PERRO re POOF RAINE A A LAL PLAS PY A IB aGd NE SLABS. ne Men te FAB fl tom ae PLP IMA DD Pah Pi npep SHORT PRACTICAL RULES ‘ FOR COMMERCIAL CALCULATIONS, INCLUDING DIVISION SIMPLIFIED AND ABBREVIATED, AND SHORT METHODS FOR MULTIPLICATION: ORIGINAL AND INGENIOUS METH- ODS IN CANCELLATION, THE RULE OF THREE, PERCENTAGE, INTEREST AND DISCOUNT, AND A SIMPLE METHOD FOR AVERAGING ACCOUNTS. TO WHICH - RXPOSITION OF PROFIT AND LOSS, DIVISION INTO PROPORTIONAL PARTS, PARTNERSHIP AND BANKRUPTCY, INVOLUTION, COMPOUND INTEREST, ANNUITIES, SINKING FUND, AND BOND COMPUTATIONS, STERLING, CHRONOLOGICAL CALCULATIONS. A SIMPLE METHOD FOR THE EX- TRACTION OF THE CUBE ROOT, AND 4 SIMPLE RULE SHOWING HOW TO DISCHARGE A GIVEN DEBT IN SEVERAL EQUAL PAYMENTS, IN A GIVEN TIME, INCLUDING PRINCIPAL AND INTEREST, AT A GIVEN RATE PER CENT. A SIMPLE METHOD FOR ADDITION, AND PRACTICAL HINTS FOR BUILDERS, ETC. BY PATRICK MURPHY. ALBANY: WEED-PARSONS PRINTING COMPANY, IgI2 Entered according to act of Congress in the year nineteen hundred and four, By PATRICK MURPHY, In the office of the Librarian of Congress at Washington. Entered according to act of Congress in the year nineteen hundred and twelve, By PATRICK MURPHY, In the office of the Librarian of Congress at Washington. PLAT IS os MATHEMATICS tiopas Sm Ind LIOKAL BUR 1 AsT On nee -= A trivial matter having led the author to make some investiga- \ tions in the fundamental rules of Arithmetic, particularly Division, —~ and having, in the course of these investigations, discovered some novel methods for simplifying and abbreviating Long Division, ‘» he has concluded to present them in this form to the public, flattering himself that they will be found, on perusal, both inter- = esting and instructive. ~ In treating the matter, the author assumes that the reader has es a knowledge of arithmetical notation and numeration, the simple 03 rules, and the principles of Division as given in most Arithmetics of the present day, and which will be essential to a proper under- standing of the subject; hence, the explanation of technical _ terms, definitions, etc., unless when necessary, will be found ; omitted. ~\ Division, we are aware, could not be changed without making, o at the same time, a corresponding change in Multiplication — one being the reverse of the other —hence, some extraordinary re- sults will be found in the contractions in Multiplication, to which attention is invited. The method for Addition will be found advantageous when- ever the person engaged in adding is liable to interruptions. , The rule for finding the weekly day, which forms the closing Dieter of this little work, is quite simple, requiring no arith- metical process beyond Division for its solution, and is easily remembered, as it requires neither monthly nor centennial ratio ~ to be added. The rule is original so far as known to the ——, | AUTHOR. ; ALBANY, January, 1886. a * + Oxctnsnrsy A 4 V1 4 PREFACE, NEW EDITION REVISED, ENLARGED AND IMPROVED. In the preparation of this work, our first attempt at authorship, Division Simplified, ete., has formed the groundwork of construc- . tion. The present edition contains the principal feature of the original work, the Vertical Line, by the use of which, most extraordinary results in arithmetical calculations are obtained, and problems solved much more readily than by the usual methods given in text-books. In the new work, we have endeavored to make the use of the line more simple and clear, explaining the most. important prin- ciples connected with it, and giving numerous examples and illustrations. Original and Ingenious Methods will be found in Cancellation, The Rule of Three, Percentage, Interest and Discount, Averaging Accounts, Profit and Loss, Division into Proportionate Parts, Partnership, Bankruptcy, and Practical Hints for Builders, which are given in the present work. In presenting this edition, the Author trusts that, by the improvements now introduced, it will be rendered worthy, in a much greater degree, of a continuance of the very favorable reception which it has already experienced from a kind public. Tort AUTHOR TO THE READER. Before attempting the simple and novel methods given in Can- cellation, the Rule of Three, Interest, Discount, etc., w thorough knowledge of our methods for Division and Multiplication will be necessary, as without the latter, the simplicity and novelty of the former can neither be understood nor appreciated. January, 1889. PREFACE. 5 IMPROVED EDITION. | Some very important improvements have been made in this edition, and fifteen pages of new matter added, which, from long experience, the Author knows will be found useful and practical. To particularize all that has been done would be tedious and unnecessary: suffice it to say that, in the present edition, a more comprehensive and lucid explanation of our Simple Method for Averaging Accounts is given; Short Rules for converting British Sterling to American Currency, and for computing Interest on British Sterling; Hints on Interest, and Other Short Methods, together with a Simple Rule for finding the face of a note, the proceeds being given. This, we will venture to say, is the first time the rule. has appeared - in print, and will be found of great practical utility. The reasons of the rules and operations (a part of arithmetical science too generally neglected both in treatises on the subject, and in teaching), are fully explained by simple and easy illustra- tions and examples; and it is hoped that the subjects will thus be rendered intelligible and attractive tothe reader. July, 1895. ENLARGED NEW EDITION. Twenty pages of useful and practical matter, carefully prepared, have been added in this edition. The article on Interest Simplified cannot fail to interest the student. The Short methods for finding Interest on Daily Balances, and for changing Commercial into Exact Interest, and the reverse; also Short Methods on Tonnage, both Net and Gross; and the Short Methods on Trade Discounts, together with numer- ous examples and illustrations, and the reasons for those Short Methods, will be found worthy the attention of the reader. July, 1900. @ We have added to the present edition a Review on Interest, with illustrations and examples, the most complete, perhaps, ever given on this subject. We have also made important changes in the body of the work, and added fourteen pages of Hints and Helps for the Studeat, which will be found both useful and interesting. August, 1904. 6 PREFACE. REVISED, ENLARGED AND IMPROVED EDITION. Several changes of a very important nature have been made in the present edition. Twenty-one pages have been thrown out, and new matter of a more useful and business character substi- tuted. Besides this, Twenty-five new pages have been added, consisting chiefly of very simple methods for multiplying together numbers of two, three, four, five and more figures, whether whole or fractional, with copious examples, illustrations and reasons. These methods, so far as known to the Author, are entirely original, and will be found exceedingly simple and practical. August, 1906. Some things of minor importance, contained in former editions, have been omitted in this edition, and the space thus gained has been filled by the insertion of valuable rules, which cannot fail to be interesting to the student. July, 1908. LATEST EDITION, ENLARGED AND IMPROVED. To this, the latest and best edition of the work, we have added a Chapter on Involution; Compound Interest, including a Table ~ showing the Amount of $1, or £1 sterling from 2% to 10%, for any number of years from 1 to 35; and a Simple Rule for the Computation of Bonds; and for Sinking Funds; also a Rule to Ascertain the Amount necessary to Discharge a Given Debt in Several Equal Payments in a given time, including both Prin- cipal and Interest, at a given rate per cent. In connection with this rule, we have given a Table showing the Amount necessary to discharge the Debt of $1, in equal pay- ments,- from 2 to 21 years, thereby facilitating the solution of such problems. Much labor and care have been given to the construction of this Table and, so far as known to the Author, it is the first of its kind given in an Arithmetical work. We have also given a Short Rule for the Computation of Paper; and a Simple Rule for the Extraction of the Cube Root. The rules are Tersely expressed; the Problems are all solved and fully illustrated, so that the Student will have no difficulty in mastering the different subjects. January, 1912. CONTENTS. PAGE Eee INLINE 5 ov. soir se pan cme stne cece ctcecs panate oa. General Principles, Land I] .............ceccee cecereces 9 PON RG cine worse os 26 es Pen tenn sd We aes fe oladine. s 10 Principal Wustrated ./..... 25... . cee es eee eee eee er eenenes 11-15 oR Mee EON SE tt. n giuccie ele e's 49 0b so .ese'e = 15 Tanne the Decimal of the Remainder. ......,....6.00.00- 17 Simple Method to find the Decimal .. ...........+-.0.00. 17 De INGIOlGS elie, fs. le kere c clelo vibe canes cece ences 18 PUTIOT DION LIUSLTALEC 6c ww eee elt e ow Seveee seme exes 19-24 Se oe Aaa > ata lals 22-5. «ys o'e dias <0 c <-6 #19 w 0s! 24 eC O LL lees oe hc pide as eS shalt els pw se 64.0 4 © Son s.9 29 Problems solved by Rule if I eetayey oe oe pees Ta ye ee 30-35 Problems solved by Rule IIT .........ccccccocees aitsrel Paes 30-47 Premarks on Rule Tl)... wc ccc enews nctcaccoesene 38-41 ORS Pa SS (TEs (2 21 a eT ad perss omits Co General Short Method for all Numbers.......... om oscar 49 Short Method to reduce Square Feet to Acres...........06- a) Gross Cost given, to find cost of One Article.............. 57 Me as a ORL fa ioe oe. ca es Ves s 8s & boar Stce oc 58 meetnods Of Proof for Division... 66... cas veces eRe Neenia 62-65 TL IROLE Tate os CWS wos eins ov alsin a ae chain ov bw weeds 65 To Divide by the Nine Digits in Direct Order ............. 68 To Divide by the Nine Digits in Reversed Order........... 69 mepoiviie byiMixed NUMbETS. 0050... wae ee Gk seed eoewe 70 meoriahcinons tor Multiplication’. 022.6005. pee SO. 71-88 To Multipy by the Nine Digits in Direct Order........ ss &2 To Multiply by the Nine Digits in Reversed Order ........ 83 Methods of Proof for Multiplication ; Figures, 9and11.... ° 88 RRO MORES TERE Par Soar edi aah kale oho. 5 ae pte wh ao. wis 26 pl lwo 0. wie w/e 89-97 ities prule.of Three”. o....5. 3 Sct hy OPE ey ere 97-106 Pee OM TIC ETOPOTHMON aiy. 4 cece die see ne ss deletes are aveten ters 107-109 erecvica ) roolems -"lron) Steel ete. ...8 oes esceccsece seca’ 109-113 ENE PR BTS ito Fs One 4 a xg Wn kb alo CR elewselawiain 6 115-121 Interest, both Commercial and Exact .... .ssseeesecee eee 122-128 martial Payments Or INGOrsements .. seine aecc ccs cuceee 129 ae T ICO iG ale ie ie i fad iets co vin eB nv seats bog e'tens ot 131-136 Averaging Accounts; Interest Methods...... ........ «- 137-142 RETRACTS A Sree tees sete 5! aise titi dias «; wade ofoie's v0 5 SOlN Ss 143-148 Divison into Proportional Parts. .............. eee ener’ ee 149 Peavitaraisi pin ss). =. bahar gs sete ete CPE nk sade? ema os) aot ove. 6 s4 ® 152 Pe SAL Ven ah PES alae 25%, Paw AREAL VA WC et «Ne eee ore ae 154 Deer terctilesee with LilUstrationses <. sickvevescesse eee. abs - 155-158 Serer Caer IHbe MOT EWP, iereisieicy pei «0 << 64 class 04'e cae 159-173 Lumber Calculations Simplified...... Se aienes aye pie eflanes aiets 172-173 MOE ICHU RICUIAUOUS) Oye c rite deeedeesestececds arte 174 Bissextile or Leap Year Explained ....cocessesscsiecvecces 176 APPENDIX. PAGE Interest Rules Tersely Stated and ene 20 ni-eis Sidiole’s ein sleaeretinamnelets 177 Special Rules — Six per cent. Method.... ... ,.....seeeceres wd ses aia pee ae 180 Interest on Running ACCounts <2 o. sive. sou - news cele theleseieierete te tne amernae 182 Important Facts to be Remembered | CaS Rae Sietele eieis, bia) cate Nenejere tele eretaaterene 184 A Simple Method for Averaging AcCounts.........ccccescsscsceec sce o-o. 185-194 Short Methods — Steel, Iron, Coal, ete s2..)2... siete 6 sone stein erate ane meets 195-200 Pounds Reduced to Gross Tons; Short Meee: seach eee a pane epee 200 Sterling — Pounds, Shillings and’ Pence. oo... se ccs sos van ele renee 201 Sterling Reduced to American Cee sio.sie oie le via seis aye isles ale eee Seamer 204-206 interest on’ Sterling’. oe seve soso ce bac Se Soeseare cab wis olerere eee 206-209 Other Short: Methods hase Ft Seisele tere tee ee roa) oslo 6 sole miaers aeenereeete ee 209-211 Hints on Interest — Savings Banks, 3/3352 wig 6 bntete'ere stew ¢ ereeeettee Japaeeee eis Interest on Monthly Payments............. celestial eileen eer weenie Seer e eee 212 Interest on Weekly Payments... «Usual » volaldcaleretelete Pusleinte 213 Simple Method to Find the Face of a Note. .6200e va 2 nes WSR ne eee 214 Interest for Months at any Rate per cent.......... ...e+eceeee Sie gharsiemiereiens 215 Interest Simplified ... = .... ... e'sinie eos 0 heels, salto eeneieeleta enema Oe ee) Important Facts, Illustrated by Exam ples... sia 8.8 sie.eiaielsoleis ee arate 218 To Change Commercial Interest to Exact Interest. 3... ee cipoastesie 219 To Change Exact Interest to Commercial Interest..,... o irelalielata oietee het ete 220 Interest on Daily Balances — Short Method....: ............ ASIC De 220 Bank Balances — Subtraction performed by Addition....... songs ate Sanletteiers 221 Pig-Iron, 2268 lbs. or 2240 lbs. — Short Methods. . oo 0's 616 Ree elton iS tenteren ceemeree To Reduce Pounds to Tons of 2268 1bS........-.sss0ccceeceeee. Siew eciersleus ei 224 To Reduce Pounds to Tons of 2240 Ibs. . bk 's ioe Wiel ete ina /ach anya teh ereter ata eae 225 Gross Tons Reduced to Net —Short Method..........c.sccceececcececeees 226 Net Tons Reduced to Gross —Short Method... ........ 00.00 ecen cee vee 227 The Net Cost being given to find the Gross Cost.........c.cceccccsce cece 223 The Gross Cost being given to find the Net Cost........ .cccc.cccccees es 228 Computing the Cost of Pounds by the Net Ton........... Sabice Reale ete veces 229-230 Gross Tons of Rails to Mile—Short Method..............cceceees BU Goer 231 Net Tons of Rails to Mile — Short Method...... Stade tateee “sik tonsietarsiniereteieiats 231 Trade Discounts —Short Methods ~.. 02. 5.2... 2. v2s cane ce cance eireemeemee=caa Interest: Reviewed. 2.0.0 90 6 lhe Fe Wess Sele veered WoL smisen 5 elle ee Reet eee 238-245 When the Rate Changes Fr equentlyoec.win cjoeaieeerieete ose wiesaaatetoen tien 245 Annual Interest. .-. scare: 6 ca.g Se celele. eralvare rebehotietelateneeseo mt arene 246 Partial Payments — Merchants’ Rule,.:.:.-..)s-0s-0.. eee alee eeae ets : 247 Hints:and Helps for the Students... ics ce cocsnl eee eee enemas Saar. 248-280 Decimal Fractions. © e660 sis ied win woes Je lw tae Se eee 254-257 To Multiply the ’Teens Together — Short Method.. . .... ..............- 258 To Multiply the Twenties, Thirties, etc., Short Method............... ... 259 To Multiply Mixed Numbers. .2:..0:.¢. 0s 83) bcd er 277-279 A Simple Method for Addition... 5... c0.0cc.0) 2202.2 ccs code sees eee Oe A Simple Method for Subtraction ................eeesee- oars sueeteaeme calc renaes 283-284 Problems with their Solutions . 2.5.52... seeccs 0 cues dele ene ERT tereeearenn 285 TAV.GLUELON ES poh Les neni eae reat ool nie erent 9 ayereSsin sere piekele of eccliel ele nectar Neer 293-295 Compound Interest, and Tables... onesie eo cleia) seal levhipie's se evens eTetele tere aaiarerer aoe acct Sinking Fund Computations srgieceret 1a'eje\ioia.oca vai clei sicte ee) eievel aia erels emilee epereeate ee eteowes 301 Bond Computations. 22 16 30s cas oe co Sere cee eee nee Ete eek tegmastirg 302-303 AMNICIES {Seek ee whi Eee ele sea eae ola cee ee 303 To Discharge a Given Debt in Several Equal Payments, Including Principal - and Interest, at Any Rate Per peace ene Te to ee acy Computations. . : ves eee . 804-307 A Simple Method for Cube Root.. ..s:ss0s+seseces anak -totee oeteene eae 808-309 DIVISION SIMPLIFIED AND ABBREVIATED. GENERAL PRINCIPLES. There are certain general principles of Division, a knowledge of which is essential to a proper understand- ing of the simplified methods given in this work. The following are some of the most important: I, Since the quotient in Division is the result obtained by divid- ing the dividend by the divisor, it is evident that the value of the quotient depends upon the relative values of the dividend and divisor. Hence, | Any change in the value of either dividend or divisor must produce a change in the value of the quotient ; but if a similar change be made in both dividend and divisor, at the same time, the quotient undergoes no change. IJ. If a number be added to the divisor, the dividend must be increased by the product of the quotient and the number so added, in order that the quotient may remain the same. Thus, 84-10 — 8...4 ' Now, if any number, say 2, be added to the divisor, 10, and we desire to divide by 12, without changing the quotient of 10, we increase 84 by 2 times the quotient 8, or by 16; then, dividing the sum, we have 100 + 12 = 8...4, the same as was obtained by dividing 84 by 10. 10 DIVISION SIMPLIFIED AND ABBREVIATED. And if the dividend be no¢ thus increased, the quotient will be diminished by the result obtained by dividing the product of the quotient and the number added by the new divisor. Thus, 84 +10 —8...4 Now, if we divide by 12, without increasing 84, we have, 84 +12—7 the same as if 2 times 8 were divided by 12, and the result sub- tracted from the quotient of 10. Thus, 84 - 10 — 8...4, less 2 times 8, or 16 divided by 12: 16+ 12 —T1...4 7 Notes. — 1. When the divisor is contained in the dividend without a re- mainder, the division is exact. 2. When the division is not exact, a part of the dividend is left, this is called the remainder and must be less than the divisor. 3. The remainder is always of the same name or kind as the dividend, being a part of it. The division of one number by another is denoted in severat ways. Thus, either of the expressions, 84 + 10 = 8...4; S4 — 8...4; or 10)84 means that 84 is divided by 10, that 8...4 the quotient is 8, and the remainder 4, fully expressed 8-4). The quotient, also, may be expressed in several ways. Thus, 875 the last expression, it may be expressed thus, 8 | 4, showing that, to divide a number by 10, we simply cut off the unit figure of es dividend for remainder. 8’/,), 8.4, or, if we substitute a vertical line for the point in Rute To divide by a number expressed by 1, with a cipher or ciphers annexed: Cut off from the right of the divi- Div1ston SIMPLIFIED AND ABBREVIATED. 11 dend, by a vertical line, as many figures for remainder, as there are ciphers in the divisor. Exam. 1. Divide 475891 by 1000. 475|891 Here, there are three ciphers in the divisor, and we cut off from the right of the dividend, three figures, 891, for remainder, and 475 is the quotient; fully expressed, 475891, Now, let a number, say 7, be added to the divisor, 1000, and let the given dividend, 475891, be divided by the sum, or new divi- sor, 1007, according to principle II. Thus: Dividing first by 1000, we get 475 for quotient, and 891 for remainder. 475|891 To finish the division, now, according to the _ 8/804 principle referred to, we multiply the quotient, 472|587 475, by 7, the figure added, and divide the pro- duct, 3325, by the new divisor, 1007, as shown in 1007)3325(3 : : : 3021 the margin, getting 3 for quotient, and 304 for 304 remainder. Subtracting this result from 475/891, found above, we get the true quotient, 472, and the true re- mainder, 587; fully expressed, 472,°8,4. The required quotient for 1007, however, may be more readily obtained as follows : First, divide by 1000; the quotient is 475, 475|891 ~ 1007 and the remainder, 891. Then, extending the 3/825 vertical line to a suitable length, multiply the 479/566 quotient, 475, by the excess, or added figure F1 7, and set the product, 3325, so that its unit 587 figure will be under the unit figure of the re- mainder, 891, and subtract the said product from 475/891. Next, multiply 3, that part of the product (now partial quo- 1 DIVISION SIMPLIFIED AND ABBREVIATED. ~ tient), to the left of the line, by 7, also; set the result, 21, under the remainder, 566, and add both; the sum, 587, is the true re- mainder, and 472, to the left of the line, is the quotient. The reason of the last process will be understood by comparing it with that which immediately precedes, in connection with the well-known principles: (a) the greater the divisor, the dividend remaining unchanged, the less . will be the value of the quotient; and (b) the less the divisor the greater the quotient. The extended vertical line, observe, divides 3325, as well as 475891, by 1000. Now, the product 3325 being divided by 1000 in the last process, while it ought to have been divided by 1007, as in the first (Prin. ID), the result, 3|325, is too large (Prin. b); and subtracting this result, which is too large, from 475|891, leaves too little; consequently, we must add the excess taken away. This excess, observe, is the difference between 3/325, in the last — process, and 3/304, the true result taken away in the first process, or 21, which is 7 times 3, the partial quotient, or that part of the product, 3/3825, to the left of the vertical line. Or the reason may be explained thus: The true result to be subtracted is 3/804; we have subtracted 3/3825; the difference is 21. Having taken away 21 too many we restore the same by ad- dition. Before giving the rule for this method of division, the following four examples are necessary, as they still fur- ther illustrate our method and explain some changes which will frequently occur when making use of this method : ne DIVISION SIMPLIFIED AND ABBREVIATED. 13 Exam. 2. Divide 3487286 by 1006. 3487|286 20/922 3466/364 126 490 In this, we first divide by 1000, as in the previous example, get- ting 3487 for quotient and 286 for remainder. Then, multiplying 3487 by 6, the product, 20922, is set under the first result and subtracted; the difference is 3466|364. In the subtraction 1 is carried, which makes 21 to the left of the vertical line, and in multiplying next by 6, we multiply not 20, but 21 (20 plus the 1 carried), setting the product, 126, under the remainder, 364, and adding; the true remainder is 490, and the true quotient 3466. Nore. — The number carried from the remainders to the quotients forms a part of the quotient to which it has been carried, and must evidently be maltiplied in with it. Exam. 3. Divide 348728654 by 1010. 348728654 3487/20 52411374 34/870 76/244 350 B45275|/894 10 904 Here, we first divide by 1000 as in the other examples, cutting off 654 for remainder. Then, multiplying the quotient, 348728, 14 DivISION SIMPLIFIED AND ABBREVIATED. by the excess, 10 (to multiply by 10, we simply annex a cipher and copy the figures), the product, 3487280, is set as in the pre- ceding examples, and subtracted (omitting the two outside fig- ures, 34, for the present, as they undergo no change by the — subtraction). Next, multiplying that part of the product, 3487, to the left of the line, by 10, also, and setting the product, 34870, as before, it is added (omitting the figures 52 for the present). : In the addition 1 is carried to 34, making 35, which is now multiplied by 10, the product 350 set as before and subtracted. The figures 3452 are now brought down, giving 345275 for quo- tient. In subtracting 350 from the result above it, 1 was carried and taken from 76, giving 75; and if actually expressed, said 1 would be set under 76. It is evident, therefore, that such a figure is a partial quotient, the same as 3487 and 34, and must be treated accordingly; hence, 1 is multiplied by 10, also, and the product 10 added to 894, giving 904 for the true remainder. Exam. 4. Divide 3509615 by 1012. 3509/1615 49/108 3467/507 504 1011 In this we divide by 1000, as in the other examples, and mui- tiply by the excess, 12, getting 8467 for quotient and 1011 for remainder. In adding the remainders 504 and 507, it would appear that we ~ ought to have carried 1 to the quotient’s place, multiplied said. 1 by 12, as in the preceding example, and subtracted; but we observe that the sum, 1011, is /ess than the divisor 1012, and,% therefore, 1011 is the final remainder. DIVISION SIMPLIFIED AND ABBREVIATED, 15 Exam. 5. Divide 7423520 by 1040. 7423/5210 + 1040 296/92 126 60 11|88 7138|48 48 In this, we cut off the cipher from the right of the divisor, 1040, ~ and also the cipher from the right of the dividend. Then, divid- ing 742352, the remaining part of the dividend, by 104, the re- maining part of the divisor, as in the preceding examples, we get 7138 for quotient, and there is no remainder. In multiplying 296 and 11 by 4, the excess in 104, it must be borne in mind that 1 has been carried from the remainder 92, in subtracting, and 1 also from the remainder 88, in adding, and that we have multiplied 297 in the one case, and 12 in the other, by 4, getting 1188 and 48, Nore.—It need scarcely be remarked that the numbers to the right of the vertical line are remainders, and those to the left quotients; that is, partial remainders and partial quotients, till the final quotient and remainder are found. Route II. From the foregoing examples and illustrations we derive the following: Rute. To divide by a number which is a little in excess of 100, 1000, 10000, etc. : I. Divide, first, by 100, 1000, 10000, etc., as the case may be, using a vertical line to cut off the remainder, as pointed out in Rule I. (Call 100, 1000, etc., in such cases, the approximate divisor.) Il. Hatend the vertical line to a suitable length. and multiply the figures to the left of said line by the excess jigure of the divisor, setting 16 DIVISION SIMPLIFIED AND ABBREVIATED. the product under the first result, so that writs will be under units, etc., and subtract. . Ill. If a figure or figures of the suid product eatend to the left of the vertical line, multiply such figure or figures, also, by the excess figure of the divisor, and include in the multiplication, the figure car- ried, if any, from the remainder to the quotient, in the subtraction ; set the result to the right under the difference already found, and add ; and so on, as long as possible, subtracting and adding alternately, and including in each multiplication the figure carried from the remainders to the quotients in subtracting and adding. Note.—When the last product figure, or carried figure, to the left of the line, is brought to the right by multiplication, the division is completed, the quotient being to the left, and the remainder to the right, of the vertical line. The reason why we subtract and add alternately will be understood from the following, in connection with the principles of Division already laid down. By referring to example 3, page 13, it will be seen that, instead of dividing 3487280 (the product of the quotient and the figure added, or the excess 10) by 1010, the vea/ divisor, we have divided by the approximate 1000, thereby getting too large a quotient; subtracting that which is too large leaves too little, as has been already explained; consequently, something must be added to rectify. The true result to be added would be that found by dividing the product of 3487 and 10, or 34870, by the real divisor, 1010. Instead of dividing 34870 by 1010, however, we prefer to divide it by the approximate divisor, 1000, getting 84/870, which is added. | The result thus added being too large (having divided by a smaller divisor than the real one), we obtain too much by the addition, and, therefore, we must subtract next; and so on, till | the parts of all the products to the left of the line are brought to — the right by multiplication; whence, the reason of multiplying by the excess figure of the divisor, and setting the products as DIVISION SIMPLIFL:ED AND ABBREVIATED. 17 directed by the rule is evident, and the reason for subtracting and adding alternately is shown. Or the reason might be expressed in general terms as follows: Dividing in each case by a quantity which is too small, gives a result: in each case too large. Subtracting too large a quantity leaves too little — adding too large a quantity gives too much — we must subtract and add till the proper correction is made. Notre.— Each new correction becomes less than the previous one, till it finally disappears. To Frnp tur DeEcrMat. Should it be desirable to continue the division into decimals, in example 8, the process, according to the usual method for Long Division, would be to annex a suitable number of ciphers to the ‘remainder, 904, and keep on dividing by 1010 till the required number of decimals is found. Thus, suppose we required three decimals in said example: 1010)904000(. 895 S080 9600 9090 5100 5050 Annexing three ciphers and continuing the division we get .895+ SimepLE Meruop or Finping THE DEcIMAL. To find the decimal from the remainder by the simplt- jied process: Annex, or conceive to be annexed, as many ciphers to the remainder as there are decimals re- quired, and continue the division as in the first part of the process. 2 18 Division SIMPLIFIED AND ABBREVIATED. Thus, taking the remainder, 904, inexam- 904]... + 1010 ple 38, again, we conceive three ciphers an- 9/040 nexed, and continuing the process, asin the gy4i960 first part of the example, we get .895+ as 100 before. » BORER. Norte.— This will be found more fully explained in a subsequent part of the work. (GENERAL PRINCIPLES. III. If a number be taken from the divisor, the dividend must be diminished by the product of the quotient and the number sub- tracted, in order that the quotient may not be changed. Thus, 84 + 12 —7 Now, if 2 be taken from the divisor, 12, and we desire to divide by 10, without changing the quotient of 12, we subtract from 84, the dividend, 2 times the quotient 7, or 14; then dividing the difference, 70 (84 — 14), by 10 (12 — 2), we have: 70+10—7 the same as: 84 + 12 -- And if the dividend be not thus diminished, the quotient will be increased by the result obtained by dividing the product of the quotient and the number subtracted, by the new divisor. Thus, 84 + 12 — Now, if we divide by 10, without diminishing 84, we have: 84 + 10 = 8...4 the same as if 2 times 7, or 14, were divided by the new divisor, 10, and the result added to the quotient of 12. Thus, 84 -+- 12 — 7, plus 2 times 7, or 14, divided by 10: 14+ 10 =—1...4 Seaee hae ee DIVISION SIMPLIFIED AND ARBREVIATED. 19 From a due consideration of the foregoing principles, the following illustrations will be readily understood : If 9216487 be divided by 1000, as in example 1, Rule I, the re- sult will stand thus: 9216|487 Suppose, now, we subtract any number, say 2, from the divisor 1000, and divide the same dividend, 9216487 (without diminish- ing it) by the difference, 998, according to the foregoing general © principles. Thus, 9216|487 18/468 9934\955 Dividing first by 1000, the quotient is 9216, and the remainder 487. To finish the division, now, according to Prin. III, we mul- tiply the quotient 9216 by 2, the figure subtracted from the divi- sor, or the difference between 1000 and 998, and divide the pro- duct. 18432, by the new divisor, 998, by Long Division, as in the margin, getting 18 for quo- 998)18432(18 tient and 468 for remainder. This quotient 998 © and remainder is now added to 9216|487 found ~ 8452 above, and the division by 998 is completed, T984 9234 being the true quotient, and 955 the re- 468 mainder. The required quotient for 998, however, can be more simply found as follows : 9216/4857 ~ 998 18/432 36 9234|955 20 DIVISION SIMPLIFIED AND ABBREVIATED. Dividing first by 1000, the quotient is 9216, and the remainder, 487. Then, extending the vertical line to a suitable length, the quotient, or figures 9216, to the left of the line, is multiplied by 2, the difference between 1000 and 998, and the product, 18482, set so that its unit figure will be under the unit figure of the remainder, 487, and the other figures in the corresponding places. Then, 2 times 18 (to the left of the line), or 36, is set in proper position under the remainder, 432; the results are now added the same as in simple addition, giving 9234 for the true quotient, and 955 for remainder; fully expressed, 9234958, The reason of this simple process will be understood | by comparing it with that which immediately precedes, in connection with the principle: the greater the divisor, the dividend not being changed, the less the quotient. Dividing first by 1000, which is greater than the real divisor, 998, the quotient obtained is too small and requires an addition. The true result to be added, it will be remembered, is the quo- tient, 18, and the remainder, 468, obtained by Long Division, by dividing 2 times 9216, or 18432 by 998, as shown in the process immediately preceding. Instead of dividing 18432 by 998, however, as in the said pro- cess, we prefer, here, to divide it by 1000. The extended vertical line, observe, performs such division by merely setting the pro- duct 18/432, as shown in the margin. Since 18432 has now been divided by 1000, while it ought to have been divided by 998, the result found, or 18|482, is likewise too small, and must have an addition also. The result required to be added is evidently the difference between the true result, 18/468, found in the first process, and 18]432, found in the second. The difference is 36, or 2 times 18, that part of the product, 18/432, to the left of the vertical line. | The reason of the simplified process might be expressed in gen- DIVISION SIMPLIFIED AND ABBREVIATED. 21 eral terms as follows: Dividing by a quantity which is too large, in every case, gives a result which is too small, and we have to keep on adding till the proper correction is made. To Fino tHe DEcIMAL. To find the decimal for the remainder, 955, true, say to four places, the usual method would stand thus: 998)9550(.9569++ 8982 5680 4990 6900 5988 9120 8982 By the simplified process, we simply annex, or conceive to be annexed, four ciphers and continue the division as in the other part of the work, thus: 9550]... 19) 100 38 .9569| Conceiving four ciphers annexed (three of them represented by dots), 2 times 9550, or 19100; then 2 times 19 are placed in proper position, as in the first part of the process; then, adding the results (rejecting that part of the decimal to the right of the line, being less than 5), we have .9569, as found by the long process. Before giving the rule for this interesting method of division, the three following examples are necessary to 22 DIVISION SIMPLIFIED AND ABBREVIATED. still further illustrate our method, and explain a few simple changes which will frequently occur when making use of this method: Exam. 6. Divide 4789365 by 998. 4789|365 33/523 231 7 4823/126 Dividing first by 1000, in this, the quotient is 4789, and the remainder, 365. Then, multiplying 4789, the partial quotient to the left of the line, by 7 (the difference between 1000 and 998), the product, 33523, is set in proper position, as has been explained in the previous illustration. Next, that part of the product (now partial quotient), to the left of the line, or 33, is multiplied by 7, also, and the product, 231, set in position. All the figures to the left of the line having now been multiplied by 7, the next step is to add the several results. Before doing so, we make a short mental examination of the partial remainders, or numbers on the right of the line, to ascertain whether any thing is to be carried to the figures on the left; and we find 1 is to be carried to 83, making 34, which ought to have been multiplied by 7, instead of 33, giving 238 as the true result to be added instead of 231. The same result is obtained, observe, by allowing 231 to remain as it is, and add 7 times the 1 carried, or 7, placed in proper position. The quotient, then, is 4823, and the remainder 126, fully SADE 48231234. DIVISION SIMPLIFIED AND ABBREVIATED, 23 Exam. 7. Divide 641458207 by 9930. 64145|820|7 + 993)0 449/015 3/143 eo, 645979997 1/9930 64593|0067 In this we first cut off the cipher from the right of the divisor, and also one figure, 7, from the dividend, which will be the last figure of the remainder. ‘Then, dividing the remaining part of the dividend by 993, as in the last example, we get 64597 for quotient, and 999 for remainder. The 7 cut from the dividend is now brought down, making the remainder 9997; this remain- der, being greater than the divisor, 9930, contains said divisor once more, 1 is added to 64597, giving 64598 for the true quotient and 67 is the remainder. Exam. 8. Divide 478498963 by 9991. 47849|8963 + 9991 43/0641 387 47892|9991 Here, we divide first by 10000, getting 47849 for quotient and 8963 for remainder. Then, multiplying the figures to the left of the vertical line by 9 (the difference between 10000 and 9991) and adding, as in the other examples, we have 47892 for quotient and 9991 for remainder. The remainder, being equal to the divisor, contains it once, and there is no remainder, but 1 is added to the quotient, giving 47893 for the true quotient. 24 DIVISION SIMPLIFIED AND ABBREVIATED. Roe III. From the foregoing examples and illustrations we have the following: Rue. To divide by a number which is a little less than 100, 1000, 10000, etc.: | I. Dwide jirst by 100, 1000, 10000, etc., as the case may be, using a vertical line to separate the quotient from the remainder. (Call 100, 1000, etc., in such cases the approximate divisor.) : Il. Hxtend the vertical line to a suitable length and multiply the figures vo the left of the line by the difference between the approximate and the true divisor (this difference is called the complement), setting the product under the jirst result, so that units will be under units, etc. Ill. [fa part of the said product extend to the left of the line, mul- tiply such part, also, by the complement, setting the product as before, . and so on till the figures on the left of the line are exhausted. IV. Add the several results; the true quotient will be on the left, and the remainder on the right, of the vertical line (not forgetting to multiply the figure carried, if any, from the remainders to the quo- tients, setting the product in proper positior and adding it, as shown in example 6). Arithmeticians have given a simple rule for dividing by 100, 1000, ete. (Rule I of this work). We have now established, in addition to said rule, two very important and equally sumple rules, namely: first, for such num- bers as are a little in excess of 100, 1000, ete. (Rule IJ), and second, for such numbers as are somewhat less than 100, 1000, ete. (Rule ITI). A knowledge of these two simple rules, together with the principles of Division, will now enable us to simplify Long Division to an extraordinary degree, and, as a mat: DIVISION SIMPLIFIED AND ABBREVIATED, 25 ter of course, problems in other branches of Arithmetic where Long Division has to be used. ‘l'o do so, we draw very largely on the following simple principle: Multiplying or dividing both dividend and divisor by the same number, does not change the value of the quo- tient. Thus, 15 + 8 —5, and 4 times 15 divided by 4 times 3, or 60 + 12 — 5 Again, 4812 — 4, and the half of 48 divided by the half of 12, or 24 6 —4 ; PROBLEMS. From a due consideration of the foregoing rules, prin- ciples and illustrations, the student will readily under- stand the following problems: Exam. 1. If 26 building lots be sold for $90350, what is the average price of each lot ? To solve this problem, we simply divide the price by the num- ber of lots. Now, it is evident that if 4 times as many lots were sold for 4 times the money, the average price of each lot would still be the same; and as we can more readily divide now by 4 times 26, or . 104 (Rule II), we divide 4 times the money, $361400 by 104, thus: 90350 + 26 3614/00 ~ 104 26 DIVISION SIMPLIFIED AND ABBREVIATED. Multiplying both dividend and divisor by 4, we take the pro- ducts for a new dividend and new divisor. Dividing first by 100, the quotient is 3614; this is next multi- plied by 4, the excess, and the product, 14456, subtracted. Then 145 (144 plus the 1 carried in subtraction) is multiplied by 4, also, and the product, 580, added. Finally, 6 (5 plus the 1 carried in addition) is multiplied by 4, and the product, 24, subtracted; there is no remainder, $3475 being the quotient, or the average price of each lot. Exam. 2. If 196 tons of iron cost $7 252, what is the price of a ton? To solve this we divide 7252 by 196, the quotient is the price. Now, it is evident that if half 196 tons be bought for half the money, the price per ton will be still the same; and as we can more readily divide by 98, half of 196, than by 196 itself, we di- vide half the money, or $3626, by 98; thus: 36/26 + 98 12 98 Here we make use of Rule III. Dividing first by 100, the quo- tient is 36, and the remainder 26, Multiplying 36, the quotient, _ then, by 2 (100 — 98), the complement, and setting the product, 72, under the remainder, 26, we add. ‘The remainder, here, being equal to the divisor, 98, evidently contains said divisor, once more, and hence the true quotient is $37, the price per ton (see exam. 8). Exam. 8. If 3834 suits of clothes cost $12408.10, what is the price per suit ? Here we sce that the divisor, 384, is nearly one-third of 1000, DIVISION SIMPLIFIED AND ABBREVIATED. a0 and bearing in mind that we have now an easy method for dividing by the numbers at either side of 1000, etc., whether a little more, or a little less, we divide 3 times the price by 3 times the number of suits; thus: 12408.10 + 334 37/224 .30 + 1002 14 15|030 30 Multiplying both dividend and divisor by 3, we get $37224.30, and 1002. Dividing first by 1000, we get 87 for quotient, and 224.30 for remainder; we next multiply the quotient, 37, by 2, the excess figure of the divisor, and set the product, 74, under the unit figure of the dollars, 224. Then, subtracting 74 leaves a remainder of 150, evidently dollars, to .be still divided by 1002. Now, besides $150 of a remainder, there is a remainder of 30 cents, also, or 15030 cents in all, to be divided by 1002. The di- vision of the cents is performed the same as the dollars, first dividing by 1000, and subtracting 2 times 15, or 30. The result is then $37.15, the price per suit. Exam. 4. If 47 tons of iron cost $1703.75, what is the price per ton ? Here, we see at a glance that 2 times 47, or 94, is a simpler divisor than 47 itself, so we multiply the cost and the number of tons each by two, and operate with the products, thus: 34/07.50 + 94 2\04 12 36|23(50 1/38 a 24/94 28 DIVISION SIMPLIFIED AND ABBREVIATED. Here we make use of Rule III, dividing the dollars first by 100, and adding for the complement (6) we get $36 for quotient, and a remainder of $23 to be still divided by 94. Now, there is a remainder of 50 cents also, making 2350 cents in all for the remainder, to be divided by 94; the division is performed . same as on the dollars, first dividing 2350 by 100, and adding for the complement (6), and we get 24 cents and 94 remaining; this remainder being equal to the divisor, its value is evidently 1, making 25 cents. The price is, then, $36.25 per ton. Exam. 5. If 202 boxes of cigars cost $875.40, what is the cost of 1 box ? 875.40 + 202 4/37.70 + 101 4 33/70 Dividing the number of boxes and the cost; that is, the divisor and dividend, each by 2, the new divisor becomes 101, by which we divide according to Rule II, and we get $4.33, the required cost. 3 In dividing the remainder, $33.70, or 38370 cents, by 101, we first divide by 100; the next step is to multiply 33 by 1, set the product, 33, under 70 and subtract, but we see by inspection that in doing so the next remainder would be less than 5, or less than half a cent (the remainder would be .37),-and without proceeding farther, we see that 33 is the correct number of cents; the answer is, therefore, $4.33. Exam. 6. If 721 boxes of cigars cost $3121.93, what is the cost of a single box ? 4 ane eee ar | 445.99 + 108 12 Be DIVISION SIMPLIFIED AND ABBREVIATED. 29 Here, we divide both dividend and divisor by 7, and the new divisor becomes 108. ) After dividing $445 by 108, there is a remainder of $83.99, or 3399 cents; to divide 3399 by 103, we would first divide by 100, this would give 33 for quotient, and 99 for remainder; the next step would be to multiply 33 by 8, and subtract the product, 99, from the remainder, 99, which would give 0 for remainder; so we see without proceeding farther in the example, that $4.33 is the correct answer. Exam. 7. If 816 hats cost $3468, what is the cost of 1 hat ? 3468 = 816 ~4/33.50 + 102 8 95. In this we divide the terms by 8, and the new divisor becomes 102. The rest is now plain; the answer is $4.25. REMARKS. Remarks on Rule I].— The attentive student cannot fail to see, at this stage of the work, to what an extraor- dinary extent Long Division may now be simplified. Confining our remarks here to Rule I, without further reference, for the present, to that equally, if not more, important Rule III, which will be fully considered here- after, we see that: If we take the number 800, for instance, every number between it and 900, differing by 8, can be simplified when used as a divisor. Thus, 808, 816, 824, 832, 840, 848, 856, 864, etc., divided by 8, will give 101, 102, 103, 104, 105, 106, 107, 108, etc., for new di- visors, to which Rule IT is applicable. 30 DIVISION SIMPLIFIED AND ABBREVIATED. Again, if the numbers between 8000 and 9000, 80000 and 90000, etc., be taken, a similar relation will be found to exist. Thus, 8008, 8016, 8024, 8032, 8040, 8056, etc., divided by 8, wiil give 1001, 1002, 1003, 1004, 1005, 1007, etc., for new divi- sors. And the same is true with regard to the numbers from 200 to 300, 300 to 400, 400 to 500, 500 to 600, 600 to 700, ete. Thus, 308, 306, 809, 312, 315, 318, etc., divided by 3, would give new divisors 101, 102, 108, 104, 105, 106, etc.; and the same may be said in regard to 3003, 3006, 3009, 3012, 3015, etc.; 30003, 30012, 30018. And 909, 918, 927, 936, 945, 954, 968, 972, ete., divided by 9, give 101, 102, 103, 104, 105, 106, 107, 108, etc., for new divisors, to all of which Rule II is applicable; and so with the other series. Hence, When the last figure or figures of the divisor are a multiple of the Jirst figure or figures, the division can be simplified. Notre.— When one number is contained in another an exact number of times, the less is said to be a measure of the greater; and the greater is called a multiple of the less. Taking the divisor, 816, in the last example, for instance, 8, the first figure, is contained without remainder in 16, the two last figures. 8 isa measure of 16, and 16 is a multiple of 8. And when the first figure or figures of the dwisor is a multiple of the last figure or figures, Rule II is also applicable, as illustrated in the following: Exam. 8. If 2001 citizens of New York pay an annual _ tax of $704352, what is the average tax to each? 3 704352 + 2001 852(176 = 10004 176 DIVISION SIMPLIFIED AND ABBREVIATED. 31 To solve this problem, we divide the whole tax by the number of persons, thus: Dividing the dividend and divisor, each, by 2, we have 352176 to be divided by 10004. Dividing by 1000, we get 352 for quo- tient, and 176 for remainder. The result thus found being too large, we multiply the quotient, 352, by the excess, 4, and divide the product by 1000, and subtract the result. This result is found by simply taking half of 352, or 176, and setting it in proper position under the remainder, 176; the vertical line performing the divi- sion by 1000; and the answer is $352, the average tax. Or thus: 7043520 + 20010 3852/1760 + 10005 © 1760 Annexing a cipher to both dividend and divisor; in other words, multiplying each by 10, the new divisor becomes 20010. Then, dividing by 2, we have 3521760 to be divided by 10005, ac- cording to Rule II, and we get $352, as before. Exam. 9. If 804 building lots be sold for $738476.34, what is the average price of each lot ? 1384|\76.34 + 804 ~923)0954-+ + 1004 4/615 918/480 95 5054 Here, we divide both dividend and divisor by 8, and the new divisor becomes 1004, by which we divide as in example 8. (It being immaterial whether we divide first by 8 and next by 100, or 32 DIVISION SIMPLIFIED AND ABBREVIATED. divide first by 100 and then by 8; we have here divided first by 100, thereby getting the position of the vertical line at once.) Multiplying the quotient, 928, by 4, we get 461.5, and dividing this by 100 gives 4.615, which is subtracted. In subtracting, 1 is carried to 4, to the left of the line, making 5; this is multiplied by the excess, 4, also, and the product, 2.5, divided by 100, giv- ing .025 as result, which is added. The quotient is then $918.50, the average price of each lot. Or thus: 738476384 + 804.. 923|Uy95|4 & 1005]0 4I615 918]480 25 5054 Moving the decimal point two places to the right, in both divi- dend and divisor (this multiplies by 100), and then dividing each by 8, the new divisor becomes 10050, by which we divide, as in example 5, page 15, and we get $918.50, as before. In applying the foregoing methods, recourse may be had to any process whereby the divisor can be reduced toa simple one; always bearing in mind that, whatever change is made tn the devisor a similar change is to be made in the dividend to preserve the relationship. By a simple divisor is meant 10, 100, 1000, 10000, etc.; or any number near to these, as 101, 1002, 10009, etc.; 91, 92, 994, 99997, etc. Exam. 10. Divide 369940704 by 77784. Here, we divide the terms each by 7 and 369940704 + 77784 multiply the results by 9, getting 100008 for 52848672 11112 a simple divisor. The quotient is 4756. 756138048 ~- 100008 38048 Division SIMPLIFIED AND ABBREVIATED. 33 Exam. 11. Divide 251076872 by 6668. In this, we add one half the divisor and one 251076872 + 6668 half the dividend to each respectively and 125538436 3334 the divisor becomes 10002. 8766115308 + 10002 The quotient is 37654, the remainder, 1$$93, 715322 being equal to 1, which is added to 37653, and 3765319986 there is no remainder. 16 10002 Nore.— It need scarcely be remarked that the remainder in division must be always less than the divisor. In making use of the present methods, however, the remainder frequently comes out equal to the divisor; in such cases 1 is added to the quotient, as in the example, and there is no remainder, the division being exact. Sometimes the remainder will come out greater than the divisor; in these cases the remainder is divided by the divisor, the result is added to the quotient already found and the difference between said greater remainder and the divisor, will be the final remainder, as shown in example 7, page 23, to which the student is referred. (See example, page 250.) Exam. 12. Divide 653134680 by 8888. 653134680 + 8888 Here, we add an eighth to each term and 81641835 1111 obtain 9999 for a simple divisor. . 7347716515 + The quotient is 73485, the remainder being de ote equal to 1, which is added to the quotient. y 73484|9999 NotEe.—If there be a remainder and it is desirable to find the equivalent decimal, annex as many ciphers as there are decimal places required and continue the division, as in the first part, illustrated in the following: Exam. 18. Divide 47032938 by 12501 to five decimal places. In this, the terms are multiplied by 8and 470382988 + 12501 the divisor is 100008, by which we divide. 37§9/63504 + {00008 The quotient is 3762, and the remainder 33408 30096 to which we annex five ciphers for the number oa of decimal places required, and continuing the 3340800000 division as in the first part, we obtain the deci- 2/67264 mal .33405, making the quotient 3762.33405, . 83405 |382736 correct to five decimal places. Note 1.-- By annexing ciphers to each successive remainder the division can be carried to any desirable length. ‘ 2. lf there be decimals in the dividend annex them to the remainder instead of ciphers, supplying the deficiency by ciphers, if necessary, to correspond with the number of decimal places required, as illustrated in the following : 3 34 Diviston Smp.iFIED AND ABBREVIATED. Exam. 14. Divide 183170503.6482: by 24995 to six decimal places. Here, the terms are multiplied by 4 and the divisor becomes 99980, a simple 183170503.6432 + 24995 one, the complement being 20. Multi- 7326 82014.5728 99980 plying 7326 by 20, the result is set to 1146520 . the right beginning at the units’ place; 40 next, 20 times 2 (1 plus 1 carried in 7828|285745/72800 adding) are 40, set in proper position, o7 eee, and by addition, we find 7328 for quo- “395509158640 tient, and 28574 for remainder, to which 5728 is annexed, also two ciphers to make six places to correspond to the number of decimal places re- quired. Multiplying 285745, now, by 20 and 57, also, by 20 and adding, we obtain .285802, true to six places, leaving 88840 for re- mainder. The quotient is 7328.285802+. And by annexing ciphers to the remainder, and continuing the division, any required number of decimal places may be obtained. Nort.—Itis worthy of remark, that any part or any multiple of a simple divisor can also be simplified when used as a divisor. Thus, if 44 of 99980= 3332624 be used as divisor. To divide by 383326% it is multiplied by 3 to get 99980: and if 4g of 99980=1249714, or 12497.5 be used, multiplying either number by 8 gives 99480. Again, if one-half of 12501=625014, or 6250 5, be used ar divisor; multiplying by 2, and the result by 8, will give 100008; if 4 times 12501=50004 be used: annexing a cipher to the latter we obtain 500040, and dividing this by 5, gives 100008, a simple divisor, and so of other numbers. (For further illustrations, see page 250.) Exam. 15. Divide $5296765 among 3001 persons. 765, the result, 5296|765, is the quotient for ———|__ ; 1000. Then, dividing this by 3, we obtain $1765)988+ | 1765(|588+ (the decimal may be continued tf de- ee sirable), the quotient for 38000. To rectify the result which is too large, 3000 being less than the true divisor, we subtract the 1-8000 part: of 1765, or eee and we obtain $1765, the required quotient. To get the 1-3000 part of 1765, remove the decimal point three places to the left and divide by 3. (1.765+3=.588+.) _ Here, we cut off the three right hand figures, . 5296/7653 + 8001 58 Division SmpiiFIED AND ABBREVIATED. 35 | Exam. 16. Divide 21187060 by 12004. In this, we divide the terms by 4, and the 21187060 ~ 12004 divisor becomes 3001, a simple one, by which “52961765 = 3001 we divide as in the foregoing example, the 76515884 quotient is 1765. . 588-4 From the foregoing examples and illustrations it will _be seen that the division can be readily simplified when the divisors run as follows: 1202, 1203, 1204, 1206, 12012, etc.; 12002, 12003, 12004, etc.; 1402, 1407, 14014, etc. ; 1602, 1604, 1608, 16016, etc. Also such as 1212, 1414, 1616, 2424, 3636, 4848, 5656, 6464, etc. If it were required to divide, for instance, by 2408, 24008, 240008, etc., we would divide by 8 and get 301, 3001, 30001, etc., for new divisors, and the division would be as in the last example, and so with the other numbers. To divide by 3636, for instance, we would first divide by 6, ‘getting 606; dividing this in turn by 6 we get 101 for new divi- sor, and so with similar combinations, always bearing in mind that Whatever operation is performed on the divisor to simplify it, the same operation must be performed on the dividend also. Prosiems Sotvep sy Rote ILI. Exam. 1. If 792 milch-cows be bought for $59400, what is the average price of each ? 59400 + 792 74125 + 99° 74 99 Bb DIVISION SIMPLIFIED AND ABBREVIATED. To solve this problem, we divide the whole cost by the number of cows, thus: A slight inspection of the divisor, 792, shows that by dividing it by 8, we obtain a simple divisor, 99, by which we can readily divide according to Rule III. Dividing both dividend and divi- sor, then, first by 8, we next divide the new dividend, 7425, by the new divisor, 99, getting 74 for quotient and 99 for re- mainder. The quotient is, then, $7482, or rather, $75, the re- quired price. Now, as the mental eye cannot always readily tell whether such a number as 792, for instance, is a multiple of some particular number, as 8, the following will be found a much more rapid way of simplifying such num- bers when presented as divisors: Let us take 800, and call it the approwimate divisor, in S connection with 792, the veal divisor, arranging them as [792 shown in the margin, and using periods, or dots, instead ate of the ciphers in 800. The difference of the divisors is 8. Examining the process, now, in the foregoing example, it will be seen that we first divided by 8, and then by 100, in other words, we divided by 800, using the component factors, 8 and 100 (8x100=800), getting 74 for quotient, and 25 for remainder. This result is too small since we have divided by 800, instead of 792. To rectify, we have to multiply the quotient, 74, by 8 (the difference of the divisors), and divide the product, 592, by 800. (Gen. Prin. III, page 18.) Now, multiplying 74 by 8, and dividing by 800, is the same as multiplying by 1, and dividing by 100 (¢8> being equal to ;4,), and that is exactly what was done in the example; we simply added +17 of the quotient, 74, that is, .74, or rather |74, the vertical line being used instead of the point. The price is, then, $7432, or rather $75. DIVISION SIMPLIFIED AND ABBREVIATED. 37 It will be seen upon examination, also, that we have simply di- vided the eighth part of the total price by the eighth part of the number of cows, that is, we have divided $7425 by 99. To divide by 99, we use 100 for approximate divisor, and 1 (100 —99) is the complement; so we add 47 of the quotient, .74, or |74. Notse.— Complement, in the language of Arithmetic, is what any number wants of being a unit of the next higher order. In the above example, 99 wants 1 of being 100, and we call 1 the complement. If 93, 993, or 9993, 7 would, in each case, be the complement, etc. Exam. 2. If the taxes paid by 6986 persons amount to $528351.18, what is the average tax to each ? (eee 528/351.18 + 6986 CTH Seis. a 14 150 62/874 124 998 Taking 7000 for approximate divisor, in this, and arranging the divisors, as in example 1, we find their difference to be 14. To divide by 7000 now, it is immaterial whether we divide first by 7, and then by 1000, or divide by 1000 first and then by 7%, the result being the same in either case. We will choose the lat- ter course. Cutting off three figures from the right of the dollars divides by 1000. Then, extending the vertical line, we divide by 7, getting $75 for quotient and $478.74 for remainder, or $75|47874 as quotient for 7000. The result obtained being too small, having made use of too large a divisor, we have to multiply the quotient, 75, by 14 (the difference of the divisors), and divide the product by 7000. Multiplying 75 by 14, and dividing by 7000, is the same as multiplying by 2 and dividing by 1000 (744, = 7,255); and as 38 DIVISION SIMPLIFIED AND ABBREVIATED. the vertical line performs the division by 1000, all we have to de is, simply to add 2 times 75, or 150, placed in proper position. There is now a remainder of $628.74, or 62874 cents, in all, to be still divided. Now, as the remainder in Division is always a part of the dividend, it is evident that 62874 is part of the new dividend, $75|47874; consequently, in dividing 62874 cents, there is no need of dividing again by 7 (the original dividend having been divided by 7), so we simply cut off three figures, 874, from the cents, and multiply 62 (to the left of the line) by 2, as was done with $75, and setting the product, 124, to the right, we add. The quotient is now $75.62, and a remainder of 998. Since the © original dividend has been divided by 7, it is evident that 998 is only the seventh part of the true remainder. Multiplying 998 by 7 gives the true remainder, 6986, which contains the true divisor once; 1 is added to 62 cents, making $75.63, the required aver- age tax. REMARKS. Remarks on Rule III. —If the student have followed the thread. of our reasoning thus far, he will readily see, on analyzing the fore- going example, that we have simply divided the seventh part of the dividend by the seventh part of the divisor; that is, we have divided $75478.74 by 998 (6986 + 7). To divide by 998 we -use 1000 for approximate divisor, and add the complement, or 2 times the quotient; 2 times $75 in the first case, and 2 times 62 cents in the other. This gives $75.62 and a remainder of 998 for quotient; fully expressed, $75.62238, or rather $75.63. From a due consideration of the foregoing results we see that: When the divisor is such that the difference between it and the next higher number ending in ciphers is divisible by the significant part-of such higher number, the complement, or multiplying Jigure, will be the result obtained by dividing said difference by the said significant part. Referring to problem 1, for instance, where we have taken 800 for approximate divisor, in connection with 792, the true divisor, we see that the difference is 8, and that said difference contains 8, DIVISION SIMPLIFIED AND ABBREVIATED. 39 the significant figure of 800, once; 1 is the complement or multi- plier. | | Again, referring to problem 2, we see that 7, the significant figure of 7000, is contained 2 times in 14, the difference of the divisors; 2 is then the complement or multiplier. For brevity we will henceforth call the multiplier in such cases the key. The student’s attention is now invited to the following: Tak- ing 800, as was taken in the remarks on Rule II, we see that every number between that and 700, differing by 8, can be readily sim- plified when used as a divisor. Thus, 800, 792, 784, 776, 768, 760, 752, 744, 736, etc. Again, if we take the numbers between 8000 and 7000, 80000 and 70000, 800000 and 700000, etc., a similar relation will be found to exist. Thus, 8000, 7992, 7984, 7976, 7968, 7960, etc.; 80000, 79992, 79984, 79976, 79968, etc. ; 800000, 799992, 799984, 799976, 799968, etc. Suppose it were required to divide by 7968, for instance, we would take 8000 for approximate divisor, and both divisors would stand thus: We would now divide by 8000, as in-problem 2, and 4 (82 + 8) is the key. And if the divisor were 79968, we would use 8. eK 79968 32 where 80000 is the approximate divisor, and 4 also the key; and so of the other numbers. And if the numbers between 900 and 800, 700 and 600, 600 and 500, 500 and 400, 400 and 300, etc.; also between 9000 and 8000, 40 DIVISION SIMPLIFIED AND ABBREVIATED. 90000 and 80000, etce.; 8000 and 7000, 80000 and 70000, etc., be compared, a similar relation will be found to exist. If it were required to divide by 2982, for instance, the divisors would stand thus: Din 2982 18 3000 is approximate and 6 the key. Reviewing the numbers, now, which can be readily simplified, when presented as divisors, let us take any series, say from 700 to 800, placing those to which Rule II is applicable in a column to the left, and those to which Rule III is applicable in a column to the right, as below, and the key for each in a column between, thus: Rule IT, ; Key _ Rule III. 700 800 107 1 792 714 2 (84 721 3 776 728 ee 768 735 5 760 742 6 752 749 7 744 756 8 736 763 9 728 770 10 720 TTT 1a 712 784. 12 704 (91 13 696 798 14 etc. To divide by 756, for instance, we would apply Rule IT; first dividing both dividend and divisor by 7, we would then divide DIVISION SIMPLIFIED AND ABBREVIATED. Al the seventh part of the dividend by the seventh part of the divi- sor, which, in this case, would be 108. The approximate divisor for 108 would be 100, and 8 the key. And to divide by 736, for instance, it would stand thus, Se2 756 64 800 would be the approximate divisor, and 8 the key (64-+8), and so of the other numbers. When the difference between the real and the approxi- mate divisor is /ess than the significant part of the latter, we indicate the division by the fractional form; illustrated in the following: Exam. 3. It required $563650.88 to pay a certain army, giving each man $23.92; how many men were in that army ¢ — 94.. 563650/88 + 2392 140912/72 = 8 —= 3 93485|45334- 78|28334. 26 23563|99|66 33 To solve this, we have to find how many times $23.92 is con- tained in $563650.88. Nore.— The arithmetical student need scarcely be told, that operations on decimals are performed as on whole numbers, due attention being paid to the proper position of the decimal point. 42 DIVISION SIMPLIFIED AND ABBREVIATED. Moving the decimal point in each two places to the right, in other words, calling both cents, throws off the decimals, and we have whole numbers at once. i To divide by 2392, we take 2400 for approximate divisor; the difference is 8, which does not contain 24, the aie part of 2400, so we indicate such division by the fraction 8, which is equal to 4, and this is the key or multiplier. We now divide by 2400, first by 100 by means of the line, and next by 24. To divide by 24 we use the component factors 4 and 6 (4x6 = 24), dividing first by 4 and then by 6, and we get 23485 for quotient and .4533+ for remainder, 3 being repeated without end. We have now to add 8-2400, or =45 part of the quotient, 23485, or, what amounts to the same thing, the ;1, part divided by 3 The hundredth part of 23485 is 234.85, and the third part of the latter is 78.2833+, 3 being repeated without end. Next, the =, part of 78, or .26, is added. Now, since the vertical line performs the division by 100, right through, all we have to do is simply divide the partial quotients, 23485 and 78, each by 3, setting the results in proper position. Adding the several results now, we get 23563 for quotient and .9966+ for remainder, 6 being repeated without end. To find the correct decimal, we continue the process, simply cutting off 66 and ee a third of .99, or .83, placed in proper position (that is the 45 part of .99, or .0033, this being the value of |33 as it now stands in the work). The remainder is now .9999-+, or .9 repeated to infinity, which is equal to1. Adding 1 to 23563 gives 23564, the number of men, and there is no re- mainder. | Norte. — To show that .9999, etc., is equal to 1: If we take the digit 9 as divisor, and any one of the remaining eight digits as dividend, and express the value of such division in the language of decimals, we will, in every case, obtain a repetition, without end, of the figure of the dividend. Thus, 4 — .11111, ete. ; 4 — .7777, etc.; and DIVISION SIMPLIFIED AND ABBREVIATED. 43 Conversely, .11111, etc., = 45 .7777, etc., = 45 and on the same principle, .9999, etc., is equal to 1; that is, .9999, etc., = 3, or 1, as in the foregoing example. Hence, To find the value of a decimal repeated to infinity: Set down the repeated figure or figures for numerator, and 9, or as many nines, for denominator, as there are figures repeated, and the result is a common fraction equal in value to the decimal. Thus, .3333, etc., = 3, or 4; and .9696, etc., = 24, etc. Exam. 4. Suppose $5083763.16 were to be divided among a number of persons, giving each $398.32; how many persons would receive that sum $ ee 50837|6316 + 39832 76254 12709/4079 ~~ 198 —49 318 53/3778 2996 fia bated 12763|0125 Here we take 40000 for approximate divisor, The difference of the divisors, 168, divided by 4, the significant part of 40000, gives 42, the key. Cutting off four figures, 6316 (one figure for each cipher in 40000), from the right of the dividend, and then divid- ing by 4, gives the quotient for 40000, or 127094079. We have now to add 42 times 12709. To multiply by 42, we use the component factors 6 and 7 (6x 7=42). Setting 6 times 12709, or 76254, a little to the left, as shown in the margin, we multiply the latter in turn by 7, setting the product, 533778, in proper position, as the rule directs. Next, we add 42 times 53, or 2226, placed in proper position; thus: 6 times 53, or 318, is set to the left, as in the margin; then 7 times 318, or 2226. A short inspection, now, shows that in adding the remainders, or numbers to the right of the line, 1 is to be carried to the left, or quotient’s place; this 1 is also multiplied by 42, and the pro- 44 DIVISION SIMPLIFIED AND ABBREVIATED. duct set in proper position. Addition now gives 12763 persons, and .0125 remaining, which is equal to $5. Notr.—To find the value of the remainder, .0125, in such examples as the foregoing, it must be carefully borne in mind, that the remainder, in Division, is always a part of the dividend; and that whatever operations are performed on the dividend and divisor, by way of preparation to simplify the division, the reverse of these operations is performed on the remainder, to find the true remainder. In the foregoing example, the dividend has been divided by 40000, conse- quently, .0125 is only the 40000 part of what it ought to be. To restore it to its proper value; that is, to the same denomination as the dividend, we multiply it by 40000. To multiply .0125 by 40000, we simply move the deci- mal point four places to the right, which gives 125, the product for 10000; multiplying 125, then, by 4 gives 500, or the product for 40000, the true | remainder. Now, since the dividend in the example is cents, the true remainder, 500, is also cents. The value of .0125 is, therefore, $5. When the divisor can be conveniently resolved into component factors, we proceed by successive division; illustrated in the following: Exam. 1. Divide 661740804 by 17946. Taking 18000 for approximate divisor, we see that the 18000 difference between that and the real divisor is 54, which 17946 contains 18, the significant part of the approximate divisor, 54 without a remainder; and this being the case, the divisor, 17946, will also contain 18 without a remainder. Now, the factors of 18 are 3 and 6 (®8xX6=18). Divid- 38{17946 ing 17946 by 8, and the result by 6, we obtain 997, a 6) 5982 simple divisor. 997 The factors of 17946, now, are 3, 6 and 997, 3|661740804 by which we divide in succession as shown 6/220580268 in the margin. The quotient is 36878, and 3676 ae the remainder, 997, which contains 997 once th ah et more, 1 is added to the quotient, making 330 36874, and there is no remainder. (See exam. 369731097, 8, page 23.) Nots.—If 18 be taken from 18000, then from the remainder and from each successive remainder, the numbers thus found, viz., 17982, 17964, 17946, 17928, 17910, 17892, etc., each differing by 18, can be treated similar to the example shown, if presented as divisors. (For further illustrations, see examples from page 249 to 253.) Division SIMPLIFIED AND ABBREVIATED. 45 Exam. 2. Divide 35372903712 by 240168. In this, a slight inspection shows that 4|240168 168, the right hand figures of the divisor, 6} 60042 is a multiple of 24, the left hand figures; 10007 and the factors of 24 are 4.and 6. Divid- ing 240168 by 4, and the result by 6, we 4135372903712 obtain 10007, a simple divisor. The fact- 4 8843225928 ors, now, of 240168 are 4, 6 and 10007 by ry which we divide as shown in the margin, BETDS 10856 set D007 ; ; ; 103) 1709 The quotient is 147284, the remainder a olnis notes 10007 being equal to 1 (12997). (See Rule 147283) 9279 II, page 15, and exam. 2, p. 18.) 728 10007 Norr.— It will now be readily seen that, when the divisor is not too large, and when the last figures are a multiple of the first figures, said first figures being easily factored, the foregoing method can be applied; such numbers as: 24168, 82192, pape 3200192, 48240, 480240, 72648, 720648, 4590, 45090, 450135, 450180, 450225, etc., etc. GENERAL SHorT Mrtuop ror ALL Numbers. When the foregoing methods cannot be conveniently applied, the ‘following short method, cutting off 50% of the usual work, can be used, and may, perhaps, be preferred in all cases; illustrated in the following: Exam. 1. Divide 17653762 by 3658. Here, we take 17653 for the first partial 3658)17653)762(4826. dividend, and drawing a line to the right of 3021/7 3, we find 4 for the first figure of the quo- 95/36 tient. We now multiply 3658 by 4; but 22)}202 instead of writing down the product and ~ 1954 subtracting, we simply add enough to each product, as we proceed, to give the figure of the partial dividend. Thus, 4 times 8 are 82 and 1 (setting down 1) are 33 which gives 3 of the dividend; 3 to carry; 4 times 5 are 20 and 3 are 23; and 2 (re- quired to make 25) is set down under 5 of the dividend: 4 times 6 are 24, and 2 (carried from 25) are 26, set down 0, the figure of the dividend being 6; carry 2; 4 times 8 are 12, and 2 are 14, and 8 is set down to make 17, The number 8021 is the remainder, to which the next figure, 7, is brought down; giving 30217 for next dividend. (Continued on page 46.) 46 Division SIMPLIFIED AND ABBREVIATED. Now, 3558 is contained 8 times in 80217: Say 8 times 8 are 64, and 3 (set down to 3658)17653|762(4826 make 67): carry 6; 8 times 5 are 40, and 6 a a are 46, and 5 (set down to make 51): carry 92902 5; 8 times 6 are 48, and 5 are 53, and 9 (set mae 0,7 | down to make 62): carry 6; 8 times 3 are 24, and 6 are 30, nothing set down, the number above being 80; the remainder is 953 to which 6, the next figure of the dividend is brought down, giving 9536 for partial dividend. The next figure of the quotient is 2: say 2 times 8 are 16, set down 0, the figure imme- diately above being already 6: carry 1; 2 times 5 are 10, and 1 are 11, and 2 (set down to make 18): carry 1; 2 times 6 are 12, and 1 are 12, and 2 (set down to make 15): carry 1; 2 times 3 are 6, and 1 are 7, and 2(set down to make 9): Proceeding thus, we find the next figure of the quotient, 6, and the remainder is 254. NotrE.— If it should be necessary to continue the work into decimals, annex ciphers as in the usual method, and proceed as before. When the divisor ends in ciphers we would proceed as in the following : Exam. 1. Divide 6569437124 by 998000. 6569/437|124 + 998|000. 13]138 ee 6582|601124 Here, we cut off the ciphers from the divisor, and as many fig- ures (124) from the right of the ‘dividend for the last figures of the remainder. Then, dividing the remaining part of-the divi- dend by the remaining part (998) of the divisor, we get 6582 for quotient and .601 for remainder. The figures cut from the divi- DIVISION SIMPLIFIED AND ABBREVIATED. 4% dend are now brought down, and the true remainder is 601124. The manner of finding the correct decimals of the remainder has been already pointed out. The decimal here found is correct to two places (.60). When the digits of the divisor are all the same figure we would proceed as in the following: Exam. 1. Divide 3009581015 by 77777. 30095|81015 T7777 429440145 11111 38694/61305 + 99999 38694 99999 Norr, — It will be readily seen that, when the digits of the divisor are all the same figure, a succession of 1’s is obtained by dividing the said divisor by one of its digits; and multiplying this quotient of 1’s in turn by 9, gives a succession of 9’s by which we can readily divide according to Rule III. Dividing 77777, in this example, by 7 gives 11111, and multi- plying this by 9 gives 99999, a simple divisor. Performing the same operations on the dividend gives 3869461305 for a new divi- dend. Dividing this by 99999 gives 38694 for quotient and 99999 for remainder, which, being equal to the divisor, adds 1 to the quotient, making 38695, and there is no remainder. Notes. — By referring to the process in the foregoing example, we see that the number of figures to be cut from the right of the dividend in such cases is equal to the number of digits in the divisor, thus giving the position of the vertical line before commencing operations on the dividend. Hints ror THE STUDENT. From the foregoing example it will be seen that, when the divisor rs such that, being divided by a measure, or 48 DIVISION SIMPLIFIED AND ABBREVIATED. exact divisor, we obtain a repetition of any of the nine digits, the division can be at once simplified. This will be made clear by the following observations: Let us take the digits 2, 3, 4, 5, etc., repeated two times, three times, etc., or any multiple of such repeated digits, and examine them carefully for a few moments, thus: 22x 6—=—182 222x6—1332 2999 x 6 — 13339 88x5—=—165 333x5—1665 3333 x 5 — 16665 44x4—176 444x%4—1776 444059 ey Bx T= 885 555x7—3885 5555 x 7 — 38885 66x38—-198 666x3—1998 6666 x 3 = 19998 © [x= 154 TI D2—1584 FIT = Bee 88x3—264 888x3— 2664 8888x 3 — 26664 99x3—297 999x3—2997 9999 x 3 — 29997 Suppose, now, it were required to divide by any of the above numbers, say 132, 1332 or 138332; we see by inspection that if these be divided by 2, 3, 4, 6 or 12, we obtain a repetition, in every case, of a particular digit: 66, 666, 6666; 44, 444, 4444; 33, 338, 38333; 22, 222, 2222; 11, 111, 1111; any one of which can be treated the same as 77777 in the last example. And the same is true of the remaining numbers, 165, 176, 385, etc. : Again, if any multiple of those multiples be used as a divisor, such as 132 x 3 = 396 1332 x 3 — 3996 165 x 38 = 495 1665 x 38 = 4995 297 x 3 — 891 2997 x 3 = 8991 etc., the same treatment is applicable; but for most of such mul- tiples more rapid methods have been already pointed out, DIVISION SIMPLIFIED AND ABBREVIATED. 49 To divide by 396, 495, 2997 or 8991, for instance, we would make use of the next higher numbers ending in ciphers, as 400, 500, etc., for approximate divisors, getting the key at once, as has- been already explained. (See remarks on Rule III.) Note. — Dividing the divisor by one of its digits, when the figures are the same, and then multiplying by 9, we need scarcely remark, is the same as” multiplying first by 9, and dividing by the digit after. To divide by 1332, for instance, we first divide by one of its measures, or an exact divisor, say 2, getting 666, and this in turn by 6, getting 111, which is multiplied by 9, getting 999 for asimple divisor. But dividing by2 and then by 6 is dividing by 12(2 X 6=12). Now, by first multiplying 1332 by 9, and dividing after, the process will be more readily performed; thus, 1332 x 9 = 11988. Taking 12000 for approximate divisor now, in connection with 11988, and arranging them as has been already pointed out, they willstand asin j9 . the margin, the key being 1, andthe division being the same as above; 11988 that is, by 2, 6 and 1000; or, by 1000 first, and then by 12, as inthe 12 last arrangement. In such cases, then, it is perhaps preferable to multiply first by 9. When the digits of the divisor are all the same figure and a cipher intervenes, as 10101, 20202, 3030308, ete., we would proceed as in the following: Exam. 1. Divide 8045261828 by 80808. 8045261828.. + 80808.. 301480/920972 + 7999992 37685/115121+ : 37685 152806 Here, we see at a glance, that, if the divisor, 80808, be multi- plied by 11, we get 888888, by which we can readily divide as in the previous example, first dividing by 8, and then multiplying by 9. Now, multiplying first by 11, and afterward by 9, is multiply- 4 50 DIVISION SIMPLIFIED AND ABBREVIATED. ing by 99. So we prefer to multiply 80808 by 99 first, vieetting . 7999992 for a new divisor. To multiply by 99, we conceive two ciphers, represented by dots, or periods, annexed to the divisor, as seen in the margin . this multiplies it by 100, giving 80808.. as the result. From this we subtract 80808, without setting down the latter, however, but simply setting down the difference, 7999992, thus: 8 from 10 _ (represented by the first dot to the right) and 2; carry 1 to 0 (to the left of right hand 8), 1 from 10 (the second dot), and 9; carry 1 to 8 (the second or middle 8), 9 from 18, and 9; and so on, all from the expression 80808. . | Going through a similar process with the dividend, that is, subtracting 30452618 from 30452618.., without setting down the first (being already contained in the last expression), we get a new dividend which is divided by 7999992; 8000000 being the approxi- mate, and 1 the key, giving 37685.152806 for quotient, true to six places of decimals (bearing in mind that the lines is used for the decimal point). Note.— When two or more ciphers intervene, as 1001001, 7007007, etc., we multiply by 999, 9999, etc. (short method), always one 9 more than the’ number of intervening ciphers, to simplify the division. (For short method of multiplying by any number of o 8, see Contractions in Multiplication, page 74.) When the figures of the divisor are repeated; thus, 212121, 323232, etc., we would proceed as in the last’ ease, as illustrated in the following: Exam. 2. Divide 16299801882 by 212121. : 16299801882... + 212191.. 8)1613680|386318 + 20999979 7)537893|462106 76841/923158 76841 999999 DIVISION SIMPLIFIED AND ABBREVIATED. 51 A moment’s glance at the divisor, in this example, shows that, if it be divided by 3 or 7, we get 70707, or 30303, and the divi- sion cap be performed as in the previous example. Multiplying both dividend and divisor, then, by 99, the new divisor becomes 20999979, with which we use 21000000 as ap- proximate divisor, the difference being 21 and the key 1. The quotient is 76842, there being no remainder. Nort. — There aretwo ways for finding the value of the remainder .999999. First. Reverse the process performed on the dividend; that is, divide by 99 and multiply by 21 (3 x 7). Dividing 999999 by 99, we get 10101, and multiplying this by 21 gives 212121, which contains the divisor once; .999999, therefore, is 1. Second. Annex ciphers and continue the division, as has been already pointed out, and 9 will be repeated to infinity; but it has been already shown that .9 repeated to infinity is equal to 1. Whenever, therefore, .9, .99, etc., is remainder in such cases, add 1 to the quotient and there is no remainder. Orner StmpLtE Meruops. It must be carefully borne in mind that the two prin- cipal rules laid down in the beginning of this work are the entire secret to our methods, a little practice enabling us to bring any ordinary number used as a divisor within one or other of those rules. Hence, the student need not expect to understand what follows unless he has thor- oughly posted himself on the groundwork. To still further assist the student to master Simplified Division, we subjoin a few more examples illustrating our methods, which, if rightly understood, will prove ex- tremely simple and interesting. Exam. 1. If 73 musical instruments cost $12802. 74, - what did one instrument cost at that rate ? 52 DIVISION SIMPLIFIED AND ABBREVIATED. From several methods of simplifying the division by 73, we select the following, as, perhaps, the most simple and expeditious that can be given. 1280274 ae 426758 2433-+-+ 42675 .8 243-+- 4267.58 24-++ 175|3975.38 + 10001 175 38]0088 | 38 Annexing two ciphers (represented by dots) to 78, in other words, multiplying it by 100, we get 7300. To multiply the divi- dend by 100, we simply call the dollars cents, and we have 1280274 cents to be divided by 7300. To divide by the latter, we set under it, one-third of itself, one-tenth of that third, and one- tenth of that tenth, and adding, we get 10001 for a simple divi- sor. Going through a similar operation with the dividend. we get 1753975.38 for new dividend, which being divided by 10001, gives 175 for quotient, and .3800 for remainder, to which decimal 88 is annexed, and continuing the process, we find 38 is the cor- rect decimal, or cents, in this case. The price, then, is $175.38, Notse.— If the division of 7300 be continued, we find the 7300 decimal to be .383, etc., as shown in the margin, 3 being nee repeated without end. And in adding a tenth, ete., 8 will be 24333 4 repeated in like manner, so that when the results are added, 9000.999 we get 10000.999, etc. Butit has been shown (note, page 42) that .999, etc., is equal to 1; therefore, 10000.999, etc., = 10001. Hence, To divide by 73 then: Multiply the dividend by 100 (simply con- ceive two ciphers annexed), and under the product write one-third of itself, one-tenth of that third, and one-tenth of that tenth, and divide the sum by 10001. DIVISION SIMPLIFIED AND ABBREVIATED. 53 Exam. 2. The annual income of a certain American citizen is $1538475 ; what is his income for one day ? 1538475 + 365 3076950. + 730. 10256500 1025650 102565 4215|4215 4915 To divide by 365 we double it, then double the dividend, and we have 3076950 to be divided by 730. Annexing a cipher (dot) now, to both dividend and divisor, we get 7300 for new divisor, by which we divide as in the previous example, and we get $4215, the income for one day. Exam. 3. If the aggregate annual tax paid by 73000 tax payers be $31220640, what is the average individual tax ¢ 312/20640 + 73000 104/06880 24333 10/40688 2433 1/04068 943 497|\72276 + 100010 4270 68 Here, we set under both dividend and divisor one-third of each, one-tenth of that third, and one-tenth of that tenth, and adding, we have 42772276-++, to be divided by 100010. The quotient is found by Rule II to be $427.68, the average tax, Norr. — We see by inspection that 68 is the correct decimal, or number of cents; for if we complete the subtraction and continue the process we 54 DIVISION SIMPLIFIED AND ABBREVIATED. get .68006, to which we annex 80, the figures omitted in the division by 10, and we have then .6800680 to be divided by 100010, and the result is .68, as found in the example. Lo divide by 73000, then, we have the following simple rule: Under the dividend write one-third of itself, one-tenth of that third, and one-tenth of that tenth, add the four lines together and divide the sum by 100010. Norz. — By continuing the decimals, in the division of 73000 by 8 and 10, it will be seen that 3 is repeated to infinity, and the sum of the four lines 1s 100009.9999, etc.; but this has been shown to be equal to 100010. (See note to example 3, page 42.) It is optional whether the vertical line be drawn before taking the parts or when commencing to divide by 100010, as, know- ing the divisor, we know the number of figures to be cut from the right of the dividend, namely: five. This simple and expeditious method of dividing by 73000 will be of great value in solving problems in interest, for days, at any rate per cent, on the basis of 365 days to the year, (Explained in the article on Interest.) Exam. 4. If 732 building lots be valued at $356484, what is the average price of each lot? 356484. + 732. 1188280. 2440 118828 244 487|1948 + 10004 1948 Here, we simply conceive a cipher annexed to both dividend and divisor, and under each result set one-third of itself, then one-tenth of that third, and adding we have 4871948 to be di- vided by 10004. The quotient by Rule II is $487, the average price of each lot. Division SIMPLIFIED AND ABBREVIATED. 55 A knowledge of the methods here given, will be of particular advantage in operations where the divisor is not subject to change, as,forexample 43560 sq. ft. to an acre; 5280 fect to a mile; 625 sq. 1. in a pole; 7.92 in. to a link; 243, or 24.75 cub. ft. to a perch of masonry; 2240 lbs. to a gross ton; 144 articles to a gross, &c. In problems on Percentage the methods will be found extremely valuable where the divisors run like 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 108, 104, Xe. Exam. In 76824763 sq. ft. how many acres, true to five places of decimals ? Norr.— Here, we have to divide by 48560, and 4.0 |4856.0 a slight inspection shows that the component factors 11/1089 — are 40, 11 and 99 (40 x 11 x 99 = 43560.) 298-6 Pointing off one figure and dividing 7682476.3 by 4 divides by 40; next, we divide by 11, 1920619.075 getting 174601.734. This is now 1746)/01.734 divided by 99 by simply pointing off _ 17/46 two places for 100, and adding for cen 17 the complement 1, once 1746 and 1763|647|34 once 17 set in proper position, the 647 result is 1763|/64734 6 65387 To find the complete decimal, now, we divide 64734 by 99; simply adding once 647 and once 6; the answer is 1763.65387 acres true to five places of decimals. Notet.—To divide by 5280; the component factors are 60, 8 and 11; make use of successive division, decimally. To divide by 625, multiply by 4 and the result by 4, and divide by 10.000 (625 x 4 = 2500, and 2500 x 4= 10000.) bearing in mind to multiply the dividend also, to preserve the relation. For 792 take 14, of the dividend and divide by 99 (792+ 8 = 99.) To divide by 2434, or 24.75, multiply by 4 and divide by 99 (2434, or 24.75 x 4=99.) The component factors of 2240, are 40, 7 and 8 (40 x 7 x 8 = 2240.) and for 144 multiply by 7 and divide by 1008 (144 x 7= 1008.) (See rule and exam. page 225.) i 56 Division SIMPLIFIED AND ABBREVIATED. i In many cases where, at first glance, it may appear difficult to simplify the division, a short inspection of the divisor will saggest amethod, ‘Take, for instance, the following Exam. Divide 49862538 by 7854, true to seven places of decimals. Here, a short inspection 49862538 > 7854 4532958 714 of the divisor shows that 9065916 -1428 it is divisible by 11, giving 6346/1412 + 9996 2/5384 714; doubling this gives Seal eee 6348/6804000/0000 1428; the sum of the three 2721/6000 1|0884 numbers gives 9996 for a 4 .6806722/6888 simple divisor. Dividing the dividend by 11, in like manner, doubling the result and adding, we have 63461412 to be divided by 9996. Dividing now by 10000, and adding for the complement 4 times 6346 and 4 times 2 we get 6348/6804. _ To get seven places of decimals, now, we annex seven ciphers (one for each decimal required) and dividing 68040000000 by 9996, as in the first part of the example, we get 6348.6806722, as required. Notrre.—It need scarcely be observed that the division may be carried to any length by simply annexing as many ciphers as there are decimals required, dividing in each case by 9996, the new divisor. Suppose it were required to find five more decimal places; annex five ciphers and divide 688800000 by 9996. (See examples on pp. 58 and 59.) — oe Division SIMPLIFIED AND ABBREVIATED. 5% The following simple rule will be found useful when the cost per gross is given to find the cost of a single article: RuLE.— Multipiy the cost per gross (up to $142) by 7; point off three figures for decimals, and the result is the cost of a single article, near enough for practical purposes. Exam. If a gross of padlocks cost $23.50; what is the cost of 1? Answer 16c. Thus: 23.50 x 7 = 164.50 and pointing off three places we have .16450, or 16 c. for business. Exam. What is the cost of 1 article at $19 per gross ? Answer 13c. Thus: 19 X 7 = 1838, and pointing off three places we have .138, or 13c. for business. Reason.—To divide by 144 we multiply by 7 to get 1008 for a simple divisor; and multiplying 19, also by 7, to equalize, we have 133 + 1008; and we simply divide by 1000 instead of 1008 which gives the result near enough when the cost does not exceed $142 per gross. Notr.—If greater accuracy be required set 8 times 133 three places to the right and deduct, thus: 133. . 1064 and, the result is correct to four places, viz. .131936 1319 And if it should be required to find the result correct to six places, we annex six ciphers, or one cipher for each decimal required, and divide by 1008. Exam. Divide $19 by 144, true to 6 places of deci- mals. Multiplying both terms by 7 we have 1338 =+- 1008 true to 6 places of decimals, thus: Annexing 6 ciphers we apply 133000,000 ~- 1008 1064;000 rule 2 (page 15) and we get 131986,000 __—*851e 1381944, true to 6 places -181944/512 DECIMALS. It has been already remarked that operations on deci- mals are performed as on whole numbers, due attention being given to the point, or characteristic of the decimal. To illustrate, we will take a few of the most Boge numbers used in Practical Mathematics. Exam. 1. Divide 4985.7192 by .7854. 49857192 +. 7854 7122456 1129 6474/96 + 102 129|48 6345/48 2158 6348106 06 Moving the decimal point in both dividend and divisor four places to the right; in other words, multiplying each by 10000, we have whole numbers. A moment’s inspection of the divisor, now, shows that it is divisible by 7, giving 1122, which, in turn, is divisible by 11, giving 102 for a simple divisor. 7 Dividing the dividend, now, by 7, and the result by 11, we have 647496 to be divided by 102. Here Rule II is applicable, and we get 6348 for quotient. DECIMALS. 59 Or thus: 49857192 7854 71229456 1122 6474960 1020 6345/4608 - 9996 2153880 8 6347/9996 Under the divisor set one-seventh of itself, or 1122; then con- ceiving a cipher annexed to this, we have 11220; dividing this by 11, we get 1020, which is set under 1122, and adding the three numbers together, we get 9996 for a simple divisor. Going through a similar process with the dividend, we get 63454608 to be divided by 9996. Here, Rule III is applicable, and we get 6348 for quotient, the remainder, 9996, being equal to 1. Hence, the Rue. To divide by .7854: Below the dividend set one-seventh of itself, and one-eleventh of that sevent., setting the latter one place farther to the left than its proper position ; add the three numbers together, and divide the sum by 9996. Exam. 2. Divide 4985.7192 by 3.1416. 49857192 + 31416 12464298 + [xd4 1780614 1618740 1586/3652 + 9996 6344 ‘19996 Dividing both dividend and divisor in this by 4, we get .7854 for a new divisor, by which we divide according to the foregoing rule, and we get 1587 for quotient. 60 DECIMALS. Exam. 3. Divide 496.258 by .07958, true to four places of decimals. Here, we move the decimal point in 49695800 + 7958 the divisor five places to the right, and we “69031995 + 9943 have 07958., or rather 7958, for a new di- 321565 .75 visor. Then, as there are only three places 168 vf decimals in the dividend, we annex two 6935/9587[5... ciphers to fill the deficiency, and, moving 501331. 75 the point five places to the right, we have 6935 9638 49625800 for a new dividend. Norte.—If 8000 be taken as approximate for 7958, andarranged thus: 8... 514 (42 + 8) isthe key. From this we conclude that if 7958 be divided 7958 by 8, a divisor will be obtained lacking only 51/ of being some power of 10 (in this case 1000). ‘Dividing both dividend and divisor, then, by 8, the new divisor becomes 994%, the complement being 54, as shown in note. Here, Ruie III is applicable, 54 times 6203, or 32/565.75; then 5} times 32, or |168, being set in proper position and added, giving 6235 for quotient and |958.75 for remainder. To find the decimal, now, true to four places, the vertical line is drawn to the right of the fourth figure, 7, of the remainder, and two ciphers (dots) annexed to the remaining part (so as to ‘give three places, |[5.., to the right of the line, as in the first part of the example); then, setting 5} times 9587, or 50|331.75, in position, as in the first part, and adding, we get 6235.9638, for quotient, true to four places of decimals (1 being allowed for the part of the decimal cut away). “Exam. 4. A perch of masonry contains 242 or 24.75 cubic feet. How many perches of masonry in 86006.25 cubic feet ¢ | 86006 .25 + 24.75 3440|25|00 + 99/00 34/40 34 3474/99 DECIMALS. 61 Multiplying both dividend and divisor by 4, in this, we have 344025 to be divided by 99. Here, Rule III is applicable, 100 being the approximate divisor and 1 the key. Dividing by 100, and adding once each partial quotient; that is, the 1-100 part, we get 347422, or rather 3475, the required number of perches. Notr. —If we used the fractional form, 2484, we would multiply by 4 also, 4 times 2434 being 99. Hence, the Rute. To divide by 24%, or 24.75; Multiply the dividend by 4, and divide the product by 99. Exam. 5. Divide 478932673 by 16667 to five places of decimals, 478932673 + 16667 28735|960388 ~ 100002 57470 38564). . 28735.38567 Here, we simply multiply both dividend and divisor by 6 (the divisor being nearly one-sixth of 100000), and the new divisor is 100002, by which we divide, getting 28735 for quotient and 38568 for remainder. To find five places of decimals, now, a vertical line is drawn to the right of the fifth figure of the remain- der, and as many ciphers conceived to be annexed as there are figures cut from the right of the new dividend, namely, five; then, continuing the multiplication by 2, as before, we get .388567, the number of decimals required. The quotient, then, is 28735 . 38567. Upon examination, however, it will be found that the decimal in the present example is not only correct to five places, but it is correct to nine places, viz.: .385672286, and the correction may be made to any required number of decimal places by continuing the process as above. Hence, 62. METHODS OF PROOF. To find any particular number of decimal places we have the following simple Rue. To the right of as many figures of the remainder as there are decimals required draw a vertical line, and to the figures on the right of said line, if any, annex, or conceive to be annexed, as many ciphers as will make the number of places (both figures and ciphers) equal to the number of figures cut from the right of the dividend at Jirst, and continue the process as in the first part. Nore.— Should the remainder not contain a sufficient number of figures for the number of decimals required, annex ciphers to fill the deficiency, and proceed according to the rule. If, for instance, seven places of decimals were required in the last example, instead of five, the line would be drawn two places farther to the right (fill- ing up with ciphers, or dots to represent them); then, annexing two more dots to the right, so as to make the correct number (five) to the right of the fine, proceed as before. METHODS OF PROOF. There are two principal methods of proving Division: First, by multiplication: Multiply the quotient by the divisor, and to the product add the remainder, if any; the result, if the work is correct, will be equal to the dividend. Second, by casting out the 9’s. This method of proof, which is very easy and convenient in practice, and gen- erally preferred by the experienced arithmetician, is given in connection with the following: : | METHODS OF PROOF, 63 Exam. How many times is 83 contained in 2869086 ? 2 2869086 + 83 3 84429/032 + 996 137/716 548 4 , " 34567|300 + 12 — 95 rem. 3 To divide by 83, we multiply both dividend and divisor by 12, and the new divisor is 996, to which Rule III is applicable. The quotient is 34567, and the remainder, 300, which is divided by 12, to get the true remainder, 25. Proor. Commencing with the divisor, 83, we add its digits from right to left, rejecting 9 from the sum; thus: 3 and 8 are 11; 9 from 11 and 2; the remainder, cr excess, 2, is now carried to the left and reserved. The digits of the quotient are next added, and 9 rejected, in like manner; thus: 7 and 6 are 13, 9 from 13 and 4; then this 4 and 5 are 9, and taking 9 from this, leaves 0; next, 4 and 3 are 7, from which 9 cannot be taken; 7 is carried to the left, as shown in tlie margin. Multiplying the excesses, 7 and 2, now, gives 14; the digits of this are added, in like manner, making 5, which, being less than 9, is added to the digits of the remainder, 25, from left to right; thus, 5 and 2 are 7, and 5 are 12; then, 9 from 12 and 3, the excess, is carried to the right and reserved. Finally, adding the digits of the dividend, from left to right, also; omitting 9 whenever it occurs, and rejecting 9 from the sums, as often as they make 9, or more, we get an excess of 2, which is equal to the excess found from the remainder, and the work is supposed to be correct. Rute. (1) Cast the 9’s from the divisor and quotient ; set the ex- eesses to the left of the work and reserve them. (2) Multiply said excesses, and cast the 9’s from the product ; add the excess here found, 64 Merruops oF PRoor. to the remainder, and cast the 9’s from the sum, reserving the excess to the right of the work. (8) Cast the 9's from the dividend, setting the excess to the right, also; if both excesses on the right be equal, the work is presumed to be correct. Norrs.— 1. Should the excess from the divisor be 0, it is evident we need not go over the quotient, as the result found by multiplication would also he 0. The excess from the remainder, if any, in that case, will be the same as that from the dividend; and if there be no remainder, the excess from the dividend must be 0. 2. The divisor, 996, and its corresponding dividend, might be taken in proving the work in the foregoing example, but in that case, the correspond- ing remainder, 800, must be taken, instead of 25, 300 being the true re- mainder for 996. Try it, the excess will be 0. 8. It may be well to remark, also, that when the division is continued into decimals, the true remainder, and not the decimal, must always be taken when using this method of proof. 4, It is hardly necessary to say that, should there be a misplacing of figs ures (an occurreuce, however, which is very rare), this nvethod of proof will fail, as it is evident the swms of the digits will be the same, regardless of their local positions. When the division is performed by the simplified methods, the proof can be obtained by the short methods for multiplication commencing at page 71; as illustrated in the two following examples: Exam. 1. In 497786 square inches, how many square reer. To divide by 144 (sq. inches in a sq. foot) - both numbers are multiplied by 7, giving 34841521008. 3484|152 + 1008 The quotient is 8456, and the remainder, 504, ee ere To prove the work, 8456 is multiplied by 1008, 3456 eS and the remainder, 504, is added. This is per- 504 formed by simply setting 8 times 3456, or 27648, 27648 three places to the right, under the remainder, “3484152 and adding; this gives 8484152, the dividend. (See example 2, page 71.) On Smee Drvisors. 65 Exam. 2. Divide 3174473 by 497. Doubling both numbers, we have 6348946 to be divided by 994, a simple divisor. The quotient is 6387, and the remainder, 268. To prove the work, 6387 is multiplied by 994, eyeey + 994 and 268 added. The process is performed by 38 se setting 6 times 6387, or 38322, three places to the 6387968 right, under the remainder, and subtracting; this 33/999 gives the dividend. (See example 3, page 72.) $348946 Nots.— The proof is to be always taken before reducing the remainder toa decimal. ON SIMPLE DIVISORS. Divisors may be simplified by any process that will make them 10, 100, 1000, 10000, etc., or that will make them a little more, ora little less, than these; as, 101, 102, 1003, 10007, etc.; or 91, 92, 98, etc.; 991, 992, 993, 9989, 9999, etc. Norge. — Composite numbers, when not too large, can be readily divided by using their component factors. This method is called successive division. A composite number is one that may be produced by multiplying together two or more numbers. Thus: 18 isequal to6x3; or 9x2; or3x3x2, Exam. In 12872 cubic feet of earth, how many cubie yards, or loads ? In this, the component factors of 27 (cub. feet in a cub, yd.) are 3 and 9 (3 X 9=27). We divide first, : ate ms by 3, and the result by 9, to get 476.74 cubic yards, aie Fil or loads. Nore. —Jn making calculations where the divisor is constant, and is a composite number, the desired results are more easily obtained by successive division than by the usual long methods; thus: 2240 pounds toa gross ton, the factors are 40, 7 and 8 (40 & 7 & 8 = 2240). Again, 5289 feet to a mile; the factors are 60, 8 and 11 (60 « 8 & 11 = 5280); and 43560 square feet to an acre; the factors are 40, 11 and 99 (40 « 11 & 99 = 43856V) ; and by the short methods already established, divison by 99 is the simplest part of the process. (See page 5d.) 5 66 On Simpie Divisors. The following suggestions will aid the student in obtaining simple divisors : To divide by 114: 11.5, or 115, multiply by 8; the results are 92 and 920, simple divisors. . 124: 12.5, 1 25, 125 or 1250, multiplied by 8 will give 100, 10, 1000 and 10000. 13: 184, 14.25, 1.325 or 1825, multiplied by 8 will give simple divisors. 13}: 13.5, 1.85, 185 or 1850, multiplied by 8 will give simple divisors, 14: The component factors are 2 and 7; or multiply by 7 the result is 98. 144: 14.25, 142.5, 1425 or 14250 multiplied by 7 give simple divisors, 144: 14.5, 145, 1450, etc., multiplied by 7 give simple divisors. To diviace by 14} or 14.25; multiply by 7, the result is 993; use 100 for the approximate divisor, and add for the quarter, etc. Exam. Divide 109185 by 145.. In this, both numbers are multiplied by 7, and we have 764295+1015. Buta slight inspection shows that this can be still further simplified. 109185 + 145 By setting 145 one place to the right, under ae aS. 1015; and the given dividend one place to the 75313765 10005 right under the new dividend, and subtracting 3765 in each case, we get 10005 for a simple divisor. aaa The required quotient is 753. 164: 16.5, 165, 1650, 16.6, 166, 1666, etc.; 167, 16.7, 1670, 1675; 168, 16.8, 1680, etc.; and 163, all multiplied by 6 give simple divisors; thus: 164 x 6 = 99; ete. 17: Multiplied by 6 gives 102; and 174, or 17.28x4=—69. (See pages 250 and 251.) ; 174: 17.5, 175, 1750, etc.; multiplied by 6 give simple divisors. 17$: 17.75, 177.5, 1775, 17750, etc.; multiplied by 8 x 7 give simple divisors, Thus: 17? or 17.75 x 8 = 142, and 142 x 7= 994, a simple divisor. . 18: The factors are 3 and 6; or 18 X 6 = 108, a simple divisor. 18}: 18.25, 1825, 18250, etc.; multiplied by 4 give 78, 730, 7300, 73000, etc.; and the teeta of dividing by these is given at pages 52 aod 53. On Srmpce Drvisors.: 67 183: 18.75, 1875, 18750, etc.; multiply by 4and to the result add } of itself. Thus: 183 x 4= 75; 4 of which is 25, and this added to 75 gives 100; etc. 19: Multiplied by 5; the result is 95: 194, 19.25 or 1925 multiplied by 5 and a fifth of the number itself added; thus: 1925 k 5 = 9625 + + of 1925 = 10010. . It must be carefully borne in mind that, whatever change is made in the divisor, to simplify it, a similar change must be made in the dividend to preserve the relation; and that when the divisor can be simplified, any multiple, sub-multiple, or aliquot part of it can also be simplified. (See page 250.) And now, if the numbers from 19 to 100, be taken and examined, the subject will be found not only interesting and instructive but it will be found that division by the majority of these, their multiples and sub-multiples, will be exceedingly simple. For instance, most of the twenties multiplied by 4; the thirties by 3; the forties and fifties by 2, etc., will give simple divisors. A few more examples of a practical nature, before closing this chapter, may be found helpful. Exam. In 1269 pounds of oil; how many gallons, 74 pounds to the gallon ? In this, we simply add athird; and atenth of the j result is the number of gallons, and the decimal] of 439 i a gallon. In other words, by adding a third to 74 it 172.810 becomes 10, a simple divisor. Exam. In 12134 feet; how many perches of 164 feet ? 12134+163 Here, both numbers are multiplied by 6, and we 7281/04 99 have 72804+99. The answer is 73533, or reducing 7/28 the fraction to a decimal, .3939, ete. pea U 73539 . Exam. In 34682487 feet; how many miles of 5280 feet? We give the solution of this two ways: First: By using the factors (60 x 8 xX 11=5280). 3468248 /7_ Cutting off one figure, and dividing by 6, divides Yee pal by 60. Then an eighth of this, and one eleventh —$56816528 of the eighth. «668 On Smmpre Divisors. Second: Here, we take the component fac- tors, 80, 11 and 6, or 80 and 66 (66 « 80 = 5280) and divide, first, by 80, by cutting off one figure and dividing by 8. ‘hen, to divide by 66, we add half to get 99, a simple divisor. The remainder of the process is clear. The answer is: 6568.6528 miles. The deci- mal may be carried to any desired length, by annexing ciphers to the remainder and divid- ing by 99. Exam. In 76824763 square feet; 3468248 .7 433031 .0875+66 216765 .5487 33 ~ 6502/96. 6312-99 65)02 66 6568/6463|12 64/63 64 .6528)39 how many acres, true to five places of decimals, 43560 sq. ft. to the acre ? The component factors of the divisor, in this, are 40, 11 and 99 *(40 x 11 x 99 = 48560). The solution is left for the student. Should he fail to see his way clear, he is referred to page 59, where the explanation is given. And when the digits follow in their natural order, the division can be simplified, as illustrated in the following: Exam. Divide 15241578750190521 by 123456789. 15241578750190591 + 193456789 1219392630001524168 987654319 123456787|87654322|01 + 99999999]09 1/23456787 2 123456789/1111111101 PAVIA OL Here, we first multiply both dividend and divisor by 8, setting the product under each respectively, one place to the left of units, and adding, in both cases, we take the results for a new dividend : ; j 4 : . ; 3 On SIMPLE DIVISORS. 69 and new divisor. Cutting off 09 from the divisor, now, and 01 from the dividend, we divide the remaining part of the dividend by the remaining part (99999999) of the divisor, 1 being the key, and we get 123456789 for quotient, and 11111111 for remainder, to which 01, cut from the dividend, is annexed, and the re- mainder is then 1111111101. The result thus obtained is too large, being the quotient for 9999999900; to correct, we subtract 9 times (09 cut from the divisor) the quotient, 123456789, or 1111111101. There is no re- mainder, 123456789 being the quotient. Notes.—1. Multiplying the terms by 8, setting the products one place to the left, and adding, we need scarcely observe, is multiplying by 81. 2. The complement of 9999999909 being 91, the quotient may be obtained without cutting off the figures as was done in the example, by adding 91 times the quotient. In that case there would be no subtraction. And on examination, it will be found that is what we have actually done, because to multiply by 91, we have used the short method, namely, multiplied by 100, and subtracted 9 (100 -—9= 91). Hence, the Rue. To divide by the nine digits in their natural order : Set 8 times the dividend under itself one place to the left, and divide the sum by 9999999909, or by eight 9’s, 09. And when the digits are in a reversed order, we would proceed as in the following: Exam. Divide 975461057789971041 by 987654821. 975461057789971041 + 9876543821 78036884623 19768328 7901234568 8)7901234568/0987654321 + 80000000001 987654321 |0123845679V125 0123456790125 In this, we multiply the terms by 8, setting the products one place to the left, as in the last example, and adding, the new di- visor is 80000000001, the key being 4. ROCF. On SIMPLE DIVISORS. Dividing, first, by 80000000000 (simply cut off ten figures from the right of the dividend, by the vertical line, for the ten ciphers, and divide by 8), we get 987654321 for quotient and |01234567890125 for remainder. From the result thus found we subtract 4 of the quotient, set in proper position; that is, 1-80000000000 part of 987654321, or .01234567890125. There is no remainder, 9876543821 being the required quotient. Hence, the Rue. 70 divide by the nine digits written in reversed order: Set 8 times the dividend under itself one place to the left, and divide — the sum by 80000000001, or by 8 followed by nine ciphers and 1. Nors. — To get 1-80000000000 of the quotient, 987654321, is to divide the latter by 80000000000, and to do so we simply cut off the ten ciphers and divide by &: But we find that the number to be divided (the quotient) does not contain ten figures to be cut off, to correspond with the number of: * ciphers, so we prefix a cipher to supply the deficiency; the number to be divided then is .0987654821, and the eighth part of this is .01234567890125, which is equal to the remainder from which it is subtracted, giving 0 for remainder (all of which can-be seen at a glance after a little praptive with our methods). Mixep Noumpers. A Mixed Number consists of a whole number with a fraction annexed; as, 344, 1064, 2494, etc., and division by these will be found as simple as viet numbers 5 thus : To divide by 2494, or tts equal, 249.25: Multiply by 4 and the new divisor is 997, a simple divisor (not forgetting to multiply the dividend, also, by 4). 2514, or its equal, 251.50: Multiplied Bye 4, gives 1006 for a sim- ple divisor. 1672, or its equal, 167.8383, etc. (3 being repeated), whick may be written 167.84: Multiplied by 6, gives 1007 for a simple divi- sor; and so of other mixed numbers. ea had SHORT METHODS FOR MULTIPLICATION, Before entering into the more important of our short methods for Multiplication, we deem it proper to give, at the commencement, a-few of the more simple, with which, no doubt, some of our readers are already familiar, but, a knowledge of them is essential to all, to fully understand the several cases which follow. To multiply by 1, followed by any number of ciphers: ‘Simply annex to the multiplicand as many ciphers as there are in the multiplier. Exam. 1. Multiply 475891 by 1000. 475891000 Here, there are three ciphers in the multiplier, and we simply annex three ciphers to the multiplicand to find the product, 475891000. Exam. 2. Multiply 475891 by 1007. 475891... 9331237 479229237 2 SHoRT METHODS FOR MULTIPLICATION. Here, we first multiply by 1000, as in example 1, using periods, or dots, instead of ciphers, and to the product add 7 times the multiplicand. Exam. 3. Multiply 475891 by 993. 4(DB91 Ss 3331237 472559763 Here, we first multiply by 1000, as in the two previous exam- ples, and from the product subtract 7 times (1000 — 993) the multiplicand. To multiply by 21, 31, 41, etc.: Simply set the product by the tens under the multiplicand, in proper position, and add, thus: Exam. 4. Multiply 6853 by 71. 68538 x 71 47971 482563 And if ciphers come between the two digits of the multiplier, proceed in the same way, only move the product as many places to the left as there are figures in the multiplier; thus: 6853 x 7001 A971... 47977853 Composite numbers which can be readily factored should always be used, so as not to need addition in the multiplication; thus: Exam. 5. Multiply 3684 by 42. 3684 x 42 22104 154728 SHort METHODS FOR MULTIPLICATION. Tou Instead of first multiplying by 2 and next by 4, and adding the partial products, we prefer to use the component factors 6 and 7 (6 x 7 = 42). Multiplying first by 6, we get 22104, and multi- plying this in turn by 7, we have the product, 154728, without addition. As we shall make use of subtraction to a large extent in the short methods for Multiplication, we may be per- mitted here to make a slight digression to say a few words on SUBTRACTION. The usual method in Subtraction is to place the less number below the greater, with units under units, etc.; but it will be often found of greater advantage to reverse this order, by having the less placed above the greater; or, by having both the less and the greater contained in one number. Both the latter methods will be used in the following short methods for Multiplication, but in the one case the words upper and lower will be interchanged throughout the rule, and in the other we shall point out the two numbers when contained in one. Exam. 6. Multiply 6847 by 4193. 42.. 6847 x 4193 47929 7 Q7T5T4.. 28709471 Here, instead of multiplying by 4193 and adding the four par- tial products, we take 4200 for approximate multiplier. The dif- ference of the multipliers is 7, by which we first multiply. getting 47929. Now, we see that 7 is one of the factors of 42; dividing v4 SHort METHODS FOR MULTIPLICATION. 42 by 7 gives the other factor, 6. It is evident, now, that by multiplying the product of 7, or 47929, by 6, we get the product for 42, or 287574, and that by simply moving this number three places to the left of units, it represents the product of 4200. Then subtracting the wpper number from the lower, in other words, taking the product of 7 from that of 4200, gives the pro- duct for 4193 (4200 — 7 = 4198). And if the multiplier were 41993, 4199938, etc.; 3493, 34993, etc.; 3495, 84995, etc., we would proceed in a similar manner. using 42000, 420000, etc.; 38500, 35000, etc., for approximates; and so on with other numbers of a like nature. Exam. 7. Multiply 4876 by 999. 4376... x 999 4371624 Here, we multiply by 1000, by simply conceiving three ciphers, represented by dots, annexed; the result is 4376000, or rather, .4376..., as seen in the margin. Now, if from this we take once the multiplicand, 4376, the difference will be the product of 999 (1000 —1= 999). It will now be observed that the product of 1000, and the product of 1, are both contained in the one expres- sion, 4376..., and that by simply taking 4376, the product of 1, from the whole, 4376.. , the product of 1000, the difference is the required result; the subtraction being performed without setting down 4876 a second time. REMARK.— It may, perhaps, be well to remark, that it is imma- terial in Multiplication which factor is taken as multiplier,. or which as multiplicand; or what position the partial products hold with reference to the multiplicand, whether they be placed to the right or the left; provided they are assigned the proper posi- tion with respect to each other. SuHort MrtTHops FOR MULTIPLICATION. V5 Exam. 8. Multiply 3984 by 3476. cae 7 3984 x 3476 16 138904... 55616 13848384 A short inspection of the factors here, shows that by taking the multiplicand, 3984, as the multiplier, the process can be shortened. Taking 4000 for approximate, the difference of the multipliers is 16. Multiplying 3476 by 4, and annexing three ciphers (dots) gives the product of 4000. Now, we observe that 4, the signifi- cant figure of 4000, is contained 4 times in 16, the difference of the multipliers, and that by multiplying 13904, the product of 4, by 4 (found by dividing 16 by 4), and setting the result, 55616, in proper position, it represents the product of 16. Subtracting this, now, from 13904...; that is, taking the product of 16 from that of 4000, gives the product for 3984 (4000 — 16 — 3984). To multiply by any number from 11 to 19. Exam. 9. Multiply 743586 by 11. 743586 x 11 8179446 Here, we simply set down first the unit figure, 6, of the multi- plicand, then add the figures from right to left, carrying when necessary as we proceed, thus: 6 and 8 are 14; 8 and 5, and 1 carried, are 14; 5 and 3, and 1 carried, are 9; 3 and 4 are7; 4 and 7 are 11; 7 and 1 carried are 8. Exam. 10. Multiply 7468 by 17. 7468 x 17 126956 76 SHorRtT METHODS FOR MULTIPLICATION. Here, we multiply by 7, the unit figure of 17, adding the figures of the multiplicand as we proceed, thus: 7 times 8 are 56; 7 times 6 are 42, and 8 (the unit figure) and 5 (carried) are 55; 7 times 4 are 28, and 6 (the figure to the right of 4) and 5 (carried) are 39; 7 times 7: 49 and 4, and 8 (carried) are 56; 7 (the last figure) and 5 (carried) are 12. Norte. — In such cases it is better to always add the figure of the multipli- cand first, adding in the figure carried after. Exam. 11. Multiply 7258 by 1013. [208.21 Mae 94354 7352354 Here, we first multiply by 1000, and next by 18 (short pags and add. Exam. 12. Multiply 13986 by 3684 AE: Fass 3684 x 13986 SIS (6S 1k 51524424. Taking the multiplicand in this (seeing that it is near 14000) for the multiplier, the process can be shortened at once. Multi- plying 3684 by 14 (short method), and then annexing three ciphers (dots) we have the product of 14000, or 51576... Now, the difference of the multipliers is also 14, and it is evident that if 14 times the multiplicand, 3684, be taken from the product of 14000, the difference will be the product of 18986. Now, the product of 14, and also of 14000, are contained in the one expres- sion, 51576...; all we have to do, then, is to subtract 51576 from 51576..., and the difference is 51524424, the required product. (See example 7.) Suort Metruops FoR MULTIPLICATION, 7? Exam. 13. Multiply 35982 by 3286. 36... 35982 x 3286 18 59148 118296... 118236852 Here, we see tha. 35982 is near 36000, so we take these two as the multipliers. Their difference is 18. Multiplying 3286 by 18 (short method), the product is59148. Multiplying this by 2 gives the product of 36 (18 x 2 = 36); annexing three ciphers (dots) gives the product of 36000, or 118296... Now, if from 36000 we take 18, the difference is 35982. Subtracting 59148, the product of 18, from 118296..., the product of 36000, gives 1182386852, the required product. Exam. 14. Multiply 46782 by 27985. Oa 46782 x 27985 654948 45 1309896... 1309194270 If the difference of the multipliers, in this example, was 14 in- stead of 15, wecould proceed as in the last. Here, we set 14 times (short method) the multiplicand under itself, and we see at a glance that if both numbers be added the result is the product of 15 (14 + 1). We donot add, however, but set 2 times 654948, the product of 14, four places to the left, having first drawn a separating: line, and conceiving three ciphers annexed, we have the product of 28000, or 1309896... Now, if from this product, the sum of the two numbers immediately above it be taken, the difference will be the product of 27985 (28000 — 14 + 1). 78 SHort METHODS FoR MULTIPLICATION. Notre.—It is hardly necessary to say to the arithmetical student, that, in taking the two partial products which go to make up the product of 15, from. that of 28000, the addition and subtraction go hand in hand, thus: Com- mencing at the top; 2 and 8 are 10, from 10, and 0. Carry 1 to 8, 9 and 4 are 13, from 20 (in this case) and 7; 2 and 7, 9, and 9 are 18, from 20 and 25 2 and 6, 8, and 4 are 12, from 16 and 4, etc. Exam. 15. Multiply 24975 by 5487. 24975 x 5487... 135925. 185789075 Seeing that 24975 is near 25000, we take these two numbers for multipliers; we see by inspection that the difference is 25. We multiply by 25000, thus: Conceiving two ciphers annexed to 5437 multiplies that number by 100, and taking one-fourth of the result gives the product of 25, or 135925. To this we annex three ciphers (dots), and we get 185925..., the product of 25000. Now, taking 25 from 25000, gives 24975, that is, 135925... minus 135925 (both contained in the one number), gives 135789075, the required product. When the quotient obtained by dividing the difference of the multt- pliers by the significant part of the approximate multiplier, consists of two or more figures which can be readily factored, we would pro- ceed as in the following: Exam. 16. Multiply 59832 by 78481. Oe ae 59832 x 8431 168 98 470586. ...1882344 13176408 4692683592 Taking 59832 and 60000 for the multipliers, in this example, we find cheir difference to be 168, which contains 6, the signifi- SHort Metuops For MULTIPLICATION. 19 cant part of 60000, 28 times. Multiplying 78431 first by 6, and annexing four ciphers (dots) we have the product of 60000. Now, 28 times 6 is 168, and if 168 be taken from 60000 it leaves 59832; in other words, if 168 times 78431 be taken from 60000 times that number, or 470586...., which we already have, the difference will be the product for 59832. To get 168 times the multiplicand, 78431, we take 28 times the product of 6. To multiply by 28, we use its component factors, 4 and 7 (4 X 7 = 28), setting 4 times 470586 a little to the right, as shown in the margin; then 7 times that product, or 13176408, is set under the product of 60000 and subtracted, giving 4692683592, the required product. When one part of the multiplier, or of the multipli- cand, 7s a multiple of another part, illustrated in the following: Exam. 17. Multiply 47634 by 16128. 47654 x 16128 762144... 6097152 768241152 A glance at the multiplier, 16128, in this example, shows that 128, the three last figures, is 8 times 16, the two first, in other words, 128 is a multiple of 16. Multiplying first by 16 (short method), and conceiving three ciphers annexed to the result, gives the product of 16000; then, ‘setting 8 times 762144 in proper position, and adding, we get the required product. Nore.—If the muitiplier or the multiplicand had been 12816, we would proceed in the same way, first multiplying by 16, as in the example; then setting 8 times the product of 16, three places to the left and adding. And if one of the factors were 1612832, 3212816, 1283216 or 1632128, the process would be equally simple; multiplying first by 16, then 8 times the product 80 SHortT METHODS FOR MULTIPLICATION. of 16, and next, 2 times the product of 16; taking care to place the partial products in proper position with respect to one another. And so of other numbers similarly combined; such, for instance, as 812, 8012, 815, 3015, etc.; 412, 416, 424, 4024, etc.; 642, 6042, 756, 7056, etc., etc. Division REVERSED. Many extraordinary contractions in Multiplication may be obtained by simply reversing our new methods for Di- vision. A few examples will illustrate. Exam. 1. Multiply 756 by 334. TOOT: 1512 252504 Here, we simply annex three ciphers (dots) to 756, and under the result set 2 times 756, or 1512. The sum of these two num- bers, without actually setting down the result, is evidently 757512, and we divide both numbers, considered as one, by 3, getting 252504, the required product. Reason: To divide by 334, we would multiply by 3, getting 1002; divide first by 1000 and subtract 2 times the quotient. In the multiplication we reverse the process, multiplying first by 1000, adding 2 times the multiplicand, and then dividing by 3. And if one or both of the factors were 3334, 83384, etc., the process would be equally simple. Exam. 2. Multiply 1668 by 478. 4783894 797304 Here, we simply set 8 times 478, or 3824, to the right and di- vide by 6, and we have 797304, the product. SHort MetuHons FoR MULTIPLICATION. 81 Reason: To divide by 1668, we would multiply it by 6 (being nearly the sixth of 10000), getting 10008 for new divisor. To divide by 10008, we first divide by 10000, then subtract 8 times the quotient. To multiply by 1668 (which in the example we have made the multiplier), we reverse the division by first multiplying by 10000, adding 8 times the multiplicand, and then dividing by 6. And if one or both factors were 16668, 166668, etc., the process would be equally simple. Exam. 8. Multiply 7342 by 8334. 61188228 Setting 8 times the multiplicand, or 58736, five places to the right, in this, and looking on both numbers as one (that is, as being added), we divide by 12 to get 61188228, the required product. Reason: To divide by 8334, we multiply it by 12, getting 100008 for divisor; reversing the division, we have the multipli- cation. Exam. 4. Multiply 142858 by 37684. 142858 ) 37684...... 1000006 226104 5383460872 In this we take 142858, as the multiplier: To divide by this number we multiply it by 7, getting 1000006 for divisor, as shown’ in the margin. To multiply by 142858, we reverse the division by multiplying 37684 by 1000000, adding 6 times 37684, or 226104, and dividing by 7. \ aa fas 82 SHort METHODS FOR MULTIPLICATION. And the numbers 1428, 1429, 14286, 14287, 142857, etc., will — be found equally simple, and so with other numbers, Exam. 5. Multiply 123456789 by 123456789. 198456789.......... 11234567799 9)1934567878765432201 9)13717420875 17114689 15241578750190521 In this we simply conceive ten ciphers annexed to the multipli- cand, and from the result we take 91 times (short method) the multiplicand, or 11234567799; then dividing the difference by 9 and 9, in succession, we have the required product. This is the reverse of division by the nine digits in direct order, (See example, page 68.) Ruue. Zo multiply any number by the nine digits in direct order: Annex, or conceive to be annexed, ten ciphers to the number to be multiplied; from the result take 91 times the said number and divide the difference by 9 and 9 in succession. Exam. 6. Multiply 987654321 by 987654321. U8iGo482 1 chaste 79012345680987654321 8779149520 1097389369 9754610577899 7 LO4L Conceiving ten ciphers annexed to the multiplicand, in this; multiplying the result by 8, and adding the multiplicand, we divide the sum by 9 and 9, in succession, as in the preceding example, and we have the required product. | This is the reverse of division by the nine-digits written in re- versed order (see example, page 69). SHort METHODS FOR MULTIPLICATION. 83 Rue. To multiply any number by the nine digits in a reversed order: Annex, or conceive to be annexed, ten ciphers to the multiplicand; to 8 times the result add the multiplicand, and divide the sum by 9 and 9 in succession. Nore.— An unlimited number of such examples could be added, but, for the student who has carefully read the preceding pages of this work, enough has been given by way of suggestion, To square any number of two digits we give the following simple Rowe. (1) Add to the given number the difference between it and the next higher number ending with a cipher. (2) Subtract from the given number the number which was added. (3) Multi- ply the sum and difference thus found, and to the product add the square of the figure added. Exam. 1. What is the square of 47? First, add 38, making ..... 50 Second, subtract 3, making 44 Their product is......... 2200 Now add 3 times 3, or... 9 2209, the square. Or, if more convenient, subtract from the given number the difference between it and the next Jower number ending with a cipher; then add to the given number that which was subtracted, and proceed as above, thus: Exam. 2. What is the square of 91? First, subtract 1, making. . 90 Second, add 1, making... | 92 (ete eee Their product is......... 8280 AMS Xl OFime wad A 1 eee 8281, the square. 84 SHortT METHODS FOR MULTIPLICATION. The process is founded on the principle that: Zhe product of the sum and difference of two numbers, is equal to the difference of their squares : If 47 + 3 be multiplied by 47 — 3, the product 1s equal to 4% squared minus 8 squared; that is, (47 +8) x (47 —8) = 47733 Now, 47 + 3 is the same as 50, and 47 — 3, the same as 44; therefore 50 x 44 is equal to (47 x 47) minus (3 x 3), that is, when 50 is multiplied by 44, and the square of 3, or 9 added to the product, the result will be the square of 47; thus: 50 x 44 — 2200; and 47? 3? = 2209 — 9; that is, 9900 — 2209 — 9 Now, it is a well-known axiom that: if equals be added to equals, the sums will be equal. If, now, 9 be added to the two last expressions, which are equals, the sums will be equal, thus: 2200 + 9 = 2209 —9+9; subtracting 9 and adding it at the — same time leaves the last expression 2209; and adding 9 to 2200 makes it 2209, which is the square of 47. Norz.— A little practice will enable a person to readily square any number consisting of two digits mentally by this simple method. This method will be found convenient in many cases where the number to be squared consists of three figures. Exam. 3. What is the square of 1092 109? = 100 x 118 + 81 = 11881, at sight. Numbers consisting of three figures may be readily squared on the same principle, thus: SHorT Metuops For MULTIPLICATION. 85 Exam. 4. What is the square of 432? 400 x 464 = 185600 459? == 4 30 x '34'== * 1020 Q2x 29= 4 186624 Subtracting 32 gives 400, adding 32 gives 464; the results are multiplied, giving 185600. The square of 32 is added next; sub- tracting 2 gives 30, adding 2 gives 34; the results are multiplied, giving 1020. Finally, the square of 2 is added, making 186624, the square of 4382. Exam. 5. What is the square of 371 ? 442 x 300 — 152600 Bila + 12%. 10:—+. 5040 1x L= 1 137641 Adding 71 makes 442, subtracting 71 makes 300; their product is 132600. Adding 1 to 71 makes 72, subtracting 1 makes 70; their pro- duct is 5040. Adding the square of 1 gives 137641, the square of 371. Rute. To square any number of two figures: (1) Multiply the units by the units and set down the unit figure of the product. (2) Multiply the sum of the units by a single figure of the tens, set down the unit figure of the product. (8) Multiply the tens together to com- plete the square, carrying as usual. Exam. What is the square of 74? Ans. 74=74 X 74 = 5476. Here, say 4 times 4 are 16; set down 6, and carry 1; then, 7 times 8 (4 + 4) are 56 and 1 are 57; set down 7, and carry 5: next, 7 times 7 are 49, and 5 are 54 completes the square. (See page 271.) 86 Saort Mergops ror MOULtIPLicatTIoN. Rue. To square numbers of two figures ending in 5: (1) Multiply the 5's together and set down the result in full. (2) Add 1 to either Jigure of the tens and multiply the other by the number thus increased. Exam. What is the square of 75? Ans. 75°= 15 + 75 = 5625. Say 5 times 5 are 25; set down in full; add 1 to either 7 and say 8 times 7 are 56 to complete the square. Reason: 7 xX 75 = 80 xX 70 plus 5 X 5 = 5600 + 25 = 5625. The rule is applicable in many cases to three, four or more figures, thus: 125. 185, 145, etc., 295; 895; 39995, etc. Exam. What is the square of 695? Ans. 695? = 695 X 695 = 483025. In this say 5 times 5 are 25; set down in full; add 1 to 69 and say 70 times 69 are 4830 to complete the square. And so with the others. and similar nuinbers. And the same rule-can be applied to any two figures, whose units when added make 10; the tens in both numbers being alike; as 47x 43; 76 x 74; 58 X 52, etc., and in many cases to three, four, or more figures, as 172 x 178; 127 x 123; 196 x 194; 193 x 197; etc., 292 X 298, 293 x 297; 394 x 396; 4998 x 4997; ete. Exam. Multiply 76 by 74. Ans. 76 X 74 = 5624, Say 4 times 6 are 24; set down in full; add 1 to either 7, and say 8 times 7 are 56; for the reason that 76 X 74= S80 X 70 plus 6 X 4 = 5600 + 24 = 5624. Exam. Multiply 397 by 393. Ans. 397x393 = 156021. Say 3 times 7 are 21; set down in full; add 1 to 39 (either factor) and say 40 times 39 are 1560 to complete the product. (See examples on page 87.) NotE.— When the units are 1 and 9 the second figure of the product will always be a cipher; as: 71 x 79 = 5609, 691 x 699 = 483009. Say 9 times1 are 9, then 0, 8 times 7 are 56. Say 9 times 1 are 9; then 0; and 70 times 69 are 4830 completes the work. Exam. Multiply i27 by 124. Ans. 127 x 123 = 15621 Plus 127 multiplied by 1 = 127 15748. Note, — Multiply as if the units made 10, and add or subtract for the difference. emer eo ~~ es +) Sica Sa bint, eel! Sport Mrruops ror MULTIPLICATION, 87 Hence, in all cases where the two units figures make 10, and the other figures are alike, the foregoing rule is applicable, as illustrated in the following ExAMPLES: 127 < 123 = 15621: Here, because 3 and 7, the unit figures, make 112 x 118 10, and 12 and 12 are alike, we simply say 3 47 X 45 times 7 are 21, setting it down in full; then, 84 x 86 adding 1 to either 12, calling it 18, we say 12 134 x 136 | times 13 are 156, which completes the product, etc. 15621. And tf one or both factors contain a fraction, the process will be found equally simple; thus: 24 x 264 = 636: In this, 6 and 4 make 10, and the 2’s are alike; 284 x 22 say one half of 24 is 12, tocarry; then, 6 times 4 36 X 34} are 24, and 12 are 36; set down in full; now add 1 48 x 424 to 2 and say 3 times 2 are 6; this completes the etc. product. 241 & 261 = 6431: Here, we say } of 24 is 6, to carry: then, 6 123 x 184 times 4 are 24, and 6 are 30; now, 3 times 2 are O2t X 3885 6; making 630, or 24 x 261: to this is added } as of 26}, or 134, making 643}. Note.— When the unit figures are 1 and 9 the second figure of the product will be a cipher; thus: 61 x 69 = 4209, By this simple rule, then, such numbers as 21 < 29, 22 x 28, 23 X 27; 31 x 39, 82 x 38, 85 xX 35; 47 x 43, 48 x 42, 58 x 52; and up in the sixties, seventies, eighties, etc., may be multiplied together without effort. Also, 191 «199, 192 x 198, 1992 x 1998, 1202 x 1208, 1807 « 1303, 798 x 792, 6993 x 6997, etc. (See examples, page 260 ) Exam. What will 198 hats cost at $1.92 each? In this, 2 and 8 make 10, and the 19’s are alike. Say 2 times 8 are 16; set down in full, then, add 1 to 198 x 1.92 19 and say 20 times 19 are 380; this makes se 16, $380.16 the cost. 88 Meruops oF PRooF. First: by casting out the 9’s: cast the 9’s out of both factors, as in the proof for Division, page 63, and reserve the excesses. Multiply these excesses together and from the result reject 9, also. Now, find the excess of 9’s in the product, and if this be the same as found from the factors, the work is generally correct. In the annexed example, the excesses of the factors are 4 and 3; their productis12. Taking 43976... .4 9 from 12 leaves an excess of 8. Then, reject- 14682. ...3 t 12 ing the 9’s from the product, the excess is 3, 6353878232 3 also. (See page 63, and note 4, page 64.) Second: casting out 11’s: commence at the units of the multipli- cand and add all the digits in the odd places; then, all those in the even places, and from the former, increased if necessary, by 11, subtract the latter, and reserve the excess. Proceed in like manner with the digits of the multiplier, and reserve the excess. Multiply these two excesses and take the even from the odd (increased by 11, if necessary) and reserve the excess. Finally, in a similar manner, find the excess in the product, and if this be the same as that from the factors, the work is generally correct. In the annexed example, the sum of the digits 43276. ..2 in the odd places of the multiplicand, is 12, and 14682. ..8 t 16 that in the even, 10; the excess is 2. Now, in 6353782382. .... 5 the multiplier, the digits in the odd places make 9, and in the even, 12. Increasing 9 by 11 we get 20 from which 12 is taken, leaving an excess of 8. Then, 8 times 2 are 16, and taking the even from the odd in this, we say 1 from 6 leaves 5, the excess. Or, 11 from 16 leaves 5. Finally, the digits in the odd places of the product make 22, and those in the even, 17; the excess is 5, also, the same as found from the factors. Norg. — In all cases where the odds are less than the evens the odds must be increased by 11. Those figures can be applied to prove Addition, Subtraction and Division. The 11 is more reliable than 9, and may be used with advantage as a check figure to test the correctness of the footings of the Cash Book, Journal, etc. By applying both 9 and 11 to the same operation, an almost abso- lute certainty of its correctness would be obtained. CANCELLATION. Arithmetical calculations are frequently abbreviated in a wonderful manner by what is called Cancellation, that is, rejecting equal factors from numbers bearing to each other the relation of dividend and divisor; or, from mul- tiplier and divisor in operations requiring multiplication and division. Exam. 1. Multiply 476 by 312 and divide the product by 416. 476 x 312 1428.. 5712 + 416 104 357 is the required result, found by the methods already estab- lished. 50 CANCELLATION. Bringing cancellation to our aid, the same result is much more readily found as follows: 4 8 416 476 ~—s-_ «28:19 104 1428 * 104 B57 In this, we see at a glance that the multiplier, 312, is divisible by 3, giving 104; and the divisor, 416, is divisible by 4, giving 104 also. Now, multiplying 476 by 312 is the same as multiplying it by the factors, 3 and 104. Next, dividing by 416 is the same as dividing by the factors, 4 and 104. Multiplying 476 by 104, and dividing it at the same time by 104, will not change the number. Rejecting the common factors, 104, then, we simply multiply 476 by the remaining factor, 3, getting 1428, which is divided in turn by the remaining factor, 4, giving 357, as found by the long process. Exam. 2. Multiply 468 by 800 and divide the product by 792, 192 468). . 8he 4168 4 472|72 Dividing 792 and 800, in this, each by 8, we get 99 for divisor and 100 for multiplier. Annexing two ciphers (dots) to 468 mul- tiplies by 100; then to divide by 99, according to the simplified method, we divide by 100 and add 1-100 as often as possible. Now, to divide by 100 we cut off the two ciphers (dots) which were annexed to multiply by 100. From this it will be readily seen that the position of the vertical line, in all such cases, is to the right of units, or through the decimal point. 2 0 CANCELLATION. 91 Exam. 3. Multiply 423 by 14 and divide the product by 728. 728 493 14 104 846 9 32 14 Dividing 14 and 728 each by 7, in this, gives 2 for multiplier and 104 for divisor. Multiplying 423 then by 2 and dividing by 104 we get 8|14—8.134+. (See p. 17.) Exam. 4. Multiply 51 by 73 and divide by 49. aU ete O leat (ol: 98 102 1/46 © Doubling 49 and 51, in this, the multiplier is 102, and the divi- sor, 98. From what has been said in example 2, the position of the vertical sine, here, is to the right of 738. Adding 2 times 73, now, placed in proper position, multiplies by 102. Then divid- ing by 98, we get 75|96=75.979+. (See pp. 19 and 21.) On examining the process it will be seen that we have added 2-100 of 73, or 1/46; and in dividing by 98 we have added 2-100 of 74; then 2-100 of 1, or 1/48 and |02. Now, it is evident that the process can be abridged, thus: 49 51 731. 98 te a ae 992 CANCELLATION. Adding 4-100 of 73, adds for 2 in 102, and also for 2 in the | division by 98. Then 2 times 2, or 4, is added, which finishes the division by 98, and we have 75|96 as before. Exam. 5. Multiply 204 by 68 and divide the product by 214. 214 68 204: 107 1/36 103 69|36 4/83 64/53 35 88 Dividing 214 and 204, each, by 2, gives 107 for divisor, and 102 for multiplier. The result is 64/88=64.82+. In this it will be seen that 2-100, or 1]36, has been added; then 7-100, or 4|838, subtracted, which is evidently the same as sub- tracting 5-100 (the difference), so we simply subtract 5-100 of 68, or 3/40, as shown in the margin. This 68 adds for the 2 in 102, and subtracts for 7 in 107 _38)40 Then 7 times 4 (3 plus the 1 carried in subtracting 64/60. 40), or 28, is added, to complete the division by 107. 28 The result is 64/88, as before. 88 Exam. 6, Divide 748 times 416 by 412. 412 748 416 103 T4808 __ it THOT Dividing 412 and 416 each by 4, the divisor is 103 and the multiplier 104. Now, we have to add for 4 in 104, and subtract for 3 in 103; their difference is 1, so we simply add 1; that is, " CANCELLATION. 93 1-100 of 748, or 7/48. This completes the multiplication by 104, and subtracts for 3 in the divisor. But as the addition exceeds the subtraction, it is evident that 7, which is in reality a part of 748, immediately above it, has got to be multiplied by 3 (in 103), and the result, 21, subtracted. Taking 21 from 48 gives the remain- der, 27, and adding 7 to 748 gives the quotient, 755. Exam. 7. What would be the result of taking $32.64, 416 times, and dividing the result into portions of $31.04 each ¢ 3104 416 3264 “St a is eal 437|43 Taking $31.04 and $32.64 as so many cents, throws off the decimals, and the multiplier becomes 3264, and the divisor, 3104. A moment’s inspection, now, shows that each is divisible by 4, giving 776 and 816. These, in turn, are divided by 8, giving 97 and 102. Then, adding 5-100 of 416, or 20|80, completes the multiplica- tion by 102, and adds for 3, the complement of 97. Next, 8 times 20, and then 3 times 1 (carried in adding the remainders 60 and 80), set in proper position, completes the division for 97. Adding the several results, we get 437/43. Nortr.— The value of the remainder, .43, is found by multiplying it by 4, and the product by 8 (the factors by which we divided in the example). The product is 1376, evidently cents in this case, or $13.76, which is not large enough to make a portion ($31.04). The answer, then, is, $31.04 could be taken 437 times, and $13.76 would remain. Exam. 8. A hatter exchanged 172 men’s hats which cost him $3.35 each, for a number of boys’ hats, at $1.68 each; how many of the latter did he receive ? 94 CANCELLATION. REMARK.— In simplifying arithmetical operations, the relation 7 which the numbers sustain to each other, must be carefully borne in mind. If, for instance,the numbers sustain to each other the relation of multiplicand and multiplier, whatever operation is performed on the one, to simplify the process, the reverse is performed on the other, that the product may not be changed; and if the relation be that of dividend and divisor, whatever operation is performed on the one, the same is performed on the other, that the quotient may not be changed. | To solve the above example, we multiply $3.35 by 172, and divide the product by 1.68; thus: 6 8 6 168 335 172 1008 1005 1032 “344... 1720 845/720 2/760 342/960 24 984 Here, 885 and 172 bear to each other the relation of multipli- cand and multiplier, and 168, that of divisor to each of the others, or, to their product. To divide by 168 we multiply it by 6, getting 1008, a simple divisor; and multiplying 172 also by 6 we get 1032. Next, to multiply 335 by 1032, we take the former for multiplier and mul- tiply it by 3 to get 1005, a simple multiplier. Multiplying 335 by 3, divides 1032 by 3, giving 344. The terms now are 1008, 1005 and 344. Multiplying 344 by 1005, and dividing the product, 345720, by 1008, according to our established methods, we get 342/984=342.976. CANCELLATION. 95. Bringing Cancellation to our aid, we can still further simplify the foregoing process, thus: 168 335 172 1008 1005 ~ B44)... 1/032 342/968 16 984 It will be observed that, in the first process, 172 was multiplied by 6, and the product divided by 3, which, by cancellation, is simply multiplying by 2. Here, then, we simply multiply 172 by 2, getting 344. Next, adding 5 and subtracting 8 is subtracting 3, in this case 38-1000. So we subtract 3 times 344, placed in proper position, or 1|032. Now, since the subtraction exceeds the addition, it is evident that 1, or rather 2 (1 plus the 1 carried in subtracting), has yet to be multiplied by 8 (in 1008), and the pro- duct, 16, added to complete the division by 1008. The result is 342/984, as before. Exam. 9. How many pairs of men’s shoes at $3.34 a pair can be got in exchange for 346 pairs of women’s, worth $2.494 a pair? 334 9494 346 1002 998 1088 95915. 1/037 958/463 To solve this, $2.494 is multiplied by 346, and the number ow times $3.34 is contained in the product is the required number. Multiplying 334 cents by 3 gives 1002 for a simple divisor, ama 96 CANCELLATION. nultiplying 2494 cents py 4 gives 998 for a simple multiplier. Multiplying 334 by 3 multiplies 346 also by 3, giving 10388; and multiplying 2494 by 4 divides 346 by 4; or, what amounts to the same thing, 1038 is divided by 4, giving 259.5. The terms, now, are 1002, 998 and 259.5. To multiply by 998 we use 1000 and subtract 2 (that is, 2 times the multiplicand), and to divide by 1002 we use 1000 also, and subtract for 2; in other words, we ~ subtract 4-1000 of 259, or 1/0386. In multiplying by 998, however, it must be borne in mind that — 259.5 has been multiplied by 2, while in dividing by 1002, only 259 has been multiplied by 2. In multiplying 259 by 4, therefore, 1 is added (found by multiplying the decimal .5 by 2, the comple- ment of 998), making 1|037 to be subtracted; the result is 258)46 pairs. Exam. 10. How many acres of land at $32% an acre can be got in exchange for 176 acres, worth $48.50 an acre ? 3 p) B94 176 48.50 7 88 97 264 Multiplying 324 by 3, gives 97 for divisor— multiplying 48.50 by 2, gives 97 for multiplier, also; and, being common factors, we reject both. Now, multiplying the divisor, 324 by 3, multi- plies the dividend, 176, also, by 3; and multiplying 48.50, one of the factors of the multiplication, by 2, divides the other factor, 176, by 2. So we simply divide 176 by 2, and multiply the result, 88, by 3, to get 264 acres, the required number. Or, add 88 to 176, because multiplying by 3 and dividing by 2 is multiplying by 1}. Nors. — Other examples, snowing methods of abbreviation, might be — added without limit, but those which are given will be sufficient by way of suggestion. RULE OF THREE. “The Lule of Three” (so called because there are al- ways three numbers given to find a fourth) has reference to that part of Simple Proportion usually taught in arithmetic, to find a fourth proportional to three given numbers. The resolution of this problem is the most important result of the theory of Proportion. On account of its great utility and its extensive application by merchants, accountants and others, it has been called the Golden Rule. A thorough knowledge of this problem will enable the student to understand, with little effort, questions in Percentage, Interest, Discount, Commission, Profit and Loss, and other branches of arithmetic having the principles of Proportion for their basis. Proportion is the equality of ratios. Ratio is the relation which one number bears to another of the same kind, with regard to size or comparative value. Thus, since 8 is double of 4, and 10 of 5, the ratio of 8 to 4 is equal to that of 10 to5. The four numbers, 8, 4, 10 and 5 are proportionals and constitute a proportion, or analogy. In this relation they are usually written thus: mes + 4:2 10: 5;-or, simply, 8's 4:33 10 3:5. < ° 98 RULE OF THREE. The first expression is read, as 8 is to 4 so is 10 to 5; and the second, 8 is to 4 as 10 7s to 5, meaning that whatever relation 8 has to 4, the same relation exactly 10 bears to 5; in other words, whatever number of times 8 is greater or less than 4, 10 is as many times greater or less than 5. | Now, 8: 4is a ratio, meaning that 8 has a certain relation to 4; 10:5 is also a ratio, and both being equal, they might also be written thus: 8 A085 and may be read as in the other cases, or, the ratio of 8 to 4 equals the ratio of 10 to 5. The two numbers, 8 and 4, are the Zerms of the first ratio, the first number is called the Antecedent, and the second, the Conse- guent ; 10 and 5, are the terms of the second ratio, 10 being the antecedent, and 5 the consequent. The value of a ratio is the quotient obtained by dividing the antecedent by the consequent; thus: the value of 8 : 4 is 8, or 2; and of 10: 5, 4,2, or 2. Since there are two terms in a ratio, and two ratios in a pro- portion, there must be at least four terms in every proportion. : The first and fourth terms of the proportion, or the outside num- bers, 8 and 5, are called the Hztremes ; and the second and third terms, or the middle numbers, 4 and 10, the Means. And when four numbers constitute a proportion the product of the means will always be found to be equal to the product of the extremes. Thus, in the proportion: 8: 4 32510 5 4 multiplied by 10 equals 8 multiplied by 5. Hence, any three terms of a proportion being given, the fourth can be readily found. Thus, if an extreme be wanting: Multiply the two means together and divide the product by the given extreme ; the quotient will be the required extreme, and if a mean be wanting: RULE OF THREE. U9 Multiply the extremes together and divide the product by the given mean ; the quotient is the required mean. Exam. 1. What is the fourth term of the proportion UNS er Solution: 24 x 4 = 96; then, 96 + 16 = 6, the required term. The proportion is now 16: 4 :: 24: 6. Proof: 24 x 4 — 96, the product of the means, 16 x 6 = 96, the product of the extremes, Exam. 2. Complete the following proportion: A ee De ea Solution: 16 x 6 = 96; then, 96 + 24 — 4, the second term. In this, two extremes are given, and one mean, to find the other. The extremes are 16 and 6; they are multiplied together and the product, 96, divided by 24, the given mean; the quotient, 4, is the other mean. From the foregoing principles and illustrations we de- rive the following: Rue. (General Rule.) To find a fourth proportional to three given numbers : I. Arrange the three given numbers in a line, in succession, setting the one which is of the same kind as the required term, the third in order. Il. Lf, by the nature of the question, the required term is-to be greater than the third term, put the greater of the other terms in the second place; but if the required term is to be less than the third, then put the less of the other two in the second place. Ill. Multiply the second and third terms together and divide the product by the first; the quotient wili be the required term. 100 RuLE oF THREE. Notrs.— 1. If the first and second terms be not of the same denomination, they must be reduced to such. 2. If the third term be a compound number it must be reduced to its lowest unit. Exam. 38. If a person is to receive $840 for 9 months’ services, how much ought he to receive for 90 days at | the same rate ? 9 x 80 = AO Oe 840 Bit ieee Bee The answer must be money, so we put $840 for the third term, and from the nature of the question the answer is to be less than $840; the fourth less than the third, therefore, the second must be less than the first. The second term, then, is 90 days and the first 9 months. Now, the first and second, although being of the same kind (time), are not of the same denomination; multiplying 9 by 30 gives 270 days, and the statement is: 270 days is to 90 days as $840 is to the required term. A glance at the two first terms now shows that each is divisible by 90, and consequently the ratio is as 3:1. So we simply divide 840 by 3, and the quo- tient, $280, is the answer. ee 840 x 90 — 75600 i8 840 x 1 — 840 ROOF 1 280 x 270 = 75600 280 X 3 = 840 Norse. — It may be well to remark that, in abbreviating operations in pro- portion by cancellation, the first and second, or the first and third terms (but never the second and third), are to be operated upon, Exam. 4. If a pole 6 feet 3 inches high casts a shadow 74 feet long, how high is a ae whose shadow is 120 feet long? fet. ft? ft. in. Tt : 120323 6.8 N 240 TH 1200 5 100 feet. NN a eee RuLE oF THREE. 101 In this, height is required, therefore, 6 ft. 3 in. height is put in the third place; and from the nature of the question, the answer is to be more than this third term, consequently, the second term is greater than the first; 120 is, therefore, put in the second place. Reducing the first and second terms, now, to the same denomi- nation, halves, and the third term to its lowest unit, inches, we have: 15 : 240 :: 75. Bringing cancellation to our aid, now, 15 and 75, the first and third terms, are each divided by 15, giving 1 and 5 (the 1 is omitted, as it does not affect the process, and the first term dis- appears). Multiplying the second and third terms, next, that is, 240 by 5, gives 1200, which is of the same denomination as that to which the third has been reduced, namely, inches. Dividing 1200 by 12 (inches in a foot) we have 100 feet, the height of the steeple. 240 x 75 — 18000 1200 x 15 — 18000 PRooF: or, or, 120 x 64 — 750) 240 x 5 — 1200 100 x 74 = 750 { 1200 x 1 — 1200 The product of the means equal to the product of the extremes. Exam. 5. If a contractor be paid $64.35 for excavating 148 cubic yards, or loads of earth, how much ought he to receive for excavating 335 loads, at the same rate? fé 3 148 : 335 :: 64/35 TO01 1005 21/45 150}15. 600 75 102 Rue OF THREE. The answer to this must be more than $64.35; hence, 335 isthe second term of the proportion. 4 To multiply by 335, we multiply it by 8, to get 1005, a sinaple q multiplier. Multiplying 335 by 3, divides 64385 by 3, giving 2145, to be divided by 148. To divide by 143, it is multiplied by 7, giving 1001 for a sim- q ple divisor. Multiplying 148 by 7, multiplies 2145, also, by 7%, giving 15015, The terms, now, are; 1001 « 1005.22. 150215 Here, we are to multiply by 1005, and then divide by 1001. Now, to multiply 150.15 by 1000, we simply move the point three places to the right, getting 150150; and to divide this by 1000, we move the point three places to the left, getting for the result, 150.150, or, as shown in the example, 150/15., the dot represent- ing the cipher. Hence, we simply draw the vertical line through the decimal point of $64.35, divide by 3, and multiply by 7, to get 150.15. Annexing a cipher (dot) to 150)15, now, we simply add the 4-1000 (5—1), or .600; the required result is $150.75 (see Cancella- tion, examples 5, 6, 7 ad 8). Exam. 6. A builder is paid $678 for 168 perches of — masonry ; how much ought he to receive for 3374 perches, — at the same rate ? | o- ORD 3 168 : 3374 :: 678 1008 1018)" Tabpiees $1361/384 In this, the answer is to be money, therefore, $678 is put in the RuLE oF THREE. 103 third place; and, as more will be paid for 3374 perches than for 168, the larger of the two is put in the middle. Multiplying 168 by 6, gives 1008 for a simple divisor; and multiplying 8874 by 8, gives 1012 for a simple multiplier. Now, multiplying 168 by 6, multiplies 678, also, by 6; and multiplying 3374 by 3, divides 678 by 3. But, multiplying 678 by 6, and then dividing by 3, is simply multiplying it by 2. Multiplying 678, then, by 2, gives 1356. The analogy, now, is: 1008 : 1012 :: 1356; and the product of the second and third terms divided by the first, is the answer. To multiply 1356 by 1012, we annex three ciphers (dots), and to the result add 12 times 1356. To divide by 1008, we cut off the three ciphers, and subtract 8 times 1356. Now, adding 12, and subtracting 8, is adding 4; so we simply add 4 times 1356 (in this case 4-1000), or 5424, placed in proper position. The multiplication by 1012 is now completed, but the division by 1008 is not. We have still to subtract 8 times the partial quotient, 5, or rather, 8-1000, which is .040. Subtracting 40, now, from 424, we get .384, or 38 cents; and adding 5 to 1356, we have $1361.38, the answer. Exam. 7. A drygoods dealer bought at auction 315 yards of silk for $352.80, and at the close of the auction he bought, at private sale, 214 yards more of the same material at the same rate; how much did he pay for the latter 2 a y, B15 : 214. 327352/80 105 107 117/60 235/00 4\70 0) 239/70]. . Tht a 104 RcuLE oF THREE. ihe The answer here is to be less than $352.80, so 214 is put for the second term. Dividing 315 by 8 gives 105 for simple divisor, and _ 214 by 2 gives 107 for simple multiplier. Dividing 315 by 8 di- vides 352.80 by 3, also, giving 117.60; and dividing 214 by 2 multiplies 352.80, or rather, in this case, 117.60 by 2, giving 235.20, The terms now are : 105 : 107 :: 235.20, and the product of the — second and third divided by the first gives the answer. We have now to add the 7-100, and then subtract 5-100; this is add- ing 2-100. The 2-100 of 235 is 4/70. Next, we have to subtract 5-100 of the partial quotient, 4, to finish the division by 105; this is found to be .20. It will be observed, now, that 20 is to be added to 70, and at the same time 20 is to be subtracted; so we simply set down 70 and add 235 and 4, getting 239/70. To find the correct decimal, or cents, now, a vertical line is drawn to the right of 70, and two ciphers (dots) annexed. To this result 2-100 is to be added, and 5-100 subtracted, as in the other part of the process; this is the same as subtracting 3-100. The 3-100 of 70 is 2110. Next, 5-100 of 2, or .10, is to be sub- tracted; or, simply, 2 altogether is subtracted and we get 68, the correct cents. The answer is $239.68. Exam. 8. If 3168 bushels of wheat cost $3200, what will 317 bushels cost at the same rate ? 32... 8168 : 317| :: 3200 39 3\17 3 320/20 In this, 317 bushels will cost less than 3168 bushels, therefore, 317 is put in the second place. Then, we simply draw the vertical line to the right of 317, and setting that number under itself, two RULE oF THREE. 105 places to the right; then 3, under the last result, two places to the right, and adding, we get $320.20, the required cost. Reason: Taking 3200 for approximate divisor, in connection with 3168, we find the key to the division, 1, that is, 1-100, as has been explained in the article on Simplified Division. Multiplying 317, now, by 3200, and dividing the result by 3200 does not change 317. The divisor is not 3200, however, but 3168, so we must add the 32-8200, or rather its equal, 1-100 of 317; and this is done by simply setting 3|17, and then |08, in proper position, as seen in the example. Note.— The process will be made clear by dividing 3168, the true divisor, and 3200, the third term of the analogy, by 32, using the factors 4 and 8. It will then be seen that we have simply multiplied 317 by 100, and divided the product, 31700, by 99, according to Rule III of Simplified Division. Exam. 9. If 1372 bushels of oats cost $617.40, what will 1470 bushels cost at the same rate ? 14.. 1372 : 1470 :: 617/40 —98 “jos 30/8700 648|27 12/96 26 661/49)00 My 1470 bushels will cost more than $617.40, therefore, 1470 is made the second term of the proportion. To divide by 1372, we use 1400 as approximate in connection with it, and the key to the division is 2 (2-100). A glance at 1470 shows that it is divisible by 14, giving 105 for a simple mul- tiplier. Dividing the second term, 1470, by 14, divides the first term, 1372, also, by 14, or rather the approximate, 1400, giving 100 for approximate divisor. Multiplying $617.40, now, by 105 106 RuLE oF THREE. (simply set 5 times that number two places to the right, and add), and dividing by 100, we get 648/27. Then, adding the 2-100 of the partial quotients, we get $661.49. Continuing the division, we find the correct decimal, or cents, to be .50, .98 being equal to 1. The answer, then, is $661.50. Nors.— If 1372, the real divisor, be divided by 14, instead of the approxi- mate, 1400 (using the factors 2 and 7 to divide by), we find the new divisor is 98; and that we have simply multiplied $617.40 by 105, and divided the product by 98. _ Now, since the key to the division is found by dividing the difference of the divisors, or 28, by 14 (the significant part of 14..), the key thus found, in all such cases, wéil be the difference between the approximate (new) divisor, found by cancellation, and the real (new) divisor, Thus, 28 + 14 = 2, the key; and14.. + 14= 100, the approximate (new) divisor; then, 100 —2 = 98, the real (new) divisor (all of which can be seen mentally). Exam. 10. If 2328 yards of silk cost $2460, what will 310 yards cost at the same rate? | Dass 2328 : 810]/:: 2468, 79 17/05 mee $327|56 Notr.— The real (new) divisor in this example is 97, found mentally, thus: | 24,. + 24 = 100, approximate (new) divisor; 72 + 24=8; then, 100—3= 97. Dividing the first term of the proportion by 24, divides the third term, 2460, also, by 24, giving 1024 for multiplier. Multi- plying 310, now, by 1023, and dividing by 97, we get $327.56, the required cost. Setting 54 times 310, or 1705, in proper posi- tion adds for 3 in the divisor, 97, and also for 24 in the mul- tiplier, 1024, and completes the multiplication. Then, 3 times 17, or 51, placed in proper position, is added to complete the division by 97. (See remarks on Rule IIT, page 38.) RuLE oF THREE. 107 Comprounp PRoporrTIOoN. Proportion is often of such a nature as to require a compound and a simple ratio, and is then called Com- pound Proportion ; but a knowledge of simple propor tion will enable us to solve compound with equal facility. A compound ratio, as its name denotes, consists of two or more simple ratios, and can be reduced to a sim- ple ratio by the following: Ruiz. To reduce a compound ratio to a simple one: Multiply all the antecedents together for a new antecedent, and all the consequents together for a new consequent. Exam. 1. If it cost $3600 to supply 30 men with pro- visions for 24 days, the rations being 20 ounces per day, how much will it cost to supply 20 men for 36 days, the rations being 18 ounces per day ? In this example, it will be observed, there are three pairs of terms or couplets, namely, 80 men and 20 men, 24 days and 36 days, _ Noting men, days. 02. 20 ounces and 18 ounces; and there 99 __ 94. 99 __ $3600 is a single term, $3600. Arranging 90 — 36-.—18 — (2) the couplets, as shown in the mar- gin, which we will call ‘noting the question”; setting the single term last, and under it, the interrogation point, to indicate that the corresponding term is wanting, we are prepared to set the antecedents and consequents of the respective ratios in their proper places. The answer to the question being money, $3600 is put for the third term in the following statement: 108 Rue oF THREE. 30 : 20 :: 8600 24: 86 20: 18 Tt00. : 12960 :: 3600 4 E2540 alee To arrange the other terms in their proper places, now, we go back to the noting, and, taking the first couplet, men, in connec- tion with the single or third term, we proceed by question and answer, thus: If it cost $3600 to supply 30 men, will it cost more or less to supply 20 men? The answer is evidently less; 20 is - therefore put in the second place, and 30 in the first. Next, if it cost $3600 to get 24 days’ supply, will it cost more or less to get 36 days’ supply ? More; therefore, 36 is the second term of the next ratio and 24 the first. Now, if it cost $3600 to supply 20 ounces of rations, will it cost more or less to supply 18 ounces ? Less; and, therefore, 18 is the second term of the third ratio. The compound ratio now consists of three simple ones, namely: 30 : 20, 24 : 36 and 20: 18. Multiplying the antecedents together, now, we get 14400 for a new antecedent, and the consequents together, and we get 12960 for anew consequent. The proportion, incomplete, now stands: 14400 : 12960 :: 83600; that is; three terms to find the fourth. — Multiplying the second and third terms, and dividing the product by the first, as in simple proportion, will give the fourth propor- tional, or the answer. Before proceeding with said multiplication and division, how- ever, we see by inspection that the process can be abridged by Cancellation, 3600 being a measure, or exact divisor, for 14400. Dividing each of these numbers, then, by 3600, gives 1 for the third term and 4 for the first, and we simply divide the second term, 12960, by 4, getting $3240, the required fourth proportional, or the answer, PRACTICAL PROBLEMS. 109 Proor: § 12960 x 3600 — 46656000 * 714400 x 3940 — 46656000 ue 12960 x 1 — 12960 ae 3240 x 4 — 12960 The product of the means equal to the product of the extremes, taking either pair of ratios. Nore. — It is almost needless to remark, at this stage of the work, that, before multiplying the antecedents and consequents together, as in the example, recourse may be had to cancellation, dividing the first and second terms, or the first and third, by any numbers that wili reduce or eliminate those terms. PRACTICAL PROBLEMS. Proportion may be greatly abridged by the use of aliquot parts. P An Aliquot Part of a quantity is such a part as, when taken a certain number of times, will exactly make that quantity. Thus, 4 is an aliquot part of 12, 24 of 10, 123 and 334 of 100, etc. Exam. 1. What is the cost of a cargo of iron weighing 259166 pounds, at $36.16 per gross ton? There are 2240 pounds in a gross ton, and by proportion, the statement is: 2240 : 259166 :: 36.16 Multiplying the second and third terms, and dividing by the first will give the solution, which, by the usual method, is quite tedious. Now, by assuming as the price per ton, some measure, or hee 110 PRACTICAL PROBLEMS. } aliquot part, of 2240, as, for instance, 82, the foregoing problem can be abridged, thus: 70 i "99-40: 2591616 3: 3 3702/3871 — price @ $32 00 462'796 — “© & 4.00 11/569-—= * as 10 5/785 a Sees O05 INS ae 01 $4183/688 — “ “ $36 16 Tf the price per gross ton were $32, the statement would be: 2240 : 259166 :: 32, and by cancellation, this becomes — 70 : 259166 :: 1; 82 being contained in itself once, and 70 times in 2240. Monae by 1, now, and dividing by 70, gives the price of the given number of pounds at $32 per ton. To divide by 70, we cut off the cipher, and also one figure, 6, from the right of the dividend (this divides by 10), then dividing by'7, we get $3702.37, the price at $32 per ton. Next, we take aliquot parts of this for $4.16, the difference between the real and the assumed price, thus: $4 is 4 of $32, therefore, the price at $4 will be the eighth part of that at $32, or, $462.79+ (the remain- ing decimals being mills, etc.). Then, 10 cts. is 7, of $4, 5 cts. is 4 of 10, and 1c. is4of 5. The sum of the several pare quo- tients, or, $4183.68, is the price at $36.16. Exam. 2. What is the cost of a cargo of iron weighing 276388 pounds at $23 per gross ton ? 2'7638|8 . 8948/40 = price @ $32 ~O8710 PCOS 66 $8 128)38875 = “ “ J POS8791. = «. * B98 PRACTICAL PROBLEMS, 111 Here, we simply divide the given number of pounds by 70, as in the preceding example, and we have the price at $32 per ton. The difference between this and the real price, $23, 1s $9, for which we take aliquot parts, thus: $8 is + of 32, and $1 1s 4 of $8. Adding the price at $8 and $1, gives the price at $9, which, being taken from the price at $32, gives that at $23, or $2837.91. Norse. — The subtraction of the two-numbers is performed at a single operation, as pointed out in short methods for Multiplication. (See note to example 14, page 78.) Exam. 3. What is the cost of 24840 pounds of coal at $3.50 per gross ton ? The statement of this problem by Proportion would be: 9240 : 24840 :: 3.50 | and the product of the second and third terms, divided by the first, would be the answer. We will here assume $28 as the ie per gross ton, and we have: 8.0 ii 9340 : 2484/0 :: 28 810150 — price @ $28 pes UR aC 1 ' By cancellation, 2240 and 28 become 80 and 1; 28 being con- tained in itself once and 80 times in 2240. Dividing then by 80, as in the two preceding examples, we get $310.50, the price at $28 per ton. Now, $3.50 is 4 of 28; therefore, the eighth part of the price at $28 is the price at $3.50. Dividing $310.50 by 8, we get $38.81, the required cost. Nore. —It may be well to remark, in connection with this method of ab- breviation, that, although there are other aliquot parts, or measures, of 2240, the nnmbers 32 and 28 will be found the most convenient for practical PB yy PRACTICAL PROBLEMS. purposes. The price per gross ton will always decide which of the two the more desirable to be taken as the assumed price. FREIGHT. When the price per gross ton is less than a dollar, we would proceed as in the following: Exam. 4. What is the freight on 23800 pounds of mer- chandise at 96 cents per gross ton ? 23/8010 8/40 = price @ 32 ©. $10/200— * + * 96e., If the price per ton were $96, instead of 96 cents, in this example, we would cut off the right hand cipher, and divide by 7, as in the other examples, getting the price at $32. But cents are the hundredths of dollars, so we cut off two more figures from the right, and then proceed as in the other examples. In all such cases, then, three figures are cut off from the right, and the process will be as pointed out in the foregoing examples. $3.40 is the freight at 32 cents, and 3 times this, or $10.20, is the freight at 96 cents. Oats. What is the cost of 24160 pounds of oats at 58 cents per bushel, the bushel consisting of 32 pounds? This problem, by Proportion, would be: é 32 : 24160 :: 58 and the product of the second and third terms, divided by the first, would be the answer. PRACTICAL PROBLEMS. * Lie Assuming that the price per bushel is 382 cents, instead of 58, the statement would be: 32 : 24160 :: 82; and this by cancellation would be: 1; 24160 :: 1. The price of the given number of pounds, then, at 32 cents per bushel is 24160 cents, or $241.60; that is, 1 cent per pound, and the process thus: 241/60 — price @ ae 433/20 = “ 420 90 66 66 15}10 437190 — “ & | lewl® At 1 cent per pound, or 82 cents per bushel, the price is $241.60. Doubling this gives the price at 64 cents per bushel, which is 6 cents per bushel too much. Then 4 cents is 4 of 82; dividing $241.60 by 8 gives the price at 4 cents. Next, 2 cents is } of 4; taking 4 of $30.20, the price at 4 cents, gives $15.10, the price at 2 cents, and deducting both from $483.20, the price at 64 cents, gives $437.90, the price at 58 cents, or the answer. Or thus: 241/60 32 120/80 16 60/40 8 15|10 2 437/90 58 The process here is self-evident and requires no analysis. Note. — The four last problems will suggest the method of abbreviating commercial calculations by the use of aliquot parts. Wheat, buckwheat, barley, etc., may be treated in like manner, PERCENTAGE, Per cent. from the Latin per centum, signifies by the 100. In business transactions, it means a certain part of every 100. Thus, 2 per cent. means 2 of every 100, and may signify 2 cents of every 100 cents, 2 dollars of every 100 dollars, 2 yards of every 100 yards, ete. The character, %, is used in business transactions to represent the words per cent.; thus 2% means 2 per cent. In Percentage, five quantities are concerned, namely : The Base, Rate per cent., Percentage, Amount, and Difference. The Base is the number on which the percentage is computed. The Rate per cent. is the part of 100 taken. The Percentage is the fourth proportional to 100, the rate per cent., and the base, taken in the order mentioned, The Amount is the base plus the percentage. The Difference is the base less the percentage. Case I. Given, the base and rate, to find the percentage. Exam. 1. What is 4% of $350. 350 x 4 $1400 PERCENTAGE. 115 To solve this, we multiply the base, $350, by 4, the rate per cent., and divide the product, 1400, by 100; simply cutting off two figures from the right; $14 is the percentage. Reason: The foregoing is simply a question in proportion, ex- pressed thus: If $4 be allowed on $100, how much ought to be allowed on $350, at the same rate? And the analogy is: as $100, base, is to $350, base, so is $4, the percentage on $100, to the corresponding percentage on $350. Thus: 100 : 850 3: 4 er, as $100 is to its percentage, 4, so is $350 to its percentage. Thus: 100 : 4:3: 3850 In either case, the product of the second and third terms divided by the first gives the fourth proportional. Hence, Rue. To jind the percentage: Multiply the base by the rate per cent. and divide the product by 100. Exam. 2. A.’s salary is $2500 a year; if he spend 10% for board, 6% for clothing, 5% for books, and 9% for other purposes, what are his yearly expenses ? -Nore.— When several rates refer to the same base, they may be added or subtracted, according to the nature of the question. Thus: 10% + 6% + 5% +- 9% = 80%; then 30% of $2500 equals $750, his yearly expenses. Case II. Given, the percentage and base, to find the rate. Exam. 1. What per cent of $350 is $14? The question, fully expressed, is this: If $14 be allowed on $350, how much ought to be allowed on $100 at the same rate ? 116 PERCENTAGE. And the analogy is: As $350, base, is to $100, base, so is $14, the percentage on $350, to the corresponding percentage on $100. Or, as the given base is to its percentage, so is $100, considered as a base, to its percentage. Thus: 350 : 14 :: 100 Multiplying the second and third terms, now, and dividing by the first, we get the rate on $100, or the rate per cent. Hence the Rue. Multiply the percentage by 100 and divide the product by the given base ; the quotient is the rate. Thus: 1400 + 350 — 4% In the example, the percentage is 14 and the base 350. Annex- ing two ciphers to 14; that is, multiplying i¥ by 100, the third term of the analogy, we have 1400; and dividing this by 350 gives 4, the rate per cent. Exam. 2. A merchant, failing, owes $14300, and his assets are only $10725; how much on the dollar can he pay to his creditors ¢ _ 143/00 107 25] vs 1001 T5|UT5 [ts In this, $10725 is the percentage, and $14300 the base. An- nexing two ciphers (dots) to the percentage, we have 1072500 to be divided by 14300. Cutting off the ciphers from both numbers and multiplying the remaining parts each by 7, we have 75075 to be divided by 1001. The quotient is 75%; that is, 75 cents on the 100 cents, or the dollar. PERCENTAGE, 117 Case III. Given, the percentage and the rate, to find the base. Exam. 1. $14 is 4% of what number ¢ The question, fully expressed, is this: If $4 be allowed on $100, on what sum ought $14 to be allowed? And the analogy is: As $4, the percentage on $100, is to its base, 100, so is $14, the given percentage, to its corresponding base. Thus: 4:100:: 14 Hence the Rue. Multiply the percentage by 100 and divide the product by the given rate ; the quotient is the base. Thus: 1400 + 4 — $350, the base. Casse LY. Given, the amount and rate, to find the base. Exam. 1. What number increased by 4% of itself is equal to 364 4 ANALYsIS: Since the amount is the base plus the percentage, it is evident that 104 is the amount of 100 (considered as a base). The question now is: If 104 be the amount of 100, of what num- ber is 364 the amount ? ~ And the analogy is: As 104, the amount of 100, is to its base, 100, so is 864, the given amount, to its corresponding base. Thus: 104: 100 :: 364 Multiplying the second and third terms now, and dividing the product by the first, will give the base. Hence the 118 PERCENTAGE. RuiE. Multiply the given amount by 100 and divide the product by 100 plus the rate; the quotient is the base. ‘Thus: 364|.. + 104 14/56 349/44 60 350\ 04 04 Annexing two ciphers (dots) to 364, we have 36400 to be divided by 104. The quotient is 350, the required base (Rule II, Division). Exam. 2. A merchant increased his stock in trade ‘by - 127% of itself, and then had $3800; how much had he at first ? , BSHO. 2-54 ULZ 3420|000 O00 + 1008 27/360 $3392/640 224 864 In this, $3800 is the amount, and 12 the rate. Annexing two ciphers (dots) to 3800, we have 380000 to be divided by 112. Multiplying both by 9, the divisor becomes 1008. The answer is $3392.86. Case VY. , Given, the difference and the rate, to find the base. Exam. 1. What number diminished by 4% of itself, is equal to 336 ? PERCENTAGE, 1192 ANALysIs: Since the difference is’the base less the percentage, it is evident that the difference of 100 (considered as a base) is 96. The question, now, is: If 100 be the base and 96 the difference, what ought to be the proportional base for the difference, 336? And the analogy is: As the difference, 96, is to its corre- sponding base, 100, so is the given’ difference, 336, to its corre- sponding base. Thus: 96 : 100 :: 336 Multiplying the second and third terms now, and dividing the product by the first, will give the base. Hence the Rue. Multiply the given difference by 100 and divide the product by 100 less the rate; the quotient is the base. Thus: 336]... + 96 13\44 52 349/96 Annexing two ciphers (dots) to 336, we have 33600, to be di- vided by 96. The quotient is 3493%, or rather 350, the remainder being 1, APPLICATIONS OF PERCENTAGE. The five rules of Percentage now established can be readily applied to problems in Commission, Stocks, Profit and Loss, Taxes, Insurance, etc. A few examples will suflice. 1. A merchant in Albany remits to his agent in Chicago $850.75 for the purchase of grain. The remit- tance includes commission at 24%; how much will the agent expend for grain, and what will be his commission 4 In this, $850.75 is the amount, and 24 the rate. Now, what is the base ? erry: os aoe 120 PERCENTAGE. To this, Case IV is applicable. Thus: 850|75 + 1025 21/25 29/50 525 025 Anatysis: In this, we have to multiply first by 100, and then divide by 100 (in 1024), consequently, $850.75 undergoes no change and we simply draw the vertical line through the decimal point and make the proper correction for 24. To multiply 850 by 24, we conceive a cipher annexed and divide by 4 (4° — 24), or, 75 of 850 as it stands, gives 21.25, which is placed in proper position and subtracted. Then, 7/5 of 21, or .525, is placed in position and added, and finally, the 71, of 1 (carried in adding |50 and 525), or |025, is subtracted. The result is $830, the base of com- mission, which is the sum to be expended for grain. Then, 24% of $830 — $20.75, or $850.75 — $830 — $20.75, the com- mission. . Note.— Commission is charged only on what is expended or collected by a person acting in the capacity of agent. 2. An agent sold real estate on commission at 3%, and returned to the owner, as net proceeds, $2425; what was the price received for the property, and what was the commission ? In this, the net proceeds, $2425, is the difference and 8 the rate. Now, what is the base ? Here, Case V is applicable.* Thus: 2425|.. + OF 72|75 2)16 6 $2499197 PERCENTAGE. 121 Annexing two ciphers (dots) to 2425, we have 242500, to be divided by 97. The quotient, by simplified division, is $2500 (the remainder, .97, being 1), the required base, or what the prop- erty sold for; whence, by subtraction, we obtain the commission, $75. Or, 38% of 2500 = $75. Note.— From a due consideration of the foregoing article, the student cannot fail to appreciate the great advantage of a thorough knowledge of the principles of proportion (given under the head of ‘‘The Rule of Three” in this work) én all cases where percentage is concerned ; for, if a set rule should be forgotten, as frequently happens, a knowledge of proportion will enable us to recall it without difficulty. If, for instance, we should forget the set rule for the last problem, we would reason thus: If, from property which sold for $100, I receive $97 as the net proceeds, from what sum ought I to receive $2425 as net proceeds? And the analogy is: 97: 100:: 2425, which gives the rule at once. EXAMPLES FOR PRACTICE. 1. What is 123% of $5600? Ans. $700. 2. What per cent. of $720 is $21.60? Ans. De 8. 18 is 25% of what number ? Ans. 12. 4, What number increased by 15% of itself is equal to 644 ? Ans. 560. 5. What number diminished by 10% of itself is equal to 504% Ans. 560. Notr.— By solving those five examples by Proportion, the reason of the set rules will be impressed upon the memory. INTEREST, In computations in Interest there are five quantities to be considered, namely: the Principal, the Interest, the Amount, the fate and the Zime. Principal has reference to money (or its equivalent) lent by one person to another, on condition that the bor- rower pays a certain sum to the lender for the use of the money. [Interest is the sum paid for the use of the principal, — and is calculated on the basis of $100 as a standard prin- cipal, and one year as the time. Amount is the principal and interest together. . Rate, or Rate per cent. per annum, is the sum allowed for the use of $100, for a year, per cent. meaning by the 100, and per annum, by the year. The Time is that agreed upon by the parties to the transaction. The most important problem in computations in interest is that in which the principal, the time and the rate are given, to find the interest or amount. Exam. 1. What is the interest of $376 for 2 months at 6 per cent 2 | INTEREST. 123 ANALYsIs: This, in substance, is a question in compound pro- portion, and when fully expressed reads thus: If $6 be paid for the use of $100 for 12 months, how much ought to be paid for the use of $376 for 2 months, at the same rate ? The statement, by proportion, would then be: THOS 316-226 Te 2 A moment’s glance at the terms of this proportion shows that the process can be abridged by cancellation; thus, dividing 12 and 6 each by 6 they become 2 and 1; next, dividing 2 thus found, and 2, the second term of the second ratio, each by 2, they dis- appear, and the proportion then is simply: TOs 30 10s Multiplying the second and third terms, in this proportion, and dividing the product by the first, we get $3.76, the fourth pro- portional, or the interest for 2 months. From these facts we see that, when the rate per cent is 6, and the time 2 months, or 60 days (30 days to a month), the interest of any principal will be just as many cents as there are dollars in the gwen principal. Taking this for our basis, it is evident that the interest for any number of months, at 6%, will be the interest for two months multiplied by half the given number of months, thus: The interest of $376 for 10 months, at 6%, is $3.76 multiplied by 5 (half of 10 months), or $18.80; in other words, the interest for 2 months multiplied by 5 will be the interest for 10 months. The interest of $376 for 3 years and 4 months, at 67, is $3.76 multiplied by 20 (half of 40 months, the number in 3 years and 4 months), or $75.20. 124 INTEREST. And the interest of $376 for 3 years, 4 months.and 15 days, at 6%, is the interest for 3 years and 4 months plus + of $3.76 (the in- terest for 60 days), 15 days being } of 60, thus: _$3/76 — the int. of $3876 for 2 mos., or 60 days, at 6%. 75|20 — - 3 years, 4 mos., or 40 mos, yy ee ““ «“ 15 days. $76|14 — e : 3 years, 4 mos., 15 days. From this the interest at any other rate per cent. can be readily obtained by the method of aliquot parts; thus, if the rate were 7%, add 4 of the interest at 67, and if at 5%, deduct the 4; if at 8%, add 4 (2 being + of 6), and if at 4%, deduct 4; if 9%, add 4, and if 3%, take the 4; 74%, add 4 (14 being + of 6), and if 442, deduct the 4, ete. Nore. —It need scarcely be remarked that, in computing interest, the business method is to reject the cents of the principal when less than 50, and when 50 or more, add $1. Exam. 2. What is the interest of $239.97 for 1 year, 5 months and 17 days @ 6%? $2|40 — the interest for 2 mos., or 60 days. 20/40 = = “1 year, 5 mos., or, 17 moa: 60 = “ 6 15 days. 08 == % e 2 days.. $21/08 — . 1 year, 5 mos., 17 days. Here, we call the principal $240; the interest on this for 2 months, or 60 days, is $2.40. Multiplying this by 8} (half of 17 months, or 1 year and 5 months) gives $20.40. For 17 days we take aliquot parts of 60, thus: 15 days = 4 of 60; the $ of $2.40 is 60 cents; then 2 days — 4), of 60; the {; of $2.40 is 8 cents, INTEREST. 125 or .08; and by addition we get $21.08, the interest for 1 year, 5 months and 17 days. Nott. — Should it be desirable to retain the cents of the principal, it will not, of course, affect the process, the figures to the right of the line being simply decimals, or cents and mills. From the foregoing principles and examples we derive the following Easy Meruop or Computine InTEREst oN ANY SuM, FOR ANY Timz, AT ANY RATE PER CENT., oN A Basis oF 360 Days To THE YEAR. GENERAL RuLeE. (1.) Draw a vertical line through the given prin- cipal, two places to the left of units; the result is the interest for 2 months, or 60 days, at 6%: (2.) multiply this by half the months in the given time (reducing the years, if any, to months): (8.) add, for days, such parts of 60 days’ interest, as the days are aliquot parts of 60; the result will be the interest for the given time at 64%, the dollars being to the left, and the cents and mills to the right of the vertical line. Then, for any rate other than 6%, add or subtract the proper proportions, as pointed out in the foregoing analysis. Nore.— If the number of days be not an aliquot part of 60, say 25 days, for instance, then say 20 days — ¢ of 60, and5— ¥f of 20; the two results, when added, make 25. If 27 days, then 30 = 1g of 60, and 3 = qs of 30; the difference is 27 (30 — 3), ete. The foregoing is also the rule generally used for Bank or Busts ness Discount, as it is called. Exam, 3. What is the bank discount on a note drawn at 3 months for $480.23 at 64 ? 126 INTEREST. 4|S0 = the interest for 2 mos,, or 60 days. 2)40 == et eT m0., OF 30 ave. : 2 FS bs 8 days (grace). $7|44 = . i 93 days Deducting the interest (discount), $7.44, from $480.23, gives the present worth or the net proceeds. Exam. 4. What is the interest of $468 for 1388 days at 6% % 4168 9136 13/8 + 3]0 12d — } of 60 93 “Als 6d = 4 of 12 468 $ 10/76 Dividing 138 days by 80, as shown in the margin, we get 4 months and 18 days, the process is then according to the rule. Now, since the interest of any sum of money for 60 days, or 2 months, at 6%, is as many cents as there are dollars; that is 1%. of the principal, the interest for 6 days is a tenth of that for 60 days, or 2 months, and for 1 day, a sixth of that for 6 days; for 600 days, or 20 months, the interest is ten times that for 60 days, or 2 months and for 6000 days, or 200 months it is ten times that for 600 days, or 20 months; in other words, any sum of money will double itself, or the interest will equal the given principal, in 6000 days, 200 months, or 162 years, at 6%, simple interest, thus: The interest of $5480 for 6000 days or 200 months = $5480. ck ‘* $5480 for 600 days or 20 months = $548.0 ee « $5480 for 60 days or 2 months = $54.80 es é«° $9480¢ for Go days! s.06 ts eee = $5.480 a ‘+ 5480 for 4 days. ieee eee et Oe And from this the interest for any time can be readily found, as illustrated in the following: : 4 q ; INTEREST. 127 Exam. What is the interest of $5498.70, for 3 months and 10 days, at 6 per cent.? In this, counting 30 days to the month, the time is 100 days, and we simply cut off one figure from the dollars to get the interest for 600 days; then a sixth of this, or $91.645, is ’ the interest for 100 days, or 3 mo., 10 da. $049|8 . 70—600da. $91} 645 —100 ** Exam. What is the interest on a note for $7560, dated Jan. 31, 1906, and payable May 5th, following, at 6 per cent.? Counting Feb, 28 days, Mar. 31, April 30 and May 5, the time is 94 days; and the interest for 100 days is $126 from which $7.56, the interest for 6 days is deducted; this leaves $118.44, the interest for 94 days. Or, The interest for 60 days=$75.60; for 30 days $37.80; for 3 da. a tenth of this, or $3.78 and for i da. a third of this, or $1.26, making, when added, the required interest. $756/0 —600da. 126) =-100 © Too Ons $11s|44— 94 °° $75|50=60 da. 37/80=30 << SSS oo 1 26= 1 cé $118 44—94 da Exam. What is the interest of $5840, for one year, 8 months and 15 days, at 7 per cent.? In this, 1 yr., 8 mo. = 20 mo. and at 6%, the interest is $584. Then, the interest for 15 da. at 6% is one fourth of 60 days’ interest, or of $58.40; this gives $14.60 which is added, making $598.60. for the given time at 6% To this we add the interest at 1%,or one-sixth of that at 67, $99.766, which gives $698.366, the interest at 77. $584|0 =20 mo. 14/60 =15 da. $598|/60 —67_ y9|766=17 $6981366 =77 Nots. —If the rate were 5%, $99.766, or one-sixth of 6% would be deducted; if 4%, a third of 6% or 2% would be deducted; if 744% a quarter of that at 6% would be added 144 being a quarter of 6, etc. (See page 124.) 128 INTEREST. RuLE. To compute interest at any rate per cent. for any number of days, on the basis of 360 days: Multiply the principal by double the rate; multiply the result by the number of days; cut off fiwe jigures from the right, counting from the decimal point always, and add a third and a siath of that third. Exam. What is the interest of $50000 for 14 days, at 34 per cent.? In this, multiplying by 7 (double 33) and then by 14, is the same as multiplying by 98 (7 x 14) at once. $50000 x 98 Multiplying 98 by 5, and annexing the ciphers, we 49100000 get $4900000. Cutting off five places, and adding s pee a third and a sixth of that third, we get the required . $68|05500 interest, to five places of decimals, And to compute interest at any rate, for any number of days, on the basis of 365 days, in other words, exact interest : Ruue. Multiply the principal by double the rate; multiply the result by the number of days; cut off five figures from the right, counting from the decimal point always, and add a third, a tenth of that third and a tenth of that tenth. Exam. What is the interest of $256800 for 1 day at 2 per cent. (865 da.) ? Here, the time being only 1 day, we simply multi- $256800 x 4 ply by 4 (double the rate); then, cutting off five ee pe places, and adding a third, a tenth of that third and aie 1384240 a tenth of that tenth, we get the interest, correct to 3494 five places of decimals. $14/07264 Norsk. — This rule will be found useful in finding the interest on daily balances. (See page 220.) Both the foregoing rules will be found more fully explained in the Appendix (pp. 177 and 178) together with the reason of the rules, PaRTIAL PAYMENTS. . 129 PARTIAL PAYMENTS OR INDORSEMENTS. The United States Courts have decided that, I. “The rule for casting interest when partial pay- ments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. Il. “If the payment exceeds the interest the surplus goes toward discharging the principal, and the subsequent interest is to be computed on the balance of the principal remaining due. III. “If the payment be less than the interest the sur plus of interest must not be taken to augment the princi- pal, but the interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied toward discharging the principal, and the interest is to be computed on the balance as aforesaid.” — Decision of Chancellor Kent. EXAM. $1000. Arnany, N. Y., May 1, 1885. Two years after date I promise to pay to A. B., or order, one thousand dollars, with interest, value received. Cru: On this note were indorsed the following payments: Jan. 1, 1886, received $150. Sept. 1, 1887, received 5. Jan. 1, 1888, received 200. 130 PARTIAL PAYMENTS. What remained due May 1, 1888? Process : | Principal on interest May 1, 1885........ ..... ee $1000 Interest from May 1, 1885, to Jan. 1, 1886 (8 mos.) @6 2, 40 Amount ee $1040 First. payment, Jan. 1) 18860-54000 2 aes vin ghee oes 150 New principal...... $890 Interest from first payment to Sept. 1, 1887 (20 mos.).. 89 Second payment, Sept. 1, 1887 (less than interest). $75 Interest on $890 from Sept. 1, 1887, to Jan 1, 1888 (4 mos.) 17 80 Amount ...... staea- SOGGH ER Third payment.January 1, 1888°.5..4 4-2 $200 Sum of second and third payments ........ Spi oS 275 New principal...... $721 80 Interest from Jan. 1, 1888, to May 1, 1888 (4 mos.)... 14 44 Balance due May 1; 1888 . 2. °.. 5520 Ses eee $736 24 Hence the Rue I. Compute the interest on the given principal to the time of the first payment, and, if less than the payment, add it to the prin- cipal and subtract the payment from the amount, the difference will be the new principal. Il. But if the payment be less than the interest, let the account stand (noting the payment on the document) tal the next payment, when, if the sum of the payments shall equal or exceed the interest then due, add the interest to the new principal and subtract the sum of the payments from the amount; the difference will be the new principal, with which proceed as before. Nots,— This rule applies to bonds, mortgages and other obligations bear- ing interest. BANK DISCOUNT. Bank Discount is an allowance made to a bank for the payment of a note before it becomes due. The money received for the note when discounted is the Proceeds, and is equal to the face of the note less the discount. The Face of a note is the sum made payable by the note. Three days, called Days of Grace, are allowed on a note, after the time it is nominally due, before it is legally due. Thus, a note drawn on April 8, at 2 months, would not be legally due till June 11. ~The person by whom a note is signed is the Maker or Drawer of the note. The person in whose favor, or to whose order the note is payable, is called the Payee; and the folder is the owner of the note. The Maturity of a note is the expiration of the days of grace. To J/ndorse a note is to have the payee or holder write his name across the back of it. 182 BANK DISCOUNT. Note. — An indorsement makes the indorser liable for the payment of a note ifthe maker fails to pay it when due. The following is the usual custom of borrowing money at banks: The borrower presents a note, either made or indorsed by himself, payable at a future specified time. Interest is calculated on the face of the note for the time it has to run from the date of discount , this interest is deducted from the face and withheld by the bank in consideration of advancing the money before the note matures. Hence the Rue. To find the discount and the proceeds of a note: I, Find the interest on the face ofthe note for three days more than the time specified ; this interest is the discount. II. Subtract the discount from the face of the note; the differ- ence is the proceeds, Notr. — When a note is given in settlement of an account, the business method isto add the interest to the debt and draw the note for the full amount. But sometimes notes are drawn promising to pay ‘‘ with interest’ (not in- cluding the interest with the debt). In such cases the amount, that is, the debt and interest together, is the face of the note, or sum made payable, and must be made the basis of discount. EXAM. — $1260.15 AuBany, N. Y., Jan. 17, 1889. Ninety days after date I promise to pay to the order of Case & King, one thousand two hundred and sixty and ;4%, dollars, at the Exchange Bank, for value received. JOHN JONES. Suppose Case & King discounted the foregoing note on Febru- ary 4, what would be the discount, and what the proceeds ? Anauysis: In the first place we find the day of maturity, thus: Counting 90 days after January 17, we find the nominal day to be April 17, thus: subtracting 17 from 31 leaves 14 days for January, then February 28 days, March 31 days, and 17 days of April make BANK DISCOUNT. 133 90. To this we add 3 days grace, and the day of maturity is April 20. Interest is computed on the note, now, from the date of discount, February 4 to April 20, at 6%. The number of days from February 4 to April 20 is 75, and the interest of $1260.15 for 75 days at 6% is found thus: 12|60 — interest for 60 days. 3|15 = A t Daas $1575 Tg’ The discount, then, is $15.75, and the proceeds $1260.15 minus $15.75, or $1244.40. Norse. —For business method of computing interest, see note, page 124; also general rule for interest, page 125. Although the foregoing rule is that which is employed in actual practice, it is founded on a principle radically false, and always gives the discount too large, and conse- quently the present egret too small, °Y the interest of the true discount. The true present worth of any debt is such a sum as would, if lent at interest at the assigned rate, amount to that debt at the time at which it would have been due. Hence the Rute. To find the true present worth of a debt: As $100 plus its interest for the given time, and at the given rate per cent., is to $100, so is the debt to its present worth. Thus, the true present worth of the note in the last example is found as follows: The interest of $100 for 75 days at 6% is $1.25; adding this to $100, and proceeding according to the rule, we have the following analogy: $101.25 : $100 :: $1260.15 : $1244.59, the present worth. 134 BANK DISCOUNT. Subtracting $1244.59 from $1260.15, the face of the note, gives $15.56, the true discount, or 19 cents less than was found by the bank method. And the interest of $15.56, the true discount, for 75 days at 6%, is found to be 19 cents. The reason of this rule: $101.25 is the amount of $100 for the given time and rate, and $1260.15 is the wmount of $1244.59 for the same time and rate; that is, the amount of $100 is to the prin- cipal which produced that amount as any other given amount is to the principal corresponding to the said amount. It is evident, also, that we might use the amount of $1, and $1 itself, or the amount of any sum whatever, and that sum itself, for the two first terms of the analogy; but it is generally more easy to use $100 and its amount. : Rute. To jind the face of a note, the proceeds being given: As the proceeds of $100 is to $100, so is the proceeds of the note to its face. Exam. 1. For what sum must a’note be drawn at 4 months to net $750 when discounted at 6% ? Norz.—The days of grace having been abolished in the State of New York, are not taken into account in the following examples. ANALYSIS: The interest of $100 for 4 mo. at 6% = $2. Deducting this from $100 gives $98, the proceeds of $100. Then, As $98: $100: :$750: x ="“X"™ _ 9765.30 the sum for which the note must be are : that is, the proceeds of a note whose face is $100, ts to that face, as the proceeds of any other note ts to its face.. The division by 98 is performed by the 750/00 — simplified process : Dividing first by 100, 15/00 2 times 750 = 1500, and 2 times 15 = 30; are set 30 in proper positition and added; the sum is $765.30 —- $765 30 (For a more simple method to find the face of a note, see p. 214.) Banx Discount. 135 RULE.—To find the interest corresponding to a given rate of bank discount : I. Assume $100 as the amount or face of the note, and find the discount and the proceeds of that amount for the time the note has torun. Then, II. Apply the following proportion : As the given time : 1 year : the principal : $100 the interest ; the rate. Exam. What rate of interest is paid, when a note payable in 60 days is discounted at 2% a month ? ANALYSIS: 2% per mo. = 24% per year. The discount of $100 for 60 da., or 2 mo., at o4y = $4. Deducting this from $100 gives 896, the ihenede! Here, then, having assumed $100 as the amount; $96 is the prin- cipal, $4 the interest and 2 mo. the given time, and we have the following proportion : As 2 mo. : aC aeeat : $96 : $100 ; Reducing this to a simple proportion, we have: 192 :1200:: 4 Multiplying the second and third terms, now, and dividing by the first, we have: “1” 4 = 4800 + 192 = 25¢, the rate. Or, having found! the discount and the proceeds of $100; draw a vertical line and set the given time and the proceeds on the left; : $4 : the rate. and on the right sct 1 year, $100 and 2 mo.|12 mo the discount. Divide the product of: $96) $100 the numbers on the right by the product $4 of those on the left, and the quotient is 192/4800 (25% the rate. NotTe.—The reader need scarcely be told that the numbers on both sides may be cancelled, if desirable; in this case they can be cut down to 100 + 4 = 25%. 136 Banx Discount. ID Pag Mal ‘ rt. a AN 4 | b, ay RvuLE.—To find the rate of bank discount corresponding to a given rate of interest: Assume $100 as the proceeds of the note and find the interest and the amount of that for the time the note has torun. Then, . As the pe ly at : : the interest : the rate. the principal : $100 Exam. A broker buys 30 day notes at such a discount that his money earns him 24% a month; what is the rate of discount ? ANALYSIS: 24% per mo. = 30% per year, The interest of $100 for 30 da. or 1 mo. at 30% = $2.50. Adding this to $100 gives $102.50, the amount, Here, then, $102.50 is the principal, $2.50 the interest and 1 mo. the given time, and we have the following proportion: As 1 mo, : 12 mo. os $2 50 : the rate. $102.50 : $100 and this now becomes: $102.50 : 1200 : ; $2.50 == 700*250_soq000 + 10250 = 29147, the rate. Or, having found the interest and the amount of $100; draw a _ vertical line and set the given time and the amount on the left; next, set 1 year, $100 and the 1 mo.|12 mo. interest, on the right. Divide the $102.50) $100 product of the numbers on the right $2.50 by the product of those on the left; 102501800000 (29447 the quotient is the rate. Notge.—It may be well to remark that the proportion given in connection with the two foregoing rules, is applicable to those problems in Interest where the principal, the interest and the time are given to find the rate. AVERAGING ACCOUNTS. The rule which determines the just time to pay, in one payment, several debts due at different times, is called Equation of Payments, or Average. Exam. 1. On January 1, 1888, Jno. Dwyer bought of Wm. Prior A bill of drygoods amounting to $300 @ 2 months, A bill of hats amounting to 400 @ 3 months, And sundries amounting to 500 @ 4 months, at what time may the whole be paid without loss to either party ? Nore. — The time which elapses before a payment is due is called the Term of Credit; and each item of a book account draws interest from the time it is due, which may be either at the date of purchase, or after a speci- fied time of credit. To find the average date of the foregoing account, we will as- sume that each item was due at the date of purchase; in other words, that they were cash transactions. On this assumption, it is evident that the purchaser would owe the merchant, at the end of each term of credit, not only the 138 AVERAGING ACCOUNTS. amount of each bill of goods, but also the interest on each amount for the time, thus: Amount of account $300, int. for 2 mos. @ 6% = $3.00 66 400, 66 3 66 66 pa 6.00 as es 500, (4) a = 10.00 $1200 ; $19.00 The transactions were not for cash, however, but on time. The purchaser, therefore, is entitled to the $19 interest; in other words, he is entitled to hold the $1200 for such time after January 1, us it would take that sum to give $19 interest at 6% per annum. The question, now, is, in what time will $1200 principal give $19 interest, if $100 principal give $6 in 12 months? The propor- tion, or analogy, for this problem, would be: $1200 : $100 :: 12 mos. OD and this, by cancellation, becomes: es gee BS eee | Then, ee the second and third terms, and dividing by the first, we have 19 + 6 = 34 months, or 3 months and 5 days. If, then, the purchaser paid the entire $1200 of account, 3 months and 5 days after January 1, there would be no injustice on either side. . Counting 3 months and 5 days from January 1, we find the average date to be April 6th. Exam. 2. On January 1, 1889, a man gave 3 notes, the first for $500 payable in 80 days; the second for $400 payable in 60 days; the third for $600 payable in 90 days. What is the average term of credit, and what the equated time of payment ? AVERAGING ACCOUNTS. 139 Process. Interest on $500 for 30 days @ 6%-— $2 50 “ AHO bao Ce es, 224-200 % GOO) 4*5.- 90 Be = 9 00 $1500 | G15 50 Here, the interest of the several payments for the respective terms of credit is $15.50, and the sum of the payments, $1500. Now, in what time will $1500 give $15.50 interest, if $100 give $6 in 12 months? The following is the analogy: $1500 : $100 :: t& mos. Moe Li DO: 2, And, by cancellation, 6 disappears and 12 becomes 2. Then we have $15.50 multiplied by 2, or $31, which is multiplied in turn by 100, giving $3100; dividing this last, now, by $1500, gives 2; months, or 62 days, the average term of credit; and the equated time of payment, March 4. Hence the following Ruue. Find the interest on each payment for its term of credit at 6 per cent., and add the results ; multiply double the interest thus found by 100, and divide the product by the sum of the payments; the quotient will be the average term of credit in months. Notse.— If the sum of the payments be not contained in the product as _ found above, multiply said product by 80, then divide, and the quotient will be the average term in days. The foregoing is called a Simple Equation, or Simple Average, having reference only to one side of the ac- count; and the terms of credit begin at the same date. When both debits and credits, or both sides of an account, are to be considered, and the terms of credit begin at different dates, the process is called a Compound Equation, or Compound Average; as illustrated in the following Ledger account: 3 ‘ 8 | v CO | ScF 89 | IT |; 00 | 008 89 | IL |} 00 | 008 ACR a hie eed 00 | 0 00 | S&T 0 3 96 ‘AON iho 00 | 08% Pes 2 ” CL 390 10 | F 00 | 08% ns “s {. 4ded Gg 13 00 | 008 sre Le og ydeg |! 0g | 9 00 | 0gE mo SOE 7 Oe o> f eg | 8 00 | 00¢ skep 901 | 93°AON 0} | ZI “SN || 09 | F 00 | OF skBp CIL | 93 °AON 0} | G “SNY & as) . 3 “ysoloqyUT]| “SJUNOUTY ‘OUILY, “qselojzuy!) “syunowWYy “OUL I, cal > ! y of 60 days, or 12 cents, making $9.72, the interest oe 243 days. Again, the interest of $240 for 316 days, for instance, would be 5 times 60, plus 16 hgh (31/6 + 6/0 =5..16). For 16 days we would say 10 = t of 60; and 6 = ao: the sum of the results is 16 days’ interest, etc. Bivorie ANID COss. Profit and Loss are commercial terms, having reference to the gain made, or the loss sustained, in the course of business. Gains and losses are usually estimated at some rate pet cent. on the money first expended or invested. Note.—It should be particularly remarked, that, by the gain or loss per cent, is to be understood the sum that would be gained or lost at the given prices, not on a hundred dollars’ worth sold, but on a@ hundred dollars laid out tu Jjirst cost, and in charges, if there be any. Ruel. The first cost and the selling price being given to find the gain or loss per cent.: As the first cost is to the gain or loss on that cost, so is $100 to the gain or loss per cent. Exam. 1. If tea be bought at 40 cents and sold at 50 cents per pound, what is the gain per cent. Ais 1) To) 4/0) L00|0 25% Here, 50 — 40 — 10 cents, the gain on 40 cents; then, as 40 cents (the first cost) : 10 cents (the gain on 40 cents) :: $100 (re- garded as first cost) : 25, the gain on $100. 144 PROFIT AND Loss. Rue II. To find how a commodity must be sold to gain or lose a certain rate per cent.: As $100 is to the gain or loss on $100, or per cent., so is the first cost to the gain or loss on that cost; and from this and the first cost, the selling price will be found by addition or subtraction. Exam. 2. How must tea which cost 60 cents per pound be sold to gain 20%? LOO:2° 20.52.60 ——— eee 12/00; then, 60 + 12 — 72 cents. Here, $100 (regarded as first cost): $20 (the gain on $100) .. 60 cents (the first cost) : 12 cents, the gain on 60 cents; then 60 plus 12 — 72 cents, the required selling price. (By cancellation, We have 60 + 5 = 12.) Or, say: As $100 is to $100 plus the gain, or minus the loss per cent., so is the cost to the selling price. 'Thus, taking the same example, we have: 1003 "1296.45260 72|00 Here, what cost $100 is sold for $120; and what cost 60 cents © will be sold, in proportion, for 72 cents. Rue III. Zo jind the first cost from the gain per cent., and the seluing price: As $100 plus the gain, or minus the loss per cent. is to $100, so is the selling price to the first cost. Exam. 8. Sold 12 musical instruments for $1500, and gained 25%, what was the first cost of each ¢ 125 :. 100 :: 1500 8 150000 1000 1200|000 + 12 — $100 PROFIT AND Loss. 145 Here, the analogy is simply this: As the selling price, $125, is to $100 (considered cost), so is $1500, the selling price, to the cost price ($1200). ‘ To divide by 125, the dividend and divisor are multiplied by 8, and the new divisor is 1000, by which we divide, getting $1200, the cost price of 12 instruments. Dividing this by 12 gives $100, the cost price of each. QUESTIONS WITH THEIR SOLUTIONS. 1. In closing the Ledger at the end of a year, the Dr. side of the Mdse. account is $38750, and the Cr. side $46500 ; what is the gain per cent. ? The Dr. side of Mdse. represents the purchases, or cost price, and the Cr. side, the sales, or selling price; the difference of the two will be the gain or loss on the ac- count, Sales. Purchase. 46500 — 38750 — 7750 gain. SOP Tous Se 00 ) TT5000 (20% 775000 Here, we find the gain on the account to be $7750; then as the cost price is to the gain on that price, so is $100 to the gain per cent. (202). 9. A merchant sold 24 musical instruments for $125 each; he gained 25% on half, and lost 25% on the re- mainder; did he gain or lose on the transaction, and how much ? 10 a ANALYsIS: 12 at $125 each — $1500; then $100 plus the gain, $25, or $125, the selling price, is to $100 (considered cost), as $1500, the selling price, to its corresponding cost; thus: 146 PROFIT AND Loss. 195-100-322 1500 8 150000 1000) 1200/000 Then, $1500, the selling price, minus $1200 cost = $300 gain | on the first half. Now, 100 — 98, or, 75 : 100 :: 1500 4 150000 3/00) 600000 $2000 cost price. Then, $2000 cost minus $1500 selling price = $500 loss on the remaining half. The loss on the transaction is.... $500 The gain f 6. eS eee ae Therefore he lost ...... | ore 2 8206 3. A music dealer paid $1500 for 12 musical instru ments which he wished to sell at a profit of 20%; what must he charge for each instrument? i 100 : 120 :: 1500 pga This is a simple question in Compound Proportion, in which — $100, considered as cost, is to $120, the selling price, as $1500 PROFIT AND LOSS. 14% cost is to its corresponding selling price; and next, as 12 articles is to 1 so is the price of 12 to the price of 1; the said proportion, by cancellation, becoming: 10s 12271500 and the fourth proportional $150, or the price which must be charged for each instrument. Nore. — From this we see that when the number of articles is 12, and the gain per cent. desired to be made is 20, we get the price of a single article, including 20 per cent., by simply dividing the price of 12 by 10. And to di- vide a number by 10 wesimply cut off one figure from the right of the divi- dend; in other words, move the decimal point one place to the left. Hence, When goods are bought or sold by the dozen we can readily tell what each article must be sold for so as to make 20 per cent. profit on the sale; and from this, by the method of aliquot parts (see page 109), we can get the price at any rate Pia cent., as illus- trated in the following: 4. If a dozen silk hats be bought for $54, what must each be sold for to make 402 profit ? 100 3140" 2354 ~ 45/60 + 12 $6|30 The process, by the regular method, would be: As $100, con- sidered cost, is to $140, the selling price, so is $54 cost, to its corresponding selling price, $75.60. Dividing this by 12, gives $6.30, the price at which each hat must be sold to make 402. Or thus: Dividing $54 by 10, gives $5. 40, the price of one hat, including 20% profit, or the same as if the selling price were $120 ($100 being considered cost). This lacks 20¢ of the desired profit. 148 PROFIT AND Loss. ) 20% is 4 of 120; now, by simply adding 4 of $5.40 to itself, we get the price at 40%, thus: : $5.40 — 120 90 = 20 $6.30 — 140 Nortz.— A little practice will enable a person to solve such questions men- tally by this process. 5. Suppose every thing as in the last question, only 10% was made by the sale, instead of 40; what was the sell- ing price of each hat? $5.40 — 120 45 — 10 $4.95 — 110 10% is 5 of 120; subtract, the selling price is $4.95. Questions of the following nature will be found useful, and, from the examples and illustrations already given, will be readily understood : 6. The population of Albany was 69422 in the year 1870, and 90905 in 1880; what was the rate per cent. of the increase during the interval ? ANALYSIS: By taking the difference we find 21483, the increase of population. Then, As 69422 ; 214838 :: 100 : 30.94+, required rate. 7. Between 1850 and 1870 the population of Albany increased by 36.76 per cent., and in the latter year it was 69422; what was it in 1850 ? 136.76 : 100 :: 69422 : 50762, nearly, the population in 1850. | DIVISION INTO PROPORTIONAL PARTS. Rute. To divide a given quantity into parts which shall be propor- tional to given numbers: As the sum of the given numbers is to any one of them, so is the entire quantity to be divided to the part corresponding to the number used as the second term of the proportion. Exam. 1. Proof spirits are composed of 48 parts of alcohol, or pure spirit, and 52 of water; how much of each is contained in 40 gallons of proof spirits? (48 + 52 — 100); then, 100 : 48 :: 40 : 19.20 alcohol. 100 : 52 :: 40 : 20.80 water. Or, having found 19.20 alcohol, deduct it from 40; the differ- ence is 20.80 watcr, as found by the analogy. Exam. 2. Suppose a train to start from Albany to New York, going at the rate of 20 miles an hour, and another at the same time from New York to Albany, going 30 miles an hour; where will they meet, the distance be- tween the two places being 145 miles ? 150 Division INTO PROPORTIONAL PARTS. (20 + 30 — 50); then 50: 20 :: 145 : 58 miles from Albany. Or, by taking : the train from NewYork: 50: 80 :: 145; 87 miles from New York. The operation is proved to be correct by adding the results together. Exam. 8. Divide $9500 among father, mother and son in such a manner that the father’s share may be one-half greater than the mother’s, and the mother’s one-half greater than the son’s. ANALYysIs: Here the parts are evidently 1, 14 and 24, or 4, $ and 9. Then . (4+6+9—19) 19:4 :: 9500 : 2000, son’s share. 19 : 6 :: 9500 : 8000, mother’s share. 19 : 9 :: 9500 : 4500, father’s share. Exam. 4. A quantity of flax seed being converted into oil, the result was found to be 329 pounds of oil and 649 pounds of cake; how much oil is that to the bushel, and what per cent.; a bushel of seed being 56 pounds, and a gallon of oil 7% pounds? (329 + 640 — 969); then 969 : 329 :: 56: 19.013, the number of pounds of oil to 56 pounds, or, to the bushel. Now, 19.0138 + 74 __ 2.535, or 24 gallons nearly. Next, 969 : 329 :: 100 : 33.952, or 34% nearly. Exam. 5. Pure water is composed of oxygen and DIVISION INTO PROPORTIONAL PARTS. 151 hydrogen, in such proportions that the weight of the former is to that of the latter as 15 to 2. Required the weight of each contained in a cubic foot, or 1000 ounces, avoirdupois weight of water. (15 + 2—17); then 17: 15 :: 1000: Bey oz. oxygen. 17: 2:: 1000: 11744 oz. hydrogen. 7000 01 aie: 15000 + aly 900/00 + 102 18 00_ 882), TOT Si = i To divide the product of the second and third terms of the first proportion, or 15000, by 17, we multiply both by 6 (seeing that 17 is nearly one-siath of 100). Then dividing 90000 by 102 ae II, Division), we get 882,38 ounces oxygen, the fraction qoz being equal to =5;. Now, 1000 — 88275 = 11744 ounces hydrogen. Or thus: 120|00 + 102 2/40 117/60 6 66 + 6 == 11, that is, 11, Multiplying 17 and 2000 (2 x 1000), each by 6, we have 12000 to be divided by 102. The quotient is 117, and the remainder 66, which is divided py 6 (by which we Ma aa and the true remainder is 11, that is, when fully expressed, 44; or, ,6 = i. a. oe _—- a PARTNERSHIP. CasE I. The gains and losses of partners in business may be ascertained as in the last rule by the following proportion : As the whole stock is to the whole gain or loss, so ts the stock of any partner to his gain or loss. Exam. 1. A.and B. forma copartnership; A. furnishes $5000 and B. $7000 as capital; they gain $960; what is each man’s share of the gain ? (5000 + 7000 — 12000); then, 12|000 : 5000 :: 960 ~ 4800/000 $400, A.’s gain. B.’s gain is found by the same analogy, using 7000 for the sec- ond term. Exam. 2. Two brothers, John and James, purchase a house jointly for $25000; John contributed $10000 and James $15000 of the purchase-money. They let the house for the yearly rent of $2000; what share of the rent is each to receive ? PARTNERSHIP. 153 25000 : 2000 :: 10000: 800, John’s share. 25000 : 2000 :: 15000 : 1200, James’ share. $2000 Case II. To find each partner’s share of the gain or loss when their capital is employed for wnequal periods of time. Rue. Multiply each stock by the time of its continuance in trade; then, using the products as stocks, proceed according to Case I. Exam. 1. A. and B. form a partnership. A. contrib- utes $3500 for 12 months, and B. $4500 for 9 months. They gain $1600; what is the share of each? 3500 x 12 — 42000 4500 x 9 — 40500 82500 Then, as $82500 : $1600 :: $42000 : $814.54, A.’s gain. 82500: 1600:: 40500: 785.46, B.’s gain. The reason of the process will be evident from the consideration, that $3500 for 12 months is equivalent to 12 times that for 1 month, that is, to $42000; and $4500 for 9 months is equivalent to 9 times that for 1 month, that is, to $40500. Hence, if these increased stocks be employed, it is evident that, since the times are then to be regarded as equal, the process will be the same as in Case I. Norte.— It need scarcely be remarked that the times, in all such operations, must be of the same denomination. If, for instance, one was 12 weeks, and the other 9 months in the foregoing example, the 9 months should be re- duced to weeks, or the 12 weeks to months. BANKRUPTCx The estate of a bankrupt may be divided among his creditors by the following analogy: As the sum of all the claims on the estate ts to the value of the whole estate, sv is the claem of any creditor to his dividend or share. Exam. 1. A bankrupt owes A. $350, B. $650 and C. $1500. His whole estate is worth only $1500; want is the share of each creditor ? ($350 + $650 + $1500 — $2500) Then, as $2500 : 1500-:: 350 :°210, A.’s share. 9500 : 1500:: 650: 390, B.’s share. 9500 : 1500 : - 1500 « 900, C.’s share. $1500 — the whole ontater Nots. —In the division of a bankrupt’s estate, it is usual first to find how much on the dollar he can pay ; that is, how much the creditors will receive for each dollar of their respective claims. Thus, resuming the same ex- ample, we have this analogy or proportion: 25/00 : 1500 :: $1, or 100 cents. 1j00° 1500|00 60400 The sum of all the claims is to the whole estate as $1, or 100 cents, to the proportional part, corresponding to a dollar, which is found to be 60 cents, or 60 per cent, Then 60% of $350 — $210, A.’s share, as before. And B.’s and C.’s can be found in like manner. USEFUL RULES. Rue I. To find the rate at which a given principal will gain a certain interest in a stated time: As the given principal : $100 Prk teed tine meet ae :: the interest : the rate. _ Exam. If $5000, invested for 1 year and 6 months, gain $525; what is the rate per cent.? The problem fully expressed is this: If $5000, in 18 mo. gain $525; what will $100 gain in 12 mo.; and for this we have the following proportion : As $5000 : $100 : : $525 : the rate. 18 mo. : 12 mo. Reducing this to a simple proportion, we have the following : 90000 : 1200 :: $525 : x, or the rate. Multiplying the second and third terms, now, and dividing by the first, we get the rate, thus: ca = 630000 + 90000 = 7%, the rate. Hence, if we draw a vertical line and set the given principal and the stated time on the left; and on the right, set the given interest, $100 and 5000/525 1 year (or 12 mo. as the case may be); then divide ee ee the product of the numbers on the right by the 90000|630000(7. product of those on the left; the quotient is the rate. Norge. —It may be well to remark that, in all cases where the terms will admit, the work can be cut down by cancellation, 156 Usrerut Rutes. Roz Il. To find what principal, in a stated time, will gain a cer- tain interest, at a given rate per cent, : As the given rate : $100 the stated time : 1 year t : : the interest : the principal, Exam. What sum of money must be invested to gain $525 in 1 year and 6 months, at 7 per cent. ? The problem fully expressed is this: If $100 gain $7 (7%) in 12 months; what sum will gain $525 in 18 months; and for this we have the following proportion : het : de : : $525 : the principal. And by reduction to a simple proportion, we have the following: As 126 : 1200 : : 525 : x, or the principal. Multiplying the second and third terms, and dividing by the first, we have: 10 <5 _ 630000 + 126 = $5000, the principal. 126 Or, drawing a vertical line, setting the given rate 155 and the stated time on the left; and the given in- 18/100 terest, $100 and 12 mo. on the right, and dividing the os 12 product of these on the right by the product of those 126|630000 on the left, we get the principal. $0000 Rue Ill. To find the time in which, at a given rate per cent. per annum, a given principal will produce a certain interest : As the principal : $100 the rate : the interest :: L year : to the time. Exam. How long will it take to have $5000 gain $525, at 7 per cent. per annum, simple interest ? The problem fully expressed is this: If $100, in 1 year, gain $7 (7%) how long will it take $5000 to gain $525; and for this we have the following proportion : Usrerut Rv ues. 157 As ie ie :: l year : the time required. Reducing this to a simple proportion, we have the following: As 35000 : 52500 :: 1 : x, or the time. And we have, now, simply, 52509 to be divided by 35000. Cut- ting off the three ciphers in 35000 and three places from 52500 (this divides each by 1000, and does not alter the proportion) we have 52.5 to be divided by 35, as shown in the margin. To divide by 35, we use the component factors, 5|52.5 5 and 7 (5 X 7 = 85) the quotient is 1.5 years, or 710.9 | tcyT-,. 0-0, 1.5 years Now, the interest on $5000 for 1 year @ 74, is $350; and the given interest is $525, and if 525 be divided by 350 the quotient is the time; hence, the rule may be stated as follows: Divide the given interest by the interest on the principal for 1 year, at the given rate, and the quotient is the time in years and decimals. Rue IV. To find what principal, in a stated time, will increase to @ given amount,.at a given rate per cent. per annum: This is the same as finding the true present worth of a debt, and the rule given on page 133, will answer here, also, viz.: As $100 plus its interest for the given time, and at the given rate, is to $100, so is the given amouut to the principal sought; illustrated in the following : Exam. What principal, in 1 year and 6 months, at 7 per cent., simple interest, will amount to $5525? Or, in other words, what is the true present worth of a debt of $5525, due in 1 year aad 6 months, at 7 per cent.? Here, the interest of $100 for.1 year and 6 months, at 7%, is $10.50. Then, $100 + $10.50 = $110.50, the amount of $100 for the given time and at the given rate; and applying the foregoing rule, we have the following proportion : £58 Usrrut Ru es. As $110.50 : $100 :: $5525 : x, or the required principal. Throwing off the decimal, now, in the first term, we have the following: 11050 : 10000 :: 5525 = “WY <*> — 55250000 + 11050 = $5000, the required principal, or, the true present worth of the debt. Questions like the following, and which are of a useful kind, are of the same nature as those regarding the interest of money :‘ Exam. If the population of a city was 240000 in the year 1896, and 300000, in 1906; what was the rate per cent. of increase during the interval ? By taking the difference of these we find 60000, the increase of population. The question now is: If 240000 gain 60000; what will 100 gain; and for this we have the following proportion : As 240000 : 60000 :: 100 : x, or the rate; that is: Oa = 6000000 + 240000 = 25 per cent., the rate. Exam. Between 1896 and 1906, the population in- creased 25 per cent., and in the latter -year it was 300000; what was it in 1896? In this, 100 + 25= 125: Then, we have 125 : 100 :: 800000 : x; 100 x 300000 ——— = 30000000 + 125 = 240000, the popu- 125 lation in 1896. (See examples, page 148.) and this gives us: Exam. If $25000 be invested in property which rents for $3500 a year, and on which $500 are paid in taxes; what rate of interest does the investment pay ? Here, $3500 — $500 = $3000, the income from $25000. Now, $25000 : $3000 :: $100 : rate; and we have “~~ <" — 300000 + 25000, or, cutting off three ciphers from each, we have 300 + 25 = 122, the rate. PRACTICAL HINTS FOR BUILDERS, Lumber and sawed tember, as plank, scantling, etc., are usually estimated in Loard Measure, hewn and round timber in cubic measure. A board foot is 12 inches long, 12 inches wide and 1 inch thick ; in other words, it isa square foot 1 inch thick. In board measure all boards are assumed to be 1 inch thick. Notre. — Lumber 1 inch thick or less is sold by surface measure, and in the trade is denominated boards. Ifmore than 1 inch thick it is called plank, and is computed at 1 inch thickness, or standard thickness; that is, the product of the surface measure in square feet multiplied by the thickness in inches is the number of feet of lumber at board measure. Rue. To jind the number of feet, board measure, in a board or plank: Multiply the length in feet by the width in inches and divide the product by 12; the result is the number of feet at 1 inch, or the standard thickness. Next, multiply the result thus found by the thickness of the plank in inches; the product is the number of feet of standard thickness, or board measure. Exam. 1. How many feet board measure in a piece of pine lumber 16 feet !oug, 9 inches wide and 1% inches thick ? | 160 PRAcTICAL HINTS FOR BUILDERS. ANALYsiIs: 16 X 9 = 144; then, 144 + 12 = 12 feet at 1 inch thick. Then, 12 x 14 = 15 ft.; or simply add 4 of 12, or 3 and we have the number of board feet = ‘15 feet. If the plank were 14 inches thick we would add the $ of 12, or 6, making 18 feet, board measure; if 24 inches thick, it would be 24 times 12, or 27 feet, etc. Note. — When the piece is 12 feet long and 1 inch thick, or less, the sur- face feet will be the same as the width in inches; thus, a buard 12 feet long, 7 inches wide and 1 inch thick is 7 feet, surfaee measure. In the trade it is customary to have 1} inch lumber resawed into boards about 4 inch thick, commonly called panel-stuff, which is bought and sold by surface measure as if it were inch, or stand- ard thickness; but the price is reduced accordingly. For in- stance, 1} inch pine lumber, worth $50 per thousand feet, would, when resawed, be sold for about $32 to $35 per thousand feet. Hence, questions of the following nature are frequently asked: Exam. 2. Whether isit more advantageous to buy 1000 feet of 14 inch pine at $50, and have it resawed into panel-stuff, paying $2 for the sawing, or to buy the same quantity already resawed, at $35 per thousand ¢ 1000 feet of 14 inch @ $50, and $2 for sawing — $52. In 1000 feet of 14 inch measurement there are 800 feet at 1 inch, or surface measurement, found thus: 1000 feet of 14 Jess one-fifth (4) 200 800 feet, There are 5 quarters in 14, therefore, 1 quarter (4) is the fifth of 5; deducting this leaves the 1 inch, or surface. Now, since Practical HINTS FoR BUILDERS. 161 every piece of 1} inch makes 2 pieces when sawed, 800 feet is doubled, giving 1600 feet of panel. Then 1600 feet at $35 per thousand gives $56, or $4 more than $52, as found above. It would be more advantageous, therefore, to buy at $50 and pay $2 for sawing. Exam. 3. A carpenter wishing to get 1000 feet of 1- inch pine boards dressed to ¢ inch, and not getting the quality suitable for his purpose in the market, concluded to take 14-inch lumber and have it resawed ; how much of the latter did he require ? To solve problems of this nature, it must be borne in mind that the surface is the same regardless of the thickness. But 1000 feet surface is 1500 feet, standard measure, at 14 inch thick, and since each piece which goes to make 1500 feet will make two pieces when resawed, it follows that half the quantity will make 1000 feet surface. Dividing 1500 feet then by 2 gives 750 feet of 14-inch lumber, the required quantity. Proor: 750 feet 14-inch lumber, less 4 250 (1 half inch = 4 of 3 half inches, or 14 inch) 500 feet at 1 inch thick, and doubling this gives 1000 feet surface; that is, 750 feet of 14 inch resawed. Or the problem might be solved thus: take half 1000 feet sur- face and we have 500; to this add one-half and we have 750 feet of 14, inch as before. Adding one-half of 500 feet to itself, we need scarcely remark, is multiplying 500 feet surface by 14, the thickness. SouTHERN PInp. Southern pine, commonly called ‘‘ Georgia pine,” or ‘‘ yellow pine,” comes in various lengths and widths, and it not unfre- quently happens that builders requiring a quantity of flooring or 1) 162 PrRacticAL HINTS FOR BUILDERS. ceiling will give a hurried order to the dealer for such quantity; — and the dealer, in delivering the same from the mill, where it has — probably just been dressed, will not spend the time to get the standard measurement, but, instead, will merely take the lengths of the pieces, or the lineal feet, converting the same at leisure into standard, or. board measure. For such emergencies we give the following simple Rue. To jind the lineal feet for any surface and to convert the same into standard or board measure: Multiply the surface to be covered by 12, and divide the product by the width of the board or plank; the quotient will be the lineal feet. Next, reverse the process; that is, multiply the lineal feet by the width of the board or plank (in the rough, or before being dressed), and divide by 12; the result is surface feet. Then add for the extra thickness, if more than 1 inch thick, and we have the standard, or board measure. Exam. 1. How many lineal feet of 14 by 4 inches (face) of Southern pine will cover 2550 feet surface 2 9550 x 12 + 4 — 7650 lineal feet. Reason of the rule: When a board or plank is 12 feet long it will cover as many feet surface as there are inches in the width of the face of such board or plank. Now, if a piece 4 inches on the face and 12 feet long cover 4 feet surface, how many feet in length will cover 2550 feet surface? And for this we have the following proportion, or analogy: 4: 2550 :: 12; that is, 4 feet surface is to any given surface (2550 in this case) as 12 feet (the length corresponding to 4 feet surface) is to the length, or lineal feet, corresponding to 2550 feet surface. Multiplying the second and third terms and dividing by the first, we get 7650 feet in length, or lineal feet. Next. how many feet, standard, or board measure, in 7650 lineal feet of 14-inch flooring, 4 inches on the face ? PRACTICAL HINTS FOR BUILDERS. 163 A piece which gives 4 inches (face) when dressed, is usually 43 inches in the rough; that is, before being dressed, and, as it és customary to buy and sell at what the lumber measures in the rough, we multiply by 44 instead of 4, thus: ' 7650 < 44 = 2868 feet, the surface, 12 then adding} (the thickness over linch)or 1717, we find the standard, or board measure to be 3585 feet. Hence, in MAKING ESTIMATES for flooring, ceiling, etc., the calculations for the material should be made on the measurements in the rough. Exam. 2. How many feet of “ Georgia pine ” flooring, 24 inches on the face and 14 inches thick, required to cover 3 floors 40 x 386 feet ? 40 x 86 x 3 = 4820 feet, the surface to be covered. Now, a piece of flooring 24 inches on the face and 12 feet long (regardless of the thickness) wil cover 24 Jeet surface. But a piece vf flooring 24 inches (face), 12 feet long, was 3 inches (face) in the rough, or 3 feet surface, and at 1} inches thick it was 33 feet, board measure. The question now is: if 3% feet, board measure, cover 24 feet surface, how many feet, board measure, will cover 4820 feet sur- face? And the proportion is: 21 + 4390 :: 38 Or, reducing the first and second terms to the same denomina- tion, halves, and the third to quarters, or fourths, ~ we have 5 : 8640 :: 15; and by cancellation this becomes 1 : 8640:: 3. 164 PRACTICAL HINTS FOR BUILDERS. Multiplying 8640 by 3, now, gives 25920 fourths, or quarter feet; which, being divided by 4, gives 6480 feet, board measrre, in the rough, or the number of feet to be paid for. From this, it will be seen that, when the material is narrow, it takes about 14 times the surface to be covered, for the required number of feet, board measure, at 14 inches thick. Exam. 3. A builder requires 4 inch pine ceiling, 24 inches on the face, to cover 10000 feet surface. He can buy ceiling suitable for his purpose for $30 per thousand feet, surface measure, or, 14 inch by 6 inch pine, from which to make such ceiling, for $35 per thousand feet. Which is the more profitable, the cost for making each piece of ceiling from the 14 inch being 23 cents? AnaLysis: A piece 24 inches on face, 12 feet long (regardless of thickness), will cover 24 feet surface. Dividing 10000 feet by 24, or using their doubles, 20000 + 5, gives 4000 pieces at 24 feet each. Buta piece 24 feet surface, dressed, was 3 feet in the rough, and, therefore, 4000 pieces at 3 feet each, would make 12000 feet surface measure, or what has to be paid for. This, at $30 per thousand, is $360, the cost. Next, a piece 14 by 6 inches and 12 feet long contains 6 feet and the quarter of 6, or 74 feet, standard measure. It will take 1000 pieces of 14 < 6 (each piece makes 4 pieces of ceiling when milled) to make 4000 pieces of ceiling. Now, 1000 pieces at 74 feet each makes 7500 feet, standard or board measure, and at $35 per thousand this gives $262.50. Adding to this the milling of 4000 pieces at 24 cents each, or $100, we have $862.50, the cost. It is more profitable to buy the ceiling already made, in this case. _ Norz. — From the three foregoing examples it will be seen that, by taking 12 feet in length as a basis of calculation, estimates for flooring, ceiling, etc., ean be readily made. PRAcTICAL H1InNTs FOR BUILDERS. 165 Roor ELrvaAtions, ETC. By the ‘‘ pitch ” of the roof is meant the ratio which the height of the ridge above the level of the roof-plates bears to the span, or the distance between the supports or studs on which the roof rests. The usual pitches are the Common or true pitch, in which the rafters are three-fourths of the width of the building; the Gothic pitch, in which the length of the principal rafters is equal to the width of the building; the Pediment pitch is when the perpen- dicular height is 2 of the width. There are also the 4 pitch, + pitch, % pitch, etc. Rue. To jind the length of rafter for any particular pitch of roof: To the square of the perpendicular height of roof add the square of half the width of the building, the square root of the sum is the length of the rafter. Nore.— The method of extracting the square root is given in almost any common-school arithmetic. It need scarcely be remarked that the rafters for the Common pitch, and also for the Gothic, are obtained without the aid of this rule. SHINGLES, Latu, Ere. A ‘‘shingle” is 4 inches wide and from 16 to 18 inches long. But shingles are seldom made of a uniform width; they vary from 2 to 10 inches, more or less, and are put up in bundles, or bunches, containing 250 shingles each (not by count but on an average of 4 inches to a shingle). Hence, there are 4 bunches to 1000 shingles. Since a shingle is reckoned at 4 inches, it is evident that the number of shingles required to cover a roof will depend on how much of the shingle is ‘‘laid to the weather.” Thus, if 6 inches be laid to the weather, a shingle will cover 24 square inches (4 x 6); and by dividing 144 square inches (1 square foot) by 24, we find it will take 6 shingles to cover 1 square foot. ig Bed em 166 PRACTICAL HINTS FOR BUILDERS. Again, if laid 5 inches to the weather, a shingle will cover 20 square inches (4 x 5); and dividing 144 by 20, gives 74 shingles to the square foot. Hence, it will be seen that, by multiplying the number of square feet to be shingled, in the one case, by 6, and by 74 in the other, we get the number of shingles required at 6 inches and 5 inches, respectively, to the weather. } Now, shingles are generally laid from 5 to 54 inches to the © weather, and for practical purposes the following simple rule will be found sufficiently accurate: Multiply the number of square feet to be shingled by 7; the product is the number of shingles required, nearly . Exam. How many sawed pine shingles required to cover a building 50 feet in length and 36 feet in width, the roof being of the common or true pitch ? ANALYSIS: In the true pitch the rafter is ¢ of the width of the building; % of 36 = 27 feet, the length of the rafter. Then, 27 doubled and multiplied by 50 will give the surface, or the num- ber of square feet to be shingled, thus: 27 X 2 X 50 = 2700 feet, the surface of roof; and 2700 X 7 = 18,900 shingles, the number required. LatH. Notr. — It is customary among the dealers to make use of the singular form, lath; as, ‘‘Have you got any lath?”’ ‘‘How many lath will cover 1000 feet?” etc. A lath is 4 feet long, 14 to 1$ inches wide and about % inch thick, usually made from pine, spruce or hemlock. Lath are seldom, if ever, counted in bunching; they are gen- erally put into a gauge, or measure, which contains about 100 pieces, more or less, and tied in a bunch; hence, 10 bunches make 1000 lath. on ae PRACTICAL HINTS FOR BUILDERS. 167 The surface of a lath 4 feet, or 48 inches long, and 14 inches wide, is 48 X 14, or 72 square inches, and of 2 lath, 72 xX 2, or 144 square inches: therefore, 2 lath, 14 inches, set edge to edge, will cover a square foot. Hence the following simple Rute. To find the number of lath required to cover any surface: Multiply the number of square feet to be lathed by 2; the product is the number required. Exam. How many lath will be required for a room 24 feet long, 20 feet wide, and 9 feet 6 inches high ? ANALYSIS: The length of the four walls is (24 + 24 + 20 + 20) 88 feet; then 88 x 94 (the height 9 feet 6 inches) — 836 square feet, or surface of walls. Next, 24 x 20 (the ceiling) = 480 square feet or surface of ceiling. Putting the two surfaces ~—— EY GOES til al SO eee Poh Ten rar cena 1316, ..the number of square feet to be lathed. Doubling this, we have 2682, the number of lath required. Nores.—1. Lath are usually set about 1¢ inch apart, to allow for the “‘clinch’’?; hence, when 11g inch lath are used, a deduction of about one- tenth, or 1 bunch in every 10, may be made. Thus, in the foregoing example, 2632 less 263 (345 of 2632) — 2369, would be nearest the true re- sult. But for 144 inch Jath no deduction is necessary. 2. Allowance must, of course, be made for doors, windows, ete. CoNSTRUCTION AND CAPACITY OF Bis, ETC. The Standard Bushel of the United States is a cylindrical meas- ure 184 inches in diameter and 8 inches deep, and contains 2150.42 cubic inches. (The capacity, 2150.42, is found by multiplying the square of the diameter, 184, or 18.5, by .7854, and the pro- duct by the depth, 8 inches.) Since a cubic foot contains 1728 cubic inches, and a standard bushel contains 2150.42 cubic inches, a bushel is equal to 14 cubic feet, nearly (2150 + 1728 = 1}, nearly), the proportion being 1 to 14, or 4 to 5, nearly. Hence, 168 PRACTICAL HINTS FOR BUILDERS. To jind the capacity of a bin in bushels: Add 4 of the quantity in bushels to itself; the sum will represent the capacity of the bin. Thus, what must be the capacity of a bin to contain 160 bushels of wheat? 160 + 40 G4 of 160) = 200, the number of cubic feet, or capacity of bin. And To jind the number of bushels contained in a bin: Deduct 4+ of the capacity of the bin from itself; the re- mainder will represent the number of bushels. Thus, how many bushels of wheat in a bin of 200 cubic feet capacity? 200 — 40 (4 of 200) = 160 bushels of wheat in a bin of 200 | cubic feet. Hence, Any two dimensions of a bin being gwen, the third dimension can be found, thus: (1) Increase the number of bushels by 4 of itself; the result will represent the number of cubic feet contained in the bin. (2) Di- vide the contents in cubic feet by the product of the two dimen- sions, and the quotient will be the other dimension. Exam. 1. What must be the depth of a bin to contain 280 bushels, its length being 10 feet and its width 5 feet ? ANALYSIS: 280 + 70 = 350; then 350 + 50 (10 x 5) = 7 feet, the depth. Exam. 2. What is the value of a bin of wheat 20 feet long, 12 feet wide, and 5 feet deep, at $2 a bushel? PRACTICAL HINTS FOR BUILDERS. 169 ANALYsIS: 20 x 12 x 5 = 1200; then, 1200 — 240 (4 of 1200) = 960, the number of bushels in bin. 960 at $2 = $1920, the value. ; Exam. 3. A coal bin is 12 feet long and 6 feet wide; how deep must it be to contain 12 tons of chestnut coal, the contents of a ton of chestnut coal being 38 cubic feet? ANALysIs: 38 x 12 = 456, the contents of 12 tons; then 456 + 72 (12 x 6) = 64 feet, or 6 feet 4 inches, the depth. Rue. To find the capacity of a vessel or space in gallons: Divide the contents in cubic inches by 231 for liquid gallons, or by 268.8 for dry gallons. Exam. 1. How many gallons of water will a cistern hold that is 4 feet by 5 feet, and 6 feet deep ? Anaxysis: (4 X 5 X 6 X 1728) + 231 — 89734 gallons capacity. Exam. 2. What must be the depth of a cistern that is 6 feet long and 54 feet wide to hold 462 gallons of water ¢ 52 x 231 ANALYSIS: iaresis = 1.87 feet, the depth. Exam. 3. A cellar 40 feet long, 20 feet wide and 8 feet deep is half full of water. What is the cost of pumping it out at 6 cents a hogshead ¢ Anaxysis: (40 X 20 X 8 X 1728) + 231 — 47875.32 gallons; then 47875.32 + 63 (gallons in a hogshead) = 759.92 hogsheads; and 6 times 759.92 — 4559.52 cents. Dividing this by 2 (half the cellar) gives $22.80, the cost. 1,0 PRACTICAL HINTS FOR BUILDERS. Masonry. Masonry is estimated by the cubic foot and by the perch, also by the square foot and the square yard. A perch of masonry is 164 feet long, 14 feet ie and 1 foot high. Multiplying these three dimensions together, we find there are 242, or 24.75, cubic feet in a perch of masonry (164 x 14 <1 = 24%), or (16.5 x 1.5 X 1 = 24.75). Notr. — When stone is built into a wall without mortar or filling an allow- ance of 234 feet is made, and 22 cubic feet make a perch. Rute. To find the number of perches of masonry ina wall: Divide the number of cubic feet in the work by 24%, or 24.75; the quo- tient will be the number of perches. Nots.— Brick-layers and masons, in estimating their work by cubic meas- ure, make no allowance for the corners of the walls of houses, cellars, etc., but estimate their work by the gzrt, that is, the entire length of the wall on the outside. Joiners, brick-layers, and masons, make an allowance of one-half the openings or vacant spaces for doors, windows, etc. Exam. 1. At $4 a perch what will be the cost of build- ing the walls of a cellar 374 feet long, 26 feet wide, 9 feet deep and 2 feet thick ? ANALYSIS: 75 + 52 = 127 feet, the girt, or outside measure; then, 127 x 9 x 2 = 2286 cubic feet, the solid content of walls. Next, 2286 + 24.75 = 92.36 perches. Multiplying this, now, by 4, gives $369.44, the cost. Nore.— To divide by 24.75, or 2484, we make use of the simplified method, as pointed out at page 60, example 4, to which the reader is referred. EXCAVATING or digging is measured and paid for’by the cubic yard, and a cubic yard of earth is called a load. Nore. — In a lineal yard there are 3 feet. Cubing 3 (3 X 3 x 8 = 27), we get 27 cubic feet for acubic yard. Hence, a box 9 feet long, 8 feet wide PRACTICAL HINTS FOR BUILDERS. id and 1 foot deep will contain a load of earth (9 x 3 « 1 = 27). And any two dimensions of a box to contain a load being given, the other dimension can be found, thus; Divide 27 by the product of the two given dimensions; the quotient is the other. (See example 1, page 168.) Exam. 2. What is the cost of digging the cellar in the last example at 50 cents a load? ANALYSIS: 874 X 26 X 9 = 8775 cubic feet of earth in cellar. Then, 8775 ~ 27 = 825 loads; and at 50 cents a load it is $162.50, the cost. Brick-work. Rute. Zo find the number of bricks in a wall: Multiply the number of cubic feet in the work by the number of bricks ina cubic foot; the product is the number of bricks required. Nore. — About 22 common bricks make a cubic foot when laid. Exam. 1. How many common bricks in a wall 70 feet long, 20 feet high, and 12 inches thick ? Anatysis: 70 x 20 x 1 = 1400 cubic feet in wall; then 1400 x 22 = 30800 bricks. In estimating brick-laying by the square yard, the rod, or by the square of 100 feet, the work is understood to be 12 inches, or 14 bricks thick, which is called standard thickness, Rue. To reduce brick-laying to standard thickness: Multiply the superficial content of the work by the number of half bricks in thickness, and divide by 3. Notge.— The superficial content is found by multiplying the length by the height. ded Exam. What is the cost of 75680 feet of black walnut at $75.75 per thousand ? In this, we have at sight the cost at $100; then, at $25, it is a fourth, or $1892, which $7568|0 = $100 is deducted, leaving $5676, the cost at $75. _ 1802/0) ae Now, 75 cts. is a hundredth part of $75, or ye we $75 m5 $56.76, which is added; this gives $5732.76, $5733\76— $75.75 the cost at $75.75. PracticaLt Hints ror BuILpERs. 173 Exam. What is the cost of 75684 feet of lumber at $32.58 per thousand feet ? Here, we cut off one figure to get the cost at $100 ; then, one-fourth of this, or $1892.10, is the cost at $25 ; now, $5 isa fifth of this, or one-half of the cost at $10, got from the top line ; next, $2.50 is half of $5; and for 8 cts. we take 8 times the cost at 1¢, that is, .75684, got from the top number: say 8 times .757, making use of three figures only, and allow- ing for those rejected; the result is $6.056. Or thus: Cutting off two figures gives the cost at $10, always; then, 3 times 10 are $30; now, $2.50 is a fourth of $10; and finally, 8 times .757 gives $6.056; the sum of the several results gives $2465.786, the cost at $382.58, the same as before. $7568/4 —$100 1892it “= + 25 878/42 = 5 189|\21 = 2.50 6|056= .08 $2465|786= $32.58 $756/84 —$10 2270/52 = 30 189/21 = 2 50 6\056=- .08 $2465|786 = $32.58 Exam. What is the cost of 75684 feet, at $37.75 per thousand ? In this, the cost at $25 is one-fourth of that at $100, or $1892.10; then, $12.50 is half of $25; and 25cts. is the hundredth part of $25; the hundredth part of $1892.1 is $18.92; the sum of the several results is $2857.07, the re- quired cost. Or thus: Here, we have the cost at $10, at sight, and 3 times this is the cost at $30; then, $5 is half of $10; now, $2.50 is half of $5, and 25cts. is a tenth of $2.50; the sum of the several results is the cost at $37.75, as before, $7568|4 =$100 189211 — 25 946|/05= 12.50 18/92= 25 $2857|07= $37.75 $756 84=$10 3970 50-2 30 8378/142— 5 189/21— 2.50 18 92— OD $2857|07=$37.75 CHRONOLOGICAL CALCULATIONS. To find the weekly day for any given date from the year 1600, New Style (N. 8S.) for any year or century thereafter, we give the following simple RuLE |. Subtract the centuries from the given year. Il. To the remainder add one.fcurth of the given year; also the number of days from January 1, up to and including the given date, and 1 for every fourth century. III. Divide the sum by 7; the remainder, counting Sunday 1, will be the weekly day. Notes. —1. For leap years add 1 to the centuries before subtracting always, counting 29 days in February; then proceed according to the rule. 2. When 7 is contained in the sum without remainder, Saturday is the weekly day. 3. In taking the fourth part of the given year, the remainder, if any, may be rejected; also, in taking one-fourth of the centuries. Exam. 1. On what day of the week did the 4th of July, 1775, happen ? SoLUTION.— Subtracting 17 centuries. from 1775 Jan. 31 da. the given year, we get 1758; to this we add 17 Yeb. 23 one-fourth of 1775, rejecting the remainder, 1758 Mch.31 “ next 185, the number of days from January 1, 443 April30 “ up to and including July 4th, and 4, 1 for 185 May 31 ‘* every fourth century in 17 ee rejecting 4 June 30 ‘ the remainder. 7)2390 July 4 “ Dividing the sum, 2390, by 7 gives a re- 341—3 185 mainder of 3; then counting Sunday 1, Mon- day 2, Tuesday 3; we find that July 4th hap- pened on the third day of the week, Tuesday. CHRONOLOGICAL CALCULATIONS. 175 Exam. 2. On what day of the week will Washington’s — birthday (Feb. 22) happen in 1912? In this, 1912 being a leap year, we sub- 1912 less (20) 19 +1 tract 20 from-the given year, the remainder ~ 7899 is 1892. Adding to this 478, the fourth part 478 of 1912; then 53 days, the number from Jan- 53 uary 1 to February 22, and 4 (19 + 4) 1 for 4 every fourth century in 19 centuries, and 9427 dividing the sum by 7, we obtain a re- Dill mainder of 5; the fifth day of the week, 346...5, Thursday. Thursday. To find the Day of the Month on which a Particular Day oy the Week will happen. Exam. 2. The presidential election occurs on Tuesday after the first Monday in November; what day of the Month will it be in 1912? To solve a problem of this kind, we have to find, first, on what day of the week November 1 will happen. SOLUTION.— 1912 being a leap year we sub- 1912 less 20 tract 20 (19+ 1) from the given year, as in the 1892 foregoing example, next, adding to the re- 478 majnder one-fourth of 1912, also the number of 306 days, 306, from January 1, up to and including — 4 November 1; and 4 (19 + 4) and dividing by 7, 7)2680 we get a remainder of 6; the sixth day of the 389.~6 Frid week, Friday, on which November 1 will happen; Shoat ish toute consequently election day will be the following Tuesday, or Nov. 5. Exam. 3. Abraham Lincoln was born in Kentucky, February 12th, 1809, on what day of the week did it happen ? 1809 — 18 = 1781 SoLuTion.— Deducting 18 from the given year 452 leaves 1781; then proceeding according to the rule 43 we find the remainder to be 1; the first day of the 4 week, Sunday. 7) 2290 On what day will it happen in 1912? Ans. Monday. 176 CHRONOLOGICAL CALCULATIONS. Exam. 3. If your birthday be Dec. 29, 1900, on what day of the week will it happen ? 1900 — 19 1881 475 363 4 2728 a 9.... Saturday. Subtracting 19 from 1900, we get 1881. Adding the required numbers according to the rule and dividing by 7, we find the day to be Saturday, the division being exact. BISSEXTILE OR LEAP YEAR, Nore. — The solar year, or the time required by the earth to go once around the sun, is 365 da. 5 h. 48 min. 48 sec. The common year is 365 days. Hence, 1 solar year is 5h. 48 min. 48 sec. longer than 1 common year. 4 ‘* years are So As oe ode 4 = years. 100 66 eG oe 24 da. 5 bb 90 G6 66 66 100 e 400 Ge Ge 6s 96 66 21 66 20 &e 6% 6 400 be ee or 97 days nearly. Hence, If 97 days be added to every 400 years, in other words, if 97 leap years be reckoned in every 400 years, the calendar will be only 2h. 40 min. in advance of true time; or about | day in 4000 years. To be more explicit, let us take the time between the years 1600 and 2000, a period of 400 years. It is clear from the foregoing that we cannot reckon every fourth year in these 400 a leap year as that would give 100 leap years, while the correct number is 97, a difference of 3 years. To distribute those 97 days, then, among 97 years, every fourth year in we 400 is reckoned a leap year, except the centennial years 1700, 1800 and 1900, ence, A fF Every year that is exactly divisible by 4 is a leap year, the centennial years excepted ; the other years are common years. Il. Ever y centennial year that is exactly divisible by 400 is a leap year ; ; the other centennial years are common years. The year 1900, for example, is a common year, because, although exactly divisible by 4, it is not exactly divisible by 400. The year 1904 isa leap year, be- ing exactly divisible by 4; and the years t600, 2000 and 2400 are leap years, being exactly divisible by 400. APPENDIX. INTEREST RULES TERSELY STATED AND EXPLAINED. GENERAL RuLe.— To jind the interest of a given sum for any number of days, at any rate per cent., on the basis of 360 days to the year: Multiply the principal by double the rate, and the product by the days, cut off five tigures from the right, counting Jrom the decimal point always, and add a third and a sixth. Exam. What is the interest of $3765 for 8 days at 3242 $3765 9125900 758 1255 $3/1375 Doubling the rate, 33, we get 74; multiplying this by 8, the days, gives 60 ; then 60 times the principal is $225900. Cut off five figures to the right, and add } and 1 of that third; the interest is $3.1375. Note.— When the principal is not large enough to cut off five figures, prefix a cipher or ciphers, to make five, and proceed according to the rule. Thus, if the principal were $65, in the example, 65 x 60=$3900; prefixing a cipher we have .03900, five decimal places, and adding a third and a sixth of that third we get .05446, the interest of $65 for 8 days at 3% 4%. Whenthere are cents in the principal there will, of course, be seven places of decimals. The reason of the rule will be understood from the following : The problem fully expressed is this : If $100 earn $33 in 360 days, what will $3765 earn in 8 days? And the solution, by compound proportion, would be as follows : $100 : $8765 :: $382 360 days : 8 days | Va = ee 178 Interest Routes TERSELY STATED AND EXPLAINED. Reducing this to a simple proportion, we have the following: 36000 : 30120 :: 32 Doubling the rate, now, doubles the first term of the proportion, and we have 72000: 3u120 :: 7s. Multiplying now by 73, double the rate (we have already multi- plied by the days, 8), and dividing by 72000, solves the problem. Instead of dividing by 72000, in the usual man- 72000 ner, we prefer to make use of a more simple divisor, 24000 4000 and that we get by adding a third and a sixth of that third, making 100000, as shown in the margin. 100000 Bearing in mind now, that whatever operation is performed on the divisor to simplify the division, a similar operation must be performed on the dividend; hence we add to 225900 a third and a sixth of that third, having cut off five figures from the right which divides by 100000. Mernop or Exact INTEREST. GeneraL Ruie.— To find the interest of a gwen sum for any number of days, at any rate per cent., on the basis of 365 days to the year: Multiply the principal by double the rate, and the product by the days, cut off five figures from the right, counting from the decimal point, and add a third, a tenth and a tenth. Exam. What is the interest of $275000 for 50 days, at 2% ¢ $275000 550(00000 183/33333 18/33333 11833338 $753/50000 7530 42 Interest Rurtes Tersery STATED AND EXPLAINED. 17Y Doubling the rate, 2, gives 4; then 50 days multiplied by 4 gives 200. Multiplying the principal, now, by 200, cutting off five figures from the right, and adding a third, a tenth of that third and a tenth of that tenth, we get $753.50. This gives an excess of 10 cents in every $1000 of the interest, or 1 cent in every $100, so we bring 10 times the dollars, or 7530, to the right, and subtracting, we get the correct interest, $753.42. The reason of the rule will be understood from the following: The problem fully expressed is this: If $2 be paid for the use of $100 for 365 days, how much will be paid for the use of $275000 for 50 days, at the same rate? And the solution by compound proportion would be as follows: $100 : $275000 :: $2 365 days : 50 days Reducing this to a simple proportion, we have the following: 36500 : 13750000 :: 2 Doubling the rate now, doubles the first term also, and we have: 73000 : 13750000 :: 4 Multiplying the middle term now by 4, and dividing the pro- duct by 73000 solves the problem. Instead of dividing by 73000, however, we prefer 73000 to use a more simple divisor, and that we get by 243334 adding a third, a tenth of that third and a tenth ° 24334 of that tenth, making 100010, as shown in the 2434 margin. 100010 — Performing a similar operation on the dividend 55000000, we add its third, a tenth of that third and a tenth of that tenth, getting 75350000 for new dividend, to correspond with 100010, the new divisor. Cutting off five figures divides by 100000 (which, by the way, may be done at the beginning or end of the process), and rejecting 10 times the quotient, or the dollars, $753, 180 Inrerest Rutes Tersety Sratep anD EXPLAINED. makes the required correction. (See example 3, page 53; also rule, page 54.) Nore.—If the principal be not large enough to give five decimal places, prefix a cipher, or ciphers, to make five, as in the rule for 360 days. If we had to use 366 days (leap year) as divisor, the complete divisor would then be the double of 366, or 732, multiplied by 100, or 73200, and by adding to this a third and a tenth of that third, the new divisor would be 100004, a simple divisor. (See example 4, page 54.) SPECIAL RULEs. Although the foregoing rules are general, there are special rules for certain rates which will be found shorter for business purposes. Foremost among these we would place the Six Per Cent. Meruop. Norrt.—This method has been already fully explained in the chapter on Interest, commencing at page 122, where we have made proportion and cancellation the groundwork of the rule, but for persons who may not have a knowledge of these subjects, the following will be found, perhaps, more simple and clear: Rue. Zo jind the interest of any sum for one year at any rate per cent.: Multiply the principal by the rate per cent. and divide the product by 100. Exam. What is the interest of $475 for one year at 6%? Multiplying by 6, the rate. and dividing by 100, we $475 get $28.50, the interest. 28[50 Suppose, now, we wished to find the interest of $475 for 2 months, or 60 days, at 6%, on the basis of 360 days to the year, we simply take § of $28.50 (2 mos. or 60 days being ¢ of a year), which gives the required interest, $4.75, or a cent for every dollar of the principal, showing that we need never figure 2 mos., or 60 days’ int. at 6%, only call the dollars of the principal so many cents, in other words, 1% of the principal is always the interest for 60 days at 64. InterEst Rutzes TrErRsELY STATED AND EXPLAINED. 181 Taking this for our basis, it is evident that the int. of $475 for 6 days at 6%, is a tentn of 60 days’ interest, $4.75, or ATS and for 3 days, the interest is half of 6 days’ interest, or .2875 Again the interest of $475 for 600 days, 20 mos., or 1 yr. 8 mos. at 67%, is 10 times the int. for 2 mos. or 60 days, $4.75, or $47.50 | and for 40 mos., or 3 yrs., 4mos., it is 2 times $47.50, or _ $95.00 The rule will be found equally simple when the number of days is not a measure or a multiple of 60 or 6. Thus, if it were required to find the int. of $475 for 13 days at 6¢. Moving the decimal point two places to the left, gives the in- terest for 2 mos. or 60 days, and moving it three places, gives _ the interest for 6 days, at 6%, always. Now the interest of $475 for 6 days is .........202.42- 475 raaor l2idays, iis 2 times: :475, Or Joie. . eels ile se .950 Petts Peay itwisys Of) (ATD NOT, oc ys wees Geb wens 79 Pate nerinverest £0T 19 GAYS OF &. soc soc ic dng eats dine _ $1.029 And the interest of $475 for 117 days, at 6%, would be found thus : SMEeresGICOr (CAV 1S, aes, inc aw noes ctnescinacs $4/75 And for 120 days, it is 2 times $4.75, or.......... oe AOU Pomardays, the int: is half of :475, or. ... 0... eee ewes 2375 Which is deducted, leaving the int. for 117 days or.... $9\2625 Exam. What is the interest of $420 for 279 days, at 6% ? Dividing 279 days by 60, it is contained 4 times, with a remain- der of 39. (See note 3, page 142.) - The interest of $420 for 60 days is........... wcccceee . $420 And 4 times $4.20 is the int. for 240 days, or.......... 16,80 The interest for 30 days is half 60 days’ int. or ......... 2,10 And 8 times 21 cts. (the int. for 3 days, or {, of 30) is 9d. or 63 Giving the interest of $420 for 279 days............... $19/53 Now since the interest of $420 for 279 days is the same as the interest of $279 for 420 days, the foregoing example can be sim- -plified by taking the dollars for the days, thus: 182 Interest Ruutes Tersety STratep AND EXPLAINED. The interest of $279 for 60 days, at 6%, is............4. $2|'79 And for 420 days, it is 7 times 60 days’ interest, or...... $19/53 (This method can be always adopted when it ismore convenient to take the dollars for the days, Nott.—It need scarcely be remarked that should the principal contain cents, it will not affect the process, the figures to the right of the line being simply decimals. Exam. What is the interest of $345.60 for 1 y. 10 mos. 26 d., at 43% % Moving the decimal point one place to the left, gives the interest for 20 months, or for 1 y. 8 mos., at 6%........ $84/560 The interest for 2 mos, is a tenth of that, or........... 3/456 And for 20 days, the interest is a third of 2 mos., or . 1/152 The interest for six days 18... .. .. 5. ..3s Sen 345 - Giving the interest of $345.60 for the given time, at 6%, 39|513 Deducting } of the interest at 6%, $39.518, or .......... 9/878 Gives the interest at 4397, or. . 22. ...0 ssc nse see $29 1635 Nore.—The difference between 6% and 4147 is 114; and 1 is 4 of 6. (See page 124, also rule, page 125.) INVEREST ON RUNNING ACCOUNTS, Instead of finding the interest on each item separately, as is frequently the case with persons having to deal with such matters, it will be found much more simple and expeditious to proceed as follows: Exam. What is the interest of $3000 for 23 days; $4500 for 10 days; $5000 for 19 days; $2000 for 26 days; and $4000 for 17 days, at 6% ? ? Multiply each principal by the number of days respectively; cut off three figures to the right, from the sum of the products; divide by 6, and the result is the interest at 62. InterEst Routes Trersety STATED AND EXPLAINED, 183 Process, $3000 X 238 == 69000 €4500 X 10 = 45000 $5000 X 19 = 95000 $2000 X 26 = 52000 $4000 xX 17 = 68000 329|000 $54(833 Note.— For any rate other than 6%, add or subtract the difference, as | pointed out at page 124. ! Exam. Find the interest on the following at 62 : Process. $2700 for 3 y. 5 mos, 12 days. 8100 13500 32400 $1500 for 5 y. 7 mos. 8 days. 7500 10500 12000 $4500 for 7 y. 9 mos. 13 days. 31500 40500 58500 $3800 for 1 y. 3 mos. 20 days. 3800 11400 76000 $5000 for 2 y. 0 mo. 28 days. BOOGIE pa HAS ee 140000 $7500 for4 y. 1 mo. 0O days. 30000 7500 909.00 834.00 318.900 $5454.00 | $417,00 $53.15 Adding the three results thus found, we get the required interest, $5924.15. Rue I. Multiply the principal by the years, months and days, setting the product of each respectively in a separate column, as shown in the example, and add. IL. Point off two figures to the right in the sum of the yearly col- umn, multiply by the rate, and the result is the interest for the years at the given rate. 184 Inrerest Ruutes Trersety STATED AND EXPLAINED. III. Point off two figures in the monthly column and take half ; and for the daily column point off three figures and divide by 6 ; the result in both cases will be the interest at 6% from which the interest at any other rate is readily found by addition or subtraction, as already pointed out. | NotEr.— Reason of the rule: The interest of $2700 for 3 years is the same as the interest of $8100 for 1 year; the interest of $2700 for 5 months, the same as $13500 for 1 mo.; and of $2700 for 12 days the same as $32400 for 1 day, at any rate per cent.; consequently the interest of the totals, $90900; $83400 and $318900, for 1 year, 1 mo.and 1 day respectively, is equal to the interest of the several principals for the given time and rate. Pointing off two figures from the right of the dollars in any Brincipat and multiplying by the rate, gives the interest of that principal for1 year at the given rate. Pointing off two figures from the right of the dollars in any principal, gives the interest for 2 mos. at 6%, and half of this is one month’s interest. Finally, pointing off three figures gives the interest for 6 days at 6%, and therefore the interest for 1 day is 1-6 of that. ImporTANT Facts to BE REMEMBERED. (1.) That the interest of any principal for 6000 days, at 6%, is equal to the princi- pal itself ; in other words, any sum of money will double itself at 6%, sim- ple interest, in 6000 days, 200 months, or 16% years. (2.) Moving the decimal point one place to the left in any principal, gives the interest of that principal for 600 days, at 62. (8.) Moving the point two places gives the interest for 60 days ; and (4.) Moving it three places gives the interest for 6 days, at 6%: thus: The interest of $4765 for 6000 days, at 6%, is $4765 € ‘* $4765 “ 600 days, ie $476.5, or $476.50 a ‘© $4765 ** 60 days, OS BAGS Se ** $4765 “* 6 days, ri $4.765 And from this simple basis the interest for any time and rate can be easily found, and often by a choice of two methods: (See exam. page 181), .ExaM. Find the interest of $2000 for 119 days, at 64. First Method. Second Method $20|00= 60 days’ interest. Take the days for the dollars, and the 40\00=120 ** 3 Ee dollars for the days: Be chy oh rs Spat 6d. \ Int. of $119 for 6000 days is $119. BES E2000 os Nate or Sears ‘Spar Pee 7 be $39/67=119 ‘“ [ int.=33e. | A SIMPLE METHOD FOR AVERAGING ACCOUNTS, In averaging, there are two kinds of equations, Simple and Compound. A Simple Equation has reference to one side of an account only, which may be either a debit or credit. A Compound Equation has reference to both sides of an account. SimpLeE EQuATION. If one person owe another, on Jan. 1, $300 payable in 4 months, $500 payable in 6 months, and $400 payable in 103 months; at what time may the whole be paid without loss to either party ? Process. $300 x 4 — 1200 500 x 6 —= 3000 400 x 10% — 4200 ale Sr The interest of $300 for 4 months equals the interest of $1200 for 1 mo.; the interest of $500 for 6 mos. equals the interest of $3000 for 1.mo.; and that of $400 for 104 mos. equals the interest of $4200 for1 mo. And since the interest of $8400 for 1 month equals the interest of $1 for 8400 months, $1200 = 7 months. 186 A Srmerte Meruop ror AvErRAGING ACCOUNTS. The time, therefore, is 7 months from Jan. 1, or Aug. 1. Rute. Multiply each payment by its term of credit, and divide the sum of the products by the sum of the payments; the quotient will be the time to be counted forward from the date at which the credits begin. . Exam. What is the equated time of payment for the following bill? New York, Jan. 1, 1893. EDWARD JONES. To JAMES FRENCH SOR. 1892. June 5, To Cash ....... 41 240 123 Ese 16, 6) 96 $124 8|0)117/6 124)196(1 mo. — 124 39 —— 72 30 124)2160(17 2108 Arranging the time in months on the margin, we find 0 mo. for April, 1 mo for May and 8 mos, for July Multiplying each item of the bill now by its respective date, and adding the results, we get $1176 for 1 day. Dividing this by 30 (days in a mo.) gives $39 for 1 mo., which is carried to the right, opposite the last date of April. Then $34, the amount for May, is multiplied by 1 mo. and $41, for July, is multiplied by 3 and.the results set under $39 and added, making $196 for 1 mo. Dividing 196 now by $124, the amount of bill, we get 1 mo. and 17 days. Counting 1 mo. and 17 days forward from April 1, inclusive, we get May 17th for the average date; and counting 60 days (term of credit), forward from May 17, gives July 16 for the due date. (See note, page 187. To impress more thoroughly upon the mind of the student, a knowledge of this important subject, it may be well to give the solu- tion of one more example, as follows: 194 A SimpLe Meruop ror AVERAGING ACCOUNTS. John King wishes to settle his account on Dee. 1, 1895; how much does he owe, charging interest at the rate of 6%? 1895. Dr. Mo. Q May 7. Tomdse. $25 175 aes Reh ak aioe 30 270 nS LE A Seana 20 300 82 Oy VULY SL ene eee 5 9) Meta Uae oe ey oF 280 210 meat Bet be TO 60 720 BOcte Vega Se 8 24 en Uh ea | $72 520 360 (e ibaat Ler 12 )- 180 $252 3)0)247\4 $652 82 Cr. 1 June12, By Cash. ae $70 600 45 cine Fae mle ia had 20 300 70 B COct, ctoestai 100 100 Teale yeas rorcnsal aieare aaa a 00 80) BT $52 B&B 52)113/2 mo. 104 a 30 52)270(5 days. 260 SOLUTION.— Counting May 0, July 2, Oct. 5 on the Dr. side, and June 1 and Oct. 5 on the Cr., we have the time arranged at sight on the margin. Proceeding now according to the rule, we get $652 for 1 mo. on the Dr. and $765 on the Cr. side. The balance of account is $52, and is on the Dr. side; the balance of interest is $113, and is on the Cr. side. In other words, the balances are on the opposite sides of the account, aud therefore the time, 2 mos. and 5 days — (found by dividing the balance of interest by the balance of account), is to be reckoned backward from May 1, giving Feb. 238d for the average date. Interest is now charged on $52 from Feb. 23d to Dec, 1 (date of settlement), at 6%. The time is 9 mos. and 6 days; the \nterest, $2.39, making the balance due, $54.39. : SHORT METHODS. Exam. What is the cost of 1278456 pounds of iron at $24.812 per gross ton (2240 Ibs.) ? Lone METHOD. SHort METHOD. 1278456 12784516 24.812 18263 657 639228 9131/828 319614 ’ 4565|914 278456 456/591 10227648 5|'707 5113824 2/853 2556912 1 426 cae alate Le ai 322 $14164/322 2240 9328 Dividing the number of pounds 8960 by 70, gives the price at $32.00 a 1 of $32“ ‘“ 16.00 14408 aor t ties Bee a 8. 00 13440 zy of $8 us ns . 80 9681 atthise ie oot 01 8960 4 of this a4 cé 4 7217 ; = 5 4978 Adding we get price at $24.81 4480 (See exam. 1, page 109.) 4980 4480 et Note.— The division by 2240 can always be simplified by cutting off one figure from the right of the dividend, counting from the decimal point, and taking a quarter, one-seventh, and one eighth. thus: 196 ‘Suort Muruops. Cutting off one figure and dividing by 4, een a divides by 40; then one-seventh of that 1133141577 quarter, and one-eighth of that seventh, be- "$14164|329_ cause these numbers are factors of 2240 (40 . x7 x 8 = 2240). Exam. What is the cost of 259356 pounds of iron at $13.75 per gross ton % 25935)6 3705|085 926|271 463/136 115/784 57/892 238/946 $1592/029 Cutting off one figure and dividing by %, gives the PPICE Abo. LEM wicis seve ew ee © 5 208 90 'e ene $32. One-fourth of $32.18 the price at... ... 4c... eee 8. One-half of $8 * EMP Sr es ee 4, One-fourth of $4 $$ §Oe i Saw de «eee arerpe Ie One-half of $1 os SS Sta ae ee ee 3:0 ole eee .50 And one-half of this RE Bese © 8 0.9-a\e\e eieneeee 25 The total.sum is the price at.......*..sss«es evens $13. 75 Or thus: Cutting off one figure, and dividing 925935]6 by 8, divides by 80, and gives the 3241195 price at $28 per gross ton always. 16201975.. $931 .568 Half of $28 is the price at $14, or 281946 25c. more than the given price. To $1592 029 get the price at 25c. we set one-seventh of the price at $14, or $231. 568, a little to the right of the work, as shown in the margin; this is the price at $2, and an eighth of this, or $28. 946, is the price at 25c. Subtracting this fram the price at $14, gives the price at $13.75. Suort Mernops. 197 Notst.— From the foregoing examples, it will be readily seen how easily the cost of any number of pounds, at $14, 16, 21, 24, 25, 28, 32, 34, 36, or in fact any price per gross ton, may be found. (See note to exam. 3, page 111.) Exam. What is the cost of 562800 pounds of iron at $37.50; of coal, at $3.75; and freight at 3874c¢. per gross ton ¢ | IRON. CoAL. FREIGHT. 56280/0 5628/00 $562|800 8040|00 —$32.00]| 804/00 —$3.20 80/400 = .32 1005|00 = 4. 100j00°== 40 10/(050== 4 goli25 = 1 LoL eos CLO Aiol2=—= 1 125|625— .50 12/562—= 5 1/256 = cota ————————|/ |__| —. SS eee $9421|875—= $37 .50|| $942|\187— $3.75 $94/218 = .374 Cutting off one figure and dividing by 7, gives the price at $32 per gross ton always; cutting off two, and dividing by 7, gives the price at $3.20, or a tenth of $32; and cutting off three figures, and dividing by 7, gives the price at 32 cents, or the one- hundredth part of $32; and the aliquot parts are the same. Or, we could have proceeded in either case as if the price were $37.50 per ton, and take one-tenth or one-hundredth at the finish. Oats. Exam. What is the cost of 4760 pounds of oats at 593,c. per bushel (82 lbs.) ? 47/160 = .32e. 23/80 — .16 1190 — 8 D975 == 2 14487— 1 1 185:=== fr 3 — ts 198 SHort Meruops. Assuming the price at 1c. per pound, we have the price at © once, at 2c. per bushel; then 16c. equals 4 of 32; 8c. half of 16; — 2c. a quarter of 8; 1 a half of 2; 2, or its equal, 3, is an eighth - of 1c.; and 54 is half of that. Exam. What is the cost of 300 bags of oats, 79 pounds net to the bag, at 25fc. per bushel ? $237|00 — 32¢. 118|50 —16 59/95 — 8 7/406 —= 1 3I703— 4 1s51— 2 995 — 4 $191|635 — 252 300 multiplied by 79, gives the number of pounds, 23700, and at 32c. per bushel, the price is $237.00. Half that is the price at 16c., or $118.50; at 8c. the price is half that, or $59.25, and 1c. is one-eighth of that, or $7.406; §, or its equal, 4, is half of 1c. or $3.703; 2 is half that, or $1.851, and 4 is half that, or .925, making $191.635, the price at 2d{c. Corn. Wheat, buckwheat, barley, etc., may be treated in like manner, by assuming the price at a cent per pound, or at as many cents per bushel as there are pounds to the bushel. Exam. What is the cost of 2940 pounds of corn at 676. per bushel (56 Ibs.) ? $29/40 — 56c. 4/20 = 8 1105 — 2 sone] $35/18— 670, SHort Meruops. 199 Cutting off two figures we have the price at 56c. per bushel ; one-seventh of which is the price at 8c.; 2c. is a quarter of 8; and ic. is half of 2, giving $35.18, the price at 67c. And if the quantity should be given in bushels and pounds, the same method of solution can be employed, as illustrated in the following: Exam. What is the cost of 364 bu. 27 lbs. of oats at 428c. per bushel ? 364/27 eer 29/12 116)75 — 32a 291187 — 8 7|296 — 91 oa 456 — 8154/1601 498 | | bo jm oojao Multiplying the bushels by 8, and setting the result two places to the right, we get 2912; multiplying this, in turn, by 4, and adding at the same time the 27 lbs., we get 11675 lbs., or the number in 36423 bu. The remainder of the process needs no further explanation. Notr.—To reduce pounds to bushels, we should never make use of long division when the number of pounds to the bushel is a composite number. Thus, how many bushels in 11675 lbs. of oats ? 11675 2918 ....3 364....27 200 SHort Mertuops. Simply use the factors, 4 and 8 (4 X 8 — 82). Dividing first — by 4, we get 2918, and a remainder of 3. Next, we divide by 8, and we get 364 bu. and a rem. of 6; this rem. is multiplied by the first divisor, 4, and the rem. 3 added, making 27 lbs. How many bushels in 14763 lbs. of corn ? 14763 2109 263....35 Here the factors are 7 and 8 (7 x 8 =56). Dividing by 7, we get 2109; dividing this by 8, we get 263 bu. and a rem. of 5, which is multiplied by 7, giving 35 lbs., or the complete rem. Hay. Exam. What is the cost of 1860 pounds of hay ¢ at $17.50 per ton (2000 lbs.) ? 18/60 2/325 $16|275 At 1c. per pound it is $18.60, which is $20 per ton. The dif- ference between $17.50, the given price, and $20 is $2.50 which is 4 of $20. Deducting 4 we have the price at $17.50. - Pounps to Gross Tons. To reduce pounds to gross tons we have the following simple Rule: Cut off one figure from the right of the pounds, and take a quarter, a seventh and an eighth; the result will be gross tons and BS decimal of a ton. Exam. How many gross tons in 79952 lbs. of iron ? SoL.— Cutting off one figure, and taking a quar- 7995/2 ter, 4 of that quarter, and { of that seventh, we 1998 8 285|5428 have 35.6928+ gross tons, (See exam. page 196. —35 16098 STHRLING. Pounps, SHILLINGS AND PENCE. Rue. Jo reduce shillings, pence, etc., to the decimal of a pound : Divide the pence ana farthings (having reduced the farthings to a decimal) by 12; to the result thus found prefix the shillings, and divide by 20. Exam. 1. Reduce 18s. 22d. to the decimal of a pound. Reducing $ to a decimal we have 2% =........ ar tale 2.75 mividing this’ by 12,:we have: ..i.0...5....5 wath swe .22916' Prefixing the 18s. to this result, we have ............ 18 .22916' mruing now Dy. 20, we Nave; i...» cies deinceise alec + £.9114583’ Note.—To divide by 20, move the decimal point one place to the left and take the half. Exam. 2. Reduce 19s. 103d. to the decimal of a pound. Seeeretial. , AHO LOR 1S CQUAL ow. weiss ss ares Coe eas 6% 10.5 Dividing this by 12 and prefixing 19, we get........... 19.875 mre TA LE FOS G WC TOb eo ace ars)s «oases ce 'ae,e o's s £.993875 Reverse Rote. To find the value of the decimal of a pound steriing to the nearest farthing: (1) Take % of the number expressed by the jirst two figures of the decimal for the shillings of the result. (2) Diminish the number expressed by the remainder, with the third Jigure of the decimal annewed, by sz of itself, what remains will be the Sarthings in the rest of the required value. 202 STERLING. Exam. 1. What is the value of £.99375. Taking a fifth of 99, the first two figures of the ,99375 decimal, we get 19 shillings with a remainder of 4, 19s. 104d. to which 3 (the third fig.) is annexed, making 48, this is diminished by 34, leaving 42 farthings or 104 d. Reason: .99875 is more nearly equal .994 than .993, and .994 = .95 + .044, and the value of .95 is found by multiplying by 20 and dividing by 100, or simply ¢ of 95 = 19s. Then the value of £.044 or £;44, by diminishing the denominator by s we have 960 (farthings in a pound sterling), and by diminishing the numerator 44 by »~, we get 42 nearly, hence £44, is nearly equal £& #2, or 42 farthings, Exam. 2. What is the value of £.8525. Taking 4 of 85 we get 17s. and the third figure 1%s. 04d. is 2 farthings, that is 3 or 4. To Divine Pounps, Suinuines AnD PrncE By 100. To divide pounds, shillings and pence by 100, in other words, to take 1% and consequently any per cent. of sterling money, we give the following simple RuLe.— Hor the pounds of the quotient take the pounds of the dividend, except the last two figures, which are to be divided by 5 for shillings ; from the remainder, with half the shillings annexed, reject sz part and regard what remains as farthings. Exam. Take 1% of £8947..138..8d. £89.. 9s. .64d. STERLING. £03 Here by cutting off 47, the last two figures of the pounds, we have £89; and + of 47 is 9, the shillings required, and the re- mainder is 2. This remainder with 7, the half of 14s. (because 13s. Sd. is more nearly 14s.) annexed becomes 27, from which 1 is rejected (nearly its 3), we have 26 farthings or 64d.; the answer is £89 9s. 64d. Note.—In rejecting the 1-23 always take what is nearest the true result. The reason of the process will be understood from the preceding rule. Exam. What is 4% of £89..16s..4d.? Dividing 89 by 5, we get 17, the shillings, and a remainder of 4 to which 8 (the half of £89 16s. 4d. 16) is annexed, making 48 farthings, from this 2 (nearly s,) is rejected, leaving 46 far- 17s. 114d.. things, or 114d. This is 1¢ of the given sum. Lost ls: LOG: Then 4 times this is £8 11s. 10d., or 4% of the given sum. Reason of the rule: Dividing £89 by 100, we get £0, leaving a remainder of £.89. If this be multiplied now by 20 (shillings in a pound), and the result divided by 100, we get 17s. But multiplying by 20 and dividing by 100 is the same as dividing by 5; so we simply take a fifth of £.89 for the 17 shillings of the answer. This leaves a remainder of £;4,, or £.04. Now 16 shillings — £46 or £,8,, writ- ten decimally, £.8 is still to be divided by 100; this gives £.008. We have now £.04, the remainder left in dividing 89 by 5, plus £.008, which make, when added, £.048, or, when expressed frac- tionally, £;43,. Diminishing the denominator of this fraction, now, by ss part of itself, we have 960; and diminishing the numerator also by a similar part of itself, we have 46 nearly ; hence, £,4%, is nearly equal to £,4°,, or 46 farthings, since £5351 farthing. 204 STERLING. STERLING REDUCED To AMERICAN CURRENCY. The first three figures of the decimal of a pound will be sufficient for all practical purposes in reducing Sterling to American currency; and to find those three figures, the following simple rule is given, which will be found preferable, perhaps, to bee given for finding the decimal in full, at page 201. Rue. —(1.) Take half the number of shillings for the first ‘a of the decimal, tf the shillings be even ; and if odd, half the shillings will be the first two figures of the decimal. (2.) Reduce the pence and farthings, if any, to farthings, by multiplying by 4, and vf the result consists of only one figure, set tt in the third decimal place, but if of two figures, set them in the second and third places, adding 1 ¢f the number of farthings iz between 12 and 36, and 2, if between 36 and 48. (3.) Prefix the pounds to the decimal thus found, and multiply by the rate of exchange. Exam. What must be paid in New York for a bill on London for £32 18s. 23d., the rate of exchange being P4.874 ? SoLutTion.— The shillings being even, half OPERATION. the number, or .9, is the first figure of the deci- O48 GIT mal. There are 11 farthings in 23d., and this 4.875 number forms the second and third figures of $160 .441125 the decimal. We have now £32 18s. 2?d.— £32.911, and this multiplied by $4.875 = $160.44. Notre.— The multiplication by the rate of exchange can be frequently shortened if we assume $5 asa standard rate. In that case, at $1 per pound sterling, the cost of £32 18s. 234d. or its equal, £32.911, would be $32.911; at $2, two times that, and at $5 it would be five times $32.911 or $164.555. Now multiplying by 5 is the same as multiplying by 10 and taking half, so that if we move the decimal point in £32.911 one place to the right, thus, £329.11, and take the half, we get $164.55, or the value of £32.911 sterling, im American currency, at $5 per pound sterling. And since the difference between $5 and $4.8714 is SHORT METHOD 124% c. or % of a dollar, we a the value at $4.873¢ by 329/11 subtracting x of 82 91Lor A. of 164.55 (4 of $1 being 164/55 the same as 3 |, of $5). sia a a ae ~—— STERLING. 205 Again, since the value of £32.911 at $1 per pound sterling, is $32.911, the value at 10 c. would be the tenth of that at $1, or $3.2911, and at 1 c. the value would be a tenth of the latter, or $.32911; at 2c. it would be twice $.32911; at 3% c. three and a half times $.32911, etc.; so that, had the rate of exchange been $4.88, $4.89¥, or $4.91, we would simply add to $160.44 (the value at $4.8734), once $.32911, or 33 c.; twice $.32911, or three and a half times $.32911 ; 1,2and 3¥% being the differences between $4.8734 and the three mentioned rates. If the rates were less than $4.8744, we would, of course, deduct the difference. To illustrate further this simple method let us take the following: Exam. What is the value of £612 17s. 9d. at $4.86 2 SoLuTION. — Taking half the 17s. decimally, we have .85 for the first two figures; 4 times 9d.— 36 farthings and 1 make 37 to be set in the second and third places, making ... .. Ret PeOOk the three decimal figures required. We have now £612.887 to be multiplied by the rate, $4.86} which gives the required value, in American currency. SHorT METHOD, Moving the point in £612.887, or, which is the 6128/87 same in effect, drawing the vertical line one 80641435 _ place to the right, and taking half, we have ie ak $3064.435, the value at $5. Then 7, of this S064 (simply a fourth set in proper position) is $76.61, 11539 which being deducted from $8064.435, would — WH |——"_ give the value at $4.874. But the rate is $4.86, $2983|229 or a difference of } of a cent, which is also to be deducted, 3 =} and +; half of $6.128 (value at lc.) is $3.064, and half of this is $1.532, the value at 1 of a cent. Taking these three results now from $3064.435, we have $29838.229, the value at $4.86}. Nortr.—In subtracting the three items from $3064.435, the process is per- formed by addition,-thus 2 and 4 are 6, and 9 (setting down 9 at the bottom), are 15 (the top figure); carry 1 to 3; 4 and 6 are 10, and 1, are 11 and 2 at bottom, are 13, etc. 506 STERLING. Or we might proceed as follows: Suppose the rate — had been $4.883, what is the value of £612 17s. 9d.? SoLution.— The difference between $5 (our standard rate) and $4.883, is 113 cents. £612 17s. 9d.= £612.887, and at $1 per £1,the value= $612.887 And at $5 per pound aie the Values 2, ce.e ea At 10c. per £1, the value is ;4, of that at $1.. = 61/288 And at 1c., the value is +5 of int at 100,27 seen 6/128 At éc., or its equal, tc., the value is half 1le.. ... = 3/064 And {ce. is a quarter of $, or } of Ic..... eer 766 $2993] 189 These four items make up the value at 113c., and deducting them from $5, or rather from its value, $3064.435 (by the method of addi- ee ea ee a a ae ee tion, as pointed out in the foregoing note), we get $2993.189, the © : value at $4.882. The reason for taking half the shillings, etc., to find the first three figures of the decimal, will be understood from the following: 16s. =£48, 55%, ifs, or -3, all representing the same value differently expressed: therefore 17s. = £47 or 785)... 0.2.2... wee eee ae And 9d. reduced to farthings =£,25,; now if we add to the de- nominator 950, +1 =z Of itself, or 40, we get 1000, and by increas- ing the numerator 86, by a similar part of itself, we make it 37 nearly ; so that £35, is nearly =£335 .-: sees eeees o eye sey See Consequently lis. 9d. expressed decimally...... ........ .... = .887 In reducing 9d. we simply say 4 times 9 are 386, and 1 are 37; and lis to be always added when the number of farthings is nearest to 24, or when the number is between 12 and 36, as stated in the rule. InrerEst on STERLING. Rute I. To find the interest of any sum for 1 year, at any rate per cent: Multiply the principal by the rate, and divide by 100. Exam. 1. What is the interest of £1 sterling for 1 year at 5% per annum 4 a ee ee STERLING. 207 SoLtution.— Multiplying £1 by 5, the rate, and dividing by 100, we have £;3,, or £.05, for the interest. But £35 = £55, or 1 shilling, or its equal, 12 pence; and since the interest of £1 for 1 year, or 12 months, at 54%, is 12 pence, the interest of £1 for 1 month is one-twelfth of 12 pence or 1 penny, and consequently the interest for 2 mos. is 2 pence; 3 mos. 3 pence, etc. Hence the following: Rue Il. To find the interest of any number of pounds sterling for a given number of months, at 5% per annum: Take the pounds as pence and multiply by the months. Exam. 2. What is the interest of £42 for 7 mos. at 5%? SoLuTion.— Calling 42 pounds 42 pence, or 3s. 6d. and multiply- ing by 7, the number of months, we get £1 4s. 6d. the interest. So likewise the interest of £42.. 10s. for 8 months at 52%, is 424 pence, or 3s. 63d. X 8= £1.. 8s. 4d. | And from this the interest at any other rate may be easily derived. Thus, to find the interest at 4%, simply subtract from £1 8s. 4d. one- fifth of itself, or 5s. 8d., and we have £1 2s. 8d., the interest of £42 10s. for 8 mos. at 4%. Had the rate been 6% we should have added one-fifth of the interest at 5%; or we might employ the following: Rue Ill. To find the interest for months at 6% per annum: Multiply the principal by half the number of months, and divide the result by 100. Exam. 1. What is the interest of £439 16s. 8d. for 1 year and 8 months, at 6% per annum ? OPERATION. SoLutTrion.— 1 y. 8 mos.= 20 mos. £439 16s. 8cde Multiplying £439 16s. 8d. by 10, half of 20 mos., 10 we get £4398 6s, 8d., and dividing this by 100, - £4398 6s. Sd. we get £43 19s. 8d., the interest required. £43 19s. 8d. To divide by 100, we set down £43, leaving a remainder of 98. Taking one-fifth of 98, we get 19s. and a remainder of 3. To this remainder we annex half of 6s., or 3, making 33 from which & is rejected, leaving 32 farthings, or 8i., nearly. (For short method of dividng by 100, see rule page 202; also reason of rule page 208.) 208 STERLING. Exam. 2. What is the interest of £756 14s. 10d. for 5 months, at 54% per annum. This example is solved by both the 5% and 6¢ rules as follows ; First Method, 5 per cent rule. SOLUTION. — £756... 14s. 10d... ese eee 7563d. nearly Multiplying this by 5, the number of mos. we have 37833d. Dividing this by 12 (pence in a shiliing) we have.. d1ds. 33d. Dividing the latter by 20 (shillings in £1) we have £15 1ds. 33d. the interest at 5%. Now there are 20 quarters, or A4ths in 5%; therefore 1 quarter, or 32 is 54, of 52. Dividing £15 15s. 38d. by 20 we have............. 15s. 94d. Which is added making the required interest..... £16 11s. 1d. Second Method, 6 per cent rule. Multiplying £756 14s. 10d. by 5, we have ........ £38783 14s. 2d. And dividing this result by 2, we have........... £1891 17s. 1d. This multiplies by 24, or by half of 5 mos..... Dividing this by 100 (short method), we have..... £18 18s. 44d. Which is the interest at 62. The difference between 6% and 54 is 3%. There are 24 quarters in 6%; therefore 3 is 4 of 24 quarters,-or of 6%. Taking 4 of £18 18s. 44d., we get the interest at 2%....... -...ceedscorees = £2 Ws) Brae Which is taken from the interest at 6%, leaving ... £16 11s. 1d. the required interest. Rute tv. To find the interest of any sum for any number of days at any rate per cent: (1.) Reduce the shillings and pence, if any, to a decimal by the rule given at page 204. (2.) Multiply the principal by double the rate, and the result by the number of days, cut off five figures from the right, counting always from the decimal point, and add «a third, a tenth of that third and a tenth of that tenth. Nots.— In computing interest on sterling for days, the basis is 365 days to the year. STERLING. 209 Exam. 1. What is the interest of £648 15s. 3d. from June 2 to Nov. 25, at 54 per annum ? SoLutron.— There are 176 days from June 2 to Nov. 25, Reducing 15s. 8d. to a decimal, we have the principal = £648.762 Multiplying this by 10, double the rate, then by 176, Ne CR lass A layne ek Fifi oes os «ens 2 vba sce £11/41821.12 Cutting off five owes from the ee and taking} ee et Gate Wie iy Gi oe oe ae eaye'e Ge vis-e oh scene ne 3|80607 Taking a tenth of this third, we get.......c.csecee 38060 And taking a tenth of this tenth, we get ........... ° 3806 Adding the four results now, we get .........s..e06 ~ £15 64294 150 Rejecting 10 times 15, or 150 from .64294, we get.... 64144 _ Finally, the value of £.64144 is 12s. 10d. nearly, 12s. 10d making the required interest £15 12s. 10d. (See rule, page 201.) Exam. 2. What is the interest of £8000 for 75 days at 4% 2 SOLUTION.— £8000 K 600... ..-. ccc ccc eceee = £48/00000 Multiplying the principal by 8, and by 75, or by 600 16 (75 X 8), at once, we get £4800000. Cutting off five 116 places to the right, and adding a third, a tenth of 16 that third and a tenth of that tenth, we get £65.76000. —@pelsannn Rejecting 10 times 65, or 650 from this, we have £65)76000 £65.7535. Then taking one-fifth of .75, we get 15s., 650 and the third figure of the decimal now is simply 7535 3d., making the interest £65.. 15s. 03d. lds. 02d. (See General Itule, page 178 ; also reason of the rule, page 179.) OruER SHort Meruops. The six per cent rule of interest already explained in this work, can be applied to the computation of other matters, such as, Wheat, Clover Seed, Potatoes, etc., where the number of pounds to the bushel is 60. 210 ; Orner SHorr. MEtTuHops. Exam. What is the interest of $1860 for 1 y. 9m. 5d. at 6%? SoLuTion.— In 1 y. 9mo. 5d., there are 635 days. The interest of $1860 for 600 days....... Se = $186)0 Half $18.60 (60 days’ int.) is the int. for 830 days... = 9/380 One-sixth of $9.30 (80 days’ int.) is 5 days.... ... = 1/55 Making the interest of $1860 for 685 days... .... = Bi0G|s8 Or, taking the dollars for the days and the days for the dollars, we have the interest of $635 for 1860 days, thus: SoLuTIon.— The interest of $635 for 600 days. = $63)5 Three times 600 days’ int. is the int. for 1800 days. = ae One-tenth of the top line, or $63.50 is 60 days.... = 6/35 And the two results, when added, is the int. for —_——/|— 1860 days os... acs cwat, es hies una er os $196/85 (See example, page 181). Suppose now we change the foregoing problem into the Sollowing : Exam. What is the price of 1860 lbs. of clover seed, at $6.35 per bushel (60 lbs.) ¢ SoLuTron.— At 1c. per lb., or at 60c. per bushel, the price of © 1860 lbs. would be 1860c. or $18.60. And the price of 1860 lbs. at $6, or 600c. per bushel= $186|0 At 30c. per bushel, it is half the price at 60c. or halt $19/60.0.0 00 wna ee ee = 9/30 And at 5c. the price is one- an of 30, or of $9.30..= 155 Making the price of 1860 lbs. at $6.35 ..... pears = $196 85 The same as was found by the first solution of the interest problem. And uf the problem be reversed so as to read: 685 lbs. of clover seed. at $18.€0, or 1860c. per bushel, it can be solved same as the second solution of the interest problem. The student will please try it. OtTxeER SHort MeErsops. 911 W HEAT. Exam. What is the cost of 16940 lbs. of wheat at 87ic. per bushel (60 lbs.) ? SOLUTION. — Cutting off two figures from 16940 gives the cost at 1c. per pound, or at 60c. per bushel... ....... = $169\40 One-third of the cost at 60c. gives the cost at at 20c.= 56I47 At 6c. the cost is a tenth of the top line, or of 60c...= 16/94 At 1c. the cost is a sixth of that at 6c.............. = 9/82 And at 4c. it is the eighth part of 1c .............. = 35 Making the total cost i,0¢iii les es $245 198 Porartogs. Exam. What is the cost of 749 lbs. of potatoes at 474c. per bushel (60 Ibs.) ? Sou.— Cost at 60c. .. II $7.49 | Or thus: Tan eee rs 3 >, | Cost at 6c. per bushel.= _—$.749 trans 3 15K, — 1.87 | And 8 times 6c.=48c..= $5.992 ¢ of PCa aac...) — dl | Less, of .749= jJc.= 62 ee Alize ...= $0.93 | And the cost at 47ic.= $5.93 Hints on Inrersst. Savings banks allow interest on deposits for a certain fixed term, generally 3 mos. or 6 mos., and calculations may be simplified in many cases by multiplying the time and rate together and dvniding the result by i, which _ will give the rate for the given time, thus: “ee Exam. What is the interest of $872 for 3 mos..at 4% per annum ? Sovurion. —*<* = 1%, and 1% of $872 = $8.72, the interest; the process being, of bins performed mentally . 912 | Hints on Interest. And if the rate were 34% per annum, we would deduct one-eighth of itself from $8.72, and the remainder is the interest for 3 mos. at 342. Exam. What is the interest of $648 for 3 mos. at 334 per annum ? SoLuTion.— The interest at 4% is simply 1% of $648.... = $6.48 And at 4%, the difference between 47% and 34%, itis} of4%. = .81 Making the interest for 3 mos. at 8334 per annum........ $5.67 Again, the interest for 3 mos. at 2% per annum, is half that at 4¢; at 5¢ it is 4% plus + and at 8% it is twice 4%, etc. And the same rule will be found to hold good in many other cases. it Exam. What is the interest of $648 for 1 year and 5 mos. at 7% per annnm ? SoLUTION.— 1 y. 5 mo.=17 mo.; 9147-102 less +. The interest of $648, at 17 = $6.48, and at 102 it is... $64.80 From this we deduct 54, of 1% or of $6.48, equal...... , 4 Leaving the interest for 1 y. 5 mos., at 7% per annum, $64.26 And so with other problems of a similar nature.. Montaty Payments, 6%. When interest is to be calculated on monthly payments, we have the following simple | RULE. (1) Add 1 to the number of months. (2) Multiply the sum by half the number of months, and the result by the payment, and divide by 2. (8) Lf the payment be in dollars, point off two decimal places, and if there are cents, point of four places. Exam. Suppose a person to pay into a Building and Loan Association, $3 a month for 15 months; what is the amount due at the end of the time at 6% interest ? Hints on INTEREST. 213 SoLUTION.— 15 + 1 = 16; then mSEesS, or OES? $1. 80 And adding the principal, $3 K 15 ............. a 45.00 We have the amount, that is, the prin. and int.... = $46.80 And for WEEKLY Payments 6%. We have the following Route, (1) Add 1 to the number of weeks. (2) Multiply the sum by half the number of weeks, by 7 and by the weekly payment, and divide by 6. (8) If the payment be in dollars, point off three decimal places, if there are cents, five places. Exam. Suppose a person to pay in 40c. a week for 10 weeks; what is the amount due at the end of the time at 6% interest ? SoLuTion. —10-+ 1=11; then -2*47*-.... = & 032 And adding the principal, 50c. x 10........ a RE 5.00 Dweenave tho AMOUNt........sser-eee es Mee eee we ees $5 .032 And from this the interest at any rate other than 6% can be found by aliquot parts, as has been already pointed out. Reason of the rule: If the payments were $1 per week, and the problem solved by the usual method, we sliould have multiplied $1 by 10; $1 by 9; $1 by 8, and so on down to the last payment; and adding the results, we get $55, the principal for 1 week. But we see that by adding 1 to 10, and multi- plying by 5, we get $55 more readily; and 7 times $55 or $385 is the principal for 1 day, at $1 payments. Now the payment is 50c.; multiplying $385 by .50, we get $192.50, the principal for 1 day at 50c. payments. And the interest of $192.50 for 1 day at 6%,is found by multiplying by 1 day, pointing off three figures from the decimal point, and dividing by 6, and we have .1920+6= .03208 +, or .032, as shown in the example, for the interest. The same line of reasoning is applicable to the monthly rule. Multiplying $3 by 15, 14, 13,12, and so on down tothe last payment, and adding the results: we get $360, the principal for 1 month. But this is more readily found by adding 1to15, and multiplying by 7¥, or, which is the saine in effect, 15 by 8, the half of 16; then by 3, and we have $360. The interest of $360 for 2 mos. at 6% is simply 1% or $3.60; and for 1 month it is half, or $1.80, hence the reason for dividing by 2. 914 Hints on INTEREST. A SmpLe Mernop. To find the face of a note, the proceeds being given: Rup. (1) Find the interest of the proceeds for the given time and vate. (2) Find the interest of that interest, and so on, till the interest ts so small as not to affect the result. (3) Add the interest thus found to the proceeds, and the sum is the face of the note. — Exam. For what sum must a note be drawn at 3 mos. to net $2500 when discounted at 6%. SoLUTION.— $25.00 is the interest for 2 mos. on........ $2500.00 12.50 . ‘s- ToTO; Making. . . $37.50 a << 3 mos. The int. of $37.50 for 2 mos. = 88c. “ 1 m6. 19¢. Making the int. for. 3 mos. = 5%7c.; and the whole int.= 38.07 Adding $38.07 interest to the proceeds, we get.......... $2538 . 07 the face of the note. Norte.— The days of grace being abolished in the State of New York, are not taken into account. If grace be allowed, add the interest for the days of grace. : Proof: The interest, or bank discount, for’ mos. at6Zon $2538.07 (the sum for which the note is drawn), is found to be 08.07 and deducting this from $2538.07, we have the net proceeds. $2500. Exam. 2. Having discounted a note in bank, I am credited with $1500 as the proceeds, what was the face _ of the note, the rate being 4% and the time 90 days? SOLUTION. —The interest for 90 days, or 8 mos. at 4% on $1500 is 1% or $15; and on $15, the int. is 15c. making... .... 15.15 and adding this tu the proceeds, we have the face....... $1515.15 Proof: The interest of $1515.15 for 3 mos. at 4% is 1% or 15.15 and deducting this from the face, we have the proceeds.. $1500.00 Pt oe Ee Hints oN INTEREST. — a We) From the foregoing hints, we derive the following simple rule for computing InreREstT FoR Monrtus. Rute. 70 find the interest for months at any rate per cent: Mul- tiply the principal by the product of the months and rate, and divide by 12. Or, multiply by the months, then by the rate, and take haif the result, and one-sixth of that half. Or, multiply the months and rate together and divide by 12; the result is the rate for the given time ; multiply the principal by this rate. Exam. What is the interest of $480 for 5 mos. at 5%? 480 x 25 $480 K5 x 5 SoLuTIon.— BAS 2 $10.; Or, — = $10. Or, 5 times 5 = 25, divide by 12 = 257; then $480 x 2, = $10. Or, because $480, the principal, is divisible by 12, we have 40 x 25 = $10, the required interest. This rule, after a little practice, is so simple that, in numerous cases it will not be necessary to use pencil or paper in computing interest, thus: What is the interest of $72 for 8 mos. at 344? Ans. 8 times 3} is 28; then 28 multiplied by 6 (42) = $1.68 the interest. Or, 28 + 12= 23, and $72 x 24% = $1.68; and if there be YEARS AND MoNTHS Reduce the years to months and proceed according to the rule. Note. —In conclusion, it may be well to remark that, in relation to the foregoing rule, the rate may be changed for the time and the time for the rate, and very often to great advantage; thus, the interest of any sum for3 mos. at 4% is the same as for 4 mos at 3%; for 6 mos. at 5%, the same as for 5 mos. at 6%; for 3 mos. ana 15 days at 3%, the same as for 3 mos. at 336%, 8 mos. 15 days being equal to 344 mos.; etc. Suppose it were required to find the interest of $320 for 3 mos. and 15 days, at 3%. Tosolve this, we reverse the problem so astoread: $320 for 3 mos. at 34%. Then the interest for 3 mos. at 4% on $320, is simply 1%.. ........... $3.20 and at %% itis an eighth of 4%, or 40c. deducted.................. 40 giving the interest for 3 mos. and 15 days at 3%........... Siscacesdne: €o\ Oas0U INTEREST SIMPLIFIED. If 360 days be divided by the rate per cent, the interest of any sum of money for the number of days thus found, at the given rate, will be equal to one per cent of the principal. Thus if the rate be 44%; dividing 360 by 44 gives 80 days; now the interest of any sum, say $1768 for 80 days, at 444, is found to be 1% of that sum, or $17.68; and for 8 days the interest is one-tenth of the latter, or $1.768; for 800 days, the interest is ten times $17.68, or $176.8, and for 8000 days it is ten times the last interest, or $1768. In other words, the interest of any principal for 8000 days, at 442, is 100 per cent, or equal to the given principal. Hence we have at sight: The interest of $1768 for 8000 days, at 447=$1768. ce 6é 800 66 cé = 176.8 66 ce 80 ce ce = 17.68 ¢é “cc 8 €é ce = 1.768 And the interst for 1 day: i! Serer And from this simple basis the interest for any number of days is easily found. An example will make this clear. Exam. What is the interest of $3765.47 for 92 days at 44%? Here we have at sight, $37.6547 80 days’ interest; Next, we have 3.76547= 8 “ And half of this, or 1.88273= 4 < Making $43.3029 —92 « IntTEREST SIMPLIFIED. 217 From the foregoing illustrations and examples the following will be readily understood : “Rate. 360 days. 4% 90 ‘ hence, 9000, 900, 90, 9and 1 day. 44g 80 ** Sram SOUO ABOU; 80,78 Foal Bigewee sos | SES TZ00, TAU A725 Bees 6% ee a ee GO00 7600760,-6 < and by 6x 40x54 x60x2 as 2 x 60. cancellation Wax7x5 Net Tons or Raits to THE Mite. Rue. — To find the number of net tons of rails to a mile of rail- road: Multiply the number 1.76 by the number of pounds of rail to the yard. Exam. How many net tons of rails to a mile of road at 60 pounds to the yard ? Here we have 1.76 * 60 = 105.60 net tons. Reason. — The process in full would be 1760 yds. x 60 x 2, di- vided by 2000; or WX O*? = THEO S' = 1.76 x 60. There are 1760 lineal yards toa statute mile, or, 8 fur, X 40 per. x 54 yds. = 1760 yds. 232 Discounts. TRADE Discounts; SHort MeErnops. Manufacturers and wholesale dealers usually allow to the trade or retail dealers, a reduction from the fixed or list prices of some kinds of merchandise. This reduction is called a discount, or a trade dis- count. In some lines of business several discounts are allowed, Discounts are taken off in succession. Thus, 25%, 15% and 10% off; or, as it is generally expressed in business, 25, 15 and 10 off, means first a discount of 25%, then 15% of what is left, and finally, 10% of the remainder. Norts. — The profit on goods is less when 25, 15 and 10 are allowed, than if 50 per cent were allowed. It is immateria] in what order the discounts are taken as it will not affect the result. Exam. 1. If goods be listed at $65 with 40% off, what is the net cost ? Usual Method. Short Method. $65 X< .40 = $26 $65 x .60 = $39 net. then 65 — 26 — $39 Instead of multiplying the list-price, $65, by .40, and deducting the result, it is much shorter to multiply by .60, the difference between 1.00 and .40, which gives the net at once ; or, as it is said in the trade 40 off is 60 on. And this short rule will hold good for any series of discounts. Nors. — Since any per cent. is some number of hundredths, it is properly ex- pressed by a decimal fraction ; thus 5 per cent. = 5% = .05. It is scarcely necessary to remark that the _list-price represents 100 per cent. 100 =100% = ad =1.00=1. Now, if from this we deduct 40% = .40, the difference will be 60% = .60 ; so we multiply the list-price by .60 straight to get 40% off. 3 x q Discounts. 935 TrapvE Discounts; SuHort Meruops. Exam. If guods be listed at $3 with 30, 20 and 10 off, what is the net cost? $3 4 Or thus: 7X 8X .9 =.504 2 1 3 8 $1.512 1.68 9 $1.512 Here we deduct, at sight, 30, 20 and 10, each, from 1002 getting 70%, 80% and 90%; or, when decimally expressed, .70, .80 and .90; or, .7, .8and .9 (the ciphers not affecting the significant figures in the decimal expressions), Multiplying by .7, .8and .9 in succession, we get $1.512 the net. Or, multiplying .7, .8 and .9 together, we get .504 = 50;4% which is the net discount equal to 380, 20 and 10; and multiplying this net by the list price we get the net cost. Norte 1. To find the net rate of discount equal to several rates: Deduct each rate from 100, multiply the differences together and the product is the net rate. Thus, 60%, 25% aud 10% off, is equal to 27% net, or to 73% off (.40 x .75 x .90 = .27); or, (.4 x .73¢ x .9 =.27). Nore 2. It should be carefully borne in mind that the cipher, or zero, having no value, is used in combinations of figures to fill places where no value is to be expressed, and thus to make the other figures occupy those places in which they will express the intended values. Hence, a cipher will not affect the value of a number unless it be placed between some significant figure and the decimal point. Thus, .7, .70, .700, .7000, etc., are all equal in value. But .07, .007, .0007, are entirely different, the local value of 7 being changed by the cipher, or ciphers coming between that figure and the decimal point. Briefly, then, the use of the cipher is to keep the significant figures in proper position with reference to the decimal point. The use of the decimal point is to mark the place of units; and whether ex- pressed or understood, its position is always to the right of units. Itis a matter of the utmost importance to have a correct knowledge of making a proper use of the decimal point in our calculations. 934 Discounts. TrapE Discounts; SHort Meruops. Given the net cost and the discount to find the list-price. RuLE. — Divide the net cost by the net discount. Exam. If the net cost of goods be $1.512, and the dis- counts are 30, 20 and 10 off, what is the list-price ? Multiplying .7, .8 and .9 together we get .504, the net discount. Then $1.512 + .504; or, moving the decimal point three places to the right in these numbers, in other words, multiplying each by 1000, to throw off the decimals, we have $1512 + 504 = $3, the list-price. Reason. — Since the net cost is found by multiplying the list-price by the net discount, we simply reverse the rule here, viz.; divide the net cost by the net discount to find the list-price. OtrueR Snort Meruops. In marking goods, merchants generally take a rate per cent. that is an aliquot part of 100, as 50, 334, 25, 124, 84, etc.; and instead of multiplying by the net discounts, in such cases, to find the net cost, it will be found preferable to make use of the method of aliquot parts, as in the following Exam. If goods be listed at $240 with 334, 25 and 5 off; what is the net cost ? Here, instead of deducting 334, 25 and 5, each, from 100, $240 and multiplying by the net discounts, .662, .75 and .95, we 80 take +, + and =, off in succession (834 = 4 of 100; 26=4 160 and 5 = 4), thus getting the net cost, $114. 40 120 6 $114 Notr.— To find the aliquot parts of 100 divide it by 2, 3, 4, 5, ete. Thus, 100 + 6 = 16%, and 16% is 1-6 of 100, etc. Discounts. | 935 Trape Discounts; SHorrt Meruops. Odd rates of discount will be found equally simple in many cases where the inexperienced calculator has to make use of long methods. Exam. If goods be listed at $148 with 474 off; what is the net ? Here instead of multiplying by .524, we say 50 is a half, $148 and 24 is 4, of 50, or 7, of 100; and adding, we have mul- 74 tiplied by .524. 3.70 $77.70 Nore. — And if the rate were 52) off, instead of 4714, the same figuring would answer, only instead of adding 3.70 we deduct (100— 471% = 6214); (50 + 264 = 5236) ; (50—23¢ = 4714). And if the rate were 3714, we would say 25= \& and 124% = 1% of 25 (25 + 1256 = 3744); 27% off would be 25= ¥ of 100 and 2¥ a tenth of 25; and so of other rates. | Exam. If goods be listed at $420 with 65% off; what is the net ? Here we say 65 off equals 35 on and multiplying $420 by $42.0 .85 we get $147 net. 105. $147 For short method we take 10 and 25; 10% = $42, at sight, and 254 = + of 420, or 105 which is added in proper position, one place to the left. , To make the process clearer it must be borne in mind that $420 SPREE AMIGA LF Gh) Gih 2 otal custo Stal eis min wale ay C6-See Mx once = $420 eer OE LOS 5. 5 0c's/e s waina a « Aah EEO ie eee = 105 Ree ite EE LOU. os.d5 a0 Valea vax hese seetc gee ececess = 42 Making 354 (25 + 10); or (10 + 25 = 85). ...ccceeereee = $147 236 Discounts. TrapE Discounts; SHort Meruops. Exam. What is the net cost of goods which are listed at $6.40 with a discount of 624% off? Here we have 100 — 623 = 374, the net $6.40 discount. We now say 25=}.......... e ebetence = Pele and 124 = 4 of 25, or.40f 100 2... 6.24.04 eee 80 Adding both results we have multiplied by 374 $2.40 Or, simply take 4 of $6.40 = 80c. and three times 80c. gives $2.40, the net cost, 874 being 3 of 100. Norsz.—If the rate were 371% off, then the net would be 6214, and to multiply by 6244 we would say 50 = 4, and 124=14 of 50. Or, take 14 and multiply by 5, 6214 being 5 of 100. Exam. If goods be listed at $54 with 674 off; what is the net cost ? Here 100 — 673} = 8234, the net discount $54 X .34 which is equal to 80 + 24. To multiply by .382} $16.2 we multiply first by .30, and for 24 we add 1.35 gy of $54, or 4 of $5.4 ($5.40), 24 being 2, $17.55 of 100, or + of 10. Or, by looking on .32} as .34, simply multiply by .384 bearing in mind that the quarter in such cases is in reality 7, of the list price, or } of its tenth. Norre.—If the rate were 3214; then the net would be 674, and to multiply by .67144 we would multiply first by .60, and to the result add its 14, 714 being & of 60. Or, take 50, 5 and 1214 (50 + 5 + 12446 = 67.) And other odd rates of discount will be found equally simple. If the rate were 5714 off, for instance, the net would be .4244. To multiply by .424¢ look upon it as .414, bearing in mind that the quarter is 7, asin the foregoing exam- ple. If the net were 5734 we would resolve it into 50, 5 and 24; 50=4;5= 5 of 50, and 244 = 1% of 5. pied wea. Discounts. 237 TravE Discounts; SHort Merruops. Exam. If goods be listed at $80 with 40, 10, 10, 5, 5, 74 and 3 off; what is the net cost ? Since the rates of discount may be taken off in any order without affecting the result, we begin here with 7}.and find the net rate as follows : Deducting 74 from 100 we get 100 73 924, and express it decimally, .925 44-Oft;) ==. 925 2775 To get 3% off we simply set 3 times oes = eee reo 4486 .925 two places to the right and sub- Ga Soe BHO 4261 tract. Then 34, of the result is sub- 5 * = 80978 1D ae ese tracted, and 34, of what is left re- 10. Goda 6 40 ‘© = .3938558x80 spectively, for the two 5’s, giving $31 .48464 .80978. Next, 10% is taken off by subtracting each left hand figure from the one immediately to the right, beginning with the tens, or second figure in .80978, thus: 7 from 8 leaves 1; 9 from 17 leaves 8; carry 1 to 0; 1 from 9 leaves 8; 8 from 10 leaves 2; 1 tocarry taken from 8 leaves 7, the result is .72881. And from this number the second 10% is got in like man- ner, giving 65593. Then 40% is taken off by multiplying .65593 by .6, the net equivalent to .40; the product, .393558 is the net rate. This is now multiplied by 80, the list price, and we get $31.48 +, the net cost. Nots.—The net rate, .393558 is nearly equal to .39;°,5,, or 393%, and is the product of .9214 x .97 x .95 x .95 x .90 x .90 x .60, the multiplication being per- formed much more simply by the process given above. Ry taking 3954,% from 100 we get 603%% which is the rate of discount off equivalent to the seven rates, 40, two 10's, two 5’s, 744 and 3 off. (See page 238, also note 1, same page.) INTEREST REVIEWED. Since the interest of $100 for1 yearat1% = a ae | the interest of $100 for 100 years at. 1% $100. in other words, money doubles, or the interest ) will equal the principal in 100 years, at 1%, simple interest. If, then, 100 years be divided by the rate per cent. it will give the time, in years, when money doubles at that rate, simple interest. Thus, any sum of money at 2%, will double itself in 50 years; at 4%, in 25 years; 5%, in 20 years; 6%, 162 years, &c. Hence, taking 100 years and 14 for the basis, and reducing the 100 years to months and days, we have the time in which money doubles at 1% = 36000 days = 1200 mo. = 100 years. “A fe 6° 2:9 -= 18000:°.*. = - 600 eee ws es “© 247 = 16000 ‘“*. = 60355 ee ee cs ‘63% = 12000. -** 2 540023 ree & ff “ 38¢ = 9600 ‘“ = 820) 5 ee a a “44% $9000 -“ = 3000 ee &¢ As ‘| 44¢ = 8000“: = (2662 ae * a ‘5G = .7200- 8. == 22a e fs £5 666.9 = 6000: . >= 200) i See ae aS 98% ==. 4800 =. £ se 2 AGN ee Be af "8 +=. 4000. 2% == 100 eee ‘s ef ‘94 = 4000.) “ + =).130) See = . © 10:'9:=.. 8600) “te 120 Sea ue “f "12-4 = -3000.. 4%" = 100 te ee &e. &e. &c. &e. Nore.—-It will be seen, on examining the foregoing, that the time is pro- portional to the rate, and vice versa; thus, the interest of any sum of money for 18000 days at 2%, is equal to the interest of that snm for 9000 days at 47; the interest at 9% for 4000 days, is equal to that at 44%% for 8000 days, or 2144, for 16000 days; and the interest of any sum for 4800 days at 714%, is equal to the interest for 9600 days at 3%%, &c Interest REVIEWED. 939 A careful analysis of the matter given on the foregoing page, will now enable us to compute interest at any rate, and for any time, with ease and rapidity. Todoso, we must keep in view the basis, or the time in which money doubles at 1%, simple interest, viz., 36000 days, 1200 mo. or 100 years. Take any particular rate, say 24%, for example, and suppose it were required to find the interest of $5764.50 for 16 days, at that rate. . Here we simply point off three $5 7645 figures to the right, counting from the decimal point, and we get $5.7645, the required interest, true to four places of decimals. Reason. —Since money doubles in 36000 da. at 1%, simple interest, it will double at 214% in 16000 da.; 214 being contained 16000 times in 36000. Hence, The interest of $5764.50 for 16000 da. at 214% = $5764.50 one tenth of this is the interest for 1600 ‘‘ os = $576.45 one tenth of the latter is the int. for L6G 2s ne = $57 645 and a tenth of this last is the int. for ‘Gass ey = $5. 7645 We have here, now, a basis to find the interest for any number of ‘days, at Q1¢, If 40 days, take + of 160 days’ interest, or of $57.646; the interest is $14.41; if 80 days, the interest will be half of $57.645, or $28.82 ; if 32 days, it will be two times 16 days’ int. or $11.529; if 10 days’ interest be required, take 4 of 160 days’ interest, then 4 of that half; or, take 1 of 160, then 1 of that fourth ; the result will be 10 days’ interest. If the time be 8 mo. 10 da. or 100 days, we have 6 times 16 da. equal 96, plus 4 da., or} of 16; and if 4 mo. or 120 da., we have 160 da. less a quarter of that, or 40 days’ interest, the difference will be 4 mo., or 120 days’ interest, &c. And if one day’s interest be required at 24%: point off three figures from the right, counting from the decimal point, the result is 16 days’ interest always. Then take 4 and } of that $ (2x8=16.) or, take } and } of that } (4x4=16.) 940 IntTEREST REVIEWED. Again, suppose it were required to find the interest on a loan, say, of $300,000 for 1 day, at 334. By referring to page 238, we see that 3? is contained 9600 times in 36000 days. ~ Now, since the interest will equal the principal in 9600 days, at 33%, simple interest, always : We have, here, at sight, the interest for 9600 da. = $300000 one tenth of this, or the interest for 960 “* = .$30000 one tenth of the latter, or the interest for 96 ‘* = $3000 and x, of this last is the interest for a $31.25 Hence, RuLE.— To find the interest of any sum for 1 day, at 38%: Move the decimal point two places to the left, in other words, take 1% of the principal, and the result is 96 days’ interest, always. Divide this by 96 for 1 day’s interest. Exam. What is the interest of $376860.48 for 1 day at 33% ? Taking 1% of the principal we have $3768 .6045 = 96 da. int. one twelfth of $3768.6048 gives $314.0504 = 8‘ * and one eighth of $314.0504 gives $29. 2563 =" 1" Notrs.—1. It will be observed that, in dividing by 96, to get 1 day’s interest, we have made use of 12 and 8, the component factors of that number — (12 x 8= 96.) And in all such cases where the divisor can be readily factored this method of division is to be preferred. Thus, to get 1 day’s interest at 714¢, we would divide 1% of the principal by 48 (the significant figures in 4800 days, or the time in which money doubles at 7147, simple interest), making use of the factors, 6 and 8; and if 5%, we would divide 1% of the principal by 72 (the sig- nificant figures in 7200 days, the time in which money doubles at 5%, simple in- terest), using § and 9, or 6 and 12, the factors of 72, &c. 2. If we examine carefully the foregoing example, it will be readily seen how easily the interest for any number of days may be obtained at 334%. If the time were 48 days, for instance, instead of multiplying 1 day’s interest by 48, we would simply take 44 of 96 days’ interest, or 6 times 8 days’ interest. 32 days’ interest is 4 of 96 days, or 4times 8 days’ interest; 33 days would be 82 plus 1 day, and 81 days would be 32 minus 1 day. By adding 1 day and 8 days’ interest we have 9 days, and by deducting 1 from 8 we have 7 days, &c. Interest REVIEWED. 241 It will now be seen that, when the rate is an exact divisor of 36000 days, or 1200 mo., the computation of interest can be made simple and interesting. And this being properly understood, com- putations will be foun‘ equally simple when the rate is not an exact divisor. Take for instance, the following Exam. What is the interest of $75684 for 1 day, at 23% ? Since 234 is not.an exact divisor of 36000, we take 8% which is contained 12000 times in 36000 ; and since the interest of any sum for 12000 days at 3%, is equal to the principal, the interest for 12 days is =,)55 of the principal; and this is found by simply moving the decimal point three places to the left in the given principal always. Here, then, we have 12 days’ interest.......... eo $75 . 684. and +, of this is the interest for 1 day at 3%........ = %6. 307 Now, 37 is 7 of 3%, and we deduct 7y............. = 525 making the interest for 1 day at 23¢..... ..... ee — $5. 782 Note.—If the rate were 344%, we would, of course, add as instead of subtract- ing ; and if 314% then } of 3% would be added, &% being 4 of 3, &c. Hence, Rue. Zo find the interest of any sum for 1 day, at any rate per cent. : Divide 86000 cays by the rate, if it be an exact divisor, divide the yiven principal by the result, and the quotient is 1 day’s interest. Tf the rate be not an exact divisor, take the nearest rate which is an exact divisor, find the interest at that rate, and add, or subtract, for the difference. If the rate be 47, for example, 4 is contained 9000 times in 36000 and we divide the principal by 9000 to get 1 day’s interest. If 414%, the divisor is 8000; if 67, the divisor is 6000; and if 5%, it is 7200, &c. Tn dividing, point off as many figures from the right, in the given principal, as there are ciphers in the divisor, cownting from the decimal point alivays. - Thus, to divide by 9000, point off three figures and divide by 9; for 6000, point off three and divide by 6. In dividing by 4500 for 8%, point off two figures and divide by 45, using the factors 5 and 9; first taking + of the principal, ther tof that fifth, making nse of short division in all such cases as pointed out in the example given on page 240. 249 INTEREST REVIEWED. The 6% method is used in most of the Banking Institutions of the country (when interest tables are not used), and is taught in the majority of Commercial Colleges, as the shortest method of comput- ing interest, a very excellent method, indeed (given on p. 180 of this work.) In adhering to this method, however, a good deal of unnecessary . labor has to be gone through in many cases where a knowledge of the foregoing methods will frequently give the interest, at sight, without figuring, at all. Take, for instance, the following problem at 447, and compute the interest on the basis of 64. Exam. What is the interest of $8000 for 149 days at 41%? Nots.—As a general rule 60 days’ interest is taken as the basis of calcula- — tion, and the solution, by the 6% method, would be something like the following: - Pointing off two places, we have.... $80.00 = G0 da. int. multiplying this by 2, we have........ . $160,000 3426s one third of 60 days’ interest, or...... 26.66°es 1 200 ieee one tenth of 60 days’ interest, or...... 8.00 == 6 en and one half of 6 days’ interest, or.... 4.00) = 8 eee giving the interest for 149 da. at 6% = $198.66 = 149“ « From this we deduct } to get.......... 49 .66 44%; 14% being 1 of 67................ $149.00 = int. at 444. Note.—Too many figures even for the 6% method. Better point off one place, and we have $800.0 = 600 days’ int. at 6%; 150 da. = 4 of 600; less 1day, or tof | $8; the difference is 149 days’ int. at 6%. Then 1 off is 442. _ But either method is too long in this case if it be borne in mind — that the interest of $8000 for 149 days, is the same as the interest of $149 for 8000 days, and that 8000 days is the basis ; in other words, | INTEREST REVIEWED, 243 that the interest will equal the principal in 8000 days, at 447, simple interest, always. Hence, The problem can be reversed so 2s to read: What is the interest of $149 for 8000 days, at 45%? The answer is at sight, namely, $149. Notr.—This method of reversing the problem can always be used when it is more convenient to take the dollars for the days. If, for instance, it were reguired to find the interest of $1000 for 149 days, at 44%, we would reason thus: The interest of $1000 for 149 days being = $149 for 1000 SG. , and since the interest of $149 for 8000 da. =35 . $149 4 of this will be the interest for 1000 ‘* = $18.625 And, in like manner, the interest of $2000, $3000, $4000, $5000, $6000, $7000, $9000, $10000, $12000; or any multiple, as $20000, $25000, $48000, &c., or any part, as $4, $40, $400, &c., can be readily obtained. To make this important reethoe more clear let us take another Exam. What is the interest of $4500 for 5 mo. 17 da. at 8%? In 5 mo. 17 da. there are 167 days. Now, since money doubles in 4500 days, at 8%, simple interest, and that the interest of $4500 for 167 da. = that of $167 for 4500 da. the required interest is at sight, viz., $167. Notre.—If the principal were $2250, the interest would be % of $167 = $83.50; if $9000, the interest would be twice $167, or $334; if $1500, the interest would be &% of $167, or $55.67 ; and if $15000, the interest would be 10 times that, &c. And if the rate were 7%, we would, in this case, deduct } of the int. at 8%. Since 7 is not an exact divisor of 36000; interest can be computed at 6% and 4 added, or at 8% and } deducted, whichever is most convenient. 944 InTEREST REVIEWED. Should the time be given in months; years and months; or years, months and days, it will be found preferable, in many cases, to take the time in which money doubles, in months, as the basis of calcu- lation, instead of days. Thus: Exam. What is the interest on a bond of $5000 for { mo, Oda. jal oy? Notr.—lInstead of taking 7200 days, the time in which money doubles at 5%, for the basis of calculation, we prefer to take 240 mo , its equivalent ; and since the interest of $5000 for 240 mo. at 5%, is $5000, the interest for 24 mo. is +5 of that, or $500. Here, then, by pointing off one place $500.0 = 24 mo. in $5000, we have 24 mo. interest, or $500 $126; + == Gee 6 mo. =} of 24 mo., + of $500 = $125 $20.83 =="2 bes 1mo. =tof 6mo.; tof $125 = $20.83; $6.25 = Qda. and for 9 days, we take 4 of $50, or 72 da. $152.08 interest, which is always at sight by point- ing off two places in the principal, $5000. Or, to get 9 da. interest we could take 6 da. =+4o0f 1 mo. and 3 da.=4of 6. Or, better thus, perhaps : Reducing the time to days, we have 7 mo. 9 da. = 219 da. and since the interest of $5000 for 219 days, is the same as the interest of $219 for 5000 days, the problem can be reversed so as to read: The interest of $219 for 5000 da., at 5Z, and the solution be obtained as follows: The interest of $219 for 7200 da., at 5%, being.... = $219 10 times this will: be:for “72000. ; <<’ -.vae.4 eee ee = $2190 Now, since the interest for 72000 da. = $2190 zs of this interest will be that for 6000 << = $182.50 4 of this will be the interest for 1000 ‘“ = $30. 42 and the difference will be for 5000 « = $152.08 — sf See eae InteEREST Rui VIEWED. Y45 Rue. When the rate of interest changes frequently : Find the rate for 1 day, and compute the interest on the given principal at that rate. Exam. What is the interest on a loan of $35000 made on Jan. 1, and paid off on the 15th, the rate of interest changing as follows: Jan. 1 24% 3 2% 6 3% Pir ake 9 3% 10 35% 12 2% 15 paid off. Here $35000 bears interest from the Ist. to the 3rd, or two days at 24%; next, 3 days at 2%, &c. Now, 237 for 2 days = 54 for 1 day, Xe. Hence the process : 24 X 2=5 Seo a Bee == 8 34 xX 2=7 3 ea ye er 3, <2 16 07% for 1 day. We now compute the interest on $35000 for 1 day at 872, or, what amounts to the same thing, the interest for 37 days at 12. And since the interest of $35000 for 36000 days at 1% == $35000 sé 6é 66 66 ce 36 days “é Lad a $35 6é ce ee 6é 6< 1 day cé¢ 66 = 97 giving the interest required ‘* 837 days a ey $35.97 Norz.—In getting J] day’s int. take } of $35; then % of the result, making use of the factors, 6 and 6 (6 x 6 = 36.) ANNUAL INTEREST. Rue. When interest is payable annually : _ I. Compute simple interest of the principal from tuts given date to the date of settlement. II, Add to this the interest of each year’ s interest from the time of its accruing to the date of settlement. Exam. What is the amount due, in 3 years, 6 months, © on a note of $8000, interest payable annually, at 5%? $8000 x .05 = $400 = Interest for 1 year. $400 x 34 = $1400 *s ‘« 35 years. $400 x .05 x 2h = $50 ‘* on the 1st year’s interest te | $400 x .05 X 1$= 30 “ $6 OSes $400 x .05x 4= 10 i «6c. Oras 8000 Principal. $9490 In this the interest of $8000 for 3% years, at 5% Next, the interest of $400 ‘“* 24 - then ¢< 66 66 ee (a3 14 6s ce finally ce 66 “6 anf 66 q year 66 Making the entire interest................. 414 = $90. Tt II “6 oe $1400 50 30 10 $1490 Notr.—It will be seen that the interest of $400 for 244 years, 14 yrs. and& yr., is equal to the interest of $400 for 44% yrs., their sum. Thus: $400 x .05 x Pei | Aly FeASY MCR N'TS, A Partial Payment is payment in part of a note, mortgage, bond, or other obligation. An Indorsement is an acknowledgment of payment, written on the back of the note, mortgage, &c., stating the time and amount of the payment made on the obligation, NotrE.—The United States Rule, given on page 129 of this work, has been adopted by nearly all the states of the Union, to secure uniformity in the method of computing interest where partial payments have been made on bonds, mortgages and other obligations. It may, however, be proper, here, to give the reader a knowledge of the method generally used in business for sete tling notes and interest accounts, and is called Tne MERCHANTS’ RuLE.—I. Compute the interest of the principal from its date to the time of settlement, and add wt to the principal. Il. Compute the tnterest of cach payment from its respective date to the time of settlement and add this interest to the payments. Ul. Krom the amount of the principal take the amount of the pay- ments, and the remainder will be the,balance due. EXAMPLE. $5000 Albany, N. Y., Aug. 1, 1902. Sixty days after date I promise to pay to A. B., or order, five thousand dollars, with interest, value received. BOOZ. Indorsed as follows: Nov. 1, 1902, $500; Feb. 1, 1903, $500; April 1, 1908, $1000; July 1, 1903, $2000. How much was due Aug. 1, 1903? Process: Principal = $5000. Payment = $500 lyr. int. at6% = 300 9 mo. int. at 6% = 22.50 $5300 Payment =a O00: Deduct 4057.50 |, 6 mo. he at 6% = ae, Wiener. aymen = P Balance due = $1282.50 4 mo, int. at 6% = 90). Payment = 2000 1 mo. int. at 67 = 10 $4067.50 948 Hints anp Hewp ror tHe STUDENT. . . .865)1709749265(4684244.5616 ...730| 2497 3074 In long operations in division, it or where the same divisor is to 2 be frequently used, zt will be 8. .19095 1549 found advantageous to form a 4..1460} »§ gg2 table of the several products of 5. .1825 1626 the divisor and the nine digits. 6. .2190 1665 Thus, to divide 1709749265 by 7. .2555 sete: 365, say to four places of deci- 8. .2920 600 mals, we look in the table and 9. .38285 2350 find that 1460, the product by 4, is the nearest below 1709; we place 4 in the quotient, and, without setting down 1460, subtract it from 1709, setting down the remainder 249, to which 7 is brought down making 2497. We then see in the table that 2190, the product by 6, is the nearest below 2497; 6 is put in the quotient, and 2190, the product by 6, is the nearest below 2497; 6 is put in the quotient, and 2190 is subtracted from 2497 and the remainder is 307, to which 4 is brought down, etc. To divide by 365, however, we prefer to proceed as follows: Annexing a cipher, and doubling the re- 17097492650 sult, we have 84194985300. ‘T'o this we add 34194985300 its third, one tenth of that third and one 11398328433 tenth of that tenth: the sum is 46847129860, 1129832843 which we divide by 10001, according to rule 113983294 II, page 15. - To divide by 10001, we simply cut off 4684712/9860+10001 four figures to the right for 10000; then deduct once 4684712, the figures on the left of the line, and to the result add 468, found on the left of the line, also; the quotient is 4684244 .5616, true to four places of decimals. (See rule II, page 15.) Reason.— By annexing a cipher to 865 and doubling 3650 the result; then adding to this its third, one tenth of 7300 that third and one tenth of that tenth, we obtain 10001 for - 24334 .0616 a simple divisor, thus establishing a simple method for 2431 dividing by 3865, always. (See examples pp. 51, 52 244 and 58.) aes Hints AND HELPs FOR THE STUDENT. 249 By simple divisors we mean 10, 100, 1000, 10000, etc., or any number which isa little more, or a little less, as 103, 1008, 1040, etc., 89, 989, 91, 94, 995, 9996, etc., also, the multiples, submultiples and aliquot parts of these numbers. Divisors may be simplified by any process that will make them 19, 100, 1000, 10000, etc. Thus, to divide by 74, 7.5, 750, 7500, etc.: add to each its third, and the numbers become 10, 10, 100, 1000, 10000, etc. Exam. 1. In 2592 pounds of oil, how many gallons, 74 pounds to the gallon ? Here, instead of dividing 2592 by 74, we add to each its 2592 third, and we have 3456+10, which gives 345.6, the re- 864 quired number of gallons. 345.6 Exam. 2. If a gross of articles cost $237.60, what is the cost of a single article? Instead of dividing by 144 (articles in a gross) $1/663.20+-1008 we multiply it by 7, getting 1008 for a simple 8 divisor, then, multiplying $237.60, also, by 7, 651520 we have 1663.20+1008; and the answer is $1.65. ; 520 (See the rule and exam. page 57.) ede Exam. 3. Divide 35886432 by 9524. By setting half of each number, in this 35886432 9524 example, one place to the right, and adding, © 17943216 4762 3768. 5236 And if 4762 were the divisor; 2 times that ee number set one place to the left, and added, would make 10002. Exam. 4. Divide 3229184 by 476. Setting 2 times each term one place to the 3229184 476 left, and adding, we get 67812864+-9996; the 6458368 952 quotient is 6784. (See exam. 8, page 23.) And if the divisor were 952; half that number set one place to the right, and added, would make 9996. 6781 |2864-- 9996 2\7124 8 6783 9996 250 Hints AnD HeEtp For THE STUDENT. Let it be carefully borne in mind that, when a divisor can be simplified, any multiple, submultiple, or aliquot part of it, can also be simplified. Thus, 167 multiplied by 6 gives 1002 for a simple divisor. Now, if we take the multiples, 334, 501, 668, 835, 8350, 1169, 1836, 18360, 1508, 15080, etc., we have: 384x*8=—1002; 501*2=1002; and by adding to 668 and 835, one half and one fifth of each, respectively, we get 1002. ‘To simplify 1169, 1886 and 1508, we deduct from each, one seventh, one fourth and one third respectively, to get 1002. And if we take the submultiples, 834, 552, 413, 332, 278, 288, 20%, 183, etc., we multiply by 2, 3, 4, 5, ete., to get 167, which, in turn, is multiplied by 6, to get 1002, the same simple divisor for all the numbers. (See page 70.) Again, many numbers, which, at first glance, may appear difficult to simplify, will be found, on slight inspection, to be easily managed, and frequently by a choice of several methods. ‘Take for instance, 69, 690, 6900, etc., and their multiples, 1388, 1380, etc., 276, 2760, 27600, etc. Exam. Divide 50950980 by 690.. We give the solution of this in four different ways, as follows: First: In this, we add tothe terms one half 50950980 690 25475490 345 76426470 1035 of each, respectively; then, setting said half one place to the right under each result and 25475490 345 subtracting, we have 733789210+10005; and oe est the quotient is 73842; the remainder being 7384119820 equal to 1 ; 185 10005 50950980 690 Second: Here, we add to the terms a third of 16983660 230 each, respectively, and we have 67934640 to be pti ees divided by 920. Cutting off the cipher from each, 434 72 we have 6793464+-92; and the quotient is 78842. - (See exam. 6, 7 and 8, p. 23.) ae 73841'92 Nots. —It is well to remember that, whether we use the method of. addition or that of subtraction, the figures on the left of the line, only, are to be operated upon; andif any figure be carried from right to left, over the line, in either case, such figure must also be operated upon. In this last example we multiply 5 (2 plus 3 carried) by 8, to get 40; and in the other example, 37 (86+1) is multi- plied by 5 to get 185. ae Pe Sy eae Hints AND HeELrs ror THE STUDENT. 251 es Third: Here, we take 700 for approximate 509509180 + 690 divisor; the difference is 10. Cutting off two figures and dividing by 7, divides by 700. T2787|118 10 We next add +o, or yo of each partial 1039 oe quotient, till all these quotients are exhausted 14 843 (divide by 7 and set each result one place to se 7 the right) using the fractional form for the eae aati remainders. The sum of the several results 73841|984 = 73842 is 73841 | 894, which is equal to 73842, the required quotient. : Dividing 50950980 by 690, it is scarcely necessary to say, is the same as dividing one seventh of the one by one seventh of the other, or dividing 72787112 by 984 (690+7=984) and, therefore, the re- mainder 987 being equal to the divisor, 1 is added making 73842. Again, in the addition, 1 is carried from the figures on the right, to those on the left of the line, and if written down, the carried figure would be set under 14: the carried figure, therefore, (1 under- stood) forms a partial quotient the same as the numbers immediately above it; and , of this figure, .013, must be added. Now, since 98; is the simplified divisor, 13 (100—984) is the com- plement, or multiplier: but multiplying by 13 is multiplying by 19; and since we have already divided by 100, by means of the line, and now dividing by 7, we have added the +, or +5, in each case. 509509/80 ; sath Pega 72787|11.428571 Fourth: The process, in this, is the same as 1039'81.428571 in the last example, except that the remainders 14/84, 285714 are treated decimally instead of fractionally. s 428571 73841|98 .571428 The remainders, in the present instance, run into circulating decimals, or periodicals, and if carried out, their sum would be -571428, repeated. Now, if 4, the fractional part of 984, be reduced to a decimal it will be found to be equal to .571428, also, repeated. The divisor 984, therefore, is equal to 98.571428; and since the remainder is equal to the divisor, 1 is added to the quotient, making 73842, as in the other cases. 952 Hints AND HELPS FOR THE STUDENT. From the examples and illustrations given, it will now be readily seen how to divide by 29, 39, 59, 79, 390, 399, 599, 59990, etc. Again, if we take such numbers as 200, 2000, 800, 30000, 400000, 700, 7000, etc., and deduct the significant figure from these numbers, and also from the successive remainders, each remainder thus found, can be simplified when used for a divisor. Take for instance, 7000, and deduct 7 from it, and also from each successive remainder, and we find that 69938, 6986, 6979, 6972, 6965, 6958, etc., can be simplified. If, for eae 6986 were presented for aimee we would com- pare it with 7000, as shown in the margin. Here we see that 14 is the difference between 6986 and {6 7000, and that 7, the significant figure of 7000, is contained 6986 in 14, an exact number of times; and this being understood, —_—— 7 will be contained in 6986, also, an exact number of times: 998 and so we divide 6986 by 7 to get 998 for a simple divisor. Exam. Divide 76874626 by 5994, exact, to six places of decimals. ‘In this, we see at a glance that 6 76874626 6... is the difference between 5994 and : 37.6’ +999 5994 6000; hence 6, the significant figure ” ue 999 in 6000, is contained in 5994 giving —_jag55/ag9 GEaiaGG 999 for a simple divisor. We then 262/666 divide the dividend by 6, getting 263 12812437.6’ which is divided by 999. . 262929|595 In dividing by 6, we treat the remainder decimally, and obtain the repetend .6’, in other words, 6 is repeated to infinity, and in get- ting the decimal part of the quotient, 6 is repeated six times for the six decimal places required. The quotient is 12825.262929, true to six places of decimals. Nore. — And if the divisor were 6 less than 5994, that is, 5988, we would divide 5988 by 6, getting 998 for a simple divisor; and so with the other numbers differ- ing by 6. Thus, 5982-+-6=997. ete. Hints Ano HeEtps For THE STUDENT. 253 When one part of the divisor is a multiple of the other part, which, in itself is a simple divisor, the division can be simplified. Take, for instance, such numbers as, 96192, 960192, 960120, etc., 93372, 930372, 48144, 480144, 36108, 360108, 24096, 240960, etc. Suppose, now, it were required to divide by any one of these, say 930372. Arranging the numbers as follows, the component factors will be readily seen: 96 96 93 93 96192 960192 93372 930372 1002 10002 1004 10004 Exam. Divide 30207318096 by 930872. 93 A moment’s glance, here, shows that 3020731] 8096930372 1208} 2924 10004 372, one part of the divisor, is 4 times 93, eee 3019523} 5172 the other part, in other words, 93 and 4832 10004 are component factors of 930372 10004 (93>< 10004=930372). We now divide, first, a be nga by 10004, getting 3019524 for quotient, 147/91 which in turn, we divide by 93, and the 10 sh quotient is 32467 | 93=32468. 39467 93 Again, if these numbers be reversed such as, 19296, 192960, etc., the division can also be simplified. To divide by 192960, for instance, we would take half, or 96480, and make use of 96 the component factors 96 and 1005, as shown in the margin, 96480 dividing, first, by 1005, and the result found, by 96: or, 1005 divide, first, by 96 and next by 1005. And if 48144 were presented for divisor, we would divide 48 by 48, using the factors 6 and 8 (6X8=48); then divide the 48144 result by 1003: and so with the others, and similar numbers. 1003 Or, doubling 48144 gives 96288, the component factors 96 being 96 and 1003. (See exam. and note page 79.) 96288 1008 Notr. — This method of using the component factors, in connection with the simplified methods given in this work, will be of great advantage when the divisor is not subject to change. (See page 55, exam. and note.) 254 Hints AND HELps FOR THE STUDENT. DECIMAL FRACTIONS, OR DECIMALS. A Decimal Fraction, or a Decimal, is a fraction having 10, or some power of 10, such as 100, 1000, 10000, etc., for its denominator. 3 9 4756 j Thus, #, zs. azodsoo 4436, are decimals. In the notation of decimals, the denominator is usually omitted, the value of the fraction being expressed by pointing off as many decimal places in the numerator as there are ciphers in the denominator. Should there not be a sufficient number of figures in the numer- ator, the deficiency is to be supplied by prefixing ciphers. Thus, 335, +30, rovooo 4338, expressed in the notation of decimals, are written, respectively, .8, .09, .00003, 4.756. Hence, conversely: The denominator of a decimal thus expressed, is unity, or 1, followed by as many ciphers as there are figures in the decimal. Thus, .57 is 75%, .004 is e475, and .00063 is +yS2q5- It will be readily seen from this notation, that the figure im- mediately to the right of the decimal point, is tenths, the next, hundredths, the third thousandths, etc. Thus, since 376 is equivalent to Bee a the fraction .376, or pins is equivalent to #oy>5-+2rh00 +1000 OF otra tri: The values of figures in decimals, as in whole numbers; are in- creased in a tenfold degree by removing the decimal point one place towards the right hand, and are diminished in a like degree by removing the point one place to the left. Thus, in the decimal .003, by removing the point one place to the right we have .03, which denotes ;3,, or 7395, and is therefore ten times the given fraction, .0038, or +29; but by removing the point one place to the left we have .0003, or +5255, which is only a tenth part of .003, or zp, OF T0000: Hence, A decimal is multiplied by 10, if the point be removed one place Hints and HeEtps ror THE STUDENT. 255 to the right; by 100, if two places; by 1000, if three places, etc., and, conversely, a decimal is divided by 10, if the point be removed one place towards the left hand, by 100, if two places; by 1000 if three places, etc., vacant places, when there are such, being supplied in both cases by ciphers. Thus, .6845x10=—6.345, or 6,343,;; 6.345 x 100=—634.5; 6.31000 — 6300. Also, 78.48+10=7.848; .784+100=.00784; 7.3+100= .073, etc. Hence, The value of a decimal is not changed by annexing a cipher to the end of it, nor by taking one away, a cipher not affecting the value of a number except when placed between a significant figure and the decimal point. (See note 2, page 233.) Thus, .50=.5=.500—.5000; each being equivalent to one half. But if a cipher be placed between the significant figure and the decimal point, the value of the number is changed at once. Thus, 50., 500., 5000., .05, .005, .0005, ete. From the foregoing view of the nature of decimals, it will be seen that there is, in every respect, the closest resemblance between them and whole numbers; and hence all operations on decimals are performed exactly in the same manner as those on whole numbers, due attention being paid to the position of the decimal point. This last circumstance, indeed, requires the utmost care when making our calculations, as the point is the characteristic of the decimal; and, from what precedes, it is evident how much depends on its | proper position. Rute I. To reduce a common fraction to a decimal, in other words, to divide a smaller number by a larger: First, annex a cipher to the smaller number and divide by the larger, and to the significant figure or cipher found in the quotient, prefix a point. Then, if there bea remainder, annex ciphers, and continue the division till nothing remains, or till the result consists of as many figures as may be deemed necessary. 256 Hints ann Hewps ror THE STUDENT. Exam. Reduce 515, to a decimal, in other words, divide 15 by 2560, expressing the quotient decimally. Here, by annexing a cipher 15 becomes 2560)15000(. 005859375 150, in which 2560 is not contained; and 12800 therefore a cipher is placed in the quotient. Annexing another cipher, 150 becomes 1500, in which 2560 is not contained, and another cipher is put in the quotient. After this, the division proceeds in the usual way, a cipher being added each time; and the quotient, or decimal is found to be .005859375, or 728) %3eo0 : Which by reduction to its lowest terms would become »}3,, the given fraction, thereby proving the work to be correct. (The work is left for the student to perform.) etc., etc. Nots.—It may be well to remark that, when the fractional form of the decimal is used, the point is always omitted, the numerator of the fraction consisting of the significant figures of the decimal, while the denominator will always be 1, followed by as many ciphers as there are places in the decimal, including signifi- cant figures, and ciphers, if there be any. RuLeE Il. To divide decimals: If the number of decimal places in the divisor and dividend be not equal ; make them equal by annexing ciphers to the one having the least number. Then reject the points and divide as in whole numbers, and if the divisor be contained tin the dividend, the figure, or figures found in the quotient will be a whole number. Having used the last figure of the dividend, annex ciphers, if there be a remainder, and continue the division till nothing remains, or till the number of figures considered necessary is found in the quotient. The part of the quotient thus obtained, will be a decimal, If, after rejecting the points, the divisor be greater than the dividend. the work will proceed according to the rule given on page 255, Exam. 1. Divide 2738.5 by 78.54. Here, by annexing a cipher to the dividend, we make the number of decimal places in both numbers equal. Then rejecting the points, we have 273850 to be divided by 7854; and we find the integral part to be 34. Annexing ciphers, and continuing the division we get the decimal part .86758, etc. The quotient, therefore, is 34.86758, etc. (For simple method of dividing by 7854, see p. 56.) Hints and Henps FoR THE STUDENT. 257 Exam. 2. Divide .1342 by 67.1. In this, we equalize the number of decimal places in both numbers, by annexing three ciphers to the divisor; and rejecting the points, we have 1342 to be divided by 671000. Then, the divisor being greater than the dividend, we proceed according to rule I, page 208, The required quotient is .002. MULTIPLICATION OF DECIMALS. Rue. — To multiply decimals: Multiply the factors as in simple multiplication and point off in the product as many places of decimals, as there are in both factors ; supplying the deficiency, when there is one, by prefixing ciphers. Exam. 1. Multiply 66.3 by .582. Here, we multiply 663 by 582 as whole numbers; the product is 385866. Then, because there are three decimal places in one factor, and one in the other, there must be four decimal places in the product 38.5866. Exam. 2. Multiply .14 by .6. Here, a cipher must be prefixed tothe product 84, as there are two places of decimals in one factor, and one in the other. The product, therefore, is .084. Notes. — Recourse should be had to our short methods, whenever possible, in multiplying decimals, as well as whole numbers, as illustrated in the following: Exam. 3. Multiply 79.96 by 79.94. Treating the factors, in this, as whole numbers, 4 and 6 make 10, and the other figures are alike: say 4 times 79.96 6 are 24; set down in full; then, add 1 to 799 making 79.94 500, and say 800 times 799 are 63920 to complete the 639.2024 product. Pointing off four places now, we have 633.2024. (See page 87.) 258 Hints AND HELPS FoR THE STUDENT. The student’s attention is called to the remaining pages of the work, which are devoted exclusively to short methods. These methods, so far as known to the author, are entirely original, and will be found both interesting and practical : To Mu.ttiety THE ’TrEns ToGETHER. Rue. (1) Multiply the units by the units and set down the unit figure of the product. (2) Add the figure to be carried, if any, to either factor, and to the result; add the unit figure of the other factor, dropping the 1 from that other factor, always. Nore. —It is immaterial to whick factor the carried figure is added, provided the 1 is dropped from the other, as illustrated in the following: EXAMPLES : A X16 272: Here, say 6 times 7 are 42; set down 2, and carry 19 x 18 4 to 16 are 20, and 7 (in 17) are 27: Or, 4 to 17, are 16 x 143 21, and 6 (in 16) are 27, making 272; the 1 being etc. dropped from the opposite factor in either case, 18 X 154 = 276: In this, say 4 of 18 is 6, to carry: then, 5 times 174 x 18 8 are 40, and 6 are 46; set down 6, and carry 4 to etc. 18, are 22, and 5 (in 15) are 27: Or, 4 to 15, are 19, and 8 (in 18) are 27; making 276. The reason for dropping the 1 from either factor will be understood from the following : If the parts of the numbers be multiplied together, instead of the numbers themselves, as shown in the margin, we have, first, 10 x 10 = 100; then 10X6= 17=10+7 60; next, 7x 10= 70 and finally, 7x6—=42. Add- 16=10+6. ing the several products thus found, we get 272, or sh 16 times 17. And here, it will be observed that, 70 having set down 2, the unit figure in 42; we add, or 4/2 carry 4 to 6 and 10 (16) leaving 7 (the 1 being dropped) O718: to be still added. Or, carrying 4 to 7 and 10 (17) leaves 6 (the 1 peing dropped) to be added. And when the ’teens are taken in a reversed order, 12, 18, 14, 15, 16, 17, 18 and 19 will become: 21, 31, 41, 51, 61, 71, 81 and 91. each ending in 1. ee ee Hints AND HELPS FOR THE STUDENT. 259 Now, Any two figures ending in 1 can be multiplied together by the following simple rule: (1) Set down the unit figure. (2) Add the tens and set down the unit figure of the sum. (8) Multiply the tens together, carrying as usual; thus: vi X 81 = 7371; In this, set down the unit figure 1; then add: bx. 01 8 and 9 are 17; set down 7, and carry 1: now mul- ox. 11 tiply; 8 times 9 are 72, and 1 are 75 completes the etc. product. - j The reason for adding in the second part of the process will be understood if the numbers be multiplied together, 91 in the usual way, as shown in the margin. Here, we see 81 that, having set down the unit figure, the figures 8 and 9 Eh are repeated in the work, and added, which can be done "371 without setting them down a second time. To Mottipty tHe Twenties, Turrties, Fortizs, Eve., ToGETHER. The Twenties, Thirties, Forties, Fifties, Sixties, Seventies, Eighties and Nineties, when written in the natural order, can be multiplied together by the following: Rue. (1) Multiply the units by the units. (2) Multiply the sum of the units by a single figure of the tens. (3) Multiply the tens by the tens; carrying as usual; thus: 23 X 24 = 592: Here, we say 4 times 3 are 12; set down 2, and 29 X 27 carry 1: then, 2 times 7 (4+ 8) or, 7 times 2, are 14, lee and 1 are 15; 5, and carry 1: next, 2 times 2 are 4, ee. and 1 are 5 completes the product. The factors in these exampies, are written in their natural order, and here it may be remarked that, when thus written, the tens in each set of numbers are alike, and the reason of the rule will be understood from the following: 260 Hints anp HeELps FoR THE STUDENT. Exam. Multiply 73 by 72, as shown in the margin: Here, having set down 2 times 8, we have next, 2 times 7; 73 then, 7 times 8, or 3 times 7; that is, 5 (2 + 8) times 7; and TR finally, 7 times 7, plus the figure carried. 5256 37 X 38 = 140€: Say 8 times 7 are 56; 6 and carry 5: now 8 times 32 x 36 15 (8+ 7) are 45, and 5 are 50; 0, and carry 35: etc. then, 3 times 3 are 9, and 5 are 14; making 1406. 43 x 48 It will be seen that the rule is applicable to all the 42 x 45 numbers down the margin; but in all cases where the etc. unit figures equal 10, while the tens are alike, the process 62 x 69 can be still further simplified by the following Rule: 67 x 64 (1) Multiply the units.together and set down the product etc. in full. (2) Add 1 to either of the tens and multiply the other by the figure thus increased ; as: %6 X 74 = 5624: In this, the units, 4 and 6, make 10, and the 7’s 48 X 72 are alike. Say 4 times 6 are 24; set down in full; etc. now, add 1 to 7, and say 8 times 7 are 56 to com- plete the product. 81 X 89 = 7209: When the unit figures are 1 and 9 they give 61 x 69 only one figure when multiplied, and in such etc. cases a cipher is set in the second place, as 81 x 89 = 7209. (See page 87, and note.) The rule is applicable to numbers of three figures, also; and in many cases, to four, or more figures; thus: 126 x 124 = 15624: Say 4 times 6 are 24; set down in full; add 123 X 127 1 to 12 and say 12 times 13 are 156; making 15624. 398 x 392 = 158016: Say 2 times 8 are 16; add 1 to 89, and say 40 493 x 497 times 39 are 1560; making 156016. 7994 x 7996=63920024: Say 6 times 4 are 24; set down in full; 3998 xX 3992 then, add 1 to 799 and 799 x 800 = 639200 etc. completes the product. And the rule can be extended to all numbers of a like nature. Having thus shown how the twenties, thirties, etc., can be mul- tiplied together when written in the natural order, let us take them, now, in a reversed order: Thus, reversing 23, 24; 382, 37; 48, 49; 57, 58; 73, 76; 87, 89; 92, 98, etc., we have Hints AND HELPS FOR THE STUDENT. 20s 82, 42; 23, 73; 84, 94; 75, 85; 37, 67; 78, 98; 29, 89, etc., and here it will be observed that the wnits, in each set of numbers, are alike. Now, To multiply any two figures by any other two, whose units are alike, we give the following Rule: (1) Multiply the units by the units. (2) Multiply the sum of the tens by a single figure of the units. (3) Multiply the tens by the tens; carrying a3 usual; thus: 32 X 42 = 1344: Say 2 times 2 are 4; set down 4; then, 7 (4+ 3) 53 X 48 times 2 are 14; 4 and carry 1: next, 4 times 3 are ov X 67 12, and 1 are 13 completes the product. The reason is plain: Take 74 x 64, as shown in the margin, 74 xX 64 = 4736: Having multiplied the units together, 83 x 93 and set down 6, we have 4 times 7; then, 14 94 x 74 6 times 4, or 4 times 6; that is, 4 times 13 hen: etc. (6 + 7) plus the 1 carried; then 6 times 7; 4736 plus 5 And the rule can be applied with equal facility to numbers of three figures, whose tens and hundreds consist of the ’teens and whose units are alike; thus: 174 XK 164 = 28536: In this, we say 4 times 4 are 16, 6, and carry 1: 163 x 183 then, 4 times 33 (16-+17) are 182, and 1 are 133; 148 x 158 set down 38, and carry 18: finally 16 times 17 123 x 133 (short method, page 258) are 272, and 13 are 156 x 156 285, which completes the product. Norr. —It will be observed, on examination, that the method made use of in multiplying 174 by 164, is the same as that in multiplying 74 by 64. And if the three figures be such that the hundreds in both factors are alike, while the units and tens consist of the ’teens; as, 417, 416, 318, 319, etc., we have the following Rue. To multiply any two numéers of three figures together, whose hundreds are alike, and whose units and tens consist of the *teens: (1) Multiply the ’teens together and set down the units and tens of the product, carrying the hundreds. (2) Multiply the sum of the “teens by a single figure of the hundreds; set down the units and tens of the product, and carry the hundreds. (3) Multiply the hundreds together to complete the product; thus: 962 Hints ANnpD HELPs FoR THE STUDENT. 417 x 416 = 178472: In this, 16 times 17 are 272 (short method); 115 x 114 set down 72, and carry 2 hundred: then 4 times 817 x 8138 33 (16 + 17) are 182, and 2 are 134; set down 215 « 215 34, and carry 1: next, 4 times 4 are 16, and 1 etc. are 17 completes the product. Should the tens in these numbers be a cipher, instead of 1, pro- ceed as follows: (1) Multiply the units together and set down the pro- duct in full. (2) Multiply one figure of the hundreds by the sum of — the units, setting down the product, also, in full. (8) Multiply the hundreds together and set down the product in full; thus: 804 < 803 = 645612: Here, we say 3 times 4 are 12; set down in 602 x 608 full; then, 7 (8 + 4) times 8 are 56; set down, 705 < 705 also, in full; and 8 times 8 are 64 completes etc. the product. In multiplying numbers of this nature together, it is well to remember that two figures are always set down for the product of the units; and two, also, for the product of the hundreds by the sum of the units. When the product of the units is only one figure, a cipher is put in the second place; and when the product of the hundreds by the sum of the units consists of three figures, two only, are set down, the third, or hundreds being carried to the next pro- duct; illustrated in the following : | 803 x 803 = 644809: Here, 8 times 3 are 9 (one figure) set down 704 x 702 two, 09; then, 6 (8+ 8) times 8 are 48; set 801 x 809 down in full; next, 8 times 8 are 64 completes etc. the work. 906 « 907 = 821742: Say 7 times 6 are 42; set down in full; then, 904 x 908 13 (6 + 7) times 9, or 9 times 18 are 117 (three 807 < 807 figures); set down two, 17, and carry 1 hundred; 709 « 708: finally, 9 times 9 are 81, and 1 are 82. The rule can be applied with equal facility to numbers of four figures, also; when the tens in both numbers are ciphers and the third and fourth figures consist of like ’teens; thus: 1103 « 1106 = 1219918: In this, say 6 times 3 are 18, and set 1204 « 12038 down in full; then, 9 (8+ 6) times 11 are 1305 « 1804 99; set down in full; next, 11 times 11 are etc. 121 completes the work. Hints anp Hexps ror THe Srupenr. 263 Or, commencing with the left hand figures: we have 11 times 11 are 121; set down in full; then 11 times 9 (8 + 6) are 99; set down in full; and 3 times 6 are 18 completes the work as before. It may be proper to remark here, also, that in multiplying numbers of this nature together, two figures, or a significant figure - and a cipher to make two, are always set down for the product of the units. And two figures only, are set down for the product of the teens by the sum of the units. Should the product give three figures, the units and tens are set down, and the hundreds are carried to the next product; thus: 1903 x 1903 = 3621409: Here, 3 times 3 are 9 (one figure) set down 1804 « 1802 two, 09; then 6 (8+ 3) times 19 are 114 1701 « 1709 (three figures); set down two, 14, and carry 1608 x 1601 1 hundred: finally, 19 times 19 (short etc. method) are 361, and 1 are 362 completes the product, And when the foregoing numbers are reversed they can be mul- tiplied together with equal facility; thus: 906 and 907 become 609 and 709, the units in both numbers being alike, while the hundreds are unlike: 609 x 709 = 431781: Here, 9 times 9 are 81; set down in full; 907 x 807 then, 9 times 13 (6+ 7) are 117 (three figures); 406 < 706 set down two, 17, and carry 1 hundred; next, etc. 7 times 6 are 42, and 1 are 43 completes the product. And any two numbers of four figures whose units and tens consist of the ’teens, the third figure being a cipher, while the fourth in each is alike, can be multiplied together by the following: RuLE. (1) Multiply the "teens together (short method) and set down the product in full. (2) Multiply the sum of the ’teens by one figure of the thousands and set down the product, also, nm full. (3) Multiply the thousands together to complete the product; thus: 4017 x 4016 -= 16182272: In this, say 16 times 17 are 272; set 83014 « 3018 down in full; then, 4 times 33 (16+ 17) are 5015 x 5015 132; set down, also, in full; next, 4 times 4 etc. are 16 completes the product. 264 Hints AND HELrs FoR THE STUDENT. Fra Or, commencing with the left hand figures, say 4 times 4 are 16; set down in full; then, 4 times 33 are 132; set down in full; and 16 times 17 are 272 completes the product, as before. If, when the sum of the ’teens is multiplied by one figure of he thousands, the product gives only two figures, then a cipher is set down to make three places, always; thus: 2019 x 2019 = 4076861: Here, 19 times 19 are 361; set down in 3014 x 3016 full; then 2 times 38 (19 +19) are 76 (two 3011 ~« 8015 figures), set down three, 076; next, 2 times etc. 2 are 4 completes the product. From the foregoing examples and illustrations, the student will find no difficulty, now, in extending the rules to the twenties, thirties, forties, etc., thus: To multiply any two numbers of three figures together when the hundreds are alike, and whose units and tens consist of the twenties, thirties, ete.: RULE. (1) Multiply the twenties, thirties, etc., together and set down the two first figures of the product, always, carrying the others. (2) Multiply the sum of the twenties, etc., by a single figure of the hundreds and set down two figures, only, of the product. (8) Multiply the hundreds together, adding the carried figures to complete the product. Exam. Multiply 823 by 824. In this, we say 24 times 23 are 552 (short method, page 259) set down 52, and carry 5 hundred: then, 8 8/23 ) AY times 47 (23 + 24) are 376, and 5 are 881; set down 81, __ 82d f and carry 3 hundred: next, 8 times 8 are 64, and 3 are 675152 67 completes the product. And if ciphers come between the hundreds and the twenties, thirties, etc., the process will be still more simple; thus: Exam. Multiply 8028 by 8024. Here, we have 23 & 24 = 552 (short method) which is set down in full; then, 8 times 47 (23 + 24) are 3876, Ba which is set down, also, in full; next, 8 times 8 are 64 Ta DRERE 64376552 completes the product. Hints anp HELps For THE STUDENT. 265 Exam. Multiply 80023 by 80024. In this, we have 23 x 24 = 552; set down in full; ‘ then a cipher: now, 8 times 47 are 376; set down in aaa full; then a cipher: next, 8 times 8 are 64 completes 6403760559 the product, Exam. Multiply 8094 by 8093. In this, we have 93 times 94 (short method) = 8742; 80194 set down 742, and carry 8: then, 8 times 187 are 1496, 80/93 1 187 and § are 1504; set down 504, and carry 1: next, 8 65504742 times 8 are 64, and 1 are 65 completes the product. igh Exam. Multiply 80004 by 80093. Here, 94 x 98 = 8742; thus: 3 times 4 are 12; 2 and carry 1: then, 7 (8+ 4) times 9, or 9 times 7 are 63, and 1 are 64; 4 and carry 6: next, 9 times 9 800/94 l 487 } 800/93 ) are 81, and 6 are 87; set down 8742 in full; now, 8 6414968742 times 187 are 1496; set down, also, in full; finally, 8 times 8 are 64 completes the product. Exam. Multiply 1034 by 10388. In this, 38 times 34 are 1292 (short method); four figures, set down only three, 292, and carry 1: then, 10134) , 72 <1 plus 1 carried are 73 two figures, set down 10 38 7 three, 073; next, 1 multiplied by 1 completes the . 1073292 — product. REMARK. — In multiplying numbers of this nature together, it will be observed that, when the factors consist of three figures, only two are set down for the product of the units and tens; and two figures, also, for the product of their sum by the hundreds. When one cipher intervenes, three figures are set down in each case, and when two ciphers intervene, four figures, and so on, according to the number of ciphers. When less than the required number of figures in the product, a cipher is set down, as 073, in the last example. 266 Hints anp HE.LpPs FOR THE STUDENT. Exam. Multiply 160042 by 160048. Here, 42 x 48 = 2016; set down in full; then, 16 1600/42 t 90 times 90 are 1440; and 16 times 16 are 256 completes 1600/48 the product. 25614402016 The process will be found equally simple tn cases of three figures where the hundreds are not alike, the units and tens in both factors consisting of any figures; and, also, under similar conditions, when ciphers intervene; thus: Exam. Multiply 518 by 314. In this, 14 times 18 (short method) are 252 (three fizures); set down two, only, 52, and carry 2: then, 3 times 18 are 54, and 2 are 56 and 5 times 14 are 170; now, 70 and 56 are 126; set down 26, and carry 1: next, 3 times 5.are 15, and 1 are 16 completes the product. Exam. Multiply 518 by 304. Here, 4 times 18 are 72; set down in full; then, 3 times 18 are 54 and 5 times 4 are 20; the sum of both is 74; set down in full; now, 3 times 5 are 15 completes the product. Exam. Multiply 473 by 604. In this, 4 times 73 are 292 (three figures); set down ~ two, 92, and. carry 2: then, 6 times 73 are 488, and 2 are 440 and 4 times 4 are 16; the sum of both is 456; set down 56, and carry 4: now, 6 times 4 are 24, and 4 are 28 completes the product. Exam. Multiply 4073 by 604. Here, 4 times 73 are 292; set down 92, and carry 2: - then, 6 times 73 are 488, and 2 are 440 and 4 times 40 are 160; the sum of both is 600 ; set down in full; next, 6 times 4 are 24 completes the product. 5{18 2 56 _ 8144 70 162652 5/18 ) 54 3/04 § 20 157472 4I73 ) 440 6045 16 285692 ere 440 6104 160 2460092 Hints anp HEeEtps FoR THE STUDENT. 267 Exam. Multiply 7034 by 6023. In this, 23 times 34 are 782; set down in full; "0184) 204 then, 6 times 34 are 204 and 7 times 23 are 161; the he a5 161 sum of both is 865; set down in full; now, 6 times 7 42365782 are 42 completes the product. Before proceeding with our examples, it will be of advantage to the student to become thoroughly familiar with the following method : To multiply any two digits by any other two, in a@ single line; illustrated in the following : Exam. Multiply 73 by 82. In this, say 2 times 73 are 146; set down 6, and carry 73|146 14: then, 8 times 8 are 24, and 14 are 38; 8 and carry 3: 82 now, 8 times 7 are 56, and 3 are 59 completes the product. 5986 Exam. Multiply 67 by 42. Here, 2 times 67 are 134; set down 4, and carry 13: then, 67 4 times 7 are 28, and 13 are 41; 1 and carry 4: now, 4 times 42 6 are 24, and 4 are 28 completes the product. 2814 Notgz.— When the student has become familiar with the process, it will not be necessary to set down the product in the margin, as shown in the first example, And when ciphers intervene, proceed as follows : Exam. Multiply 607 by 402. In this, 2 times 7 are 14; set down in full; then, mul- tiply crosswise: 2 times 6 are 12 and 4 times 7 are 28; ae now add both products; 28 and 12 are 40; set down in shot full; and 4 times 6 are 24 completes the product. Exam. Multiply 7003 by 4006. Here, 6 times 3, or 18, is set down, then a cipher to make three places; now, 6 times 7 are 42 and 4 times 3 7003 are 12; 42 and 12 are 54; set down, then a cipher to 4006 ' make three places, also; now, 4 times 7 are 28 completes 28054018 the product. 268 Hints AnD HE.LPs FoR THE STUDENT. REMARK.—In multiplying numbers of this nature together, it will be observed that, when only one cipher intervenes, two figures are. . set down for the product of the units; and two figures, also, for the result found in multiplying crosswise; and when two ciphers inter- vene, three figures are set down in both cases, and so on, according to the number of ciphers. . When the products are less, a cipher, or ciphers are set down to make the required number, as in the fore- going example; illustrated also in the following: Exam. 702 by 504. In this, set down 4 times 2, then a cipher, 08, to make 702) 10 two places; then, 4 times 7 are 28 and 5 times 2 are 10; oat 98 28 and 10 are 38; set down in full; and 5 times 7 are 35 353808 completes.the product. And any number of three digits can be multiplied by any other three, in a single line as follows: Exam. Multiply 724 by 348. Here, by the method for multiplying two digits by sas 83 any other two, given on the preceding page, we have 3/48) 836 24 x 48 = 1152; set down 52, and carry 11: then, 3 201952 times 24 are 72, and 11 are 83 and 7 times 48 are 336, both set on the margin to the right; now, the sum of both numbers is 419; set down 19, and carry 4: next, 8 times 7 are 21, and 4 are 25 completes the product. If ciphers intervene proceed as follows: Exam. Multiply 7024 by 3048. In this, 24 X 48 = 1152: set down three figures, 152, and carry 1: then, 3 times 24 are 72, and 1 are 73 a rete A and 7 times 48 are 336; their sum is 409; set down in 51409152 full; next, 3 times 7 are 21 completes the product. Hints Anp Hers FoR THE STUDENT. 269 GENERAL SHORT METHOD. The following short method can be applied in all cases: Exam. 1. Multiply 78 by 63. When the multiplier consists of two figures: - Here, say 3 times 8 are 24; set down 4 for the first 78X63 figure of the answer one place to the right of the 93 multiplicand; carry 2, then, 3 times 7 are 21, and 2 are A01k 23; set under the units and tens of the multiplicand, as shown: hext, 6 times 8 are 48, and 3 (add downwards) are 51; set down 1 and carry 5; 6 times 7 are 42, and 5 are 47, and 2 (adding downwards) are 49 completes the product. Exam, 2. Multiply 3478 by 74. In this, 4 times 3478 are 13912; set down 2 for the first figure of the answer, and 1891 under the multi- 3478 X74 plicand as shown: then, 7 times 8 are 56, and 1 are "7907 hue 57; set down / and carry 5; 7 times 7 are 49, and 5 557372 are 54, and 9 (add downwards) are 63. Proceeding, : we obtain 257372, the product. Exam. 3. Multiply 2356 by 347. When the multiplier consists of three figures: ’ Here, we set 2, the first figure of the answer, two 2356347 places to the right, and 1649 is set as shown in the “Teo ee margin: next, 4 times 6 are 24, and 9 are 33, which 943 gives the second figure of the answer, and 3 to carry; 817532 then 4 times 235 plus 3, gives 943, which is set in proper position, making the second partial product 9483, the unit figure 3 being the second of the answer. Now, 3 times 6 are 18, and 7 (4+3 adding downwards) are 25; set down 5 and carry 2; 3 times 5 are 15, and 2 are 17, and 10 (6+4 downwards) are 27. Proceeding, we obtain the product, 8175382. 970 Hints anp Hetps ror THE STUDENT. Exam. 4. Multiply 15673 by 5432. When the multiplier consists of four figures : In this, 6, the first figure of the answer is set three 156735432 places to the right, and the remain‘ng part of the 3134 product, 3134, is set as shown: next, 3 times 3 are 9, A702 and 4 are 13, gives 3 for the second figure of the 6269 ’ answer and 1 to carry; and 4702 set down as shown: 85135736 then, 4 times 3 are 12, and 5(8+2) are 17, gives 7 for the third figure of the answer, and 1 to carry: and 6269 is set in proper position. Finally 5 times 3 are 15, and 10 (1+0+49) are 25, this gives 5, the fourth figure of the answer, and 2 to carry. Pro- ceeding thus, we obtain 85135736, the complete product. Exam, 5. Multiply 2346 by 3404 Here, the product by 4 is 9384; 4 is set three places 2346 x 8204 to the right, for the first figure of the answer,and — 938 938 as shown: then, the next figure of the multiplier 469 being a cipher, the 8 is brought down for the second 7516584 figure of the answer. Now, 2 times 6 are 12, and 3 are 15, gives 5 for the third figure, and 469 is set in proper position as shown. Finally 3 times 6 are 18, and 9+9 are 36, gives the fourth figure of the answer, and 3 to carry. Proceeding, we obtain 7516584, the complete product. Nore.—It will be observed that the unit figure of the answer is set to the right of the multiplicand as many places as there are figures in the multiplier less one, and that the partial products are one less than the significant figures of the multiplier, always. Exam. 6. What is the cost of 347 lbs. of sugar, @ $5.32 per hundred ? UsuaL METHOD. SHORT METHOD. 347 3475.32 5.32 69 694 105 1041 $18 . 4604 17385 $18. 4604 Hints anp HeELPps FoR THE STUDENT. 271 SQUARING OF NUMBERS. Exam. What is the square of 73 ? In this, say 3 times 3 are 9; then, 6 (58 + 3) times 7, or 7 73 6 times 6 are 42; 2, and carry 4: now, 7 times 7 are 49, and 4 13 are 538 complete the square, (See pages 259, 260 and 261.) D829 Exam. What is the square of 417 ? Here, the units and tens consist of the ’teens and the hundreds are alike: 17 times 17 (short method, page 258) are 289; set down 89, and carry 2: then, 4 times 34 ae I 34 (17 + 17) are 136, and 2 are 138; set down 38, and carry 173889 — 1: now, 4 times 4 are 16, and {1 are 17; this completes the square. ‘Exam. What is the square of 824 ? In this, the units and tens consist of the twenties, and the hundreds are alike: 24 times 24 (short method, page 259) are 576; set down 76, and carry 5: then, 8 als times 48 are 384, and 5 are 389; set down 89, and carry 678976 53: now, 8 times 8 are 64, and 8 are 67 completes the square. REMARK.— It may be well to remark, here, that, in squaring numbers of three digits, only two figures are set down for the pro- duct of the units and tens; and fwo, also, for the product of their sum by the hundreds, the remaining figures being carried in both cases. Exam. What is the square of 1236 ? Here, 36 X 36 = 1296; set down 96, and carry 12: then, 12 times 72 are 864, and 12 are 876; set down 76, 3 ms 73 and carry 8: now, 12 times 12 are 144, and 8 are 152 1527696 completes the square. When a cipher intervenes any number of three figures can be squared at sight; thus: 272 Hints AND HELPS FOR THE STUDENT. Exam. What is the square of 808 ? In this, say 8 times 8 are 64; set down in full; then, 808 double 64; 2 times 64 are 128; set down 28, and carry 1: 808 now, 8 times 8 are 64, and 1 are 65 completes the square. 652864 Exam. What is the square of 12034 ? Here, the units and tens consist of the thirties, and we have 34 X 34 = 1156 (short method); set nown 156, 120134 68 and carry 1: then, 12 times 68 are 816, and 1 are 817 120|84 (three figures) set down in full; now, 12 times 12 are 144817156 144 completes the square. (See pages 265 and 266.) Exam. What is the square of 12734 ? In this, we first multiply as if the 7’s were ciphers, 12/7 34 that is, 12034 x 12034, as in the previous example. 12/7 at 68 To this we have to add 7 times 68 (84 + 34); 7 times ase 7 and 7 times 24 (12 + 12); that is, 700 times, 7 being 173376. in the hundreds’ place; thus: 7 times 8 are 56; 6 and 162154756 carry 5: 7 times 6 are 42, and 5 are 47; 7 and carry 4: 7 times 7 are 49, and 4 are 53; 3, and carry 5: now, 7 times 24 are 168, and 5 are 173; the sum of both products is the square. (See page 270.) Exam. What is the square of 70342 ? Here, 42 x 42 = 1764; set down in full; then, 7 times 84 are 588 (only three figures) ; set down four, 703142 0588; now, 7 times 7 are 49; the result is the pro- 703|42 t 84 duct of 70042 by 70042, to which is added 8 (that is, 4905881764 — 300) times 3| 84, or 1152 and 8 times 14 (7+ 7) or 6 421152. . (3 + 3) times 7 are 42; the sum of both products is 4947996964 the square. The following will be found interesting and practical: Any number of three figures can be multiplied by any other number of three figures when the hundreds in both are alike, and whose units and tens, when added, make 10, 20, 30, 40, 50, 60, etc., 100, 110, 120, 1380, 140, etc., as follows: | Hints AnD HeELPs FOR THE STUDENT. 273 ' Exam. Multiply 742 by 708. < In this, 42 and 8 make 50, and the hundreds are alike: 5 Omit the cipher in 50, and set 5 above the 4: now, say 8 7142 times 42 are 336; set down in full; then 7 times 75 are 7|08 525 completes the product. 525336 Exam. Multiply 518 by 512. 3 Here, 12 and 18 make 380, and the hundreds are alike: 5/18 Say 12 times 18 are 216; and 5 times 53 are 265. 0/12 265216 REMARK.— In multiplying numbers of this nature together, three Jigures must be always set down for the product of the units and tens; when four figures are obtained, the fourth is carried; and when only two are obtained, a cipher is set down in the third place; illustrated in the following: Exam. Multiply 489 by 471. In this, 71 and 39 make 110; set 11 above 43: now, 111 by the short method, page 267; we have 39 X 71 = 2769; A\3 set down 769, and carry 2: then, 4 times 51 are 204, and att 2 are 206. 206769 Exam. Multiply 748 by 702. Here, 48 and 2 make 50; set 5 above the 4: now, 2 times 5 48 are 96 (only two figures); set down three, 096; then, 7 ae times 75 are 525 completes the product. 525096 And when a cipher intervenes in both factors, the process will be found equally simple; but in this case four figures are set down for the product of the units and tens: if only three are obtained, a cipher is set in the fourth place; illustrated in the following: Exam. Multiply 3071 by 3089. In this, 89 and 71 make 160; set 16 above 07: now, 1/6 71 x 89 = 6319 (short method); set down in full; then, 3 ot i times 316 are 948 completes the product. 9486319 ite 974 Hints ano HxExLps FoR THE STUDENT. Exam. Multiply 702+ by 7016. Here, 16 and 24 make 40; set 4 above the 2. now, 16 Ao times 24 are 384 (three figures) set down four, 0384; 70|24 then, 7 times 704 are 4928 completes the product. Or, 70}16 having set down 384, say 70 times 704 are 49280. 49280384 When the units and tens make an even 100; set down their pro- duct and make four figures always, setting down a cipher or ciphers when four are not obtained; then add 1 to either eu of the hundreds, and multiply by che other; thus: Exam. Multiply 988 by 912. In this, 88 and 12 make 100: say 12 times 88 are 1056; 9/88 set down in full; now, add 1 to either 9, and say 10 times 9)12 9 are 90 to complete the product. 901056 Exam. Multiply 792 by 708. Here, 92 and 8 make 100: say 8 times 92 are 736; set 7,92 this down, then a cipher to make four places; now, 8 708 times 7 are 56 completes the product. 560736 Exam. Multiply 1299 by 1201. 12/99 In this, 99 times 1 are 99; set this down, then two 12/01 ciphers; now, 12 times 13 are 156 completes the product. 1560099 Exam. Multiply 5938 by 5962. Here, 62 and 88 make 100, and the hundreds and thousands in both factors are alike: now, 62 times 38 59138 (short method) are 2356; set down in full; then, adding 59/62 1 to either 59. say 60 times 59 are 3540 to complete the 35402356 product. | And if a figure be changed in either factor, the product is obtained with equal facility; thus: Hints AnD HE.Lps For THE STUDENT. 275 Exam. Multiply 5988 by 5964. 59/38 In this, we multiply as if 64 were 62, as in the pre- 59/64 ceding example, and to the result thus found, 2 times 39402356 5938, or 11876, is added for the required result. 11876 39414232 Exam. Multiply 5938 by 5762. Here, we assume 57 to be 59, and multiply as in the 59138 preceding examples, and from the result, 2 (that is, 57/62 200) times 5938, or 1187600, is deducted; the difference 35402356 is the required product. And if the multiplier were 11876.. 6162 instead of 5762, we would add 1187600. 34214756 Exam. Multiply 15947 by 8953. We multiply in this as if 159 were 89, and we have a a 8947 « 8953 = 80102491, to which is added 7 (that is, 80102491 62671... 7000) times 89538, or 62671000, to get...............6. 142773491 A knowledge of the foregoing methods, with a little practice, will now enable us to obtain extraordinary results, without much mental effort, in many cases where, by the usual methods, it would require considerable labor. Take, for instance, the following: Exam. Multiply 120342 by 120358. In this, 58 and 42 make 100, and the remaining 1203/42 figures are alike: 42 x 58 = 2436 (short method, page 1203/58 267); set down in full; now, adding 1 to 3 (either 14484122436 one); we have 1204 X 1203: say 3 times 4 are 12; set down in full; next, 7 (8 + 4) times 12 are 84; set down in full; and 12 times 12 are 144; set down in full, completes the product. (See page 262.) 276 Hints anp HELps FoR THE STUDENT. Exam. Multiply 82341 by 2359. Here, we have, first, 8 (that is, 80000) times 23859, or 18872, set five places to the left of units; then, oe 2 2341 X 2359 = 5522419, which is set in proper posi- eae tion, and added; this gives the required product 5599419 41 x 59 = 2419; then adding 1 to 23, ‘we have 94242419 24 X 23 = 582. 7854 The following method for multiplying any number by 7854, or .7854, so much used in practical mathematics, will be found prefer- able to the usual method: multiply 84628 by .7854, First, multiply by 7 and set down the result a second time, one place 34628 ..7854. farther to the right; then, double 949396 be the latter number and set down the 242396 7854=5 4 40 result a second time, also, each one 484792 14 484792 lace to the right, and add the re. 5" place to the right, and a ee” 37196. 8312 sults. The veason is shown on the margin. 7854 Numbers ending in 5, when not too large, can be readily multiplied together, or squared, by the following: Rue. (1) Multiply the units together and set down the product in. full. (2) Multiply the remaining figures together and to their product add half their sum; thus: 165 x 45 = 7425: In this, say 5 times 5 are 25; set down in full; 245 x 65 then, 4 times 16 are 64, and 8 (half of 16) are 72, 145 x 85 and 2 (half of 4) are 74 completes the product. etc. When the sum of the numbers is odd, drop 1 to make the number even, then add half this even number, and set down 75, always, for the first two figures, instead of 25; thus: 175 & 45 = 7875: Here, the sum of 17 and 4, or 21, is odd, and 195 x 65 75 is set down, instead of 25; now, 4 times 17 are 135 x 85 68, and 10 (half of 20) are 78. Or, add 8 (half 16) etc, and 2 (half of 4) as you proceed. Hints anp HELPs FoR THE STUDENT. OTT Now, the square of 695, or 695 x 695 = 483025: For squaring numbers ending in 5, the rule 195 x 195 given on page 260 is preferable, since the units 230 X 285 equal 10, and the other figures are alike always. etc. Here, we say 5 times 5 are 25; set down in full; then adding 1 to 69, we say 70 times 69 are 4830 to complete the work. The reason for adding half the sum will be understood if we take any two numbers ending in 5, say 85 and 45, and multiply their parts together, as shown in the margin, Here, it will be seen that the partial products, when added, make 4825, or 85 x 45, and that, 80 + 5 omitting the ciphers, we have 4 times 8, or 382, =i Dae plas 6 (the half of 4+8) plus 5 times. Anda 00+ 30 similar mode of reasoning will show why 75 must 3200 a0 : 35 be set down, instead of 25, when the sum is odd, The rule given on page 260, for multiplying together numbers of two or three figures whose units equal 10, the otber figures being alike, can be applied to mixed numbers whose fractions make 1, and whose whole numbers are alike; thus: 84 X< 84 = 724: In this, the two fractions, when added, make 1, Tk X TF and the 8’s are alike: say half by half is a quarter 6 X 64 (4); then, add 1 to 8 and say 9 times 8 are 72. etc. (8;.= 8 5, and 8 5 X 8.5 = 72.25, or 724.) Rue. (1) Multiply the fractions together and set down the result. (2) Add 1 to either whole number and multiply the other by the number thus increased. mi62.< 165.— 27245: Here, the sum of 3 and 3 is 1, and the whole 142 x 144 numbers are alike: now, 3X $=}, which is etc. set down; then, 16 times 17 (short method) are 272. And any two mixed numbers, each having the fraction 4, can be multiplied together by the following Rule: (1) Multiply the fractions together and set down the result. (2) Multiply the whole numbers together and to their product add hulf their sum; thus: OF x 74. = Tt: In this, 4X 4=4, which is set down; then, 7 183 x 64 times 9 are 63 and 8 (half of 7 + 9) are 71; the pro- etc, duct is 71}. ‘ 278 FRACTIONS. When the sum of the whole numbers is odd, add half the next lower even number, and set down 2 for the fraction, always, instead of +: thus: 174 & 44 = 783: Here, 17 and 4 are 21, an odd number, set 194 x 64 down 3; now, 4 times 17 are 68, and 10 (half of 20) etc, are 78, This is the application of the rule given on page 276, for multi- plying together numbers ending in 5. GENERAL RULE. To multiply ang two mixed numbers together : (1) Multiply the whole numbers together. (2) Multiply the whole number of the multiplicand by the fraction of the multiplier. (8) Mul- tiply the whole number of the multiplier by the fraction of the multi. plicand. (4) Multiply the fractions together; and add the four products. Exam. Multiply 1434 by 168. In this, 16 times 14 (short method) are 224, which is pasadena | set down; then, ¢of 14=—103; next, 4 of 16=8 and 104 4X $= 2: adding the four products, now, we get 242%, the required product. Pagar The work can be done mentally, in most cases, without setting ~ down the several products; thus: multiply 12% by 83. Here, we say 8 times 12 are 96, and 4 (4 of 12) are 100; 122 x 84 and 6 ( of 8) are 106; then, } x 4= ,3, which gives... ot LOG ie Now, in actual business transactions, where the fraction of a cent cannot be taken (the rule being to reject the fraction when less than half a cent, and when a half, or more, to add a cent) the following simple method will answer all practical purposes: take 144 lbs. of tea, @ 16% ¢ per lb. In this, we say # of 14, to the nearest unit, is 11; and 11 8 3 of 16 is 8; then, 16 times 14 are 224, and 19 (11 + 8) 1434 x 163 are 248, or $2.48. $2.48 The correct answer is 2.42%, as found above; but $2.43 for business. FRACTIONS. 279 Exam. What is the cost of 244 yds. of cloth, @ 233 ¢ per yard? Here, # of 24 to the nearest unit is 18; and the half 18 12 of 23 to the nearest unit is 12; then 23 times 24 (short 245 X& 23% method) are 552, and 80 (18 + 12) are 582. $5.82 Note. — In multiplying by each fraction, always take what is nearest the true result; thus, the 4 of 23 is nearer 12 than 11, etc. When either factor is an aliquot part of 100, 1000, etc., or is con- veniently near an aliquot part, the multiplication can be more easily performed by division. The aliquot parts of a number are obtained by dividing it by 2, 3, 4, 5, 6, etc., as shown in the margin, where the aliquot parts of 100 are given. RuLE. To multiply by an aliquot part of 100: If there be no decimal in the mulliplicand, annex two ciphers and divide by 2, 3, 4, etc., as the case may be. If there be decimals, or cents, move the decimal point two places to the right and divide as directed. If a mixed number, reduce the fraction to a decimal. © oS HW UU a ad Tl POs Colm aaa Sater saecoptnc|m SDD ES? SUP Go ae i=) es|ao p= — Oo | al — Exam. What is the cost of 162 yds. of cloth @ $1.25 per yard? Here, 162 being } of 100, instead of multiplying in the usual way we move the decimal point two places to the $125 right in $1.25, this multiplies by 100, and gives $125; $20.83 now, + of this gives $20.83 the required cost. Or, since 125 is 1 of 1000, by reducing the fraction in 16} to a decimal, we have 16.666’ to be multiplied by 125. Now, moving the point three places to the right, we have 16666. cents, and 3 of this gives $20.83 as before. 280 FRACTIONS. Computing THE Cost oF COMMODITIES. In computing the cost of commodities, such as dry goods, etc., when the commodity contains a fractional part, the process can be simplified by reducing the fraction toa decimal and applying the method of aliquot parts. Take for example 4883 yards of goods at any particular price. By reducing 3 to the decimal of a yard, we have 4883 yds. = 488.75 yds. Now, the cost of 488.75 yards at $1 per yard = $438.75 and at 10 cents per yard, the cost is one-tenth of that = 43.875 at 1 cent per yard, the cost is one-tenth of thatat10cents= 4.3875 and at 3 cent, the cost is half that at 1 cent = 2.1937 And font this basis the cost at any given price can be easily ob- tained, illustrated in the following examples : Exam. 1. Find the cost of 4383 yds. of silk at $1.37} per yd. In this 4883 yds. =438.75, and at $1 per yd. the cost is $438.75 Now, 374c.= 25c., plus 124c.; and 25c.=40f $1= .... 109.69 and at 12ic. the cost is one-half that at 25c.= .... 54.84 making the total cost $603.28. $603.28 Exam. 2. What is the cost of 86% yds. at $1.384 per yd. Here we have 863 yds. = 86.875 yds. and proceeding as $86 .875 in the foregoing example, we obtain the three first items 21.718 which give the cost at $1.374; and the cost at 1c. is 10.859 found from the first item ($86.875) by moving the point . 868 two places to the left; this gives .868 making the cost $120.32 $120.32. Exam. 3. What is the cost of 67% yds. at $1.874 per yd.? In this, 675 yds.= 67.625; and at $1, the cost= .... $67 .625 Now, the difference between $1.874 and $2, is 124c., so 185.25 we multiply by 2 to get the cost at $2 per yd. and 8.453 from this we deduct one-eighth of the cost at $1, or $126.797 $8 .453, the cost at 124c. (4 of. $67.625 = $8.453) to get the cost at $1.874. Note.— To reduce the fractional part to a docintal annex ciphers to the numerator, or suppose them annexed, and divide by the ‘denominator. ADDITION. The following method for Addition requires no “ ¢ar- rying,’ and will be of advantage to accountants when liable to be interrupted in their calculations : 3829 . 25 768 .50 4687.49 2823 .35 7547 .28 3760.82 675 . 64 1846 .35 3785.10 23270 .48 6453.3 29723 .78 Commencing at the top of the left-hand column and running downwards, we find the sum to be 23, which is set down in full, in the usual manner; then, without carrying, we find the sum of the next column to be 62; 2 is set in its proper place under the column, and 6 to the left, one line lower; the sum of the next coiumn, without carrying, is 47; the next 50; then 34, and finaily 38. Thesum of the two results thus obtained is the required sum. Nortg. —It is preferable to write the numbers of each sum in the natural order, thus; 62, write down 6 first and then 2, etc. 289 7 ADDITION. Instead of taking one column into consideration, as in the foregoing, let us take two columns, as in the follow- ing example: 7633.42 3893.87 6324.73 39250. 78 5 6 ee 39756 .78 Commencing at the bottom of the left-hand column and run- ning up, we find the sum to be 34; call this 340; now run down the next column, starting with 40, bearing in mind the 8 (300), by taking hold of the third finger of the left hand; the sum of the two columns is 392, which is set down in full. Next, take the third column from the left, without carrying, the sum is 49, call this 490 and run down the fourth column starting with 90; tue sum of both columns is 550, which is set down in full, 50 in proper position, and 5, that is 500, to the left and one line lower; the sum of the next two columns is 678, which is set down as shown, and the total sum is 39756.78. Note.-- If the number of columns be odd, first add the left hand column, then two columns at a time, thus: 764.54 The left-hand column is 26, the sum of the two next, 123.45 298; and the sum of the two last, 324. 876.43 23 Hints AND HeE.Lps For THE STUDENT. 283 The following rule for Subtraction will be found simple and practical : RULE. — Add to the minuend, first, what the untt figure of the subtrahend wants of being 10; and for the succeeding figures, add what each wants of being 9. Drop 1 immediately to the left of the last figure of the subtrahend always. Exam. 1. From 74721 take 39864. Here, we say 6 and 1 are 7; 3 and 2 are 5; 1 and 7 are 8; T4721 4is4; 6 and 7 are 13; drop the 1, and the remainder is 39864 34857. (7)84857 Exam. 2. From the sum of $687463 and $2346; take $6942. $687463 + $2346—$6942=$682867. In this, we say 8 and 6 are 14, and 3 are 17; 7 and 1 to carry: 1 and 5 are 6, and 4 are 10, and 6 are 16; 6 and 1 tocarry; 1 and 8 are 4 and 4 are 8; 3 and 2are 5, and 7 are 12; 2. and 1 to carry; 1 and 8 are 9; but dropping 1 immediately to the left of the subtractive number, we set down 8; then setting down the last figure, 6, the remainder is $682867. Exam. 3. Received $756.47; $982.34; $765.26; and paid out $476.29; what is the balance on hand? In this example, we set the subtractive number under $756 .47 those to be added, and perform the whole process by 982.34 addition, adding mentally, what the first figure of 765. 26 $476 29 wants of being 10; and what each remaining Cesanaron 476.29 figure wants of being 9. The sum of the last column git oS is 30; but dropping 1 from 3, immediately to the left of $2027.78 $476.29, makes it 2. The reason of dropping 1 to the left of the last figure of the sub. trahend, always, will be understood from the following : 984 Hints AND HeELps FoR THE STUDENT. Any number, which, when added to another, makes 10, 100, 1000, 10000, etc., is called the complement of that other. Thus, 434+57= 100; 43 is the complement of 57; and 57 is the complement of 43. The complement of $476.29 is $523.71, both numbers, when added, making $1000. Subtracting either number, now, from 1000, is the same as adding. its complement and dropping the 1000. Thus, subtracting $476.29 from $1000 leaves $523.71; eee as adding $523.71 $1000 $1000 and subtracting $1000, or dropping the 1 in 416.29 ek 1000. $523.71 ()528.71 And this being understood, any number can be subtracted from another, by adding (mentally) the complement of the number to be subtracted, dropping 1 to the left of that number, always. (See page 221.) Exam. 1. From 547632 take 876. In this, we say, 4 and 2 are 6; 2 and 8 are 5; 1 and 6 547632 76 546756. are 7; here we drop 1, and set down 6; then 4, and 5. Exam. 2. From 370234 take 18547. Here, we say, 3 and 4are 7; 5 and 8 are 8; 4 and 2are 6; 3702384 1; 8 and 7 are 15; carry 1 to 3 is 4, but here 1 is to be 18547 dropped, and we set down 3, 851687 Prosi~ems With TuHerr So.vutions. 985 The following problems, with their solutions, are given for the purpose of showing the student how they and similar problems may be solved. PAPER PROBLEMS. Prop. 1. What is the cost of 187 sheets 25 lb. paper, 500 sheets to the ream, @ 18c per lb.? For the solution of this an similar problems, we give the follow- ing simple RuLE.— Multiply the number of sheets by twice the weight of the ream, and the result by the price ; .or by twice the price, and the result by the weight, whichever is most convenient, and point off five decimal places, always. Solution. —187 X 50 X 18 = $1.68300; or, 187 XK 25 x 36 = $1.68300; for business $1.68. The reason for pointing off five places: If we proceed in the usual way we multiply the number of sheets by the weight and the result by the price, and divide by 500. Now, by doubling the weight, or the price, we double 500, also, making 1000 for the divisor. Three places are cut off for the ciphers in 1000, and two for .18 in the multiplication, making five places in all. Note. —If the ream consist of 480 sheets, proceed according to the rule, and at to the result four per cent. of itself (.04) thus, $1.68300 x .04 = .06732; then, 1 683 067 $1.75 Reason: The difference between 500 and 480 = 20, and when ue divide by 500 instead of 480, we obtain a result too small by go's or rfp-= .04. Pros. 2. 28 lb. paper is listed to be sold at 10c. per lb. by the ream, but for a quantity purchased less than a ream, 25% extra is charged; what is the cost of 264 sheets on these terms ? Solution.— 264 Xx 28 K 20 = $1.47840; then 25% = + and we have $1.47840 plus 4= .36960 $1.84800 r, By adding the percentage to the number of sheets, the weight, or the price, whichever is most convenient, we obtain the same result, thus, 264 plus + or 66 = 330, and we have 330 x 28x 20 = $1.84800; or, by adding 25% to 28 making it 35, we have 264 x 35 X 20 = $1.84800. 286 Proptems Wiru THEIR SOLUTIONS. Paper PROBLEMS. Weights of paper equal in thickness to 24 x 36. Rue. — To find the weights of paper equal in thickness to 24 x 86: (1.) Draw a vertical line, and on the left of said line, set the given. dimensions ; and on the right set the required dimensions, and the given weight. (2.) Divide the product of the numbers on the right by the product of those on the left; the result is the required weight. Exam. If 24x36 weigh 70 pounds; what will 40x 48 weigh ? aE NS Solution. — 24|40 It is scarcely necessary to re- ue 36/48 mark that recourse may be had to 38 70 Beep ec in eo As —_——————- such cases, thereby shortening the — 641184400(165.5 process. thuss . lai 155.5 Dividing 24 and 48 each, by 24, we obtain 1 and 2; and dividing 36 and 40 each, by 4, we obtain 9 and 10. We have now 10 x 2 X 70 +9 = 155.5. PrrcenracE PROBLEM. If. $30,000 be invested in property which rents for $250 per month, and on which $750 are paid in taxes; what rate of interest does the investment pay ? Solution. — Here, $250 per mo. is $3,000 per year, out of which $750 are paid in taxes leaving $2,250 income on the investment. Now, if $30,000 in one year give $2,200 income, what income (rate) ought $100 give? And for this we have the following proportion: As $30,000 : $100 : : $2,250 :x, or the rate on $100; and we have $225,000-+$30,000—747%. (See Useful Rules, pages 155 to 158.) * RuLE. — Divide 100 times the income by the investment. Tae Porunation PRospiem. If the population of Albany was 90,000 in the year. 1900, and 100,000 in 1910; what was the rate per cent increase during the interval ? Solution. — Here 100,000 — 90,000 = 10,000, the increase of popu- lation during the interval. And we have: As 90,000 :100 :: 10,000 :x = 1,000,000 + 90,000 = 115%. (See page 158 ) RULE. — Divide 100 times the increase of population during the interval, by the population of the earlier date; the result is the rate ‘per cent. PRoBLEMS witH THEIR SOLUTIONS. 287 Tue Grant PROBLEM. General Grant sold 2 horses for $198 each, he gained 10% on one and lost 10% on the other, what was the result of the sale? Solution.— In this, rule 111 of Profit and Loss, page 144, is appli- cable: To find the first cost from the gain per cent., and the selling price. . Now, what cost $100 was sold for $110 and we have to find what the horse cost which was sold for $198. As the selling price, $110 : $100 cost, : : $198 selling price to x, or the cost, and we have 100 x 198 + 110 = $180, the first cost. Selling price $198, cost price $180; the result was a gain of $18 in the first place. Next, what cost $100 was sold for $90; and we have the following As $90 : $100: : $198 : x = 19800 + 90 = $220, the cost in the second place, showing a loss of $22; then $22 — $18 = $4 loss, the result of the sale. (See problem 2, page 145.) Tue Sauipe PrRoBuem. A ship springs a leak 40 miles from shore, and admits 3% tons of water in 12 minutes; 60 tons will sink her, but the pumps throw out 12 tons per hour. Find the average rate of speed to bring her to shore before sinking. Solution.— The ship admits 33 tons in 12 min., she will admit 5 times that in 60 min., or 183 tons per hour (83X5=183 tons) but the pumps throw out 12 tons per hour: 183 — 12 = 63 tons, the quantity admitted per hour. Now, if 63 tons be admitted in 1 hour, how long will it take to admit 60 tons? The proportion is as follows: As 63:60: :1:x=60 + 63 = 88 hours. Next, if in 83 hours the ship has to go 40 miles, what must be the speed per hour? As 8§h.;:1h.: :40m.:x= 40 + 88 = 4} miles, Proof: 83 x 43 = 40 miles. 288 PROBLEMS witH THEIR SOLUTIONS. Tue Ort PROBLEM. A quantity of flax seed being converted into oil, the result was found to be 658 lbs. of oil and 1276 lbs. of cake; how much oil is that to the bushel; what per cent. ; and what is the value of the oil at 10¢ per gallon; a bushel of seed being 56 lbs., and a gallon of oil 74 Ibs.? Solution. — Here, the rule given on page 149 is applicable, thus, 658 + 1276 = 1934, the whole number of pounds obtained from the seed. Now, as 1934 : 658: : 56: x; and we have 658 x 56+1936=19.05 lbs. to the bushel. . Erne this by 74 lbs. to the gal. we have 19.05 + 71 = 2.54 gal. 24 gal. nearly to the bushel. ‘eed we have 1934: 658: :100:x, or the percentage, that is, we have 65800 + 1934 = 84¢ nearly. Finally, dividing the number of pounds of oil, viz., 658 by 7% gives the number of gallons; 658 + 73 = 87.7 gals. nearly, and at 10¢ per gal. the value is $8.77. Nots. — To divide by 74, add one-third and take one-tenth. (See exam. 1, p. 249.) THe CHICKEN PROBLEM. Do figures lie? Let us see. Two women sold 30 chickens each and agreed to divide the proceeds equally. One sold her chickens 2 for $1, getting $15 for 80 chickens and the other sold hers 3 for $1, getting etd, for her 30 chickens. This made $25 for 60 chickens. The merchant being asked to divide the money, said: You sold yours 2 for $1; and yours, 3 for $1, that is 5 for $2; well 5 into 60, 12 times and 12 times 2 are $24; $12 each. But the women received $25; how can this be explained? Solution. — 2 for $1 = 1 for 50¢ 3 for $1 =1 for 33k¢ 2 for 834¢ or 1 for 413¢ average price, Now, 60 @ 412¢ = $25. PROBLEMS WITH THEIR SOLUTIONS. 289 Tuer Contract PROBLEM. Four men contracted to do a certain job of work for $8600; the first employed 28 men 20da., 10 h. a day; the second, 25 men 15 da.,-12 h. a day; the third, is men 25 da.. 11 h. a day; and the fourth, 15 men 24 da., 8 h.aday. How much should each contractor receive ? Solution. — 28x 20 10=5600, the number of hrs. for 1st contractor. 25X15 12=4500, =‘ Py ‘fone oad exe 4050, 4 a % ye aard as 15X24x 8=2880, ‘“ i 4th ss 17930, total number of hours for $8600. And here the rule given on page 149, for division into proportional parts is applicable; the question resolving itself into this: _ If $8600 be paid for 17930 hours’ of work; what should be paid for 5600 h., 4500 h., 4950 h., and 2880h.? For this we have the following: 1st. As 17930 : 5600: :8600 : a ENS = $2686.00 2nd. As 17930 : 4500: : 8600 : Se = 2158.39 8rd. As 17930 : 4950 ; : 8600 : age eee = 2374.24 4th. As 17980: 288: :8600:x esi 1381.37 $8600.00 Tur Coat PRoBLeEm. A boy agreed to work for a mechanic 20 weeks, on condition that he should receive $20 and acoat. At the end of 12 weeks the boy quit work and received $9 and the coat. What was the value of the coat ? Solution. — The boy’s loss in 8 weeks (20—12) was $11 (20—9). If he had worked the 8 weeks he should have received $11 and $9. Now, if for 8 weeks he should have received $11; what should he receive for 12 weeks? KaG 1olers hee r= At = $16.50 he should have received $16.50; but he received ee 9.00 $9, the coat, therefore, was worth the difference $7.50. £7.50 290 PropieMs witH THEIR So.urions. MarxinG Goops. Short methods for trade discounts have already been given from page 232 to 237. In connection therewith, it may be well to give here.a few examples on the mark- ing of goods, calling attention to the fact that losses may be sustained when supposed profits are being made. Exam. 1. Suppose goods are marked to sell at 40% above cost, and we offer a discount of 20%, our profit is not 20% but 12%; for the reason the discount is not cal- culated on the vost of the goods but upon the marked or asking price, which includes the first cost and the per cent. of profit. Solution.— What cost $100, we have marked at $140 and on this we allow a discount of 20%, or }...... 28 showing a profit of only 12¢. $112 Again, suppose, without full consideration, we offered a discount of 30% on the foregoing, instead of 20%; what would be the result? Here; what cost $100, we have marked to sell for............ $140 and 380% of this is found to be $42, which on being deducted... 42 shows a loss of $2. ; $98 RuLE.— To mark goods so as to allow a discount on the asking price and have them net a certain profit: Divide the net, or desired price by 100 less the discount. Exam. 2. What must be the asking price of goods, to allow the purchaser a discount of 20% and net the manu- facturer $25 per dozen ? Solution. —Here, we have 100%—20%=802; decimally expressed, .80. Dividing the desired price, $25, by .80, we have 2500 + 80 = $31.25. Or, Applying the rule given under the head of percentage Case V, page 118; the question may be asked thus: What number diminished by 202 of itself is equal to 25? And for this we have the following proportion : } As 80:100: : 25: x = 2500 + 80 = $31.25 Proof: 20¢=iand } of $31.25= 6.25 deducting $6.25 we obtain. ..... $25.00 Marxina Goons. 291 Exam. 3. What must be the asking price of a piano costing $350 to allow a discount of 25% to the buyer and still gain 20% above cost ? Solution.—To realize the desired profit in this, 202, or } of $350.00 the cost, is added to itself; this gives $420, or the net price 70.00 at which the manufacturer wishes to sell the instrument. $420.00 Now, what sum diminished by 252 of itself is equal to $420? As $75 : $100: : $420 : x = $42000 + 75 = $560.00 the asking price. Hence the following : RuLE.— Add the desired profit, or rate per cent. to the cost priee, multiply this by 100 (annexing two ciphers) and divide the result by 100 less the rate of discount. Proof: Marked price $560 Less 25% discount 140 Net selling price $420 Less added profit 70 Cost $350 RULE.— 70 find what rate per cent. of discount may be allowed on the asking price to realize cost: Subtract the cost from the asking price, annex two ciphers, and divide the result by the asking price. Exam. 4. If the cost of a piano be $350 and the ask- ing price $560; what rate per cent. of discount can be offered to realize cost ? Solution. — $560 — $350 = $210; then $21000 + 560 = 3744¢. Case 2 of percentage, page 115, is applicable here: Given the per centage and base to find the rate. Deducting the cost from the asking price gives the percentage. In the present example, 560 is the base and 210 the percentage; and the problem may be resolved into this: If the percentage on 560 be 210; what is the percentage on 100? And for this we have the following proportion : } As 560 :210: :100:x = 21000 + 560 = 38734. _ This explains the foregoing rule. Proof: 374% of 560 = $210 and $560 Less 210 Cost $350 _ 999 Marxina Goobs. RULE.— To tell quickly what a single article should sell for, when commodities are bought by the dozen, to gain a certain per cent. of profit: Divide the cost per dozen by 10; the result will be the selling price including a profit of 20% always; and from this any per cent. of profit may be readily obtained, illustrated in the following examples: Exam. 1. If hats be bought for $35 a dozen; what must be the selling price of a single hat to make a profit of 20%? Ans. 1, of $385 = $3.50. The reason for dividing by 10 to obtain a profit of 202, will be understood from the following proportion: As $100 considered as cost, is to $120, selling price, so is $385 cost to its corresponding selling price; and as 12 articles is to 1, so is the cost of 12 to the selling price of 1; thus: Reducing thistoasimple As100:120: :385:x proportion, and cancelling 12 ee 1200 and 120, we obtain 1200 : 120: :35:x 10 to 1 as 85 tox, and divid- 10: 1: :35=385+ 10= $3.50 ing 85 by 10, we get the selling price of a single hat including a profit of 20%. (See exam. 3, page 146; and note page 147.) Exam. 2. If shoes be bought for $32 a dozen pairs; what should be the selling price of one pair to realize a profit of 50%? Solution.—To make a profit of 20%, we take one-tenth of $32=$3.20 and this represents 120%; but we desire a profit of 50%, ora selling price of 150%; the difference is 30% (150—120) and 30% is one-fourth of 120%, so we add to $3.20 one-fourth of itself, .80 or 80¢; this gives the selling price of a pair, $4. $400 Proof: 12 pairs cost $82.00 Added profit, 50%.... 16.00 Selling price of 1 pair $48.00 + 12 = $4 Nots.— To make 60% profit, add % of itself to the 20% (160—120=40) and 40% is 4% of 120%; to make 30%, add og (120— 120 = 10) and 10% is 149 of 120%; for. 8314%, add (13314 — 120 = 184) and 1314% is 1% of 128%, ete. And when the desired profit is less than 20%, deduct, thus: To make 1624%, deduct 14, (120— 11634=314) and 314 is 146 of 120%; to make 15%, deduct 14 (120—115=5) and 5% is 144 of 120%, etc. tING OME UL TLON. Involution is the process of raising a number to any proposed power. A Power of a number is either the number itself, or the product arising from using the number a certain number of times as a factor. The first power of a number is the number itself. Thus, the first power of 5 is 5. When the number is used twice as factor, the result is the second power, or the square, of that number; when three times, the third power or cube; when four times, the fourth power, etc. Powers are denoted by a small figure placed above and to the right of the number, to show how many times it is taken as a factor; thus, 5! = the first power of 5. 5? = 5 X 5 = 25, the second power, or square of 5. 5°? =5 xX 5 X 5 = 125, the third power, or cube of 5. 5¢=5 X5 X5 X 5 = 625, the fourth power of 5. The small figure above and to the right of the number is called the Index, or the Hxponent, of the power. ; Notr.— The number of multiplications is one less than the numbar of factors As the first power of a number is the number itself, its index or exponent is generally understood. PRINCIPLES I. A number is raised to a given power by taking it as a factor as many times as there are units in the exponent; thus, ei x Xo "1D. Il. The product of any two or more powers of the same number, is equal to the power indicated by the sum of their exponents thus, Pao 419 0) (0 XO KD) = OPT 8 SS BS. Nore.—It frequently happens in the multiplication of decimals, or in raising a decimal to a proposed power, that a greater number of decimal figures is obtained in the product, than is necessary for practical accuracy. This may be avoided by making use of the sclcwsne contracted method for the multiplica- tion of decimals: RULE. (1) Count off, after the decimal point in the multiplicand, (annexing ciphers, if requisite) as many figures of decimals as it is necessary to have in the product. (2) Below the last of these, write the unite figure of the multiplier. and write the other figures in reversed order. (3) Then multiply by each figure of the multiplier, thus in- verted, neglecting all the figures of the multiplicand to the right of that figure, except to find what is to be carried; and let all the partial products be so arranged, that their right hand figures may stand in the same column. (4) Lastely, from the sum of the partial products, cut off the assigned number of decimal places. 294 | INVOLUTION. Exam. Multiply 7.24651 by 81.4632, so that there may be three decimal places in the product. CONTRACTED METHOD. COMMON METHOD. 7.24651 Here 1, the unit figure of the multi- 7.24651 2364.18 plier, is set below 6, the third decimal 81.4632 579721 ~+figure of the multiplicand; 8, the figure "71449302 9247 which precedes 1, is put after it; 4, the 94 |73952 9898 figure which follows it, is set before it, A34'7906 435 etc. Wethen say, 8 times 5d are 40, and1 9898 604 99 (carried from 8 times 1) are 41: 1 is then WOABI51 { set down ane 4 carried, oo the ces of 517972018 san ao, the work by 8 proceeds in the usual Way.