— SS SS GZ ~ eee ))} WY Mi Wy (G Hi Hh Wy) WH) (« Cc (G a“ ) ) ) ) ) y)) a. oe , )) )) i |) ( { UNIVERSITY OF RNC i A cK ) LIBRARY _ } | yi) 1) a) We K( Hl | z Pt ae a Lynn ee l (( ) — )) /; 7) ) 7 \ ‘ a KN ui i ) ) ) \ _ \ (« a )) )) yyy Wy Wy He )) Hy Peg SNS Na = goin oe: SE See a — SS NS itm See Saree ee Se ee SE em ee ERI IU ERNE LOO ETON, | SOS ee Se RS SID Se OE I Soe LL = — EE , / SS SS SS LSS a So Sts See ROTTS Se ZO ge 2. i roam FEE Ea ORC FE yO i SE ae ee —_——_ a FE Dae cata Ce eo / Se Dre NS a : od el ge a te ee eg OO aE ee ee LOOT Is A Tir AL LD Er K\ ee nT gee —— ae Ze FETE BoC > A a ath cat PET EE TR See a SU SS See aa SLA —= i) WY / ))))) ) )))) Wy BOE Wh SS eS AN \\ )) Wh OEE —— )))))))) y WH, Wi «a ( ce ) a " cl ») )) Y cq ie Ce )))) c : ))) ) ))) \ a )) “ ( Cl REVISED AND. ENLARGED RAPID CALCULATOR, FOR The Use of Public Schools, Business Colleges, Teachers, Bankers and Private Students. A Complete Work on Rapid Calcu- lation, CONTAINING ALL THE SHORT METHODS KNOWN TO THE EDUCATOR OR BUSINESS MAN. NE-Wsy OR Re Grey U.S. A. Second Edition, January 1, 1904. Third Edition 1907. Te 3 ; og eae hoe 3 a ~ eR BY C. W.. ROBBINS 3 ey a8 oo a ee N Commerce. KOGA CSBtechert 1,35 PREFACE. In this age of steam and electricity, the author confidently believes that any text-book on arithmetical calculations that saves time, lessens the possibility of error, and teachers labor-saving methods, will be wel- comed by the intelligent teacher, business man, or private student. In this work all of the short methods worthy of notice, known to the teacher or business man, are given. Some of these methods, in the judgment of the author, are very much better than others, but they are all given that the student may select the one that suits his pe- culiar fancy. Thorough and clear explanations of each method have been given that the private learner, as well as the pupil in the school, may get a clear understanding of every method in the book without consulting a teacher. It can truly be said that the average pupil as he graduates from most schools is not prepared to take a position in business and perform calculations with any degree of satisfaction. But little time is devoted to practice in our average school on any subject after the pupil has mastered the principles so that he can solve problems under that rule. Hence, he finds himself leaving school poorly prepared for business, as he can neither solve problems, add nor multiply rapidly or ac- curately. He is therefore compelled to enter the school of business, and there learn what he should have learned in his earlier school days. The thoughtful and progressive teacher finds him- self at a loss to teach this subject, as he would like to teach it, as he fails to find anything that will render him the necessary assistance. To arrange a work that 124715 A _ RAPID CALCULATOR REVISED. will be a guide and an invaluable companion to the in- telligent and progressive teacher, student or business man, a work, which if carefully studied and practiced, will lead the thoughtful and industrious pupil to a knowledge of rapid, accurate figuring, that will be of inestimable value to him all his life, is the aim of the author. RAPID CALCULATOR REVISED. 5 ADDITION Rapid addition depends upon a process of group- ing. There is just as much reason for pronouncing each letter of a word when reading, as there is for considering each figure separately when adding. Forty years ago, the pupil was taught to read by spelling out each word separately, but what would you think of a person who advocated such a method to-day? Why do we still continue spelling out the figures in adding the same as our forefathers spelled out the words in learning to read? We do not say “c plus a, plus t,” equals cat. Then how absurd to say 5 plus 4 plus 2 equals 11. It is much easier to learn the combination of fig- ures than it is to learn the combination of letters. Then why do we add to-day by the same method that Noah probably used in computing the number of tim- bers in the ark? The practice of seeing and adding a single figure at a time is no help to swm reading, which is the only true basis on which we can hope to acquire any great speed in addition. Instead of looking at 2 and 3 as separate and dis- tinct numbers, we must see their sum, 5, with as much ease and rapidity as we prononuce the word hat from glancing at the letters which form it. Let no one suppose that he will ever attain the highest results in addition so long as he continues to add one figure at a time. 6 RAPID CALCULATOR REVISED. We believe that the methods of addition herein presented will at once commend themselves to every live teacher, and we know they will be welcomed by the accountant for their simplicity and rapidity. There are several excellent methods of adding, but all of them are based on one foundation, namely, sum reading or grouping. The pupil should thoroughly familiarize himself with the following table, remembering that the one thing to be learned is, that { is simply another form of 2, and that ? is simply another form of 7, etc. RAPID CALCULATOR REVISED. a TABLE No. 1. 1 1 1 2 sep 5 ie iby pe 4. 3 ee oS Tha eee Nese Seles 6. 5. 4 bea Fae feed sO Tee Sgn cs Fee e Geta Os" D EXPLANATION. In this table we show 20 different combina- tions of numbers to pro- duce results from 2 to 9 inclusive. In practicing on addi- tion tables, the pupil should not add the num- bers, but merely glance at them and call their sum: this tis imperative. The complete mastery of these tables will lay the foundation for _profici- ency in addition. Do not hesitate in passing the eye over these num- bers, but firmly fix your attention upon your work and run over them very rapidly. TABLE No. 2. LO sto AS O aie hee Oe Do ea a ae 1 ine Peles epee © Bie cA OSB G 2 Bae a eee ete © Ass BG ~ Daebek 5 aman 6. en egy | er 3 aay 6 Git Qos Fie sie 9, 3 8 9 9 9 RAPID CALCULATOR REVISED. 10 11 12 13 ‘14 15 16 17 18 EXPLANATION. This table shows the combinations of numbers from 10 to 18, inclusive. After the pupil is thoroughly familiar with these tables, he should practice on the numbers, naming only the sum of the units, thus: This practice is abso- lutely necessary to secure readiness in adding the units and counting the tens. If the pupil has thoroughly learned the first table, when this one is mastered, he will know the sum of any two numbers at sight, and will be in possession of the fundamental princi- ples of addition. If the RAPID CALCULATOR REVISED. 9 pupil knows the sum of any two numbers at sight read- ily, with a little practice, he will know the sum of any four. If the pupil has learned that 2 and 3 is 5 and that 4 and 8 is 7, and that 7 and 5 is 12, he knows the sum of 2 and 3 and 4 and 3 by simply looking at them; for, it is evident that, if he knows the sum of the first two and the sum of the second two and the sum of the numbers that are the sums of the first two and second two, he will know the sum of the entire four numbers. GROUPING BY THREES. The following table contains all groups of three figures. These should be committed to memory so that the pupil can instantly call their sum without performing an addition. To learn to group. three figures at a time is not at all a difficult task, and the pupil once in possession of this knowledge has pro- ceeded far on the road to rapid addition. TABLE No. 3. | EY pat pad pa RAPID CALCULATOR REVISED. 10 TNA Hvis NN A alos odds QUAD © rio 09 dA r4r19 10% 11 12 13 tH 10 OD kor Kor COM 26 or or aor’ | 27 RAPID CALCULATOR REVISED. 13 GROUPING EXERCISE. The pupil should now take columns and, passing the eye quickly from top to bottom, group enough fig- ures to make ten or more, calling only the unit figure. No addition should yet be attempted. 5) 4 qe Be EXPLANATION. 3 Zrie rT. 4 Bt 5 Beginning at the top 8 6 of the first column, we 4 have 5 and 6, which in 6 2 reality makes 11, but the 3 pupil should call only the 5) At 1 units’ figure 1. 3, 2 and 2+ 5 2) 7 makes 12, but call it 8) 2. 8 and 6 makes 14, but-callt:4.—-5, 2 and 8 3] makes 15, but call it 5. 0 5 3 and 7 makes 10, but 7 Al 6 call it 0. 6 and 9 makes 9) Thabutcailit a. Sool-and 6 6 makes 15, but call it 5. 5 4,2 and 7 makes 13, but 9 2) call it 8. Always group 2, enough figures to make 8 A‘ 1 at least ten, and call only 1 5 a the units’ figure. 6 14 RAPID CALCULATOR REVISED. wh Si a age Ol ee The pupil should group numbers similar to these, and practice on them until he has no trouble in instan- taneously naming the sum of enough figures to make ten or more. This practice is absolutely indispensable if we wish to secure the best possible results. While practicing on these exercises, the mind should be cen- tered wholly on the units; the tens will take care of themselves. (The teacher will here give the class a few prob- lems in addition for the purpose of impressing pre- vious ‘principles. ) | METHOD OF ADDITION BY DROPPING THE TENS. It has long since been decided by accountants that all methods of addition by which we can hope to attain any great speed must have for their foundation GROUP- ING. But the process of adding and retaining large numbers in the mind is, with many, a very laborious task, and one which in some cases is totally impracti- cable. To obviate this difficulty and reduce addition to its simplest form, we here present .a method which is based upon a process of grouping by tens. By this method of addition the longest column can be added with the same ease as a short one. The largest calcu- lation that you will ever be compelled to make, by this method, is, to add nine and nine together, which we call ezght, no attention being paid to the tens. No col- umn of figures can possibly contain a calculation which will be more difficult to perform than the one above mentioned. This brings expert addition within the reach of everybody. Heretofore it has been thought that only the man with a giant faculty for numbers could rapidly add column after column of figures, and write their results with wnerring correctness. A ten- year old child can be taught by this method to add a column of figures of marvelous length in an almost incredible space of time... Remember, however, that RAPID CALCULATOR REVISED. 15 while we are changing addition from a laborious task to a pleasant recreation “There is no excellence with- out great labor,” and it is with this as with everything: else; practice is the essence of success. Many methods of addition are advocated which are entirely beyond the grasp of the ordinary individ- ual, and only suited to experts, but this one is within the reach of everybody, and by its use the twelve-year- old school boy can vie with many experienced account- ants and bank cashiers. The following cut illustrates how the tens are held on the fingers when adding: ey ~ BOF — Ss 150 SF FI fii f / A ih Nh i he) GnbukGrv) Zi! j\"" l WW ZZ ( ; whi CN Sonbir ous 7 SS om ares : ; fi y ( bf EGA Go | FO WN . iar , = a pL SLO JIC, "\ £ou As will be observed, fifteen tens may be retained on the fingers. If the column should make more than 150, you may commence at the first, and go over the fingers the second time, and if necessary even a third or a fourth time. We make a record of the tens by simply placing the end of the thumb on the first joint of the first fing- er for one ten; on the first joint of the second finger 16 RAPID CALCULATOR REVISED. for two tens; on the first joint of the third finger for three tens; on the first joint of the fourth finger for four tens; by straightening the thumb for five tens; by placing it on the second joint of the first finger for Six tens, ete. CO OT ho Ve eee ees esl ee ee io) otherwise count but one ten. EXPLANATION. * ff Commencing at the bottom of this column, we mentally see 1 (8, 3) and place our thumb on the first joint of the first finger. To the 1 that we now have we add 5 (9, 6) which gives us 6 and we place our thumb on the first joint of the second finger. To the 6 we add 5 (8, 7) which gives us 1, and we place the thumb on the first joint of the fourth finger. It will be ob- served that this time we moved two places. Whenever the sum of the units is ten or more, you must count two tens, To the 1 we add 2 (3, 5, 4) which gives us 3, and we straighten the RAPID CALCULATOR REVISED. Le thumb. To the 3 we add 5 (9, 6) which gives us 8, and we place the thumb on the second joint of the first finger. To the 8 we add 5 (7, 8) which gives us 3, and we place our thumb on the second joint of the third finger. (It will be observed that the sum of the units here gives more than ten.) To the 3 we add 4 (5, 2,7) which gives us 7, and we place our thumb on the second joint of the fourth finger. We now write the 7 at the bottom of the column for the units’ figure of the answer. By looking at the position of the thumb, you will observe that we have 9 tens; hence the result to the column is 97. We now give a column in which the process of grouping is carried to a greater extent than in the one previously explained: EXPLANATION. Having given quite a lengthy explanation in the preceding example, we. will omit a portion here. Beginning at the’ bottom of this column we mentally see 6, to which we add 3, which gives us 9; skipping over the 8, 7 and 5, for they simp- ly make two tens, we add 5 to the 9, which gives us 4, to which we add 8, which gives us 2 for the units’ figure of the ans- swer. By observing the position of the thumb, you will notice we have 8 tens; hence the result is 82. ’ € 3 —~—< CO iw) Or a ergo @ % eas) a) WORTNAONUTNH WORD + ror) Se ey Sas, sae co | (o/) iw) 18 RAPID CALCULATOR REVISED. Add the following numbers, retaining the tens on the fingers: | 1892 3221 423 100 5 4563 56 1893 17 7891 1359 465 356 3265 267 3276 78 5432 15 8945 492 9891 999 1234. 1379 1234 653 9765 pie 623 5678 9875 3223 _ 5981 9087 4105 4671 358 5634 65 6006 495 1218 1111 333. 7215 7654 866 495 9634 3322 5963 99 519 7845 7877 63 673 6005 123 1895. 4321 3478 4567 1900 6743 5795. 890 654 8976 3881 976 3112 - 1231 8462 5A 5485 1892 6421 3219 6666 - 17 7399 9876 187 418 8888 998 422 RAPID CALCULATOR REVISED. 19 METHOD OF ADDITION BY DROPPING THE TWENTIES. In adding by the method of dropping the tens on the fingers, some persons find it difficult to move the fingers as fast as they are able to group the numbers. To such, we respectfully recommend the method of dropping the twenties on the fingers. In adding by this method, the fingers are used in exactly the same manner as in the method previously explained, with the exception of the fact that, when the thumb is placed on the first joint of the first finger, it indicates one twenty; on the first joint of the second finger, two twenties, or forty; on the first joint of the third fin- ger, three twenties, or sixty, etc. Fifteen twenties, or three hundred, may be re- tained on the fingers, as the following cut will illus- trate: In adding by this method, we will begin at the top of the column, and add downward, though the pupil 20 RAPID CALCULATOR REVISED. may begin at either the bottom or the top. This is simply a matter of convenience, and will not affect the speed. EXPLANATION. We begin at the top and first see 1 (7, 4) and adding 1 (8, 3) to this we have 2 and one twen- ty which we record by placing the thumb on the first joint of the first fin- ger. Looking down still farther we next see 2 (5, 7), whieh added to the 2 we have, gives us 4. We next havel (6, 5), which added to 4 gives another twenty and 5. 5 plus 8 (9, 4) equals 8, plus 1 (3, 8) gives us another twenty and 9; plus 3 (7, 6) gives us still another twenty and 2: plus 3° (4,,.9)-eqtia =e. plus 8 (5, 8) equals one more twenty and 8; plus 3 (7, 6) equals still an- other twenty and 1. RAPID CALCULATOR REVISED. PA We find now that our thumb is on the second joint of the first finger, which indicates six twenties, or 120, which added to one gives us the result 121. HOW TO KEEP RESULTS TO EACH COLUMN SEPARATELY. In adding, the result to each column should us- ually be written separately. Write it in light lead pen- cil figures to the right-of the numbers added and it need never be erased. Add the following: To add these numbers, 465 we begin at the top of 338 first column, and _ by [2A grouping, say 4 (5, 8, 637 Al) OL (ASG Deel) ee 584 (fai aol, aad Woy ex- 976 amining the fingers we 825 find 3 twenties recorded, 387 which gives us 62 as the 420 Oe result to the first column. T47 59 We now begin at the top 886 69 of the second column and » ellen BaMecCm CG Mie or on Ole ek 6992 Ciebe; Oe) loc See8), and we have 53 as the re- sult to the second column, plus 6 to carry gives 59. We next begin at the top of the third column and Bays 14,135,117, 6, 5), 2: (9,8), 4 (3,,4,-7, 8), and we have the result 64, plus 5, to carry, gives us 69. 22 RAPID CALCULATOR REVISED. Commencing at the bottom and reading up we now have the result to all the columns 6992. Do not be discouraged if you cannot use this me- thod the first time you try, but practice diligently and you will be rewarded many times for your trouble. Remember that this is an age of steam and elec- tricity, and the RAPID ACCOUNTANT has no time to throw away adding by the old methods; and, if you wish to be quick, drop all single figure methods at once and master one of the methods herein presented, and ad- dition will be rendered simple, easy and rapid to you. This subject is the corner stone of all calculations. RAPID CALCULATOR REVISED. 23 CIVIL SERVICE METHOD OF RETAINING RE- SULTS TO DIFFERENT COLUMNS WHEN ADDING. The following method of retaining results is very useful to clerks who are compelled to stand at an office window and add when several individuals are asking them questions, or likely to ask questions at any mo- ment: EXPLANATION. 2643 5782 . By this process, the 6958 47 result to the second col- 4737 | 29 umn is written one place 6528 36 to the left of the result 4279 nee | to the first coulmn; the doar result to the third col- 30927 umn, one place to the left of the result to the second column, etc. When through, add the partial results, and you will have the result to the entire number of columns. GROUPING METHOD OF ADDITION. Such persons as can instantaneously tell the re- sult of numbers like 27 and 18, 45 and 23, 89 and 16, etc., may find no great advantage in retaining the tens or twenties on the fingers. For the benefit of this class, we here give what is known as the Grouping Method, based upon a scientific method of sum reading. 24 RAPID CALCULATOR REVISED. In learning to add, or in teaching children to add, the sign (++) should never be used while practicing; for the reason that we are to see and think only the sum. We are not to calculate, but merely to see and to read sums the same as we pronounce words when reading. 3 719 7? 6 63 EXPLANATION. 5 2 Beginning at the bot- 4 tom of this column, we 3 see 2 (8, 4) and adding 4 2 (5, 7) to this we have 24; adding 10 (8, 2) to 9 this we have 34; adding D 15 (6,9) to this we have 6 49 49. 8 0 2 34 In making this last named addition, do not 7 try to add 15 to 34 by the 2 ordinary process. Think 5} 24 of it as the number 15, but center your mind on 4 the five, and think what 2 5 and 4 makes; then call 8 the result 49. 79 Now center your mind on the units’ figure 4 of the-next group (4, 3, 2, 5), and think what 4 and 9 will make, then call the result 638. Now center your mind on the units’ figure 6 of the next group (6, 7, 3,) thinking what 6 and 3 will make, and call the result 79. If you wish to make a success of this method of adding, you must center your mind on the units’ figures, and not be afraid to work. RAPID CALCULATOR REVISED. 25 Remember practice is the golden key that unlocks the vault of knowledge. Add the following numbers by the grouping me- thod, also by the method of dropping the twenties on the fingers, retaining the results of the columns; and finally by the Civil Service Method. 4567983 3497658 8735215 1930256 2653149 6958732 7294631 9786320 5462934 9365287 5432167 3785645 2591034 8963451 8264537 6728553 7853296 3459781 3625149 4563145 7284536 5938765 9784632 3165988 4264881 1236547 4577112 1234567 9897965 9922333 9756312 4243659 8546729 6543210 8631297 3927475 8996554 4567823 6789342 3264876 6759432 1234567 5796831 5678919 9876543 9564234 3459256 2109875 7326845 12138145 7878955 26 RAPID CALCULATOR REVISED. CIPHER METHOD OF ADDITION. This process of addition, like all other rapid pro- cesses, has for its basis, grouping; but, by no other process, do we group in such a peculiar manner as by the Cipher Method. The successful application of this method lies in the pupil’s ability to pick the tens, twen- ties, and thirties out of the columns instantaneously. The numbers, by this method, are not added in the or- der in which they are written in the column, but we skip from place to place, selecting those numbers which will give a cipher termintaion. This may, to the un- practiced eye, seem to be a very awkward and clumsy - process; it seems that figures would often be left out entirely; but it is wonderful what marvelous results will follow intelligent, well directed practice. Before attempting to add by this method, the pupil should thoroughly familiarize himself with the fol- lowing tables: Every group of two figures amounting to ten: Every group of three figures amounting to ten: ~~ ee et CO me DO ea Ww pt OT No _ CG —- me WwW Ww opye) 1 hS)ia=s SS) | INO? 925 -Or RAPID CALCULATOR REVISED. Zt Every group of three figures amounting to twenty: Septet he Os OD.) Sie Oe Ob hoortsiace OFe P26 tit Oro Tae ibe 6G Every group of four figures amounting to thirty: GC © H 00 69 © Oris O Or Os © © oon ca PROCESS OF ADDING. Daonownmwm won o each =) EXPLANATION. Beginning at the - bot- tom of this column, we first take 8 and 2 which gives us 10. We next take 7 (skipping the 5) and 38 which gives us 10 more, or 20, adding in the 5 we now have 25, to which we add 6 and 4, which gives us 35; now adding 8, 7 and 5, we have 55 to which we add 3, 2 and 5, which gives us 65; next adding 2 and 8 we have 75, to which we add 7, which gives us 82, the result toe -othe=;) columnies Alt 28 RAPID CALCULATOR REVISED. is by no means necessary that the pupil follow the exact method of grouping that we have indicated nor, indeed, is it necessary that any two pupils should group by the same method, but remember that you must group. | It is marvelous how many cipher terminations can be made from a few mental transitions of figures. In adding by this method, the pupil should name only the figures which occur in the third column; namely, 25, . 35, 55, ete. Every figure is to be seen and used, but not called. HOW TO ADD SEVERAL COLUMNS AT ONCE. This is a process of reading numbers, and when once acquired, very few additions are performed. This will be found a very useful method to bank cash- iers, tellers, and all persons who are engaged in a busi- ness in which there are many short columns of figures to be added, especially columns containing a good many ciphers. For the ordinary individual, we do not believe that this method can vie with the single column method in adding long columns of solid figures. By — this process, we always read two numbers at once, but | not more than two; while, by the single column process, we often read three or four numbers at once. Many persons suppose that, if a man adds two columns at a time, he must necessarily be much more rapid in addition than the man who adds one column at a time, but such does not necessarily follow; for the man who adds one column at a time may read it off in groups of four figures each, while the man who adds two columns at a time only reads in groups of two fig- ures each; hence will it be seen that the man who reads four figures as one will be more rapid in addition than RAPID CALCULATOR REVISED. 29 the man who reads two figures as one, other things be- ing equal. This method of addition, however, fills a long felt want, and is much superior to any single col- umn method for adding short columns of figures, such as we usually find on deposit tickets, etc. Below are given a few exercises for practice: Mince ahi toe EXPLANATION. Aout 3 126 36 Begin at the left and lg OR, 2, read the amounts at a 2 PEP ape aos glance without adding them in the _ ordinary way. Remember, these numbers are to be read, not added. Simply glance at them, and call their results, 59, 96, 68, 97, ete. The ambitious pupil will make for himself, hundreds of such exercises as the above, and practice on them until he can read them as easily and quickly as he can read the words in a book. By this method the pupil should always commence at the upper left hand corner, and read to the lower right hand cornev. It will be observed that, in the ex- amples given above, the sum of none of the digits makes as much as ten. They are, however, quite as readily disposed of when they make ten or more, as the following examples and explanation will illustrate: 30 RAPID CALCULATOR REVISED. EXPLANATION, 48 39 58 262° 47-39 All there is to do to read these numbers it to look one place ahead, and when the pupil sees two numbers that make ten ‘or more he should eall 65 43) 3526 the digits to the right 1p AD one more. Sometimes it Se may be necessary to look two places ahead, but oc- casions of this kind only occur when the figures one place ahead give 9, and those that are two places ahead give 10 or more. In the above exercises, with- out noticing the right hand digits, the first number will be read as 64, but observing that the sum of the digits on the right makes more than 10, we call the sum of the left hand digits one more, or 7; hence we have 74. When the pupil goes to call the right hand digit, it is not at all necessary that he should know the sum of the | left hand digits, but he simply wishes to know whether _ their sum is ten or more. This fact can be known at a glance. No attention is paid to carrying as is given in the customary method. In this method of addition, only two digits are combined vertically at the same time. The pupil should now write a large list of numbers similar to the following, and practice on them till he can read them rapidly and accurately. Don’t add, but read, and remember that you should always begin at the left and read to the right. | ] 245678 456782 514236 154231 7) ca a 76145 34218 674513 711231 ! 4 ioe reat - ete AE ep fas ff i) Un} if % A og K R . Ms ‘ ; ee » , : f eZ *? 15. = aa FeLi SSAA SESSA: Yy, Z| xy - ef ry if i] vy ty h' 4, wy js ‘ v) ' | RAPID CALCULATOR REVISED. 869845 414316 45236 41321 Nh { | 31 32 RAPID CALCULATOR REVISED. PROCESS OF ADDING SEVERAL COLUMNS ' AT ONCE. EXPLANATION. The pupil should di- A Oe vide these numbers into Tera | sae eey § syllables. Instead of yee en Dre no calling 2 and 6 twenty- Seu tee g six, they should be read Dnt Gh ay LOR ae “two. six’ (2. G0 ee ee 2 Sar aM AU eles should be read “three 6. ee Oral eae one” (38...1). This pre cess of dividing numbers PAE a into syllables will doubt- less, to the casual obser- ver, appear to be a thing of little impor- tance, but, to the one who is performing the addition, it will be found to be a most powerful auxiliary. Beginning at the upper left. hand corner of the col- umn, and reading the first two numbers, we have 5. .7. These numbers are not to be added, but read. Now, conceiving the 5 and 7 to be placed above the 4 and 2, we read 9. .9, now conceiving the 9 and 9 to be placed above the 3 and 0, we read 12. .9; again conceiving the 12 and 9 to be placed above the 5 and 4, we read 18. .3, eonceiving the 18 and 8 to be placed above the 2 and 3, we read 20..6; conceiving the 20 and 6 to be placed above the 6 and 5, we read 27..1, which means 271, the result of the two columns. RAPID CALCULATOR REVISED. 33 This method of addition needs no commendation from us. Its superiority over the methods ordinarily presented in the school text-book will be appreciated at a glance. Notice the comparision between the two methods; by the old method, to add 24, 36 and 42 we are taught to say that 24 plus 30 equals 54, plus 6 equlas 60, plus 40 equals 100, plus 2 equals 102; while by this method, we have their sum by simply glaneing at them twice. Reading the first two-numbers we have 60, which placed above the 42 and read, gives us 102. The old method of adding two columns may be a very nice thing with which to adorn the pages of some theoretical text- book, but when applied to practical work, it will be seen that the method above presented is superior to it. Any one who can add the above numbers as we have ‘directed, even slowly at first, has only to perse- vere, and the day is not far distance when he will as- tonish himself and friends with his speed and accu- racv in business calculations. The pupil, however, must center his mind on his work. Don’t be afraid of the numbers, but get into the midst of them, and make them talk to vou. Add the following problems by the method pre- viously explained : o4 RAPID CALCULATOR REVISED. 34 4G 52 33 25 15 ST ete enn) 2) Wt 31 eae 15 B40 eee 45 50 60 40 10 11 63.68, "2%. 56 = 2a 20. 42° 18 = 4a 54-0). AB O* 2h 31 “02° - B20 3 1B a 23° 26. 22° 90 an 44. "34. “85. 52. Let us now consider the process of adding THREE COLUMNS at a time. — $. 2.31 4.20 1.22 3.42 $ 11.15 EXPLANATION. Place your pencil on the decimal point in the second number between the 4 and 2, and reading say, $6.51... Now dropping the pencil to the decimal point between the 1 and 2, say $7.73; dropping the pencil to the decimal point between the 38 and 4, say $11.15. Do the work rapidly, without hesitation or repetition. The important thing is, to learn to con- centrate the attention. 2 RAPID CALCULATOR REVISED. _ 35 We will now give a problem containing FOUR COL- UMNS. EXPLANATION. $ 23.40 | 5d LG Place your pencil on 42.02 | the decimal point be- 50.23 tween the 1 and 2, and —__—— say $54.61. Now drop $146.86 the pencil to the decimal point between the 2 and 0, and say $96.63; now drop the pencil between the 0 and 2, and say $146.86, and you have the result to the four columns. You have accomplished a wonderful thing, and a thing which the majority of people look upon as being nearly impossible. You have added an account con- taining figures four columns wide and four deep all at once. 3 With a little practice by this method, you can soon astonish your friends and associates as well as com- mand the respect and admiration of any business com- munity. Below we give a few examples tor practice. The pupil should not stop when he has added these, but he should write out exercises for himself, and practice on them until he can read them without hesitation. S 4.25 3.20 4.00 20.10 3.50 4.80 3.25 4.00 25.60 $ 12.00 8.40 4.25 6.20 4.00 18.25 13.40 20.60 $ 214 $ 500 640 200 125 6435 $ 30.40 18.00 5 6.75. 50.00 10.20 30.40 11.26 30.00 5.35 A,20 45 .65 30,20 - 40.70 ao 4") S 50.10 34.20 45.00 64.20 5.80 540.20 RAPID CALCULATOR REVISED. $ 34.20 40.00 61.21 42.30 20.50 8.45 6.20 3.40 } fy $ 15.00 5.00 4.25 65.20 30.45 420.35 RAPID CALCULATOR REVISED. ot THE DOT METHOD OF ADDITION. This will be found to be a very easy and accurate method for beginners. It will assist in grouping num- bers in tens, and relieve the mind of the strain that addition usually causes. We especially recommend it for children, or, in fact, to anyone who has trouble in adding correctly long columns of figures. 38 RAPID CALCULATOR REVISED. EXPLANATION. This method of addi- tion is somewhat similar to the method of adding by dropping the tens on the fingers. We group enough figures to make ten or more, and place a dot for the tens, re- taining the units in the mind. Beginning at the bot- tom, and grouping the first three figures, (4, 5, 6) we have 15. We place a dot for the ten, and mentally call the units’ figure 5; adding the 5 to the two figures above (3,7), we have another ten and 5 units. We place a dot for the ten, and add the 5 to the two figures above (4, 8) which gives us one ter and 7 units; placing a dot for the ten, and adding the 7 tothe figure9 above, we have one ten and six units; placing a dot for the ten, and adding the 6 to the figure 5 above, we have one ten and 1 unit; placing a dot for the ten, and adding the one to the next two figures above (6, 3), we have one ten and no units; ‘placing a dot for the ten, and adding the two figures above, we have 7 units. We write 7 for the units’ figure of the answer. Now counting the dots, it will be observed that we have 6 which indicates six tens, or 60; hence the answer is 67. Sl MAA waIR WOU wp oo RAPID CALCULATOR REVISED. 39 HOW TO ADD HORIZONTALLY. The accountant will sometimes find it necessary to write several items on the same line and add them horizontally. To rearrange these items so as to add perpendicularly would be a vast amount of unneces- sary work. This method will be found to be espec- ially valuable for making extensions on tax books, etc. a ds Or 9 DO bo ol | co} mono co | DO OTL Co me | ror! EXPLANATION. To add numbers horizontally, we begin at the left and add to the right since the eye is more accustomed to moving from left to right than from right to left. Commencing at the top number on the left we add the units’ figure, 5, of the first number to the units’ figure, 2, of the second number which gives 7; then to the units’ figure, 6, of the third number which gives us 13; then to the units’ figure, 1, of the fourth number which gives us 14. Writing down the 4 and adding the 1 to the tens’ figure, 4, of the first number, we have 5, which added to the tens’ figure, 4, of the second num- ber, gives us 9; which added to the tens’ figure, 2, of the third number, gives us 11; which added to the tens’ figure, 2, of the fourth number gives us 13. Writing down the 3 and adding the 1 to the hundreds’ figure, 2, of the first number, we have 3, which added to the hun- dreds’ figure, 3, of the second number gives us 6, which added to the hundreds’ figure, 4, of the fourth number, 40 RAPID CALCULATOR REVISED. gives us ten; hence the answer to the first row is 1034. We now add the second, third, and fourth rows by a similar process; then adding their several totals per- pendicularly, we have 7831. This method of addition is very practical, and the pupil should make himself thoroughly familiar with it. Instead of adding the numbers horizontally, the pupil may apply the veading method of addition if he so desires. By this method, we would conceive the first number, 245, to be placed above the second num- ber, 342, and read the result 587; now conceiving the 587 to be placed above the third number, 26, we read the result 613; now conceiving the 613 to be placed above the fourth number, we read the result 1034. This last named method is very much more rapid than the first. The only objection to it is the difficulty which some experience in its application, though any one who has thoroughly mastered the methods of addi- tion previously explained will have no trouble in adding these numbers by the reading process. HOW TO PROVE ADDITION. The best and most certain proof of addition is, to add the columns in a reverse manner, and, by the good accountant, this is by far the most rapid process, for, in other methods, very few facilities for grouping are offered, while, by adding columns in a reverse man- ner, the calculator is not deprived of his grouping me- thod. We, however, present a method, which is a modi- fication of the old method of casting out the nines, for the benefit of any one who may wish to use it. RAPID CALCULATOR REVISED. 41 4562—17=—8 | 3150=-20=—0 oe 14386—20—2 | | agen 19,. 1-49-10, 140-1 8694—27—0 A782—21-—3 5986=—-28—1 Bos50e-219, 1--9=-10, 1-00—1 EXPLANATION. — Beginning at the upper left hand corner and ad- ding’ horizontally the 4, 5, 6, and 2 together, we have 17 which we write in a column to the right of the num- bers. Now adding the 7 and 1 together, we have 8 which we write in a column, for the remainder, still farther to the right; adding the 3, 7, 8 and 5 together, we have 23, which we write under the 17. Now ad- ding the 2 and 3 together we have 5, which we write in the column under the 8, and thus we proceed until we have disposed of all the figures of the numbers to be added. When the sum of the digits of any of the remainders is 9, the 9 should be rejected; for ex- ample, the digits of the fourth number from the top when added make 18. 1 and 8 added together gives us 9; we reject the 9 and write a cipher in the third col- umn. Now adding perpendicularly the 1, 3, 2, 5 and 8, we have 19, which we write to the left of the num- bers; adding the 1 and 9 we have 10, and immediately adding the 1 and 0 together we have 1, which we write to the right of the 19. Now adding the figures of the answer, 3, 9, 5, 0 and 2, we have 19; adding the 1 and 9 together, we have 10, and immediately adding the 1 and 0 together we have 1. The figures of the an- 42 RAPID CALCULATOR REVISED. swer, when added and reduced, give the same remain- der as the figures of the numbers to be added; hence the addition is correct. By the ordinary method of casting out the nines, when the sum of the digits of any of the numbers pro- duces a number that is greater than 9, we divide by 9 and retain the remainder, but, by this method, when we have a number composed of more than one digit, we add its digits together, and, if the first addition does not give a number composed of only one digit, we add the digits of the remainder or successive remainders, as the case may be, until we have a number composed of one digit only. It takes a good many words to explain this pro- cess, but when once understood, it is very quickly done. It may be shortened somewhat by applying the fol- lowing method: Remember, however, that when you have a 9 or any two or three figures that will make a 9, such figure or figures should be omitted. Commencing at the upper right hand corner and observing that 4 and 5 makes a 9 we omit them; now adding the 6 and 2 we have 8,-+8 the first figure of the second row, which gives 11; adding the digits of 11 together, we have 2,7 gives us 9 which we drop; ad- ding the 8 and 5 together we have 13, and adding the digits of 18 we have 4, which we add to the first figure of the next number, and thus proceed with the remain- der of the numbers, obtaining the same final remain- der as by the process previously explained. RAPID CALCULATOR REVISED. 43 SUBTRACTION. There are several methods of subtraction, all of which are comparatively simple in their nature. The method, however, of subtraction by addition is, per- haps, the simplest process for the beginner, and in- deed, when once mastered, it will doubtless be found to be the most rapid method for the accountant. By this method, we begin at the r7ght and write in order the numbers under the line, which added to the subtrahend will produce the minuend. If at any time one of the digits of the subtrahend is larger than ° the digit above it in the minuend, mentally add 10 to the figure in the minuend; then find what number ad- ded to the figure in the subtrahend will produce the figures in the minuend. When 10 has been added to any figure in the minuend, the next figure in the sub- trahend to the left must be called one more. Be Gre eae ott EXPLANATION. Peer a 95.4. een Beginning at the right, i fg ee iar. Ba we say that 3 added to 4 will produce 7; hence we write 3 for the first figure of the answer. The next figure of the subtrahend, 9, is larger than the figure of the minuend, therefore we mentally add 10 to the figure of the minuend, which produces 13; we then find that 4 added to 9 will make 13; hence we write 4 for the next figure of the answer. Now calling the next figure of the subtrahend one more, or 4, we find that 1 added to it will produce 5; hence we write 1 for the next fig- ure of the answer. 4 added to 2 will produce 6; % 44 RAPID CALCULATOR REVISED. hence we write 4 for the next figure of the answer. 1 added to 1 will produce 2, hence we write 1 for the last figure of the answer. Pupils will calculuate much more rapidly when taught by this method than when taught by the ‘“‘bor- rowing process.” Having learned the sum of any twe numbers at sight, with a little practice, they will read- ily discern what number added to another will produce a given number. We beg to say at this point that we are not pre- tending to teach a method by which we may explain all the minutia of subtraction, but we are presenting methods to secure rapid and correct work, believing this to be of far more importance to the ordinary school boy than all the technical terms, or high sounding ex- planations that have been employed since the time of Adam. There is no objection, however, to the teacher’s giving any explanation of subtraction that he may see fit, for such explanation will not materially retard the pupil in learning this method. The BUSINESS MAN, however, does not care so much for explanations as he cares for accurate and ready answers. We now beg to call your attention to a method of subtraction similar to the above which begins at the left to subtract instead of the right. Divo: boda Omek EXPLANATION. iA be Bed | — By this method we sim- Lue 28) oe ply begin at the left and write for the answer the numbers which added to the subtrahend wil! produce the minuend. When we look one place ahead, and see a figure in the subtrahend which is larger than the figure above it in the minuend, we must call the figure in the subtrahend to the left one more. RAPID CALCULATOR REVISED. 45 In beginning this subtraction, we look one place ahead, and see that the 4 in the subtrahend is larger than the figure above it in the minuend; hence we must call the first figure of the subtrahend 2 instead of 1. 1 added to 2 will produce 3; hence we write 1 for the first figure of the answer. Now we wish to know what number added to 4 will produce a 3, and we find it is 9; hence we write 9 for the second figure of the answer. 2 added to 5 will produce 7, and we write 2 for the next figure of the answer. 3 added to 2 will produce 5, hence we write 3 for the next figure of the answer, etc. EXAMPLES FOR PRACTICE. vag Wig ety 65 {ga Sees, Seo Ob Neiad = G er italteve ia » 4 4 J al be aD Bw or 4.2 Te eee 96754 33 Dy Bere a SO -4r 10 tos be9 Beer re os a 711468 Geb ou ece oo ea. 7. ee GLO Ove Od eon o. Dela OO) do Se2e9 55 PROOF OF SUBTRACTION. The following method for proving subtraction will, doubtless, be found of some utility to those who ex- perience any difficulty in subtracting correctly. This method is similar to our method for proving addition ; hence very little explanation will be necessary. 46 RAPID CALCULATOR REVISED. 29AG7-——22 219-4 > 8954==26, 2-+4-6—8 EXPLANATION. 14513147 12 4S Adding the figures of A+ 9—13,—8—5 the minuend together and reducing them to a num- ber composed of one digit, we have 4; adding the figures of the subtrahend together and reducing thera to a number composed of one digit, we have 8; adding: the figures of the difference together and reducing them to a number composed of one digit, we have 5. The figures of the subtrahend, when reduced, give a larger remainder than the figures of the minuend; hence we must add 9 to the remainder (4) given by the minuend; this gives us 18. Now subtracting 8 from 13 we have 5. The two final remainders are the same; hence the subtraction is correct. ~ RAPID CALCULATOR REVISED. A7 MULTIPLICATION. There is perhaps no part of arithmetic, nor indeed is there any part of mathmetics, that offers greater facilities for securing short methods than does multipli- cation. Our intention, in treating this subject, is to give all possible methods that possess merit; hence some of the methods herein presented are only useful in exceptional cases, while, on the other hand, the ma- jority of them are methods which the calculator can carry into the counting room or bank, and find them of ~ value every day of his life. They are methods which no time nor change nor chance can possible eliminate from the fields of commercial calculations. For further evi- dence of their superority and usefulness, we have only to point to the vast army of young men over this coun- try, who, by their knowledge of this subject, are en- abled to do from two to three men’s work. We, there- fore, earnestly advise the youny man, who wishes to keep apace with the rapid march of time, to master these methods thoroughly. He should study them until they are as clear before his mind as the HORS of the noonday sun. HOW TO MULTIPLY ANY TWO NUMBERS, NEAR ONE HUNDRED, TOGETHER. oof eae We first subtract the ST: a § numbers from 100 and multiply the remainders ee 3 LO together, which gives us the first two figures of the result, then subtracting crosswise, 6 from 87 or 13. from 94 we have the last two figures of the re- sult. AS RAPID CALCULATOR REVISED. The last two figures may also be found by adding 87 and 94, always dropping the left hand 1from the 87 sum, thus: 94 1&1 HOW TO MULTIPLY NUMBERS OVER ONE HUNDRED TOGETHER. 114 is 14 more than Wye Sea ee 100, and 105 is 5 more FOR evo than 100. Multiplying ae the excesses together we Lia 10 have 70 for the first two figures. of the result, then adding we have 119, which gives the total result 11970. _In adding always cross out the top 1. Jt must not be considered. er aes 19 times 18 gives 342. 118. .18 We put down the 42 and PAT 9S Le carry the 8 to the next ee column, adding as usual. TAQ 2 In multiplying numbers either over or under 100 together the product of the excesses or complements should fill two places. If they only fill one place prefix a cipher. If they fill more than one place carry the third figure to the next col- umn. RAPID CALCULATOR REVISED. Ag HOW TO MULTIPLY ANY NUMBER BY BLEVEN. D425 el 37675 Write down the first figure of the multipli- cand, 5. Add the first to the second and write that down and so on to the last, after which write the last figure of the multiplicand; carry when > necessary. The process will be like this: 7 (3, 4) {: 5 (2,4) (5,2) 5 MULTIPLICATION OF NUMBERS WHOSE TENS’ FIGURES ARE ALIKE AND WHOSE UNITS’ FIGURES ADD TEN. We first say 4X6 is 24 for the first two figures of the result; then calling 7 one more, or 8, we say $< 7-56, and we, have the result 5624. EXERCISES. 3634 - 95x95 46 4b Tots 84 86 63 < 67 ap Yee 85 85 54 56 15X75 38 X< 32 58x52 65 65 ri! rare 9298 4347 83 « 87 63 67 50 RAPID CALCULATOR REVISED. THE LOWELL MULTIPLICATION RULE. These numbers. are 63 multiplied by the preced- 3 29 ing rule. We say 9 Le times 3 is 27, and write 1827 it down for the first two figures of the answer. Then calling the next figure of the multiplier one more, or 38, Wwe say 38 times 6 is 18, and write it down for the next two figures of the answer. HOW TO TELL WHEN THIS RULE MAY BE USED. In the above example, the complement of 9 is 1; that is 9 is one less than 10, and 1 times 6 is the same as 3 times 2. This rule applies to all numbers that will stand this test. Multiples of 11 may be multiplied by numbers whose digits add 10, by the above method. oo IS A multiple of 11, 30 and the sum of the digits 5 46 of the multiplier is 10; — hence we say 6 times 3 1518 is 18 for the first two figures of the answer. Calling 4 one more, or 5, we say 5 times 3 is 15 for the next two figures of the answer. RAPID CALCULATOR REVISED. 9b 6438 96x 66 3626 9339 32 X< 66 86 47 31x68 Ae aa ME 8429 84 67 8855 EXERCISES. 2816 TAX 382 82 49 93 482 2617 87x 464 69 <27 34702 6944 42384 24< 42 49173 68 36 37172 23% 44 61391 6438 98371 22 46 147 48 2148 14677 MULTIPLICATION OF NUMBERS ONE OF WHICH IS OVER AND THE OTHER UNDER ONE HUNDRED. 114. .14 2K pe 3 111 Multiply the excess and complement figures together, and subtract 42 110 58 MULTIPLICATION their product from. the sum of the two numbers, always setting the pro- duct of the excess and complement figures two places to the right of the sum of the numbers. The first 1 in the multiplicand should not be considered. OF NUMBERS OVER ANY NUMBER OF HUNDREDS. 306. .6 409. .9 125154 We multiply the excess figures together, which gives us 54 for the first two figures of the an- swer. Then multiply- 52 RAPID CALCULATOR REVISED. ing crosswise, we have 27 (39) again multiplying crosswise, (64) we have 24; adding the two products we have 51, the next two figures of the result, 34 equals 12 for the last two figures of the result. | MULTIPLICATION OF NUMBERS UNDER ANY NUMBER OF HUNDREDS. We find the comple- | 4 3 ment figures of 396 and p06... 493 to be 4 and 7, which pL Sean tl multiplied together gives 5 28 for the first two ——_—— figures of the product. 1952 28 Then calling each of the hundreds’ figures one more, we multiply crosswise, 4x7 plus 5448. This we subtract from 100, which leaves us a remain- der of 52 for the next two figures of the result. Multiplying the increased hundreds’ figures to- gether we have 20 (54). From this we subtract 1, which leaves us 19 to complete our product. If the swm of the products of the hundreds’ and the complement figures, produces more than 100 sub- tract from 200 and deduct 2-from the product of the hundreds’ figures instead of 1. 9 912 plus [xX Tesh0, 893 27 which we subtract from 688. .12 200 to obtain the remain- Tee der 43. -9X7=63, from pes ra Be which we deduct 2 to ob- ‘O13 84 tain the number 61. RAPID CALCULATOR REVISED. 53. MULTIPLICATION OF NUMBERS OVER ANY NUMBER OF HUNDREDS BY NUMBERS OVER ONE HUNDRED. The excess figures are A713. .73 fovdnduler re bo MeO 1 AO We put down the 76 and — carry the 8 4 x12+8 SZ LG | (carried) +3) = Cunits’ figure) —=59. We put down the Yandcarry the 5. 7-512 which gives us 2, and 1 tocarry. 1--4==5 which completes the re- sult. No attention is paid to the multiplier after com- mencing to add. MULTIPLICATION OF NUMBERS NEAR FIR LY. We first subtract these 43... 7 numbers, separately from ay dae! bs 50, and multiply the re- ate mainders together for Pe 291 the first two figures of the answer, thus: 438 1s 7 less than 50, and 87 is 13 less than 50. 13 times 7=91, which we write for the first two figures of the answer; now subtracting crosswise 13 from 43 we have 30, and dividing by 2 we ~ have 15, which we write for the last two figures of the answer. We niay also subtract 7 from 37, which leaves 30, and then divide by 2 for the last two fig- ures (15) of the answer. MOLTIPLICATION OF NUMBERS IN THE THENS. ) ere The product of the ex- iY pees cess figures is 72. We ee put down the 2 and ea, carry the 7 to the next column and add. 749+ 824. Put down the 4 54 RAPID CALCULATOR REVISED. and carry the 2. 2+1=—8, which gives the result 342. Omit the upper left hand 1. HOW TO MULTIPLY ANY NUMBER OF NINES BY ANY OTHER NUMBER. 999999 Consider the last figure 435682 of the multiplier 1 less ——_—________— and write down the mul- 435681564318 tiplier, 485681. Nowsub- tract the multiplier less one from the multiplicand and write the remainder 5643818, down. HOW TO MULTIPLY ANY NUMBER BY NUM- BERS IN THE TEENS. Multiply 31234 by 18. By this process, we 31234 multiply, in regular or- 13 der, the figures of the multiplicand by the right 406042 hand figure 3, of the mul- tiplier, and to their pro- duct add the figure of the multiplicand which stands next to the right; also add — the figure to carry, if any. Beginning at the right we multiply 3 by 4 which gives us 12. Writing down the 2 we have | to carry; multiplying the 8 by 3 we have 9, plus 1 (to carry) plus 4 gives us 14, writing down the 4 and carrying the 1, we multiply 2 by 3, which gives us 6, to which we add 1 (to carry) and 3 which gives us 10; writ- ine down the 0 and carrying 1 we multiply 1 by 3 RAPID CALCULATOR REVISED. Sh which gives us 3; to which add 1 (to carry) and 2 » which gives us 6; multiplying 3 by 3 and adding the figure to the right we have 10; writing down the 0 and carrying the 1 which we add to the 8 we have 4 for the last figure of the answer. This process is very simple, and, with a little practice, wonderful speed may be attained. The above numbers may also be multiplied as fol- lows: 3125413 By this process we 93702 multiply the figures of the multiplicand by _ 3, 406042 and write their product one place to the right of the multiplicand; then adding we have the result. HOW TO MULTIPLY ANY NUMBER BY ONE HUNDRED AND ELEVEN. 5342 Beginning at the right, 111: we place the first figure of the multiplicand, 2, 592962 down for the first figure of the answer; adding the 2 to 4, we write their sum, 6, for the second ate of the answer; adding 2, 4 and 3 together, we have 9 for the third figure of the answer; adding the 4, 3 and 5 together, we write 2 for the fourth figure of the answer, and have 1 to carry; adding the 3, 5 and 1 (to carry) together, we have 9, which we write for the fifth figure of the answer. We now write the last figure of the multiplicand, 5, for the last figure of the answer. Osc: RAPID CALCULATOR REVISED. The process will stand as follows: 2 == |2 244 = |6 ree Wie = |9 41815 =1]2 3+5+1 (carried)— |9 5 ak MULTIPLICATION BY ALIQUOT PARTS. It is hardly possible to estimate the untold advan- tage that aliquot parts are to the accountant. To multiply any number by 10. annex a cipher. 100 annex two ciphers. 1000 = annex three ciphers. 10000 eannex four ciphers. : @ annex a cipher and divide by 9. 1+ annex a cipher and divide by 8. annex a cipher and divide by 7. 3 13 annex a cipher and divide by 6. 2+ annex a cipher and divide by 4. ot annex a cipher and divide by 3. 81 annex two ciphers and divide by 12. 384 annex two ciphers and divide by 3. 25 annex two ciphers and divide by 4. 50 annex two ciphers and divide by 2. 124 annex two ciphers and divide by 8. 142 annex two ciphers and divide by 7. 162 annex two ciphers and divide by 6. 374 multiply by 3, annex two ciphers, and divide by 8. . 621 multiply by 5, annex two ciphers, and divide by 8. ~] RAPID CALCULATOR REVISED. * — 662 multiply by 2, annex two ciphers and divide by 3. 75 annex two ciphers and deduct | of the product. 15 annex one cipher and add $ the product to itself. 112} annex two ciphers and add ! of the product. 1534 annex two ciphers and add 4 of the product. 831 annex three ciphers and divide by 12. 125 annex three ciphers and divide by 8. 1663 annex three ciphers and divide by 6. 250 annex three ciphers and divide by 4. S004 annex three ciphers and divide by 3. 375 multiply by 3, annex three ciphers, and divide by 8. 8531 multiply by 5, annex three ciphers, and divide by 6. 875 annex three ciphers and subtract 1 of the pro- duct. : EXAMPLES. 1. Multiply 564 by 24 4) 5640 1410 This is the same as multiplying by 10 and dividing by 4. If we multiply by 10, we have taken the multi- plier 4 times too often, because 10 is 4 times the multi- plier, 24. 2. Multiply 6346 by 25 4) 634600 158650 3. Multiply 266 by 34 3) 2660 ———} 8863 58 _ RAPID CALCULATOR REVISED. 4. 5. 6. i: 8. a: 10. Multiply 365 by 13 6) 8650 6081 Multiply 868 by 11 8) 8630 1078% Multiply 629 by 38} 3) 62900 209662 Multiply 2314 by 124 8) 231400 28925 Multiply 3424 by 8} 12) 342400 285331 Multiply 2649 by 142 7) 264900 Multiply.3418 by 163 6)341800 569663 RAPID CALCULATOR REVISED. 1h bs 18. 14, 15. Multiply 4642 by 62} 4642 5 8) 2321000 290125 Multiply 8432 by 75 4) 843200 210800 652400 Multiply 6346 by 1124 8) 634600 79325 713925 Multiply 6542 by 125 8) 6542000 817750 Multiply 4963 by 875 8) 4963000 620375 4342625 60 RAPID CALCULATOR REVISED. MULTIPLICATION BY NUMBERS NEAR ALI- QUOT “PARTS: After the pupil is thoroughly familiar with the process of multiplying by aliquot parts, with a little practice, he can also multiply by numbers near aliquot parts. ae EXAMPLE. 1. Multiply 2416 by 248. 4) 2416000 EXPLANATION. ee 248 is near 250, so we 604000 first multiply 2416 by 2416 2 =, 4882 250; this we accomplish eer 8 by annexing three ciph- 599168 Ans. ers to the multiplicand and dividing by 4, which gives 604000. It will be ob- served that we have taken the multiplicand twice two often, as 250 is 2 more than 248;.hence we multiply 2416 by 2, which gives us 4832. This we _ subtract from 604000, which leaves us a remainder of 599168, the product of 2416 by 248. 2. Multiply 4368 by 127. 8) 4868000 EXPLANATION. 546000 127 is near 125, so we 4368 x 2=— 8736 first multiply 43868 by 125; this we accom- 554736 plish by annexing three ciphers and divid- ing by 8, which gives us 546000. 127 is 2 more than 125, so our multiplier is 2 too small; hence the product will be too small by two times the multiplicand. RAPID CALCULATOR REVISED. 61 Multiplying the multiplicand by 2 we have 8736, which we add to 546000 for the complete product, 554736. 3. Multiply 32519 by 997. 32519000 EXPLANATION. 32519“ 3— 97557 | _— 997 is near 1000, so we 32421443 first multiply by 1000; this we accomplish by annexing three ciphers to the multiplicand. 997 is 3 less than 1000, so we multiply the multiplicand, 32519, by 8, and subtract it from the product obtained by multiplying by 1000; this leaves us the correct result, 32421443. 4, Multiply 428 by 243 4) 42800 10700 of 428— 214 hw) 10486 EXPLANATION. 241 is near 25, so we first mlutiply 428 by 25: this is accomplished by annexing two ciphers and dividing by 4, 244 is 4 less than 25, so we have taken our multiplicand 4 times too often. t of 428 is 214. This we subtract fr om 10700, and we have the product of 428 by 241; namely, 10486. 62 RAPID CALCULATOR REVISED. HOW TO MULTIPLY NUMBERS TOGETHER WHEN THE TWO LEFT HAND FIGURES ARE ALIKE AND THE RIGHT HAND FIGURES ADD TEN. Multiply 146 by 144. 146 EXPLANATION. 144 We first multiply the 21024 two right hand figures, 4 and 6, together, which gives us 24 for the first two figures of the result; then calling the two left hand figures of the multiplier one more, or 15, we say 15 times 14 is 210 for the three left hand figures of the re- sult, or after adding 1 to 14, we may say 15 times 4 is 60, putting down the 0 and carrying the 6; then multi- plying 1 by 15 and adding 6 (to carry) we have 21, which completes the product. EXAMPLES. 1 36h4 1 4 5 Lori aet | Gees age bs. fe ga ay 3 Aveo tale 17°36 Eek O03 Lae ae £3.54 Lipa t Leas 15.4.6 RAPID CALCULATOR REVISED. 63 HOW TO MULTIPLY TWO NUMBERS OF TWO OR THREE FIGURES EACH, WHEN THE LEFT HAND FIGURE OR FIGURES ARE ALIKE AND THE SUM OF THE UNITS APPROXIMATE TEN, MORE OR LESS. 46 EXPLANATION. 45 Saka In problems of this 2024 kind, we call the right 46 hand figure of the mul- ee tiplier such a number as, 2070 when added to the right hand figure of the multi- plicand, the sum will be 10. In this problem we con- sider 5 one less, or 4; multiplying 46 by 44 we have 2024. Our multiplier is one too small; hence the pro- duct is 1 time the multiplicand too small, therefore we add 46 to 2024, which gives us the correct result, 2070. Multiply 127 by 122. VPAare EXPLANATION. ve ceo Calling the right hand 15621 figure of the multiplier Li At | one more, or 3, We Say aaa 3 times 7 is 21, which we 15494 write for the first two figures of the product; then multiplying 12 by 3 we have 156 for the next three figures of the product, hence the product of 127 by 123 is 15621; but our multiplier is one less than 123, therefore we have taken the multiplicand once 64 RAPID CALCULATOR REVISED. — $$$ too often; hence subtracting one time the multiplicand, or 127, from 15621, we have the product of 127 by 122° namely, 15494. 84 78 yi 71 TOT: 5616 87 78 7308 5538 HOW TO MULTIPLY TWO NUMBERS OF TWO FIGURES EACH WHEN THE TENS’ FIGURES ADD TEN AND THE UNITS’ FIGURES ARE ALIKE. 48 EXPLANATION. 68 Biel In problems of this 3264 kind, we first multiply the units’ figures together, which gives us 64 for the first two figures of the an- swer. We next multiply the tens’ figures together, and to the product add one of the units’ figures (6«4 +8,) which gives us 32 for the next two figures of the answer. When the product of the units’ figures fills but one place, a cipher must be written in the tens’ place. EXAMPLES. ep) O99 CO op) re o 00 ww Lolita) Bho ey) | RAPID CALCULATOR REVISED. 65 HOW TO MULTIPLY NUMBERS WHEN THE UNITS’ OR TENS’ FIGURES “ONLY ARE ALIKE. Multiply 64 by 34. 64 D4 2176 Multiplying the units’ figures together, we have 16. We write the 6 for the first figure of the answer and carry the 1. Now adding the figures that are not alike (8+6) we have 9, and multiplying one of the figures that are alike (4) by 9 we have 36, +1 (to carry) gives us 37. We write the 7 for the tens’ figure of the answer, and carry the 3; now multiplying the 3 and 6 together and adding in the 3 (to carry) we have 21, which we write to complete the product. Multiply 74 by 79. TA 79 5846 Multiplying the 9 and 4 together we have 36; we write the 6 for the first figure of the answer, and carry the 3. Now adding the figures that are not. alike (4+-9) we have 13. Multiplying one of the figures that are alike (7) by 18 we have 91, +3 (to carry) gives us 94; write the 4 for the tens’ figure of the an- swer and carry the 9. Multiplying the two left hand figures together and adding in the carrying figure (7<7+9) we have 58, which we write for the two left hand figures of the product. 66 RAPID CALCULATOR REVISED. Remember, by this method of multiplication, the pupil should always add the figures that are not alike, it matters not whether they occur in the units’ or tens’ place. HOW TO MULTIPLY BY SUCH NUMBERS AS THIRTY-ONE, FORTY-ONE, SEVENTY-ONE, ETC. Multiply 2345 by 31. 2345 «31 7035 72695 EXPLANATION. By this process we multiply the figures of the mul- tiplicand by the left hand figure of the multiplier, placing the first figure of the product under the tens, and add the product obtained to the multiplicand. PROBLEMS. 1, .4562x 41. 2. 302461. 3 24831. 4, “0491. 5 345 81. HOW TO MULTIPLY BY ANY NUMBER, ONE PART OF WHICH IS A FACTOR OR MULTIPLE OF THE OTHER PART. Multiply 3124 by 244. EXPLANATION. 3124 244 We first multiply 3124 by the right hand figure 12496 (4) of the multiplier 74976 which gives us 12496. The left hand figures | 762256 (24) of the multiplier RAPID CALCULATOR REVISED. 67 are 6 times the right hand figure (4) of the multiplier; therefore 6 times 12496 must be the same as 24 times 3124; so we multiply 12496 by 6, writing the result one place to the left of the previous one. Adding the two partial products together we have the complete pro- duct, 762256. . Multiply 4121 by 624. A121 EXPLANATION. 624 ee We first multiply 4121 24726 by 6, which gives us 98904 24726. (The last figure on of the product must al- 2571504 ways be written under the figure you multiply by.) The right hand figures (24) of the multiplier are 4 times the left hand figure (6) ; therefore 4 times 24726 is the same as 24 times 4121. So we multiply 24726 by 4, which gives us 98904. (This product must be written two places to the right of the previous. one.) Now adding our partial products together we have 2571504 for the complete product. Multiply 231423 by 36945. ie 231423 EXPLANATION. 36945 toast 2a ee alae We first multiply the 2082807 multiplicand by the 8331228 figure 9 of the multiplier, 10414035 which gives us 2082807. oo = 2a ieee The two left hand figures 8,549,922 735 (36) of the multiplier are 4 times 9; therefore 36 times the multiplicand is the same as 4 _ times 68 RAPID CALCULATOR REVISED. 2082807. So we multiply this latter number by 4, which gives us 8331228. (This number is written one place to the left of the 7, or directly under the figure 6 of the multiplier.) We now observe that the two right hand figures(45) of the multipler are 5 times 9; therefore 45 times the multiplicand is-the same as 5 times 2082807. So we multiply this number (2082807) by 5, which gives us 10414035. - (This product should be written two places to the right of the 7, so that the last figure. will fall directly under the figure 5 of the multiplier.) Now adding the partial products _ to- gether we have the complete product, 8549922735. a ITT ons - HOW TO MULTIPLY WHEN THE MUL- TIPLIER CAN BE RESOLVED INTO EASY FACTORS. Multiply 2342 by 28. EXPLANATION. 2342 4 The factors of 28 are epi 4 and 7; multiplying the 9368 multiplicand by 4 we 7, have 9368, and multiply- ing this product by 7 we 65576 have the complete pyro- duct, 65576. _ By this method we only use two lines of figures exclusive of the multiplicand, and have no addition to perform. By the old method it takes three lines of figures exclusive of the multiplicand, and there is one addition to perform; hence we have saved a line of figures and an addition. RAPID CALCULATOR REVISED. 69 HOW TO MULTIPLY ANY NUMBER OF MIS. CELLANEOUS FIGURES BY ANY NUMBER OF LIKE FIGURES AS it!,.7777, 888, ETC..- Multiply 5123 by 4444. 5123 4444 22766612 3x4 12 B--2<4-.1 (curried) == 2|1 ee oe Ae (curried )—=<2}6 3494145x4,42 (carried) — 4]6 21445 %4,+4 (carried) = 3|6 4-75 x4,--3. (carried) = 2/7 BGA -earried ) -===.2>-2 EXPLANATION. By this method we multiply the first figure of the multiplicand by one of the figures of the multiplier, then add the first figure of the multiplicand to the se- cond figure of the multiplicand and multiply by one of the figures of the multiplier; we continue adding hori- zontally and multiplying by one of the figures of the multiplier until the number of figures to be added horizontally is equal to the number of figures contained in the multiplier. In our horizontal additions after we have reached the left hand figure of the mulitplicand, we decrease the number of figures to be added horizon- tally, one at each successive multiplication. We first multiply 3 by 4, which gives us 12. We write down the 2 and carry the 1. Now, adding the 3 70 RAPID CALCULATOR REVISED. and 2 together, we multiply by 4 and add the 1 (to earry), which gives 21. Write down the 1 and carry the 2. Now adding the 3, 2 and 1, we have 6, and mul- tiplying by 4 we have 24,+-2 (to carry) gives us 26. Now, adding the 3, 2, 1 and 5 together we have 11, and multiplying by 4 we have 44,42 (to carry) gives us 46. We write down the 6 and carry the 4. The pupil will now observe that in our grouping we have arrived over the last figure of our similar fig- ures, So we commence dropping our dissimilar figures, or the figures of the multiplicand, one by one as we group. Now, adding the 2, 1 and 5 together we have 8, and multiplying by 4 we have 32,+4 (to carry)=—836. We write down the 6 and carry the 3. Now, ad- ding the 1 and 5 together we have 6, and multiplying by 4 we have 24,+3 (to carry) gives us 27. We write down the 7 and carry the 2. Now, multiplying 5 by 4 and adding 2 (to carry) we have 22, which we write down for the last two figures of the product. SIMULFANEOUS, OR CROSS MULTIP- LICATION. This method of multilpication is of inestimable value to the business world, and the one who thorough- ly masters it is enabled to appear as a prodigy in num- bers and lead those who are unacquainted with the pro- cess by which he works, to suppose that he is pos- sessed of some supernatural power. It is, however, so simple in its nature that the ordinary school boy will have no trouble in readily comprehending and applying it? It has stood the test of the counting room for many years, and, never before during its history, has it re- ceived such high encomiums as at the present day. RAPID CALCULATOR REVISED. fg a By this method we can multiply any two numbers together and place the result in one straight line. Some have said that it is too difficult, but with ten minutes’ practice, we have seen boys multiply numbers of three and four figures eaclt by numbers of two figures each, in much less time than they could possibly do it by the ordinary method. If the pupil, who has thoroughly mastered one of our systems of addition, will now place himself in possession of this method of multiplication, he has only to launch his bark upon the mighty sea of of business, and, with a little perseverance, it will soon be laden with the golden gems of success. No person can be proficient in handling numbers without a thor- ough knowledge of this process. | The operations of this method are based upon the following principles: 1. Units multiplied by units produce units. 2. Tens multiplied by units produce tens. 3. Units multiplied by tens produce tens. 4. Hundreds multiplied by units produce hun- dreds. 5. Tens multiplied by tens produce hundreds. 6. Units multiplied by hundreds produce hun- dreds. Thousands multiplied by units produce thou- sands. 8. Hundreds multiplied by tens produce thou- . sands. 9. Tens multiplied by hundreds produce thou- sands. 10. Units multiplied by thousands produce thou- sands. 11. Ten thousands multiplied by units produce ten thousands. 12. Thousands multiplied by tens produce ten thousands. 13. Hundreds multiplied by hundreds produce ten thousands. 14. Tens multiplied by thousands produce ten thousands. 72 15 RAPID CALCULATOR REVISED. . Units multiplied by ten thousands produce ten 16. Lt 18. 19; 20. 21. thousand. Hundred thousands multiplied by units pro- duce hundred thousands. Ten thousands multiplied -by tens produce hun- dred thousands. ; Thousands multiplied by hundreds produce hundred thousands. . Hundreds multiplied by thousands produce hundred thousands. Tens multiplied by ten thousands produce hun- dred thousands. Units multiplied by hundred thousands pro- duce hundred thousands. So on for the higher numbers. EXPLANATION. 36 To multiply these num- A438 bers together, we first Bias say 3x6 equals 18, and 1548 put down the 8 and re- tain the 1 in the mind. We now say 3x8=—9 plus 1 (to carry) equals 10+46=—34. We put down the four and retain the 3 mentally. We now say 4X3—12 and 12+:3 (to carry) ==15 which completes the answer, 1548. FIGURES| FIGURES CARRYING | PRODUCT 6><3 == 118 38,11 (carried) +6X4— 3]4 84,18 (carried)—= 1 5 ont w RAPID CALCULATOR REVISED. : Multiply 488 by 267. 438 267 116946 CARRYING | PRODUCT FIGURES | FIGURES 8<7—5|6 37,15 (carried) +8 x 6=7|4 4Xx7,+7 (carried) +3x6+8x2—6|9 46+6 (carried) +2<3==3|6 AX2+38 (carried)—1]1 EXPLANATION. Beginning at the right-we first multiply 8 by 7 which gives us 56; write the 6 for the first figure of the answer and carry the 5. 37,+5 (carried) +86 74; we write the 4 for the second figure of the answer and carry the 7. AX7,1+7 (carried) —35, and 35+3 x6+8%2—69. We write the 9 for the third figure of the answer and carry the 6. We are now through multiplying by 7 aS we have multiplied the last figure of the multi- cand by it; therefore it is omitted, and we next multi- ply 4 by 6, which gives us 24,+-6 (carried)—30, and 30+-3 x 2—36; we write the 6 for the fourth figure of the answer and carry the 3. 4x2,-+-8 (carried)—11, which we write for the last figure of the answer. This rule is especially valuable in extending bills. In fact, no one who does this class of work can afford to be without a knowledge of it. 74 RAPID CALCULATOR REVISED. LIGHTNING MULTIPLICATION. This is simply a modification of cross multiplica- tion, and while it is, perhaps, no better than cross multi- plication, it is far simpler and more readily compre- hended by the ordinary individual. This method mul- tiplies any number by any other number. It matters not how many figures the numbers may be composed of, or what the figures are. The superior advantages of this method will, by the good accountant, be appre- ciated at a glance. To the young man just starting in life, the value of it can hardly be estimated. Multiply 4867 by 231. 4367 132 1008777 py 6x14+7x*K3= 2|7 3x1+6«3+7X2,+2 (carried) 3]7 , EXPLANATION. The first thing is to, write down the figures of the multiplier in a reversed order. Now multiplying the right hand figure (7) of the multiplicand by the left hand figure (1) of the multiplier, we have 7, which we write for the first figure of the answer. Multiplying the next figure (6) of the multiplicand by 1 we have 6-73-27. We write the 7 for the next figure of the ~I oO RAPID CALCULATOR REVISED. answer and carry the 2. Multiplying 3 by 1 we have 3,+2 (to carry). which gives us 5+-638+-7x2=37. We write the 7 for the next figure of the answer and have 3 to carry. Multiplying 4 by 1 we have 4+3 (to carry )=7,+3x3+6x2-—28. We write the 8 for the next figure of the answer and carry the 2. We have now multiplied every figure of the multiplicaid by the figure 1 of the multiplier; hence in our future multi- plications, we omit the 1. Multiplying the 4 by 3 we have 12,12 (to carry) —14,+3x2—20. We write the 0 for the next figure of the answer and carry the. 2. We now omit multiplying by the 3 for the same reason that we omitted multiplying by the 1. Multiplying 4 by 2 we have 8+2 (to carry) which gives us 10, which we write for the last two figures of the answer. This method is so simple in its nature that a child ten years of age will have no trouble whatever in com- prehending and readily applying it. The thing to be remembered is, that you commence with the left hand figure of the multiplier, and mulitply every figure of the multiplicand by it, following up with the figures of the multiplier in regular order, as you proceed to the left with the multiplication by the left hand figure of the multiplier. No figure of the multiplicand is mul- tuplied by more than one figure of the multiplier at the same time. Study this method carefully, and you will be more than paid for your trouble. Its usefulness and rapidity will only be fully unfolded to you after you have thoroughly mastered it. Don’t be afraid that you cannot learn it. It does not require a giant intellect to multiply by this method; as we have _ previously stated, a child can learn it. Remember ‘Perseverance is the price of success.” 76 RAPID CALCULATOR REVISED. SLIDING METHOD OF MULTIPLICATION. This method is the same as the one previously ex- plained, with the exception that the figures of the mul- tiplier are written on a separate piece of paper, and slided to the left above the figures of the multiplicand as we proceed with the multiplication. It is even simpler for the beginner than the previous method, though there is no need of the advanced ‘pupil going to the extra trouble of sliding the figures; for, by him, it will be readily comprehended without this labor. Multiply 632 by 541. 145 632 2 We first write the figures of the multiplier in a re- versed order on a separate piece of paper, and place the paper so that the figure 1 of the multiplier is just above the first figure of the multiplicand. Now multi- plying the right hand figure of the mlutiplicand by the left hand figure of the multiplier we have 2, which we write for the first figure of the product. Now slide the paper with the multiplier on it one place to the left, and it will appear thus: 145 632 S12 | Now 3x14+24=—11. Write down the right hand 1 for the next figure of the product and carry the left hand 1. Now slide the multiplier one place farther to the left, and it will appear thus: RAPID CALCULATOR REVISED. 77 145 632 912 Now 6X1+3x4+2™5,-+1 (to carry)—=29. Write down the 9 for the next figure of the product and carry the 2. Now slide the multiplier one place farther to the left, and it will appear thus: 145 632 1912 Now 64+35,+2 (to carry) =41. Write down the 1 for the next figure of the product and carry the 4. Now slide the multiplier one place farther to the left, and it will appear thus: 145 632 . 841912 Ans. Now 6X5,+4 (to carry)—34, which we _ write down for the two remaining figures of the product. By this method it will be observed that you place the left hand figure of the multiplier over the right hand figure of the multiplicand, and proceed toward the left until the multiplicand passes from under the multiplier, multiplying each time the figure or figures of the multiplicand which fall under any figure or figures of the multiplier. 78 RAPID CALCULATOR REVISED. MULTIPLICATION TABLE. It would be much better if the pupil would learn the multiplication table up to twenty times nine instead of up to twelve times twelve. Those wishing to attain the utmost proficiency in rapid calculation should thor- oughly master the following table: ipeps) 4h 5) 6) 7) 3. te 4] 6 | 8 eee ee f Bes 5.6 (9 be oe eet oT | i £7 8] 12 116 1.0 te el ee 36 | r 5 1 d04 16 £20 | oe 0 45 | 6 | 12118 | 24) 80} 86 | 42 | 4e 9 7| 14/281 |28| 85 | 421) 49) 560 sae 8 [16] 24; 321 40] 48) 56) 64 9) 18 | 27 136) 45) 64) 6381.7 10.| 20130 | 40] 50/601. 70) 801. onm | | 24 | | | 32 | | | | | | | | | | | | | Bae 114] 22 {33 | AA bb | 666.| 77 | 885 eee | | | | | | | | | | | | | | | | | | | | | | | | | | | 12;,| 24:1 36-[ 48° |< 60 | 727] 84 96 ee [718s] 26-| 39: (52 | 65 1978 | 91 | 10a ee 14/28 | 42 7 56 | 70 | 84° | 98 | 21a ee 155|-30 | 45 | 60 | 75 | 90 | 105 (Aap e 16)| 82. | 48] 64 | 80] 96-] 112 | 128) ee 17;| 84) 51 | 68 | 85 | 102 | 119 | 186 eae 18/36 | 54 | 72°| -90 | 108] 126 | 1447 eee 19°[ 38 | 57 (76. | 95 | 114 | 133) 152 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 | SQUARING NUMBERS. We here present all the best methods known for squaring numbers, and many of them will be found to be very useful in performing business calculations. RAPID CALCULATOR REVISED. tte TABLE OF SQUARES AND CUBES. Numbers | Squares Cubes Numbers Squares | Cubes | | 1 eae ete 1 26 676 17576 2 4 | 8 Pl fai SN Pre 19683 3 | 9 at ZO ale 134 21952 4 | 16 64 29 841 | 24389 Fig Amar a5, 125 30 900 27000 6 147.36 216 31 961 29791 af AQ 343 O2 1024 | 32768 8 | 64 512 33 1089 39937 Oe, 81 729 o4 1156 539304 Ee LOO.” fe 1000 OD 1225 42875 ae 121 1331 36 1296 46656 12 | 144 1728 oO” 1369 50653 Ee LOO 2197 38 1444 | 54872 tase = 196 2744 5) | 1521 59319 15 | 225 3075 40 | 1600 64000 LGe 206 | 4096 41 | 1681 68921 tie ceO0e, 1 tc 4 Le 42 | 1764 | 74088 18 | 324 | 5832 43 1849 | 79507 19 361 6859 44 | 1936 85184 20 | 400 8000 45 | 2025 91125 Z5ge ps 44) 9261 46 2116 97336 22 | 484 | 10648 AT 2209 103823 2a.:\),.529 | 12167 48 | 2304 110592 24 | 576 | 18824 AQ 2401 117649 25 | 625 | 15625 50 | 2500 125000 80 | RAPID CALCULATOR REVISED. SQUARES AND CUBES—Continued. Numbers | YAIANNAQaITgNaH NWR USO ONG OSs Squares | PA 2600" i 2704 2809 2916 3025 31386 3249 0364 D481 5600 5625 5476 5329 5184 5041 4900 AT61 4624 4489 4356 4225 4096 53969 3844 ol21 Cubes 157464 166375 175616 | 185195 | SOS EES 205379 | 216000 | 421875 405224 | 389017 313248 oo7911 343000 328509 .pl4452 200763 287496 | 274625 262144 250047 238328 ea 00ST 1 182651, 140608 | 148877 Numbers Squares 76) £5 ae 78 79 80 | 81 82 | 85 34 85 86 | 87 88 | 89 90 91 | 92 93 94 Ora 96 OF 98 99 200 ue 5776 5929 6084 6241 6400 6561 6724 6889 7056 7225 7396 7569 7744 7921 8100 S98). 8464 8649 8836 9025 | 9216 9409 9604 9801 . Cubes 438976 456533 474552 493039 512000 531441 551368 STL787 592704 614125 6536056 658503 681472 . 704969 729000 753571 778688 804357 830584 857379 884736 912673 941192 970299 1000000 RAPID CALCULATOR REVISED. ae Sh HOW TO SQUARE ANY NUMBER ENDING IN FIVE. We first multiply 5 by 145 5, which gives us 25 for 145 the first two figures of the product; then call- 21025 ing the 14 in the mul- tiplier one more, or 15, we say 15 times 14—210 for the next three figures of the product. HOW TO SQUARE ANY NUMBER ENDING IN TWENTY-FIVE. EXPLANATION. To square any number 625 ending in 25, first square 625 its left hand digit and to this product add one-half 390625 of the digit before it was squared (omitting frac- tions should any occur), and to this annex 0625 if the left hand digit is even, and if odd annex 5625. We first multiply 6 by 6, which gives us 36; now taking one-half of 6, or 3, and adding to 36, we have 39 for the first two left-hand figures of the product; now to this annex 0625 and we have the complete pro- duct. If the left-hand figure of the number’ to’ be squared is odd, then the fourth place from the right will be 5 instead of 0, thus: 525 525 275625 $2. RAPID CALCULATOR REVISED. Ee Multiplying the two left-hand together we have 25. Now taking the one-half of 5 we have 243. We drop the 4 and add the 2 to the 25, which gives us 27 for the two left-hand figures of the product. We now annex 5625 to complete the product. In the first example the process is virtually the same as multiplying 61 by 61, and, in the second ex- ample, the same as multiplying 5} by 54. To multiply 64 by 61 the process would be as follows: EXPLANATION. We first say 6 times 6 64 is 36; now taking the 4 61 of 6, or 3, and adding to — 36 we have 23: 1+ multi- 39, plied by 4— 4. Now if we wanted the square of 625 instead of the square of 6}, we would simply extend the ;!; in the pro- duct four places to the right, thus: 16) 10000 0625 This we annex to the right of the 39 and we have the square of 625. HOW TO SQUARE ANY NUMBER ENDING IN SEVENTY-FIVE. 875 875 765625 RAPID CALCULATOR REVISED. 83 EXPLANATION. To square any number ending in 75, first multiply the left-hand digit by a number one higher than itself, and to this product add 4 of the left-hand digit if even, and if odd add 3 of the number next higher than the left-hand digit; to this annex 5625 if the left- hand digit is even, and if odd annex 0625. To square 875, we first multiply the 8 by one more than itself, or 9, and to this product we add the 4} of 8, or 4, which gives us 76. Now annexing 5625 to the left of 76 we have the result, 765625. When the left-hand figure of a number ending in 75 is odd, the fourth place in the answer from the right will be a cipher instead of 5, and we take 4 of the left-hand figure after it has been increased by 1 instead of tak- ing it before it has been increased by 1; thus: T15° 600625 Multiplying the left-hand figure by one more than itself, or 8, we have 56; to this product we add 4 of the left-hand figure after it has been increased by 1 (7+1-2), which sum equals 60; this we write for the first two left-hand figures of the answer. . Now an- nexing 0625 we have the resuit, 600625. This is based upon the same principle as squaring fractions ending in 3. For example: 84 RAPID CALCULATOR REVISED. To square 8? we first multiply 8 by one more than itself, or 9, which gives us 72; to this product we add + of 8, or 4, which gives us 76. Now squaring ? we have ;%; hence our complete product, 763%. We can now obtain the square of 875 by extending the ;% four places to the left and annexing it to the right of the 76. 16) 90000 5625 HOW TO SQUARE ANY NUMBER BE- TWEEN TWENTY-FIVE AND FIFTY. To be proficient in this work, the squares of num- bers as high as 25 should be thoroughly memorized. What is the square of 27? 27 oR O 62 ns 2Oe 23? —529 - 529 -+2—729 Ans. In problems of this kind first subtract 25 from the number to be squared, and immediately subtract the remainder from 25. Then square the second remainder and add to its hundreds’ figure the first remainder. In the above example, 27--25—2. 2 from 25 leaves Za. squaring 23 we have 529, adding the first remainder (2) to the hundreds’ figure, we have 729, or the square Of 2a HOW TO SQUARE ANY NUMBER BE- TWEEN FIETY AND SEVENTY -FIEs What is the square of 64? RAPID CALCULATOR REVISED. 85 64—25 = 39 64—50 = 14 £427 196 196-439 —4096 We first subtract 25 from 64, which leaves 39. We now subtract 50 from 64, which leaves a remainder of 14. Squaring the last remainder (14) we have 196, and to the hundreds’ figure we add our first remainder (39), which gives us 4096, or the square of 64. Re- member that under this rule we first mentally subtract 25, then 50 and from the number to be squared, retain- ing both remainders. HOW TO SQUARE ANY NUMBER BE- TWEEN SEVENTY-FIVE AND ONE HUNDRED. What is the square of 86? 86—75 = 11 25—11 = 14 Lt 1OG 196-+ (86°—14) —7396 In problems of this kind we first subtract 75 from the number to be squared, and then subtract the re- mainder from 25. Square the second remainder and to its hundreds’ figure add the original number minus tne second. remainder. $6——75—11, and 25—11—14. 14 squared equals 196. 86—14—72. ‘72 added to the hundreds’ figure (1) makes 7396, or the square of 86. MULTIPLICATION BY SQUARING THE TWO NUMBERS. The product of any two numbers is equal to the square of the mean minus the square of one-half their difference. The mean is a number as much greater than the less as it is less than the greater. 86 RAPID CALCULATOR REVISED. Multiply 18 by 14. 18 | 162 — 256 14 (Lof 4)? = 4 252 252 The mean of 18 and 14 is 16, and the square of 16 is 256. The difference between 18 and 14 is 4, and $ of the difference is 2. The square of 2 is 4; subtract- ing 4 from 256, we have 252, the product of 18 by 14. HOW TO SQUARE ANY NUMBER OF NINES. Wuat is the square of 9999? 9999? 99980001 - To square any number of nines, we first write a 1, then write as many ciphers as you have nines, less one, write an 8, then write 9 as many times as you have nines less one. HOW TO SQUARE ANY NUMBER OF SIXES. This may be accomplished instantaneously by con- ceiving the sixes of the multiplier to be nines, and the ~ sixes of the multiplicand to be fours, then multiply the nines by the fours by the process explained in a pre- — vious portion of this work for multiplying any number of nines by any other number. What is the square of 666? 666, Gos 666, 444 443556 We first write three nines for the multiplicand, and three fours for the multiplier. We call the right ‘7 RAPID CALCULATOR REVISED. 87 hand figure of the multiplier one less, or 3, and write 443 for the first three left hand figures of the product. Subtracting the multiplier less one from the multi- plicand we have 556 for the three right hand figures of the product. HOW TO SQUARE ANY NUMBER OF THREES. This may be accomplished by conceiving the threes of the multiplicand to be nines, and the threes of the multiplier to be ones. What is the square of 33333? 33390, 99999 DOd00, ipaeal 1111088889 HOW TO MULTIPLY ANY NUMBER OF SIXES BY THE SAME NUMBER OrCEMREES:. Change the sixes to nines and the threes to twos, and proceed as before. Multiply 6666 by 3333. 6666, Bao 3333, Bone 22217778 HOW TO SQUARE ANY NUMBER OF ONES. Count 1, 2, 3, 4, etc., up to the number of ones to be squared, then count back. 88 RAPID CALCULATOR REVISED. What is the square of 11111? 1 3s Bs Be 123454321 HOW TO MULTIPLY FIVES BY TWOS. Change the fives and twos to ones, and proceed as in the previous example and annex a cipher to the right. Multiply 5555 by 2222. 5d50== 1111 MA oa iif at Ek 12343210 LIGHTNING INVOLUTION. By a thorough mastery of this method, the pupil can square any number either large or small almost in- stantaneously. This method is new, and is much to be preferred to any other method of involution for all kinds of numbers that we now know of. The major- ity of methods for squaring numbers have too many ex- ceptions to render their use practical for any one who is not constantly engaged in this kind of work. This method applies to all kinds and classes of numbers, and when once learned, it is not easily forgotten. What is the square of 36? 362 1296 CARRYING|PRODUCT FIGURES |FIGURES 6? 3/6 (3x2) x6+8 (carried) —3/9 3?+8 (carried) —12 RAPID CALCULATOR REVISED. 89 We first square the first figure (6) which gives us 36. We write down the 6 and carry the 3. Now doub- ling the second figure and multiplying by the 6 (8x2 <6) we have 36, to which we add 3 (to carry) which gives us 39. We write down the 9 and carry the 3. Now squaring the last figure (8), and adding in the 3 (to carry) we have 12 which we write for the last two figures of the answer. | What is the square of 43521? 43521? 1894077441 CARRYING|FIGURES FIGURES |FIGURES 2 1 (22) x1 |4 (52) X1+42? —==1|4 (ax2) xt (5x2) x21 (carried)==217 (4X2) x1-- (82) x2-+-5?-+-2 (carried) —4 7 (4K2) x2+ (8x2) x5+-4 (carried) —5/|0 (42) X5-+3?--5 (carried) —=5/4 9 8 (42) X3-+5 (carried) =2 421-2 (carried) = 1 EXPLANATION. We first square the right hand figure (1), which gives us 1, which we write for the first figure of the answer. Now doubling the second figure(2) and mul- tiplying by the first figure (1) we have 4, which we . write for the second figure of the answer. Now doub- ling the third figure (5) and multiplying by the first figure (1) we have 10, to which we add the square of the second figure (2) which gives us 14. We write the 4 for the third figure of the answer and carry the 1. We now double the fourth figure (3) and multiply by the first figure (1) which gives us 6. We now double 90 RAPID CALCULATOR REVISED. the third figure (5) and multiply it by the second figure (2) and add 1 (to carry) which gives us 21. This we add to the 6 previously obtained which gives us 27. We write the 7 for the fourth figure of the answer and earry the 2. We now double the fifth figure (4) and multiply by the first figure (1) which gives us 8; we now double the fourth figure (8) which we multiply by the second figure (2) which gives us 12. We now square the third figure (5) and add 2 (to carry) which gives us 27; now adding these last three products to- gether we have 47. We write the 7 for the fifth fig- ure of the answer, and carry the 4. We now double the last figure (4), and multiply by the second figure (2), which gives us 16. We now double the fourth figure (8) and multiply by the third figure (5) which gives us 30; now adding these last two products to- gether we have 46+4 (to carry) 50. We write the 0 for the sixth figure of the answer and carry the 5. We now double the last figure(4) and multiply by the third figure (5) which gives us 40. Now squaring the fourth figure (3) and adding 5 (to carry) we have 1440-54. We write the 4 for the seventh figure of the answer and carry the 5. We now double the last figure (4), and multiply by the fourth figure (3) which gives us 24 to which we add 5 (to carry) which gives us 29. We write the 9 for the eighth figure of the answer and carry the 2. Wenow. square the last figure (4) which gives us 16 to which we add 2 (to carry) which gives us 18; this we write to complete ~ the answer. HOW TO SQUARE ANY NUMBER UNDER ONE THOUSAND. This method is here published for the first time, and it is to this that we are indebted for our ability to square numbers mentally. Any one with good natural ability and a fair amount of practice, can attain a pro- ficiency in squaring numbers by this method that will RAPID CALCULATOR REVISED. 91 excite the wonder and admiration of any audience. By a thorough mastery of it, the lightning calculator may adorn his cap with the most brilliant plums of _ approbation, and fill his coffers with the golden laurels of success. What is the square of 148? 148° 43°—= 2304 A8x2—= 96 1 ioe b 21904 We first square 48 (by the method for lightning involution), which gives us 2304. We now muliply 48 by 2 which we write two places to the left of the first figure of our partial product, under the 23. We now square the last figure(1) which we write one place to the left of the square of 48; now adding we have 21904. What is the square of 248? 248°? 43°— 2304 48x 4192 2A 61504 92 RAPID CALCULATOR REVISED. ——————— SSS = 29687616 | 143== 2044 ~~ SS S ES ~ SS SSS Sa = SS = aS S > SSS Tae y Lo D i. f SRS Vi y LEN SS SSS SS SS SS WES WSS BOSS The above cut represents a lightning calculator squaring numbers mentally by the rule on previous page. The calculator has his back to the board and calls the answers without seeing or making a figure, while a bystander writes the results as called by the calculator. This is not a picture to tickle the fancy, but is one that represents an actual ocurrence, and its. sight should fill every one with a desire to master the rule by which audience after audience has been amazed and mystified. We first square 48, which gives us 2304. We now multiply 48 by 4 which gives us 192, the right hand figure of which we write in the hundreds’ plac2, letting RAPID CALCULATOR REVISED. 93 the others follow to the left. We now square 2 which we write in ten thousands’ place; now adding we have 61504. If the number to be squared is between 100 and 200, multiply the two right hand figures by 2 for the second partial product; if between 200 and 300, multi- ply by 4; if between 300 and 400 multiply by 6; if be- tween 400 and 500 multiply by 8; if between 500 and 600 multiply by 10; if between 600 and 700 multiply 12; if between 700 and 800 multiply by 14; if be- tween 800 and 900 multiply by 16; if between 900 and 1000 multiply by 18. The number to multiply by may always be found by doubling the hundreds’ figure of the number to be squared. What. is the square of 843? 843° 1849 7088 710649 43°—= 1849 4316 = 688 8°—64 710649 We first square 43 which gives us 1849. As this number is between 800 and 900 it is evident that we must multiply by 16. We first multiply the first figure (3) by 16 which gives us 48; we put down the 8 under the third figure of the square of 48 (1849) and carry the 4. Multiplying the next figure (4) by 16 we have 64-4 (to carry)=-—68; we put down the 8 and carry the 6. Now squaring the last figure (8) we have 64+6 (to carry)=70 (see first illustration). Now adding we have 710649. 94 RAPID CALCULATOR REVISED. In multiplying by 16 it will-be found more con- venient to use our short method explained in a previous part of this work, but for fear the pupil has not master- ed that method yet, we have multiplied by the process of ordinary multiplication. HOW TO SQUARE ANY NUMBER UNDER TEN THOUSAND ENDING IN : TWENTY-FIVE. We present this method because it is simpler than the method of lightning involution. It is here pub- lished for the first time so far as our knowledge ex- tends. A thorough understanding of this method will materially assist the pupil in performing the seeming- ly mircaulous involutions hereinafter explained. What is the square of 8625? EXPLANATION. Squaring the 86 (by the process of multiply- 8625 ing numbers near 100 8625 together) we have 7396; ————_ we now take 4 of 86 and 7396 write it under this. Now 43 annex 0625 commencing 0625 one place to the right of SEE the above mentioned par- 74390625 .tial products; adding we have 74390625, the square of 8625. HOW TO SQUARE ANY NUMBER UNDER ONE HUNDRED THOUSAND ENDING IN TWENTY-FIVE. What is the square of 843625? RAPID CALCULATOR REVISED. 95, EXPLANATION. First squaring 43 we 843625 have 1849. Now doub- 843625 ling the left hand figure (a (8) of the multiplier we 1849 have 16; multiplying the 7088 fourth figure (3) of the 421 multiplicand by 16 we 5625 have 48. We write the ———_—_—— 8 two places to the left 7110705625 of the first figure of the square of 43 and carry the 4. We now multiply 4 by 16 which gives us 64,---4 (to carry) —68; we write down the 8 and carry the 6. Now squaring the last figure (8) we have 64,16 (to carry)—70. We now take 4 of 843 which gives us 421 (we omit the frac- tions). We now annex 5625, commencing to write it one place to the right of the above mentioned partial products; now adding (see illustration) we have 7110- 0705625. PROOF OF MULTIPLICATION. This method is based upon the principle that the excess of nines in the product of the two numbers equal the excess of nines in the product of the excesses of those numbers. 26482—8. Excess in multiplicand. 285. Si “multiplier. 607936—4. $ “product. 8 x<5=—40!-. 4--0=-4; ">. Final excess. Reducing these numbers by the method explain- ed for addition we find an excess of 8 in the multipli- cand, an excess of 5 in the multiplier, and an excess of 4 in the product. 96 RAPID CALCULATOR REVISED. Multiplying the excess (8) in the multiplicand by the excess (5) in the multiplier we have 40; this reduced gives us a final excess figure of 4, which is the same as the excess obtained from the product; hence the work is correct. Below we present a few bills which should be ex-— tended by the short process for multiplication previous- ly explained: DRY GOODS. SEDALIA, Mo., April 10, 1892. Blackford & Son, Clinton, Mo. Bought of C. D.: Minter @& Go. | : Yd. Bibon @ 162¢ Sa ad ee Pr, Fur: Gloves “(@:$2.507 2. vee Doz Collars "@ 8232.2 ae Yd. Doeskin’ @799¢ 22 34 “Velvet Ribbon. (@ $°.22403%. * Gingham @ "250" ae “MSN (0.2 15e.8 ee ” inen(@ “Sil Geen tia ae DRUGS. $ 2! $146|72 SEDALIA, Mo., August 26, 1892. Messrs. Taylor & Co., Bought of Ott Bros. WW Wty NOB bo Doz. Hair Brushes @_$3:50. 23: “Tooth Brushes @ | “Tooth Powder @ $1.80:. 2. > Wace Powder @ 32° 3-7 Boxes English Chalk @ $.75...... Doz. Cuticura S0ap @W S242 15 oe $ 7) 7| 3/60 1/50 2/25 6 $27|35 RAPID CALCULATOR REVISED. 97 CLOTHING. SEDALIA, Mo., August 25, 1892 Charles Williams, St. Louis, Mo., Bought of Blair Bros. 3 SYEREECY (Pah Bs ee sean an ne | $ 54 5 Pog uats. (a $206 vere a Sk ok 60 6 Pease: (Derba wee ne ie ote ee wk es eS dee eOVErCcoals :(@) -S1B.. ao. eed neh ole | 182 Giereer Doze Collars) @). bo. 2. ee: oe acs “He 40 BOOTS AND SHOES. | Sedalia, Mo., December 1, 1907. John Smith, Kansas City, Mo., Bought of Wm. Courtney. fee eres Kid=shoes? @ $2.257 2.2.0... $ 9 eee rs... Groat shoes -@» 72.0005... 35 < 5 Bam eerie oH ats. @erp2.00 0... 5 Set ees || 20 CLOTHING. Sedalia, Mo., Nov. 26, 1907. Mr. B. W. Johnson, Ft. Scott, Kas., Bought of St. Louis Clothing Co. — ES LT RUASERE (73:2 3 be egret Sa Se ete || $234 2 (OVELCOAUS aA =. ©. Leena este so. 36 Besse @ $225- Peles. | 18 reels OuUil se (SCG 2 eee wets. Ko oleh ens | 5 18 Mandkerenierss @SosCe se oe 6) 14 IBS U DMA BOR iets Giri. oe ake ss | 6|72 5 Dea HEU REI Miami. won eere soe [3245 98 RAPID CALCULATOR REVISED. DRY GOODS.. SEDALIA, Mo., July 14, 1892. Brill & Co., Lincoln, Mo., Bought of H. H. Marean. Case Nop ae 338 Pieces Cotton, . 30°40. 526). 40% “26 see ; 26 AL @e oo eee 4) 16 |75 No. 134 , 12 | Pieces Merrimac Prints, 42-5600 (20. AG a toe 292-45" 36". 19 aia @. SOS. OP Oe ea ae No. 126 | 15 | Pieces Delaine, 34... 29)» 367 243 62h se 31) 46" 83" 27 eae 42. .637 41 @° 164¢c_ 3a, RAPID CALCULATOR REVISED. 99 DIVISION. This subject does not offer the facilities for short cuts that Multiplication offers, neither is it a subject of such great importance to the business man as the subject of Multiplication. We do not know that all the short cuts in Division, herein presented, will be found of ‘practical utility to the ordinary school boy, still there are many of them that possess virtues worthy of the recognition and use of every accountant in Amer- ica; HOW TO DIVIDE BY ALIQUOT PARTS, To divide any number by 12 multiply the dividend by 9 and point off one decimal place 1} cé 6é ia3 ¢é 8 6é é é és 9? ”? 13 éé “<é “é ec i ¢é é é sé 9? %? 12 cé é é a9 6 “< 66 ¢é 6é 9? 9? 24 <¢ &é “é (79 4 “cc é 66 ¢ 9? ”? 34 é “cc “é “ec 3 “é 6é 6é éé 99 9? 5 éé ¢ c¢ 66 LY, cc ¢é 6é “é 99 9? 64 éé 66 éé éé 16 6é 6é ¢¢ two 9? bP] 81 “é 6é ¢¢ é¢ 12 6é 66 ‘é ‘i oe 9? 9 124 ‘é ¢ cé (79 8 “é 6é ing ¢¢ +B] 9 142 ce cé sé 6é ff é éé ¢é ée bP ] 9? 16% ce é “é é 6 “< é é é +B] 9 18? multiply by 16 divide by 3 and point off two deci- mal places. 224 multiply by 4 divide by 9 and point off one deci- mal place. . 100 RAPID CALCULATOR REVISED. 25 multiply the dividend by 4 and point off two deci- mal places. 314 multiply the dividend by 16 multiply the product obtained by 2 and point off three decimal places. pale 334 multiply the dividend by 3 and point off two deci- mal places. 374 multiply the dividend by 8 divide by 3 and point off two decimal places. 624 multiply the dividend by 16 and point off three decimal places 66% multiply by 3 divide by 2 and point off two decimal places. 75 multiply by 4 divide by 3 and point off two decimal places. 83! multiply by 12 and point off three decimal places. 874 multiply by 8 divide by 7 and point off two decimal places. 88 multiply by 9 divide by 8 and point off two deci- mal places. 125 multiply by 8 and point off three decimal places. | 1331 multiply by 3 divide by 4 and point off two deci- mal places. 16632 multiply by 6 and point off three decimal places. 225 multiply by 4 divide by 9 and point off two deci- mal ‘places. 250 multiply by 4 and point off three decimal places. 333} multiply by 38 and point off three decimal places. 375 multiply by 8 divide by 3 and point off three de- cimal places. RAPID CALCULATOR REVISED. 101 625 multiply by 16 and point off four decimal places. 8334 multiply by 12 and point off four decimal places. 875 multiply by 8 divide by 7 and point off three deci- mal places. 7 15 multiply by 2 divide by 3 and point off one deci- mal place. 35 multiply by 2 divide by 7 and point off one deci- mal place. 45 multiply by 2 divide by 9 and point off one deci- mal place. 1. Divide 245 by 13. 245 9 220.5 Answer. Multiplying the dividend by 9 we have 2205, and pointing off one decimal place we have 220.5, the quo- tient arising from dividing 245 by 13. 2. Divide 543 by 13. 543 7 380.1 3. Divide 5424 by 124 5424 8 433.92 4. Divide 12342 by 16% 12342 6 740.52 102 RAPID CALCULATOR REVISED. 5. Divide 321 by 18% 321 16 3)5136 ny 17.12 6. Divide 21542 by 25. 21542 4 861.68 7. Divide 2134 by 314. 2134 16 34144 2 od 68.288 8. Divide 3421 by 334. 3421 3 102.63 - 9. Divide 5426 by 663. 5426 3 2) 16278 81.39 RAPID CALCULATOR REVISED. 103 10. Divide 4828 by 75. 4328 4 3) 17312 57.702 11. Divide 21432 by 125. 21432 8 171.456 12. Divide 421545 by 625. 421545 16 674.4720 PECULIAR CONTRACTIONS. 1. Divide 84240 by 24. 30) 84240 4 | 2808 702 3510 To divide any number by 24, divide by 30 and add 1 of the quotient to itself. In the above problem, we first divide 84240 by 30 which gives us 2808. 30 is 6 more than 24 which is 3% or 4 more; therefore our quotient is 1 too small. 1 of 2808 is 702 which added to 2808 gives us 3510. 2. Divide 549 by 734. 3)549 183 —_——————__—_ 73.2 104 RAPID CALCULATOR REVISED. To divide any number by 74, add 4 of itself and point off one decimal place. In the above problem 4 of 549 is 188 which added to 549 gives 732; pointing off one decimal place, we have 73.2. This process is _ the same as dividing by 10 and adding 4 of the quo- tient. 3. Divide 438560 by 54. 6(/)) 4356) 9|726 | 803 80632 Ans. We first divide by 60 which gives us 726. 60 is 6 more than 54; therefore our divisor is + (6= 5 of 54) too small. + of 726 is 802; adding 802 to 726 we have 8062. HOW TO DIVIDE BY FACTORING THE DIVISOR. Divide 245184 by 24. 6) 245184 4) 40864 10216 Ans. It will be observed that the factors of 24 are 4 and 6; so dividing by 6 and then dividing by 4 is the same as dividing by 24. This is much preferable to divid- ing by long division, and whenever possible the divisor should be factored. Do not divide by the process of long division when you can use the process of short di- vision just as well. Some pupils continually divide by 11 and 12 by long division, when, if they know the mul- tiplication table up to 12 times 12, they might just as easily divide by short division. In dividing by short division do not write the successive remainders to be earried to the right thus: RAPID CALCULATOR REVISED. 105 8) 4 5°4°4 568 This method of division is a positive injury to the pupil using it, and should be checked at once. No teach- er Should permit even the smallest pupils to indulge in this injurious practice. CONTRACTED METHOD FOR DIVIDING ALL KINDS OF NUMBERS. EXPLANATION. 432) 645840 (1495 By this process we ee multiply the figures of 2138 the divisor by those of the quotient, and mental- 4104 ly subtract their result from the dividend, writ- 2160 ing down the remainder only. By examining the 0 figures of the dividend and divisor, we see that 432 is contained into 645 one time. We write the 1 for the first figure of the quotient; we now multiply the first figure (2) of the divisor by 1, which gives us 2, which we immediately subtract from the figure 5 of the dividend, which leaves a remainder of 3. We write the remainder under the 5 of the dividend; now multiplying the figure 3 in the divisor by 1 we have 38, and subtracting from the 4 in the dividend, we have a remainder of 1, which we write beneath the 4. Now multiplying the figure 4 in the divisor by 1 and sub- tracting from the dividend, we have a remainder of 2, which we write beneath the 6. We now bring down the next figure (8) of the dividend; by examining the numbers we find that 432 is contained in 2138 four times, so we write 4 for the next figure of the quo- tient. Now multiplying the first figure (2) of the 106 RAPID CALCULATOR REVISED. divisor by 4 we have 8, and subtracting.from 8 we have a remainder of 0, which we write immediately beneath the 8. Now multiplying 3 by 4, we have 12; we subtract the 2 from the 3, which gives us a remain- der of 1, and carry 1 of the number 12 to the next or- der. Multiplying 4 by 4 and adding 1 (to carry) .we have 17. Now subtracting we have a remainder of 4. In performing any of the subtractions, if the fig- ure in the minuend is smaller than the figure to be subtracted, we must carry 1 in addition to whatever other carrying figure we may have. The number 17 will give us 1 to carry, and in our subtraction the figure 7 is larger than the figure 1 above it, so we must carry — 2; now subtracting 2 from the figure (2) above it, we have a remainder of 0. We bring down the next fig- ure (4) and we find that 432 is contained into 4104 nine times. Multiplying the first figure (2) of the divisor by 9 we have 18, and subtracting we have a remainder of 6 with 2 to carry. Multiplying 3 by 9 we have 27, +2 (to carry)—=29; now subtracting we have a re- mainder of 1 with 3 to carry. Multiplying the next figure (4) by 9 we have 36,+3 (to carry)=39; now subtracting, we have a remainder of 2 with 4 to carry, and subtracting 4 from 4 we have a remainder of 0. We now bring down the next figure (0) and we find that 432 is contained into 2160 five times with a remainder of 0; hence our quotient is 1495. Below we present the same problem worked out in a little more abbreviated and convenient form. In the following example the remainders are written ver- tically over the figures of the dividend. EXPLANATION. 242 111 By examination we 306 find that 432 is contained 432) 645840 into 645 one time. We ——_—_- multiply the first figure 1495 (2) of the divisor by 1, and subtract. from 5; RAPID CALCULATOR REVISED. 107 which gives us a remainder of 3; we write the 3 above the figure 8 of the dividend. Now multiplying the sec- ond figure (3) of. the divisor by 1 we have 3 and sub- tracting from 4 we have a remainder of 1, which we write above the figure 3. Now mutiplying the third figure (4) of the divisor by 1 and subtracting we have a remainder of 2, which we write above the 1; hence, reading vertically, we have a remainder of 213, and our next minuend is 2138; 432 is contained into 2138 four times, so we write 4 for the next figure of the quotient. Now multiplying the first figure (2) of the divisor by 4 we have 8, and subtracting from 8, we have a remainder of 0; we write the 0. directly above the second figure from the right of the dividend. Now multiply the second figure (3) of the divisor by 4 we have 12, and subtracting 2 from 3 we have a remainder of 1, which we write directly above the 0 in the second remainder. Multiplying the last figure (4) of the divisor by 4 we have 16,-++-1 (to carry) —17, and subtracting, we have a remainder of 4, which we write directly above the figure 1 in the second re- maider; we have 2 to carry and subtracting 2 from 2 we have a remainder of 0, and thus we proceed until all the figures of the dividend have been disposed of. PROOF OF DIVISION. This method of proving division is based upon the principle that the excess of the nines in the dividend equals the product of the excesses in the divisor and quotient plus the excess in the remainder. 243) 253456 (1043 243 1045 972 736 729 7 108 RAPID CALCULATOR REVISED. Divisor. 243. ,— 0, excess in divisor Quotient. 1043525 8,0 SS owe Sauer. Remainder. {eae 7, oS yemamaer. 0<8-++7,—7, final excess. Dividend*253456;.. > — 7, excess in dividend. EXPLANATION. Reducing these numbers by the method explained. for addition, we find no excess in the divisor, an ex- cess of 8 in the quotient, an excess of 7 in the remain- der, and an excess of 7 in the dividend. Multiplying the excess (0) in the divisor by the excess (8) in the quotient we have 0, to which we add the excess (7) in the remainder which gives us 7, the same excess that we have in the dividend; therefore the work is correct. RAPID CALCULATOR REVISED. | 109 FRACTIONS. In the business world, calculations involving frac- tions are, generally speaking, of a comparatively simple nature, still a thorough understanding of them is one of the chief requisites of an accountant; hence, while we have explained a great deal about them in previous portions of this work, we feel that a special chapter on this subject is, not only appropriate, but very es- sential in a book of this kind. HOW TO ADD FRACTIONS. What is the sum of 3 and ?? EXPLANATION. 2, ee & | 4 9-+-38— 15 = 15%. By this method we find the product of the numer- ator of the first fraction and the denominator of the second fraction, to which we add the product of the numerator of the second fraction and the denominator of the first fraction for the numerator of our answer. We next find the product of the denominators of the fractions to be added, for the denominator of the an- swer. In the above example we have arranged the frac- tions to be added, vertically; so we first find the pro- duct of the means (3 and 3) to which we add the pro- duct of the extremes (2 and 4) which gives 17 for the numerator of our answer. We now multiply the upper 110 RAPID CALCULATOR REVISED. mean (3) by the lower extreme(4) which gives us 12 for the denominator of our answer; hence the result is Find the sum of 4, 4, and. 4 ) 4s 2xK4=8, +1x5 t 2x95 515 4 pe 2 ) 10>< 220, --1asc3 — fs 10«3 ie la re Ss = We first write the fractions vertically. Now ad-_ ding the 4 and 4 together (by the method previously explained) we have +3. We conceive the +3 and 2 to- gether (by the method previously explained) which gives us 32 ,.or 143. ANOTHER METHOD OF ADDING FRACTIONS. This method is preferred by many, to the preced- ing one. We give both, however, and the accountant may make his own selection. RULE. Multiply each numerator by the product of all the denominators except its own, and find the sum of these products which will be the numerator of the answer. Multiply the denominators of the fractions, to be added, together for the denominator of the answer. Find the sum of 3, 3? and 4. EXPLANATION. 2x 45540 Multiplying the num- oxo 45 erator of % by 456" we 4x4 3—48 have 40. Multiplying the a numerator of 2 by 3x5 133 we have 45. “Mule eee a 60 ing the numerator of 4 5 42% or 21%. Ans, by 43 we have 48. Ad- ding 40, 45 and 48 to- RAPID CALCULATOR REVISED. Vy, gether, we have 133 for the numerator of our an- swer. Multiplying the denominators of the frac- tions together (3x45) we have 60 for the denom- inator of our answer; hence our result is 74°, or Pde oa HOW TO SUBTRACT FRACTIONS. From 3 take %. EXPLANATION. pe By this process we 3 peas first write the subtrahend ( oats NG or number to be subtract- 3 ed, and immediately be- ie Bec neath it we write the | 4 minuend. We now mul- tiply the means togeth- er from which we subtract the product of the ex- tremes, for the numerator of our remainder. We now multiply the upper mean by the lower ex- treme for the denominator of our remainder. In the above problem, multiplying the means (8 and 3) together we have 9, and multiplying the ex- tremes (2 and 4) together we have 8; now subtracting the product of the extremes (8) from the product of the means (9) we have remainder of 1 which we write _ for the numerator of our answer. Now multiplying the mean (3)-by the lower extreme (4) we have 12 which we write for the denominator of our answer; hence our result is +. ANOPHER METHOD FOR SUBTRACTING FRACTIONS. RULE. Multiply the numerator of the minuend by the denominator of the subtrahend, and from the product subtract the numerator of the subtrahend multiplied by the denominator of the minuend; the remainder will be the numerator of the answer. Multiply the denomi- at RAPID CALCULATOR REVISED. nator of the minuend by the denominator of the subtra- hend for the denominator of the answer. EXPLANATION. From ¢# subtract %. Pea Multiplying the num- 2<6==10 erator of the minuend an (4) by the denominator 2 of the subtrahend (38) a oslo we have 12. Multiply- 4 Ans. ing the numerator of the subtrahend (2) by the denominator of the minuend (5) we have 10. Sub- tracting 10 from 12 we have 2 for the numerator of our answer. Multiplying the denominator of the minuend (5) by the denominator of the subtrahend (3) we have 15 for the denominator of our an- swer ; hence our result is ;. HOW TO MULTIPLY FRACTIONS. Find the product of }«4#x2xH. SASS Oey SC NS ee ee 4 $ 6. 8 16 EXPLANATION. Arrange the fractions in a parallel line, and drop equal factors from numerators and denominators. — Multiply the remaining factors of the numerator to- gether for the numerators of the product, and the re- maining factors of the denominators together for the denominators of the product. ANOTHER METHOD FOR MULTIPLYING FRACTIONS. This method is the same as the preceding one with the exception that a perpendicular line is used in- stead of a parallel line. Some prefer the perpendicu- lar line. The student, however, may use either. RAPID CALCULATOR REVISED. 113 Find the product of 3x#x?x«3x#@. EXPLANATION. Ay Draw a perpendicular Ag line and write the num- 3615 erators of the fractions 21 ead b on the right hand side of — Ans. the line, and the denomi- SoC B= D4 nators on the left hand side; then cancel and multiply the remaining factors. together on _ the right hand side of the line for the numerator of your answer, and multiply the remaining factors together on the left hand side of Uae line for the denominator of your answer. HOW TO DIVIDE FRACTIONS. RULE. Invert the terms of the divisor, and proceed by the method given by our first rule for the multiplication of fractions. Divide # by ¢$. 49 9Q ak a ANS. a $2 10 EXPLANATION. Inverting the terms of the divisor (3) we have 2; multiplying + by 2 we have ;%,. Divide ?Xix +3 by «$x 13. $f 8 8-7 5 7 Se, yd ne: eax 48 Idd G2 18 16 | ANOTHER METHOD FOR DIVISION OF FRACTIONS. Divide 3 by 3. 114 RAPID CALCULATOR REVISED. 2 EXPLANATION. See Write the divisor di- 9 SO Rema we rectly under the divi- pes See dend, and multiply the fe extremes together for yas the numerator of your answer, and the means together for the denominator of your answer. In the above problem multiplying the extremes (2 and 4) together we have 8, which we write for the numerator of our answer. Now multiplying the means (3 and 3) together we have 9 which we write for the 8 denominator of our answer; hence the result is § . yy Divide 3X4Xyo by 3X#X#. Page ae dee a 375 “ids | 6 ! 5 — 48 or O23 Ans. 8a ht ee Ss neni Pier Fe fe . EXPLANATION. We first write the fractions of the divisor (3,2 , 2,) immediately beneath those of the dividend. Now omit- ting equal factors from the means and extremes we have a factor 5 remaining uncancelled in the means which we write for the denominator of our answer. Multiplying the remaining factors in the extremes (4 and 7) together we have 28 which we write for the numerator of our answer; hence our result is = or 52. Fractions may also be divided by writing the numera- tors of the dividend and the denominators of the divisor on the right hand side of a perpendicular line, and the denominators of the dividend and the numerators of the divisor onthe left hand side of this line: Cancel and multiply the remaining factors together on the right hand side of the line for the numerator of the quotient, and multiply the remainders together on the left hand RAPID CALCULATOR REVISED. 115 side of the line for the denominator of the quotient. By this method the preceding problem would be stated thus: D O 2 A 10)7 4 4 i) b 2 2 Z | — — 25 or ai! HOW TO MULTIPLY MIXED NUMBERS. Multiply 124 by8t. 124 8 123xK4i= 35 124x8= 9832 1013 Ans. EXPLANATION. Commence with the fraction of the multiplier, if there be a fraction in the multiplier, and multiply the entire multiplicand by the first whole number of the first partial product of the answer. Next multiply the entire multiplicand by the first whole number of the multiplier, and write the result for the second partial product of the answer, and continue multiplying the multiplicand by the whole numbers of the multiplier: until the multiplication is complete. Add the several partial products together, and the result will be the complete product. | In the above problem, we first multiply 4 by 4 which gives us ;4. We next multiply 12 by + "which gives us 3; hence our first partial product is 34, We next multiply 4 by 8 which gives us 2%. We write 116 RAPID CALCULATOR REVISED. down the 3 and carry the 2. Multiplying 12 by 8 we have 96,12 (to carry) gives us 98; hence we have 982 for the next partial product. Adding 3,4, and 982 we have 1012 for our complete product. Multiply 256842 by 181. 256842 131 256842 x 13—3339002 3381815 Ans. HOW TO DIVIDE MIXED NUMBERS WITHOUT REDUCING TO IMPROPER FRACTIONS. Divide 569414 by 431. 431) 56944 (131248 434 1239—1364 1293 6341—715 431 2875 = 2875 2848 Serre tie) 431 EXPLANATION. By inspecting the first two figures (56) of the dividend, we find that the divisor (431) is contained in- to them 1 time, so we write 1 for the first figure of the quotient. Now subtracting 431 from 56 we have 123. We bring down the next figure (9) of the dividend, and our minuend appears thus, 1239. The fraction 4 RAPID CALCULATOR REVISED. 117 belongs to the same order as the 2; and to get it in the order with the 9 we must multiply it by 10. 10 times 3—71, so we have 74 to add to 129 which gives us 136}. A31 is contained into 1361 three times; 3 times 431— 1292 and subtracting 129% from 1361 we have a re- mainder of 6%. We now bring down the figures (44) that occupy the next order in the dividend, and our new minuend appears thus: 6241. The 2 in the min- uend belongs to the same order as the 6; to get it in the order with the 4 we must multiply it by 10. 10 times 374, and 74 added to 644—71%. 431 is con- tained into 712 one time; subtracting 431 from 712 we have a remainder of 28,4, or expressed as the fraction- ; 2855 al part of the divisor, This reduced to a simple A314 fraction = 242, hence our result is 131 248. VALUABLE CONTRACTIONS FOR BUSINESS MEN. Find the cost of 64 yards of goods at 174 cents per yard. EXPLANATION. 174 By this process we 64 first multiply the whole number of the multipli- 1736-102 cand by that of the mul- Lic4e 9 . tiplier. We next multi- Gras 2 ply the whole number SEES in the multiplicand by $1.13 Ans. the fraction in the mul- tiplier, and if this pro- duct contains a fraction under } we omit the fraction; but if the fraction is 3 or more, we omit the fraction and call the pro- duct one more. Next multiply the whole number in _the multiplier by the fraction in the multiplicand (omitting the fraction in the product as previously explained). Now multiply the fraction in the multi- plicand by the fraction in the multiplier and if their product is less than 4 omit it; if 4 or more call it 1. LESS RAPID CALCULATOR REVISED. In the above problem we first multiply 17 by 6 which gives us 102, and we write this for the first partial product. We next multiply 6 by 4 which gives us 2; this we write for a third partial product. The product of 4 by 4 is 4; as this is less than $ we omit it. Adding our partial products together we have 113; hence our answer is $1.13. The result to this problem, if multiplied out by the ordinary method of arithmetic, will be $1.122; which is virtually $1.13. This process will give practically the same result as that produced by the ordinary pro- cess,-and we are sure that business men can use it to advantage. Find the cost of 115% yards of goods at Tic per “yard; | EXPLANATION. 1152 ee 74 Multiplying 115 by 7 we have 805 which we Jb x 7=—805 write for our first par- PISS 29 tial product. — Multiply- TxA oo ing 115 by 4 (disposing aX1i= 0 of fractions as previous eas ly explained) we have $8.39 29 which we write for our second partial pro- duct. Multiplying 7 by 2 (disposing of fractions by method previously explained) we have 5 which we write for our third partial product. 2 multiplied by 4 gives us a product which is less than 4, so we omit it. Now adding our several partial products together we have 839; hence our answer is $8.39. In multiplying a whole number by a fraction, first multiply by the numerator and divide by the denomina- tor, thus: 14 multiplied by 3 equals 14x2~+3=—94. In exceptional cases this method of multiplication RAPID CALCULATOR REVISED. 119 may make a difference of one cent in the product, but ordinarily the result will be the same in business as that produced by other processes. By this method a twelve-year-old. school boy can be taught to tell men- tally and almost instantaneously the result of such problems as the following: 1. Find the cost of 154 yards of goods at 51¢c per yard. 2. Find the cost of 124 yards of goods at 84c per yard. 3. Find the cost of 144 yards of goods at 24¢ per yard. : 4. Find the cost of 18 yards of goods at 62 per yard. How to multiply similar numbers together whose fractions add one. Multiply 84 by 84. EXPLANATION. 4 Call the figure in the 3 multiplier one more and multiply the multiplicand (24 by it, and write the pro- duct for the whole num- bers of the answer. Now multiply the fractions to- gether and write their product for the _ fractions of the answer. In the above problems we call the figure (8) of the multiplier one more, or 9, and multiply the multipli- cand by it which gives us 72. Now multiplying 4 by 4 we have 1; hence our result is 724. Multiply 62 by 6}. 63 EXPLANATION. 16% Multiplying 6 by 7 (6 423, increased by 1) we have 42, and multiplying 2 by 4 we have ;';; hence our answer is 42,3. 120 RAPID CALCULATOR REVISED. Multiply 14% by 141. 14: 15 dd 2104; HOW TO MULTIPLY ANY TWO NUMBERS THE DIFFERENCE OF WHICH IS ONE AND THE SUM ON WHOSE FRACTIONS IS ONE. Multiply 81 by 73. . EXPLANATION. 4 By this process in- 3 crease the figure of the — larger number by one Tx9=263 and multiply by the 1—— 13 smaller number, and write their product for 6342 the whole number of the answer. Next subtract the square of the fraction of the larger number from 1, and write the remainder for the fractional part of the answer. In the above problem we increase the larger num- ber (8) by 1, which gives us 9; multiplying 9 by 7, we have 63, which we write for the whole number of the result. Now multiplying 1 by 1 we have ;., and sub- tracting ; from 1 we have a remainder of 13, which we write for the fractional part of our answer; hence the result is 6313. | HOW TO MULTIPLY ANY TWO NUMBERS TO- GETHER WHOSE FRACTIONS ARE SIMILAR. PROcESsS.—First, multiply the whole numbers to- gether, to which add the product obtained by multiply- ing the sum of the whole numbers by either of the frac- - tions. Next add the product of the fractions and you will have the complete result. RAPID CALCULATOR REVISED. 12i Multiply 64 by 83. EXPLANATION. 63 84 We first multiply 6 by 8, which gives us 48; this 6x« 8=—48 we write for our first ——- partial product. Now 6+8xi= 7 adding 6 and 8 together 4xj= 4 and multiplying by 4, we have 7, which we write 554 | for our second partial product. Multiplying 4 by 4, we have i, which we write for our third partial product; adding our several partial products togeth- er we have 554, the complete product. Multiply 94 by 64. 91 EXPLANATION. 64 ) a We first multiply 9 by 9x 6—54 6, which gives us 54 for eae our first partial product. 9+6x4i= 5 Now adding 9 and 6 to- 4xX4=— +} gether and multiplying by 4, we have 5 for our 594 second partial product. Multiplying 4 by 4 we have 2 for our third partial product; adding our sev- eral partial products together we have the complete product, 593. In solving practical problems the pupil should omit writing the several partial products, and add them together mentally. HOW TO MULTIPLY NUMBERS ENDING IN TWENTY-FIVE, FIFTY AND SEVENTY- FIVE TOGETHER. PROCESS.—Reduce the 25, 50 or 75, to a fraction of hundreds, and reduce to its lowest terms, and then —— eee Fy ¥ _ * on ae ois ee ere rental ALDL LI OOOO A aw a ee eee baal ———— wa ww 127 RAPID CALCULATOR REVISED. - ———_—" multiply by our process for multiplying mixed num- bers together and carry the fraction in the result four places to the right. Multiply 675 by 425. EXPLANATION. 62 hundreds. 41 hundreds. In this problem 75-=% (hundreds) and 25==i (hundreds) hence we have 62 (hundreds) to be multiplied by 44 (hun- dreds. ) Multiplying 62 41 by the method of multiplying mixed numbers to- gether, we have 29,4, (ten thousands.) 5 carried four places to the right —0625, and this annexed to 29= 290625, the product of 675425. 29,', ten thousands. a= 0625 675 « 425—290625 Table of sixteenths and their equivalents in deci- mals: j,==.0625 6,375 44—.6875 a —,125 — 4375 —— 15 f= 1875 5 t¢—=-8125 4,25 = .5625 re—= 875 §,—=.38125 19—— 625 43.9379 HOW TO MULTIPLY BY NUMBERS CONTAIN- — ING FRACTIONS NEAR A UNIT. Multiply 12 by 83. EXPLANATION. 12 82 Our multiplier (82) is near 9 so we first multi- 12><9=—1.08 ply the multiplicand | Pe fee a (12) by 9 which gives us ER 108, 8% is 4 less than 9, 106 Ans. so we have taken our multiplicand 4 times too often. 4 of the multiplicand (12)—2 and subtracting 2 from 108 we have 106, the product of 12x83. RAPID CALCULATOR REVISED. 123 Multiply 18 by 73. 18 14 18x 8144 18--4— 4} 1394 Ans. TABLE SHOWING THE PRICE OF A SINGLE ARTICLE WHEN THE PRICE PER DOZEN IS GIVEN. Per Doz. Per Piece Per Doz. Per Piece. 25¢ — ds $ 6 —— 50e 50¢ me Ale = Sy — 58ie 75¢ = 6ic $ 8 = 662c¢ $1 — Sie $ 9 = 75¢e $2 ==: ..16%¢ $10 ee 83i¢ $3 Se OC $11 = 912¢ $4 = 334¢ $12 == + 100c $5 Sah ake Te SHORT METHOD FOR MULTIPLYING DECI- MALS RESERVING ANY NUMBER OF DECIMAL -PLACES. To multiply 2.45839742608479 by 4.2532967429 reserving three decimal places, or correct to three decimal places. Begin at the decimal point in the multiplicand and count to the right as many decimal places as you wish to reserve in the product; draw a_ perpen- dicular line to the right of the last figure count- ed, and write the multiplier so that this line separates the whole number and decimal. Com- mence to multiply with the left hand figure of the multiplier, multiplying the figure of the multiplicand that is as far on one side of the line as the figure of the multiplier in use, is on the other side. Do not write 124 RAPID CALCULATOR REVISED. this result in the partial product, but add its tens to the product obtained from the next figure of the multipli- cand, which sum you write to the left of the perpendi- cular line. After multiplying all the figures to the left in the multiplicand by this figure in the multiplier, you proceed in like manner with the other figures of the multiplier till there is nothing more to multiply, al- ways observing that you begin with that figure of the multiplicand as far on one side of the line as the figure of the multiplier in use is on the other side. Add and you have the result. By carrying the tens we mean to carry 1 for any- thing from 5 to 14 inclusive, or 2 for anything from 15 to 24 inclusive and so on. It will be seen in the example below that all fig- ures, to the right of the first one used in the multipli- cand, are dropped. We first multiply the 3 2.458 | 39742608470 in the multiplicand by 4,| 2532967429 the 4 in the multiplier. — We do this to find the 9.833 carrying figure. 492 You must always mul- 123 tiply one place more than 7 you wish to retain so as —______—_—. to obtain the correct fig- 10.455 ure to carry. We now multiply 8 by 4 which gives us 32, and 1 (to carry) makes 33. We now continue multiplying by 4 by the usual method. We now go one place to the right of the line and multiply by 2. 8x2=—16 which gives us 2 to carry. 5x2+2 (to carry) —12. We put down the 2 and con- tinue the multiplication as usual. 5<5=25 which gives us 3 to carry. 45-43 (to carry) =—23. Put down the 3 and continue the multiplication. 4x*3—12 which gives us 1 to carry. 23-41 (to carry) =7. Adding we have the result, 10.4554. RAPID CALCULATOR REVISED. 125 In multiplying by this method as you go to the right of the line in the multiplier, drop places in the multiplicand equal to the number you have used in the multiplier. CONTRACTED METHOD. $48.241/321 .|0431135 — ORDINARY METHOD. 1930 145 $48.241321 5) .0431135 $2.080 Ans. 241206605 144723963 48241321 48241321 144723963 192965284 2.0798521929335 or $2.08 Ans. In business if we have five mills or over, it is call- ed a cent; hence the results by the contracted and ordi- nary processes are practically the same. Every one should thoroughly master this method. Remember this is an age of steam and electricity and we have no time to spend solving problems by long, tedious processes. HOW TO MULTIPLY BY NUMBERS CONTAIN- ING DECIMALS THAT ARE ALIQUOT PARTS. RULE. Reduce the decimal to common fractions, and pro- ceed by the method explained for multiplying common fractions together, and at the conclusion of the multi- plication, if desired, reduce the fraction in the an- swer to a decimal. 126 RAPID CALCULATOR REVISED. Multiply 8.25 by 4.75. EXPLANATION. 8 Ae 8.25 is equal to 8}. 4.75 is equal to 42. Mul- 39-3,-or 39.1875 tiplying 81 by 43 by 3 method previcusly ex- plained we have 39,°, or, reducing the ;3; to a decimal, we have 39.1875. Multiply 16.625 by 4.25 EXPLANATION. 162 .625 is equal to §; .25 41 is equal to 1, so we mul- tiply 168 by 44, ‘which 7024 gives us 7034 Ans. Multiply 18.125 by 5.33834. EXPLANATION. 125 18. equaleoe, 333} is equal to 4, so we 963 multiply.. 18$— by = | which gives us 963. Ans. CONTRACTED METHOD FOR DIVISION OF DECIMALS CORRECT TO ANY NUMBER. OF DECIMAL PLACES DESIRED. EXAMPLE BY CONTRACTED METHOD CORRECT TO THREE — DECIMAL PLACES. 23.479624388462978) 9.28 |4967849687759784692478964 (395 704 Ans. a correct to 224 3 decimal 214 places. Jt, 12 1 RAPID CALCULATOR REVISED. Lea After writing the divisor and dividend as in ordi- nary division we place a dot over the 8 in the hun- dredths’ place in the dividend because hundredths’ denomination results from multiplying 10 (a unit of the highest denomination found in the divisor) by .001 (a unit of the lowest denomination required in the quotient.) (10.001—.01, denomination to be mark- ed in dividend.) After marking the 8 in the dividend with a dot over it, we draw a vertical line to the right of it and drop all figures in the dividend to the right of this line. We now see that 4 in the tenths’ place in the divis- or will fall under the marked figure 8 of the dividend, hence we place a dot over the 4 and drop all figures in ‘the divisor to the right of 4. We now find how many times the contracted divisor is contained in the con- tracted dividend, and place it in the quotient (3). We now say three times 7 is 21, and add the 2 to carry, to three times 4 (the marked figure) which gives us 14. We write the 4 beneath the 8 and carry the 1. 38 times 2 is 6, add the 1 and we have 7 to write under the 9. Subtract, and we have 224 for a new dividend. We now cancel the 4 from the divisor and use it only in carrying its tens to the next higher order at the next ‘multiplication. After placing a dot over the 3 in the unit’s place in the divisor, we find that the divisor 23 as it now ‘stands, is contained 9 times in the dividend 224. We now multiply 4 by 9, which gives us 36. We carry 4 for 36, because it is more nearly 4 tens than it is 3 tens. 9 times 8 is 27, and adding the 4 carried, we thhave 31. Write the 1 under the 4 in the dividend and carry the 3. Nine times 2 is 18 and 8 carried is 21, which we write under the 22 in the dividend. Sub- tract and we have 13 remaining for the next dividend. We now cancel the 3 in the divisor and place a dot over the 2. We see that 2 is contained 6 times in 13, but we must take into account the tens of the next lower order, and by so doing, we find that 5 will be our next 128 RAPID CALCULATOR REVISED. figure in the quotient, as 5 times 4 is 20 and 8 times 5 is 15 and 2 carried is 17, and 17 is more nearly 2 tens than 1 ten, hence we have 5 times 2 is 10 and 2 carried is 12. We now cancel the last figure in the left of the divisor, point off three places in the quo- tient, and find our answer is .395, correct to three decimal places. In cases where absolute accuracy is required, re- serve one more place than you really wish to retain, and the answer will be absolutely correct. COMMON METHOD. That the pupil may see how much time is saved by the contracted method, we give below the same problem as given above, worked by the common method of division. When the problem involves dollars and cents in the answer it will be seen that the exact number of cents will be found by reserving as many as three dci- mal places and the problem can be solved in one-tenth the time that it will take by the common method. RAPID CALCULATOR REVISED. 129 EXAMPLE BY COMMON METHOD. 23.479624388462978) 9.28496784968775978469247 8964 (.895447886902+ 7 04388873165388934 2 24108053314886638 2 11316619496166802 127914338187198364 117398121942314890 105162162448834746 93918497553851912 112436648949826349 93918497553851912 185181513959744372 164357370719240846 208241432405035264 187836995107703824 204044372973314407 187836995107703824 162073778656105838 140877746330777868 - 211960323253279709 211316619496106802 64370375711290764 46959248776925956 17411126934264808 130 RAPID CALCULATOR REVISED. INTEREST. Under this head we give a large variety of the best interest methods known to the world at the pre- sent day. We do not claim to give every method, but we do claim to give the best methods, and we believe that no book published prior to this, contains so great: a variety of good interest methods. This is a subject of great importance to everybody, and the value of a thorough understanding of it can hardly be overes- timated. If you have trouble in finding the interest accurately and readily on any sum of money, place yourself in possession of some of these methods, and the calculation of interest will be rendered rapid, easy and accurate. By these methods the problems that were regard- ed as an arduous task by our forefathers are the de- light of the twelve-year-old school boy. The processes of computing interest explained by our school text- books forty years ago, however good they might have been in their time, are not able to keep apace with the rapid improvements of this century. So far as the counting room is concerned, their death knell was sounded several years ago, and it is only a question of a very short time when they shall be obliterated from the pages of our school text-books, and sink to their final resting place in the graveyard of oblivion. RAPID CALCULATOR REVISED. 13] a | THOUSAND-DAY METHOD FOR COM- PUTING INTEREST. This method, though, perhaps too new to be in general use yet, has come to take its place in the fore- most ranks of interest calculations. By this process it is just as easy to find the interest on an odd princi- pal, as an even principal. It matters not what a tangled labyrinth of days, per cent., principal, and time you may have, the thousand-day method is the omnis- cient prince that will successfully untie the most diffi- cult Gordian knot of interest calculations. 132 RAPID CALCULATOR REVISED. - This method is based on the fact that any prin- cipal will dcuble itself in one thousand days at 36 per cent. We first find the interest on the principal for the given time at 86 per cent, after which, by the methods hereinafter explained, we may find it for any per cent. PROCESS.—Interest on any sum of money for one thousand days equals the principal; interest on any sum of money for one hundred days equals 4 of prin- cipal; interest on any sum of money for ten days equals sia Of principal; interest on any sum of money for one day equals z,55 of principal. : From the above we deduce the following: To find the interest on any sum of money for one hundred days, move the decimal point one place to the left; to find the interest on any sum of money for ten days, move the decimal point two places to the left, to find the interest on any sum of money for one day, move the decimal point three places to the left, thus: the interest on $324 for one thousand days equals $324; the interest on $324 for one hundred days equals $32.4; the interest on $324 for ten.days equals $3.24; the in- terest on $324 for one day equals $.324. Any number of days is composed of so many thou- sands of days, so many hundreds of days, so many tens of days, and so many units of days. By this method, we first find the interest on the principal for the number of thousands of days; next for the number of hundreds of days; next for the number of tens of days; next for the number of units of days, and adding we have the interest for the whole time. In solving practical problems, do not move the decimal point, but leave it in its original place, and con- ceive the decimal point to be moved by moving the finger above the figures of the principal. RAPID CALCULATOR REVISED. 133 Find the interest on $348.20 for 4 months and 3 days at 36 per cent. $34|82 —interest on principal for 100 days. 6|964—interest on principal for 20 days. 1|044—interest on principal for 3 days. $42.828—interest on principal for 123 days at 36 per cent. EXPLANATION. 123 days is composed of 1 hundred, 2 tens, and 3 units. We first find the interest on $348.20 for 100 ' days by moving the decimal point one place to the left, which gives us $34.82. We next find the interest on $348.20 for 10 days by moving the decimal point two places to the left which gives us $3.482; multiplying by 2 we have the interest for 20 days, which is $6.964. We then find the interest on $348.20 for one day by moving the decimal point three places to the left, which gives us $.3848; multiplying this by 3 we have the interest for 3 days, $1.044. Now adding we have the interest for 123 days at 36%: namely, $42.828. HOW TO FIND THE INTEREST AT DIFFERENT RATES. After obtaining the interest at 36% to find it at 6% divide by 6. 9% divide by 4. 8% multiply by 2 and divide by 9. 10% divide by 6, move the decimal point one place to the right and divide by 6 again, or divide by 3 and deduct ¢. 7% divide by 6 and add + of itself. 74% divide by 6 and add + of itself. 4% divide by 9. 1% divide by 36. 2% divide by 18. 134 RAPID CALCULATOR REVISED. le ae no ee a 3% divide by 12. 34% divide by 9 and deduct +. 44% divide by 8. 5% divide by 6 and deduct 4of the quotient. 54% divide by 6 and deduct +5. 84% divide by 4 and deduct +; . 11% divide by 3 and deduct +. 12% divide by 3. 18% divide by 2. 20% multiply by 5 and divide by 9. Find the interest on $246.12 for 123 days at 9 per cent. | 24|612—Int. for 100 days. 4/922—Int. for 20 days. \738—Int. for 3 days. 4) 30272 $7.568 Ans. By this method it will be found that reserving three decimal places will give a result sufficiently ac- curate for all practical purposes. The pupil should be instructed to draw a perpendicular line and place his finger in such a position over the principal as will in- dicate the interest for one thousand, one hundred, ten days, or one day, as the case may be; then commence his multiplication three places to the right of his fin- ger, writing the decimals on the right hand side of the perpendicular line and the whole numbers on the left. If at any time there are not three places to the right of the pupil’s finger, he should commence multi- plying as far to the right as he has places, observing that the first figure in the product should be placed no farther to the right of the perpendicular line than its first factor is to the right of the pupil’s finger. - RAPID CALCULATOR REVISED. 135 2. Find the interest on $2413.02 for 213 days at 6 per cent. hull) First step, | 4) $2413.02 482/604 We place the finger above the figures of the multi- plicand between the 1 and the 3, which shows our in- terest for 100 days, namely, 241.802; multiplying this by 2 we have the interest for 200 days, namely, $482, 604. We write the decimals to the right of the perpen- dicular line. Second step, i $2413.02 482'604 Int. teh sees by Ist step. 24|130 We now place the finger above the figures of the principal between the 4 and the 1, which gives us the interest for 10 days, namely, $24.13, which we write as previously directed with the decimals to the right of the line. Third step, i $2413.02 482|604 Int. produced by Ist step. ~ 24|130 “ $f ““2nd_ step. 7|239 We now place the finger above the figures of the principal between the 2 and 4, and we have the inter- - 136 RAPID CALCULATOR REVISED. est for 1 day, namely, $2.413+. This we multiply by 3, which gives us the interest for 3 days, namely, $7.239. This we write for our third partial result, as previously directed. Fourth step, $482|604 Int. produced by 1st step. 24113022" ‘* 20 Sten. Tivo se : “8d wep. 6)513|973—Total Int. at 36 per cent. $85.662-+ —Int. at 6 per cent. 3. Find the interest on $240 for 9 months and 27 days.at 9 per cent. Nine months and 27 days are equal to 297 days. To reduce months to days, seat by 3 and annex a cipher. $24 =Int. on $240 for 100 days. 3 _ $72| Int. on Prin. for 300 days. W2es 6 : 3 days. é 4) 71|23—= 66 éé ¢é éé 297 cé $17.82 Int. at 9 per cent. EXPLANATION. Two hundred ninety-seven days is near 300 days, so we first find the interest on the principal for 300) days which gives us $72; this is for three days too | much, so we subtract the interest on the principal for ~ 38 days (.72) from $72, which leaves us $71.28, the in- RAPID CALCULATOR REVISED. 1387 terest for 297 days at 36 per cent. Now dividing by . 4 we have $17.82, the interest on the principal for 297 days at 9 'per cent. oa A thorough understanding of multiplication by aliquot parts is a necessary adjunct to performing cal- culations rapidly by this method of computing inter- est. 4. Find the interest on $2416.18 for 249 days at 6 per cent. 4) 2416.18 604 45S Int. for 250 days. AB re ee Ie sah BOUL) Oem. oie DAD Ol. 3" $100.27 Ans. 5. Find the interest on $200 for 314 days at 4 per cent. 31|4 2 9) 62|8—Int. on $200 for 314 days. $6.98— Ans. EXPLANATION. The interest on $200 for 314 days is the same as the interest on $314 for 200 days, so we consider the dollars as days and the days as dollars, and moving the decimal point one place to the left, we have the in- terest for 100 days, $31.40. Now multiplying this by 2 we have the interest for 200 days, $62.80, and divid- ing by 9 we have the interest at 4 per cent., $6.98. When the dollars are expressed in even hundreds 138 RAPID CALCULATOR REVISED. or thousands, and the days are odd, consider the dol- — lars days and the days dollars, and proceed as prev- iously directed. 6. Find the interest on $6000 for 241 days at 8 per cent. . 241 6 1446—Int. on $6000 for 241 days at 36 per cent. 2 9) 2892 $321.38+ Ans. 7. Find the interest on $250 for 634 days at 10 per cent. 4) 634 6) 158.5—Int. on $250 for 634 days at 36 per cent. 6)26.417+ —Int. at 6 per cent. $44.0380 =—Int. at 10 per cent. EXPLANATION. Interest on $250 for 634 days is the same as the interest on $634 for 250 days; 250 is + of a thousand, and the interest on $634 for one thousand days would be $634; for 250 days it would be + of $634, or $158.50. Dividing this by 6 we have the interest at 6 per cent, $26.417. Dividing the interest at 6 per cent by 6 and moving the decimal point one place to the right we have the interest at 10 per cent, $44.03. The pupil should spare no pains to make himself a thorough master of this method for computing in- terest. It would be impossible for us to enumerate here its untold advantages. It is not only particularly valuable in performing all kinds of interest calcula- tions, but is equally useful in solving problems in ac- counts current and equation of accounts. All ac- RAPID CALCULATOR REVISED. 139 counts current may be shortened at least one-third by the application of the thousand-day method of com- puting interest. ANOTHER THOUSAND-DAY METHOD. HOW TO FIND THE INTEREST BY THIS THOUSAND-DAY METHOD. EXAMPLE.—Find the interest of $432.69 for one year, four months and seven days at 9%. Solution: 43 |269 A492 173 | 076 38 | 942 865 4)212|883 Int. at 36 per cent. 53|22075 Int. at 9 per cent. EXPLANATION. Reduce the time to days if not given in days. If the days consist of three figures, draw a line one place to the left of the decimal point in the principal and tet the line represent a new decimal ‘point; if tie days consist of two figures draw the line two places to the left of the decimal point; if of only one figure, draw it three places to the left of the decimal; if the days con- sist of four figures, draw the line through the decimal point; if the number of days consists of five figures, move the line one place to the right of the decinial point. Now, this line represents the decimal point in the solution of the problem. Write the days below the principal and to the right of the line, and draw a nori- zontal line under the days. Begin the multiplication by multiplying the third figure of the principal to the right of the line, by the 140 RAPID CALCULATOR REVISED. first figure of days to the right of the line. Set units’ figure of the product in the third place to the right of the line in the partial product, carrying the tens to the next higher order until you multiply all of the figures of the principal by the above named figure in the days. For the next partial product you will begin multiply- ing by the second figure of days to the right of the line, multiplying the second figure of the principal to the right of the line, always carrying to this partial pro- duct the tens that would occur by multiplying the fig- ure of the next lower order in the principal. Write the units’ figure of this product in the third place to the right of the line and proceed as with first multi- plication. Next begin with the third figure in days to the right of the line, multiplying the first figure in the © principal to the right of the line, setting units’ figure of this product in the third place to the right of the line, and proceed as above. So continue until you have used all the figures in the days. Observe that you use the next figure to the right in days as multiplier at each successive multiplica- tion, always beginning with the next figure to the left in the principal, from the one you began with in pre- vious multiplication. Never neglect to multiply, how- ever, the next lower figure in the principal than the one you are going to begin with, carrying the tens to the product of the first figure that you are to multi- ly. Add all these several products together and this will be the interest for the time at 36%. Now, for one per cent; divide this by 6 and the quotient by 6; for 2% divide by 9 and that quotient by 2; for 3%, di- © vide by 12; for 4%, divide by 9; for 5%, divide by 9 and add + of the result to itself; for 6%, divide by 6; for 7%, divide by 6 and add + of the result to itself; for 8%, divide by 9 and multiply the result by 2; for 9%, divide by 4; for 10%, divide by 6 and divide the result by 6 and multiply the last quotient by 10; for 11%, divide by 3 and subtract ;4 of the result from this; for 12%, divide by 3; in short, take as many 36ths of the answer as you wish to find per cent in- _ terest. RAPID CALCULATOR REVISED. 141 THE TWELVE PER CENT MONTHS AND TENTHS METHOD. To find the interest on any principal at 12 per cent move the decimal point of the principal two places to the left, and mutipy by the time in months and tenths. a Divide any number of days by 3 for tenths of months, thus: 18 days equal ;°; of a month; 24 days equals +3; of a month. Find the interest on $243.12 for 4 months and 3 days at 12 per cent. EXPLANATION. 4|.1 9.4913 Three days equal .1 of a month, hence, our 9725 time is 4.1 months. Now 243 moving the decimal point ee of the principal two $9.968 Ans. places to the left we have 2.4812, which we multiply by 4.1 by the short method for multiplying decimals together. This gives us $9.968 or practi- cally $9.97. After finding the interest at 12 per cent. to find it at 6 per cent divide by 2. 8 per cent multiply by 2 and divide by 3. 9 per cent multiply by 3 and divide by 4. 10 per cent multiply by 5 and divide by 6, or de- duct 4. 15 per cent add t. 20 per cent multiply by 5 and divide by 3. Find the interest on $521 for 2 months and 9 days at 8 per cent. 142 RAPID CALCULATOR REVISED. $5.21—Prin. with decimal point moved two places to the left. 2.3 —Time in months and tenths. $11.983—Int. at 12 ‘per cent. 2 3) 23.966 $7.99 —Int. at 8 per cent. - SIX PER CENT MONTHS AND TENTHS METHOD. To find the interest on any principal at 6 per cent move the decimal point of the principal two places to the left and multiply it by one-half the number of months and one-sixth the number of days, written as tenths of months. In taking one-half the number of months, if you have an odd month reduce it to days, add in the number of days that you have, take one- sixth, and write the quotient as tenths of months. Find the interest on $324 for 4 months and 18 days at 6 per cent. $3.24 —Prin. with the decimal point moved two places to the left. | 2.38 —One-half the number of months and one-sixth the number of days written as tenths of months. $7.452—Int. at 6 per cent. EXPLANATION. We first take one-half the number of months (4) which gives us 2. Now dividing the number of days (18) by 6, we have 3. which we write as tenths of RAPID CALCULATOR REVISED. 143 months, hence, our multiplier is 2.3. Now moving the decimal point of the principal two places to the left, we have $3.24, and multiplying this by 2.3, we have $7.452, or $7.45. After finding the interest at 6 per cent to find it at 7 per cent add } of itself. 74 per cent add } of itself. 8 per cent add } of itself. 9 per cent add + of itself. 10 per cent move the decimal point one place to | the right and divide by 6. Find the interest on $425.23 for 1 year 4 months and 17 days at 6 per cent. $4,252 |3 8| 283 34018 850 340 13 $35.221 or $35.22. EXPLANATION. 1 year and 4 months equals 16 months, and this divided by 2 gives us 8. One-sixth of 17 days equals .2838-+; an- nexing this to 8 we have 8.2838+. Now moving the decimal point of the principal two places to the left we have $4.2523, and multiplying by 8.283 by the short method for multiplying decimals to- gether we have $35.221, or $35.22. Remember that in the answer 5 mills or over should be called a cent but anything under this amount is not considered. 144 RAPID CALCULATOR REVISED. 7. A Wo [BB ee yf ges Ai VoL \ aE ena WOM AY] Ze CK (sgl aN . a! 4 fs i KL Oi) 4 S| ke i/ = 2 rad i Hy) | LZ SS SS3—~s - wy 7 a ~~ Bg RS ee SSSS Fo. C2 HN = MM I oor? F oN i - ‘. qf Ko y he 4 ; SW SIXTY-THREE, NINETY-THREE, AND THIRTY- THREE DAY METHOD. This method is particularly valuable to bankers, and we believe that for calculating interest for any of the above named days, this is the shortest method in existence. The principles governing the calculations of interest by this method have been known for a long time, but never before so far as our knowledge extends have they been reduced to a scientific process so that the work may be performed almost automatically. This method is by far shorter than the use of interest tables, and the high encomiums which it has received from the banks all over the country speak far more for its merits than it would be possible for us to speak on _ cold paper. The basis of this method for computing interest is 6 per cent. However, after it is found at 6 per cent it may readily be found at any other per cent. RAPID CALCULATOR REVISED. 145 HOW TO FIND THE INTEREST FOR SIXTY- THREE DAYS. Move the decimal point of the principal two places to the left, then divide by 2, writing the result immediately under the principal one place farther to the right than it would ordinarily be written, then add and you will have the interest for 63 days at 6 per cent. Find the interest on $2438.12 for 63 days at 6 per cent. ) | $2|4312—Int. for 60 days. | 1215— 6é 6é 3 66 $2.55 =Int. for 63 days at 6 per cent. EXPLANATION. _ Moving the decimal point of the principal two places to the left we have $2.4312; dividing this by 2 we have $1.215+; we write this one place farther to the right than it would ordinarily be written or, in other words, we divide by 10, which gives us $.1215. Now adding this last number to $2.4312 we have $2.55, which is the interest at 6 per cent. HOW TO FIND THE INTEREST FOR NINETY- THREE DAYS. Move the decimal point of the principal two places to the left, then divide by 2, writing the result immediately under the principal. Now write the re- sult that was obtained by the dividing by 2 down again one place farther to the right than it was writ- ten before and add the three numbers together for the interest for 93 days at 6 per cent. Find the interest on $548 for 93 days at 6 per cent. $5|/48 —Int. for 60 days. ] TA — “ce sé 30 6c“ 274— 6é “< 3 66 $8.494—Int. for 93 days at 6 per cent. 146 ~ RAPID CALCULATOR REVISED. EXPLANATION. To find the interest on $548 for 93 days we first move the decimal point two places to the left which gives us the interest for 60 days ($5.48). Now taking the one-half of $5.48 ($2.74) we write it immediately beneath the $5.48. We now write the $2.74 down again one place farther to the right than it was writ- ten the first time. Adding, we have $8.494, the in- terest on $548 for 93 days at 6 per cent. HOW TO FIND THE INTEREST FOR THIRTY- THREE DAYS. Move the decimal point of the principal two places to the left and divide by 2, writing the result immediately under the principal. Now write the re- sult that was obtained by dividing by 2 down again one place to the right and add the two quotients to- gether, but do not add in the principal. This is just the same as the process for 93 days. with the excep- tion that for 93 days we add all three of the numbers and for 33 days we simply add the two lower numbers. Find the interest on $548 for 33 days at 6 ver cent. $5|48 ———- Ait 4-s=Int. 1or:30 days 274— ce 66 3 (74 $3.014—Int. for 33 days at 6 per cent. EXPLANATION. Moving the decimal point of the principal two places to the left we have $5.48; taking one-half of this we have $2.74. Now taking one-tenth of $2.74, which is the same as writing it one place to the right. we have $.274. Now adding $2.74 and $.274 we have $3.01, which is the interest on $548 for 33 days at 6 per cent. RAPID CALCULATOR REVISED. 147 After having obtained the interest at 6 per cent to find it at 10 per cent move the decimal point one place to the right and divide by 6. 7 per cent add $ of itself. 5 per cent deduct ¢ of itself. 4 per cent deduct } of itself. 8 per cent add 4 of itself. 12 per cent multiply by 2. 18 per cent multiply by 3. 20 ‘per cent multiply by 10 and divide by 3. Find the interest on $2418 for 63 days at 8 per cent. lo 8 Int for 60 days at 6 per cent. 2 09— 6é sé 3 “é 1a “cc «é rey WA) 389— “< ‘é 63 «6 “é 6 é cc 8 463— “eé éc 63 iad cc 3 é ‘é $30,002 6é ‘6 63 “<é é 8 “ce “é TWO HUNDRED-MONTH METHOD OF COM- PUTING INTEREST. A Thousand Years as a Day and a Day as a Thous- and Years. This method of computing interest is similar to the thousand-day method, and will be found to be a very valuable acquisition to interest methods. ‘his process is based on the fact that any sum of money will double itself in 200 months at 6 per cent; there- fore the interest on any principal at 6 per cent for 200 months is equal to the principal. The interest on any principal at 6 per cent for 20 months is equal to 4 of the principal. The interest on any principal at 6 per cent for 2 months is equal to ;j> of the principai. The interest on any principal at 6 per cent for 6 days is equal to =, cf the principal; hence to find the in- terest on any sum of money at 6 per cent for 20 months, we move the decimal point one place to the 148 RAPID CALCULATOR REVISED. left. To find it at 6 per cent for 2 months, we move the decimal point two places to the left, and to find it at 6 per cent for 6 days, we move the decimal point three places to the left. For example, the interest on $246.12 for 200 months at 6 per cent is equal to $246.12; the interest on $246.12 for 20 months is equal to $24,612; the interest on $246.12 for 2 months is equal to $2.461-++ ; the interest on $246.12 for 6 days is equal to $.246+. Find the interest on $826.15 for 1 year 10 months and 18 days at 6 per cent. $82|615=Int. for 20 months at 6 per cent. 8|261— (7 66 = 66 66 6 66 2|478—= “ “ 18 days at 6 per cent. SIO DAS ee 1 year 10 months and 18 days at 6 per cent. EXPLANATION. One year and 10 months are equal to 22 months. We first find the interest on the principal ($826.15) for 20 months; this we accomplish by moving the dec- imal point one place to the left, which gives us $82.615 We now find the interest on the principal for 2 months by moving the decimal point two places to the left, which gives us $8.261+. We now find the interest on the principal for 6 days by moving the decimal point three places to the left, which gives us $.826+. Multi- plying $.826-+ by 3 we have the interest for 18 days, — or $2.478. Now adding the above partial results to- gether we have the total amount of interest, $93.354. Find the interest on $746 for 3 years, 5 months, 13 days at 6 per cent. RAPID CALCULATOR REVISED. 149 $149\|2 =Int for 40 months. yp Bre aS Sea | ‘ . iaeoas Pe I days. : : tFd— AE SE day. $154.546— “ “ 3 years, 5 months, 13 days at 6 per cent. EXPLANATION. 3 years, 5 months, and 138 days are equal to 41 months and 13 days. The interest on the principal ($746) for 20 months is $74.60. The interest for 40 months is twice $74.60 or $149.20. The interest on the principal for 2 months is equal to $7.46. The in- terest for 1 month is one-half of $7.46, or $3.73. The interest on the principal for 6 days is equal to $.746. The interest for 12 days is equal to twice $.746, or $1.492. The interest for 1 day is equal to 4 of $.746, or $.124. Adding our partial results together we have the total interest, $154.546. Find the interest on $2418 for 9 years, 6 months and 21 days at 6 per cent. $1209| =Int. for 100 months. On aS EY eS a alli 8A 48 36 a é é 4 6é hiznae see = 418 “dans, a 209— 66 ¢é 3 6é BhemOrtzos oh 9 years, 6 months, 21 days at 6 per cent. We have now given some of the most difficult in- terest problems possible, and, as the reader has seen, they all readily yield to the 200 month-method. After finding the interest at 6 per cent the inter- est may be found at any per cent by the methods ex- plained for finding the interest at any rate under 63, 93, and 33 day method. 150 RAPID CALCULATOR REVISED. Find the interest on.$46.13 for 8 months and 24 days at 8 per cent. - $1/844—Int. for 8 months. 1342202 Gels ave 3) 20282 26 oo 8 months, 24 days at 6 per cent. |676—= 6<é cé 8 66 24 6é 46 2 6é $2.704— 6 6% 8 66 24 6é $6 8 6é We now proceed to demonstrate the truthfulness of the announcement made at the heading of this -method, ‘‘A thousand years as a day.” Find the interest on $120 for 1000 years at 6 per cent. | $120 _ 60 $7200 Ans. Find the interest on $120 for 1 day at 6 per cent. 6) .120 .02 Ans. NotTEeE.—The above principle is based on 360 days to the year. If you solve above problems by the 1000 day method counting 365 days to the year, you will get +; less than above answers. THE TWO PLACE INTEREST METHOD. . Moving the decimal point in any principal, two places to the left, will give the interest at RAPID CALCULATOR REVISED. 4 per cent for 720 days. *74 ‘per 1 ‘cc 6“ 260 6“ *Q 14 6“ ‘é PA0 73 *81 7, “cé“ é 180 ‘eé 8) 24 ““ 6“ 144 6“ *O4 3 ne eas bt Ba 10 e341 Be sir SRB yee es AMOS 4 66 6“ 90 ‘“ *1] *Ad ing 6é 85 ¢é *115 5 sé a9 F lyie “cs 12 #54 iad 6é 65 6é 15 6 66 6c“ 60 eé 18 #61 ‘<“ 6“ 55 6“ 20 aT 6“ 6“ 51 66 24 * Approximately. cent for ‘ ‘79 45 151 48 days. Find the interest on $120 for 90 days at 6 per cent. Le Z0==(ntator- 60: days. GOs. Cha D Opes eo $1.80—= ‘< E90. 3 EXPLANATION. Moving the decimal point of the principal two places to the left, we have $1.20, the interest for 60 days at 6 per cent. Ninety days is one-half more than 60 days, hence, the interest for 90 days must be one- One-half of $1.20 is $.60, and $.60 plus $1.20 equals $1.80, the interest for half more than it is for 60 days. 90 days at 6 per cent. Find the interest on $180 at 8 per cent for 50 days. $1.80—Int. for 45 days. a, tee 5 $2:00— “c“ ‘é 50 “cc 6é 6é 152 RAPID CALCULATOR REVISED. EXPLANATION. Moving the decimal point of the ‘principal two places to the left, we have the interest for 45 days ($1.80.) Fifty days is 5 days more than 45 days, or more, hence, the interest for 50 days must be #, or more than it is for 45 days. One-ninth of $1.80 is $.20, and $.20 added to $1.80 gives us $2.00, the inter- est for 50 days at 8 per cent. cD|4 col Find the interest on $624.14 at 9 per cent for 120 days. EXPLANATION. $6 | 2414 3 The interest.o7 $18.723 Ans. $624.14 for 40 days at 9 per cent is $6.241-+. The interest for 120 days is three times $6.241-+, or $18.723-. BANKERS-MONTH METHOD OF INTEREST. This method will be found particularly valauble to Savings Banks, Loan Associations, and, in fact, to every one who has interest to compute for an even number of months. The interest at 24 per cent per annum is equal to 2 per cent a month. 18 1d 12 7: ‘““ ‘é 1 ‘“ 7; 10 66 ‘“ ‘ 5 ““ “c 9 66 ‘“ ‘“é 3 ‘“ fT: ‘6 rz ‘6 2 ‘“c ‘““ 3 ~ ~ n~ nw n~ nw I nn nw nw nw Sal el Csi bole bo] RAPID CALCULATOR REVISED. 153 To find the interest by this process multiply the principal by the product of the rate per cent per month by the number of months, and point off two places. 1. Find the interest on $842 at 4 per cent for 6 months. EXPLANATION. 842 2 Four per cent per an- num is equal to 4 per $16.84 Ans. cent a month. Multiply- ing 4 by 6, the number of months, we have$,or2. Now multiplying the principal ($842) by 2 and moving the decimal point two places to the left, we have $16.84, the interest on $842 at 4 per cent for 6 months. 2. Find the interest on $524 at 12 per cent for 7 months. EXPLANATION. $524 : i Twelve per cent per annum is equal to 1 per $36.68 Ans. cent a month. One per cent a month is equal to 7 per cent for 7 months. Multiplying the principal ($524) by 7 and moving the decimal point two places to the left, we have $36.68, the interest on $524 at 12 per cent for 7 months. 3. Find the interest on $2518 for 4 months at 18 per cent. EXPLANATION. 2518 6 Eighteen per cent per - annum is equal to 14 per $151.08 Ans. cent a month. Multiply- ing 14 by 4 (the number of months), we have 6. Now multiplying the princi- 154 RAPID CALCULATOR REVISED. See. pal by 6 and pointing off two decimal places, we have $151.08, the interest on $2518 at 18 per cent for four months. 4. Find the interest on $65 for 3 months at 8 per, cent. EXPLANATION. 65 2 . Eight per cent per an- mee es num is equal to % per $1.30 Ans. cent a month. Multiply- ing % by 3 (the number of months), we have 2. Now multiplying the princi- pal ($65) by 2 and pointing off two decimal places we have $1.30. Ans. # CANCELLATION METHOD. Draw a perpendicular line and place the principal, rate per cent and time in days on the right side of the line, and the number 360 on the left side. Cancel, and move the decimal point in the result two ‘places to the left and you have the interest at the given rate. EXAMPLE. Find the interest on $400 at 9 per cent for 73 days. 40010 3609 A) 73 $7.30 CONTRACTED INTEREST DIVISORS. 6 per cent = == 60 10 per cent = 36 9 = 40 12 e == 30 8 - == 45 ~—%4 cee ae _ Place the principal and the time in days on the RAPID CALCULATOR REVISED. Hes). right side of the line and the divisor opposite the rate desired on the left side and proceed as directed in pre- vious rule. EXAMPLES. Find the interest on $240 at 9 per cent for 81 days. ZAQ « 7) The divisor, 40, is ta- ‘81 ken from the table. $4.86 Ans. Find the interest on $213.14 at 74 per cent for 3 months and 6 days. 1213 14 Ag _Three months and 6 W. days equals 96 days. The ae divisor, 48, is taken from $4.26+ Ans. the table. LIGHTNING INTEREST AND TIME RULE. By this method we multiply the time in months and tenths by ;4 of the principal, and point off one decimal place for the interest at 10 per cent or two places for the interest at one per cent. | A note for $240 was dated in the year 1875, May 15. It was ‘paid in the year 1878, June 21. Find the interest at 10 per cent. EXPLANATION. Oe One E Fey eiel yal We reject the century ——_—_____—-- figures, and write down 37.2 the figures of the year. 20 , June is the sixth month, so we write a 6 in our $74.40 Ans. minuend. May is the fifth month, so we write 156 RAPID CALCULATOR REVISED. a 5 in our subtrahend. We now take 4 of the number of days for tenths of months; 21 days equals .7 of a month, and 15 days equals .5 of a month. Now sub- tracting .5 from .7 we have .2. Subtracting 5 from 6 we have 1; subtracting 75 from 78 we have 3. 3 years are equal to 36 months, +1 month—87 months; hence our time is 37.2 months. ;, of 240 is equal to 20. Multiplying 37.2 by 20 and moving the deci- mal point one place to the left we have $74.40, the in- terest on $240 for 3 years, 1 month and 6 days at 10 per cent. , If we wish the interest for any per cent other than 10, we would first find the interest at 1 per cent which, in the above problem, would be $7.44, and multiply this by the rate per cent. A note for $300 was dated in the year 1884, June 18. It was paid in the year 1886, August 12. Find the interest at 10 per cent. EXPLANATION. BG; SA 12 . 84 6.6 18 Performing our. sub- —_____—_- tractions as in the pre- 4) 25.8 vious problem, we find ——. the difference in time to $64.50 Ans. be 25.8 months. + of 300 equals 25, and mul- tiplying 25.8 by 25, and. moving the decimal point one place to the left, we have $64.50. Ans. A note for $600 was dated September 27, 1887. It was paid July 15, 1892. Find the interest at 10 per cent. EXPLANATION. 92S fe =15 We find the difference Shae Oe in time to be 57 months ———___—. and 6 tenths. Multiplying 57.6 this by +4 of the princi- 50 pal and moving the deci- on mal point one place to $288.00 Ans. the left we have $288. Ans. RAPID CALCULATOR REVISED. 1o% *- This method of computing interest is well worth the careful study of anyone, and it is doubtful whether a more convenient method can be found by which to find the interest at 10 per cent in partial payment ex- amples. of the quotient that was obtained by dividing 160 RAPID CALCULATOR REVISED. by 5, and add these three last quotients to the quotient that was obtained by dividing by 5, and we will have the accurate interest. | Find the accurate interest on $600 at 7 per cent for 42 days. 600 7x 42—176400. 5)17.6400—product of the principal, rate per cent and time in days with decimal point moved four places to left. 4)3.528 ==+ of 17.64. 882 =} of 3.5238,” 5) .38528—8.528 with decimal point moved one place to left .0705=1 of .3528. $4.8333—accurate interest on $600 at 7 per cent for 42 days. EXPLANATION. Multiplying the principal (600) rate per cent (7) and time in days (42) together, we have 176400; mov-. ing the decimal point of 176400 four places to the left,, we have 17.64. Taking + of 17.64 we have 3.528. Now taking + of 3.528 we have .882._-Now writing 3.528 down again with decimal point moved one place to left © we have .3528. Taking } of .3528 we have .0705; ad- ding 3.528, .882, .3528 and .0705 together we have -$4.83-++, the accurate interest on $600 for 42 days at 7 per cent. RAPID CALCULATOR REVISED. 161 TIME TABLE. | FROM TO Jan | Feb | Mar} Apr| May | Jun | July | Aug | Sep | Oct | Nov} Dec Jan OS. 12 1 2 3 4 5 6 7 8 9 10 11 Days | 365 31 59 90 120 151 181 212 | 243 273 | 304 334 Jan | Feb | Mar| Apr| May)! Jun | July | Aug | Sep | Oct | Nov} Dec Feb Mos. a6 | 12 1 2 3 4 5 6 7 8 9 10 * | Days} 334 365 28 59 &9 120 150/181) 212 242 | 273 303 Jan | Feb | Mar| Apr| May | Jun | July} Aug) Sep | Oct | Nov| Dec Mar Mos. 10 ll 12 1 2 3 “4 5 6 7 8 9 Days | 306 | 337 | 365 31 61 92 122 153 184 | 214 | 245 | 275 Jan | Feb | Mar Apr | May} Jun | July} Aug} Sep | Oct | Nov} Dec Avr Mos. c 10 11 12 1 a 3 4 5 6 7 8 Pr.) Days | 275 | 306 | 334 | 365 | 30 61 91 122 | 153 | 183 | 214 | 244 Jan | Feb | Mar| Apr| May | Jun | July} Aug | Sep | Oct | Nov| Dec Ma { Mos. 8 9 10 11 12 1 2 3 d 5 6 7 ¥ Days | 245 | 276 | 304 | 335 | 365 | 31 61 92 123 | 153 | 184 | 214 Jan | Feb | Mar| Apr! May | Jun | July| Aug | Sep | Oct | Nov| Dec June | Mos. 7 8 9 10 11 12 1 2 3 4 5 6 Days | 214 | 245 | 273 | 304 | 334 | 365 30 61 92 | 123 | 153 | 183 Jan | Feb | Mar| Apr| May | Jun | July | Aug! Sep | Oct | Nov| Dec Tai { Mos. 6 7 8 9 10 11 12 a ye 3 4 5 ¥\ Days| 184 | 215 | 243 | 274 | 304 | 335 | 365 | 31 62 92 | 123 | 158 Jan | Feb | Mar| Apr| May| Jun | July | Aug | Sep | Oct | Nov} Dec Au hae 5 6 7 8 9 10 11 12 1 2 3 4 8. Days| 153 | 184 | 212 | 243 | 273 | 304 | 334 | 365 | 31 62 92 | 122 Jan | Feb | Mar| Apr/ May | Jun | July} Aug] Sep | Oct | Nov| Dec Sept | Mos oa 5 6 7 8 9 10 11 12 1 2 3 P*-1 Days | 122 | 153 | 181 | 212 242 | 273 | 303 | 334 | 365 | 30 61 91 Jan | Feb | Mar| Apr/| May| Jun | July} Aug! Sep | Oct | Nov| Dec Oct { Mos. 3 4 By) 6 7 8 9 10 11 12 1 2 *‘ Days} 92 123 | 151 182 | 212 | 243 | 273 | 304 | 335 | 365 31 61 Jan | Feb | Mar| Apr] May | Jun | July} Aug} Sep | Oct | Nov! Dec Nov { Mos. 2 3 4 5 6 7 8 9 10 11 12 u *\ Days} 61 y2 | 120 | 151 | 181 | 212 | 242 | 273 | 304 | 334 | 365 30 Jan | Feb | Mar| Apr]! May] Jun | July | Aug| Sep | Oct | Nov| Dec Dec { Mos. 1 2 3 4 5 6 7 8 9 10 11 12 vo Days 31 62 90 121 15L | 182 | 212 | 2438 | 274 | 304 | 335 365 EXPLANATION. This table shows the time in months, and the time in days from any day in one month to the correspond- ing day in any other month. If we wish to find the teme from March 10, to June 10, we move the finger down the first left hand column until we come to March. Now move the fin- 162 RAPID CALCULATOR REVISED. ger to the right until it is under June, and we find the number of months is three and the number of days is 92. If the day to which we wish to reckon be either greater or less than the date reckoned to in the table, we should add or subtract the difference as the case may be. EXAMPLE. How many days from July 6 to December 15? Moving the finger down the first column to July, and then moving it to the right till it rests under Decem- ber, we find the number of days from July 6 to De- _cember 6, to be 153. The difference between Decem- ber 6, and December 15, is 9 days; therefore we must add 9 days to 153 days, which gives us 162 days, the difference in time between July 6 and December 15. How many days from November 24 to August 14? Moving the finger down the first column to No- vember, and then moving it to the right till it rests un- der August, we find the difference in time between November 24, and August 24, to be 273 days. August 24 is 10 days more than August 14, so we must sub- tract 10 days from 273 days which leaves us 263 days, the difference in time between November 24, and Aug- ust 14. ; How many days from January 30, to August 31? We find the number of days from January 30, to August 30, to be 212. August 31 is 1 day more than August 30, so we must add 1 day to 212 days, which gives us 213 days, the difference in time between Jan- uary 30, and August 31. No person who has to count time can afford to be without this table. It may be used to a great advan- tage in the process of averaging accounts. and is espec- ially useful in solving problems in partial payments. RAPID CALCULATOR REVISED. 163 INTEREST LAWS IN THE UNITED STATES. ~--These laws were compiled from information furnished by the Attorney-Generals in the different states and territories. > States 2 ae oe . o ise] 7 oOo © z ‘oy ke wi} 2s Penalties for Re and eer — oR . ae = : oN = i i es . 7 N\ { \ WN ip 4": cat we = NRG SI Mh FETE NO NAN ~~ Set, 6 oN = - TC a es : \ Aw! LC dimensions, 5, 4 and 3. Cancelling and multiplying our remainders together we have 48, the number of bushels of beans a box will hold that is 5x48 feet. How many bushels of apples in a bin 10 feet long, 74 feet wide, and 6 feet high? 288 bu. Axs. RAPID CALCULATOR REVISED. 181 At 25 cts. a bu. what is the corn in the shuck in a crib 1087 worth? EXPLANATION. If the price per bushel is given, find what part of a dollar it is equal to, and write the numerator of the fraction on the right hand side of the line, and the de- nominator on the left hand side. 25 is equal to 4 of a dollar, so we write the numerator (1) on the right hand side of the line, and the denominator (4) on the left hand side of the line. For corn in the shuck, we use a 10 on the left hand ~ side of the line, and a 3 on the right hand side of the line. We write these immediately beneath the 4 and 1. We next write the dimensions of the crib on the right hand side of the line. We now cancel and multiply our remainders together which gives us 42, the value of the corn in a crib 1087 at 25c per bushel. The following rule for measuring articles by stricken measure, though not quite so short as the first one, is, perhaps, a little more accurate: - To find the number of bushels of shelled corn, oats, wheat, etc., that a bin will contain, multiply the number of cubic feet by .8034. How many bushels of oats in a bin that is 205 «4? 205 4—400 1 8034 | 400 321.400 Ans. 182 RAPID CALCULATOR REVISED. HOW TO FIND THE NUMBER OF BUSHELS OF WHEAT, CORN, ETC., IN ANY ROUND BIN. RULE. Square the diameter, multiply by the height, and multiply this product by .6311. ee =—7 Ss oy =; SS =o Deets) —— SS as ANS SS SSS ts RE oes SS SSS LIS Ae, A NS J oe SR wie G Se /) a Wee OS. -k Hy HN es = RAN SS Wana’ Nies awe AS How many bushels of oats in a cylindrical vessel 10 feet in diameter and 6 feet high? EXPLANATION. 10? 6—600 Squaring the diameter 6511 (10) we have 100; mul- 600 tiplying this by 6 we ee have 600 the number of 378.6600 Ans. cylindrical feet the ves- sel contains. Now mul- tiplying .6311 by 600 we have 378.66. Ans. RAPID CALCULATOR REVISED. 183 HOW TO FIND THE NUMBER OF BARRELS OF CORN ON THE COB IN ANY BIN. Multiply the number of cubic feet in the bin by 8, and point off two places; and the result will be the num- ber of barrels of corn it contains. This allows 5 bush- els to the barrel. How many barrels of corn in a bin 1064? 10x 6x 4=240 8 -——. Moving point two places to left, 19.20 Ans. EXPLANATION. Multiplying the dimensions together we find the number of cubic feet in the bin to be 240. Multiply- ing 240 by 8 and pointing off 2 places we have 19.2, the number of barrels of corn the above named bin con- contains. If you wish to reduce barrels to bushels, multiply by 5. This process will produce the same result as is given by most of our revised arithmetics. Those, how- ever, who prefer a rule that will more accurately con- form to arithmetics that use the exact: number of cubic inches in a bushel for determining results to problems of this kind may multiply cubic feet by .0804, thus: How many barrels of corn in a bin that is 2011 | <6? : 20 11 6==-1320 .0804 105.600 440 106.040 Ans. 1 84 RAPID CALCULATOR REVISED. At $.90 per bushel what is the wheat worth in a bin that is 10x6x4? 19\9 o|4 10 6 4 864—5—$172.80 Ans. EXPLANATION. $.90 is equal to ;°, of a dollar, so we write a 9 on the right hand side of the line and 10 on the left hand side. We BT, i} opapT ieiibae DIFADIM a PEED vA Ly HHA 35) HY, 04 1] 7 wil NO A Wheat is measured by stricken measure, so we place a 4 on the right and a 5 on the left of the line. We now write the dimensions of the bin in feet on the right hand side of the line. Cancelling the two tens and RAPID CALCULATOR REVISED. 185 multiplying the factors left uncancelled on the right of the line together we have 864. Dividing this by 5, the remaining factor on the left of the line, we have 172.8, hence our answer is $172.80. HOW TO CALCULATE THE VALUE OF GRAIN, FRUITS, ETC , WHEN THEIR WEIGHTS ARE GIVEN. Place the total number of pounds on the right hand side of the line and the number of pounds in a bushel on the left. Write the price per bushel as a fractional part of a dollar, and place its numerator on the right hand side of the line, and the denominator on the left. What is the value of 2400 pounds of wheat at 80c per bushel? 40° 60 2400 5 4 LG. Yee USP EXPLANATION. We first write the total weight of the wheat (2400) on the right hand side of the line. By refer- ring to the table on following page, we find the number of pounds in a bushel of wheat to be 60, so we write this on the left hand side of the line. 80c is equal to 4 of a dollar, so we place a 4 on the right hand side of the line, and a 5 on the left hand side of the line. Cancelling and multiplying our remainders together we have $32. Ans. Find the value of 640 pounds of oats at 20c per bushel. 32 640 20) 5 4 $4. Ans. RAPID CALCULATOR REVISED. 186 the ear in 70 pounds of corn alue of 21 bushel. Find the v at 30c per 10—$9.30 Ans. 93— RAPID CALCULATOR REVISED. 187 WHIGHT OF PRODUCE PER BUSHEL, AS ES- TABLISHED BY LAW OF CUSTOM. 2 Articles. 2 E ce. ef sa = 2 = Vy O La oe yy OSU ei Rack GEASS Tale ae ee en 60 60 60 | 60 Mir imeertnGer te oir ec eg Ip celts! atae ee 36 56 06 56 RB TEIBCRY oo. tie. cece ee cas 70 | 68to70 | 70 70 ersten Fett bien, ii oo ere hk |, os 30 10 Pl AS De RE So ocak hee eos 56 BG ef eee 56 ELPA te Sa ger jacana sR s, mes on 32 32 32 EAE Se 4 pS OR A a ea 48 48 Ye Aided pe: P ifishwieetatoes 40g oe. hee 60 | 60 60 | 60 Dumerls Potatoes eye. oe week 56 50 eee OC) EST Re aie alle eS 60 60 oe ote kaso penns «eect cits nk. we 46 46 2 Fas 46 a) Sr RC es RE PC pias eae 20 20-* fe 20: lovers NeClset . MP ee Ts ok 60 60 med BO Mamotiypecd*.. eo, . cee teow Cs 45 45. ah 45 Diindariane eed. ow suie ces 5 48 | 48to50|....| 48 LAOS i Toe ACR a a O80 Se 4.4, 44 wee | 44 Plae DEEW AN en Swe oo ris 536 56 Pe OG TTL Po ope Pot bees a ae 30 50 MWS ecw. Red-top Seed or Herd’s Grass.... | 14 14 gf cnce sy aaiavaees Osage Orange Seed............ 36 OO eh Oe eee PUOM MUM GE & act'a a wie leiei aie «ck cea 42 45 eee Kentucky Blue Grass Seed ...... 14 14 Se ions PONArGM Grass. fo oo aes Sew 14 14 Bed | PEAS RNC Weal Gee mre ct eGo 0s a oe 50 4 ete iota er en eek Gn Bf oUtooO! Of iimas PORTO NIONVOeESE cs ies othe cee 3 | 28 28 ee Pe sae 188 RAPID CALCULATOR REVISED. WEIGHT OF PRODUCE PER BUSHEL — Continued. 2 pa wv = fe rs) Articles. 3 a S te Ben in ee op O re O Pens cs Se ss See 60 60 mee Split Peas 2: 33a eee GO 2 . viene Dried Apples.530 5.3 ee ee 24 25 ee Dried 'Penches.< 3 hea ee 33 oo . es Malti 60335. So ee eee 38 34 ee Dall kk sawes eee BU ea) Re, Coals. a eee SOs eee 80 Peanuts, dry Southern.......... OL Vs sae Coelion Seed... a ee Sen At ara Paranips.=3.S oceans a See eS eee gmmon Lurnips:< 240. tn Sees: Tae De arias 2 ee eee 5b Mee eee Rutabagas Sit ate ded ait win inne Yoke teneta aie hs ee Green Peas, unshelled.......... 0 |) eae Green Beans, unshelled ......... OO" hee Green: Anples 22 ce: 7 os eg MOI ca Green: Peaches 05 4S see yo pee eee oe, Green: Pears 4.0 Sn ee AP yes Coke x25 pao a0 eo eee 40 189 RAPID CALCULATOR REVISED. | | i e. | | Ki ae tae j | 3 ' Pa | a mm | & ; i] 2 melee a) | Ela;el? é/e/./8/3 Commodities. E 3 ald 3 E r 8 g a P E E Cats z be i 8 f E SIEIEISIELEIS( S/S 1518/8 /o]Flelels|2lalslelery elo} rs Def yey) OF fie srt ort emt de | PD ie Oe ge Poe lm: | peealinn CW Coiled ee I emer ieee | Oo Aa. | eae) peak Beane Barley ...... ..../50|. ./48/48/48/48/32|. . |46)48/48/48] . .|48/48/48/ 48/46/47 46|48|45 Beane vice Arete eoleoleoleo|-|".|_.|-.|..leol. 11. BBOIE el hic bails onteestent Bituminous Sere Meleolzolsolsolee | Saat ec Wael Soak MRI |e ec leet Blue Grasa; peed =. .|. 1. . | L444 Pare eet. WO Bee Dec aed bate Bes BRP re Rain De) Buckwheat .... ..|40/45/40160/52/52|..|. .|46/42/42/52). ./50/50/48). ./42/48]. .|46/42/42 WESTON SB CANS 0: oe chien |e |SO/DOILO|s ail sre [eaters ace) ail Oeil ate ls Ube carte Clover Seed .....4..1..160145/60160/..|..|..160/60/60|. .|. .|64/60/60/60) . _. {es}. 2160160 IOTIOd “A DDIEN: jh. ake cli ete | aki tere lee | ck lees Oaks ee etme iene | seen ake to rt extagen Dried ee a poms PHP baer areips bedcamarrl cab lak aorta Plax oSeed ice. che als los (DOL O4/B6/ BGI hel. or sa bes (DG) 5105/00 1Bb O6I- ee weg aiips. Hemp Seed ......./52/. .144/56/44/441. at me Aa ; PCAN AOL Tete aie 5, 56|52/48/56|56|56).. 56|56|56|52! . 456 |58| 56/5656 |56|56)56 Indian Corn in ear j..|. .|70/50/68} . aaa Sch Lavin Aoi aie an ee lias ; Indian Corn Meal .|32]. .!48/62|. 50 m4 50 50]. Lt lineelives +f oe 50 _ ,|28/32|58/35/33 32|30}30)382|382|385 30]. 32/32|32/34/32| . ORCS on hua AY, ANa AOPIBUVOL DD ach ace | Oe nitaene (0 (uae wei et pena ie lit 150}. COTTON er ee Maur. 50 Potatoes .... ...../54/60/60/46/60/60 Pat ee eo . .{60/60]. 60|. 60/60 60|60 Tevet e ance ps oho pee 156|56/56|56 _ |BGIBGIBGIBG|56I. . puke SASS Gee OS a ‘aba Pat Gee aff RL ean : al Dar ae ee eae galeoleol 2 Al anes . (561. i |: os Timothy Seed ee is las 45|45|45]. Ni ae he Abia, ti 144]. a ‘a WV NIGH Gorin. utes ee 1 lee 60 . |60 60|60|60}. . 60 60/60 60 tel WARS Wheat Bran Mate alelo'(20) eol20/20/8% | true Sea PAU Ree Rane io Foie g ‘hen 190 RAPID CALCULATOR REVISED. HAY MEASURE. Measuring is a very unsatisfactory method for es- timating the weight of hay, but there are times when it is impracticable to use the scales and a close approxi- mation to the true weight will answer the purpose. It is estimated that timothy hay in stacks of ten feet in height measures about 500 cubic feet to the ton; clover between 600 and 700 cubic feet; new mown hay, about 675 cubic feet; fine hay well settled, 450 to 500 cubic feet. Wy aa a : 28 TS ow om oe an rere ee ioe TO FIND THE NUMBER OF CUBIC FEET IN A CIRCULAR STACK multiply the square of the circumference by .04 of the height. RAPID CALCULATOR REVISED. 191. How many cubic feet in a circular stack whose circumference is 20 feet, and height 15 feet? 20°—400 . 15x .04-= .6 240. cubic feet. EXPLANATION. Squaring the circumference of the stack (20) we have 400. Multiplying the height of the stack (15) by .04 we have .6. Multiplying 400 by .6 we have au Ans. TO FIND THE NUMBER OF TONS OF HAY IN A MOW. Divide the product of the length, height and width, in yards, by 15 if well packed; if shallow and the hay recently stacked divide by 18. How many tons of hay in a mow 10 yards long, 5 yards wide and 4 yards high? 10«5x4=200 20—200 decimal point moved one place to the left. 63—1 of 20 134 Ans. EXPLANATION. ) Multiplying the dimensions of the mow, 10, 5 and 4 together, we have 200. Dividing this by 15 we have 134. Ans. To divide any number by 15 move the decimal point one place to the left and subtract 4 of the quo- tient thus obtained from itself. 192 RAPID CALCULATOR REVISED. TO FIND THE NUMBER OF TONS OF HAY IN SQUARE OR LONG STACKS. Divide the product of the length, width and half the heighth, in yards, by 15. How many tons of hay in a rick 12 yards long, 5 yards wide and 8 yards high? 125 (4 of 8)==-240; 240—-15—=16, B (12° EXPLANATION. Taking 4 of the height (8) we have 4. Maultiply- SSk& i i . AAP isin 107 NY Yell ANN SOUT WIRY Ae ae RW Gs Wah \ \\ \ ae tt . A | i \\ A NEAR , is BROS DRUMS CAA TIL OPP Ve oe La ing 12, 5 and 4 together we have 240, and dividing 240 by 15 we have 16. Ans. Shorten by cancellation. “s RAPID CALCULATOR REVISED. 193 TO FIND THE NUMBER OF TONS OF HAY IN A LOAD. Multiply the length, width and height, in yards, together and divide the product by 20. How many tons of hay in a load 5 yards long, 3 yards wide and 2 yards high? EXPLANATION. 5x<3x2—30 eae Multiplying the dimen- ee sions 5, 3 and 2 together 1.50 Ans. we have 30, and divid- ing 30 by 20 we have 14. Ans. To divide any number by 20, multiply by 5 and point off two places. To find the number of tons of hay in a circular stack, multiply the square of the circumference in yards by 4 times the height in yards; point off three places and subtract 4 of the quotient from itself. How many tons of hay in a stack whose circumfer- ence is 15 yards and height 6 yards? 152225, 225<24(4x6)—5400 5400 with decimal point moved three places to left equals 5.4. EXPLANATION. Squaring the circumference (15), we have 225. Multiplying 225 by 24 (4 times the height), we have 5400. Pointing off three decimal places in 5400 we have 5.4. One-third of 5.4=—1.8. 5.4 —1.8=—3.6 Ans: Remember that no claim is made to the absolute accuracy of these rules, so do not criticise them too harshly because they may not give exactly the same answer as that produced by the rule current in your locality. They are based on the best information we can obtain as to the general customs prevalent through- out the United States. 194 RAPID CALCULATOR REVISED. AL pie 2m of A eas] SSSSS3 \ { = Se x NN SSS) | Ss SSOw ——— = SS Be SHA ta (eee ee iy < a SE EH thes ea rire U ° Ly SSE SS 4 Rea =< Past Byes ee —_S Ls a0 mess )\ At { | Ae % 7 Ms ai 4, Vee io Se Pa: ag eee SS ——S- aa ——= apesecg es m vat LOG AND LUMBER MEASURE. This is a subject so vast in its nature that we might with perfect propriety have prepared a special work on it, and therefore while we do not claim that this treatise is exhaustive, we do claim to present a great variety of the shortest and best rules on this sub- ject known to the world. Lumber is measured by the board foot, and a board foot is a piece of lumber 12 inches long, 12 inches wide and 1 in. thick. . In calculating the number of board feet in any board, anything under an inch thick is computed the same as an inch. ; | To tell the number of feet of lumber in planks, scantling, etc., draw a perpendicular line, and place the number 12 on the left hand side of this line; on the right hand side, place the length of the piece of lum- ber, in feet, and its width and thickness, in inches; then cancel and multiply the remainders together for the number of board feet. | 4 195 RAPID CALCULATOR REVISED. How much lumber in a board 12 feet long and 8 hes wide. Inc 6 12 8 8 feet. Ans. How much lumber in 20 scantling 8 feet long, inches wide and 2 inches thick? 1? Ans. At $2.25 per C what is the lumber in 24 boards 18 feet long and 10 inches wide, worth? 160 feet. ~ it Hc i a i ~ 196 RAPID CALCULATOR REVISED. EXPLANATION. $2.25 equals $24, or % 9 $2. We place the numer- 4 ator(9) on the _ right 1010/94 hand side of the line and 12|18° the denominator (4) on 10 the left hand side of the sR Ene line. If the price is $8.10 Ans. given at so much a hun- dred, place 100 on the left hand side of the line. If at so much a_ 1000, place 1000 on the left hand side of the line. In this problem, we write 100 on the left hand side of the line. Now place the 12 on the left hand side of the line and write the dimensions of the lumber on the right hand side of the line, as previously stated; can- cel and multiply your remainders together, the re- sult will be the answer in dollars and cents. TABLE. The following table for telling the number of feet of lumber in planks, joists, scantlings, etc., from 8 to 30 feet long will be found the shortest cut possible to produce correct answers: To tell the Number of Feet in a Piece of Lumber 1 in. thick and 8 ft. long, multiply the width in inches by 2 and divide by 3; 1 in. thick and 10 ft. long, multiply the width in inches by 5 and divide by 6; 1 in. thick and 12 ft. long, multiply the width in inches by 45 1 in. thick and 14 ft. long, add 4 of the width in in- ches to itself; 1 in. thick and 16 ft. long, add 4 of the width in in- ches to itself; 1 in. thick and 18 ft. long, add 4 of the width in in- ches to itself; RAPID CALCULATOR REVISED. 197 1 in. thick and 20 ft. long, annex a cipher to the width in inches and divide by 6; 1 in. thick and 22 ft. long, multiply the width in inches by 11 and divide by 6; 1 in. thick and 24 ft. ie multiply the width in inches byi2* isn. thick and 26 ft. long, multiply the width in inches by 24; 1 in. thick and 28 ft. long, multiply the witdh in inches y 23; 1 in. thick and 30 ft. long, annex a cipher to the width in inches and divide by 4. How many feet of lumber in a plank 1 in. thick, 8 in. wide and 24 ft. long? 8x2—16 Ans. | By referring to the above table we notice that for a plank 24 ft. long we multiply the width in inches by yA How many feet of lumber in a board 1 in. thick, 6 in. wide and 14 ft. long? 6)6 1 7 Ans. By referring to the above table we notice that for a board 14 ft. long we annex } of the width in inches to 6.. 4 of 6=1; 1+6=—7. Ans. How many feet of lumber in a board 1 in. thick, 10 in. wide and 30 ft. long? 4)100 25 Ans. By referring to the above table we notice that for a board 30 ft. long we annex a cipher to the width in inches and divide by 4. 10 with the cipher annexed= 100, and 100 divided by 4—25. Ans. 198 RAPID CALCULATOR REVISED. HOW TO TELL THE NUMBER OF FERT OF LUM- BER THAT CAN BE SAWED FROM A LOG. Measure the diameter in inches at the small end. Take half of the diameter, in inches, and multiply it by the diameter. Multiply this product by the length of the log in feet, and divide by 12; the result will be the number of feet of smooth lumber, board measure. Example: How many feet of lumber can _ be sawed from a log 24 inches in diameter and 30 ft. long? 24 Diameter. 12\12 Half, Diameter. 30 Length of Log. 2430720 ft. Ans. The above rule does not allow anything for the guage of the saw. If you wish to know how many feet of square- edged boards can be sawed from a log, allowing for guage of saw, divide the above result by the thickness of the board plus the guage of the saw. For example: _JIn the above problem, if the boards are 1 inch thick and + inch is allowed for guage of saw, 576 feet © (720-14) of lumber could be sawed from the log. If the boards are 2 inches thick, we would divide 720 by 21 (4 for guage of saw) which would give 320 feet of lumber if measured on the surface, but as the lumber is 2 inches thick, we must multiply this result by 2, which gives us 640, the number of feet of two- inch boards, board measure. It will be observed that more board feet of two- inch boards can be sawed from a log than one-inch boards; for in sawing two-inch boards, less lumber is wasted by the guage of the saw. RAPID CALCULATOR REVISED. 199 DOYLE’S LOG RULE. This rule has been in use for several years and has met with the universal favor of lumbermen in all parts of the United States. It is a very close approx- imation to a scientific rule. It may favor the buyer in small logs, and the seller in large ones, but logs are often crooked and no rule averages a more nearly correct result than this one. RULE. From the diameter of the small end of the log, in inches, subtract 4. The square of the remainder will be the number of board feet yielded by a log 16 feet in length. After finding the number of board feet in a log 16 feet long, For a log 8 ft. long, take 4. For a log 10 ft. long, take &. For a log 12 ft. long, deduct 4. For a log 14 ft. long, deduct }. For a log 18 ft. long, add }. For a log 20 ft. long, add 4. For a log 22 ft. long, multiply by 11 and divide by 8. For a log 26 ft. long, multiply by 18 and divide by 8. For a log 28 ft. long, multiply by 7 and divide by 4 For a log 30 ft. long, mlutiply by 15 and divide by 8. How many feet of lumber can be sawed from a log 20 ft. long and 28 inches in diameter? 28 4 24, 24°—576. 4)576—number of feet in a log 16 ft. long. 144—1 of 576. 720—number of feet in a log 20 ft. long. 200 RAPID CALCULATOR REVISED. EXPLANATION. Subtracting 4 from the diameter of the log (28) we have 24. Squaring 24 we have 576, the number of feet of lumber in a log 16 feet long. By referring to the table, we find that for a log 20 feet we add + to the - number of feet in a log 16 feet long. i of 576—144. 144576—720. Ans. The following rule, though a modification of Doyle’s rule, produces the same result, and, for many problems, is much shorter. RULE. From the diameter of the log at the small end, in inches, subtract 4. Square 4+ of the remainder and multiply by the length in feet. How many feet of inch boards can be sawed from a log 20 feet long and 28 inches in diameter? 28 4 . EXPLANATION. —— Subtracting 4 from 4)24 the diameter (28) we have 24. 4 of 24 is 67—36 equal to 6; squaring 6 we 20 have 36, and multiply- — ing 36 by 20 we have 720 Ans. 720. Ans. How many feet of inch boards can be sawed from a log 30 feet long and 36 inches in diameter? 36 4 EXPLANATION. 4) 32 Subtracting 4 from 36 a we have 32. 4 of 32 is 8?—64 8; squaring 8 we have 30 64. 64x30—1920. Ans. RAPID CALCULATOR REVISED. 201 ——=- SSS — 2 Ti HOW TO FIND THE NUMBER OF CUBIC FEET IN THE LARGEST SQUARE PIECE OF TIM- BER THAT CAN BE SAWED FROM A ROUND LOG. RULE. Draw a perpendicular line, place the diameter in inches, one-half the diameter in inches, and the length of the log in feet on the right hand side of the line. Place 144, or two 12’s on the left hand side of the line, then cancel; multiply the remainders together, and the result will be the number of cubic feet. How many cubic feet in the largest square piece of timber can be sawed from a round log 20 feet long, and 24 inches in diameter? 94° =-diameter of log in inches. 12:12 == half of diameter of log in inches. 12:20 =Tlength of log in feet 40 Ans. 202 RAPID CALCULATOR REVISED. HOW TO TELL THE NUMBER OF CUBIC FEET IN ROUND TIMBER. RULE. Multiply the square of the circumference at the middle of the log, in feet, by 8 times the length, and point off two places in the answer. How many cubic feet in a log 6 feet in circumfer- ence and 20 feet long? 6°—86 , 20 EXPLANATION. 720 Squaring the circum- ference of the log (6) we have 36; multiplying 36 57.60 Ans. by 8 times the length (160), or by 20 and 8, and pointing off two places we have 57.6. Ans. RAPID CALCULATOR REVISED. 203 ZN 12. was Wednesday : Wh \ ! Sa tore eraA 5 i ——— —— =— Se ‘ ; [ Brees : | eAseias “hy , \ tht CeetHH 1S ene 2 25 B22 ee ‘ i Wi (A MET Ee a get TN ; _& tt Salute ee ae i | ae ae 4 es 4 ON pes SN ete a ee eee aie - eel HOW TO TELL THE DAY OF THE WEEK. By the following method you can tell immediately what day of the week any date transpired, or will transpire from the commencement of the Christian Era for a term of three thousand years. MONTHLY EXCESS FIGURES. Ratio of June is Ratio of September is Ratio of December is Ratio of April is Ratio of July is Ratio of January is Ratio of October is Sn Be Nia Been NOTE—January and February are one less in Leap years. The monthly excess figures of the other months are not affect- ed by Leap year. 204 RAPID CALCULATOR REVISED. Ratio of May is Ratio of August is Ratio of February is Ratio of March is Ratio of November is Se CENTENNIAL TABLE. The ratio to add for each century will be found in the following table: The ratio of 200, 900, 1800, 2200, 2600, 3000 = 0. The ratio of 100, 800, 1500 2 ea ee ae tes sl. The ratio of 700, 1400, 1700, 2100, 2500, 2900 i is Bs The ratio of 600, 1300 wis ka kn ae tes, The ratio of 500, 1200, 1600, 2000, 2400, 2800 i Fe 4, The ratio of 400, 1100, 1900, 2300, 2700. is 5. The ratio of. 300,:1000...... <2 eee is 6. RULE. Reject the century figures, annex a cipher to the figures of the year, and divide by 8 (rejecting the frac- tions should any occur), and to the quotient thus ob- tained, add the day of the month, the ratio for the month, and the ratio for the century, and divide.the sum by 7. The remainder will indicate the day of the week. A remainder of 1 indicates Sunday. A remainder of 2 indicates Monday. A remainder of 3 indicates Tuesday. A remainder of 4 indicates Wednesday. A remainder of 5 indicates Thursday. A remainder of 6 indicates Friday. No remainder indicates Saturday. In the year 1784 what day of the week was July 15? RAPID CALCULATOR REVISED. 205 8) 840 105 EXPLANATION. 15—day of month. Striking off the two 2—ratio 10G July. left hand figures, we 2—ratio for 17, cen- have a remainder of 84. [tury. Annexing a cipher to 84 7) 124 we have 840. Dividing ees 840 by 8 we have 105 to 17 with 5 Rem. which we add the day of Thursday. Ans. the month (15), the ex- -cess figure for July (2), and the excess figure for 1700 (2) which gives us 124; dividing 124 by 7 we have 17 with a remainder of 5. The remainder 5 indicates the fifth day of the week, or Thursday. In the year 1841 what day of the week was Sep- tember 24? 8) 410 es EXPLANATION. 51 24 Striking off the two 4 left hand figures and an- ae nexing a cipher to 41, 7)76 we have 410. Dividing ee 410 by 8 and rejecting 10 with 6 Rem. fractions, we have 51, Friday. Ans. to which we add 24 the day of the month and 1, the excess figure for September, which gives us 76. Dividing 76 by 7 we have a remainder of 6 which in- dicates the sixth day of the week, or Friday. In the year 1877 what day of the week was Aug- ust 11? NOTE—In finding the day of the week for the present cen- tury, no attention is paid to the centennial ratio as it is 0. 206 RAPID CALCULATOR REVISED. 77=two right hand figures of the year. 19=1 of 77 omitting fractions. ll—day of month. 5—excess figure for the month. 7)112 16 with 0 Rem. Saturday. Ans. EXPLANATION. Instead of annexing a cipher to the two right hand figures of the year and dividing by 8 we may divide the figures of the year by 4 (rejecting fractions should any occur), and add the quotient to the figures of the year. In the above problem, we strike out 18 . which leaves us 77. +4 of 77 equals 19. The day of the month is 11, and the excess figure for August is 5. - Adding 77, 19, 11, and 5 together we have 112; 112 divided by 7 leaves a remainder 0, which indicates the last day. of the week or Saturday. In the year 1835 what day of the week was Nov- ember 9? . 8) 350 7) 43 6 with 1 Rem. 9—day of month. : 6—excess figure for Nov. TAG 2 with 2 Rem. Monday. Ans. RAPID CALCULATOR REVISED. 207. EXPLANATION. This process, though virtually the same as those previously explained, is a little more convenient for performing mental operations. Annexing a cipher to the figures of the year (35) we have 350. Dividing 350 by 8 we have 438. Dividing 43 by 7 we have 6 with a remainder of 1; to the remainder (1) we add the day of the month (9) and the excess figure of the month (6) which gives us 16. Dividing 16 by 7 we have a re- mainder of 2 which indicates the second day of the week, or Monday. : By this method, the pupil may divide by 7 as of- ten as he pleases, as all that is necessary is, for him to retain the remainder paying no attention to the ‘ques tients. What day of the week was January 10, 1840? 8) 400 | 50 9—day of month. 2—excess figure for Jan. in Leap year. 7) 61 8 with 5 Rem. Thursday. Ans. NOTE—Years divisible by 8 after a cipher is annexed are Leap years. Also years divisible by 4 are Leap years. ‘This does not include centennial years. 208 RAPID CALCULATOR REVISED. HOW TO TELL The Day of the Week, The Day of the Month, The Month of the Year That A Person Was Born on, Also His Age in Years. Ask a person to write down the number of the day of the week, the number of the month, the day of the month on which he was born, as one number; then multiply the number by 2, add 5 to the product, mul- tiply the sum by 50, add the age in years to the product and subtract 365 from the sum. Now ask the person to tell you the remainder, and you add the number 115 to the remainder. ©The first figure of the number that you now have will indicate the day of the week; the second one or two figures, the month; the third one or two figures the day of the month; the fourth one or two figures, the age in years. ; A person was born on Monday, May 16, 1842. © 2516 2 Multiply. EXPLANATION. 5032 ; 5 Add. Monday is the second eee day of the week so we 5037 write down a 2. May is 50 Multiply. _ the fifth month, so to the right of the 2 we annex eee Add age ' a 5 which gives us 25. : The day of the month is. 251900 16, and annexing 16 to 365 Subtract. the right of 25 we have Fe HORSES 2516. Multiplying 2516 251535 by 2 we have 5032. Ad- 115 Add. ding 5 to 5032 we have 2,5,16,50 5037. Multiplying 5037 ee ele by 50 we have 251850. © o ® e Be tee If a person was born in BP the year 1842, in the ee year 1892 he would be oe 50 years old. Then ad- = ding 50 to 251850 we have 251900. Subtracting 365 from 251900 we have RAPID CALCULATOR REVISED. 209 251535. Adding 115 to 251535 we have 251650, the first figure of which indicates the day of the week; the second figure the month; the third and fourth figures the day of the month, and the fourth and fifth figures the age in years. If the day of the week on which a person is born is not known, it may be found by the rule for telling the day of the week, or if desired, THE MONTH, THE DAY OF THE MONTH, AND THE AGE MAY BE TOLD BY THE FOL- LOWING METHOD: PROCESS. Write down the number of the month and the day of the month as one number, multiply by 2, add 5, mul- tiply by 50, add age in years, subtract 365, and add 115 to the remainder; the first one or two figures thus found will be the number of the month; the next one or two, the day of the month, and next one or two the age in years. A person was born June 4, 1850. 64 2 Multiply. EXPLANATION. 128 June 1s’ ‘the sixth 5 Add. month, so writing down 133 the number of the month and the day of the month 50 Multiply. as one number we have 6650 64; multiplying by 2 we 42 Add age. have 128, adding 5 we 6692 have 133; multiplying by 865 Subtract. 50 we have _ 6650, to or which we add 42, the age 6327 in years, and we have 115 Add. 6692. Subtracting 365 6,4,42 from 6692 we have 6327. SE & Adding 115 to 6327 we Be ee have 6442. The first sais figure (6) indicates the 5 months; the second fig- 5 ure (4) indicates the oe day of the month; the third and fourth figures indicate the age in years. 210 RAPID CALCULATOR REVISED. HOW TO FIND THE COST OF SINGLE ARTICLES WHEN PURCHASED BY THE GROSS.* Multiply the cost per gross by 7, and move the decimal point three places to the left. Find the cost of a single knife if purchased at $60 per gross. 7 $.420 Ans. By moving decimal point three places to the left. Find the cost of a single pen if purchased at 80c @ gross. $.80 7 $.00560 Ans. By moving decimal point three places to the left. MECHANICS’ IRON RULE. To find the weight of wrought iron by measure- ment, multiply the thickness and width in inches by the length in feet, annex a cipher to the product and divide by 3. Find the weight of a bar of wrought iron 20 feet long, 2 inches wide and 4 inches thick. 20«2« 4—160. 160 with cipher annexed equals 1600. 3) 1600 5334 Ib. Ans. EXPLANATION. Multiplying the dimensions, 20, 2, and 4, together we have 160. Annexing a cipher to the 160 we have 1600; dividing 1600 by 38 we have 5334. Ans. *NOTE—The answer given by this process is a smail frac- tion too much, but is sufficiently accurate for all practical pur- poses. RAPID CALCULATOR REVISED. i one For cast iron deduct ;; from the weight of wrought iron. For steel add x4 to the weight of wrought iron. For copper add + to the weight of wrought iron. For brass add 75 to the weight of wrought iron. For lead add 4 to the weight of wrought iron. Find the weight of a bar of lead 12 feet long, 4 in- ches wide, and 3 inches thick. 12x4x<3=144 144 with cipher annexed equals 1440. 3) 1440 2)480—weight of wrought iron. 240—4 of weight of wrought iron. 720—weight of lead. EXPLANATION. Multiplying the dimensions 12, 4, and 3, together, we have 144. Annexing a cipher to 144 we have 1440; dividing 1440 by 3 we have 480, the weight of the bar if it were wrought iron. By referring to the table we notice that we add 4 of the weight of wrought iron to itself to find the weight of lead. 4 of 480 is 240 and 240+480—720. Answer. WOOD MEASUREMENT. A cord of wood is 8 feet long, 4 feet high, and usually 4 feet wide. By custom, however, anything from 34 to 4 feet is usually considered a sufficient width for a cord of wood. To find the number of cords of wood in a pile, draw a perpendicular line and place the length and the height on the right hand side of this line. Place Sse PC Zee eT fd —— = a LL 47d > Unddsgates Unt, Sei (2% the numbers 8 and 4 on the left hand side of the line. Cancel and multiply the remainders together, and the result will be the exact number of cords. How many cords of wood in a pile 32 feet long and 12 feet high? 30% 4 12 12 Ans. At $3.50 a cord find the value of wood in a pile 40 feet long and 16 feet high? EXPLANATION. 27 ‘ $3.50 is equal to 3. hag ; We place the numerator A1¢* on the right hand side pau ae aes of the line, and the de- $70 Ans. nominator on the left. We also write an 8 and 4 on the left side of the line, RAPID CALCULATOR REVISED. . 2138 and place the length and height of the pile of wood on the right hand side of the line. Cancelling and multiplying our remainders together we have 70. Ans. wy nity ia Yi iiswe pe “|b Some 4 fee wang all > ATT Meat nat apy {Tn ining pe A : a Sag sles = = Lt 3 f (lees ce, Ve MMA age elo Wy WMA ee 214 ; RAPID CALCULATOR REVISED. HOW TO TELL THE NUMBER OF GALLONS OF WATER IN ANY TANK, BOILER, OR CYLIN- DRICAL VESSEL OF ANY KIND. Square the diameter in feet, and multiply by the length in feet to find the number of cylindrical* feet the vessel contains. Multiply the cylindrical feet by 6 and deduct 4 of the cylindrical feet from the product obtained by multiplying by 6. The result will be the number of gallons, very nearly. How many gallons of water in a boiler 4 feet in diameter and 20 feet long? 4? —16 16« 20320 EXPLANATION. 320 Squaring the diameter 6 (4) we have 16. Multi- eas plying 16 by 20 we have 1920 320, the number of cylin- 40—=4 of 320 drical feet in the boiler. —— Multiplying 320 by 6 we 1880 Gal. Ans. have 1920. 4 of 320 is 40. 1920—40—1880. HOW TO TELL THE NUMBER OF GALLONS IN ANY RECTANGULAR VESSEL. This rule is here given for the first time, and, so far as the author’s knowledge extends, is decidedly the shortest and simplest rule in existence. NOTE—A cylindrical foot is the solid contents of a square, which would exactly hold a cylinder 1 foot long, and 1 foot in diameter The actual contents of the cylinder is found by mul- tiplying cylindrical feet by .7854. RAPID CALCULATOR REVISED. 215 RULE. To find the number of cubic feet in the vessel, an- nex a cipher, after the cipher is annexed deduct + of the number from itself. The result will be the number of gallons very nearly. How many gallons in a vessel 20«8x4? 20*8x4—640 6400—640 with cipher annexed. 1600} of 6400 4800 Ans. EXPLANATION. Multiplying the dimensions 20, 8, and 4 together we have 640. Annexing a cipher to 640 we have 6400. 4 of 6400 is 1600; 6400 minus 1600—4800. Ans. The result produced by this rule is sufficiently ac- curate for all practical purposes, but those who desire to obtian a still more accurate answer may deduct .02 of the cubic feet from the answer given by the above rule. The solution would then stand as follows: 6400—640 with cipher annexed. 16001 of 6400. 4800—Number gallons produced by first rule. 12.8—640 multiplied by .02. 4787.2 Ans. EXPLANATION. We find the 4800 by the preceding rule. Now multiplying the number of cubic feet (640) by .02 we have 12.8; subtracting 12.8 from 4800 we have 4787.2. Ans. 216 RAPID CALCULATOR REVISED. The above problem worked out by the ordinary process of arithmetic would appear as follows: 20x 12—240 8x12— 96 AN 12S =) As 240 96 1440 2160 23040 48 184320 92160 231) 1105920 (4787+ Ans. 924 1819 1617 2022 1848 1740 1617 123 HOW TO TELL THE NUMBER OF GAL- LONS IN A BARREL. RULE. Multiply the square of the mean diameter, in in- ches, by the length in inches, and this product by .0034. Divide the sum of the head and bung diameters by 2 for the mean diameter. coal RAPID CALCULATOR REVISED. ! & ] 8 f] R ini iy i SW ( nn fe | \ a — Rast Uy ru {| oIiserere: ot b HALA) ff : , ma ery | How many gallons in a barrel, the bung diameter of which is 22 inches, the head diameter 18 inches, and. the length 26 inches. ~ 22—bung diameter. 18—head diameter. 2)40 sum of bung and head diameters. 20 mean diameter. 20°—400 26 10400 .0034 35.3600 Ans. 217 218 RAPID CALCULATOR REVISED. EXPLANATION. Adding the bung diameter (22) and the head di- ameter (18) together we have 40. Taking 4 of 40 we have 20; 20 squared equals 400. 400 multiplied by 26 equals 10400; 10400 multiplied by .0034—35.36, the number of gallons in the barrel. 49h HM UY 44 | 5 | 5 jor Aue 1, fT fT ASS no nS oe no ag NAME SIZE. |Se_(e". |S". |e, [S= ; Koslbos | hon | hos |ho'R ES FS AS FS “esl oes | a4 | oad |oa8 | | A ba oo Ad ree Baltimore.... |8444232|6.5 |13 |19.5 |26 [82.5 New York.... (8 X34x24/734 [14% [214 |284 [853 Maines aos. 2. 74X382X23/7.2 |14.4 [21.6 |28.8 [36 Michigan.... |8443%23|64 [122 |19 [|254 [31% Hire ye oo: [94x48 22/62 [1384 [20 |263 |334 Common..... I8 x4Ex2 |8 {16 |24 |82 {40 Pressed Brick |84<44X234/6.48|12.96/19.44|/25.92/32.4 How many Baltimore brick will be required to build a 13 inch wall 40 feet long and 20 feet high? 40 20—800 EXPLANATION. 194 Multiplying the length 800 (40) by the height (20) of the wall we have 15600 Ans. 800. By referring to the table we find the number of Baltimore brick in a 13 inch wall to be 194 and 800 multiplied by 194 gives us 15600. Avs. 220 RAPID CALCULATOR REVISED. How many fire brick in a 18 inch wall 60 feet long and 12 feet high? 60 12—720 EXPLANATION. 720 Multiplying the length 20 and height of the wall together we have 720. 14400 By referring to the table we find the number of fire brick in a 13 inch wall to be 20. Multiplying 720 by 20 we have 14400. Ans. STONE WORK. In some localities, 164 cubic feet are allowed for a perch of stone; in some, 22 cubic feet, and in others 243 cubic feet are allowed. RULE. To find the number of perches in a wall 14 feet thick, allowing 164 cubic feet to the perch, multiply the length of the wall by the height and divide by 11. How many perches in a wall 24 feet long, 10 feet high, and 14 feet thick? 24 10—240 EXPLANATION. 11) 240 Multiplying the length 21> Ans. by the height of the wall we have 240, and 240 di- -vided by 11 gives us 21% Ans. If the wall is 2 feet thick, first find the number of perches in a similar wall 14 feet thick and add 4 of itself. If 21 feet thick add 4 of itself. If 24 feet thick add 3 of itself. How many perches in a wall 604 feet long, 4 feet high, and 2 feet thick? RAPID CALCULATOR REVISED. 221 SNR wc a ZZ ST Vie < = : N ree MIC, gE _ Te +> a 604 4 EXPLANATION. 11) 242 Multiplying the length by the height of the wall Paes we have 242; 242 divided 74=4 of 22 by 11 gives us 22, the number of perches in a 291 Ans. wall 604 feet long, 4 feet high, and 24 feet thick. As the above wall is 2 feet thick, we add i of our an- swer to itself. 4 of 22—74; 74+22—291. Ans. If 22 cubic feet are allowed to the perch, draw a perpendicular line, and write the length, height, and thickness of the wall in feet on the right hand side of this line. Place the numbers 2 and 11 on the left hand side of this line and cancel. 222 RAPID CALCULATOR REVISED. How many perches in a wall 20 feet long, 4 feet high, and 2 feet thick? : 20 y) If 242 cubic feet are 4 allowed to the perch, di- rt vide the product of the 9 length and height by 11 for a wall 21 feet thick. 11) 80 Deduct 4 of the answer if the wall is 14 feet 73An: thick; deduct ¢ if the wall is 2 feet thick, and add 3? if the wall i: 21 feet thick. How many perches in a wall 34 feet long, 9 feet high, and 2 feet thick? 34 EXPLANATION. 9 : — Multiplying the length 11)306 (34) by the height 9 we have 306. Dividing 306 2 ae by 11 we have 27 ;% the 3834 number of perches ‘in a wall 34 feet long, 9 feet 24 = Ans. high and 21 feet thick. For a wall 2 feet thick we must deduct ; tof 275% equals 3-4; and 3+; from 27% leaves 24 = Ans ~ We may also find the a eae of verthes ina wall, allowng 243 cubic feet to the perch, by multiplying the cubic feet by .0404, How many perches in a wall 20 feet lone 5 feet high, and 2 feet thick? 20*5x2—200 .0404 200 8.0800 Ans. This last rule in an approximation, but one which is sufficiently accurate for all practical purposes. RAPID CALCULATOR REVISED. 223 TY SN R aN y ; a Za HOW TO FIND THE NUMBER OF TONS OF BI- TUMINOUS COAL THAT MAY BE PLACED IN A RECTANGULAR BIN WHOSE DI- MENSIONS ARE GIVEN IN FEET. EXAMPLE. How many tons of bituminous coal will a rectan- gular bin hold whose dimnsions are 68x10? Solution: 6 EXPLANATION. 4 |$° To the left of a vertical 40110 line place 40; to the right, 12 Tons. the dimensions in feet, the 40 being the divisor and the product of the dimensions, the dividend. Cancel when possible. The answer will be tons. 224 RAPID CALCULATOR REVISED. HOW TO FIND THE NUMBER OF TONS OF AN- ~ THRACITE COAL THAT CAN BE PLACED IN A RECTANGULAR BIN WHOSE DIMENSIONS ARE GIVEN IN EBERT. EXAMPLE. How many tons of anthracite coal will a bin hold 5x8 X12? Solution: At 5 EXPLANATION. 235 Multiply together 27 8 and the three dimensions —— in feet, and point off, 1080 three places. The an- 12 swer will be tons. 12.960 Tons. SHORT METHOD FOR EXTRACTING CUBE ROOT. This process is new and comparatively simple in its nature. Numbers are of two kinds; either perfect cubes or surds, thus: 64 is a perfect cube; it is the cube of 4. But 65 is a surd or imperfect cube. We first present a method applying to “perfect cubes, and next, a method applying to surds. The pupil should familiarize himself with the cubes of num- bers as high as 9. The following table may be nelD for this purpose: Ii cteet oxox 2s DXB xX O==26 4x4 4—64 5b 5==125 6x6 6—=216 TXTXT=848 5 BS Kee Ie 95<9 9-29 RAPID CALCULATOR REVISED. 225. In extracting the cube root, we should begin at the right hand side and divide the numbers into periods of three figures each. The last period, however, may not contain three figures. But it should never contain more than three figures. It may contain either three figures, two figures or one figure. If the first period on the left is between 1 and 8, the root figure is 1; if between 8 and 27, the root figure is 2; if between 27 and 64, the root figure is 3; if between 64 and 125, the root figure is 4, and so on. In a perfect cube, the root figure of the right hand period may always be unerringly selected. If the last figure of the right hand period is 1, the last figure of the cube root of the number is 1. If the last figure of the right hand period is 2, the last figure of the cube root of the number is 8. If the last figure of the right hand period is 3, the last figure of the cube root of the number is 7. If the last figure of the right hand period is 4, the last figure of the cube root of the number is 4. If the last figure of the right hand period is 5, the last figure of the cube root of the number is 5. If the last figure of the right hand per- jiod is 6, the last figure of the cube root of the number is 6. If the last figure of the right hand period is 7, the last figure of the cube root of the number is 3. If the last figure of the right hand period is 8, the last fig- ure of the cube root of the number is 2. If the last figure of the right hand period is 9, the last figure of the cube root of the number is 9. From the foregoing explanations, it will be ob- served that we may, by inspection, find the cube root of any perfect cube of two periods. Extract the cube root of 42875. 42875 (35 oo EXPLANATION. We observe that the left hand period is between 27 and 64, so its cube root must be 3. The right hand 226 RAPID CALCULATOR REVISED. period ends with a figure 5. The only number cubed that will produce a 5, is 5; hence, the second figure ~ of the cube root of 42875 is 5. Extract the cube root of 166375. 166375 (55. EXPLANATION. By inspection we find that the left hand period of this number is between 125 and 216, so its root figure is 5; as the last figure of the next period ends with 5, its root figure is 5; hence the answer is 55. © Extract the cube root of 185193. 185193 (57 EXPLANATION. The first period is between 125 and 216, so its root figure is 5. The next figure ends with a 3, so its root figure is 7; hence the answer is 57. Let the pupils write the cube root to each of the following numbers choosing at sight the final root figure according to the kind of figure the number ends with: | ,/488976 ,/.000024389 V ci V ,/614125 ,/.091125 Ai ,/262144 ,/.000002744 V we ,/843192 ,/.000021952 Ve : Weenie | ,/.912673 V RAPID CALCULATOR REVISED. . | 227 EXTRACTION OF THE CUBE ROOT OF PER- FECT CUBES OF THREE PERIODS EACH. Extract the cube root of 14886936. 14886936 (246 | 2319/68 EXPLANATION. As the first period is between 8 and 27, its root figure must be 2. We now square 2 and multiply it by 8 for our trial divisor (12). We now bring down the first figure of the second period which gives us 68 for a dividend. Our trial divisor (12) is contained into 68 four times ;* hence the next figure of our root must be 4. As the number ends with 6, its last root. figure must be 6; hence our answer is 246. Extract the cube root of 48228544. “48228544 (364 OT 32 8—27|212 | | EXPLANATION. As the first period is between 27 and 64, its root figure must be 8. Subtracting 27, the cube of 3, from *NOTE—Our trial divisor (12) is contained into 68, five times in reality, but, in extracting the cube root, it is usually necessary to deduct at least one time from the actual number of times the trial divisor is contained in the dividend. The trial divisor may always be found by taking triple the square of the root figures already found. 228 RAPID CALCULATOR REVISED. —_ = 48 we have 21, and bringing down the first figure of the next period we have 212. Squaring the root figure already found (3) and multiplying by 3 we have 27. 27 is contained into 212 six times. (The pupil should be careful not to get the second root figure too large.) As the last figure of the number is 4, the last figure of the root of the number must be 4; hence our answer is 364. | Extract the cube root of the following: ,/44361864 Verse ene ,/184217728 \/ ,/105154048 Vien Shea _/129554216 \/, ,/12812904 V ,/153990656 Vv ; ,/3881420489 V HOW TO EXTRACT THE CUBE ROOT OF PER- FECT CUBES OF MORE THAN THREE PERIODS. Extract the cube root of 1879080904. 1879080904 (1234 1 1?x3=364| 879 ere: joe 12? 8—=432|1510 EXPLANATION. By inspection we find the root of the first period to be 1, the cube of which we subtract from the first RAPID CALCULATOR REVISED. 229 period. Squaring the 1 and multiplying it by 3 we have 3. We now bring down the first figure of the next period which is 8. Our trial divisor (8) is con- tained into 8 twice, so the second figure of our root must be 2. Multiplying the first figure of the root (1) by the second figure (2), we have 2, and multiplying this by 3 we have 6, which we-write one place to the right of our trial divisor (8). Squaring the second figure of our root (2) we have 4 which we write one place to the right of the 6, which gives us 364 our complete divisor. We now bring down the remaining figures of the second period which gives us 879. Mul- tiplying 364 by 2 we have 728; subtracting 728 from 879 we have 151. Squaring 12 and multiplying the product by 3 we have 432 for our trial divisor. Bring- ing down the first figure of the third period, and an- nexing it to our remainder we have 1510. 432 is con- tained into 1510 three times; hence the _ third figure of our root must be 3. As the last fig- ure of the number is 4, the last figure of the root must be 4; therefore our answer is 1234. Extract the cube root of 8421182563625. 8421182563625 (20345 8 2?38—12/4 2.0? 381200) 4211 13600 1200) 611 | EXPLANATION. By inspection we find the root figure of the first period to be 2. Squaring the 2 and subtracting the result from the first period, we have a remainder of 0. 230 RAPID CALCULATOR REVISED. Bringing down the first figure of the second period we have 4. Squaring the first figure of the root (2) and multiplying by 3 we have 12. 12 is larger than 4, so our next root figure must be 0. We now bring down the second period and with it the first figure of the third period. Squaring 20 and multiplying by 3 we have 1200 for our trial divisor. 1200 is contained into 4211 three times, so our third root figure must be 8. Multiplying 1200 by 8 and subtracting from 4211, we have 611. We now divide 611 by our previous trial divisor (1200) less the right hand figure. 1200 with the right hand place cut off is equal 120. 120 is contained into 611 four times, so our fourth root figure must be 4. The last figure of the number is 5, so our last root figure must be 5; hence our answer is 20345. After the third root figure is found, divide the re- mainder by the previous trial divisor less the right hand figure, for the fourth root figure. Extract the cube root of 40047321597. 20047321597 (34213 3°27 eae 3°xX8—27 |13047 8xX4X8= 386 | 42— 16|12304 oe 3076| | 34? 3=3468|7433 16936 3468497 347 — Pia0 © RAPID CALCULATOR REVISED. 231 Extract the cube root of the following numbers: _/2176782336 7 | 3/12521107822861 V 3/999183710672625 V HOW TO EXTRACT THE CUBE ROOT OF SURDS. The main difference between extracting the cube root of surds and perfect cubes, lies in the selection of the last figure. In extracting the cube root of a surd, it is impos- sible to select a final root figure, as there is no final root figure to select. Extract the cube root of 2.875087681. 2875087681 (1.4224 1 peak 12X8=8 |1.875 436|1744 | 142 38—588|1310 |1176 pee 588] 134 118 |_—— | 16 Zon RAPID CALCULATOR REVISED. EXPLANATION. By inspection we find the root figure of the first period to be 1. Cubing 1 and subtracting from the first period we have a remainder of 1. Bringing down the first figure of the second period we have 18. Squaring our first root figure (1) and multiplying by 3 we have 3 for a trial divisor. As our dividend is 18 and our trial divisor 3 our next root figure must be 4. We now bring down the next two figures of the second period which gives us 1875. Multiplying the first root figure (1) by the second root figure (4) we have 4, and multiplying 4 by 3 we have 12, which we write one place to the right of our trial divisor (8). Squaring the second root figure (4) we have 16, which we write one place to the right of the 12; adding we have 486 for our complete divisor. Multiplying 436 by 4 we have 1744. Subtracting 1744 from 1875 we have 181. We now bring down the first figure of the next period which gives us 1310. Squaring the root figure already found (14) and multiplying by 3, we have 588 for a trial divisor. 588 is contained into 1310 twice. Multiplying 588 by 2 and _ subtracting 1176 from 1310, we have 134. We now _ divide 134 by our trial divisor (588) less the right hand figure. Though the right hand figure is not consider- ed as a part of the divisor, the carrying figure obtain- ed therefrom must be considered in finding the sub- trahend. The right hand figure of .588 is 8; _ this multiplied by 2 will give a carrying figure of 2, so multiplying 58 by 2 and adding in the carrying figure we have 118 which subtracted from 134 leaves 16. We now have our result correct to three decimal places. It may be carried still further by dividing the remain- der (16) by 58 less the right hand figure. © RAPID CALCULATOR REVISED. Zoo Extract the cube root of 7 correct to four decimal places. 7.600000000000 (1.9129-+ 1 8 |6000 27 81| 651/5859 Ba rose 1083 Cua 1084/3827 217 10$3|110 98 alt 12 EXPLANATION. By inspection we find the root of the first place to be 1; cubing this and subtracting from 7 we have a remainder of 6. We bring down the first figure of the next period which gives us 60. Squaring 1 and mul- tiplying by 3 we have 8 for our trial divisor. We now find the next root figure of our dividend to be 9. Mul- tiplying the first root figure (1) by the second root figure (9) we have 9, and multiplying this product by 3 we have 27 which we write one place to the right of our trial divisor. Squaring the 9 we have 81 which we write one place to the right of the 27. Adding we have 651 for our complete divisor. We now bring down the next two figures of the second period which 234 RAPID CALCULATOR REVISED. gives us 6000. Multiplying our complete divisor (651) by 9 and subtracting the product from 6000 we have 141. We now bring down the first figure of the third period which gives us 1410. Squaring 19 and multiplying the product by 3 we have 1083 for our trial divisor. 1083 is contained into 1410 one time with a remainder of 327. We now divide our remainder (327) by 1083 less the right hand figure. (We obtain a carrying figure of 1 from the right hand figure of our divisor (8).) Multiplying 108 by 2 and adding in the carrying figure we have 217. Subtracting 217 from 327 we have 110. We now divide our remainder 110 by our trial divisor 10838 less the two right hand fig- ures. The two right hand figures (838) when multi- plied by 9 will give us a carrying figure of 8. Thus, 9 times 3 is 27, this gives us a carrying figure of 3. 9 times 8 is 72 plus 8 (to carry) equals 75. 75 gives us a carrying figure of 8. Now multiplying our divisor by 9 and adding in the carrying figure (8) we have 98. 98 from 110 leaves 12. It will be observed that our remainder is greater than our divisor, but this is frequently the case in the extraction of the cube root of surds. This result is cor- rect to four decimal places, as verified by logarithms. RAPID CALCULATOR REVISED. 235 MISCELLANEOUS. SIZES OF BOXES OF DIFFERENT MEASURES. A box 24 inches by 16 inches by 28 inches will contain a barrel. A box 26 inches by 104 inches by 8 inches will contain a bushel. A box 12 inches by 10 inches by 9 inches. will contain a half bushel. A box 8 inches by 8 inches by 42 inches will con- tain one gallon. A box 9 inches by 5 inches by 3 inches deep will contain a half gallon. A box 4 inches by 4 inches by 44 inches will con- tain a quart. USEFUL TABLE. 1 cu. ft. of White Pine, seasoned, weighs. 0010S. BA Sie “ White Oak, eo ee iy eres ae “ Water seeueeD #2, Os Ls ** Earth, loose ed eh rane § 15 sarees 5 ae * Earth, compact, Af ig ewe 2A ae tek “ Brick Sere Lane hts ** Clay ges 3 ea ba aaa: Lio “ Clay with stones Sree LOW) ae | Pe “ Cast Iron vera Maes ite at: BeOIUS IIAKC Ss Phi. tt eats Chk eRe 1 ton LS ace * Gravel, before digging, make..... Avante PATE te eS *“ Gravel, after digging, ee ier NSsveta See oo * Earth he Pe ete tis Tees As are * Sand a dt Lge 1 “ contains j of bu., or nearly 7 gal. 236 RAPID CALCULATOR REVISED. 2 cu. ft. of Ear Corn make 1 bu. of Shelled Corn, nearly. 1 “* ‘* Hard Coal contains 48 to 54 Ibs. 1 bu. of Timothy contains about..... 41,823,360 seeds Lo er Olover “ yes 16/400 96057 1 aye e Svese 68886905. 1 te Dats 2 os 6 640 0CG Pe ee Wheat at dW oe ee DOE ZOU eee 1 * contains 1 cubic foot or 2150 cubic inches. 1: ounce. Pure Gold: is: ‘worth... A ..¢23....es $20.672 1.” Gon’ Gold is) 2 a ee 18.60 i ee sf Silver iss st ee 1.22 1 Silver dollar can be coined from .9 oz. of pure silver. Amer. Coin Gold has 900 parts pure and 100 par augy Eng. 66 6é ¢é oo 6¢ ¢¢ ey Amer,, °°“: Silver.“ 900. “" =“) *§. 106 ae “ Eng. 6é 66 6é 37 6é é 6¢ 3 99 : 99 1 sq. Chain (Gunter’s) is equal to ...... 16 sq. rods 10 sq. Chains (Gunter’s) are equal to........:. 1 acre 1 section of land contains 640 acres. | 1 acre is 4,840 sq. yds. 48,560 sq. ft. 6,272,640 sq. in- ches. TABLE OF RAILROAD WEIGHTS. 3 The Weights Given in the Table Below May Be Used When It Is Not Practicable to Weigh the Articles Enumerated. PER BUSHEL. Apples, rob ag |<) 9 Geiecu ane ie ee catier Coie Sener 24 lbs. per bu. PTOCN is ee 56 * Barley. eos oo aes eee ee 48 $e Beans, whites... 5a ee 60S os 8 CQStON i Ore ae ee AG2 - ‘SBYane Seo Ui Se ie eee eee 200 Sea Buckwheats 2. 2s eae eee ee Bots - Charcoal 25-3 tee ee Spee Macca Rates Zone ys Clover. ‘Seed: 3s cer ee ee 60" Coal 05 scjco et eee ee 807% - RAPID CALCULATOR REVISED. 237 aes GR oe Se eo AO Ibs. per bu. SRPIPRTIETICE fe0/ se. Sots uc le Sond vb ee as bb 7a ¥ Pen CAN rs Sion, 8, a texeeercte aki. os Bes SANE eka _ ROUTE Belen htt ie Bee Se a kg wd uae Aire he NPE SOR OR ESIC 5 oc isk yo 8's abate ao aves Age ss ite. (Ste aS 9 tie oa a ae em e Bele OV OTE Se oo aiecs bvecacs Ngee, oA ede 5 2: TOES we Eica caer Fz) 9 ea eo 45 roe cS e VRC Lee er ee Ks 450 ¥ " ere COOTENALG oy. st: eee: oho os peetOU LOD eae ns 5 aces e's ae its : es pepe LOG Viel ola. eo ore we AD ie “ STE ORC ret ar e ie saan sore «oes ooh 1g Te ~ NSN Gs oe Re Scrat os we Ros le SOs - MIL AEISAT ICN ocr icicle o> Fe cine aan 38 sf SLO MIGR ES WO. SiS Sica soe ek ees BD os SONAR, Rua Sytee eenth | 58 fal Saga ee ea a APA ig = DUETIGS TTD ops Tiles en ES ag eR ae a OOF MeILATGES -OITIEN © ona es soa ae: GUT. a * WRC eaters hr ashe eto 50g “ PEROT oS MS ni a gan ae ear Mi ace 60 “ $ PABLOT INU ETL AIY 2... 7hc Soe Pr ke se os Pe: ve ibid Ns ba [ora boo ere a Oo she: os Py Mee Titers Me to yon Sk ea ak OG ire . PSTN ECU ier ke ee = oer er phesit obese if TB The a0 1 ies sae weit! Bre CSS ot pee eee DO pire ¥ CA) PEE Bi 2 ga er ta Geen aie ee aaa aa 605s a PER BARREL. ere Ole. ts. eae BOL yiee Sy ne on 390 lbs. per bbl. PMemerscelrand EOTUCr <3 6. ces cc we SD U sre agree aie ae a fo api Ad Rol Milesdacat ely 6é ¢é ¢é éé Bee ©: : 100 ce 4 99 i | an fa ei Aneta) ON ok ee PYODIGS © OTCON Se eek es ce og wok ew LOD Sea peras., Beat ye) Pie ee pet 7. ce 330i a Meet 238 RAPID CALCULATOR REVISED. Cider otis So ee ee ee 400 lbs. per bbl. Corn: Meal. 23.203. an sata eee 215° Goes Eos. ious hc eee ee 200° 33 ee Bish ® scp voce. 0s wee Coaeadee hater ea 300 >< eure fs ag ante te wae Ss Ge ga gc ee 160: . ‘Sp aee Os OE atte alee eee ee 80 eee PEA A Neer eres ctu ss SSL Os 40 “ per kit BIOUY ache Pere ne ae eae 196. 4 oe High WiMes 3:5... genres aa ene 390: eee Liquors, not otherwise specified ..... 400 es DLOIASSES 2 mood? Pace date «| oan meee aia 515. See Oeics sith cae Me es ieee eee ace 360 eee Let) Cee WME Meron hy De Lie 330. “i= Resin iaif4: 5 cua (te cum ete eee a 300. ‘te Salt. PNG.) vari. ee ee 810. Se Viewer 445086 0a oe 400. “oo Spirits: Turpentine: +..5 ois oe eer 360°: "ae Water (hime... 0.0... 3 Se eee oe 300: Whiskey n.8% <2. 20. Uy Se eee 390 “eee ESTIMATED WEIGHT OF LUMBER AND OTHER ARTICLES Amount LIGHT LUMBER. Weight, for car lbs load, feet. Pine, Poplar, Basswood, Butternut, Spruce and Hemlock, thoroughly seasoned, per 1,00 ft). .4.....2<5% 2,500 10,000 Black Walnut, Ash, Maple and Cherry, seasoned, per 1,000: ft. 24.28 4,000 6,000 Gum and Cottonwood, seasoned, per L,OO0! Dt ce ee ee ee 3,000 8,000 MEDIUM LUMBER. Pine, Whitewood, Basswood, Buttonwood, Hemlock (green), Spruce (green), per L000 te ee ss ee 4,000 6,000 ~ RAPID CALCULATOR REVISED. 239 Black Walnut, Maple (green), Cherry (green), Oak (seasoned), Hickory (seasoned), Elm (seasoned), Cot- tonwood (green), per 1,000 ft... 4,500 6,000 HEAVY LUMBER. Oak (green), Hickory (green), Elm tere) pert 000 9th aie 3s ee: 5,000 6,000 Oak (part seasoned), Hickory (part seasoned), Elm (part seasoned), per MeO OME Leet SNe Myc ete eae ss 4,500 6,000 HOOP POLES. Bee OtlOOULCAIMS ike hehe eer tee ee Load 4 ft. high Becsoned 25-LOOu CAT. . 2 freee soe ewe Load 5 ft. high STAVES AND HEADING. Green 28-foot car ..... UP Nicaea eee Load 4 ft. high MAANGNCU: ZOsLOUU CAD ies wee cee tess Load 5 ft. high BARK. A MMeEPCOt DOT. CONG 2 Di ccs io vhs cog os 3500 7 eds. COEENEC (Wa WMD Y 4p O79 Ok 00 a6 CoP ay A a 1500 9 cds. SHINGLES. Ba re ISO FLO mt Sirs, cog at ees ares flak he 135 70,000 Greenport at OU Oa ust cde st sy. cad, 180 90,000 LATH eee TUL MM a ce he hs 5 paca u've ees 500 50,000 BRICK. Common, per car load, each ......... 4 6,000 Baraenen car load, Gach ....\.).... ss... «6 6 4,000 240 RAPID CALCULATOR REVISED. LIME, COAL AND WOOD. Lime and Coal, per bushel ......... Coke, per bushel... eee Wood (soft, dry), per cord. .3...3. (soft, green), per cord....... (hard, green), per cord ..... (hard, dry), per cord ....... 66 66 66 SAND, STONE, ETC. mand, per: cubic Varden. ni ee Gravel, per cubic yard Stone, undressed, per cubic yard.... Marble, per cubic foot.....4. sie Slate; sper =cubie fo0ts.. 2. cee Ice, per-cubic footie 2. oe a oeee ee © @ 6 Page ean a eae tect al. ols ators. v 6,0", eles claip'< qe 6's eS pe wste's wale 3) See RTP T WL WO Sara ray dicts! aca nike Foon wa oco(ern helk.a.we ce eel «ele elo ae 7 Deer ERT METS VE TOUS: 5 5/2 soa. sees. e ht cie'siter¢ Lies’ w ob A on. ele reeubae & y Reem CeO TCI SO 1, 5 clave «an lo ste sack elehs o's de a 6s wisi ont are eue eis 13 Method of Addition by dropping the Tens.............. 14 Method of Addition by dropping the Twenties.......... 19 How to Keep Results to Each Column Separately...... 21 Civil Service Method of Retaining Results of Different Mar DTASEUMR Pee AY PLETE seh CLATS as vie ilar cins ov ei oy no asvate a'a'p 9/948 23 Groumner method. Of AGGITION: 0. ow cco pcan eee to ee ole Table of ‘Squares and -CubeS =i. sar.. suk css eee 79 How to Square Any Number Ending in Five iiss anaes 81 How to Square Any Number Ending in Twenty-five .... 81 How to Square Any Number Ending in Seventy-five.... 82 How to Square Any Number Between Twenty-five and WLC ye iewkins Valle awh one ok Ae aren aah te 84 How to Square Any Nuimnber Between Fifty and Seven- LY-TV Geos wings ee PS eek ere we 6b a ag ee eet 84 RAPID CALCULATOR REVISED. 243 Page How to Square Any Number Between Seventy-five and PISMRE UTTER OUP) ohe oy ohare ao Gok o Re Fae ake ak Olea ee ws 8d Multiplication by Squaring the Mean of Two Numbers... 85 How to Square Any Number of Nines ................ 86 How to Square Any Number of Sixes ............... 86 How to Square Any Number of Threes ................ 87 How to Multiply Any Number of Sixes by the Same PUNE TOL chk LRGOCR crn gies a ea; Soy - Shere ohare o's 2 oes, a ho 87 How to Square Any Number of Ones ................. 87 PAW CLOPNLILIDLY Gd! LVOSs DY EL WOS iis 6 a kes crete aces coc 88 ly cola eyo UO Tet ied a aha oo Gell, ee pap a> SaOR SS a0 ea ee a ee 88 How to Square Any Number Under One Thousand.... 90 How to Square any Number Under Ten Thousand End- Peeep ttre WORRY IVC 8. Ge re Cae ee See ears oS cece s 94 How to Square any Number Under One Hundred Thous- mand bndine in Twenty-0ve ii. dst s 00 cata oes 94 Pata MOREE yA ITER PLL AULOII ero cele so Oe oie otic sauce 0 és,a wieele® s 95 Pe Cea OSM RTL wapn) Oh reales te gles os 2.5 uo wtehe Sen y Wee ote 96 LORRI er a) el eg PE vinta pic koe Oaks A oes via Gee a hes 99 POWs LIVie) DY AMOGUOL) FALLS. -..cka cca s chee sesces 99 ea Tee A TEE ISL GUS S airir ais oy eeie o's, « ais ETAS cals eck seed guna «6 103 How to Divide by Factoring the Divisor ............. 104 Contracted Method for Dividing All Kinds of Numbers... 105 ROM SCMRIE LINAC eet tee emer e hs Coren eet | a ota etalevadsless wees 3.0 wee 107 te ELC ISS See sie: enone oe oo kao ahd pa ts Conger 109 Fay ORAM EE AOEI ON cette die Sty. she aie wis no be aad sco e ws ahs 109 Another Method for Adding Fractions ........... wee FLO TRE TE UT eT A CUIOIL Scio sp alas ert eo boniera'c wreipie 45 ae sen we | Another Method for Subtracting Fractions .......... rie Bice Vio CLL UO PACLIONS cr acclies sa aes cash see oa cdn yl2 Another Method for Multiplying Fractions .......... 112 PIO WRU L VIC SE POCTIONS. case slahoclere aie cles eats erated fave rye es Another Method for Division of Fractions ............ 113 How to Multiply Mixed Numbers .................... 715 How to Divide Mixed Numbers Without Reducing to Im- Samet pire RE EYL CLOTS cea are lai’ wie aed aed ateWe a Bho bs 416 Valuable Contractions for Business Men ............. ta How to Multiply Similar Numbers Together Whose Frac- ies eee Ete SENET ee toes) bg Gs one aha a aie 3 le o'a thao ep Pee adi 119 How to Multiply Any Two Numbers the Difference of Which is One and the Sum of Whose Fractions is CUE Neils | One See ate pe ear aE ear h e ieat dats 120 244 RAPID CALCULATOR REVISED. Page How to Multiply Any Two Numbers Together Whose Fractions: are Sc & oleae de pike) ee ee 130 Thousand-Day Method for Computing Interest ........ 131 How to Find the Interest at Different Rates .......: 133 Another -Thousand-Day: Method ........i. :.. yf ere onicloate ele Sat iets EK cra etetes 8 e308 203 OREGON OT BDIC Sake ce eee ds nok See weeds ale F siea) atie vip eieters 204 How to Tell The Day of The Week or Month And The Month in Which a Person Was Born .......... ws... 208 How to Find The Cost of Single Articles When Purchased PRE ESAT CIM er, erate e <<. une % Side 'e « Alejens ale ase claw pes b's ’ 210 PLEA ns UAT COL CU Lakes wiesc.s's suciely aise eke Fu os Oo vues gO ts 210 WOOD MEASUREMENT eoeoev ee? a eh Tae SP eS 3 E.G O.0 DB lor'e 2-2 295? 19'S 211 2A6 RAPID CALCULATOR REVISED. Page How To Tell The Number of Gallons of Water. in Any Tank, Boiler, or Cylindrical Vessel of Any Kind ...... 214 How to Tell the Number of Gallons in any Rectangular Vessel Sa ee a 214 How to Tell the Number of Gallons in a Barrel........ 216 BRICK WORK 2.02 oe, soe -abisaeey 0k a eee 218 STONE WORK eco eee cle aim a ois alee wie ee wb 3s Some 220 MBASUREMENT. OF COAL ° oc dicvec sd eweu pec eae 223 How to Find the Number of Tons of Bituminous Coal that May be Placed in Rectangular Bin Whose Di- mensions are given in Feet. ....0: 0.45 cceww sue een 223 How to Find the Number of Tons of Anthracite Coal that Can be Placed in a Rectangular Bin Whose Dimensions are Given in Feet... . 0... .s.ss. 5 enews 224 SHORT METHOD OF EXTRACTING CUBE ROOT ...... 224 Extraction of the Cube Root of Perfect Cubes of Three Periods Each: si..e ise sees Sea ae | ee ee How to Extract the Cube Root of Perfect Cubes of More Than Three. Periods 20.0... 2.4... 0602 228 How to Extract the Cube Root of Surds .............. 231 MISCHLLUANHEOUSs sug ae oe eee hie eee ee eee Sizes of Boxes of Different Measures ................ 235, Usetul Pable \arcie tt ease t oo a Ue aes ae ee 235 Table: of Railroad. Weights J.....-2.¢., ae E5236 Estimated Weight of Lumber and Other Articles ...... 238 a Ca Ly a a ) Wy We )) YH WW — ~ i Ki \ i 7 )) (( (( (( a a ((( @ CG a fi Wy a ( Cc ) { KC ID Tes Tie LS \\ = ZS ) SS LE EEE SSS SE (( ((( ) AK H(( WH (Ct I) | ) NN )))») Wy, Yu Wy ) “( 7 ff ) AN \\ \ ))) LE a \ Sa Sele )) Wy it Hit ) ( A ca )) })) Wi Same Ui 7 )) ) Fe EEO OES oS POM ES Rie RS YT DER SBI FAS Se ins aD ————E—_ aS CC ee SS SSS pF tat Serene es FSS Sa eee SS SR aa oe SZ Se SSS SS aes PIE )) Wy ee / WH CC cq — ) ) on “ >) 0) ( ao ) } )) oo (Kt K ( ( NK (« ( ((( ( (( oe (( LO Sa NN > — Lee He Hi ii Hil Wy i (( ca « cc yy), cl A " \ ») | )) / Wy) y; cc ( » (( »)) )) yy)» « ))) a as aw gan, amet EE nN See —— SE IS = es eee Za ao oe Bee = Ke Se See Say es ee SS es ee SS SN ; —— Sa EGS = ibe SSS Se HH ((( \ ) | a WH K( | NN Wi a ) i) \ a ))) 7 )))) KK \ o \ ] )) 7) )))) Wy Wy a | a )))) yy NN AK ; i yy ) if . i) ))) )) SS Wi UNIVERSITY OF ILLINOIS-URBANA 513.92R53R C001 RAPID CALCULATOR 3RD ED. NEW YORK HI TH 30112 017102739