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Un dea symbolea suivants apparaitra sur la darnlAre image de cheque microfiche, selon le caa: le symbole — »• signifie "A SUIVRE". le symbole V signifie "FIN". Mapa. plataa, charts, etc.. may be filmed at different reduction ratios. Thoae too large to be entirely included in one expoaure are filmed beginning in the upper left hand comer, left to right and top to bottom, aa many framea aa required. The following diagrama illuatrata the method: Lea cartaa. planches, tableaux, etc., peuvent Atre filmte A dea taux da rMuction diff Grants. Lorsque le document eat trop grand pour itre raproduit an un saul clich*. il est film* A partir de Tangle sup^rieur gauche, de gauciie k droite. et de haut an baa, an pranant la nombra d'imagaa n^cassaira. Las diagrammes suivants illustrent la m^thoda. 1 2 3 1 2 3 4 5 6 "15 ''^ ARITHMETIC /or « h Cnlkg^B mi Irjnmls, ADAPTED TO THE DECIMAL SYSTEM OP OUKBENCY, ft FEOM THE ARITHMETIO OP BARNARD SMITH, ESQ., M. A., Fallow and Tutor of St. Pat«r>s Collage, Gmbiirlg% BT THE KEV. A, P. MORRIS, M. A., Oxpoed, Principal, And Mr. JAMES RAVEN, late Miihematioal Master, OP THE GRAMMAR SCHOOL, HAMILTON., HAMILTON : PRINTED AND PUBLISHED BT GILLESPT & ROBERTSON. 1860. f»»PA^^^ r 't ■ The acknowledged superiority of the Treatise on Arithmetic, by Mr. Barnard Smith, over any that have preceded it from the English press, induced the compUers of the present work to undertake its adaptation to the Decimal System of Cunenoy which is now ili use in this Province. The excellence of the original consisted in its clear explanation of principles and in the numerous examples worked out under each Rule, which. enabled the Student to see at a glance the appliisation of the reasoning in each case. In the words of the late learned Dk. Peacock : « The Rules in this Arithmetic are stated with great clearness. The Examples are well selected, and worked out with just sufficient detail, without being encumbered by too minute explanations; and there prevails throughout it that just proportion of theory and practice which is the crowning excellence of an elementary work." These characteristics have been retained in the present work- and while its adaptation to the Decimal System has been made complete, care has been taken to retain sufficient examples of the English ^rrenoy to enable the Student to become thoroughly familiar with that method of computing money, rendered so necessary by the. commercial relations of the two countries. The present opportunity, is taken of thanking Mr. McCallum, Principal of the Central School, Hamilton, for a valuable addition to the article on the Double Rule of Three, and for several suggestions which have been adopted in this book. Hamilton, O.W. \ Itth January, 1860* )• A. P. MORRIS. f ,. TABLE OF CONTENTS. •»»' II Amthmetio. _ . „^ Definitions, Notation and Numeration, i Addition, ". ^ Simple Addition, , * * g Subtraction, g Simple Subtraction, ,• j Multiplication, 2i Kmple Multiplication, ***' 12 Division,..." jiy Simple Division, , , , ^j Greatest Common Measure, .* 24 Least Common Multiple, 27 Mitcellaneous Questions and Examples, 31 Fractions, ^ 34 Vulgai;Fractions,.... ^ 35 Addition of Vulgar Fractions, 45 Subtraction of Vulgar Fractions 47 Multiplication of Vulgar Fractions, 49 Division of Vulgar Fractions, 51 Miscellaneotu Examples toorked out, , , 53 Miscellaneous Questions and Examples, 55 Decimals, ' ^^^ 59' Addition of Decimals, g2 Subtraction of Decimals, 54 Multiplication of Decimals, , , , ^ 05 Division of Decimals," , . . . ^ Vulgar Fractions expressed as Decimals, ^0 Circulating Decimals, 73 Miscellaneous Questions and Examples, 73 vi TABU OF OONTINTS. ^^wtaaal Coinage^ .... Concrete Numbers, TableofBngliahOoins,. ....,'*..*.'* ' ^^ Measures ofWeight,. length, ..;!.7.'.;;; 92 " Surface, ^4 " SoUdity, 95 " Capacity, .* 96 " Number, 9t " Time, .*.*.*.*.'.*.'.'.' ?^ Beduction, Compound Addition, " "• " Subtraction,....'.'.*.*.* ^^^ " Multiplication,....'.*.'.' ^^^ " Division,,... "" Ill 113 Reduction of Fractions, " l^edmals,.... '.'.'.'.'**.' ^^^ " Currencies, ..*.!!.*.'.'!!!![[ ^^^ Practice, •' V"^^^ Square and Cubic Mea^;; 0^;; M:a*t;;i;^t;on,' Duod'e'cLnli; .* .* ." m Miieellaneous Examples worked oui, . Miteellanem, Questions and £xample», .....'..........'. igQ Rule of Three, Double Rule of Three I'^l Interest^ .'**** * * 190 Simple Interest^ ..., 206 Pormuteforthe^i'mputetionif'^tere^t;; ??,^ Compound Interest, 211 Present Worth and Discount * ^^^ Present "Worth, .. 217 ' 218 TABLI OF COKHNTS. Discount, Stocka, Brokerage and Commission,* .*!!.'.*!!*. i 1 i [ ] ooi Partial Payments— Ponns of Notes, . Jtl Profit and Loss, * ^^ Division into Proportional Parts, .*.'.'.*.*.'.*.* nil Fellowship or Partnership, .'.'.' J,J* Simple Fellowship, [[[[ ^ Compound Fellowship, .*.'.'!.*.'.*.'.'!! * ^^' Equation of Payments, ....*.!'.**.! **" Application of the term Per Cent, .'. „f? •Example* and jQuettioM on Average 339 Square Root, Cube Root, .....*.'.*.*..*... " '*^ Progression, * ' * ^^^ Arithmetical Progression,. '..*.'!.*.'!.*. ^ Geometrical Progression. . . .•.«*. jj.. „ ..261 MwellaneoM QueBtions and Example*, jgg ▼H -«•» ,^. n f ARITHMETIC. DEFINITIONS, NOTATION AND NUMERATION. Artiolb 1. By a Unit is meant a single object or thing, oonsidered as one and undivided. ^ »«mwTJu 2. Number is the name by which we signify how many objeott or thmgs are considered, whether one or more. When, for instance. number of the things referred to will be one, two, three, or four, according to the case; and so one, two, three, four, and the rest are called numbers. 8. Numbers are considered either as Abstract or Cokoreti. Abstract numbers are those which have no reference to any particular kind of unit; thus, five, as an abstract number, signifim hve units only, without any regard to particular objects. Concrete numbers are those which have reference to some particular kind of unit; thus, when we speak of five hours, six yards seven horses, the numbers five, six, seven, are said to be concret^ numbers, having reference to the particular units one hour, one yard, one horse, respectively. 4. Arithmetic is the science of Numbers. Jti ^}^ *^""i^ers in common Arithmetic are expressed by means of the figure 0, commonly called zero or a cypher, which has no value in itself, and nine significant figures, 1, 2, 3, 4, 5, 6. 7. 8. ^ which denote respectively the numbers one, two, thrU, four five! SIX, seven, eight, nine. These ten figures are sometimes UUed JJioiTs ; but this name is often improperly limited to the nine sijgnmcant figures above mentioned, which are then called the nine The number one, which is represented by the figure 1, is called jS^^When any of these figures stands by itself, it expresses its ^^\TL^' 'TT ""^'"J • ^^""^^ ^ expresses nine abstract unite, or mne particular things : but when it is followed by another fig^e 2 "ABITHMETro. together with ten times foSS ™d .^""^ **""*« "!»« ""its. soon by a tenfold inoreCforSS^.l"?** ™il» "^"'^ = ""d jae value which thus beloCfo aZLT^-**^ «*«« fo^ows it position 01 place, is oaUed itsXl! va^ ™ consequence of its « ^ffi "'' ''"^' *«- « «-!"- - ;.,«^, ^^, „„, „^„ ^v^sK^i^^'^t'ti^terr^L^r^^^^ - P«-«ne /fiTi *'!i':*y-«'g''t, thirty-ninT- we'th!f r"™' *'""y-«'^. thirty sinL^U^^r^rrr?-^'^''^^^^^^^^^ «e,t number to thiris,^:;S-^-r-th^^^^^^^^^ o! DEFINraOKS, HOTAWOH AND NUMERATIOK. 8 expresses ffhen it is times its line units, lore '. and follows it. ice of its and also proceed units to of units lousands sands of 3reds of sands of >f units, lions of 30 on to lits ; or, ogether 12, 13, f units units; lixteen, ^,25, ' units ^1 nine y-two, seven, 4,35, r-one, hirty- ), 60 ; the s, or 1^ one hundred of units, together with no tens of units, together with no units ; or, as it is briefly read, one hundred. By pursuing the same system in higher numbers the figure occupymg the fowth place from the right hand will represe * so ?"T S?u ^f ^^^>M Express the following numbers in figures : (1) Sixty-three ; eigkty-one ; ninety-nine ; forty ; thirteen. ARIIHMSTIC. «»d ninety-thrT' '■'°'""-^«J «<» two thousand, five hSndSl ^'^"r St°S;>„:^^ t'ttef -5 -""o-^. one millions, fiftj.four thousand and pl^h.-i ^^ ^^^J fifty-four tbrteen millfons, twentrthL^^^^^^ hundred and ^ (7) Two bill ons ; nine bnW fl ^f^"? ^ *"^ *"^«e. 10000001- S^J^?f2 5 418254. ADDITION. of the same denomination ZallZn^^ ^f^*" <*"«'■«'« ""mbers ft fa <>'«IK.undAdditio; wh nTr t"' "''P'"*^' •" ooBorete numbers of l,e 1^"^?'? \''* *»''«" '»«<'*«'■ denominations of that kind- «,n^I, ^^?,f.> >"" "f different years, months, •".» j...- ' "Pounds, shiUinsrs. and no«, ■ ^ ™" '^^' '' o'' 8»"ons, quarto, ana" pints! '' ADDITION. ven hundred seventy ^ne; and nine. thirty-three seventy-six ive hundred illions, one ^; fifty-four undred and :iiree. >usand and ninety-four nbers : 5 ). I is equal ^ether is t ^pouiro. together umbers ogether ifferent 12. The sign +, plus, placed between two or more numbers signifies that the numbers are to be added together : thus 2+5+7 signifies that 2, 5 .- . 7 are to be added together, and denotes their sum. The sign =, bqual, placed between two numbers, signifies that the numbers are equal to one another. The sign , vinculum, placed over numbers, and the sign ( ), or [ ], called a bracket, enclosing numbers within it, are used to denote that all numbers under the vinculum, or within the bracket, are equally affected b y all numbers not under the vinculum or within the bracket : thus 2+3 or (2+3) or [2+3], each signify, that whatsoever is outside the vinculum or bracket which affects 2 in any way, must also affect 3 in the same way, and conversely. The sign . •. signifies * therefore.* SIMPLE ADDITION. 13. Rule. Write down the given numbers under each other, so that units may come under units, tens under tens, hundreds under hundreds, and so on ; then draw a straight line under the lowest line. Find the sum of the column of units ; if it be under ten, write it down under the column of units, below the line just drawn ; if it exceed ten, then write down the last figure of the sum under the column of units, and carry to the next column the remaining figure or figures; treat each succeeding column in the same way, and write down the full sum of the extreme left-hand column. The entire sum so marked down will be the sum or amount of the separate numbers. 14. Add together 5469, 743, and 27. Proceeding by the Rule given above, we obtain 5469 743 27 6239 The reason for the Rule will appear from the following considerations When we take the sum of 7 units and 3 units and 9 units, we get 10 units and 9 units, or 19 units; we therefore place the 9 units under the column of units and carry on the 1 ten units to the next column, viz. the column of tens. 6 ! ^^tranano. Now the sum of l f^M o i. 3 tens, or 18 tens • ITL. ^^' f ^^^* «^^ 6 tens ,« in f 2f te«« and camr'ortST^^/^ *h« » tenrSnder fh ^? ^"^ , down at ft]l W^"^?'* »«1>' We been worked f k +1?; +700+40+1 Now aiding the oo,un««:'::tith ^foo+noo'+m'tjt"" Ex. n. ^ ^ I7 (3) 234 ^ 753 ~ J45 1899 (S) 12 35 56 80 183 494 687 656 336 (4) (6) 1721 3333 5046 2754 654 321 804 509 (7) 750 36 1843 _661 3190 (8) 4789 2346 3857 5005 w 10 tens and ler the column fc column, viz. mdreds, is IQ fore place the yy on the 1 thousands. > thousands ; ids, and the bus. eaaiM Ajamnov. 10+9) and so on) adding the "ng them Me will in t) 654 321 804 4789 2346 3867 5005 (9) 9102 (10) 84670 479 5437 8776 29 901 21904 (11) 1790621 206803 353 9003766 (12) 256783 21003 5734 40036 21 100001 (13) 627432 543201 678641 548200 868759 345678 (14) 892764 93687 9482 100 152346 11 (15) 1807353 298743 5987 760003 247 50705 423578 (16) 117064 92973 827569 351 777777 65656 r,.i}7L ^^^ together 7384, 326, 6780, and 57: also 6740 ft74fi T065591234, 87«, 9183460789. and 5000 ;^f SS22' 73600, 27978462, 333, 5875896006, 48!nB32^6^I^ 80082148379, 18846, 1118868673, and 688M^5. ^^^' fhi*?*^ Add together one thousand, four hundred and e«hty. Z!!'i.'T°i'"''^'"!^"°'l"^'*y-^; thirty-nine; forty thoulmi s^T- fift/J^""'*?"^-'^""'' "^^ thousand, eiiht hiidred^d sixty ; tilty thousand and seven. (21) Add together the foUowing numbers : fifteen thousand ^nZl,-?'^'^ ""1 ""''r " ' ^°" "^"-J^ »°d nine ; two TS and th.rty.four thousand and fifty ; four mUlions, three thoS (22) Add together the following numbers • twentv-two millions, SIX hundred thousand, five hundred and t^fiTe four ; one hundred and eleven millions, six hundred and fifty thousand and fifty ; three hundred and twentv.«iv S.!?''^!!.^^ Cred'knStn S'^' ''^ f^-^^^ ^'^^^ ^^ nunored and ten millions, one thousand seven hundred and ten • one billion, three hundred thousand and five. ' ASlTHlfSTZO. SUBTRACTIOK. The number found bv anKfro^V- .? ^^®**®^ """^ber. from th, greater is"<^,'i I'tmS """"'" "^ '-» ""-be™ SIMPLE SUBTRACTION. j»d^ hundred, .„d ,o on , .henT^rst^^, S' it'ta^ " Am KSbt rftrr^el^jr- "^r <^'^^ 'o- which Stands immediately o^er t Z\lf^^ "^ """ W^' «"» Ae hne just drawn, units under unite tif.,'5,"'*"'"'"^'''- Wow hut If the units in any figure in th. i^ t«M "ider tens, and so on : of «D te in the figure K it, «dd ten T T ""^ '^ """"'"^ then take the number of units in fLi ^ J*'^ "PPe"" %ire, and "the upper figure thus ?nor1^"ed iT^K"*"™ '■'<"» tSnumW Wore, and then carry one toTe LCfi * "-emainder down a. ne entire difference I remainder so mfc °f *' ^°^«' «"«■ dltferenoe or remainder of the gfven number ""' '"" ''* "^ ?J>- at. Subtract 4938 from 6123 P«K»eding by the Rule given above, we obtain 5183 4938 -that the remainder is one hundred eighty-five (186). mrn^ritun^r"^'^'^'^"^^^^^^^^ units to the 8 nnL ^J™" *™« ."■»<», we therefore ad^ ,© «^%8 ».ts fr^ml3t^;-: We'nr le^ft/te-^^^ SUBTRACTION. 9 place 5 under the column of units : but having added 1 ten units to the upper number, we must add the same number of units (1 ten umte) to the lower number, so that the difference between the two numbers may not be altered ; and adding 1 ten units to the 3 ten unite m the lower number, we obtain 4 tens or 40 instead of 3 tens or oO. Again, we cannot take 4 tens from 2 tens ; we therefore add 10 tens or 1 hundred to the 2 tens,which thus become 12 tens or 120 • and then taking 4 tens or 40 from 12 tens or 120, we have 8 tens or 80 remammg ; we therefore place 8 under the column of tens : but havmg added 1 hundred to the upper number, we must add 1 hundred to the lower number for the reason given above : and *??"^,l?"''1''^^ ^ *^® ^ hundreds in the lower number we obtain 10 hundreds or 1000 instead of 900. ""^^oer, we Again, we cannot take 10 hundreds from 1 hundred, and we therefore add hundreds or 1 thousand to the 1 hundred, whiTh 1000 from 11 hundreds or 1100, we have 1 hundred or 100 left- we therefore place 1 under the column of hundreds : but having added 10 hundreds or 1 thousand to the upper number, we 3 add 1 thousand to the lower number for the reason given above • and addii^ 1 thousand to the 4 thousands in the lowefnumberwe obtam 5 thousands or 5000 ; • "umwer, we 5000 taken from 5000 leaves • therefore the whole difference or remainder is 185. 21. The above Example might have been worked thus * miffing down at full length the local vSlues of the %ure8 : ' ^ ^ 5123= 5000 + 1004- 20 +3 = 4000+1000 +100+ 20 +3 =4000+1000+100+10+10+3 =4000+1000+110+13 (collecting the first 10 with the 100, and the second 10 with the 3 ) 4938=4000+900+30+8. llZf2' ''^^''^^"^^^"g <^he columns, thousands from thousands &c we get the remamder or difference "u»anas, ozc. =100+80+3=185. ^^ mte. The truth of all results in Subtraction m^rr h. ^.^--M »~ - -" tae less numoer to tne diiference or remainder -'If thi« «nm "^ "i"°i?«f B w (1) 663 m go (6) 4236 3089 Ex. in. ^^ampies in Simple Subtraction. (^> f.^ (3) 704 iii 483 (4) (6) 80502 38672 806 720 (7) 46095 28736 (8) 655655 123456 (18) How much grater is 1643862i9lhan 48476798. " '• i22222'>^OOOttan7W070077, , (14) Take two thousand JST'^''*^'^^"^^^^^^^^^ rfx ; tfcee thousand Z^th? f^r' *"°°' '^'^ f^o^sand and ninety forty-four. """ *«'"' ^o^ seven thousand, nine huS (16) Subtract five hundrTT..'"""^^*<»'S8''i five hundred by D or lo one ff.' ^'^^ 5^ i" ' <"■« hundred by C • ^ All other numbers fl2r;Zi^?f°"^ by M or CP,. ^ ^ ' <*"aoter,subjectto^arii'^r"'""^^^^^^^ for 8 : LV for 55 ; LxlvH for 77 c6xr . oj /^ ^°' « 5 ^^ Secondly Who.,. 1, '"'^ " > "^^-S-I for 811. whole exp/ekion deno^^wJT^'^'^^by one of &,. value ti,e for 100-10, or 90. '°'^^®5 ^L for 60-10, or 40 • XC - -^"f^H, —tance. to Stood for-500^;: ^^r^! MULTIPUOAYION. tl and so forth. And every C prefixed and q annexed to CIq, increased the value of the latter tenfold ; for instance, CCIm stood 10000 ; CCCIooo for 100000 ; and so forth. Fourthly ; A line drawn over a character or characters increased the value of the latter ^thousandfold ; for instance, V" stood for 6000; C for 100000; IX fcr 9000; and so forth. It follows then that either XXXXVI or XLVI will represent 46 • and that either M.DCCC.LIV, or CIq-IqCCCLIV. orlDCCCLini will represent 1854. Ex. IV. (1) Express in Roman characters, thirty ; forty-eight ; fiftv-nine • 222 ; 6000 ; 1843. ^ b > J > (2) Express in words, and also in Arabic figures, the, values of XXIII; LXIX; CCXVIII; Tl; CLDaU; MC. MULTIPLICATION. 23. Multiplication is a short method of finding the sum of any given number repeated as often as there are units in another given number : thus, when 3 is multiplied by 4, the number produced by the multiplication is the sum of 3 repeated 4 times, which sum is equal to 3+3+3+3 or 12. The number to be repeated or added to itself, is called the Multiplicand. The number which shows how often the multiplicand is to be repeated or added to itself, is called the Multiplier. The number found by multiplication is called the Product. The multiplicand and multiplier are sometimes called * Factors,' because they are factors or makers of the product. 24. Multiplication is of two kinds. Simple and Compound. It is termed Simple Multiplication, when the multiplicand is eitiier an abstract number, or a concrete number of one denomination. It is termed Compound Multiplication, when the multiplicand contains numbers of more than one denomination, but all of the same kind. 25. The sign x , placed between two numbers, signifies that the numbers are to be multiplied together. \l 12 ARITHMBTIO. 8«. ae following Table ought to be learned correctly : line, when multiplied bv 2 t'J' ' ' ^' *"• "' '2, in the first under the resp^^e Tumber^* oTX' iC-^-^ ^^'''g P'^-^ multiplication of which they aril ./h^tw J r ^\'"'^' *«'» *« products when the %ur^"nU,eV^J^''"* '>'«*'«•'« several multiplied by 3, and so on ^'" ''"* "« respectively fi»d 1>W m Jy^hmfe.,^' """i multiply 6 shillkgsTv 4 i . t'"'°«' SIMPLE MULTIPLICATION. "nf^u^^-tenfte'^te^sS^^^ rWr^^"-^' ""i-^ the multiplicand, beginning wS,c,^?; ?*«i"Piy each figure of plaoeofthemultiplief (by me^fiju " m' V *« %ui-e in thi units- set down and car^y as n ffion X.^^V- V"'*^"'«P««»ti°»h multipUcand, be^nning ^iTZ u^l\ i"P'r'«'''''«"^e»''*e place of tie multiplierfpladng L ««; / *' "«"« " "-e tens' the tend of tl,« i!_» .i' *^ .8 "■* "i^t figure so nK(oi„„j ...j place of the multiplie;rptog"fte"fi«;''/ *** ''«'""« ™ the'ten;" "■- * - • e line abive, £ex? fit"' ^T '". "^^'"^^ "idei- Krt !i "-- """lextngure under theh.m^.„j , «a Ji uie same way with «inl, ...".j. -""''' ""a so o^"'p/c^'?,« T^^' «•« next figure Snd;r"tl.e"h3Td-''""'' oceed ai .« ^^e way with each suc^d^'g" fig^Tf 8IMPL1 IflTLTIPUOATIOir. 18 the multiplier. Then add up aU the results thus obtained, by the rule of Simple Addition. Note. If the inultiplier does not exee«d 12, tha multiplioaUon can be effected euily m one line, by means of the Table given albove. 28. Ex. Multiply 7654 by 397. Proceeding by the Rule given above, we obtain 7654 397 53578 68886 22962 3038638 The reason for the Rule will appear from the following considerations. When 7654 is to be multiplied by 7, we first take 4 seven tmies which by the Table gives 28, i, e., 8 units and 2 tens • we therefore place down 8 in the units' place and carry on the 2 tens : again, 5 tens taken 7 times give 35 tens, to which add 2 tens, and we obtain 37 tens, or 7 tens and 3 hundreds ; we put down 7 in the tens' place, and carry on 3 hundreds : again, 6 hundreds taken 7 times give 42 hundreds to which add 3 hundreds, and we obtain 45 hundreds, or 4 thousands and 5 hundreds ; we put down 5 in the hundreds' place, and carry on the 4 thousands : again, 7 thousands taken 7 times give 49 thousands, to which we add the 4 thousands thus obtaining 53 thousands, which we write down. * Next, when we multiply 7654 by the 9, we in fact multiply It by 90 ; and 4 units taken 90 times give 360 units, or 3 hundreds, 6 tens, and units : therefore, omitting the cypher we place the 6 under the tens' place, and carry on the 8 to the next figure, and proceed with the operation as in the line above. When we naultiply 7654 by the 3, we in fact multiply by 300 ; and 4 multiplied by 300 gives 1200, or 1 thousand, 2 hundreds tens, and units ; therefore, omitting the cyphers, we place the first figure 2 under the hundreds^ nlace. and nrn^^ii ao before. Then adding up the three lines' of 'figures'^which we nave just obtained, we obtain the product of 7654 by 897 14 ■^•JWDiifia >wn at fu "^^m f^ '"Z*^* J^^^^ ^^e been workpH f k * ^ the W: values of thefigu^^^^^ '^""' P"*^^ = 7 avr. »xIOO+ »xlo+ 7 '«wxiooooo+ xlOOOOO + 8x100000 + 1 y 10000 + 1x100000+1x10000+8x1000 + »xlO0O+ 9x100 + 7x100 + 1x10 ^<»«»07i^,^B0500T2^riSSr:^ _^2xl0+8 =8000000x1000000 + 2x10000 + 1^^^^^^^^ _^^T*xiuuoo + l0xl000+7xlonnj.i7r~r:c: =»oooooo+^;7I^55^;n:i^^;— ^^^ 30. Jfthe multiplier or mii]f!,.i- i. / «-«)0, and 670 be multiplied by Sm"wS """"'P''*"* 6200 °^^ 686^ f^ He reason u clear • f** • >. 1824000 70mul«plied,,.orknt7lra^^^£« .te^ n= uiuiuplj, by the 6 we takeUe ed thus, imtting annpLB iruLviPLKUTioir. 15 ixio+as 8 8 1x104^8 '200 5^,. 14000. her place, tiplier we \ multiply rom that instead of take the .r^^i^il"^-^ « times, when we multiply by the 2, we reaUy take the multiplicand, not 20 times, but 200 raes. ^ ^?* ^^®5.S^° numbers are to be multiplied together, it it n matter of indifference, so far as 2x4=2+2+2+2=8, 4X2=4+4 =8; theiofore the results are the same, that is, 2X4=4x2. • That 'he product of one number multiplied by another will be equal to the product of the latter multiplied by the formeT may perhaps appear more clearly from the following mode of shS this equality m the case of the numbers 3 and 5. «°ewmg 8=1+1+1; •'^''^=^iVV+Vj"'^^''^''^^"'^^-^^-*-^^+<^+^+i)+(i+i+n +1+1+1 + 1 + 1 + 1 }.=ie. +1+1+1 + 1 + 1 + 1^ .. ^.n^'}'"^^ '' -f "* *^^ o«.^ from left to right, there are 3 ones U.ken 5 times; if we regard them taken from top to bottom, we Imve 5 ones repeated 3 times; and the number of ones in each ca^e IS the same; t. .3x5=6x3: and so in the case of any Uo other numbers multiplied together. ^ 33. IJe truth of all results in Multiplication maybe proved bv using the muhiplicand as multiplier, and the^muftS as ^rn^T'^i 'It Pu'"^"'' thus obtained be the samel the product found at first, the results are in all probabUity true 34. We have hitherto confined our attention to products formed by the multiplication of two factors only. Products may ho^S t'JrmedT '^' "^-^^'^^^^''on of thre'e or more fSs ; tMsIs termed Continued Multiplication: thus, 2x3x4 denotes the fhi«^ t? r ^'f ^*'?^'^l^y ^' ^°^ the product tJiis obtained to be ^}^±!'.^^^'^''i'^ i^^ ?«"ti«^«d product of 2, 3, and 4 : we may 16 (1) 534 4 2136 (5) 2057 ^ 7 (9) 35976 11 ABITHlfETIO. Ex. V. Examples in Simple MulHplieation. (2) 673 3 (6) 57409 8 (10) 91525 12 (13) 87298 46 (17) 840607 80 (14) 16097 59 (18) 175 189 (3) 2867 6 (7) 2745638 9 (11) 257 53 771 1285 13621 (15) 296897 (16) 69284 (4) 7492 6 (8) 5763 n 63393 (12) 96843 17 83 (19) 6298 769 (21) 25607 5004 (22) 78847 8803 90 6235560 (20) 5423 603 16269 32538 3270069 (23) Find the product of 234578 by 18 bv 29 anri «T«« i. ko ?^»24846 by 67, by 95, and also by ^0 ; 2^846067 b v 20^ h ' 1008, and also by 907; 8409631 by 21711 bv 700q\! «uV and also by 7980. "J' ^wn, by 7009, by 8435, (26) Multiply six hundred and fifty thousand and ninety bv three chousand and eight ; also seventy-six millions, eight Sousknd nine" ""^ --vHvc, uy ume miliions, nine thousand and DIYISIOK. 17 (27) Find the continued product of 12, 17, and 19 ; of 378L 3782, and 3783 ; and of 6565, 6786, and 9898. (28) Multiply 20470 by 1030, and 2958 by 476, explaining the reason of each step in the process. DIVISION. 35. Division is the method of finding how often one number, called the Divisor, is contained in another number, called the Dividend. The result is called the Quotient. 36. Division is of two kinds, Simple and Compound. It is called Simple Division, when the dividend and divisor are, both of them, either abstract numbers, or concrete numbers of one and the same denomination. It is called Compound Division, when the dividend, or when both divisor and dividend contain numbers of different denominations, but of one and the same kind. 37. The sign-r, placed between two numbers, signifies that the first is to be divided by the second. 38. In Division, if the dividend be a concrete number, the divisor may be either a conc^-ete number or an abstract number, and the quotient will be an abstract number or a concrete number, according as the divisor is concrete or abstract. For instance, 5 tons taken 6 times give 30 tons, therefore 30 tons divided by 5 tons, give the abstract number 6 as quotient ; and 30 tons divided by 6, give the concrete number 5 tons as quotient SIMPLE DIVISION. 39. Rule. Place the divisor and dividend thus : divisor) dividend (quotient. Take off from the left-hand of the dividend the least number of figures which make a number not less than the divisor ; then find by the Multiplication Table, how often tBe first figure on the left-hand side of the divisor is contained in the first figure, or the first two figures, on the left-hand side of the dividend, and place the figure which denotes this number of times in the quotient : multiply the divisor by this figure, and bring down the product »..^ ■^^x■^vl.a^uv 11; iixjiii wic uuuiDer whicu was caKcu on at the left of the dividend : then bring down the next figure of the dividend, and place it to the right of the remainder, and proceed as before ; if the ii ARnmoitic. B^^p'^^^^^^t^nT^r'.^^^^r^^,^ the down, and the quotiSt if Vhpr!i^^'*^^''^^®«'^*fa"s brought determined, or ?f tSbe Tr:m:Mer""T'^^''^ ^''^ ^« '^-^ remainder will be thus determined ' ^"°*^'"* ^^ ^^^ g^f JJ-titeThi^^^^^^^^ --ber Which stands above 52 J' W wmainder ]?e ffreater Xn *f *??*.'^ ^^'°°« of smaller value- 40. Ex. Divide 2338268 by 6758 Proceeding by the Rule given above we obtain 6768)2338268(346 20274 31086 27032 40548 4054 8 Therefore the quotient is 346 represent two millionMhre" hundr^r.^u!''" "'^ the dividend and two hundred. "''^ """^ thirty-eight tliousand, In working, we properly place th7f 1 ^^S"*^ '^' convenience *esubtr,5t theVo^t thus found 1-1 '^ " """"' ''^^''' S108, which represenrthree huZin J! ''''^" * remainder of himdrcd. Bring down the 6 bv fh. p T '?? *ousand, and eight »0, but the cy/herTs omUtTfor ,W«; *'' \^''''"^ « 'ens or number now^VesenT thwe Ldrld r/''°^" «'«*«'' = «h« hundred and sixty • M^i! "nndred and ten thousand eiirht fl«8x40=27mo: weoLtrt"°'"*'r'' *" «'"«»' '" this, Td «ibtract the 27032 'frZ T' ^^yP""*' *' «ie end as before, and -^d. is 4054, whieh- ^r=i^^T:,^trCe^ 1 can eiaailybe SJWfl^ DIVIS^N. and forty. Bring down the 8 by the Rule, and the number now represents forty thousand, five hundred and forty-eicht: 6758 is contained 6 times exactly in this number. '^ *» Therefore 346 is the quotient of 2338268 by 6758. 41. The above example worked, without omitting the cyphers, would have stood thus : jt*'^^'^ 6758)2338268(300+40+6. 2027400 310868 270320 40548 40548 hence it appears that the divisor is subtracted from the dividend 300 times, and then 40 times from what remains, and then 6 times from what then remains, and there being now no remainder 6758 IS contained exactly 346 times in 2338268. ' folS^s ^"^"^^ ""^ ^^^ ^^''''^ "^^^^"^^ ""'^^^ ^^""^ ^®^° ^^^^ ^ 2338268=2027400+270320+40548 6758)2027400+270320+40548(300-1-40+6 +270320 +270320 +40548 +405 48 42. Ex. Divide 56438971 by 4064 4064)56438971(13887 4064 15798 12192 36069 32512 35577 32512 30651 28448 2203 «^'»fX,t5n,''"^"'"'"^ '" ^^^' 13887 Ume8,with the remainder 2203. II 20 ! 1 ABITHMBTIO. fig?rSfroml'ri^^^^^^ and as „.any ao cutoff at the riXhan^^^^^^ f *^"'" ^^« ^^P^ers remaining figuri accordinrto t^^ then proceed with the remainder afnex r/gteVcu^^^^^^^ fy.''. ^'i ' ?"^ *^*^^ ^^^ remainder. ^ ^^ "^^'^ ^® dividend for the total Ex. Divide 637623 by 3400. Proceeding by the Rule, 34,00) 5376,23 (158 34 197 170 275 272 ^^efore, 8400 u, contained in 587523, 158 time^ with remainder S4SS'lM tTmTsIUtriSderfoo"/ '"<"' /'''^«« ~"'«- contain 8400 at all thlTS^. '^,'"^- J ^l*^ «' ^3 does not remainder 300+23, or sls.^ '^'^^ '"'"^'"''^ ''^ 158, with t«Siu wfth"ej;hif "PP"" ''"'' «'• <«"- •■«1 dividend both '^^^^l^Zi^Zi^l^^^^Z^Z^'^. -PfT^d into A number which can be semmfo/l i«t„ r » greater than unity, or which to othl ^ °" respectively multiplying together two otmn^ u '""''^^> '« Produced by ««» uVty,*is I "oMwsrruS*^^^^^^^ 6 which =8 X 3, 8 wWoh =2 X aT^ ^''"' * ''''''<''" =2x2, because they are composed or .^nsisUTthT^^''^ "r*"^'^ -nora numbers, each o/which U ^eaZ Sfantr'"' "' '^^ "^ saittetitt ^:t:ir-"^/!r,«^-'«' t^» -Uy, are prime to each other. ^* ' *""' '"^ iiumoers 3, 5, 8, 1 1, are SIMPLE DIVISION. m process can be 45. When the divisor is a composite number, and made up of two factors, neither of which exceeds 12, the dividend may be divided by one of the factors in the way of Short Division, and then the result by the other factor : if there be a remainder after each of these divisions, the true remainder will be found by multiplying the second remainder by the first divisor, and addini to the product the first remainder. Ex. Divide 56732 by 45. 9 45 56732 6303—5 1260—3 the total remainder is 9 x 3 + 5, or 27 + 5 = 32. Therefore the quotient arising from the division of 56732 by 45 IS 1260, with a remainder 32 over. The reason for (he above Rule is manifest from the following considerations. 6303 is 5 times 1260 together with 3, and 56732 is 9 times 6303 together with 5, . or is 9 times (5 times 1260+3,) together with 5 or is 45 times 1260+27-|-5, ' or is 45 times 1260+32. 46. The accuracy of results in Multiplication is often tested by the following method, which is termed " casting out the nines" • add together all the figures in the multiplicand, divide their sum by 9, and set down the remainder ; then divide the sum of the Igures m the multiplier by 9, and set down the remainder : multiply these remainders together, and divide their product by 9, and set down the remainder : if this remainder be the same as the remainder' which results after dividing the product, or the sum of the digits in the product, of the multiplicand and multiplier by 9 the sum is very probably right ; but if different, it is sure to be wrong. This test depends upon the fact that "if any number and the sum of Its digits be each divided by 9, the remainders will be the same. Ihe proof of which may be shewn thus : 100=99+1, where the remRindpr munf. ha nn^ "'It^^^Ha^ '•n'^ *t-- " •' digits in 100, viz. 1, be divided by 9, since 99 is divisible bv 9 without a remainder. ^ n Similarly, A9JTHMSTI0. 200=2X99+2, 300=3x99+3, 400=4x99+4 600=5x99 + 5* / &c.=&c. Also the number 532 = 500 + 30-f-2 = 5x100+ 3 x104-2 . . =5x99+5+3"x9+3+2- whence it appears that if the parts SxiOO^viJi a c ,. . make up the entire number hrpnl / • ^ J ^ }^' *"^ ^' which will be 5, 3, 2 respect"vX 'and th.rpl t."^ ^^ ^' ^^" ^^^^ainders is divided by 9,^tTiri^^^ '^' remainder, when 532 divided by 9: ^ ^^"^ '^°'^' ^« when 5+3+2 is muTSpIie'ity I7' *'^ *^^^ '^^^^' '^' ^^« ^^^e as an example 533 533 57 3731 2665 Now 30381 533=9x59+2=531+2 T,. , . 57=^9 X 6 + 3= 54+3 57t^t:ir9rilft a^:^^^^^^^^ -r^^^-^ that 531 x is left after dividing he prXcfoVssI^^^^^^^^ remainder which same as the remainder wK U 1.1 t ^? ?2 ^^ ^' "'"s* ^^ the and 57 by 9. '' ^^^ ^^^^ ^^^^^^"g the product of 2 Again, since the product of 57 and 2=^54+^^ vo ^ .l product of 54 and 2 when dividprlX. o r^+^)^^» and the therefore the remainL which is tft ^IrV'T' T ^^^^^i^^er, 633 and 57 by 9 must brth. L after dividmg the product of dividing the UCt of 3 ' n'/'o**^" remainder left after product^f theTn:iL:r?wh1ch\?e'^^^^^^^ ^-^^^ 't multiplicand and multiplier res^ely b^^^^ "' ^^^ . JV'^ ^? dividing either 30381. or th« s"m of its ^\~'-^- ' • - - 15, oy ., the remainder left is 6, and 3"x 2 dtvitedT^X SIMPLE DrQir><3 T'U^ lQ, ' & " -.""iiz«. Aixc iOot, ui visuf yfi\i be tne greatest common measure required. Um^ GREATEST COMMON MEASURE. <|5 Ex. Required the greatest common measure of 475 and 589. Proceeding by thd Rule given above, 475) 589 (1 475 114) 475 (4 456 19) 114 (6 114 therefore 19 is the greatest common measure of 475 and 589. Reason for the above process. Any number which measures 589 and 475 also measures their difference, or 589-475, or 114 Art (50Y also measures any multiple of 114, and therefore 4 i 114 or 456 Art. (50) ; ' * and any number which measures 456 and 475, also measures their difference, or 475—456, or 19 • ?S"q "^""?:i^^^ greater than 19 can measure the original numbers 589 and 475 ; lor it has just been shown that any number which measures them must also measure 19. Again, 19 itself will measure 589 and 475. For 19 measures 1 14 (since 1 14=6 x 19) • therefore 19 measures 4 x 114, or 456, Art. (50) • therefore 19 measures 456+ 19, or 475, Art. (50) • ' therefore 19 measures 475 + 114, or 589 ; therefore since 19 measures them both, and no number greater than 19 can measure them both, ^ 19 is their greatest common measure. 52. To find the greatest common measure of three or more numbers. '' Rule. Find the greatest common measure of the first two numbers; then the greatest coihmon measure of the common measure so found and the third number ; than that of the comSon measure last found and the fourth number, and so on IV w re^uH ""'''"'' '" '"""^ "^^^ ^' "^^ greatest comm"on mea^uTe" 26 ▲RITHMKTIO. Ex. Find the greatest common measure of 16, 24, and 18. Proceeding by the Rule given above, 16) 24 (1 16 ■ , ■ ' 8) 16 (2 16 therefore 8 is the greatest common measure of 16 and 24. Now to find the greatest common measure of 8 and 18, 8) 18 (2 15. 2) 8 (4 therefore 2 is the greatest common measure required. Reason for the above process. It appears from Art. (50) that every number, which measures lo and 24, measures 8 also ; therefore every number, which measures 16, 24, and 18 measures 8 and 18 ; » » , therefore the greatest common measure of 16, 24, and 18, is the greatest common measure of 8 and 18. But 2 is the greatest common measure of 8 and 18 • therefore 2 is the greatest common measure of 16, 24, and 18. Ex. VII. 1. Find the greatest common measure of 16 and 72. 55 and 121. 272 and 425. 825 and 960. 176 and 1000. 689 and 1573. 30 and 75. 128 and 324. 394 and 672. 775 and 1800. (12] 1236 and 1632. -{151 63 and 99. 120 and 320. 720 and 860. 856 and 936. 6409 and 7395. 5210 and 5718. 2484 and 2628. 4067 and 2673, ^^«« , ^ . 1729 and 5850. (18 2023 and 7581. (20) 468 and 1266. (21 3444 and 2268. (23) 5544 and 6552. (24 J2S^ *?^ ^^®^^- (26) 80934 and 1 10331; i^Q'.^^^^^Lo P®) 1^536 ^^d 23148. 42237 and 75.5ft2. /9A\ ^qk^xa «>.j i%r^o,rw/^i 10353 and 14877. (32) 271469 and 30599. MABT OOmiOB MULTIPU, ^ 2. Find the greatest common measure of 805 ?S? Vh .I?a "^ ^' "34, and 1347. OUD, 1311 and 1978. (6) 28 84 lfi4 nn^ ^ <^h»* ^^'""^on measure, and place the quotients so obtained and the undivided numbers in a line beneath, separated as before. Proceed in the same way with the second line, and so on with those which follow untU a row of numbers is obtained in which there are no /riiumbers wtiich have any common measure greater than unity Then the TerilWr^' ^" ^'^ ^'^''^'^ ^"^ ^^« ""-^- - the lltt line will be the least common multiple required. vIZ\llh^tj?t- ^' ^°"r^ advantageous to begin with the lowest rrj^l ? ^ **/ 4^^"^^' and to repeat this as often as can be dona • and then to proceed with the prime numbers 3. 6, Ac. in the same wy ' Ex. Find the least common multiple of 18, 28, 30, and 42. Proceeding by the Rule given above, 2 2 3 7 18, 28, 30, 42 9,14, 15, 21 9, 7, 1 5, 21 3, 7, 5, 7 .u . . ^' *' ^' * therefore the least common multiple required, =2x2x3x7x3x5 = 1260. Reason for the above process. Sinc€l8=2 X 3 X 3; 28=2 x 2 x 7: 30=2 x .^ y .^. zt9=o v 3 v. 7. it IS clear that the last common multiple of 18 and'28 must contam as a factor 2 x 2x 3 x 3x V ; and this factor ItseTf is m ARITHlimO. evidently a common multiple of 2 x 3 x 3, or 18, and of 2x2x7, or 28 ; now the least number which contains 2x2x3x3x7 asa factor is the product of these numbers ; therefore 2x2x3x3x7 is the least common multiple of 18 and 28 : also it is clear that the least common multiple of 18, 28, and 80, or of 2x2x3x3x7 and 30, or of 2 x 2 x 3 x 3 x 7 and 2x3x6 must contain as a factor 2x2x3x3x7x6, and this factor itself is evidently a common multiple of 2 x 3 x 3 or 18, 2 x 2 x 7 or 28, and 2x3x5 or 30 ; hence it follows as before that 2x2x3x8x7x6 is the least common multiple of 18, 28, and 30 ; again, the least common multiple 0f2x2x3x3x7x6 and 42, or of 2x2x3x3x7x6 and 2x3x7 must contain 2x2x3x3x7x6 as a factor, and this factor, as before, is evidently itself a common multiple of 18, 28, 30, and 42 ; now the least number which contains 2x2x3x3x7x6 as a factor, is the product of these numbers. Therefore this product, or 1260, is the least common multiple required. Note 2. The above method is sometimes shortened by rejecting in any line, any number, which is exactly contained in any other number in the same line ; for instance, if it be required to find the least common multiple of 2, 4, 8, 16, 10 and 48 ; the numbers 2, 4, 8» 16, since each of them is exactly contained in 48, may be left out of consideration, and 240, the least common multiple of 10 and 48, will evidently be the least common multiple required. tl (3) 7 and 15. (6) 333 and 504. (9) 4662 and 5476. 5415 and 30106. Ex. VIII. 1. Find the least common multiple of 16 and 24. (2) 36 and 75. 28 and 35. (5) 319 and 407. 2961 and 799. (8) 7568 and 9504. (10) 6327 and 23997. (11) (12) 15863 and 21489. 2. Find the least common multiple of 12, 8, and 9. (2) 8, 12, and 16. 6, 10, and 15. (4) 8, 12, and 20. 27, 24, and 15. (6) 12, 51, and 68. 19, 29, and 38. " - (8) 24, 48, 64, and 192. 63, 12, 84, and 14. (10) 5, 7, 9, 11, and 15. 6, 15, 24, and 25. (12) 12, 18, 30, 48, and 60 (14) if, i'^^ 14, ana isju 1 [3 5 7 9 m 10, ana r^. -(15) 54, 81, 63, and 14. (16) 24, 10, 32, 45, and 25. I Tj MISCBLLANKOTO QUESTIONS. S] l^A^di^ ?il„«3, and 14. 81 18, 24, 72, and 144. oo^ o«^' ^*' ^^» ^3, and U. 225, 255, 289, 1023, and 4095. Ex. IX. several steps. ^ *' *"^ S^^e the reasons for the Fifd^JIICl;^ 'Xi^;--'^»g of 90^0909, and of 90909. employed. a'Herence, and explain fully the processes ii'fS&'stanfS^S.^^/r "' ^''^ ^^"^ to (4) A person whole 4™ 73 w^ 37 ^e.*" 'If'"? «"-• his e d^t son ; what is & son's all ^""^ "''' *' '^^ »*'"•* of anf«iS,t'!ri"g?'- 'h^t.nns "vincnlu.', -bracket'. country 1 i„ 50 ; i„ how manv vpL" -Jf^u* '" ^^' «»d i" the "* l«v. « , . „^; J~ 'J"S» "lii g... 3« f., ». ,„a„^ (1) Define Miilfinij^o^j^^ . , t^*. . , of Wo numbers /hrsamrrwhr- ^""r '^^ *« Product performed. '*"* '" whatever order the operation is 32 ARITHMETIC. (2^ The Iliad contains 15683 lines, and the iEneid contains 9892 lines ; how many days will it take a boy to read through both of them, at the rate of eighty-five lines a day 1 (3 Explain what is meant by the greatest common measure . and by the least common multiple of two or more numbers -and Iw that the product of two numbers is the product of thejr leas common multiple into their greatest common measure. Fmd the least common multiple of 12, 16, 21, 52, and 70. ■ (4) Explain the meaning of the sign-, and find the value of r'7854-49l3f X 8-(203'74-12530)-?-53-6 4- (396456-2364H556. ^ (S^At a game of cricket A, B, and (7 together score 108 runs; B and C together score 90 runs, and A and C7 together score 51 runs ; find the number of runs scored by each of them. IV (1) Define Addition, and Subtraction. What is meant by a prime number ? When are numbers said to be prmie to each other? Give examples. ,,. . ^ , Explain the rule of carrying in the addition of numbers ; exemplify it in the addition of 3864, 4768, and 15938 (2) There are two numbers of which the product is 3'73625 the greater number is 875 ; find the sum and difference of the "73^ A father was 21 years old when his eldest son was born ; how old will his son be when he is 50 years old, and what will be the father's ase when the son is 50 years old '? . . ,. _, . " (4) WritTin figures one hundred millions,on8 hundred thousand, one hundred and Ine; and in words 1010101010. Express m figures M.DCCC.XL. -4. ? r, t?:„;i fVio ^(5) When are numbers said to be 'composite 1 ^ -F^^d the greatest number which can divide each of the t^o numbers 849 , Ind 1132 ; also the least number which can be divided by each ot them ; explaining the process in each case. (1) Multiply 478 by 146, and test the result by casting out the nines In what cases does this method of proof faiH Divide 4843 by 99, and prove the correctness of the operation by any tpst vou olease. ... . ^, _-ji_..i. "(2^ What number multiplied by 86 will give the same pruuuc. as 163 by 430 'i MISCELLANEOUS QUESTIONS. gS (4) A gentleman dies, and leaves his property tljus • 10000 »^d1^nofi;L{:'l7' '^ ''"""^ toVs'elLtZ.;on^ To^o'Sst t^h' oF rr;Vo^ir„r375?d" r^: ^ rwSipr:Li"oft ~'^- ^'»' -o- o^ (5) The quotient arising from the division of 9281 bv a cerfaiin number is 17, and the remainder is 373. Find the divLr. V. dmsor terminate with cyphers. aonagep, it the (3) The remainder of a division is 97, the quotient efl-i tmH th- d.v«or 91 more than the sum of both. WhruTdl^WeZ^* nt;Kfl mltdi^^rVr ^' -^"^ -"'^ ^ ^^ 1- ' his^^el^?ril'T*°' ''^°f ^^ '" ^' ^ ^° »0"« and a daughter • 6 s age equals the sum of the ages of his children ; two veaiisinM his ^e was double that of his eldest son ; the sum of tKes S the ather and the eldest son is seven tim^ as ^ZL ^t^f th. youngest son ; find the ages of the children. ^ °^ *' 34 ARITHMXTIO. FRACTIONS. 58. If 1 represent any concrete quantity, as for instance 1 yard, it is divisible into parts : suppose tiie parts to be equal to each other, and the number of them 3 ; one of the parts would be denoted by ^ (read one-third), two of them by f (read two-thirdsY three of them or the whole yard by | or 1 ; if another equal portion of a second yard divided in the same manner as the first be added, the sum would be denoted by a ; if two such portions were added, by f ; and so on. Such expressions, representing any number of parts of a unit, that is, of the quantity which is denoted by 1, are termed Broken Numbers or Praotions • we may therefore define a fraction thus : ' 5i>. Dbp. a Fraction denotes a part or parts of a unit ; it is expressed by two numbers placed one above the other with a line drawn between them ; the lower number is called the Denominator, and shews into how many equal parts the unit is divid^*^ ; the upper is called the Numerator, and shews how man^ of such parts are taken to form the fraction. Tnus I denotes that the unit is divided into 6 equal parts, and that 5 of these parts are taken to form the fraction : so, if a yard were divided into six equal parts, and 5 of them were taken, then denoting one yard by 1, we should denote the parts taken by the fraction f Again, J denotes that the unit is ivided into 6 equal parts, and that 7 such parts are taken to form the fraction ; for instance, in the example before us, one whole yard would be taken and also one of the equal parts of another yard divided in the same manner as the first. 60. A fraction also represents the quotient of the numerator by the denominator. Thus, I represents 5-r 6 ; for we should obtain the same result, whether we divide one unit into 6 equal parts, and take 5 of such parts (which would be represented by f ) ; or divide /ve units into 6 equal parts, and take 1 of such parts, which would be equivalent to ^ pa -t of 5 units, i.e, 5-r6 : hence | and 5-r-6 will have the same meaning. 61. When fractions are denoted in the manner above explained they are called Vulgar Fractions. Fractions, whose denominators are composed of 10, or 10 multiplied by itself; any number of times, are often denoted in a VULGAR FBAOTIONS. FflcTioN™^^"^ J a°h the numerator of the fraction by it, * 'mumpty tne Thus f X 3=f Reason for the above process, vl'^^l. %^^2:^ a-^ Z^ of those samara ■eaoh''«re.'' ' *" *' '"* "'''"' "^ "^ P""^ ^^ «>« 36 ARITHIIBTIO. Ex. X Multiply ^ separately by 3, 9, 12, 36. Multaply i^ separately by 7, 15, 21, 45. Multiply i separately by 4, 5, 7. Multiply ^^j separately by 25, 64. Thus,|.-r3=— ?— =^. 7 7x3 21 ^ Reason for the above process. In the fraction f , the unit is divided into 7 equal parts and 2 rvf those parts are taken ; in the fraction ^, the uniV ifd vided fntf 21 equal parts, and 2 of such parts are'teken: but sinTeTchpart m tiie latter case is equal t^ one-third of each part in^e forme* S£l%rrT^''l-Pr^ are taken ?n each case T Clear that /j- represents one-third part of ^, or f divided by 3. Ex. XI. Divide I separately by 2, 3, 4, 6, 10. JJivide jVy separately by 1 1, 20, 25, 45. Divide \^ separately by 6, 9, 12. Divide \ separately by 1, 7, 13. _ ^7?: ,.^ ^^, numerator and denominator of a fraction be bofh mJaDuibv^Tr*?-"" ^--l "^fr'^'-^tor of the fraction * be Sme vX fs5 "^ ''"' ^ ^' "^''^ '= °^ *« Reason for the above process. In the fraction f the unit is divided into 7 equal parts and 2 of ^iTquCa^ Sid S-' "f *'V"^*^^" ^ **^^ '-^ ^^^"^'^^ in'o *i equal parts, and six of such parts are taken. Now there are 3 times as many parts taken in the second fraction asThere are [n S\ p" ttl-r^^ ^,-!.3 P-ts in the second fraction arfon^^^^^^^^^^^^^ • to 1 part in the first fraction; therefore the 6 parts taken in the Setr/T^'T. ^^"^ *^^ ^ P^^^ ^k- inLfir^t Action; _ 67. Hence it follows that a whole number mnv b» ^o"™r»»d mto a vulgar traction with any denommator, by "multiplyh^ Ae VULOAR riuonoNs. 87 number by the required denominator for the numerator of the fraction, and placing the required denominator underS ; for 6=1 and^ to convert it into a fraction with a denominator 5 or 14, we "~T~"r>r5~5'' ^6_ 6xl4 84 1 ~1 X 14~ 14'- Ex. XII. (1) Reduce 7, 9, and 11, to fractions with denominators 3 7 and For if the numerator of the fraction 4 be multinliVrl h^r a *k V^r^^Sl^^S^Jiltt^^ylT^ into 8 e^ual 3 units : also 4 signifies that nnitv t ^^ -5^. ^^ ^^^ivaJent to m me case ot tiie first traction are equal to 1 oart in ih^ n/aT 4^ i are equal. ' °''' "" °***'" """"^^ *^« fra«ti°ns V and the resulting fraction iL fy. *^ ^^ denominator by 2, th^XJ "^u"*"' "^t ^'^ ™'' ^ divided into 8 equal parts and that 3 of such parts are taken ; and A signifies ZttTlr^ mm ARITHHBTIO. the same value as lilr or ^ 16 ' 16' nuZer^" '^''''''' "" '"'^'P'' f'^'^"^ ^ « ^^ole or mixed Ex Reduce V and V to whole or mixed numbers! oy the Kule given above, V =5, a whole number ; Measonfor the above process. Q. ^^ 25 5x6 5 *« ''i'!,^ * ''g"'fies that the unit is divided into 5 equal parts and Again, ??=?»±_5_6il5+5 6 6 6 with equals^ together with |, that is,=5 together with |, by what has been said above ; or as it is written, 5|. Ex. xm. numbILT ** '"°"°'^ ™P""°P'' '■'^"°"' "« "^^^ »' whole f 17) VtW. 70. Tb^erfwcc a mixed number to an improper fraction, liuLS. Multiply the integer hy the denominator of the fraction, VULOAR FRACTIONS. 39 >le or mixed and to the product add the numerator of the fractional part • the result wi 1 br, ae required numerator, and the denominator of the fractional part the required denominator. ""^^na^r ol tne Ex. Convert 24 into an improper fraction. Proceeding by the Rule given above, Reason for the above process. ^ 2| is meant to represent the integer 2 with the fraction 4 added But 2 is the same as ^ or ^ ; and therefore 24 must be Uie Zr/i>! 'r'^'f ^^ ^ ^' ^" V ; for V denotes that unity is ^tl:Zl:Z^''''''''^' ^^^^-^^^^ suchpart^togefhe^ Ex. XIV. Beduce the foUo—ixed „u„.ters to improper f^«o„s : ;5 Uh. ^UI;v ?Uli \t\U 9 2003f (10)857|i. (} 57^?' rS Itl!' Nl57|H. h8Jl7i4 ll9i427X. i^i 1m|||: 71. To rAoe a compound fraaUm to it> eguivaUnt simpU/raclion Kuufc Multiply the several numerators together for th« ro^rLitsisLr""'""^ *"« — ^ ^-=^-- Ex. Convert ^ of f into a simple fraction. Proceedmg by the Rule given above, lofI=i2iI=2i 5 8 5x8 40* Reason for the above process. hy^i- \&Zl^^T !*^?/^*.^^* P^r* «f «nity which is denoted oy ^ thus If unity be divided into 8 equal parts and 7 of th^^^ be taken and if each of these be again LidKto 5 equaf ptts and 3 of each set of parts be takenfthen each of the parts wfu be one-fortieti. partof the original unii, and the number rpartrSken --""' "— - 21 3x7^ will be 3 X 7. or 21 ; the result therefore is-- or 40' 6x8' that is ^ ARITHMITIO. 6 8 5x8 «fr^«'nnf^? 7'^"'''"^ compound fractions to simple ones, we may strike out factors common to one of the numerators and one of the denommators; for this is in fact simply dividing the numerator and denommator of the fraction by the same number. Art.X) Thus|of2TJjoflTV=fofHofH, _ 3x25xl6 _ 3x5x5x4 x4_ 4 5x12x15 5x3x4x3"x5"~'3 frn^l'Kr'"^*''!''"''®^.^*^y'^'^^^"g ^"* the factors 3, 5, 5, 4, from the numerator and denominator. '»>"*» Ex. XV. Reduce the following compound fractions to simple ones • (1) foff (2) lof^i^. (3) fof^ (4) 4 of 11 (5) lof^of7. (6) I of^ of I of ^ of 28 ' (7) Aof2iof4oflOf (8) ^ofl2|of4off off of9. (9) /joffof^^off ofT?^of2of3rV (10) 4ofiof|of70f of^oflT-VofM?. 72. Dbf. a Fraction is in its lowest TBRMS,when its numerator and denommator are prime to each other. Note. When the numerator and denominator of a fraction are arl?"Tv. ^^\ ^*^f.' they have (Art. 44) a common factor greater than unity. If we divide each of them by this there results a fraction equal to the former, but of which the terms, that IS, the numerator and denominator are less, or lower than those of the original fraction; and it maybe considered to be the same fraction m lower terms. When the numerator and denominator are prime to each other, that is, have no common factor greater than umty, It is clear that its terms cannot be made lower by division of this kind, and on this account the fraction is said to be m its LOWEST TERMS. 73. To reduce a fraction to its lowest terms. Rule. Divide the numerator and denominator by their greatest common measure. ^ Ex. 1. Reduce f A|| to its lowest terms. . First, find the greatest common measure of 6465 and 7335. VULOAK FRACTIONS. 6465) 7336 (1 6465 41 870) 6466 (7 6090 376) 870 (2 120) 375 (3 360 16) 120 (8 therefore IS* is the greatest common measure. 16)6466(431 16)7336(489 rL. 60 ii m_ _y ^ 135 therefore the fraction in its lowest terms=||^. Beason for the above process. Ex. 2. Reduce ^o ^^ j^^ j^^^^^ ^^^^^^ ^'^^It a'^'a'"'^ n»i«erator and denominator by 10 -il, dividing numerator and denominator by 3. " Ex. XVI. . Reduce each of the following fractions to its lowest terms • (1) 4' (2^ i« (5) if (6) / t5. (3) if (8) WV. ▲RITHMXTIC. ( w am. 10) m 11 HI. (2a) Miff. 27) enf a if 8 0"' wiVA a common 74. 7b rtfrfww fractions to equivalent ones aenomtnator, thi^wm k!';"!? *^« ^^^«<^ «,«™»"o» "^"Jtiple of the denominators : th 18 wi 1 be the common denominator. Then divide the common multiple so found by the denominator of each fraction, and muTp?? Wonrt^"/r fr"^ ^^'^ '^' """^^^^^^^ -^ ^^e faction S belongs to it for the new numerator of that fraction. Note 1 If the given fractions be in their lowest terms, the above rule will reduce them to others having the leasi common denominator; \f the least common denominator be require7Se fJI bf ~d. ''^"''^ '" '^''' ^"^''' termslefoii the comLn^dl"omin1t^^^ ^*' '^' "'' '^"^'^^^' ''^^^^^"^ ^^^^ * Proceeding by the Rule given above, 2 2 2 3 12, 16, 24, 33 6, 8, 12, 33 3, 4, 6,33 3, 2, 3,33 1, % 1,11 therefore least common multiple =2 x2x2x3x2xll =558; therefore the fractions become respectively, 5 X 44 220/ . 528 ' T2ir44=528r^^" " 44 'X 9x33 297/. 528 ^528^"^^ 242/ . ^528(^"^'^' 272 / . 528 16x33 11x22 24x22 17x16 12 16=^3), 24 = 23). eriTtc^ ( Since 16^ Od a common VULGAR FRACTIONS. *||| or the fractions with a common denominator are • Reason for the above process. The least cominon multiple of the denominators of the iriven fractions will evidently contain the denominator of any one of the fractions an exact number of times. If both the numerator and denominator of that fraction be multiplied by that number the value of the fraction will not be altered (Art. 66) and the denominator will then bo equal to the least common multiple of all the denominators. If this be done with all the fractions, they will evidently be, in like manner, reduced to others of the same value and having the least common multiple of all the denomina- tors for the denominator of each fraction. m£''ll?; '^fj!^ denominators have no common measure, we must then multiply each numerator into all the denominators, except its own, for a new numerator for each fraction, and all the denouiinators together for the common denominator. de^minftor"^^ *' ^' ^' ^^ ^^^^^^^^^"^ fractions with a common The least common multiple of the denominators therefore the fractions become 1x7x9 ^63^ 2v5x9^90 1x5x7 35 5x7x9 315 7x5x9 315 9^5177=816 or the fractions with a common denominator are J5JL JLO aw(\ SlSl. 7 IS* 3TS5 "'"^ 3TS- Ex. XVII. Reduce the fractions in each of the following sets to equivalent fractions, having the least common denominator : '0 i,f,andf. (3) f, |,andf. (5) f fV^ and \\. y) f » jh and \\, (9) h A, and ||. OO |,f, f,andi. [3 |,||,and,VV '5) ^, A, If , ,\ and ,V (19) H, H, H, T* J, anr^ .^ I, and 1. (2) f, 4) i' and ,^. (6) f |,andf (8) A, tV, and if 10) f, f,iand,V ;i2) I, iViV^andf ,'4) iVf H,and^|. }^) i J, i, A, ^V, and 11. (20) fi,ii,||,andW. 44 ABITHMETIO. into the same numbeTof eau J^^^^^^^ ""^ jt ^'^"'^^^ "^ ^^'^^^^ tors will show us how many of S I ^f **"" respective numera! or which is the gre^l'TeVo^.^eK n^^r^ ^o T ^^ ^ i^irst, to find the least common multiple of the denominators ; 2 3 5 27, 24, 6, 16, 6 27, 12, 3, 15, 5 9, 4,1, 5,5 ^» 4, 1, 1, 1 therefore the least common denominator =2x3x5x9x4=::1080; therefore the fractions become 5x40^200, 11x45 ^495, 1080 3x216 27x40 1080 24x45 4x72 _288, i080 5 X 180 _ 900 , 6x180 l08b 648 15x72 1080 517216=1080' S"L^lt;^^^^^^^^'^^^--^' H the next,,V the next, , ^ Ex. XVIII. i, Compare the values of (1) f, f, and j\. (2) h h f , and I l3) i off, /j, and A of 4. A» i%, H, andff; h tV> 115 tV» and f ^. f » 3i» A> tVj and |f y, 3i and f of 9f . |> ih H, I, and f |. fi^A< ^' ^" >"V» a^i, and ■^. a. Find the greatest and least of the fraot,inn« K^} h 1^., f, h and i. (2) j^,y^;T|, ^^, ,„d «. ii le between reduced to se then we o obtained, v^e numera- each case ; on. ninators ; 900 , 1080 the next, \h ADDITION OF VUI^AR FRACTIONS. ADDITION OF VULGAR FRACTIONS. 46 75. Rule. Reduce the fractions to equivalent ones with their aru^rrT-^'"""^^"^'^^^^^^ ^" the Sew numerators tog^^^^^^^^^ and under their sum write the common denominator. Ex. Find the sum of j^, -, and ^. Proceeding by the Rule given above, First, find the least common multiple of the denominators : 3 5 15, 21, 35 5, 7,35 1, 7, 7 1, I tt frtdot iTclr "" ""^'P^^ = 3 X 5 X 7=105 ; therefore 7 x7 105 10x5 50 21x5" 49+60+48 105 147 16x3 48 35x3"~r05 , 49 15x7 therefore their sum = 105 105 "35 Reason for the Rule. In each of the equivalent fractious, we have unity divided into ? equal parts, and those fractions represent respectively 49 50 and 48 of such parts ; therefore the^ sum of the fractions must represent 49+50+48 or 147 such parts, that is, must be — Note 1. If the sum of the fractions be a fraction which is not in Its lowest terms, reduce it to its lowest terms ; and if the result be Ml improper fraction, then reduce it to a whole or mixed number: thus iii=±|=ii| : the same remark applies to all results in Vulgar Fractions. ^^ Note 2. Before applying the Rule, reduce all fractions to their lowest terms, improper fractions to whole or mixed numbers and compound fractions to simple ones. ' Note 3. If any of the given numbers be whole or mivo^ numbers, the whole numbers may be added together as in simple addition, and the fractional parts by the Rule given above 46 ABPPHMBTIO. Ex. Find the sum of |» 3^, 10}, and -• = 13+14^1 + 1. Now to find the sum of |- + y+l+^. First, find the least common multiple of the denominators • 5 14,15 ,5,11 4 3 1 i 1 therefore the least common multiple' fj,. f .u . =2x5x4x3x11=1320; therefore the fractions become 3x 165 _ 496 8 X 165 i320' 2x264^ 528 , 14x88 ^1232, 15x88 1320 9 X 60 540 , 22x60""l320 5 X 264 1320 therefore the sum of the fractions ^495J2232j-528 + 5^ 2795 _559, 1320 -I^ "~264 ( <^^v^<5»ng numerator and denominator by 5, ) =2 V therefore the whole sum=13+25VT = 15vU^ Ex. XIX. 1. Add together, (1) f ^d 4 (4) I and ^ (7) ^j and /y. (10) VV and VV- 2. Find the sum of 10 f f » and j-V 3) i, f , and Jy. f5) f , f , and j\. (2) f and f (5) tV and ^y (8) ^, and f i. (11) 3f aud7f '2 and^ I and If . , , and 24. (12) 4f and 9^. \^i t, f , and ^. f A» and ^. I, f,and f (4) T^j, h and /,. (6) #, 4, and i. ,(^) i'i»i,and f (iO) i. 2|, and 13tV. SUBTRACTION OF VULGAR FRACTIONS. 47 }2) iof^of|,6i,and^. 14 jl, f • - - (20) 387^, 285^, 394^, and f of 3704. 8. Find the value of ^H + 6|i-f i^ + rof |4+i| + ^ of2j. i4-|-3A+H+7tv+^V+fof|. 5^ + 1 off of 3^ + 9^+^ of I of 4. f ofl2 + f of| + 3f of Uof4-,4_4-+|of3*of J ofl l. 270? + 650/,l5000|4-53f +ii *^ * ^^» °' ^'^• i of J + ^V of (l + |f) + |i+|J of jl + i}. SUBTRACTION. 76. Rule. Reduce the fractions to their least common denommator take the difference of the new numerators, and place the common denominator underneath. . r ^« Ex. Subtract - from --* 2 8 nn^L'"''^^^'"!!'^ ^^%'^l^^ given above, since 8 is clearly the least wuTbeirdl^ ' o y 23 Product =2! X 3? X lol x 2of x 5^, 2 8 8 9 23' ^5 27 81 184 124 2'''8''8''"9~''23' ^ 5x9x3x9x9x8x23x4x 31 2x2x4x8x9x23 ~' (or cancelling, i. e, dividing numerator and denominator by the product of the common factors 9, 8, 23, 4,) product ■ — -,=9410^, 2x2 DIVISION OF FRACTIOUS. 51 tor by )y the ;or by >y th? )y the Multiply ^byf. A 3* by 2f . ^ , ^ of f by 5f of 3. Ex. XXI. f by if. f * by H il! ^ibyioff (9) |byf n by |. I2by f of5. - - - (10 If of 3a. by IJ^ of 4* of a. H of Vj of V^ by ^V of 37H of 3^ of ,V I of 2,V of V, of 3^ by A of If '^ SyV of 3^ of .jfj of 34 by ^f y of ^j of 1| of 19. 2. Find the continued product of 3) IH. 2| of IJ^, ^f, J/y, SjV of 49, and ^Y 4) f 2^ 3Vr, 5/^, and 6,iy. 5) 1*1%, if, tV3> Uf and 4 of U. ifi) Hh UH, Uh nh and liif DIVISION. 78. Rule. Invert the divisor, i. e. take its numerator as a denominator and its denominator as a numerator, and proceed as in Multiplication. Ex. Divide 1 by ^ . Proceeding by the Rule given above Reason for the rule. 11 '5 11^3 33* If— be divided by 3, the result is jA_„r|,(Art.65). A^^^ ^^"^* ^^ ^ ^^^^^ ^^^ ^"^*^^' ^^' ^" ^^ber words, is only one fifth part of the required quotient, since, instead of dividing by 3 we have to divide by f , which is only one-fifth part of 3 ; and the quotient of ^j divided by f must therefore be 5 times greater than if the divisor were 3. Hence the above result ^ must be multiplied by 5 in order to give the true quotient. j.nuroiuie, the quotient =— x & — "^ ^ " = ' ' 33 33 33 d2 ARITHMETIC. Note 1. Before applying this rule, mixed numbers must be reduced to improper fractions, and compound fractions to simple ones, as in the following Examples : Ex. 1. Divide 4^ by ^f. 13_^11 3 4 4J-f-2f. 13 4 -3^0 .52^^19. 33 33 Ex.2. Divideloflbyi^of7. 4 8 -^ 16 lof-I-^-of7 =3x7_^ 15x7 ^^3x7 4 8 ' 16 * 4x8 ' 16X1 4x8 _3x7xl6 _3x7x4x4 ^J_. 10 16x1 15x7 4x8x15x7 4x2x4x3x5x7 Note 2. C!oMPLBx Fractions may by this Rule be reduced to simple ones. \i 1 1 f^ Thys, 21 2 4 2 7 2 7 =__ y — * 4 5 10 Or thus, ^f=i= fx4x2 , 21 i |x4x2 multiplying the numerator and denominator of the complex fraction by the product of the denominators of the simple fractions, 20^^10 1i=-i'=l=_?. i? ^ 1_ 3x3 _3 30 30 V 2"^ 1=2^30 2x3x10 20* Again, Or thus, ZI= 30 I t x2xl Vx2xl 60 .3^ '20' Again 30_30_ V _30_^9_30 ^ 2_ 3 x 10 x 2 4^ f f 1 ' 2 1 9 3x3 20 =6|, Or thus, 30 "I: Sl a Vxlx2_60_20_ t xlx2 9 3 6|. 1. Divide, 3byf I Ex. XXII. 2) f by 1. 5) U by 31. (3) f by i-f . (6) n by ||. lust be > simple xl x7 luced to jomplex ■actions, 3 JO- DIVISION or FRAOTIONa. 53 4loi lto%tin by ^7] '' "' * "' ''■ ' •> «THby e^oflH. (14) f of f of 4 of 1 .\ by yv of ^fy of j| of 1^. 2. Compare the product and quotient of 2J by 3t. 3. Reduce to simple fractions tlie following complex fractions • 0)1 (^)i-l (3)?ii (4)4 (6) 5f (7) 13A i If (5) m 111 ^1 4 79. Miscellaneous Examples in Fractions worked out. Ex. 1. What number added to J4- Vv will give 2i ? This question in other words is the following :'« What number will remam after i+J^ has been subtracted from 2} 'i " Now, 2i-(i4-J^^) = 21-1-/^ = V-l-A Therefore the number required =|. JSTote. It will be remembered, that all quantities within a vin- culum are equally affected by any sign placed before the vinculum. 7 I^"V\*^7\''''^ expression -(i + Vv) means that the sum of J and /^ has to be subtracted from 21; whereas-i + J^- would SX tuTt."^ "'"''''^ '""^ ''' -" »^^" ^ ^ ^^ Ex. 2. What number subtracted from 14f will leave la for a remamder? " *^ ** lor a Number required:^ 14f—l|=::(14+l + |)—(i + i + |\ =(14+u)_(2+n)=i4-2+(V~f)=12|. Or thus, Hf-l3 = i33_^ = ,o7_| = ,^i = 12^^ Ex. 3. What number multiplied by If will produce 14a i This question in other words is the following: "IfMa K« divided by If, what will the quotient be 1" * But HI = V =59 8 _ 59x2 118 Therefore the number required = Ex. 4. What number dividea -^ 1 . IX 11 *- iT produce 10,^^^ 54 ▲RITHMBTJO. This question in otber words is the following : " What is the product of If and lOyV^" The product of If and lO/r = V x YT' = 4' = V=14f. Ex. 5. Reduce the expression (£j+_^ — _ o{—\ x U ^ ^7 10^ 18 7/ ^ to Its simplest form. \7 10^ 18 7/^ 4 \^ V 18'<7/ ^ Ex. 6. Simplify the expression llili- = Ul + i ^ Si 4i 6+4+3 12 2^ 3i 4i if 120+90 + 70 l + ^f + i ill 315 _ii_jj^315_ 13x3x105 _105_j - tH 12 286 3x4xl3x22"88" " Ex. 7. Divide 3J— ^ of ^V by 21i + t\r+4i of 5. 3i— iofi-V = V~l = -^VV"* = W- 10^ 3 259 =2^7+211 = 21+21+^ + 1 = 21+21+^=43^=^ therefore the quotient required ^109^259 _ 109 6 36 ■ 6 36^259 Ex. 8. Simplify the expression 109 ^ 6x6^259 = i^- 1554 iof 1 13 1 + 1 Now* 1 _ 1 i+i ,+ S+i 3 + i ^ 1 1+^ ^ 39 39+4 _39 4 hat is the jti + 90 4-70 315 05 \S =1H = w. 5 _259 "" 6; 109. 1554 43; MIS0BLLANE0U8 EXAMPLES IN VULGAR PRA0TI0N8. 55 If 1 13 43 48- therefore Ex.9. Si„..pUfy l^i+l of ^-m ^l^. The expression = ill.l^A_ll .306 { = ill4.il? 2 I 228 ( 4 "^ 38 ""T f ^ 305 ^5 5 2 ) 228 19 Y'^ 5 j" 2 ) 228 305 (the least common multiple of 4, 38, and 3, = 38 x 2 X 3) - jlL2LJJ_ x3+175x2x3--2x38x 2) 228 I 38x2x3 " } x^ _ i 627+10 50--152 ^ 228 i 228 5 ^ 305 _ 1677~1 52, 228^ 1525^ - 228 305 "305 _ ., ^ Ex. XXIII. Miscellaneous Questions and Examples on Arts. (58—79). s^nlo^f^^® a fraction ; what is the distinction between a Vulgar and a DecimaUraction ? How many different kinds of Vu£ fractions are there 1 Give an example of each kind ^ 2. Find the sum and difTerence of |i of 7^, and If divided by %:slSiV""'''''''''*''"^^-"^ (1) [|+| of 5^] X [K|+3f]. (2) 3,1, of 34-^^v of 9. (3) ||^?i+i (4) Ijx 4 ^x41- 1 r 4. Shew that the fraction |l|±^ lies between the greatelt and least of the fractions |, 4. and -ft. 5. The difference of two numbprd i« !»; 4 . +u^ " 20a : find the smaUeTnumber ^ ' *' «'""*'' "™^" 66 ▲RITHMXTIO. II. 1. If the numerator and denominator of a fra(bion be both multiplied or both divided by the same number, the value of the fraction is not altered : prove this by means of an example. 2. What number subtracted from 41 i leaves 19^? and what number multiplied by 2j\ of ^ produces 3^ of | ] 3. When is a fraction said to be in its lowest terms 1 Reduce the frrctions fff || and aVyVW *<> th^ir lowest terms.- 4. Simplify (1) % M^-iA. (2)3Jof5iof}-iofA. (3) (A+A)-{3-i)xa+J). (4) A ofgof ^ . 6. Divide the product of 2y^ and 2f by the difference of 2f and 2i.^ Explain why it is necessary in the addition and subtraction of fractions to reduce the fractions to a common denominator. III. 1. Shew by an example that multiplying the numerator of a frac- tion by any number, is the same in effect as dividing the denominator by that number, and conversely. 2. Simplify (1) 275i+62^+1031i + iof4150|.>/(2) fa-MjV xie^-M^f LJ±.^. + L (4) 4i-3f 3-2X 3i 9^2+4|. ^ ^ 4-} + 3i i=3i: 3. Which is the greater, i of 4 or i of 5? and by how much? 4. Divide the sum of the fractions ^ and j\ by the product of y\ and If ; and reduce the result to its lowest terms. 5. What number is that, from which if you deduct f — ^ and to the remainder add the quotient of ^ divided by 2^, the sum will be lyV 1 IV. 1. Define a vulgar fraction; an improper fraction; and the terms numerator and denominator of a fraction. Prove by means of an example the rule for the multiplication effractions; and multiply the sum of | of i and IJ by the difference of yV a^^ } • * (3) sh sa wl be both e of the e. od what terms. 3 'of 2f on and jommon )f a frao- ling the f-^Hf much 1 )duct of } and to am will md the lication bv the QUK8TI0NS ilND IXAMPLIS IF FRACTIONS. 67' 2. Reduce to their most simple forms the following expressions • (3) HH|. (4) ^j of (l + 5i) + # of ^ of (^~2f)-.l (5) Vlizdi . 3. What number added to ^ of (1-| i— jl4.i\ m-keq .y slfgj ; and multiply the sum of i, 1^, and f by the difference of ^ and ,=V, and divide the product by II by U4 2. Reduce to their simplest forms (1) (3) (4) (6) (i-f)-(i-«. (2) l^IZlA. fofH-i|ofH+?ofg- Hi-H |of/y-^ofyV (^> 2^x 1 4i 3i + 6A 3. What is meant by the symbol a ? ohoTi'""^ *^ ^f ^^ ^'"*^*^^" ^^'""^ »<^<^ed to the sum of 4 i and « shall make the result an integer. *' *' ^*» 4 t/^^ S! '""" ""S^ir ^'1*^^'* ^^ ^^^«^ <>^the fractions f, Jl^ turns "^^ ' ""^ *^' ^^^'' *^^ ' ^"^ *he difference of th^e wW £?"" H^ I ^^ ^ ^®^^^» ^*^ gives his son X of his share- what portion of the estate has he then left ? * ' 58 ▲RITHMBnO. VI. - VI. fr«;;fnn« *! f ® ''"^'l K ^^^''''' ^°d subtraction of vulgar fractions ; and prove them by means of an example. ^ 2. Simplify (1) fof^-^of^+foflH. (2)'M^.3!. (3) {4xfxl3^}-f [ix^+40). U)^~-Hl fhf ;« f^^^^e a^r^W ^/i«>^c/, and compound fraction. ^Exnkin the method of reducing a compound fraction to a simple one ^ Ex. |of 6 ofyVrofli. 4. Shew by means of an example how a fraction is affpnfo^ if the same number be added to its nLerator a?d denominl^^^ '' 5. Multiply 3x by 3,V, and divide-^^^y 11^, and find the 6 'wL^n^^'T *^L'T ^"^ ^^^^^^"«« <>f <^hese results. b. What number added to ^i+m ™iii nro^lnp*. qaaT J a u number divided by 2,V will produce^? ' 3||i? and what VII.J 1. Shew from the nature of fractions that a. +4=f a. that i nf 4=H; and that f-r^=i.i. 3^t-?i , inat f ot 2. Simplify (1) (3) i. 2^. 6i + T^ 3f (3iof4|)-(8j_i.)of(3j_i.) (2) 2*+3|+A+,>, + 6iJ. (4) GVof3J) + (|-.jj)_/^i_^\^(2_^) 3. Simplify t£Lt£g. and talce the .e,«lt fro„> the sum of K>|,3A,7f|. ^^,■"^1*1 V ^*^®5 *' *' ^' ^"^ *' s<*toact the 8um from 2 mnltinlv the result by f of J J of 8, and find what fraction tW^is of 99 ^ which was the remainder of the score and 4 cTT^ i . \ B tm.. as many as the other Whl^'^^f %Z' ±Z I'^f'^'i rwis, and the score of each man ? ""' " ""' " "" """'"'''' ""^ . DZOIMALS. DECIMALS. 69 80. It has been stated that figures in the units' place retain their intrinsic values, while those to the left of the units' place increase tenfold at each step from the units' place ; therefore, according to the same notation, as we proceed from the units' place to the right , every successive ^figure would decrease tenfold. We can thus represent whole numbers or integers and fractions under a uniform notation by means of figures in the units' place and on each side of It ; for instance, in the number 5673-2412, the figures on the left of the dot * represent integers, while those on the right of the dot denote fractions. The number written at lergth would stand thus, 5x 1000+6X 100+7 xlO+3+t^+^±^ + ^_i+„2^ ihe dot IS termed the decimal point, and all digits to the right of It are called Decimals, because they are fractions with either 10, 100, or 10 X 10, 1000, or 10 x 10 x 10, &c, as their respective denommators. ^ 81. It may here be observed, that, when a number is multiplied by Itself any number of times, the product is called a Power of that number; being called the second, third, fourth, &c. power, according as the number is multiplied once, twice, three times, &c. by Itself, that is, according as it is employed twice, three times, &c. as a factor. ' 82. It will be seen from what has been said, that Decimals are m tact fractions having either 10 or some power of 10, for their denominators. For this reason also they are called Decimal (Becem^ten) Fractions, in contradistinction to Vulgar Fractions wnich,as we have seen, are represented by a different notation! and not limited m their denominators to 10, or powers of 10. 83. From the preceding observations, it appears that First, •2345=-2.+_L. JL ._A_ 10 100 1000 10000* Now the least common multiple of the denominators of the fractions is 10000 : therefore, reducing the several fractions to equivalent ones with their least common denominator, we get 3 100 4 10 5 •2345=1x^^0^ 10 1000 100 100"^IOOO 10^ 10000 _2000+300+40+5 2345 10000 10000* I' ARITHMSTIO. Secondly, -00324 =^+~^+ 8 2 ,,, , 1000"^ 10000 "^lOuOOO (the least common multiple of the denominators is 100000) 10000 _0_ 1000 3 100 2 10 4 ^ 10 ^ 10000 __300+20+4 100 1000 324 1000 "" 100 "^10000 ""lO'lOOOOO + 100000 100000. Thirdly, 56.Sl6=5xl0+6+r%+-l^4.~.J> ^(the^least common multiple of the denominators is 1000) 1 10 6 .5x^^1000 6 1000 8 100 1 1000 "■ 1 ''looo-'io^Ioo-'loo^io 1000 ^500004^6^00j^O+10+6 _ 56816 1000 — - "looo- romLtt^^ '-' 0000) 4 0000 30) omposed a vulgar or those in either as many for the r of 10 le form We see that ;327 -0327, and -3270 are respectively equivalent t^rnr^V vl'^*'"^' ^ ^r^ numeraJor%nd tL first and third of which have also the same denominator, while the denommator of the second is greater. Consequently, -327 is equal to -3270, but -0327 is less than either. The value of a decimal is therefore rot affected bv aMxina cyphers to the right of it ; but its value is decreased byVS cyphers : which effect is exactly opposite to that which is produced by affixing and prefixing cyphers to integers. ^ 86. Hence it appears that a decimal is multiplied by 10. if the decimal point be removed one place towards the n^A/hand ; by 100, If two places; by 1000, if three places; and so on and conversely, a decimal is dMed by 10, if the point be removed one place to the left hand; by 100, if ^ places; by 1000, T'^m places ; and so on. Thus 5-6 xlO=i|.x 10=56. 5'6 X 1000=f & X 1000=5600. 5-6-M0=f^x-i^::,_5_^_^.56. 5'6-5-1000=f A X tttV— Aa.-— '"^^fx .»,• • fif ff™tfg« arising from the use of dear .als consists in this,vi2 : that the addition, subtraction, multiplication, and division ofrfmma/ fractions, are much more easily performed than those of .t^/^ar fractions ; and although all vulgar fractions Cnno? be reduced to finite decimals, yet we can find decimals so near thdr h^l!/!"'V / ' ''T ^"'^"S ^'^'^ "«^"g t*^« ^^"^«^ instead of the vulgar fraction is not perceptible i«. Ex. XXIV. 1. Convert the following decimals into vulgar fractions • •1; -3; -31; -311; 31111; 31111111. WUermsV^' ^''"''™^ '^'''"'^^' '"^^ "^"^S*" ^'^"^^^"' ^ *^«^'' •5; -25; -35; '05; -005; -0256; 000256; -00008125. 3. Express as vulgar fractions in their lowest terms : -075 ; -848 ; 302 ; 3434 ; 3434 : -034.^4 • -Af^nnAf; . oqa.>iaa , ARITHMETIC. 4. Express as decimals, tV ; A ; j\ ; VA ; ^h ; rir^ir ; m^\ ; sw ; rUlh ; r/irVir ; •7 separately by 10, 100, 1000, and by 100000 • •006 separately by 100, 10000 and by 10000000 • •0431 separately by 100, and by 1000000 • o nn?! f P^'^^^^y ^y 10» 1000, and by a million ; 9-0016 by ten hundred thousand, and by 100. 6. Divide •51 separately by 10, 1000, and by 100000 • •008 separately by 100, and by a million • ' 5-016 separately by 1000, and by 100000 • 378-0186 separately by 1000, and by a million. foJfl ^^P^e««/ccording to the decimal notation, five tenths ; seven- tenths ; nineteen hundredths; twenty-eight hundredths ; five thousandths ; ninety-seven tenths ; one millionth ; fourteen and four-tenths; .wohundred and eighty,and four ten-thousandths ; seven and seven-thousandths ; one hundred and one hundred-thousandths ? one one-thousandth and one ten-millionth ; five billionths. 8. Express the following decimals in words : •4; -25; -75; -745; 1; -001; -00001; 23-75 • 2-375 • aqT*^ . •00002375; 1.000001 ; -1000001 ; -00000001 '' ^' ^1 1 ADDITION OF DECIMALS. 88. Rule. Place the numbers under each other, units under unite, tens under tens, &c., one-tenths under one tenths, &c. ; so that the decimals be all under each other : add as in whole numbers, and place the decimal point in the sum under the decimal point above. Ex. Add together 27-5037, -042, 342, and 2-1. Proceeding by the Rule given above, 27-5037 /042 342- 2.1 371-6457 Mie, The same method of explanation holds for the [| t s h a ADDITION OP DECIMALS. 68 fundamental rules of decimals, which has been given at length in explammg the Rules for Simple Addition, Simpllsubtraction, and the other fundamental rules in whole numbers. Reason for the above process. If we convert the decimals into fractions, and add them together as such, we obtain wgotuor 27-5037+042+342+2-1, 275037 42 10000 342 21 1000"^ 1 "^lo; (or reducing the fractions to a common denominator). 275037 lOOOlT "^Toooo 420 3420000 21000 10000 "'"Toooo" 3716457 „, = -T0000 = 371-6457, (Art. 84). Ex. XXV. 1. Add together : (1) -234, 14-3812, -01, 32-47, and -00075. (2) 232-15, 3-225, 21, -0001, 34-005, and -001304. {^^) 14-94, -00857, 1.5, 5607-25, 530, and -0057. 2. Express in one sum : (1) •08 + 165 + l-327+'0003 + 2760-H-9. (2) 346 + -0027+-25 + -186+72-505+'0014+-00004 (3) 6-3084+-006+36-207+-0001+364+008022. (4) 725-1201+34-00076 + -04+50-9 + 143-713. (6) 67-8125+27.105+17-5 + -000375+255+3-0125. 3. Add together : pr^vLh'eS;'"''' ■''''*^'' '•"««*""'«• "•'' --^ '-5; ""I J'vUr&ur'''"' '''■^' ^-"""^^MOOO, and 3-4073; Jlu^*' ■"'^' "'*'"'^'^' '"^"^^^^ ""'J 4957-5 ; and verify (4) Five hundred, and nine-hundredths ; three hundred and seventy-five ; twenty thousand and eighty-four, and seventv eieht hundred thousandths; eevenmiIlions,tlotiu,a;d.andtwrhlHrd 04 ARITHMETIO. SUBTRACTION OF DECIMALS. 89. Rule. Place the less number under the greater, units under unite, tens under l.^.s, &c., one-tenths under one^teniJ^ &o ! t\TH2%P.^rV".^^'"PP^^^^ ^f necessary i^the^p^^^^^^^^^ the right of the decimal : then proceed as in Simpleffitrac'ion Ex. Subtract 5-473 from 6'2S. Proceeding by the rule given above, 6-23 5-473 •757 Reason for the above process. . 6-23 — 5473 = ~ 100 6' iOOO 6230 6473 1000 1000 _ 757 " 1000 = -757, Art. (84.) Ex. XXVI. o hr^^^^ ***® difference between 2-1354 and 1-0436 • 7.ftqf; c,«^ 56 oVj 27 148 Z„' LT"^ 'f ■" ^13'«0683 ; 35-009876 from 00 078, 27 148 from 9816; and prove the truth of each resdt. W^T*'^""'*'* the difference between seven and seven tenths • =■ fouTrthr";: "^f^ ""^^ r^" "•""»""'« ; also between tvyty: ^ luur -f- ^n« ^unuredtiis j j...a verify each result. .■■'.> t MULTIPLICATION OP DECIMALS. 65 MULTIPLICATION OF DECIMALS. 90. Rule. Multiply the numbers together as if they were whole numbers, and point off in the product as many decimal places in both the multiplicand and tLe multiplier ; if there are not figures enough, supply the deficiency by prefixing cyphers. Ex. 1. Multiply 5 34 by -21. Proceeding by the Rule given above, 5-34 •21 534 1068 1121i Now the number of decimal places in the multiplicand + the number of those in the multiplier =2 +2 =4 • therefore product= 1*1214. Ex. 2. Multiply 5-34 by -0021. 5-34 •0021 534 1068 11214 We must have 6 decimal places in the product ; but there are only 5 figures: and therefore we must prefix one zero, and place a pomt before it thus -011214. , t "^ Reason for the above process. 6-34x-21=??fx^ 100 100 _11214 10000 = 1-1214. Again 5-34 X 0021=5?! x-?l_ 100 10000 _ 11214 1 nnnAAA i www =•011214. ^ ARITHMETIC. Ex. XXVII. 1. Multiply together : (1) 3-8 and 42 ; -38 and -42 ; 38 and 42 ; -038 and -0042. (2) 417 and -417; -417 and -417 ; 71956 and -000025. (3) 2-052 and -0031 ; 4-07 and -916 ; 476 and -00026. 2. Multiply (proving the truth of the result in each case ) (1) 81-4632 by -0378. (2) 2735 by 7-70071 (3) -04375 by -0764. 3. Find the product of (1) -0046 by 7-85. (2) -00846 by -00324. (3) -314 by -0021 (4) -009 by -00846. (5) -009207 by 6056. (6) -00948 by 29 • proving the truth of each result. ' of fi2^\"4 *!|)'l2TnHT9^n P^'^r' ^^ \ •^^' •^^^' ^"d 100; also ot 1^ 1 2, 012, and 120 ; and prove the truth of the results 5. Find the value of (1) 7-6 X -071 X 2-1 X 29. (2) -007 X 700 X 760-3 X -00416x100000. DIVISION OF DECIMALS. .JLo ."f ''*'• ,^^*?/f ""^^^^r of decimal places in the dividend exceeds the number of decimal places in the divisor. '"''^'^'^^ Bulb. Divide as in whole numbers, and mark off in \\^ quotient a number of decimal places equal to the excess of th! number of decimal places in tL divid^snd over the number tf decimal places in the divisor; if there are not figuL suffic^^^^^^ prefix cyphers as in Multiplication. ^ sufticient, Ex. 1. Divide M214 by 5-34. Proceeding by the rule given above, 5-34) -1-1214 (21 1068 534 534 Now the number of decimal places in the dividend —the mimbAr of decimal places in the divisor=4-2=2 ; therefore the quotient =-21. . \ DIVISION OF DIQIMALI. e7 Ex. 2. Divide -011214 by 68-4. 53-4) -011214 (21 1068 534 534 Now the number of decimal places in the dividend — the number of decimal places in the divisor =6-1=5; therefore we prefix three cyphers, and the quotient is -00021. Reason for the above process. 11214-f-5-34 ,11214_^534 10000 ' 100 10000^634 = ^^^^^ 100 ^ 534 ^10000 21 1 X ;r- 1 100 (<.?««« 11214 _o, , 100 smce -— — — = 21 and .. = 534 10000 = ii =-21 100 -011214-r-53-4 100 y Again, _ 11214 _^534 ^ 11214 10 ^ 11214 1000000" 10 1000000^534 X 10 =~ 1 = 21 1 ^100000 100000 534 1000000 00021. 92. Secondly. When the number of decimal places in the dividend IS less than the number of decimal places in the divisor. Rule. Affix cyphers to the dividend until the number of decimal places m the dividend equals the number of decimal places m the divisor ; the quotient up to this point of the division will be a whole number; if there be a remainder, and the division be carried on further th-^ f^mirp" i- iV - - -v this point wffl be decimals ^^ 'i"""™' *^' 08 if ARITUMXTIO. Ex. Divide 1121-4 by -SSi. Proceeding by the Rule given above, •634) 1121-400(2100 1088 534 00 Rtason for tht above process 1121-4-T-534=ll?lf--i34 10 1000 1000 ^11214 1000 _ 11214 10 534 53r ^ TO"'^^^ ^ 100=2100 DiSn of nt' ^y'T^ "^i^takes in the proof of examples in 10 iaa\ Deo ma.s, always contrive in the process to seoarate 10, 100, &c. in the tvro fractions from the other figures asTthe fnZeTe'tT "k^' th' r" "^^^^ *^ ^^^^^ the'mSi;n:at[o: 11 xnere De tens left m the denominator ; nor, if there be tPn«a Ipff m the numerator, to effect it until the last step of l^Verat^^^^^^ Ex. Divide 172-9 by -142 to three places of decimals. •142) 172-900000 (1217 605 142 309 284 250 142 1080 994 860 852 800 710 90 Here we must affix 5 cyphers to 172-9; for if we affix two ,^1^?!;^. ^„^;-l«' *^«. <^--on up to th^t poinl w7l ^fv'e 'Z ^.t.5.»x i^art .u ,ne quotient only, and therefore as the quotient DIVI8I0IT OP DECIMALS. 69 is to be obtained to three places of imals, we must affix three cyphers more, that is, we must affix ve altogether. Reason for the above process. 172'9-r-142 ^1729^ 142 ^ 1729 1000 _ 1729 100000 10 1000 142 "IT "142 ^ToW" _ 172900000 1 — X 142 1000 Now l^^^ = im606...from above; therefore the result = ^^^^^^^- = 1217-605. 1. Ex. XXVIII. Divide, (proving the truth of each result by Fractions :) 10*836 by 5-16, and 34-96818 by -381. •02. :. by 1-003, and -02916 by -0012. •oOGdl by 27, and 1-V7089 by 4-735. by -1, by -01, and by -0001. 31-5 v. '. and 5-2 by -32. 32r. 0025, and -03217 by 6250. 4-63636 h, ^1-34, and 15*4546 by -019. •429^' ^^{ by 59-64, and 2147-04 by -036. • 12-6 by -0012, and -065341 by -000475. 3-012 by -0006, and ^^^3916669 by 541-283 130-4 by -0004 and bv ' and 46-634205 by 48^ 1-69 by 1-3, by -13, b; 3, and also by -013. •00281 by 1-405, by 14 ^5, and by -001405. 72-36 by 36 and b -0036 and -003 bv 1-6. 6725102-3544 by 7089, an., hy -7089. 10363-84-75 by 39^-25, and =09844 by -0046. 816 by -00u4, and -00196106528 "^ w 2-38645 18368830 5 by 2315, 231-5, .a v 2315 ^^•■/V^^^J *^>«-V -^5, rxa b^ \}\)\}{jv .u. b84-1197 by 1200- 1, .ad also by -0120021. "-5. 7Q ▲RITHMITIO. (1) 325 by 8-7 ; -02 by 1-7 ; 1 by -013 (2) -009384 by -0063 ; 51846734 by 1-02 (3) 7380-964 by 023 ; 6-5 by 342 ; 25 by 19 (4) 176432-76 by 01257 ; 74571345 by 65354962 (6) 37-24 by 29 ; 0719 by 27-53. 3. Find the quotient (verifying pich result) of ' (1) 0029202 by 157, and by i-57. (3) (7^ of j + ^l) by -0005; of 31-008 by 4^ of U of tWV; •7575byl6i. 93. Certain Vulgar Fractions can be expressed accurately asDecimah Ex. 1. Convert | into a decimal. 5 I 3-0 . -6 rhere is one decimal place in the dividend and none in the divisor ; therefore there is one decimal place in the quotient Note, In reducing any such fraction as j\ or ^^^ to a decimal we may proceed in the same way as if we wJre reducing | tS care however m the result to move the decimal point 'one plac! further to the left for each cypher cut off. ^ 8 -6 =•«' Thus 8 50 =■<>«' fr tlf^JA. ^Tt:^l 5' ^^ *«n by 10. 100, &c.. accordiD XO UU. U 00, 0210. g DIVISION or DI0IMAL8. 71 Ex. 2. Reducejjto a decimal. 16) 5-0000 ('3125 48 20 16 40 82 80 80 or thus (4 1.25( 2500 •3125 -.^^=•3125 3 8 Ex. 3. Convert--- and ^t^^^;^ 512 51200 Now 512=8x64=8x8x8 3-000 into decimals. 8 8 8 •375000 •046»75000 •00585d375 or^-^ is equivalent to -005859375, and^is equivalent to •00005859375. Ex. 4. Convert | + 3i + 25«V+6tVV into a decimal. f+3i + 2A+6,VV=114-| + i+^,+^.,V 5I_3^ 8 I mo 4|9-00 •^ •''''' "2^ .•.tV=*225 5 5 5 •125 110 220 •440 •088 therefore i=.6, |=.125, /, =.225„V^=-088 ; therefore the whole expression =ll + '6+l25+-225+*088 = 12^038. li^ 7S ARITBMETIO. ^ Ex. XXIX. 1. ixeduce to decimals : 2. Keduee to decimals ■ " W I+-061. (5) j..,_,. (6)f,,Hl. ""'-rof^otU. (8) 5,J^+.75offof7i. ,^ ,„ second power o{ 10. ^ """ M.W;»»«roflO. .n^ If"; 10x10x10x10 .JiMpomrono. and 80 on, similarly of other numbers. into*a d'SmaraCLtllX';'" *° <""^.^^\'' ^"'««- fr"""'"" affixing cyphers to the Zml.! *"""!! *" "^ '°''«^' '«"•"« »""1 the fraction: nowl0=2x5 «n^ ^ ' ^ ?^ denominator of which 10 can be broken ud ' twi "' T '\^ ""'^ ^^"'""^ "^ lowest terms, if Z denZinatortr' V"^ '^' '^'"'"o" '' '" "« factors 2 and 5, orone ofTem ' r "' <""»P°'«*0'«*J-s, from must come roSnd agtror tur"or t"""""^' f' f ""« '^'«» always affix the mmehauL m fL° ^^^/T****^ = '*"" ^^"^ ^e whenever any foZer X «„!'''! J!!'^?.""' ""^'y " «ypher, recur. Wow'when we diviHAKr ""'"'\''"^ quotient will also ow Men we divide by any number, the remainder must VULGAR FRACTIONS EXPRESSED AS DECIMALS. 78 always be less than that number, and therefore some remainder must recur before we have obtained a number of remainders equal to the number of units in the divisor. 95. Pure Circulating Decimals are those which recur from decimafJ''"'"^' *^"^ '^^^^"•' '^'^^'^W'"' ^^^ P"^® circulating Mixed Circulating Decimals are those which do not begin to recur, till after a certain number of figures. Thus -128888 , -0113636 , are mixed circulating decimals. The circulating part, or the part which is repeated is called the rBRioD OR Repetend. Pure and mixed circulating decimals are generally written down only to the end of the first period, a dot being placed over the hrst and last figures of that period. Thus -3 represents the pure circulating decimal 333.,. .'^9 -3036... *^^? '639639... •1?? mixed -1388... '^^136 -0113636... 96. Pure Circulaling Decimals may be converted into their equivalent Vulgar Fractions by the following Rule. Rule. Make the period or repetend the numerator of the fraction, and for the denominator put down as many nines as there are figures in the period or repetend ; this fraction, reduced to its lowest terms, will be the fraction required. Note. The fraction is only reduced to its lowest terms for the sake of exhibiting it in its simplest form. It is not of course actually necessary so to reduce it. ^ ^ Exs. Reduce the following pure circulating decimals, -3 -87, •857 J 42, to their respective equivalent vulgar fractions. Procef ding by the rule given above, .31 '^-■9~F . . 27 3 V '*'--99~~ll' •867142 = ^-^^ = 95238 ^ 6 x 15 873 _ 6 999999 mill -i X 15873 ~ r ' l\ i I 74 ABITHMBTIO. Let the circulating decimal -SSSS... be represented by a symbol « ; then ^^.3333 ^ J J' ui therefore lo times a:=10 times -3338... „ • =3-3333... (Art. 86). Jew 10 times. ^ ^^^^^^^J^^^^^l^y 1 time x, will leave 9 times x, ^^^ 3-3333...-^-3333=3.3333... ' — -3333... =3 or 9 times a;=3 therefore 1 time x, that is x or -3333... =i=i. ThTn' ^^* ^^^ c^^^^lating decimal •2727- be represented by x. aj=: -272727 100 times a? =100 times '2727 TV. !• inn . =27-2727... (Art.' 86). Therefore 100 times ^^^^^ed by^l ^^^^ ., will be equal to ^. or 99 times a;=27 ; 27 ^ therefore x or •2727...=qq= -~ yy i 1 Next, let the recurring decimal -857142 be represented by x T*»«°. ^=-857142857142... ^ and' weTafe''" ^'' '''' ^^"''' '" '"'^ P'"'^' ^^ ^"^^^^^^ 1^00000, 1000000 times a?= 1000000 times -875142 =875142-875142... ; tiierefore 999999 ic= 857 142, 857142 or X- 999999' 6 which fraction, reduced to its lowest terms = A.^''^ \ ^^ ^^^^""^ '" ^^^^ ^^«® is t<^ multiply the recurrinff NoU 2. Thej^ow;m of numbers are often expressed by placing CIRCULATING DECIMALS. 76 a small figure (equivalent to the number of factors and called the 7im:::ZT:is:l'^ ^^^"^ ^^ ^'^^ ^^^^^^^^^ of thenu.nber, Thus 10 X 10, or the second power of 10 is expressed by 10*, 10 X 10 X 10, or the third power of 10 . 10»' 10 X 10 X 10 X 10X10, or the fifth power of 10." V and so on. » . 97. Mixed Circulating Decimals may be converted into their equivalent Vulgar Fractions by the following Rule. Rule Subtract the figures which do not circulate from the figures taken to the end of the first period, as if both were whole numbers ; make the result the numerator ; and write down as many mnes as there are figures in the circulating part, followed fo'rl7eZr;t:r'"^ ^^^ '^""-^^ ^^ ^^^ nonlir^cmatingpart Exs. Reduce the following mixed circulating decimals, -14, •0138, '2418, to their respective equivalent vulgar fractions. Proceeding by the Rule given above .14="-! '« •0138= 90 138—13 90' 125 9000 9000 1 =— , in its lowest terms. 2418-2 2416 •24.1 ft = —JZ • ^^^^ 9990 ^9990 ^ 1208 4996* coL'z:!:^ ''"' ""''' ""'"' "^^^^'* -^^^^ ''' /^^^--^ Let the mixed circulating decimal be represented by x in each 01 the above cases. ^ First, let a: = -1444... ^ If, by multiplication, we change the decimal in s-uc^ « rv.or,«ov. luat ine non-circulating part is rendered a whole number, and also change it so that the non-circulating and circulating parte to the 76 ARITHMBTIO. end of the first period are rendered a whole number, and then subtract the first result from the second, we shall get rid of the circulating part. Thus, multiplying first by 10 to get the 1 out as a whole number, and then by 100 to get the 14 out as a whole number, we have 10 times a;=10 times •1444... = 1-444... 100 times a;= 14-444... ; therefore 100 times a;— 10 times a; =14-444...— 1.444... or 90 times a;= 14.444... —1.444... therefore Next, let = 18 13 00* a: =-013888... Xz=' Here there are three places in the non-recurring part, and one in the recurring part ; therefore multiplying first by 1000, and then by 10000, we have 1000 times a;=1000 x -013888... = 13-8888... ; and 10000 times x=l 38-8888... ; therefore subtracting, as before, SOOO times x = 138—13 = 125 ; therefore Next, let X' 125 9000 Jl_ 72* x= -2418418... Now we have one place in the ion-recurring part, and three places in the recurring part ; therefore multiplying first by 10, and then by 10000 we have 10 times a;=2-418418... therefore therefore 10000 times a;=2418-418418... ; 9990 times x=2418— 2 = 2416 ; 9.4.1 A 1'>C8 x= 9990 - 4995 OIROULATINa DKCIMALS. it Ex. XXX. 'X ^:_^®^"*^® ^,^®. following vulgar fractions and' mixed numbers circulating decimals : K'-/ 9 ) II > 3T > T* (3^ 3.2.3.1 ' 7_9Aa 33.3. 4 95 > 14 • 6 1} \S20 f _.. y V , ^ ■ 1 7 3 3a7 ? ^OTToir* 15-5-3- ^"sa 3* (2) U ; - - - W 2M^3 ; iVyJ'^ ; 2Hf III. 2. Find the vulgar fractions equivalent to the recurring decimals • (1) -7 ; -07 ; -227. (2) -583 ; -135 ; -263 ; (3) -00185 ; 3 024; -01230. (4) -142857; -397916; -382142857 (6) -307692; -6307692; 27857142. (6) -342753; -03132132; 8-02083. (7) 85-60806 ; 3'642857i ; 127-00022095. 98. The value of the circulating decimal -999... is found by Art (96) to be f or 1 ; but since the difference between 1 and -9=-l* between 1 and -99=^01, between 1 and -999=-00l, &c., it appears that however far we continue the recurring decimal, it can never at any stage be actually=\. But the recurring decimal is considered =1, because the difference between 1 and -99... becomes less and less, the more figures we take in the decimal, which thus m fact, approaches nearer to 1 than by any difference that can be assigned. In like manner, it is in this sense that any vulgar fraction can be said to be the value of a circulating decimal ; because there is no assignable difference between their values. 99. In arithmetical operations, where circulating decimals are concerned and the result is only required to be true to a certain number of decimal places, it will be sufficient to carry on the circulating part to two or three decimal places more than the number required : taking care that the last figure retained be increased jy 1, if the succeeding figure be 5, or greater than 6 - because, for instance, if we have the mixed decimal -6288 and and stop at -628 it is ^V.r that -628 is less, and -629 is greater than the true value of :iiG -lecimal : but -628 is less than the true •000111^ "•* '"^^ ^^^ '' ^''^^' ^^^"^ *^'^ *'"" ^*^"« b^ Now -000111... is less than -000888... Therefore -629 is nearer the true value than '628. ' ! ii [ I 78 ARITHMETIC. Ex. 1. Add together -33, -0432, 2-345; so as to be correct to 5 places of decimals. •3333333 •0432432 2-3454 546 2-722031T Ans. 2-72203. Ex. 2. Subtract -2916 from -989583, so as to be correct to 6 places of decimals, •9895833 •29 16667 •6979166 Ans. -69791. Note. This method may be advantageously applied in the Addition and Subtraction of circulating decimals. In the Multiplication and Division, however, of circulating decimals, it is always preferable to reduce the circulating decimals to Vulgar Fractions and having found the product or quotient as a Vulgar fraction, then, if necessary, to reduce the result to a decimal. Ex. XXXI. (1) Find the value (correct to 6 places of decimals) of -1. 2-418+ 1-16 + 3-009+-7354+24042. 2. 234-6+9 928+ -0123456789+ -0044+456. Zf' _?"*5— 3; and 7'72-6045 ; and 309— 94724. (2) Express the sum of f|, |^«, and j\, and the difference of i»y'j and 4^^, as recurring decimals. (3) Multiply 1. 2-3 by 36 ; -7575 by 366. 2. 406 by 62 ; 825 by -36. 14.)' DivMe^"^ ^^ ^ ' ^^^ ^^ ^' ^* ^'^^^ ^^ *^^^^ ' ^^^ ^^ '^^' 1. 19502 by 4 ; -37592 by 05. 2. 54 by •if ; 13-2 by 5-6. 3. 411-3519 by 58-7645 ; 216595 by -04 ; -6559903 by 48-76. Ex. XXXII. Miscellaneous Questions and Examples on Arts. (80 99). I. (1) Define a Decimal ; and shew how its value is affected by rect to 5 )3. rect to 6 by -84. by 5-6. 48*76. )9). Jtedby MISOBLLANBOUS QUB8TI0NS. lyo affixing and prefixing cyphers. Reduce -0625, and 3-14159 to fdS."'^"p"" ^'^ '^^'^^^^^ b«*-- sorandni:: not^tnlL^;^^^^^^^^ ^^-i.., do (5)^pIify_1.^2U72f + 316^ + 2.875. 2. .026649-2^. ' 5+-5 ^ITS 10* ^' { '18+ 009} -r 016. ^^^ ^'""'^"iS: ^^ I?fr J ^«^"«« the quotient to the form 1'0714285. Divide 91-863 by 87-56. 11. foui'LSiZir. '■'^T"'^^°''"^^«" ^""^'^ "•ousand **"" lerToWF *s ft aecimtU. ■ * .S^l ^^*^ the effect as regards the dedmfil point of muItinlvinT and dividing a decimal by any giten wwer oi 10 Wr ,. .5' ^ in words the meaning of 397008-40ti.^ ^ dj p,„ ^l"* t^ f?, ■2^m'iS6'?"'"f'"^' 'y ^'^ ^"' g'- "•« -- of i, •0003 X -004. -^^vme o/^/a»8 oy 1620; find the value of • i^g— ' reduce ^V+y f ^~ Jj3„ ^ a decimal. (5) Simplify, expressing each result in a decimal form 1- TifhoOf^. 2. (2^ + 6)^(31—1). lF> 4 5 3. *-4+f 7-875+^--| ^' ^3 «V« + lTB^oo+5eJ^^ + 2-000875. 80 ARITBMBTIO. h (6) Find a number which multiplied into 3132*458 will give a product which differs only in the 7th decimal place from 1S2S'6S7% III. (1) Divide 684-1197 by 1200*21, and also by -0120021 ; and 594*27 by *047 to three places of decimals, jand explain fally how the position of the decimal point is determined in each of the quotients. (2) Simplify, expressing each result in a fractional and decimal form, •015x21 ^ 3^—04 1. 2. 035 • "' 5—0625* 3. | + *14+f ofl-C784. 4. Q-~|)x(f+H). (3) What is meant by a * Recurring Decimal' ? What kind of vulgar fractions produce such decimals ? State the rules for reducing any recurring decimal to a vulgar fraction. Multiply 5*81 by -4583, and divide 1*13 by -006132. Is /^ reducible to a recurring decimal 1 r (4) Shew that if lyV* ^fj^ 3^^, 4^ be added together, (1) as fractions, and (2) as decimals, the results coincide. (5) A man walked in 4 days 60 miles ; in each of the three first days he walked an equal distance, in the fourth day he walked 13*95 miles ; find the amount of his daily walking. (6) A person has '1875 part of a mine, he sells -17 part of his share ; what fractional part of the mine has he still left 1 IV. -1 (1) State the Rules for the Addition and Subtraction of Decimals. Add together 1*23, *123, -0123, *00123, and 123; and find the vulgar fraction corresponding to the -result. Find the fraction equivalent to 31*457457, and subtract it from the fraction y/. (2) Write down in figures the number, three millions six thousand and five. Also write down in words the signification of the same figures when the last is marked off as a decimal. (3) Compare the values of 5 x *05, 1-5 x *75, and 2*625-^5. (4) Find the product of *0147i47 by *333 ; and the quotients of •12693 by 19-39 ; of 132790 by *245 ; of -014904 by S-^% ; of 61061 by 3-05 ; and of 6106*1 by 305000. MISCELLANEOUS QUESTIONS. 81 (5) Shew that the decimal '90437532 is more nearly represented by -90438 than by -90437; and find the value of II -J— ^ 1 ) 4 ^^^ ( 5""3x5'"^53r5^~7x5"^ + ^^* 3 "239 accurately to 5 places of decimals. (6) A person sold -15 oi -i. estate to one person, and then ^ of the remainder to another person. What part of the estate did he still retain ? 4- V. (1) Express ^(6^+21—3), f HI, and also the product of 8* and (3^— A) of 4 as decimals. (2) Simplify 4-255 X -032 „,.,...' 1. •00016 3. (tV of" 35}— 31) + (2-5625+71). 4. 593-r 1-78 x-36-r 072. (3) State at length the advantages which decimals possess over vulgar fractions ; what disadvantages have they ? Shew whether ^ or f f f is nearer to the number 3*14159. (4) Find the value of 1 + 1 + _!l_ + 1 1x2 of decimals ; and also of X ( l-A+3x4^J_ 102 ■ - --^ 1x2x3 3x4x5 + &c., to 7 places x2x3 10«j 103 \^ 102 1x2 10* 1 expressing it (1) as a decimal, and (2) as a fraction. (5) Find the Earth's equatorial diameter in miles. Supposing the Sun's diameter, which is 111-454 times as great as the equatorial diameter of the Earth, to be 883345 miles. (6) In what sense is a vulgar fraction said to be the value of a recurring decimal 1 Explain how a sufficient degree of accuracy may be obtained in the addition and subtraction of circulating decimals to any given number of decimal places, without converting the decimals into fractions. Tit TTI* 1 . 1 & sum of decimals. i-^, 4-163, and 9-457, correct to 5 places ™ ARITHMXTIO. VI. (1) Prove the Rule for Multiplication of decimals by means of^the example 404-04 multiplied by -030303. Multiply -345 by ^ ; and divide -04813489963 by -6593, and by -006593. (2) Explain the meaning of 1\ and 7'; and find what vu] ar d^fferenci! '^"'''^'"' "" '^' '""" "^ "^'^ ^"^ ^'^^ divided by the (3) Reduce to their lowest terms 1??1®» ^^^ ?^^^- 1033-2 ^^ 5-7980 (4) Shew that :2!i^?l^Z::5?5x;025_2 •376—- 025 -5 ^^'^ ***»* 7+Xe=3-14169 nearly. Reduce -1293131 to its equivalent vulgar fraction. (5) What decimal added to the sum of lA, f,andi5 will make the sum total equal to 3 ? "' ^' =** div^dLd^^ ^"''*'^''* ^"^'"^ ^^* ^""^ *^^ ^^^^'°" '^^^ fi«d the D1CIM4L oonrAox. by means y -345 by 93. at vui ;ar ed by the ) t I H will find the DECIMAL COINAGE. 100. Table of t nadia ^nd United States icncy* 10 dimes, 100 cen^ o' 1000 mils, make 1 dollar « $?; currerovL'lT/T '^- P'^^-^'ng ^ble that this system of ^aUtL> lot' ?^'*.'""''*^J "" '^ " '« soLranged, coino^^h i.S °^/ ^or^ ^^?«d ^ the unit of money in Canada and Stef^ a'S T"^^ \ ■^^'''''*'' the gold coins current in the United States, and those of silver, current in Canada. 8i ARITHMETIC. inflJ^Z^^:^^'r"''" regards gold .nd silver onf if ^ r.^i f r£^> »i^ r rr -^"^ ">- ^<'"'" ^e »m of money expended, and express thelnswer in a faujtS (8.) A farm of 125 acres is sold for 15824. How muph ;« th^t Llw^lt ^F^,"^ -werasai^tion, and&e"^'^;:^' n. (1.) Find the value, in dollars, of ^ $5f + 5|c+5|m . 8. 80016c+ gin 1-25 2. 4. •006 •05^ -6 Ian?? *lftSSl'i'^^/ tradesman realized 11536 from the sale of some lana , 91856-15 from his stock in trade • ^SQn-^^ a.^^^ t- ^ and carriage ; and $59.63 from otherekt w^^^^^^ '^"'^^ after paying $2563.758 to his creditors anH 125^^0^!^^^^^ expenses of the auction? fi^i&SO for the (3) Find the difference between $ (•1001+3-572) $(7-356+-101) AA' -:, ,. '^^^ and ooi and divide the remainder by 1|. * (f ). A,?^®^<^^»t possesses a fortune of $206 060 and wi«»,«a f^ start his three sons in life ; to the eldest he give^ 1 oT hir^^^^ !2„^^.^!??!^lf-i-^ ^ the third ^, find L vaL of 1^^^^^^^ o«».^ , oxau wuat remains of the father's fortune. " " h and silver ine dollars )1 of pears 31. Find fractional Jch is that difference •36, what nXOIMAL OOINAOX. ) of some tiis horse s possess for the ishes to fortune, jMI f\rto?a U (5) Find the value of 12-8 of 2-27 1136 __ ^ $(4-4 - 2-83) ^f 6;SofJ 1-6 +2-629 2-25 nf ISn f f °l,l®°i*"' 'J" furnishing his house, had to buy 95 yards of carpet for his drawing room at 12-75 a yard ; 87 yards for his dmmg room at $1-625; 166 yards for his study a/dbX^ms tTeti^rr ''''^ f ?^' P^^y*^^> ^^ ^^ whole cS^Tf carpeting the house, and the average price of the carpet per yard. III. (1.) The estimated coat of the Great Eastern steamship was $5,000,000; what would be the value of tH of ^ of the vessel ? And supposing her shares rose in the market from $5 to f 7 ^b per share, what would be the increase on 51 -2 original shares ? (2.) A farmer goes into town with $15-75 in his pocket to pav his grocer's bil which consisted of the following items: 7U7f root at 48c per lb. ; 5 ounces of spices at 7+c per ounce • what money will he have remaining after paying the debt 1 ' 4t^mVo?9Tl^^^^^ '''""" ^'''^ '' "-^^ '' 1-1^' ^^ (4) A speculator bought 500 acres of land |br $9874 • and 250 acres for $647|. He sold 487^ acres for $1246^ How m^h Ser 'to ' -'"^'^"^ \"^ \' "^^^^ ^"«* ^^ ««" it per a^,raS neither to gain nor lose by the transaction ? , « per^lOoT-fee^^^^ *^' ^"^"^ ""^^^^ ^""^ ""^ oak planking at $12-76 4^375 of $?.30.*^"^ ""^^'^^ of $3-8677083+5-8 of $2-4114583 IV. hundred and (1) Read $6'00101 : and ftYnpAss in fimn-oc ««,,o« ""--rA/- ^^^ toee thousand and nine 6oH^'-^;^,n^;^^ r^^Zp^ ^■ dunes m oents and mUs, and 706884 mUs in dollars! 90 ARITHMXTIO. f» gallon * '' ""• ' *'"' ° °85 gallons of molasses at »0-668 «(i+^*S»+X)'"''"'^ "'^ ''°"" ""^"P""^ ''y "^ '"' e^^o 1 b^uihdf IJso 0?^ :' ra."" *''*'*-'''^' fi^-^ ""o -' of ^fl^K^''^' 'l** ""'"* ofs"ga>- » quarter when -75 nuarters no»t m76, »d what must be paid for ^ quarters of^upTtThe' paf 4^5^;^^ '^r^Zt'Ue<^t 'x:^f-o:t\£' market to the value of «35,-16 per share ; ie thenTowZ What n^A liA TvtAl^A^ l*-- iL_ i .. . _ _ 2.-4- 7 So ^® fit!"® ^^ ^^'^ transaction ? and find the value of 1^-11 nf a share at the advanced price 1 ^^^^ ^*^ * toTtJ^^a^^tfrn TdS'r"'*T"'«°\" E"«'-d difficulties in the wav renL t?? °"^\ "*"' "'^ P'««'i<»' should be cr^ulj^ S fl^ • ' ""Pf^y "wt ite introduction some'A'^^t^tinJ^^^Z'w^wV^SCr? '"""•'"" « place in a decimal scale of c^y'^^Srst/j^Ti''* been most fevorablv receivod for rt.V J;'' 5 , * *'"«'' ^as IK>und sterling for l^lf 'H'.I^tlt^^' *^" "^ 10 florins i pound £i CorresfL ^ ^'^^^»^+"''™>='^l »«• *=• S"-^ U)rrespondmg m arrangement to - . _ra. „ari„ TTiix aittvi- m a corresponding degree. 76 lbs. of 30 lbs. of at 10-653 will give iie cost of rters cost ir at the w^hich he le in the it. What England practical eduction use for cupying lich has ikes the OONORETB NUMBERS. 91 Canada 3e only >; thus, a. [ual to being other, CONCRETE NUMBERS. TABLES. 103. Our operations hitherto have been carried on with regard only to abstract numbers, or concrete numbers of one denomination. It IS evident that if concrete numbers were all of one denomination • It, for instance pounds were the only units of weight, yards of length, years of time, and so on, such numbers would be subject to the common rules for abstract numbers. Again, if the concrete numbers were of different denominations, and those denominations differed from each other by 10 or multiples of 10, then all operations with such concrete numbers could be carried on bv the rules which have been given for Decimals. But generally with concrete numbers such a relation does not hold between the different denoniinations, and therefore it is necessary to commit to memory tables, which connect the different units of weight K:i^^^^^^^^^^ *^^^'^^^' '"^^ ^^^--* -its ^'IlZ':Z^:^roTZl.:''''' "^^* ^^^^-^ of these tables, ■ TABLE OP ENGLISH MONET.; 2 Farthings make 1 Half-penny. 2 Half-pence 1 Penny. 12 Pence i Shilling. 20 ShiUings l Pound. Pounds, shillings, pence, and farthings were formerly denoted by £, ., d and q respectively, these letters being the fiJst letters names of certain Roman coins or sums of money. £sd areS the abbreviated forms for pounds, shillings, and pence reVSy. ^ILrf."^ ^. P?."" ^''^^'^^ 1 ^^''^^S i denotes a STennj i denotes three farthings ; shewing that one ferfhinc fw. fi,iu°"Z» I* ABITHMETIO. The following coins are in common use in England: COPPER corns. A Farthing, the coin of least value A Half-penny =2 Farthings. A Penny =4 Farthings. '* SILVER coma threepenny ) „ „ piece ... f =3 Pence. Fouxpemiy. 5 _ piece ... y * ^ence. A Sixpence . . =6 Pence, A Shilling... =12 Pence. A Florin =2 Shillings. A Half.Crown=2 Shillings and 6 pence. A Crown =5 Shillings. aOLD COINS. A Half.Sovereign=10 Shillings. A Sovereign =20 Shillings. Note. The office at which coin pass or become current for legal The following coins have been in use at various periods in England, but with the exception of the first two, which are used under different names, they are now obsolete : BILVKR COINS. A Groat. A Tester = 4 Pence. = 6 Pence. GOLD COINS. £ 8. d. A Noble =0 6 8 An Angel =0 10 AHalf-Guinea. .. =0 10 6 A Mark or Merk. =0 13 4 A Guinea =1 1 A Oarolus =1 3 A Jacobus =1 6 A Moidore =1 7 is made and stamped, so as to money, is called the Mint, MEASURES OF WEIGHT. TABLE OF TROY WEIGHT. 104. ^is table derives its name probably from Troves in France, the first city m Europe where it was adopted. It seems to have been brought thither from Egypt. It has also been derived ivomTroy-novant the monkish name for London, It is used in weighing gold, silver, diamonds, and other articles of a costly nature; also in determining specific gravities; and generally in philosophical investigations. » s '= ""J' *" The diiferent units are grains (written grs.), pennyweights (dwts) ounces (oz.), and pounds (lbs. or ibs.),and they are connected Zs ' 24 Grains. ....... make 1 Pennyweight . . 1 dwt. 20 Pennyweights 1 Ounce.. 1 oz 120unces 1 Pound '.llb.'orib. frnm fl ^AA^^\'tt'^''' ""^ ""^^8^^, a grain of wheat was taken S ."S'^^^ff ^^^"^."'/"^ ^«^"g ^«" ^^i«d» ^as used as a weight, and called a ^grain\ « »d » Note % Diamonds and other precious stones are weighed by a 9 t( Ig a] si is sc tb or ar mi of un hu ho^ OBI d. 8 6 4 MIASUB18 OF WBIOHT. 15T?V ^|\?^** weighing about 8^ grains. The term * carat ' applied to gold has a relative meaning only; any Quantity of pure Thl^' -J^f^^oT^ \'^^ "^""^ ^^^^^ metal, \eing supposed to be divided into 24 equal parts carats); if the gold be pu?e, it 18 said to be 24 carats fine; if 22 parts be pure gold and 2 Srti alloy, it 18 said to be 22 carats fine. b *^^ * pans Standard gold is 22 carats fine : jewellers' gold is 18 carats fine. TABLE OP APOTHECARIES' WEIGHT. 106. Apothecaries weight only differs from Troy weight in the subdivisions of the pound, which is the same in both. This table 18 used in mixing medicines. The different units are grains (crs \ scruples (3.), drams ( 3 •), ounces ( I •), pounds (lbs. or ft>s.) and they are connected thus : '* ^ 20^ Grains make 1 Scruple. . 1 sc. or 1 3 a.Scruples 1 Dram ... I dr. or 1 3 . fe8 Drams 1 Ounce. . , 1 oz. or 1 ? 12 Ounces 1 Pound . . 1 lb. or Ift .' TABLE OP AVOIRDUPOIS WEIGHT. 106. Avoirdupois weight derives its name from Avoirs ( goods or chattels,) and Foids (weight). It is used in weighing all heavy articles, which are coarse and drossy, or subject to waste, as butter ineat, and the like, and all objects of commerce, with the exception of medicines, gold, silver, and some precious stones. The different units are drams (drs.) ounces (oz.) pounds (lbs.) quarters (qrs ) hundredweights (cwts.) tons (tons.) and they are connected thus : 16 Drams make 1 Ounce i oz 16 Ounces 1 Pound i ib' 25 Pounds 1 Quarter i qr. 4 Quarters , 1 Hundredweight 1 cwt 20 Hundredweights ... 1 Ton i ton. 1 lb. Avoirdupois weighs '7000 grains Troy ; 1 lb. Troy weighs 5*760 grains Troy; therefore 1 lb. Avoirdupois =|M^ of lb. Troy =1 <^^ of 1 lb. Troy = i.J|ofllb. Troy =14 oz. 11 dwt. 16 grs. Troy =1 lb. 2 oz. 11 dwt. 16 grs* Troy. Note. In England, 28 lbs. =1 qr. or 112 =1 cwt. In Liverpool however, a new weight has lately been introduced, called the CENTAL, which corresponds with the hundredweight of 100 lbs. "^ ▲RITHMITIO. MEASURES OF LENGTH. TABLE OF LINEAL MEASURE. 107. In thio measure, which is used to measure distances, lengths breadths, heights, depths, and the like, of places or things : ' 3 Barley Corns (in length) make 1 Inch, which is written 1 m 12 Inches i Foot ifv* JJeet 1 Yard, [.'.'.".'lyd SFeet 1 Fathom i S Ji^f^8 1 RoC7. There will therefore be 9 cubic feet in the solid figure DL. Similarly if another lineal foot HN^ were taken, and a plane NO were drawn parallel to J3X, there would be 9 cubic feet contained in the solid figure HO. Similarly, there would be 9 cubic feet in the solid figure NE. Therefore, there are 27 cubic feet in the solid figure DE, or in 1 cubic yard. The following measurements may be added : A Load of rough Timber = 40 cubic feet. A Load of squared Timber= 50 cubic feet. A Ton of Shipping = 42 cubic feet. A Cord of Wood =128 cubic feot. Note. A pile of wood 4 feet wide, 4 feet high and 8 feet long makes a cord, — 1 foot in length, of such a pile, is sometimes called a cord foot. It contains 16 solid feet ; consequently 8 cord feet make 1 cord. m TABLES — ^MEASURES OP CAPACITY. W MEASURES OF CAPACITY. TABLE OP WINE MBASUEB. pviiVii''.*?'' uT"""®' ^^ 7^'"'^ ^^^^ »°^ ^" liquids, with the exception of malt liquors and water, are measured : 4 Gills make. . 1 Pint. 2 Pints 4 Quarts . . . 10 Gallons . . . 18 Gallons . . . 42 GaUons . . . 2 Tierces 63 GaUons . . . 2 Hogsheads 2 Pipes . 1 Quart . . . . 1 Gallon. . . . . 1 Anker. . . , . 1 Runlet . . . . 1 Tierce. . . . 1 Puncheon 1 Hogshead 1 Pip9 ITun . Ipt . 1 qt, 1 gal. 1 ank. 1 run. 1 tier. 1 pun. Ihhd. 1 pipe. 1 tun. TABLE OP ALE^AND BEER|MEASURB. milured :*^'' "'''^"''' ^^ ""^'"^ ^"^™*^* ^^^"°^^ ^''^^ ^^^r are I ^^^^ make 1 Quart . . . . l qt. ^Q"f« IGallon...!^ ,^^^°^« lFirkin....lIr: ll^^T^ 1 Kilderkin . 1 kil. 36 Gallons x Barrel "« bar 1| Barrels or 64 Galloift. . .* 1 Hogshead' .' 1 bhd • 2Hogsheads i Butt i butt. 2^^"8 iTun itun. TABLE OP CORN OR DRY MEASURE. 113. In this measure, by which all drv onrY^ry^r.Ai4.•^ 2 Quarts make i rottie . . . - i pot. ?gSr-------------i^"::::Jpk. I ^"^^^ make 1 Pottle . . . . l 2Jo«=les IGallon....! gal. tl^''l\- IBushei ll'^'^f 1 Strike.... istr. 4 Bushels i ron^u t !: __ 2 Coombs or 8 Bushdis ; .' i I Quart"; *. ! ! T" JQ^iarters 1 Load . . . i load 2 Loads or lO Quarters. . . 1 Last. ...;.' i ^I* 1 bush, str. ooomb. ARZTHMXHO. TABLE OF COAL MEASURE. 114. In this measure, which is not much used now, as coals are sold by weight : 4 Peeks make 1 Bushel. 3 Bushels. ... 1 Sack. 36 Bushels 1 Chaldron. 1 MEASUEES OF NUMBER. MEASURES OF TIME. TABLE OF TIME. 116. * 1 Second is written thus 1* 60 Seconds make 1 Minute. . 1' 60 Minutes 1 Hour . . .1 hr. 2f Hours 1 Day ....1 day. ^I>ays 1 Week...l wk. TABLE OF NUMBER. llo. 12 Units. . . .make 1 Dozen. 12 Dozen i Gross. ' 20 Units 1 Score. 120 Units 1 Long Hundred. 24 Sheets of Paper 1 Quire. 20 Quires 1 Ream. 10 Reams l Bale. * A year is divided into 12 months, called Calendar months ^l«r^I''^?f^^.^'' 1?"^^ of which are easily remembered by means of the following lines : ^ Thirty days hath September, April, June, and November: February has twenty-eight alone, And all the rest have thirty-one: But leap-year coming once in foiu*, February then has one day more. -4 «?ay, or rather a mean solar day, which is divided into 24 equal portions called mean solar hours, is the standard unit for the measurement of time, and it is the mean or average time which elapses between two successive transits of the Sun across the meridian of any place. The time between the Sun's leaving a certain point in the McUphcmd. Its return to that point consists of 365-242218 mean soUir days or 365 days, 5 hours, 48 minutes, 47i seconds, very nearly, and is called a ^olar year. Therefore the civil or common year, which contains 365 days, is about ith of a day less than the solar year; and this error would of course in time be very considerable, and cause great confusion. Julius Csesar m order to correct this error, enacted that every 4th year should consist of 366 days : this was called Lea'n or mssextiie year, hi that year February had 29 days, the extra day bemg called * the Intercalary' day. TABLES — IMPERIAL STANDARD MEASURE. 99 oals are ME. 18 1" .1' .Ihr. .1 day. . 1 wk. Qonths, red by ato 24 for the which 3s the in the mean , very mmon i than ) very every extra But the solar vear contains 365-242218 days, and the Julian year contains Sr: ^5 or 365^ days. Now 365-25 -365-2421 18=-0.07782. Therefore in one year, taken according to the Julian calculation, the Sun would have returned to the same place in the Ecliptic •007782 of a day before the end of the Julian year. Therefore in 400 years the sun would have come to the same place in the Ecliptic -007782X400 or 3-1128 days before the end of the Julian year ; and in 1257 years would have come to the same place, -007782 x 1267 or 9-7819, or about ten days before the end of the Julian year. Accordingly, the vernal equinox which, in the year 325 at the council of Nice, fell on the 21st of March, in the year 1582 (that is 1257 years later) happened on the 11th of March ; therefore Pope Gregory caused 10 days to be omitted in that year, making the 15th of October immediately succeed the 4th, so that in the next year the vernal equinox again fell on the 21st of March ; and to prevent the recurrence of the • error, ordered that for the future in every 400 years, 3 of the leap years should be omitted, viz. those which complete a century the numbers expressing which century, are not divisible by 4 • thus 1600 and 2000 are leap years, because 16 and 20 are exactly divisible by 4; but 1700, 1800, and 1900 are not leap years because 17, 18, and 19 are not exactly divisible by 4. ' This Gregorian style, which is called the new style, was adopted in England on the 2nd of September 1752, when the error amounted to 11 days. The Julian calculation is called the oUityle : thus Old Michaelmas and old Christmas take place 12 days after New Michaelmas and New Christmas. In Russia, they still calculate according to the old style, but in the other countries of Europe the new style is used. Sir Harris Nicolas in his Chronology gives the dates at which the new style was adopted in different countries. Of course it was almost immediately adopted by most of the Roman Catholic courts of Europe, 117. TABLE OF ANGULAE MEASURE. 1 Second is written 1 see. or 1". 60 Seconds make 1 Minute 1 min. or 1' 60 Minutes 1 Degree 1 deg. or 1°. 90 Degrees 1 Right Angle .1 rt. ang. or 90' 100 ARITHMETIC. i«Pq«A f®''®'''^ Of every circle is considered to be divided into 360 equal parts, each of which is often called a degree as U subtends an mtgl of P at the cenfre.of ,the circle ^ ' 118. An Act of the Imperial Parliament "por, AsdeRTAiNiNo and I^I^aT"" Pr^^^^y «^ Weights and Measures," iS U^ngland, came mto operation on the fir.^. of January, 1826. It is thereby enacted, of^^^wll^^nf^!!,*'*^ '^^^^r^ Y^'^of 1760, then in custody ^WW J^ J ut ?''"'® ^^- Commons, shall be the Imperdl ^tatuiard Yard, (the brass being at the temperature of 62° by St^l'i ?i ' thermometer) ; and that this Imperial Standard Yard shall be the unit or only standard measure of extension, wherefrom or whereby a 1 other measures of extension whatsoever, whether InZr.^- ^T^' 'T'^^^^\ ^''^^^^^' '^^^^ be divided, c;)mputed' W /S'° ' thnty-sixth part of this yard shall Now the length of a Pendulum vibrating seconds in the latitude h^^f^^'^'V^T'^'' ?-:? ^\^^^ level of the sea, is found to o? iTer rc£ ^ch"' '-'' '' ^"^^ ^"^^^^> ^^^ ^^^^ ten-thousandths viJ'^lf^i!?-. ^^1 ""^"t^ ''^ recovering the Imperial Standard Yard should it be lost. In lact, the brass Standard Yard of 1760 SLtrfnIsV"''"' "^'"^ '^ *'^ ^'' '' *^^ «^-« ^^ ?hT„n>^^' **!?f ^^ *^.' T^^y ^^ *^« «^^^ <^ffi««r' «hall LnLue wei^s sLlI^:''^ • f^'^'^ of Y^iff^^, from;hich all other weights shall be derived, computed and ascertained ; that 5760 ST^J^ \^ conteined in the Imperial Standard Troy Pound, and 7000 such grains in the Avoirdupois Pound. Nowthe weight of a cubic iwcA of distilled water is 262-458 T^ Vt tTl^ ^^^"^ ** '^ '^'^''^ -^ ^^^ thermometer S* '^•jxf .^^^^® *^® ^^*ns of recovering the Imnerial mSwlnT^'^'^'^^' ^^l^^*- I"^^«^' thelrassweTgUof 1758 was destroyed or lost at the above-mentioned fire. Thirdli/ ; That the Standard Measure of Capacity for Liauids ^^^TF. }^. F,'!:'^±^l''''^^^.-!1. 3!«3ght of distilled wate'r, RBDUOTION. 101 3 divided ree, as it KING AND RES," in 6. custody Imperial ' 62° by ard Yard lerefrom whether mputed, trd shall latitude found to isandths Standard of 1760 [ouse of ' of the sontinue 11 other It 5760 Pound, J62-458 IOmeter tnperial sight of Liquids Gallon;' water, >meter, Now this weight fills «77-274 cubic inches, therefore the Imperial Standard Gallon contains 277*274 cubic inches. The Imperial Bushel consisting of eight gallons, will consequently be 2218-192 cubic inches. o , i j REDUCTION. 119. Reduction is the method of expressing numbers of a superior denomination in units of a lower denomination, and conversely. Thus £1 is of the same value as 240rf., and £21 as 5040c?., and conversely ; and the process, by which we ascertain this to be so, is termed Reduction. First. To express a number of a higher denomination in units of a lower denomination. Rule. " Multiply the number uf the highest denomination in the proposed quantity by the number of units of the next lower denomination contained in one unit of the highest, and to the product add the number of that lower denomination, if there be any in the proposed quantity; repeat this process for each succeeding denomination till the required one is arrived at." Ex. 1. How many pence are there in £23. 15s. i * Proceeding by the Rule given above, £23 . 15s. 20 460+16 or 475s. 12 5700rf. or £23. 15s.=5700d. Reason for the process. There are 20 shillings in £1. Therefore there are (23 x 20)s. or 460s. in £23, and so there are 460j.4-15s., or 475s. in £23. 15s. /.i^^*^?^vT*^® **^®^® *^® ^^ P®'^^® ^° **•; therefore there are (475 X 12)rf., or 6700rf. in 475s. i. e. in £23! 15«. 102 ARITHMBTIO. Ex. 2. Beduoe 2 tons, 7 cwt, 8 qrs., 24 lbs. into lbs. tou owt. qrs. ibi. 2 . 7 . 3 . 24 20 40+7 or 47 cwt. 4 188+3 or 191 qrs. 25 955 382 4775+24 or 4799 lbs. miles fur. per. yds. 106 . 6 . 25 . 2* 8 ^ 848+6=854 fur. 40 34160+25 per. =34185 H 170925 170921 188017^ + 2J yds. =188020 36 1128120 564060 6768720 in units of a higher denomination. ^w^""' " ?'!i^® ?^ ^''^'' ""'"^^ V the number of units which connect that denomination with the next higher, and the remamder, if any, will be the number of surplus units of he lower denomination. Carry on thi« Droce»" tillNr— — \ " denomination required.'"" ^ ' ^"" "^''"^ *' '°® RBotronoir. lod 'longs, 25 pe^'?^' How; many pounds and shillings are there in 6700 Proceeding by the Rule given above, 12 I 5700 2,0 I 47,5 £28. I6s, In dividing 475 by 20 we out off the and 5 by Art. (43.) Beason/or the above process, „f^'l'*P*°"*,o* ''■'"'18! ^o<^foK in any given number of ^^I?l\71^ ^^ pence th„e j^ j jyuing, so that in 6700rf. or (IZ X 47o)a. there are 475s. Again, since 20*=£1 ; therefore in any given number of shillinffs for every 20 shillings there is £1. smuings, Hence, in 475^., or (20x23+15),. there are £23, and 15,. nfif^^'k®'""^ ^""^ f ^^ *^°^® ^»^«« i« tl»e converse of the h?w!5 V""'"'*? ^ ""^ t""^ '"'"^^ ^^^°«^ V «i<^J»«r of them may be tested by workmg the result back again by the other rule. m In this Example It will be convenient to bring the inches to &^ v'a'"''-^ ^' ^"^^.'^^ *^ P^^^- I^ ^ J«^%ard there hSf-yards. * "" * ^ ® ***®''® "^ ^* ^^^^ °'' ^^" are 5^ yards or eleven ition in >f units uid the of the at the 11 4,0 8 272668—1 90889 -1 1 4 • 15148 — 1 half-yard or 18 inches. 137,7-17 po. 34-2 fur. therefore the answer is 4 miles, 2 fur., 17 po., 22 in. J) 104 ARITHMSTIO. mllM ftir. polM In. Proof 4 . 2 . 17 . 22 B 34 furlonffs. ' _iO I36a+17 =1377 poles. 1377 11 15147 half-yards. 18 121176 15147 22 272668 inches. Ex. 3. How many grains of gold are contained in 9 lbs., 11 oz., 13 dwts., 20 grs. 1 Prove the result. * Iba. OE. dwti. grs. . 11 . 13 . 20 12^ 108+11=119 oz. in 9 lbs., 11 oz. 20 2380+ 13 or 2393 dwts in 9 lbs., 1 1 oz„ 13 dwts. 24 9572 4786 or 57432+20 57452 grs. in 9 lbs., 11 oz., 13 dwts., 20 grs. Proof (6 2,0 12 57452—0 I 14363—5 ) 20 grs. 239,3 219—13 dwts. 9 — 11 oz. therefore in 57452 grs., there are 9 lbs., 11 oz., 13 dwts.', 20 grs. RIOUOTION. 105 Ex. XXXVI. (1) Reduce (verifying each result) ; * « 1. £57 to pence ; and 613 guineas to farthings. 2. £15 128. to pence ; and 5000 guineas to pence. 3. £83 159. 6|(f. to farthings ; and £393 0«. ll^d. to half-pence. (2) Find the number of pounds in 5673542 farthings, and prove the truth of the result. (3) Reduce the following, verifying the result in each case : 1. 59 lbs., 7 oz., 14 dwts., 19 grs., to grains ; and 37400157 grs. to lbs. 2. 56332005 scrs. to lbs. Troy ; and 536 lbs. to drams and scruples. i 3. 7 tons, 15 cwt., 2 qrs., 16 lbs. to ounces ; and 7593241 drs. to tons. 4. 5838297 oz. to tons ; and 33 tons, 17 cwt, 3 qrs., 23 lbs., 15 drs. to drams. 5. 171bs., 2 § , 2 3 to grains ; and 34678 grs. Apoth. to oz. Troy. 6. 3 m., 7 fur., 8 po. to yards ; and 573 miles to inches. 7. 1364428 in. to leagues ; and 74 m., 3 fur., 4 yds. to inches. 8. 4 lea., 2 m., 2 in. to barleycorns ; and 50 m., 3 po. to yards. 9. 7 fur., 200 yds. to chains ; and 6 cubits, 1 span to feet. 10. 84 yds., 1 qr. to nails ; and 56 Eng. ells, 1 qr. to nails. 11. 83 Fr. ells, 3 qrs. to nails ; and 73 Fl. ells, 1 qr. to nails. 12. 35 ac, 2 ro. to poles ; and 56 ac., 2 ro. to yards. 13. 3 ro., 37 po., 26 yds. to inches ; and 3 ac, 30 po. to feet. 14. 15 ac, 3 ro. to links ; and 50000 po. to acres. 15. 29 cub. yds. to feet ; and 158279 cub. in. to yards. 16. 17 cub. yds., 1001 cub. in. to inches ; and 26 cub. yds., l9 cub. ft. to inches. 17. 563 gals, to pints ; and 365843 gills to gallons. 18. 5 pipes, 1 hhd., 35 gals, to pints ; and 487634 gills to tierces. 19. 760 bus., 3 pks. to quarts ; and 2 qrs., 1 coomb, 3 pks. to gallons. 20. 3659712 pints to loads ; and 7 Ids., 1 qr., 2 bus. to pecks. 21. 56 reams, 19 quires to sheets; and 52073 sheets of paper to reams. 22. 36 wks., 5d., 17 hrs. to seconds ; and 1 mo. 6i 23 hrs., 59 sec. to seconds. ^ loe ▲BITHIIITIO. (4.) How many barrels, gallons, quarts, and pints are there in 18r6381 half-pints 1 (5) One year being equivalent to 365 days, 6 hours, find how many seconds there are in 27 years, 245 days. (6) From 9 o'clock, p.m., Aug. 5, 1852, to 6 o'clock, a.m., March 3, 1853, how many hours are there, and how many seconds ? (1) In 5972 cords of wood, how many cubic feet, and cord feet ; and in 76267 cubic feet, how many cords. (8) In the United States there are 3,250,000 square miles, in British America there are 2,450,000 square miles, and in the British Islands there are 1 15,000 square miles; how many acres do they collectively contain *? COMPOUND ADDITION. 120. Compound Addition is the method of collecting several numbers of the same kind, but containing different denominations of that kind, into one sum. Bulb. "Arrange the numbers, so that those of the same denomination may be under each other in the same column, and draw a line below them. Add the numbers of the lowest denomi- nation together, and find by reduction how many units of the next higher denomination are contained in this sum. Set down the remainder, if any, under the column just added, and carry the quotient to the next column : proceed thus with all the columns." Ex. 1. Add together £2. 4*. 74<^., £S. 5*. lOirf., £15. 15s., and £33. 12«. U^d, Proceeding by the Rule given above, £ *. d. 2 . 4 3 . 5 15 , 15 . lOj . 33 . 12 . 11^ :54 . 18 . 51 Reason for the above process. The sum of 2 farthings, i farthing, and 2 farthings, =5 farthings, = one penny, and 1 farthing j we therefore put down ^, that is, a COMPOUND ADDITION. 107 one farthing, and carry 1 penny to the column of pence. Then (l + ll+10+7)rf.=29rf.=(12X2+6V. or 2 shillings, and 5 pence ; we therefore put down 5(/., and carry on the 2 to the column of shillings. Then (2+12+ 15 + 5H-4)».=385.=(20xl + 18)«.=£l.andl8i4 we therefore put down 18»., and carry on the 1 pound to the column of pounds. Then ( 1 + 33+ 15 + 3+2) pounds equal £54. Therefore the result is £54. 18*. 5;|^a. Note. The method of proof is the same as that in Simple Addition. Ex. 2. Add together 34 tons, 15 cwt, 1 qr., 14^ lbs. ; 4& tons, 3 cwt., 18^ lbs. ; 18 tons, 19 cwt., 3 qrs. ; 7 cwt., 6| lbs; 2 qrs., 19 lbs. ; and 3 tons, 7^ lbs. tons 34 42 18 3 owta. 15 3 19 7 Ans. 99 • 6 qri. 1 3 2 Iba. 14i H 19 • 15H Proceeding in this case as in ordinary fractions we have (A+?+i+i) lbs. =(A+H+fV+TV)lb8. =Hlbs. =lHlb8. we therefore put down \\ lb. and carry one to the column of lbs, and then proceed as in the former example. X. 0)1 6 5 8 2 «. 7 11 8 d. 6 3 4 8 1. 11 tons. owt. qn. Ibf. (4) 16 . 17 . 2 . 23 13 . 10 . . 20 17 . 15 . 2 . 19 84 . . 3 . 22 11 . 11. 1 .11 Ex. XXXVII. X. B, i, (2) 33 . 16 . 3f 67. 0. 7^ 73 . 19 . 10^ 29. 9. 9i 47 . 16 . 8] OS. dn. Bc. gw. (5) 22 . 3 . 2 . 19^ 56 . . 1 . lOf 3.2.2. 11 15 . 6 . 1 . 9i 79 . 4 . 1 . 10 (3) 528 . 14 . IIA 854. 19. 4 578 . 18 . 9| 507. 0. 04 859 . 14 . 11^ M. ro. po. (6) 82 . 2 . 24J 18 . 3 . 14tV 20 . 1 . 27 56.0. OA 45.3.30 (7) Find the sum £28. 14*. 6|fl?., £27. 18«. 4^-^, £79. 12*. M £19. 18*. lOjd., and £85. 14s. 3^rf. ; also of £678. 10». 2d., £325. m Ito ARnHMino. 6*. 5rf., £487 18#. 9rf., £607. 0«. lU, and je779. 10«. 8A ; also of £i.Us.§ld, H.O'- T}d.;£12.16*.0jc/.;X10.O».0irf.;£1.7«.6|rf, and €14. 1^. T ,e^. ; also of £20. 16*. 5rf|., £14. 16#. Ofrf., £5. 1S# R^lu £. X 19*. Ijrf. %iid £18. 3*. 4^d, and prove the result itx eiftch case. («) Add together 2 lbs., 9 oz., 1 dwt , 28^ grs. ; 8 lbs., 6oz., 4 dwts., 20 2rs. ; 1 lb., 10 oz. 5 dwts., 12f grs. ; 14 lbs., 11 oz., 14 dwls., 19 gra. ; and 2l lbs., 8oz., 18 dwts. 11^ grs. : verify the result. (9) Add U>g«^^her 3 drs., 2 scr., 19 jrs. ; 2 drs., 2scr., 11 grs. ; 7 drs. 17 grs. ; t) drs. 1 scr., 9 grs. ; and 5 drs., 1 scr., 13r''j grs. : explain the process. (10) Find the aggregate of 18 lbs., 14 oz., 6 drs. ; 9 lbs., 6oz., 15 drs. J 46 lbs., 9 oz., 8 drs. ; 9 lbs., 15 oz., 4 drs. ; and 14 lbs., 12 oz., 12 drs. ; also of 1 cwt, 2 qrs., 24 lbs., lOf oz. ; 11 cwt, 18 lbs., 9^ oz. ; 13 cwt, 3 qrs., 17 lbs., 14^ oz.; 7 cwt, 1 qr., 20 lbs., 9 oz. ; and 19 cwt, 2 qrs., 19 lbs., 14 oz. (11) Find the sum of 11 yds., 2 ft., 9 in. ; 27 yds., 1 ft., 3| in. ; 36 yds., 2 ft., 10^ in. ; 48 yds., 2 ft, 11 in.,; and 51 yds., 1 ft., 8^ inJ|; also of 26 m., 7 fur., 23 po., 3 yds. ; 22 m., 5 fur., 27 po., 5i yds. ; 37 m., 4 fur., 3^ yds. ; 86 m., 6 fur., 38 po., 3,^. yds. ; IRd 25 m., 1 fur., 29 po., 2^ yds. (12) Find the sum of 43 yds., 2 qrs., 3 na.; 37 yds., 2 qrs., H na ; 23 yds., 3 qrs., 2 na. ; 41 yds., 2 qrs., 2^ na. ; and 38 yds. 2 qrs. 3 na. ; and of 11 Eng. ells, 2 qrs., 3 na.; 13 Eng. ells, 2 qrs., 14 na. ; 39 Eng. ells, 4 qrs., 2 na. ; 37 Eng. ells, 4 qrs., B^j na. ; and 79 Eng. ells, 3 na. : and prove each result (13) Find the sum of 25 ac, 2 ro., 16 po. ; 30 ac, 2 ro., 25 po. ; 26 ac., 2 ro., 35 po. ; 63 ac, 1 ro., 31 po. ; and 34 ac, 2 ro., 29 po. : also of 5 ac, 2 ro., 15 po., 25a sq. yds., 101 sq. in. ; 9 ac 1 ro., 35 po , 12| sq. yds., 87 sq. in. ; 42 ac, 3 ro., 24 po., 23f sq. yds., 57 sq. in. ; 12 ac, 2 ro., 6 po., 13J sq. yds. , 23 sq. in. ; and 17 ac, 24 po., 30 sq. yds., 113 sq. in. : explain each orocess. (14) Find the sum of 3 c yds.. 23 c. ft., 171 c in. ; 17 c. vos., 17 c ft., 31 c in. ; 28c yds., 26 c ft., 1000 c in. ; an 23 h., 69 m. and 11 mo., 1 w 68 m ; also, 7 yrs., 28 w., 8 s. ; 26 yrs., 5 w. 6 d. ; 58 yrs., 6 d 28 h., 59 s. ; 43 ^,, 23 h., 50 m., 12 si. , and 124 yrs., 14 w., 19 h., 37s. » ». , «iu (17) When ^ was born, A'b ag( was 2 yrs., 9 mo., 3w., 4 d. ; when £was born, ^'s age was 13 yrs., and 8 d. : when D wa^ born, CTs age was 9 mo., 2 w., 8 d., 28 h. ; when S was bom, D's age was 6 yrs., 11 mo., 23 hrs. ; when Fw&s bom, JE"s age was 7 yrs.j 8 w., 5 d., 15 h. What was ^'s age on /"s 5th birui-day ? COMPOUND SUBTRACTION. ,.J^^- Compound Subtraotioh is the method of finding the diflerence between two numbers of the same kind, but containing different denominations of that kind. RuLB. "Place the less number below the greater, so that the numbers of the same denomination may be under each other in the same column, and draw a line below them. Bei in at the right hand, and subtract if possible each number of the )wer line from that which stands above it, and set the remainder underneath But when any number in the lower line is greater than the number above it, add to the upper one as many units of the same denomination as make one unit of the next higher denomination- subtract as before, and carry one to the number of th ^ next ^gher denomination in the lower line ; proceed thus throughout Ex. 1. Subtract £88. 18*. S^d. from £146. 19*. 6\d. Proceeding by the Rule given above, 146 88 19 18 £58 • • 9^ Reason for the above process. Sincere?, is greater than irf.. we add to 4-rf. 4 farthin«« /^^• i penny, thus raising it to 5 farthings : and when 2 fartiiincs^are subtracted from 5 farthings, we have three farthings left- we no ARITHMETIC. therefore place down fd: and in order to increase the lower number equally with the upper number, we add one penny to the o pence. ^ •' Now 9 pence cannot be taken from 6 pence ; we therefore add 12 pence or U to 6 pence, thus raising the latter to 18rf. : we take the »fl?. from 18. and put down the remainder 9d ; then adding 1«. to 18«. the latter becomes 19*. : 19*. taken from 19». leave no remamder : we then subtract 4)88. from £146., as though thev were abstract numbers. It is manifest that in this process whenever we add to the upper line, we also add a number- of the XrJ ^ "" ^^"®' "'' ***** *^® ^""^^ difference is not Q ^""'i^;, ^®"^'*^* ^^^ ^^'•' ^^ ^^-^ 16f dwts., from 144 lbs, 8 oz., 14^ dwts. » lb. 144 . 8 dwts. 14f 106 . 11 . \QfL 87 . 8 . 17f f is greater than ^, therefore we add 1 or f to 1, which makes it y . Now V-f=v_6=^. We must repay the dwt. by adding 1 dwt. to the 10 dwts. Art. (76.) Ex. XXXVIII. £ a. d, (1) 149 • 4 • 63 86 • 13 ' 2^ fa. po. yds; (4) 14 • 34 • 5 1 • 38 • 4 £ $. (2) 309 • 13 • 11^ 119 • 19 • 10^ ae. ro. po. (5) 63 • 1 • 29^ 57 • 2 • 38f owt. qr. lb. OB. (3) 63 • • 18 • 1 58 • 1 • 12 • 10 qrs. bu8. pk, gal. (6) 64 • 3 • 1 • 8 • 5 • 3 • 1 (7) Subtract £456. 15*. llfo?. from £534. 13*. lUd.i and prove the result. ^ ' (8) Find the difference between the following numbers, and verify the results : . ° ' 1. 426 lbs., 8 oz., 1 /^wt., 7grs., and 38S lbs., Soz., 11 dwts., 21 grs. ' 2. 5836 lbs., and 4976 lbs , 7 oz., 13 dwts., 19 grs ?• oni?^' ^,uo«7U. «f5. #*i«. -. iV. \^l 3) 14471bs.l0oz. 4dwts.+1000.(4) 262 tons, 10 cwt.+ 2400 cwt.-T-2400. OOMPOITHD DiyiBION. (5) 26380 mo. 3 w. 6 d.^260(K). '""^ 21 ac., 3 ro., 17 perches x -02; „.„. 24 ac, 3 ro., 10 perches x 112, and x 11-2. 117 [Of Auoov mo. o w. o a.-r»OOUO. }?| oi !^' I !^'* ]1 p®'^!?®" ^ ;^,^; a°? ^^'''s. 3«. x -0507. 127. When the divisor and dividend are both compound numbers of the same kind, Rrlb. «' Reduce both numbers to the same denomination : divide as in Simple Division, and the result will be the answer required. Ex. How often is 6s. 8|rf. contained in £15. 8«. 9rf. ? Proceeding by the above Rule, 5 • 3J 22 63 4 255 £ 16 • 20 318 12 3825 4 18 d. 9 15300 255) 15300 (60 1530 Therefore 60 is the answer. Reason for the above process, 5«. 3ffl?.=255 farthings, £15. 18*. 9rf.=15300 farthings; and 255 farthings subtracted 60 times from 15300 farthines leave no remainder. ® Ex. XLII. £160. 4*. 8|d-r£l. 10*. Ud, £401 4*. 3rf.-r£2. lis. 6\d. 44cwt., 2 qrs., 11 lbs. 4-1 cwt., 2 qrs., 17 lbs. ^loJ?^*' ^ "^""-^ y<^«-' 2 qrs., 1 na. 9487 bus., 2 pks.-r-143 bus. 3 pks. 1416 ac, 2 ro., 16 po.-r-4 ac, 3 ro., 27 po. [8) 617 Ids., 1 qr.-T-12 qrs., 1 pk. m. 118 ▲RITHMino. 128. We shall now add some examples of the Multiplication and Division of numbers, comprising different denominations, but of the same kind, by mixed numbers. In the case of Multiplying by a mixed number, it will generally be found advantageous, first to multiply by the integral part, and then to add to the result thus obtained the result given by multiplying by the fractional part. Thus, for example : Multiply je2. 6«. 8rf. by 8/^. (£2,Qs.Sd.)xS=£7. (£2. 68, Sd.) X 7 £16.6*. 8rf. «, ,„ ^^ io "= 10 — =^^- ^^^' ^"^ Therefore (£2. 6s. Sd.)x3^=£7+£i, 12,. Sd.=^£S. I2s, 8d. In Division it will be found advantageous to reduce the mixed number to an improper fraction. Thus, for example : Divide £89. 17». e^d. by 19J. 19J = V. Now £89. 17,.6K-V=^^5^;^^^^*=£4.n,.0He/. Ex. XLIII. rr (3j (5 (7 8 £40. Us. i. X 57,V 3 ro., 35 po., 27^ yds. x 81f 597 cwt., 2 qrs., 8 lbs. -i- 13|. 571 yds., 2 qrs., 1 na.-T-234. 4 mi., 3 fur., 37 po., 4| yds x 5f REDUCTION OF FRACTIONS. £20. 18«. 2if^d-M2ff 84 tons, 13cwt.,31bs. x 23f. 6491 yrs., 8 mo. -^375yV. 129. To find the value of a fractional part of a numher of one denomination in terms of the same or lower denominations. Rule. Multiply the given number by the numerator of the fraction, and divide the product (if possible) by the denominator ; if there be a remainder, multiply the numerator of the fraction which remains by the number of units connecting the given denomination with the next lower denomination, and divide the product by the denominator ; if there still be a remainder, proceed ». .vs. «T/ «.« »--t- -jiriiiv? TTcejr ao rt mil wic uxs\t i'tjiuaiiiuer, ancl so on till you come to the lowest denomination. The compound number RIDUOTION OP nUOTIONS. 119 formed of the integral pwts reserved from the successive quotients. %1 IfTi' '' ***' »^V'<^"««on, will be the valueTeS mte. If the given number comprise different denominations of the same kind : reduce the different denominations toX lowest denomination involved, and the above rule may be then aVnlild ? or the value may be found by the method shewJ [n Art ft ' Ex. 1, Find the value of -^ofjCl. Proceeding by the Rule given above, 8 8*2 35 and-g-of 1*.=— — (;?.=6(?. ; therefore the value required=17«. 6d. Beason for the above process. —of £1 is the same as 1 times— -Qf £j and-g-of£l=-g~=-; therefore 7 times-|-of XI =7 timesf =^*=i7i..=i75. 6d. ^i+'i of /on,!^' '"^"^ "^ ^ "^ ^^^^^^ "^ ^1+^ ^n off of yOf £15=£~-y-=:£y=iJ4^^ ^__2x20 40 * 7 7 *«=-;y-«.=545., ly *• — 5X12 60 therefore f of £15=^4. 5*. 84c?. 3 _3x20 60 ^T— ~~7~«. = Y». =8|*., 120 ARITHMBTIO. 4 4x12 48 y *. ^—fj — fl?.=yrf. =6f d, therefore £S^=£S. Ss. Q^d. ioffoff ofjei=|of£l. 20 =2^«. — mt S» 6 6x12 72 -y-».=— 7y — a.=-yrf.=10^rf. therefore ^ of 4 of f of £1=2». lO^cf. 2 3 2 2x12 g-of -y-of 1*.= yof ls.= — = — d. 24 therefore required value=£4. 5*. S^d.+£S. Ss. 6^d.+2s. lOid. -{■S^d.=£7. 17s. 6^d. Ex. 3. Find the value of f of a bushel — ^ of a peck. ^ of a bu8.=S|_4 pk3.=|0p^^ _2^ pj^^^^ I pk.= ?|i qts.= ^^ qts. = l^ qts, ; therefore - of a bus. =2 pks., 1| qts., y ^ of a pk. =i^ qts. = ^ qts. =54 n ts.; therefore required value =2 pks., 1^ qts.— 54 qts. =1 pk., 4,V qts. Ex. XLIV. (1) Find the respective values of 1. I of 4*. 7d. ; If of £1. 25. 9d.; ^x H of 21s. ; 1 of | of 9s. lO^d. "" "1 J ^'^ "TO. ^Cc. , yy vi «%,-x. xia. ui*. j |- Ul jy Ol JV/6', U», J y'lj- 01 100 guineas. RlDUOnON OF rRAOTIONB, 19] 128, oa. 4. T^ of a cwt. ; ? of a lb. Avoird. ; | of a mile ; f of an acre. 6. A of a mile ; tV of a day ; | of a yard ; ^ of 3 cwt, 1 qr., 14 lb. 6. 7f of a lb. Avoird. ; If of a lb. Troy ; 2f of a gal. ; 4/^ of an acre. '" 7. 8/y of a hhd. of beer ; ^ of a tun of wine ; 6** of a bus. 8. 2^ of a load ; 3^^ of a cub. yd. ; 9,«, guineas. 9.lofi of lOf hrs. ; £^ ; ^of ^ of a moidore. ^^' ^H ^^ ^^^' ^'' ^^'^' J ^ of M of 12i of ^ of X2 X ^. 11. f of l'cwt.x5|; I off of lib. -rf 12. 19f of 5ac. Iro. Ijpo. ; 2f of 8 yds. 2ft. 2VVin.~8f. (2) Find the values of 1. i of £1+^ of a guinea+3*. 2d. 2. i of £l + f of 2«. 6rf.-f I of U 3. lofXl + f ofU + J^d 4. * of lib. + 1 of loz. + ^ of Idwt. 6. I of 5 acres + * of Sro. + 7 po. 1 + 4 ^' TT" ^f ^tons + W of 4cwts. + 1 of Iqr. 7. J of 3 cords+^ of 3 cord feet. 8. f of 2U.+i off of !£.-} off of 5*. + l of i of U. ^2 4 11 + ^ ^* 3f ^^f=f of 2 guineas+3|-| ^^ ^^• 10. f of a ton+f of a cwf+f lb. 11. f lb. Troy + 1 lb. Troy— | oz. Troy. 12. /^ of a mile--| of a fur.+ A-po. 13. yf,. cub. yds.+2| cub. ft. 14. f of a qr.+f of a bus.— ^ of a qr. \a L'^not'-'F P""-' ^^y^^'-^i of 6 mi., 3 fur., 37 po., 4^ yds. 16. 7| of 365^ d. X 3/^ of f wks.+f of 5| hrs. ^ ' * J^ * ^1 ?^ ^^•' ^ ''^- ^^ po., 2f yds.— I of 6 ac, 2 ro., 17 po. 25A yds. *^ 130. To reduce a number, or a fraction, of any denomination, to a fraction of another denomination. Bulb. « Re.duce the given number, or fraction, and also the number or fraction to the fraction of which it is to be reduced, to 1 1S2 ABITBIfinO. their retpeotive equiTalent ralues in terms of some one and the same denomination : then the fraction of which the former is made the numerator, and the latter the denominator, will be tiie firaotion required." Ex. 1. Reduce Sa. 5d. to the fraction of £1. Proceeding by the above Rule, 3«. 5d.=4l pence, £1=240 pence; therefore fraction required =^|V. JReaton for the above process, £1, or unity, is here divided into 240 equal parts ; and 41 of such parts being takf>n, the part of unity, or £1, which they make up, is represented by ^\. Ex, 2. Reduce f of £1 to the fraction of 27^. fof£l=20 times f of 1«. 5x20 ^ 8 5x5 - 2 '• V. 27«. =275. 5x5 lired 2 5x5 .27 ""27 ~ 2 • 1 5x5 1 25 therefore fraction required ~ 2 ^27~54 For 27*. is divided into 27 equal parts ; and f of £1 is divided into y of such parts ; therefore the part of unity, or 27«., which the V 25 latter represents, '^^^^'^' Ex. 3. What part of | of a ton is 2| of 1^ of i^ of a cwt ? ^ of 1^ of ^ of a cwt.=f of A of 4 of a cwt. — f X I X I cwt. • ^ ofa ton=V cwt. |x^ x| Therefore fraction required = — ■— — 3 =fx|x|x^ ""T"hr' Eioucnov OF nuonovi. Ex. XLV. I» (1) Reduce 1. 6rf. to the frtu3t on of 1*. ; ^ 3*. 4^^ to the fraction of £1. iL £1. 8*. 4d. t(> the fraction ot iJQ. 6#. 8dL j and 2*. 0*rf. to the fraction of 10*. Qd, * 8. £18. 7*. 6d. to the fraction of £2; and 6*. 7#rf. to the fraction of 7*. dd. ^ 4. 3 qrs., 19 lbs. to the fraction of a ton; and 6U lbs. to the fraction of 4 oz. 5. 3 ^rs., 4 lbs. to the fraction of 2 owt ; and 5 oz.. 2| dra. to the fraction of a grain. ,u \^ ^.^•» ^liP^- ^ ^^ fraction of an acre; and 26» sq. yds. to the fraction of 2 acres. • -i ^ ,AA ^^^y^^-y ^ ft., 6 in. to the fraction of a mile : and 6 cub. ft., 100 cub. m. to the fraction of a cubic yard. 8. 2 qrs., 2^ na. to the fraction of an Eng. ell ; and 8 h., 8 m. to the fraction of a day. A « ^''} ?• ^^ *^® fraction of 9 ao., 2 ro. ; and 1540 yds., 2 ft., 9 in. to the fraction of 2 miles. ^ * 10. 5i lbs. to the fraction of a cental ; and 8| feet of wood to traction of a cord. the fraction oVa\t'eL ''"''"" "'' '^' ^'' ' "^' ^ ^^' '* ^^ ^ 1 ^\a '^^^•' ? *^*y^'''' ^'^^ m. to the fraction of a day; and 1 ro. 20 po. to the fraction of an acre. ^ 13. 4 bush., 2^ qts. to the fraction of a load ; and 8 quires 7 sheets to the fraction of a ream. i i ^ Af o^; ^* ?o'?if*'' ^ ^® ^^^^^^'^ of £2^ ; and 2i cwt. to the fraction ot z tons, 12 lbs, to Z friz of J° "" '■""'"°" °^ ^' ■"""'^ ' ■""* *""^ 8uine« JLlon pl4"-iJ^ *' '"^''°" °' '^ *"• ''^ "• »"" H y-''- 1° (2) Reduce fractiL^o^f^/ *° *^^ fraction of £1; and | of a farthing to the fro!;- ^ ^^^o ,^7*-' *° ^® ^''*^*'^° ^^ 1 ^^- ; and ^ of 2 lbs. to the traction of 24 tons. Q 9 of 2^1 «. t ui a 10. TO me iraction of a cwt. ; and | of a yd. to the fraction of a mile. ' j . w uuc 194 ARITHICBTIO. penny of a mile 4. s^-g of £1 to the fraction o the fraction of a yard. 5. f of 4 of lOs, 6d, to the fraction of 2$. 6d, ; and 1 oz. Troy to the fraction of 1 oz. Avoirdupois. 6. ^ of a pole to the fraction of a league ; and 3| furlongs to the fraction of 2| miles. 7. f of 7^ of 16^ yards to the fraction of a furlong ; and 4 of ^ of a guinea to the fraction of 2s* M, 8. f of 16». O^d, to the fraction of 17« 6c?. ; and -^l-g of a lb. Troy to the fraction of a pennyweight. 9. I of a lb. Avoird, to^the fraction of 2 lbs. Troy ; and | of 2*. 6i. to the fraction of £1. lis. Qd, 10. i of French ell to the fraction of a yd. ; and f of a crown to the fraction of ^ of Is, Qd, 11. M of a sq. in. to the fraction of a sq. yd. ; and ^ of a yd. to the faction of an English ell. •036 12. j^Vt 0^ * y®*" *o ***® fraction of a day ; ^ |giy5 to the fraction of a farthing. (3) 1. What part of £9 is ^ of VV of 2s. 6d 1 2. What part of a second is tttuV^t o^ * ^^V ^ 3. What part of f of a league is i of a mile ? 4. What part of ^ guineas is 5| of t«j of £7 1 5. What part of 3 weeks, 4 days, is ^ of 5| sec. ? 6. What part of | of an acre is 25,«i- po. ? 7. What part of 3V of a min. is j^-g of a month of 28 days? 8. What part of ^ of 4 tuns of wine is 3^ hhds. 1 9. What part of 3 fathoms is iSf of ^ of a pole ? 10. Which is the greater, ^V of a day, or 4 of an hour and by how much 1 REDUCTION OF DECIMALS. 131. To reduce a decimal of any denomination to its proper value. Rule. " Multiply the decimal by the number of units connecting UiX^ XX'Cii^V x\.f TTt \jx \A\^-tx\^^iixm»%.x»^£* tt *v»" ^^^^' ^-. » • j j-- decimals as many figures in the product, beginning from the right hand, as there are inures in the given decimal. The figures on RlDXTOnOir OF DlOnCALS. 125 {he left of the decimal point will represent the whole numbers in the next denomination. Proceed in the same way with the decimal part for that denomination, and so on." Ex. 1. Find the value of '0484 of £1, Proceeding by the Rule given above, £ •0484 20 •96805. 12 ll-6160rf. 4__ ^ , , 2-4640y. therefore the value of -0484 of jBI = 1 ld2-4640j. jReason for the above process. 484 •0484of£l=T^of£l. 10000' 9680 116160 d. 10000 10000 ^^'^•■*'Iooo^- =lU+2|,Vffy. Ex. 2. Find the value of 13-3375 acres. Aoreg. 13-3375 4 1-3500 ro. 40 14.0000 po. ' therefore the value is 13 ac, 1 ro., 14 po. i I 199 ARITHMITIO. Ex. 3. Find the value of •972016 of£l. Ist method. '972917 20^ 19-4583405. 12 6-500080rf. 4 therefore the value is 19* S^oT. nearly. 2nd method. •972916 of £l=2!??l±=»^„f.l Art ,97. 875625 / 467 \ "900000°^^^=" \^ 480 ^^^) «. 467 iVb^^. The latter is generally the better course to adopt. Ex. 4. Find the value of;|^ of 3| tons— 3405 of If qrs. =24 cwt., 3 qrs., 18f lbs. __/3402 5 \ ■"\^9990^3^^^ )^^^-» 21x25\ -^f— jibs. =145*^ lbs. therefore the value of the expression =24 cwt., 3 qrs., 18| lbs.-14Albs. ==24 cwt., 3 qrs, 4yWIbs. = 1 ton, 4 cwt., 8 qrV, 4y^lbs. { Bifiuonoir Of inoncAij. m Ex. XLVL j (1) Find the respective values of I 1. -45 of £1 ; -16875 of £1 ; -87708 of £1 ( 2. -875 of a lea. ; 2-5384375 of a day ; -6 of I lb. Trov 3. 6-156510416 of £4; -046875 of 3 qrs., 12 lbs ^' \ 4. -85076 of a cwt. ; -07325 of a cwt. ; -045 of a mile i 5. 4-16525 of a ton ; 3-625 of a cwt. : -05 of an acre ) 6. 3-8343 of a Ib.Troy ; 2-46875 of a qr. ; 4-106 of 3 cwt, 1 or.. 21 lbs. » 1 » 7. 3-8375 of an acre ; 3-5 of 18 gallons. 8. -925 of a furlong ; -34375 of a lunar month. 9. 5 06325 of £100 ; 3 8 of an Eng. ell. 10. 2-25 of 3| acres ; 2-465 of 25 shillings. 11. 1-605 of £3. 2s, 6d, ; 20396 of 1 m^ 530 yds 12. 4-751 of 2 sq. yds., 7 sq., ft. ; 20005 of £63. Ot, 34.rf 18. 2-009943 of 2 miles ; 1005 of 15 centols. * (2) Find the respective values of 1. -383 of £1; -47083 of £1; ' 4694 of lib. Troy. 2. -6746 of 27«. ; -138 of lOs. 6d. ; 2-6 of 5oz. Troy. 3. -142857 of 6 ounces ; 3*2095328 of 3 drams. 4. 4-05 of 1^ sq. yds. ; -163 of 2J miles ; 4-90 of 4 d., 3 hrs. 5. 3-242 of 2i acres ;:22?l?of 2^ of 2-5 days. •5681 A iliof "^*^ *5® difference between -77777 of a pound and 8*. shiint * '^^^'^ ''^^^^ "^^ * P""""^ ^""^ ^"^^^ ^^ * (4) Find the respective values of the following expressions : 1. -68125 of £1 + -375 of 13*. 4ef.+ -605 of £3. 2». 6d. 2. 31 of 5 cwts— 40972 of 5 lbs+2-75 of 5 oz. 3. 18-731 of a mile+17-505 of amile+f of ayard. 4 . 2-81 of 365^ days+5-75 of a week— ;J of 6f hours. 5. ^ of r\ of 3 acres— 2-00875 square yards+-0227 of Si square feet. » 128 AEUHMETIO. ^ (5). Which is the greater, -0231 of a ton, or 'lO of a cwt? 132. To reduce a number or fraction of any denomination^ to the decimal of another denomination. Bulb. " Reduce the given number or fraction, to a fraction of the proposed denomination ; and then reduce this fraction to its equivalent decimal." Ex. 1. Reduce 13*. 6\d. to the decimal of £1. 13». e\d.=lQ2\d=^^^d. £l=240flf.; 649 4 649 therefore the fraction=g|^=QgQ- 960) 64900 (-67 5760 7300 6720 580 We may work such an example as the above more expeditiously by first reducing ^d. to the decimal of a penny, which decimal will be '25, and then reducing 6-25cf. to the decimal of a shilling bj dividing by 12, which decimal will be -520833, and then reducing 13-520833«. lo the decimal of £1 by dividing by 20, which process gives '67604166 as the required decimal of £1. The mode of operation may be shewn thus : 4 12 2,0 100 6-25 13-520843 -6700416 Ex. 2. Reduce 3 bjish., 1 pk. to the decimal of a load : and verify the result. 4 8 5 100 3-25 -40625 -08125 therefore -08125 is the decimal required. % load : and RXOnOTIOSr OF dxoimals, •08125 Id. 6 n W •40625 qrs. 8 3-25000 bush. 4 1-00000 pk. therefore '08125 of a load=3 bus., 1 pk. ♦l.«^i'3f i^i *°g«*h«^f Jf a rood and ^ of 5 perches. Reduce the result to the decimal of an acre : , -, 2X40 f of 1 ro.=— ^perch. =2x8 perch. = 16 perches ; f of 5 perches, =^ perches, = 3^ perches ; therefore the sum=19^ perches. Now to reduce 19f perches to the decimS ofan ^^' acre: 4 4,0 3- 19-75 •49375 therefore the decimal required =-49375 acre. Or it may be worked thus : 19^ perches,= If perches. 4 — —-T-^fcO ac. 4 _79 100 =•49375 ac. Ex. 4. Express the sum of -428571 of £15, { of -Lof | of £1. 12,. and 4 of 3rf., as the decimal of £161' •42857i of £15=4||fii of £15. =^of '£15=£y ' =£6. Qs. Q^d. ; 180 ▲RITBMBTIO. I of i. of I of £1. 12».=i of T^ off of 82#. = V«.=2». 3^rf.; 4of 3rf.=Vrf. = 14c?; therefore the siMn=£6. 89. 64(f.+2«. 3}d.+Hd, =JE6. 11*. 2,0 10 11 6-55 therefore the decimal required = '655. •655 !: ^^^ Ex. XLVn. (1) Reduce 1. 6». 4td. to the decimal of £1 ; and 8*. S^d. to the decimal of £1. 2. 4*. l^d. to the decimal of £l ; and 15*. ll^d. to the decimal of £1. 8. 10*. O^d. to the decimal of £1 ; and 5*. 8f cf. to the decimal of £5. 4. £3. 1 1«. 9f rf. to the decimalof £1 ; and also to the decimal of £2. 10*. 5. 2 oz., 13 dwts. to the decimal of 1 lb. ; and 4 lbs., 2 so. to the decimal of 1 oz. 6. 2 qrs., 21 lbs. to the decimal of 1 ton ; and 3 cwt, 3 oz. to the decimal of 10 cwt. 7. 2 fur., 41 yds. to the decimal of a mile ; 1 fur. 30 po. to the decimal of a league. 8. 2 sq. ft., 73 sq. in. to the decimal of a square yard ; and 3 ro., 20 po. to the decimal of an acre. 9. 14 gals., 2 qts. to the decimal of a barrel ; and 3 qrs., 8 pks. to the decimal of a load. 10. 4 days, 18 hrs. to the decimal of a week; and 11 sec. to the decimal of 5 days. 11. 1^ guineas to the decimal of £1^ ; and 1 lb. Troy to the decimal of 1 lb. Avoirdupois. 12. ^ inches to the decimal of 2^ miles ; and 20^ lbs. to the decimal of 3^ lbs. iA n* —I-.- X. xU. J.-: ,.1 ^.T 01 ^ . »«.J OIVl ~n1a 4-f\ tX%A O. Of pKS. WJ W«3 UOUlUiill Ul Oj qra. ; »UU « « j gaxis. w VIJ.T3 decimal of 1^ qts. MDUOTIOW Of 0URR1H0II8. 181 the decimal of 1 owt., 2f qrs. * * 16. 8 wk^, 5j d. to the decimal of 5^ hrs. j and 1 min., 21 sea to the decimal of ^ of a lunar month. •» *♦ ■•o- ^^* ^ J!^-""" f° r^o®, decimal of 19 sheets ; and 3* acres to the decimal of 3^ sq. yards. •^ wto 17. 33 vds. to the decimal of a mile ; and 7*. 8AMJL<2. tn & the decimal of 10*. Qd, ' »iVWr«. to * (2) Reduce 1. i of 68 to the decimal of 21». ; and 6f cwt. to the decimia 01 a ton. '• ^Mim':ltfft"^^''^ ^" ^ oMi^«n^ t^ yds. to the decimal of ^ of 2 miles; and * of 8^ sq. yds. to the decimal of 2 acres, 1 ro. 4. I of 4| hrs to the decimal of 365^- days ; and 9tV of il pecks to the decimal of 3^ qrs. J^ » *^^ »Tt oi ff 5. 3 lbs., 6 oz. IVoy to the decimal of 10 lbs. Avoird. • and i oz. Avoird. to the decimal of i oz. Troy, ' (3) Express f of a crown+| of a shilling as a decimal of 7# ^(4) Express f of half-a^rown+.4 of a shilling as a decimal* of (5) Add together f of a day, f of an hour and 4 of 6 hoi,r« . and express the result as the decimal of a week * * thiteu? In'^i^""" ^' * ^^ * -^^ ^'i-'- perch a. de^L^'f^tU""'''*'''^^'-' ^^' '^'^'^ ^' -- to the REDUCTION OF CURRENCIES. 133. Pounds, shillings and pence, the denominations of the PnirH.i. ionetary system, are still in use in varinnc ^o.^^" .?- - -^^ - i- • — --«- ^-«ivo Ox tiug v>onnnent. however, m America, has depreciated in monetary The value of comparison with the pound sterling, owing to the great amount of 102 ARITHMETIC. paper circulation. TLls depreciation in va'.ue has been greater in IL districts than in others, ai.d consequent y ^^e P^'^-'f ^^^^ has various values. The respective equ.valents ot tl.e pound m dollars and cents will be seen in tht. followuig table : HALIFAX OR CANADA CURRBNCT. £ I in Canada, and Nova Scoiia • ~'^- ' and*l=5«=£i. HBW YORK CURRENCY. £1 in New York, Ohio, N. Carolina — - =>2i» ! and |1=8«.= £|. NEW ENOLANb CURRENCY. £1 in New England States, Kentucky, Virginia, Tenne88ee=|3i, and|l = 6«.=£TV PENNSYLVANIA CURRENCY. £1 in New Jersey, Pennsylvania, Delaware, Maryland .-=t2|, ^ and |1=7j. 6i -»^|. GEORGIA CURRENCY. £1 in South Carolina and Gfeorgia -Wf» ^- *^ and $1=45. 8(f.=£5V The value of the pound sterling in Canada is $4.87 or in Canada currency £1. 4.. 4^: Its value in the United Sutes is fixed by the Stelt 14.84! but its real value as compared with the gold corns of that country is 14*866. ^^ 134 The following table shows the value of the principal foreign coisindoltoandlents, as fixed by the Senate of the Umted States: 1 Pound Sterling or Sovereign ,$ 4-84 1 Guinea^ English J'JJ 1 Crown « *oq 1 Shilling piece "--J^g 1 Franc, French "• ^ 1 Doubloon, Mexico ^?-J" 1 Specie Dollar of Sweden and Norway a • "» 1 Specie DoUar of Denmark. ^-"J 1, Thaler of Prussia, andN. States of Germany... 0.69 1 -CT^-i^/xf A«atriftT»TSTnmraandCitVOf AuRSDurg. W.«JO Ducat 1 Ounce' of Sicily ...•..•.*.• .......«..' 2.4S> 135. To reduce any Currency to dollar e and centt. Rdlb. Express the sum given, dt-cimally, in the principal unit of the curr^n-y, and iiwltiply by the dollars and cents equivalent to that uuiL ' < > .V ? i,^. r Ex. 1. Riduce £5. 12*. 6rf. sterling to dollars and cents. Proceeding by the above rule : £^. Ue 6fl?.=£6625 (Art. 132 ) 5 025 487 39375 45000 22500 127 393'/ 5 therefore £5. 12*. 6rf.=|27- 393 + Beason for the above process. Since £i=$4'87, therefore £5 625=|(5-626x4'87) = $27-39375. Note,-— To reduce dollars and cents to any other currency it Wi of course simply be necessary to divide by the number of dollars and cents contained in the principal unit of thatourfen^y. Ex. 2. Reduce,$75679'80 into sterling money ; lOw into Fwnch 75679-80-r4-87 =£15540 therefore $75679-80=£(5540. again 1 Franc=-lS6 ' ' ' ^' * therefore $76679-80-r-^^6=No. of Franjjs An J7M7$-80 =40Q$80/r Francs. ^ Ex. XLVin. l)Jleduce 1;he following Sterling money to ^olla)rs and centi ; irst takmg the pound sterling at |4:87, and secon,dly atj^||4!8^. 1. £i7f,and€n5. 10*. ^. • £618. 15^*., ind £459^ l^. 6|i :'.\t '" •t^ ARXTBMZTXO. (2) Reduce to sterling money at I487, to the pound sterling. 1. $84,738 and 18484.485 2 1369 02425, 1923890 and $796076f . s! X18. Hi. 3rf. ; and £196. 15f. lOld. Canada currency. (8) Beduce : 1. $7928-04 to Canada currency. 2. $8037 06 to New York currency. 8. $684-125 to New England currency. (4) Reduce to dollars and cents. 1. je85. 17». lOrf., Canada Currency. 2. £^. 19«. 3(?., New York Currency. 3! £15. 7». Qd., Georgia Currency. (5) Reduce, 1. 715 francs to dollars and cents. 2* 18-6 Austrian Florins to Canada Currency. 8. 189-5 Mexican Doubloons to dollars and cents, and also to sterling money. (6) Find the value, in dollars and cents, of 1 £75 sterlings £16. lis. 3rf., Canada currency + £18. 5». Georgian currenc/+£1087. 7.. 6rf., New York Currency. " 2. 71 guineas + 15-5 Prussian Thaler8-2-8 Crowns. 3. $71687-03 + £\^ Canada Currency + IH Danish dollars-£4i sterling + ^ of ^ of f of $0-05. m What would be gained by changing £198. 17^5. sterling in Canada, instead of the United States'? (8) Which is the greater 987^ Mexican Doubloons, or $16692-50, and by how much? (9) Find the value in dollars of W of £1 Bterling +_^/[^l ^^ Canada Currency+xV of £1 New York Currency + £tV of 1 Danish Thaler + £j\ of 1 Ducat of Naples. (10) Find the difference between f of £25 Georgia Currency, and I of £25 Canada Currency. FRACTIOI. 196 PRACTICE. 136. p«r. An Aliquot Part of a number is such a part as. when taken a certain number of times, will exactly make up that 18 &T ' ^ '"* *"" ^''^^^^^ ^*'"* of 12, 6 an aliquot part of Practice is a compendious mode of finding the value of any number of articles by means of Aliquot Parts, when the value of an unit of any denomination is given. exlm^ r^* ^^^ Practice will be easily shown by the following iJjt;^'"^ ^^^ ''*'"® ^^^^ ^^*^-' 3 V»'> 5 lbs. of Sugar at 156.75 per cwt. 6 »« The method of working such an example is the following : 1 cwt. costs $56.75 .-. 84 cwts. cost $ (5675 x 84) =14767-00. 2 qrs.=| of 1 cwt. .-. the cost of 2 qrs. =i the cost of 1 cwt. =1 28-376. 1 qr.=J of 2 qrs. .•. the cost of 1 qr. =^ the ccst of 2 qrs. =| 14'1875. 5 lbs.=j of 1 qr. .•. the cost of 5 lbs. =} the cost of 1 qr. =$ 2-8375, therefore by adding up the vertical columns, the cost of 84 cwts. 3 qrs. 5 lbs. =$4812*39. The operation is usually written thus : 2 qrs.=^ of 1 cwt. 1 qr.=^ of 2 qrs. 61bs.=^oflqr. 156*75 = value of 1 cwt 84 22700 45400 4767-00 =value of 84 cwt. 28-375 =vaIueof2qrs. 14-l875=valueof 1 qr. 2-8375= value of 5 !bs. I48l2-3U00= value of 84 cwts. 3 qrs. 5 lbs. ik iW ARffsmxV. Note, — It will generally be found more oonvenlent to expreH 'rt. and lbs. in the denomination of lbs. and take aliquot parts of le cwta., thus: 2 qr». 15 lbs. =66 lbs. of which 501b8. =iof a cwt.,10 lbs. = ^of 501b8.,and 81b8.=r|of lOlbi. 137. The following are examples of Practice in Canada Currency : Ex. 1. Find the value of 3825 things at £2. 17«. ^d. each. The operation is as follows : 10#. = I of £1 6*. = I of 10». . 2*. = i of 10». (.-.Ukeiof £1912. 10«.) 4rf. ^ i of 2a. i 3. Find the value of 72 cwts., 3 qrs., 7 lbs., of sugar, at 135-60 per cwt. ^ ° ' -4. Find the value of 9 yds., 2 ft., 10 in., at $1-75 per yard. ^4 6. Find the value of 15 oz., 6 dwts., 17 grs., at |1'25 per oz. 6. Find the value of 7 cwts., 1 qr., 15^ lbs., at $12-37 per cwt. ^^if 7. What will, the rent of a farm, consisting of 196 acres 3 roods, 30 poles, amount to at $4*84 per acre ? ' 8. What would be the cost of putting up 230 rods, 24 yards of fencing at |0-75 per rod ? ' * ^ 9. Find the cost of 18 tons, 16 cwt., 3 qrs., 16 lbs., at ft378-63 per ton. 10. Suppose a steamship require 45 tons, 15 cwts., 2 qrs. of coal per day ; what would she cost a year for fuel, the coal beinc f 4*50 per ton ? '6 11. Find the value of 48 sq. yds., 8 sq. ft., 114 sq. in., at I3-381 per sq. yd. and of 139 cords 25 feet of wood at $3-25 pei cord. 12. Find the cost o ,o7 mings at £1. 18s. Qld. each ; and of 30 cwt., 2 qrs., 14 lbs., at £l. 17*. S^d. SQUARE AND CUBIC MEASURE. CROSS MULTIPLICATION, DUODECIMALS. 138. Definitions : (1) A Parallelogram is a four-sided figure, of which the opposite sides are parallel. , (2) A Kectangle is a right-angled parallelogram. (3) m Area of a figure is the c^uantity of surface contained m It : and is estimatAH niiTr>oT>inoiiTT u,^ «u» i /• -. parts of a time it contams a certain fixed area, which is assumed for its measuring unit. mm m ARTrHMXSIO, (4) A Solid is that which hath length, breadth, and thickness. (5) The Cafaoitt, or Volums of a solid, is the quantity of space, comprehending length, breadth, and thickness, which it contains or takes up. (6) The word Content is also frequently used to denote length, area and capacity or volume ; the length of a line being called its linear content ; the area of a figure, its superficial content ; and the capacity or volume of a body, or of a portion of space, com- prehending length, breadth and thickness, its 3olid content. (7) A Parallblopipbd is a solid contained by six quadrilateral figures, whereof qycij opposite two are parallel. (8) A RxoTANOXTLAR Parallblopipbd is one in which the several angles of the quadrilateral figures, which contain it, are right angles. 189. By reference to the Tables, Arts. (109, 110), and the observations upon them, we see that, in the sense there indicated, length multiplied by length, produces area, and area multiplied by le^th produces capacity ; the units in the products in these cases difierins in kind from the units in the factors ; thus, a rectangular ai ea, whose adjacent sides are 4 and 3 feet respectively, is divisible into (4 X 3) or 12 equal squares, as shewn by the accompanying figure, the length of a side of each square 12 3 4 being one linear foot. The rectangular area in tins case is said to be the product of the two adjacent sides, represented respectively by numbers, the units in the numerical pro- duct being no longer linear feet, but square feet. Similarly, if the adjacent edges of a rectangular parallelopiped be 3, 4, and 2 feet, respectively, the capacity of the solid is equivalent to 24 cubes, each containing one cubic foot ; and thus the capacity of the parallelopiped is correctly expressed by the product of the three adjacent edges represented respectively by numbers, the units in the numerical product being no longer linear feet, as in the factors, but cubic feet. Perhaps the readiest way of working examples in square and cubic measure is that of reducing all the different denominations to the same denomination ; and proceeding as in the examples subjoined. 1 2 3 1 5 9 2 6 10 3 7 11 4 8 12 of 21 SQUARB AOT> OUBIO lIKiSUBB. 139 d thickness. quantity of s, which it note length, ig called its mient; and jpace, com- uadrilateral which the tain it, are 0), and the e indicated, altiplied by these cases rectangular , is divisible iompanying 3 4 ~3~ 7 11 4 8 12 larly, if the and 2 feet, io 24 cubes, city of the ►f the three the units in I the factors, square and ainations to B examples Ex. 1 . Find the area of a rectangular court-yard, 17 ft. 6 in. lomr, and 13 ft. 4 m. broad. ^* The area=(17 ft. 6 in.) x (13 ft. 4 in.) =17^ft.xl8ift. 700 ^ =— sq.ft. =238^ sq.ft. =25 sq. yds., 8 sq. ft., 48 sq. in. Ex. 2. Fmd the expense of paving a floor, whose length is 33 ft. 2 in., and breadth 18 ft., at $1-50 per square yard. Area of floor=(33 ft. 2 in.) x 18 ft. =83^ ft. X 18 ft;. = I -^ X 18 I sq.ft. j sq. yds. / 199 18 = (t"^9 199x2 6 sq. yds. Therefore cost, which=cost per yard x number of yardi is= J 1.50x1?!^) =$99.50. r^f^^'J' ^^\^^y square yards, feet and inches will remain out of 400 square feet of carpeting, after coveruig the floor of a room ^1 ft. 9 m. long and 16 ft. 11 in. broad? Area of the floor=(2l^ x 16|^)sq. fL ^/87 203\ . 29 X 203 4x4 sq. ft. . 5887 " 16 sq. ft. 140 I . ABITHMETIO. Therefore the number of square feet of carpet remaining after covering the floor=400— 5?§Z 6400—5887 16 513 ■"16 =32 sq. ft., 9 sq, in. =3 sq. yds., 5 sq. ft., 9 sq. in. Ex. 4. Find the capacity of a cube, of which each edge is 1 ft. Sin. Capacity =len^th x breadth x height, =(l|xlf xl|)cubicft. \3 3 3/ =— — cub. ft. 27 =4 cub. ft., 1088 cub. in. Ex. 5. A block of marble is 2 yds. 1 ft. 3 in. long, 1 ft. 7 in. broad, and 2 ft. thick ; find its solid content, and its value at |2'25 per cub. ft. Its content=(2 yds. 1 ft. 3 in.) x (1 ft. 7 in.) x 2 ft. ^ =7ift.xl/,ft.x2ft. =(7jxlTVx2)cub. ft. =(t^8^0^"^-^- _ 29xl9 ~ 2 X 12 And $2-25=$2^= cab. ft. =22 cub. ft.: 1656 cub. in. Therefore its value= r I 2£ y ??iil? \ _1^_*ri .ar«i ^V* 2x12/ "32""'"*""^ iQvuti AHD cDBio mistno. 141 &.fl fi. ^""», •?*" feet multiplied by linear feet give square feet, It follows that square feet divided by linear feet give lin^ feet Similarly, square yards divided by'linear yards |ive W yards and so on Again, since linear tU mul«plied V Bq"S^ feet give cubic feet, it follows that cubic feet divided ^iS i.^'^?* \ ^^ '""^ *'°°* ■* concerned, length and breadth or W "^^^je^^ O' '"•«»d* ""l height, have to be mSl^ togetter ; but never length and breadth and height to be So fed ^tZ: ' ^^ """y *"'^ P'""* ^^' soUd 3tt .J'%\ ^^^ ?* 'i'P'T °^ '»T>e' , 1043 ' 36 • =|28ff Ex. 7. What must be the length of a beam, the end of which is II ±^'L'l?!lVt''^ -1^^ -",*-t may'be the same asThat 4'¥Tb:3T9 'i;7ardT2 ft: iTiL^'r '"^"^ — P-tively ^* 142 ABITHMBTIO. Content of 1st beam=(length of beam in linear ft. x H x H) cub. ft. 2nd..-.=(4i x3| X 12f ) cub. ft. By the question, .-. (length of beam in linear ft. x 1^ x li) cub. ft.=(4i x 3f + 12|) cub. ft. ; .-. ('length of beam in linear ft. X J. x±Vub. ft (-y=7 in. Then 9 ft. x 7 ft., or rect. J-ff,^ (7= rectangular figure ACJSB, which contains 63 sq. ft., and 9 ft. x 7 primes, or rect.-45,^2)=rectangular ^ E figure ADFB, which is y^th part of 63 sq. ft. t\Zl\rfV^^''l ^'^ }^ ^'' ^^ ^^' each^Ji), it follows that SureTc^A "^^"^ ^"''''' '"'^ = ^^^^ i» rectangular Ex. 2. Multiply 17^ft. 8 in. 6 pts. by 12 ft. 6 in. 3 pts. 17 12 8' 6 6' 3 207 8 6 . 7 . 9V 4.3, 0" 10" 6 »f 216 . 6 6 Hf . 10" . 10 . 6 \ / fi (\ =216sq.ft.+ I To+,-T7+T7; +- ^ ^ \^ 12^ 144^ 12 X 144^ 12 x 12 xlii =216sq.ft.+ (l44+12-^+i2xl|^^^ 10 =216sq.ft.+72sq.in.+^ Igsq. in.+j^q. in, ■ i i mi'i w u i Mi ' p ' f i I f : I 146 JLRITHMITIO. Ex. 8. Find by cross multiplication the capacity of a cube, whose edge is 2 ft. 8 in. ; and prove the truth of the result by vulgar fractions. * ft. • 2 . 8' 2 . 8 5.4 1.9.4" 7 . 1 2 . 8 14 . 2 . 8 4 . 8 . 10 . 8" 18 .11 a8cub.ft.+ (Tj^+ 6 . 8" 6 . 8 \12^ 144^ 1728 ^jcub. ft. ,o , ,, .1584+72+8^ , A = 18 cub. ft. + , cub. ft. =181111 cub. ft. =18 cub. ft. 1664 cub. in. Proof by vulgar fractions. Content= (2| x 2| x 2|) cub. ft. /8 8 8\ , - __512 ~\3 ^ 3 ^3/ ft-= 27 cnb. ft. =18 cub. ft. 1664 cub. in. 141. In the examples of Cross Multiplication we see that a decimal and duodecimal scale of notation is employed, the figures of the feet being expressed and multiplied in the ordinary way ; whereas in other places the number 12 is always used instead of 10- Cross Multiplication is not, therefore, properly termed Duodecimal Multiplication or Duodecimals ; because, although the different denominations are connected with each other by the ^„^v^^r 1*>. atlll thp. differfint diffits of those denominations are connected with each other by the number 10. L SQUARK AND OUBIO MBASURK. 147 Ex. L. I 1. Find the area of a rectangular board, whose sides are 2 ft. 9 in. and 10 fl. 4 in. respectively. V 2. A room is 7 ft. 3 in. long, and 13 ft. 10 in. broad ; find the area of the floor in feet and inches. A 3. Find the number of square feet and inches in a rectangular piece of ground 9 ft. 3 in. by 3 ft. 5 in. V 4. The floor of a room, which is 15^ ft. wide, contains 91 sq. yards ; find the length of the room. \ 5. A rectangular plot of ground 26 ft. broad, contains 92 sq yds. 4 sq. ft. ; find its length. ^' ;^ 6. Find the breadth of a room, whose length is 224 ft. and whose area is 397^ ft. 7C ^f?-. How many planks 12 ft. 6 in. long, and 8i in. wide, will floor a room 50 ft. by 16 ft. ? >8. Find the area of a square building, whose side is 26 yds 5 in. ^ * 9. An area, measuring 30 ft. 6 in. by 8 ft. 9 in., is to be paved • what will it cost at the rate of $1-13 per square foot % ' *< 10. Find the cost of a slab 5 ft. 7 in. long, and 3 ft. 8 in. broad at $0-75 per sq. ft. * 11. Find the area of a floor which measures 18 ft. 6 in. by 12 ft. 3 in., and the expense of carpeting it at $0'80 per sq. yd. 12. What will be the expense of painting the surfaces which measure respectively as follows % ' n 23 ft. 6 in. by 20 ft., at $1-25 per sq. yd. (2) 14 ft. 3 in. by 11 ft. 11 in., at $M0 per sq. ft. (3) 13 ft. 6 in. by 8 ft. 9 in., at $2-05 per sq. yd. 13. Work by Cross Multiplication each of the following examples and prove the truth of each result by Vulgar Fractions : ^ » [1) 18 ft. 9 in. X by 14 ft. 7 in. (2) 23 ft. 8 in. x by 16 ft. 9 in. (8) 27 ft. 6'. ^\ X by 5 ft. 3'. (4) 22 ft. 8^ in. x 16 ft. 7^ in. " 4 ft. 6'._5*. X by 9 ft. 4'. 7". m\ 7.'i ft. 71 in yKv 5lft ft »— ... . . . ~j .j.^, nf. «i in (7) 5 yds. 2* ft. 2 in. 3 pts. x by 5 yds. 11 in. 7 pts, 148 llRITHMXTIO. I i 14. How many yards of carpet ^ yd. wide will cover a room 40 ft. 8 in. by 24 ft. 6 in. 15. What length of paper f of a yard wide will be required to cover a wall 15 ft. 8 in. long by 11 ft. 8 in. high i 16. Find the cost of a carpet f yard wide at $0*83^ a yard, for a room 20 feet by 1 8. 17. Find the expense of carpeting the following rooms : (1) 12 ft. 4 in. long, and 12 ft. 6 in. broad, with carpet | yd, wide, at $2*75 a yard. (2) 29| ft. long, and 14| ft. broad, with carpet | yd. wide, at $1*50 a yard. (3) 15 ft. 6 in. long, and 12 ft. in. broad, with carpet 24 in. wide, at $1*85 a yard. (4) 26 J ft. long, and 18 ft. broad, with carpet £ yd. broad, at 10-87^ a yard. (5) 19 ft 7 in. long, and 18 ft. 11 in. broad, with carpet 25 in. broad, at $1*25 a yard. 18. Find the content, and (when required) the cost, of the following : (1) A piece of timber, whose length, breadth and thickness are respectively 54^ ft., 5 ft., and 2 ft. 5 inches, at $0.03 a solid foot. !2) A cube, whose edge is 1 ft. 8 in., at $0*12 a solid inch. 8) Digging a cubical cellar, whose length is 12 ft., at |0'25 a solid yard. [4) A cistern 6 ft. deep, having a square bottom, of which each side is 2^ ft. [5) A wall 1000 ft. long, 10^ ft. high, and 2 ft. 1^ in. thick. [6) A cube, whose edge is 13 ft. 7'. 1", 19. Find the number of feet and inches in the floor, and the number of cubic feet and inches in the volume of a room 23 ft. 10 in. long, 18 ft. 4 in. broad, and 11 ft. 3 in. high. 20. Find the length of paper, |^ths of a yard wide, required to cover the walls of a room, whose length is 27 ft. 5 in., breadth ^ 14 ft. 7 in., and height 12 ft. 10 in. 21. What would be the cost of painting the four walls of a room, wuose iuugui is tM: lu o lu^f ureoui/U i«i i\i, c xu., mux msi^ikh 11 ft. 6 in., at $ MO a square foot? i! •QUARS AMD CUBIC MBABltRX. 149 r walls of a 22. Find the expense of painting the walls and ceilings of each of the first two, and the walls of each of the last two of the fol- lowing rooms: (1) A room whose length is 16 ft. 8 in., breadth 15 ft. 9 in and height 14 ft., at $0-30 a sq. yd. '* (2) One whose length is 15 ft., breadth 10 ft;., and heicht 9 ft;. 9 m., at $0-37^ a sq. yd. * (3) One whose circuit is 41^ ft., and height 8 ft. 5 in., at I0*36 a sq. yd. (4) One whose circuit is 72 ft., and height 10^ ft., at 10-25 a sq. yd. And find also the expense of papering the walls of the first two of the abov rooms with paper 1 ft. 9 in. wide at the following prices—the first at $1.00 a yard, and the second at |0-37^ a yard. 23. The length, breadth and height of a room are 7 yds. 1 ft 3 in 5 yds, 2 ft. 9 in and 4 yds. 6 in., respectively. What length of paper two feet broad will be required to cover the walls, and what will It cost at $0-05 per yard 1 , « wuau *,% Supposing the cost of a carpet in a room 25 feet long at f 1.50 a square yard, to be $37-50, determine th^ breadth of the room. 25. In a rectangular court, which measures 96 ft. by 84 ft there are four rectangular grass plots, measuring each 22i ft, by 'i8 ft • find the cost of paving the remaining part of the court at $0'17 De- square yard. ^ 26. If a piece of cloth be 94^ yds. long, and H yds. broad, how Droad IS a piece of the same content whose length is 74a yds J 27. How many square feet and square inches remain out of 313 sq. ft. of carpeting, after covering a room 16 ft. 9 in. by 12 ft !f? ^^ ^^® ^^^^® ^^ ^^^ requisite carpeting at 10-75 a 28. On laying down a bowling-green with sods 2 ft. 6 in. lone by 9 in. wide, it is found that it requires 75 sods to form one strip extending the whole length of the green, and that a man can lay down one strip and a quarter each day : find the space laid down m 8 days. o-*^- ^.?'^^^°' ^*^*^'^'^^®^ ^®"g^^ is i^l y how ^any bushels will be required for the consumption of 979 men for three months and a half? md also centimes nc=100 ^ntimes. d to one jvereigns overeign. id. b.Troy perform a srform the ays, I3f)days, ~ — days, on of 979 mSOBLLANEOUS BXAMPLBS WORKED OUT. 155 858 men in 6 months consume 234 bushels, 234 .*. 1 man in 1 month consumes^^s — sbus.. .•. 979 men m 1 month consume ^.-^ "ft" bus., .Q^Q • Q, .u / 979x234 7\ .-.979 men m S^months consume I x - J bus.or 155f bus. Ex. 10. If 5 men or 7 women can do a piece of work in 87 days ; in what time will 7 men and 5 women do a piece of work twice as great ? 5 men=7 women, 7 .*. 1 man =-g- woman, 49 .*. 7 men=^women, / 49 \ 74 .'. 7 men and 5 women = I ■5~+5 ) women =-r- women. Now by the question, 7 women in 37 days do the piece of work, .'. 1 woman in (37 x 7) days does . 37x7^ .*. 74 women m ^. days do.. 74 . 37x7x5^ .*. -g-women m — =g — days do 74 . 37x7x5x2 .*. -g-women m ;^ or in 35 days do twice as much. Ex. 11. A bankrupt owes three creditors A^ £, and CI250, 1330, and $420 respectively, and his property is worth $125 ; how* much will each creditor receive, and how many cents in the dollar? Debts amount to $(250 + 330 + 420), or $1000. If the bankrupt has $1, he pays_i_ part of his debt, $125 i^ part of debt. lAAA * * vvv part of debt * 156 ARITHMETIC. .'. A gets 131.25., B gets $41.25., and C gets $52.50. He pays \ of $1., or $-125., in the dollar. Ex. 12. Gunpowder being composed of nitre 15 parts, charcoal 8 Pftrts, and sulphur 2 parts ; find how much of each is required fornie cwt. of powder. The whole number of parts=(15-f 3+2)=20. Of every 20 parts, 15 3 3 2 1 _or —is nitre, — is charcoal, — or— is sulphur. 20 4 20 20 10 ^ 3 .'.--of 16 cwt., or 12 cwt.=quantity of nitre required, rrtof 16 cwt., or 2f cwt.= ... charcoal •TT^of 16 cwt., or 14 cwt.= sulphur 10 •' ^ Ex. 13. Of a certain dynasty, \ of the kings are of the same name, \ of another, ^ of a third, and -^ o{ & fourth, and there are 5 besides : how many are there of each name ? Representing the whole dynasty by unity, or 1. — =number of kings of one name, o J_— ,. of a second 4 2-= ..of a third ._-. is- of a fourth .--. 12 ^^^i+T'^i'*"ii"S* .-. whole dynasty-l?, or l-^,or i == no. of remaining kings in it « He pays charcoal required red. the same there are UBgtinit MI8C1LIAH10U8 BZAlfPLlS WORKXD OUT. l^ But by the question, 84 °^ "^^^y* o^^of the whole dyna8ty=5 ; .*. 1, or the whole dynasty,=5 x ~ =24 • ■■■ "^InTi of M "^ ''' """"• " °' '^ '^^' « °^ ** 3"^ J ■^** ^1* ^^ *^° ^^ * P^®*'® <^^ ^ork in 5 days, J5 can do it in 6 days, and can do it in 7 days ; in what lime will A £ and f r1-"^ '!' 'Kl""^^ ^ ^^'^ ^ ^i'^d -1«° i« what time i and ^ working together, A and (7 together, and ^and <7tog1ther could respectively finish it & » ^" ^ logecner, Representing the work by unity or !. In one day A does -. part of the work, j^does. 6 ' .-.-.... ..Cdoes— 7 » .-. time in which Ai-~£i- C would finish the work =J3!yital of $"15 1 6. What is the difteif^.nce between simple ana compound Practice 1 Required the pri> ol cwt., 3 qrs 16 lbs. a^ jA. 6s. Old. per quarter, by practice. 7. Determine the expense of papering a room 12 ft. high, measuring 20 ft. by 15 ft., at the rate of $'05 per squar- /arrj. 8. In the civil year, 97 days are intercalated into 400 years ; what is the average length of the year 1 irice of 7| tat $1-75, f his rent"? B can do it ^ can do it. , whilst 12 I will they currencies respective sper ncre; I. of spices d Multipli- ;he suii) of ig $4-87 to tions ; and I chain .f jxpre.*^ the quarts, and /e^ id a scliool 3 $90 in 7 ►f r-l5? compound i«. Old. per 2 ft. high, ar /ard. 400 years ; Mli LLANBOUS QUI8TI0NS AND EXAMPLES. 163 9. If 15 horscs, and 148 sheep can be kept 9 days for $75, what sum will keep 10 horses and m. sheep for 8 days, supposing 6 horses to eat as much as 84 sheep ? J i tt 6 ^^ 10.^^, and C are three workmen : A can do half a piece of work m 3 hours, being twice as much as B can do ; and A,£ and C7 can together do.the whole in 2^ hours. Shew that C can do in 5 hours as much as JB can do in 9 hours. 11. Find the sum and difference of 7 tons, 13 cwtb. 2 qrs 15 lbs 8oz., and 9 tons, 3 cwts. and divide the sum by 95. 12. What are the different uses to which Troy weight and Avoirdupois weight are respectively applied ? Express 56 lbs. Avoirdupois m lbs. dec, Troy. ^ IV. 1. Expluiii how whole numbers are represented in the decimal or common system of notation. Multiply 729 by 37, and explain the process. » ^ r " 2. Add together the fifth of a shilling, two-sevenths of a crown of £25 ''""^'^ ^ ^"^'^^^ ' ^^^ ^^^^^^ ^^® ^^^"^* *° ^^® decimal 3. Two persons gained in trade $375 ; one having put in $500 rrceiveV ' ""^"^ ^"'^ ""^^^^ P'°^^ ^"^^^ '^'^ ^'''^^^ 4. Taking the circumference of a circle at 3| times its diameter hnd the cost of a marble column of two feet breadth and five vards height, marble being $5^ per cub. ft. (Area of circle = 4 circumference x semi-diameter.) * lo^i ^^^ certain number of men can throw up an entrenchment in 12 days hen the day is 6 hours ,_ug, in what time will they do It when the 'ay is 8 hours long? ■' 6. Boug. 3 loads of wood : the first was 8 ft. long, 4 ft. wide and 3 f . h.gh : the second 7 ft. long, 4 ft. wide, and 2 ft. high ' the thud was I ft ong, 3 ft. wide, and 3 ft. high. How mfn^ solidft mt- whole? How many cord ft., and ho manjTco^ds? 7 Reduce $2375^ -' -edish dollars to dollars and cents the exchange, •ngatll-Oo. rdoUar. Andfindthevalueofl 000 000 rupees at )3i ea h. ?vwv,vuu 8. xi.e !_ iicr- used for rolling a bowling-green, being 6ft d ^n m circi mference, by 2 ft. 3 in. wide, is observed to make 12 • 164 ▲BITHMETIO. revolutions as it rolls from one extremity of the green to the other ; find the area rolled when the roller has passed 10 times the whole length of it. 9. Divide $1400 among '-4, B and (7, in such a manner that as often as A gets |5, B shall get $4, and as oflon as B gets $3, C shall get $2. 10. A fr idulent wine-merchant sells as brandy a mixture of brandy and rum at $10^ a gallon, which is the proper price of his brandy, that of his rum being $5^ a gallon. Supposing one-third of the whole mixture to be rum, ascertain how much a gallon he gains by his dishonesty. 1 ] . How many revolutions will a wheel, which is 4 yards in circumference, make in 3 miles'? 12. Find the difference between £1000 Canada currency, and £1000 Georgia currency ; also between £596^ sterling and £596j^ Canada currency. Express the answers in dollars and cents. V. 1. Divide 550974 by 1472 ; find the quotient and remainder. Explain the operation, and prove the result. 2. Shew that the value of a fraction is not altered by multiplying the numerator and denominator by the same number. 3. Express the fractions -^^ -f, and /^ by corresponding fractions having the same denominator, and find the sum. 4. Eight bushels of wheat are consumed annually by each person in Canada; if wheat.be at $1*25 a bushel, and the population 2,500,000, what is the value of a quarter of a year's consumption"? 5. A certain number of men mow 4 acres of grass in 3 hours ; and a certain number of others mow 8 acres in 5 hours : how long will they be mowing 1 1 acres, if all work together *? 6. The depth of water in a cistern whose base contains 1344 sq. in. is 3 fl. 9 in. Find (1) the number of cub. f.. of water in it, and (2) the depth of the same quantity of water in another cistern whose base contains 1088 square inches. 7. If a main can do a piece of work in 8^ days by working 6 hours a day, how many hours a day must he work to finish it in 5 days ? XL t .U^v:ii XJi. X J. 1 1 amsu a piece of work in 17 days, how many days will it take 11 men and 7 women to finish it? the other ; I the whole ner that as gets $3, C mixture of price of his g one-third I gallon he 4 yards in rrency, and and £596^ cents. remainder. nultiplying ig fractions Bach person population asumption? hours; and w long will QS 1344 sq. water in it, ther cistern working 6 finish it in III \i uays, nish it % MISCELLANEOUS QUESTIOIfg AND EXAMPLES. 1^5 »,il* ^r^"^ 'f ^^f ^ ^°- ^^"8 ^y ^S ft- « in. Wide, an rt high ; It has two doors, each 8 ft. high by 3 ft. 9 in wide L\\ windows, one 5 ft. by 7 ft., the other two 5^ft by 4 I each Wha^ wjn^cost to paper the room with paper a yar'd wide at 10 cen J 10. A loaded truck weighs 4 tons, 3 qrs. 1 lb. the truck itself weighs a ton and a half, and it contains 758 equal packles; find the weight of each package. ^ ^ ' \u }}'• ^^Ujan^. Shakspear was born April 23rd, 1564. How lone IS It since his birth to January )st, 1860. 1 ^ firsf Di^cr^^'vS^ t*^ ^'/^' M ''' ^ P^'""'' n^«««"ring as follows : hrst piece 37 yds 3 qrs. 1 nail ; second piece 41 yds! U Flemish ells; third piece 43 yds. 1^ English ells. ^ * J^^emisft VI. 1. Multiply £10. Ms. ^d. Canada currency by 764 ; and find by Practice the value of 8764 things at £10. 17*. 6frf each. 2. A bankrupt's assets amounted to 1.542. and his orprliM^o received 12^ cts. in the dollar : find the amount of his debts is twi^tE nS"^""^^-' \^'" ""f'"""^ ^^^^ ^ y^'^ nieasure which whlust^telrgthf '^^ ^'"^' ^^^^^^^ '^ ^^ ''^ ^-^^ 1-g^ i ont'^tTot'fi^t^^f"*^^^^^ ^f ^^I^l "^^^"'^ ^^' ^^S<>' ^"d «oW ! vnrH w^ u^i-t ^.y^'^' ^"^ *^« remainder at a loss of $4 a yard. How much did he gain by the operation ? " ^ ^' *t 5 Estimate the cost of a dish of almonds and raisins consisting of SIX ounces of almonds and three quarters of a pou^of raists^ supposing almonds to be 37^ cts, and raisins 17 cts. a pound * 6 If 5 cwt 3 qrs. 14 lbs. cost $30 per cwt, what will be the m'^r "^ '^' '"'' "^ '^' ^*^°^^ ^^' ^''^ reduced by na!; t^f^T ^"^' ^^ ''''^- ^ ^"'- 2^ ^b«- *^^ sugar $109-60, and Drodfc^mat *^' ^'^'''"'' ^'''''''' ^'^'' Multiplication and Find the cost of papering a room 20 ft. lontr ifii ft h^o^A .„j 166 ARITHMETIC. 9. If a snail, on the average, creep 2 ft. 7 in. up a pole during 12 hours in the night, and slip down 16 in. during the 12 hrs. in the day,- how many hours will he be in getting to the top of a pole 35 ft. high ? 10. The profits of a tradesman average $135*75 per week, out of which he pays 3 foremen, 10 clerks, and 5 assistants, at the rate of $10, $5, and $4, per week respectively ; His yearly outgoings for rent, &;c., amount to $3640*621. Find his net annual profit. 11. A traveller walks 22 miles a day, and after he has gone 84 miles another follows him at the rate of 34 miles a day ; in what time will the second traveller overtake the first *? 12. Divide 100 acres, 3 roods, 8 rods, of land, between four persons, A, J?, (7, 1), so that A shall have } of the whole, B I of the remainder, (7 ^ of what then remains, and D the rest. How much will each one have 1 VII. 1. What is meant by a fraction"? Find the value of | of ^ of 13 tons, and then express the result as the fraction and decimal of 237 tons, 10 cwts. 2. By what number must £5. 6s. 3i£?., be multiplied, in order to give as product £85. Os. 4d. ? Divide £34. 13*. into 3 parts, one of which shall be twice, and the other 4 times as great as the third. 3. If a year consists of 365-242264 days, in how many years will its defect from the civil year of 365|^ days amount to 1 day 1 4. If 15 men take 17 days to mow 300 arres of grass, ho^v long will 27 men take to mow 167 acres ? 5. If 20 men can perform a piece of work in 12 days, how many men will accomplish another piece of work, which is six times as great, in a tenth part of the time ? 6. I am owner of f of f of i of a ship worth $6549 and sell ^th of the ship ; what part of her will then belong to me, and what will it be worth ? 7. A bankrupt owes $900 to his three creditors ; and his whole property amounts to $675 ; the claims of two of his creditors are $125 and $375 respectively ; what sum will the remaining creditor receive for his dividend ? 8. Shew that a cistern 13 ft. 4 in. long, 8 ft;, broad, and 5 ft. 3 in deep holds just twice as much water as another which is 7 ft. long, 6 ft. 8 in. broad, and ft. deep. MISCELLANEOUS QUESTIONS AND EXAMPLES. 167 .1. ^V^'^i^P^^^^*^® calender as now in use. On June 21 of 1851 10. What will be the cost of 10? bales of cotton, each bale weighing 3 cwt 8qrs., at *10-68i per cwt.? Also C how many sohd feet there are in a stick of timber 40 ft. 9 in long, 1ft 11. A person counts on the average $7000 in an hour • what sum will he count in 67 days, if he work 9 hours a day " ' * £^?^^f^^^^^■ ^'- ^^i* '**'""§' *° <^«"»''a currency; and *55. 18«. 6*, Canada currency, to sterling money. ■''»'" ' JIJ.A. 1. Find the value, at $16 per oz.,of 13 lbs.9 )zs. 3 dwts. of gold 2. How many times will a pendulum vibrate in 24 hour, which vibrates 5 times in 2 seconds. ' 3. A crystal palace has a glazed area of 28,000 sq. vds. • find atloaHsTftj '"'^'' '^^^'" '^''' ^^^' ^^^ "^^^ ^^^-^^ '^'t *hh^^ '''■^'^'^u ""^^^'^^^^ '>^ a debt of $950, a dividend of $0-63 in whole? '"^^^^"^"^'J "^^^^ ^«^« the creditor receive on the 5. Reduce 12 ft., 4^ in., to the fraction of a mile, and find the corresponding decimal. ' ® 6. If 29,040 copies of a paper be printed, each copy consisting of 3 sheets, and each sheet being 3^ ft. long by 2 ft. broad how many acres will one edition cover ? ' ^ 7. A man has an income of $800 a year; an extra taxis est«b lished of 14 cents in the dollar, while l duty of 3 centrperTb is taken off sugar; what must be his yearly consumption^ofsu^a? that he may just save the extra tax 1 ^ ^ 8. If ^ can do as much work in 5 hours as B can do in 6 hours or as (7 can do in 9 hours, how long will it tak. to clpfete a piece of work, one-half of which has been done by A 3f 12 hours and B working 24 hours ? vvuxMug i^ fJh^'''^ ^^® "T^^^ of shillings and pence which are equivalent to the recurring decimal -3333 . _ . . of a pound. H"^ vaien i 168 ARITHMETIC. 10. Explain how the Statute defines " a yard," with reference to a natural standard of length. Find the corresponding linear unit, when ftrf acre is one hundred thousand square units. 11. A Liverpool merchant has bills falling due in New York and Montreal: what number of sovereigns will he have to transmit, supposing the New York bill to amount to $7260, and the Montreal one to $9740 ? - 12. A merchant has 10 pieces of cloth, of equal length, and together containing 575 yards, 2 qr. 2 na. What is the length of enih piece 1 IX. 1. Explain the process of Long Division. Reduce 9^^V+-^Hf— ^^ its equivalent whole number. 2. Shew how to convert any proper fraction into a decimal. Reduce f and y^^j to the decimal form. 3. State what kind of vulgar fractions can be expressed in finite decimals ? Can the quantity ^~} — ^\ be so expressed ? How many cents should be given in exchange for ^—^ of a dollar. 4. If two-thirds of an academic term exceed one-half of it by 131 days, how many days are there in the whole term 1 5. How many barleycorns will reach round the earth, supposing the circumference of it to be 25000 miles ? 6. The length of a rectangular field, which contains 1 ac. 1 ro. 15 po., is 453 yds. 2 ft. 3 in. ; find its breadth. 7. A butler concocts a bowl of punch, of which the following are the ingredients : milk 2^ qts., the rind of one lemon, 2 eggs, 1 pint of rum, and half-a-pint of brandy. Compute the value of the punch, reckoning milk at $'05 a quart, lemons at $-60 a dozen, eggs at $-02 each, rum at $3'50 per gallon, and brandy at $5'00 per gallon. 8. A Cochin China hen eats a pint of barley and lays a dozen eggs, while a Canadian hen eats half-a-pint of barley and lays five eggs. Supposing the eggs of the Canadian hen to be half as large again as those of the Cochin China, which is the more economical layer ? ference to lear unit, BW York have to 260, and tgth, and ength of r. nal. I in finite ±i of a tbyl3i, ipposing ic. 1 ro. allowing . 2 eggs, value of a dozen, it $5-00 a dozen nd lays half as le more MISCELLANBOUS QUESTIONS AND EXAMPLES. 169 9. If 72 men dig a trench 20 yds. long, 1 ft. 6 in. broad, 4 feet deep, m 3 days of 10 hours each, how many men would be required to dig a trench 30 yds long, 2 ft. 3 in. broad, and 6 feet delp, in 15 days of 9 hours each? , . '^* 10. A person engages to build 100 rods and 10 ft., of stone fence- at one time he builds 17 rods, 5 ft., at another 37 rods, 15 ft* How much still remains to be built ? » * • 11. Five Chinamen divide 100 lbs of rice as follows ; the first .akes I of 1 of the whole ; the second takes ^ of J of the remainder ; the third takes ^ of ^ of the second remainder; the fourth takes i of ^ of the third remainder; and the fifth had what was left. How much did each receive? 12. Add together | of 4 degrees of 69^ miles each, and 4 of 4 furlongs. ' X. 1. Explain the rule for the addition of decimals ; add together f and '061 ; subtract -003 from -02; and divide -0672 by -006. 2^ Subtract ^ of | from | ofj\, and multiply the result by | 8 3 The fore wheel of a carriage is 10 feet in circumference, and the hind wheel is 16 feet; how many revolutions will one make more than the other in 100 miles ? . ti ¥^ "^^ ®*'"" ^^^'"^^ ^" "^^ ^^y^^ how much will 10 men earn m 11| days? 5. A person expend?^ $72 in the purchase of cloth, hew much can he buy at the rate of $0'64 a yard 'i 6. What is the cost per hour of lighting a room with ten burners, each consuming 4 cub. in. of gas per second; the price of gas bemg $1-50 for a thousand cubic feet 1 *,1;^y^?^ ^? *^® ""^^"^ ^^ ^ ^^^- ''' bushels, 3 pecks of wheat, at $r37^abushen ' ^[ T^ bushels, 2 pecks of malt cost $100, what is the price per 8. What length of paper, # of a vard wide, will be ror^i,4^-,i *^ cover a wall 15 ft. 8 in. long by 11 ft. 3 in. high? * " " '" 9. " Define a Rectangular Parallelopiped." no ARirHMSTlO. A block of wood, in the form of a rectangular parallelopiped, measurej along its edges 18^ feet, 5| feet, and 3 feet, respectively ; determine its value on the supposition that a cubical block, measuring 11 inches along the edge, is worth $0*50. 10. If 36 men, working 8 hours a day for 16 days, can dig a trench 72 yards long, 18 wide, and 12 deep, in how many days will 32 men, working 12 hours a day dig a trench 64 yards long, 27 wide, and 18 deep? ^ 11.-4 received ^ of a legacy, B ^,axid C the remainder. Now it is found that A had $80 more than £. How much did each receive ? 12. Suppose a man consumes f of a day in sleep, | of a day in eating, | hr. each day in amusement, f hr. each day in idleness. How many days, of 10 hours each, has he for work in the course of the year ? XI. 1. Express ,yg. as a decimal ; and thence find its value when unity represents $300. 2. A Township containing 2456 acres is rated on a 'rental of $30700 ; a rate of $0*5 in the- dollar being levied, what on the average is the charge per acre ? 3. Find the price of 2 tons, 16 cwt., 17 lbs. of sugar at $1-25 a Ton. 4. If 1 cwt. of an article cost $70, at what price per lb. must it be sold to gain j\ of the outlay ? 5. Find in inches and fractions of an inch the value of 0000355 P 36 f a mile. Explain the process employed. 6. Express £17. 18.?. S}d. sterling money, and £19. 13c" 4 144. The relation of one number to another, in respect of magnitude, is called Ratio ; and when the relation is considered in the first of the above methods, that is, when it is estimated by the difference between the two numbers, it is called Arithmetical B >- - • but when it is considered according to the secxjnd method, tJ4*t u% when it is estimated by considering what multiple, part or r>ar?>, one number is of the other, or, which is seen from above to be the same thing, by the fraction which the first number is of the second, it is called Geometrical Ratio. Thus, for instance, the arithmetical ratio of the numbers 12 and 3 is ft • whii^ *i,«i» geometrical ratio is ^ or 4. In like manner the arithmetical ratio of 7 and 13 is 6, while iheir geometrical ratio is .^, 172 ARITHMETIC. 145. It is more common, however, in comparing one number with another to estimate their relation to one another in respect of magnitude according to the second method, and to call that relation so estimate*' by the name of Ratio. According to this mode of treatment, which we shall adopt in what follows, " Ratio is the relation which one number has to another in respect of magnitude, the comparison being made by considering what multiple, part, or parts, the first number is of the second, or how many times or parts of a time, or both, the second is contained in the first." 146. It is plain that, for any two numbers, the fraction in which the first is numerator and the second denominator, will correctly express the ^nultiple or part, or both which the first number is of the second, or the number of times or parts of a time, or both, of a time the second is contained in the first Thus, if we take the numbers 12 and 3, the fraction 'J, which is equivalent to the whole number 4, shows the multiple which 12 is of 3, or the number of times 3 is contained in 12. And again if we take the numbers 7 and 13, the fraction -f^ will express the part or parts which the number 7 is of 13, or will express the part or parts of a time that 13 is contained in 7 : for 1 is one thirteenth part of 13, so that 7 must be 7 thirteenth parts of 13, that is, -j'^ths of it ; and 1 is contained 7 times in 7, so that 13 must be contained only y'^ths of a time in 7. We conclude therefore that the ratio of one number to another may be estimated and expressed by the fraction in which the former number is the numerator and the latter the denominator. 147. The ratio of one number to another is often denoted by placing a colon between them. Thus the ratio of 7 to 13 is denoted by 7 : 13. As we have shown that the ratio of one number to another may be expressed by the fraction in which the former is the numerator, and the latter the denominator, we see that 7 : 13 is=T\. The two numbers which form a ratio are called its terms ; the first number, or the number compared, being called the first term, or the Antecedent, and the second number or that with which the former is compared, the second term, or the Consequent of the ratio. 148. If the two numbers to be compared together be concrete, they must be of the same kind. We cannot compare together 7 days and 13 miles in respect of magnitude ; but we can compare 7 days with 13 days ; and it is clear that 7 days will have the same relation to 13 days in respect of magnitude, which the RULl OF THRBB. 178 I number 'espect of t relation mode of 10 is the agnitudb, , part, or times or 3t." in which correctly ber is of or both, ) take the ;he whole imber of imbers 7 hich the ^ime that so that 7 and 1 is ily T^ths > of one I fractio!! itter the loted by to 13 is » number e former see that re called tig called r or that , or THB 3oncrete, gether 7 compare have the bich the number 7 has to the number 18, so that the ratio of 7 days to 18 days will be the same as the ratio of the abstract number 7 to the abstract number 13, and may be expressed by the fraction ^. If however the concrete numbers, though of the same kind, be not in the same denomination of that kind, it will be convenient to reduce them to one and the same denomination in order to find their ratio. Thus, if one of the numbers be 7 days and the other be 18 hours, the ratio of the former to the latter will not be that of 7 to 18, but that of 7 days to 13 hours, that is, 168 hours to 18 hours, which will clearly be the same as that of the abstract number 168 to the abstract number 13, and so will be expressed not by y\, but by ^. , We see, then, that 7 days : 13 hours is the same as 168 : 13, and that each is= ^jO. Thus it is plain that when the numbers are concrete, we may reduce them to one and the same denomination, and then, in considering their ratio, treat them as abstract numbers. 149. Proportion is the equality of two ratios ; so that, when the ratio of one number to a second is equal to the ratio of a third number to a fourth, proportion is said to exist among the numbers, and the numbers are called Proportionals. Thus,^the ratio of 8 to 9 is equal to that of 24 to 27, for the former ratio is f, and the latter ratio is f 4, which is also equal to |. The ratios being equal, ^Ui/^iortion exists among the numbers 8, 9, 24, 27 ; and thus those numbers are proportionals. 150. When proportion exists among four numbers, that is, when the ratio of the first to the second is equal to that of the third to the fourth, tv is iiropOitioii or equality is often denoted by writing down the two ra* that—— ^ ^ ' 24~8- 154. If only three of the numbers in a proportion be given, we can by means of them find the fourth, and the method or Rule by which it may be found, is one of great importance in Arithmetic. We have seen that proportion exists among the numbers 8, 9, 24* 27. If the first three numbers only were given, and we were required, by means of these, to find the fourth, the method or rule to be adopted ought to determine a number to which 24 would have the same ratio, as 8 to 9 ; or, which is seen from the last article to be the same thing, it ought to determine a number which will have the same ratio to 24, which 9 has to 8 ; this number being of course 27. Almost all questions which arise in the common concerns of life, so far as they require calculation by numbers, might be brought within the scope of the Rule of Three, which enables us to find the fourth term in a proportion, and which, on account of its great use and extensive application, is often called the Golden Rule. 165. The Rule of Three, then, is a method by which we are enabled, from three numbers which are given, to find a fourth which shall bear the same ratio to the third as the second to the first, that is, shall be the same multiple, part, or parts of the third as the second is of the first ; in other words, it is a Rule by which' when three terms of a proportion are given, we can determine the fourth. As most of the practical cases in which this rule is made use of relate to concrete numbers, we shall express the Rule with especial reference to such cases, adding, however, a short direction for cases in which abstract numbers only are concerned. 156. Rule. " Leaving out of consideration superfluous quantities, find, out of the three quantities which are given, that which is of the same kind as the fourth or roqjjired quantity ; or that which is distinguished from the other terms by the nature of the question : place this quantity as the third term of the proportion. " Now consider whether, from the nature of the question, the fourth term will be greater or less than the third ; if it be greater, then put the 'larger of the other two quantities in the second term' and the smaller in the first term ; but if less, put the smaller in the second term, and the larger in the first term. 176 ▲RITHMITXO. « Take care to reduce the first and second terms to one and the same demoruination, and also to reduce the third so that it may be wholly in one denomination ; remei )enng, however, that it the quantities involved be all of the sa ,ie ^d, it »« """^^^^Jjy to reduce all the three terms to the same de. mm tion, but onl> the first and second terms to one and the same denomination, and the third to a single denomination, wb" h will not necessarily be the same as the former. When the terms have >een P operly reduced, multiply the second and third together an i ^^jvide by the first, treating all three as abstract numbers. The ^^^ ^ 't"l \^^^ the answer to the question, in the denomina .on to which the third term was reduced." If the case be one in which abstract numbers only are concerned, the question itself will show at once which of ^he ^^^^ers will form the third term of the proportion: the second and first will be determined as above explained ; and then the answer . > t^ 4"f «^^^^ will be found by such multiplication and division as ar ', directed in the Rule. The arrangement of the given terms in the manner luentioned at the V.:pnnini of the Rule, is commonly called stating ^.^f 9«*«*^">;- S^LiLiss a word or two, or a letter, or a symbol, will be added to '4 ]M;'esent the fourth or required term. Nolc 1. The process denoted by the above Rule may often be much abbreviated by dividing the first and second, or the first and Trd terms, (but never the second and third) by any number jhich liU divide each of them without a remainder, and using the quotients instead of the numbers themselves. For 9 • 12 : : 21 : 28 is the same as V'7=if which is the sam as /=U which is the same as 3 : 4 1 : 21 : 28, which represents ?he first proportion after its first aud second terms have each been divided by the same number 3. . , . , A eain 9 : 12 : : 21 : 28 is the same as T^r=ii, which is the same asS, which is the same as 3 : 12 : 1 7 : 28, which represents llrst 'proportion after the first and third terms have each been ^' Again! 9? 12 : : 21 : 28 is the same as ^^\h }>."t this is ^^ the fam; as f =,>, wbich^ - 1^.^^ rld'a^d'tMrd tt Zr^Z^^^^^' ^^ |-is> equal to ^, and of course 9 ; 4 : : 7 : 28 is not a true proportion. RULI or THRXB. 177 id the t may that if essary it only >n, and rily be operly by the will be le tiiiA d earned, ill form will be uestion ected in loned at uestion, e added »ften be first and jr which }ing the the sam [presents ach been the same [presents ach been is is not 8, which rd terms 0/r> ^^ method just stated (/ but as the distinctir writers on Arithmet The Rule of Thrt or less requires less Note 2. Although we ^ ^id in the Rule, multiply the second and ^^d terms together . u divide their product by the firrt? ; it wiw c found in most cu advisable not to perform the actufd multipL -ation until we have Jiscovered, by putting the expression in the form of a fraction, whether there be any factor or factors common to the numerator and denominator, and if so, have rejected such factor or factors. 157. It may be proper to observe that the Rule of Thre^ is applicable in two * iferert kinds of cases, according to which it is called the Rule of Three )'rect or the Rule of Three Inverse. The is applicable to both kinds of cases ; een the two is commonly noticed by til be right to show in what it consi'^te. irect is that in which more requires more, or, in other words, in which » greater numb"5r requires a greater answer, or a less number a less answer. Thus in the question, *• If 4 acres of land cost $550, find the cost of 15 acres, after the same rate." The 15 acres being more than the four acres, will require a larger sum than $550 for their purchase, and so, in this case, more requires more. Again in the question, " If 15 acres of land cost $1160 find the cost of 4 acres, afler the same rate," the four acres being less than the 15 acres, will require a less sum than $1160 for their purchase, and, therefore, in this case, ^ ;ss requires less. Such cases belong to the Rule of Three Direct. The Rule of Three Inverse is that in which more requires less, or less requires more : or, in other words, in which a greater number requires a less answer, or a less number a greater answer. Thus ' the question, " If 4 men can mow a certain meadow in 3 days, hud the t .ae in which 6 men ought to mow it," the six men being more lan tht four, should perform the work in less time, and so, in this case, more requires less. Again, in the question, " If 6 men can mow a certain meadow in 2 days, find the time in which 4 men ought to mow it," the 4 men being fewer than the 6,will require a longer time for performing the work, and therefore, in this case, less require more. Such cases belonjr to the Rule of Three Inverse. Mule of Three Direct Ex. 1. Find the value of 37 yards of jilk, when 25 yards cost $150. IMAGE EVALUATION TEST TARGET (MT-S) 1.0 I.I i^l2i2 us IL25 III 1.4 I 2.0 1.6 / Hiotographic Sdoices Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 873-4503 \ iV A •S^ \ :\ <^ '^ "W k^ 178 ABITHMSTIO. 1. There are here three given quantities, 25 yards, 37 yards, and $160, and we 'have to find a fourth which will be the price of 37 yards. It is manifest that three given quantities, 25 yards, 37 yards, $150, and the required sum, must form a proportipn, because the 25 yards must have the same relation in respect of magnitude to the 87 yards, which the $150 (cost of 25 yards) has to the required sum (cost of 37 yards). Proceeding then by Rule (Art 157) we observe that the $150 is of the same kind as the required term, viz. money ; we make that the third term of the proportion ; and since the required sum (cost of 37 yards) must necessarily be greater than $150 (cost of 25 yards), we make 37 the second term, and 25 the first. We have thus the first three terms arranged as follows : . 25yd8 :37 yds.:: $150. . And the entire proportion will be as follows : 25 yds. : 37 yds. : :$150 : required cost. The first and second terms are in one and the same denomination, and require no reduction ; therefore the proportion is, 25 yds. : 37 yds.: :$150 : no. of dollars jn required sum. And by our rule we must now treat the numbers as abstract, multiply the second and third together, and divide by the first 150 37 25 i; 1050 450 5 5 5550 1110 222 The quotient 222 gives the number of dollars required. The above process is usually written down as follows : ydi. yds. $ 25:35::i50 37 1050 450 jaULX OF TRRXX. 179 Reason for the above process. We have the cost of 25 yards given, viz. 1150, in order to enable us to find the cost of 37 yards. It is manifest that the required sum must have the same relation in respect of magnitude to 1150 which 37 yards have to 25 yards ; that is, the ratio of the required sum must be equal to that of 37 yards to 25 pards. Now the ratio of the number of dollars in the required sum to $150, is the same as that of the abstract number which indicates how many dollars the required sum contains to the abstract number 150j and may (if the former number be called the required number) v J 1- xi. /• ^- required number be expressed by the fraction -^ — r^jr . required number 37^ •*• 150 ""25' required number 37 X 150= — X 150, or 150 required number X 150 37 x 150 150 or required number: 150x37 25 37x150 26 , (Art. 66.) or: 25 This result shows that if we arrange the three given terms, 25 yards, 37 yards, and $150 in the following manner : ydB. yd». 25:37::$i50, and then consider the numbers to be abstract, as if they had been written 25: 37:: 150, we shall obtain the abstract number which will show us how many dollars there are in the required sum by multiplying the second and third terms totrether and dividincr the product bv the first-. • an/i then by treating this number as concrete, that is, as so many dollars, we have the required answer in dollars. h:^ The reason for the procees may also he sho^n s» follows : The cost of 26 yards is $150 ; .'. the cost of 1 yard is-jrr- dollars ; , . /ISO \ , „ 150x87^ „ .% the cost of 37 yards is ( 05 x 37 1 dollars= — ^ — dollars ; or if we arrange the numbers in the form ydi. yds. 25 : 37 : : ii50, tapA then treat them as abstract numbers, multiply the second and t^ird together, and divide the product by the first, the quotient will give the number of dollars in the required sum of money. Ex. 2. If the tax on $565 be $35*60, what will be the tax on $35701 The $35-60, being of the same nature with the sum required,mu8t be placed in the third term in the proportion ; and as the required tax must clearly be greater than $35*60, we must place $3570 as the second, and $565 as the first term. $565 : $3570 : : $36*60 : the required tax. 3570 249200 17800 10680 565) 12709200 (224*94y»TV 1130 , 1409 fc llSO 2792 2260 5320 5085 2350 2260 90 .*. the required tax is $224*94^. V 00 tl 75. .-. (3} x 364) days : 1 day : :$20375 : daily income, whence, daily income=$15-99||f Ex. 4. Two places, A and J?, are distant from each other 324 miles by railway. A trkin leaves A for B at the same time that a train leaves B for A ; the trains meet at the end of 6 hours, the train from -4 to ^ having travelled 16 miles an hour more than the other. How many miles did each travel an hour ? Each train is supposed to run with uniform speed : whcii the trains meet, the whole distance must have been passed over by them. . . .-. 6 hrs. : 1 hr. :: 324 miles : miles passed over by both trains in 1 hr. whence, miles passed over by both trains in 1 hr.=:54 therefore by question, (54 -16) -1-2, or 19 = miles travelled per hour by one train, and therefore 54—19, or 35=miles travelled per hour by the other. Ex. 5. An English gentleman, after paying an income-tax of 7rf. ' in the £, has £248. 10*. 8c?. left ; what was his gross annual income % For every 19*. 5c?. which he now has, he had £1. before he paid his income-tax ; .-. 19ft M. : £248. 10?. 8rf. : : £1; • r^uitwl income, s before g ran 7 ; by the rds, And els: rtnight ; vings at pposing f$6726 ler 3i24 le that a jrs, the re than lidti the >ver by in 1 hr. led per avelled c ofTrf. icome? e, BULS or TBBIB. Ex. UI. 9k^^^^ 1. If 4 yards of cloth cost 112, wh&t will 06 yards of the same doth cost? 1^2. If 9 yards of cloth cost $45, hoT macy yards oan be bousht for $225? ^^ ^ 7. If 7 bushels of wheat be worth $7-70, what will be the value of 3 bushels of the same quality 1 M 4. The rent of 42 acres of land is $68, how many acres of the same quality of land ought to be rented for $273 1 6. If the cost of 72 tons of coals be $540, what will be the cost of 42 tons? 6. How much must be given for 13 articles at the rate of £S, 16s. 6rf., Canada currency, for 6 articles ? 7. How long will a person be saving $194*26, if he put by $1*75 per week ? 8. IM a number which shall bear the same ratio to 9, which 20 does to 15. |j{, 9. If 2 cwt, 3 qrs., 14 lbs. of sugar cost $28*90 what quantity of the same quality of sugar can be bought for $86*70 ? // 10. If 3 cwt., 3 qrs. cost $33*75, what will be the price of 2 cwt, 2 qrs. ? iC 11. Find the value of 23 yds., 1 ft. of doth, supposing 4 yds., 81 in. of the same quality to cost $17*50.^ J */ 6 12. What will be the school tax, at 3 cents in the doUur, on $25710? 13. 5 gallons, 2 pints of parafine oQ cost $5*88, what will 75| gallons cost ? 14. What is the tax upon $872*25, when $1170 is rated at $315? 15. If one bushel of malt cost $1*23, how much can I buy for $897*62^? y 16. Find the price of 2 tons, 3 cwt., 14 lbs. at $15*11 per Quarter. 17. If 9 acres of land sell for $230-62J, what should 6 acres bring at the same rate ? " 16. Find the amount of a servant's wages for 215 days at $1*25 a day. z / ^ h c^ ARITHIIETIO. 19. A ba^ikrupt's debts amount to $10550, and his assets to •1056 ; how much in the dollar can he pay 1 20. A cistern is filled with water, by 2 pipes, in 3 hours 25 minutes ; in what time would it be filled by 5 pipes of like size ? ■5!. A bankrupt pays |-075 in the dollar, and his assets amount to 1950 ; find the amount of his debts. 22. The velocity of a locomotive on a railroad is 36 miles an hour, how far does it move in 30 seconds ? ,23. If I of a bushel of wheat cost Iff, what will $f purchase? 24. If l^lbs. of indigo cost 1384, what will 49-2 lbs. cost ? 26. Find a fourth proportional to the numbers 3, 3*75, and 40. 26. If 10 men can mow a field in 12 days, in how many days will 15 men mow it ? 27. If a man walk 62 miles in 3 days, in how many days will he jralk 80 miles 1 ^ ^ 28. How many yards worth 13-25 a yard must be given in exchange for 935^ yards worth 14*60 per yard 1 29. If 7 times | of ^ of an estate be worth $15000, what is » of I of it worth? ' P5. Find the price of 2 tons, 16 owt,, 17 lbs. of sugar at 10-25 for 2^ lbs. . ' 31. If a person travelling 19 hours a day perform a journey in 24 days, in what length of time will he perform the same journey if he travel 16 hours a day ? 82. If 3f oz. Avoir, cost $2 87^, what will 30^ lbs. cost ? 33. How many men must be employed to finish a piece of work in 15 days, which 5 men can do in 24 days ? 84. If 356 ac., 3 ro., 39^ po. be rented at $1713-585, what is the rent of 2 acres ? 35. The governor of a besieged place has provisions for 54 days, at the rate of 1^ lbs. of bread to each man per day, but is desirous* to prolong the siege to 80 days, in expectation of succour ; in that case, what must the ration of bread be ? 86. If 27 bush. 2 pks., cost $65, what is the price of 16^ bush.? 87. How many yards of drugget, an ell wide, will cover 40 vds, of carpet f yd wide? *" RULE OF THBIB. 187 assetfi to hours 25 ke size 1 s amount miles an irchase ? JOStI , and 40. any days lays will given in vhat is f at $0*25 urney in journey b? of work what is 54 days, lesirous in that ^bush.? 40 vdfl- 88. A merchant, owning | of a vessel, sells | of his share for $500 ; what is the whole vessel worth ? / 38. A field is 121 yds. long and 86 yds. broad ; what will be its value at $80 an acre 1 I 40.. If the price of 1 lb. of sugar be $0-0625, what is the value •of -75 of a cwt. 1 41. If 3| shares in a mine cost $22'645, what will 28* shares cost ? 42. If 34^ yards of cloth cost $53|, how many yards can be bought for $35yv '? i43. Find the rent, at $3 an acre, of a rectangular field, whose ides are respectively 50 chains 40 links, and 56 chains 25 links. 44. In what time will 25 men do a piece of work which 12 men can do in 3 days? 1 45. If -3 of 4-5 cwt. cost $11*55, what is the price per lb. ? 46. When $| will buy I dwt. of gold, how much gold can be bought for $8000 ? ^ 47. If a piece of building land, 375 ft. 6 in. by 75 ft. 6 in., cost $118, what will be the price of a piece of similar land, 278 ft. 9 in by 151 ft.? 48. A servant enters on a situation at 12 o'clock at noon, on Jan. 1, 1859, at a yearly salary of $175, he leaves it at noon on the 27th of May following; what ought he to receive for his services ? 49. A was owner of j\ of a vessel, and sold ^j of f of his share of l^is share for $*^o j what was tiie value of ii of f of the vessel? SO.' A exchanged with B 60 yards of silk, worth $1-44 a yard for 48 yards of velvet ; what was the price of the velvet a yard ? ' I 51. If it require 30 yds. of carpeting, which is ^ of a yard wide, |o cover a floor, how many yards of carpeting, which is U yds. Wide, will be necessary to cover the same floor ? 52. If by a leak of a ship % enough water run in to sink her in 4 hours, how long will she float ? ^JX A watch is 10 minutes too fast at 12 o'clock (noon) on Monday, and it gains 3". 10" a day ; what will be the time by' the watch at a quarter past 10 o'clock a. m. on the following Saturday ? m ARITHHFTIOa 84. Theciroumferenceofa circle is toitidiameteraa 8-1416 • !• r"n^"*/®®* *"^ ^^^^^^ ***® circumferenoft of a circle whose diameter is 22^ Uaet, (66. If $100 in 12 months, bring an interest of |7, what wUl be the mter est of $76 for the same time 1 6rt. If the carriage of 3 cwt. cost $2| for 40 miles, how much ought to be earned for the same price for 26f miles ? h.^L!{ ^ ^^^f ^u ^rin^. ^^ * fortnight, what must my income be that I may lay by $200 in a year 1 (^8. If VV of a city lot be sold for $500, what would A of the swne lot sell for at the same rate ? "^ *fe/ '"J!' *!?''*'^.' ^^»5\^«»ghs 1 lb. 10 oz. 10 dwts., cost $^•66 ; what is the value of the silver per ounce ? RO. A man, working 7^ hours a day, does a piece of work in » flays ; how many hours a day must he work to finish it in 4^ days ? 61. Suppose sound to move 1100 feet in a second, how manv S« t'T.V' r^'^^'J^ ''}''^ ^^^htning is observed 16 se2 motk^ofliSf '' ""^ ''^^''^"'""' ^"^^ °''^' ^"' '^" 62. How much did a person spend in 64 days, who with an annual income of $8180, is 900 dollars in debt^X^eld of a year » ^•.^^2?i^®°' ^? ^^™e°» an «^»,„, — in that denomination to which the third term was reduced. months. Id be in uld gain )f Three, a capital ^[uestion. 000 by the iwo first i terms jduct of '■■ snswcr « DOUBLX RULB OF THBBl. 195 Or thus : * A capital of $2000 gains $50 in 3 months, ' ^ tl $50 in (3X2000) months, II ..-$1 in (^^) months, 13000 ....II in (Ji^) months, 1- 13000 ... |175in(^2i^?^2il25'|n,onths V 50x3000 J^^^^^* / 2000 X 175 X3\ ^A 3000X50 j'^Q'^th«> that is, if we arrange the given quantities as follows, $3000 : 12000 ) . .o $50 : $175 [ • -^ "*' we obtain the required time in months by multiplying the two first terms togother for a final first term, the two second terms together for a final second term ; and then dividing the product of the second and third terms by the first term. Ex. 2. If a tradesman, with a capital of $2000, gain $50 in 3 months, what sum will he gain, with a capital of $3000 in 7 months ? ' The $50 in the supposition corresponds to the quantity sought in the demand. Make this $50 the third term. Then taking the capital of $2000 in the supposition, and that of $3000 in the demand, and considering them with reference to the gain in the third term, we observe that if the amount of capital be iucreased so also will be the gain in a given time, and thus the fourth term' would be greater than the third ; therefore we place the $2000 as the first term, and the $3000 as the second. Again, taking the 3 months in the supposition, and the 7 months in the demand and considering them in like manner with reference to the gain in the third term, we observe, that as the time is increased so also will be the gain from a given capital, and thus the fourth term would h greater than the third ; therefore we place the three months" J^'l first term and the 7 months as a second. We thus obtain the following statements : $3000 : $3000 ) . .^-^ Sm :7m ] • ••^"* Proceeding according to our Rule, we obtain the following statement : 2000x8 : 3000x7 :: 5o, , . , . , . J 11 3000 X 7 X 60 and the required sum m dollars- 3 X 7 X 60 2x8 2000x3 =7X25=176. The answer is therefore $175. Ex. 3. If 7 horses be kept 20 days for $14, how many will be kept 7 days for $28 ? The 7 horses in the supposition correspond to the required quantity (number of horses) in the demand. Make this the third term. Then, taking the 20 days in the supposition, and the 7 days in the demand, and considering them with reference to our third term, we obsferve that if the number of days be diminished, the number of horses which can be kept in them for a given sum of money will be increased, and thus a fourth term would be greater than the third ; we therefore place the 7 days in a first term, and the 20 days in a second. Again, taking the $14 in the supposition, and the $28 in the demand, and considering them with reference to the third term, we observe that if the sum be increased, the number of horses which can be kept by it in a given time will be increased also ; so that here also a fourth term would be greater than the third ; we therefore place the $14 in a first term, and the $28 in a second. We thus obtain the following statements : 7 days : 20 days ) . . ^ , $14 : $28 f • • ^ ^^''®®^' whi^, by our Rule, will give the following single statement : 7 X 14 : 20 X 28 :: 7, and thus, the required number of horses = 20 X 28 X 7 7x14 The answer is therefore 40 horses. =40. following Y will be required the third ihe 7 days our third ished, the m sum of )e greater erm, and ' pposition, reference iased, the e will be e greater 1, and the Its: tent: 7 DOUBLE &UUi OW THREE. IW ^:^ ^^J 8®* ® ^* ^®^8**^ ^^ ^^^^ for erf. when wheat is 15*. a bushel, what ought a bushel of wheat to be when I Met 12 oz of bread for 4d.1 o ^ The price of a bushel of wheat is required ; to this the 16». in the supposition corresponds. Place this as the third term. Then takmg the 8 oz. in the supposition and the 12 oz. in the demand, and considenng them with reference to the price in the third term we observe that the greater the weight of bread we obtain for a given sum the less will be the price of a bushel of wheat, and so a fourth term would be less than the third; we therefore place the 1^ oz. as a first term, and the 8 oz. as a second term. Again taking the 6c?. in th^ supposition and the 4rf. in the demand! we consider that the less we pay for a given weight of bread, the less will be the price of a bushel of wheat, so that here also a fourth term would be less than the third ; therefore we place the 6d. as a lirst term, and the 4rf. as a second. Thus we have the following statements : ^ 12 oz ;. : 8 oz. I . . I : 4d. f • • 158. which, by our Rule will give the following single statement : 12x6 : 8x4 :: 15, and thus, the required price will be 8x4x15 8x15 12x6 3x6 4x5 20 =-g-«. ^Y*' = 6». 8rf. Ex. 5. If 20 men can perform a piece of work in 12 days, find the number of men who could perform another piece of work 3 times as great, in }th of the time. ' The first piece of work being reckoned as 1, the second must be reckoned as 3. The 20 men in the supposition must be taken as the third term. Ihen, taking the piece of work (represented by 1) in the supposition, and the piece of work (represented by 3) in the demand, we observe that if the work be increased, the number of men to perform it ma given time must be increased, and we therefore place the 1 as a first term, and the 3 as a second. Again, taking the 12 days m the supposition and the V days in the demand, we observe that i. lu^ !i..,.«,jv,x VI viojo we uiiuiuisuea, tne numoer oi men required to perform any given work will be increased, and therefore we 198 ARITHMSTIO. place the y days as a first term, and the 12 days as a second term, inus "we nave the following statements : 1*3 ) y days : 12 days [•■•^^ ^®"» which, by our Rule, wUl give the following single statement : V: 3x12.: 20, and thus the required number of men will be 3x12x2 3x12x20x5 ~V = 12 =300. Ex. 6. If 252 men can dig a trench 210 yards long, 3 wide, and u P'm1°oo ^^^^ °^ ^ ^ ^'^"^^ ^^«*^ ; i» ^ow many days of 9 hours each will 22 men dig a trench of 420 yds. long, 5 wide, and 3 deep ? The first trench contains (210 x 3 x 2) cubic yds. =1260 cubic yds. T*»e second (420x5x3) cubic yds. =6300 cubic yds. On the supposition therefore that 252 men can remove 1260 cubic yds. of earth in 55 hours, we have to find in how many hours 22 men can remove 6300 cubic yds. TJe 55 hours correspond to the quantity sought. Make this the third term. Then, takmg the 252 men in the supposition, and the 22 men in the demand, we observe that if the number of men be diminished, the number of working hours in which a given work can be performed will be increased, and we therefore place the 22 men as a first term, and the 252 men as a secondf Again taking the 1260 cubic yds. in the supposition and the 6300 cub' yds. m the demand, we consider that if the number of cubic yds be increased, the number of working hours in which a given number of men can perform the work will be increased also? and therefore we place the 1260 cubic yds. as a first term, and the tWOO cubic yds. as a second. Then we have the following statements : 22 men : 252 men \ . . ^^ , 1260 cub. yds. : 6300 cub. yds. [••55 hours, which, by our Rnle, will give the following single statement : 22x1260: 253x6300:: 55, y fi as w an m( in in roi I 801 OOUBLI RULI or THRU. 199 working hours and thus the required time 252x6300x 55 22x1260 _ 252x 5 X 55 22" working hours =3150 working hours _3160, —■ ~g~"days of 9 working hours =350 such days. Ex. 7. If 560 flag stones, each 1^ feet square, will pave a court- yard, how many will be required for a yard twice the size, each flag-stone being 14 in. by 9 in. ? Superficial content of each of former flag-stonei =(4XU) sq. ft. = (iX|) sq. ft.=f sq. ft. Superficial content of each of the latter flag-stones =(ll X Vu) sq. ft. = (f X J) sq. ft.=| sq. ft. Considering the first court-yard as 1, and therefore the second as 2, our statements will be i sq. ft. : f sq. ft. ) . . 1 • 2 f • • °"^ nag-stones, which, by our Rule, will give us the following single statement : f : f x2 :: 560, and thus the required number of flag-stones =(f x2x560)-^f =(f X 560 X f ) __9x 560x8 ^^^^ - =2880. 2x7 Ex; 8. If 10 cannon, which fire 3 rounds in 5 minutes, kill 270 men m an hour and a half, how many cannon, which fire 5 rounds m 6 mmutes, will kill 500 men in one hour? The first lO cannon, firing f of a round in a minute, kill 270 men m I hours. It is required to find how many cannon, firing * of a round in a minute will kill 500 men in one hour. ~ iLl ".' Ti — T " , ""Pi^ooiwon uorrcSpouG to the quantitv sought m the demand. We makp his the third term. Then w$ JAnmtMfKi. taking the I of a round in the supposition and the * of a round in the demand, we observe that if the part of a round which is fired in a minute be increased, the number of cannon for effecting a certain slaughter would be diminished ; and therefore we place the I of a round as a first term, and the | of a round as the second. Again, taking the 270 men in the supposition and the 500 men in the demand we observe that an increase in the number of men killed would require an increase in the number of cannon- and therefore we place the 270 men as a first term, and the 500* men as a second. Again, taking the | hours in the supposition and the 1 hour in the demand, we consider that if the time in which a certain number of men are killed be diminished, the number of cannon would be increased ; and therefore we place the 1 hour as a first term and the ^ hours as a second. Sur statements will therefore be, f round : ^ i^Sanii ) 27^ men : 50« men V : : 1« cannon, 1 hour : I hours ) which, by our Rule, will give us the following single statement: f X270X1 : ^x5eox|:: lo, or 5 X 45 : 3 X 50 X 3 : : 10, .*. required number of cann(in= 3 x5»x3x 1# 5x45~ =2«. Ex. 9. A town which is defended by 1200 men, with provisions enough to sustain them 42 days, supposing each man to receive 18 oz. a day, obtains an increase of 200 men to its garrison ; what must now be the allowance to each man, in order that the provisions may serve the whole garrison for 54 days ? The 1400 men will belong to the demand : for the question is what must be the allowance to each man, when the garrison is increased to 1400 men, in order that the provisions may last 54 days. The 18 oz. must clearly, according to our Rule, be the third term. Taking the 1200 men from the supposition, ^d the 1400 men from the demand, we consider that if the number of men be increased, the allowance to each must be diminished, in order that the provisions may last a given time ; and we therefore place the 1400 men as a first term, and the 1200 men as a second. Again, taking the 42 days in the supposition and the 64 days in" the DOUBLX RULl 07 THRU. round in ih is fired fTecting a we place e second. ) men in p of men ion ; and 300 men n and the which a iniber of hour as 3nts will 201 )ment : I. ovisions eeive 18 a ; what hat the istion is, rison is last 54 le third e 1400 men be ier that lace the Again, in the demand, we consider that if the number of days during which a S^usTh« T"' ""l T'^'^'f ^' 1"^^^^^^' '^' alloWce K tn must be dimmished; and we therefore place the 54 days as a H? tWefore ^ '' '''' " ' ^^^^'^^ ^^'^ Our staZe^ It 1400 men : 1200 men ) ...^ 54 days : 42 days | • • 1° oz. which, by our Rule, will give us the following single statement : 1400X54 : 1200x42:: 18, .-. required allowance =i^^^l2iL® -lo ,, 1400x54 ^2.— l2oz. so that 12 oz. will be the answer. y.. \^f' ^1 'I ^^" ^"f t^o»s in Proportion, simple and compound m av be solved by simple ratios, without the lengthened pSrilril^ pursued m such cases; and when canceling is^rnXd^i^ operation is very much shortened. As the vadous temsTinrl, nf ratios, antecedents and consequents, perfect and imnLTif • ^ °^ and falling, have already heen^xplain'ed To 2de7^^^^^^^ IS unnecessary. The problem under ^onsideratbnshoS^^?^ Tl^rirr'^ ^^ '^' P"P^^' ^"^ '' ^" «"^h questions c^sis? of perfect ratios, except one, and as the terms of each nftil ratios consist of the same kinds of quantities, dl we have to do ! to place them together in vertical rolnmna ah .^ °^ ^^ t Jratios .houw\e asM Xefo Z o thl'l^m' S^"'"'!,' warL^atrtr;*;^"^''"''"^"'^ of theproUem. AnVxT/l^ samelSnTdaysf " '' "'^'' """ "^^ ««" ^"^^ "o th. Here we first put down the imper- fect ratio X : 8 using X (or any other mark) for the term in which X : 8 4 : 12 which, after cancellation, will stand as follows- 12x2 1 — = 24 men. it is deficient, next we ask the question, would 4 days reni,ir« more or less men than 12 davs? fha onoJ^^ ;^^^Jl°y^ require put down the other as a rising'ratio Tfil '" **"""^*^^^> ^^ ^« 2b If ARITHMETIC. », (8.) Ne,t let ui »aJ «, .ftor nrtk to the question :-If 8 men in iouM^Jritl 1h ^'-'t'^^'i r" « fi«Hhow many men work LZ '^i '"''• *"'''"'« « '"'"" " ^"^ ' An.. 3o! Tl,e X : 8 ('raperfeot ratio, men required). 4 : 12 ' 8: 10 The imperfect ratio x ; 8 as before, we ask, using the term of demimd with reference to the term wanting in the imperfect r^tio would 4 day, require more or less men thSn 13 drsTn Sch to perform the same amount of labor ? the answer isf 4 days would require more men (it is evident the shorter the time the more S 4*. .',™?J""P'°y'''') ""«" *« "»«» i^ » rising one, anZut down 4 : 18; likewise we ask would 8 hours pe? day requC r"«o 8 "• 10 ""^.l'"" '" ""r "/'^ !,»'"- »'«™''t-« "rising of 1 J- • ^*° T '^'^•' ""^ »' »" "'6 antecedents or factors ^«| o't t^^^-5rd;: ^"tTib'turT'2h7;iT X 20 4 8 8 16 12 10 (Imperfect ratio, men requjred). which after cai7:,ellation will stand as follows : 6 x 4=24 Ans. ms is the same as before, except the second ratio, would 16 aci .. r^uire^more ct less men than 20 acres? lo.s, 'a falling rpt'n iniM ^!-/''"''T^''f P^^^^^"' ^^^^'^g '^o ^ewer than seven ratios is just as easily solved as any of the preceeding, and so of anv question so far as the r .tement is concerned. ^ nhll^^'^r^'^^^' '"^ ^ ^ "« " '^' 14 ^^"^s long can compose 20 Bheets of 24 pages m mm .. 3(: ^ 50 lino, in a p^e and 40^^ in • line ; m how manj c^j of 7 hour, long caTircompS eempose a volume to ha nrinfofi i»i fv, ^i--fj t^ompositorg c a DOtTBLI RlJLl OP THREl. ) men in ny men iO. The uired). term of ct ratio, ^hich to s would le more and put require a rising factors equents n mow ay how a day? ujred). SOS 1 Ans. 5 acres ' rntin tios is f any >se20 etters sitors ug40 X : 16 7 : 14 10 : 5 20 : 40 24 : 16 60 : 60 kt U r 4^ .' 50 which after cancellation will stand as follows : 16 x 2=32 Ana. \V fiiid by inspection the imperfect ratio and put it down x • 16 • Wen we ask, would 7 hours require more or less days than 14 ? more,arismgratio7 : 14; would 10 compositors require more itZ'Ir^ ^^'" ^ compositors? less, a falling ratio 10 '5 wonid 40 «neets require more or less than 20 sheets 1 more a rismg ratio 20 : 40; would 16 pages require more or e. Jhan 24 pages less a falling ratio 24 : 16 ;Vould 60 hL Squire iZr """ ^'unnT **^^" ^""^^''^ more a rising ratio 50 ^So ^o f^' r?"-^ ^^ *'"''' ''^"^^^ '"^^^ ^' !««« ^han 40 letters* the; would require more, a rising ratio, 40 : 50. ^ It is eddent that if the value of x be substituted for it in these Xtt?*^ ''^'' will just cancel; this must always be so Inf!^ *i ! T Py* ^^^^ correctly, as the product of all the antecedents equals the product of all the consequents. Ex. LIU. .!^ a '^ "'^'^ T "■? *P ^ ^""^^ ^^ ^^ *»<^"^s, how many men wUl reap 15 acres m 14 hours ? / 2. If 3 men earn £15 in 20 days, how many men will earn 15 g^ccis m 9 days, at the same rate ? A^. ^^.J.Vu'"'^' ^** ^? ^"'*'^^' ^^ ^^^'^ ^" 42 days, in how many days will 1 horses eat m bushels ? ^ 4. If 800 soldiers consume 5 sacks of flour in 6 days, how many wm consume 15 sacks in 2 days 1 j , »uy J^Jl ^liJ'^^j^^'^ b® consumed by 6 horses in 13 days, what quant ty will 8 horses eat in 11 days, at the same rate ? ^n /oV'"''"®' T PJ'?"^^ ^^®^ ^^^^« ^^ Q -' 2 po., be $390, what would be the rent of another farm, containing 26 ac, 2 ro. 23 do If 6 acres of the former be worth 7 acres of the latter ? ' 19. If 1500 copies of a book of 11 sheets require 55 reams of paper, how much paper will be required for 5000 copies of a book ot Zd sheets, of the same size as the former ? 20. If 5 men can reap a rectangular field whose length is 800 feet and breadth 700 ft. in 3^ days If 14 hours each; fnhow m^y &Vrld?r960t r ' ""^ "'^ ^ '''' "'^^^ ^^^^^^ ^^ '''' fl. If a thousand men besieged in a town with provisions for 5 weeks, allowing each man 16 oz. a day, be reinforced with 500 men more, and have their daily allowance reduced 63. oz • how long will the provisions last them ? "" ' ' 22. If 20 masons build a wall 50 feet long, 2 feet thick, and 14 " feet high, m_ 12 days of 7 hrs. each, in how manv davs of 10 h's ^cn ^wiii ou masons build a wail 500 feet long, 4 thick, and 16 30, what id., how nonths ? months, hat will $385 ; mg will 12 men is come 2 hours ^ing 16 rking 8 3 work 0, what 23 po., ams of a book 00 feet many s 1800 s for 5 ;h 500 ', how md 14 ' knd 16 OOtrBLZ RVXX 6f THRKI. S05 23. If 10 men can perform a piece of work in 24 days, how many men will perform another piece of work 7 times as great m one-fifth of the time ? ^ ^ 24. If 125 men can make an embankment 100 yards lonff, 20 feet wide, and 4 feet high, in 4 days, working 12 hours a day, how many men must be employed to make an embankment 1000 yards long,^16 feet wide, and 6 feet high, in 3 days, working 10 hours a 25. What is the weight of a block of stone 12 ft. 6 in. long. 6 ft. 6 m. broad, and 8 ft. 3 in. deep, when a block of the same stone 5 ft. long, 3 ft. 9 in. broad, and 2 ft. 6 in. deep, weighs 7500 lbs. ? 26. If 100 men drink $200 worth of wine at |M0 per bottle how many men will drink $720 worth at $1-20. per bottle, in the same time, at the same rate of drinking? 27. If 5 horses require as much com as 8 ponies, and 15 quarters last 12 ponies for 64 days, how long may 25 horses be kept for $205 when corn is $2*75 a bus. w ^LfX y^^- ^^ cloth, which is 18 in. wide, cost $236-50, ^3^7111 119^ yds- of yard-wide cloth of the same quality cost? 29. 124 men dig a trench llO yds. long, 3 ft. wide, and 4 ft. deep, m 5 days of 11 hours each; another trench is dug by half the number of men in 7 days of 9 hours each ; how many feet of water is it capable of holding ? 30. If the 5 cent loaf weigh 3-35 lbs. when wheat is at $1-75 a bus., what ought to be paid for 47^ lbs. of bread when wheat is at $2*40 a bus. 1 31. A pit 24 ft. deep, 14 sq. ft. horizontal section cost $80 to dig out ; how deep will a pit be of horizontal section 7 ft. by 9 ft which costs $40-50? ^ 32. The value of the paper required for papering a room supposing it f yard wide, and 6 cents a yard, is $10.75 : what would It conae to if it were 2 feet wide and 5 cents a yard ? 33. 7 men working 16 days can mow a field of corn 1320 yards long and 880 wide ; what will be the length of the side of a field 1320 yards broad, which 4 men can mow in 42 days ? /J,\/i'^ j\ ^ ?1: ^ohT ^^ ?^^ ^''"S, 21 feet broad, and 8 inches thick, ^-^T I wcigus i^ov/ ius ; wuat must be ihe length of another beam of the ' ' same material, whose breadth is 3^ feet, thickness 7^ inches anj>ff^'^/t weight 2028 lbs ? * > « y - / ^ j 209 l&ITHMSTIO. 35. If 12 oxen and 35 sheep eat 12 tons, 12 owt. of hay in 8 days, how much will it cost per month (of 28 days) to feed 9 oxen and 12 sheep, the price of hay being $40 a ton, and 3 oxen being supposed to eat as much as 7 sheep ? 36. If 1 man and 2 women do a piece of work in 10 days, find in how long a time 2 men and 1 woman will do a piece of work 4 times as great, the rates of working of a man and a woman beinff as 8 to 21 ^ 37. A person is able to perform ^ journey of 142*2 miles in 44 days, when the day is 10*164 hours long; how many days will he be in travelling 505*6 miles, when the days are 8*4 hours long 1 38. If the sixpenny loaf weigh 4*35 lbs. when wheat is at 5*75«. per bushel, what weight of bread, when wheat is 18*4«, per buchel, ought to be purchased for 18*13*. ? 39. If a family of 9 people can live comfortably in England for 1560 gumeas a year, what will it cost a family of 8 to live in Canada in the same style for seven months, prices being supposed to be I of what they would be in England ? . INTEREST. 165. Dbf. Ii^sREST is the sum of money paid for the loan or use of some other sum of money, lent for a certain time at a fixed rate ; generally at so much for each $100 for one year. The money lent is called the Principal, The interest of $100 for a year is called the Rate per Cent. The principal + the interest is called the Amount. Interest is divided into Simple and Compound. When interest is reckoned only on the original principal, it is called Simple Interest. When the interest at the end of che first period, instead of being paid by the borrower, is retained by him and added on as principal to the former principal, mterest being calculated on the new principal for the next period, and this interest again, instead of being paid, is retained and added on to the last principal for a new principal, and so on ; it is called Compound Intirist. hay in 8 )d 9 oxen en being ays, find •f work 4 lan being les in 4^ '^s will he long ? at 5*75«. ir buchel, jland for ) live in supposed ) loan or t a fixed Cent. interest SiMPLB of being i on as i on the instead >al for a 6IMPL1 XNT1BB8T. SIMPLE INTEREST. 207 166. To find the interest of a given turn of money at a aivenrate per cent, for a year, RuLi. « Multiply the principal by the rate per cent., and divide the product by 100, as in (Art. 126)" Mte 1. The interest for any given number of years will of course be found by multiplying the interest for one year by the number ot years ; and the interest for any parts of a year may be found ^^the mterest for one year, by Practice, or by the Rule of Note 2, If the interest has to be calculated from one given day to another, as for instance from the 30th of January to the 7th of JJebruary the 30th of January must be left out in the calculation, and the 7th of February must be taken into account, for the borrower will not have had the use of the money for one day till the 31st of January. ^ MteB. If the amount be required, the interest has first to be tound for the given time, and the principal has then to be added to it. Ex. Find the simple interest of $250 for one year at 5 per cent per annum. '^ Proceedmg according to the Rule given above 1250 5 $12-50 therefore the interest is $12*50 Measonfor the Process, The sum of $100 must have the same relation in respect of magmtude to $250 as the simple interest of $100 for a year has to the simple interest of $250 for a year ; and thus the $100, $250 15, and the required interest must form a proportion. (Art. 149). We have then $100 : $250 : : $5 : required interest, vrnon/>£> «>Ai^iii«<\;/1 i»<^^^>>».4. 250X5 10^(Art. 156). which agrees with the Rule given above. '.i) n»t m ,i mmu,t«- m ABlTdMtTXO* Examples worked Ex. 1. Find the simple interest and amount of $460.14 for 1 year, 10 months, at 3^ per cent. $46014 138042 115035 1495-455 4 ) $46014 115035 Int. for 1 year Int. for 6 mos., or ^ of 1 year = Int. for 4 mos., or | of 1 year = =$1495455 7-477275 4-98485 .•. Int. for 1 year, 10 months =$27-416675 .-. Amount = $460-14+ $27-416675 = $487556675. Note. In examples like the above we may reckon 12 months to the year ; but if calendar months are given', the interest will then be best found by the Rule of Three; as for instance in the following example: Ex. 2. Find the simple interest and the amount of $500, from June 15, 1843, to Aug. 27, 1843, at 4^ per. cent. ? 500 4-5 2500 2000 $2250-0 $22-50= interest for 1 year. The number of days from June 15 to Aug. 27 =15+31+27 = 73, Hence, 365 days : 73 : : $22-50, whence it may be found that interest required =$4-50 ; .-. amount=$500+$4-50=$604-50. Questions on Commission, Brokerage and Insurance, these charges being usually made at so much per cent, amount to the same thing as jfinding the interest on a given amount at a given BJUrjJt QirT9BXBT. m 4 for 1 nths to ill then in the K from f 1 these to the i given rate for one year, and may therefore be worked by the Rule given above for Simple Interest. There is, however, one case of Insurance which it may be 'well to notice by an example worked out. Ex. If goods worth 11200 be insured at |16 per cent., to what amount must they be insured, so that in case of loss the party insured may recover the value of the goods and the premium ? If they be insured at their actual worth the premium paid will be lost, since the insurer will get $1200 only. But if every ($100—16.), or $84, be insured for $100, then, m case of loss, the value of the goods $84+ $16, {the premium paid) will be recovered. Thus we have $84 : $1200 : : $100 : sum which is required to be insured : whence, sum required to be insured=$1428-571^. Note. If it be required to find the interest on a sum of money in £. 8. d. it will be necessary to reduce the given sum to the decimal of a pound, and then to proceed under the ordinary rule. Ex. LIV. 1, Find the simple Interest fl) On $85 for 1 year at 5 per cent. (2) On $310 for 1 year at 4 per cent. On $1000 for 1 year at 4t^ per cent. On $475 for 3 years at 5 per cent. On $936-75 for 2 years at 4 per cent. On 556-50 for 6 years at 5 per cent. On $945 for 2 years at 4 per cent. On $198 for 1 year at 3^ per cent. ^ On $2361 for 2^ years at 3 per cent. (10) On $98 for i year at 2| per cent. 2. Find the amount 11) Of $1000 for 2 years at 41 per cent. (2) Of $2833i for 41 years at 3 per cent. Of $1050-625 for 6 years at 4^ per cent. Of $i39| for 3| years at 5^ per cent. Of $1895-35 for ^ years at 2f per cent. 2o (5) 210 ▲RITBMSTIO. i f6) Of £1534. 6«. Zd. for 1| years at 3| per cent (7) Of jWII. 10*. for I year at 4} per cent. [8) Of $1595-125 for 5f years at 3J per cent. 3. Find the simple Interest and Amonnt Of $375 for 3 years, 8 months, at 3^ per cent. Of $446^ for 3 years, 3 months, at 6 per cent. Of $220 for 7 months at 3f per cent. Of £243. lOff. for 2 years, 5 months, at 4| per cent. Of $10 for 117 days at 3^ per cent. Of $684 for 1 year 11 months, at 1^ per cent. Of $500 from March 16, 1850, to January 23, 1851, at 5|- per cent. Of $7290-56 for 2 years, 35 days, at 7f per cent. Of £34. 10*. from August 10 to October 21, at 6^ per cent. "4. What is the annual cost of insuring $40000 worth of property at ^ per cent. % . . 5. What must be the suni insured at 4^ per cent, on goods worth $19100 so that in case of loss the worth of the goods and the premium may be recovered ? 6. At 7^ per cent, what will be the cost of insuring property worth $500, so that in the event of loss the worth of the goods and the premium of insurance may be recovered ? 167. In all questions of Interest, if any three of the four (principal, rate per cent., time, amount) be given, the fourth may he found} as, for instance, in the following examples. £x. 1. Find the amount of $225 in 4 years at 8^ per cent, simple interest. Interest for 1 year=$ — Tno =$ — 4 =$?2= $7-875 o .% Interest for 4 years=$(7-875 x 4) =$31 -50. Amount = Principal + Interest =$225 + $31-50 =$256-50 cent. ^ 1851, at 21, at 6^ f property on goods ^oods and property the goods the four ourth may per cent., SIMPLE INTSRBST. %\\ Ex. 2. In what time will $225 amount to $256.50 at 84 per cent ? ■ '^ *n$!^.^'^^""^^^^==^^^*^^ ^^'«^ ^3 <^*»« in<=erest to be obtained on f325 m order that it may amount to $256-50. But Int. of $225 for 1 year=$7'875 ; which must have the same relation m respect of magnitude tp the $31-50, as the one year has to the required time. .'. $7-876 : $31-50 :: l year : required number of years, whence required number of years =4. Ex. 3. At what rate per cent., simple interest, will $225 amount to $256-50 in 4^ years? In other words, at what rate per cent. will $226 give $31-50 for interest in 4 years, or ??i!^, or $7-875 • « - 4 m one year ? Then $225 : $100 : : $7-875 : required rate per cent., whence required rate per cent. =3^. Ex. 4. What sum of money will amount to $256*50 in 4 years, at 3^ per cent., simple interest ? $100 in 4 yrs. at ^ per cent, amounts to $ 1 00+$(34 x 4) or $1 14 ; and this $114 must be to the $25650 as the $100 is to the required sum of money ; ^ .-. 1114 : $256-50 : : $100 : required number of dollars, whence required number of dollars =$225. 168. The student who has made sufficient progress in Algebra to understand the following equations, will find it advantageous to- remember the formula M=P+Pnr, by which all examples in Interest may be easily worked, and which may be explained aa follows : " r Let P represent the principal (in dollars). n — . — number of years. »• interest of $1 for 1 year. ■3/" *- amount of $P in n years. Then $1 : $P : : r : interest of $P for 1 year or interest of $P for 1 year=Pr. »years=Pm. 212 AftlTHMSnO. (1) .-. w» = P + Pm. (3) = P (1+rn.) and in — P=Prn, m— P (4) also -p7-= w apd since w=*P (1 +r».) (6) .-. l+rn P. Wherefore having any three of the quantities P. m. r. n. given, the fourth may easily be found. Note. — Care must be taken not to misunderstand the meaning of r which is the interest of $1, not of $100. 169. The following formula, deduced from the rule under Art. 166, will be found applicable to all cases of interest when three of the four quantities (time, rate, interest, principal) are given, and from its simplicity will be easily understood and remembered. (1) (2) (3) 100 .'. prt. = t X 100 . «_*xl00 ,\p — — — rt iXlOO and r = (4) and t = pt ex 100 pr The following examples will illustrate this formula : What is the interest of $100 for 16 yrs. 8 mo., at 6 per cent ? (1) i =$ ^QQx^> iig|==$ggg=$ioo. ^ ^ 100 3 What is that sum, the interest of which at 7 per cent for 5 yrs. 4 mo. is $728 ? (2) ^^1 728x^^^ 104 X 100 ^^10400 ^ 1=$1950. siMPLi nnraiuBflT. At what rate per cent, will $600 gain |800 in 5 years 1 ^ ' 600x5 How long will 11800 be in gaining $702 at 6 per cent 1 (4) .=Z^^2i2^=ll^=6l=6 yrs. 6 mo. ^ ' 1800x6 18 ^ -^ 218 Ex. LV. 1. What sum will brmg $250 in 4 years at 5 per cent, simple interest ? 2. At what rate per cent, will $540 amount to $734 in 9 years, at simple interest 1 3. In what time will $850 amount to $402*50 at 3 per cent simple interest ? '4. At what rate per cent, will $752*60 amount to $9970 in 8| years, at simple Interest *? 5. In what time will $729 amount to $1250*85 at 7^ per cent, simple interest ? 6. At what rate will $729*37^ amount to $5695 in 25 years at simple interest 1 • 7. What sum will produce for interest $75 in 2^ years at 6^ per cent, simple interest 1 8. What sum will amount to $529*30 in 3^ years at 6| per cent, simple interest ? 9. What sum will amount to $38*5 in 3 years at 4 per cent, simple interest 1 10. In what time will $1275 amount to to $1549*25 at 3f per cent, simple interest? 1 1. At what rate per cent, simple interest, "will $936 amount to $1157*75, in 4-J years? 12. In what time will $125 double itself at 5 percent simple interest 1 13. What sum will amount to £425. I9s. 4|rf. in 10 years at3| per cent simple interest, and in how many more years will it amount to £453. lis. Id. 1 214 ARITHMBTIO. 14. What sum of principal money, lent out at 5 per cent, per annum, simple interest, •will produce in 4 years the same amount of interest as £250, lent out at 3 per cent, per annum, will produce in 6 years 1 16. W4iat time will be required for $2,500 to gain $470, at 8 per cent ? 16. In what time will $284-75, at 5| per cent, give $18.75 interest? ' . .17. What principal will in 7 years and 6 months, at 8 per cent, amount to $2600 ? 18. The interest of $120 for 2 years 9 months and 12 days is $18*86, what is the rate per cent ^ 19. What is the interest at 8| per cent, of $384*25 from January 12th, 1859, to April 4th, 1860 ? - 20. What is the amount of $S7*05 for 2 years 3 months and 20 days, at 7 per cent ? 21. What principal will, in 4 years and 9 months, at 8 per cent., give $19*38 interest 1 22. The interest of $248 for 2 years 1 month and 20 days is $29*194, what is the rate per cent 1 COMPOUND INTEREST. 170. To find the Compound Interest of a given sum of money at a given rate per cent, for any number of years, RuLB. "At the end of each year add the interest of that year, found by Art. (166), to the principal at the beginning of it; this will be the principal for the next year j proceed in the same way as far as may be required by the question. Add together the interest so arising in the several years, and the result will be the compound interest for the given period." The reason for the above Mule is clear from what has been stated in Arts. (165 and 166). Sz. Required the compound interest and the amount of $720 for 8 years at 5 per cent. COMPOUND INTZRXST. 916 Proceeding as in Simple Interest for the 1st year ; , $720 , 5 $3600 $720=1'* principal, 36 =!•» interest, by addition, $756 =2~* principal, of which find in. at 5 per cent, 5 $37-80=2°* interest, $756=2"* principal, 87-80 =2'^ interest, by addition, $793*80=3'* principal of which find in. as above. 5 $39-6900=3'* interest, $793-80= principal for 3'* year, 39-69 = interest for 3** year, by addition,.-.$833.4y = amount of $720 in 3 years at 5 per cent The compound. interest for that time . c=sum of interests for each year, =$36 + $37-80+$39-69=$l 13-49. If the rate per cent, be a factor of a 1 00, the following method, by practice, will be more concise than the preceding. Thus : 5=,V of 100 5=,V of 100 5=jV of 100 $720 =!•» principal, 36 =!•* interest, 756 =2°* principal, 37-8 =2"* interest. 793 8 =3'* principal, 39 •69=3'* interest. 833 49= amount in 3 years. I^ote 1. It is customary, if the compound interest be required for any number ot entire years and a part of a year, /^for instance for 6| years), to find the compound interest for the 6th year, and then take ^ths of the last interest for the |ths of the 6th year. 916 ABITHMSnO. Note 2. If the interest be payable half-yearly, or quarterly, it is clear that the compound interest of a given sum for a given time will be greater as the length of each given period is less ; the simple interest will not be affected by the length of each period. Ncte 3. In working sums in compound interest it will generally be fouud advisable to reduce to decimals any vulgar fractions that may occur in the time given, or the rate per cent. 171. Any example in CSonopound Interest may be worked by remembering the formula, M=P(l-{-r)n. Let P denote the principal (in dollars), n number of years, • r interest of $1 for 1 year," M amount of $P in n years. Since at the end of the first year, $1 amounts to ($l+r), and since the same proportion holds for each successive year, we obtain $1 : (ii+r)::$p : $P(i+r), or the amount of IP in 1 year is $P(l+r). Similarly, |1 : (ei+r) :: $P(l+r) : $P(l+r)(l+r), or|P(H-f)», or the amount $P in 2 years is $P(1 + r)\ Similarly, the amount of $P in n years is |P(1 +r)", orif=P{l+r)"; In which equation any three of the quantities Jf, P, r, n being given, the fourth may be found. Ex. Find the amount of $200 in 2 years, at 4 per cent., Compo'md Interest. Here P=8200, w=2, r= 4^, ^=-04 ; .-. M= P(l+rY = $200 (104)» = $216-32 Ex. LVI. 1. Find the compound interest of $2000 in 2 years at 4 per cent, per annum. 2. Find the amount of $600 in 3 years at 3| per cent., allowing compound interest. PRISIITT WORTH AHD DISOOUITT. an terly, it en time 88 ; the eriod. snerally ons that ked by r), and 6 obtain 1+r)', n being jr cent., 8. Find the oompuur i interest -^^ ^270 in 'i yt;%ri, at 3 per cent 4. Find the amount of $690 lot 3 years it 4^ per cent, compound interest. <■ . ' ' 5. Find the amount of 1230*75. for 3 years, at 6 per cent., compound interest. 0. Find the difference in the amount of $416*50, put out for 4 years at 2^ per cent., Ist at simple, 2nd at compound interest. 7. Find the compound interest of $130 in 3 years at 4 per cent, (interest being payable half-yearly). 8. What will $1760*75 amount to in 2| years, allowing 4 per cent, compound interest ? 9. A person lays by $230 at the end of each year, and employs the money at 6^ per cent., compound interest ; what will he be worth at the end of 3 years 7 10. Find the difference between the simple and compound interest of $416| for 2 years at 6| per cent. 1 1 . What is the difference between the simple and the compound interest of £13,333. 6«. Sd., for 5 years, at 5 per cent ? 12. Find the amount of $180 in 3 years at 4^ per cent, compound interest. 13. What sum of money, put out at compound interest for 2 years at 5 per cent., will amount to $100? 14. What sum at 10 per cent, compound interest, will amount in 2 years to $264*60 1 15. ^ and B each lend £256 for 3 years, at 4^ per cent per annum, one at simple interest, the other at compound interest ; find the difference in the amount of interest they respectively receive. PRESENT WORTH AND DISCOUNT. at 4 per Allowing 172. A owes B $600, which is to be paid at the end of 6 months from the present time ; it is clear that, if the debt be discharged at once, (interest being reckoned, we will suppose, at 4 per cent, per annum,) B ought to receive a less sum of money than $600; in fact, such a sum of money as will, being now put out at 4 per cent, interest, amount to $600 at the end of 6 months. The sum which B ought to receive now is called the Present Worth of the $600 2d 218 ARITHMBTIO. diue 6 •months hence, and the sum to be deducted from the $000, in consequence of immediate payment, which is in fact the interest of the Present Worth, is called the Discount of the $600 discharged 6 months before it is due. Dbf. We may therefore define Present Worth to be the actual worth, at the present time, of a sum of money due some time hence, at a given rate of interest ; and we may define the Discount of a pum of money to be the interest of the Present Worth of that sum, calculated from the present time to the time when the sum would be properly payable. PRESENT WORTH. 173. Rule. Find the interest of $100 for the given time at the given rate per cent., and state thus : $100+ its interest for the given time at the given rate per cent. : given sum, ; ', $100 I present worth required." Ex. 1. Find the present worth of $600, due 6 months hence, at 4 per cent, per annum. Proceeding according to the above Rule, Interest of $100 for 6 months at 4 per cent, is $2. .-. $102 : $600 : : $100 : required present worth, whence, required present worth =$588 '2338^. The Reason for the above process is clear from the consideration, that $100 in 6 months at 4 per cent, interest would amounjb to $102, and therefore $100 is the present value of $102 due 6 months hence : and consequently we have 1'* debt : 2°* debt : : P* present worth : 2"'^ present worth. Ex. 2. Find the present worth of $838, due 1 9 months hence, at 6 per cent, simple interest. Since the interest of $100 for 19 months, at 6 per cent. =${Hx6)=$V=$9i .'. $109^ : $838 : : $100 : required present worth, whence, required present worth =$765 '29+ Ex. 3. What is the value, at 16 years of age, of a legacy of $1000 payable at 21 years of age, allowing simple interest at 4 per cent. ? Since $100 at 4 per cent, simple interest will in 5 years amount DISOOUST. 219 to $120, therefore the present worth of $120 due 5 years hence will at that rate be $100. Hence $120 : $1000 : : $100 : required valued whence, required value =$8S3^. DISCOUNT. 174. Rule. Find the interest of $100 for the given time at the given rate per cent., and state thus : $100+its interest for the given time at the given rate per cent. : given sum : : interest of $100 for the given time at the given rate per cent. : discount required." Ex. 1. Find the discount of $600, due 6 -^^onths hence, at 4 per cent, per annum. Proceeding according to the above Rule, The interest of $100 for 6 months at 4 per cent. =$2 ; therefore proceeding according to the Rule, $102 : $600 : : $2 required discount, whence, required discount =$11 'TOy"^. The reason for the above process is clear from the consideration, that $2 is the interest for 6 months, at 4 per cent., of $100, the present worth of $102 due at the end of that time; and consequently we have 1"* debt : 2"* debt .* ! discount on l'*debt : discount on 2~*debt. Ex. 2. Find the discount on $1000, due 15 months hence, at 5 per cent, per annum. The interest of $100 for 15 months at 5 per cent. =$6-25. .-. $106-25 : $1000 : : $6-25 : required discount, whence required discount =$58*82i nearly. Ex. 3. Find the discount on £127. 2« for half-a-year at 5 per ' cent. £I00f : £127yV :•: £f : required discount; whence required discount=£3. 2*. Note 1. Discount =ffiven sum less Present. Worth • Proaanf Worth=given sum less Discount. Note 2. Bankers and Merchants in discounting bills calculate 220 ARITHMETIC. interest, instead of discount, on the sum drawn for in the bill, from the time of their discounting it to the time when it becomes due, adding three dats of grace, which are usually allowed after the time a bill is nominally due, before it is legally due ; which is of course an additional advantage. When a bill is payable on demand, the days of grace are not allowed. Note 3. If a bill, without the days of grace, should appear to be due on the 31st of any month which contains only 30 days, the last of that month, and not the first day of the next, is considered as the day on which the bill is due. Thus a bill drawn on the 31st of October, at 4 months, would be really due, adding in the days of grace, on the 3rd of March. Also bills which fall due on Sunday, are paid in England on the previous Saturday. Ex. A bill of $1000 is drawn on Feb. 16th, 1851, at 7 months' date ; it is discounted on the 8th of July at 5 per cent. What does the banker gain by the transaction f The bill is legally due on Sept. 19 ; and from July 8 to Sept. 19 are 73 days. The interest of $1000 for the time=$10. The true discount = 9j90j^ .'. the banker's gain $09 fV, . Ex. LVII. 1. Find the Present worth of [1) $233i^ due 1 yearhence,at 5 pr. ct. pr. ann.,simple interest. 2) $252.9675 6^ [3) $676f ..6 months 3 4) $284-90 ..6 3^ 5) $460-50 ..7 4 6) $390 ..7 3i._.. 7) $672 ..8.. 7^ 8) $1261-05 ..1 1 9) $35 ..4 4^ (10) $1250 ..3 6 (U) $2110 .11 5 12) $275^ .15 4 [l3) $918 ..4years 5 .14) $500 .19 months 5^ hh\ $800 .20 years. „.-.5^ (16) $2197 ..3 years 4 ....compound interest the STOCKS. 221 2. Find the Discount on (1) $63^ due 4 months hence, at 4 per cent, per annum, [simple interest. (2) $1380-375.. 9 3 ,. (3) $107-25 ..6 5 . (4) $125-50 ..3 ...3^ :5) $487 ..5 ...3^ 6) $340 ..5 4 7) £3640 .10 44 .. (8) £813 9». ..11 4I ..... [9) £250 15s. .17 months ...5 (10^ $55 146days 4f (11) A bill of $649 is dated on June 23, 1853, at 6 months, and is discounted on July 8, at 3^ per cent. ; what does the banker gain thereby 1 (12) Find the true discount on a bill drawn March 17, 1859, at 3 months, and discounted May 2, at 5J per cent. (13) Find the simple interest on $545 in 2 years, at 3^ per cent. , per annum ; and the discount on $583-15, due 2 years hence, at the same rate of interest. Explain clearly why these two sums are identical. (14) Explain the difference between Discount and Interest. Five volumes of a work can be bought for a certain sum, payable at the end of a year ; and six volumes of the same work can be bought for the same sum in ready money : what is the rate of discount 1 (15) A tradesman marks his goods with two prices, one for ready money, and the other for one year's credit allowing discount, at 5 per cent. ; if the credit price be marked £2 9s., what ought to be the cash price 1 STOCKS, BROKERAGE AND COMMISSION. 175. Stocks are Government funds, the Capital of Banks and other Joint Stock Companies. Brokerage is the allowance paid to a Broker, or dealer in stocks or bills of exchange for tranaacting business. This allowance in 222 ▲RITHMETIO. 1 England is £^ or 2s. 6d. per cent. The rate of allowance varies, however, on this continent. Commission is an allowance made to an agent for the sale, purchase, or car^ of property, on the money employed. The capital df a company, or money paidin^ is divided into shares^ which are owned by Stockholders. The original cost of a share is its 7)ar value. If it sell in the market for more than its original, X j is said to be above par, or at an advance ; if it sell for less, it is behw par, or at a discount. The original cost of a share is usually |100, though it is sometimt, $25, $50, $500, <&c. The rise and fall in stocks is a per cent, on the par value. Thus a share, whose par value is $i^^, at 16 per cent, advance, will ^^'"g \H of its original cost ; at 16 per cent, discount it will bring ^u* t\V of its original cost. The profits of these companies are every year, or every half-year, divided among the Stockholders. The amount so paid out is called ?, dividends The actiiai cost of $100 share in any company, when, for instance, the $100 share is worth in the market $94^, will be $94j+the brokerage. The actual sum received on such share where the brokerage is 1 per cent, will therefore be ($94|— ^) $94. All examples in Stocks depend on the principles of Proportion ; thoc3 cf most frequent occurrence will be now explained. Ex. 1. Required the sum which will purchase $1500 worth of mining shares, at $82 ; the $100 share. in this case $100 stock cost $82 in money ; .*. $100 I $1500 : : $82 money ; required sum of money. whence, required sum of money =$1200. Ex. 2. What amount of Railway stock will $4050 purchase, the marke^ value of the $100 share being $90 ? In this caje $90 money will purchase $100 stock ; .-. $90 : $4050 : : $100 stock ; required amount of stock. whence, required amount of stock =$4500. "Ry- 3. If I invp.st. ^\^*I0 in r minincr nnrnnnnv nauinnr 9 nav cent.,the stock being $93^ per $100 share,and pay $| for brokerage, what does it cost me 1 II st 11 Ci q ri e fi a STOCKS. 238 Every $100 stock costs me ($93^ + |), or $93f ; .-. $100 stock : $1520 stock : : $93f : required sumof money ; whence, required sum of money=$1419-30. Ex. 4. What sterling money shall I receive for £1920 13*. 4d, in the 3^ per cent, government stock at 98|, brokerage being £| per cent. 1 . £100 stock realizes £(98-J— i)=98a ; /. £100 stock : £1920| stock : : £98f : required sterling money ; whence, required sterling money=£1896. 13*. 2c?. Ex. 5. If I invest £7927 10«. in the 3 per cent, government stock at 94f, what annual income shall 1 receive from the investment 1 For every £94f I get £100 stock, and the interest on £100 stock is £3 ; therefore for every £94f of money I get £3 interest ; /. £94f : £7927 10*. : : £3 : required annual income ; whence, required annual income =£252. Hfote 1. If it be required to find the income arising from a certain quantity of stock, it is merely a question of simple interest N^ote 2. It may be noticed in the above examples, that when the question was simply to find the amount of stock, or money realized by sale of stock, the 3, 4, or other rate per cent, never entered into the statement ; and when the question was simply to find income arising from any sum invested in the funds, then the $100 never entered into the statement. Note 3. All questions of the transfer of stock from one kind to another, belong to the Rule of Three inverse. Note 4. The par value of all English Government stock is £100 per share. Ex. LVIII. 1. Find the quantity of Railway stock, the par value being $100 per share, purchased by investing : 11 ) $2850 paying 3 per cent, at 75. 2) $712 paying 3^ per cent, at 89. [3) $504 paying 6 per cent, at 96. (4) $883-50 paying 4 per cent, at 93. (5) $3741 paying 3| per cent, at 87. (6) $500 paying 3 per cent, at 83^. 224 ARITHMXTIO. 3. Find the yearly income arising from the investment in government securities, par value the same as preceding. n $1008 paying 6 per cent, at 84. [2) $5580 paying 6 per cent, at 93. 3) $1138-50 paying 4| per cent, at 92. 1 4) $1638 paying 4^ per cent, at 93^. 5) $2000 paying 6 per cent, at S^. (6) £3500 in the 3 per cent, consols at 94|-, brokerage | per cent. 8. Find the quantity of Bank stock purchased by investing (1) $800 paying 4 per cent, at 75^. (2) $4311 paying 3^ per cent, at 85f. (3) $2000 paying 3^ per cent, at 94. (4) $2353 paying 3 per cent, at 90f , brokerage | per cent. (5) $3277 paying 4 per cent, at 105^, brokerage \ per cent. (6) $10000 paying 3^ per cent.''at 99^, brokerage ^ per cent. 4. (1) I bought 27 shares of Railway Stock, at 13 per cent, discount and sold them again at an advance of 2 per cent. How QMich did I gain by the operation 1 The par value was $100. (2) A gentleman paid a broker J per cent, to invest $19278 in government funds. How much was the brokerage ? (3) A person in England transfers £1000 stock from the 4 per cents at 90, to the 3 per cents at 72 ; find the alteration in his income. (4) Which is the better investment, $1896 in city debentures, paying 6 per cent, at 87, or in Railway shares, paying 6^ at $89 ? (5) A lady has a bequest of $10500 ; she paid an agent 2^ per • cent, commission per annum to take care of the money for her. How much did the commission amount to 1 (6) A factor sells 43 bales of cotton at $375 per bale, and charges 2 per cent, commission. How much money must he pay to his principal ? (7) What is the value of 50 shares of Bank stock, 12 per cent, below par ; the par value of which was $200 per share ? (8) Find the income produced by $12600, and invested in a blanket factory, paying 7^ per cent., the stock being at the time of purchase $85 per $100 share. „ PJLRttAL PAYKnCNTfl— lO&MS Of K0TI8. fM „ PARTIAL PAYMENTS— FORMS OF NOTES. 176. When notes, bonds or obligations receive only partial payments or endorsements, the following Rule is that usually adopted : Rule. The rule for casting interest, when partial payments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. If the payment exceed the interest, the surplus goes towards discharging the principal, and ' the subsequent interest is to be computed on the balance of the principal remaining due. If the payment be less than the interest, the surplus of interest must not be taken to augment the principal, but interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surj^us is to be applied towards discharging the principal ; and interest is to be computed on the balance, as aforesaid. Ex. 1620 Hamiltok, Nov. 1, 1857. For value received, 1 promise to pay Samuel Thompson, or order, the sum of six hundred and twenty dollars on demand, with interest. THOMAS JONES. The following endorsements were made on this note : — 1858, Oct. 6, received 161-07 ; March 4th, 1850, $89*03 ; Dec. 11, 1859, 1107-77 ; July 20th, 1860, $200-50. What is the balance due, Oct. 15, 1860, allowing 7 per cent, interest? 2i 236 ABITHMETK). The amount of note, or principal, is $620'000 Int. on the same to Oct. 6, 1858, at 7 per cent., is . . 40386 Amount due on note, Oct. 6, 1858, is 660*386 The first endorsement is 61 0*70 599-316 Interest from Oct. 6, 1858, to March 4, 1859, is 17247 Amount due March 4, 1859, is 616568 The second endorsement is --. 89'030 527-538 Interest from March 4, 1859, to Dec. 11, 1859, is 28-414 555-947 The third endorsement is 107 770 448-177 Interest from Dec. 11, 1859, to July 20, 1860, is 19-085 467-262 The fourth endorsement is - 200-500 266-762 Interest from July 20, 1860, to Oct. 15, 1860, is 4-409 ^ws. 271171 Ex. LIX. $850 Toronto, May 1, 1856. 1. For value received, I promise to pay Timothy Snooks, or order, three hundred and fifty dollars, with interest, at 6 per cent. M. BROWN. Dec. 25, 1856, there was endorsed $50 ; June 30th 1857, $5 ; Aug. 22, 1858, 115 ; June 4, 1859, $100. How much was due April 5, 18601 $143^^(^ London, C. W. Aug. 1, 1857. 2. For value received, I promise to pay Henry Toms, or bearer, one hundred and forty-three dollars, and fifty cents, on demand, with interest. S. JONES. Deft. 17. 1857. there was endorsed $37-40 : Julv 1st. 1858, $7-09 ; Dec. 22, i859, $13-13 ; Sept. 9, 1860, $50-50." How much remains due Dec. 28, 1860, the interest being 7 per cent. '? pRom Aim LOSS. 987 I PROFIT AND LOSS. 177. Dbf. All questions in Arithmetic which relate to gain or loss in mercantile transactions, fall under the head of Peofit AMD Loss. Examples in Profit and Loss are worked by the principle of Proportion : various examples will now be worked out by way of illustration. Ex. 1. If a cask of wine containing 84 gallons cost 1412.60, what is gained by selling it at $6 per gallon 1 The gain = selling price less first cost ; the selling price=l(6x84)=$504; therefore the gain =1504—1412 '50 =$9 1-50. Ex. 2. A ream of paper cost me 15-25 what must I sell it at, so as to realize 20 per cent. 1 The reasoning in this case is, If $100 gain $20, or produce $120, what will $5-25 produce % .-. $100 : $525 : : $120 : required amount in dollars, whence, required araount=$6*30. Ex. 3. A man buys 33 geese for $30-50 ; at how much per head must he sell them to gain 10 per cent, on his outlay I In this case, $100 : $30-50 : : $110 ; selling price of the geese in dollars, whence, selling price = $33-55. .-. selling price of each goose = $'|-^«=$1-01|. Ex. 4. A person buys shares in a railway when they are at £194, £15 having been paid, and sells thom at £32. 9*. when £25 has been paid : how much per cent, does he gain? He buys each share at £19^, and he afterwards pays upon it £(25— 15V or £10 -, therefore at the time he sells, he has paid on each share £29. 10«. ; therefore by selling at £32 9*. he gains on each £29. 10.. ; therefore by selling at £32. 9jr. he gams on each £29. 10s. which he has paid (£32. 9*.— £29. 105.)=£2. 19*. ; whence, gain per cent. =£10, or gain is 10 per cent. / ▲ItlTBMXTXO. Ex. 5. What was the prime cost of an article, which when sold for $12, realized a profit of 20 per cent. 1 Here what cost £100 would be sold for $120 ; .-. $120 : $12 : : $100 : prime cost in dollars, whence, prime cost =$10. If the above example had been, •' What was the prime cost of an article, which when sold for $12, entails a loss of 20 percent. V* then $80 : $12 : t $100 : prime cost in dollars, whence, prime cost = $15. Ex. LX 1. Bought 6 cwt. 3 qrs. 14 lbs. of cheese at $9*50 per cwt., and sold it again for $11*20 per cwt. What was the gain upon the whole 1 2. If 6 cwt. 3 qrs. 14 lbs. be bought for $47 and sold for $65-80 what is the rate of gain per cwt. ? 8. Find the total value of 43 articles at M. 6«. Sd. each, 67 at £11. 8«. 4rf. each, and 4 at £13. 15s. 4rf. each. What is gained or lost by selling them at the rate of 3 for £28 ? 4. A person buys 400 yards of silk at $500 and sells 300 yards at $1*50 a yard, and the rest, which is damaged, at $-75 a yard ; find how much per cent, he gains or loses. 5. A grocer buys 2 cwt. of sugar at 10 cents per pound, and 4 cwt. at 9 cents ; he sells 3 cwt. at 9^ cents per pound ; at what rftte per pound will he be able to sell the remainder so as neither to gain nor lose by the bargain 1 6. If a commodity be bought for $6.50 a cwt. and sold for 9 cents a lb., find the rate of profit per cent. 7. Bought goods at 10 cents per pound, and sold them at $7.50 per cwt. ; what is the gain or loss per cent. 1 8. An article which cost 3«. 6rf. is sold for 3*. lO^d. ; find the gain per cent. 9. Goods were sold at $12, at a profit of 22f f per cent. ; what was the prime cost 1 10. If a tradesman gain $5*50 on an article which he sells for $22, what is his gain per cent. ? 11. A man sells a cow for $24*60, and loses $18. per cent, on what the cow cost him ; what was the original»cost ? PROVR Aim LOBS. 12. By selling an article fur $5. a person loses 5 per cent. ; what was the prime cost, and what must he sell it at to gain 4| per cent. ? I • 18. The cost price of a book is |^ ; the expense of sale 5 per cent, upon the cost price ; and the profit 25 per cent upon the whole outlay ; find the selling price of the book. 14. If by selling an article for $25*50, 8 per cent, be lost, what per cent, is gained or lost if it be sold at $38 ? 15. I bought 500 sheep at $2*10 a-head; their food cost me 27| cents, a-nead : I then sold them at $2'425 a-head. Find my whole gain, and also my gain per cent. 16. A person having bought goods for $40 sells half of them at a gain of 5 per cent ; for how much must he sell the remainder so as to gain 20 per cent on the whole ? 17. A person has goods worth $30 ; he sells one-third of them so as to lose 10 per cent. ; what must he sell the remainder at so as to gain 20 per cent, on the whole ? 18. I buy ^ house for $5000, and sell it Immediately at a profit of 30 per cent ; what do I receive, supposing the expenses of the sale to be 5 per cent 1 19. The prime cost of a 76-gallo& cask is $23'625, but 18 gallons are lost by leakage ; 9 gallons of water is then mixed with the remainder, and it is sold at $'375 a gallon. Find the whole gain, and also the gain per cent 20. A stationer sold quills at lis. sterling a thousand, by which he cleared f of the money ; he raises the price to 13«. 6d. What does he clear per cent, by the latter price ? 21. A person sold 72 yards of cloth for $43*50 ; his profit being the cost of 11*52 yards; how much did he gain per cent 1 26. A person expends $3000 in railway shares at 15^ per cent, discount, and sells them at par ; what does he gain by the transac- tion, and what per cent. 1 27. A wine-merchant bought 14^ pipes of wine, which having received damage, he sold for $1120 ^f , thereby losing 20 per cent ; find the cost of the wine per pipe, and the selling price of it per gallon. 28. A farm is let for £96 and the value of a certain number of quarters of wheat When wheat is 38«. a quarter, the whole ▲RZTHiaTXO. rtnt is 15 per cent lower than when it is 66#. a quarter. Find the number of quarters of wheat which are paid as part of the rent 29. A man having bought a lot of goods for $150, sells ^rd at ft loss of 4 per cent. ; by what increase per cent, must he raise that selling price, in order that by selling the rest at the increased rate, he may gain 4 per cent, on the whole transaction? 80. A person bought a French watch, bearing a duty of 25 per cent, and sold it at a loss of 5 per cent. ; had he sold it for $3 more, he would have cleared 1 per cent, on his bargain. What had the French maker for it 'i DIVISION INTO PROPORTIONAL PARTS. 178. To divide a given number into parts which shall be proportional to certain other given numbers. This is merely an application of the Rule of Three ; still it may be well to state a general Rule, by which examples which come under the above head may be worked. RuLB. State thus : "As the sum of the given parts ; any one of them : : the entire quantity to be divided : the corresponding part of it" * This statement must be repeated for each of the parts, or at all events for all but the last part, which of course may either be found by the Rule, or by subtracting the sum of the values of the other parts from the entire quantity to be divided. Ex. 1. Divide |40 among A, B. and (7, so that their propor- tions may be as 7, 11, and 14, resp ^tively. Proceeding according to the Rule given above, 32 : 7 : : 40 dollars : ^'s share, 32 ': 11 : : 40 dollars : B'8 share, whence A's share=$8'75, and ^'s share= $13*75. C's share may be found from the proportion 32 : 14 : : 40 dollars : C's share ; whence C's share=$17*50 ; or by subtracting $875+ $13-75, or $22*50 ^rom $40, which leaves $17*50, as above. ruLowtHip OR PAknruuiHip. 281 Th« rtaaonfor thi above proem is dear from the consideration, that 40 dollars is to be divided into 32 equal parU, of whioh A is to have 7 parts, SI 1, and C 14. w Ex. 2. Divide $11000 among 4 persons, A, B, C, D, in the proportions of |, |, |, and }. Sum of shares=^^ ; .*. H '. i • • 111000 : A'b share in dollars, whence ^'s share =$42854 Similarly, ■'% JB's 8hare=$28674, C'b 8hare=$2142f. D'b 8hare=$17144. Ex. 8. Divide $45000 among A, B, C7, and />, so that A'» share : B'a share 1 : 1 : 2, S'a : (7*5 : : 8 : 4, and d \ D't : : 4 : 5. In this case, B's 8hare=2 A^s share, 3 Cs share=4 B's share, 4 jD'* share =5 CPs share ; .'. we have (Ta share=|-6'* 8hare=f -4'« share, and i)'s share=| CPs share= y A^s share ; .*. At share + B's share + ^' re+D'* ahare =^'« ..arex(l+2ff+V), =9 ^'» share j .-. -4'« share=|5000, 5'«=10000, C?'*=$13333i, i>'«=$16666|. , FELLOWSHIP OR PARTNERSHIP. 179. Dbp. Fellowship or Partnership is a method by which the respective gains or losses of partners in uiy mercantile transactions are determined. Fellowship is divided into Simple and Compound Fellowship ; ip the fnriinAr the S"m* f^^ TTinnpv nnf. in \\\r t.lip !BAV*».rg,l Dartnftrg continue in the business for the same time; in the latter, for different periods of time. — fm AlUtBlllTlO. SIMPLE FELLOWSHIP. 180. Examples in this Rule are merely particular applioations of the Rule in Art. (178)) and that Rule Idierelbre applies. Ex. 1. Two merchants, A and B^ form a joint capital ; A puts in $240, and B $860 : they gain $80. How ought the gain to be divided between them ? $(240+360) : $240 : : $80 : A'» share in dollars, whence, A^s 8hare=$32, and .*. B^a share=$48. HFoie, The estate of a Bankrupt may be divided among his creditors by the same method. Ex. 2. A bankrupt owes three creditors, A^ B, and C7, $175, $210, and $265, respectively ; his property is worth $422'50 : what ought they each to receive? $650 : $175 : : $422^ : A's share, $650 : $210 : : $4221 : BU share, whence A^s share=$113j ^'* share=£l36| ; .-. CPs share=$172i. . I COMPOUND FELLOWSHIP. 181. RuLX. " Reduce all the times into the same denomination, and multiply each man's stock by the time of its continuance, and then state thus : As the sum of all the products \ each particular product .* \ the whole quantity to be divided \ the corresponding share." Ex. 1. A and B enter into partnership ; A contributes $3000 for 9 months, and B $2400 for 6 months, they gain $1150 : find each man's share of the gain. Proceeding by the Rule given above, $(8000x»+2400x6) : $(8000x9) :: $1160 \A'8 share of gain, or $41400 : $27000 ! ! 1150 : A^b share of gain, and $41400 : $14400 : : $1150 : Ba share of gain ; whence, As share=:$750, and B^a 8hAre^$400. BQUATIOK OF PAYMENTS. 233 tiona puts o be . I his 175, •50: tion, and :the {000 find hare The reason for the above process, is evident from the considera- tion, that a stock of $3000 for 9 months would be equivalent to a stock of 9 times $3000 for one month ; and one of $2400 for 6 months, to one of 6 times $2400 for 1 month : henbe, the increased stocks being considered, the question then becomes one of Simple Fellowship. EQUATION OF PAYMENTS. 182. Dkp. When a person owes another several sums of money, due at different times, the Rule by which we determine the just time when the whole- debt may be discharged atone payment, is called the Equation of Payments. Mte. It is assumed in this Rule that the sum of the interests of the several debts for their.respective times equals the interest of the sum of the debts for the equated time. Rule. ** Multiply each debt into the time which will elapse before it becomes due, and then divide the sum of the products by the sum of the debts ; the quotient will be the equated time required." Ex. 1. A owes B $100, whereof $40 is to be paid in 3 months and $60 in 5 months : find the equated time. Proceeding according to the Rule given above, then (40 x 3+60 x 5)=40+60) x equated time in months, whence, equated time=4^ months. The reason for the above pi^^cess^ in accordance with our assumption, is clear from the consideration that the sum of the interests of $40 for 3 months, and $60 for 5 montiis, is the same as the interest of $(120+300), or $420 for 1 month; if therefore ^ has to pay $100 in one sum, the question is, how long ought he to hold it so that the interest on it may be the same as the interest on $420 for one month. The statement therefore will be thus : $100 : $420 : : l month : required number of months; whence, required number of months=4| months ; which is evidentiy the equated time of payment, and agrees with the result obtained by the Rule given above. the end of 9 months : Ex. 2. A owed B $1000, to be paid 2t 2S4 ▲BITHMBTIO. he pays however |200 at the end of 3 months, and $300 at the end of 8 months : when was the remaider due 1 In this case, (2^0 X 3 + 300 X 8 + 500 x number of months required) = 1000 X 9, or 500 X number of months required=6000 ; whence number of months required=12. Ex. LXI. 1. A company of militia consisting of 72 men is to be raised from 3 towns, which contain respectively 1500, 7000, and 9500 men. How many must each town provide 1 2. Divide £17*5875 into two parts which shall be to each other as 5 : 16. 3. Divide 4472 into parts which shall be to each other in the ratio of 3, 5, 7, 11 ; and also 1500 into parts which shall be in the ratio of |, f and 4. 4. A bankrupt owes A $2501, B $203^, and (7|141| ; his estate is worth $421i|^ ; how much will A, B, and receive respectively 1 5. A mass of counterfeit metal is composed of fine gold 15 parts, silver 4 parts, and copper 3 parts : find how much of each is required in making 18 cwt. of the composition. 6. Two persons have gained in trade $720; the one put in $2200 and the other $1800 ; what is each person's share of the profits ] 7. In a certain substance there are 11 parts tin to 100 of copper. Find the weight of tin in a piece weighing 24 cwt. ? 8. A man leaves his property amounting to $13,000 to be divided amongst his children, consisting of 4 sons and 3 daughters ; the three younger sons are each to have twice the share of each of the daughters, and the eldest son as much as a younger son and a daughter together ; find the share of each. 9. Two persons, A and B, are partners in a mercantile concern, and contribute ^12000 and 420000 canital resnpo.t.ivelv ? A \a tr>. .--.-- — J- — — J. . — J J — — _.^ have 10 per cent, of the profits for managing the business, and the remaining profits to be divided in proportion to the capital I APPLIOATIOirS OF THK TERM PER OEHT. 235 contributed by each ; the entire profit at the year's end is 18000 ; how much of it must each receive *{ 10. Divide $100 amona A, jB, C and Z>, so that 5 may receive as much ss A ; Caa mua corn-rent equally apportioned between wheat and barley ; it is valued at |930 a year when the average price of wheat is $M0 a bushel, and that of barley $*75 a bushel ; find the rent when wheat rises to the average price of $1'50 per bushel, and barley to that of ll'OO per bushed. First we must find the number of bushels of wheat and barley at the given rent of $930. S930 — - — =$465 the sum to be raised by each kind of grain ; ' 465 110 465 •75 ■4:22j\ bushels of wheat ; =620 bushels of barley ; .'. rent in lattfer case=(422y«r x H+620 x 1) =|1254tV Ex. 5. A person's average annual income from 1830 to 1850 was $1550. In 1830 his income was $1680. and in 1851 his income was $1625, what was his average annual income from 1831 to 1851 (inclusive)? ^ \; 8QUABB ROOT. 241 His total inoome from 1831 to 1851 inclusive -$1550x21 + 11625—11680. =132495 I ? $32495 .♦. his average income= — sT" =$1547,^-. Ex. LXIII. 10 1. In 1845 the rental of an estate amounted to $18697, in 1846 to $17292, in 1847 to $2013550, in 1848 to $20078-75, in 1849 to $18582, in 1850 to $24048-25, in 1851 to $21631 ; find the average rental of the 7 years. 2. The number of bushels of grain exported from a country in 11 successive years were 2679438, 2958272, 3030298, 3474302, 2243151,2327782,2855525,2538234,3206482,2801204,3251901; find the average exportation during that period. 3. In a class of 23 children, 8 are boys, 15 girls. The age of the boys— 4 of 8, 2 of 11, 2 of 12. Of the girls— 5 the average age of the boys, 4 of 9, 2 of 10, 4 of 13. Find the average age of (a) the boys, (b) the girls, (d) the whole class. 4. There are 25 children on the register of one class in a school. 19 have been present at one time or other during the week. The sum of days on which the children have attended is 84|. What is the average number of days per week attended by each child ever present during the week, there being no school on Saturday or Sunday 1 Give the answer in decimals. 5. In a school of 7 classes, the average number of days attended by each child in Class I. is 4*5 ; Class 11., 4 ; Class III., 3 9 ; Class IV., 4-1 ; Class V., 3-6; Class VI., 42 ; Class VII., 3-3. Find the average number of days attended by each child in the school. 6. Divide $960 among three persons in such a manner that their shares shall be to each other as 5, 4, and 3, respectively. hiR SQUARE ROOT. 184. The Square of a given number is the product of that number multiplied by itself. Thus 36 is the square of 6. 2o 242 ▲RITHMBTIO. The square of a number is frequently denoted by placing the figure 2 above the number, a little to the right. Thus 6* denotes the square of (J, so that 6' =36. 185. The Square Root of a given number is a number, which when multiplied by itself, will produce the given number. The square root of a number is sometimes denoted by placing the sign v/ before the number, or by placing the fraction ^ above the number, a little to the right. Thus V 36 oi (36)i denotes the square root of 36 ; so that ^/ 36 or (36)i=6. 186. The number of figures in the integral part of the Square Root of any whole number may readily be known from the following considerations : The square root of 1 is 1 100 is 10 10000 is 100 1000000 is 1000 &;c. is &c. Hence it follows that the square root of any number between I and 100 must lie between 1 and 10, that is, will have one figure in its integral part ; of any number between 100 and 10000, must lie between 10 and 100, that is, will have two figures in its integral part; of any number between 10000 and 1000000, must lie between 100 and 1000, that is, must have three figures in its integral part ; and so on. Wherefore, if a point be placed over the units* place of the number, and thence over every second figure to the left of that place, the points will shew the number of figures in the integral part of the root. Thus the square root of 99 consists, so far as it is integral, of one figure ; that of i98 of two figures ; that of 176432 of fhree figures ; that of 1764321 of fwir figures ; and so on. 187. The following Rule may be laid down for extracting the square root of a whole number : RuLB. " Place a point or dot over the units' place of the given number, and thence over every second figure to the left of that place, thus dividing the whole number into several periods. The number of points will shew the number of figures in the required root. (ArtT 186.) Find the greatest number whose square is contained in the first SQUARt ROOT. 348 ing the denotes :, which placing ^ Skhove otes the Square om the between le figure 10, must integral [lust lie s in its ied over second number i root of i98 of i432i of ting the le given : of that is. The required the first period at the left ; this is the first figure in the root, whioh place in the form of a quotient to the right of the given number. Subtract its square from the first period, and to the remainder bring down the second period. Divide the number thus formed, omitting the last figure, by twice the part of the root already obtained, and annex the result to the root and also to the divisor. Then multiply the divisor, as it now stands, by the part of the root last obtained, and subtract the product from the number formed, as above mentioned, by the first remainder and second period. If there be more periods to be brought down, the operation must be repeated." Ex. 1. Find the square root of 1369. 1369 (37 9 67 469 469 Afler pointing, according to the Rule, we take the first period, or 13, and find the greatest number whose square is contained in it. Since the square of 3 is 9, and that of 4 is 16, it is clear that 3 is the greatest number whose square is contained in 13; therefore place 3 in the form of a quotient to the right of the given number. Square this number, and put down the square under the 13; subtract it from the 13, and to the remainder 4 affix the next period 69, thus forming the number 469. Take 2x3, or 6, for a divisor ; divide the 469, omitting the last figure, that is, divide the 46 by the 6, and we obtain 7. Annex the 7 to the 3 before obtained and to the divisor 6 ; then multiplying the 67 by the 7 we obtain 469, which being subtracted from the 469 before formed, leaves no remainder ; therefore 37 is the square .oot of 1369. Reason for the above process. Since (37)' =1369, and therefore 37 is the square root of 1369 ; we have to investigate the proper Rule by which the 37, or 30+7, may be obtained from the 1369. Now 1369=900+469=900+49+420 =(30)'+7='+2x30x7 =(30)''+2x 30x7+7" where we see that the 1369 is separated into parts in which the 30 and the 7, together constituting the square root, or 37, are made m ARmiMlTIO. distinctly apparent. Treating then the number 1869 in the following form, viz. (30)*-f 2 X 30 X 7+7' we observe that the square root of the first part or of (30)', is 30; which is one part of the required root. Subtract the square of the 30 from the whole quantity (30)'-f-2 x 30 x 7 + 7', and we have 2 X 30 X 7 + 7'remaining. Multiply the 30 before obtained by 2, and we see tb it the product is contained 7 times in the first part of the remainder, or in 2 x 30 x 7 ; and adding the 7 to the 2 x 30, thus making 2 x 30-+-7 or 67, this latter quantity is contained 7 times exactly in the remainder 2X30 x 7 + 7 or 469 ; so that by this division we shall gain the 7, the remaining part of the root. If we had found that the 2 x 30 + 7 or 67, when multiplied by the 7, had produced a larger number than the 469, the 7 would have been too large, and we should have had to try a smaller number, as 6, in its place. The process will be shewn as follows : (30)«+ 2x30x7 + 7' (30+7 (SO)" 2x30 + 7 2x30x7+7' 2x30x7 + 7' This operation is clearly equivalent to the following : 900+420+49 ( 30+7 900 60+7 420+49 420+49 This again is equivalent to the following : 1369 ( 37 9 67 469 469 which is the mode of operation pointed out in the Rule. Note 1. The reasoning will be better understood when the student has made some progress in Algebra. Note 2. The divisor obtained by doubling the part of the ri)0t already obtained, is often called a trial divisor^ because the SQUARS ROOT. quotient first obtained from it by the Rule in (ArL 187), will sometimen be too large. It will be readily found, in the process, whether this is the case or not, for when, according to our Rule, we have annexed the quotient to the trial divisor, and multiplied the divisor as it then stands by that quotient, the resulting number should not be greater than the number from which it ought to be subtracted. If it be, the quotient is too large, and the nun^^ber next smaller should be tried in its place. Not9 3. If at any point of the operation, the number to be divided by the trial divioor be less than it, we then affix a cypher to the root, and also to the trial divisor, bring down the next period, and proceed according oo the Rule. Ex. 2. Find the square root of 74684164. ioo4164 ( 8642 04 i2x8«=16} {2x86=172} }2X 864=1728} 166 ! iOGS 1724 17282 724J 6896 34564 34564 Therefore 8642 is the square root of 74684164. Ex. 3. Find the square root of 7 lo905 12350625. 71690512350625 (8467025 64 {2x8=16} {2x84=168} {2x846=1692} (2x8467=16934) \ 1693402 4233506 (2 X 84670=169340) J 13386804 164 769 656 1686 11305 10116 16927 118912 118489 16934045 84670225 84C70225 .'. 8467025 is the required square root. S46 ARITHMBTIO. 188. Again, since the square root of '0] is *1 •0001 is -01 •000001 is -001 •OOOOOOOlis -0001 5; -5. Ex. LXIV. 1. Find the square roots of [I) 289; 576; 1444; 4096. (2) 6561 , is) 98596 ; 37249 ; 1 1664. (4) 998001 ; 978121 ; 824464 29506624; 14356521 ; 5345344. 236144689 ; 282429536481 ; "^82475249. 295066240000 ; 4160580062500. 2. Find the square roots of (1) 167-9616; 28-8369; 5764801. (2) -3486784401 ; 3915380te9. (3) -042849; -00139876; 00203401. (4) 5774409; 5-774409. (5) 120888-68379025 ; 240398012416. 3. Extract the square roots of (1) 16; 1-6; -16; -016. (2) 2356; -1; (3) -0004; -00*^81; 379-864.(4) 20i ; 153^ 31 5*04 (5) f ; A;3i ; ^ («) :o2i ' ^^^ ' ^'^ ' *^ ' to four places of decimals in each case > ^re the root does not terminate. 4. Extract the square root of -0019140625 and reduce the result to the corresponding equivalent fraction in its lowest terms. 5 Find the side of a square field equal in area to a rectangular field 700 yards wide and 2800 yards long. 6. A rectangular field measures 223 yards in length, and 120 yards in breadth ; what will be the length of a diagonal path across it? 7. Find the length of the side of a square enclosure, the pavmg of which cost JS27. Is. 6c?. at 8d. per sq. yard. 8. The hypothenuse of a right-angled triangle is 51 yards, and the perpendicular is 24 yards, find the base. 9. A ladder, whose length is 91 feet, reaches from the extremity of a path 35 feet wide, to a point in a building on the other side, which is within 9 inches of the top of it ; find the height of the building. 10. Extract the square root of -0050722884, and find within an inch the length of ^a side of a square field, the area of which is 2 acres. 2h 25<^ ARITHMBTIO. 11. Two persons start from a certain point at the same time, the one goes due east at the rate of 12 miles an hour, and the other due north at the rate of 9 miles an hour ; how far are they distant from each other at the end of six hours 'i 12. A ladder 36 feet long will reach to a window 28 feet from the ground, on one side of a street ; and if the foot of the ladder be retained in the same position, will reach to a window 20 feet high on the other side. JB'ind the breadth of the street. 13. A society collected among themselves for certain purposes a fond of $45-9375 : each person paid as many cents as there were members in the whole societ|p Find the number of members. 14. The area of a circular lake is 295066*24 square yards, how many yards are contained in the side of a square of equal superficies? CUBE ROOT. 192. The Chbb of a given number is the product which arises from multiplying that number by itself, and then multiplying the result again by the same number. Thus 6x6x6 or 216 is the cube of 6. The cube of a number is frequently denoted by placing the figure 3 above the number, a little to the right. Thus 6' denotes the cube of 6, so that 6''=6 x 6 x 6 or 216. 193. The Cube Root of a given n'lmber is a number, which, when multiplied into itself, and the result again multiplied by it, will produce the given number. Thus 6 is the cube root of 216 ; for 6x6x6 is=216. The cube root of a number is sometimes denoted by placing the sign \^ ^before the number, or placing the fraction ^ above the number, a little to the right. Thus 1^^216 or (216)i denotes the cube root of 216; so that ^216 or (216)i=6. 194. The number of figures in the integral part of the cube root of any whole number may readily be known from the following considerations : The cube root of 1 is 1 1000 is 10 1000000 is 100 1000000000 is 1000 &ic. is &c. CUBK BOOT. 251 Hence it follows that the cube root of any number between 1 the S^^^^^^^^ number, and thence over every third figure lo tCleVof that place, the points will shew the number of figures in the integral part of the root. Thus the cube root of 677 consists, so far as it is integral, of one figure ; that of 198999 of two figures ; that of 134198999 of three figures ; and so on. 195 The following Rule may be laid down for extracting the Cube Root of a whole number : RuLK " Place a point or dot over the units' pla<5c of tfie given number,' and thence over every third figure to ohe left of thjjt nWe thuVdividing the whole number into severa periods. The tnmievoi^s^m shew the number of figures in the required root. (Art. 194.) :. . . ^ 4. "Find the greatest number whose cube is contained in the first periS at fne feft ; this is the first figure in the root, which place m the form of a quotient to the right of the given number. "Subtract its cube from the first period, and to the remainder bring down the second period. « Divide the number thus formed, omitting the la^t two figures by 3 tiSes the square of the part of the root already obteined, and annex the result to the root. « Now calculate the value of 3 times the square of the first figure in thVrloMwhich of course has the value of so many tens) +3 tinSs the product of the two figures in the root + the square of he n figure in the root. Multiply the value thus found by the second tore in the root, and subtract the result from the number formedls ab^ve mentioned, by the first remainder and tJ.. .econd peTod If thero be more periods to be brought down, the operation must be repeated." Ex. 1. I'ind the cube root of 15625. 25S \\ ARITHMBTIO. 15(525 (25 2»=8 g y^ 03=12 3 x(20)«=3x 400=1200 3x20x5= 300 5'= 25 1525 Multiply by 5 7625 7625 ' W( wl fr< ' re AH 7625 After pointing, according to the Rule, we take the fi st Period, or 15, and fio.) the greatest number whose cube is ctmtaSnod in it. Since the . Annex the 5 to the 2 before obtained; and calocate the value of 3x (20)«+3x20x5+5', which is 1525; SLn?t '^.'°^ ^^*^^ ^y ^ "^^ ^^^^^'^ *^^^S' which being subtracted from 7625 before formed leaves no remainder, therefore 25 is the cube root required. Reason for the above process, Since (25)«=15625, and therefpre 25 is the cube root of 15625 • we have to investigate the proper Rule by which the 25, or 20+5' may be obtained from 15626. ' ' > Now 156^5=8000+7500+125 =8000+6000+1500+125 =(20)'+3x(20)»x5 + 3x20x5»+5\ where we see that the 15625 is separated into parts in v ?ch the 20 and the 5, together constituting the cube root, or 2'> are made distinctly apparent. Tre.f.*yg then the number .-S-J5 in the {20)»+3x(20)»x5+3x20x5'+5'', OUBK BOOT. 253 we observe that the cube root of the first part or of (20)' is 20 ; which is one part of the required root. Subtract the cube of the 20 from the whole quantity, and we have 3 x (20)» x 5 +3 X 20 x 5" + 5' remaining. Multiply the square of the 20 before obtained by 3, and we see that the product is contained 5 times in the first part of the remainder, or in 3 x (20)' x 5 ; and adding 3 times the product of the two terms of the root + the square of the last term of the root, thus making 3 x (20)»^+3 x 20 x 5 + 5", we see that this latter quantity is contained 5 times exactly in the remainder 3 X (20)" X 5 + 3 X 20x5' +5', so that by this division we shall obtain the 5, the remaining part of the root. The process will be shewn as follows : (20)'+3 X (20)« X 54-3 x (20) x 5» + 5* (20 +5 (20)" divisor=3 X (20)', 3x(20)''x5+3x20x5»+5» 3x(20)^i?_ *^^~37(20)» ~'^' .•.{3x(20)''+3x20x6+5'i x5= 3x( 20)'x5+3x20x5'+5' This operation is clearly equivalent to the following: 8000+6000+1500+122 (20+5 8000 3 X (20)'=1200, and Hn=5 (1200+300+25) X 5= 6000+1600+125 6000 + 1500+125 This again is equivalent to the following : 15625 (25 8 3x2''=3x4=l2, and].f= 3x(20)»=1200 + 3x20x5= 300 + 5'= 25 1525 5 7625 7625 7625 which is the mode of operation pointed out in the Rule. Note 1. The reasoning will be better understood when the student has made some progress in Algebra. 854 ARITHMinO. ^il'''* ^d'^^int'y'''''" ^^""^ ^* Obtained according to the Rule given in (Art 195) is sometimes called a trial divisor, because the aTve'Exa'lfel^^ be t.o large, as was th^'cTL'tt wrioii J . ' '^*'''!^ "^^ "^^ "^"«^ *^y a 8»°a"er number. dTlKrnnl^ '''''''^" whether-the number obtained from the division IS too large or not, because if it be too large, the quantity which we ought to subtract from the number formed by a remainder and a period will turn out in that case to be larger than that A'^A^Au l^u** ?«/ point of the operation, the number to be thr rl { '^" '''u ^™"\^^ ^''' '^"^ i* 5 ^^« ^^ a cypher to the root, two cyphers to the trial divisor, bring down the ne^ period, and proceed according to the Rule. . Ex. 2. Extract the cube root of 95443993. trial divisor=3 x 4'=48 3X(40)»=4800 3x40x5= 600 5' = 25 5425 5 95443993 (457 ' =64 31443 314 -^goes 6 times, but 6 will be found too large; try 5. 27125 27125 trial divisor=3 X (45)'=6075 I 4318993 Now 45 has the valiite of 450 ; .*. 3 X (450)«=607500 3x450x7= 9450 7«= 49 616999 7 4318993 43189 ^75* ^^®® ^ times, and we are led to conclude that 7 is the figure, be- cause 7'=343, and 3 is the final figure in the remainder. 4318993 Therefore 457 is the cube root reijuired. ihe Rule iause the se in the number, irom the (Quantity mainder lan that 1st try a r to be pher to he next , but 6 large; is, and nclude re, be- d3 is n the CUBS ROOT. 255 Ex. 3. Find the cube root of 223648543. 223648543 (607 6»=216 trialdivisor=3x6'' =108 trial divi9or=3 x (60)»=10800 3x(600)''=1080000 3x600x7= 12600 7"= 49 1092649 7 7648 76 is not divisible by 108; 7648543 7648543 bring down the next period and affix to the root; HHf goes '^ times, and 7 seems likely to be the figure required; since 7' =343 and 3 is the final figure 7648543 in the remainder. Therefore 607 is the cube root required. 196 Aeain, since the cube root of '001 is 'I, the cube root of -000000001 is -001, n the subject of pomtmg, the rule for extracting the cube root of a decimal or dLlt^T 'iiT ''fv!"^ P*'*^ ^^ ^ ""'""^^ ^"«^t,er and partly of a decimal will be the sar c u3 tkir. before given (Art. 19M for finding the cube root of a whole number. As tjie decimal notation 18 only an extension or continuance of the ordinary integral notation pJ^eTwni t!Z^Z^l:\^ — ^^^- ^-° ^- ^^e may find the cube root of each separately ; and ^tllwerw^^^ thus be^obtained .. a vulgar fraction ; if U, we can firsWeduce ^llTT^ "^ ^ ^?r^' ^^ ^ * ^^«^« """^ber and deciLl and then find the r .ct of the resulting number. The answer wH thus be obtained either as a decimal, or as a whole numberTnd decimal according to the case. Also a mixed number may be reduced to an improper fraction, and its root extracted in the lame way Ex. 4. Find the cube root of 48228*544. 3x(30)»=2700 3x30x6= 540 6' = 36 3276 6 3x3»: 48228544 (364 3'' =27 =27 21228 19656 3x(36)''=3888 3x(360)'=388800 3 X 360 X 4= 4320 4'== 16 393136 4 lPr>P,<5 1572644 1572544 1572544 Therefore 36*4 is the cube root required. OTOIF '^OOT. * 267 Ex. 5. Find the cube root of *t j007 to three places of decimals. •000007000 (019 1 3xl'=3" 3 X (10)'= 300 3x10x9=270 9^= _81_ 651 9 5859 600 5859 141 Ex. 6. Find the cube ror of f to three places of decimals. • 8»=! 555555555... 555555555 ('822 512 3x8'=192 3x(80V=19200 3x8v ^ = 480 ■^*-^-- 4 19684 2 S9368 43555 39368 i. (82y=2017! ..x(8. )''=2017200 3x820x2- 4920 2"- 4 I 4187555 2022124 2 404424S 4044248 143307 200. Hig> . roots than the squs^^ and cube can sometimes be extracted b) means of '-e Rules foi uare and cube root ; thus the 4th root is found bv akinc he sq re root of the square root ; uiie 6th root by ta il he square loot of the cube root, and so on. 2i 258 ARITHMXTZO. !llll Ex. LXV. 1. Find the cube roots of (1) 172b ; 8375 ; 29791. (2) 54872 ; 1)0592 ; 800768. (8) 681472; 804857; 941192. 4) 2406104 ; 69426581 ; R365427. 5) 251289591 ; 2887262 ; 48228544. 6) 17173512 ; 259694072 ; 926859375. [7) 27054086008 ; 219865827791. 2. Find the cube roots of •389017 ; 82-46.759 ; 95448'998 ; •000912678 ; *001906624 ; '000024389. 8. Find the cube roots of (1) 8, -3, 03. (^)5j (3) 405, Vff ; 7i; 3-0041 5. (4) -0001; 250 686' 1257-728 44-6. 16884 to tiiree places of decimals, in those cases where the root does not terminate. 4. Find the cube root of 233*744896, and also the cube root of the last-mentioned number multiplied by -008. 5. The cost of a cubic mass of metal is £10481. 1*. Ad. at 10*. hd, a cubic inch. What are the dimensions of the mass ? 6. A cubical block of stone contains 50653 solid feet, what is the area of its side 1 7. A cube contains 56 solid feet, 568 solid inches ; find its edge. 8. Find the cost of carpeting a cubical room, whose content is 21717-639 solid feet, with carpet 21 inches broad, at $1*25 a yard. 9. A cubical box contains 941192 solid inches ; find the cost of painting its outside surface at 6 cents a square foot. 10. If the solid content of a cube be 37 ft. 64 in., shew that its surface will be 66 ft. 96 in. 11. The edge of a cubical vessel is 2 feet long: what is the length of the edge of another cubical vessel containing 3 times as much? 12. Find the 4th root of 43046721 ; and thr 6th root of •000000304096. ▲RITHMBTIOAL PBOOBBSUOIV. 259 res. 8 not otof 10*. \t is jdge. nt is rard. St of it its the ssas t of PROGRESSION. ARITHMETICAL PROGRESSION. 201. A series of numbers that increase or decrease by a common difference is said to be an Arithmetical Progression. When the terms are constantly increasing, the series is an Arithmetical Progreaaion Ascending ; when constantly decreasing, the series is an Arithmetical Progression Descending. Thus, 1, 3, 5, 7, 9, &o., is an Ascending Arithmetical Progression; 10, 8, 6, 4, 2, is a Descending Arithmetical Progression. The first and last terms of a Progression are called the extremes ; the intermediate terms are called toe means. The terms of an Arithmetical Progression may be fractional, thus : i, 1, Ht 2, ^, 3, &c., having a common difference of ^ ; h h 1» H) 1|» ^> ^» ^t» ^» ^*^'' ^^^^"^ * common difference of \. From the nature of an Arithmetical Progression it follows that the sum of the extremes is equal to the sum of any other two terms equally distant from them, or to twice the middle term, if the number of terms be unequal; this will be evident from inspecting the following Progression : 1, 3, 5, 7, 9, 11, 13. 18 5 7 13 11 9 7 9 11 13 5 3 1 14 14 14 14 14 14 14 In Arithmetical Progression there are five terms to be considered. 1, the first term; 2^ the last term; ^, the common difference; 4* the number of terms ; 5, the mm of all the terms. These quantities bear such a relation to each other that any three of them being given, the remaining two can be found. Since there are five terms and only three of them necessary to be known, it follows that there are twenty distinct cases in Arithmetical Progression. We shall, however, notice but two of the most important, and refer the student to Algebra for the others. THB LAST TERM. 202. To find the last term, when the first term, the common differenf 1 and the number of terms are given. «*y, 260 ARITHIISTIO. Rule. To the first term add the product of the common difference into the number of terms, less 1. Ex. What is the last term of an Arithmetical Progression, whose first term is 4, the common difference 3, and the number of terms 5 1 Proceeding by the rule given above, 44-(3x4)=16. Heason for the process. It is clear that the second term of an Ascending Progression is equal to the first, increased by the common difference ; the third is equal to the first, increased by twice the common difference ; the fourth is equal to the first, increased by three times the common difference ; and so on, for the succeeding terms. In a decreasing Progression the last term is equal to the first term, less the same product, thus : What is the last term of a Progression, whose first term is 16, the common difference 3, and the number of terms 5 ? 16— (4x3) =4. THE SUM OF ALL THE TERMS. . 203. To find the sum of all the terms, when the first term, the last term and the number of terms are given. Rule. Multiply half the sum of the extremes by the number of terms. Ex. The first term of an Arithmetical Progression is 2, the last term is 50, and the number of terms 17 ; what is the sum of all the terms 1 Proceeding' by the rule given above, 2+50 ,^ ^^^ —^x 17=442. Reason for the process. The sum of the extremes of an Arithmetical Progression being equal to the sum of any two terms equally distant from them, it follows that the terms must average half the sum of the extremes. LXVI. 1. What is the 100th term of an Arithmetical Progression,whose first term is 2, and common difference 3 ? I GEOMETRICAL PROGRESSION. 261 2. What is the 50th term of an Arithmetical Progression, whose first term is I, the common difference ^1 3. The first term of a decreasing Arithmetical Progression is 4680, the common difference is 3, and the number of *^erms 120. What is the last term 1 and the sum of all the terms '? 4. A tapering board, 6 inches wide at the narrow end, and 1 2 feet long, is found to increase \ an inch for every foot in length. What is the width of the wide end ? 5. A person travels 25 days, going 11 miles the first day, 135 miles the last day : the miles which he travelled in the successive days form an Arithmetical Progression. How far did he go in the 25 days 1 6. A person makes 12 monthly deposits in a Savings Bank ; the first deposit consisted of $25, the second of $30, the thi: 1 of $35, and so on, in Arithmetical Progression. How much did he deposit in 12 months ? 7. What is the sum of an Arithmetical Progression whoso first term is 7, and last term 11131 8. A man has in his orchard 34 rows of trees ; in the first row there are 20 trees, in the second 24, in the third 28, increasing in Arithmetical Progression. How many trees are there in the last rowl 9. A merchant bought a certain number of pieces of cloth, the prices of which increased by $2 ; the first piece cost $3 and the last $43. How many pieces did he buy % , GEOMETRICAL PROGRESSION. 204. A series of numbers which succeed each other by a constant multiplier is called a Geometrical Progression. The constant factor by which the successive terms are multiplied is called the ratio. When the ratio is greater than a unit, the series is called an ^r\anfkA 9/4««/»/v7 T^fv»f\r%v*sooinf^ * wliPii Ipfia tlion n. unit. t.liA series is called a Descending Geometrical Progression. Thus, 1, 3, 9, 27, is an Ascending Geometrical Progression, whose ratio 262 ARITHMETIC. 18 3; and 1, t, tV, tV i8 a Descending Geometrical Progression whose ratio is J-. o > In Geometrical, as in Arithmetical Progression, there are five terms to be considered : I, the first term ; 2, the last term • 6 the common ratio ; 4, the number of terms ; 5, the sum of all These quantities are so related to each other that any three beinc given the remaining two can be found. We will demonstrate two of the most important, leaving the student to pursue the remainder in Algebra. I THE LAST TERM. 205. To llnd the last term, when the first term, the ratio and the number of terms are given. Rule. Multiply the first term by that power of the ratio which IS expressed by the number of terms less cie. Ex. Find the last term of a Geometrical Progression, whose first term is 2, the ratio 3, and the number of terms 8. Proceeding by the rule given above, 2x3^=4374. Tlie reason of the process is rendered evident by the following consideration : ° If the first term be 3 and the ratio of the progression 5, the series will be r o , 'j 3, 3X5, 3x5x6, 3x5x5xft&c. or3, 3x5, 3x5» 3x5" &c. in which it is plain that the last term (3 x 5") will always be equal to the first tei :n (3) multiplied by that power of the ratio (5) which IS expressed by the number of terms less one. THE SUM OF ALL THE TERMS. 206. To find the sum of the terms of a Geometrical Procession when the fu-st term, the last term, and the ratio are given. OXOMETRIOAL PROORESSION. 263 * i It Rule. Divide the difTerence between the first term and the last term multiplied by the ratio, by the difference between the ratio and a unit. The following examples will explain the process : Take the progression of which the ratio is 3, 4, 4x3, 4x3»,4x3»,4x3* multiplying each term by the ratio, we obtain the series 4+3, 4+3», 4+3», 4X3*, 4X3" The sum of which is 3 times the sum of the first series. If the first series be subtracted from the second, the remainder will be 4 X 3*— 4 which must be twice the given series. But this remainder is the difference between the first term (4,) and the product of the last term (4 x 3*), multiplied^by the ratio (3) while 2 is the difference between the ratio and a unit. The sum of the series will therefore be found according to*the Rule. 207. If the number of terms of a decreasing Geometrical Pro- gression were infinite, that is increased without limit, their sum would be equal to the first term divided by the difference between the ratio and a unit. Thus, if the series 9, 3, i, ^^ &;c . in which the ratio \ were continued to an infinite number of terms ; the last term would be diminished without limit, that is it would be 0, and by the preceding proposi- tion the sum of all the terms wouid be (9-0xi)-^(l--i)-9-rf=13^. On this principle the value of a Repeating Decimal may be computed. (Art. 96.) For example, the Rcpetend -4444 is equal to i*o+To7+TTsVir+ TToTTT *"^d so on without limit. Now this is a decreasing Geometrical Progression, in which the ratio is i^y, the sum of all the terms is therefore iV"^-i^ir=i=|. Ex, LXVIL 1. Th« first term of a Geometrical Progression is lOO, the ratio 864 ▲RITHMBTIO. m of the progression is |, and the number of a terms 7 ; what is the last term ? and the sum of all the terms ? 2. The first term of a Geometrical Progression is 5, the ratio is 4 and the number of term 9 ; what is the last term ? 8. A person travels 5 miles the first day, 10 miles the second day, 20 miles the third day, and so on, increasing in Geometrical I'rogression : if he continues to travel in this wEy for 7 days, how lar will he go the last day ? ^ ^ » 4. A ball falling from the height of 10 feet, by its elasticity bounds 5 feet, and again falling, bounds 2l feet, and so on continuing to bound ^ as high as it fell. What will be the whole Gistance made in the successive falls before coming to a state of rest? and what the whole distance made by its successive bounds ? . ^' '?^^L^oo ^^T" ,^^ ^ Geometrical Progression is 4, the last term is 78732, and the ratio is 3. What is the sum of all the terms ? 6. What is the sum of an infinite number of terms of a i^eometrical Progression whose first term is 1000 and ratio 1 1 7. A man commences bu3iness with a capital of $20 and doubled it m every 2 years. What was his capital at the end of 25 years ? 8. What is the sum of the infinite series 1, |, :|, J-, &c. 1 9. What is the sum of the infinite series -V ?^ 9 jj Q l^\l^ ^^ ^^^^l""! "^"^^^ r^"^ '^^^ ^* 1 «e»* for the first yard, 3 for the second, 9 for the third, and so on, what would be the price of the last yard? and what would the whole amount to? 1 iW"^^ '^ ^^^ '•'"^ ""^ ^^^ '""^'-^^ «^^i^« J' h h &c., also of MISOELLAJTBOUS QUESTIONS. 266 Mscellaneous Questions and Examples on preceding Arts. \ .^'°o,*J*®, ^®?S*^ ^^ ***® interior edge of a cubical bin which contains 310 bushels of wheat. (A bushel fills 2218 192 cubic inches.) iA?'n^^*^ ^® ***® ^**^^® ^*^"® ^^^i y^^' of cloth at $.3* a yard, 10| lbs. of tea at $-75a lb., and 43 bush, of com at $-70 a bushel \ Uivide the sum among 4 people in the proportions 1, 2, 3, 4. 3. Assuming only the definition of a vulgar fraction, prove that the numerator and denominator of any vulgar fraction may be multiplied or divided by the same integer without altering its value. (a) What fraction of a pound sterling is 4^|--10f |+9^--^|S^^ (b) Find the value of i x V of f i+ f Si+gir) x e^\ 4. A.debt is to be discharged at the expiration of 6^ months, 1 IS paid immediately, and i at the expiration of 3 months : when ought the remainder to be paid ? 5. If 10 men or 15 boys can reap 20 acres of com in 6 days working 14 hours a day, how many boys must be employed to assist 3 men to reap 6 acres in 1| days of 8 hours a day 1 6. What is the height of a closet 8^ ft. by 6| ft. which will exactly contain 12 boxes ^ ft. long, 3^ ft. wide, 2^ ft deep ? 7. Two lines are 41-06328 and -0438 of an inch long respectively. How many lines as long as the latter can be cut off from the former ? What will be the length of the remaining line 1 8. Explain the method of extracting the cube root of a number. J md the area of tli -^ j face of a cube which contains 733626753859 cubic inches. 9. Sharfes in ^ certain Railway pay 63-25 dividend per annum. How much must I give for them to get 5 per cent, for my money ? A person having bought 20 shares at this price sells them when thev have risen ^7 c^anh ar\i\ >>nxra rv.iv.i*«,. c*^^^\. «4. nn i m per cent. Fmd the change in his income. > r .r s y 260 ▲IlITHMETIO. JO. What sum of money wUl amount to 1845 in 2 years at 4 per^cent compound interest, and what will it amount to in 2 mtt oJ}' \ ?!fi?^u*i ^®"^ '^^ ^"^^els of corn at a profit of 8 per cent and 37 bushels at a profit of 12 per cent. • if he had so?d for tie'o^ni ^" ""'""^^^ ^'^^ ^^^^ ^^« **^« P"«^ h« P^'^d are apportioned as follows ; 41 per cent, to pay the working IXTuLFllrT- !^ '^'l *h-hareholders^a^iyfden7:t ^^^^ rate ot d^per cent, on their shares; and the remaindei $15000 IS reserved ; find the paid-up capital W the company miLns^'^fiThJLrfr'/"' ^^^r^ '^''^ ^""^^^d ^l^o"«««d mimons, tivo hundred and seven thousands, three hundred and sixty four ; and in writing 236045978213428. """^^^^ ^^^ 2. When the pound sterling was worth 24 francs, 75 centimes a traveller at Dover received 15.. for a Napoleon (20 francs^ Of how much was he cheated ? ^ «"" v^v irancsj. ut Fi5 W^P ^""J ^y firs^Principles to calculate values by Practice. Find by Practice the value 750 articles at £5. 8s 4rf each and fuXc" f ' '"' ' ^"- ''^ '''' '' ^^- '- ''' P- cwMCa^^^^^ FrLion.^'tm%r""^ '''""" ' ^"^^^'^ ^^' ^ ^^^^'"^^ (a) 2xv'l+^;TVl-i 5x Vl+Lxv/1— 1 2-6 "8-7* Wl(6H.f)^-ffyg-i|x-05of5. + ^.. (c?) -0576 X 1-97 + •142857-r24 + -0454864. its ddt^il^^^^^^^^^^ ''P'"'""' " '^"^^^ ^" y^'^'^ ^^^ '^' ^^«gth of 5. A and ^ can finish a piece of work in li days, A and O in ears at 4 a 2 more >f 8 per had sold received he paid »in year working id at the 115000, housand ced and bimes, a s). Of ractice. h ; and Canada decimal MISCELLANEOUS QTTBflTIONS. 267 2 days, and B and Om 3 days If $6 be paid for the piece of work, what are a day's wages of each workman ? 6. A tax of $530 is to be raised from 3 towns, the numbers of m..abitants of which are respectively 2500, 3000, and 4200 How much should each town pay, and each person in it ? 7. If 15 men or 40 boys do a piece of work in 12 days, how many days would 10 men and 20 boys take to do a piece of work 7 times as great ? r "v*«. 8. Define Interest and Discount. Show that the Interest and 9 The breadth of a room is 14 ft. ; the cost of papering the walls at $-05 a square yard is $4 ; and that of carpeting the room at $-225 a square yard is $5.60. Determine the height and length of the 10. It is observed that 20 men, all of equal strength, build a wall 16 feet high, 30 feet long in 60 days, and 35 others, also of equal strength, build a wall 20 feet high, and 40 feet long in 64 cWs-^ ^^'^ ^»*i<>of the strength of the men of the two 11 A person :^.s 200 shares in a certain Railway, for which he gives $100 per share. When they are pa-ing $2 per cent, he sells them all at $46 per share, and invests the process in City Debentures at 92, ^ar],:. 6 per cent. Find the alteration m his income. " 12. Explain the method of pointing in the Extraction of the cube root of decimals. Find the square root of l^^and the cube root of 423564-751. igth of C in III. 1. Shew from first principles how to divide one fraction bv another. ^ Prove that the fraction ?±| is greater than f and less than f . Simplify M"^'' 9x5 __llj 14x3 15 268 ARITHMETIO. ^; 2. Express (a) (i+f)^+(i+f >.+(i+f )rf. as the decimal of £1. (5) 5^ cwts. as the decimal of a ton, and 2^ qrs. as the decimal of a cwt. 3. A man contracts to perform a piece of work in 60 days, and immediately employs upon ic 30 men ; at the end of 48 days the work is only half done ; required the additional number of men necessary to fulfil the contract. 4. A person increased his capital annually ^rd part, and at the end of 4 years, one year's interest thereon at 4^ per cent, amounted to $270. What capital did he start with? 5. A can do a piece of work in 12 hours, B in four hours, and C in 3 hours. -4, B and C all work together for half an hour, when A leaves off. How long will it take B and C to finish the piece of work 1 6. Explain the method of pointing in extracting the square root of a whole number, and also of a decimal. (a) The surface of a cube is 86*94 square feet, find the length of its edge. (6) Given that the square of 10129 is 102596641, find the square of 101293 without going through the operation of squaring. (c) Given that the square root of 105625 is 325, find that of 10582009. 7. If 144 men can dig a trench 40 yds long, 1 ft. 6 in. broad, and 48 ft. deep, in 3 days of 10 hours each ; how long must another trench 5 ft. deep, and 2 ft. 3 in. broad be, in order that 51 men may dig it in 15 days of 9 hours each % I 8. If a cubic foot of marble weigh 2*716 times as much as a cubic foot of water, find the weight of a block of marble 9 ft. 6 in. long, 2 ft. 3 in. broad, 2 ft. thick, supposing a cubic foot of water to weigh 1000 oz. / 9. Find the equated time of payment of $200 due 14 months' hence, and of $300 due 19 months' hence ; and determine the present value of the whole sum, (supposed to be due at the equated time) allowing 3^ per cent, simple interest. i 10. A publisher wishes to net $*70 for each copy of a work ; A what price should he put upon it so as to be able to allow the trade 30 per cent, discount 1 \ 11. What is the present worth of $840, due 3 years and 6 months hence, at 6 per cent. ? t M|iiMiMiai«ii •MfMH MIS0ELLANIOU8 QUESTIONS 269 P \i 12. Which is the greater v/ 2 or ^ 3 1 Find the cube root of 50 30-912 65536 IV. 1. A stationer bought 40 reams of paper at $-625 a ream, and 60 reams at $-775 a ream ; find the whole cost, and the average price per ream, and if the whole be sold at $-75 a ream, find the profit. 2. Express in figures thirty-four and two thousandths, and by it divide 2825565-2. What alteration must be made in the quotient if the decimal point in the dividend be moved eight places to the left? ^ 3. Three men, working 9 hours a day, take 16 days to pave a road 315 yds. long and 30 ft broad ; how many days will four men, two of whom work 8 hours, and two 10 hours a day, take to pave a road 1575 yds. long, and 35 ft. 6 in. broad ? 4. The areas of two cubes are respectively 5359-375 and 5-359375 cubic feet ; find the difference of the lengths of their edges in inches. 5. A and B agree to divide their travelling expenses in the proportion of the numbers 7 and 5. A pays in the whole $25-80, and B pays in the whole $15-85. What is the one to pay and the other to receive in order to settle the account ? 6. When are four quantities said to be in proportion ? Shew by means of your definition that $191*625 : $31-50 : *. 365 days ; 60 days and deduce the method of working the following question : If 3 workmen earn between them $191-625 in a year, in what time will they earn $31-50 ? 7. The Discount on a sum due one year hence at 5 per cent, per annum interest is $15. What is the sum 1 8. If 8 variegated silk scarfs, measuring each 3 cubits in breadth and 8 in length cost 100 nishcas ; what will a like scarf 3^ cubits long and ^ a cubit wide cost in terms of drammas, panas, cacinio, and cowry shells ? 1 nishca=16 drammas, 1 dramma=16 panas, 1 pana==4 cacinis, 1 cacini=20 cowry shells. 9. A person invests a sum of money in 50 casks of sugar each containing 11 cwt. 3 qrs. 2 lbs, at $3-50 oer ewt of 28 lbs. what price mujt he sell them at after 6 months to realize the same interest ^ he might have had for his money at 4| per cent. ? ■w** — y 270 ARITHMETIC. 10. It is agreed that the rent of a farm shall consist of a fixed sum together with the value of a cei t,ain number of bushels of wheat ; when wheat is $2 a bushel the rent is $250, when wheat is $2-25 a bushel the rent is $260, what will the rent be when wiieat is $2-50 a bushel 1 11. ^ and B can do a piece of work in 1» days ; B and C in 15 days and A and C in 25 days ; they all work at it for 4 days ; A then leaves, and B and C go on for 5 days ; B then leaves : In how many days will C finish the work '\ 12. A ship's hold is 99 ft. long, 40 ft. broad, and 5 ft. deep, how many bales can be stowed in it each 3 ft. 6 in. long, 2 ft. 8 in broad, and 2 ft. 6 in deep, leaving a gangway of 4 ft. .>road % 1. (a) The French metre 'being 39-37 in., how many yards are there in 36(10 metres % (6) 3 versts being ^2 miles, in what time will a man travel over 2500 versts at the rate of 10 miles an hour 1 2. State what fractions produce terminating decimals, and what produce recurring decimals. E^ri; > n the reason. _ Reduce to decimals the vuli;. ,..j,;^ions f, f^, f-?^ aind add them ; and divide their sum by #gi®5:41 to tw© decimal places. 3. A silversmith purchases a ij^rge dish weighing 80 »z., and forms it into 2 dozen of dessert spoons, and one dozen of table- spoons. If the latter weigh 28 oz., what is the weight of each dessert-spoon, and what is its value at W of ^'^^ P^i' grain 1 ^ 4. Add together | of f of £2. 5*., f of 3 guineas, -27 of £1. 18s. 6(^., and 2*154 of £2. 15s., and reduce the result to the fraction of 25 guineas. . 6 How much may a person who has an annual income ot $840-50 spend per day, in order to save $63-966| m. after paying a Tax of $-16 in the dollar 1 . 6. Find the square root of H:l-^-5^, and the cube root of -03. 7. If a piece of work can be finished in 45 days by 35 men, and if the men drop off by 7 at a time at the end of every 15 days, how long will it be before the work is finished ? e, uivide jDioyQ^ auiuiig -c-i, Xj, v una ^, -•' u,,— -- T'^'i B's share :: 6 : 5; ^'s share : C's share M 2 : 3; and Cs share ! D's share >: 4:3. i I %«iri i| 1 MI301LLAKE0US QUESTIONS. «n 9. What is the ' st of paper for the wall^ '' a room 30 ft. long, 16 ft. broad, .nd V. ^. high, the paper 1 » yds. wide, and its price 4^ & »ts per ^ .rd 1 What woul.. the cost for a room twice as loi , twice ab broad, and twice . high, the paper twice as wide, and costing twice as much per yard as before 1 10 If whc 25 ! r -ent. is lost in grinding wheat, a country has to iniport ten million quarters, but can maintain itself on its own produce if only 5 per cent, be lost, find the quantity of wheat grown hi ^>if^ country. n Hovvr many flag-stones, each 5-76 ft. on^ and 4*15 ft. wide are required for paving a cloister whicl es a rectangular court 45-77 yds. long and 41-93 yds. ; the cloister being 12-45 ft. wide ■? ^ ^^ . , 12. ' man wishing to invest $10 m Grovernment bonds bearing interest at 6 percent, inquires tii . price of the stock, and finds it to 1)" 86 per cont.; he delays the investment however, until bonds 1 -o risen to 87. What effect has the delay on his income 1 VI. ] Explain our decimal system of f^r^thmetic, and how it is that we are enabled with digits to express any number, however ^''2. If 12 men or 18 boys can do f of a piece of work in 6i hours, in what time will 11 men and 9 boys do -ua rest? 3. Express -0025 by .. simple fraction, and if^- + 6y\— ^i by 4 Explain what is meant by Compound Interest. What is the differe^ice between the Simple and Compound Interest of $345-50 for 2 \ears at 35 per cei, 5 Define discount. If the di? .'unt on $226-33, due at the end of a year and a half be $12-80 hat is the rate of interest 1 6 Show iiow to divide 4 things of the same size and material among 3 children. A, B, and C7, by merely breaking one of the four and so that ^'s share shall be f + | + i of a whole one ; £ s share 3. + /„ + 2^0 of a whole one ; and C's share the remainder. 7 A grain of pure gold can be drawn out into a wire 550 feet long • find the cost of a wire of the same thickness which ^= uld extend round the earth, assuming the circumference of the earth to be 25 000 miles, and the value of gold to be $21-25 per oz. troy. ». (a) Jt -d Jlto C, 1^ or jD, auu O — .*g- Oi j-^, uiiki tiiv iwvxv x}i ^= :JkE!3 ^ "A.^. IMAGE EVALUATION TEST TARGET (MT-S) 1.0 Ui IM 12.5 ^ 1^ 12.2 1.1 i.*^ n^ L25 iU 11.6 — 6" '#' o 7 PhotDgraphic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4S03 >%^%^ €^1^. ^ ^^ ^ / ^"T-sajs^ ■w» 9VPi P"P "iPii 272 •r: I (b) Simplifyi ''iofH . . . (c) Divide 10-88e by 61-6 ; and 1083-6 by 6-16, and also by •00516, and prove each result by vulgar fractions. 9. A shopkeeper buys \ cwt of tea at 4«. 2d per lb., and mixes it with tea which cost him 2*. llrf. per lb. Mow much of the latter must he add to the f, and also by per lb.) and iiow mucb of [ the mixture ? [aid down on ber of acres income, and 9 income. ^th of the le one who is turned by a )r29. Wbat rs? k holding 18 aye been its z. mt^ning 30 a line, how there are 50 cubical box, . in 16 days, 03. 10«., the ►rmer? >k by selling it at $1 -761 )K, a w. PRINCIPIA UTINA, AN INTRODUCTION TO THE UTIN UNGFAGE, i^ By CHARLES D'URBAN MORRIS M. A. Hector of Trinity School, New York, formerly FeUow of Oriel'coUege, Oxford. NEW YORK : PUBLISHED BY MASON, B30THERS, 6 AND 1 MerOBB STHlBiiPP Meboeb Street. 1860. SCHOOL BOOKS. • •»- The Irish National Series of Schooi BooIk, As approved by the Board of Pivblic Instruction of Canada, KE-PUBLISHED AT THE "SPECTATOR" OFFICE. <•■■> HEAD LINE COPY BOOKS, PLAIN AND FANCY JOB PRINTING, BOOK-BINDING LITHOGRAPHING; &c., &c. 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