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Un daa symbolaa suivants apparattra sur la damlAra imaga da chaqua microficha, salon la caa: la symbola ^^ signifia "A SUIVRE", la aymbola ▼ signifia "FIN". plataa. charta. ate., may ba fMrnad at d i ffarant reduction ratios. Thoae too large to be entirely included in one expoeure ara filmed beginning in the upper left hand comer, left to right and top to bottom, aa many framae aa required. The following diegrama illuatrate the method: Lee cartea, planches, tablaeux. etc.. peuvent Atre filmAe A dee taux da rAduction diff Arents. Lorsque le document est trop grend pour Atra reproduit en un soul cllehA. il est flimA A partir da I'angle supArieur gauche, do gauche A droite. et do hieut en bee. an pranant la nombra d'imagea nAcassaire. I.aa diagrammes suivants illustrent la mAthode. 1 2 3 1 2 3 4 5 6 * ^ ■!fti. '■•.It *•< [//am .'i* -1' !'hm€tic.\ 1— ^ TBI CANADIAN AEITHMETIC. DESIGNED FOR SCHOOLS AND ACADEMIES • • «■ > IN BRITISH AMERICA, IN FOUK PARTS. COMPILED FROM THE B^ST AVTHORa, BY A TEACHER. PICTON, C. W. PRINTED BY JOHN DOUGLASS, 1845. « ,.'» J;- 1«^: »i m PREFACE. Among the many Arithmetics extant it is uniYersally conceded that there is none that is in every respect adapt- ed to the wants of Schools in this Province. There is scarcely a School in which there are not several different works on Arithmetic in use. This is very peiTplexing to the Teacher, as vnell as a positive disadvantage to the pupil. It would not only be of gi-eat advantage to the pupil, but a great saving of money to parents to have a uniformity of books in our Common Schools — for it fi^equently occurs, that when a pupil is removed from one School to another, a new set of books must be purchased, and the old ones laid asidft as useless. Every school room should be fur- nished witli a black board, about four by six feet m size, and all the pupils siiould be classed and exercised upon it daily, which cannot conveniently be done if each pupil uses a diflerent author. Much time is wasted in many Schools by teaching individually instead of in classes.— While a class is reciting and exercising on the black board it affords an excellent opportunity for the Teacher to ex- plain to the whole class at the same time any part that they do not under.^tand. When a class is called up to re- cite, while one of tliem is solving the question given out by the Teacher, on the board, the rest of the class should solve the same question on their slates at the same time, and it should not be dismissed until each one of the class thoroughly understands it. When all the pupils in a School use the same author, and are classed and exercised in this way, it excites a spirit of emulation and interest, which is a powerful incentive to action. Many of the Arithmetics in use are American, and consequently are filled almost entirely with questions in Federal money, to the exclusion of the currency of this Province ; othen again are better oafeulated for those whp are considerably ll!Wll!l!l«*«l,.J._iJ* tr PREFAOS. IT advanced in the science, than to the capacities of the ma- jority of pupils in our Common Schools. In many Arith- metics a mechanical method is presented of performing certain operations (tccording to m/e, without assigning any reason for such operations. Thus in Subtraction, the rea- son why one is carried, and ten borrowed; or in Multipli- cation, why the figures are placed in a certain method; or in Division, why Multiplication and Subtraction are per- formed, is never explained or illustrated. To the pupil they are a sort of cabalistical process, which he finds will Mng the right answer^ and this is all he can know from any thing he learns from the book. Others again are too sparing of examples, especially in the simple rulesj in consequence of which the pupil is hurried through them without understanding the first prin- ciples. Hence his ambition is fettered, and he can see no beauty in a science in which obscurity is behin'^ him, and impenetrable darkness before him. The compiler of the following work has enlarged upon and given numerous examples in the different rules, m proportion to their piactical importance in the business transactions of life. At the commencement of each rule, one or more examples have been wrought at length, and the method of operating has been clearly and fully ex- plained. No term is used until it has been defined — and the examples under the various rules are mostly of a prac- tical nature. The arrangement of the rules is that which appeared to the compiler the most easy and natural. After Reduction and the Compound Rules, those which are of the most practical importance are first introduced. It is believed tliat this arrangement will be found of great ad- vantage, especially to those who have not an opportunity yo go through all the rules of arithmetic. The intercourse and trade between this Province and the United States is so extensive, it is important that every one should understand the currency of that country — therefore a short article on Federal money has been intro* duoed immediately after the compound rules. In the arU^ PBEFACB. V ele on the subject of Proportion, the useless and perplexing distinction of direct and inverse proportion has been die- carded, and a single rule, which is simple and general, has been adopted. The solutions of some of the most difficult questions are intended to ai^sist such as may study the science without the aid of a Teacher; and perhaps they may be of advan- tage to some teachers who have had but little experience in their prof jssion. To a business man, a practical and correct knowledge of Arithmetic is of the first importance; hence the public should receive with indulgence every attempt to improve this interesting arid important department of instruction. In a popular treatise on a subject which has engaged the attention of many Authors endued with talents of the highest order, much originality cannot be expected. This work, compiled from the best authors, is respectfully sub- mitted to the inspection of the public, with the simple re- quest, that they will examine critically and impartially be- fore they condemn it. Prince Edward Seminary, 7th mo. (July,) 184^5. ADDRESS TO PUPILS. Mt Young Ff4ends, This book is compiled expressk ly for your benefit, and is intended to assist you in acquiring a knowledge of one of the most useful, and if properly pursued, most interesting branches of science. Arithmetic is not pronounced a dry and difficult study by any one who pursues it un- derstandingly ; and that it may be perfectly under* stood by all who resolve to do so, there is no rea- son to doubt. It is to be hoped that very few, if any of you, will be satisfied short of this ; for it is not only a great waste of time and labor, but is also doubly perplexing to spend months, perhaps years in a dull monotonous drilling upon a subject, for want of a proper application, while the student who per- Heveringly removes each obstacle which he meets, will progress far more rapidly as well as pleasantly, and at the end will enjoy the gratification of hav- ing gained the object of his pursuit. I trust, that in the following pages you will find no difficulty which may not by a little sober re- flection be easily removed. An exs^mple in each case is worked out and explained, so that the most intricate may, by the application of your mindf with a little assistance from a teacher, be render- ADDRESS TO P„p,M., «<« quite plain and easy anrf , ^ constant ruJe never to pass If" '^"''^^ ""^ko it a "'O'-oughly understand^ it 'a^^.P"' ^'«"°« "•ethod, your progress mf;' /'"'°"S'' ^'^ this :t^j-----e-nrsLTa,r;^^^ omemor.. no answers 7',^°.''° •=°'»'»'««d 'earned, as you proceed .' .J° '"' "»oroughly yo"-- boots homeland stud^ ,1 ° '^"""P'"^- Take 'ong winter evenin." ^ ""'° P'"''^ during the '"•ended as an introduction {' ^^'" ^"'-"e » e^ensi.o field of se,o„ce U "'"' ''"P°«''« and "•ough acquaintance with a.-ffr'"'? ^''""'' « ">o- P'iepared to ad vance^th "■""'"= J'"" ^"1 be .«'«h '"-'-gorating potveT^J, T ""'^ ="«'="'y «nd 'ea' knowledge. °B°To.d; ""'P'"'°'''»ath;mat. parent difficulties a "hi '0'°"™^''' then by ap. '^-y be, as they often havoT'"'"""'"^"'- ^hey punted, and when o„ee ° J"' '^'''^P'^'^'y «"- ' "tead of perplex; p^ov! T '"" ""^^ *'"- i«- ' your minds. A to I I *■""' "famusement to ^.^''n aptly comtS't.nfr'l'" this study Z the country of an en my r'""'"'""^ *''""'«'» -r--.«^erewillheW;Sh^;i2L- ADDRESS TO PUPILS. lake it a without by this rather ^tly iiK go are mitted >ughly Take igthe metic ipline to fit me is : and tho- 1 be and |nat- ap- hcy jur- in- to las gh n- id behind with but a poor prospect of success. I have read an anecdote of a lad who boasted that he had been through the arithmetic and could perform any sum in it. Some person gave him the following question : " How much will sixty pounds of beef amount to at three pence per pound, provided it is three fourths fat?" After pondering awhile, he gave it up in dispair saying, if it were not for the fat he thought he could do.it. So much for going through the arithmetic without thinking. You must not be afraid of thinking. It is the very thing that will strengthen and improve your minds, while at the same time it will make you the conquerors of the field before you. The only thing wanting with most young persons, is a determiri' ation to obtain knowledge. There is time enough- wasted by nearly every one between the ages of fifteen and twenty, to acquire a good English edu- cation. Many of the most eminent men, whose names are justly honored, have attained the elevat- ed stations which they occupy in society and in the world, by their own unaided and untiring ex- ertions. Close application, joined with unconquer- able perseverance, will overcome all obstacles ; one after another will give way, until eventually, you will find yourselves standing upon the hill of science. May you then my young friends, deeply possess your minds with an esteem for that which is truly worthy of your time and attention. Seek enjoy- ADDRESS TO PUPILS «^l' render you 1,30^ ''''' ''"<* t'-uo pietv '^'V happy, is .h^o deS „f "'■"'^^ ''°«ored and Your friend, \ THE AUTHOR. /t: ,".m>; ARITHMETIC. Arithmetic is the science of numbers. It con- tains the following live principal rules of opera- tion, viz: Numeration, Addition, Subtraction, Multiplication and Division. NUMERATION AND NOTATION. Numbers arc expressed by certain characters called figures. There are ten of these characters, viz: 1, 2, 3, 4, 5, G, 7, 8, 9, 0— the last of which is called a cipher, or naught. The nine others are called significant figures or digits.* They are also called Arabic characters, because they were first introduced into Europe by the Arabs. A unit iii a wholo thing of any kind. Thus, if the number be feightfeet, one foot is the unit; if it be four pounds, one pound is the unit, &c. If the figure 1 stands alone, it represents one unit ; figure 2, two units ; figure 3, three units, and 80 on. • From the F-.itln word dicilnit, a fingpr— Iheir number being equal to that of the (ingerB on both our Imndt. QuKKTioKs — Whxt \» /.riihmeticl How many principal rulei doea it contain? What are thfyT Ih>w are nunibpra exprett* edi How many oF \\\e»n charart<>ra are tberel What fe the Uat oF there charactpra cnlledl \y hat are the nine othera called? From what ia tii? word dijrit derived, and wliy are these charao. tera ao called? Why nre they rolled Arabic choractera? Whtl Ian unit? Give exii'nplpn. What doea thn fijiure I reprflMnt libcn ilanding atone? Figure 2? Figure 31 47 5? &o. N. f '■ ri: f f i i r, 12 In tr» Av^^. i . If we wish, "" *™ DOTATION. if ^ '""" e"» We must wr;*^ * . ' ^» ^^us, lo • if towarW. it, . ^ '"^ removal of -, fi °' Hence 1 « called a un-, r ■ " '""**• *]I7* • * • Writ IVumi figur( g numbers 'eration is the r by figures •es. -"ToKSt--^ »•'-• on. of H.f,itdT" h""' "•* "»°y "? .h."" "" '^J''"'- W>' KVMBRATION AND DOTATION. i NUMERATION TABLE. S i i S3S SQ H S V] ^ 2 PQ g 3^ 11 i=i vm o o?5 10 o ^ to E-t O »TC3 s 5-c s = w o 'f w .2 '2 99 c S 5 5i=S ttt^cy Ke^h Shpq Ol «= 2 C '3 te&H^ We-h Whh3. ^ « • 2 5 • 3 4 1 *■ ■ 5 6 4 7 1 3 i 5 -^ 13 2 1 9 8 7 6 2 8 5 4 1 4 3 1 2 3 4 7 9 . , , .354 6 7 1 1 1 ! 1 14 3 7 6 7 5 4 3 2 ] L 14 3 1 6 2 5 . 4 3 ' 1. 5 6 7 8 4 1 3 12: 3 4 5 1 3 7 8 1 9 \ 2 ' 1 5 7 ( 3 10 1 2 1 1 .45: 3 7 1 12 18 9 1 6 7 1 3 5 1 2 1 3 4 16 7 9 1 2 4 8 9 1 12 4 1 2 73 1 4 5 4 3 4 5 1 6 7 --94] I 2 3 4 3 5 - I1 1 1 2 1 4 7 8 9 The words at the head of the above table, units, tenst hundreds, ^. arc applicable to all numbers* and must be committed to memory by the pupil. * Tliii t&ble if formed occordin]; in the French nr.ethod of numeration. The Eliighah method givei six placee to thou« •andi, &o. QuEvrioiii.— Repeat the numtralton table. Hew may ib* itading of figurci be facilitated? \f 1 k I' ; 1 ^ '1 Hi i ) f '•'. f ff I# VVMBBATIOir AiND NOTATION* In order to faciiitate the reading of figures they are often separated into periods of three figures each, counting from the right hand towards the lefl. The first period is called the period of imits — the second, thousands — the third, miUions—^the fourth, biliions-^the fifth, trillions — the sixth, quad- rillions, &,c. Quad. 457, Tiil. Bil. 623, Mfl. 589, Thous. 801, Units* 213. EXERCISES IN NOTATION. Write four in figures. Write twenty one. Write seventy five. Write one hundred and one. Write six hundred and seventy nine. Write four thousand and twenty eight. Write nine thousand nine hundred 6l nine. Write ten thousand. Write one hundred thousand. Write one million. Write one hundred million. Write four billions. Write two hundred and eighty five thousand se- ven hundred and nineteen. Seven hundred and thirty six thousand one hun- dred and fifty six. Seven million one hundred and sixty one thou- sand nine hundred and six. Three million seven hundred thousand six hun- dred and seventv four. Twenty seven million fifty four thousand three hundred and ninely six. One hundred and eighty two million three hun- dred and seventy five thousand nine hundred and nine. QuEmom.— What it the first period call«d1 What ii tb« te* ooftd««lled1 Thttbirdt Thefuurtht Thefiftht ThtaiiUit KVMBRJLTION AND NCTTATION. 115 WMAS Nil^MEHATION. Before the k troduction of the Arabic figures, a method of expressing numbers by Roman letters was employed. As this method is still m use, it is important that it should be learned. The letter I stands for one ; V, for five ; X, for ten ; L, fifty ; C, C, one hundred; D, five hundred; M, one thousand. As often as a letter is repeated its value is re- peated. When a less number is put before a great- er, the less number is subtracted from the greater. But when the less number is put after the greater ^ it is added, EXAMPLES. In IV, the less number, I, is put before the great- er number V, and is to be subtracted, making the number four. In VI, the less number is put after the greater, and it is to be added, making the number six. In XL, the ten is to be subtracted from fifty. In LX, the ten is to be added to fifty. ROMAN TABLE. I II III IV V VI VII VIII IX X XX XXX XL L One Two Three Four Five Six Seven Eight Nine Ten Twenty Thirty Forty Fifty LX Sixty LXX Seventy LXXX Eighty xc Ninety c One hundred cc Two hundred ccc Three hundred cccc Four hundred D Five hundred DC Six hundred DCC Seven hundred DCCC Eight hundred DCCCC Nine hundred M One thousand A line drawn ovtr a number increaiei it a thouiand timtt, tbui, X ezprtiiei ten tliouaandp and XlT twenty thoutand. U5 ■ r 1 SIMPLB ADBmOlf. SIMPLE ADDITION. If four apples are added to 5 apples, hovr many in all? Every one will answer, 9. ' Here a single apple is ihc unit ; and the number 9 contains as nianv units as the two numbers 4 and 5 ; and the operation by which this result is ob- tained is called addition.' Hence, addition is unit' ing several numbers in one. The number which is obtained by uniting several numbers into one is called the sum, or sum total. In the above exam- ple, 9 is the sum of 4 and 5 added together. Tiir: SIGNS. One stra'ght Tme crosi-irg nnothcr at r'jrht nrgle?, thus -f- is called plu?, which s'gnifics more. When placed he- twcen two n\imher!*, it denotes that tlicy are to be added together, thus 2-[-4-|-3 denotes that 2, 4< and 3 arc to bo added together. Two short parallel lines are called the rgn of equality. Thus 2-f3-f 4=:9 and 4-f-5-|-6= 1 3. When placed be- tween two ij umbers it denotes that they arc equal to each other. Note. — Bpfore adding larjre num'aors the pup'ils should lie able to add small numbers incnlully, oiid iioi by couniing their finx gera, or somelhing else, as mmiy do. By bning thorough al ibo commenccine.it much time and labor will be saved. The pupils should now be cliitispd, >ind conibinuiions like Ihs lullowing siiould be propounded until the nature of additioa is well understood. Questions. — What iflnddiiinn? VVhtit it ihe number cnlled which is obtained by uniting oeverul numbers into oneT What is the sum or4 ond &T WltHt i!< the nign uf additinn? What ia itcalledT What does it •ignifv'? When placed between two numbers, what does it denote 7 Give an example. What is the sign of equality! When plaecd between two numberi whii itoea it showf wttm(im9f^^-mmt»mm SIMPLE , 1 and are how many t 2and 1 ? 3 and 2 1 .5 ^nd, 3 ? 8 and 4 ^ APBITION. 17 14> and 3 are how many 1 15 and 4 T 16 and 5 17 and 6 ^ ,18 and 7 5 and 4- 6 an^ 3- 7 andi5- 8 and 6- 9 and 5- 10 and e- lla^i4?- )2and6- 1 ? ? 7 .? 19 and 8- 20 and 9- 9 and 4- 8 and 7- 7 and 9- 6 and 7- 5 and 8- 4 and 9- 4-4-9=ho\v many? 8+7+2= how many 1 5+4+3 = how many 1 6+5+4= how many ? 2+0+4+6 =: how many ? 9+3+2+1= how many? 10+4+6+2=hovv manyl n+3+l+0=how many ? 12+5+6+4 = how many ? 134-6+5+4+2= how many? 14+7+6+5+4+1 =how many? RULE FOR ADDITION. , Write the numbers to be added — units under bnlts, tens under tens, hundreds unde** hundreds, &c., and draw a line underneath. Add each column separately, beginning with the right hand column. When tlie sum of any column is not more than nine, write it down under the co- lumn ; but when it is more than nine, write only the right hand or unit figure under the column, - ' ■ , . . , 11 I !■ QuBBTioNs.— How do we write down ibe nufnb«r« far add i« tionT Where do we befriii lu addl When the gum of «ny co« lumn ii not more Ihan 9 what do we dot When, it exceeded what do wt dol What da we set down at the last coluinnt b3 18 ana carry (he ]g{^ u j 4 6 3 8 unite, tens utLZttr"' ''H^'' »"''« „ 2 1 6 the rale. w» .t ^ "• '« Erected in « 3 2 9 right hand „!S? '"*" commence aTtho _» 2 1 2 fin*d,L"l„7Lne':;11''l"''. and ^e nowtt'Zw*the en^rj? ^"9 "»<""'«» «» 14 , (2) 22345 PRol „" """"6 '^ *« "J*. 'III add the. «:: n^u y Vtl.' 1^^- 78 *he work may bo ref^^TZ^^T "^^ Sum, 99755 ^mio fiooA 99755 Q»MTiojr.-.How do ^^ P'ove tddiUoiif 8IMPLX ADDmoir. (3) 22321 (4) 23432 (5) 110331 (6) 143450 41332 42212 224212 467089 12123 13124 103123 356748 13220 21101 220320 910310 88996 99869 657986 1877597 140670 (8) 23456 (9) 456780 (10) 541012 596704 54321 134108 134167 86034? 12345 120212 34160 104539 67890 967342 5603 210110 30102 710011 414 121401 87549 81216 21 234561 (12) 1345601 (13) 5430161 (14)^ 123003 3413215 241678 54 456784 1014494 34124 671 341612 3742121 9671 3416 172310 34167 540 12467 416789 841 31 91511 432111 21 9 87894 10 PRACTICAL EXERCISES. 1. Find the sum of 4+8+6+10+12+16+18+26» +81-fl3. Ans. 2. What is the sum of 100+101+141+106+91-(. -[-104-4. Ans. 3.* Five bovs commenced playing marbles — ^Willet had 18, Levi 21, ^ohn 11, William 10, and Thomas 9 ; how nany marbles had they all 1 4. Charles bought apples as follows : at one time 5221^ «t another 7540 ; then 1368, then 5648, then 7300: how ■any did he buy in all? Ans. 27077. '%0 'if&'n PLC ADDinoir. 5. A man had ^1 sheep in qm field, 104 in apothe?*, 9^ lAj another, and 164 in another. How many iri-all? is. /f he populaiibh of MontiJBal 19 45,000, of Quebec 3d0^6p0, of Toronto 20,000, of ISkingston 8,000; what is thdiehlire population of the above named cities? . Ans. 1030OO. ^. t'rom the creation of the world to the Deluge was 1656 years ; thence to the building of Solomqn's temple 1344 V^ars; thence td the birth of Christ 1004 years. HdW ol'd is the world the present y^ar? . , ,>; , ; 8. A drover paid 300 dollars for 200 sheep, 525 dollars for 250 sheep, and 1000 dollars for 504 sheep ; bow ma-^ ny did he buy, and what did the whole cost! , , . . , Ans. 954 sheep, and they cost 1S[!24 dollars. 9. St. Puul's Cathedral in London cost 800,00Opound3 sterling, the Royal Exchange 80,000 pounds, the Mansion House 40,000 pounds, Blackfriara bridge 152,840 pounds, We^rhinster bridge 389,000 pounds, and the Monument 13^,000 pounds ; what is the amtount of these ,$ums t Ans. 1^74,840 pounds. 10. The British dominions are ;estirnated in squairp miles as follows: England and Wales 57,812, Sco^ond 82,167, Ireland 31,874, Islands in the Pritish Seas 332$ Colonies and Dependencies in Europe ^4, in Asia 1,204,664, in Africa 200,723, m North America 754,577, in the West Indies 77,552, in South America 52,400, and in Austra* lasia 500,000 ; what is the whole amount ? Ans. 2,912,225. 11. The populati6n of the British Empire and Colonies is as follows: England 14,995,000, Wales 911,000, Scot- land 2,'6f28,tK)0, Ir^l^hd 8,^66,d(]l5, ^orth America 1,580, 000,West Indies and South America 845,060, Afcca 300,- OOO, Asiatic 124,641,000 ; Ceylon, Ghin-tndia &c. 1,400- QOOj CW^^hic^ 575,0l60; (^uery, the entare pojpiMonI Jim, l56i4l,0Q0. 15. Th^pfoWaii'dn of UdpibW is 1^500,006, Ifcches- ^ 182,OO0i iivteobl' 165,0(3^^ ^-^v-'^o. A m a»^. Leeds 123,000, Bi^Stbl 1 17,000, ^-.i'^'^^^i^i^B^^^s^^mi mm SIMPLE SUBTRACTIOir. 12 lor- irich 61,000, Sheffield 59,000, Hull 54,000, Nottingham 51,000, Portf-mouth 50,000, Cambrit'g^ 2*1,000, Osford 21,000, and York 25,000 1 how many inhabitants in alii Ans. 2,650,000. . 13. Accordirg to the census of 1842, the population of Canada West wos as follows: Eastern District 27618, Ottawa Dist. 7386, Johnstown Di^t. 31839, Bathurst Di£t. 21083, Dalhousie Dist. 15681, Prince Edward D'ltU 14396, Midland DiA. 34438, Victoria Dist. 5214, New- castle Ditt. 30425, Colborne Dis?t. 13265, Home and Sim- coe Dists. 83294, NiogiiraDist. 34348, Gore Dist. 44232, Wellington Dist. 11418, Brock Di:.t. 17315, Talbot Dist. 10193, London Di^.t. 29357, Huron Dist. 6515, Western Dist. 21498. What was the entire population of Canada West at that time ? Ans. 459,818. SIMPLE SUBTRACTION. 1. If 4 apples are taken from 6 apple- how many will remain ? 2. If 3 pence arc taken frcm 8 pence, what will re- main? In the first example tvyo apples will remain, and the two is called the dillcrcnce bctwccjn 4 apples and 6 apples. Ilenco subtraction is taking a less numbei' from a greater. The gj eater of the two numbers is called the minuend^ and the less the sub' trahendy and the difference is called the remainder, A short horizontal line ( — ), is the s'gn of sub- traction ; it is called minus, which is a Latin word, signifying less; it shows that the number after it is to bo taken from the one before it. Thus, 0—4=5, and is read 9 minus 4 is equal to 5, or 9 less 4 equals 5. QoKKTioNs. — VVIial id suUraclioiil What is tlir |!re«ter num- ber called? Whut is the lens number called? What in the dif. ference called 7 What is the siKn of etihtractinn? What if ii •illedt What does the term signify T What does it shPWT _^.^*v many / Q f — now many ? 8--4==howmany? r^— now many ? io^~?°^^ many ? iti=:how many ? 13_Q ^'^'"any? Ti ^^^''ow many ? Jp f7howmany? I8Z7 K^'^'^any? 21-70 .°'^'"«"y^ ot f^-Jo^^niafly? icJ=howmany? Write down'tL""" «t»'-«''<^"o''- " """"" "amder direct/y below "' '"'' ^"'^ the «. It- S ot S/y'"*^^ "PP- «- - smaller *e upper figure bit i^thfr'^ "" '° ^^ nd ny? ny? ay ? ly ? ny? ly? ly? ly? »y? »y? 3r the * tens> hand > Jow- e re- id to 1 to act- BIM;P|.S SUBTRACTIOIV. 2S 4561 EXAMPLIB8. From 4567 Subtract 1314 3436 subtrabe^Bd. 2187 minuend. 3243 1249 remainder. In the first example, we begin at the right hand and sub- tract each figure in the lower line from the * Take 132941 Add the remainder and subtrahend to- gether ; if the work is right, the sum will Rem. 290711 be equal to the minuend. Proof, 423652 Addition may be proved by subtracting continually from the amount the several numbers which were added to pro- *' ^■' I ■ 11 I ■ _ , _ „ ,^ - — . — I, , . I. M— I I I— ■ _ I. ■ ■■■ ■ - I II ■■■11^ QoBSTioN8.--lt an equal amount be add***] to the minuend and ihe attbtrahend. is the remainder affected? How do we prOT« iobtraelionl How may addition be proved by aubtracViont ^ . «t-(..,*,rw...WH* ( ; t; L. U SIMPLE SUBTRACTXOir. ducc it, and if the work is right therc^ will be no remainder. ILLUSTRA.TION. — 5,4, 7=16: proof, 16 — 7=9, and 0^-4=5, and 5—5=0. W (5) From 3621531 Prom 91041234 Taiie 1841675 Take 70134165 . 1779856 (6) 31.5701341 U 1870 134 (9) 74)6789 946340 (7) 12301014 9109417 (10) 1043456 141897 20907069 (8) 841397 198745 01) 10101010 1010101 •l 12. From 41078912 take 19416781. 13. From 72416714 tako 13741010. 14. From 91012412 take 9178743. SIMPLE MULTIPLICATION. 1. A boy g:ve.s8 apples to eacli of 3 companions? how many does he give to tliem all ? 2. If 1 bushel of apples coi:t 9 ponce, how many pcnco must I pay for 4 bushels ? 3. If one orange costs 3 pence, what will 4 omngcs costi The answers to the above quoslions may be ob- tained by addition ; but the operation may be much facilitated by a rule called Multiplication. In the first example the numljcr 8 apples is repeated 3 Gmes, we may therefore add 8 three times to itself, thiv 6, 8, 8=24 ; or we may say 3 times 8=24, and so of tb* Moond and third examples. tmmmmm- f- SIMPLE MULTIPLICATION. 25 Multiplication is a short method of repeating one number as many times as there are units in • another. The number to be repeated is called the multi- plicand. The number which shows how many times the multiplicand is to be repeated is called the muUi' plier. , The answer is called the product, because it is the sum produced by multiplication. The multiplier and multiplicand taken together are called factors. Sign. — Two short lines crossing each other in the form of the letter X are the sign of muitiplicatioa ; thus, 4x2 =8, which means that two times 4 are equal to 8, or 4 times 2 are S. Note to Pupils. — I hope you will so far consult your own interest, as to commit the following ta- ble perfectly to memory before you attempt to proceed farther. No progress can be made with- out it. if yo^ apply yourselves perseveringly to the task, you can soon accomplish it ; and if you should proceed no farther, you will find it of great advantage to you through life. — — . ,«_^ QuBSTiom.—- What ia muItiplioationT What ia th« number callad which ia to be repeated'' What ia the multiplier 1 What ia the anawer calted! Why? What are the multiplier and multiplicand taken together, called f In the firal example, which ia the multiplicand? The muUipiier? The pcoduct?-* What ie the eif n of Multiplication? 26 ; i SIMPIE MVL Mi : I \ i TIPIICATION. SIMPLE MULTIPLICATION. 27 ^^«^2il, ^^ "^ "^ i sssigf. ^^ "^ — ^»Sg{:>^ »^ "^ iM 1 ■^ »-» 1-^ I "5 weopo nULE FOR MULTIPLICATION. L Set down the multiplier under the* multipli- cand, so that units shall fall under units, tens under tens, hundreds under hundreds, &c. and draw a line underneath. When the multiplier does not exceed 12, begin at the right hand of the multiplicand, and multiply each figure contained in it by the multiplier, set- ting down and «airying as in addition. 2. When the mulliplier exceeds 12, multiply by each figure of the multiplier separately ; first by the units, then by the tens, then by the hundreds, &c., being carefial always to place the first figure of each product directly under the figure by which you multiply. Add up the several products, and their sum will be the product sought. EXAMPLES. I. Multiply 145 by 3 ; that is, find 3 times 145. The answer to this example might be obtained by adding 145 together 3 times ; thus, 145+145+145=435, or more readily by multiplying 145 by 3. OPERATION. We first write the multii)lier un- der the multiplicand, and then say 3 times 5 are 15, or 5 units and 1 ten-*- we write down the 5 units only, and reserve the ten to be added in the ten'H place 5 we then say 3 times 4 are 12, and 1 more makes 13, that is 3 tens and one hundred. We now write down Multiplicand 145 Multiplier 3 Product 435 QoESTioNi.— How do wo let down numbers for multiplying? Where do we begin to muUiply? How do we multiply? When the multiplier exoeedi 12 now do we proceed? how do we place the products? What do we do with the eeveral produoti? What ii their aum? ff-j t il • i ■ s • r I 3d *« 3 ten, in th '""'" ^"'''n'wCATro^. It is nl!i,„ e ' "''"«i is 145 aT' *" "* find »he muSr ' "?^°/ times as S^ ''own the JJu^Micand 365 r^^\^^rioN. ^""••P«- _f4« si.^"o?||;-P/-^^^^^^^^^^^^^ eon. Product ^^ ''«'H and place .'hi " "T *''« 2 hun- -3790 under the figClf^ ?':'^"« Arecfly Con«eq„c„.,,,.,.3,„^ ,, ""'"'^P'-'^""" 73000 ">"« be the enZT,^ *** Products ^ ^n'te the mnu: i- pRoop. - i ^^e hundreds for ^'^Sandlare ^^i so we find t^mes repeated. ? that Muiti^ ^^* and that ^^ down the are units in ler. '«^%her coa- * write it un- s under units, '^e then mul- the 6 units, rthe2hun- luct directly y^Y', lastly cts together, 2190 14600 ct is 73000 89790 246. K5 is the P^'er, and »re alike, ^^ How ' be used samples. SIMPLE MULTIPLICATION. 29 (3) 4372543 2 (4) (5) 427365 5729385 3 4 8745086 • (6) 4567145 . 5 (7) (8) 3541678 13456789 6 7 (9) 3451248 8 2314521 9 (11) (12) 314156 1213451 10 12 13. Multiply 14. « 15. « 16. « 17. « 18. « 19. « 20. « 480 by 1324 by 3648 by 3725436 by 12765235 by 537467 by 673526 by 346726 by 36 Product 17280 45 « 59580 72 « 265248 i 43 « 160193748 \ 275 « 3510439625 .i 367 « 197250389 2674 « 1801008524 i 3426 « 1187883276 i CONTRACTIONS IN MULTIPLICATION. I. When the number is 1, and any number of ciphers after it, as 10, 100, 1000, &c. We have already learned that a cipher placed on the right of a number, changes the units place into tens, tin? tens into hundreds, &c. Hence, When the multiplier is 10, 100, or 1 with any number of ciphers annexed, place as many ciphers to the multipli- cand as there are ciphers in the multiplier, and the multi- plicand so increased will be the product required. QuKtTioNs.— How 18 a number affected by placing a cipher on the right of it? When the multiplier it 10, 100, 1000, &c. how do we proceed? o8 ■ ft 30 k 1 II : I * 111 SIMPLE MULTIPLICATION. EXAMPLES. U Multiply 341 by 10 Ans. 3410. 2. « 475 by 100. Ans. 47500. 3. « 3159 by 1000. Ans. 3159000. 4. « 2346 by 10000. Ans. 23460000. II. When there are ciphers on the right hand of one or both of the factors. RULE. Neglect the ciphers and multiply by the signifi- cant figures only ; then place as many ciphers to the right hand of the product, as there are in both of the factors. EXAMPLES. (1) 365 40 • (2) (3) 4567 46000 300 340 Ans. 14600 Ans. 76400X24 7532000X580 21200X70 4871000X270000 7496430X695000 1370100 184 138 4. 5. 6. 7. 8. Ans. 15640000 Ans. 1833600. Ans. 4368560000. Ans. 1484000. Ans. 1315170000000. Ans. 5210018850000. III. When there are ciphers standing between signifi- cant figures of the multiplier they may be disregarded, EXAMPLE. 1. What is the product of 12318 multiplied by 7004 ? Questions. — When there are ciphers u.; Ihe right hand of one or both the factors, >vhat is the method of proceeding? How many ciphers should be placed at the right handof the productt When there are ciphers standing between significant figures of the multiplier how should we treat them. •mmimm Ahs. 3410. Ans. 47500. ^ns. 3159000 ns. 23460000. hand of one or / the signifi. y ciphers to are in both 46000 340 isT 138 15640000 >. >ooo. ). >oooooo. !850000. en signifi^ 'regarded. ^y 7004 ? land of one ogf How e product? t figures of SIMPLE MULTIPLICATION. 12318 7004 M Operation. •> 49272 86226 86275272 2. What is the product of 506 multiplied by 302 ? Ans. 152812. 3. Multiply 154326 by 3007. Ans. 464058282! IV. When the multiplier is a composite number. A composite number is the product of two or more num- bers, which are called the components or factors. Thus, 4X3= 12. Here 12 is a composite number, and 3 and 4 are the factors, 8X^=40 ; 40 is also a composite number. BULE. When the multiplier is a composite number, multiply by each of the factors in succession, and the last' product will be the entire product sought. EXAMPLES. 1. Multiply 365 by 16. The factors of 16 are 4 and 4, or 2 and 8, or they are 2 and 2 and 4 j for 4X4j=16 and 2X8=16 j also, 2X2X4=16. 365 365 365 4 8 2 1460 4 Product 5840 2920 2 730 2 5840 1460 4 Product 5840 Questions.— Whet is a composite number? Give an example. What are the factors? When the multiplier is a composite number how do we multiply? What will the last product be? 32 ■4^ I. m SmPlfi MVLTlPUCATtOif. *• Midtiply 8423675 by 64 ^'^ /ns. 38604360. 5. Multiply 5246789 by 81 ^'^- 839115200. 6. Muluply 4103413 bv 100 ^M. 424989909. ' ^ Ans. 410341300. 1 Wk . ''**C"CAr, EXERCISES. 4 .„d71"" ""'"'»' » *=•«. the facto™ of which a« 9 5 4. Suppose a man wem f 7^ i o^"^' ^^^ houre. a«;il ^^ «>- n-be, of which 1 1^?*! Z^l Sj-Jp '-'''> -.Sin a ^iie, howln^f r^^„ 8- What wiU 194 chests oftoa costl?,^'^?*'*' ''''<'«• . "'°^^^*'5doUareacheslt »• A man sold a farm «/.-» • • . -Ans. 14550 . •- - ac. , what ^^c^:izz':r' "" ''^^- ^Ju' ^"PP^'se a book to contnin ^A""^' ^^^^ *^«"a»^. "*'*TO«i^w»^jiL 6IMPLE DIVISION. BS A.ns. 274032. s. 38604360. . 539115200. . 424989909. 410341300. lich are 9, 5, the other 45 miles a day ; how far will they be apart at the end of 12 days ? Ins. 912 miles.. 13. The component parts or factors of a certain num- ber are 4, 6, 8, 9, 12, and 14 ; what is the number ? Ans. 290304. 14. A dollar contains 5 shillings ; how many shillings in 299 dollars 1 Ans. 1495 shillings. 15. One pound contains 4 dollars ; how inany dollars in 1540 pounds! Ans. 6160 dollars. 109 trees in on each tree, contain ? )138 apples, 'n days in a 8 hours. I a day, how > days, how '60 men. 14^, and 21 . 31752. ny rods in 00 rods, irsachest? 14550. 'or 19dol- dollars. lines on my letters 57500. of 58000 consisting miles, id travel 3s a day, i SIMPLE DIVISION. John has 25 apples, and wishes to divide them equaUy among 8 bo^^s. In this example, giving each boy 1 would take 8 and leave 17 ; giving each 1 a second time would take 8 and leave 9, and giving each 1 would again take 8 and leave 1. 8 has been sub- tracted from 25 3 times ; hence 8, is contained 3 times in 25 and there is 1 Operation. 25 8 lat remainder 17 8 2nd remainder 9 over, g 3rd remainder 1 By continued subtraction we can always find how many times one number is contained in ano- ther, and likewise what remains when it is not contained an exact number of times. We can however arrive at the same result by a much shorter and more expeditious method called Division. - ~ -^ QuEstiDNs.— When John divides 25 apples equally amongf 8 boys, how many does he give to each? How many times does 85 contain 8? How many remain? By continued subtraction what can we find? By what other method may we arrive al the same result? Bii ' tf 'ant «„,„,. "* ^«e Quotient There are th ®^^^'^®* Joilounna- ;. ^ P'^^cedtne if ;= * . » '"us -:. ;» .1, ^ ten oier^f 'V *"« 24^4-B o*°> *vided Bv !.. °^« 2 "n 2 how„ """^^ femffiaS »""'"''"" |;»8 « " !*m32 ;; « ? «« in 10 2 in J2 2 in 14 5 in 1^ 2 in IS (e tt (t « ivide 840 by 24. "^^^^^^»on. Operation. T^. . J^ividend ,i:,"" """^^^"^P'e we set fhn 120 "ght oi the dividend a*, tht VIUE8TI0N8. — VVhnr. 1 1 j- ■^ -. ' *"'•"• " *''« operation died. SIMPLE DIVISION. 39 product is 72, which is placed under 84 and subtracted irom it and 12 remains. We now bring down the next figure of the dividend, and conclude that 120 will contain 24 5 times ; we therefore place 5 in the quotient and mul- tiply as before. Thus we find 840 contains 24 35 times. 2. Divide 2756 by 26. Operation. In this example we Dividend. ' say 26 in 27 once, and Divisor 26)2756(106 Quotient, place the 1 in the quo- 26 ticnt, multiply 26 by 1, subtract and iDring down 156 the 5 ; we then say 26 156 in 15, times and place the in the quotient, wo -:l then' L'lii.'g down the 6, and find that the divisor is contained in 156, 6 times. Note 1. If the remainder at any tiiu i I'C greater than the divisor, the quotient figure must be increased. Note 2. If the product of the divisor and quotient fi- gure exceed that part of the dividend directly above it, the (juotient figure must be diminished, 3. Divide 30d60 by 84. Operation. Proof. 84)30660(365 365 Quotient. 252 84 Divisor. 546 504 1460 2920 420 420 30660=thcdividend. QuRiTtoNi. — If llie remainder at any lime be greater than lh« diTiiur, wlint mu«t be dune'f If the product ot the diviaur and quotient figure exceed that part of the dividend direotl/ above it, what inuat be doneY m .»{ ' * ill i M ^ 40 SIMPLE DiVlSIOir. 4. Divide 25648 by 21. Operation. 21)25648(1221 21 Proof. 1221 Quotient. 21 Divisor/ 46 42 44 42 1221 7 remainder. 25643 =:thc dividend. 28 21 5. 6. 7. S. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. '^ S' Ans. 724GG7— 3 rem. Ans. 192i()5— 11 rem. Ans. 4192G1 Ana. 234. Ans. 2S35 '^-\ Ans. 17359— Ansj. 3283.. Ans. 1345. •1 rem. 7 rem. Divide 9420674 by 13. Divide 3271916 by 17. Divide 9643007 by 23. Divide 6318 by 27. Divide 9G414by34. Divide 9373S7 by 54. Divide 147735 by 45. Divide 145260 by 108. Divide 24167 by 125, and prove tlie operation. Divide 3410S by 87, Divide 10416 bv 140, Divide 541678 by 341, Divide 674160 by 410, Divide 940161 by 365, Divide 141041 by 1341, 20. Divide 8416759 by 2140. CONTIIACTIOWS IN DIVISION. I When the divisor is a composite number. RULE. When the divisor is a composite number, divide the dividend by one of the component parts, and a a a a a a a u a 6i a n SIMPLE DIVISION. #4J the quotient arising from that division, by the oth- er ; the last quotient will be the answer. EXAMPLES. 1. Divide 54}30 apples equally among 15 boys. Operation. In this example 5 and 3 are the 5)5430 component parts or factors of 15. — First divide the apples among 5 boys, 3)1086 1st quo. and we find they will have 1086 ap- pies a piece. Then let each one of 362 quo. sought, these boys divide 1086 among 3 boys, and they will have 362, and the whole number of parts will be 15. 2. Divide 18576 by 48= 4X12. Ans. 387. 3. Divde 9576 by 72= 9X 8. Ans. 133. 4. Divide 19296 by 96=12X 8. Ans. 201. To obtain the true remainder, when factors have been used as divisors, multiply the last remainder by the first divisor y and to the product add the first remainder. 1. Divide 4967 by 32=4X8. Operation, 4)4967 Ans 155^^-. 2. 3. 4. 5. 8)1241 — 3, 1st remainder. 155 — 1X4+3=7 the true remainder. Divide 956789 by 7x8=56. Ans. 17085|J. Divide 4870029 by 8X9=72. Ans. 67639f^. Divide 674201 by 10X11 = 110. Ans. 6129,Vo« Divide 445767 by 12X12=144. Ans. 3095-jV4-. QcxsTioNii.— When the divisor is a composite number, how 11 tlie diTiiion performed] What will be the answer? How may the true remainder be obtained when factors have been used as divisors? d2 tfmm SIMPLE DlVISIOi^i II. When the divisor is 10, 100, 1000, &c. ftULE. Cut off as many figures from the right hand of the dividend, as there are ciphers in the divisor ; the other figures of the diviuend will be the quo^ timtf and the figures cut oflf will be the remainder. EXAMPLES. 1. Divide U364, by 100. Ans. 143,^0 Operation. In this example there are two Os in the 1|00)143|64». divisor; therefore we cut off two figures from the right hand of the dividend, and the quotient is 143, and the remainder 64». 2. Divide 24367 by 10. * Ans. 24 10 ^> la 45 la 6? 6 — 3> XM 3> 14 7? 15 T» It should be constantly borne in mind, that if the numerator and denominator he divided hy the same number, the value of the fraction remains the same. Questions. — What are the numerator and denominator taken together called? If a mf?!oii be divided into 8 equal parts, and 4 of them be given to Charles and the remaining 4 to Henry, what is each. boy's share of the melon? What are 4-8 equal tot What is the difference in value between 4 8 and J? What if this operutien of changing fractions called? How may a frac- tion be reduced to lower terms? If the numerator and deno^ minator be divided by the same number what effect has it on the ▼alu« of the fraction? It the numerator and denominator in the fraction 5-15 be both divided by 5, what will the ezpreision be? Will its yalue be changed? ^! SUPPLEMENT TO MULTIPLICATION. 47 ogether 3 and 4 other 4 3h boy's boy has loes not ■educing iividing number 7iainder, linder and 7> 15 T» latif the the same \he same. lator taken parts, and to Henry, 3 equal tol What if nay a frac- and deno^ nfl it on the rninator in expreiiion SUPPLEMENT TO MULTIPLICATION. You have learned how to multiply by integers, or whole nnmbers. It '^ .ecessary that you now learn how to mul- tiply by ^ mixed number, which is made up of a whoU number and a fraction : thus, 2}, 4^, Gj^, 9^, &c. Multiplication has been defined to be the repeating of one number as many times as there are units in another. Hence multiplying by 1 is taking the multiplicand once^ multiplying by 2 is taking it twice, multiplying by 3 is tak- ing it three times, &c. Multiplying by a broken number is taking a part of tho multipUcand as many times as there are hke parts of a unit in the multiplier. Multiplying by ^ is taking one half o^ihQ multiplicand. Multiplying by \ is taking one fourth of it. Multiplying by -\ is taking three fourths of it, &c. To multiply a whole number by a mixed num- ber observe the following nuLE. First multiply the multiplicand by the whole number of the multiplier ; then multiply it by the numerator of the fraction, and divide that product by the denominator ; add the two products toge- ther, and their sum will be the entire product of the mixed number. QoESTioNB. — What is a mixed number? Give an example! How has multiplication been defined? What is multiplying by 1? by 2? by 3? What is multiplying by a broken number? Multiplying by \ is the same as what? by ^7 by ^7 Repeut the rule for multiplying by a mixed number? h^ %. 1-. u f- 43 SUPPLEMENT TO MULTIPLICATION. EXAMPLES. 1. Multiply 48 by ^. Operation. When the numerator is .1, as in this ex- 48 ample, there is no necessity of multiplying 1 . by it, for it siipply repeats the multiplicand. 2)48 24 Product. 2. Multiply 36 by 4^ Operation. 36 36 4 3 144 4)108 27 , 27 Product of 36 Xf 171 Product. In this example we first multiply 36 by 4, and the pro- duct is 144 ; we then set 36 down in another place, and multiply it by f, and find the product to be 27 ; the sum of these products are the entire product of 36 multiplied by 4f . Multiply 39byi 96 by J- » • 3. '4. 5. 6. 7. 8. 9. 10. 11. 12. 150 by -I- 69 by 2i 100 by 41- 48 by 23- 55 by 5^ 346 by 2t 100 by 6 1 240 by 18f Ans. 13. Ans. 24. Ans. 30. Ans. 161. Ans. 425. Ans. 132. Ans. 286. Ans. 939|. Ans. 666|. Ans. 4453f . PRACTICAL EXERCISES. 1 . There are 16^ feet in a rod, and 320 rods in a mile : how many feet are there in a mile 1 Ane. 5280. his ex- tiplying lUcand. the pro- ce, and Ithe sum ultipUed I a mile : 5280. PRACTICAL EXERCISES^ 49 It. What is the cost of 8} tons of hay at 10 dollars per Ion? Ans. 87^ dollars. 3. If 4f- bushels of wheat make 1 barrel of flour : how many bushels will it require to 250 barrels? Ans. 1200 bushels. 4. There are 69^ miles in one degree : how many miles in 360 degrees ? Ans. 25028 miles. 5* What is the cost of 25 horses at 15^ pounds each? Ans. 387^ pounds. Note to Pupjls — You have now been through the fundamental or foundation rules oi Arithmetic, and I hope are able to solve all the questions thus far under standingly, and can also answer all the questions at the bottom of each page. A thorough knowledge of the foregoing rules will greatly facil- itate your future progress. I will now present you with a few , PRACTICAL EXERCISES. Involving the principles of the preceding rules. 1. 240+670+81+14+9+2=howmany? Ans. 1016. 2. 10460148— 1341009=how many ? Ans. 9119139. t, 340084 X4005=how many? Ans. 1362036420. 4. 2674236-^634=how many ? Ans. 4218^8V 5. ■^•^iH^=howmany'? Ans. 49. 6. 25+18+10— 13x9-7-3 = how many? Ans.Jl20. 7. 100+75x4— 2-;-4=hovvmany? Ans. 174i. 8. 12X11+141—13+10x2= Ans. 52. Note. — A line, or vinculum, drawn over several num- bers, signifies, that the numbers under it, are to be taken ai one whole number ; thus 4+3 X 8 — 2=42. 9. 10. 5+6—4x9— 6=how many? 10— G+4X7— 3+l=how many? Ans. 40. Ans, 5|. 9—5 Question. — Wiiui dues a line, or vinculum, druwit over sev- •ral numbers signify} E iS ^1 Hi H' it !f ? '.I .»{ 50 PRACTICAL EXERCISES. n. From the sum of 100, 84, 75, 18 and 31 subtract the sum of 10, 9, 8 and 2. Ans. 279. 12. The sum of two numbers is 1045, one of the num- bers is 109, what is the other 1 Ans.'936. 13. If the minuend be 1460 and the subtrahend 1390; what is the remainder] 14. If the minuend be 1460 and the remainder 70 ; what is the subtrahend 1 15. If the subtrahend be 1390 and the remainder 70 j what is the minuend i 16. What is the product of 537467 and 367 ? Ans. 197250389. 17. If the divisor be 84 and the dividend 30660 ; what is the quotient ? Ans. 365. 18. If the quotient be 365 and the divisor 84 ; what is the dividend ? 19. The factors of a certain number are 86972 and 1208 ; what is the number 1 ^Ans. 105062176. 20.- A farmer sells a horse for 18 pounds, 5 cows for 4 pounds a piece, 6 oxen for 5 pounds each, and 100 sheep at 1^ pounds a head : How much did he receive for them all ? Ans. 218 pounds. 21. A gentleman sells 150 bushels of wheat at^ (one fourth) of a pound per bushel, 25 tons of hay at 3 pounds per ton ; and takes in part payment a wagon at 10 pounds, and a yoke of oxen worth 20 pounds, and the rest in cash : How much money did he receive 1 Ans. 82^ pounds. 22. Sold a ship for 11516 pounds, and I owned (three fourths) of her ; what was my part of the money ] Ans. 8637 pounds. 23. A gentleman left his son, sixteen thousand sixteen huhdred and sixteen pounds, and his dau|;hter one half as much : Query, the daughter"'s portion . Ar s. 8808 pounds^ 24. The circumference of the earth is 25000 miles : how long would it take a man to travel around it, supposing him to travel 4JJ miles a day ? Ans. 595^ days. 25. What wiil 12| yards of cloth come to at 6 dollars a yard? Ans. 76f dollars. PRACTICAL EXERCISES. 51 26. The Bible contains 3 1 173 verses ; how many vers- es must be read in a day, that it may be read through in 365 days? Ans. 85^^ 27. What is one-eighth of 59876 1 Ans. 74-84 1 . 28. What is four fifths of 64870 1 Ans. 51896. 29. Write down four thousand six hundred and seven- teen 5 multiply it by 12, divide the product by 9, and add 365 to the quotient, then from that sum subtract five the -i- sand five hundred and twenty-one, and the remainder will be 1000. Try it and see. 30. If a man's wages amount to 625 dollars iu b'l weeks, how much may he spend a week and save 261 dollars 1 Ans. 7 dollars a week. 3 1 . How many times can 24 be subtracted from 1416? Ans. 59. 32. Sold 5505 pounds of butter at 12 pence per poumi, and took my pay in molasses at 36 pence a gallon ; how many gallons did I receive 1 Ans. 1835 gallons. 33. How many barrels of flour, at 8 dollars per barrel will it take to buy 62 horses at 95 dollars each I Ans. 736 1 . 34. How many men must be employed to do a piece of work in 1 day, that 11 men can perform in 18 davs? Ans. 198dcy.s. 35. Thomas and Joseph are studying arithmetic. Tho- mas is 322 examples in advance of Joseph, but Joseph performs 55 examples in a day, and Thomas 41. In how many days will Joseph overtake Thomaf^i Ans. 23 days. 36. A. B. and C. made up a purse of .500 pounds. A. put in 75 pounds, B. put in three times as muiih: How much did C. put in"? Ans. 210 pounds. 37. What number must be subtracted from 294106 in order that the remainder shall be 14230 1 Ans. 279876. 38. In 30416 dollars ; how many pounds. Ans. 7604. 39. In 8940 shillings j how many doUajs. Ans. 1788. 40. A farmer purchased a farm, for which he paid' 1850 dollars. He sold 50 acres for 60 dollars ner acre, mmmm WW J WI ■mainaanHr- 52 PRACTICAL EXERCISE!^. 'it 'J ' and the remainder stood him in 50 dollars an acre : ho^ much land did he purchase. Ans. 351 acres. 41. The mariner's compass was discovered in Eng- land in the year 1302 : how long since that time*? 42. A gentleman wishes to distribute 1200 apples a- mong 5 boys : he gave the first boy one third of the whole : the second one fourth, the third one fifth the fourth one sixth ; and the fifth the remainder: how many apples did each boy receive. C 1st boy 400, I 2nd « 300, Ans.^ 2d " 240, I 4th « 200, I 5th " 60. .43. A speculator bought 536 acres of land at 10^ dol- lars per acre : he sold A. 100 acres at 12 dollars an acre, Ji. 150 at 11 dollars per acre, C. 145 at 13 dollars per a- cre ; he was obliged to sell the balance at 8 dollars per acre : did he gain or loso* by the purchase, and how much ? also, what was the average price that he obtained per acre. Ans. he gained 235 dels. Average price 10^5 6 tlollars. Note by the Printer. — The following exercises weit) inadvertently omitted by the compositor, and should have been inserted immediately after the examples in subtraction. •» - PRACTICAL EXERCISES IN SUBTRACTION. 1. From fifteen million take fifty six thousand, and what will remain. Ans. 14,944,000. 2. A man paid 150 dollars for a good horse and sold liim for 175 dollars ; how m.uch did he gain 1 Ans. 25 dol. 3. What number is that which, taken from 5487, leaves 999? Ans. 4488. 4. Columbus discovered America in the year 1492, how many years since ] 5. Queen Victoria was born in the year 1819 succeed- ed to the throne of England in 1837 ; what was her age p.: the latter period ? : how acres. L Ettg- iples a- whole : ih one ►les did 400, 300, 240, 200, 60. 0^ dol- an acre, 3 per a- lars per 1 much ? )er acre. 35 dols. dollars. ges weix) lid have traction. nd, and 144,000. ind Bold 1. 25 dol. 5487, 4488. 1492, jucceed- ;r a|(e et PRACTICAL EXERCISES. 93 6. Subtract one from one million. Ans. 999999. 7. From one hundred million two hundred and forty •even thousand, take one million four hundred and nine. Ans. 99246591. 8. The number of inhabitants in the U. States, in 1830, was 12,840,540, in 1840 they amounted to 17,069,957. What was the increase in 10 years? Ans. 4,229,417. 9. The author of this book was born in the year 1814, What is his age the present year, 1845. 10. Canada was ceded to Britain, by the French, in 1763, hov/ long since that time? 1 1 . What number, together with these three, viz. 130 1 , 2561 and 3120, will make ten thousand. Ans. 3018. 12. What number must be added to 175 to make 1101. Ans. 926. 13. There are two numbers, whose dilTerence is 10, 101 ; the greater number is 100,000 : Query the less. Ans. 89,899. 14. A man dies woith 10,104 pounds, he leaves to his daughter 4,115 pounds, and the remainder to his son : what was the son's portion ? Ans. 5989. 15. The sun is ninety five millions of miles from the earth, iind the moon is two hundred and forty thousand miles, how much farther in the sun from us than the moon ? Ans. 94,760,000. 16. The minuend being 135 and the subtrahend 100: what is the remainder? Ans. 35. 17. The subtrahend being 100 and the remainder 35 j what is the minuend ? Ans. 135. 18. The minuend is 135, and the remainder 35 ; what is the subtrahend. Alls. 100. 19. The sum of two numbers is 100, and one of th« numbers is 35 : what is the other? Ans. 65. QuEHTioNg. — When the minuend and lubtrahend are giYcn, how do we find the remainder? When the aubtrahend and r«. mainder are given, how do we find tha minuendl When tht ium of two number! and one of the numberi are given, how do we find the other? i2 ■M-ia ■pSp > ( 54 PRACTICAL EXERCISES, 20. The greater of two numbers is 100 and their differ^ enee 35 ; what is the less number ? Ans. 65. 2 1 . The less of two numbers is 65, and their difference 35 ; what is the greater 1 Ans. 100. Questions. —When we have the greater of two numbers and their difference, how do we find the less number? When we have the leis of two numbers and their diiferlsnce given, how do we find the greater? E^D OF PART I. I '' i--^}'- •' £« m CO]\jIPOUND NUMBERS. A number expressing things of the same kinfl is called a simple number. For example, ^0 horses, 4 pounds, 8 bushels, are each of them simple num- bers, A compound number, expresses things of differ- ent kinds ; thus 5 pounds, 8 shillings and 10 pence is a compound number, also, 1 year 2 months 2 weeks and 3 days, 2 bushels 3 pecks 4 quarts, are compound numbers. When numbers have differ- ent names as above they are mostly called differ- ent denominations, TABLES OF COMPOUND NUMBERS. QuciTioRi.— What do simple numberi ezpreiiT What do coinpoand numberi eipreiil Give an example of a aimple number, of a compound number. What ii underitood by diff- erent denominationif Are 8 poundi and 4 pounda of the lame denomination? Are 6 pounds and 4 shilliDgsT Are seyeral litnnbers of diffcreat ^^oominatiooi oflea oonii«oted tuget^ft? Oite an example? " ' PENCE TABLE. d, s, d. s, d, J 20=1 8 2=24. 30=2 6 3=36 4.0=3 4 4=48 50=4. 2 5=60 60=5 6=72 70=5 10 7=84 80=6 8 8=96 90=7 6 1=108 100=8 4 10=120 110=9 2 11=132 120=10 12=144 mtti gfjgggat mmtm r^ . : TABLES OF COMPOUND NUMBERS^ ENGLISH MONEY. The denominations of English Money, guineas, pounds, shillings, pence, and farthings. 4f farthings marked /ar. make 1 penny marked d. 12 pence make 1 shilling b. 20 shillings ^ make 1 pound £, 21 shillings make 1 guinea. Note. — Farthings are generally expressed in fractions of a penny. Thus, for 1 farthing we write ^. for 2 far- tl^ipgh ^d. and fpr 3 farthings ^d. TROY WEIGHT. i Gold, silver, jewels and liquors, are weighed by this weight. Its denominations arc pounds, oun- ces, pennyweights, and grains. 24 grains, (gr.) make 1 pennyweight marked pwt. 20 pennyweights " 1 ounce, oz. 12 ounces " 1 pound, — : — lb. .. APOTHECARIES* WEIGHT. This weight is used by apothecaries and physic clans in mixing their medicines. Its denomina- tions are pounds, ounces, drachms, scruples, and grains. 20 grains, (grs.) make 1, scruple. 3 scruples " 1 drachm. 8 drachms " 1 ounce. * 12 ounces " 1 pound. QuEBTioNS.— What are the denominations of English Moncyf Repeal the table. How are farthings generally expressed! How many pence in 6 farthings? In 8? In 107 'in 121 How ma. ny shillinss in 14 pence? In 161 In 187 In 337 In 271 In S21 In 441 How many pounds in 24 shillings! In 30 shit* lingsl In 361 In 401 What articles are weighed by Troy Weightl What are its denominational Repeat the table. What is the use of Apothecaries' Weightl What are itf df nf* minationsl Recite the table. TABLES OF COMPOUND NUMBERS. 61 AVOIRDUPOIS WEIGHT. By this weight are weighed all things of a coarse drossy nature, as tea, sugar, grain, hay, tallow, leather, medicines, (in buying and selling) and all kinds of metals except gold and silver. The de- nominations are tons, hundreds, quarters, pounds, ounces, and drachms. 16 drachms (drs.) make 1 ounce, marked oz. 16 ounces " 1 pound " lb. 28 pounds <* 1 quarter " qr. 4 quarters make 1 hundred weight, marked cwt. 20 hundred weight make 1 ton. a T. Note. — In this weight the words g?-oss and neit are used. Gross is the weight of the goods, together with the box, cask, bag, bale, &c., which contains them. Nett is tlie weight of the goods only? or what remains after deduct- ing from the gross weight, the weight of the box, bag, cask, &c. A hundred weight is 112 lbs., as appears from thd table. LONG measure. Long measure is used in measuring distances, or other things, where length is considered without regard to breadth. Its denominations are degrees, leagues, miles, furlongs, rods, yards, feet, inches, and barley corns. J barley corns, {bar,) mak e 1 men, marked m. 12 inches a 1 foot. ^< ft. 3 feet i( 1 yard. « yd. 5^ yards or 16.^ feet i( 1 rod, perch or pole, " rd. 40 rods it 1 furlong. " fur. 8 furiongs or 320 rods (( 1 mile. « mi. 3 miles a 1 league, « L. 60 geographical or ) 69 i statute miles, ) a 1 degree, "deg.or^ QuRiTioNB. — What are weighed by Avo'sdupois Weight?-- Whai are its denominations? Repeat the table. What ii meant by gro^a weight? What by nett? What is a cwt? When ia long niMaure Uf«d1 What are ita denominatiuna? Repeat the tabU^ 58 TABLES OF COMPOUND NUMBERS. '^1 360 degrees make a great circle^ or circumference! of the eartl^* ^foTE. — A hand is a measure of 4 inches, and is use^ to ineasiire the height of horses. A fathom is 6 feet, and is nibstly used to measure the depth of water. CLOTH MEASURE. By this measure all kinds of cloth, tapes, dec, are measured. Its denominations are ells French, ells English, ells Flemish, yards, quarters, nails and inches. 2^ inches (in.) make 1 nail, marked na. 4 nails « 1 quarter of a yard " qr. 4 quarters (6 1 yard, « yd. 3 quarters (i 1 Ell Flemish, « E. FI. 5 quarters cc 1 Ell English, « E. E. 6 quarters « 1 Ell French, « E. Fr. LAND OR SaUARE MEASURE. This measure is used in measuring land, or any thing in which length and breadth are both consid- ered. The denominations are miles, acres, roods, perches, yard«, feet and inches. 144 square inches (sq. in.) make 1 square foot, Sq. ft. 9 square feet " 1 square yard, Sq. yd, 30^ square yards " 1 square pole, P. 40 square poles " 1 rood, R. 4 roods, or 160 sq. rods " 1 acre, A. :; 640 acres " 1 square mile, IM. Note. — The Surveyor's or Gunter's chain is used in measuring land. It is 4 rods, or 66 feet in length, and is divided into 100 links. QvBtTioifs. — What is a hand in measure, and for what ig it uiedY What ia a fathom, and for what is it used? What are met^it^red by Cloth Measure? What are its denominations? K^p^at the 'table? For what is square measure used? Wh>^ are the deoominations. Repeat the table. For what is the Sur- ▼eyor'a or Gunter'i chain used? What is its length? How is it diftdedt TABLES OP COMPOUND NUMBERS. 59 SOLID OR CUBIC MEASURE. Solid or cubic measure is used in measuring timber, stone, wood, earth, and such other things as have length, breadth and thickness. Its deno- minations are tons, cords, yards, feet and inches. 1728 Bolid inches (S. in.) make 1 solid foot, S. ft. 27 solid feet " 1 solid yard, S. yd. 4 et of round timb'^'* *' 1 ton, T. 50 .. t of hewn tirnbt •* 1 ton, T. 128 solid feet « 1 cord of wood, C. Note. — A pile of wood 8 feet long, 4 feet wide, and 4 feet high makes a cord. And what is called a cord foot of wood is 16 solid feet : that is 4 feet in length, 4 feet in width, and 1 foot in height, and 8 such feet, that is 8 cord feet make 1 cord. WINE MEASURE. Wine measure is used in measuring all liquors excepting beer and ale. Its denominations are tuns, pipes, hogsheads, barrels, gallons, quarts, pints, and gills. 4 gills (gi.) 2 pints 4 quarts 31^ gallons 63 gallons 2 hogsheads 2 pipes or 4 hhds. make 1 pint, 1 quart, 1 gallon, 1 barrel, 1 hogshead; 1 pipe, 1 tun, « marked pt. « qt. (C gal. u bar. a hhd. it pi- a tun. Note. — A gallon wine measure, contains 231 cubic inches. Questions.— How is Solid or Cubic Measure used7 "What are its denominations? Recite the table. What is the lenfflh, breadth and heicht of a cord of wood? What is a cord foot? How many cord feet make a cord? What is the use of Wine Measure! What are its denominations? Repeat the table.— Hbw many solid inches in a gallon, wine measure? f 60 TABLES OF COMPOUND NUMBERS. n St V 'J } ALE OR BEER MEASURE. This measure is used in measuring ale, beer and milk. Its denominations are hogsheads, barrels, gallons, quarts and pints. 2 pints (pt.) make 1 quart, marked qt. 4 quarts " 1 gallon, « gal. 36 gallons " 1 barrel, " bar. 54 gallons " 1 hogshead, " hhd. Note. — A gallon, beer measure, contains 282 cubio inches. DRY measure. Dry measure is used in measuring all dry arti- cles, such as grain, fruits, roots, salt, coal, &c. Its denominations are chaldrons, quarters, bushels, pecks, quarts and pints. 2 pints (pt.) make 1 quart, 8 quarts " 1 peck, 1 bushel, 1 quarter, 1 chaldron. Note. — A gallon dry measure contains 268f cubie inches. A Winchester bushel is 18^ inches in diameter, 8 inches deep, and contains 2150| cubic inches. time. Time is naturally divided into days, by the re- volution of the earth upon its axis, once in 24 hours ; and into years, by the revolution of the earth round the sun once in 365 days, 5 hours, 48 min- utes, and 48 seconds ; this period is called a Solar QuRsTioNs. — What is the use of Ale or Beer Measure? Wiiat are ita denominations? Repeat the table. Ilow many cubia inches in a gallon^ beer measure? For wliat is Dry Measurt ■sedf What are its denominations? Repeat the table. Ho«r many inches, solid, are conlamed in a gallon, dry measure? Ho«r many inches in diameter is a Winchester bushel? Ho«r many inches deep? How many cubic inches does it contain? 4 pecks 8 bushels 36 bushels a marked qt. it pk. bu. (6 it 9 qr. ch. TABLES OF COMPOUND NUMBERS. 61 * year. In order to keep pace with the solar year in our reckoning; we make every fourth year to contain 366 days, and call it Leap year. The denominations of time are years, months, weeks, days, hours, minutes and seconds. 60 seconds (sec.) make 1 minute, marked m. 60 miautes " 1 hour, « hf. 24* hours " 1 day, *< da. 7 days « 1 week, " wk. 4 weeks " 1 month, " mo. 13 months, lday,6hrs. " 1 Julian year" yr. The year is also dividedinto 12 calendar months, which contain an unequal number of days. 2 3 4. 5 6 7 8 9 10 11 12 Names. No of 1 days. January, 31 February, 28 March, 31 April, 30 May, 31 June, 30 July, 31 August, 31 September, 30 October, 31 November, 30 December, 31 Total. , 365 Note. — When any year can be divided by 4t with- out a remainder, it is called leap year, in which the 2d month (February) has 29 days. Questions, — How is time divided into days? How into jearsf What is a solar year'{ Why is ever}' fourth year called Leap year? What ore the denominations of time'' Repeat the table. How many calendar months in a yearl Name them, and tht number of days ia each. How many days has February in tb« Leap yenr^ 9t TABLES OF COMPOUND NUMBERS* B07 OIRCUJiAR MEASURE OR MOTION. This measure is used in estimatimg latitude and JoDgitude, and also in measuring the motions of the heavenly bodies. Every circle is supposed to be divided into 360 equal parts, called degrees. 60 seconds (") make 1 minute, marked ' 60 minutes " 1 degree, " ® 30 degrees ** 1 sign, « b. 12 signs, or 360 o « i circle « c. TABLE OP PARTICULARS. 12 units make 1 dozen. 12 dozen "'^ " 12 gross or 144 dozen " 20 units « 24 sheets of paper <' 20 quires ** BOOKS. A sheet folded in two leaves is called a folio. 1 gross. 1 great gross. 1 score. 1 quire. 1 ream. four eight twelve eighteen <6 « a quarto, or 4 to. an octavo, or 8 vo. a duodecimo, or 12 mo. an octodecimo, or 18 mo. n REDUCTION. Reduction is changing the denomination, or name of a number without altering its value. The reducing of a denomination of greater vsL]ue into a .1* ■ Questions.— For what is Circular Measure used? How is er- ery circle supposed to be d i vided 1 Recite the tabLe. How many units make a dozcnl How many dozens make a gross? A great gross? How many units make a score? How many sheets of paper make a quire? How many quires a ream. VVhen a sheet is folded into two leaves what is it called? Im four letTes? In eight? In twelve? la eighteen? REDUCTION.' 03" denomination of less value, is called reduction de- scending, and is performed by multiplication, — The reducing of a denomination of less value into- one of greater value is called reduction ascending, and is performed by division. It ia evident, if we wish to change 8 feet to inches, we must multiply l)y 12, because 12 inclies make 1 foot thus, 8X12=96. If we wish to change 96 inches to feet it is also plain that we must divide by 12, thus 96 -r 12 =8 feet. Hence it follows that, reduction descending and ascending pceiprocally prove each other. RULE FOR REDUCTION. I. To reduce from a higher to a lower denomi- nation 3IuUiphj the greater denomination by that number which is required of the less, to make one of the greater, adding to the product, as many of the less denomination, as are expressed in the giv- ei: sum. And so on through all the denominations. II. To reduce frorr a lower to a higher denom- ination. Divide the less denomination by that number which is required of the less to make one of the next greater ; the quotient will be of the same name as the greater denomination, and the remain- ders will be of the same denomination with the dividend, and are to be put as a part of the answer. Questions. — What is reduction? What is reduction descend- ing? How is it perforinedt vVhat is reduction ascending! How is it performed? Mow do we change feel to inches? li this reduction ascending, or descending? How do we change inches to feet? Is this descending or ascending? How is re- duction proved? How do we reduce from a higher to a laww denomination. r^ 64 REDUCTION. ^ EXAMPLES IN ENGLISH MONET. 1. In £2 9s. 6d. 3far. how many farthings 1 £ 2 20 8. 9 d. far. 6 3 49 shiUlings. 12 594 pence, 4 Ans. 2379 farthings. In this example we first reduce the pounds to shillings by multiplying by 20, and add 9 shillings to the product — making 49 shillings. We then reduce the 49 shillings to pence, by multiplying by 12, and add the 6 pence to the product — making 549 pence ; we then reduce these pence to farthings, by multi- plying by 4, and add the 3 farthings to the product, making 2379 farthings. This is called reduction descending, because we have changed a higher denomination to a lower. 2. Reduce 2379 fartliing to pounds. 4)2379 12)594..3 far. 20)49..6 pence. £2,. 9shilllings. £ s. d, far. Ans. 2 9 6 3 This example is the rs- verse of the first example. — We first reduce tlie farthings to pence by dividing by 4, and the quotient is 549penc3 and 3 far. (or quartcs's mark- ed qr.) over. We then re- duce the pence to shillings by dividing by 12, and the quotient is 49 shillings and 6 pence over. We next reduce the 49 shillings to pounds, by dividing by 20, and find there is 2 pounds and 9 shil- lings over. 3. In 35 pounds how many shillings? Ans. 700a. .Qaxas-ioNa. — Mow are pounds reduced to shillings? Shillingt to pence? Pence to farthings? How are farthings reduced to pence? Fence to shillings? ahillings to pounds? REDUCTION. 05 rooa. ingst 4. In £145 how many shillings? Ans. 29000. 5. In J6376 how many shillings and pence ? Ans. 7520s. 90240(L 6. Reduce Je56 14s. 6d. to pence. Ans. 13614d. 7. In 165 pounds 13 shillings, how many farthings t Ans. 159024qr8. 8. In 128s. how many pounds? Ans. J£6 8s. 9. Reduce 1046 petice to pounds. Ans. £4> 7s. 2d. 10. In 6169 pence, how many pounds? Ans. £25 143. Id. Jl. In 180960u. how many pounds. A.m. £754. 12. Reduce £967 14s. 7d. to pence. Result 232255d. 13. In 135764qrs. how many pound??? Ans. £141 8s. 5d. 14. In 48 guineas how many pence r An.- 12096d. 15. How many pence, shillings, and pour J ^ , are there in 17280 farthings? Ans. 4'^20a. 360.3. £78. . 16. Reduce 369936 farthings to gunti s. Ans. 367 guineas. 17. Reduce 878 guineas to pence and the:?e pence to pounds. Fact. 2212r)6d. and £921 18a. 1^ In 525 dollars at 5s. each how many pence? Ans. 3l500a. 19. In 126000 farthings how many dollars ? Ans. 525 dollars. EXA?,irLES IN THOY WEIGHT. 1. In 271b. lOoz. 13dwts. of gold, how many grains? Ans.* 160632. 2. In 81b. Ooz. 7dwt. 2gri^. how many grains? Ans. 46250. 8. Reduce 158262gii:. to pounds. Ans. 27lb. 5oz. 14dwt. 6grs. 4. Reduce 376457grs. to pounds. Fact. Seib. 4oz. 5dwt. 17grB. QousTioNs. — How are (lounds reduced lo ounces, in Troj Weight? Ounces to pennyweights? Pennyweights to jjraintt Grains to pennyweights? Pennyweights to ounces? Ounotif to pounds? f3 66 I REDUCTION. 5. In 3751b. lOoz. 16grs. how many grains? Ans. 2164816gre. 6. In 167597dwts. how many pounds? Ans. 6981b. 3oz. 17dwts. 7. In 251b. 9oz. lOdwt. how many grains? Ans. 148560gra, 8. Reduce 97645745grs. to pounds. Result 169521b. 4oz. 12dwt. 17gr8. EXAMPLES IN APOTHECAniES WEIGHT. 1 . In 171b. how many ounces, drachms, and scruples ? Ans. 204oz. 1632drs. 4886scru. 2. In 1332005 grains, how many pounds? Ans. 2311b. 3oz. 5grs. 3. In 51b. of drugs, how many parcels, each 16 drachms. Ans. 30 parcels. 4. Reduce 271b. 3oz. 2drs. to grains. Result 157080grd. 5. In 245 parcels of drugs each lOoz. 3drs. 2scru. how many pounds? Ans. 21^1b. 6oz. 2drs. Iscru. EXAMPLES IN AVOIRDUPOIS WEIGHT. 1 . In 1 5 tons, how many hundred weight, quarters and pounds? Ans. 300cwt. 1200qrs. 336001bs. 2. In 67200 lbs. how many tons ? Ans. 30 tons. 3. In 9cwt. 51bs. how many ounces ? Ans. 16208OZ. 4. Reduce 20571005 drachms to tons. Ans. 35T. 17cwt. Iqr. 231bs. 7oz, 13dr. 5. One of the stones in the walls ot Balbeck is said to have weighed 683T. 8cwt. Query, its weight in pounds ? Ans. 15308161b8. 6. In 57T. lOcwt. 3qrs. 141bs. how manv drachms? Ans. 3299788S dfi. 7. Reduce 4768768 drachms to tons. Ann. 8T. 6cwt. Iqr. Slbii* Questions. — In Apothecariei Weight how ere pounds reduoe4 to giaiii'!.1 Grains to pounds? In Avoirdupoii weight, how Af9 tons reduoed to draohmsl Druchmi to tonsT REDUCTION. (J7 8. A merchant would put 109cwt. Oqrs. l^lbs. of raisins into boxes, containing 261bs. each ; how many boxes will it require 1 » Ans. 4)70 boxes. EXAMPLES IN LONG MEASURE. How many barley-corns will reach round the globe, it being 360 degrees ? Operation. 2)360 o 69i 180 3240 2160 25020 miles. 3020 500400 75060 8006400 rods, 16^ 4003200 48038400 8006400 132105600 feet. 12 1585267200 inches. 3 Ans. 4755801600 bar. In this example, we first multiply 360 ® by 69^, the number of miles in a degree ; then by 320, the number of rods in a mile : then by 16^, the number of feet in a rod j then by 12 the number of inches in a foot, lastly by 3, the number of barleycorns in an inch, — which producer the answer, 4755801600 barleycornt^ 68 reduction; t. Reduce 4755801600 barleycorns to degrees. Ans. 360 degrees. This example is the re- verse of the first example. — We first divide by 3, the no. Operation. 3)4755801600 12)1585267200 in. 132105600 feet. 2 33)2642 1 1200 i feet. 320)8006400 rods. 25020 miles. 2 of bar. in an inch ; then by 12 the number of in. in a foot — ?nd find tlie quotient to be 132105600 feet, which ore to be reduced to rods. — We cannot easily div'ide by the mixed number 16^ on account of the fraction ^. — We overcome this difficulty, thus, 16.', feet=33 half feet, and 132105600 feet =264- 211200 half feet, which di- 139)50040i miles. viJcd by 33 gives 8006400 rods ; we reduce these rods 360 dcgrecn. to miles by diviiling by 320, the number ofroJs in a mile, and to degrees by dividing by 69^ miles=139 half miles, and 25020 milcs= 5001-0 hnlf miles, which divided by 139 gives 3 GO degret'ij for tho quotient. Note. — When the divi^^or is a mixed number we may reduce it to halvc^!, ihinls, fourths, &c. and the dividend to tho same 5 then divide, r.nd the quotient will be the an- swer. 3. How many inches arc in 273 miles ? Ans. 17297280in. 4. In 34594560in. luAv many miles'? Ans. 546m. 5. Reduce 2m. Ifur. 8p. 3yds. 2in. to inches. Ans. 136334 in. 6. Reduce 2280060 barley-corns to miles. Ans. 11m. 7fur. 38p. 2yds. 2ft. Questions. —In long measure, how arc degrees reduced U btrleycornsi Barieycorni to degrees! REDUCTION. 69 1. X. 3. i. 5. 6. 7. yards. 8. 7. Required the number of revolutions a wheel IStL 4in. circumference will make in running 150 miles. Ans. 43200. EXAMPLES IN CLOTK MEASURE. In 15yds. 3qr. Ina. how many nails ? Ans. 253na. In 1012 nails of cloth, how many yards? Ans. 63yds. Iqr. Reduce 73 ells Flemish to quarters. Ans. 219qr8. In 1752 nailsj how many ells Flemish 1 Ans. 146 ells. How many ells English are in 1408 nails'? Ans. 70 E. E. 2qr8. In 47656 nails how many ells Eiiijlish ] Ans. 2382 E. E. 4qrB. Reduce 475 ells English to nails and these again to Result 9500na. aiul 593yds. 3qrs. In 4 boles of cloth, each 12 pieces, and each piece 24 ells English, how many yards and ells Flemish 1 Ans. 1140yds. 1920 E. Fl. EXAMPLES IN LAND OR SQUARE MEASURE. By the table of long measure wc learn that 3 feet in length make 1 yard, also by the table of square measure, tliat 9 square feet make 1 s(juare yard, that is 3 feet in length and 3 feet in breadth. To make this plain let the email square represent 1 square foot. It uill be perceiv- ed that the large square contains 9 such squares, c^ 1 ^ ^ 1 V t A square is the space included between 4 equal lines, drawn perpendicular to each other. Each line is called a side of the square. If each side be 1 foot, it is called a square foot, if 1 yard it is called a square yard : if 1 rod a square rod, &c. (|uK8TioNs. — In clulli nieaiurt how do we reduce ydn to iiich«>if Ella English lo incheil £Ut Fiemtsh to inches? Ells French to inches? Inches to yards,&f.? 1 1^ ' i 1 yd. =3 feet. *•* "1 • II 0» r • 70 REDUCTION. Note. — ^When the length and breadth are given, w«* find its square contents, by multiplying the length by the breadth, 1; Required, the contents of a board 14 feet long and 2 feet wide. Ans. 28 feet 2. In 24 ac. 2 r. 26 p. how many perches ? Ans. 3946 perchee.- 3. Reduce 365 ac. 3 r. 13 p. to perche?. Ans. 58533 perches.- 4. In 267464 perches, how many acres I Ans. 1671ac. 2r. 24p. 5. If 2599200p. be divided into 25 equal tracts, how many acres will be in each? Ans. 649ac. 3r. 8p. 6. In 37456 square inches, how many square feet ? Ans. 260 sq. ft. 16 sq. in. 7. In 14972 square rods, how many acres t Ans. 93a. 2r. 12p. 8. In 3674139p. how many square miles! Ans. 35m. 563a. Ir. 19p. EXAMPLES IN CUBIC MEASURE. In long measure we consider length only : in square measure, length and breadth are both con- sidered : cubic or solid measure has regard to length, breadth and thickness. A cube may be represented by a solid block having 6 equal sides^ Suppose we have a number of blocks, containing a cubic foot each, if we lay 9 of these upon the floor, thoy will cover a square yard, and as each block contains a cubic foot, the solid contents of the square yard will be cubic feet. Then if we Questions. — How many feet in lermth make n yard ? How many square (cct niuke a square ymtl 7 Wliiit is a square ? What is eiK'l) tine called? If eucii side be I foot, what is it called) If each side be I yard what ia it called? When the length and breadth are given how do wp find the contents of a •quareT How do we reduce miles tu inches in Square Measure? lachei to miles? REDUCTION. 71 •cover tliis layer with another of blocks, contain- ^ ing also 9 solid feet, the pile will contain 18 solid feet ; if we pile on yet another layer the pile will contain 27 solid feet, or one cubic yard, and as the pile is 3 feet in length, 3 feet in breadth, and 3 feet in depth, its solid contents are found by mul- tiplying its length and breadth^ and depth together, 1. How many cubic inches are there in a brick, that is 8 inches long, 4 inches wide, and 2 inches thick 1 Ans. 64. 2. How many solid feet in a pile of wood 25 feet long, 4 feet wide, and 6 feet high ? How many cords 1 Ans. 600 feet and 4 cords and 88 feet. 3. In 14 tons of hewn timber, how many solid inches t Ans. 1209600. 4. Tn 12 cords of wood, how many solid feet and inches? Ans. 1536 feet and 2654208in. 5. In 4608 solid feet of wood, how many cords 1 Ans. 36. EXAMPLES IN WINE MEASURE. 1. In 5 tuns 1 hogshead of wine, how many gallons t Ans. 1323. 2. In 10584 quarts of wine, how many tunsl Ans. 10 tuns, 2hhds. 3. In 24bhds. ISgals. 2qts. how many pints? Ans. 12244. 4. In 1789 quarts of cider, how many barrels? Ans. 14bbls. 25qts. 5. In 58 barrels of wine, how many gallons, quarts tnd pints? Ans. 1827gals. 7308qts. 14616pta. QoESTioNfl.— In lonjr measure what do we coniider? What in iquare measure? To what has cubic meaoure regardl How may a cube be rcpresenied? How many cubic feet in a cubit yard? What io its length? What is ilfi breadth? What ii itt depth? How are its solid contents found? In cubic measura how are tons reduced to inches? How are cords reduced to inches? Inches to cords? To tons? How are tuns reduced to gilli, in wine measure? Gills to tuns? i i i ■^ i L 72 REDUCTION. ir EXAMPLES tin ALE OR BEER MEASURE. 1. Reduce 4<7bar. 16gals. 4) .<^) . £ B. d. £ s. d. £ e. d. 72 16 4| 31 11 11 146 19 9i 84 10 6^ 7 8 310 17 lOi 12 12 9 19 18 9 4G1 10 llj 28 8 1 41 16 1 19 8] 41 19 34 59 6.1 61 1 1 50 1 8 17 10 9 39 11 10} Questions. — In simple numbers wlien do we rarry 1? In compound numbers vvlien do we carry 1? In passing from pence lo siiillings what do we carry? In passing from Bhillings to pounditt COMPOVND ADDITION. 77 TROY WEIGHT. n •i lb. 11 114 223 Operation. (1) oz. pvvt. In adding up the column of grains, we find their sura to be 47 = 1 pwt. 23gr. Wc set down the 23 and carry the Ipwt. to the coUimn of pennyweights ; we find their sum to be 4<2pwt. =2oz. 2pwt. Carrying 2 to Sum, 350 5 2 23 the ounces, we, find their sum to be 29oz.=21b. 5oz. W« then carry the 2 to the column of pound:^, and find their f*\\m to be 350. 8 9 10 18 6 17 19 16 18 In lence 13 to (2) lb. oz. pwt. gr. 16 10 14, 16 23 8 7 20 62 10 18 15 83 1 5 7 76 3 12 3 252 10 18 13 ^^\ Al POTIl )h. oz. dr. scru • R«*. 24 7 2 1 16 17 11 7 2 19 36 G 5 7 15 9 7 1 13 S) 3 4 1 9 16 10 3 2 17 lb. 273 342 657 724 (3) oz. pwt. 1 9 10 9 7 15 3 13 10 18 gf- 14 5 21 17 23 APOTIIECAHIES WEIGHT. (2) lb. oz. dr. scru. gr. 12 11 6 1 15 4 9 7 12 9 10 1 2 16 4 8 1 2 19 9 1 10 4 9 1^ 1 6 >A 46 1 4 1 18 «9 78 COMPOUND ADDITION. AVOIRDUPOIS WEIGHT. (^>.. cwt. qr. lb. 2 3 27 1 1 17 4 2 26, 6 1 13^ 3 3 15 6 2 16 * 26 2 (2) :^ U). oz. dr. T. cwt. qr. lb. oz. dr. 24 13 14. 91 17 2 24 13 14 17 12 11 19 9 0.17 10 12 26 12 15 14 13 2 4 9 11 16 8 7 47 11 3 19 14 5 24 10 12 69 1 12 11 12 12 77 19 3 27 15 11 122 7 7 320 12 2 11 1 1 4. A farmer sold 4 loads of hay, weighing as follows, viz : 12cwt.3qrs. 161bs., 13cwt.2qr3. 121bs. 13oz,, 15cwt. Iqr. 211bs. lloz., 16cwt. Iqr. 241bs. 8oz. ; what was the weight of the whole in tons? Ans. 2T. 18cwt. Iqr. 191bs. LONG MEASURE. (1) (2) L. mi. fur. f#k yd. ft. in. bar. 16 2 7 39 90 2 11 2 327 1 2 20 155 1 9 1 87 1 15 327 7 1111 50 2 1 2 432 2 4 35 624 1 5 2 CLOTH MEASURE. m (2) ^T. (^) yds. qrs. na. E.E. qr. na. £.F. qrs. na^ 71 3 3 44 3 2 84 2 1 13 2 1 49 4 3 7 1 3 16 1 6 2 3 76 2 42 3 3 84 4 1 52 2 3 57 2 2 7 53 2 2 49 2 2 61 2 1 9 2 3 251 3 254 2 2 285 2 COMPOUND ADDITION. 7^ ^ LAND OR SQUARi: MEASURE. dr. 14 12 11 5 12 11 >W8, JWt. I the (1) (2) Sq. yd. Sq. ft. Sq. in. M. A. R. P. 97 4 104 1 700 3 37 22 3 27 6 375 2 25 105- 8 2 7 450 1 31 37 7 127 11 30 25 263 16 27 277 -38 SOLID OR CUBIC MEASURE. n. (^> ^(^) (3) T. ft C. ft. . feet, inches. 41 43 3 122 .. 13 1446 12 43 4 114 16 1726 49 6 7 83 3 . 866 4 27 10 127 14 . 284 108 19 27 62 48 866 WINE MEASURE. •gal. (1) qt. pt. &' hhd. (2) gal. qt. pt. (3) tun. hhd. gal. qt. 39 3 1 3 42 61 3 1 34 2 34 2 17 2 1 2 27 39 2 19 1 59 1 24 3 1 9 14 1 28 2 2 1 19 1 1 2 9 2 1 19 32 2 8 3 16 24 1 1 37 3 11 1 40 2 1 1 4 3 19 150 2 1 100 24 1 139 3 22 3 1^ 80 COMPOUND ADDITION. DRY MEASURE. (1) (2) ch. bu. pk. qt. ptc ch. bu. pk. qt. pt. 27 25 3 7 1 141 36 '3 7 2 59 21 2 6 3 21 32 2 4 1 2 12 7 1 m 9 10 3 9 182 10 4413 M. 22 3 7 1 259 12 1 .riJJ r\ . TIME. - i ;i) .: -^ ^ (2) - V' rao. wk. da. Iir. ^^k. ua. hr. m. BCf. 4 11 3 6 20 8 8 14 55 57 3 10 2 5 21 10 7 23 57 49 5 8 1 4 19 '20 6 14 42 10 101 9 3 7 23 6 5 23 19 59 55 8 4 G 17 - 2 2 20 45 48 172 2 14 4 ' 50 4 1 41 34 t P- V CIRCULAR MOTION. :\\ (1) (2) » & o t >r s. o I M tt r 3 29 17 14 11 29 59 59 n: 1 C 10 17 . 40 10 yds. 3qr., and the lilth 27yd. 2qrs. ; how many yards did he buy ? . Ans. 191yds. Iqr, 6. If 1 ci-loin contaiuo 25hhJ. 27g.il. 3qt. ; a second 37hhd. 2(1gal. 2(it. ; a third 3r)!ilul. C)^r.\[. l<|t. ; and a 4th 4Dhhd. 15gal. 3qt. : ^^ hat quantity will tlicy joijitly contain? Ans. 143l'ihd. lilgal. Iqt. 7. A man bought 3 piles »>f ^vood, which contain ai fdlovvs, viz. 37 cords, 51 feet; 14 cords, 120 feet; 19 eords, 95 ; how many cords did he purchase ? Ans. 72 cords 10 feot. w*f 82 COMPOUND SUBTRACTION. COMPOUND SUBTRACTION. Compound subtraction like that of simple num- bers, is taking a less number fr^hn a greater , but of difierent denominations. The principles invol- ved are the same as those already explained in simple subtraction, except that it takes different numbers to make a unit in the next higher denom- ination as explained in compound addition. RULE. Write the less numbers under the greater, pla- cing those of the same denomination directly un- der each other. Begin with the lowest denomina- tion, and take successively the lower number in each denomination from the upp&r, and write the remainder underneath as in subtraction of simple numbers. If the lower number be greater than the one a- bove it, add as many units to the upper number, as make one of the next higher denomination, sub- tract the lower number therefrom, and add 1 to the next higher denomination in the subtrahend, and so on through all -he denominations. Proof — as in simple . ubtraction. EXAMPLES. 1. Subtract £29 19.s. 8d. from £36 15s. 7d. Placing them according to the rule, they stand thus : Operation. £ 8. d. 36 15 7 29 19 8 "6 ISTT In the above example 8d. cannot bo taken from 7 ; we therefore add as many units of tliis denoinination to 7, as are re- quired to make a unit of the next higher, that in 12, which added to 7 make 19, Questions.— What is Compound Subtraction? How doea it differ from Simple Suhiruotion'' Repeat the rule. If the lower number be greater thuii the one above it what id to be done? — How is Compound {Subtraction proved? I. ■ COMPOUND SUBTRACTION. 88 bo [any re- ler, |19, |ait Iwer 1? — subtract 8 from 19, and 11 remains. As 12 pence have been added to the minuend, an equal quantity must be added to the subtrahend, we therefore carry 1 shilling to the 19, which makes 20. As this cannot be subtracted from 15, we add to the 15 as many of this denomination as make one of the next higher ; that is 20, which added to 15 makes 35. Subtract 20 from 35 and 15 remain. As 20s. have been added to the minuend, 1 pound must be added to the subtrahend of the next higher denomina- tion, making it 30, which subtracted from 36 leaves 6. (1) £ s. d. qr. From 346 16 5 3 Take 128_17 4 2 Rem. 2f7~T9 19 1 Proof, 346 16 5 ~3 £ s. d. qr. From 476 10 9 1 Take 277 1 7 7 1 Rem. 198 13 2 3. Borrowed ^£29 16s. and paid £IQ lis. 9d. : how much remains due? Ans. £13 4s. 3d. 4. From ^£397 16s. 6d. ; take jei44 lis. 4d. 3qr. Ans. '^"■3 5s. Id. Iqr. 5. How much does £?s 1 7 6s. exceed £ 1 78 18s. 5id.? Ans. £138 7s. 6ld. 6. From eleven pcimds take eleven pence. Ans. XIO 19s. Id. 7. From one hviihirNl pounds, Inke four pence half- penny. . Ans. £99 19 7^1. (8) lb. oz. pwt. gr. From 273 Take 98 10 18 21 Rem. i^^ J^^ 3 Proof, 273~~6~T~0 (J>) lb. oz. dr. scr. gr. 40 7 3 2 7 4 5 2 8 2* 16 2 5 2 19 5 H ■Map v^ %^ 84 COMPOUND SUBTRACTION. (10) T. cvvt. qr. lb. oz. 7 14 1 3 6 2 6 3 4 11 5 7 T~26"T1 ^ (12) yd. qr. na. 47 3 2 27 2 3 (11) Deg. m. fur. pol. yJ. ft. in. Ixu 54 48 7 28 4 2 2 2 36 25 6 39 2 8 () 22 23 29"~3"~2" ~6~'2' 20 3 (13) A. JR. P. 36 3 28 28 2 39 8 ~0"~2 9 (14) T. ft. m. 44 44 884 39 39 982 ~5 4.1630 (;5) T. hlul. gal. qt. pt. gi. 88 3 55 3 48 62 2 2 40 2 56 1 2 (16) ch. l)u. pk. qt. 47 34 3 6 20 35 7 26""35 2 7 /,'! '(. ■ (17) yr. mo. vvk. da. lir. mi. sec. 250 4 3 6 22 55 49 220 8 5 23 59 55 29~ 9~ 3'~0"22 55 "54 (18) 8. '^ ' " 9 23 45 54 3 7 40 56 GT5~4V 58 PRACTICAL EXF.aCI3E3 ' *; * IN COMPOUND ADDITION AND SUBTR ACTION. 1. From Olio pound take one flvi'thing. An>\ 193. Hd 3f;ir. 2. What mm luhhd to iJ17 I Is. S^d. will make £ 100? 3. Lent a friend at one time JG13 I'Js. 8d. ; at : notlicr jG35 lO.s. 6(1. ; he paid tiie at one time, £10 lOs?. . Id. ; at another, JG9 3,-!. 6d. ; how much '-^i., iiris dsie ? An . £29 J 2s. 9(1. 4. From M years, 6 monlii>', 11 day^, taiio 10 yean", 6 months 29 day^^. Aih-^. IJmo. J2da. PRACTICAL EXERCISES. 85 2 2^ 5. A gentleman paid his 3 laborers as follows : to A. he gave ^11 13s. 6d. ; to B. 13s. 6d. more than to A. ; and to C. he gave as much as A. and B. both ; how much did the gentleman pay out? Ans. i^-iS Is. 6. A merchant bought ITcvvt. 2qr. 141b. of sugar, of which he sells 9cwt. 3qr. 251b. how much of it remains unsold? Ans. Tcvvt. 2qr. 171b. 7. From a piece of cloth which contained 52yd3. 2na. a tailor was ordered to take 3 suits, each 6yds. 2na. : how much remains of the piece 1 Ans. 32yds, 2qi;. 2na. 8. The revolutionary war broke out jjetvveen Great Britain and the Uunited States, April 1.9th, 1775, and a general peace took place January 20th, 1783 ; how long did the war continue? Ans. 73T. 9mo. Id. 9. A merchant buys two hogslieaJs of sugar, onu weighing 8cwt. 3qr. 211b., the other 9cwt. 2qr. 61b. ; he sells two barrels, one weighing 3cwt. Iqr. 211b.. 14oZm the other, 2cwt. 3qr. 151b. 6oz.; how much remains mi hand! • Ans. 12cwt. 261b. 12oz. 10. A boy sets out upon a journey, and has 200 miles to travel ; the first day he travels 9 leagues, 2 miles, 7 fur- longs, 30 rods; the second day 12 leagues, 1 mile, 1 fur- long ; the third day 14 leagues ; the fourth day 15 leaguesj 2 miles, 5 furlongs, 35 rods ; how far had he then to travel f Ans. 141. Imi. Ifur. ISrd 9(1. COIMPOUND MULTIPLICATION. To multiply a compound number by a simple one, is tu repeat the compound number as many times as there arc units in the multiplier. I. I When the multiplier does not exceed 12. RULE. Write down the compound number and set the multiplier under the lowest denomination. Multi • Questions. — What is multiplying a compound number by a simple one? When Ihe multiplier does not exceed 12 what is the rule ? ': 1} 'It \ n ■J .' ■ ' ( i\ ul- '!t 86 COMPOUND MULTIPLICATION. ply the several denominations in succession by the multiplier ; reduce the results to the next higher denomination, and set down the excess as in addi- tion; proceed in the same way through all the denominations, and set down the entire product when you come to the last, EXAMPLES. 1. Multiply £3 9s, lad. by 4. Operation. £> s. d. 3 9 10 4 Product, 13 19 4 Place the numbers as the rule directs, and then say 4> times lOd. are 40d.=3s. 4d. ; we set down the 4d. in the lower line ; then 4 times 9s. are 36s. and 3 shillings to carry make 39s.=j61, and 19s. over — set down the 19s. ; then 4 times jG3 are ^612, and JGI to carry make X13, and the product of i33 9s. lOd. by 4=^613 19s. 4d. ^ X s. d. 2. Multiply 111 6 .^'.n*i>I' qr. 2 by 5 5 (3) T. cwt. qr. lb. oz. 9 3 27 12 7 Product, £7 17 8 2 *^* 3 9 3 26 4 'W 4. What will be the cost of 5 yards of calico, al 3s. 9d. per yard ? Ans. 18s. 9d. 5. A man bought 7 sheep at 9s. 6d. a head ; what did they coat him ? Aoi. £3 6s. 6d. §, What will be the cost of 9 hats at 9s. 9d. each ? Ans. £4 7s. 9d. 7. Bought 9 pieces of shirting, each containing 28yds. 2qrs. 2na ; ho ." many yards in all 1 Ans. 257 yds. 2qrs. 2na. 8. W'r -t is the cost of 12 bushels of wheat at 98. 6d. per bushel { Ans. Je5 14s. a. What will 12 horses come to, at je29 16s. 8d. each 1 Ans. £S58. COMPOUND MULTIPLICATION. 87 II. When the multiplier exceeds 12 and is a .compo- eite number. RULE. Multiply the compound number by one of the component parts or factors, and then that product by the other factor, and the last product will bo the answer. EXAMPLES. 1. What will 16 yards of broadcloth come to at £J^ 33. 3d. per yard ? Operation. ie2 3 3 r: 4 In this example 4 and 4 are the fac- tors of 16, because 4X4=16. b 13 \ 4 3s. 9d. did 6d. 9d. rds. [na. 6d. I iMi >hi 58. Ans. 34 12 -^ "' \ 2. What will 20 barrels of flour come to, at JGI 6s. 3d. each ? m Ans. £1'o 5s. 3. What must a man pay for 36 sheep, at 9s. 4d. each? Ans. J616 16?. 4. What ia the weight of 48 pipes of wine, each weigh- ing 18cwt. 2qrs. 141bs. Ans. 894cwt. or 44T. 14cwt. 5. In 21 pieces of cloth, each containing 24yds. 2qr8. 3na. ; how many yards? Ans. 518yds. Iqr. 3na. 6. How much water will be contained in 96hhds. each containing 62gal, Iqt. Ipt. Igi. Ans. 55991gal. 7. What is the cost of 120 dozen of candles at 5s. 9d. ,. per dozen ? Ans, ^£34 108. III. When the multiplier exceeds 12, and is not a com- posite number. QuKSTioN.— When the muUiplier exceeds 12 and ii a cornpo- «ile number what is the rule] 88 COMPOUND MULTIPLICATION. ^ •> •S9fi.=z£l las. lOid.: :«S .8(1. RULE. Multiply the simple number by each of the de- nominations separately, and reduce each product to the highest denomination named. Th^n add the several products together, and their sCim will be the answer sought. EXAMPLES. 1. Multiply £5 3s. 8d. by 13. Operation. In this example we 13 13 first multiply the 13 3s. 8d. by 8d., and the pro- _. dnctisl0'id.=:8s,8d. then the 13 by 3s. and the product is 39s.=jeiips.5 then the 13 bv £5, and the product is £65. We then add the se- veral products toge- ' '- ther for the answer, Alls. (7 7 8 as #10 rule directs. 'What is the'cos-jt of 139 oxen at £6 8s. 9d. eac^^- Operation. 139x9a.=]251d.=£ • 5 4^. 3d. 139X8s.=ni2s.-je 55 12s. 139Xi26=^831=:je834 ' 13 5£ J. 19s. s 8d. i) Ans. XS94 IG 3 3. What is the worth of 47 pounds of butter at 9^d. i>cr pound ? Ans. ^1 17s. 2^d. 4". Wiiat will 19 yards of cambric come to, at 1 Is. 6d. per yard ? Ans. ,i^lO 18s. 6d. 5. What is the cost of 52 pounds of tea, at 5s. 9d. per pound? Ans. iei4.19s. Question. — When the multiplier exceeds 13, and is not a com- posite number, how do we proceed? V? COMPOUND DIVISION. 89 6. What is the weight of 17 hogsheads of sugar, each weighing 8cwt. 3qrs. 141bs. Ans. 150cwt. 3qrs. 141bs. 7; If one yard of cloth cost 7 shillings and 10 pence, what will 65 yards cost? Ans. ^£25 9s. 2d. 8. What is the cost of 46 bushels of wheat at 4s. 7^d. per bushel? Ans. JGIO lis. 9^d. 9. What is the cost of 1 17c\vt. of raisins, at ^1 2s. 3d. per cwt. Ans. X130 3s. 3d. 10. In 357 hogsheads of sugar, each 15cwt. 3qrs. 171b9. how many cwt. ? Ans. 5676cwt. 3qrs. 2 libs. 11. If a steamboat in crossing the Atlantic goes 211mi. 4fnr. 32rd. a dayj how far will she go in 15 days? Ans. 3174 miles. ence=4^ ^ c\ 92 COMPOUND DIVISION. 7. A man paid £2 10s. for 15 bushels of com ; what did he pay per bushel 1 Ans. 3s. 4d. Si Bought 36 bushels of apples for £2 14s. ; what was that per bushel ? Ans. Is. 6d. 9. Sold 81 barrels of flour for £U'7 16s. 6d. ; how much was that per barrel 1 Ans. JGI 16s. 6d. 10. Bought 64 gallons of oil for ^630 8s. ; what did it cost per gallon ? Ans. 9s. 6d. 11. Bought 144 reams of paper for J696 ; what did it cost per ream? Ans. 13s. 4d. 12. 176 men consumed in a week 13cwt. 3 qrs. lib. 6oz. of bread; how much did each man consume 1 Ans. 81b. 12oz. 2dr. 13. If 232 bushels 3 pecl^s 7 quarts of wheat be put into 105 bags, how much will each bag contain ? Ans. 2bu. 7qt. 14. Divide 18 gallons equally among 144 soldiers. Ans. 1 pint each. *» PRACTICAL EXERCISES IN COMPOUND MULTIPLICATION AND DIVISION. 1. In 15 loads of hay, each weighing 1 ton 3cwt. 2qr8. how many tons 1 Ans. 17T. 12cwt. 2qrs. 2. What will 24 barrels of flour cost at £2 12s. 4d. per barrel ? Ans. je62 16s. 3. The Pri nee of Wales receives a salary of JB 1 50,000 a year; how much is that per day 1 Ans. je410 19s. 2d. 4. Bought a dozen silver spoons, which together weighed 31b. 2oz.l3pvvt. 12gr.''; how much silver did each •poon contain ? Ans* 3oz. 4pwt. llgr. 5. Bought 17cwt. 3qr9. 191b8, of sugar, and sold out 900 tiiird of it ; how much remains unsold ? Ans. llcwt. 3qr8. 22lb8. 6. If 168 bushels 1 peck 6 quarts of wheat he put into 95 baga^ how many bushels in each ? Ans. 4bu. 3pkf. 2qtfl. i ;gg«m>tmiiiii,,-inmii„i ifjt mm PRACTICAL EXERCISIS* 93 7. In ^ pieces of cloth, each measuring j^7|^yards, how maay yards? Ans. 971yds. Iqr. 8. If a man's wages amount to JS257 2s. 5d. in 12 months, what is that per month 1 Ans. £2t 8s. 6|d. 9. A privateer took a prize of JE30,000, of which the owner took one thirds and the officers one fourth ; the re- mainder to be equally divided among 125 seamen ; how much must each seaman receive? Ana. iSlOO. 10. A certain gentleman lays up every year J6294 12s. .6d., and spends daily JBl 12s. 6d. ; what is his in- cjome? Ans. ^£887 15s, 1 1 . If lcwt.*of sugar cost £S 10s. what is it per pound ? Ans. 7idt *. PRACTICAL EXERCISES IN THE FOUR C0MP0UI»JD RULES. ♦ ^^ 1. Find the amount of forty pounds nine fihiUlngs, six- ty-four pounds and nine pence, ninety-five pounds nine- teen shillings, and seventeen shillings fourpence half-penny. Ans. £201 6s. l^d. 2. Received of 4 men the following sums of money, viz : the first paid me £37 lis. 4d. ; the second j625 168. 7d. ; the third, JG19 14s. 6d., and the fourth as much as all the other tliree, lacking 193. 6d. I demand the whole sum received. Ans. J6165 5s. 4d, , 3. Borrowed JSIOO, and paid in part as follows: at 1^e time je21 lis. dd. ; at another time Jei9 Hs. 4id. ; at anotlier time 10 dollars, at 6s. each, and at another time two guineas, at 28s. each, and two pistareens, at 14id. each ; how much rem:vins due, or unpaid 1 Ans. £52 128. Sjd. 4. How many days are there in 15 years of 365 days ^hrs. 4Sm. 51sec. each? Ans. 5478da. 15hr. 12m. 45sc«. 5. A club in Quebec, consisting of 25 men, joined Anr a lottery ticket of J610 value, which came up a prize of H P^lA^TICAL EXI^CISES. Jt ^: iS4000* I wish to know what each man corv^nbuted, and what eacl| man's share came to. - Ans. Each contributed 8s. ; each share dS 160. .mi. - ' ' 6. Three Merchants A. B. & C. have a ship in com- Sany. A. has f, B. f, and C,\ and they receive for •e\^t £228 16s. 8d. It is required to divide it among the owners, according to their respective shares. Ans. A's share jei43 Qa^ 5d. B's;e57 4s. 2d. and C'a £29^128. Id; T 7. A man paid J667 4s. for a pile of wood containing 64 cords ; he sold 30 cords for ^629 16sr.'f for how much must he sell the remainder per cord, so as not to lose 1 ifc. Ans. £1 2s. 8. A gentleman jpurchased of a silver^ith, 2 dozen silver spoons, each weighing 3oz. 4pwt. Igr. ; 2 dozen of tea spoons, egch^ weighing 15pvvt..l6gr. ; 3 tankards, each weighing 22oz. 14pwt. He sold him old silver to the amount of 61b. lOoz. 3pwt. j how much remained to bo paid for 1 Ans. 61b. 9oz. 12pwt. 9. A farmer has 6T. 8cwt. 2qrs. 14lbs. of hay to bo removed in 6 equal loads ; how much must be carried at each load? - Ans. IT. Icwt. Iqr. 211bs. ' 10. A merchant had JG19H8 to begin trade with ; for 5 yeani' together, he cleared JG1086 a year; the next 4 years he cleared je2715 10s. 6d. a year j but the last 3 years he was in trade he had the misfortune to lose one year with another, je475 4s. 6d. a year ; what was his real fortune at the endjaf 12 years ? Ans. ^633984 8s. 69. 11. What will Icwt. of cheese come to at 2id. j at 2|d. } at 3d. ; at 2d. ; at 3^d. per pound ? *" Answers in course : £1 3s. 4d. ; £1 5s. 8d. ; £1 88. ; 18s. 8d. ; £1 12s. 8d. 12. Out of a pipe of wine a merchant draws 12 bottles, each containing 1 pint 3 gills ; ho then fills six 5 gallon demijohns : then he draws off 3 dozen bottles, each con- taining 1 quart 2 gills ; how much remained in the cask? Ans. 82gal. Ipt. * JT^ ■^^ BILLS OF PARCELS. 96 I I \ ^ 13. A privateer takes a prize worth JS 12465, of which the owner take# one. half, the officers ^ne fourth, and the remainder is equally divided among the sailors, who are 125 in number ; how much is each sailor's part 1 Ans. je24 18s. T^d. 14". A printer uses one sheet of paper for every 16 pa- ges of an octavo book ; how much paper will be necessa- ry to print 500 copies of a book containing 336 pages, al- lowing 2 quires of waste paper in each ream. Ans. 24 reams, 5 quires. 12 sheets. *'* ■ ak- BILLS OF PARCELS. Note. — In keeping accounts, making out bills of par- cels, w<» draw an oblique line and place the shillings cm the left hand of it and 'ttie pence on the right, to designats the price of one article: thus, 2/6, 4/9, whidh are read 28. 6d. and 4s. 9d. Picton, October 14th, 1845. 1. Joseph Faithful, -^^ Bought of James Trust, 8 yards of Calico at 1/3 9i « « « 1/ 10 « Cloth «5/6 # 50 lbs. of Sugar, « 6id. 25 « of Rice «« 3^4. 30 «< of Codfish "4id. Total Cost, £6 Os. l^d. K^ingston, Sept. 19th, 1845. Otis Smith, Bought of Richard Good, 15 yards of Satin, at 9/6 of Silk, « 17/4 ofFine Cloth," 19/8, of Cambric, « 3/2, of Velvet, « 27/6, of Sheep's grey, at 6/3 18 » 12 (( 16 (( 13 (( 23 « t^ Total Cost, je62 2«. Sd. 9(5 BILLS OF PARCELS. Queljec, June 18tli, 1845. 3. Caivin Hawley, Bought oi" John Punctual, 54 bushels of Wheat, at 4/9, %. 75 « of Oats, « 1/8, 25^ « of Com, « 2/6, 4iton3 of Hay, « 40/6, je31 7s. 6d. i Montreal, July 24th, 1845.* * 4. Samuel Edwards, Bought of Andrew White, 90 yards of Broad Cloth, at 8/4, 100 a 66 « 10/ 6$ 112 c< Satinet, «^3/7^, 126 «' 66 «« 12/111, 144 « 66 « 19/11, 162 6( U " 9/3, 70 6( of Bombazine, " IQ/'Tir 198 66 of Italian Silk, " 16/^, 132 (( « « 8/11, 66 (( 66 « 16/11^ # ■# Ans. £752 14s. l^d. Toronto, Nov. 19tli, 1845. 5. William Clark, Bought of John Black, & Co. 92 yards of Satinet, at 3/5^, 94 " of " " 6/9i, 102 « ofDurant, ^ 1/8, . ^, 104 Silk Vests, « 6/7, 106 Leghorn Hats, « 11/9^, 114 pieces Nankin, « 8/3|, 1 16 pounds of Thread, « 9/1 1 J, Ana. je257 17s. 8d. i Id. r- i. 8d. FEDERAL MONET* 97 FEDERAL MONEY, Federal Money is the national currency of the United States. Its denominations are,->^ihe mtVA the cent, the dollar and eagle, TABLE OF FEDERAL MONET. 10 mills arc equal in value to 1 cent« 10 cents are equal to 1 dime. 10 dimes, or 100 cents, are equal to 1 dollar, 10 dollars are equal to one eagle. In this table, 10 units of either denomination, make one unit oi the next higher denomination, and this is the same way that simple numbers in- crease from the right to the left. Therefore, the denominations of federal money may be added, sub' tractedj multiplied and divided by the same rules that have already been given for simple numbers* From the table it appears — First, that cents may be cJianged into mills by multiplying by 10, or by annexing a cipher. Thus, 5 cent8=50 mills. Second, that dollars may be changed into cents by multiplying by 100, or an- nexing two ciphers, and into mills by annexing three. Thus, 8 dollars=800 cents, or 8000 mills. This character, $, placed before a number shows the number to express dollars, thus $14, is 14 dollars. When , dollars and cents are written in one sum, they are separa- ted by a point, thus, $10.45 : to be- read 10 dollars and 45 cents. Take notice, there must be two places of figures for cents : therefore if the cents#>e4ess than 10, a ciplier Questions. — What is federal money? What are its denomi- nations^? Repeat the table. How many units ofeither denom> inalion make one of the next higher? How may federal money be added? subtracted? multiplied? and divided? How may cents be changed to mills? Dollars into cents? How into mills? How many cents in 8 dollars? in 9? in 10? What character stands for dol!:irs? When dollars and cents are written toge* ther, hoAT are they separate^? How many places must there be for cents? 96 FEDERAL MONEY. f must be placed on the left hand of the cents ^ thus, 84 dol- lars and 4 cf^nts is written, $84.04. Federal money is generally read in dollars and cents. — Cents are reduced to dollars by dividing by 100, or, which is the same thing, by cutting off the two right hand figures, mills to dollars by cutting oft' three right hand figures. iEXAMPLES. 1. How many cents in 89 dollars ? Ans. 8900 cents. 2. How many cents in 468 dollars? Ans. 46800 c^nts. 3. How many cents in 48 and 19 cents? Ans. 4819 cents. 4. How many dollars in 8643 c^nts ? Ans. $86.43. 5. How many dollars in 1903 cents ? Ans. $19.03. 6. How many dollars in 6489 mills? Ans. $6.48.9. 7. What is the sum of $34.25, S18.04, $142, $176. 81 and $0.58. Operation. In %^Tlting federal money for addition be careful to place dollars under dollars, 34.25 18.01< 142.00 176.81 .58 cents under cents, &c. Then add the same as in simple numbers. $371.68 8. Add together 36 dollars, 7 dollars and 45 cents, 86 cents, 130 dollars and 6 cents, and 340 dollars 1 cent. Aii«. 514 dollars 38 cents. 9. Add together 46 dollars 9 cents, 100 dollars 7 cents, 99 dollars 75 cents, 451 dollars 99 cents, and 1 dollar 1 ct ' • Ans. 698 dollars 91 cents. 10. What is the expense of one quarter's schooling, allowing $19 for board, $9 for tuition, $3.75 for books, and 92 cents for stationary? Ans. $32.67. Questions. — When the cents ore less than 10 what ii to be done? How is federnl money };onerally read? How are cents reduced to dollars? How are tiuli& reduced to dollars? How do we set down federal money for addition? How do we add tederal money? ^. - V FEDEBAL MONET. 99 g t . ' 4 Operation. 319,00 47,56 11. A man paid $75.41 for ahorse, $54.04 for a yoke of oxen, $21 for a cow, $7.41 for 4 sheep, $1.50 each for two pigs, and $64 for a wagon ; how much mo- ney did he pay out ? Ans. $224.86. 12. From 319 dollars take 47 dollars and 56 cents. When either of the sums in Federal money presented for subtraction has no cents ex- pressed, tlie places of cents may be si^pplied by two ciphers j then proceed as in simpU? $271,44 numbers. 13. Subtract 654 dollars from 783 dollars and 48 cents. « Ans. $ Subtract $31,12 from $5390. Ans. $ Subtract 42 cents from $51. Ans. $ Subtract 7 cents from $1. Ans. $ Subtract 5 cents from $754. Ans. $ Subtract 4 cents from $4 Ans. $ 14. 15. 16. 17. 18. 19. A man's income is $3000 a year ; he spends $187,50 ; how much does fie save ? ^^ Ans. $2812,50. 20. How much muf.t be added to $40,17 to make $100. Ans. $59,83. 21. How much is 18 times 4 dollars 72 cents'? Operation. $4,72 is the same as 472 cents ; we there- 4,72 fore multiply it as 472 cents, and the product 18 is 8496 cents. To change these cents to dollars we divide them by 100 ; thii> .:f done by pointing olf two figures for a remainder. — The quotient, 84, is dollars, and tlie remain- der, 96, is cents. 3776 472 $84,96 Note. — If we wish to obtain the value of several arti- cles in Federal money, the articles may be mulUplied by the price of one, or the price by the aiiicles^ and the pro- QuESTioNs. — When either of the sums presented for subtrac- tion has no cents expressed, what do we do7 How do vi^e then proceed? How do we obtain the value of several articlei ii^ federal money? In what denomination will the anawer be? fiiw 100 FEDERAL MONEY. dud will be the answer in the lowest denomination men* tioned in the price. 22. What must be paid for 6 pounds of tea, at $1,20 per pound? Ans. $7,20. 23. At $1,05 per pound, what is the value of 5 chests of tea, each chest containing 64< pounds ? Ans. $336. 24f. What will 55 yards of cloth come to at 37 cents per yard ? Ans. $20,35. 25. What will 300 bushels of wheat come to at $ 1,25 per bushel ? Ans. $375. 26. What is the value of 9704 oranges, at 3^ cents each ? Ans. $339,64. 27. What will be the •ost of 47 barrels of apples at 1| dollars per barrel] - Ans. $82,25. 28. If 637 dollars be divided equally among 24 men, what will each man receive 1 Operation. 24)637(26 dollare 48 157 144 1300(54 cts, 120 100 After dividing the dollars by the number of men, it appears from the quotient«rand remainder, that each man can have $26, and still 13 dol- lars will remain undivided. We change $13 to cents, by annexing two ciphers, and then divide the cents by the number of men, from which it appears each man will have 54 cts., and 4 cents will remain. 96 -^ Rem.' 4 cents. 29. A man bought a piece of cloth containing 72yds. for $252 : what did he pay per yard 1 Ans. $3,50. 30. A farmer purchased a farm containing 725 acres, for which he paid $18306.25 : what did it cost him per acre? Ans. $25.25, QvuTioif.— 'In dividing dollars, if a remainder occur, what if to be done in order to divide the remaUiderl FEDERAL MONEY. 101 15. !■ •31. A farmer receives $840 for the wool of 14?00 sheep : how much does each sheep produce him 1 > Ans. $0.60. 32. At $954 for 3816 yards of flannel, what is that a yard 1 Ans. $0.25. 33. Bought 72 pounds of raisins for $8, what was that a potind ? ' Ans. 11^ cents. PRACTICAL EXERCISES IN FEDERAL MONEY. 1. Bought 1 barrel of flour at 6 dollars 75 cents, lOlbs. of coffee for 2 dollars 30 cents, 7 pounds of sugar for 92 cents, 1 pound of raisins for 12A cents, and 2 oranges for 6 cents ; what was the whole amount? Ans. $10.15.^. 2. A man having j|)500 lost 83 cents : how much had heleft? * Ars. $499.17. 3. How much must be added to $16.82 to make $25? 4. If I pay 22 cents a gallon for 72 hogsheads of mo- lasses, each hogshead containing 63 gallons, and then sell the whole for $936, how much do I losel Ans. $61.92. 5. From 3 dollars take 7 cents. . Ans. $2.93. 6. A man dies leaving an estate^f ,^33000 to be equally divided among his four children after his wife shall have taken her third. What was the wife's portion, and what the part of each child 1 A $ $11000 wife's part. ^"^* I $ 5500 each child's part. 7. A person on settling with his butcher finds that ho is charged with 126 pounds of beef at 9 cents per pound j 85 pounds of veal at 6 cents per pound : 6 pairs of fowls at 37 cents a pair : and three hams at $1.50 each : how much does he owe himi -^ Ans. $23,16. 8. A farmer bargains witli his tailor for a new coat ev- ery six months, a new vest every three months, and three pairs of pantaloons a year : the coats to cost $29.50 each, tlie vests $3 a piece, and the pantaloons $12 a pair : at tho end of two years how much did he owe him. Ans. $214. 9. A farmer has six ten acre lots, in each of which he pastures 6 cows : each cow produces 112 pounds of biU- i2 1 102 FEDERAL MONEY. ter, for which he receives 18^ cents per pound : the ex- pences of each cow are 5 dollars and a half: how mucli does he make by his dairy? Ans. $547.92. 10. A mart lets out 2000 sheep with the condition that he is to have three fourths of what they produce after deducting the expenses of shearing : they yield 4f pounds of wool a head, which is sold at 47^ cents per lb. The expense of shearing is one tenth of the whole : what does the owner of the sheep receive ? Ans. $25.65; END OF PART 11. ■ni^Pffi «,7::;v... ■M'-^ ^y* *fe PART III. SIMPLE INTEREST. Interest is money paid for the use of money that has been borrowed. The sum of money lent is called the principal. The sum paid for the use of money is called interest. Amount is the prin- cipal and interest added together. Per annum signifies by the year. It is customary to pay a cer- tain sum for every hundred pounds, dollars &c. — ^ In this Province six pounds a year is paid for the use of every hundred, and in England five pounds for every hundred that is borrowed. These rates are established by law and are called legal interest. Usury is taking more interest than the law aU lows. The expressions six per cent, seven per cent, &c., signify that six or seven pounds or dollars are paid for every hundred borrowed. Per signi- fies for, and cent, is the abbreviation of centum, the Latin word for hundred. Rate per cent,, then signifies rate by the hundred. In all notes on interest if no particular rate per cent, is mentioned, it is always understood to be legal interest that is promised, in this work Qper cent, will be understood when no r|ite per cent.' is mentioned. QufisTioNs, — What is iaterettl What is the principal?— What is the siiiin calle4'1^nich is paid for the use of the princi- pal? What U the aipountl What is meant by per annum? What is legal interest? What is usury? What is the meaning^ of per cent, t W hen np. rate per cent, is mentioned what inte^y nt is undei^atood? 104. SIMPLE INTEREST. CASE I. To find the interest of any given sum for one or mote years. RULE. I. Multiply the principal by the rate per cent, and di- vide the product by 100 and the quotient will be the inter- est for one year. II. When the number of years exceeds one, multiply the interest for one year by the number of years : the pro- duct will be the interest for that number of years. EXAMPLES. 1. What is the interest of ^6150 10s. 6d. for one year, at 6 per cent ? We first multiply the principal by the rate per cent, and divide the product l)y 100, by cutting off the two right hand figures in the pounds — We then multiply the remaining right hand figures by the next in- ferior denomination, and divide by 100 OS before, and so on through all the denominations. qr.2|24 Rem. -. ' '> »•■ :' '''" '\' And. £9 Os. 7d. 2qr. • ' • • 2. What is the interest of jei55 for one year? Ans. j£0 6s. 3. What is the interest of .€236 10s. 4^d. for a year at 5 per cent ? Ans. JGll 16s. 6d, 4i, What is the interest of £2 128. 9.id. for a vear 1 Ans. £0 3s. 2d. QoESTioMs. — How do we find ttit* interest on ony given eum tbr one year? How do we find the interest for several years? Operation, del 50 10s. 6d. 6 £9 03 3 20 s.0|63 12 d.7 56 4 • 1 t * m. \ SIMPLE INTEREST. 100 5. Required the interest of j£200 10s. 4d. for 3 years at 7 per cent. Ans. J842 2s. 2d. 6. What is the interest of Jei50 8s. 3d. for 5 years at 6 per cent ? Ans. je45 2s. 5^ . 7. Required the amount of j654j7 15s. at 5 per cent. \yev annum for 3 years. Ans. ^£629 18s. 3d. 8. What is the amount of J6500 for 5 years at 5 per centi Ans. jt;625. 9. What is the interest of J6325 12s. 3d. for five years at 6 per cent ? Ans. ^697 13s. 8d. 10. Required the interest of ;€855 17s. 6d. for one year at 5| per cent per annum. « Ans. £4i9 4s. 3d. *11. What is the interest of JE246 18s. for five years at 4^ per cent per annum 1 Ans. ^£52 9s. 3^d, CASE II. To find the simple interest of any sum of money, for any number of years, and parts of a year. I. Find the interest on the given sum for 1 year. 11. Multiply the interest of one year by the given num- ber of years, and the product will be the answer for that time. III. If there be parts of a year, as months and days? work for the months by the aliquot or even parts of a year? and for the days by the aliquot parts of a month, or for the days multiply the interest of one year by the number of days and divide the product by 365. NOTE.- an aliquot part, as i^, \, |, &c. m An exact or even part of a quantity is called If we wish to find the in- terest of any sum for any part of a year it may be done by dividing the interest for a year by that pa»t. For example, for six months, which is half a year, we divide the yearly interest by 2, for 1 month by 12, &c. and the same in re- gard to parts of a month. By a little exercise of the mind it is plain that one month is -^\ of a year ; 2 months the {• of a year ; 3 months the 4 of a year ; 4* months the |, and 6 montlis the ^ of a year, &c. And so of the days, al* ( I : 106 SIMPLE INTERfiST. lowing 30 days to the month ; 1 day is -^\ of a month, 2 ^*ys -iV> 3 days -i\, 5 ds^s |, 6 days -J^, 10 days i, and 15 dajni i- of a month, &c. It is sometimes mor^ convenient to take parts of parts. ^ EXAMPLES. 1. What is the interest of JCISO I63. 8d. for 4> y^ears, 7 months and 20 days at 6 per cent ? Operation. £ s. d. 150 16 8 6 First find tlie interest for 1 year and then for the number of years. Then because 6 mo. are the one-half of a year, we divide the interest of one year by 2, and the quo- tient is the interest for 6 mo. We then di- 9|05 vide the interest of 6 mo. by 6 for the inter- 20 est of one month, and the interest of 1 mo. — by 2 because 15 days are the ^ of 1 mo. 1|00 and the interest of 15 days by 3 because 5 days are the i of 15 days. Wlierefore the sum of tliese several numbers must be the interest for the whole time. 6 rao .= iyear. * . is f of 6 mo. is i of 1 mo. is -|^ of 15 da. • (Vhat is the inl £9 Is. 4. = Interest 1 = do. 6= do. 1= do. 6 2=do. 6 0^=do or 1 year. 1 mo 15 da. 5 da. 36 4 4= 10 15 7 2 for 4 do. for 6 months, for 1 month, for 15 days. . for 5 do. 2. ^ 41 19 7 25 Ans. tercst of £ 100 10s. for 1 1 months If QoKiTioNi.--Huw do we find the intcreit for monthit How for dayil By what other method may the interest for daye be foundl SIMPLE INTEREST. 107 Operation. jeiOO 10s. 6 '6 mo* 4>mo. Imo. h We fiiid the interest for 6 mo. by dividing the interest — ^ '" for 1 year by 2, theft for 4 ' 603 00 Intfor 1 yr mo. by dividing the interest for 1 year by 3, and for 1 mo. 301 10 do. 6 mo. by dividing the interest for 4» 201 00 do. 4 mo. mo. by 4, then add the sever 50 05 do. 1 mo. al sums together ; 6 mo. 4» mo. and 1 mo.= ll months. je5,52 15 20 8.10,55 12 We mjiy if we please add the several sums logether be- fore dividing by 100, as in the above example, which is frequently the better way. d.6,60 qr.2,40 Ans. £5 10s. 6d. 2qrs, ^ 3. What is the interest of £250 16s. Sd. for 28 days? £ s. 4 15 1 Int. for 1 vr. 28 mimb. of dN^s 365)421 8(jei 365 56 SO ( Over,) First, find the interest for 1 year, muhiply it by the no. of days, and divide the pro- duct by 365 as the rule di- rects. QuE8TioNi.-.-Wliat part of a yeor is 1 monlliT 2 monthst— 3 monlhi? 4 monthil 6 months What part of a month ii a day? 2 dayi? 3 dayt7 5 days? G liaya? 10 daya? 15 dayat i ; 108 SIMPLE IITTERBST. * '. ! f 1128(3a. 1095 33 12 396(l(i. 3«5 31 Ans. jei 33. Id. ■#c 4| 4. What is the interest of £^1 17s. 8d. for three months? . Ans. 17s. 4|d. 5. Required the interest of JS174 10s. 6d. for 3 years and 6 months? Ans. je36 13s. 6. Of je 150 16s. 8d. for 4 years and 7 months 1 Ans. £41 9s. 7d. 7. Of je75 83. 4d. for 5 years and 2 months 1 Ana. £23 7s. 7d. 8. What will £3000 amount to in 12 years and 10 months? Ans. £5310. 9. What is the interest of £257 5s. Id. for 1 year and three quarters, at 4 per cent ? Ans. £18 Os. l|d. 10. What will £279 13s. 8d. nmount to in 3 years and a half at 5j per cent, per annum ? Ans. £331 Is. 6d. 11. What is the interest of £71 3s. 11 id. for 1 year, 5 months and 25 days? Ans. £6 Gs. 11. 12. What is the interest of £397 9s. 5d. for 2 years 3 months, at 3^ per cent ? Ans. £31 6s. 13. Required the interest of £300 for 2 years, 7 mo. and 20 days ? Ans. £47 10s. 14. What is the interest of £80 19s. lid. for 3 years, 10 months and 10 days ? Ans. £18. 15s. 3^d. ^ - .illiy <#' RATIO. 109 15. What is the interest of je320 10s. 8d. for 2 years, 10 months and 20 days ? Ans. £55 1 Is. 2d. 16. .Required the interest of j£ 500 for 3 yean, 11 mo. and 11 days, at 3^ per cenU? Ans. £69 Is. G^d. 17. What is the interest of X 100 for for 46 days ? Ans. 14s. 9d. 2qr. 18. What is the interest of Je25 18s. 9d. for 9 years 9 months and 9 days at 5 per cent 1 ^ ^ Ans. X12 13s. 6d. Iqr 1- RATIO. The word ratio means relation ; and when it is asked what ratio one number has to another, it means in what relation does one number stand to another ? Thus, when we say the ratio of 1 to 2 is J, we mean that the relation in which 1 stands to 2 is that of one-half to a whole. * Again, the ratio of 3 to 4 is J, that is, 3 is j of 4, or stands in the relation of | to the 4. So also the ratio of 4 to 3 is i ; for the 4 is 4 thirds of 8, and stands to it therefore in the relation off. The relation of 12 to 24 is J, and the relation of 24 to 12 is 2. When therefore we find the ratio of one number to another, we find what part of one number ano- ther is, , Then the ratio of 4 to is f ; that is, 4 is 4 sixth of 6. The ratio of one number to another QuEBTioNB.— What if the meaning of the word ratio? When it is a«ked what ratio one number bean to another, what ia meant! What ia the ratio of one to 2T What ia the relation of 12 to 24? Of 24 to 12? When we nbtiUn the ratio of one number to another what do we find? What pait of 6 ii 4? Sl'l ttftO RVLE OF THREE^ N If! r:- f ^ Ml ( n ! >• i" H then may always be expressed by a fradiion, in which the first number is put fornumerator, and is called the antecedent, and the second number is pxi for denominator, and is called -the consequent. Thus the ratio of 8 to 4 is f or 2. That is, 8 is twice 4, or stands to 4 in the relation of a dupli- cate or double. RULE OP THREE, OR PI^OPORTION. When quantities have the same ratio, they afe ^aid to be proportional to ear.h other. Thus the ra- tio of 2 to 4 is J, and the ratio of 4 to 8 is J ; that is, 1 has the same relation to 2 that 4 has to 8 ; therefore these numbers are called proportionals. Again, 4 is the same portion or jyart of 8 that 10 is of 20 J wherefore these numbers are called pro- portiondls. A proportion then is a combination of equal ratios. Points are used to indicate that there is a pro- portion between numbers. Thus, 4 : 8 : : 9 : 18 is read thus, 4 has the same ratio or relation to 8 that 9 has to 18. Or more briefly, 4 is to 8, as 9 to 18. The fobrth term of every proportion may be found by multiplying the second and third terms together and dividing their product by the first term. For example, if the first three terms of a proportion are 3, 9, 12, the fourth is 36, for 9X124-3=36. Questions.-— Mow may the ratio of one number to another be expressed? Which number is put for numerator? What is put for denominator? What is this called? When are quan- tities said to be proportional to each other? Give exampieii. What are used to indicate that there is a proportion between numbers? Give an example on the slate. How is it read? IIow may the fourth term of any proportion be found? If the first three terms arc 2, 4 and 6, what is the fourth? ■ «>-&U>"l!-"llMl'UjlK4t."JM J^ RULE OF THREE. The first and fourth terms oltk proiportion are called the two extremes, and the second and third t&^ms are called the two means, „ in every proportion the product of the two ex- :?; iremes is equal to ihe product of the two means, Eor example, in the proportion 4 : 12 : : 8 : 24 4X24=8X12=96. The Riile of Three takes its name from the clr- eumstance that three numbers are always given to find di fourth, which shall bear the same proportion to one of the given numbers as exists between the * other two. * To find the fourth term when three are given we have the following general RULE. I. Write that number for the third tenn which is of the same kind with the answer or number sought. II. Then consider from the nature of the ques- tlon; whether the answer required must be greater or less than this third te»* n ; if greater, write the greater of the other given numbers for the second term, and the less for the first ; but if the required answer must be less than the third term, set down the less of the other two numbers for the second term, and the greater for the first, III. If the first and second terms contain differ* ent denominations they must be reduced to the » - ', Questions. — What are the first and fourth terms of a propor- lion called? What are the second and third terms called? In every proportion what is the product of the two extremes equal to? Give an example. From what does the Rule of Three take its name? In stating a question what is the first thing to be done? What is the next? If the required answer must be greater than the third term how must we place the other two numbers? How if the answer is to be less than the third term? t. I. . i 'M [\\ [I I k f» H 1 llli RULE Ot THRE£4 fame denomination, nnd the MiVe^ to the /ot/^es/ de- nominatioB hientioned in it. Then multiply the second and third terms toge- ther, and divide the product by the first term, and the quotient will*be the fourth term or answer sought of the same denomination as that to which the third term was reduced. Note. — The same rule is applicable, whether the given quantities be integral, fractional or decimal. Proof. — Divide the product of the extremes by one of the mean terms, and if the work is right the quotient will be the other mean term. ' , EXAMPLES. 1. If 5 pounds of butter cost 75 pence; how much will 13 pounds cost ? Operation. As 51b. : 13lb. : : 75»l. 13 225 75 In this example it is plain that the answer must be mo- ney ; we therefore write the 75 pence as the third term. It is also plain that the price of 13 pounds is greater than . the price of 5 pounds, that is, 5)975 the required answer is greater than the third term ; we 195 pence, therefore set down the 13lbs. for the second term, and the 511)3. for the first term ; we then multipiy the second and third terms together, and divide by the first, as the rule di- rects, and the quotient is the answer in the same denomina- tion as the third term. 2. If a footman perform a journey in 21 days, when QuEbTioMs.— If the first and second terms contain different denominations what must be done] Hnw must we reduce the third term? Which two do we multiply together? By whai do we divide the product? Whtt will the quotient bt? Of whai denomination? m -*,**■ ' RULfi OF three; 113 the (lays are 15 hours long, in how many days of 9 hours can he perform the same journey 1 Operation. Here the term similar to the requireil hrs. hrs. days, answer is 21 days, which is conse- 9 : 15 : : 21 quently the third term. Then, since 15 the footman will not travel so far in a day of 9 hours as in one of 15 hours, 9)3 15 the required answer mut^i be greater than the third term ; the 15hrs. must there- Ans. 35 ds. fore be the second Icvni and 9hrs. the first. Operation of Proof. In this example the terms 9, 15 9x35-1-15=21 or and 21 arc given, and the lai^t 9x35=315—21 = 15 term 35 ii^ lound. Then the product of the extremes 9, 35, it^ 315; this being divided by either of the means gives the other. 3. If 3cwt. of sugar cost £9 2?. OJ. what wilHcwt. 3qr3. 261bs. cost at the same rate 1 Operation. cwt. cvvt. qr. lb. £ p, J, 3 26 9 2 20 12 28 19 28 3361b. 5581b. 182 12 . 2184.1, 558 336)1218672(3627il. 12)3627 20)302 3d. jC15 2s. 3d. An». QuESTioK.— How is the Rule of Three prcved f k2 lU RULK OF THRE£. First state the question as the rule directs. We then re- duce the first and second terms to pounds, then the third to petice, which is the lowest denomination named in it. We then multiply the second and third terms together and divide by the first, and the quotient is pence, which we reduce to pounds, &c. 4. If a man's wages be JG75 10s. for 12 months, what is -that a month ? Ans. £6 5s. lOd. 5. If one pair of gloves cost 4s. 6d. what will 19 do- zen pairs cost ? Ans. jG516s. 6. At lO^d. per pound, what is the value of a firkin of butter weighing SGlbs.? Ans. £2 9si 7. If a man spends 3s. 4d. per day, how much is that a year? Ans. JG60 16s. 8d. 8. What is the value of 2cwt. of sugar at 5cl. a pound ? Ans*:^4 13s. 4d, 9. If 240 sheep yield 6601bs. of wool^ hovv many pounds will be obtained from 1200 sheep 1 Ans. 33001bs. 10. If 4^ tons of hay will keep 3 cattle over the ,vin- ter, iiow many tons will it take to keep 25 cattle ti: same time? , Ans. 37 i tons. 11. Bou^it 8 chests of sugar, each 9cwt. 2qrs. ; what do they come to at £2 5s. per cwt. ? Ans. JB171. 12. Hovv much butter can you buy for JG23 2s., at 9d. l^er pound ? Ans, 5cwt. 2qrs. 43. If 30 barrels of flour will support 100 men for 40 days, hovv long would it subsist 25 men ? Ans. 160 days. 14. A owes B jB679 6^., but compounds with him by paying 3s. 4d. on the pound ; hovv much does B receive for his debt? Ans. Jeil3 4s. 4d. 15. A goldsmith sold a tankard for £H 12s. at 5s. 4d. per oz. ; what was the weight of the tankard ? r Ans. 21bs. 8oy. 5pwt 16. If 2cwt. 3qrs. 2 libs, of sugar cost £6 Is. 8d. what cost 354cwt. ? Ans. £73. 17. If a man spend 7d. pe> day for bitters, how much is that in a year ? Ans, JCIO 12s, 1 |d. RVI.E OF THREE. 115 18. If 90 busheli of oats will feed 40 horses for 6 days, how long will 450 bushels last them ? Ans. 30 days. 19. If 5cwt. 3qrs. 141bs. of sugar cost £6 Is. 8d., what will 35cwt. 281b|i. cost 1 Ans. je36 lOs. 20. What is the cost of 3cwt. of coffee at 15d. per pound? Ans. je21. 21. If 3 quarters of a yard of velvet cost 7s. 3d. how many yards can be bought for i£13 15s. 6d. ^^ Ans. 28yds. 2qrs. 22. Sold a ship for j£537, and I owned f of her ; what was ray part of the money ? Ans. j£201 ?»-. 6d. 23. If a staff 5 feet long cast a shade on level ground 8 feet, what is the heiglrt of that steeple whose shade at the same time measures 181 feet? Ans. 113^ feet. 24. Bought 50 pieces of kerseys, each 34 ells Flemish, at 8s. 4d, per ell English ; what did the whole cost? Ans. ;e425. 25. Bought 200 yards of cambric for JC90, but being damaged, I am willing to lose £1 10s. by the sale of it ; what must I demand per ell English ? Ans. 10s. 3|d. 26. If an ingot of gold weighing 91b. 9oz. 12pwt. bo worth J£470 8s. what is that per grain ? Ans. 2d. 27. Bought 4 bales of cloth, each containing 6 pieces, and each piece 27 yards at JG16 4s. per piece ; what is the value of the whole and the cost per yard ? Ans. je388 16s. at 12s. per yd. 28. What will be the cost of 72 yards of cloth at the rate of Je5 12s. for 9 yards? Ans. i244 16». 29. A person^s annual income is JC146 ; how much ian that per day ? Ans. 89. 30. If 3 paces or common steps of a person be equal to 2 yards, how many yards will 160 paces make ? Ans. 106 yards 2ft. 3 1 . How many yards of carpeting that is 3 feet wide, will cover a floor that is 27 feet lopg and 20 feet broad ? Ans. 60 yards. 32. What is the cost of 6 bushels of coal at the rate of £1 14s. Od. the chaldron f Ans. 5s. 9d. 116 RULIf 01' THREE. ii i! 33. When hens are 9 shillings a dozen, what will be the price of 6 dozens of eggs at 2 pence for 3 eggs? Ans. 4)8. 34. If 6352 stones 3 feet long will complete a certain quantity of wall, how many stones of 2 feet long will raise the like quantity ? Ans. 9528. 35. If a person can count 300 in two minutes, how many can he count in a day ? Ans. 216000. 36. A garrison of 536 men have prrvisi» ' * *"''*" l '''' **''''*^' '5|t» PRACTICE. ^ PKACTICE. ' " * Practice is a short ibethod of finding the aa*^ swers to questions iji the Rule of Three, when the first term is a unit or one. It has acquired its name firom its d^iiiy use among merchants and business men, it being an easy method of working, where the price of a unit is given to find the price of a quantity. For example, if one yard of cloth cost ten shil" lings, what will 40 cost ? This question may be easily solved by the rule called Practice. If the cloth had been £l per yard, the cost of 40 yards would have been £40 ; but since it is on- ly a part of a pound per yard, the whole cost will be the same part of £40 that the cost of one yard is of one pound, that is J^of 40. Hence the cost is i of £40, or £20. One number is said to be an aliquot part of an- other when it forms an exact or even part of it. For example, 4d. is ah aliquot or even part of a shilling. So is 5s. of one pound ; it is one fourth part, being contained in 20s. 4 times. TABLE op ALIQUOT PARTS. jParts of jei.lTaftrof Is. rPaflsoTl TonJParts of acvvtT cwt. qr. lb. 10s. =i 6d. = \ 10 = h 2 or 56 =i, 6s. 8d. — j, 4d. = ?, 5 ^^4 1 or 28 ^•i 5s. —\ 3d. ~"? 4 — 1. — ~ 5 14 = 1 4s. =i 2d. — -L 6 24 g Parts of a qr. 3s. 4d. =1 1 ;,d. — J. 2 "^"l 141bs. = ^ 2s. 6d. = I Id. — _1_ 7 =i ls.8d.=-jV 4 =^ ^ H ~^ K QuBSTioNa.— What is Practice? From what has it derived its Dtiniel When iton« nuniber said to be an aliquot part of an>« otherl Mention tome oF Ihe aliquot pajrta of £1. Of a ihilling. Of a ton. Ofaowt. Ofa quarter, dec. ■>- »■ PRACTICE. 119 ^ C AS5 I. Wben the price of one yarcli pound, &c. is less than a penny. - RULE. Find the value of the given quantity at Id. a yard, pound, &lc. and divide it by that even part, and the quotient will be the answer in pence. — Biit if the price be not an even part of Id., as 3qrs., take parts of parts, and add the^results together For the answer in pence. rr EXAMPLES. 1. What is the value of 4528 oranges at ^d. each ? Operation. Ii|il4528 1132d.=:e414s.4d. It is evident that the price of 4528 oranges at Id. is 4528d. It is also plain that at |d. the price would be ^ as many pence as there are oranges ; we thero- forc divide by 4 and the quotient is the answer in pence. 2. What is the value of 4528 eggs at ^d. each ? ^d.=3qr. the greatest even part of which is 2qr. or ^d. ; wc there- fore divide by 2, and tlie quotient is the price at 0., and becatiee ^d. is \ of id. we divide this quotient by 2 for the price at Jd. ; wc then add the quotients together and the sum is the price at ^d. Operation. i.im4528 i|2264 vaUieat^d. 1132 value at ^d. 3396 value at ^d. Ans. £U< 3s. 3. What is the value of 5704 lemons at Jd. t Ans. £o 18s. lOd. 4. Wliat is the value of 6813 do. at id. 1 Ans. Jei4 3s. id. ■ ' r ill 11 I <• QuEBTioNf. — When the jir'ico of a yd. lb. &c. ig nn even part of a penny what is firat to be donel Wliat will the quotient beT If it be not an even part of Id.? What part of ^ is |t When the price of u unit ii I of id. how do w<{ divide t •■«i .ts^','*»*««*-'. I I 1*1' t 120 PRACTICE. 5. What is the value of 9424 do. at |d t Ans. £29 08. 6. What is the value of 1487 do. at |d. 1 Ans. £4f 12s. ll|d. CASE II. Whw the price of one yard, pound, &c. is less than Is. RULE. Find the value of the quantity at Is. a yd. &c., then take such part or parts as the price is of Is. ; add the quotients together, and their sum will be the answer in shillings. EXAMPLES. 1. What is the value of 371 libs, of butter at 7|d. per pound 1 Operation. 3711 Tt is plain that the price of 3711 lbs. at Is. is 3711s. then because 6d. is i of Is. we divide by 2 for the price at 6d., then by 6, because 1 is ^ of 6d., then by 2 again, because 2qrs. is ^ of Id. ; again by 2, bo- cause Iqr. is ^ of 2qrs. ; we then 2396s. 8d. Iqr. add their quotients as the rule di- rects. 6 1 JL ■2 X 4 1855 6 309 3 154 7 2 77 3 3 jeil9 163. 8d. Iqr. 2. Wliat is the cost of 862 yards at 2d ? Ans. £1 3s. 8d. 3. « « 749 « 4d.f Ans. jei2 9s. 8d. 4. « « 113 « 6d.] Ans. £2 16s. 6d. 5. « «< 899 « 8d.? Ans. £29 19s. 4d. QvBiTiONi.— When the price of a yard, pound, &c. ii leu than 1 ihilling what ia the firat thing to be done? What part or part! ofthia do we Ukel What vill tne aum of the quotients be? X 4 ns. £29 98. I 12s. ll|d. ?£j?5 than Is. a yd. &c., e is of Is. ; am will be r at 7|d. per rice of 37 11 len because by 2 for the ), because 1 ain, because n by 2, be- !. ; we then the rule di- PRACTIC£. 121 £1 38. 8d. jei2 98. 8d. £2 16s. 6d. P29 19s.4(l. , &c. !■ leis What part quolientrt bel 6. What is the costof ^4>7 yards at 3^d.? Ans. £31 68. 2Ad. 2456 7. 8. 9. 10. 11. 12. u M U u tc a u u M u u « « 4id. ? Ans. £43 98. lOd. 3271 « 7d, ? Ans. £95 8s. Id. 2759 « 8id. ? Ans. £97 14s. 3^d. 5272 « 9d. 1 Ans. £197 14s. Od. 3254 « 10id.l Ans. £142 7s. 3d. 7972 « ll|d.? Ans. £390 5s. lid. CASE III. When the price is any number of shillings under 20, or an even part of 1 pound. RULE. Multiply the quantity by the price for the an- swer in shillings, or Find the value at £l per yard, &c. and then take parts, or parts of parts as the case may require, and the quotient or sum of the quotients will be the an- swer in pounds. EXAMPLES. 1. What will 129^ bushels of oats cost at 2s. 6d. per bushel ? £ s. C2s. 6. Ill 129 10 value at £1 per bushel. Operation < ( Ans. £16 3s. 9(1. value at 2s. 6d. Because 2s. 6d. is j^ of a pound we divide the price at £1 by 8, aud it is evident that the quotient will be the price at 2s. 6d. QuRiTioNi.— When the price it any part ofa pound how may the anawer be fbund? How ii it done otherwiie? In what de« nomination will the quotient or quotients bel L •*<• y\ ■i ^oIjA it r • > 122 PRACTICE. 8. What is the value of 527 yards at 4s. 1 u u 1271 u Ans. jei05. Sa. 58. T r Ans. £817 15s. 4. a «. 2710 « 6s. t Ans. X813 Os. 5. « « 191 « 8s.'? 76 Ss. 6- it « 600 « ISs.? . . . An8.je390 Os. 7. u « 1075 « 168.? Ans. jeSOO Os. 8. u « 2150 « 19s.? Ans. 2042 10s. 9. t( « . 543 « 6s. 8d. ? - Ans. 181 0.^. 10. it « 127 « 33. 4d.? Ans. je21 3d. 4d. n. (( « 461 CASE IV. « Is. 8d.? Ans. de38 8s. 4d. When the price is pounds, or pounds, shillings, pence, and quarters. RULE. Multiply the given Quantity by the pounds, then work for the shillings oy case 3d. for the pence by case 2nd, and for the quarters by case 1st, add the several quotients together and the sum will be the answer. EXAMPLES. 1. What is the cost of 680 acres of land at £3 9s. 1^. per acre? •f QvitTioNf.— Kepeat the rule for performins; the operttion when the priO€ is In poundi, or poundi, shillingf, pence, tnd qutrtcre. PRACTICE. 12a .■Li. «/8* is. 6d. Id. 2qr. i 4 680 5 3 2040 170 136 1 17 6 2 -i 1 Operation. 16 8 8 4 value at £Z per acre, do at 5s. per do. do at 4s. per do. do at 6d. perdo. do at Id. perdo. do at ^d. per do. An!?. ;e2367 5 do at Je3 9s. 7Ad. We first multiply the price by the pounds, then it is ev- ident that the price of 680 acres at jGI per acre would be J6680, then as 5s. and 4s. are even parts of 20s. we take the \ and the ^ of 680 for the price at 9j<. — then because 4s.=48d. and 6d. is \ of48d. we take ^ of the price at 4.-:. for the price at 6d., then \ of that for the price at Id., and i of the price at Id. for the price at 2(^1^?. and their sums is the answer sought. 2. What is the value of 124 acres at £'i 5s. ^^iiA Ans. ^£406 7s. 24k«'4Ft«^^ ^f PRACTICE. CASB V. When the price and quantity are of several denominations. RULE. .\ Multiply the price by the integers, or whole numbers in the given quantity, and take parts for the rest from the price of an integer ; which added together will be the answer. EXAMPLES. 1 . What cost 5 cwt. 3 qr. 14»lbs of raisins at £2 1 ls.8d.? 2qt Iqr. 141b. Operation. £ s. d. 2 11 8 We first multiply the price by the 5 cwt. then because 2 qr. is ^ cwt. 5 we divide the price of cwt. by 2, and the quo- 12 18 4 cost of 5 cwt. tient is the cost, of 2 qr. ; 1 5 10 do. of 2 qr. then as 1 qr. is ^ of 2 qr. 12 11 do. of 1 qr. we take ^ the price of 2 6 5^ do. of 1 4 lb. qr. for the price of 1 qr.; also, because 14 lb. is ^ 1 An«. JB15 3 6i whole cost. qr. we take ^ the price of one quarter, and the quotient is the price of 14 lbs. 2. What is the cost of 5cwt. Iqr. of sugar at JE2 17s. per cwt. 1 Ans. jei4 19s. 3d. 3. What is the cost of 14cv^i. 3qr. 71b. of beef at 13s. Sd. per cwt.? dClO 2s. S^d. 4. At jei 4s. 9(1. per cwt. what is the value of 17cwt. Iqr. 171bs. cheese 1 Ans. £21 10s. 8d. .5. At £3 17s. 6d. per cwt. what is the value of 25 cwt. 2qr. 14Ib. tobacco! Ans. je99 5s. lljd. 6. Bought 78cwt. 3qr. 121b. of currants at Je2 17s. 9d. per cwt.j what did I give for the whole? Ans. £2^.1 :4s. QuKtTioNi When the price and quantity are Qf ^everi^l ^^t nominationi, what b the rule? PRACTICE. 125 ieoominations 's, or whole ike parts for «^hich added atje211s.8d.? multiply the e 5 cwt. then qr. is icwt. the price of and the quo- costof 2 qr. ; [".is ^ of2qr. the price of 2 price of 1 qr.; 3e 14 lb. is ^ 1 i the price of 4 lbs. at £2 17s. €14 19s. 3d. f beef at 13s. £10 2s. S^d. lueofl7cwt. e21 10s. 8d. value of 25 99 5s. ll^d. £2 17s. 9d. k ^2^. / :4s. Jf aeverald^. PRACTICAL EXERCISES. 1. What is the cost of 650 pigeons at ^d. ea(ih.1 Ans. 13s. 6id. 2. What is the value of 245 ducks at ^d. each, t Ans. 10s. 2id. 3. Bought a box of oranges containing 525, at |d. each; what did they coct me»? Ans. £1 12s. 9|d 4. What is the value of 120 lb. of rice at 3d. per lb.1 Ans. £\ 10s. 5. Bought 8012 lb. chalis at 2|d. per lb. Ans. £91 16s. Id. 6. How much will 3906 lb. of beef come to at7id. per lb, 1 Ans. £122 Is. 3d. 7. What will 1847 yards of cloth come to at 5s. 8d. per yard 1 ^ Ans. £523 6s. 4d. 8. If an ell of Holland cost 4s. 6d. what is the value of 5 pieces each 12 ellsl Ans. £13 10:?. 9. What is the value of 1234 yards of muslin at Is. 1 1 Jd. per yard ? Ans. £122 2s. S^d. 10. What cost 287 bushels of wheat at 17s. 6d. per bushel I Ans. £251 2s. 6d. 11. How much will 47 tons of hay amount to at £6 6s. 8d. per ton. Ans. £297 13s. 4d. 12. Sold 26 acres of land for £11 14s. per acre ; what is the amount ? Ans. £304 4.^. 13. If 1 yard of cloth co t £1 19s. 4d. how much will 1677 yards come to. Ans. £3298 2s. 14. Sold 16cwt. 2qr. 17ib. of sugar at £2 15s. lid. per cwt. what was its value 1 Ans. £46 1 Is. Id. 15. Sold 56cwt. Iqr. 171b. of sugar at £2 15?. 9d. the cwt.; what does it come to ? Ans. £157 4s. 4id. 16. What will 51 acres of land be worth at £3 2s. 2d. per acre. Ans. £ 1 58 lOs. 6d; 17. What will 4 E. E. 3qr. 2na. of broadcloth cost at £2 38. 8d. per yard ? Ans. £12 16s. 6id. l2 M0i f n ''' ' i ' > : % 126 COMMISSIOjy AND BROKERAGE. COMMISSION AND BROKERAGE. Commission is an allowance made to a Factor or person engaged in buying and selling goods for another. A Factor is an agent who transacts bu- siness for his employer. Brokerage is an allowance made to dealers in money or stocks. The allowance made is generally a certain per cent, or rate per hundred on the monies paid out or received, and the work is the same as casting the interest on the same sum for one vear. EXAMPLES. 1. If I employ a factor to sell goods for me to the va- lue of £257 D 17s. 6d.} what must I pay him at 4 per cent.l Operation. £ s. d. Here the work is the same as sim- 2575 17 6 pie interest, we multiply by the rate 4 per cent, and divide by 100. 103,03 10 20 0,70 12 . 8,40 4 1,60 Anfc'.^eiOS 0.^ 8id. 2 . My correspondent semis me word that he has bought goods to the amount of JC 1286 on my account ; what will his commission come to at 2^ per cent. 1 Ans. £32 3s. Questions.— What is Commission] What is a factor? What li Brokerage? What is the allowanoe generally made? To what is the work similar? COMMISSION AND BROKERAGE. 127 3 dealers in mt; what will Ans. £32 3s. 3. A facto? sells land to the amount of JS25,500 and is to receive 2i^ per cent, commission; bow much must \ pay over to his principaU Ans. £24iS62 10s. 4. What is the commission on £34t96 at 6 per cent. ? Ans. je263 15s. 2id. 5. A gentleman sent a broker £3825 to be invested in stock, the broker is to receive 2 per cent, on the amount paid for the stock ; what "wagjjie value of the stock pur- chased ] Operation. 100 o waybe 2,00 100 102 : 100:: 3825 3825 As the broker is to re- ceive 2 per cent., it follows that JB102 of the money received by him, will pur- chase JG 100 of stock : there- fore lOO/.added to the com- mission is to 100, as the - given sum to the stock which it will purchase. 102)382500(3750 Ans. Proof. -Commission on 3750 at 2 per cent, is £75 and 3750 +75=3825. 6. A factor receives £708 15s. and is directed to pur- %- ciiase steel at £45 per ton : he is to receive 5 per cent, on the money paid : how much steel can he purchase ? Ans. 15 tons, 7. A broker bought 200 shares of bank stock for A. — He paid £197 per share, and he is to receive one fourth per cent, on the money he received ; how much must A. pay for the stock ? Ans. £39498 10s. 8. A bank fails, and has in circulation bills to the a- mount of £267581. It can pay only 9^ per cent. : how much money is there on hand? Ans. £25420 3s. lOjd. 9. A merchant shipped to his agent in Montreal 1000 barrels of flour, which was sold at 20s. per barrel : what ^ 128 INSURANCE. did he obtain for the flour, and what commission did he pay at 1| percent? . ( He received ^6982 10s. for the flour. ( And paid £17 10s. commission. ■s 't • INS URANC E. An Insurance Co»fJp(d^ a \^(xdy of men who in return for a certain compensation, promise to pay for the loss of property insured. The written engagement they give is called a Policy. The sum paid by those who own the property, to the Company who insure it, is called Premium. It is reckoned at so much per cent, on the value of the property insured. EXAMPLES. 1. What will be the premium for insuring a ship and cargo from Quebec to Amsterdam, v^jaed at J637800, at 4^ per cent. ? Ans. £1701. 2. What would be the premium for the insurance of a house valued at ^£5500, against loss by fire for 1 year at ^ per cent. ? Ans. £21 10s. 3. What would be the premium for insuring a ship and cargo, valued at JG37500 from Montreal to Liverpool, at 3^ per cent. ? Ans. £1312 lOg. 4. What would be the insurance of a steamboat from Kingston to Toronto, valued at JEMOOO, at 1^ per cent? Also, at I per cent? At \ per cent? At \ per cent? At \ per cent. ? Answers in course, £210, £105, £70, £46 13s. 34d. £35. QuEbTioMS. — Wlial is an Insurance Company? What is the written engagement which they give called? What is the Prc<« "^^uunt? How ii it reckoned 7 nission did he Is. for the flour. >mmi8sion. .4 DISCOUNT. 129 5. What is the insurance on a store and goods, valued at £27000, at 2^ per cent] At 2, at 1|, at |, at i, at \, at ^, and at -J- per cent.? ^ Answers in course, £607 10s., £540, J^405, £202 lOs., £135, £67 10s., £54, £45. of men who 1, promise to The written Hey, the property, ed Premium, \ the value of 'ing a ship and d at £37800, Ans.£1701. le insurance of e for 1 year at \ns. £27 lOs. ring a ship and > Liverpool, at 3. £1312 lOs. teamboat from 1 ^ per cent ? ^er cent ? At ■A6 13s. 3id. What is the hat is the Prct DISCOUNT. Discount is a deduction made from a debt, for paying it before it is due. If, lor example, I owe a man £300 two years hence, and am willing to pay him now, I ought to pay only that sum which, with its interest, would in two years amount to £300. The sum which, in the time mentioned, would by the addition of its interest, amount to the sum which is due, is called the present worth'. The question then is in the above example, what sum together with its interest at a certain per cent, would in two years amount to £300? This is found by the following RtJLE. Find the amount of £100 or dollars, for the time and rate proposed in the question. Then, as the amount found is to the amount given, so is £100 or dollars to the principal or present worth re quired. The present worth, subtracted from the whole debt will leave the discount. £XAMt>LES« 1. A debt of £372 is due 4 years hence ; what mo- ney paid down will discharge it, allowing 6 per cent, per annum discount? QucsTiOKs.— What ia Piacount? What ia called the present worth of a note, or debt? Repeat the rulel When the preient worth is found, how do we And the discountt ^ - 130 Iv DISCOUNT. 100 6 Operation. 600 4 2400 100 - 124 amount of £100 for the time. Then as 124 : 372 : : 100 100 124)37200(300 present worth. 372— 300=72 discount. It is evident that JGIOO is the present worth of jC124 due 4 years hence, because JCIOO amounts to JC124 for the time and rate given ; hence L124 bears the same relation to L372 as LlOO to the discount on L372. Note. — This method of computing discount is the cor- rect one, but the mode generally adopted at the banks is to compute the interest on the whole note to be discounted in a manner which produces a small excess, and, deducting tliis interest, advance the remainder to the holders — thus virtually charging interest not only on the sum advanced, but on the part withheld. This when the sum is small is a trivial error, but in a large one the error is sometimes ponsiderable. For example, the interest of L500 at 5 per cent, for 12 years, exceeds the discount of the same sum for the same time and at the same rate by LI 12 10s., a sum too great to lose. 2. What sum in ready money will discharge a debt of L925 due one year and 8 monthti i.ence at 6 per cent. ? Ans. L840 18s. 2d. 3. What is the present worth of 1*600 due 4 years hence, at 5 per cent. I Ans. L500« LOSS AND GAIN. 131 le. present worth. worth of £124, tojei24d. 4. Bought 441b. at L6 12s. and sold it again for LH 10s. 6d. : what was the profit on each pound ? Ans. lOj^d. 5. Bought a hogshead of wine at Ll 5s. per gallon, and sold it ibr LIS : was there a loss or gain ? Ans. Loss of 15s. II. To know what is gained or lost per cent. RULE. First find what the gain or loss is by subtraction, then, as the price it cost : is to the gain or loss : : so is £100 to the gain or loss per cent. QoBiTioir.— What i« the Rule to find what it gainad or loai per oent.t LOSS AMD GAIN. 138 is., at lA 5s. wmucU'was be profit on a ien on the 75 I yards at 28. rd to gain LI ^8. lid. Ans. und but could er cwt : did I t,L2 lls.4d. again for L8 Ans. lO^d. IS. per gallon, 1 . Loss of 15b. ent. subtraction, in or loss : : gained or loii EXAMPLES. I. A boy bought a knife for 2s. and sold it again for 2§. 8d. ; what did he gain per cent, or in laying out LlOO ? It is plain that the boy Operation. Ss. 8d. 2 Od. gains 8d. in selling his knife; that is 2s. gained 8d. We then say, if 2s. gain 8d. what will JGIOO gain ? because the gain on ]e 100 must be in the same proportion as the gain on 2 shillings. Bought sugar at S^d. per lb. and sold it again at £i\ 8d. gain, s. d. jC 2 : 8 : : 100 : je33 6s. 8d. Ans. 2. 17s. per cwt. ; what did I gain per cent. ? Ans. je25 19s. 5jd. 3. At l^d. profit on a shilling, how mucli per cent. ? Ans. £12 10s. 4. If I buy 12hhds. of wine for £204 and sell the same again at JG14 17s. Gd. per hhd. do I gain or lose, and what per cent. ? Ans. I lose 12^ per cent. 5. At 5s. profit on a pound, how much per cent ? Ans. 2.5 per cent. G. If by selling pepper at lO^il. per pound there are 2d. lost on each : ro(|uired the loss per cent. Ans. 16. III. To know how a commodity must be sold to gain or loae so much j)cr cent. RULE. As £100 : is to tiio price it cost : : so is £100 with the proiit added, or loss subtracted, to the gain- ing or losing price. EXAMPLES. 1. If I buy Irish linen at 2s. 3d. per yard, how must 1 sell it per yard to gain 25 per cent. ? As JCIOO : 2s. 3d. : : £125 to 2s. 9d. 3qrs. Ans. QuEgTio!«.--Whrti in tlie rule for finding how a commodity must be sold, to gain or ioie so much pci c«Mit. 7 n Mi I III ■ ■ II — T Questions. —If the answer exceed 100, what is the exoeiiT If it be less than lUO, what ii the deficiency? What ii Equa^ tion of Caymental 1 136 EQUATION OF PAYMENTS. Is* t :l ii f!l m The interest of jei2 for 3 mo.=int. of £1 for 36 mo. do of i:;i8 for 4 mo.=int. of £^ for 72 do do of Je20 lor 9 mo. = int. of ^1 for 180 do £50 288 Now it appears that it will be the same to him to hav'c £1 for 36, for 72, and for 180 months, as it would to have the 12, the 18, and the 20 pounds for the number of months specified. He might therefore keep £1 just 288 months, and it would be the same as keeping the ^650 for the numher of months specified. But as the whole sum of money lent was £50, he may keep this only one fiftielh of the time he might keep £1. Therefore if 288 months be divided by 50, the quotient will be the equated lime of payment, which is 5^ ^ months. RULE. Multiply each payment by the time before it be- comes due, and divide the sum of the products by the sum of the payments ; the quotients will be the mean time. EXAMPLES. 2. A owes W £600 : JC200 is to be paid in 2 months, iJ200 in 4? months, and £200 in 6 months ; what is the mean lime for the payment of the whole 1 Operation. We here multiply each sum 200x2=400 by the time at which it bc- 200 Xl— 800 comes due, and divide the sum 200X6= 1200 , of the products by the sum of the payments. 2400-; 600=4nio. Ans. 3. A man ov\es me £300, to be paid as follows ; \ in 3 months ; ^ in 4 months, and the rest in 6 monUis ; what is \\\c mean time fi)r payment ! Ans. 4.^ months. 4. A merchant lias due him £300 to be paid in 60 day<», £500 to be paid in 120 days, and 750 to be paid in Hepeat the Kule fur Kqualion ofFaymentf, FELLOWSHIP. 137 180 days ; what is the equated time for the payment of the whole 1 Ans. 137i^ days. 5. A owes B J61200, ^ is to be paid in 6 months, \ in 8 months, and the rem^'nder in 10 months j what is the equated time for the payment of the whole ? Ans. 7^ months. FELLOWSHIP. The Rule of Fellowship is a method of ascer- taining the respective gains or losses of individu- als engaged in joint trade. The money, or value of the nrticles employed in trade is called the Capital or Slock* The gain or loss to be shared is called the Dividend. It is plain that each man's gain or loss should be in proportion to his share of the Stock. Hence the following nuLE. As the whole stock is to the whole gain or loss, so is each man's share to his share of the gain or loss. PROOF. Add all the separate ])rofits or shares together ; their sum should be equal to the whole profit or stock. EXAMPLES. 1. A and B buy certain merchnndise amounting to £160, of which A pays JEPO, and B je70 ; they gain by the purchase je32 ; what is each one'^ share of the profits ? £00+£1<^=£1SO, tl.cn i^ ICO : 32 | »» ; ^{^ ^;s share. 2. Three merchants make a joint stock of j£1200, of QuBtTiona.^What \» Fellowship? What it the Capital or 8tockt What ii the dividend? Repeat the rule. ]]ow ii it proved 1 m2 138 DOUBLE FELLOWSHIP. 1! M ml Ml ^1 m which A put in £240, B je360, and C £600 ; and by trad- ing they gain JG325 ; what is each one's part of the gain ? Ans. A's part £65, Wa £97 10s., C's £162 10s. a. A bankrupt is indebted to A £21 1, to B £300, and o C £390, and his whole estate amounts only to £675 10s. which he gives up to those creditors ; how much must each have in proportion to his debt 1 Ans. A must have £158 Os. 3^d., B £224 13s. 4Ad. and C £292 16?. 3|d. DOUBLE FELLOWSHIP. When several persons who are joinejd together in trade employ their capital for different periods of time, tlie partnership is called DoubleFellowship. For example, suppose A puis £200 in trade for 4 years, B £300 for 3 years, and C £100 for 1 year; this would make a case of double Fellowship. Now it is evident that there are two circum- stances which should determine each on?'s share of the profit ; first, the amount of capital he puts in ; and secondly, the time which it is continued in the business. Wherefore each one's share should be proportional to the capital he puts in, multiplied by the time it is continued in trade. Hence the following RULE. Multiply each man's stock or share by the time it was continued in trade ; then. As the sum of the several products is to the whole gain or loss, so is each man's particular pro- duct to his particular share of the gain or loss. Questions.— What is Double Fellowship? What two circom- ttanoet should determine each one's share of the profits? Re* peat the rule. TAHE AND TRET. 139 3 ; how much EXAMPLES. 1. A and B enter into partnership. A puts in J6840 for 4 months, and B puts in ^6650 for 6 months ; they gain d6300. What is each one's share of the profits ? A's stock de840x4<=3360 B's stock i2650x6=3900 je7260:300 "\ 3360:;ei3816s.lOd 3900:^6161 3s Id. 2. A put in trade JG50 for 4 months, and B ^660 for 5 months ; they gained JG24. How is it to be divided be- tween them ? Ans. A's share ^69 12s., B's £14> 8s. 3. C and D hold a pasture together, for which they pay £54? J C pastures 23 horses for 27 days, and D 21 horses for 39 days. How much of the rent ought each one to pay ? Ans. C £23 5s. 9d. j D £30 143. 3d. 4". A, B and C hold a pasture in common, for which they pay £19 per annum. A put in 8 oxen for 6 weeks, B 12 oxen for 8 weeks, and C 12 oxen for 12 weeks. What must each pay of the rent ] Ans. A must pay £3 3s. 4d. j B, £6 6s. 8d. j and C, £9 10s. TARE AND TRET. Ta7'e and Tret are allowances made in selling goods by weight. Draft is an allowance on the gross weight in fa- vour of the buver or importer. It is always de- ducted before the Tare, Tare is an allowance made to the buyer for the weight of the hogshead, barrel or bag, &c., con- taining the commodity sold. QoESTioHt — What are Tare and Tretl What ii Drat\! What is Tare? Li ' 1. I 1^ li w '; UO TARE AND TRET. Gross weight is the whole weight of the goods, together with that of the hogshead, barrel, bag, &c., which contains them. Suttle is what remains after a part of the allow- ances have been deducted from the gross weight. Nett weight is what remains after all the deduc- tions are made. y All the questions in this rule may be worked by the Rule of Three. ,, ^, EXAMPLES. 1. What is the nctt weight of 112cvvt. Sqrs. 121bs. of tobacco ; tare on the whole, 6cwt. 3qrs. 201bs. ,, cwt. qrs. lbs. 112 3 12 gross weight, 6 3 20 tare. Ans. 105 3 20 nelt weight. 2. If the tare be 411)s. per cwt. what will the tare be on 6T. 2cvvt. 3qr. 14!bs. Ans. 4cwt. Iqr. IS^lbs. 3. What is ithe nett weight of 3 casks of indigo, each weighing 4c wt. 2qrs. 14lbs. gross j tare on each cask Icwt. Oqr. 121bs. { Ans. lOcwt. Sqrs. 61bs. 4. What is the nctt weight of 20 hogsheads of sugar, weighing in all 246cwt. 3qis. 7Jbs. gross, tare 161bs. per cwt. ? Ans. 21 Icwt. 2qrs. Gibs. 5. What is the nett weight of I32cwt. Iqr. 201bs. gross, tare 143 and three fifths ? What is their difierence ? 12. What is tlie diircrence between six eighths and three eighths? What is their sum ? What id the sum of five eighths and two eightlis? What is their difference? What is the sum of four eighths and four eighths ? What is their difference ? 13. What is the sum of four twelfths and eiglit twelfths ? What is their dilYerence ? What is the difier- ence between five twelfths and Reven twelfths? What is their sum ? 14. How many halves in one ? How many thirds ? Fourths? Fifths? Sixths? Sevenths? Eighths? Ninths? Tenths? Elevenths? Twelfths? 15. How many halves in two? How many fourths? How many eighths ? How many elevenths ? How many twelfths? 16. How many fourths in four? In five ? In six? In seven 1 How many fifths in four 1 How many se- venths ? How many tenths ? 17. How many elevenths in three ? In six how many ? In nine ? In ten ? :^^i! XdAR FRACTIONS. 143 s? How many hs in three units 1 ny tenths in two? How many four- il to one seventh 1 To five sevenths ? One half of one ? Of one fifth ? id one half? Of 1 and one fourth ? ■ fourth and three and two fifths? lum of two fiftlirf e? six eighths and !iat is the sum of their difference? eighths ? What fifths and eiglit 'hat is the diller- Jlftlis? What is w many thirds ? ighths? Ninths? V many fourths ? ^ ? How many five? In six? How many se^ six how many ? * 18. How many whole units in two halves? In three halves ? In four halves ? In five halves ? In six halves ? In eight halves ? In nine halves ? 19. How many units in three thirds ? In four thirds ? In five thirds? In seven thirds? In nine thirds? In eleven thirds ? 20. How many units in four fourths ? In five fourths ? In eight fourths ? In nine fourths? In eleven fourths? In sixteen fourths? 21. How many units in ten tenths ? In fifteen tenths ? In twenty tenths? In twenty four tenths? In thirty three tenths ? 22. How many eights is one fourth equal to ? What is the sum of one fourth and one eighth ? Two fourths and two eighths? One fourth and five eighths? Three fourths and two eighths ? 23. How many tenths are two fifths equal to ? What is the sum of one fifth and eight tenths ? What is the sum of one sixth and one twelfth ? Of one sixth and ten twelfths ? Before proceeding farther the pupil is requested to re- view carefully what is said of Fractions on the 4'4 ?5 6 3 1 o; An Improper Fraction is one in which the numerator equals or exceeds the denominator. They arc called im- proper fractions because they are equal to or exceed unity. When the numerator is equal to the denominator, as f , the value of the fraction is equal tol. If the numerator i i IT ■a ■;.-;1) lU VULGAR FRACTIONS exceed the denominator as f , the value of the fraction is greater than 1, as the following : .1 A £. yj 3) 5> 7> 3J XJL ia iA 8 J a J 7 • A Simple Fraction is a single expression, as |. It may be either proper or improper. The following are simple fractions : 1 '2 A JI 45 5 J 5 5 6 5 .5. J2L Si 35 45 65 A Compound Fraction is a fraction of a fraction, or se- veral fractions connected by tlie word of, as the following : ^ of ^, i of i, I of i of 10, i of I of 12. A Mixed J^umber is composed of a whole number and a fraction, as the following : ^•U *45 "g5 ^y A Jlf/a:cc/Fradio?i is one whose numerator or denomi- nator is a mixed number, as 4.^ 10. A whole number may be expressed fractionally by writ- ing 1 below it for a denominator. Thus, 2 may be written ',-, and is read 2 ones, 4 " « f, « " 4 ones. 7 « « -I, « " 7one:s. But 2 ones are equal to 2, 4< ones are equal to 4, 7 ones to 7, &c. Therefore the value of a number is not altered by placing 1 under it for a denominator. Questions. — What is a fractioikl How niatsy kiiula of Vulgar Fractions are there? What are Ihey ? What is a Proper Frac- tion? Is its value greater or less than 1? Give an example of a proper fraction. What is an Improper Fraction? Why is it called improper? When is Us value equal to 1? When is it greater than I? Give an example of an improper fraction' What is a Simple Fraction? Give an example. What is a Compound Fraction? Give an example. What is a Mixed Number? Give an example. Is five-eighths a proper or impro- per fraction? What kind of a fraction is eight-fourths? Wluit is its value? What kind of a fraction is nine-eighths? What is ita value? What kind of a (raciion is one-half of one-third? What kind of a fraction or nunber is 4 Ihrcie fourths? 7 on^'- eeventh? 9 two-thirds? ' the fraction is VULGAR FRACTIONS. 149 You have learned (page 4«5) that the denominator shows into how many equal parts a unit is divided, and the numerator shows how many of the parts ary expressed by the fraction. You have also learned that the numerator and denomi- nator taken together are called the terms of the fraction, and that dividing both terms by the same number does not change the value of the fraction. tor or denomi- ionally by writ- REDUCTION OF VULGAR FRACTIONS, CASE I. To reduce a fraction to its lowest terms. RULE. Divide the numerator and denominator by any number which will divide them both without a re- mainder, and those quotients again in the same way until there is no number greater than 1 that will divide them both without a remainder. EXAMPLES. 1. Reduce -^^ to its lowest terms. Operation. 6) 6 1 Ans. 6) 12 """2 Here it will be seen that the frac- tion is in the lowest terms, as no number greater than 1 will divide the numerator and denominator. — It will also be seen that its iermt only are altered, not its valv£. QoESTioNs.— How may a whole number be expressed fraction* allyl Does this alter its vdlue? Give an example. What doei the denominator of a fraction show? What does tlie numera. torshowl What are the numerator and denominator taken to- gether called? If both terms be divided by the same number does it change the value of a fraction? Repeat the rule for re- ducing a fraction to its lowest terms. '4il 1*6 ^vuiAji WBihcniffs, 3. Reduce -1-^ to its lowest terms. Operation. 2>m 2) &2t=2) 26_13) 13 1 2) 312'^ 2) 156 2) 78"" 13) 39*^3 ^** "*"' 4. Reduce ff*^ to its lowest terms. Ans. I^.. '•"'n* ^vf-' "flW^ 5. 6. Reduce if4f to its lowest terrtis. Reduce -Hif to its lowest terms. 8 b' Ans. |. Ans. |. (GJREATEST COMMON DIVISOR. There is another method of reducing a fraction to its lowest terms, which 's often preferable to the above, viz : dividing the terms by their great- est common divisor". In the first example above, 6 is a common divisor of both tferms of the frac- tion -i*a- : it is also their greatest common divisor. Any number greater than 1 that will divide two or more numbers without a remainder is called their common divisor ; and the greatest number that will so divide them is called their greatest common divisor. The greatest common divisor of two numbers is found by the following RULE. Divide the greater number by the less, then the diviflor by the remainder, and continue to divide the last divisor by the last remainder until nothing remains. The last divisor will be the common divisor sought. iV I ar of the ivrtt num- bers 13^ and 165. ' ^ p Operation. * Proof. 136)165(1 15)135(9 135 135 15)165(11 165 30)135(4 120 Greatest common div. 15)30(2 30 Let us try if the \em number 135 is the greatest common divisor. It will exactly divide itself, but will not divide 165 without a remainder ; we divide it therefore, by this remainder, and find still a semainder of 15. We divide the last divisor by this remainder, and nothing is left. Therefore 15 is tlie greatest common divisor of 135 and 165. 2. What is the greatest common divisor of 323 and 475? Ans. 17. 3. Required the greatest common divisor of 2310 and 46261 Ans. 6. 4. What is the greatest common divisor of 1092 and 1428? Ans. 84. ,i 5. What is the greatest common divisor of 1197 and 'f 805? Ans. 7, Note. — To find the greatest common divisor of more than two numbers ; find the conmion measure of two of them as above, then find the greatest divisor of this com- mon measure and a third given number ; and so proceed to the last. The pupil may now reduce the following fractions to their lowest terms by dividing both terms by their greatest common divisor. ...«►■ -•*»:«»*~' 148 YULGAR FRACTlQNSi ■■'H'l U i EXAMPLES. 1. Reduce -^^ to its lowest terms. Operation. yfe first find Uio 70)175(2 ^ greatest common di- 1*0 % visor of 70 and 175 ^ , . ^ to be 35 ; then re- 35)70(2 35) 70^_2^ juce -^, to its low- "70 35) 175 5,Ans. est terms by dividing at once by this num- ber. 2. Reduce f ^ to its lowest terms by the last method. Ans. ^. 3. Reduce ^l^f to its lowest terms by tho last method. Ans. ^. 4>. Reduce -jVA" *o *^s lowest terms by both methods. Ans. 4« 5. Reduce -^f^^ to ?t^ lowest terms by both methods. Ans. I. 6. Reduce + JJ ^ to ita lowest terms by both methods. Ans. ^. CASE II. To reduce a mixed number to lis equivalent improper fraction. RULE. Multiply the whole number by the denominator of the fraetion, to this product add the numerator, and place their sum over the give^i denominator. EXAMPLES. . ■ 1. Reduce 12"} to its cfjuivalcnt improper fraction. l2xS=96-f3=99. Ans. \\ It is plain that multi- plying 12 by tho de- 1 whole thing is equal to S eighths nominator 8, makes it — hence, 12 whole things equal 96 eighths. Then 3 96 eighth{>. cigths added make V . 2. Reduce 45}^ to an improper fraction? Ans. ^l^. VULGAR FRACTIONS. 149 e first find tlio est common di- of 70 and 175 i 35 ; then re- ;rms by dividing ice by this num- he last method. Ans. ^. tho last method. Ans. ^. by both methods. Ans. 4. by both methods. Ans. \. )y both methods. Ans. ^. livalent improper denominator ihc numerator, denominator. Dpcr traction. I plain that multi- g 12 by the de- untor 8, makes it ighth:^. Then 3 i added make V • >nl Ans.J^p. 3. How many 24th8 in 365r/,- 1 Ans. a|^. 4. Reduce 1 92 f^ to its equivalnt improper fraction 1 Ans-HV^- 6. Reduce 240 jYs to its equivalent improiier fraction. Ans. -af 11^. 6. Reduce 876 1^^ to its equivalent improper fraction. Ans. ^i^y^. «! « M CASE III. To reduce an improper fraction to its equivalent whole or mixed number. RULE. ^ Divide the numerator by the denominator, the quotient will be the whole number ; and if th^-e be a rena' flf^' place it over the given denominator. ^, EXAMPLES. Operation. 1. Reduce ^^ to its equivalent whole 7)t8 or mixed number. . — Cf Ans. 2. Reduce A,^ to a whole number. Ans. 84-|-7=12. From the above examples we may perceive the truth of the following principle, viz : — The value of every fraction is equal to the quotient arising from dividing the numerator hy the denominator, 3. Reduce \*-/- to a whole or mixed number. Ans. 12-,V. 4. Ill V^ of bushels, how many bushels? Ans. 5^. 5. If I give I of an orange to each of 12 children, hovr many orangos do I give ? Ans. 3. QuKttTioNit. — ilnvv nany eighllis in 12 M-hole numbers? How liiiiiiy eigliihii in 12 an'i Ihn>ceit{htht7 Kf>penl the rule for re- iiuuing a inixed number lo itt equivalent improper fraction?— Uuw many wlii>le nunit>ers in 48 leventliiT and liow many ■««• vntlia oveit Repeal the rule for reducing an improper rraotion lo iii c<)nivalenl wliql« or niijied number. v2 'I 1 life 150 . 6. VULGAR FRACTIONS. RcdfUce ^f to its whole or mixed number ? 7. Reduce ^iVs^ and Vu^V ^ their equivalent whole or mixed numbers. Ans. 24 and 8|^^f . 8. Reduce f |, ^4^jf , ff ^ and ^4^0^ to whole or mixed numbers. Ans. If, 19, 5f and 1-^^. . CASE IV. To reduce a whole number to an equivalent fraction having a given denominator. RULE^ Multiply the whole number by the given denom- inator, and set the product over the said denomin- ator. EXAMPLES. 1 . Reduce 8 to a fraction whose denominator shall be 5. Here 8x5=40 ; therefore ^ is the required fractfon, for 40-4-5=8, according to case III. 2. Reduce 18 to a fraction whose denominator shall be 8. • Ans. -4^- 3. Reduce 125 to a fraction whose denominator shall be 15. Ans -Lfp. 4. Reduce 135 to a fraction whose denominator shall bo 175. Ans. -^ffp. CASE V. To reduce a compound fraction to its equivalent simple one. aULE. I. Reduce all mixed numbers to their equiva- lent improper fractions by case II. ■• "-— — - ■ .^1 .. ■ . - ■■ I --^ - ■ — — . - -. — ■ - I -,.- , ■ ^ „ _.. , ^^, .■■■-■■■ Questions.— To what, is ihe value of every fraction equal?— - Repeat the rule for reducing a whole number to an equivalent fraction having a given denominator. How many whole num. beri in 63 BevenlhsT In 06 eighthil In 100 tenths? How ma- ny 8tha in7uniti? How many lltha in 6? How many Otha in 71 If the denominator be 5 what fraction do we form of 9? Of II? Of 18? VULGAR FRACTIONS. l&l valent fraction livaient simple II. Then multiply all the numerators together for a numerator, and all the denominators together for a denominator ; their products will form the fraction required. EXAMPLES. 1, Reduce -^ of ^ to a simple fraction. Ans. -j^. 4 '^ 2 If I of 1 is I, then i of ^ must be half as much, or ^ of 1 . 2. Reduce ^ of -^ of ^ to a simple fraction. Ans. -^\-, 3. Reduce | off off to a simple fraction. Here |XiX7=-iW=ii=f by reducing the fraction to its lowest terms, as shewn in case I. Or, by cancelling or drawing a perpendicular line after the 3'8 and 6'a in the numerator and denominator, thus, 3| /S 6, /S 7 — T By cancelling the 3's we only divide both terms by 3 ; and in cancelling the 6's we divide by 6. Hence the value of the fraction is not affected by thus cancelling like figures, which should always be done when the numerator of one is like the denominator of another. 4. Reduce f of |^ of -j\ to a simple fraction. Here ^V^^V-®- — '^^^ — ^-s — -8- — ^ Arm. Or ^ V**'va-'- — -*J- — a. Ana .'?, Reduce | of -,»„- of |f to a simple fraction. Ans. -iVb . 6, Reduce -^.^ of |f of-,^/ of 20 to a single fraction. Ans. «^i=23V. Reduce 1 of | of|5 o(-^.- of f to a smgle fraction. Ans.f^. Required the value of i || of 33 J in a single frac- Ans. 15-,Pj. How many apples are f of | of f of -^^ of | of 40 7. 8. tion. 9. apples ? Ans. 20. QuESTiuNi «-How do we reduce a compound fraction to a aimple one? When there are like figures in the numerator and denominator, what do we do with ihemT Does this alter the value of the fractioni What is half of one.third? What is one- third of one.fif\h? What is one third of threetweAhs? two. third! of vixninthit three-fourths of eighteleventhiT four-fiftba offivetwelfthtt 152 VULGAR FRACTIONS. n'«j CASE VI. To reduce fractions of different denominators to equiva- lent fractions having a common denominator. RULE. I. Reduce compound fractions to simple ones, and whole or mixed numbers to improper frac- tions. II. Then multiply each numerator by all the ' denominators except its own, for the new numera- tors, and all the denominators together for a com- mon denominator ; the common denominator pla- ced under each of the new numerators will form the several fractions sought. EXAMPLES. 1 . Reduce |, \ and f to a common denominator. - 1X3X7=21 the new numerator of the let. 1X4X7=28 « « 2nd. 4X4X3=48 « « 3d. 4X3X7=84 the common denominator. Thsrefore W'i^\ and if are the equivalent fractions. It is plain that this reduction does not alter the value of the fractions, for the numerator and denominator of each are muitiplied by the same number, and by reducing each to its lowest terms, we should have again the original frac- tion*? For --1 — ' ^ft — ' nn4oJ. 48J48 ****" 4 h ' i CASE VII. ^ To reduce a mixed fraction to a siim>le one. , RULE. » Multiply the numerator and denominator of the given fraction by the denominator of the fraction annexed, to the product of the numerator adding the numerator of the annexed fraction, and the products will be the terms of the fraction required. Note. — In the application of this rule it should be con- sidered that a fraction multiplied by a -^hole number equal to its denominator, produces a whole number equal to its numerator. EXAMPLES. 422 1. Reduce _I® to a simple fraction. 49 42|X8+7==343 numerator. 49X8 1^92 denominator. S. Reduce — Mo a simple fraction. 46 =J Ans. 8 Ans. \. „, r»i QoESTioNft.-«How may a fraction, wlinse numerator or de* nominator is a mixed number, be reduced to a airnple fraction f What doet a fraction, multiplied by a whole number equal !• itf denominator, produce f mmon denomi- 4 5 5 4 6 *'*" it* lator of f , f , ^ fi-JQ. 3.0. and^<^ r of -A-j 1^4- and -, ^%\ and in, ist common de- > 4 2 «"" 4 U. • quivalent frao- itor. AJi .3.3. and 4-*-. 4 8 > 4 8 **"" 4 k • one. ninator of the f the fraction iraior adding tion, and the ion required. should be con- \h()!e numher e number equal VULGAR FIUC170RS.. •2 Ans. 8 Ans. ). lumerator or de« ■irnple fraction Y number equal !• 34 3. Aeduce — to a simple (ractioo.. 45| 73 4. Beduce to a simple fraction. 131f 157 Ans^l. Ans. |. The following cases relate to fractions of different deno-^ minations. The next two cases are th^ reyerse of each other. CASE VIII. To reduce a fraction from a lower to a higher, denomina- tion. RULE. I. Consider how many units of tjhe, given de- nomination make one of the next higher, and place 1 over that number forming a second fraction. II. Proceed in the same manner from the second denomination to the third, and so on to the denomination desired. III. Connect the several fractions thus foirmed by the word of, making a compound fractiop, then reduce the compound fraction to a si.nple one by Case V. EXAMPLES. 1. Reduce f of a penny to the fraction of J61. &. of -J- of -J'-=— J^— =-J— of JCI. Arifi Here the given fraction is f- of a penny ; but one penny of a shilling, and one shilling is ^\ of a pound ; hence f of a penny is equal to f of VsT o^ «\- of a ^= 2"i¥» Therefore the reason of the rule is evident. 2. Reduce f of a barleycorn to the fraction of a yard. Ans. -s-f 7 of a yd. Operation. | of ^ of -^\ of | =^|t. Questions.— To reduce a fraction from a lower to a higher denomination, what is the first step? What ii the secondT The third? Are i of a £ and of i of a £ the saniie, or different denominations^ One-fifth of a £ and i of a shiltins? } of a day and i of an hour? One-seventh of a week and | of a month? i of a toot and § of a rod? ^ of an inch and two'fifiha a yard? O is -i\ 158 VyLGAR FRACTIONS, Three barleycorns make an inch, we therefore first place 1 over 3 ; as 12 inches make a foot, we next place 1 over 12, and as 3 feet make a yard, we plaf3 1 over 3. 3. Beduce ^ of a farthing to the fraction of a shilling. ' Ans. -g^,-. .,. 4. Reduce f of an ounce Troy, to the fraction of a ;pound. Ans. t/,-. 5. Reduce f of a pound Avoirdupois, to the fraction of a cwt. i t ; Ans. ■^^, . 6. Reduce ^ of a farthing to the fraction of a J8. Ans. ^^^ £. 7. Reduce ^ of an ounce to the fraction of a ton. "**•"*'• 7 1 6 a * • ,, ^,8, Reduce \^ of a minute to the fraction of a day. ^ ^^' Ans. -xVaV« ^ day. . ! »■ .... CASE IX. To reduce a fraction from a higher to a lower deno- mmation. RULE. I. Consider how many units of the next lower denomination make one unit of the given denomi- nation, and place 1 under the number forming a second fraction. II. Proceed in the same manner with the de- nomination still lower, and so on to the denomina- tion desired. III. Connect the several fractions thus formed, making a compound fraction, which reduce to a simple one by Case V. EXAMPLES. I. Reduce f of a JG to the fraction of a penny. Ans. ^ifM. Operation. \ of Y of Y=^d. QuKaTioHi.— In reducing fractions frum a higher to a lower denomination what is the first step? What the secondl Third? VULGAR FRACTIONS. 159 I lower (leno- Hero ^ of a j6 is -^ of 20 shillingg ; but 1 shilling is equal to 12 pence,'hence -f of a £=z\ of Y of ^="^7 ^ The reason of the rule is therefore apparent. 2. Reduce f cwt. to the fraction of a pound. Ans. Af^lb. 3. Reduce t/^- of a pound Troy, to the fraction of an ounce. Ans. ^oz. 4. Reduce -/^ of a week to the fraction of a da -. Ans.^| day. 5. Reduce ^j]-^ of a hogshsad to the fraction of a gal- lon. Ans. I gal. 6. Reduce | of a tun to the fraction of a gill, Ans.^Ji-=^8-<^ gill. 7. Reduce —g^T of a day to the fraction of a minute- Ans. mm. 8. Reduce -^-J^ of a furlong to the fraction of a foot Ans. ^ I ft. ' I i jl :. CASE X. To find the value of a fraction in whole numbers of a lower denomination. RULE. I. Reduce Ihe numerator to the next lower de- nomination and divide the result by th<; denomi- nator. II. Reduce the remainder, if there be one, to the denomination still less, and divide again by the denominator, and so proceed to the lowest deno- mination. The several quotients placed in order, Will be the value of the fraction in the different denominations. Questions. —To find the value of a fraction in the lower de« nominations of a whole number what is the first thing to be donel What is the nexll How do we reduce an integer to a fraction ofa given denomination? To find the value of a fraction in the lower denominations oi a whole number what is the first thing 10 be done? i mi 160 VULGAR FRACTIONS. EXAMPLES. 1 . What is the value of '^ of a J6. Operation. 2 20 3)40 13s . 1 Remainder. 12 We first reduce the numerator from the denomination of pounds to that of shillings. Dividing by the deno- minator gives 13s. and 1 over. Re- ducing this to pence and dividing a» before gives 4d. 3)12 4d. Ans. 13s. 4«d. 2. What is the value of i^,; of a day ? Ans. 7h. I2m. 3. Find the value of | of an acre. Ans. 3R. 20P, 4. Find the value of -,V of a cwt. Ans. Iqr. Tibs. f). What is the value of | of a hogshead ? Ans. 52gals. 2qts. 6. What ia the distance of -^% of a mile? Ans. 7fur. 8p. 7. Reduce ^^ of on ell English to its proper value. Ans. 1yd. Oqr. 3na. 8. Reduce | of a mile to its proper quantity. Ans. 4fur. 22rds. 4yd.s. 2ft. lin. 2|bc. '.">' • . CASE XI. To reduce an integer to a fraction of a given denomina- tion. RULE. ' Reduce the number to the lowest denomhiation mentioned in it; then if the reduction is to bo QoBtTioN.— How do wt rtduce an integer to a fraction of » given denoniinaUonf ^k. VULGAR FRACTIONS. 161 made still lower, proceed as in Case IX, but if to a higher denomination, proceed as in Case VIII. .f;'<:^' EXAMPLES. 1. Reduce 4tl. 2qrs5. to the iVaction cfa shilling. Ans. '1. Operation. We first reduce the given 4 nomination ; and secondly, to sjt common denomi- nator. CASE I. When the fractions are of the same denomination and have a common denominator. _ RULE. Add the numerators together, and place their sum over the common denominator ; then reduce the fraction to its lowest terms, or to its equivalent mixed number. EXAMPLES. 1. Add i, f , I, and ^ together. Ans. -i^i'=2|. Operation. l--|-2+3+5=n. Hence -^J-=their sum. It is plain that as all the parts are fourths their true sum will be expressed by the number of fourths ; that is 1 1 fourths, which equal 2^. 2. 3. 4. Add Ans. ;€^=:11, s, ,., and 4 of a JG together. What is the sum of |, §, f and I ? What is the sum of -,'Vj Aj "7a I 7> 1 7> I V> I 7> 17 """ I 7' Result, If =51 ^ Add together. and Ans. \^=2. U? Ans.V,. CASE II. When the fractions are of the same denomination, but have difiercnt denominators. RULE. Ilcuuce mixed numbers to improper fractions, by Case II, page 148 ; compound fractions to sim- ple ones by Case V, page 150 ; and all the frao- QuBSTioifii— When fraction! are of ihe tame denoiniiialion knd have a oonimon denominator, how do we find their aumt What ia the aum ofS thirdi, 4 thirda and 1 IhirdT Of 1 fourth, 2 fourtha and 6 fourthal When fraotiona have different deno- roinatora how do wu add IhemT How do we reduce fracliona to a oommon denominatort How mav ] fourth and 1 half beoddedf VULGAR TtLXCtlOVn. I6i nination and tions to a common denominator, by Case VI, page 159. Then add them as in the last article. EXAMPLES. 1. A4d f, i, and f together. Ans. ^^=4|f. Operation . Afler reducing to a com- 6X3X5=90, let numerator. mon denominator, the 4« X 2 X 5=4<0, 2nd numerator, new fractions are |-^, |^, '2X3X2=12, 3rd numerator. i^=Vf > which, redii- 2x3X5=30, the denominator, ced to its lowest terms, becomes 4j|-f, Add f , § and % together. Ans. 2-^\^ Add 4| and 9g together. Ans. 14( 2, 3. Find the lowest common denominator, and add -,^2-, f , f and -aV- Ans. l-.a^j. Note. — When there are mixed numbers, instead of re- ducing them to improper fractions, it is better to add the whole numbers and fractional parts separately, and then add their sums. 5. Add I91-, 6|, and 4| together. Ans. SO-.Ve. Operation. Operat'n. fract'l p'ts. 1 9+6+4=29 whole numbers. ^+'l+*;z=-[lf= 1-a.a.. Then 29+l-«iV=30-,Vr;> the sum. ^, Add 12i, 3|, and 4j together. Ans. 20f^. >.-•. Add -,V, ?, 45, and » of » Icgether. Ans. 6^f. 8. AM together ^ of 95, and 5 of 14. Ans. 43 1^. 9. Add 3i, 63, 8-,«;, and 659. Ans. 84^^^. 10. Add g of I of 13^ apples, ^ of i of 2^ apples, | of ? of 7i apples, and | of |^ of 3^ apples together. Result, 5f». *» CASE III. When fractions are of different denominations. QuEiTioNi.— When there art mixed numberetobeadded, what' ie Ihe beat method? When the Tractioni tre of different deaom. Inaliona what ia fint to be done? What ia nextl What part of i ponnd ia ^ of an ounce Troy weight! Then what la tba •um of thret-lwtnij.fourthi of a pound and htlf an ounce! 164 VULGAR FRACTIONS. RUIiE. : '•>•'• Reduce the fractions to the same denominatipn. Then reduce them to a common denominator and add as in Case I. EXAMPLES. -0 1. Add I of a pound Troy, ^o f of an ounce. ' Ans. 5f oz. Operation. 3 of -Li — -3-3- 3 1 "~~ 8 -2^Y = 5f ouncea. %: In this example we first reduce | of a pound to the fraction of an ounce, and find it is \^ of an oz. Then ^J^ and f reduced to a common denominator are ^J^ and f^, which added together, make ^4°/, and rechiced to a mixed number equal 5 g-- ounces. ■ Or the I of an ounce might have been reduced to tlie fraction of a pound, thu:^, f of -^^=-7^-+^= ? ^ = 7! of * pound, which being reduced by Case X, equals 5| oz. 2. Add 3 of a day to f of an hour. Result, lOhrs. 2Gmin. 3. Add 1^ of a ton, to vV o^ ^ cwt. ' Result, 16cvvt. Iqr. 23-j'g- Iby. 4. Add ^ of a week, -iV of a day, and f of an hour to- gether. Result, 4> days, 14hrs. 59 ^ min. 5. What is the sum of ? of £15, £3}, \ off of f of a £, and f of | of a shilling ? * Ans. £1 17s. 5 jd. 6. Required the sum of % of | of 3^ tons, '^ of -,»,- of ^ tons, and «- of 7,- of 5^ cwt. Ans. ST. 17cwt. ll^lb. Note. — The value of each fraction may be found sepa- rately, and their scvernl values then added. 7. Add g^ of a year, J of a week, and J of a day, 4 of a year=^ o(^l^ days=219 days, i of a week=i of 7 days= 2 days, Shrs. } of a day = ^hrs. Am. 121dayg,llhri. VULGAR FRACTIONS. IW 8. Add I of a cwt., ^^ of a lb. 13oz. and ^ of a cwt. 61bs. together. Ans. Icwt. Iqr. 271bs. 19oz. 9. Add 4 of a week, ^ of a day, ^ of an hour, and ^4*- of a minute together. Ans. 2 days, 2hrs. 30' 45", to. Add I of a yard, | of a foot, and | of a mile toge- ther. Ans. 154»0yds. 2ft. 9in. SUBTRACTION OF VULGAR FRACTIONS. We have seen that fractions cannot be added until they arc reduced to the same denomination and to a common denominator. The same is necessary before they can be subtracted. Subtraction of Vulgar Fractions teaches how to take a less fraction from a greater, CASE I. When the fractions are of the same denomination, and have a common denominator. RULE. Subtract the less numerator from the greater and place their difference over the common deno- minator. EXAMPLES. 1. What is the difference between Here 7 — 4= 3, hence, |=the difference. 2. Subtract^ from f*. 3. From m take iff. 4. From i\lf take it*|. 5. From J^J^i take ^YuV- iandf] Ans. |. .A Ans. Ans. -^,\. Ans. f^li. Ans. VsV- QuBSTioris — Con one*third oi an hour be subtracted from two. thirds of a day without reduction? Can one-fuurth of a dav ba iubtraoted from one-iizth of a day! Before subtracting fraotiona what reduction! are necessary t What does subtraction of frito- lions teachl 1.66 VULGAR FRACTIONS. Vr !|»M Hit CASE II. When fractions are of the same denomination, hut have different denominators. RULE. Reduce mixed numbers to improper fractions, compound fractions to simple ones, and all the fractions to a common denominator; then subtract ns in case I. EXAMPLES. 1. What is the difference betvveen f and 1 1 Ans. -/j. i=z^^ and |=i^ J therefore ij-^ — ^^^ = 2^ difference, 2. What is the difference between -^V and ^}1 Ans.-ja_. 3. 4. From M^take^of 19 5. 6. 7. From fl take ii^. From li take f ^. From I take '| of |. Ans. l-j\ . Remainder 0. Rem. 13-2-. 1 V Ans. |. What is the difference between | of f of 20, and | of|. Ans. ^1 . 10. From 37fi, take 3 ^ of i. Ans. 36-jVv. CASE III. When the fractions are of different denominations, RULE. Reduce the fractions to the same denomination ; then to a common denominator, and add as in case I. QucsTioNB. — When fraciions are of the same denomination, and have a common denominator, Itow do we lublract them? When they have different denominators what must be done? — When the fractions are of different denominations what is th* TMle? VULGAR FRACTIONS. 167 f ^ofaJe=f^- EXAMPLES. 1. What is the difference between ^ of a J6 and ^ of a shilling] Ans. 9s. 8d. ^ of a shilUng=^ o^ ■/o = «V of a j6. -^\-=f:l of a je=9s. 8d. difference. 2. From f of an oz. take | of a pvvt. Ans. llpvvt. 3gr:?. 3. Subtract -i\- of a lb. from ^ of a cwt. Ans. Iqr. 271bs. 6oz. 10-,Vdr. 4f, From 3| weeks take -^ of a day, and ^ of § of ^ of an hour. Ans. 3w. 4da. 12hr. 19m. 17,'sec. 5. From 1| of a lb. troy wt. take } of an ounce. Ans. lib. 8oz. IGpwt. 16gr. 6. What is the difference betwoen -j^.; of a hogshead and -^%- of a quart ? Ans. IGgals. 2qts. Ipt. ^fgills. 7. What is the difference between ^ of -,3^- of S^Ibs. Troy wt. and-^^ of ^ of 3,}lbs1 Ans. 3oz. 13pwt. H<'|grs!. 8. From the sum of f of 7 miles, '^ of J of 2^ miles, and f of -,2,7 of 3^ miles, take the sum of ^ of f of 1| miles, I of-iV of 2^ miles, and l^ of ^ of 2^ miles. " Result, 2m. 2fur. 21 J ^rods. MULTIPLICATION OF VULGAR FRAC- . TIONS. If 1 apple cost ^ of a penny what will 2 apples cost ? 3 apples 1 5 apples i 7 apples ? 8 apples ? 9 apples ? Multiply the fraction j*- by 4. -iVX4=;f=| = li Ans. Or by dividing the denominator by 4 we have -^X4= 4)i2"~3~'^»^"^' Multiplying a fraction by a whole number is increas- ing the value o( ihe fraction as many times as there are units in the multipher. This we have Reen in the above hll' •1. 168 VULGAR FRACTIONS. example, may be done, by either multiplying the nume^ rotor y or dividing the denominator. Thus f X8=V=3. ^ Or 3=?=3. 8)8 1 Hence the following general principle : If the denominator remains unchanged, multiplying the numerator of a fraction by any nurnber is multiply- ing the fraction by that number. Or, If the numerator remains unchanged, dividing the denominator by any number, is multiplying the frac- tion by that number. For, the less the denominator the greater is the size of the parts into which a unit is divided, as ^ is more than \ ; and the greater the numerator, the greater the numlter of parts expressed by the fraction, as | is greater than J. Hence, CASE I. To multiply a fraction by a whole number, we have the following RULE. ^>T:T Multiply the numerator, or divide the denomi- nator by the whole number. EXAMPLES. 1. Multiply -3/4- by 12. Ans.S-,',-. Here, JJ^^^st q^ _37X12=444=3V«. 12)Ui 12~" '* 144 144 2. Multiply ^^ by 7. Ans. 6^ QuERTioNs.-— What it multiplyinf; a fraction by a whole num* berT Repeat the principle etated above- VULGAR FRACTIONS. 169 3. Multiply -Vt^ by 9. 4. Multiph' i^ by 5. 5. Multiply f f f by 49. Ans. J-H^. Ans. 42 1 . Ans. 124-j^t. CASE II. To multiply one fraction by anotber. You have already learned that multiplying by a fraction is taking a part of the multiplicand as many times as there are like parts of a unit in the multiplier. For example to multiply 8 by | is to take ^ of 8 which is 6. Hence, when the multiplier is less than 1, we do not take the whole of the multiplicand, but only such a part of it as the fraction is of unity. Thus, if the multiplier is one half of unity, the product will be one half of the multipli- cand ; if the multiplier be { of unity, the product will be \ of the multiplicand. Hence, to multiply by a proper fraction does not imply increase, as in multiplication of whole numbers. For example. Multiply | by 5. Here % is to be taken ^ times, that is | is to be multi- plied by 2 and the product divided by 3. This residt is ob- tained by multiplying the numerator by the numerator and the denominator by the denominator. For the numerator 3X2=6, and the denominator 4X3 = 12, thus, -,".-, and as twelfths are three limes less Uian fourths, it follows that the fraction has been divided by 3 as well as multiplied by 2. Hence we have the foUowins hole nam* Qui'.sTioNK. — "Why (Joes dividing the denominator increase the value of a fraolioni Why doc» multiplying the numerator in- crease the finction? How do we multiply a fraalion by a whole number? What is multiplying by a fraction? When the niul. tiplier is less than 1 what part of the multiplicand do we talce? If the multiplier is ^ what pari of the multiplicand will the product be? If it is {1 Does multiplying by a proper fraction imply increase? Does multiplying the denominator increase or diminish the value of the Aaction? Why? in VFLGAR PRAGTIONS. RULE. Reduce mixed numbers to improper fractions, and compound fractions lo simple ones ; then multiply the numerators together for a numerator. and the denominators together for a denominator. EXAMPLES. 1. Multiply i by f. Here 1 1 X3= 3 5X7~35 35. Ans. -^^. It will be seen that -^^g- is 2. Multiply 5^ by i. only \^ of j^, and f multipli- ed by ^ ih iiOt the whole of one fifth, but only 3 7ths of it. 3. Multiply ^ of §, by | of | of |. Ans. -^4- This may be somewhat shortened by cancelling ; thus, of§' = iXi, ofi=fof^ = ^^=l-. Ans. 4. Multif, V i of ^ of if, by f of 5}. Result, f f J. 5. , Required the product of 6 by *| of 5. Ans, 20. 6. ' What is the product off of f by | of 3?^? *: Ans. If. Required the product of 7f , 2j, 3^, and f of J^? Ans. 39. What is the product of 5, %, f off and 4^? Ans. 2-/f. Required the product of 4^, | off and 18 f^? Ans. 9- Requircd the product of 14, f, f of 9 and 6f ? Ans. 540. What is the continued product of 6^, 2'|, f of '13 and A Ans. -,V. 24. "' — 7. 8, 9. 10. 11. H ( 4 v 9*- DIVISION OF VULGAR FRACTIONS. Suppose there are -^ of apples which we wish to divido c qually among 4 children ; we should take the parts ex- QuE6TioN8. — ReppBt the rule for multiplying one fractitin by another. How many ways are there of dividing a fraction by a whole number? What are they? Is there any differense in the renulta? VULGAB FRACTIONS. 171 pressed by the numej^tor 16, and divide them again into four portions, this would be dividiag the numerator by i. If we wished to take one half the parts we should take one half of 16, this would be dividing the numerator by 2, and the fraction would be f *. Again, If we have | of an apple and wish to divide it among i children, we must divide the parts again, in o ler to sham it equally; let each fourth be divided into 3 equal pait', each part will be -i\- of an apple and each one's dharo of the whole will be -(^.j of an apple. From these examples it appears that there are two ways of dividing a fraction by a whole number, viz. To divide the numerator, or, if this cannot conveniently be done, To multiply the denominator. £)2 = 1. Q.^ =_?=! Here it is plain that tv. 6 6 ^6X3 18 6 whether the numera oris divided by 3, or the denominator multiplied by 3, the re- ■ult is the same* From what has been shown we have the following gen- eral principle, viz. // the denominator remain unchanged, dividing ihe numerator by any numbery is dividing the fraction by thai number. Or, // the numerator remain unchanged, multiplying the denominator by any number, is (Uviding the fraction by that number. CASE I. To divide a fraction by a whole number. RULE. Divide the numerator, or multiply the denomi- nator, by the whole number. QuEBTioNc. — How does nuilliplying the denotninator divida or diminiBli the value of the fraction? Which is the greater^ thre«.foDrths or three-twelfths? What gen(?ral principle is •tated abovel Repeat the rule for dividing a fraction by a whoU number. M I 172 1. 2. 3. 4. 5. 6. 7. VULGAR FRACTIONS. Divide % EXAMPLES. ^ * % Ans. |. Aas. '|. Divide ff by 2, by 7 and by 14. Ans. -,\-, -,\-, -,%-. by 2, Divide ^ by 2, Divide -If by 9. Divide ^^^ by 15. Divide |||| by 19. Divide -^Ye^f by 15, Ans. •"-»»»• -i 8 5 19 ■"■"S ^ 7 5 b Arm —3 7. 9_ XXIIS. 19 8 •ff^ CASE II. To divide one fraction by anotiier. EXAMPLES. 1. Divide 7,^y by 1st Operation, 4=4X1 5 5 4 =-.8- - _1 80~'2 A 5' '8X5=40 If the divisor were 4 the quotient would be -y^,7. But since the divisor is only ^ of 4, the true quotient must be 5 times -/y-j^or the fifth of a num- ber will be contained in the dividend 80 5 times more than the number itself. In this operation we have actually multiplied the numerator of the dividend by 5 and the de- nominator by 4 ; that is, we have inverted the terms of the divisor and multiplied the fractions together, 2nd Operation. ' Since multiplying the denomina- 8 . 4_4) 8_2__1 tor by 4 is the same as dividing 20 ' 5~5W0~4~2 the numerator, and multiplying the ^ numerator by 5 the same as di- viding the denominator, we may, if we please, divide 8 by 4 and 20 by 5. Questions. — How do we divide one fraction by another? Is mulliplying the denominatnr the same as dividing; the niimerao tor? Is multiplying tiie numerator the same as dividing the de- nominator? If then we divide the numerator by the numerator and the denominator by the denommator, or if we invert the di- visor and multiply the fractions together, will the result be the same? VULGAR FRACTIONS. 178 ■I i Hence for division of one fraction by another, we hav3anie clTect as dividing the deno- minoior, 10. It is also evident that if the numerator and deno- minator be both mullijj/ied, or both divided by the same number, the value of the fraction ivill remain the same. EXERCISES IN THE FOUR PRECEDING RULES. U What is the sum of 2G^, 18^, 19], 13^ and J-jLii Ans. 93 ^\-. 2. Bought ^ of 3^ of r)c\vt. of sugar at one time ; al QL'EsTi3Ns.~Whul is a fruciioii? Whul does the denomina- tor sliuwT What does the nuincnitor show? To wiiut is the vulue of every fruction cquiil? When tlie nuuierutor is leas than the drnomlnator what iu the value uf llic fraclion? When the numerator ih equal to tlip dcnoniinator? When the numerator \» gr<;ater than tiie denominulor? llepeat the 7lh propoiilion. VULGAR PKACTIONS. 175 L2.1 H at iii)a< the Ihan tlie lion. another, \ of 5^ of 6c\vt. ; at another, ^ of y of 8c\vt, ; how much did I buy ? Ans. 20 ff^. 3. What is tlie value of f of a ton, and -'V-of a cwt. T Ans. 12cwt. Iqr. 81bs. 12-,8„-oz. 4». Bought 3 pieces of chrth ; the first contained \ of 3- of f of f yards ; the second ^ of f of 5 j and the third f of f of 8 j what did they all contain? Ans. 2yd^. 2qrs. 1^ na. 1. From I of an ounce take 1^ of a pwt. Ans. 6pwt. 15gr. 2. Take | of a day and I of '| of % of an hour from 3'| weeks. Ans. 3wk. 4da. 12hr. 19m. 17|sec. 3. From 1^ of a j£, take I of a shilling. Ar:>. Xl 9s. 3d. 4. One man bought 1 of 4i cwt. of iron, another 1 of 9 4^ 9 5i^ cwt. ; how much did one buy more than the other? H . Ans. 3 ['^ ^^^ drachms. 1. Multiply j; of a bushel by '* of 7. Ans. '3\\ bu. 2. If I own Y of «' f^'iipj ""<1 ^^^^ 3 ^^"i ^^ "'.y share, what part is it of the? whole? Ans. -^^. 3. How many miles are -^^- of 7 miles, multiplied by \-\ of 87- \- ? * Ans. 403 l miles. 4. What will be the cost of 17^ yards of cambric at 2.^ shillings per yard ? An:j. £"2 3s. 9d. 1. Iff of a yard of cloth cort 3s. what id the price per vard? Ans. 5j8. 2. Paid G6()'* pence f(M* marbles at Gd. a piece ; how many did I buy? Ans. 111|. 3. In 8.\ weeks a family consumes 1G5'; lbs. of butter ; how much do they consume a week ? Ant^. 19 /^,^j lbs. 4. If 50 bushels of wheat costiil7-|, what is it per bushel? Ans. 7s. Od. Ijgqra. » I . Ill .— — ^ Repeal the 8ll). The 9lU. Tlie lOlh. : 176 DECIMAL FRACTIONS. DECIMAL FRACTIONS. The division of the unit into tenths, hundredths, thou- sandths, &c. forms a system of numbers called Decimal Fractions, (from the Latin word decern, which signifies fen,) because they increase and decrease in a tenfold pro- portion, in the same manner as whole numbers. * The denominator of a decimal fraction is never written ; the numerator is written with a point prefixed to it, and the denominator is understood to be 1, with as many ci- phers annexed as there are figures in the numerator. — Thus : ,5 tenths is the same as -,\-, and ,75 hundredths is the same as -^^^^^y, and ,316 thousandths is the same as When a whole number and a decimal arc written togo- flier, the decimal point is placed between them. Thus: 24,6 is 24-;V; r),7l is .V,.V; ^^8,364, is 48f/i,-. Decimals decrease in a tenfold proportion, counting from tlie left to the right. Thus : 5 is only one tenth the vali» it would expror^'3 in the place of units, by taking away the (iucimal j)oii)t; and ,05 is only one tenth as much as ,5. So it is plai:) liiat tliey diaiinish in a tenfold proportion as llicy recede from the place of unit:. Ciphers placed ', &c. QuKSTiowft. — Mow lire docimnl rr;ir.li'>ns fnnnedT What is tlit Hwanin^ of (iccem? Why are Ihfy c illed docinialst IJow art deuimal rr.ictiotiFi oxprestspd? Give an exuinplo- When a wliols nuiiilipr and a decimal nre wriUen logoilier, where ia the de^'w mal point placed? Give nn example, Whiit ia ll:e value of &. wriUen aa a decimal compared ^viiii ila value in the unil'i plast? UuMT do deciuiala decreuie? DECIMAL FRACTIONS. 177 \vr art Miolt of 5, DECIMAL NUMERATION TABLE. I 8C ea c o c S o c '>^\ (V =3 J3 4 6 4^ 6 4 6 7 5 4 1 2 7 4 3 4 6 5 4 3 6 4 ti a a a is read 4 tenths. " 64 hundredths. 64 thousandths. 6754 ten thousandths. 1234 hundred thousandths. 7654 milUonths. 43604 ten millionths. Decimal fractions are numerated from the left hand to the right, beginning with the tenths, hundredths, &c., as in the above table. EXERCISES. Write upon the black board or slate. 'Tteer , and ♦hrec- tenths. Eighteen, and seventy-five luind'-r 'liuj. Five, and five thousandths. One, and one millionc!.. Five, and five tenths. Seventy-five, and nine-ten!' -. \,t ADDITION OF DECIMALS. RULE. Write tiic numbers under each other, tenths un- der tenths, hundredths under hundredtiis, &c., tlien add as in whole numbers, setting; the decimals in the sum directly under those in the numbers to bu added. QuuBTinNg. — Do ciphers placed nn the ri^ht Fmnd of decimals affect their viilun? Why nol7 Whnl is ihe vulue of ,5 written as a decimal? Ot ,50? Of ,500? To what is each one equal? f 178 DECIMAL FRACTIONS. EXAMPLES. 1. What is the sum of 37,04, 704.,3 and ,0376 t Operation. 37,04>» In this example we place those of 704,3 the same value under each otherj'tenths ,0376 under tenths, &c., then add as in whole numbers?. Ana. 741,3776 2. Add 4,035, 763,196, 445,3741 and 91,3754 toge- ther. Ans. 1303,9805. 3. Add 72,54-32,071+2,15744-371,4+2,75. Ans. 480,8784. 4. Add ,7509+,0074+,69+,8408+,6109. Ans. 2,9. 5. T'^ 9,999999 add one millionth part of a unit, and the sum will be 10. 6. What is the sum of one tenth, one hundredth, and one thousandth. Ans. ,111.. ' 7. What is the sum of 4, and 6 ten thousandths? Ans. 4,0006. 8. Find the sum of Twcnty-fiv.3 hundredths. Threa hunured and sixtv-fivc tiioui'andlhfc'. Six-tenths and nin« raillionths. ' Ans. 1,215009, SUBTRACTION OF DECIMALS. RULE. Place the numbers according to their value, then subtract as in whole numbers, and point off th« j^ecimals as in Addition. EXAMPLKS. 1. From 837,642 take 579,358. Ans. 25b,284. QuKsTioNs. — Mow d(jc» pvery cipher placed on the left of a dcciiiiiil iifft>ct iti« value/ Why aol Give example?. How are <{eciinHl (rariinnti i.uiMcriitod'' Flepf'ul the nilt> i'ur a ,97708+ 9. Bring £34 16s. 7^d. into a decimal expression. Ans. iS34,8322916+ 10. Reduce 3qr. 2na. to the decimal of a yard. Ans. ,875. 11. Reduce 1 gallon to the decimal of a hogshead. Ans. ,015873. 12. Reduce 7 oz. 19pwt. to the decimal of a lb. Troy. Ans. ,6625. 13. Reduce 3qrs. 211hs. Avoirdupois, to the decimal of a cwt. . Ans. ,9375. 14. Reduce 2 roods 16 perches to the decimal of an Bcrc. Ans. ,6. 15. Reduce 2 feet 6 inches to the decimal of a yani. Ans. j833333+ 16. Reduce 5fur. 16p. to the decimal of a mile. Ans. ,675. 17. Reduce 4^ calendar months to the decimal of a year. , Ans. ,375. 18. Reduce 109 days 12 hours to the decimal of a year. Ans. ,3. 19. Reduce 3qr. 121bs. 5oz. 1,92 dr. to the decimal of a cwt. Ans. ,86. 20. Reduce 3 pecks 6 quarts 1 pint to decimals of a bushel. Ans. ,953125 bu. 21. Reduce Scvvt. 3qr. 161b. to decimals of a ton. Ans. ,2946428+ton. DECIMAL FRACTIONS. 185 CASE III. To reduce a decimal fraction to its value. RULE. I. Multiply the decimal by that number which it takes of the next lower denomination to make one of this higher, and point off so many places for a remainder, to the right hand, as there are places in the given decimal. II. Multiply the remainder * he next inferior denomination, and cut off a re .mder as before, and so on through all the parts of the integer, and the several denominations standing on the left hand make the answer. EXAMPLES. , . 1. What is the value of ,832296 of a pound Sterling ? Operation. We firrjt multir.j/ the decimal by 20, which brings it to shillings, and after cutting off from the right as ma- ny places for decimals as in the given number, we have IGo. and the de^ cimal ,645920 over. This wo re- duce to pence by multiplying by 12, and theji reduce to farthings by mul- tiplying by 4'. ,832296 20 8,16,645920 12 d.7,751040 rar.3,004160 Ans. 16s. 7d. 3rar. 2. What is the value of ,5724 of a £ ? Ans. lis. 5d. 1,5 qr> 3. What is the value of ,85251 of a JC ? Ans. 17s. Od. 2,4qr. 4. What is the value of ,040625 of a pound Sterling ? Ans. 9:|d. .Questions. — To reduce a decimal fraction to its value, what is the first step? What is the second? ^3 L ^T^li^c IMAGE EVALUATION TEST TARGET (MT-3) ^V ^ /^^^ z 1.0 I.I 11.25 IAA128 |2.5 lu iki |2.2 Sf lift ""^ !g IAS 12.0 Photographic Sciences Corporation «■ ^ :1>^ <^ O'^ 33 WIST MAIN STRUT WIISTM.N.Y. MSM (71*)S73-4S03 ..•^ s ^ v\ ^ ►-^ 6'^ 186 RULE OF THREE. 5. What is the value of £1 ,88 ? Ans. £1 17s. 7d.+ 6. What ia the value of je,3375 1 Ans. 6s. 9d. 7. What is the value of ,875cwt.1 Ans. 3qrs. lilbs. 8. Reduce ,67457ibs. Avoirdupois to its proper value. Ans. lOoz. 12,68992dr. 9. What is the value of ,6 17 of a c wt. 1 Ans. 2qr. ISIba. loz. 10,6dr. 10. Find the value of ,76442 of a pound Troy. An}?. 9oz. 3p\vt. llgr?. 1 1 . What is the value of ,875 of a yd.? Ans. 3qr. 2na. 12. What ia the value of ,875 oCa hhd. of wine? Ans». 55gal. Ipt. 13. Find the proper quantity of ,089 ol a mile. Ans.28po. 2yd. 1ft. 11,04 in. 14. Find the proper quantity of ,9075 of an acre. Ans. 3roods. 25,2po. 15. What is the value of ,569 of a year of 365 days? Ans. 207da. 16h. 26m.24sec. 16. What is the value of ,712 of a furlong ? Ans. 28po. 2yds. 1ft. ll,04in. 17. What is the proper quantity of 142465 of a year? Ans. 51,9998725 days. What is the difference between ,82 of a day and 18. ,32 of an hour? Ans. 19h. 21m. 36sec. RULE OF THREE, OR TROPORTION. A correct knowledge of the Rule of Three is of the utmost importance, it being applicable in almost all arithmetical operations. I will now show you how the work may, in particular cases, be considerably shortened. I. When the second or third term is a multiple or an aliquot part of the first ; divide the second or third term by the first. QoRsTioN. — Wlint is the fimt method proposed lor sbortsoiog operslioos in the Rule of Threel RULE OF THREE. 187 lay, liog t. If 6 yards of cloth cost £2 3s. 4(1. what will 36 yards cost ? Opeartion. We state the question an yd. yd. £ s. d. in page 111, then because the 6{ : 3|6{ :: 2 3 4 firs^t term is an aliquot part of 6 the second, vvc divide the sc- cond by the fn>t, and multi- 13 Ans. ply tl 10 third term by the quo- tient 6, and the result is tiie Fame as thoi gh the second and third term had been multiplied, and the product divi- ded by the first term. 2. If £6 buy 24? yards of cloth, how many yards may be bought for jeil? £ £ yd.^. 6\ : 11 ::2|4.| 4X11— 41.yd.s.An8. II. When the first is a multiple of cither the Bccond or third, divide the first by the second or third. 1; If 12 yards of cloth cost JUl8 what will 4 yardi •ost? yd. yd. £ 112| : 4| :: 18 3 Then IS : 3=6 An« S. If 36 yards cost £2 3s. 4d. what will G yards cost? • yd. yd. j6 s. d. 3|6| : 6| :: 2 3 4 JEO 7 2| Ani. III. When the first and either of the other given ternDs have a common measure, divide them by it and use the quotients instead of the given num- bers. • II I - I- - QucMTioKt.— Whtt !• th« Mcond method meniionedt What k the tbirdt 1 1 188 KULE OF THREE, m^ U If 36 yards cost £3 2s, Gd, what will 24 yards cost? yds. yds. £ s. d. 12)36 : 24 :: 3 2 6 3 2 2 Operation.^ 3)6 5 r.- je2 1 8 Ans. 2. If 3 barrels of 'iflour cost 12 dollars what "Wili 16 barrels cost? bbl. bbl. Jg 3) 3 : 16 :: 12 $ 1 4X16=64 Ans. 3. If 25 yards of cloth cost £2 3s. 4d. what will 5 yards cost ? Ans. 8s. 8d. 4. If 12 hats cost 60 dollars, how much will 40 cost ? Ans. 200 dollars. f). If 30 barrels of flour will subsist 100 men for 40 (lays, how long will it subsist 25 1 Ans. 160. 6. If 120 sheep yield 360 lbs. of wool, how many pounds will be obtained from 600? Ans. 1800. 7. If a man travel 210 miles in 6 days, how far will lie travel in 40 days? Ans, 1400 miles^ RULE OF THREE BY ANALYSIS. The solution of questions by analysis consists in finding the ratio of two of the given terms, and multiplying this ratio by the other term. The ratio of two of the terms will generally express the value or cost of a single thing. EXAMPLES. 1 . If 3 barrels of flour cost 24 dollars, what will seven barrels cost. QuBiTioN*.— In what does the solution of questionB by analy* ■isconaial? What wiii the ratio of two of the terms generally express? ,^ ,,_. _. „ .' I J RULE OF THR£E. 189 5 . By dividing the 24 dollars by 3 we get the cost of 1 bbl. For, if 24 dollars will buy 3 barrels, it is plain that \ of it will buy 1 barrel. This, multiplied by 7, gives 56 dollars, the cost of 7 barrels. 2. If a family of ten persons spend 3 bushels of malt in a month, how many bushels will serve them when thero are 30 in the family 1 If 10 persons spend 3 bushels, it is plain that 1 person in the same time, would spend -,^- of 3 bushels, that is -j^^- of a bushel : and 30 persons would spend 30 times as much, that is, ?-g^=9 bushels, Ans. All the que:*tions in Proportion may be solved on general principles as above, without the formality of a statement. 3. If a field will feed 6 cows 91 days, how long will it feed 21 cows. ' Ans. 26 days. 4. If I walk 84 miles in three days, how far should I walk at the same rate in 9? Ans. 252'. 5. If 2 lbs. of sugar cost 25 cents, what will 100 lbs. of coffee cost, if 8 lbs. of sugar are worth 5 lbs. of coffee. Ans. 20 dollan. QUESTIONS INVOLVING FRACTIONS. RULE. State the question according to the directions given in page 111. Then having reduced when necessary, the similar terms to the same denomin- ation, and mixed numbers to improper fractions, invert the first term, and multiply this term thus inverted and the other two continually together, for the answer sought, which will be of the same denomination as the third term. . ^ EXAMPLES. 1. If ^ of a yard cost f of a pound, what will i®,- of an r.n„r.«ii n«-f ? |yj.=:|of I of |=;^=i ell Eng. ell English cost ? QuBtiTioNs. — May all questions in proportion be solved with, out a Btatementt Repeat the rule for operotioni in which there •re fractions. Why is the first term to be inrertedl H ii W i^i t.lJ 'iii 190 RULE OF THREE* Ell. Ell. ^e Ab i : -1^5 :: f First term inverted fX-i*5Xf=-iVo-i^=i: lOs. 3d. If qr. Ana. 2. If f of a yard cost J of a pound, what will 40| yds. come to 1 Ans. Je59 8s. 6 Ad. 3 . If f oz. co3t £\^, what will loz. cost ? Ans. £1 5s. 8. 4. A person having f of a vessel, sells | of his share for £3 12 J what is the whole vessel worth ? Ans. ^6780. 5. A merchant bought 5^ pieces of cloth, each con- taining 24<| yards at 9s. ^d. per yard; what did the whole amount to? Ans. £60 10s. Od. 3fqr. 6. What is the value of ^ of | of f of a pound, at the rate of -,\ of a JS for -f of a pound 1 Ans. £1}. 7. If i of a ship be worth J cf her eargd, valued at JSSOOO ; what is the whole ship and cargo worth ? ^ Ans. Jei0031 14s. ll/fd. 8. If -{-^Ibs. of sugar cost -^^ of a shiUing, what will of a pound cost 1 Let us solve this by analysis. First, divide -jV of a shil- ling by -^ of a lb. and the quotient is -.Vs^* which is the price of one pound ; then,f | of-iV;5-s.=f^-|^|s., the cost of fi of a pound. f ^jjs.=4d. Sflf^iqr. Ans. 9. If 7 pounds of sugar cost | of a dollar, what will V2 pounds cost ? It is plain that if 71b. cost | of a dollar, lib. will cost j- of |=a%- of a dollar, it is also evident, that 12 pounds will coat 12 times -/«- of a dollar=T/8 X I2=f |= If dol. Ans. 10. How many pieces of merchandise, at 20 Js. a pc. must be given for 240 pieces, at 12s. a piece ? Ans. 149 iVf. 11. If 16 men finish a piece of work in 28^ days, how long will it take 12 men to do the same work ? Ans. 37 1^ days. 12. If ^Ib. less by |, costs 13 |d., what costs 141bs. less by i of 2lbs.? Ans. £4> 9s, 9/5-d. 13. A merchant bought a number of bales of velvet, each contaming 129 j-f yards, at the rate of 7 dollars for ^ il COMPOUND PROPORTION. 191 0| yds. 3. 6Ad. 5s. 8. is share . £780. ch con- did the I. 3f qr. 1, at the aluea at ll-A-d. ^hat will >f a shil- ch is the e cost of qr. An?. t will 1-2 ill cost \ .inds will 1)1. Ans. ^8. a pc. 14Q _ij._ how lys, b^days. Its 141bs. t. 9.AJ. If velvet, lare for !? yards, and sold them out at the rate of 11 dollars for*7 yds. and gained 200 dollars by the bargain ; how many bales were there? Ans. 9 bales. First find what he gained in selling 1 yard ; then it will be easily ascertained how many yards he must sell to gain $200. COMPOUND PROPORTION. Compound Proportion, is a method of jierforming such operations as require two or more statings. It is common- ly called Double Rule of Three, because its operations can be performed by two statings in the Rule of Three. RULE. Make that number which is of the same kind with the required answer the third term. Then with this third term and each pair of similar terms com- plete a stating in the single rule as already taught; then, having reduced the similar terms to like de- nominations,^and the third to its lowest given de- nommation, multiply the numbers in the second and third places continually together for a divi- dend, and those in the first for a divisor ; divide the former by the latter, and the quotient will be the required answer, of the same denomination as the third term. Note. — The principles of contraction given in single proportion, are equally applicable in this rule. EXAMPLES. 1. If 225 bushels of oats be eaten in 30 days, by 20 horses, how many bushels will suffice for 50 horses 16 days? I' I .•/. Q Questions.— What ifl compound proportion? Why it it called double rule of three? Repeat the rule for compound proportion. ■ ^^'.^--^.^T-'V I t 192 COMPOUND PROPORTION. Operation, horses. 20 : 50 ^ bushels. days. I : : 225 30 : 16 J 6|00 800 800 6)180q00 Ans. 300 In this example it is plain that the term similar to the number sought is 225 bushels ; which is therefore set in the third place. Then putting the question on the number of horses, the fourth term must be more thnn the third, for it is evident that 50 horses will require more in a given time than 20 horses, hence 50 must be the second, and 20 the first term. Again, putting the question on the number of days, since a given number of horses will eat less in 16 days than in 30 ; the fourth term here is less than the third ; whence 16 is the second term and 30 the first. 2. If 20 horses in 30 days eat 225 bushels of oats, in how many days will 50 horses eat 300 bushels?'? Operation, horses. 5|0| : 2|0| o bushels. 2|215| : 3|0|01 3i 4 i^ We fir.it stat'3 this question as the rule dircct^\ Then be- cause 10 is an aliquot part of 50 and 20, we divide them by 10 and cancel them. For the same reason we divide 225 and 300 by 75, and canctil these numbers. We then di- vide the third term, 30 by 5, and cancel it, and the 6 by 3, and cancel the 6 and 3. Then — as the numbers are all canc^l- Ans. 16da. led in the first tf^rm, we multi- ply those in the second aiul third togather, and the product is the answer. 3. If a family of 8 persons expend 360 dollars in 9 months, how much will serve a family of 18 persons 12 months] Ans. $1080. 4. If 3 men in 4 days eat 151bs. of bread, how much will suffice 5 men for 12 days! Ans. 75lb&. 5. If 20cwt. be carried 50 miles for $15, how much days. :3,0 6| 2 4. 8 2 linlhat [number ^hich is 1 place. 1 on the J fourth he third, ) horses ^en time 50 must the first • of days, 16 days he third ; [)f oats, in I s question Then he- lot part of le them by For the Wide 22r) nd canctil e then di- , 30 by 5, liie 6 by 3, Id 3. Then all cancel- L we multi- iccond and lollars in 9 1 persons VZ Ins. $1080. Ihovv much .xns. 75lbft. how much CbMPOUND PROPORTION. 193 will 45cwt. cost to be conveyed 80 miles? Ans. ^Si. 6. If 12 men in 6 days mow 80 acres, in how many days will 25 men mow 250 acres ? Aris. 9 days. 7. If 5 men make 300 pairs of shoes in 40 days, how many men will make 900 pairs in 60 days ? Ans. 10 men. 8. If 144< men can build. a wall 32 feet high in 8 days, 4 in how many days can 63 men build a wall 28 feet high, of the same length? Ans. 16 days. 9. If a footman, when the days are 14 hours long, can travel 276 miles in 16 days ; in how many days can he •- travel 828 miles, when the days are but 12 hours long? Ans. 56 days. 10. If the wages of 6 men for 14 days be 84 dollars, what will be the wages of 9 men for 1 1 days ? Let us work this question by analysis. First, $84-f-14 =6 dollars, what 6 men earn in 1 day; then $6-7-6 men = 1 dollar, what 1 man earns in 1 day. Then $1X9X^1 = 99 dollars, what 9 men will earn in 11 days. ;,^ Ans. $99. 11. If 56lbs of bread be sufficient for 7 men 14 days, how much bread will serve 2 1 men 3 days ? If 7 men consume 561bb. of bread, one man i,. ihe same time would consume | of 561bs.= 'V'lbs. ; and if he con- sumes ^7^ lbs. in 14 days, he would consume -,\- of s^'' = I ^ lbs. in 1 day. 21 men would consume 21 times as much as one man ; that is 21 times ^ ^ = J-i^^ "lbs. in 1 day, and in 3 days they would consume 3 times as much ; that is. .3 5iJs.=:36lbs. Ans. ^ Or place the numbers that occupy the third and second places above a line, and the first terms below it, and can- cel them, thus, 4 3 •'!l.5i>-'*:J»TK*-'i!;fllSi.*ejijm - ■r?r 194, COMPOUND PROPORTION. A# Slid £^ I 18=:-fl7=what 1 student spends in 1 day. then -1-^X20=^^1^^ =whlit 20 students Spend in 1 day, and je|-ffQ-x30±=:what 20 students spend in 30 days =:X39 ISs. 4|i. Ans. 13. If 3 men reteive £S-^\ for 19i days' work, hwr muoh must 20 men receiv^ for 100^ dayst 89X21 X2|0| X i01=35689^^^305 p,, g . ^na 1)0|X3X 4| 117 2| ^•^- 14. If a barrel of beer last 7 persons 12 days, how much will be drank by 42 persons in a year ? Ans. 182 bar. 18gal. 15. If 32 men build a wall 36 feet long, 8 feet^high, and 4 feet wide, in 4 days, in what time will 48 men build a wall 864 feet long, 6 feet high and 3 feet wide 1 . Ans. 36 days. 16. Supposing 3 compositors to set 15^ pages in 2| hours, how many will be required to set 69 J pages in 6^ hours ? Ans. 6. 17. If 12 oxen eat 7f acres of pasture in 5^ weeks, how many oxen will be required to eat 13| acres in 10 j^ weeks? Ans. 11 oxen. 18. If 16 compositors set 150 pages of types, each page containing 48 lines, and each line 50 letters, in 3 days of 10 hours each j how many compositors will be re- quired to set 500 pages of 72 lines each, and 45 letters in a line, in 6 days of 8 hours each ? Ans. 45. SIMPLE INTEREST BY DECIMALS. In calculating interest, the rate per cent is a certain number of hundredths of the sum lent. Thus, if 1 per cent, is paid for XlOO, it is y^-jj-=,01 part of the sum lent. If 6 per cent, is paid, it is the ^§-=,06 part of the sum lent. QuRBTiuNs.— In calculating interest, how h the rate peroent. to the sum lentt If 1 per cent ii paidj what part of £100 ia iti SIMPLE INTEREST BT DECIMALS. 195 1 day. dinl Odays Ana. s, how . 18gal. eihigh, en build I J6 dayfl. s in 2\ s in 64 Ans. 6. weeks, inlOV^ I oxen, each in 3 II be Fe- tters in ns. 45. .S. certain ff 1 per im lent, the sum Iper oenl. 100 is i(« .^ For this reason all calculations in interest are properly sums in decimal multiplication. Thp rate per cent, may always be written as a decimal fraction of the order of hundredths. Thus, 1 per cent, may be written ,01 ; 2 per cent. ,02; 3 per cent. 0,3; At per cent^04« ; 5per cent. ,05 ; 6 per cent. ,06, and so on. RULE. Reduce the shillings and pence to the decimal of a pound — see page 183. Then find the interest by the rule on page 104 : after which reduce the decimal part of the answer to shillings and pence, — see page 185. EXAMPLES. I, What is the interest at 6 per cent of ;e27 15s. 9d. for 2 years ? Operation. £211 15s. 9J.=£27,7857 6 1,667250 2 je3,334500 20 8.6,690000 12 d.8,280000 4 ,qr.l,120000 We first find the interest for one year ; then multiply by 2, which gives the in- terest for two years. We then reduce to pounds, shil- lings and pence* Ans. £3 6s. 840.4- J^ QuESTioNB.— irC p«r cent, is paid, what part of £100 is it? What are all calculutiona in interesiT How may the rate per tent, alwaya be wrilteni Repeat the rule. 196 SIMPLE INTEREST BY DECIMALS. ( ■i <\ II '■\ 2. What is the interest of Jei21 8s. 6d. for 4i years, at 6 per cent, per annum 1 Ans. ^632 15s. 8d. l,36qr. 3. What is tlie interest on £61 19s. 6ci. at 6 per cent, for 3 years, 8 months, 16 days ? Ans. JG15 2s. 8^d. 4. What is the interest on £127 15s. 4d. at 6 per cent, for 3 years and 3 months ? Ans. £24i I8s. 3id.+ 5. What is the interest of £107 16s. lOd. at 6 per cent, for 3 years, 6 months and 6 days? Ans. ^ 6. What will £219 13s. 8d. amount to in 3 years and , a half, at 5\ per cent, per annum? Ans. £331 Is. 6d.-|- .7. What is the interest of £514 10s. 2d. for 3 years ap.d a half, at 4 per cent. ? Ans. £72 Os. 7|d.+ 8. What is the interest on £255 10s. 8d. at 6 per cent, per annum, for 6 years and 6 months ? Ans. £99 13s. l|d. 9. What is the interest on £53 18s. 5d. at 6 per cent. for 7 years and 12 days? Ans. £22 15s. ld.-|- Find the interc-t on the following note. £127 lOs. Toronto, Jan. 1st, 1845. For value received, I promise to pny on the 10th day of June next, to T. Lawrence or order, the sum of one hun- dred and twenty-seven pounds and ten shillings, with in- terest from da;e, at 7 per cent. S. S. SMITH. Ans. £3 19s. 24d. To find what remains due at the end of a given time, on jiotcs, bonds, mortgages, &c. when payments are made at different times. There are various methods of casting interest when se- veral payments arc made, no two of which produce ex- '^ actly the same result. The following is as simple and com- preliensive as any with wliich I am acquainted. RULE. Multiply the principal by the time it bears in- terest, before any part of the debt is discharged, Repeal the rule for casting interest ou notes &u. when partial payments are made at different times. SIMPLE IlfTBBEST BT DECIMALS. lay ftfitl from the given principal deduct the first pay- ment ; multiplying the remainder by the time be- tween the first and second payments ; from the re- mainder deduct the second payment, and multiply the remainder by the time between the second and third payments, and so proceed through all the payments. Then add all the products together, and find the interest on tl;c sum for 1 vear, 1 month, or 1 day, according as the times of payment are, vears, months, or davs, and this interest added to the last remainder will be the sum due at the end of the given time. EXAMPLES. jeeOO Kingston, May 13th, 1845, One year from date, for value received, I promise to pay Wm. Howland or bearer six hundred pounds currency, with interest at 5 per cent, per annum. H. Good., Now the maker o^ this note, H. Good, pays the 9th of JPuly ;£200, and the 17t:. of September JE 100 ; how much principal and interest is he to pay at the end of the year ? Operation. J&600X57 =34200 200 400X70 =28000 150 250X238 =59500 1^1700 5 365)6085,00 jei6 13s. 5 je250 16 a-d. 13 5-7 266 13 5-/ff Ans. In this example we first multi-i ply the principal by the time that expires before the first payment is made, which is 57 days; then deduct the payment dG200 from the principal, and multiply the remainder by the time between t!ie first and second payment, which is 70 days, then deduct the second payment J6150 and multiply the remainder by the time between the second pay- ment and the time of settle- ment, which is 238 days. — Then add the sex'eral products together and find the intecfti b2 198 SIMPLE INTEREST BY DECIMALS. If 'I I » on the Bum for 1 day at 5 i)er cent., by the rules previously given. The interest is j6 16 13s. 5r/jd., which being added to j£250, the part of the principal rennaining unpaid, will give the sum yet due, which is £266 13s. S/jd. Ans. By a little reflection this method is evident, because the several principals multiplied by their respective times, w^ill produce sums which will bring in as much interest in 1 month or 1 day as the several principals would in their les- ' pective times. jeSOO. ' ' '^ Toronto, June 11th, 1845. ' Due J. Robson or bearer, for value received, three hundred pounds cun'cncy. J. Thomas. Three montlis after date £60 was paid and endorsed on this note ; four months after that J6100, and five months jjfter that JG75 ; how much is due on the note at the end i.f 18 months? Ans. X79 15. A merchant borrows £2d0 for 2 years at 8 per cent, and Mgrees to pay as fast as he can ; now at the expiration of 9 months he paid JG80, and 6 months afterthatJG70, leaving liie remainder the full term of two yeai*s. How much ])nncipal and interest has the merchant then to pay? Ans. £127 IGs. A gives to B. on interest on first of November, ISM-, X*6000 at 4'.^ per cent. B. is to ))ay iiim at the expiration of 2 years, having liberty to pay belbrc that lime as much of the principal as he jileases. Now B. pays. The 16lh Dec. 1841, Tiie 11th of March, IS If), The 30th ditto, The 17th August, • The 12th of February, 1846, X900 1260 600 800 1048 Il SIMPLE INTEREST BY DECIMALS. 190 and If in 1 yr. 4. mo. the interest and principal on a sum at 6 per cent, amount to £61,02, what is the principal 1 We first find what will be the amount of a £, with ita interest, for the given time. This amounts to JG1,08 now, as every £ in the original sum gained ,08 of a £ interest, there were as many pounds as there are j£ 1,08 in £61,02. Ans. ^56 lOs. RULE. Find the amount of £l for the given time, and divide the sum given by this amount. EXAMPLES. What piincipal at S per cent, will amount to JG85,12 in 1 year G months? Ans. jG76. ■ II. To find the principal, when the time, rale and in- tere?-t are known. What sum at mteicj^t at 6 per cent, will gain JE10,5 in lyr. 4m o.? One pound put at interest for that time, would gain JG,08, and therefore it requires as many pounds ns there are JG,OS in jei0,5. * Ans. jei3J 5s. RULE. Find interest of £1 for the given rate and time. Divide the given interest by this and the quotient is the princii)al. EXAMPLKS. 1. A man paiil JC4,52 interest at 'he rate of 6 ])crccnt. at the end of 1 year 4 months ; what was the principal. Ans. £56 lOs. 2. A man received X20 for interest on a certain noto at the end of 1 year at tho rate of 6 per cent. ; what waa the principal? Ans. je333,3333. QuEiTioN. — IIow do wc flild thfl principal, when the time, rale, t'ld amount ore known 1 How do we find the principal when the time, rale, and interest are kiioWDl 200 SIMPLB INTEREST BY DECIMALS. III. To find the rate vvheatbe pnncipa], interest and time are known. If jC3,'^S is paid for the me oC£b4t, I year and 6 mo. what is the rate per cent. ? If this sum were at interest at 1 per cent, it is plain that it would ])roduce j€,54>. As many times, therefore, as J6,5^ is contained in j£3,78, so much more than one per cent, is the rate. RULE. Find the interest on the given sum at 1 per cent, for. the given time, by which divide tiie given in- terest : tlie quotient will be the rate at which inter- est was paid. EXAMPLES. 1. If JG2,34 i^ paid for tlic use of jS4i58 for 1 month, what is the rate per ccr>t.] Ans. 6 percent. 2. At £4>6 ICs. for tlie U3C of Je520 for 2 years, what is it per cent.? , ^ Ans. 4 A per cent. IV. Tiie principal, rate per cent, and interest being given to find the time. What is the tinio required to gain jG3,78 on JG36, at 7 percent.? ,» The ir.tjrc'oton jG36, 1 year at 7 per cent, is £?,52 ; iicnce jG3,78 y iJ2,02=: 1,5 years, the time required. UUL!]. Find the interest for 1 year on the prineipal giv- en, at ihc given rale, by which divido the given interest; tlio qnoticnt will be the time required in years and decimal p;:rts of a year. ]:XAMPL]JS. 1 . Paid £20 for the use of JGGOO at 8 per cent. ; wimt was the time? Ans. 5 mo. nearly. 2. Paid Je28.24'2 for the use of X217 i)a. at 4 per cent. what was the time. Ans. 3 yrs. 3 mo. QuKbTioNH. — How do wti liiiil lim rule, wttcii the (iniicipul iii> tcml and time arc known? Flow \% the time found, when tht principal, rale per cent, and intercit are knuwnl EXCHANGE. EXCHANGE. SOI Exchange is a rule by which the money of one state or •ountry is reduced to that of another. Par is equality in value ; but the course of exchange ii frequently above or below par. Agio, is a term used to signify the dilVereP-cCj in somo countries, between banli and current money. ' Whenever, in the commercial transactions between dis- tant places, the debts and credits are nearly eijual, bills of exchange may be conisidered as negotiable according to the intrinsic value of the nominal amount ; exchanges are then eaid to be at par. But when the debts become greater in amount than the credits, exchanges must rise or sell above par. %^ CURRENCY TABLE. British Sovereign, *. Pound Sterling, United States Eagle, coined before 1834, do. do. do. between 18IM } and 1841, ^ United States and Mexican Dollar, Half Dollars of the above, Quarter do. do. •'^ Eighth do. do. , Six-])ence do. French five frank pieces, British Crown, British half do. . ' British Shilling, British Six- pence, iL •._ QvK»Tio.Ns.— What 18 Exchange t Agio? 1 1 o 8. d. 4 4 4 4 13 4 2 10 5 1 2 6i 1 3 7i ^ 4 8 6 1 3 Oi 1 2^ 7-2*- What i» ParT What i« i ■\ M' 202 EXCHANGE. CANADIAN CURRENCY. To change Canadian Currency to Federal Money, or Federal Money to Canadian Currency. 1 dollar in this currency is 5s.=60d., and 60d.=:100 cents Federal money, or 3d. =5 cents ; hence cents may be changed to pence, or pence to cents by the Rule of Three. , EXAMPLES. ^' 1. In 40 cents how many pence? As 5 cents *, 3d. : : 40 cents : 24d. Anaw 2. In 95 cents how many pence ? Ans. 57d. 3. In 18d. how many cents? Ans. 30 cents. 4. In 54d. how many cents ? Ans. 90 cents. 5. Reduce 9s. 3cl. to Federal money. Ans. $1,85. 6. Reduce $2,50 to currency. An«. 12s. 6d. 7. Reduce £25 10^^ Cd. to Federal money. 1 pound — 4 iloHnrs ; a dollar tlierdbrc is \ of a pound ; hence there will he r.rs many dollars as there arc quarters in the sum. Therefore, Reduce ihe ^um ioihc decimal of a j}ound and multiply it by 4. ^ .. ( £25 10s. Gd.=:.je25,525. • .* . uperaiion. ^ 25,525x4=:$102,10 Ans. Chanac JCG9 15s. 5d. to Federal money. 8. 9. Reduce JGG lis. G^d. to Federal money Ans. $297,083+ Ans. $27,304. 10. Reduce £95 19s. llUl. to Federal money 11. Reduce $102,85 to pounds. Ans. $383,9875. O|».'ration. r 4)102,85 25,7125 An«. £25 14s. 3d. Here we divide by 4, because 1 dollar is \ of a pound, and the quo- tient is pounds and decimals of a po u nd. V^Uh^l'l<>.N!•. — l:i 1)0.1. iioM' niauv ueuiut In «5ii. iiuw inaiiyt How do we reduce pence to ceittsl Oenlfl Id pfnce? Now do we chaiiKe puundR, ithillingB and ppnce to Federal monvyt How Federal money to pounds, shillings, Hlq. 'M * EXCHANGE. 203 ley, or .=:100 its may Rule of d. AnSk IS. 57d. cents?. cents. 12ti. 6d. pountl ; quarters mliiply .525. llO Ans. f,083+ 127,304. 13,9875. [cause 1 I he quo- lu of a iiaiiyl [How do rt How "»■■ 12. Reduce $500,80 to Currency. Ans. £125 4s. 13. Change $118,25 to Currency. Ans. je29 lis. 3d. 14. Change $250 to Currency; Ans. M2 10s. .-A 4. a. To change Sterling money to Currency, or Currency money to Sterhng — work by the Rule of Three. ■^ EXAMPLES. 1. Reduce j£150 10s. English money, to Canadian Currency. As jei : jei 4 4 : : £150 10 to £183 2 2 Ans. 2. Reduce £183 2s. 2d. Currency to Sterling money. As £1 4 4 : £1 : : £183 2 2 to £l50 lOs. Ans. The pupil will perceive that these two e^Camples recip- rocally prove each other, and as£l Sterling is equal to £1 48. 4d. Currency, the operation is evident. 3. Reduce £500 Sterling to Currency. Ans. £608 6s. 8d. Reduce £125 Currency to Sterling nrioney. Ans. £102 143. 9i Reduce £250 10s. Sterling money to Canadian Currency. - Ans. £304 15s. 6d. 6. A person in Enp;land deposits the sum of £50 Ster- ling in the Bank of British North America, London, and receives a draft on the branch of the Bank at Toronto, Ca- nada, for the amount, which he sends to his correspondent in Canada. How much must the latter receive in Cana- dian Currency, exchange being at par — how much, if at 5 pr cent, premium ? * J £60 16s. 8d. at par. -^"^* I £63 17s. 6d. at 5 per cent. 7. A person in Kingston being desirous to remit to London £60 Canadian Currency — for what amount Brit- ish must the bill be, exchange being at par. Ans. £49 6s. S^d. QuKrnuNs. — Huw do we reduce Sterling money to Currenoyf How Currency to Sterling? END OF PART IH. , ^-r i ■ f H » r [ . rJ^^ ■ .et ( ..■t^i:i^* a'f .' ^=- i / . ' 1 ! > ! V. r *s * ,. . , * 'Jit'':-'- •■' ,( , f. n ,' I '. < . < 1i* t.t i ,^, .,^-, "■im-.. ! '5 1 .. , «'?♦"■! /• V ?ir T^' .''C| ^5l Jy .-V ' ■'^■. m ^v PART IV. ^^ ^ INVOLUTION. When a number is multiplied into itself, it is said to be involved, and tlie process is called involution. Hence, Involution teaches the method of finding the powers of number's. T\\Q product which is obtained by multiplying a num- ber into itself,, is called a Power, Thus, when 2 is multiplied into itself o;ice, it is 4-, and this is called the second power of 2. If it is multiplied in- to itself twice, (2X2X2=8) the answer is 8, and this is called the third power. The number which is involved is called the root, or first povjer. Thus, 2 is the root of its second power 4, and al- so the root of its third power 8. A power is named or numbered accov^Xmg to the number of times its root is used as ^ factor. Thus the number 4 is called the second ])owev of its root 2, because the root is twice used as a factor; thus 2X2 =z4f. The number 8 is called the third power of its root 2, because the root is used three times as a factor, thus 2X2 X2=8. The method of expressing a power is by writing its root and then placing a small figure above it, to show" the num- ber of times that the root is used as a factor. Thus the sec- ond power of 2 is 4, but instead of writing the product 4, wc write it thus 2-. QoESTioNs.— What ia involution? What doeii Involution tcachir What is a power? Whnt is the second power of 27 The third? Whut is the root or first power? Give an example? How is a power named or numbered? Give examples? What is the method of oxpresHin); a power? Express the ■oaond pow- er of 2? Th6 third? Tlie fourth? What is the Index or Ex- ^o enil For what is it used? Repeat an exampio. t 1 w \ I f I 206 INVOLUTION. The fourtli The third power of 2 is written thus 2 * power of 2 is 16, and is written thus 2*. The small figure that indicates the number of times that the root is used as a factor is called the Index, or exponent. The index is used to denote the power to which the root is to be raised, thus 5 * denotes that 5 is to be involved to the fourth power. The second power is called the square. The third pow- er is called the cube, — the fourth power is called the JBi- quadrate, — the fifth power is called the sursolid, and the sixth power is called the square-cubed. . - RULE OF INVOLUTION. Multiply the number continually by itself as many times less 1 as there are units in the expo- nent: the last product will be the power sought. EXAMPLES. 1. What is the cube of 5 ] Ans. 5 X 5 X r)= 125. 2. What is the cube of 7? Ans. 343. 3. What is the fourth power of 4? , Ans. 256. 4. What is the square of 14 ? ' , Ans. 196. 5. What is the 5th power of 2? Ans. 32. ^ 6. What is the 7th power of 2 ? Ans. 128. A fraction it^ involved by involving both numerator and denominator. 7. What is the square of ^ ? 8. 9. 10. 11. 12. 13. What is the cube of § "^ Ans. ^. Ans. rj'r, Ans. 7iV,-. Ans. 30^. Ans. 915-,^,r. Ans. 32768. What is the fourth power of f ? What is the square of 5.^ ? What is the square of 30^ ? , , Perform the involution of S''. Involve "aV, ^^, and -^ to the third power each. Ans. 14. Involve 21 1^ 15. Raise 25 to the fourth power. Ans. 390625. 16. Find the sixth power of 1,2. Ans. 2,985984. Questions.— What is the second power of a number called? The thirdi The fouithl The fifth? The crixUi? Repeat the rule of Involution. How ii a fraction involved? - _6 4 6 » 3 1 U ) Ans.' 9393931. 7 '» y ' EVOLUTION. 'j*-^ 207 EVOLUTION. Evolution is the process oC finding ihc root of any number ; that is of finding that number which multiplied into itself, will produce the given number. The Square Root is a number which being squared will produce the given number. It is expressed either by this sign ^y placed before a number, thus ^yl-, or by a fraction \ placed above a num- ber, thus, i'. The Cube Root or Third Root is a nuni])er which be- ing cubed, or multiplied by ittielf twice^ will produce the a 1 given number. It is expressed thus, ^ ; or thus 27^. All the other roots are expressied in the same manner. 4 The Fourth Root has this sign ^y put l)cforc a number, or else \ placed above it. There are some numbers whot:e roots cannot be precisely obtained, but by moans of deci- mals, we can approximaie to the number \v hlcli is the root. Numbers whose roots can be exactly obtained are callc'^ raiional numbers. Numbers whose precise roots cannot be olitained are called surd numbers. When the root of several numbers united by the sign-f- or — is indicated, a vinculum, or line is drawn from the sign of the root over the numbers. Thus, the square root of 59 — 10 is written ^/59Zl Jo. ' The root of a rational number, is a rational rout, and the root of a surd number, is a surd root. QuKBTio.Ns. — Wlisit is EvoluUon? What is the Square Root? Show how the Sqimre Root is expressed. What is the Cube Ruot? Express the fourth ro.J, (to. Can ihe roots of all num- bers be exactly obtuined? How may we approximate to the number which is the root? What are rational nunibers? When we express the root of several numbers united by the signs + o'* — > '^ow do we indicate that the numbers are all affected by the sign of the roof? Express the Square Root of 48+6 — 4. What is a rational root" -A surd root ? 1 r rr. u y^- >' J il 3t 208 EXTRACTION OF THE SQUARE ROOT. EXTRACTION OF THE SQUARE ROOT. Extracting the square root is finding a number, which multiplied into itself, will produce the given number; or it is finding the length of one side of a certain quantity, when that quantity is placed in an exact square. Thus, s/'ixi=:^16=4>, and ^49=7 for 7X7=49. RULE FOR EXTRACTING THE SCIUARE ROOT. I. Point oft' the given number, into periods of two figures each, beginning at the right hand. II. Find the greatest square in the first left hand period, and subtract it from that period. Place the root of this square in the quotient. To the re- mainder bring down the next period for a dividend. III. Double the root already found and place it on the left for a divisor. Seek how manv times the divisor is contained in the dividend, exclusive of the right hand figure, and place il iigure in the root and also at the right hand of the divisor. IV. Multiply the divisor, thus increased, by the last figure of the root, and subtract the product from the dividend. To the remainder bring down the next period, for a new dividend. But if the product should exceed the dividend, diminish the last figure of the root. V. Double the root already found for a new di- visor, and proceed as before, until all the periods are brought down. EXAMPLES. 1. What is one riilc of a square room, which contains 784 square feet? Questions. — What is the extraction of the Square Root?- ot of number iht The fifth? "Whttt is the first step in extracting the Square Root of numbers? What is the second? What the third? The four Repeat the entire rule? EXTRACTION OP THE SQUARE ROOT. 209 1 contains Operation. We first point off the number T84>(^ as the rule directs, and find the 4 root will consist of two figures, a /c?i and a unit. We then take the highest period 7 (hun- dreds) and^ find how many feet there will bo in the lar- gest square that can be made of this quantity, the sides of 20 feet, which must be of the order of tens. No square larger than 4 (hundreds) can be obtained in 7 (hundreds), the sides of which 20 feet. will be each 20 feet, because 20X20=400. These 20 feet or 2 tens being sides of the square arc placed in the quotient as the first figure of tha root. This Square may be represented by Fig. 1. We now take out the 400 from 700, and 300 square feet remain. . These are added to the next period 84, ma- king 384, which are to be arranged around the square A in such a way as not to destroy its square form ; consequently the additions must be made on two sides. To a^^ccrtain tiie breadth of these additions, the 384 muc^t be divided by the length of the two sides 20X20= 40, and as the root already found is one side, we double this root for a divisor, making 4 tens or 40, for as 40 feet is the length of these sides, there will be as many feet in breadth as there are forties in 384. The quotient arising from the division is 8, which is the breadth of the addition to be made, and which is placed in the quotient, after the ^ tens. s2 * I 310. EXTRACTION OP THE SQUARE ROOT. 20 feet. Fig. 2. 8 feet. O B C 20X8=160 8X8=64 A hO o X 20X20=400 II ^ I-* o^ o 20 feet. 8 feet. 7'84(28 root, 4 48) 384 384 It will be seen by Fig. 2, that to com • plete the square, the Q^ corner C. must be f* filled by a small ^ square, the sides of which are each equal to the vndih of B and to C, that is 8 feet. ^ Adding this 1o the 2. 4 tens, or 40, we find • that the whole length of the addition to be made around the sqr. A, is 48 feet, instead of 40. This multi- plied by its breadth, 8 feet, the quotient figure, gives the cori' tents of the whole addition, 384 feet. — As there is no re- mainder, the work is done, and 28 feet is the side of the given square. It iriay be proved by involution, thus, 28 X 28 =784, or by adding together the several parts of the figure thus, A contains 400 square feet, B « 160 « D « 160 C « 64 a 66 (( (( 784 If in any case, there is a remainder, after the last period is brought down, it may be reduced to a decimal fraction by annexing two ciphers for a new period, and the same Questions. — If there is a remainder after the last period is brought down how may we proceed? If the dividend be too emdl to contain the divisor what roust be done? How many f gures will there always be in the root? ■ - ■^fcft — ^.^^ EXTRACTION OP THE SQUARE ROOT. 211 process continued. If at any time the dividend be to./ small to contain the divisor a cipher must be placed in the root, and another period brouglit down. The pupil will find by trial, that the root always con- tains just half as ma7iy, or one figure more than half as many figures as are in the given quantity. Hence, thg propriety of pointing off into periods of two figures each, and there will always be as many figures in the roots as there are periods. 2. What is the square root of 998001] f 99'80'01)999 root. 1 81 Operation. - 1 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 189)1880 1701 1989)17901 17901 Find the square root ot 676. Find the square root of 625. What is the square root of 487204] « 63840U « 556516] « 488601] « 32761] « 69] « 83] « 299] « 892 ] a ti (6 a t i .^f 220 MENSURATION. 'T 3. The statute bushel contains 2150,4252 cubic or solid inches ; what must be the side of a cubic box that will contain the same quantity 1 Ans. 12,907 inches. 4. A stone of a cubic form contains 474552 solid inch- es ; what is the superficial content of one of its sides 1 Ans. 6084. MENSURATION, Mensuration signifies to measure, hence measuring sur- faces is called mensuration of snperjices ; and measuring of solids is called mensuration of solids. To find the area of a square or of a rectangle. RULE. Multiply the length by the breadth, and the pro- duct will be the area, or superficial content. EXAMPLES. 1. How many acres are contained in a piece of land 160 rods lo"g, and 80 rods wide 1 A rectangle is a four sided figure like a square, in which the sides are perpendicular to each other, but th© adjacent sides are not equal. 1 O 00 160 rods. 160X80=128004-160=80 acres. 2. How many aci*es are there in a lot of land 320 rods long, and 160 rods wide ? Ans. 320 acres. 3. What is the area of a square field of which the sides are each 33,08 chains 1 Ans. 109A. IR. 28P.+ 4. What is the content of a square piece of land of which the sides are 25 rods each? Ans. 3A. 3R. 25P. To find the area of a triangle. RULE. Multiply the bose by the perpendicular height and one half oi\\\e product will be the area. QuKHTiuN.— Whai KiVlenBuraliuit? Huw do we fiitd tlie area of a square or rectangle? How do we find the area of a trianglet MENSURATION. 221 C A triangle is a figure bounded by three straight lines. Thus, A B C is a triangle. A B is the B base, B C the perpendicular, and A C the hypothcnuse. EXAMPLES. 1. The base of a triangle is 40 yards and the perpen- dicular 20 yards ; what is the area? AnS. 400 square yards. 2. In a triangular field the base is 40 chains and the perpendicular 15 chains : how much does it contain? Ans. 30 acres. 3. How many acres are contained in a tr'angle whose base is 320 rods, and perpendicular 40 rods? Ans. 40 acres. To find the area of a triangle having the three sides given. RULE. I. From lialf the sum of the three sides sub- tract each side severally. II. Multiply the half sum and the three re- mainders continually together, and the square root of the product will be the area required. EXAMPLES. 1. Required the area of an oblique triangle, the three sides of which are 13, 14, and 15 rods. 13 21 21 21 21 J the sum. 14 13 14 15 6 15 — -_ _ _ — 8 7 6 126 2) 42 sum. 7 21 half the sum. 882 8 V70-56(84rods. 15 rods. QuBsTioH.— What 18 the rule for finding Uie urea of a triangle having the three sides given? t3 222 i MENSURATION. •F I 1 21. Required the area of an oblique triangle, the three aides of which are 80,120, and 160 rods, Ans. 29 acres 7 poles. A - To find the circumference of a circle when the diameter is given. rule. Multiply the diameter by 3,1416 and the product will be the circumference. A circle is the ]>ortion of a plane bounded by a curved line every part of which is equally distant from a certam point within, called the centre. The curved line AEBD is call- ed the circumference^ the point C the centre ; the line AE pass- ing through the centre a diame- ter, and CB the radius. The circumference AEBD is 3,1416 times grcnter than the diameter AB. Hence, if the diameter ^s 1, the circumfer- ence will be 3,1416. Wherefore if the «hametcr is known, the circumference is found by multiplying 3,1416 by the diameter. examples. 1 . The diameter of a circle is 93 feet ; what is the circumference ? Ans. 292,1688 feet. 2. The diameter of a circle is 20 rods, what is the cir- cumference 1 Anc. 62,832 rods. To find the diameter of a circle when the circumference is given rule. Divide the circumference by 3,1416 and the quo- tient will be the diameter, examples. 1. What is the diameter of a circle whose circumfer- ence is 78,54 feet] Ans. 25 feet. ^ ■M» ■ l-l^l— 11 I ■ .w I ■» I 11 11 I . I .. ■■ ■ ■ ..■ ■■ .1-1 ■ ■■^■^■^^^ QucsTioiia.'— How do we find the circunitiprence of a circle when the diameter ii givent How do we find the ditiineler of a oirde when the circumference ii givenf ircumference i MENSURATION. j#" t 223 2. What is the diameter of a circle whose circumfer- ence is 11752,1944. rods? Ans. 37,09 rods. To find the area of a circle when the diameter is known. RULE. Square the diameter, and then multiply by the decimal ,7854. EXAMPLES. 1. What is the area of a circle whose diameter is 5 feet? An?. 19,6350. 2. What is the area of a circle whose diameter is 8,75 feet? Ans. 60,1322 sq. feet. 3. How many acres are there in a circle of one mile diameter ? Ans. r)02A. 2R. 24P.+ To find the surface of a sphere or ball. KULE. Multiply the square of the diameter by the de- cimal 3,1416. EXAMPLES. 1. What is the surface or area of a sphere whose di- ameter is 12 feet? Ans. 452,3904 sq. ft. 2. What is the surface of a globe whose diameter is 7 ? Ans. 153,9384. 3. Required the area of the surface of the earth, its mean diameter being 7918,7 miles? Ans. 196996571,722104 sq. miles. To find the solidity of a sphere or globe. RULE. Multiply the cube of the diameter by the deci- mal ,5236 and the product will be the solidity. EXAMPLES. 1. W^hat is the solidity of a sphere whose diameter ia 17 inches. 17 X 17 X 17 X, 5236 =2572,4468 solid inches. Ana. QuEBTiuMs. — How 18 the area qI' a circle lound, when the di* •meter it known? How do we ^nd the lurf^ce of a •pheret<-« Uov do we find the tolidity of a eplier-eT 224 MENSURATION. m 2. What is the solidity of a globe whose diameter is 12 inches ? Ans. 904,7808 solid inches. To find the convex surface of a cylinder. RULE. Multiply the circumference of its base by the altitude, and the product will be the answer. EXAMPLES. 1. What is the convex surface of a cylinder, the diam- eter of whose base is 20 and altitude 50 f Operation. We first multiply the diameter by by 3,1416 which gives the circumfer- ence of the base. Then multiplying by the altitude, we obtain the convex 3,1416 20 62,8320 50 surface. Ans. 3141,6000 2. Required the surface of a cylinder, the diameter of whose base is 20 and the altitude 20 ? Ans. 1256,64. TIMBER MEASURE. I. To find the solid contents of a square or four sided stick of timber of equal bigness from end to end. RULE. Multiply the breadth by the depth, and then mul- tiply the product by the length, and the result will y\>e the solid contents. EXAMPLES. 1. A squared piece of timber 15 inches broad and 15 inches deep, and 18 feet long j how many solid feet does it contain] ^ ' Ans. 28,152. Questions. — How may the convex surface of a cylin4e;r be fouodi How is the solid contenvs oTa four sided stick of tioiber fgund, the size being equal fron;i e^d to end? MENSURATION. 225 2. What is the solid contenta of a piece of timber whose breadth is 16 inches, depth 12 inches, and length 12 feett Ans. 16 feet. 3. How many solid feet in a stick of ship timber, whose breadth ia 3 feet, and depth 2^ feet, and length 45 feet? Ans. 337,5 feet. II. To find the solid contents of a round stick of tim* ber of equal bigness from end to end. RULE. Multiply the area of one end by the length, and the product will be the solid contents. EXAMPLES. 1. What is the solid contents of a round stick of timber of equal bigness from end, whose diameter is 28 inches and length 25 feet 1 Ans. 1 06 [^^ feet. 2. Required the solid contents of a round stick of tim- ber, whose diameter is 18 inches and length 50 feet? Ans. 88,3575 solid feet. III. To find the solid contents of a tapering stick of limber, whether square or round, when one end is a point. RULE. Multiply the area of the big end by one third of its length, and the product will be the answer. EXAMPLES. 1. What is the contents of a tapering square stick of timber 24 feet 9 inches long, 16 inches square at one end, and a point at the other? Ans. 14-,*^/^- feet. 2. What is the contents of a tapering round stick of timber 30 feet long, 18 inclics diameter at one end, and a point at the other? Ans. 174-feet. Questions. — I4t)vv do we find Ihe solid contents of a round ■tick of limber of equal size from end to end? How do we find the oonlenls of a tapering slick of limber when one end is a point? 226 MENSURATION. 'it.- \ I lip 1 S( :^, 1 ; ^ ':i • : 1 1 1 ^< 1 f Uii IV. To find the solidity of a square piece of timber which tapers regularly but does not come to a point. RULE. Ist. Add together the breadths at the two ends and also the depths. 2nd. Multiply these sums together, and to the result add the products of the depth and breadth at each end. 3d. Multiply the last result by the length, and take one sixth of the product, which will be the solidity. EXAMPLES. 1. How many cubic feet in a piece of timber, the breadth and depth of the large end being 14 inches and 12 ijiches : and of the smaller, 6 and 4 inciiei?, and the length 30J feet. 14 12 16X20=320 6 4 14X12=168 — — 6X 4= 24 20 16 512 square inclic;-. But 512 sq. in.= "y-sq. ft. Then, --V X30.iXi = l8vV, solid feet. 2. Wiuit is the number of cu])ic feet in a flick of hewn timber, \vhos;e ends arc 30 inches by 27 and 24 inchosby 18, and the length 24 feet. Ans. 102 solid (eel. 3. How many cubic feet in a stick of limber whoso larger end is 25 feet by 20, the smaller 15 feet by 10, and the length 12 feet ? Ans. 3700 bolid feet. V. To find the solid contents of a tapering round stick of timber, when the small end is not a point. Questions. — To find the solidity of a square stick of timber which tapers regularly but dues not come to a point, what it the first thingr to be lioiicl What is the second step? The third! MENSURATION. 227 t the two encb RULE. Multiply each diameter by itself separately ; multiply one diameter by the other : add these three products together, multiply their sum by the length, annex two ciphers to the product, and di- vide by 382 ; the quotient will be the solid con- tents. EXAMPLES. 1 . What is the solid contents of a round stick of timber* whose diameter at the big end is 12 inches, at the small end 9 inches, and length 30 feetl Ans. 18-,^^V feet. 2. What is the solid contents of a round block of mar- ble, whose diameter at the big end is 23 inches, and small end 15 inches, ami length 34 feet 8 inches ? Ans. 68 feet+. VI. To find how many solid feet a round stick of timber of eiiual thicktiess from end to end, will contain when hewn square. RULE. Take one half of the diameter in inches and square it, this square being doubled and multiplied by the length gives the content in inches. EXAMPLES. 1. If the diameter of a round stick of timber be 18 inches, and its length 30 feet, how many solid feet will it contain when hewn square ] Ans. 33 |Q^ feet. 2. If a round stick of timber 28 feet long and 22 inch- es diameter, was hewn square, how many solid feet will it contain. Ans. 4>7jIj feeu VII. To find how many square edged boards of a given thickness can be sawn from a log of a given diameter. ■i ■ ■■ I II ■——ii— ^— ■ ■ ,., -I,,, ■.,. 1^ .^l.■,■■, I—— - ■■■■ ■■ ■- ■ -. ^ - „^ Qur.tTiuNi. — How do we find the loiidity of a tapcrinf; round ptiok oftimbcfr, when the imall end ie not a pointt How- may we. find what will be the content of a round stick of timber of equal itse from end to end, when hewn iquare? i 228 GUAGING. RULE. 1st. Find the solid contents of the log when made square by the last rule. . 2nd. Then say, as the thickness of the board, including the saw carf, is to the solid feet, so is 12 inches to the number of feet of boards. EXAMPLES. 1. How many feet of square edged boards, 1^ inchei thick, including the saw gap, can be sawn from a log 16 feet long and 18 inches diameter ? Ans. H4. 2. How many square feet of boards, 1^ inches thick^ including the saw gap, may be sawn from a log 28 feet long and 24* inches diameter 1 Ans. 384. 1 ;l. t 1 M « GAUGING. ' Gauging is finding the contents of any box, tub, cabk or otlier vessel. EXAMPLK. 1. There is a cask whose head diameter is 25 inches, bung diameter 31 inches, and whose length h 36 inches ; how many wine gallons does it contain 1 Also, how ma- ny beer gallons 1 ' The mean diameter of a cask id found hy addinn to the head diameter two thirds of tiic diftbrence between the bung and head diameters, or if tho staves are not much curved, by adding three fifths. Tfiin reduces the cask to a cylinder. Then, to find the solidity, wo multij)ly the square of tlj« mean diameter by the decimal ,7Sr)4« and the product by the length ; this will give the solid contents in cubic inches, which tlivided by 231 (the cubic inches in a gallon wine Qu^KTioKR. — What \n the rule for fituiinj; how many iiquart edged boaru* or a given thicknpss may be iiawn from a log of a given diamet^r^ VVliai is gaiginj;? How is the mean ditmettr of a cattk found? GAUGING. *229 measure) will give the content in wine gallons, and divid- ed by 282 (the cubic inches in a gallon beer measure) will give the content in ale or beer gallons. In this process we see that the square of the mean diame- ter will be multiplied ,7854, and divided for wine gal. by 231. Hence we may con ract the operation by only mul- tiplying their quotient ( V =,00345,-) that is, ,0034 (or by jS^, pointing otf 4 figures from the product for deci- mals.) For the same reason we may, for beer gallons, multiply by (^^^^=,0028, nearly,) ,0028. Hence, the following BULK. Multiply the square of the mean diameter by l!»e length ; then multiply this product by 34 for wine, or 28 for beer, and pointing oflf 4 decimals, the pro- duct will be the content in gallons and decimaU of a gallon. In the above example, the bung diameter is 31m. — 25 inches the head diameter=6in. difference, and '^ of 6=4- inches; 25in.-f-4in.=29in. mean diameter. Then 29^ =481, and 481 X36in. =30276. 3276X24=1029384. Ans. 102,9384 wine gal. 3276X28=847728. Ans. 847728 beer gal. 2. How many wine gallons in a cask whose bung di- ameter is 36 inches, head diameter 30 inches, and length 50 inches. Ans. 196,52gal. 3. How many wine gallons, and how many beer gal- lons, in a cask whose length is 36 inches, bung diameter 35in., and head diameter 30in. 1 ^^^'} 112 beer gal. Then QuRtTioNi. — How do we find Ihe lolidity of a caskf What will the product give! How are the cubic inches reduced to wine gnllontil Huw to beer gallons? How may the tiperation be contracted? Repeat the rule. 230 m DUODECIMALS. DUODECIMALS. Duodecimal is derived from the Latin word duodecimo signifying twelve. They are fractions of a foot, which is supposed to be divided into ttoelve equal parts called primes, marked thus, ('). Each prime is supposed to be subdivided into 12 equal parts called seconds, marked thus, (''). Eacli £Ccond is also supposed to be divided into 12 equal parts called thirds, marked thus, C0» ^"^ '^^ ^^ ^^ ^"7 extent. It thus appears that 1'. an inch or prime is -,^7 of a foot. 1'' a second is -^V of -iV? or jj^j- of a foot. 1''' a third is ■|\' <^^"iV o^iVj or — 7V7 of a foot, &c. Whenever there- fore any number of seconds (as 4"), are mentioned, it is to be understood as so many j\-f of a foot, and so of the thirds, fourths, &c. Duodecimals are added and subtracted like other com- pound numbers, 12 of a less order making 1 of the next higher, thus, 12'''' fourths make 1 third 1'". 12'" thirds make 1 second 1". 12" seconds make 1 prime or inch 1'. t 12' inches or primes make 1 foot. These marks are called " '" are called indices. Duodecimals are chiefly used in measuring surfaces and solids. MULTIPLICATION OF DUODECIMALS. RULE. Set the multiplier in such order that the feet thereof may stand under the lowest denomination of the multiplicand. Then multiply and carrv one for every 12 from one denomination to another, QuBiTioNs. — From what i& duodeoimal derived? What are duodeeiinulB? How are feet supposed to be divided? How are these parts airain divided? How are duodecimals added and subtracted? How man^ of one order mnke one of the next high, tr? What are the distinfruishing marks called? DUODECIMALS. 231 and set down the result, so that the lowest deno- mination in the product may stand directly under the number we multiply by. Then add as in com- pound addition. EXAMPLES. 1. A load of bark measures 12rt. 3 in. long, 4fCt. 2in. high, and 3ft. Gin. wide. Required the number of solid feet therein. Operation. 12 3' 4 2' 4-9 2 0' 6'f 51 0' 6ff 3 6f 153 1 25 6 6 3 We first place the numbers as the rule directs. We then multi- ply the lowest denomination in the multiplicand by the highest in the multiplier, which is 4 feet ; then by 2in. and place the product di- rectly under the multiplier, and find the product to be 51 feet Oin. and 6 seconds, which being multiplied by 3ft. 6in. gives the solid content of the load. Ans. 178 7' 9'' 0'^' In multiplication of duodecimals it will in all cases be found, to hold true, that the product of any two deno- minntions will always be of the denomination denoted by the sum of their indices'^ thus, the sum of the indices of S'X't' is " hence the product is 32"; and thus primes multiplied hy primes will produce seconds: primes multi- plied by seconds produce thirds j fourths multiplied by fifths produce ninths, &c. 2. How many square feet in a board 16 feet 9 inclies long, and 2 feet 3 inches wide ? Ans. 37ft. Sin. 3". 3. How many solid feet in a block 15ft. 8iii. long, 1ft. 5in. wide, and 1ft. 4in. thick. Ans. 29ft. T V 4''^ Questions. — Flow are duodeciiiiali cliieAy usedl Mow do we set down the numbers for inulliplyingT How do we multiply, carry, and set down the result? In mulliplicalion of duodeoU muls what will be found to hold true? Repeat tome examples. IK. 232 DUODECIMALS. 4. A board raeasurea 28ft. lOin. long, and 3ft. 5in. wide I required its contents. Ans. 98ft. Gin. 2". 5. In a pile of wood 176ft. in length, 3ft. 9' wide, and 4ft. 3' high, how many cords 1 Ans. 21 cords, and 7-,\rCord ft. over. 6. How many square feet in a pavement 371ft. 2' 6'' in length, and 181ft. 1' 9'' in breadth ? Ans. 67242ft. 10' 1'^ 4''' 6''''. 7. What is tlic price of a marble slab, whose length is 5ft. 7in. and the breadth 1 fo«.-- TOin. at 6s. per foot ? An;5. £3 Is. 5d. 8. What will the paving of a court yard come to at 3s. 2d. per yard, if the length be 27 feet 10 inches, and the broauth 14 feet 9 inches ? Anss. ^7 4s. 5d. POSITION. Position is a rule for finding, nn unknown number, by means of supposed numbers. It is of two kinds, Single and Double. These rules are generally given by writers on Arithme- tic ; yet most of tlie questions that are usually solved by them may be easily worked on general principles, and aU of them admit of very simple solutions, by Algebra. • SINGLE POSITION. Single Position includes the solution of questions by ono supposed number. RULE. Suppose any number, and perform on it the ope- ration indicated in the question ; then As the result of the operation, Is to the number given ; So is the supposed number, To the nprnber sought. QuKBTioNs. — What in Pusiiion? How many kinds are there? Whttl dues single Position include? What is the Ilule? V POSITION. 233 nd 3(1. 5in. m.6in.2''. Ift. 9' wide, ord ft. over. nift.2'6'' ose length is !i' foot 1 £3 Is. 5d. )me to at 3s. es, and the . £1 4s. 5d. number, by nds, Single >n Arithme- ly solved by lies, and all lebra. • lions by ono it the opo- -'4 ^ i Proof. — Perform, on the number found, the operation indicated in the question, if the work be right the result will be the same as the given number. EXAMPLES. 1. A School Master being asked how many pupils he had, answered. If I had as many more as I now have, half as many, one-third, and one-fourth as many, I should then have 148. How many scholars had he ? Operation. Here the supposed number Suppose he had 12 pupik. as many=12 as many =6 as many =4 1 1 3 I ^ 12, then as many more, \e ^, the I, and the I of 12 added together make 37, then according to the rule we aav as 37 : 148 : : 12 to 48 the answer. \ as many =3 37 As 37 : 148 : : 12 : 48 Ans. 2. What number is that, which being increased by ^, ^, and I of itself, the sum shall be 125 1 Ans. 60. 3. A. B. C. buy a quantity of cloth for JG340 ; of which A. i)ays three times more than B.and B. four times more than C ; what did each pay ? Ans. A. paid ^240, B. jGSO, and C. £20. 4. A person l)ought a chaise, horse and harness, for £60 ; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness. What did he give for each 1 Ans. harness, £6 13s. 4d. ; horse, £13 6s. 8d. j chaise, c£40. 5. A gentleman being asked how much money he had on hand, said that ^, J, -J- and -J- of his money amounted to JC114. What amouiit of money had he? Ans. JC120. 6. A Gentleman being asked his age, said, iff of the years I have lived, be multiplied by 7, and '| of them lie added to the product, the sum will be 219 ; what waa hiv age ? Ans. 45 yeans. is are iheret Question. — How ii Fusition proved? \2 2S4 POSITION. i>i " DOUBLE POSITION. Double Position is used for eolving such questions as re- quire two supposed numbers. RULE. Suppose any two convenient numbers, and pro- ceed with each of them separately according to the conditions of the question : then mark the er- rors with the signs +, or — according as the result of the operation, exceeds or falls short of the given number in the question. Then multiply the first supposed number by the second error, and the se- cond supposed number ^y the first error, and di- vide the sum of the products by the sum of the er- rors, if they are diflerently marked ; or the differ- ence of the products by the difierencc of the errors, if they are marked alike, and the quotient will be the answer. — Proof, as in Single Position. EXAMPLES. Divide iClOOO, so that B. may have twice as much as A., wanting JE80, and C. three times as much, wanting ^ilf»0 ; wiial was the share of each 1 Operation. -.^r ,^. . ., M- - A I 1 ^nc\r\ We now multiply tha First» Huppose A. had JE200 n . ^ \ ' ' , nil ortrv hi'st simposeu number then B. had 320 ^^p. v. \u ^ * 200 by the last error 150=30000, also the last supposed number 180 by the first error 30=5400. Then as the errors are marked alike, we divide the difference of the prc;- ducts 30000 — 5400 =2460 by the differ- ence of the errors 150 —30=120, and the quotient is £205 A's and Q. had 450 970 Supposed num her ) 1000 too small, hence \ the error is also. \ — 30 error. 2nd, suppose A. had iG 180 then B. had 280 and C. had 390 850 .Apposed number ) 1000 too email, hence the > errorfl are alike. ) -150 error. <4tVN»Tioi«.— l*'or what lu double puHtUun uitiUY Kepeat llie rui«, jstiona as re- s, and pro- cording to ark the er- sthe result f the given !y the first and the se- or, and di- 1 of the er- the difFer- the errors, cnt will be ion. I as much as icli, ^\antipg multiply Iha )sed number le last error )00, also the >sed number e lirst error Then as are marked divide the of the prc- )00 — 5400 y the differ- errors 150 SO, and the J £205 A's ). POSITION. 235 >eat iii« rul«, share, then 205x2— 80=330 B's share and 205x3— 150=C's share. 2. A. B. and C. bu"-'' a house which cost £500, of which A. paid n certam sum ; B. paid JGIO more than A. and C. paid as much as A. and B. both ; how much did each man pay ? Ans. A. paid £120, B. JEISO, and C. £250. 3. What number is that, which being multiplied by 3, the product increased by 4", and the sum divided by 8, the quotient will be 32 ? Ans. 84. 4. If my horse cost six times as much as it did, more £4, the sum vvould be JGlOO; what was the price of the horse? Ans. £16. 5. Two men, A. and B. lay on* equal Gums of money in trade ; A. gains £126, and B. loses £87, and A's mo- ney is now double to B's ; what did each lay out 1 Ans. £300, 6. A laborer was hired for 60 days, upon this condi- tion, that for every day he wrought he ehould receive 4s. and for every day he was idle should forfeit 2s ; at the ex- piration of the time he received £7 10s. ; how many da?a did he work, and how many was he idle ? Ans. He worked 45 days, was idle 15 day«. 7. There was a fish caught below Kingston, whose head was 9 inches long ; his tail was as long as his head and half his body, and the length of his body was equal to that of his head and tail : what was his whole length ? Ans. 6 feet. 8. A person has tvv^o horses, and a saddle worth £50 ; now, if the saddle be put on the back of the first horse, it will make his value double .that of the second j but if it he put on the back of the second, it will make his value triple that of the first ; what is the value of each horse ? Ans. one £30, the other £40. 9. Two gentlemen, A. and B. have both the same in- ^ come ; A. saves f of his, but B. by expending £50 a year 236 ANNUITIES. more than A. at the end of 4 years finds himself j£ 100 in debt ; what does each receive and spend per annum ? Ans. Each receives i3125 per annum. A. spend* jeiOO, and B. ^150 a year. It Ri^'i ,i3t . ANNUITIES. An annuity is a sum of money, payable every year, for a certain length of time, or forever. An annuity, in the proper sense of the word, is a sum paid annually ; yet payments made at different periods are called annuities. Pensions, rents, salaries &.c. belong to annuities. When annuities are not paid at the time they become due, they are said to be in arrears. The sum of all the annuities in arrears, with the interest on each for the time they have remained due, is called the amount. When an annuity is to continue forever, its present worth is a sum, whose yearly interest equals the annuity. I. To find the amount of an annuity at simple interesrt. RULE. First, find the interest of the given annuity for one year, and then for 2, 3, 4, and so on, up to the given number of years, less 1. Then multiply the annuity by the given number of years, and add the product to the whole interest, and the sum will be the amount sought. EXAMPLES. 1. What is the amount of an annuity of JBIOO for four years, computing interest at 7 per cent. ? Questions. — What is an annuity? When are annuities said to be in arrears? What is the amount? When an annuity ip to Guntinue tbrever, what is its present worth? How do we find tUe amount of an annuity al simple interest? ^ ANNUITIES. 237 The interest of J6100, at 7 per cent, for 1 year, is J67. for 2 years, 14«. for 3 years, 21. Four year's annuity, at JGIOO per year is Jei60x4>=4<00. Ans. .^442 2. If an annuity of .-670 be forborne 5 years, what will be due for the princii)al and interest at the end of said term, simple interest being computed at ^ per cent, per annum? Ans. £385 Os. 3. A house being let upon a lease of 7 years, at JG400 per annum, and the rent being in arrear for the whole term, I demand the sum due at the end of the term, simple interest being allowed at £6 per cent, per annum. Ans. JG3304. II. To find the present worth of an annuity at simple interest. RULE. First, find the present worth of each year by it- self, discounting from the time it becomes due; then the sum of all those will be the present worth. Note. — This rule depends on the principles of discount, see page 129. EXAMPLES. 1. What is the present worth of £400 per annum, to continue 4 years, at 6 per cent, per annum? 106 1 377,35849=pres't worth of 1st. y'r. ^^^ i • 100 • -400 • 357,14285= " 2d. y'r. jjg /► . luu . . ^.uu . 338^98305= « 3d. y'r. 124 J 322,58004= « 4th. y'r. £1396,06503 2. How much present money is equivalent to an an- nuity of £100, to continue 3 years; rebate being made at 6 per cent. ? Ans. £268,37. Questions. — How is the pres«?ni vvorili of on aiiiiuily at sim- ple inlereai fciuntl? On wlial principles does this rule depend? ill 'ft'^: I ii if'-:! t :;^ M ::^-»i ! If! :;!•>! I' ft 238 ALLIGATION. 3. What IS j680 yearly rent to continue 5 years worth in ready money at 6 per cent. ? Ans. ^6340 15s. 4d. ALLIGATION. The rule of Alligation teaches how to compound or mix together several simples of different qualities or prices, so that the composition may be of some intermediate quality or price. It consists of two kinds, Alligation Medial, and ,/lili' gation Alternate. ALLIGATION MEDIAL Teaches how to obtain the value, or mean price of a mixture, when the quantities and prices of the several arti- cJes are given. RULE. As the whole mixture is to the whole value, so is any part of the composition to its mean price* EXAMPLES. 1. A merchant mixes 20 bushels of wheat at 10s. per bushel, with 36 bushels of corn at 6s. per bushel, and 40 bushels of oats at 4s. per bushel ; what is a bushel of the mixture worth 1 Operation. 20 X 10= 200s. We first find the whole mixture to be 36 X 6=216 96bu., and the whole value to be 576s. 40 X 4=160 then according to the rule we say, as — 96 : 576 : : 1 bushel to the worth of one 96bu. 576s. bushel of the mixture. As 96 : 576 : : 1 : 6s. Ans, 2. A grocer mixes 601bs. of sugar at 8d. per pound with 20lbs. worth 12d. per pound j what is the value of one pound of the mixture ? Ans. 9d. QuesTioNs. — What does Alligation teach? Of what kinds does it consist? What does Alligation Medial teach? Repeat Hie rule. ALLIGATION. 239 3. A tobacconist mixed 361bs. of tobacco, at Is. 6(1. per pound, 121bs. at 2s. per pound, with 121b8. at Is. lOu. per pound ; what is a pound of this mixture worth 1 . Ans. Is. 8d. 4. A grocer mixed 2cwt. of sugar at 56.s. per cwt. and Icvvt. at 43s. per cwt. and 2 at 50s. per cwt. together ; I demand the price of 3cwt. of this mixture? Ans. JGT 13s. 5. A refiner melted togellier 8oz. of gold, of 22 carats fine, lOoz. of 20 carats fine, ]2oz. of 16 carats fine, 8oz. of 18 carats fine j what was the value of the composition ? Ans. 18 Hr carats fine. ALLIGATION ALTERNATE Is the method of finding what quantity of each of the in- gredients whose rates are given, will compose a mixture of a given rate ; so that it is the reverse of Alligation Me- dial, and may be proved by it. This is called IdUigation Alternate, because the same question frequently admits of different answers, CASE L When the prices of the several simples are given, to find how much of each at their respective rates must be taken to make a compound or mixture, at any proposed price. RULE. I. Reduce the mean price and the prices of each separate article to the same denomination. II. Connect with a line each price that is less than the mean price, with one qr more that is greater ; and each price greater than the mean price, with one or more that is less. III. Write the difference between the mean price and the price of each separate article oppo- site the price with which it is cor aecled ; then the Questions. — What is Alligation Alternuie? How may it be provedl Why is it called Alligation Alternate? How' do we find the proportional parts to be taken when the prices of the several simples are given? H 240 ALLIGATION. ill ! i' V sum of the differences standing against any price will express the relative quantity to be taken of that price. EXAMPLES. A merchant would mix wines worth 16s. ISs. and 22s. per gallon, in such a way that the mixture be worth 20s. per gallon ; hew much roust be taken of each sort] Operation. 16_ 20 \ 18— 22— 2 at 16s. 2 at 18s. 1^X2=6 at 22s. In this example 20s. is the mean price ; then as 16s. is less, and 22s.great- er than the mean price, we connect 16 and 22 ; then as 18s. is less than 20s. it must also be connected with 22s. ; then as the difference between 16s. and 20s. is 4s., and between ISs. and 20s. is 2s. the 4 and 2 must both be placed opposite 22, hcjause 16 and 18 are both connected with that number, also because 22 is connected to 16 and 18 the difference between 20 and 22 must be placed opposite those two numbers. So that 2 gal. at 168. 2 at 18j. and 6 at 223. will make a mixture worth 20s. per gallon, or any other quantities bearing the proportion of 2, 2, and 6. 2. What proportions of tea at 8s. at 9s. at lis. and at 12s. must be taken to make a mixture worth 10s. per pound. Ans. 21bs. at 8 &. 12s. and lib. at 9 and lis. 3. A goldsmith has gold of 16, of 18, of 23 and of 24 carats fine ; what part must be taken of each, so tl.jt the mixture shall be 21 carats fine ? Ans. 3 of 16, 2 of 18, 3 of 23, aud 5 of 24. 4. What portions of wine at 14s.j at 24s., at 21s., and at 10s. per gallon, must be mixed together so that the mix ture shall be woiili 18s. per gallon 1 Ans. 6 gal. at IOki., 3 at 14s. 4 at 21s. and 8gal. at 24s. .<•! CASE II. When one of the ingredients is limited to a certain quan- tity, to find the several quantities of the rest^ in proportion to the given quantity. ALUGATION. 241 any price taken of 8s. and 22s. 5 worth 20s. sort] mple 20s. is ice ; then as id 22s.great- mean price, 16 and 22 ; )e connected L6s. and 20s. t and 2 must I 18 are both is connected I 22 must be L 2 gal. at 168. /orth 20s. per proportion of it lis. and at »rth 10s. per at 9 and 11b. 23 and of 24 so tl.at the aud 5 of 24. ^ at 21s., and I that the mix • |) gal. at 10»., Icertain quan- lin proportion RULE. I. Find the proportional quantities of the sim- ples as in Case I. II. Then say, as the number opposite the sim- ple whose quantity is given, is to the given quanti- ty, so is either proportional quantity to the part of its simple to be taken. EXAMPLES. 1. How much wine at 5s., at 5s. 6d., and 6s.per gal. must be mixed with 4 gallons at 4s. per gallon, so that the mixture shall lie worth 5s. 4d. per gallon 1 Operation. 8"! . . Simple whose quantity is known. f48_ n i 60_ (^72_ 2 L 4 J- Proportional quantities. 16) 1 2 8 o Then 8:4:: 2 8:4:: 4 8 : 4 : : 16 Ans. 1 gal. .5s., 2 at 5s. 6d., and S at ih. How much wator must bo mixed with 100 gallons of wine worth 7s. 6d. per gallon to reduce it to 6s. 3d. per gallon ? Ans. 20 gallons. 3. A farmer made a mixture of wheat at 4s. per bush., rye at 3s., barley at 2s., with 12 bui«!iels of oats at 18d. per Inishel : ho*v much is taken of each sort wlien the mixture is worth 3s. 6d. . ^96 bu. of wheat ; 12 bn. of rye ; " ( J 2 bu. of barley ; and 12 of oats. v. With 05 gallons of wine at 8s. per gal. I mixed other wine at Gs. 8d. per gal. and some water: then I found it stood mo in 6.s. 4;1. per gallon. I demand how much wine and how much water I took. Ans. 95 gal. wine at 6s. 8d. and 30 gal. water. Qur.aTioN<;, — When one of the ingredicnlii ii limited to a cer« tnin quantity, how do we find tiie proportional quantities of the rcBlT VV 242 ALLIGATION. s^ '^li . CASE III. When the whole composition is limited to a ceartain quantity. RULE. I. Find the proportional quantities as in Case I. II. Then say, as the sum of the proportional quantities is to the given quantity, so is each pro- portional quantity to the part to be taken of each. EXAMPLES. 1. A grocer has four sorts of sugar worth 12d., lOd., 6d., and 4n of 2 to each succeeding figure. When the se- ries decreases, that is, when it is formed by the constant subtraction of the common difference, it is called a dess- cending series, thus, 14, 12, 10, 8, r>, 4, 2, &c. F ^ the series is formed by a continual suhlraction of 2 f» . ;.ach preceding figure. The figures that make up the series arc called the tcrm^^ of the series. The fust and /c/5/ terms arc called the ex- tremes^ and the other terms the means. From t'le above, it may be seen that any term In a se- ries may be found by continued addition or subtraction, but in a long series this process would be tedious. A much more expeditious method may be found. The ages of six persons are in arithmetical progrcsHon. The youngest is 8 years old, and the common difference ia 3 ; what is the age of the eldest? In other woids, what is the last term of an arithmetical series, whose fuiet term is 8, the number of terms 6, and the common difference 3 1 8, 11, 14, 17, 20, 23. Examining this series, we find that the common differ- ence, 3, is added 5 times, that is, one less than ihe num- ber of terms, and the last term 23, is larger than the first QuKSTioNs. — Whiil 18 an ariiliinctical lerieii, or prtigreiisicnT When i« it called on asccindinf^ lerieit When ii it called a dcicendinf; seriesl What are the lerma of the aerieiY What are the txtnmtsf What are the mtantf How may any term in a lerica be found? 244 ARITHMETICAL PROGRESSION. term, by 5 times the addition of the common diflference, tliree j Hence, The age of the elder person is 5x3+8=23 Ans. Therefore, when the first term, the number of terms, and the common difference, are given, to find the last term we have the foll^owing RULE. Multiply the common differ-ence into the number of terms, less 1, and add the product to the first term. EXAMPLES. 1. The first term is 3, the common difference 2, and the nimiber of terms 19 : what is the last term 1 «> Operation. 18 number of terms less 1. 2 common difference. 36 3 1st term. Ans. 39 last term. 2. If the first term be 4, the common difference 3, and iho ninubei of terms 100. wliat is the last term 1 Ans. 301. 3. .folin owes Samuel a certain ?um, to be paid in aritiiinelical progression ; the first payment is 6d., the num- ber of payment!"? 52, and the common dilference of the pay- nuuil^i is 12d. Whatislhc]:isti)aymcnt? Ans. £2 lis. fid. 4. A man put out jC 100, at 7 per cent, simple interest, A>hich amounted to £101 in a year, £114i in 2 years, and so on in arillnnctical progres:-^ion, with the common difler- eiice of jG7 : what was tiic amount due at the expiration of 50 years ? A ns. Je450. 5. A man bought 100 yards of cloth in arithmetical pro- fession : for the first yard he gave 4 pence and for the last 801 pence, what was tiie common increase on tho price of each yard? Question.— \V lien the tirtit lenn, the number of terms, and tho common ilifierence are given, how do we find ihe Uit teim? ARITHMETICAL PROGRESSION. 245 As he bought 100 yards, and at an increased price up- on every yard, it is evident that this increase was made 99 times, or once less, than the number of terms in the series. Hence the price of the last yard was greater than the first, by the addition of 99 times the regular increase. There- fore if the first price be subtracted from the last, and the remainder be divided by the number of additions 99, the quotient will be the common increase ; 301 — 4=297-f- 99=3, the common difference. Hence, when the ex- tremes and number of terms are given, to find the common difference, we have the following RULE. "^ Subtract the less extreme from the greater, and divide the remainder by the number of terms less 1, and the quotient will be the common difference. EXAMPLES. 1. The extremes nre 4 and 104, and the number of terms 26 ; what is the common diflerence? 104— 44-'ili— 1=4 Ans. 2. A man hns 8 sons, the youngest is 4 years old and the eldest 32, their ages increase in arithmetical progression: what is the common dill'ereiice of their ages? Ans. 4 years. 3. A man is to travel from Kingston to a certain place in 12 days : to go three miles the first day, increasing ev- ery day by the same number of miles, so that the last day's journey may be 58 miles : required the daily increase. Ana. 5 miles. Suppose we are required to find the sum of all the terms in a series whose first term is 2, the number of terms 10, and the common difference 2. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. 20, 18, 16, 14, 12, 10, 8, 6, 4, 2. 22, 22, 22, 22, 22, 22, 22, 22, 22, 22. The fir$t row of figures above, represents the given se- Qur.rnoN. — Whon the extremes and number of terms ar« given huvv (!o we find the common differehcel \v2 246 ARITHMETICAL PROGRESSION. i\ lies. The second, the same series with the oiclei- inverted, and the third, the sums of the additions of the correspond- ing terms in the two series. By examining these series we shall find that the sums of the coiTesponding terms are the same, and that each of them is equal to the sum of the ex- tremes, viz. 22. Now, as there are 10 of these pairs in the two series, the sum of the terms in boihj must ho 22X10=220. But it is evident, that the sum of the terms in one series can he on\y half as great as the sum of loth, therefore, if we divide 220 by two, we shall find the sum of the term^ in one series, which was the thing required. 220-^2=: ! 10, the sum of the series. Hence, When the extremes and number of terms are given, to find the sum of the terms, we have the following RULE. Multiply the sum of the extremes by tho number of terms, and divide the product by 2. EXAMPLES. 1. The extremes are 2 and 100, and the number of terms 22 : what is the sum of the scries? Operation. 2 first term. 100 last term. 102 sum of extremes. 22 number of terms. 2)2244=1122 Ans. 2. How many strokes does the hammer of a clock strike in 12 hours? Ans. 78. 3. A man bought 19 yards of linen in arithmstical progression ; for the first yard he gave Is. and for the last yard JCI 17s.; what did the whole come to ? Ans. £18 Is. 4. There are in a certain triangular field, 41 rows of corn ; the first row, being in one corner, is a single hill, . y ■■■ ■ ■■! 11^ I I.MII. ■ ■ II _ . I ■■! ■ — . , M.^. ■■ . I . ■ I M- ■ ■ — Ill I 1.1 I ■ ■! ■! —— ^ QvEiTioif.— When the extremea and number of lermi are giv« •n, how do we find the sum of the termsf GEOMETRICAL PROGRESSION. 24.7 one series and the last row, on the opposite side, contains 81 hills; how many hills of corn in the field 1 Ans. 1681. 5. Suppose 144 oranges were laid 2 yards disstant from each other, in a right line, and a basket placed two yards distant from the first orange, what length of ground will that boy travel over, who gathers them up singly, returning with them one by one to the basket ] Ans. 23 miles ^ ^nr. ISO vards. GEOMETRICAL PROGRESSION. ^ny series of numbers, consisting of more than two terms, which increases by a common multiplier, or decrea- ses by a common divisor, is called a Geovieirical series. Thus, theh-eries 2,4, 8, 16j .32, &,c., consists of terms, each of which is twice the preceding, jnd this is an increasing or ascending Geonu'trical seriev«-'. The series .32, 16, S, 4, 2, consiiits of numbers, each of which is one holfihc pre- ceding, and this is a decreasing or descending Geometrical series. The common multiplier or divisor is called the Ratio, and the nunibers which form tlie series are called Terms. In JirilhmciicaliMuWn Gcowc/'/zVa/ progression, if any three of the five following terms be given; the other two may be found. 1st. The first term. 2nd. The las^t term. 3cl. The number of terms. 4th. The common difierence. 5th. The sum of all tiie terms. A man bought a piece of cloth containing 12 yards, ths first yard cost 3d., the second 6, the third 12, and so on doubling the price to the last, what cost the last yard ? 3X2X2X2X2X2X2X2X2X2X2X2=3X2"= G144 Ans. Ill examining the above process, it will be seen that the price of the second yard is found by multiplying the first Qv£STioHfi. — What is a Geometrical seriesT Give an exampla of an ascending Geometrical series? DeH<:ribea descending Ge- ometrical series. What is the Ratio? What are the terms? What are the five terms, any threfl of which being given, the otlier two may be found. 248 GEOMETRICAL PROGRESSION. i payment into the ratio 2 once ; the price of the MiVrfyard by multiplying by 2 twice^ &c., and that the ratio 2 is used as a factor eleven times, or once less than the number of terms. The last term then, is the eleventh power of the ratio 2, multiplied by the first term 3. Hence, when the first term, ratio, and number of terms being given, to find the last term, we have the following IIULE. Raise the ratio to a power whose exponent is one less than the number of terms, and then mul- tiply the power by the first term, the product will be the last term. EXAMPLES. 1. The first term is 3 and the ratio 2 ; what is the 6th term? 2X2X2X2X2=2^=32 3 first term. Ans. 96 2. A man purchased 12 pears: he was to pay 1 far- thing for the firsit, 2 farthings for the second, and so on, doubling each time : what did he pay for the last 1 Ans. £2 2s. 8d. 3. A geniienion dying, left 9 sons!, and bequeathed his estate in the following manner: to his executors JG50: his youngcsit son to have twice as much as the executors, and each son to have douljletiie amount of the son next young- er: vyhat was tlie eldest son's portion? Ans. ^225600. 4. A man plants 4< kernels of corn, wliich at harvest produce 32 kernels ; these he plants the second year ; now suppposing the annual increase to continue 8 fold, what would he the produce of the IGth year, allowing 1000 ker- liels to a pint ? Ans. 2199023255,552 bushels, Tlie most obvious method of ohidhnrng \[\q sum of the terms in a geometrical series, might be hy addition,\i\xii\\\% is not the most expeditious, as will be seen. QuitTioN.— When the first term, ratio, and number of terms are given, how do wc find the laat term? GEOMETRICAL PROGRESSION. 249 A man bought 5 yards of cloth, giving 2 pence for the first, 6 pence for the second, and so on in a threefold ra- tio ; what did the whole cost him 1 2, 6, 18, 54, 162, 6, 18, 54, 162, 486. The first of the above lines represents the original se- ries. The second, that series, multiplied by the ratio 3. By examining these series, it will be seen that their terms are all alike, excepting two : the first term of the first se- ries, and the last of the second series. If now we sub- tract the first series from the last, we have for a remainder 486 — 2=484 as all the intermediate terms vanish in the subtraction. Now the last series is Mree times the first, (for it vvas made by multiplying. the first series by 3,) and as we have already subtracted once the first, the remainder must of course be twice the first. Therefore if we divide 484 by 2, we shall obtain the sum of the first series. 484-f 2=242 Ans. As in the preceding process, all the preceding terms vanish in the subtraction, excepting the first and last, it will be seen, that the result would have been the same, if the last term only had been multiplied, and the first sub- tracted from the product. Hence, when the extremes and ratio are given, the sum of all the terms may be found by the following RULE. Multiply the greater term by the ratio, from the product subtract the least term, and divide the re- mainder by the ratio less 1. EXAMPLES. 1. what is the sum of the series 1 The first term is 3, the ratio 2, and last term 192 ; Ans. 192x2—3=381—2—1=381. QuKSTioNs. — When the exrretnes and ratio are given, how may the sum of all the terms be found? If the last term be not given in the question what must be donel I 250 PERMUTATION. Note. — If the last term be not given in the question ; ''Tst find it by the rule in the last article, then proceed ay above. 2. A gentleman whose daughter was married on New Year's day, gave her one shilling towards her portion, and was to double it on the first day of every month during the year ; what was her portion 1 Ans. JB204) 15s. 3. What is the sum of the series 16, 4, 1, ^, -j\-, -/.^-, and so on to an infinite extent ? Ans. 21^. Here it is evident the last term is 0, or indefinitely near to nothing ; the extremes therefore are 16 and 0, and the ratio 4. 4j. What debt can be discharged in a year, by paying 1 farthing the first month, 10 farthings the second, and so on, each month in a ten fold proportion ? Ans.'jS115740740 14s. 9^d. 5. A man bought a horse, and by agreement was to give a farthing for the first nail, 2 for the second, 4 for the third, &c. There were 4 shoes, and S nails in each shoe f what did the horoe come to at that rate ? Ans. de4473924 5s. 3j(L PERMUTATION. Permutation is the method of finding how many changes may be made in the order in which things succeed each other. What number of permutations may be made on the let- ters A and B? They may be written A B. or B A. What number on the letters A B C 1 Placing A first, A B C, or A C B, Placing B first, B A C, or B C A, Placing C first, C A B, or C B A. From these examples it will be seen, that of two thinga, there may be two changes, 1x2=2, and of 3 thingsthero may he 6 changes, 1 X2X3=6. Questions. ^Whut is rennutalion? flow may we find what ntfVnber of change:} or perinututions may be made on uny given number of thinqs? MISCELLANEOUS EXERCISES. 251 Hence, to find the number of different changes, or per- mutations of which any number of different things are capable, Find the continued product of the natural series of numbers, from 1 to the given number, EXAMPLES. 1. Four gentlemen agreed to remain together, as long as they could arrange themselves differently at dinner ; how many days did they remain 1 Ans. 24<. 2. How many variations may there be in the position of the nine digits 1 Ans. 362880. 3. Seven gentlemen met at an inn, and were so well pleased with tlieir host, and with each other, that they agreed to tarry so long as they, together with their host, could sit every day in a different position at dinner ; how long must they have staid at said inn to have fulfilled their agreement? Ans. 110^ |^f years. 1 1 1 1 1 i t MISCELLANLOUS EXERCISES. Ans. 230. Ans. 3:^. I. 7+4— 2+3+40x5=how many ? ^- 3-1-6— 2 X4— 2 =howman> ? 2X2 3. What number is that which being divfdeJ by 10, the quotient will be 72 ? Ans. 1368. 4. What number is that which being multiplied by 15 the product will be | ? Ans. -/,7. 5. What is the difference between six dozen dozen, and half a dozen dozen ? Ans. 792. 6. What is the difference between thrice five and thirty, and thrice thirty five 1 Ans. 60. 7. What fraction is that, to which if you add f , the sum will be I? ' Ans.-^'J. 8. What number is that which being divided by |, the quotient will be 21 ? Ans. 15^. 252 MISCELLANEOUS BXERCISES. 9. What number taken from the square of 54 will leave 19 times 46? Ans. 2042. 10. If I buy 1000 Ells Flemish of linen for ^690, what must it be sold for per Ell English, to make JBIO by the purchase ? Ans. 3s. 4d. 1 i. A Fnail in getting up a pole 20 feet high, was ob- served to climb up 8 feet every day, but to descend 4 feet every night ; in what time did he reach the top of the pole 1 Ans. 4 days. 12. What number added to the 43rd part of 4429, will make the sum 240 1 Ans. 137. 13. How many bushels of potatoes, at Is. 6d. per bushel, must be g;iven for 32 bushels of barley, at 2s. 6d. per bushel 1 Ans. 53^ bu. 14. A boy bought a number of apples ; he gave away 10 of them to his companions, and afterwards bought 34 more, and divided one half of what he then had among four companions, who received 8 apples each j how ma- ny apples did the boy first buy ? Ans. 40. 15. A man married at the age of 23 ; he lived with his wife 14 years j she then died, leaving him a daughter 12 years of age ; 8 years after, the daugliter was married to a man 5 years older than herself, w^ho was 40 years of ag3 when the father died ; how old was the father at his death ? Ans. CO years. 16. There is a room 18 feet in length, 16 feet in width, and 8 feet in height ; how many rolls of paper, 2 feet wide, and containing 11 yards in each roll, will it take to cover the walls ? Ans. 8-J-f rolls. 17. How many steps of 30 inches each must a man take in travelling 54^ miles? Ans. 115104 stops. 18. How much time would a person redeem in 40 years, by rising each morning half an hour earlier than he now does? Ans. 304^ days. 19. There is a house, the roof of which is 44^ feet in length, and 20 feet in width on each of the two sides ; if 3 shingles in width cover one foot in length, how many shingles will it take to lay one course on this roof? if 3 ill leave . 2042. ), what > by the 3s. 4d. cvas ob- d 4 feet le pole 1 4 days. r 4429, IS. 137. 6d. per 2s. 6d. 53^ bu. \re away fught 34 \ among low ma- \.ns. 40. ed with aughter married years of ?r at hif? years. [1 width, et wide, ;o cover |.-f rolls. It a man }4> slcp>?. in 40 Ithan he l^ days. f feet in [ides ; if many )f ? if 3 MISCELLANEOUS EXERCISES. 25:i courses make one foot, how many courses will there be on one side of the roof? how mnny shingles will it take to co- ver one side, also t) cov\^r both .aides'? Ana. 16020 shingles. 20. Said John to Dick, my purse and money are worth jG9 2s., hut the monoy is 25 tirnss as much as the purse ; I demand how much money was in it? An<, £8 15s. 21. The third part of an army was killcil, the fourth part taken prisoner?, and 1000 fled; how many were in this army, how many killed, and how many captives? Ans. 2400 in the army, 800 killed, and 600 taken pri- soners. 22. If 3 men cnn do a piece of work in 4.^ hou'*;?, in how many hours will 10 mjn do the same work ? Ans. 1-"- hours. 23. Jacob, by contract was to serve Laban for his t\, t; daughters, 14 years; and when he had accompli -bed 11 years, 11 montlir', 11 we?ks, 11 days, 11 hours, 11 ,uii lites, how long had he to serve ? Ans. lyr. 11 mo. 3vv. 2da. 12h. 49min. 24. A man had two silver cups of unequal woig'it, ha- ving one cover to both, weighing 5oz, ; now if tho cover is put on tha less ciip, it will be double the weight of the greater cu]), and put on the greater cup it will be three times as heavy as the less cup j what is the weight of each cup? ^ Ans. 3oz. less ; 4oz. greater. 25. A man and bis wife can drink out a cask of beer in 12 days, but when t!ie man Was froiti ii'me it lasted the womnn 30 days ; how many days wouid the man be in drinking it alone ? Ans. 20 days. 26. The great bell of Moscow, ia Russia, the largest in the world, is 67 feet in circumf rence, 19 feet high, and weighs about 448000 pounds ; how many teams would it require to move this bell, if each team draw 30c\\t. ? Ans. 133^ teams. ,. 27. From each of 16 pieces of gold a person tiled the worth of 2s. 6d., and then offered them in payment for their original value, and the fraud being detected, and the pieces weighed, they were found to be worth in the whole, ao X 254 MISCELLANEOUS -EXERCISES. i I in ft I: 8 ft I' t more than 8 guineas j what was the original value of each piece? Ans. 13s. 28. Two men carry a kettle weighing 200 pounds ; ihe kettle is suspended on a pole, the bale being 2ft. Gin. from the hands of one ; and 3ft. 4in. from the hands of the other ; how many pounds does each bear ? Ans \ ll'*?^Po»n<^'^' ' I 85^- pounds. 29. A person bought 160 oranges at 2 for a penny, and 180 more at 3 for a penny ; after which he sold them out at the rate of 5 for 2 pence ; did he make or lose, and how much 1 Ans. He lost 4 pence. 30. If a person step 70 paces per minute, and 28in. each pace ; how fast is that per hour ? Ans. 1|^^ miles. 31. A wall of 700 yards in length was to be built in 29 days. Twelve men were employed on it for 11 days, and only completed 220 yards ; how mnny men must bo added to complete the wall in the required time ? Ans. 4 men. 32. There is a stone which measures 4 feet 6 inches long, 2 feet 9 inches broad, and 5 feet 4 inches deep ; how many solid feet docs it contain ? -*- Ans. GGft. 33. What is the product of 23. Gd. multiplied by 2s. Gd.? Ans. 6s. 3d. 34. I sum up lialf mankind, And add two thirds of the remaining half, And find the total of their hopes and fears. Dreams, empty dreams. Cowper. What j)art of mankind are mere dreamer this author 1 35. Whc . time, between 4 and 5 o'clock, arc the hour and minute hands of a watch exactly together! Ans. 21-,*,- min. past 4. 36. A. can mow an acre of grass in 7^ hours, B. in Sf hours, in what time will -they jointly perform it? Ans. 4 hours. according to Ans. g. MISCELLANEOUS EXERCISES. of each IS. 13s. )ounds ; Ih. 6in. mds of pound:?, pounds. penny, Id them )se5 and ti pence. d 28m. 'j- miles. built in LI days, must bo 4> men. ) inches p ; how s. GGft. by 23. 6s. 3d. PER. fding to Ans. I. he hour past 4. I, B. in b hours. 37. A captain, mate, and twenty seamen, took a prize worth j£3501, of which the captain takes 11 shares, and the mate 5 shares ; the remainder of the prize is equally divided among the sailors ; how much did each man re- ceive 1 i The captain, £10G9 ir)«. Ans. } The mate, 480 5s. I Each Bailor, 97 fj?^. 38. Divide the number 360 into 3 parts, which shall be to each other as 2, 3 and 4. Ans. 80, 120 and 160. 39. Two merchants have gained £4-50, of which A \^ to have three times as much as B ; how much is each to have ? Ans. A. ^1337 10.^. and B. £112 10:^. 40. A. and B. traded together, and gained i^lOO ; A. put in j£640. ; B. put in so much that he must rcceivo d£60 of the gain ; I demand B's stock. Ans. JGOGO. 41. The wall that separates China from Tartary was built 2000 years ago ; it crosses the InrjrpFt rivers and moi'in- tains, and is 1200 miles in length, 30 \\\t li^gh, and 24ft. broad ; how many cubic feet docs it contain ? Ans. 4r)6 1920000. 42. The surface of a middle sized man is IG squaro feet, and the skin is said to bo perforated by a thousand holes in the space of a square inch. How many pores does the human body contain, according to this calculation ? Ans. 2301000 pores. 43. Divide 97dog. .55mi. 7fur. 3r)po. 4rt. 2in. lb. by G. Ans. lGdeg.20min.7fur. 12po. 8ll. llin. 1^ b. c. 44. If a herring and a half coj^t a penny and a halt', how many can be bought for eleven pence ? 45. The entire amount of specie throughout the world is estimated at one billion nine hundred millions of dollars; how long would it take to count tiiis sum at the rale of 50 a minute ? Ans. 72 years, 108 days, 21 hours and20inin. 46. One end of a certain pile of wood is perpendicular to the horizon, the other is in the form of an inclined plane ; the length of the pile at the bottom is 64 feet, length at the top 50 feet, height 12 feet, length of the wooil 5 feet j re- 256 MISCELLANEOUS EXERCISES. quired the number of cords it contains ? Ans. 26 |f. 47. A may-pole, whose top was ])roken off by a blast of wind, struck the ground at 15 feet distant from the bot- tom of the pol^ ; what was the height of the whole may- pole, supposing the length of the broken piece to be 39ft. 1 Ans. 75 feet. 4«8. In the midst of a meadow well stored with grass, I took just an acre to tether my ass ; How long must the cord l:c, that feeding all round, He may 'nt graze less or more than an acre of ground? Ans. 39,25-f yards. 49. If a quantity of provisions ecrves 1500 men 12 \veeUi?5 at the rate of 20 our.corf a tiny for each man ; how many men will the same pi ovi,->:oiis mnintain for 20 weeks, nt the rate of Soz. a ilr.y i'ov cnc h mnn ? Ans. 2250 men. 50. A young,^r l^rotlifr lorcivcd iJrSd'OO, which vvas just y of his elder brotlier's fuitune; what wat? the father wonh? Ans. $19200. 51. ir20 men cnn peiforni a piece of work in 12 days, how mnny men will aceomuli. h three limes as much in one-nrih oi'tliLMlme? ' Ans. 300. 52. Siippo-'o that I Iiin-c -"- of a t:li!p worth .$1200 ; what \)[u\. iiavo I Kit altjr ^-c'liiig " of ', of my share, and whiit is it woith 1 . Alii.-. -, .„ leit, worlli ^ ?8(),(i()-|- 53. "\^'h;tt mimher i^^ tlu'l \vlii( ii bcii.g mu'.f.plied by j] of ,;i of 1^, the ywi.hx{ will I;o 3 '? Ans. IJ. 5-1.. J\Jy bono and saddle t( gather are worth .';»132, and my bor.^-e is wojtli 10 t.mes a;j mixli au the hailille: what id the value of tlie h(Mvo? Ans. $120. 55. A firmer belt g asked bow ir.nny >^beep be had, answered, that he b.d tliem in fiw) fields; in the fir.'-t ho had \ of Ills HiM-k, in the second -^, in the tliird \, in tho fourlii -,' , and in the f fdi 'l."0 ; bow many bad be I Ans. 1200. 5G. Sound travels about 1 142 feet in a ?econ«l. Now if the ilash of a cannon be fccn ;it the moment it is fired, and the report heard 45 i^cconds ai'ter, what di.'tance would the observer be from the gun ? Ans. 9uii. 5fur. 34rd.+ MISCELLANEOUS EXERCISES. 257 300. 11I.+ 57. In a certain orchard, ^ of the trees bear apples, \ of them bear peathegs, ■} of them plumf, 120 of them cherries, and 80 of them pears ; hdw many troes are there in the orchard 1 Ans. 240. 58. A circular fish-pond is 865 feet in diameter j what id itd circumference, and what is its area ? . ( Circumference 2717,484ft '^"^^* I Area 5S7655j915rt. square. 59. A well is to be stoned, of which (the diameter is 6 feet G inches ; the thickness of the wall is to be 1 foot 6 inches, leaving the diameter of the well within the stones 3 feet 6 inches. If the well is 40 feet deep how many feet of fetone will be required ? Ant^. ?A<''2^'^S feet. 60. A i^liip has a leak by which it would fi'.l and sink in 15 houn-, but ljy moans of a pump it could be emptied, if full, in 16 hours. Now if the pump is worked Tk m the time the leak L(glnr, how lorg before the ^bip will sink? Ans. 240 hours. 61. How many planks ir>rt. lorg anf5460, left a wife and two children, a ion and di.i^litcr, absent in a fcre'gn ceuntry. He directed that irhis^:on returned, the mother should have one third of the ovtato, a!.d the son the rtniairider: but if the da\'gbter rcluineil, hlio ^h()uld have one tbiid, and the mother the rturjiimior. Now, it to happened tliat they both rcturneil ; hov/ mutt the c^tale Le divided to i\\\Ci\ tho father's ii'.tetitlons ? Ans. Darj-hter J?^7S0, Son {f<31£0, Mother $1560. 63. A cistern containirg CO gallons of water has three unequal cocks for di^ch^rging it ; the laigoht will empty it ill 1 hour, the b^econd in 2, and the third in 3 hours ; in what time will the ci^ern Le cn)pticd if they all run trgcther ? AM 32. mm. 64. A. can do a piece of work alone in 10 days, and B. in 13 days ; m what time can they do it if they work together 1 Ane. 5^5 days, X2 268 MISCELLANEOUS EXERCISES. Ans. 65. The accounts of a certain school are as followe, viz: -j^g- of the boys learn geometry, | learn grammar, -f^; learn arithmetic, t/„- learn to write, and 9 learn to read ; what is the number in each branch 1 5 learn geometry, 30 grammar, 24 arith- metic, 12 writing, and 9 reading. 66. A stationer sold quills at 1 Is. a thoiiFand, by which he cloarcd | of his money ; but they growing ecarce he ^ai^ed the \n\cc to 13s. 6d. a thousand ; what did he clear at the lastpriccy on each JGIOO laid out? Ans. £9Q 7s. 3-;\d. 67. A water tub holds 147 gallons ; the pipe usually brings in 14 gallons in 9 minutes ; tiie tap discharges at a medium 40 gallons in 31 minutes. Now, supposing these to bo left open, and the v^nterto !;e turned on nt 2 oVIock in the mornir.g; a servant at 5 t^liuts the tap, and wishes to know in what time the tub v\iil be filled in case the, wa- ter continues to flow. Ans. The tub will be full at 3 m.in. 48 ^] \ sec. after 6. Gf^ Take ^ a square foot from -",- of an acre. Ans. IR. ISP. 5yd. 4rt. 69. Two men and a boy were cngr ccd to do a pieco of woiU; orse of the men could do it in 5 days, the other in 8 day.-*, and the boy in 10 days ; how long uould it tuko the thrve t( jn ther to do it? Ans. 2-,''- davr. 70. Afler laying out J of my money, and j of the re- mainder, I had 72 guineas left; how much hrul I at fiivt? Ans. 120 guinea;?. 71. Two persons, A. and B., fire on the opj^osite tides of a wood which is 53G yards in cireumference ; they be- gin to travel in the same direction at the san.c mement. — A. goes at the rale of 1 1 yards per minute, and B. at tb.c n»tc of 34 yards in 3 minutes. The question is, how ma- ny times the quiclcer one must go round the wood before ho overtakes the slower? Ans. 17 time?. 72. ir a person take snufi" onco in 10 minutes, and Hj^cnd I of II minute in the process of snulVing, adjusting tho box, and blowing \1\q nose, how many hours will be thus MISCELLANEOUS EXERCISES. 259 I u t-idos y Ih'- nt. — : tl:c spent in 7 years, allowing 13^ hours to a pnuff-taking day, and 365^ days to the year? . ( 2646i^-{ hour?, or the wakeful •' An*' < ? •' ' ( hours of more than 5 months. 73. Afatherdyingbequealhs loone of hits ems ~ of his Cf^tate, and to another If of the rcmcindcr ; uj^on a divi- sion the latter bequest is found to fce $200 lets than the former. Qucry^lhe Irgr.cy of earn ? An.?:. T!ic former i^^MOO, the bttu' $1200. 74". If A. can mow an acre of grat;- in 5^ h< m^., and B. cnn mow ll acres In 9^ hour?, in what time can they jointly cut S^ acres'? Ans. 22'| hour?. 7D. How many pounds of f carbon ? Anr. 9 ; ji pound?. 7(). If a horFO can draw G^ tons en a rr.llrcad which ascends 25^ fret in a mile, w li;;t wcifilit wov.ld l.e drawn if the af'ccnt wai' only If) Tt. in a mi. the I'orcc rcfiiiired to move a given wt. being as the accent per mile ? Aii!'. 10 [ -' 3 tons. 77. A person in health lias ab.out 7r) pu'jat-onp, or beats of the artery in a minute. Now, a gun bcirg fired on one t-ide of a river, an observer dircctiy i/ppotite counts nine pul-at'oriF at ids wri.-t between seeing tl,c flaph and hearing the report; whatwa; llu> brcrdfli oi'ti.c river? An^\ Im'. Ifur. lOOydt'. Sft. 78. A hare starts 40 yards be'bre a greyhound, and is /tot perceived by him till >;he has been uj) 40 seconds ; 8h« runs at the rate often miles an h( ur, and tiie d( g. on view, mrkei^i after liCr at tlio rale of IS males an hour ; how Icrg will tiie counc hold, and what t^paec will be run over from the J pot \Nherc the di g t-taitcd 1 Ans. (iO-''.^ sec, and f:30\d?. fpace. 79. If to m,y age there added be, One-half, one-third, and three timiCc, three, Six score and ten the vuxn will be ; What id my o^e, pray show it me 1 Ans. 66yni. 260 MISCELLANEOUS EXERCISES. 80. A gentleman divided his fortune among his thre« sons, giving A. £9 as often as B. £5, and to C. but £3 as often as B J67, and yet C's dividend was je2584? ; what was the amount of the whole estate ? Ans. je 19466 2s. 8d, 81. The yearly interest of Mary's money, at 6 per cent, exceeds -\- of the principal by J6100, and she does not intend to marry aily man who is not scholar enough to tell her fortune ; pray what is it? Ans. ^10000. 82. A. and B. can do a piece of work in 4 days, and B. and C. in 6 dayg, and A. and C. in 5 days : in what time can they all do it trgetherl Ans. 8-3^- days. 83. A. and B. can do a piece of work in 5 days; A. can do it in 7 days ; m how many days can B. do it ? Ans. ns days. 84«. A mnn died, leaving £1000 between his two sons, one 14}, and Ihc ether 18 years of ?ge, in ^:uch proportion that the tharc of each, being put to intercut at 6 per cent., ishould amount to the srme turn when they diould arrive at the age of 21 ; wlint did each receive ? Ans. The elder £rA.6,l;)3-\- ; the younger i:4.53,846+ 85. A. B. and C. would divide i^lOO Ijctwcen them, 80 as that B. ir.ay have £3 more than A., and C. JG^ mor« tlian B. ; how much mu.^tcach niark have? An^^ A. £30, B. .i;]3, and C. £31. 86. A. and B. undeital:c a j/iece of work for $54-, on which A. employed 3 hands 5 days^, and B. employed 7 hands 3 days ; what pait of the work was done by A., what part by B., and what was each or^e's hhare of the mo- ney ? Ans. A., vV ; B., v- ; A's monev, $22,50 ; BV, $31,50. 87. A.B. and C. traded in ccmpanv ; A. put in $500, B. $350, and C. 120 yards of dotli ; they gained $332,50, of which C's share was $120 ; what was the value of C's cloth per yard, and what was A. and B's share of the eaini C's cloth per yard, $4. Ans. < A's share of the gain, $125. -^ B's do. do. $87,50. MISCELLANEOUS EXERCISES. 261 88. There are 3 horses, belonging to 3 men, employed to draw a load of goods from Kingston to Toronto, for JE26,45. a. and B's horses together are supposed to do % of the work ; A. and C's -^^- ; B. and C's ^^^ ; they are to be paid proportionately ; what is each one's srhare of the money? C A's ^£11,5 (=f 3) Ans. \ B's ^25,75 (=-/g^) ( C's i:9,2 (=-\) 89. A gentleman left his Hon a fortune, ^'j- of which he spent in. 3 months; ^ of {; of t'^e remainder las:t?d him 9 months longer, when he had only £537 left ; what waa the sum bequeathed him by hirf father? Ans. £2082 18s. 2j2,-d. 90. . There is a square field, each side of v.hich is 50 rods ; what is the diistance between opposite corners ? Ans. 70,71+ 91. What is the area of a square field, of ^vhieh the opposite corners arc 70,71 rods apart? and wiiat is the lengtii of each fc-ide? Ans. to la>t, 50 rods, nearly. 92. A trader beirg cmbarraf:ted, owes .*•' 31C0, which the creditor requires to be immediately paid. He has goods which lie can ^c]I at auction for ca^h at IT) per cent. below the jurt\a!i.e; he cannot borrow nio:i..'y without allowing a premium of 9 j)er cent., and payii g intc^rctt at 6 per cent, per annmu on the whole. Now, .".dniitt'ng ho can fiell his goods for their vahie within a year, which will be more elligihle, to eend th':m to auction, or to borrow mo- ney on ihcKC conditions, to !;ati^^fy his creditor? Ans. To borrow, by .'^,71,64?. 93. Two travellers, A. and 13., at the distance of 59 miles from each other, Fct out in order to meet. A. begins his journey 1 hour before B., and travels 7 miles in 2 hours ; B. proceeds 8 nnles in 3 liours; how far will they travel respectively before they meet? Ans. A. 35nii. B. 24». 94. A hare starts up r;0 of its own leaps heforc a grey hound, and takes 4 leaps while the hound takes 3 ; but tho hound goes as far at 2 leaps as the hare does at 3 ; hovr . 262 MISCELLANEOUS EXERCISES. I u many leaps will the hound take to catch the hare ? Ans. 300. 95. A. B. and C. can complete a piece of work in 15 days ; A. can do it in 30 days, and B. in 40 ; in what time can C. perform it] Ans. 120 days. 9Q, A servant, having eloped from his master, travels 14« hours a day, at the rate of 3^ miles an hour ; at the end of two days a courier is sent in pursuit, who travels 9 hours in the day, at the rate of 7 miles an hour ; in what ♦••nQ will he overtake him 1 Ans. 7 o^ys. 97. A Greek epitaph, designed for the tomb of Dio- phantus, is said to have stated that he passed \ of his lifo in childhood, -^.r in adolescence ; that after f and 5 years more had been passed in a married state, he had a Son who lived to ^ his own age, and whom he survived 4 years. What then was the age of Diophantus] Ans. 84yrs. 98. Three ninson?, A. B. and C. are to build a wall. Now A. and B. can build it in 12 days ; A. and C. in 15 ; B. and C. in 20. In what time can they jointly effect it| and how long will they ^:cveral]y required Ans. Jointly, 10 days, A. 20, B. 30, C. 60. 99. ' A gentleman meeting, by accident, with a dange- rous wound, sen^5- in an hour ; they pump out -i\-, hence the water gains -i^3-=-^ig-=-^-j;^ of the ship per hour ; hence it will fill in 240 hours. Ex. 62. The mother was to have twice as much as the daughter and half as much as the son. Hence, the daughter 1 part, the mother 2, and son 4=7 parts in all, then 5460+7= $780 daughter's part, 780X2=$1560 mother's, arid 1560X2=$3120 son's part. Ex. 63. The first will empty m Imin. -^^ of it ; second rh of it ; third -^-^ of it, together -6V+T2-¥+Tii7== aV-j »" 1 min. Then -3^^,- : 1 : : Imin. : 32-i*j-min. Ans. Ex. 64- A. does in Iday -^^- of the work ; B. -^^3- ; to- gether -i\+-i\-=f3^o ; thenf 3-j- : 1 : : Iday : o^^days Ans. Ex. 65. -iV+l+A+'/.r ^vill be equal to ail the school except the 9 who read. Of the denominators of these fractions 80 is the least common multiple ; hence -3%-+! 77 +ft+i^=li J and li^=-,V the residue of the school, which is=9; then if -P^^ of a school is equal to 9, how many in the school ? It is plain there are 80 ; then -i\ of 80=5: i of 80= 30 5 3 5 1 J of 80=24; :,^,r of 80=12. i u 266 SOLUTIONS. Ex. 66, lls.X|=»what he first cleared on each thoi.- eand j hence, they cost him | of lls.=6s. lO^d. He af- terwards sold for 13s. 6d. ; then 13s. 6d.— 6s. 10id.=6s. 7id. what he cleared per thousand by the latter price. — Then, as 6s. lO^d. : : 6s. 7^d. : £100 : £96 7s. 3-,^,-d. Ans. Ex. 67. In Imin. the pipe brings -^ gal • The tap dis- charges in the same time |-^ gal. ; hence it fills in 1 min. -^ — ii==2Virgal. It so runs for 3hrs.=180min. ; hence, it fills in that time -/-^S X 180=Hf ^gal- It has then 147 — 4^=3_apgal. J then, as U : Hf^ : • 9min.: ^^^ min=63min. 48^^sec., that is, after 5 o'clock it fills in 63 min. 48f-}^4s6c. ; that is at 6 o'clock 3 min. 48^J-^sec. Ex. 69. All can do in 1 day, i+i+TV=i-^ Then, 15^ : 1 : : Ida. : 2,^^ day, Ans. tn^ 70 1 1 — " • J- of 3 — 3_ • tlion J J 1 P what was spent, and 1 — jA=H> what was left} then, as ■^ : f^ : : 72 : 120 Ans. Ex. 71. A. goes 11 yards per minute, or 33 yards in 3min, while B. goes 34 yard«< ; hence B. gains 1 yd. in 3 min. ; and to gain ^f *=268yd, he must travel 268x3= 804 minutes. Then, 3 : 804 : : 34 : 9112 distance travel- led; then, 9112—536=17 times round. Ex. 72. 13f hr.=!=Vhr. ; 365^da.=^4<2.x . then ^X "^V^X1='^H?,^^ hours. As^i^^l^^ : : | min.: iLajJLajL.t==2646|f 9 hr. Ex 73. f— f = f ; then if of ^=H J also, f — 1| = ,^9. Then, as ,\ : f: : 200 : 1400—200= 1200. Ex. 74. As 5'|hr. : 9^hr. : : 1 A. : f fA. quantity A. would mow in 9jhr. : then ff-\-l=%%^ acres, quantity A. and B. together would mow in 9^hours. ^3^ Y=22'| hours Ans. As -2-3-1 •*»-3 6 8 5.11 4 9f«lb. Ex. 75. »f ^X 4%== V^ J then, as f^ : \^ : : Ex. 76. As 15^ft. : 25|ft. : : 6^ tons : lOjfg Ans, Ex. 77. The velocity of sound is 1 142ft. per second, the distance passed over in Imin. w^ould be 1142x60= 68520 feet. But in this time the pulse beats 75 times. Hence, 75 puis. : 9 puis. : : 68520ft. : 8222ft.= lmi. 4fur. 100yds. 2ft. #> SOLUTIONS. 267 ■i k Ex. 78. Because 40 seconds is just -^^ of an hour, and the hare runs 17600 yards per hour. She must run 17600-i-90=: 1958yds. in 40sec., and would be 195««4- 4'0=235|yds. before the dog at his starting ; but as the dog gained 8 miles in 18, he would gain 8 yards in 18, or 4 in 9 ; hence, 4:9:: 235| : 530yds. run by the dog. And 1760X18=31680 : 530 : : 60min. : Imin. 0-/,-sec. Ex. 79. The meaning of this question is, that the num- ber 9 added to once his age, together with ^ and ^ of hia age, the sum shall be 130 ; or since the sum of the parts 1 and ^ and | is Vj ^'^^^ of his age is 130 — 9= 121 ; then 11:6:: 121 : 66, Ans. Or by Position. Ex. 80. As 7 : 5 : : 3 : 2|, and 94-5+2; = 16 ^ ; then, as 2| : 16^^ : : ^2584 : i^l9466 2s. 8d. Ans. Ex. 81. Suppose 200; then 2U0x6-f 100=12 in- terest; then ^„^ = 10, and 12—10=2 ; then 100—2=98, error. Again, suppose 300 ; then 300X6-4-100= 18 in- terest; then ^// = 15, and 18—15=3; then 100—3 = 97, error. Then 97x200= 19400; and 98x300=29400; then 29400—19400=10000 pounds, Ans. Ex. 82. A. and B. can do |, B. and C. ^, and A. and C. I per day ; then 4~l~i+i= el? which -^2 (because each man, by the conditions, is taken twice) =-i3./^- what all would do in 1 day; then l--—^^;^^ z=:3-^%- days, Ans. Ex. 83. A. and B. together can do i, and A. can do I alone ; then f — 5 = 3^5 what B. can do in a day ; then l-L-3«3 = 17idays, Ans. Ex. 84. Amount of £1 for (21—18) 3yrs.=i21,18.) Amount of ^1 for (21—14) 7yrs.=jei,42.) - -^i f =£2,60. Then (as they will receive inversely as the time.) As ^£2,60 : £1.42 : : JElOOO : £546,153 Ans. As £2,60 : £1,18 : : £1000 : £453,846, Ans. Ex. 85. B. has 3, and C. 7 more than A. ; 7-|-3=10, to be taken from 100= 90-f3=30 A's ; then, 30+3=33. B's, and 30+7=37 C's share, Ans. Ex. 86. 3 men 5 days=15 men 1 day, and 7 men 3 days=21 men 1 day ; 15+21=36 days; il=-^\r,A,', ii68 SOLUTIONS. ?,i=-i\-, B] then $54X-,V~$22,50, A's money, and $54X-,V=f31,50, B's money, Ans. Ex. 87. First, to find the gain of A. and B. ; C's gain being $120, $332,50—120=212,50, the gain of A. and B. together ; then, $850 : $500 : : $212,50 : $125, A's ; and $850 : $350 : : $212,50 : $87,50, B's. To find the price of C's cloth 500+350=850; then as 212,50 : 120 : : 850 to C's stock, $480-7-120=$4 per yard. Ex. 88. ■^+-i%-+i^=f|-^2, (as each man's horses are taken twice in the question)=f^; then, f% — | (A. 9nd B.)=t/„-, C's j f f, — ,%- (A's and C's)=2%-, B's ; and w^—H (B's and C's)=^, A's ; then A., will have H of $26,45=$n,50; B. will have -/ff=$5,75, and C. -/g ==S9,20. Ex. S9. I of |=g i then, je537 is |— 1=|, and jen37-^^|=jei432, tlie sum he had after he had spent -j\ «)f his! fortune, and consequently this must be -}-|- of what he hadal first; then, i51432-Mi=je2082 18s. 2-,\-d, Ans. Ex. 92. As the goods, when sold at auction, sell at 15 j)ei' cent, below their just value, a quantity worth $100 must be sold for $85. Therefore, as $85 : $3400 : : $100 : $4000, the value of the goods which must be sola at uuction to pay the debt. Debt $3400, premium at 9 per cent. $306 ; amount of debt and premium $3706. Interest on this sum for 1 year $222,36 ; amount $3928,36, which is less than $4000 by $71,64, Ans. Ex. 93. A. travels f miles, and B. 4 miles in 1 hour ; ^+1=3^^ miles distance travelled by bovh in 1 hour ; 59 — |=b:J-^j- mi.ssthe distance they are apart when B. sets out. ' ^ * X-sV=9, number of hours B. was travelling ; then i^X*,^ =35 miles A. travelled; fX9=«24 miles B. tra- velled. Ex. 94. As 2 : 3 : : 3 : 4^s3number of the hare's leaps, which are equal 3 of the hound. But the hare takes but 4 leaps while the hound takes 3. Therefore, the hound by taking 3 leaps gains ^ leap on the hare ; he must consequently take 300 to gain 50. ■■ It* SOLUTIONS. 269 . Ex. 95. "iV — Vo — 4V= iio=P^^' ^* ^^^^ ^^ 1 *^^y 5 he will therefore do the whole in 120 days. XiX. Vi, 6 riyn^7T^2 84 > 8 4 8 4 8 4 P"*^ "' his life which the 54-4^ years composed j therefore, -^\ : 9 : : 1 : 84 years, Ans. Ex. 98. The part done in a day by A. and B. Is -/^-, by A. and C. -,V> and by B. and C. -/„-. The sum of these fractions, f , including the part of each twice, is evi- dently twice the part performed in a day by A. B. and C. together. They therefore perform -i\p part of the work in a day, or the whole in 10 days. Again, -/g- — :^'y-s=rTj\j part A. can perform in a day ; -^\ — -iV= s'o part B. does in a (lay ; ,-^^, — ^., = ^f^ part C. does in a da3% Ex. 99. The reciprocal of a number is the number in- verted; hence, ,^i-.-\-^ + ^=.-Jf^ ^ then, as -,VV.r : t^ : : $100 : $39,25 and as ,7,^5 : ^\, : : $100 : $32,71, also, f\.Vo : Vf : : $100 : $28,04-. Ans. Ex. 100. 2+3+4=9, f of ^f^=£52 A's gain ; f of ^^^=^678 B's gain ; ^of ^^^^=£104 C's gain. Now -^f =13 ; -^ = 15,6 ; -"^^==8 : and 13+15,6+8=36,6. ^ C 13 ) £ ( jei560 A's stock. As 36,6 I 8 .6 1392 : I £1S12 B's do. £ 960 C's do. I Y'4 INDEX. «•- PART I. Page Numeration and Notation, - 11 Roman Numeration, - - - - 15 Simple Addition, - - - 16 Simple Subtraction, - _ . . 21 Simple Multiplication, 24 Simple Division, - - _ . 33 Fractions, _ - - - 44 Supplement to Multiplication, 47 )* , PART II. Compound Numbers, Tables of Comnovmd Numbers, Reduction, Compound Addition, - Compound Subtraction, il Compound Multiplication, - Compound Division, - Bi'ls of Parcels^, Federal Money, 55 55 62 75 82 85 89 95 97 >% PART III. Simple Interest, Ratio, - - - - Rule of Three, Practice, - - - - Commission and Brokerage, Insurance, - - - - Discount, - - - - Loss and Cain, - - - Equation of Payments, Fellowship, - - - Double Fellowship, - , - 103 109 110 118 126 128 129 131 135 137 138 INDEX. Tare and Tret, - . - . Vulgar Fractions, Reduction of Vulgar Fractions, Addition of Vulgar Fractions, Subtraction of Vulgar Fractions, - Multiplication of Vulgar Fractions, Division of Vulgar Fractions, Review of Vulgar Fractions, Decimal Fractions, Addition of Decimals, - Subtraction of Decimals, Multiplication of Decimals, - Division of Decimals, - Reduction of Decimal;?. Rule of Three, - - - - Rule of Three by Analysis, - Rdle of Three in Vulgar Fractions, Simple Interest by Decimals, Exfliange, - . . . 139 141 M5 161 165 167 170 173 176 177 17S 179 180 18^ 1S() 188 189 194 201 PART \X, Involution, _ - - Evolution, - . - Extraction of tiie Square Root, Extraction of the Cube Root, Mensuration, _ _ - Gauging, - - - - Duodecimals, - .. - Position, - - - - Double Position, Annuities, - - . Alligation, . - - Arithmetical Progression, Geometrical Progression, Permutation, - - - Miscc'llaneous Exercises, Solutions of do. xlOo 207 208 213 220 OOw 230 0«JO 231 23() 23S 243 247 250 251 263 ^p ?