 •427(5616 4499443 7-617412 7-()2.31.52 7-62;SM4 7(534607 7-640.321 7-04(5027 7-651725 7-().5;4i4 7-c,f<:vw 7-66.s7(5(5 7-67143(1 7-(WMW5 7•6.s.^7.•5.3 769 1:572 7-(597lH>2 7-702(525 7-708239 7-7i:5ii45 7-7194^2 7-72^'(i.-52 7-73(1614 7-7:5(;KS8 7-7417.53 7-747311 7-7.52.S(il 7-7.58 !()2 7-7(5.39.3(5 7-769162 7-774<)80 7-78| 7-92(5IOS| 7-931710 7-9;S7005 7-942293 7-947.574 7-9.52848 7-958114 n 7-02.S-V14 7-(;3(r,(i7 7-(>10.'«l 7-C>4a)27 7-0r)172') 7-(h')74U 7-r)rhKi71 7-8<)24r)l 7-.S()7'.)2.'i 7-813:M) 7-81ssl() 7-821294 7-82',)7.H5l 7-KUl(;'.t 7-84(i.'>;i5 7-81»ii)13 7 W 1424 7-8r)(;.s2.s 7-Mi222l 7-807613 7-8721HM 7-87,s3(;8 7-8.S373r; 7-WlM)'.>r) 7-8'.»4l47 7-8y!»7S»2 7-yo."/i2<.ti 7-1)1(1 ICDI 7-'.iir.7^Jl 7-!l211l»0| 7-y2()IOSi 7-!»31710 7lW70(»r» 7-9422(t3 7!»47.'>74 7-!)r)284S 7'lto8n4 . RQITARES, CUBES; AND ROOTS. 401 Ko, I Square. I Cube. Sq. Root, CubeRoot No. Square. Cube. Sq. Root. CubeRoot SOU 25.')()2.') 506 2.'>ii0.3(5 607 2.''.7019 .•508 2.580(54 509 2.59081 1510 2ii01()0 ^11 261121 512 2()2144 !513! 263169 6141 264196 i615' 265225 516 i 266256 617 1 267289 51S| 26K524 519 1 269,361 520 1 270400 5211 271441 522 1 272484 273529 274.576 275625 276676 277729 2787.H4 279341 280900 5311 281961 532 2S.3024 6.33' 284(J89 5341 28.51,56 523 524 ^25 526 5-7 528 629 530 1535 15,36 5,37 '5,'}8 '.'^39 640 ,541 542 545 546 2,86225 2,87296 288369 289444 290,521 291600 292681 293764 5431 294849 544 29.59.36 297025 298116 647 299209 648 649 550 551 652 5.53 554 5.55 656 557 658 659 660 661 562 663 664 665 566 567 300.304 :W1401 302500 303601 304704 305809 306916 308025 3091.36 310249 au.'wi 312481 31.3600 314721 31.5844 316969 318096 319225 320.356 32148tf 2.87,87625! 22 211.5.54216 30.32:5843 31096512 31872229122 ,32651000; 22 .3,34,328.31 22- 22' 34217728 3,5005697 :i,579()744 36590875 37.388096 38188413 38991832 3979,8;j.59 40ft)8000 41420761 42236048 43055067 43877824 44703125 4.5.531.576 46363183 47197952 4803.5.889 1.8877000 49721291 50.5087()8 514194.37 5227.3,304 53130375 5,3990650 548,541.53 .55720872 56,590819 57404000 58:540421 59220088 0010.3007 60989184 61878025 0277 1.3:56 630073-23 61500,592 t)5409149 66.37.5000 67281151 68196608 23 69112:57712,3 70031464 2,3 7095,3,875 i 23' 71879616 2.3' 72.S08C93 23' 73741112 2.3 74070.879 : 23- 75610000: 23' 70558481 : 23- 77.504328 1 2^5 781.53547 1 23' 79400144 2,3' 80,302125 2,3' 81.321496:23 82284263 1 23' 4722051 1914438 5l(;6(W)5 5:588553 .5610283 5,831796 6053091 0274170 049.5033 •071.5681 •09.56114 ■7156,3.34 •7.370340 •75901.34 •781,5715 80;5.50,S5 8251244 •8473193 •8<)9 19:5.3 .8910403 •9128785 •934IV899 ■9561806 ■97-8'2500 OOOOOOO (12172,89 0434.372 06512,52 ■0867928 1084400 1300670 15107:i8 1732005 1948270 21037.35 2.379001 2594007 2.8089:5,5 3023004 3238(J76 31,52351 3066429 .3880311 4093:HI8 4:507490 4.520788 473:5892 4946802 51.59520 5372046 5,5,S4;5,S0 5796522 0008474 02202.36 0131808 6()43191 0854386 700.5:592 7276210 7486,842 7097280 7907545 8117618 7-9ft5.371 7-908627 7-97.1873 7-979112 7-9*1344 7-989.570 7-994788 8-0OOOOO 8-00.5205 8-010103 8-01.5.595 8-020779 8-02.59.57 8031129 8-036293 8011451 8-040t)03 8-051748 8-050886 8'002018 8-007143 8-072202 8-077.374 8-0,S24,S0 8-087,579 8-092(57 8-0977.59 8-102.8.39 8-107913 8-1129."!) 8 11.8041 8-12,3090 8-128145 8-1.33187 8-1,38223 8-14.3253 8-148270 8-1.5:5294 8-1.5H.-5i)5 8-163310 8-16S.-509 8-173:502 8-178-2,89 8-183269 8-188244 8-193213 8-198175 8-203132 8-20,80.82 8-21,3027 8-217966 8-222,898 8-227.825 8-2.32746 8-2,370(51 8-242571 8-247474 8-252:571 8-2.57203 8-2(52149 8-2(57029 8-271904 8-280773 .56,8 5(59 .570 .571 ,572 57:5 .574 575 576 577 ."^78 .579 .5.80 .581 .582 .583 .584 ,5,^5 58(5 5.S7 5.88 589 590 .591 .592 593 .594 595 596 597 598 5i»9 600 601 (502 (503 (504 (505 (506 (507 (508 (509 010 Oil 012 613 614 615 616 017 018 019 020 021 (522 023 024 025 02(5 027 028 629 a'50 ^T" .322624 32:5701 .324900 .32(5011 .327 1S4 32,8,329 329476 .3.'50(525 .3.31776 a32929 .3340,84 .3,'5.5241 3:5(5400 337.501 3.38724 3.39,889 341056 .342225 343396 344569 34.5744 316921 348100 349281 .3.'>0464 .351649 352,s;56 .354025 ,3.5.5216 3-56409 .3.57(504 3.58801 360000 .30120' .362404 36;50()9 304810 30<5025 .3072.36 363449 .3(590(54 370.881 372100 37:5.32i 374544 375709 370996 378225 379456 380089 •381924 383101 3,8-1400 38,5641 3.80884 .38,8129 .3,89376 390(525 39187(5 ;393129 394.384 39,5641 396900 1,8.32504.32 1842-J(i009 18519,-5000 186169411 1,87149248 1881:52517 189119224 190109375 191102976 19210O():5:S 193l(H).5.52 194101.5.39 1951 i20W 1901-22041 197137,3(58 1981.55-2.S7 199176704 2002U1025 2012:500.50 2022(52003 203297472 2013;5()409 20.5,379000 20642.5071 2074746S8 208527857 209584584 21064 1.S75 21170,8730 212770173 21.3817192 214921799 216(HI0O0(» 217081,801 2181(57208 2192,5(5J27 220348864 2214151'25 22254.5016 22.304a543 2217.5,5712 22580(5529 22(5981000 22,8099131 229220928 2.30:54(5:597 23117.5544 2320Oa375 2;i3741896i 2.318,85113' 2.3(50290.321 237170(5.591 2:5,8328(W0| 2,3948,3001 ' 240041848 241.80/307 2 1297( tU 244.4(026 24,53 l'J;/S 24019Id8d 247(57.'9920 260:i99681 26-a69»282 26-0798724 8-21, ->• 5,35 8-2,8049:5 8-291344 8-296 1 IK) 8-.3010.'JO 8-:50.V<6.5 8-310(W4 8-31.5517 8-;5203.'55 8-;525l47 8-329954 8;5:547.55 8-3.39551 8-341,-541 8.319126 8-;5.V5905 8-3,58(578 8-3(5.3146 8-36,8209 8-3729(57 8-;577719 8-:5824(55 8-387206 8-;591942 8-390(573 8-101:598 8-406118 8-4108.33 8-41.5542 8-420216 8-424945 8-429(5.38 8-434:527 8-4.39010 8-443088 8-44,8:560 8-4.5,3028 8-4,570'.)l 8-402318 8-4(57000 8-471047 8- <762.89 8-4,S(>926 8-48.5.558 8-490185, 8-494.S061 8-4994-23! 8-.50 10:55 8-508(542 8-513243 8-517.SU) 8-522132 8-.527(tl9 8-.531(501 8-640700 8-(M£8l7 8-6i«{/» 8-6644a/' u-&6duyu< 8-56^5^ 8-6ti(MtUl| 8rBi: t 402 SQUARES, CUBES, AND ROOTS. No. A31 632 (•)33 634 635 636 637 638 639 640 641 642 643 644 645 646 647 64.S 64!) 650 651 652 GXi 654 6.'A 6,56 657 658 659 660 661 662 663 664 665 666 667 668 669 C70 671 6:2 674 67.5 676 677 678 679 680 681 682 68.3 684 6.S5 686 687 688 689 6i»<) 691 692 693 Square. Cube. Sq, noot OnbeRnot No. .398161 399424 400689 4019.')6 403225 404496 405769 407044 40,S;«I 409600 410881 412164 413449 414736 4UW25 417316 418<>09 419904 421201 422500 42:i.S01 425104 42640{) 427716 4290iift 4;?o:$36 431649 4,32964 434281 435600 436921 438244 439569 440896 442225 44;}55(; 444.S,S9 446224 447561 4489(M) 4.50241 451.58-1 45292<) 15-1276 4.55625 4.56976 4.5ii.329 459684 461041 462100 46.3761 46.5124 466489 4678'''' 469225 470596 471969 47.3.344 474721 476 im) 477481 478S64 480249 2512.39.591 '2.5-1 197134 1 2524,35963 25-1.3961021 2.536.361.37: 2.5- 1.W49 13 254840l04'2.5'l79.r>G6 2.56047875! 25- 1992063 2572.594.56,25-2190404 2.58 1748.53 ! 25-2;}8.H.58.5 2.59691072 25-2.586619 260917119 262144000 26.3374721 264609288 2t>;'.,S47707 267089984 26.8.3,36125 269586136 270840033 272097792 27;i'}.59449 274625000 27.5894451 277167808 27844.5077 279726264 25-278449,3 25-2982213 25-3179778 25-.33771S9 25-,r.74447 25-,3771,551 25-3968502 25-41 6,5;U)1 25-4,361947 25-4,5,58441 25-4754784 25-49.50976 25-5147016 25-,5312907 25-.5.5.38647 25-5734237 281Ci:r5l2.5-.5!l29678 282,300416 25 8124969 28;i593393 284,S9()312 286191179 287496000 08.8804781 i,r^llV.528 2914.34247 292754994 294079625 295408296 296740963 29,8077632 299418,309 30076,3000 302111711 .303464448 304821217 306182024 307546875 30,891.5776 3102.8,8733 31166;57.52 31.304()8;}9 3144.32000 31.5821241 317214.568 318611987 .'P''H) 1.3504 !Wi-il9125 322828,856 324242703 32.5660672 t27082769 32,8.50! tOOO 329JI.39.37 1 .3.31.37.3,^38 332812557 25-6,320r2 2.5-6515107 25-67099.53 25-6904652 25-7099203 25-7203(507 25-7487864 25-7681975 25- 7875939 25-8069758 25-826.3431 25-84.5(;960 25-8()50343 25-881,3,582 25-9036677 25-9229623 25-94224,35 25-9615100 25-980762 i 26-OOOOlK'O 26-0192^37 26-0.384'!31 26-0576::84 2t)-07&-<096 26-0959767 26-1151297 26-1.342637 26 15.3,39.37 26-1725047 26-1916017 26-2106843 26'2-297541 26-2488095 2(;-2678511 26-28687.80 26-»158929 26-3248932 0-577152 8-581681 8-58(i205 8-590721 8-595238 8-5!)9747 8-604252 8-6087.53 8-613248 8-617739 8-622.'25 8-62670G 8-631 1.S3 8-0;j.56.55 8-640123 8-644.585 8-619044 8-6.53497 8-657946 8-662.391 S-6()6<31 8-6712i;(; 8-6756i>7 8-680124 8-6845 It 8-68S963 8i;'>;«7(» 8-697/8.* 8-71121 S.8 8-70(i.-),S7 8-710983 8-715373 8-7197.59 8-724141 8-728518 8-7.32892 8-737260 8-741621 8-74.598.5 8-750,340 8-754691 8-7.590.3S 8-76.3.381 8-767719 8-772053 8-77(5383 8-78070S 8-785029 8-789346 8-79.3659 8-797968 8-802272 8-806572 8-8108(58 8-815160 8-819447 8-82,3731 8-82.S(H)9 8-83-2285 8-8.3()55(5 8-84082.3 8-84,5(K5 8-8-19344 Square. Cube. Sq. Root. CnboRooi 694 695 (596 (597 693 (599 700 701 702 703 701 705 706 707 703 709 710 711 712 713 714 715 716 717 718 719 720 731 7i)2 723 724 725 726 727 i 7281 729 730 731 732 7,^3 7,34 735 736 737 7,38 739 740 741 742 743 744 745 1 46 747 748 49 750 51 52 7,53 754 755 J6 175 4816,36 48.3025 484116 4,85.>-^)9 4,87204 488(501 490000 491401 492,-<04 494209 49,^616 497025 49,'-!i36 499849 ,5012(34 .502681 604100 .50.5.521 50(5944 50.8.3(59 509796 511225 5126.56 514089 51.5.524 516961 518400! .'=.19841 5212-4 522729 524176 625625 527076 628.529 529984 531441 .5.32900 63 (361 635.^24 5372.S9 638756 540225 541696 643169 .514644 546121 647(>00 649081 550564 6.520491 5.5.50251 5.5(5516' 55.80091 5.59.504 601001 662500 564(K)1 6()5.)04 .5670091 6(5.^.516! 57»025' 6715.36 .3.342.5.^384 1 26-3438797 3,3.5702.375 '26-362,8527! a37l53.5,36:26-38l8ll9 3.3.«608.S73: 26-4007.570' .34(H)6S.392' 26-4196896 3415.32099 1 26-4.",S6081i 34.3000000:26-4575131 ; 341472101126-4764046 .34.594840812(5-49.52826! 34742S927, 26-51414721 34,8913664 26-.5.3299,s3 3.5O402(525!26-.5.^l8361 3.51.*J.5816i26-67066O5i .3.53.393243 ! 26-6894716 i ,35-1894912 26-6082(594 1 3.5(5400829 1 26C270.5.39 1 a57911(M.)() 126-64.582.52 ;i;-.9-}2.5431i26-664.58;«l 3(TO44128!26-6.S.3328ll 3(52467097:26-70805981 3(5.399434 4 26-7207784 365.525,«75 26-7.3948;{9 367061696;26-7.581763 86.8601813:26-7768.5.57; ;J701 462.32 26-7955220 371694959'26-81417.54 37.3248000! 26-83281.57 374.S05.361: 26-8514432 376367048; 26-8700577 37793.3067; 26-888(5.593 379503424126-9072481 381078125 126-92.58240 3.826.57176126-9443.872 384240.583:26-9629375 .3.S5,S2»;i52: 26-9814751 387420489 27-000(K.>00 3S90170J0 1 27-0185122 .390617.'^91 127-0370117 392223 168 1 27-0.5549,'s5 39;K.?2837 27-073[>727 S9.544(5S»04 1 27-0924344 39706.^375l27•110a834 39.S6S82,56i27-1293l99 400315553 27-1477439 401947272! 27- 1661.5;54 40.'i;)8.3419l27-184.5.5-y 40.5224000! 27-20294 10 406869021 127-2213152 4085184,83 27-239(5769 410172-107^27-2.580263 411830784 27-2763^534 41349.36'25|27-29-16881 4151(50936 !27-3130{K)6 416832723:27-3.713007 418.508992127-3495887 420189749 :27-.3(578644 42187.5000 /7-.3861279 42.3.564751' 27-4043792 42.52.59008 '27-4226184 4269.57777' 27-4108455 42,86(^1064127-4.590604 430.3(5.8.875 27-4772(Vi3 432081210 27-4954612 8-85.35981 8-8.57,S4* 8-862iK'.>: 8-86«;;};57 8-.870.'>76 8-874810 8-.87!)O40 8-88320(1 8-8874.SS 8-891706 8-89592<) 8-900 l.'iO 8-904:J36 8-908.5.38 8-912737 8-91(5931 8-921121 8-925.'«J.S 8-9294'.«) 8-933(568 8-9.378-!;5 8-9421114 8-94618: 8-9.5034! 8-954503 8-9.5,^6.58 8-9(52,S09 8-9(5(59; 8-971101 8-97.5240 8-979.376 8-98.3.509 8-987637 8-9917(52 8-995SS.3 9-00(M)00 9-004113 9-008223 9-012.329 9-016431 9-020529 9-024624 9-028715 P-032802 9-03(58,86 9-040965 9-045'041 904911 9-0.53183 0-057248 9-061310 9-065367 9009422 9073473 9-077.520 9-081.563 9-0,86603 9-089639 9-09.3(572 9-097701 9-101726 9-10.5748 9109760 Ke. 7.57 7.58 759 760 761 762 763 SQUARES, CUBES, AND ROOTS. 403 .oot. CnboRoo 38797 28527' I81I9 )7ri7G J08tt6 S()08li 5282« ' 41472: 183(;i I lt66l)5! 947 IGj S2r)i>4| 70r)S9| ;i328ii 8(»5!)8: 07784 1 y»s;«) 81703! lW.^'i7| '552201 41754 128157 il4432 tH)577l ;8()593 172481 58240 4*872 21*375 4751 0(H)(H) 85122 0117 41185 1*727 4344 38834 ^311*9 r7439j 554 155M 294101 13152 *67C9 0203 3034 10881 WOOfl 3007 15887 ■8044 1279 '<7!*2 0184 8455 ■0004 2(V« 4512 ] Ks. 'Square. 8-8535981 8-857849, 8-862(K'5- 8-80«;.'i;}7i 8-870570 8-874810 8-879040 8-S8;<20fi 8-887488 8-891706 8-8959209 8-987037 8-991702 8-9958s;3 9-000000 9-0(*4113 9-008223 9-012329 9-010431 90205'29 9-024024 9-028715 P-0328(.)i 9-03t'>S80 9-0401*05 9-045'04l 9049114 9-053183 0-057248 9-001310 9-065307 9-069422 9073473 9-077620 9-081503 9-036003 9-089039 9-093072 9-097701 9101726 9-105748 9-109700 Cube. Sq. Root. CubeRooi tJNo.jSqil ' . i ■ Cube. Bq. Root. CubcRoot 57 753 759 700 761 762: 703 7^4 65 OG 707; 763, 7691 770; 771: 772 1 773 i 774' 775 j 77G| 7771 778! 779 j 780: 811 82! 7831 784 1 785 7861 787 1 788 89 790' 791^ 792; 793; 794: 795 1 796 1 97 793 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 «lrt 619 673049 5715041 5700811 577tWO] 579121 580044 582109 583'j9r> 585225 680756: 58.: i9 589824 591301 592900 .594441 595084 697529 599076 a»y625 602170 003729 005284 000841 608400 009901 011524 613089 014050 0102-25 017790 019309 020944 6-22)21 624 1 (*0 625081 6272Gt 028849 03)130 632025 63;J010 635209 a'i0804 63840V 640000 641001 043204 644809 640416 648025 649636 6.51249 652804 654481 450100 6.'>7721 659344 6C)0909 602596 664225 6*35S.56 607489 609124 670701 433793)93 4.35519512 437215179 43:s97Gi*00 440711081 4424507-28 444194947 41594,3744 447097125 44945.>090 45121 7'")03 4.529S4,S;}2 454750009 45653.301*0 4.-).S,314011 40(*099043 401,881*917 463081821 4054M375 40728><570 40t*{*974;i3 471*910952 472729 1. -19 474.552000 470.371*541 47S211708 480048087 48l,S90;i<*4 48:i730025 485.)8705i) 487413403 489303872 491109009 493039000 49491.3071 49079:{088 4980772.>7 500.50". 1 84 .502459875 .50i;i58.330 500201573 508109.592 510():S2;599 5120000540-i! 28-44-292.53' !23-40(*49-<9 |28-.l7W»0I7i 128-4950137 28-5131549; !2.8-.5.300,-i.52! 1 28-5 1,82013 1 •28-.50.57137I pi<-.5,-*.-«ll9| '28".00t)993 |28-6i81700 91i;J781 91 1779 i Sl-21-01 9-l-25si*5 9-129i0.i 9-l3i8.)3 9- 1:57797 9-1117,-;^ 9-11.5771 9-1197.57 9-i,5;!r;!7 9-157711 9-l(;i'"ot! 9-10,'.0,50 9-lti9.;2:* 9-17.15^5 9- 177511 9-1815(11* 82() .821 ."(22 823 824 .825 820 :S27 828 S29 830 831 8:52 h;j:5 '51 3'55 8:!0 .^57 9-1.8511:! KVi 9-189102 9- 19:5:! 17 9-1972^9 9-201229 9-2(15101 9-2it: 11)1)0 9-2i;5i)25 9-210.C>!i 9-22'* ^73 9-224791 9-22870/ 9-2321)19 9-237,528 9-2 10 1:5:5 9-2ii:;,55 9-21S231 9-252i;!t* !*-250:t22 9'2,59'.»ll 9-20:5797 9-207080 9-27 1,5.59 9-27.5t;55 9-279:}08 9-2i'5r;8 9-2S7044 9-290907 9-294707 9-29,-;024 9-;5(,r2l77 9-;5llO:52s 9-310175 9-3U(*19 9-317.800 9-:{v:i097 9-32/532 9-329,'«V5 9-.T5;5lW' 9-;5.-$7(*lV 9-3K*.S3S 9-34 KW 9'31S473 9-;}.52280 9-35009.5 .8.59 810 811 842 S4;5 814 815 -110 sl7 ,S48 ,S4'.* ,S50 ,'^51 S.52 8.53 K51 8.55 ,'-v"t) .S57 85 S .^■5J H>M) ,S(il 8t;2 803 ,S01 805 860 807 808 •09 871* ,-^71 872 87;i 874 -75 87b 877 r;^8 .87!;* 880 .8,81 882 072400 1 074011 075!-,8|l {)77:<2!)j 07:y7ii 0-^0025 682276 o;;;;'.)29 0-1,55 SI 687211 6.-(.^'!*00 000501 092'22l G9:SS89 095550 ()97225 ()9889ti 700509 70-2-241 70:5921 7(),50(*i* 707281 70,S901 710041! 712:530 7! l!(25| 7 1 ■■■710, 717 MOi 7!9!0i: 7'2'»80l! 7-2'2,-50()| 721-'.*li 72.5:101. 727009 72J:ilO 731025 7:!27:50i 7314(9 7.30101 737881 73,*!;i*0 74i:;2i 74:504 1 7447(59 740i:*(i 74,8225 749950 7510-51* 753 124 7.55 li;i 75,;: 100 75s;5ll 70i)3S4 702129 70:5870 7(5.5025 7(i7;57() 709129 770.N8i 7720 U ■'74400 770101 77/9^4 .551 3'T;000 1 28-(..35r.421 ' 55:5:^87001 !28 0.5:5(>l*7Ci 5.5.54 12243 12.8-0705 124 j .557 t( 17(57 2.^-(5^71'700' ,5,59 r;(!22i:2.>S- 70.510021 50!51.50-25i28-7228l,32' .50:5.5,59970128-74021.57 .50.500,i::831 28-7570077 1 .50700.'1.552 28-7719.-;91l 5(;972-27S9: 23-792:50011 57I787(H*0 2.8-3097200; 57;i8.5019i: 2,8-827' )7(*0i .57.59:50;5')8'-28-.8411102' ,57,8(*'}95;57'28-,8,il7,'59l! 5800.):5701:2.8-.8790,582: .5.^2K-<2875;28-.89(53(;-i')| 584277050' 28-923(;04.;! 5S037025:5 ' 28-9,301*52:5 1 .5.S^H:(IT2 2.8-9182297' 59;).5S'.»719;iS-905!9;'.7 59-27t*t'.)0(* 28-9.S27,5:55, .591S2 5:!2I '29-0O000iH)| ,590017088 •20-0 17-2.303 1 ,5'.)9')771()7 2')-()3M--.2:5 O*121l5-U-29-O5i07.Sl! 0),5!51'2-. 2',*-oos.-;-;37| 0(*5("57.i0-29-l»-^(i-)79li 0it7(;i5i-23 29-Hi:;JOI4! OiiOvKllO.: 29-1-20 (.3901 OllitOO,!!!* 29-i;57(Mili) 0!ir250iU 29-15(7.595 (!1029,505r2;*-l7l!H)13 0ISJ7i*:;os 29-lsiii»3'H» 02:',:50 177 29-200 li;37 6:i2-'5.5'-0375 29-'24(»3<:50 0-27222-:>lG -29-'2.574777l 02'(422793 29-2745()23! 0310-2,-(712 29-29lO)70! G!3-5:i9779 29-;ii*870131 (5;50O.50O:*O 2,);;-2.57,500| 03-!J773-1818 -29-,52:'0!01 00.5.-5:5-.'(;i 7 1-20-510.5731 007(i270-2-l •29-,5(i,34010 609,»2I,s75''21*-,5.'-i():;9s9 072221,370 29-.5972972 0715-201.-53 29-01 II.N58 070,-i,!01.52, 29-0310013, 079151 1:59: 29-0179:5-25; 6,-!l472(HHJ ■29-00179;59 6.8,-579784i: 29-6810142 C8til2rf968i29-09S1848 9-a51»002 9-:5tJ'i705 9-,'5'i7.505 9-,-57 1.502 9-37.50-90 9-37.^Ss7 9-:582075 9-3S01(iO 9-3902 12 9-39102!* 9-.-i:*779:i 9- 101501) 9-iO,>.559 9-109105 9-4I2S:;0 9-4i!;i5;;o 9-42i*:J87 9- 4-24 1 42 !i-42r-:'4 9-431i;i2 9- 1:!5-58-^ 0-4:;9i3i 9- M2n7') 9-4!. 1.107 94;5i),;41 9-45 1072 9-4r)7-'.00 9-4( 910.-.2I7 9'40-'.. )<;<.■. 9l72i;.-t2 9-170:5l»5 9-1.^0100 9-(>3Sl3 9-1.87.'>K8 9-191220 9-401-.(l9 S-498'515 9-.5O2:508 9-.5(V5998 !)-5()90.85 9-513:570 9.521100 9-.5-2.S079 9-.531749 9-.5.-554I7 9-,V51»0,82 9-54-2744 9-5!01(*3 !*-,5.5(!0.59 9-5.53712 9-.5.5730:5 9-501O1I 9-50 1050 9-50.S.9.S 9-5719:58 9-.575,574l 9-.579203I 9-,582840 9-5.'^;4G3 81 9-590094 s rs^^- 404 8QUABES, OUB£S| AND EOOfA. No. Square. Onba. Sq. Root CabcRoot Ko. Squar«. Cube. 8q. Root. Cubcllooi 883 779689 688463387 29-7153159 9-593716 942 887364 836896883 30-6920186 9-802804 884 781456 690807104 297321375 9-597337 943 889249 83S5<51807 .%'-7083051 9-806-271 mi 783225 693154125 29-74894% 9-6(M)955 944 891 i;w 8412323.84 30-7245830 9-809736 886 784996 695606456 29-7657521 9-604570 945 893025 843908(525 30-7408523 9-8131!t9 887 786769 697864103 29-78254.')2 9-008182 S46 894916 34(5590536 30-7571130 9-816659 888 78H544 700227072 29-7993289 9-611791 947 89.3.S09 849278123 30-7733651 9-820117 889 790321 702595369 29-8161030 9-6l5:W8 948 898704 851971392 30-7896086 9-82:5572 890 792100 704969000 29-8;«S678 9-619002 949 900001 864670;i49 30-8058436 9-827025 891 7i»3881 7073471*7 1 29-8490231 9 •0226(13 950 902500 857:^75000 30-8220700 9-8:50476 892 795664 709732288 29-8663690 9-620201 951 904401 860085:«1 30-8382879 9-833924 893 797449 712121957 29-8831056 9-629797 952 900:W4 862801408 30-8544972 9-837369 894 799236 714610984 29-899.8328 9-633:J8l 954 910116 868250(564 30-»86S!»04 9-8442.5;l 9-847692 896 802816 71"323i;i6 29-9:^2591 9-6405G9 955 912025 87098;}875 30-9030743 897 804609 72i734273 29-949958:1 9-644154 956 913936 873722816 30-9192497 9-851128 893 806404 724150792 29-9066481 9-647737 957 915849 876467493 30-9354166 9-854562 899 808201 720572699 29-9a'};i287 9-651317 958 917764 879217912 30-9515751 9-857993 900 810000 72-010<>620 »• 658468 960 921600 ^84736000 :iO-9338668 9-864848 902 813604 733870808 30-033:il48 9-662040 961 923.V21 88750:^681 31 0000000 9-8(58272 903 815409 73f„3l4327 30-0499584 9-0(i;")t)0'.> 962 925444 890277128 31-0161248 9-871694 904 817iSl6 738763261 30-0665928 9-0(59176 963 927:569 893056347 310322413 9-875113 905 819025 741^17025 30-0,832179 9-672740 964 929296 895,841344 3r()48:M94 9-8785:50 9<)6 820836 743677416 30-099a^39 9()76302 965 931225 898632125 31-0644491 9-8bl945 907 822649 746142643 30-1164407 9-()79860 966 933156 901428696 31-0805405 9-,885357 908 82440 t 748613312 3n-i33o;5a3 9-6-U416 967 9:^089 904231063 31-0966236 9-888767 909 826281 7510.S9429 30-1490269 9-686970 968 937024 9070392;J2 31-1126984 9-892175 910 828100 753.W1000 30-1662063 9-690521 969 93S9G1 909853209 31-1287648 9-895580 911 829921 756058031 30-1827765 9-694069 970 940900 912673000 31-1448230 9-898983 912 831744 758560528 .<«-19961 30-3479818 9-729411 980 960400 941192000 31-3049517 9-932884 922 850084 783777448 30-3644529 9-732931 981 962361 944076141 31-3209195 9-936261 923 851929 7863JJ0467 30-3809151 9-7:«)448 982 964324 946966168 31-336.S792 D-939636 924 853776 788889024 30-39736.83 9-7;}9963 983 966289 949862087 3r.3523308 9-943009 925 855625 791463125 30-4138127 9-743476 984 968256 9527(33904 31-3687743 9-946380 926 85747(B 794022776 30-4302481 9-746986 985 9702-25 955671625 31-3847097 9-949748 927 859329 796697983 30-4466747 9-750493 986 972196 958585256 31-4006369 9-953114 928 861184 799178752 30-4630924 9-753998 987 974169 961u04803 31-4165561 9-956477 929 863041 801765089 ;^-4795013 9-757600 988 976144 964430272 31-4:524673 9-959839 930 8649C0 804357000 30-4959014 9-761000 989 978121 967361669 31-448:J704 9-96;5I93 931 866761 806954491 30-5l2-2i)26 9-764497 990 980100 970299000 31-4642654 9-966555 932 868C24 80!»557568 30-5286750 9-767'.»92 991 982081 973242271 31-4801525 9-969909 933 870489 812166237 30-5150487 9-771484 992 984064 976191488 31-4960315 9-973262 934 87235G 814780504 30-5()14i:i6 9-774974 993 986049 97914(5657 31-5119025 9-976612 935 874225 817400375 .30-5777697 9-778462 994 988036 982107784 31-5277655 9-979960 936 876090 820025856 30-5941171 9-782946 995 990025 985074875 31-54:56206 9-983:505 937 8779(i9 822656953 30-6104557 9-785429 996 992016 988047936 31-5594677 9-"86649 938 879844 8252it3672 ;W-6267857 9-788909 997 9940119 991026973 31-5753068 9-1 5!«90 939 881721 827936019 30-64;U0b9 9-792;W6 998 996004 994011982 31-5911:580 9-9y;5.329 040 883600 830584000 30-6594194 9-795861 999 998001 997002999 31-6069613 9-996666 941 8854S1 833237621 30-6767233 9-799334 1000 1000000 1000000000 31-6227766 10-000000 CubeRookj 9-802804 9-8()G271 9-809736 9-8131119 9-816659 9-820117 9-823572 9-827025 9-830476 9-83392-1 9-837369 9-840813 9-8142.5;l 9-8476'.r2 9-851128 9-854562 9-857993 9-861422) 9-864848 9-8(38272 9-871094 9-875113 9-878530 9-8bl945 9-885357 9-888767 9-892175 9-895580 9-898983 9-902383 9-905782 9-909178 9-912571 9-915962 9-919;i51 9-922738 9-926122 9-929504 9-932884 9-936261 9-939636 9-943009 9-946380 9-949748 9-953114 9-956477 9-959839 9-963198 9-960555 9-969909 9-973262 9-970612 9-979960 9-983305 9-"S6649 9-1 •)9990 9'9y:i329 9-996666 10-000000 ANSWERS TO MISCELLANEOUS EXERCISES. 2. 3. 4. 8. 10. 11. Id. 13. 14. 15. 13. 19. Exercise 8. Sixty-seven trillions eight liundred and forty-five billions three hundiod and ninety-eight millions six hundred and seventy- eight thousand nine hundred and four. Five (juadrillions nine hundred trillions seven hundred and four billions sixty millions forty thousands, and sixty thousand six hundred and four Imndr. Iths of millionths. MVDCCLXIX. 42986U000. 5. .SG7-;31|. 6. 11\)'.n. 7. GUoOOOOVOOlG -000009. 46078900. 09 800403. •8139. 6V89O00O0. trH:r29S()Oi>00000. lOODOulOOOOOlOOl-OOOOOOOOOOOl. •0007609. * 16. Xincty trillions eight hundred and seven billions sixty millions live hundred and four thousand and thirty. Four (luintillions four quadrillions forty trillions four hundred billions sixty thousand four hundred and thirty-two, and one trillion ten billion two hundred and three million forty thou- sand five hundred and six hundredths of irillionths. 77j!r cord.s. 717 cords 91 cubic feet. 20. DCCXVIII, DCXIV^CDXCIX, CMXCIX, YMMMDCXLin, XCVMCXLIX, CLXMMMCMLXXXVI, UDXLMVCDXLIV. 21. 333, 1989, and 1000001. 25. $3-75yV, $24-58^, $71|, and $757-47^. EXERC ISE 17. 1. $18029304. 9. 92438 lbs. Soz. 2 dr. 1 scr. 2. $13999999-73. 13grs. 3. 3G497318. 10. 1G98728G02536. 4. 35857536. 11. 78990 bushels. 6. 27424500. 12. $04-97. 6. 271G33. 13. 9032 yds. 3qrs. 2na. 7. 9504000. 14. 1037957G01-5. S. 327040000. , 15. $16444-9602. 4.06 ANSWERS TO MISCELLANEOUS EXERCISES. Exercise 22. 2. 3. 4. 6. 6. 7. 8. 9. $34756-8121. $N30('.o4-y20t;. ayOSdvd. or Oyrs. 20Jdy3. 8137. $108. §29. 4onrta 10. il. J 2. 13. 14. 15. 16. •5 78 oz. 250 lbs. 10-1 5*7. 2 bush. 1 A pts. 1 pk. 1 gal. 2 qta. 1 1 2G7 days l^^l^ hours. 'Exercise 23. 1. 789011 !20'7 14. ! 2. Sixty-.scvt;n iMillions ciulit i hundred and tlihleeu tliuu- i sand four hundred and twenty, and twenty-(>no inilUou thu ly thou.siind and f'orty-bix billit)r.lhs. Seventv-two nsillions, and seven ty-two billion tiis. One billio)! one million and on« hundred, and ten tril- lion ten niiliion and one tenths of quadrillionths. 3. DCCIX, M-VC(-'CLXXVI, MXCMXCIX, LXXXVMIV, MMM():^iXLVMiiDXCVl. 4. 5397;3 lbs. 5. £y 18s. 11 fd. 6. 10837 vrs. 119 days 2houry. 7. $291 9 -50 ^Aj-. 8. $123-77. 9. {320ltO0O02( "43 -00000000501 6. 10. 1 ac\'e 1 r<.jod 3 per. 4 yds. 5 ft. 1 1 in. 11. $122(>8-;]0. 12. 54 years 19 weeks 8 days 1() hours 33 minutes. 13. 741000000, -OOTll, 741000000, •000000741, -000000000741, •00741, and 74-1. 14. IG. 16. 17. 18. 19. 20. 21. 22. 23. 24. 26. 26. 27. 28. 29. 30. 3I3 32. 33. 34. 35. 00. •0331682. 4Vrv,VV hhds. ^0750. il^i 58 aeres. ^0-501. $37. 3 lbs. Ooz. 14dwt. ISJ-grs. 29 acies roods 21 per. 14 yds. 1 5 lbs. 4 oz. $^890-38f. 1082094. 10800. $360^15. 8247-95. $132062. 1C9-49. 879-99 ,V $59^85. i5o2-12i. 1 dwt. 14grs. V CCCCCCDCCIX. •50218+. 37. 18G909G969-e&. 38. $1713-34. 39. f,2 1-1483. 40. 230^" 1. 2. 3. 4. 5. 6. 7. 8. 9. ANSWERS lO MISCELLANEOUS EXERCISES. 407 Exercise 40. I. 2. 3. 4. 6. 6. 15. 16. A-t688-16-,V 275oti inilca 1 fu'- 21 per. yd3. 1 ft. 6 in. 96. 500313 octcnary and 20222133 quinary, 1213094-982'75. LX XMXO DXnil and CCXXXMVDLXVII. 7 t2000000905000O78014-000008'720001 1 7. 277200. 8. See XLVIII Recapitulation. Sec. I., i)a^e 57. 9. 0427629770(35001-1. 10. 11. See Table, page 125. 12. $2089-51^ 13. 27. 14. Sec Recapitulation XLVIII page 57. 17. Seventy-one trillions three hundred billions one hun- dred millions two hundred tliousand four hundred and one, and seventy thousand four hundred and two tril- lion ths. Oae uurM'UHjd and thirty-four qu'idrillions nine hundred trillions one hundred and one billions one hundred thoustvnd and one hundred, and two hundred million twenty thousand and two trilliontlis. Four quadrillions seven hun- dred trillionij twenty thou- sand and S(^ven, and two hundred and seventy-eight hundredths of trilliouths. £2272 Od. S^d. 13. 2'^' x5' x3x23. 19. 87 ft. r 1" 3" 0"" 10"'" Qiliin . . .m;;/// 1 ,.11111111 20. 011436. 21. 10383. 22. 4096. 23. 11 acres 3 rds. 7 per. 19 yds. Oft. 130 in. 24. 330900. 25. Child's share, $179-41-1-; woman's, $358'82/i-; man's, $1 794-1 2, V 26. 1023 and 512. 23. 48359-8979094. 29. 72-2487-0873859. 39. 05 lbs. 7oz. Odrs. 1 scr. 31. 1, 2, 4, 7, 8, 14, 19, 28, 88, 50, 76, 133, 152, 206, 532. 1064. 32. 82^^^- yards. I Exercise 63. 1. h -iVo, 2^(J, !s\, and jljj. 10. 14A^- and -c^oV 2. ^• 11. $134-15^ 3. $4-52-3V 12. Si>28387-06i. 4. 1 ;?6' 13. 311gjy bushels. 6. Gave away ?•§ and kept ^^. 14. 1 and IvMV 6. 1^- 15. 2H bushels. 7. $212-99^1-. 16. |. 8. Longer part 72 feet and 17. ni shorter part 64 feet. 18. 5/6- and2fj. 9. 1058//o acres; $13219-68|. 1 9, $i333-33^ or ^j of the wholo, 408 ANSWERS Tc/ MISCELLANEOUS EXKRCISE8. Exercise 11. 1. -8. a. 1-4445506778. 3. 4 days 17 liours 55 min. 80 sec. 6. 156-85931270094. 6. -739157196 of a mile. 7. 16 sq. ft. 104^§ inches. 8. 1 iicre 3 roods 13 per. 22 yds. 9. lli^aiidlfo. 10. 26-783V428671. 11. 71-86193. 12. 11-546 oz. 13. 75^ yards. 14. 13-5i69583. 15. 3, 3, 1, 4, 1, and 9. 16. 476-65628119. 17. 9. Exercise 78. 2. 702000007030017-0000000004000076. 3. 1017116606-6. 10. 20790. 4. n- 11. 1375<-12 and 2049151. 6. 10-.¥o¥o. 12. 66. 6. 5044 bricks. 13. 1 day 23 hours 24 min. M^i 7. Ill sq.ft. 0' 9" 7'" 4"" 5'"" seconds. 5""" 14. 19860 lbs. 2 oz. 9^ drs. 8. ^m. 15. $158-76. 9. 12225 bush 2pk8 Ogal 2qts. 16. f , 11 mh and ^mz- 17. 7040000, -0000704^ 704000( 7-04. )0000( }, -00000000704, -0000^04, 18. Hm^ 24. 13450Hi 19. Mail's share = £66 Os. 4|d., 25. 134062^ lbs. or 18406^ gala. woman's = £33 Os. 2|d., 26. $295-69/jj^. child's =£1103. 0|d. 27. 247fr. 20. 190i;i^. 28. 6AV 21. 1,2,3,4,5,6,9, 10, 12,16, 23. 18, 20, 25, 27, 30, 36, 45, 30. 29x3x6. * 50, 54, 60, 75, 81, 90, 100, 31. 55045884 lines. 108, 135, 150, 162, 180, 32. $45-59. 225, 270, 300, 324, 405, 33. $90-96 ii. 450, 340, 675, 810, 900, 34. 3-185988. 1350, 1G20, 2025, 2700, 35. 21592f. 4060, 8100. 36. $2 1588 -90. 22. 117. 37. $142-8248. 23. Lunar month = 29 days 12 38. 293. hours 44 min. 3 seconds. 39. Hu, mi n ., m% Solar year = 366 days 5 HH. mh Mh' hours 48 min. 48 secoudfl. 40. $103-35^ i ANSWERS TO MISCELLANEOUS EXERCISES. 409 Exercise 89. 1. 2. 3. 6. 7. 8. 9. 10. 11. 12. 13. 23. 24. 25. 2o. 27. 28. 30. 31. 32. 33. 34. 36. 2 : 3. $479'30§. 7787 63«;e3|j- duodenary^ 760/0 .^Mndenary. 5-57052 oz. 3 yds. 3 (^rs. na. 0^1^ in. 82962-70. Ibush. 2pk. Ogal. 1 qt. 17: 8; 88 : 176; 17:8 and 23: 11; 6:7 and 88: 176; 1173:616. 39 per cent. 5 4 H^' 4. Greatest 21 : 27; least 9: 18. 6. 67-100565661872498. 12014313iJ^7mnary, and 14. 10,%. 15. £2 Is. 2id. nearly. 16. 8fg \:y8. 18. 52. 19. 50^. 20. -026856599989+. 21. -0778. 22. 4-32958 miles. 704876837 nonary; 10011110101000011001111010000 fetnary; 1 U46453021 «fi/> 53. 51. 55. 56. 57. 68. 59. 60. 1296. 38-395 7119^ years. 144. 355i 8} days. )};2469-71. and 4,^3, 3.^3, 2.V Each man had 60 ; A caught 50, B 60, C 70. 191 and 17763. 44-997 years. 61. A,$1B66'95!|; B,)j(1169-965; C, )i5973-(H;V3. 62. 1, 2, 4, 1429, !W58, 5716. 63. 2g«„. 64. Man's share = $91«»14*|, woman's = $4f)9 5T44, ajid chiid'o = $15yi94V- 66. 24. 66. $21-03. 67. (j!reatest9: 16; lea^tlO: 19; eomp. ratio 21 : 247. 68. 8-6818452. 69. •Ol9K)till8. 70. 2781-849813156089829057. 71. 157086 feet. 72. 85 spirits, 35 water. 73. 422-32. 74. 70 and 14. 76. 223-82460586. 76. 6-32341. 77. 58 and 28. 78. 156240. 79. 80401. 80. 228^:1617. 81. 3 and H, or 4 and 1^, or 6 and 14, &c. 82. U|. 83. 5,^1 minutes past 1 o'clock. 84. 6-585461; 3-602675; 5-187521; 2-118509; 0-196295; 1-969276. 85. $4-314. 86. X $672 and Y $1120. 87. J 7". 88. 4321. 89. 18| lbs. at4d. ; 18| lbs. at 6d.; and 74^ lbs. at 8d. 90,. 10, 22, 26. 91. 1, 8J, 16|, 24a, 32f, 40. 92. 7. 93. Apple 2d., pear 3d. -'*• 4 8 • 95. $275. 96. $124 and $1664. 97. 11000000000011-0000000011. >3-6276 ; 98. $3649-3932. 99. 2« X 3' X 7 X 11. 100. 28f 101. 117. 102. 62f gal., 83^ gal., and 140 gal. 412 103. 104. 106. 106. 107. 108. 109. 110. 116. 116. 117. 118. 119. 120. 121. 122. 123. 124. 125. 126. 127. 128. 129. 130. 131. ANSWERS TO EXAMINATION PROBLEMS. A, £194 168. l|fd. ; B, £129 178. 4W.\ C, £97 88.0^4.; D, £77 188. 6^^d. 1^1280-838. 111. l8t, '46 inchcn ; 2ik1, 157 10 hours. in. ; 3rd, "82 in. ; 4th, 41 years. 3'H9 in. 4029 days. 112. 7. 117. £4 168. 113. I2019-6B1 ; |487l 80n ; 44iV $481B-80B; |6467-7«9i 1422-2 Ibl. $1825. 1 14. 1"* 300 yrs. ; 2"" 56 827 yra Ist, $920-20; 2nd, $2760*60 ; 8rd, $6521-20. Paid each workman $28-66| ; Ist company cleared 87j^ acres ; 2nd company, 77^-^ acres ; coat of clearing, $84^^ per acre. 15 and 11. 132. 61 of each, rem. £lj'i{. $2340-00. 133. $200. 132 dava. 134. 10 per cent. A, $2180; B, $1635; C, j 125. $1388-888. $1308 ; D, $1090. 136. Is. 9d., Is. 2d., and 7d. h §i Uh tmt. S^\ll 1 137. A, $25 ; B, e,25 ; C, $50 ; 801H and 411^30^4-. | D, $100. Sum $58 0s. 8,Vi,J. ; quo- 138. -057. tienc 32414-56. 139. y^gV ; 162,Vo ; 1 \H ; i,', \ 2308. 140. 96: 17f. 141. $89H; 6107H; $14-^3; nnd $179iV 142. $15009-41. 143. 17^, 32^, 48^, and 63g ; 3& nnd 86905. 144. 38^ days. ^^hh yds- $214. 1*' 175 yrs. ; 2°* 41-914 yrs. lOi'f perches. 111104. 9, 27, 81, 243, 729, 2187, 6661. 9i 804 in. 9-834 in. 12-426 in. and 80 inches. TBI tVJS n 88. 0H<1.; 'H ; and, -57 J2 in. ; 4th, |!4R'n son ; «6467-7a9 ( Bd 56-827.vr.i 'I leared 87'^ aring, $8^%^ m. £111 ., and 7d. 25 ; C, $60 ; and 68J ; 3&