Abacus Made Easy Si 00 Mae e

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iy Dr. Mae, avidow


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THE ABACUS MADE EASYA Simplified Manual

for

Teaching the Cranmer Abacus

byMae E. Davidow, Ed.D.Teacher of Mathematics

Overbrook School for the Blind64th Street and Malvern AvenuePhiladelphia, Penna. 19151

Published ByOVERBROOK SCHOOL FOR THE BLIND


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Copyright, 1966OVERBROOK SCHOOL FOR THE BLIND

All rights reserved. No part of this book may be reproducedin any form without permission in writing from the author,except by a reviewer who wishes to quote brief passages inconnection with a review written for inclusion in a magazineor newspaper.


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ABOUT THE AUTHORDr. Davidow has been a teacher at Overbrook School

for the Blind since 1935. This gifted teacher, however, firstcame to Overbrook as a student, having lost her sight at theage of 10. She received a B.A. from New Jersey Collegefor Women, now Douglas College, part of Rutgers University.Temple University granted her a Master's Degree in 1949and then a doctorate in I960.

Here at Overbrook, Dr. Davidow was instrumentalin establishing the use of the Cranmer Abacus as a part ofthe regular curriculum. Her enthusiasm for this pioneermethod of teaching mathematics led others to follow hereagerly. In her role as coordinating teacher, she workedwith the members of the Mathematics Department and theresults Were highly successful. Hopeful that this success at

Overbrook might be experienced by many teachers elsewhere,she was encouraged to write the following manual.

We are indeed appreciative of the unstinting effort ofDr. Davidow in presenting Abacus Made Easy.

Joseph J . KerrAssistant Principal


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Digitized by the Internet Archivein 2012 with funding from

National Federation of the Blind (NFB)

http://www.archive.org/details/abacusmadeeasysiOOmaee


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ACKNOWLEDGMENTSIn the writing of this manual I have received invaluable

aid and suggestions from the many teachers and students withwhom; I have worked on the abacus.

I owe a great debt of gratitude to the administratorsof the Overbrook School for the Blind who encouraged me to delvemore deeply into the study of the abacus and who made it pos-sible for me to attend the first Abacus Institute ever held inAmerica. I appreciate the opportunity that was given to me towork with the instructors of mathematics as they were teachingthe use of the abacus to their students. In this manner bothteacher and student learned the language and method of operation,

I feel it is appropriate to acknowledge with gratitudethe work of Fred Gissoni whose text, Using the Cranmer Abacus,helped lay the foundation and groundwork which enabled me towrite this manual.

I wish to express my special thanks to the many volun-teers who gave unstintingly of their time to work directly withme in compiling this manuscript.


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Ill

CONTENTS

Acknowledgements iiForeword vIntroduction to the Tool 1Setting Numbers 3Addition 5Addition of Two Digit Numbers 11Subtraction 14Multiplication 21Treatment of the Zero 27Multiplication of Two Digits 29Division. 31Long Division 35Addition of Decimals 43Subtraction of Decimals. ............. 46Multiplication of Decimals 48Division of Decimals . 50Fractions 54


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IV

Addition, of Fractions 56Subtraction of Fractions 60Multiplication of Fractions 63Division of Fractions . . . . . 67Per Cent 69Square Root 71Conclusion 76


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FOREWORDUsing the Cranmer Abacus is an excellent text writ-

ten by Fred L. Gissoni of the Bureau of Rehabilitation Ser-vices, Kentucky Department of Education, at Lexington,Kentucky

.

In 1964 I attended the Abacus Institute at the Univer-sity of Kentucky, held under the direction of Mr. Gissoni.This was the first such institute ever conducted in America.We were also privileged at that time to have with us ProfessorPaske, an instructor of the abacus at the School for the Blindin Copenhagen, Denmark.

In the past two years, I have taught the CranmerAbacus using Fred Gissoni's text. However, as an instructorof the abacus to both teachers and students. I found it desirable

to have a simplified manual at our fingertips. During thisperiod I have made, with Mr. Gissoni's permission, adapta-tions of his text. This has been done while working withteachers and with children from second grade through highschool- It is my hope that this manual will make the teachingand the learning of the abacus more meaningful to both studentsand teachers. I shall attempt to explain in the simplest, most


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VI

concise manner how to add, subtract, multiply, divide,handle decimals, fractions, percent and square root.

The method of instruction for an individual or for aclass is essentially the same. The approach to teaching theabacus is likewise the same for a second grade student, asenior in high school, a college student, a teacher, a parent,or any interested person, either blind or sighted, who wishesto learn the abacus. When we first examine the abacus, weall learn about the tool in the same manner. We must moveslowly; the language must be simple; every step must beunderstood. The slow learner, as well as the gifted child,learns the abacus in the same manner. However, after thelearning has taken place, and the individual knows how tomanipulate the tool, then each one operates it at his individualspeed.

In teaching the use of the abacus, one must be verypatient. One must use the language level of the individual.The abacus is studied one step at a time. As one step islearned, it must be practiced again and again. Then one maymove to the next step. One takes his time and learns itthoroughly.


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INTRODUCTION TO THE TOOLThe first lesson consists of handling the abacus. One

must become acquainted with the tool. We notice that theframe is oblong in shape. There are thirteen rods, or columns.,attached to the frame. On each column there are five beadswhich travel up and down. There is a separation bar whichcuts across all thirteen columns, approximately two-thirdsthe distance from the bottom of the frame. This bar separatesfour of the beads or counters from the fifth bead.

We shall now examine the abacus in its proper opera-ting position. The abacus will always be on our desk in theposition which is about to be described. The bar which sepa-rates the one bead from the other four beads runs from rightto left. The section containing the single beads is on the upperpart of the abacus, and the four beads are toward the bottomof the frame .

First let us run our fingers along the face of thelower edge of the frame. Here we find a dot at the end of eachcolumn. In addition to these dots, we will find a short., verti-cal line between every third column. The dots are to helpus locate our place, and the lines, which are called "unitmarks" or commas in writing numbers, can also serve as a


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decimal point. These same dots and lines are found on theseparation bar between the four beads and the one bead.

Now let us place the index finger of our right handon the dot to the extreme right, on the bottom of our frameThis dot corresponds with the column to the extreme right.It is the units column. Moving from right to left, the nextcolumn is the tens column and the next is the hundreds column.Then we touch the line between the columns which separatesthe hundreds column from the thousands column. Next wehave the thousands column, ten thousands, hundred thou-sands, separation line, millions, etc. , up to one trillion.As our fingers move along the frame, the index finger of ourleft hand is always immediately to the left of the index fingerof the right hand. The position of our hands is extremelyimportant in all of our operations as we manipulate the abacus.The keyword, while learning the fundamentals of the CranmerAgacus, is patient repetition.


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SETTING NUMBERSWhen we put a number on our abacus, we do not say

we write it. In abacus language we use the word "set."When we want to remove a number or erase it, we say thatwe ^clear" it. As we move a bead toward the bar or rrset"it, it takes on a value. When we move it away from the bar,or "clear"' it, it loses its value. On a single column we canshow the ten digits from zero to 9- When there are no beadsmoved toward the bar, the column is in a zero position. Thento write 1 , we set one bead by pushing it up to the bar. To

write 2, we set another bead; then we set a third bead, and wehave 3; set a fourth bead and we have 4. Each one of thebeads below the separation bar has a value of one. The singlebead above the bar has a value of five.

Let us now write from 1 to 5 --writing each numberseparately. To set the number 1 , we would move one of thebeads on a column up to the separation bar. To set the number2. we would set 2 beads up to the bar. To set 3, we wouldmove 3 beads up to the bar. To set 4, we move all 4 beadsup to the bar. To set 5, we would move the upper bead downto the bar, However., we must be sure that there are no beadsbelow the bar on that column. To set the number 6, we would


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move the upper bead down, and move one lower bead up inthe same column. Five plu$ 1 equals 6. To set 7, we set a5 and then set 2 on the same column. Five and 2 equals 7.To set 8, we set 5 above the bar and 3 below. To set 9. weset the 5 bead and the four l's. In setting 9 we see that allthe beads on a column are moved as close to the bar as pos-sible.

Much time must be devoted to setting numbers. Aswe write our numbers with the right index finger, or thumb andindex finger, our left index finger is always resting on the

column immediately to the left of the right hand. Let uswrite 12. The right index finger sets a 1 in the tens columnwhile the left index finger rests on the hundreds column.Then both hands move to the right and the right index fingerand thumb push up or set two beads in the units column. Muchpractice is needed in writing and reading numbers. Littlechildren can write how many desks there are in the classroom;how many sisters they have, etc. , etc. Each teacher givesnumbers according to the level of her group, Older children canwrite their phone numbers and then have a classmate read them.When the student can set and read numbers quickly, then he isready to begin addition.


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ADDITION

We will start our addition by adding the number 1 toitself a number of times. We will place the index finger ofthe right hand on the extreme right column on the abacus.This is the units column. The left hand, as we mentionedbefore, is immediately to the left of the right hand on thetens column. We set the number 1 by moving one of the lowerbeads up to the bar. This is done with the index finger of theright hand. Next add 1 to this 1. This is done in the samemanner with the index finger of the right hand. Now add another1 . Now add another 1 . We now have four beads in our unitcolumn. It is permissible to use the thumb if one finds iteasier. We find that if we want to add 1 more, we cannot addit directly. There are no more lower beads to move up tothe bar. Since there are no more rfones ,r to add to the fourbeads that we have in our units column, our right index fingermoves above the bar and sets a 5. We only wanted to add 1more, but we added 5 more. We must now figure out howmany more beads were added than the one bead that we wanted.There are several ways that this can be done. Each teacherfigures out according to her grade level which method to use.One teacher might say, "I asked you to give me one cent and


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you gave me a nickel, how much change should I give you"?The student figures out that 1 from 5 is 4 and he must returnfour pennies. We can think of the four beads that we have seton the abacus as four pennies and when we clear the fourbeads, we are returning the four pennies. This is known asindirect addition. There are many different ways this can beexplained. This depends on the age of the student and hisunderstanding of numbers.

One second grade teacher refers to this indirectaddition as working with partners. The child learns that if hewishes to add 1 to 4, and there are no more beads on the bot-tom of the bar, he must go to the top of the bar and set a 5.The child learns that the partners of 5 when you add 1 to 4are 4 and 1. So the child clears the 4. If, on the other hand,we were adding 4 to 1 , the child would set a 5 and clear 1 .After the student learns the relationship of the 5 and itspartner 1, he can set the 5 and clear the 4 in one continuousmotion. We slide the 5 bead down to the bar, move our fingerdown across the bar, and slide the four lower beads away fromthe bar. By doing this we have added 1 to the 4 and we nowhave the digit 5 showing on the units column. In order to add 1to this 5, we continue as we did before. We move one lower


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counter up to the bar. We now have a 5 above the bar and a1 bead or counter below the bar. Five plus 1 equals 6. Thenwe add 1 to 6 and we have 7. Add 1 to 7 and we will have 8.Add 1 to 8 and we will have 9- We want to add 1 more to 9and we find that there are no more beads on the unit columnto add directly,

Our left index finger that has been resting on thecolumn immediately to the left of the right hand (on the tenscolumn) now comes into play. We can refer to that indexfinger as the "helping finger." When we can add no morebeads with the right index finger, the left hand comes intoplay and assists. The left hand will set one bead on the tenscolumn (push it up to the bar). This one bead has a value of10 since it is in the tens column. We only wanted 1 more, butthe left index finger gave us 10 more. Now we must figureout how many extra we have.

This can be done again in several ways: (1) We had 9,we wanted 1 more; our left hand gave us 10; let us see howmany extra we have. One from 10 is 9- We have 9 too many,so our right hand will clear the 9- We now have a ten in thetens column and a zero in the units column. This gives us avalue of 10. We see then that 9 plus 1 equals 10. (2) We can


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speak of it in terms of money. We want a penny to add to 9-We received a dime. We must give 9 cents change. Onefrom 10 is 9- (3) Again we can refer to our partners. Thepartners of 1 are 1 and 9- Here again, we have addedindirectly.

As we continue to add more I's, we add them in theunit column with the right index finger, or thumb, as mentionedabove. One added to 10 gives us the digits 11. When we addanother bead in the units column, we have 1Z. Thus, one 10and one 1 gives us 11; one 10 and two l's give us 12, etc. , etc.

We continue adding 1 until we have 44. Then with the rightpointer finger we set 5, clear 4, and we have 45. We con-tinue to add 1 until we have 49- Then with our left index fin-ger (our helping hand) we set the 5, clear 4, and with ourright index finger clear the 9- We now have 50. We continueadding 1 until we have 99- We find we cannot add 1 morewith the right hand; neither can the left hand assist us bygiving us another 1 in the tens column. So the left handreaches over and gives us a 10 in the hundreds column.(This ten has the value of ten 10's or 100.) But we now havenine extra 10's,. so with our left index finger we clear the nine10's and the right index finger clears the nine units. Although


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we wanted to add only 1 more, we added 10 more and thus had9 extra to clear.

A good addition exercise to introduce in the beginningis to add from 1 to 45. First add 1 to zero.. Begin with thehands in starting position- -the right hand on the units columnand the left hand immediately to the left of it on the tenscolumn. With the right index finger we set 1. Then with ourpincher fingers (the index finger and the thumb) we set 2more beads. We are now ready to add 3 more beads. But wesee that there is only 1 bead. So we must turn to the 5. Weset 5. How many extra did we set? Three from 5 is 2.Since we have 2 too many, we will clear 2, We check ourabacus and find that we have 6. We have learned that twosets of 3 makes 6, so we are sure that the abacus is right.We will now add 4 more. We find that we only have 3 beads,so our left hand will help us. The left index finger will set a10. What is 4's partner to make 10? The child has learnedit is 6. So he will clear 6 with the right hand. Or we may saywe gave 10, but we wanted only 4 more; how much too manydid we give? Four from 10 is 6, so we clear 6. The 6is cleared with the right hand. We now add 5. That can beadded directly. Now we add 6. We cannot add 6 directly.Here again, the left hand sets another 10. One says, 6 from


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10 is 4; so we must clear 4. We cannot clear 4 directlywith the right hand; so the right hand clears a 5 and sets 1 .Here again figuring 4 from 5 is 1 tells us we must give backa 1 . We wanted to take away 4 pennies; but we took away anickel, so we owe 1 cent. We give back 1.

We now have 21 on the abacus. We add 7. This canbe added directly by setting 5 and 2 more. We now have 28.We are ready to add 8. We cannot add 8 directly, so the lefthand sets a 10. Eight from 10 is 2; so the right hand clears2. We now add 9- This cannot be done directly. Here againthe left hand sets a 10. Nine from 10 is 1, so the right handclears 1. We now have 45.

This same exercise is continued by starting out bysetting 1 on the units column and then adding 1, 2, 3, etc.through 9. When we start with 1 on the abacus, our answer is46. When we start with 2 on the abacus, our answer is 47, etc

Following are some one -digit addition examples forpractice:2 + 2= 3 + 5= 7 + 2 =6 + 1= 4 + 5= 5 + 1 =2 + 5= 2 + 3= 3+4 =4 + 1 =


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ADDITION OF TWO DIGIT NUMBERSAddition on the abacus is done from left to right.

When addition is done from left to right, we do not have tocarry . Let us add 36 plus 27, On the tens column we set a3; then on the units column we set a 6. Since we were going

to write 36, we put our right hand on the tens column. Theleft hand is immediately to the left of the right hand restingon the hundreds column. Then the right hand moves to theright to the units column to write the 6. The left hand fol-lows the right hand, and when the right hand is writing the6 in the units column, the left hand is resting on the 3 in thetens column. For low vision students and sighted teachers,it is helpful to keep the pointer fingers on the lower part ofthe columns instead of resting the fingers on the beads. Inthis manner we do not cover the numbers we have set; thusmaking it possible to read with the eye what we have written.

We are now ready to add 27. We place our right handon the tens column and add 2. Since we do not have 2 beads,we must add 5. Two from 5 is 3, so we clear 3. Then ourright hand moves to the units column (followed by the lefthand) and we are ready to add 7. We cannot add a 7 directly,so the left hand will set a 10. Since we only wanted 7, we


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will say 7 from 1 is 3. We will clear 3 with the right handin the units column. We cannot clear 3 directly, so we clear5. Three from 5 is 2. Since we cleared 5 and had to takeaway 3, we took away 2 extra beads. We will give back 2with our right hand. The answer is 63.

Direct addition is not difficult and in the beginning,one should give the student simple problems such as 123 plus321. Here we set 1 in the hundreds column with the righthand (the left hand being immediately to the left of the righthand) . Both hands move to the right and the right hand setsthe 2 in the tens column. Then both hands move to the rightand we write the 3 in the units column. We then go back tothe hundreds column and the right hand sets the 3 in thehundreds column. Then both hands move to the right and theright hand sets the 2 in the tens column. We move again tothe right and set the 1 in the units column with our right hand,Our answer is 444. Our left hand did not have to assist theright hand. Now let us add 789 to 444. We place our righthand on the 4 in the hundreds column. We cannot add a 7 tothe 4 directly, so the left hand will give us a 10 in the thou-sands column. This 10 is equal to 1 hundreds, but we wantedonly 7 more hundreds. Therefore, we must clear 3 hundreds


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(7 from 10 gives us 3) .Both hands move to the right and we are ready to

add 8 tens. We do not have 8 tens, so our left hand willassist. The left hand will give us 10 tens or 1 bead in thehundreds column. We have added 2 tens too many, so weclear 2 tens. We now move to the units column and add 9ones. We do not have 9 units, so our left hand assists againby setting a 10. Although we only wanted 9, we add 10. Wesay 9 from 10 is 1. We then clear a 1 in the units column.Our answer is 1233.

Exercises in two- and three-digit addition:23 + 25 = 52 + 17 = 34 + 15 =

71 + 16 = 81 + 8 = 62 + 20 =

24 + 65 = 12 + 72 = 72 + 16 =25 + 23 = 653 + 230 = 825 + 162 =

907 + 52 = 425 + 23 = 873 + 34 =769 + 25 = 473 + 192 = 395 + 232 =439 + 123 = 892 + 532 =


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SUBTRACTIONSubtraction, can be introduced at the same time that

addition is being taught. For example, when we add 1 to 1,we have 2. Now let us take 1 away, or subtract 1. What dowe have left? We find that we have 1 left. Clear the abacus.Then we add 2 and 2. We have 4. Now let us subtract 2. Wehave 2 left. In setting beads and clearing, we must always keepin mind our hand position; the left hand always follows theright.

Let us set 32 and subtract 21 . We set the 3 of the 32in the tens column with the right hand. Then we move our handsto the right and set the 2 in the units column. From this wewish to subtract 21. Here again, as in addition, we work fromleft to right. We place our right hand on the 3 in the tens columnof the 32 and take 2 away, leaving a 1 in the tens column.Then both hands move to the right and the right hand clears a1 in the units column from the 2. Our answer is 1 1 . This issubtracting directly.

For every addition problem, a subtraction examplecan be given. We start by saying let us add 1 to zero. We set1. Let us subtract 1 from 1. Our answer is zero. Now to


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zero add 1 . Let us add two more. We have 3. Subtract 1 .We have 2 left. Subtract 2 and we have zero. We set 1. Weadd 2 more. Let us add 3 more. We find that we cannot add3 directly, so we set 5 and we recall that the- partner of 5 and3 is 2. So we clear 2 and our answer is 6. We now subtract 1.We clear 1 and our answer is 5 . In other words, the differencebetween 6 and 1 is 5. We could also have said that you had 3cents and I will give you 3 more. But since I do not have 3cents I will give you a nickel. Then how much money must yougive back to me? The child will return 2 cents and have 6remaining

.

To continue with the subtraction example, we took 1from 6 and have 5 remaining. Let us take away 2 more. Wecannot clear 2 directly, but we can clear 5. What is the part-ner of 5 and 2? We recall that it is 3, so we return 3. Wewanted to subtract 2 from 5. We subtracted 5. How many-extra did we take away? We took away 3 too many. So wereturn 3. Now we see that when we took 2 from 5 we have aremainder of 3 , or the difference between 5 and 2 is 3. Wenow have 3 left on the abacus. We subtract 2. This can be donedirectly. There is 1 left. We subtract 1. This can be donedirectly and our answer is zero.


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We now start from the beginning and add 1 . Then 2more. Then we add 3. We must set 5 and clear 2. We nowwish to add 4. We cannot do it directly with the right hand,so our left hand will help us. The left hand sets 10. We gave10, but we only wanted 4 more; how much too many did we give?Four from 10 is 6, so we clear 6 with the right hand. Inworking with partners, we can also ask what is 4's partner tomake 10? It is 6. We set a 10 with the left hand, clear 6with the right, and find that the sum of 6 and 4 is 10.

Now we turn again to subtraction; 10 minus 1. Sincewe have to subtract a 1, we put our right hand on the unitscolumn. Since there is a zero in the units column, we turn tothe left hand to assist us. Our left hand will clear a 10. Butwe only wanted to subtract 1 . We check to see how manyextra we took away. We say, 1 from 10 is 9- We must there-fore return 9- So our right hand will give back 9 by setting a5 and 4 more in the unit column. We now wish to subtract 2.We clear 2 directly. We have 7 left. We wish to subtract 3.We cannot subtract 3 directly, but our right index finger canclear 5. We wanted to subtract 3. So we clear 5 and set 2.We now have 4. Seven minus 3 gave us 4. The partner of5 and 3 is 2, so we gave back 2 after clearing the 5. We now


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subtract 2 more. This can be done directly. Let us add 2more to the 2. Now subtract 4. We have zero left.

We start again and add from 1 to 45 by adding 1, 2,3, 4, 5, 6, 7 , 8, and 9- We now have 45. We now subtract1 . Our right hand is on the units column and the left hand ison the tens column immediately to the left of the right hand.We cannot subtract 1 directly, so we say 1 from 5 is 4; wecleared 5 and set 4. We subtract 2 directly by clearing 2beads with the right hand. We are now ready to subtract 3.The right hand cannot subtract a 3, but the left hand can helpus. We subtract a 10. How many extra did we clear? Threefrom 10 is 7. That means that we cleared 7 too many, so wemust return 7. The right hand will give back 7. We will set5 and 2. We now have 9 in the units column and 3 in the tenscolumn. We subtract 4. This is done directly. We clear 4with the right hand. We now subtract 5. This is done directlywith the right hand. We are now ready to subtract 6 from 30.The right hand cannot subtract a 6 because there is a zero inthe units column, so the left hand must assist us. The lefthand clears a 10. We wanted to subtract 6 and we subtracted10. How many extra did we subtract? Six from 10 is 4; orthe partner of 1 and 6 is 4; so we give back 4 leaving us 24.


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Now we are ready to subtract 7. Since we have only 4 in theunits column, we must go to the tens column. The left handwill subtract a 10. We have taken away 3 too many. "We wantedto subtract 7, but we subtracted 10, so we check by saying7 from 10 is 3. That means we took away 3 too many, so wemust give back 3 in the unit column. We cannot give back 3directly, but we can give back a 5. We gave back 2 too many;so we must clear 2 beads. We now have 17. We subtract 8.We cannot subtract 8, so we clear 10. We have taken away 2too many, so the right hand returns 2 beads. Eight from 10gives us 2. We now have 9 beads on the abacus. We areready to subtract 9- We do this directly, and our result iszero.

For the next exercise, start by setting 1 on the abacus.To this add 1, 2, 3, 4, 5, 6, 7, 8, and 9- We now have 46on the abacus. We subtract in the same manner, starting with1, 2, 3, 4, 5, 6, 7, 8, 9- We have 1 left on the abacus. Thenwe set 1 more bead and start out with 2 on the abacus. Nowwe continue as before by adding 1, 2, 3, 4, 5, 6, 7, 8, and 9When we get to 47 we begin subtracting 1, 2, 3, etc. , etcThis is an excellent exercise for addition and subtraction.

Let us examine a few subtraction examples. We shall


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set 64 on the abacus. We write 6 with the right hand in the tenscolumn and the left hand is on the rod to the left of it. Bothhands move to the right to set the 4 in the units column.While the right hand sets the 4, the left hand rests on the tenscolumn. We will now subtract 46. We place the right hand onthe 6 in the tens column. To subtract 4, we find we cannotsubtract it directly, so we clear a 5 with the index finger andset 1 with the thumb. This leaves a Z in the tens place. Bothhands move to the right and we are ready to subtract a 6 fromthe 4 ones. Since we cannot take away 6 from 4, our left handmust help us. The left pointer finger will take away a 10and since we only wanted to subtract 6, we will say 6 from 10is 4 and give back 4 ones. We cannot give back 4 ones directly,so we give back 5. But we had to give back 4 and we gave back5, so we have given back 1 too many; therefore, we clear a 1.Our answer is 1 8.

Now let us subtract 89 from 182. We put our righthand on the tens column to subtract the 8 from the 8. We moveboth hands to the right to subtract the 9 in the units column.Since we cannot take 9 from 2, our left hand will help us.There is a zero in the tens column, so our left hand will go tothe hundreds column and clear 1 bead which is equal to 1:0 tens.


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Since we wanted to clear only 1 ten, we will return 9 tens inthe tens column. We now say 9 from 10 is 1 and return a 1in the units column with the right hand. Our answer is 93,

Subtraction Exercises:24 - 12 - 124 - 93 =46 - 31 = 275 - 184 =76 - 35 = 378 - 179 -

82 - 19 = 500 - 9 =

97 - 68 = 328 - 29 =


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MULTIPLICATION

Since multiplication is a rapid form of addition, theteacher can introduce multiplication whenever she feels itfits into her mathematical curriculum. The second grade

teacher may wish to demonstrate to her class that if we wantto add 2 two times, we can say 2 twos are 4. And the childcan set first 1 set of 2 and then another set of 2 and see thatthe answer is 4. The child may have learned at an earlierage to count by 2's, but now he sees how he gets 2, 4, 6, 8,etc. The mathematics becomes meaningful. The teacher thenshows the child that 2 x 1 is 2, 2 x 2 is 4, etc. , etc. Nowlet us examine how this is set up on the abacus.

The student must learn that each part of a multipli-cation problem has a name. We learn these names so that weall speak the same language. The number which is multipliedis the multiplicand. The number by which the multiplying isdone is the multiplier. These two together are called thefactors. The answer is called the product.

Let us multiply 3x6. The multiplier 3 is placed allthe way to the left on the extreme left column. Now we go tothe right hand side of the abacus to figure out where the


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multiplicand should be placed. We count up how many columnsare used in the multiplier, how many are used in the multipli-cand and add one column for the abacus. For example, inthe digit 3 there is one place; in the digit 6, there is 1 placeand we add one for the abacus giving us 3 places. From theextreme right, we count in 3 places - -first column units, secondcolumn tens, and third hundreds - -there we place our 6. Theleft hand is immediately to the left of the right hand. Withthe right hand on 6, we say 3 x 6 is one eight As we say one,the right hand moves to the right of the 6 and sets the 1 . Theleft hand follows the right hand and as we write the 1 with theright hand, the left hand rests on the 6 or the column wherethe 6 is. After we have set the 1, move to the right and setthe 8 as we say it. As the right hand has moved to the rightto set the 8, the left hand has followed it and is resting onthe 1, or the column where the 1 is. We then pick up the righthand and clear the 6. Our answer is 18.

Let us now multiply 7 x 43. We will write our multi-plier 7 to the extreme left. We must now figure out where towrite the 43 of the multiplicand. We count up our digits:1 digit in the multiplier, 2 digits in the multiplicand and addone place for the abacus. We go in 4 columns from the extreme


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right column and write, or set, 43. (The 4 is set in thethousands place, and the 3 in the hundreds place.) We nowmust check and see how many empty columns we have afterthe 3 of the 43. There should be 2- -one for the digit in themultiplier and one for the abacus. We are now ready to multi-ply. With our right index finger on the 3 of the 43 and ourleft index finger resting on the 4 of the 43 we say: 7 x 3 istwo (setting the 2 in the tens column as we say it), then wemove both hands to the right as we say one, and set it in theunits column. We pick up the right hand and clear the 3. Wenow place the right hand on the 4 of the 43 and say 7 x 4 istwo (the right hand moves to the right as we set the 2 and theleft hand follows it, resting on the 4). Then we move to theright and are ready to write the 8 of the two eight with the 2in the tens column. The left hand is now resting on the 2that was just written in the hundreds column. Since the righthand cannot set an eight with the 2 already in the tens column,the left hand can help us. It will give us 10 which is equal to10 tens in the hundreds column. Because we added 10 when wewanted only 8, the right hand will clear the 2 that is there;since 8 from 10 is 2 and we have added 2 too many, the righthand clears the 2 in the tens place. We pick up the right hand


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and clear the 4. The answer or product is 301. There shouldbe no difficulty in reading the answer because it will always beon the extreme right side of the abacus.

Let us now multiply 43 x 7. We will set the 43 asthe multiplier to the extreme left and we shall determine whereto set the multiplicand, 7. We count 2 digits for the multi-plier, one for the multiplicand and one for the abacus. Thatmeans we go in 4 columns from right to left and we write the7 in the thousands column. We put our right hand on the 7and the left hand to the left of it, and we say. 4 of the 43 x7 is two, and we move our hands to the right and write the2 in the hundreds column. Then we move both hands to theright and set the eight in the tens column, (our left hand restingon the 2 of the hundreds column) . We hold the hands in thatposition while we say: 3 (of the 43) x 7 is two, (and we set the2 where the right hand is resting, with the 8). Since we cannotwrite a 2 where the 8 is, the left hand will help us. The lefthand will give 10. Because we wanted only 2, and 2 from 10is 8, the right hand will clear the 8 on which it was resting.Then we move both hands to the right and set the 1 (of the 21)in the units column. The left hand is now resting on the zeroof the tens column. We pick up the right Ijand and clear the 7.


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We now read the product, 301. When we multiply 43 x 7, wemust remember to multiply the 7 by the digits of the multi-plier in the order of their occurrence. For example, we firstsaid, 4x7, and then we said 3x7.

We might mention here that when a zero is to be re-corded on the abacus in multiplication, we press the rod or

column gently or touch it and hold the hand there so that weknow our position. For example, let us multiply 5 x 47. Wewill set the 5 to the extreme left and count 4 columns fromright to left to set the 47. By counting up our digits we findthat we have 3 and 1 for the abacus. We place our right handon the 7 and say: 5 x 7 is three (we move our hands to theright and set the 3 in the tens place). We then say five, moveto the right and set it in the units place. We pick up the righthand and clear the 7. Place the right hand on the 4. Withour right hand on the 4 we say: 5 x 4 is two (our right handmoves to the right and sets the 2 in the hundreds place). Aswe say zero (for the zero of 20), our right hand rests on the3 which is already in the tens place and we give the 3 a littletouch as if we were recording the zero. Then we pick up theright hand and clear the 4. The answer or product is 235.

Let us multiply 4 x 32. Place the 4 to the extreme


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left; then count in 4 places from the extreme right and set the32, putting the 3 in the thousands place and 2 in the hundredsplace. We are now ready to multiply. Place the right hand onthe 2 and the left hand on the 3. "We say 4 x 2 is zero 8. Aswe say zero, our hands move to the right and our right handis on the tens column as we record the zero by touching thebar. Then both hands move to the right and our right handsets the 8 in the units column. (We must learn that wheneverwe multiply one digit by another digit, our answer will have2 digits. If it is an 8, it is zero eight.) We pick up the righthand and clear the 2. We put the right hand on the 3 and say:4 x 3 is one two. As we say one, both hands move to the rightsetting the 1 immediately to the right of the 3. We then moveboth hands to the right and set the 2 in the tens column. Wepick up the right hand and clear the 3. The answer is 128.

Multiplication Exercises:

17 x 14 = 25 x 5 =37 x 6 = 43 x 7 =63x5= 82 x 7 =49x9= 95 x 3 =56 x 8 = 71x6=


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tens place; we move our hands to the right and set the 2 inthe units place. We pick up the right hand and clear the 3.The product is 912.

Let us now multiply 4 x 3 02. We will put our righthand on the 2 and say: 4 x 2 is zero eight. Our right handmoves to the right and touches the rod on the tens column aswe say zero. Then we move to the right and the right handsets the 8 in the units column. We pick up our right hand andclear the 2. We then put the right hand on the 3 and say:4 x 3 is one, (moving to the right and setting the 1 in thethousands column) , and we say two as the hands move to theright and set the 2 in the hundreds column. We pick up theright hand and clear the 3. The product is 1208.

Multiplication Exercises Using Zeros:5 x 302 = 703x4 =902 x 3 = 204 x 7 =809 x 5 = 207 x.14 =805 x 32 = 107 x 36 =608 x 24 = 506 x 203 =


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MULTIPLICATION OF TWO DIGITSLet us take the example 45 x 67, Write 45 to the

extreme left. There are two digits in the multiplier, 45, andtwo digits in the multiplicand, 67, giving us four digits, thenwe add one more place for the abacus, making it 5 places.We count 5 rods from right to left and set 67 (the 6 being placedin the ten thousands place, and the 7 in the thousands place) .This leaves us 3 empty columns or rods to the extreme right.We are now ready to multiply.

With our right pointer finger on the 7 and the lefthand pointer finger on the 6, we say. 4 x 7 is two eight. Weset the 2 to the right of the 7 and then set the 8 to the right ofthe 2. The left hand follows the right hand. We hold our righthand on the 8 and our left hand is on the 2. Now we say:5 x 7 is three five. We write the 3 where our right hand ison the 8. We cannot write a 3 where the 8 is; so our left handadds a 10 and our right hand clears a 7, because 3 from 10is 7. Then we move our right hand to the right and set a 5 inthe units column. The left hand rests on the 1 in the tenscolumn. Pick up the right hand and clear the 7 from the 67.


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Place the right pointer finger on the 6 and say: 4 x

6 is two four. Set the 2 to the right of the 6 and add 4 to 3,We add 4 by setting 5 and clearing 1, and we hold the righthand on the 7 and the left hand on the 2. Then we say: 5x6is three zero- We write the 3 where our right hand is holdingthe 7. Since we cannot write a 3 where the 7 is, our lefthand will help us by adding 10 where the 2 is and our righthand will clear 7, because 3 from 10 is 7. Then we move ourhands to the right and set the zero where the 1 is. We recordthe zero by touching the 1 with the right hand. We pick up theright hand and clear the 6. Our answer is 3015,

Multiplication of two digits by three digits is done ina similar manner. Multiplication by three or more digits isdone in the same manner.

Exercises:34 x 29 = 98 x 56 =

103 x 42 = 235 x 74 =206 x 18 = 104 x 36 =

87 x 45 = 246 x 39 =342 x 58 = 234 x 125 =


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DIVISION

As we begin division, we must acquaint ourselveswith the names of the different parts of a division problem.The number that does the dividing is called the divisor.The number into which it is divided is called the dividend.The answer is the quotient. For example., if we woulddivide 6 into 48. the 6 would be the divisor, the 48 is thedividend, and the 8 is the quotient.

In setting up an example in division on the abacus, weset the divisor to the extreme left of the abacus. The divi-dend is set to the extreme right on the frame. The quo-tient will appear to the left of the dividend.

In short division, we have one digit in the divisorand as many digits as we wish in the dividend. We then dividethe divisor into one or, if necessary, more digits in the divi-dend. This results in a quotient figure which is entered onthe abacus. The figure which has been entered in the quotientis then multiplied by the divisor and the product resultingfrom this multiplication is subtracted from the dividend.

Let us now work an example . Our first examplewill be 792 divided by 4. Our divisor is 4; our dividend is


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792. We will set our divisor to the extreme left. We willset our dividend, 792, to the extreme right. We put ourright hand on the 7 of the 7 92 and ask: 4 will go into 7how many times, or how many 4's in 7? Four will go into 7one time. The question arises where to place the first quo-tient, digit 1. We examine the divisor, 4, with the firstdigit in our dividend. If the divisor is equal to, or smallerthan, the first digit in the dividend, we skip one column tothe left of the dividend and set the 1 in our quotient. Fourwent into 7 one time. We skip one column and set the 1 withthe right hand. Then with our right hand on the 1 and theleft hand to the left of it, we say: 1 x 4 is zero four. Aswe say zero, the right hand moves to the right and rests onthe column with a zero. Then as we say four, our righthand moves to the 7 and subtracts 4. We subtract 4 by clear-ing 5 and setting 1 with the right hand. We are then ready todivide again, and we say: 4 into 39 will go 9 times. Whereto place the 9? We examine the divisor with the first numberin the dividend. Our divisor 4 is larger than the 3 in the divi-dend. Since it is larger, we will write the quotient digit, 9,immediately to the left of the 3 in the dividend. Now, with theright hand on the 9, we will say: 9 x 4 is three six or 36. As


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we say three, the right hand leaves the 9, moves to the nextcolumn where there is a 3, and subtracts it. We move tothe right and say 6. The right hand is now where the 9 isin the dividend, and we subtract 6. This leaves 3 and weexamine the dividend and see that we have 3Z. We say:4 into 32 goes 8 times. Since 4, the divisor, is larger thanthe 3 in the dividend, we will write it immediately to the leftof the 3. Now, with the right hand on the 8 and the left handimmediately to the left of it, we say: 8 x 4 is three two or32. We move to the right and the right hand falls on the 3of the 32. As we say three, we clear it. Then we move theright hand to the 2, say two and clear it.

We are now ready to read the answer. We go tothe extreme right of the abacus and count off one column forthe digit in the divisor, and one for the abacus. Everythingto the left of that is quotient. Our quotient is 198, We cannow check the division by multiplying the divisor by the quo-tient. We place the right hand on the 8 and the left on the 9We say: 4 x 8 is three two or 32. We move to the rightand set the 3 in the tens place. We then move to the rightand write 2 in the ones place. We pick up the right hand andclear the 8, We place the right hand on the 9 and say: 4x9


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is three six or 36. As we say three, we put the 3 to theright of the 9, then move our hand to the right and set the6 where the 3 is in the tens column. We pick up the righthand and clear the 9- Then we place the right hand on the 1and say: 1 x 4 is zero four. As we say zero, the handmoves to the right and touches the column next to the 1to record the zero. We move on to the right and the righthand sets the 4 with the 3, making it 7. We pick up the righthand and clear the 1. We now have 792. This is what wehad in the dividend when we first started to divide 4 into 792.

It is a good policy to do the same problem over andover until the process is well established. It is also goodpractice to divide and then check it through by multiplica-tion. In this way we have the practice of dividing, subtrac-ting, adding and multiplying.

Exercises:255 f 5 = 357 * 7 =8199 t 9 = 2112 t 3 =2418 t 6 = 896 t 6 =3936 + 8 = 812 * 4 =9876 - 6 = 2348 s 9 =


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LONG DIVISION

Long division is the process by which a dividendmade up of any number of digits is divided by a divisor madeup of two or more digits. We set it up on the abacus in thesame manner as we set up the short division example.

Let us take an example with a divisor of 45 and thedividend of 3 015. We write our divisor to the extreme left.We write our dividend 3015 to the extreme right.

When we divide with more than one digit, we lookfor a trial divisor. Since 45 is close to 50 we say that ourtrial divisor should be 5 . I have found that when dividingfrom 43 to 49 it is a good idea to use 5 as a trial divisor.

We are now ready to check and see how many timesthe trial divisor, 5, will go into the first two numbers of thedividend. Since it would not go into the first digit of 3, wesay that 5 will go into 30 six times. We check the firstdigit of our divisor with the first digit of the dividend and wefind that the 4 in 45 is larger than the 3 in 30. So we writethe 6 immediately to the left of the 3 in 30. With our righthand on the 6 we say: 6 x 4 is two four. As we say two, ourright hand moves to the right and we subtract the 2 from the 3.


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Then our hands move to the right and as we say four ourright hand touches the column with a zero. Since we cannottake 4 from zero, we subtract a 10 with the left hand. Wewanted to subtract only 4 but the left hand cleared a 10, sothe right hand will return 6. Our right hand remains onthe 6 and the left hand is on the zero to the left of it, and

we say: 6 of the quotient times the second digit in thedivisor, which is 5, is three zero. As we say three, theright hand, which is on the 6 will subtract a 3 by clearing 5and setting 2. Then both hands move to the right as we sayzero. As we say zero, the right hand, which is now on the 1,records the zero by pressing the rod. We are now ready todivide again. We take the trial divisor, 5, into 31 and wesee that 5 will go into 31 six times. Since the 4 in the divi-sor is greater than the 3 of the 31 in the dividend, we writethe 6 immediately to the left of the 3 of the 31 . And with ourright hand on the 6 we say: 6 times the first digit in thedivisor, 4, is two four. As we say two, the right handslides over to the 3 and we subtract 2. Then both handsmove to the right as we say 4; and since the right hand cannotsubtract 4 from 1, the left hand assists by clearing a 10.Now we wanted only 4 and the left hand cleared 10. That


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means we took away 6 too many; so the right hand will returna 6. With our right hand on the 7 and the left hand on thezero to the left of it, we say: by going back to the last digitwe have in our quotient 6 times the 5 in the divisor is threezero. As we say three we subtract it from the 7 by clearing5 and setting 2. "We move to the right and with the righthand record the subtraction of the zero by pressing thecolumn where the 5 is. We are now ready to divide again.We look at our divisor and we examine our dividend. Wesee that we have 45 left. Then we say that 45 will go into45 one time. Since the 4 in the 45 is the same as the 4 inthe divisor and the second digit in each are the same, wemust skip a space from the 4 in the dividend and set the 1with the 6. We now multiply again, the 1x4 giving uszero four, the right hand moving to the right and the leftfollowing. We say zero, then move to the right and sayfour. As we say four, we subtract the 4. We move bothhands to the right. As the right hand rests in the columnwhere we cleared the 4, we say: 1 x 5 is zero 5. As we sayzero five, we move to the right and when we reach the 5 weclear it.

We are now ready to read the answer. In order to


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read our answer we must figure out how many spaces wemust go in from right to left before we read it. First wecount the number of digits in our divisor- -we have 2; thenwe add 1 column for the abacus-- 2 and 1 is 3. So we willcount in 3 places from right to left. Everything after that isquotient. Our quotient is 67.

We can now check our answer by multiplying thequotient 67 by the divisor 45.

In examining how the example is set up at thismoment, we find that the quotient, 67, is located in theproper position for multiplication. The 6 of the 67 is in5 spaces or rods from right to left. We can check and seethat there are 2 columns for the multiplier, 2 for the multi-plicand and 1 for the abacus.

Every division example should be checked bymultiplication; and every example that is multiplied shouldbe checked by division.

These are excellent exercises and the studentrequires continual practice.

Let us work an example with a remainder. Thedivision is done in the same manner, but let us see where theremainder will be and how we should treat it.


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Let us take the simple example of 4 into 1234. Weset the 4 to the extreme left. TThen we set the 1234 to theextreme right. We say: there are 3 fours in 1 2 . Sincethe 1 in the dividend is smaller than the 4 in the divisor, weplace the 3 of our answer immediately to the left of the 1.Then we say: 3 x 4 is one two or 12. As we say one, wemove the hand to the right and subtract 1. When we say two,we move the right hand to the right and subtract the 2.Then we say: 4 into 34 will go 8 times. Since the 4 in ourdivisor is larger than the 3 in the dividend, we set the 8immediately to the left of the 3. We say: 8 x 4 is three twoor 32. We move the right hand to the right and subtract 3.Then we move the right hand to the right as we say two andsubtract 2 from 4. This leaves a 2 in the units column.Since 4 will not go into 2, the 2 is our remainder. We arenow ready to check our answer, or quotient. We count offone place for the divisor, one place for the abacus --thismakes 2 places from right to left. We read our answer. Ouranswer is 308 and a remainder of 2. Any figure that would bein the last columns would be remainders.

Now let us check our answer. We will multiply thequotient by the divisor. We will put our right hand on the 8


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answer in each part of the quotient, we correct our answer aswe go along.

Let us divide 225 by 26, Since 26 is greater than 22let us use 3 for a trial divisor- We compare 3 with the 22of 225 and find that it will go 7 times. Since 26 is greaterthan 22, we set the quotient 7 immediately to the left of the22. We multiply the estimated quotient figure, 7, times 26.Seven times 2 is 14; subtract the 14 from the 22; the differenceis 8. We now have 85 on the abacus. Seven times 6 is 42,and the difference is 43, We note that the remainder, 43, isgreater than the divisor, 26. Hence, our estimate of 7 istoo low. We will take our trial divisor, 3, and divide it into4. The answer will be 1 . Since 26 is smaller than 43, weskip 1 rod and add the 1 to the 7, making the quotient 8. Wemultiply the 1 just added times 26 and subtract this from 43,leaving us 17. Our quotient is 8 and 17 remaining.

In Mr. Gissoni's text, Using the Cranmer Abacus,he refers to this as "upward correction." He also discusses"downward correction" and how to work the examples andmake our corrections. I find that if a trial divisor is intro-duced early, it will not be necessary to make a downwardcorrection

.


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Examples:

444 ft 12 =56883 - 38 =33456 - 272 =4933 - 11 =25025 - 25 =

42

9734 *. 18 =73983 - 72 =63345 ft 123 =27615 - 95 =9826 t 64 =


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ADDITION OF DECIMALSThe fundamentals of adding and subtracting decimals

are the same as those in the addition and subtraction ofwhole numbers. The comma or "unit mark" is used as adecimal point in both addition and subtraction of decimals.

When we have one, two, or three decimal places inthe example, we use the first decimal point from right toleft. Should there be more than 3 places in the decimal, wemust use the second decimal point from right to left. Thatbecomes our decimal point for that particular example.

For example, let us add 3. 1 plus 2. 04 plus 2, 005,Since the smallest decimal is thousandths, we need 3 decimalplaces. Therefore, we will put our first whole number tothe left of the first; decimal point. Going from right to left,we find our first decimal place and set the whole number, 3,immediately to the left of it. Immediately to the right of thedecimal point we will set our 1 tenth. We are now ready toadd 2. 04. We add the whole number, 2, to the 3 which is tothe left of the first decimal. This gives us 5. Next we placethe 4 hundredths in the hundredths place which is to the rightof the tenths place. This is 2 places to the right of the


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decimal place.

Now we are ready to add 2. 005. We add 2 to thewhole number 5. T hen we add . 005 by setting the 5 in thethousandths place which is 3 places to the right of the deci-mal point. Our answer is 7. 145.

Let us now look at an example with more than 3decimal places, i.e. , more than 3 digits in the decimal.

Example; 4. 0005 + 2. 006 + 1 . 000002 + 3 . 05 +10. 3 + 1. 00002.

Since there are more than 3 digits in some of ourdecimals; we must use the second decimal point, or unitmark, from right to left as our decimal point in this parti-cular example.

First we set the whole number 4 to the left of thesecond decimal point(from right to left) . Since 5 ten thou-sandths requires 4 places after the decimal point, we touchthe first three columns for the zeros, and then we set the 5.We then add 2. 006 by adding 2 to the whole number 4, makingthe whole number 6. We now go in 2 places for the zerosafter the decimal point and write the 6 for 6 thousandths. Weadd 2 and 2 millionths by adding the whole number 2 to the 6,making it 8; and the 2 millionths is set in the sixth place


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after the decimal point. We now add 3, 05 by adding the 3 tothe 8, making it 1 1 , and setting the 5 hundredths in the hun-dredths place. Now add 10.3 to the 11, making it 21 in thewhole number and set 3 tenths in the tenths, place . Then add1 and 2 hundred thousandths by adding 1 to the 21, making it22, and set the 2 in the hundred thousandths place, which is5 places after the decimal. Our answer is 22,356522.

Examples:3. 86 + 2. 06 = 6.446 + 6. 002 =4. 09 + 6.29 = 7.678 + 9. 085 =3.2+9-934= 2.3367+15.00245=67. 5932 + 84. 000945 = 7, 003456 +15. 3274 =69-93574 + 182.37095 = 21.7765 + 19-329 =


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SUBTRACTION OF DECIMALS

We have already mentioned that the subtraction ofdecimals is done in the same manner as the subtraction ofwhole numbers. However, we must determine where to placethe decimal. We follow the same procedure as we did inaddition of decimals.

Let us examine a subtraction example: 5. 354 - 2. 13Z.Going from right to left, we set the whole number 5 to the leftof the first decimal point. Then set 354 thousandths to theright of the whole number. We are now ready to subtract2. 132.

From the whole number 5 subtract the whole number2. This leaves us 3. From 3 tenths subtract 1 tenth and thisleaves us with 2 tenths. From 5 hundredths subtract 3 hun-dredths and that gives us 2 hundredths. From 4 thousandthssubtract 2 thousandths and we have 2 thousandths. Our answeris 3.222.

Let us now work a subtraction example that has morethan 3 places in the decimal: 7. 674598 - 5. 342165.

The whole number 7 is set to the left of the seconddecimal place from right to left. The next six digits in thedecimal are written to the right of the whole number 7. From


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the whole number 7 subtract 5, leaving 2. From 6 tenthssubtract 3 tenths, leaving 3 tenths. From 7 hundredths sub-tract 4 hundredths, leaving 3 hundredths. From 4 thousandthssubtract 2 thousandths, leaving 2 thousandths. From 5 tenthousandths subtract 1 ten thousandth leaving 4 ten thousandths.From the 9 hundred thousandths subtract 6 one hundred thou-sandths leaving 3 one hundred thousandths. From 8 millionthssubtract 5 millionths, leaving 3 millionths. Our answer is2.332433.

Indirect addition and subtraction examples are done inthe same manner with decimals as they are done with whole

numbers .Examples:

6.42 - 4. 01 = 9- 173 - 4. 152 =67.081-42.931= 78.403-57.132 =89-573 - 48.374 = 62.7193 - 54.0347 =

49-2339 - 19-8745 = 123.73092 - 54.39705 =784. 03964 - 522. 7432 = 548. 85357 - 528 78538 =


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MULTIPLICATION OF DECIMALS

In the multiplication of decimals, we set the multi-plier and multiplicand in the same manner on the abacus aswe set whole numbers when we learned the process of multi-plication. After we have completed our multiplication, wepoint off as many places in our product as there were placesin the multiplier and multiplicand combined. In other words, weadd up all the decimal places in the multiplier and multipli-cand and then point off that many places in our product bycounting from right to left.

For example, let us multiply 5 tenths by 37 hundredths.We will set our 5 to the extreme left. Then we count in 4 rodsfrom right to left and set the 37. Now we are ready to multiply.

With our right hand on the 7 and the left hand on the 3,we say: 5 x 7 is three five. Set the 3 as we move to theright, and then set the 5 as we move to the right. We pick upour right hand and clear the 7. We put our right hand on the3 and say: 5 x 3 is one five. Move the right hand to the right,followed by the left hand, and as we say one, set it with theright hand. As we move to the right and say five, set the 5 withthe 3. Pick up the right hand and clear the 3. Our product is


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185. Since we had one decimal place in the multiplier and two

decimal places in the multiplicand, we add them together andget three. Then we must point off 3 places in our productfrom right to left. Our answer is .185.

Let us multiply 2 tenths by 4 tenths. We set the 2to the extreme left. Then in order to know where to set the 4,

we count one place for the 1 digit in the multiplier one for the1 digit in the multiplicand, and 1 place for the abacus. Thatgives us 3 places. We count in 3 columns from right to leftand set the 4. With our right hand on the 4 we say: 2 x 4 iszero eight. As we say the zero, we move our hands to theright and press the bar to the right of the 4 and move bothhands to the right as we say eight and set the 8. Then clearthe 4. Since there was one decimal place in the multiplier andone decimal place in the multiplicand, we will point off two deci-mal places. Thus our product is . 08.

Examples:.4x9-7= 63.2 x 14.43 =.6x5.32= 2. 71 x 82. 3 =7.2x9-85= 8.6 x 54. 35 =.24 x 9- 38 = 7. 65 x 9- 326 =


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DIVISION OF DECIMALS

In our practice with division of decimals on the abacus,we treat the division as if we were working, with whole num-bers. Then we subtract the number of decimal places in thedivisor from the number of decimal places in the dividend.The difference indicates the number of places to point off inthe quotient.

However, if we do not have any decimal places in thedividend and we do have one or more in the divisor, then addas many zeros to the dividend as there are decimal places inthe divisor and divide as in working with whole numbers.Since we have changed both divisor and dividend to wholenumbers, there are no places to point off.

Let us take the example .4 into 8.4. We will set our4 to the extreme left. Our 84 is written to the extreme right.Four into 8 goes 2 times. Since the 4 in the divisor is smallerthan the 8 in the dividend, we will skip a rod, and write the 2.Then with our right hand on the 2, we say: 2 x 4 is zero eight.As we say zero we move to the right, press the column, moveon tp the right and subtract the 8.

We then say: 4 will go into 4 once. Since the 4 in the


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divisor is equal to the 4 in the dividend, we will skip a columnplacing the 1 immediately to the right of the 2 in the answer orquotient. We say: 1 x 4 is zero four. We move our hands tothe right, touch the column as we say zero and subtract the 4when we say four. Since there is one digit in the divisor andone for the abacus, we count in two columns from right toleft. Everything to the left of this is the answer.

Since we had one decimal place in our divisor, andone decimal place in our dividend, we subtract. 1 from 1 andwe get zero. We do not point off any places in our answer.Our answer or quotient is 21.

Let us use the example .3 into 96. Since we have onedecimal place in the divisor and no places in the dividend,we will add as many zeros in our dividend as there are deci-mal places in the divisor. There is one decimal place in thedivisor, so the divisor becomes a whole number and our 96becomes 960. This done by multiplying both divisor and divi-dend by 1 when there is one place in the divisor. If there aretwo places in the divisor, then we multiply both divisor and divi-dend by 100. This clears the decimal in the divisor and gives2 zeros in the dividend. Hence, there are no decimal placesin the quotient.


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We divide 3 into 9 and it will go 3 times. Since the 3in our divisor is smaller than the 9 in the dividend, we willskip a column and place the 3 and say; 3x3 is zero nine,moving to the right, touching the column for the zero, movingon to the right and subtracting the 9 as we say nine. Then wesay: 3 will go into 6 how many times? It will go twice. Hereagain the 3 is smaller than the 6 in the dividend. So we skip acolumn and place the 2 to the right of the 3 in the answer. Andwith the right hand on the 2, we say; 2 x 3 is zero six., movingthe hands in the proper manner and subtracting the 6. Sincethere is one digit in the divisor and one for the abacus, wecount in two places from right to left, and everything to the leftof these 2 columns is the answer. Our answer is 320, Thereare no decimal places in our quotient.

Fred Gissoni, in his text, Using the Cranmer Abacus,describes a completely different method for locating the deci-mals in both the product and the quotient, The writer hopesthat when the contents of this manual are fully examined byinterested people in the field, we may perfect a more simplifiedmethod of locating the decimal point.

Exercises:.205 * .5 = .189-* 3 =


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.279 r .9 =25.25 * 2.5 =6.66 t 1.8 =43.56 * .71 =

53

. 330 * 7 =

96.3 t 3 -370. 37 r .11 ^27.24 * .32 =


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FRACTIONS

A fraction is a numeral such as 1/2 or 3/4. In theintroduction of fractions we must learn to distinguish betweenthe numerator and the denominator. For example, in the com-mon fraction 1/2, 1 is the numerator and 2 is the denominator.The numerator 1 indicates the number of fractional unitstaken. The denominator 2 tells into how many equal partsthe unit is supposed to be divided.

The whole number 1 may take many forms. Forexample, 2/2 = 1 / 1 = 1 ; 3/ 3 = 1 / 1 = 1 ; 4/4 = 1/1 = 1, etc. It

may have many names. If you divide a pie into 6 parts andkeep all 6, you will have 6/6 of your pie, or all of it. If youdivide it into fourths and keep all 4, you will have 4/4 or allyour pie. Nine-ninths, 2/2, 3/3, 100/100 are all names for1 because we have all the parts into which the number is di-vided.

Any number divided by 1 or multiplied by 1 keeps itsown value. Seven t 1 is 7; 3 x 1 is 3; 1/2 x 1 is 1/2. Thiswill be true no matter what name is used for 1 .

Let us check this on the abacus. We set the numera-tor 2 to the extreme left, skip two rods and set the denomina-


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tor 2. In order to reduce 2/2 to its simplest form, we mustfirst find the greatest common factor of the numerator and thedenominator. The factor common to 2, the numerator, and to2, the denominator, is 2. We divide the common factor 2into the numerator 2 and change it to 1; we divide the commonfactor 2 into the denominator 2 and change it to 1. We have 1/1which equals the whole number 1 .

Let us find the simplest name for 4/8. Set thenumerator 4 to the extreme left, skip two rods and set thedenominator 8. What is the largest common factor of 4 and8? It is 4. Dividing 4 into the numerator 4 is 1 . We changethe 4 to 1 . Dividing 4 into the denominator 8 gives us 2. Wechange the 8 to 2. Thus, the simplest form is 1/2. This pro-cess is commonly referred to as the reduction of fractions.

Examples:3/6 = 5/10 =

7/14 = 10/16 =3/9 = 4/12 -6/15 = 8/14 =2/18 = 7/21 =


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ADDITION OF FRACTIONS

In the addition and subtraction of fractions , we dividethe abacus into three sections. One section is for the wholenumber, the second section is for the numerator, and thethird section is for the denominator.

Since a common denominator is necessary in the addi-tion and subtraction of fractions, we shall first determinewhat it will be, and then place it in the denominator section.The common denominator is placed to the extreme right.The whole number is placed to the left of the second comma,or the second unit marker from right to left on the abacus.If there is one digit in the whole number, it is placed on theseventh column from the right.

The numerator is placed to the left of the first comma,or unit marker (from right to left). The denominator, asmentioned before, is placed to the extreme right. If the de-nominators should be 3 digits or more, a system must beworked out whereby the three sections are moved to the leftand arranged in such a manner that they do not run into oneanother.

Since we generally need a common denominator in


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the addition of fractions, we shall get the common denominatorby multiplying the two denominators together. Then placethe common denominator in the denominator section of theabacus. We are now ready to work an example: 4 5/6 f5 7/9.

We must first find the lowest common denominator.One way to do this is to take the largest denominator, which is9, and multiply it by 2 . Two x 9 is 18. Then we check to seeif the other denominator 6 will go into 18. If it does, we willuse the 18; if nbt, we would multiply the 9 by the next numberhigher than 2, which is 3, etc (Many times we multiply thetwo denominators and thereby find the common denominator.)In this case it is 18; so we place the denominator 18 to theextreme right, which is the section for the denominator.

We are now ready to work the example . We take thewhole number 4 of the 4 5/6 and set it in the whole numbersection, which is to the left of the second unit marker orcomma (the 7th column) We are now ready to change the 5/6to eighteenths. In order to change 5/6 to 18ths 9 we set the 5to the extreme left, skip a space and set the 6. We divide thedenominator 6 into the common denominator 18 and we get 3,Then we multiply that 3 by the numerator 5 and we get 15.


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We set the numerator 15 in the numerator section of the abacus,which is to the left of the first unit marker or comma.

For the student who is learning for the first time howto change a fraction to a common denominator, the followingmethod might prove helpful. In this example we are changingthe 5/6 to 18ths.

The student sets the numerator 5 to the extremeleft; then he skips a space and sets the denominator 6. Withthe right hand on the denominator 6, he asks: 6 will go intothe common denominator 18 how many times? It will go 3times. He clears the denominator 6 and places his right hand

on the numerator 5 and says: 5 x 3 is 15. Then he clearsthat 5 and places the 1 5 in the numerator section. He nowhas the whole number 4, the numerator 15, and the denomina-tor 18.

We are ready to add the 5 7/9 We add the 5 to the4 in the whole number section. We now have the whole number9- We must change the 7/9 to I8ths. Nine into 18 will gotwice; Z x 7 is 14. We shall add 14 to our numerator in thenumerator section. We have 29 in the numerator. Weexamine our numerator and denominator and find that thenumerator is larger than the denominator, so, we divide the


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denominator into the numerator. Eighteen into 29 will goonce. Then we add the 1 to the whole number 9 and get a 10in the whole number section. We say: 1 x 18 is 18. Wesubtract the 1 from the 2 in the 29; and subtract the 8from the 9 in the 29- This leaves us 1 1 in the numerator.As we look at the three sections we read our answer:10 11/18.

Examples:3 1/3+4 5/6 = 4 1/5+2 3/5 =5 2/7+4 4/7 = 12 2/9 + 11 4/9 -6 3/8 + 7 7/8 = 5 2/3+4 5/6 =7 3/5+4 3/4 = 9 5/7 + 3 5/8 =8 2/5+3 4/9 = 11 5/6 + 7 4/9 =


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SUBTRACTION OF FRACTIONS

Subtraction examples are set up as in addition.Again, we divide the abacus into three sections- -whole num-bers to the left of the second unit marker; numerator to theleft of the first unit marker; and the common denominator tothe extreme right.

Let us work the example; 7 5/8 - 5 2/3.First we find the common denominator by multiplying

the two denominators 8 and 3. Eight x 3 is 24. "We will setthe 24 to the extreme right. We are now ready to write 7 5/8We will set the 7 in the whole number section which is tothe left of the second unit marker (from right to left). Thenwe change our 5/8 to the common denominator 24.

To change 5/8 to the common denominator 24, setthe 5 of the 5/8 to the extreme left. Skip a space and set the8. With the right hand on the 8 of the 5/8 say: 8 into 24 goes3 times . Clear the 8 and put the right hand on the 5 and say:5 x 3 is 15. Clear the 5 and set a 15 to the extreme left.Now we have changed our 5/8 to 15/24. Clear the 15 and setthe 15 in the numerator place which is to the left of the first

comma.


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We are now ready to subtract the whole number 5from the 7 which leaves 2. We then subtract the 2/3 from the15/24. However, since we cannot subtract thirds from 24ths,we will change the 2/3 to 24ths . Set the 2 to the extreme left.Skip a space and set the 3 , With the right hand on the 3 say:3 into 24 goes 8 times. Clear the 3 and say 8 x the numerator2 is 16. Clear the 2 and set the 16 to the extreme left. Wenow have 16/24, We will subtract the 16/24 from 15/24.

Since we cannot take 16 from 15, we will have toborrow a whole one from the whole number 2., which "willleave us a 1 in the whole number section. Since the wholenumber 1 has 24/24, we will add that tothe 15/24 and thisgives us 39/24. We are now able to subtract the 16 in thenumerator from the 39, and we have 23/24, Our answer is1 23/24,

Addition and subtraction must first be done withsimple examples with common denominators. At this momentwe are demonstrating how to work with lowest common denomina-tors. In the beginning of fractions, one should work theexamples with direct addition and subtraction.

Examples:7 4/5-2 2/5 = 9 5/7-5 3/7 =


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15. 5)56 -4 1/6 =18 11/15 - 9 8/159 4/9-7 3/4 =17 7/8 - 11 2/3 =

62

23 7/9 - 12 5/9 =8 3/5-2 4/7 =14 1/2 - 7 8/9 =13 3/7 - 12 3/4 =r


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MULTIPLICATION OF FRACTIONS

In the multiplication of fractions, we again dividethe abacus into sections. We will place our numerators onthe left side of the abacus and our denominators will beplaced on the right.

Let us take the example 1/2 x 3/4, We will set thenumerator, 1, to the extreme left. Then place the denomina-tor, 2, to the extreme right. Since we are multiplying thatby 3/4, we will place the numerator, 3, two columns to theright of the first numerator, 1 . We will then place thedenominator, 4, two columns to the left of the denominator 2.

On the left side of the abacus we now have thenumerators 1 and 3; and on the right section we have thedenominators 4 and 2.

The student learns that in the multiplication of frac-tions we multiply numerator by numerator, and denominatorby denominator. We place our right hand on the first numera-tor 1 and say: 1x3 (the second numerator) is 3. As theright hand moves from the numerator 1 to the numerator 3,the left hand is placed on the 1, and as we say 3, the lefthand clears the 1 and the numerator 3 remains.


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Then we put our right hand on the denominator, 4,

and say: 4 x 2 is 8. As the right hand leaves the denomina-tor 4 and moves to the denominator 2, the left hand is placedon the denominator 4, and as we say 4 x 2 is 8, the left handclears the 4, and the right hand sets an 8 where the 2 was.Our answer is 3 in the numerator and 8 in the denominator,

or 3/8.Let us now take as an example 3 1/2 x 1 1/7, Three

and one -half is changed to an improper fraction, or 7/2. Weset the numerator 7 to the extreme left. We set. the denomina-tor 2 to the extreme right- Then we change 1 1/7 to 8/7.We skip two columns after the first numerator 7 and set thenumerator 8. Then we place the denominator 7 two columnsto the left of the denominator 2. We check to see if we cancancel.

This example calls for an introduction to cancella-tion. We check to see if there is a number that can be dividedevenly into one of the numerators and one of the denominators.We put our right hand on the first numerator, which is a 7.Then we place our left hand on that 7 ^yhile the right handchecks in the denominator section for a number that can bedivided by 7. The right hand sees another 7. Seven will go


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into 7 one time. The right hand changes that 7 to a 1. Theright hand returns to the left hand and we say: 7 into 7 goesonce; and change that 7 to a 1 .

The right hand checks the second numerator. It isan 8. The left hand is placed on the 3 while the right handmoves to check the denominators. The first denominator is

a 1 and the second denominator is a 2. What number will goevenly into 2 and 8? Two will go into 2 and 8. Two will gointo 2 once. The right hand changes the 2 to a 1 . Then theright hand returns to the 8 and we say: 2 into 8 goes 4 timesand we change the 8 to a 4. We are now ready to multiply.

The right hand is placed on the numerator 1 as wesay: 1 x 4 is 4. As the right hand leaves the numerator 1and moves to the numerator 4, the left hand is placed on thenumerator 1, and as we say 4 the left hand clears the 1 andthe 4 remains. Then we put our right hand on the first de-nominator and say: 1 x 1 is 1 . As the right hand leaves thedenominator 1 and moves to the second denominator _l theleft hand is placed on the first denominator 1 . As we say:1 x 1 is 1, the left hand clears the 1 and the right hand re-mains on the other 1 . Our answer is 4 in the numerator and1 in the denominator, or the whole number 4-


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Examples:1/3 x 4/5 - 2/5 x 5/6 =3/4 x 7/9 = 7/8 x 4/7 =3/8 x 4/9 - 2 1/2 x 4 1/3 =1 1/7x1 1/8 = 2 1/3x1 2/7 =3 3/4 x 2 1/5 - 11/6x2 2/5 =


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DIVISION OF FRACTIONS

Division, of fractions is done in fhe same manner asmultiplication. The only difference is that we invert the divi-sor and proceed as we did in multiplication of fractions 5 i.e. ,multiply numerator by numerator and denominator by deno-

minator .Since division of fractions is the inverse of the multi-

plication of fractions 3 we can refer to the division of fractionsas multiplying the dividend by the reciprocal of the divisor.The inverse of a number is a reciprocal of a number of anyfraction. For example., the reciprocal of 2/3 is 3/2. Werewrite each division example as a multiplication exampleand then find the answer. For example j let us divide 2/3 by1/5. Of the dividend 2/3, we set the numerator 2 to theextreme leftj and the denominator 3 to the extreme right.We then multiply this by the reciprocal of l/5 ; which is 5/1,Of the 5/l ; we set the numerator 5 two rods removed fromthe numerator 2 of the dividend The 1 of the denominator isset on the third rod to the left of the 3, We then proceed as inmultiplication of fractions, by multiplying the numerators(2 x 5) giving us 1 in the numerator, and the denominators


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(1x3) giving us 3 in the denominator. Our answer is 10/3.Since 10/3 is an improper fraction, we will divide the de-nominator 3 into the numerator 10, and our answer is 3 1/3,

Examples:1/2 * 3/4 = 4/5 * 1/2 =5/9 * 2/3 = 4/5 * 2/5 -1 1/2 * 1 1/3 = 2 1/2 * 5/9 =2 1/4*1 2/7 = 11/8* 1/4 =5 2/3*1 2/5 = 3 1/3 * 2. 1/2 .=


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PER CENTSince per cent means hundredths, it should be easy

for us to write per cents as fractions or decimals. If wewere to say 15%,yOf 125, we would change the 15% to 15/100or . 15 and multiply . 15 x 125/1. This would set up on theabacus in the same manner as we set up .15 x 125. We wouldset the 15 to the extreme left and on the sixth, fifth andfourth rods from the extreme right, we would set 125 andmultiply. The answer would be 1875. We then point off twoplaces in our answer since per cent is the same as hundredths

,

The answer is 18, 75.When the per cent has a simple fractional equivalent,

such as 25% =1/4, or 33 1/3% =1/3, it would be quicker torevert to the multiplication of fractions. For example:33 1/3% of 129- Instead of multiplying .33 1/3x129, wecould multiply 1/3 x 129- This would set up as follows: thenumerator 1 of the 1/3 would be placed to the extreme leftand the denominator 3 to the extreme right. We skip tworods from the numerator 1 and set the 129- The denominator1 (of the whole number 129) is placed two rods to the left ofthe denominator 3. We check for cancellation and find that


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3 will go into 129 forty-three times (changing the 129 to 43);

3 will go into the denominator 3 one time, changing the 3 toa 1 . We then multiply the numerator by the numerator andthe denominator by the denominator, and our answer is43/1 or 43.

Examples;

17% of 215 = 23% of 368 =9% of 420 = 33 1/3% of 912 =12.3% of 47 = 87 1/2% of 168 =6. 05% of 103 = 25% of 210 =5 0% of 1232 = 62 1/2% of 1 84 =


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SQUARE ROOT

We know that the square root of a number is one ofthe two equal factors of the number. The square root of16=4 since 4 x 4 is 16, or 4 squared = 16. Though there aremany tables of squares and square roots, they are not alwaysat our fingertips. In this manual we shall discover how toextract the square root of a number by division. This is donein much the same way as in long division, but the divisorchanges with each step. Instead of doubling the divisor, inthis manual we shall halve the dividend, thereby working with

smaller numbers in our dividend.Let us extract the square root of 1225. We will

move 6 places from right to left and set 1225. We separateour digits into groups oi two from right to left starting fromthe decimal point. We will have as many digits in the squareroot as we have groups in the original number. We here havetwo groups, 12 and 25. For the first trial divisor , take thesquare root of the largest perfect square which is less thanthe first group (12).

First we will extract the square root of the 12 andsay: what is the greatest square in 12? The greatest square


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in 12 is 9 and the square root of 9 is 3. We p

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