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BE A
Human
CALCULATOR
Mere Observation - No Magic
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BE A
Human
CALCULATOR
Mere Observation - No Magic
Notion Press
Old No. 38, New No. 6
McNichols Road, Chetpet
Chennai - 600 031
First Published by Notion Press 2016
Copyright © Rajesh Sarswat 2016
All Rights Reserved.
ISBN 978-93-5206-607-0
This book has been published with all efforts taken to make the material errorfree after the consent of the author. However, the author and the publisher do
not assume and hereby disclaim any liability to any party for any loss, damage, or
disruption caused by errors or omissions, whether such errors or omissions result
from negligence, accident, or any other cause.
No part of this book may be used, reproduced in any manner whatsoever without
written permission from the author, except in the case of brief quotations embodied
in critical articles and reviews.
CONTENTS
Preface�������������������������������������������������������������������������������������������������������������������������������� ix
Acknowledgments����������������������������������������������������������������������������������������������������������xiii
1.
2.
3.
4.
5.
6.
Some Basic Techniques��������������������������������������������������������������������������������������� 1
1.1 Digital root or digit sum����������������������������������������������������������������������������� 1
1.2 Bases and complements ��������������������������������������������������������������������������� 2
1.3 Vinculum��������������������������������������������������������������������������������������������������������� 3
Addition��������������������������������������������������������������������������������������������������������������������� 7
2.1 One line addition or row addition��������������������������������������������������������� 8
2.2 Marking the carry forward technique�������������������������������������������������� 8
2.3 Addition of decimal numbers ���������������������������������������������������������������10
Subtraction�������������������������������������������������������������������������������������������������������������13
3.1 One line subtraction���������������������������������������������������������������������������������13
3.2 Subtraction by adding compliments���������������������������������������������������13
3.3 Adding and subtracting together���������������������������������������������������������14
Multiplication �������������������������������������������������������������������������������������������������������17
4.1 Multiplication of two digit numbers���������������������������������������������������17
4.2 Multiplication of three digit numbers�������������������������������������������������20
4.3 Multiplication of two numbers with
unequal number of digits �����������������������������������������������������������������������23
4.4 General method of multiplication �������������������������������������������������������24
4.5 Multiplication near a base number�����������������������������������������������������26
4.6 Some special cases of multiplication���������������������������������������������������30
4.7 Multiplication of three numbers simultaneously���������������������������33
Divisibility��������������������������������������������������������������������������������������������������������������37
5.1 Divisibility tests of 2, 4, 8, 16 etc.���������������������������������������������������������37
5.2 Divisibility tests of 5, 25, 125, 625 etc.�����������������������������������������������39
5.3 Divisibility tests of 3 and 9���������������������������������������������������������������������40
5.4 Divisibility test of 11���������������������������������������������������������������������������������41
5.5 Divisibility check for Prime Numbers�������������������������������������������������43
5.6 Divisibility check for Composite Numbers���������������������������������������46
Percentage�������������������������������������������������������������������������������������������������������������49
6.1 Equivalent fractions ���������������������������������������������������������������������������������49
6.2 Finding quicker percentage �������������������������������������������������������������������50
6.3 Flipping the Number Technique�����������������������������������������������������������51
Contents
7.
8.
9.
10.
11.
12.
13.
14.
Division�����������������������������������������������������������������������������������������������������������������53
7.1 Division by using vinculum �����������������������������������������������������������������53
7.2 Division by using complements���������������������������������������������������������55
7.3 Modified cancellation Technique�������������������������������������������������������57
Squaring Techniques ���������������������������������������������������������������������������������������59
8.1 Squaring of numbers ending with 5�������������������������������������������������59
8.2 Squaring of numbers from 51 to 59�������������������������������������������������60
8.3 General Method of Squaring a two digit number�������������������������61
8.4 Squaring near a base number �����������������������������������������������������������62
8.5 General Method of Squaring of any number ���������������������������������63
Cubing Techniques �������������������������������������������������������������������������������������������67
9.1 Cubing a two digit number�������������������������������������������������������������������67
9.2 Cubing near a base number�����������������������������������������������������������������69
Square Root���������������������������������������������������������������������������������������������������������73
10.1 S quare root of Perfect Square Numbers consisting
of 3 or 4 digits (By Inspection) �������������������������������������������������������73
10.2 S quare root of Perfect Square Numbers consisting
of 5 digits (By Inspection) �����������������������������������������������������������������76
10.3 Approximate square roots of non-perfect squares�������������������79
Cube Root�������������������������������������������������������������������������������������������������������������83
11.1 Cube Root of Perfect Cubes up-to 6 digit numbers�������������������84
11.2 Cube Root of Perfect Cubes of 7 to 9 digit numbers�����������������89
11.3 Finding approximate cube roots of non-perfect cubes�����������93
Fractions and Decimals�����������������������������������������������������������������������������������97
12.1 Addition of mixed fractions���������������������������������������������������������������95
12.2 Subtraction of mixed fractions���������������������������������������������������������95
12.3 Multiplication/Square of Mixed Fractions ���������������������������������95
12.3 Converting recurring decimals into fractions�����������������������������96
LCM and HCF����������������������������������������������������������������������������������������������������101
13.1 Finding LCM and HCF of two numbers��������������������������������������101
Checking Your Answer����������������������������������������������������������������������������������103
Algebra������������������������������������������������������������������������������������������������������107
15.
Long Division or Synthetic Division��������������������������������������������������������109
15.1 Long division ��������������������������������������������������������������������������������������109
15.2 S ynthetic division of a polynomial by a linear
polynomial of the type x ± a ����������������������������������������������������������109
15.3 S ynthetic division of a polynomial by a linear
polynomial of the type ax ± b��������������������������������������������������������111
15.4 S ynthetic division by a polynomial of degree two
and above����������������������������������������������������������������������������������������������112
vi
Contents
16.
17.
Factorization of Polynomials����������������������������������������������������������������������115
16.1 Factorization of quadratic polynomials by inspection ��������115
16.2 Factorization of cubic polynomials��������������������������������������������116
Solving Equations������������������������������������������������������������������������������������������121
17.1 Linear equations in two variables������������������������������������������������121
17.2 Linear equations in three variables��������������������������������������������124
17.3 Quadratic equations��������������������������������������������������������������������������128
17.4 Solving equations of degree more than 2����������������������������������131
17.5 Some special equations��������������������������������������������������������������������131
(a) Equations of the form ����������������������������������������������������������������������131
x+ y =a
x− y =b
(b) Equations of the form ����������������������������������������������������������������������132
ax + by =
c1
bx + ay =
c2
2
(c) Equations of the form: ��������������������������������������������������������������������132
( x + a ).( x + b) = ( x + c).( x + d ), where ab = cd
(d) Equations of the form: ��������������������������������������������������������������������133
( x + a ).( x + b) = ( x + c).( x + d ), where ab ≠ cd
(e) Equations of the form: ��������������������������������������������������������������������133
1
1
+
=
0
ax + b cx + d
(f) Equations of the form: ����������������������������������������������������������������������134
ax + b ex + f
N2
1
=
⇒ N=
cx + d gx + h
D1 D 2
where N 1 + N 2 = D1 + D 2
(g) Equations of the form: ��������������������������������������������������������������������135
1
1
1
1
+
=
+
x+a x+b x+c x+d
Where, (x+a) + (x+b) = (x+c) + (x+d)
Answers����������������������������������������������������������������������������������������������������137
vii
Preface
The roots of this book could be traced to the year 1991 when I
skimmed through a small article in a vernacular newspaper on quick
mathematics. The article was based on a technique of multiplication
evolved by Jakow Trachtenberg, the famed Russian mathematician. The
article left a deep imprint in my mind, and I wondered as to whether there
were other such techniques of quick mathematics. This incident led me
to look for more insightful articles and papers. As I was preparing for
competitive examinations conducted by various public agencies which
almost always require consummate mathematical skills for success, I
found these techniques truly useful. Little did I realize at that time that my
incessant journey into the realm of quick mathematics had begun! I had
taken to a passion that would be in my psyche for all times to come.
These techniques worked very well, and I cleared a plethora of
competitive examinations. The more I practiced the techniques, the better
I got at them. Very soon I was able to appreciate the deeper and subtle
aspects of quick mathematics. I realized that speed at solving problems
directly influenced the appreciation of the problems. Moreover, through
appreciation of mathematical problems, every great mathematician
would tell you, is the key to solutions. I also deem it essential to mention
at this juncture that the faster you get at problem-solving, the greater is
your accuracy.
I have always had this conviction that only by spreading knowledge
does one expand one’s own sphere of knowledge. Holding on to one’s
ken as a well-guarded secret doesn’t help. Such an approach makes the
knowledge stagnant and often devoid of soul. So I was very happy to share
my techniques with my family and friends. Many of my friends actually
improved their grades in mathematics, and of course, many of them
cleared difficult competitive examinations. It was then that I decided to
write this book for the benefit of all. However, I resisted my impulse of
rushing through with the project. That’s because I wanted the book to be
truly useful to all kinds of students (even for those who hate mathematics!)
Preface
cutting across continents and nations. And in order to make the book
truly useful for all, it was imperative for me to conduct more exhaustive
research on quick mathematics and the educational patterns pertinent to
mathematics in various countries and cultures.
I am so happy that after years of research and writing, this book has
finally become ready. I must say in full humility that the exhaustive nature
of my research allowed me to innovate on many of the existing techniques
and even devise my own methods. I have presented the essence of all my
hard work in this book.
It will perhaps also not be out of place to mention that this is my third
academic book. While one of my earlier works deals with the conceptual
aspects of mathematics, the one that you are currently reading lays greater
emphasis on essential appreciation of problems and quick calculations.
While I do not guarantee miracles to the students willing to improve
their problem solving abilities, I can certainly assure better performance
and success in competitive examinations and of course better appreciation
of problems and their solutions if one works hard and perseveres with the
methods so as to eventually master them.
I have read scores of books that make similar promises, but to my utter
dismay I discovered that more often than not the authors of such books
had not tested the techniques, they propound and promote in real-life
problem solving situations. However, the techniques that you shall find
in this book have been tested and used (not only by me but by countless
other people) in examinations time and again. So I have no hesitation in
calling these techniques trustworthy and truly effective.
I have also very often come across techniques mentioned in other books
which are pretty impractical and sometimes completely unusable. I must
tell you again that this book is not a package of magic. It is rather a package
of methods that if practiced and persevered with can churn up magical
results! This book could be a great resource for various competitive
examinations and students in middle and senior school. It could help
the reader in myriad ways depending upon his or her needs and scope
for practice. At the same time one could figure out as to which technique
would work for one and which would not, again depending upon one’s
set of circumstances and needs. By reading this book, the students will be
able to:
x
Preface
•
•
•
learn quicker methods by observing some simple techniques;
compare various techniques available on each topic;
know the limitations of each technique;
• s ave some precious minutes in various competitive and school
examinations by employing the quick calculation techniques;
• develop their own tools in the field of quick calculations.
I have consciously avoided writing the lengthy and often dull derivations
of the formulas and methods that will help you improve your calculative
speed and accuracy. That’s because the presentation of those derivations
and lengthy explications would only add to the volume of the book and do
little else to improve your skills.
Now I offer you this book for your reading and mathematical pleasures!
I shall warmly welcome your comments on the work… even the bitterest
of criticisms would be accepted with the same spirit.
xi
Acknowledgements
To start with, I am thankful to my Late father Shri H.S. Sarswat and my
mother Smt Maya Sarswat because of whom I have been able to achieve
the things I have always aspired for.
My wife Swati Sarswat and son Aniket Sarswat have always been great
sources of inspiration and encouragement. I spent countless hours in
my study for research and writing which rightfully belonged to them.
However, I never heard even the mildest of complaints. In fact, their
excitement about a book on quick practical mathematics being written
catalyzed this work by a degree that I could never have achieved if I had
worked in isolation.
I am also thankful to my friends Virender Singh, Yogesh Sharma,
Vijayant Arora, Parveen, Tapas Dey, Pankaj Dhamija and Soumitra Basu for
their profound support to my endeavors. I have derived untold motivation
and encouragement from their ideas and words.
I also wish to extend my heartfelt thanks to my friend Biswajit Banerjee
for being a constant source of inspiration! Biswajit being a writer himself
often demonstrates how one should pursue one’s literary passion
regardless of the circumstances one finds oneself in.
We are a large family, and I get immense support from all my family
members in my ventures. I am thankful to my brothers, sisters, cousins
and all other loved ones for their indescribable contributions to my life.
My sister Rashmi Sarswat merits special mention for being a perennial
source of motivation. Her words and actions have always acted as infallible
pillars of strength.
Last but not the least, I would also like to extend my appreciation and
thanks to Shri Utkaarsh R. Tiwaari, my senior, for his timely support that
greatly facilitated the writing of the book.
Chapter
1
Some Basic Techniques
Before proceeding further, students should understand some powerful
techniques used in mental/faster calculations:
1.1 Digital Root or Digit Sum:
Digital root of a number is the sum of all the digits of a number, continued
until there is only one digit left.
Example 1:
Digital root of 123 = 1+2+3=6
Example 2:
Digital root of 563281 is 5+6+3+2+8+1=25 = 2+5=7
Example 3:
Digital root of 2637896 is 2+6+3+7+8+9+6 = 41 = 4+1=5
The process of finding digital roots can further be simplified by using a
faster method, known as “Casting out Nine Method”.
CASTING OUT NINE METHOD
In this method, we eliminate or leave 9 or group of digits whose sum is 9. The
digits left thereafter at the end are added up for finding the digital root.
Let’s understand this method through some examples:-
Example 4: Digital Root of 93 is 3 (cast out nine to get the left out digit
3 as the digital root.)
Example 5: Digital root of 732 is 3 (cast out 7+2 = 9, 3 is the digital root.)
Example 6: Digital root of 9 7 4 3 8 2 1 is 7 (cast out 9, 7+2 and 8 + 1 and
the left out numbers 4 &3 are added to get 7 as the digital root.)
Note:
(1) F
rom the above examples, it may be seen that the value of 9 may be
taken as 0 for the purpose of calculating digital root.
Be a Human Calculator
(2) T
he concept of digital roots can be used to check the correctness
of addition, subtraction, multiplication, division and lots more. We
will learn about the same in detail in chapter 14.
Exercise 1.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Find the digital root of the following numbers:
1.
213927
2.
3148653
3.
4.
4846471
5.
43471648
6.
7.
334325
8.
8856746
9.
10.
123456
523827
567324
678905
1.2 Bases and Complements:
Bases: The numbers starting with 1 and follwed by 0’s are called bases.
For example 10, 100, 1000, 10000, 100000, and so on are called bases.
Normally, these kinds of numbers are considered as bases for doing
faster/calculation in various methods and tricks.
Numbers such as 25, 50, 250, 500 etc., which are factors or multiples of
bases are called secondary bases.
Complements: The difference of the given number from its next nearest
base is called complement of the given number.
Example 1: Consider the number 156. The next nearest base of 156 is
1000. So, the complement of 156 is 1000 – 156 = 844.
Example 2: Consider the number 73. The next nearest base of 73 is 100.
So, the complement of 73 is 100 – 73 = 27.
Example 3: Complement of 6 is 4 (Base is 10).
Example 4: Complement of 25 is 75 (Base is 100).
Example 5: Complement of 6545 is 3455 (Base is 10000).
Finding Complement of a number:
Complement of a number can be calculated by subtracting all digits of the
number from 9 and the last digit by 10.
2
Be a Human Calculator
Example 6: Find the complement of 358732?
Solution: 9 9 9 9 9 10
– 35873 2
6 4 1 2 6 8
Example 7: Find the complement of 5183?
Solution:
9 9 9 10
– 518 3
4 8 1 7
Finding Complement of a number when its last digit is zero
Find complement of the number after leaving the last digit (that is zero)
by using the method explained above. Add zero to the answer to get the
complement of the given number.
Example 8: Find the complement of 69560670?
Solution: 9 9 9 9 9 9 10
– 6 9 5 6 0 6 7 0
3 0 4 3 9 3 3 0
Exercises 1.2
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Find the Complements of the following numbers
1.
234
2.
3457
3.
523827
4.
10910
5.
989
6.
167324
7.
21975
8.
885674
9.
678905
10.
990
1.3 Vinculum:
(i) T
he vinculum is one of the most powerful methods used in mental/
faster calculations as it allows us to remove some or all digits over five
from a calculation so that only 0, 1, 2, 3, 4 and 5 are used.
(ii) V
inculum Numbers are those numbers which have one or more digit
as negative (shown by using a bar over them). The vinculum is a
horizontal line written over a digit thereby making it negative.
3
Be a Human Calculator
Conversion of ordinary numbers into vinculum numbers
We convert any number into a vinculum number by converting the
digits which are above 5 to less than 5 without changing the value of
that number.
Step-I: Digit or digits to be converted are replaced by their complements
from next higher base;
Step-II: Increase the immediate left digit by one.
Example 1: 8 = 08 = 12 (as complement of 8 is 2)
(In 12, 1 is positive digit and 2 is negative digit, so the place value of 1 is
10 and place value of 2 is -2. Therefore 12 = 10 - 2 = 8).
Example 2: 288 = 312 (as complement of 88 is 12)
as 312 = 300 -12 = 288
Example 3: 174 = 234 (as complement of 7 is 3)
as 234 = 200 -30 + 4 = 174
Example 4: 589 = 611 (as complement of 89 is 11)
as 611 = 600 -11 = 589
Example 5: 284491 = 324511 (as complement of 9 and 8 are 1 & 2
respectively)
as 324511 = 300000 - 20000 + 4500 – 10 + 1 = 284491
Conversion of vinculum numbers into ordinary numbers
Step-I: Vinculum digit or digits are replaced by their complements from
next higher base;
Step-II: Decrease the immediate left number by one.
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
12
= 08
234
= 174 (as complement of 3 is 7)
312
611
(as complement of 2 is 8)
= 288 (as complement of 12 is 88)
= 589 (as complement of 11 is 89)
324511 = 284491
4
Be a Human Calculator
Example 6:
109
= 111 = 091
1009
= 1011 = 1191 = 0991
(Better if we do, 100 – 9 = 91)
Example 7:
(Better if we do, 1000 – 9 = 991)
Exercises 1.3 (a)
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Convert the following numbers into vinculum numbers by reducing digits
more than 5 to digits less than 5:
1.
4.
7.
10.
237
10917
21975
993
2.
5.
8.
3457
889
885674
3.
6.
9.
52382
167324
678905
Exercises 1.3 (b)
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Convert the following vinculum numbers into ordinary numbers:
1.
4.
7.
10.
237
1342
34795
98765
2.
5.
8.
5372
2131
21921
5
3.
6.
9.
1326
27801
1002
By the Same Author
Avoid Silly Mistakes in
Mathematics
Amazon bestseller in Counting and
Numeration Category at
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Chapter
2
Addition
Addition is the first arithmetical operation we learn, so for the students
with good mathematical skills, no short tricks are required for doing
a faster addition. However, for students having an average or weak
mathematical ability, some techniques are required, which may help
them in computing addition of numbers at an improved pace. Before
learning these techniques, students must ensure that they are thorough
in one digit addition and are able to add the following numbers within
a fraction of second:
1+1=2
2+2=4
3+3=6
4+4=8
1+3=4
3+1=4
2+4=6
4+2=6
3+5=8
5+3=8
4 + 6 = 10 5 + 7 = 12
6 + 4 = 10 7 + 5 = 12
1+2=3
2+1=3
1+4=5
4+1=5
1+5=6
5+1=6
1+6=7
6+1=7
1+7=8
7+1=8
1+8=9
8+1=9
1 + 9 = 10
9 + 1 =10
2+3=5
3+2=5
2+5=7
5+2=7
2+6=8
6+2=8
2+7=9
7+2=9
3+4=7
4+3=7
3+6=9
6+3=9
4+5=9
5+4=9
5 + 5 = 10
5 + 6 = 11
6 + 5 = 11
4 + 7 = 11 5 + 8 = 13
7 + 4 = 11 8 + 5 = 13
3 + 7 = 10 4 + 8 = 12 5 + 9 = 14
7 + 3 = 10 8 + 4 = 12 9 + 5 = 14
3 + 8 = 11 4 + 9 = 13
6+6 =12
8 + 3 = 11 9 + 4 =13
2 + 8 = 10 3 + 9 = 12
7 + 7 = 14
8 + 2 = 10 9 + 3 = 12
6 + 7 = 13
7 + 6 = 13
8 + 9 = 17
9 + 8 = 17
6 + 9 = 15
9 + 6 = 15
2 + 9 = 11
8 + 8 = 16
9 + 2 = 11
9 + 9 = 18
7 + 8 = 15
8 + 7 = 15
7 + 9 = 16
9 + 7 = 16
6 + 8 = 14
8 + 6 = 14
Be a Human Calculator
2.1 One Line Addition or Row Addition:
We start learning addition in various columns as follows:
2134
3455
+ 6666
12 2 5 5
It is only a matter of habit that we add numbers using columns as
we have been doing it for years after years. If a question appears in the
examination like:
215.49 + 345.76 + 2134.98 = ?
The first reaction of most of the students (if not of all) would be as
follows:
215.49
345.76
+ 2134.98
It is an interesting fact that you are going to get the same answer, if
you add the numbers after re-writing them in columns as per our habit or
write the sum directly in front of the question itself by using the normal
addition. It will save at least the time of re-writing the numbers into
columns. So our first trick is (which of course is a mere observation) that
whenever you get any question of addition in the examination, don’t rewrite it on paper in columns. Instead, you should write the answer directly
in front of the question itself. It may look like:
215.49 + 345.76 + 2134.98 = 2696.23
Our procedure will remain the same as we do in normal column
addition. This technique will also save some space on your examination
question booklet, which you can use for really lengthy questions.
Summary of the discussion is-Add the numbers in whatever formats
they are given in the question (without re-writing) to save time.
2.2 Marking the Carry Forward Technique:
(1) I n this technique, we never count beyond ten. As soon as the
accumulated sum crosses 10, we reduce it by 10 and move forward
with a reduced figure, and put an appropriate mark on the number
that made our total more than 10.
8
Be a Human Calculator
(2) W
hen we move to add ten’s digits, we count the marks made while
adding the unit’s digits. These marks will be added as carry forward
when we will add ten’s digits.
(3) R
epeat step -1 and step-2 till you get the answer.
Example 1: Let us illustrate the technique with an example
2134 + 3453 + 23456 + 1234 + 2345=?
We start adding unit’s digit from left to right:
4 + 3 = 7, 7 + 6 =13 (As the sum has crossed ten, we will put a cross mark
over 6 and reduce number by 10. Now we have 3 as the accumulated sum),
3 + 4 = 7, 7 + 5 = 12 (As the sum has again crossed ten, we will put a cross
mark over 5 and reduce number by 10. Now we have 2 as the accumulated
sum). We will put 2 as the unit’s place of the answer as follows:
2134 + 3453 + 23456 + 1234 + 2345 =
2
Now we may see that there are two cross marks at unit’s places. These
are “carry forward” for addition of digits at ten’s place. We will start with
carry forward as follows:
2 + 3 = 5, 5 + 5 = 10 (cross mark 5 and start with 0), 0 + 5 = 5, 5 + 3 =8,
8 + 4 = 12 (cross mark 4 and put 2 as the ten’s place of the answer
2134 + 3453 + 23456 + 1234 + 2345 =
22
Again carry forward is 2. We will start adding digits at hundredth place
as follows:
2 + 1 = 3, 3 + 4 = 7, 7 + 4 =11 (mark 4 and start with 1), 1 + 2 = 3,
3 + 3 = 6. Put 6 as the hundredthplace of the answer.
2134 + 3453 + 23456 + 1234 + 2345 =
622
Now, carry forward is 1. We will start adding digits at thousandth place
as follows:
1 + 2 = 3, 3 + 3 = 6, 6+3 =9, 9 + 1 = 10 (mark 1 and start with 0),
0 + 2 = 2. Put 2 as the thousandth place of the answer.
2134 + 3453 + 23456 + 1234 + 2345 =
2622
2134 + 3453 + 23456 + 1234 + 2345 =
32622
Carry forward 1 + 2 (only number at ten thousandth place) = 3.
Therefore, our answer is:
9
Be a Human Calculator
2.3 Addition of decimal numbers
Example 2: 215.491 + 345.76 + 2134.9823 =?
Solution:
Step-1: Make number of places after decimal equal by adding appropriate
number of zeroes
215.4910 + 345.7600 + 2134.9823
Step-2: Add the numbers as mentioned in article 2.2 to get,
215.4910 + 345.7600 + 2134.9823 = 2696.2333
Exercises 2
Total Questions 25
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 25 questions in 5 minutes)
Add the following numbers:
1. 23784 + 73423 + 76854 + 464564
2. 75767 + 3453 + 54646 + 46644
3. 423424 + 78 + 7878 + 3453535
4. 8688 + 7575 + 242342 + 5555353
5. 5535 + 535 + 654 + 96567 + 5354
6. 685787 + 74789 + 757 + 6656
7. 64644 + 553353 + 3486 + 54646
8. 56335.424 + 75757.32323
9. 56563.3323 + 5335.44 + 68.332
10. 6464.555 + 3535.5353 + 3355.97
11. 355.535 + 3535.898 + 586.36635
12. 996.3532 + 3663.90630 + 366.323
13. 7474.366 + 4774.36643 + 66464.3566
14. 3666.646 + 363.36464 + 363.3636
15. 3663.363 + 47.6898 + 588.47 + 34667.3437
10
Be a Human Calculator
16. 63673.688 + 36.3636 + 699.557 + 356.36
17. 965.46734 + 5858.558 + 699.6795 + 3526.36
18. 8548.54788 + 234663.3467 + 7.4856
19. 547.848 + 688956.89568 + 57.97976
20. 57754.7878 + 5484.6868 + 68865.9997
21. 373.48 + 5656857.568856 + 47.754757
22. 57756.577 + 547.78678 + 547.5478
23. 7676.788 + 68965.99 + 780567.4634
24. 5477.77 + 4778448.488 + 568568.56865
25. 3466736.6634 + 74754.5685 + 575.7647
11
By the Same Author
How to Memorize Formulas in
Mathematics
Book-1
Calculus
Amazon bestseller in Calculus
Category at
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Chapter
3
Subtraction
3.1 One Line Subtraction or Row Subtraction:
As explained in the chapter addition, students have to change their
habit of column subtraction to give way to One Line Subtraction or Row
subtraction in a normal fashion:
So, instead of going for:
76543
– 23456
Go for,
53087
76543 – 23456 = 53087
3.2 Subtraction by Adding Complements:
This is a powerful method, which uses addition of complements to do
subtraction and so in the process, we end up doing no subtraction for
getting the answer. The method has two simple steps:
Step-1: If the first digit is more than the second digit, simply
subtract it.
Step-2: If the first digit is less than the second digit, add the complement
of second digit to the first digit and increase the immediate left
digit of second number by one by putting a • over it.
Example 1: 76546 – 23453 =?
76546 – 23453 = 3
•
76546 – 23 4 53 = 93
(As 4 is less than 5, we will add the complement of 5 ( i.e. 5) to 4 to get 9
and will increase next digit 4 by 1 by putting a • over it).
•
•
76546 – 23 4 53 = 093 (5 - 4 = 5 - 5 = 0)
•
76546 – 23 4 53 = 53093 ( 6- 3 = 3 and 7 – 2 = 5)
Be a Human Calculator
Example 2: 2378452 – 634625 =?
•
2378452 – 6346 2 5
=
7
•
2378452 – 6346 2 5
=
27
As 2 < 5, add complement of 5 with 2
to get 7. Put • over next number 2.
•
•
As 5 > 2 , 5 - 2 = 5 - 3 = 2
•
•
•
•
As 4 < 6, complement of 6 i.e. 4 + 4
=8. Put • over next number 4.
•
•
As 8 > 4 , 8 - 4 = 8 - 5 = 3
•
•
2378452 – 63 4 6 2 5
=
827
2378452 – 63 4 6 2 5
=
3 8 27
•
2378452 – 63 4 6 2 5
=
43287
•
As 7 > 3, 7 - 3 = 4
2378452 – 63 4 6 2 5
= 2043827
3 - 3 = 0, 2-0=2
Note: Whenever, you add the complement of a number to the lesser
number, increase immediate left digit of the first number by 1 by putting
a • mark over it.
3.3 Adding and subtracting together:
As a matter of habit, whenever we came across a question in which we
have to perform addition and subtraction together; we try to do it in
two or more than two steps depending upon the qunatity of numbers
we have to add and subtract whereas the same can be performed by
doing a single row addition/subtraction together. In these questions,
whenever we have a number to carry over, we can use “Marking the
Carry Forward Technique” and whenever, we have to subtract the
bigger digit from a smaller digit we can use “Subtraction by Adding
Complements Technique” as explained above.
14
Be a Human Calculator
Let us understand the concept through an example:
Example 3: 214789 – 065466 + 5345345 = ?
214789 – 65466 + 5345345
=
8
9–6+5=8
214789 – 65466 + 5345345
=
668
7–4+3=6
214789–65466 + 5345345
=
68
8–6+4=6
214789 – 65466 + 5345345
=
4668
4–5+5=4
1-6 + 4 = 5-6, as we have to
•
2 14789– 65466 + 5345345 subtract larger digit from
=
9 4668
smaller one, we will add
complement of 6 i.e. 4 to 5 to
get 9 and will increase the left
digit 2 by putting a • over it.
•
•
2 14789– 65466 + 5345345
=
56 9 4 6 6 8
2 - 0 + 3 = 3 + 3 = 6 and 5 + 0
+0=5
Exercises 3.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Subtract the following numbers:
1. 23784 - 13423
2. 75767 - 3453
3. 423424 - 353535
4. 8242342 - 5555353
5. 96567 - 65354
15
Be a Human Calculator
6. 685787 - 74789
7. 553353 - 54646
8. 56335.424 - 7575.32323
9. 56563.3323 - 5335.44
10. 6464.555 - 3535.5353
Exercises 3.2 (2 Minutes)
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Subtract the questions given in Exercise 3.1 by adding the complements
method and check your answer.
Exercises 3.3 (2 Minutes)
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Solve:
1. 23784 - 73423 + 76854 + 464564
2. 75767 + 3453 - 54646 + 46644
3. 423424 - 78 - 7878 + 3453535
4. 8688 + 7575 - 242342 + 5555353
5. 5535 - 535 + 654 + 96567 - 5354
6. 685787 - 74789 - 757 + 6656
7. 64644 + 553353 - 3486 - 54646
8. 56335.424 + 75757.32323 – 312.23
9. 56563.3323 - 5335.44 + 68.332
10. 6464.555 - 3535.5353 + 3355.97
16
By the Same Author
How to Memorize Formulas in
Mathematics
Book-2
Trigonometry
Amazon bestseller in Trigonometry
Category at
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Chapter
4
Multiplication
4.1 Multiplication of two digit numbers
Students should observe to learn the working of this method.
A B
× C
D
A × C/A × D + B × C/B × D
Step - 1: Multiply the unit digits (rightmost digits) of both the numbers,
A
B
× C
D
//B × D
Step - 2: Add the cross product of the digits as shown below:
A
× C
B
D
/A × D + B × C/B × D
Step - 3: Multiply the ten’s digits (leftmost digits) of both the numbers.
A
× C
B
D
A × C/A × D + B × C/B × D
Note: If the results obtained in step 1 and step 2 have more than one digit,
note down the unit place of the result and carry over the ten’s place of the
result to the left. Let us understand the process through some examples:
Be a Human Calculator
Example 1: Solve 13 × 12
Step - 1: Multiply the unit digits (rightmost digits) of both the numbers,
1 3
× 1
2
// 3 × 2
Step - 2: Add the cross product of the digits as shown below:
1
× 1
3
2
/1 × 2 + 3 × 1/3 × 2
Step - 3: Multiply the ten’s digits (leftmost digits) of both the numbers.
1
× 1
3
2
1 × 1/1 × 2 + 3 × 1/3 × 2
So,13 × 12 = 156
Example 2: Solve 14 × 13.
Step - 1: Multiply the unit digits (rightmost digits) of both the numbers,
1 4
× 1 3
//4 × 3
Step - 2: Add the cross product of the digits as shown below:
1
× 1
4
3
/1 × 3 + 4 × 1/3 ×2
18
Be a Human Calculator
Step- 3: Multiply the ten’s digits (leftmost digits) of both the numbers.
1
× 1
4
3
1 × 1 /1 × 3 + 4 × 1/4 ×3= 1712
So, 14 × 13 = 182
Example 3: Solve 28 × 35.
Step- 1: Multiply the unit digits (rightmost digits) of both the numbers,
2 8
× 3 5
//8 × 5
Step - 2: Add the cross product of the digits as shown below:
2 8
× 3 5
/2 × 5 + 3 × 8/8 × 5
Step- 3: Multiply the ten’s digits (leftmost digits) of both the numbers.
2 8
× 3 5
2 × 3 /2 × 5 + 3 × 8 /8 × 5= 63440
So, 28 × 35 = 980
Note: After knowing the procedure correctly and after some practice,
students will be able to get the answer only in one step as follow:
Example 4: Solve 78 × 47
7
× 4
8
7
28 /81 /56 = 288156 = 3666
19
Be a Human Calculator
Exercises 4.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Multiply the following numbers:
1.
41 × 24
2.
3.
4.
5.
6.
7.
8.
9.
10.
78 × 56
91 × 27
89 × 74
76 × 54
65 × 87
76 × 43
69 × 31
83 × 42
99 × 43
4.2 Multiplication of three digit numbers
Observe the following steps carefully to learn the method:
A B C
× D E F
A × D/ A × E + B × D /A × F + C × D + B × E/B × F + C × E /C × F
Step - 1: Multiply the unit digits (rightmost digits) of both the numbers,
A
× D
B
E
C
F
/C × F
20
Be a Human Calculator
Step - 2: Add the cross product of the last two digits of both the numbers
as shown below:
A
B
× D E
C
F
/B × F + C × E/C × F
Step - 3: Add the cross product of the Extreme digits of both the numbers
and to the sum add the product of two middle digits as shown below:
A
B
× D
E
C
F
/A × F + C × D + B × E/B × F + C × E /C × F
Step - 4: Add the cross product of the two leftmost digits of both the
numbers as shown below:
A
× D
B
E
C
F
/A × E + B × D/ A × F + C × D + B × E /B × F + C × E /C × F
Step - 5: Multiply the hundredth’ digits (leftmost digits) of both the
numbers.
A
× D
B
E
C
F
A × D/A × E + B × D/ A × F + C × D + B × E /B × F + C × E /C × F
21
Be a Human Calculator
Example 5: Solve 123 × 321
Step - 1: Multiply the unit digits (rightmost digits) of both the numbers,
1
2
× 3
2
3
1
/3 × 1
Step - 2: Add the cross product of the last two digits of both the numbers
as shown below:
1 2 3
×3 2 1
/2 × 1 + 3 × 2/3 × 1
Step - 3: Add the cross product of the Extreme digits of both the numbers
and to the sum add the product of two middle digits as shown below:
1
× 3
2
2
3
1
/1 × 1 + 3 × 3 + 2 × 2/2 × 1 + 3 × 2/3 × 1
Step - 4: Add the cross product of the two leftmost digits of both the
numbers as shown below:
1 2 3
×3 2 1
/1 × 2 + 2 × 3 /1 × 1 + 3 × 3 + 2 × 2/2 × 1 + 3 × 2 /3 × 1
Step - 5: Multiply the hundredth’ digits (leftmost digits) of both the
numbers.
1 2 3
×3 2 1
1 × 3 /1 × 2 + 2 × 3 /1 × 1 + 3 × 3 + 2 × 2/2 × 1 + 3 × 2 /3 × 1
22
Be a Human Calculator
So, 123 × 321 = 381483 = 39483
Note: After knowing the procedure correctly and after some practice,
students will be able to get the answer only in one step as explained
in the next example:
Example 6: Solve 789 × 456
7
× 4
8
5
9
6
7 × 4 /7 × 5 + 8 × 4 /7 × 6 + 9 × 4 + 8 × 5/8 × 6 + 9 × 5/9 × 6
So, 789 × 456 = 28671189354 = 35 9 7 84
Exercises 4.2
Total Questions 10
Ideal Time 3 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 3 minutes)
Multiply the following numbers:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
412 × 241
781 × 561
913 × 273
891 × 743
763 × 541
651 × 873
764 × 433
691 × 316
831 × 423
991 × 433
4.3 Multiplication of two numbers with unequal number
of digits:
The procedure for multiplication of two numbers with unequal number
of digits is same as explained in previous pages. The only different thing
we have to do is to make digits of both the numbers equal by adding a
23
Be a Human Calculator
zero or zeroes prior to the number having lesser number of digits. See the
following example:
Example 7: Solve 789 × 56
7
× 0
8
5
9
6
7 × 0 /7 × 5 + 8 × 0 /7 × 6 + 9 × 0 + 8 × 5 /8 × 6 + 9 × 5 /9 × 6
So, 789 × 56 = 035829354 = 441 84
Exercises 4.3
Total Questions 4
Ideal Time 1 Min
(Students are advised to practice this exercise till they are able to
solve all 4 questions in 1 minute)
1.
2.
412 × 41
781 × 56
3.
789 × 54
4.
987 × 34
4.4 General Method of Multiplication:
The general method of multiplication may be understood with the help of
following diagram:
24
Be a Human Calculator
Exercises 4.4
Total Questions 20
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 20 questions in 5 minutes)
Question
Multiply
1
312 × 223
2
2341 × 2345
5
1234 × 4321
3
4
2314 × 456
45 × 67
6
12345 × 54321
9
4536 × 78
12
23 × 456
7
8
69 × 89
673 × 54
10
567 × 678
11
231 × 23
13
2345 × 567
16
4578 × 987
19
345 × 765
14
789 × 91
15
234 × 567
17
3456 × 098
18
345 × 78
20
34 × 78
25
Be a Human Calculator
4.5 Multiplication near a base number:
(a)
When both the numbers are near the same base and above
the base:
Step-1: Find the base near the given numbers and write down the excess
of numbers from the base in front of numbers as follows:
12
× 13
102
× 104
1008
× 1007
+2
+ 3 (base = 10)
+2
+ 4 (base = 100)
+8
+ 7 (base = 1000)
Step-2: The answer will have two parts. The rightmost part of the answer
will be the product of excess of two numbers from their base. The number
of digits in this part will be equal to number of zeroes in the base.
12
+2
× 13
+ 3 (base = 10)
/ 6 (one digit)
102
+2
× 104
+ 4 (base = 100)
/ 0 8 (two digits)
1008
+8
× 1007
+ 7 (base = 1000)
/ 0 5 6 (three digits)
Step-3: The left part of the answer will be obtained by adding the first
number with the excess of second number or by adding the second
number to the excess of first number.
12
+2
× 13
+3
(12+3) or(13+2)/ 6 = 156
102
+2
× 104
+4
(102 + 4) or(104 +2)/08= 10608
1008
+8
× 1007
+7
(1008 + 7) or (1007 + 8) / 0 5 6 = 101556
26
Be a Human Calculator
(b)
When both the numbers are near the same base and below the
base:
Step-1: Find the base near the given numbers and write down the deficiency of numbers from the base in front of numbers as follows:
98
× 97
-2
- 3 (base 100)
999
-1
× 991
- 9 (base = 1000)
Step-2: The answer will have two parts. The rightmost part of the answer
will be the product of deficiencies of two numbers from their base. The
number of digits in this part will be equal to number of zeroes in the
base.
98
× 97
999
× 991
-2
- 3 (base 100)
/06 (two digits)
-1
- 9 (base = 1000)
/009 (three digits)
Step-3: The left part of the answer will be obtained by subtracting deficiency of the first number from the second number or by subtracting
deficiency of the second number from the first number as follows:
98
-2
× 97
-3
(98-3) or ( 97-2)/06 = 9506
999
-1
× 991
-9
(999-9) or (991-1) /009 = 990009
(c) When both the numbers are near the same base, one is above the
base and another is below the base
98 - 2
× 107 +7
−−
−−
(98 +7) or (107-2)/ 1 4 = 105 1 4
27
Be a Human Calculator
−−
105 1 4 = 10486 (see chapter 1, how to convert vinculum number into
ordinary number).
(d) When both the numbers are near a secondary base
The method will be similar to method 4.5 (a), (b) and (c) with only one
difference. We will choose a base and a secondary base and find the
deficiency or excess of given numbers from the secondary base. However,
the number of digits in the rightmost part of the answer will be equal to
the number of zeroes in the base. Let us understand through an example
of multiplying 48 by 42.
Base = 100, Secondary Base = ½ × 100 = 50
48
-2
× 42
-8
(48-8) or (42-2)/16 = 40/16
Now as our secondary base = ½ × 100 = 50
The leftmost part of the answer will also be ½ × 40 = 20
Therefore, 48 × 42 = 2016.
(e) When both the numbers are near different bases
53
× 102
+3 (Base 50)
+2 (Base 100)
Here, the ratio of bases = 100/50 = 2
Multiply the first row by 2, we have:
106
+6 (Base 50)
× 102
+2 (Base 100)
(106 + 2) or (102 + 6)/12
Therefore, we have 10812,
As we had multiplied the first row by 2, we will divide the interim answer
by 2 to get 10812/2 = 5406 as the answer.
Therefore, 53 × 102 = 5406.
28
Be a Human Calculator
Exercises 4.5
Total Questions 20
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 20 questions in 5 minutes)
Question
Multiply
1
102 × 103
4
96 × 94
2
108 × 109
3
98 × 99
5
996 × 994
8
996 × 1008
11
1004 × 1002
14
1009 × 991
6
7
9
10
12
13
15
16
53 × 52
104 × 96
59 × 52
109 × 107
104 × 51
1005 × 997
502 × 501
499 × 497
17
1001 × 1002
20
103 × 93
18
19
57 × 59
1006 × 994
29
Be a Human Calculator
4.6 Some Special Cases of Multiplication:
(a) Multiplication by 11 to 19:
Multiplication by 11: Write down the number by prefixing 0 before it
and add every digit to its right digit to get the answer.
To get 123 × 11, write 123 as 0123
0123 × 11 = 0+1/1+2/2+3/3+0 = 1353
Multiplication by 12: Write down the number by prefixing 0 before it
and add twice of every digit to its right digit to get the answer.
0123 × 12 = 0 × 2+1/1 × 2 +2/2 × 2 +3/3 × 2 + 0 = 1476
Multiplication by 13: Write down the number by prefixing 0 before it
and add thrice of every digit to its right digit to get the answer.
0123 × 13 = 0 × 3+1/1 × 3 +2/2 × 3 +3/3 × 3 + 0 = 1599
Multiplication by 14: Write down the number by prefixing 0 before it
and add four times of every digit to its right digit to get the answer.
0123 × 14 = 0 × 4+1/1 × 4 +2/2 × 4 +3/3 × 4 + 0 = 161112 = 1722
Multiplication by 15: Write down the number by prefixing 0 before it
and add five times of every digit to its right digit to get the answer.
0123 × 15 = 0 × 5 +1/1 × 5 +2/2 × 5 +3/3 × 5 + 0 = 171315 = 1845
Multiplication by 16: Write down the number by prefixing 0 before it
and add six times of every digit to its right digit to get the answer.
0123 × 16 = 0 × 6 + 1/1 × 6 +2/2 × 6 +3/3 × 6 + 0 = 181518 = 1968
Multiplication by 17: Write down the number by prefixing 0 before it
and add seven times of every digit to its right digit to get the answer.
0123 × 17 = 0 × 7 + 1/1 × 7 +2/2 × 7 +3/3 × 7 + 0 = 191721 = 2091
Multiplication by 18: Write down the number by prefixing 0 before it
and add eight times of every digit to its right digit to get the answer.
0123 × 18 = 0 × 8 + 1/1 × 8 +2/2 × 8 +3/3 × 8 + 0 = 1101924 = 2214
30
Be a Human Calculator
Multiplication by 19: Write down the number by prefixing 0 before it
and add nine times of every digit to its right digit to get the answer.
0123 × 19 = 0 × 9 + 1/1 × 9 + 2/2 × 9 +3/3 × 9 + 0 = 1112127 = 2337
(b) Multiplication by 9, 99, 999 etc.
Add number of zeroes equal to number of 9’s in the multiplier to the given
number and subtract the given number to get the answer.
453 × 99 = 45300 – 453 = 44847
(We have added two zeroes after the number as we are multiplying by 99)
453 × 999 = 453000 – 453 = 452547
(We have added three zeroes after the number as we are multiplying
by 99)
453 × 9999 = 4530000 -453 = 4529547
(We have added four zeroes after the number as we are multiplying by 99)
(c) Product of numbers when the sum of the Last 1, last 2, last 3, last
4 - - - digits added respectively equal to 10, 100, 1000, 10000 and
the remaining part of both the numbers is same:
This method is used for calculating products like 98 × 92, 998 × 902, 9978
× 9022 etc.
Method:
Step 1: Divide the number into two different parts separating common
part of both the numbers with the remaining part of both the numbers
whose sum is 10, 100, 1000 etc.
Step 2: Rightmost part of the answer will be obtained by multiplying the
rightmost parts of the numbers whose sum is 10, 100, 1000 as the case
may be.
Step 3: Multiply the remaining part (which is same for both the numbers)
by its successor to get left part of the answer.
31
Be a Human Calculator
Example 1: Solve 98 × 92
Solution: Visualize the given product as 9/8 × 9/2,
Here left most part is same for both the numbers i.e. 9 and sum of
rightmost part is 8 + 2 = 10.
The rightmost part of the answer will be 8 × 2 = 16
Leftmost part of the answer will be 9 × 10 = 90
Therefore, the answer will be 9016.
Example 2: Solve 698 × 602
Solution: The rightmost part of the answer will be 98 × 02 = 196
Leftmost part of the answer will be 6 × 7 = 42
Therefore, the answer will be 42196.
Some more examples:
Question
Left part of
the answer
Right part of
the answer
Answer
97 × 93
9 × 10 = 90
7 × 3 = 21
(two digits)
9021
988 × 912
9 × 10 = 90
88 × 12 =1056 901056
(four digits)
91 × 99
891 × 809
9 × 10 = 90
1 × 9 = 09
(two digits)
8 × 9 = 72
9009
91 × 09 =0819 720819
(four digits)
32
Be a Human Calculator
Exercises 4.6
Total Questions 10
Ideal Time 3 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 3 minutes)
Question
Multiply
1
899 × 801
4
3456 × 999
2
3
94 × 96
1234 × 11
5
1099 × 1001
8
2345 × 17
6
7
9
10
797 × 703
4567 × 99
678 × 9999
8765 × 19
4.7 Multiplication of three numbers simultaneously:
(i) When all the numbers are above a particular base:
101
+1
× 103
+3
102
+2
Leftmost part of the answer
= 101 + 2 + 3 or 102 +1+3 or 103 +2+1 = 106
Middle Part of the answer
= 1 × 2 + 2 × 3 + 3 × 1 = 11
Rightmost part of the answer
= 1 × 2 × 3 = 06
33
Be a Human Calculator
(As all the three numbers are more than the same base 100, the number of
digits in the middle part and right part of the answer will be 2).
Therefore, the answer is 1061101
(ii) When all the numbers are below a particular base:
99
98
× 97
-1
-2
-3
Leftmost part of the answer
= 99 - 2 - 3 or 98-1-3 or 97-1-2 = 94
Middle Part of the answer
= (-1 × -2)+(-2 × -3) +(- 3 × -1) = 11
Rightmost part of the answer
= -1 × -2 × -3 = 06
(As all the three numbers are more than the same base 100, the number of
digits in the middle part and right part of the answer will be 2).
Therefore, the answer is 94/11/ 06 = 941094
(iii) When the numbers are below and above a particular base both:
102
98
× 97
+2
-2
-3
Leftmost part of the answer
= 102 - 2 - 3 or 98 +2-3 or 97 +2-2 = 97
Middle Part of the answer
= (2 × -2)+(-2 × -3) +(- 3 × 2) = 04
34
Be a Human Calculator
Rightmost part of the answer
= 2 × -2 × -3 = 12
(As all the three numbers are more than the same base 100, the number of
digits in the middle part and right part of the answer will be 2).
Therefore, the answer is 97/ 04 /12 = 969612
(iv) When the numbers are near different bases:
102
13
× 104
+2
+3
+4
(Base = 100)
(Base = 10)
(Base = 100)
102
+2
(Base = 100)
Multiply the middle number and its excess by 10 to have,
130
× 104
+30 (Base = 100)
+4 (Base = 100)
Leftmost part of the answer
= 102 + 30 +4 or 130 +2 +4 or 104 + 30 +2 = 136
Middle Part of the answer
= (2 × 30)+(30 × 4) +(4 × 2) = 188
Rightmost part of the answer
= 2 × 3 × 4 =24 (Note)
(However, as smaller base is 10 and bigger base is 100, the number of
digits in rightmost part will be one and number of digits in middle part
will be two).
Therefore, the answer is 136/188/24 = 137904
35
Be a Human Calculator
Exercises 4.7
Total Questions 10
Ideal Time 3 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 3 minutes)
Question
Multiply
1
101 × 102 × 103
2
1002 × 1004 × 1005
5
101 × 102 × 98
3
4
98 × 99 ×97
999 × 998 × 997
6
1002 × 1004 × 996
9
1004 × 1008 × 992
7
8
10
101 × 12 × 103
1004 × 1006 × 112
991 × 997 × 998
36
Chapter
5
Divisibility
5.1 Divisibility Tests of 2, 4, 8, 16 etc./ Divisibility test of
the numbers of the form 2n:
Number
Divisibility Test
2= 2
Last digit of the given number
should be divisible by 2 or
even.
1
4= 2
2
8= 2
3
Last two digits of the given
number should be divisible
by 4.
16 = 2
4
32 = 2
5
Last three digits of the given
number should be divisible
by 8.
Last Four digits of the given
number should be divisible
by 16.
Last five digits of the given
number should be divisible
by 32.
Examples
Number
Check the
divisibility by
Solution
31890
2
21985634
2
Divisible as the last digit
of the given number is
even (0).
Divisible as the last digit
of the given number is
even (4) or divisible by 2.
(Contd.)
Be a Human Calculator
2178988
4
31276
8
21256784
16
2178932064
32
Divisible as the last two
digits of given number
(88) are divisible by 4.
Not, Divisible as the
last three digits of given
number (276) are not
divisible by 8.
Divisible as the last four
digits of given number
(6784) are divisible by 16.
Divisible as the last five
digits of given number
(32064) are divisible by
32.
Exercises 5.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Question
Check the divisibility of following
numbers by 2,4,8,16 and 32
1
998
2
1024
3
9967
4
12345
5
123464
8
576775338
6
7
9
10
9962
100400
54676
99789
38
Be a Human Calculator
5.2 Divisibility Tests of 5, 25, 125, 625 etc./Divisibility
test of the numbers of the form 5n:
Number
Divisibility Test
Last digit of the given number
should be divisible by 5 or
should be 0.
1
5= 5
25 = 5
Last two digits of the given
number should be divisible by
25 or should be 00.
2
125 = 5
3
625 = 5
4
Last three digits of the given
number should be divisible by
125 or should be 000.
Last Four digits of the given
number should be divisible by
625 or should be 0000.
Examples
Number
Check the
divisibility by
31895
5
Divisible as the last digit of
given number is 5.
2178900
25
31275
25
Divisible as the last two digit of
given number are 00.
21256000
125
2178932500
125
21985630
5
Solution
Divisible as the last digit of
given number is 0.
Divisible as the last two digits
of given number are divisible
by 25.
Divisible as the last three digits
of given number are 000.
Divisible as the last three digits
of given number are divisible
by 125.
(Contd.)
39
Be a Human Calculator
2170000
625
1272500
625
Divisible as the last four digits
of given number are 0000.
Divisible as the last four digits
of given number are divisible
by 625.
Exercises 5.2
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Question
Check the divisibility of following
numbers by 5, 25, 125 and 625
1
990
2
1025
3
9960
4
12345
5
123465
8
576775335
6
7
9
10
9960
100400
54675
91250
5.3 Divisibility Tests of 3 and 9:
Number
3
9
Divisibility Test
Sum of all the digits of the
given number should be
divisible by 3.
Sum of all the digits of the
given number should be
divisible by 9.
40
Be a Human Calculator
Examples
Number
Check the
divisibility by
Solution
32895
3
Divisible as the Sum of all
the digits (3 + 2 + 8 + 9+5
= 27) is divisible by 3.
21985650
9
Divisible as the Sum of all
the digits (2 + 1 + 9 + 8 +
5 +6 +5 = 36) is divisible
by 9.
Exercises 5.3
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Question
Check the divisibility of following
numbers by 3 and 9
1
990
2
1025
3
9960
4
12345
5
123465
8
576775335
6
7
9
10
9960
100400
54675
91250
5.4 Divisibility Test of 11:
If the difference of sum of digits at odd places and sum of digits at even
places is a multiple of 11 or 0, the given number will be divisible by 11.
41
Be a Human Calculator
Examples
Number
Check the
divisibility by
Solution
32895
11
21985650
11
(5+8+3)-(9+2) = 16 – 11= 5,
Not divisible
41985658
11
32835
11
(0+6+8+1)-(5+5+9+2) = 15
– 21= -6,
Not divisible
(8+6+8+1)-(5+5+9+4) = 23
– 23= 0,
Divisible
(5+8+3)-(3+2) = 16 – 5= 11,
Divisible
Exercises 5.4
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Question
Check the divisibility of following
numbers by 11
1
990
2
1025
3
9966
4
10307
5
2244858
8
576775335
6
7
9
10
9724
100400
54675
91250
42
Be a Human Calculator
5.5 Divisibility Check for Prime Numbers:
To understand the divisibility by prime numbers, we have to learn concept
of positive osculator and negative osculator.
Positive Osculator or One More Osculator:
If a number or its multiple + 1 = n × 10, then n will be positive osculator
of that number.
Examples
Number
Find Positive Osculator
7
7 × 7 + 1 = 50 = 5 × 10 (Ans : 5)
13
17
19
13 × 3 + 1 = 40 = 4 × 10 (Ans : 4)
17 × 7 + 1 = 120 = 12 × 10 (Ans : 12)
19 + 1 = 20 = 2 × 10 (Ans : 2)
Negative Osculator or One Less Osculator:
If a number or its multiple - 1 = n × 10, then n will be negative osculator
of that number.
Examples
Number
Find Negative Osculator
7
7 × 3 - 1 = 20 = 2 × 10 (Ans : 2)
13
17
19
13 × 7 - 1 = 90 = 9 × 10 (Ans : 9)
17 × 3 - 1 = 50 = 5 × 10 (Ans : 5)
19 × 9- 1 = 170 = 17 × 10 (Ans : 17)
Divisibility test for prime numbers using osculators:
Step-1: Find the positive or negative osculator of the given number
(whichever is small);
Step-2: Multiply the unit digit of the number by the osculator and add
to the remaining number (in case you have used positive osculator) or
43
Be a Human Calculator
subtract from the remaining number (in case you have used the negative
osculator).
Step-3: Check whether remaining number is divisible by the given number. If it is still a big number; repeat the step-2 till you get a smaller number by which we can get the divisibility by visualization only.
Number
21793
4897645
Check the
divisibility by
Solution
7
As 7 × 3 -1 = 2 × 10, its negative
osculator is 2.
2179 – 3 × 2 = 2173
217 – 3 × 2 = 211
21–1 × 2=19
(not divisible by 7)
Therefore, the given number
is also not divisible by 7.
As 29 + 1 = 3 × 10, its positive
osculator is 3.
489764 + 5 × 3 = 489779
48977 + 9 × 3 = 49004
4900 + 4 × 3 = 4912
491 + 2 × 3 = 497
49 + 7 × 3 = 70
(not divisible by 29)
Therefore, the given number
is also not divisible by 29.
29
Note: The method may sound little lengthy but it is very useful for
students who are not comfortable with tables higher than 10 and unable to do quick division. It is also useful for checking divisibility by
large prime numbers like 79, 89 etc.
44
Be a Human Calculator
Let us see some more examples,
Number
2157872
2341567
Check the
divisibility by
Solution
As 79+ 1 = 8 × 10, its positive
osculator is 8.
215787 + 2 × 8 = 215803
21580 + 3 × 8 = 21604
2160 + 4 × 8 = 2192
219 + 2 × 8 = 235
23+ 5 × 8 = 63
(not divisible by 79)
Therefore, the given number
is also not divisible by 79.
79
As 71 -1 = 7 × 10, its negative
osculator is 7.
234156 – 7 × 7 = 234107
23410 – 7 × 7 = 23361
2336 – 1 × 7 = 2329
232 – 9 × 7 = 169
(not divisible by 79)
Therefore, the given number
is also not divisible by 79.
71
45
Be a Human Calculator
Exercises 5.5
Total Questions 10
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Question
Check the divisibility of following
numbers by 7, 29, 31, 37
1
990
2
1025
3
9966
4
10307
5
2244858
8
576775335
6
7
9
10
9724
100400
54675
91250
5.6 Divisibility Check for Composite Numbers:
Number
Divisibility Test
12
As 12 = 3 × 4, the number should be divisible by 3 and 4 both.
18
As 18 = 2 × 9, the number should be divisible by 2 and 9 both.
14
15
21
24
26
28
33
As 14 = 2 × 7, the number should be divisible by 2 and 7 both.
As 15 = 3 × 5, the number should be divisible by 3 and 5 both.
As 21 = 3 × 7, the number should be divisible by 3 and 7 both.
As 24 = 3 × 8, the number should be divisible by 3 and 8 both.
As 26 = 2 × 13, the number should be divisible by 2 and 13
both.
As 28 = 4 × 7, the number should be divisible by 4 and 7 both.
As 33 = 3 × 11, the number should be divisible by 3 and 11
both.
Similarly, students may derive divisibility test for other composite
numbers.
46
Be a Human Calculator
Exercises 5.6
Total Questions 10
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Question
Check the divisibility of following
numbers by 12, 14, 15, 18
1
990
2
1025
3
9966
4
10307
5
2244858
8
576775335
6
7
9
10
9724
100400
54675
91250
47
Chapter
6
Percentage
In every QA or DI paper, students have to find percentage of a given number
on a large number of occasions. A faster method based on common sense
is therefore required to save time.
6.1 Equivalent Fractions
The first step towards getting faster percentage is to learn some equivalent
fractions as described below:
%
Equivalent
Fraction
1%
1/100
5%
1/20
10%
1/10
8%
15%
20%
25%
2/25
3/20
1/5
1/4
%
1
3 %
8
1
6 %
4
1
8 %
3
1
12 %
2
1
13 %
3
2
14 %
7
1
33 %
3
Equivalent
Fraction
1/32
1/16
1/12
1/8
2/15
1/7
1/3
(Contd.)
Be a Human Calculator
30%
3/10
50%
1/2
40%
60%
75%
1
37 %
2
1
62 %
2
2/5
2
66 %
3
1
87 %
2
3/5
3/4
3/8
5/8
2/3
7/8
Example 1: 87.5% of 800 = 800 × 7/8 = 700
Example 2: 12.5% of 80 = 80 × 1/8 = 10
Example 3: 75% of 400 = 400 × 3/4 = 300
6.2 Finding Quicker Percentage:
We can always find 1% or 10% or 100% of any number by observation
immediately. With the help of this fact and equivalent fractions given
above, we can find any % of any number with a quicker speed.
Example 1: Find 7% of 329.
Thought Process: 1% = 3.29
7% = 3.29 × 7 = 23.03
Example 2: Find 42% of 7289.
Thought Process: 10% = 728.9, 1% = 72.89
40% = 728.9 × 4 = 2915.60
2% = 72.89 × 2 = 145.78
Therefore, 42% of 7289 = 2915.60 + 145.78 = 3061.38
Example 3: Find 114% of 287
Thought Process: 100% = 287, 10% = 28.7, 4% = 2.87 × 4
Therefore, answer is = 287 + 28.70 + 11.48 = 327.18
50
Be a Human Calculator
Example 4: Find 49% of 3452
Thought Process: 50% = 1726, 1% = 34.52
Therefore, answer is = 1726 – 34.52 = 1691.48
Example 5: Find 273% of 523
Thought Process: 200% = 523 × 2 = 1046, 70% = 52.3 × 7=366.1, 3% =
5.23 × 3 = 15.69
Therefore, answer is = 1046 + 366.10 + 15.69 = 1427.79
Exercise 6
Total Questions 10
Ideal Time 3 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 3 minutes)
1. Find 23% of 489
2. Find 87.5% of 400
3. Find 49% 463
4. Find 22.4% of 3245
5. Find 25% of 484
6. Find 112.5% of 1200
7. Find 36% of 7843
8. Find 57% of 5674
9. Find 674% of 2341
10. Find 28% of 435
6.3 Flipping the Number Technique:
We know that, x % of y = y % of x but this small observation may save a lot
of time, when we apply it on many questions like:
Example 1: 48 % of 50 = 50% of 48 = 24
Example 2: 348 % of 75 = 75% of 348 = ¾ × 348 = 3 × 87 = 261
Example 3: 58 % of 25 = 25% of 58 = ¼ × 58 = 14.5
Example 4: 82 % of 20 = 20% of 82 = 1/5 × 82 = 16.4
Example 5: 81 % of 33 1/3 = 33 1/3 % of 81 = 1/3 × 81 = 27
51
Chapter
7
Division
7.1 Division by using vinculum:
Let us divide 48765 by 998, we have,
998 | 48765 | 48.86
- 3992
---------8845
- 7984
----------
8610
- 7984
----------6260
- 5988
---------272
We have learnt about Vinculum Numbers in Chapter-1. Now, let us learn
a great use of these numbers. We found ourselves uncomfortable while
dealing with division by big numbers like 998, 999, 9971, 879 etc. If bigger digits of these numbers are converted in vinculum digits less than 5,
the divisor will look like:
998 = 1000-2 or 100 2
999 = 1000 -1 or 1001
9971 = 10031
879 = 1000 – 121 = 1 1 2 1
And obviously the division with vinculum numbers will be far easy
keeping in view the small digits.
Let us have a relook on the question solved above by using vinculum
number.
Be a Human Calculator
We can write 998 = 1000-2 or 1002, division will be as follows:
−
100 2 | 48765 | 48.86
−
−
-400 8 {876 − 8 = 876 − (−8) = 876 + 8 = 884}
---------8845
−−
−−
−−
− −
-80 1 6 {845 − 16
= 845 − (−16)
= 845 + 16
= 861}
---------8610
−−
−−
-80 1 6 {610 − 16
= 610 − (−16)
= 610 + 16
= 626}
----------6260
-60 1 2 {260 − 12
= 260 − (−12)
= 260 + 12
= 272}
---------272
Isn’t it simpler as compared to normal division? Practice it further till
students are comfortable with this new method.
Let us take another example of division of 123456 by 879.
We can write 879 = 1000-121 or 1 1 2 1, division will be as follows:
−
− −
11 21 |123456 |140.45
−−−
-1 1 2 1
--------3555
− − −
- 4 48 4
---------3960
− − −
- 4 48 4
----------4440
− − −
-5605
---------45
----------
54
Be a Human Calculator
Explanations:
−−−
{234 − 1 2 1 = 234 − (−121) = 234 + 121 = 355}
− − −
{3555 − 4 48 4 = 3555 − (4000 − 484) = 3555 − 3516 = 39}
− − −
{3960 − 4 4 8 4 = 3960 − (4000 − 484) = 3960 − 3516 = 444}
− − −
{4440 − 56 05 =4440 − (5000 − 605) =4440 − 4395 =45}
7.2 Division by using Complements:
Another interesting application of techniques given in Chapter-1 of this
book is doing division using Complements of the numbers. Let us see the
following example:
Example: Divide 2402 by 99.
Solution: Here base 100, complement of 99 = 01 and this will be our
modified divisor.
Step-1: Number of digits in the divisor will be equal to number of zeroes
in the base, so divisor will be 01.
Step-2: Dividend should be divided into two columns known as Quotient
Column and remainder Column by drawing a vertical line. The number of
digits in the remainder column will be equal to number of zeroes in the
base. So, the division should look like this:
Divisor
Quotient Column
Remainder Column
0
|
|
1
2
4
0
2
Step-3: Under quotient column, find the sum of the first column as follows:
0
1
Quotient Column
Remainder Column
|
2
|
|
2
4
0
2
Step-4: Now multiply the first quotient digit 2 by divisor 01 to get 02 and
write this below next two columns as shown:
55
Be a Human Calculator
Quotient Column
0
1| 2
|2
Remainder Column
4
0
|0
|2
2
|
Step-5: Now get second quotient digit as follows
Quotient Column
0
1| 2 4
0
|2
4
Remainder Column
|0
|2
2
|
Step-6: Now multiply the second quotient digit 4 by divisor 01 to get 04
and write this below next two columns as shown below:
Quotient Column
0
1| 2
4
0
|2
4
Step-7: Get the remainder as shown below:
Quotient Column
0
1| 2
4
0
|2
4
Thus, the quotient is 24 and remainder is 26.
56
Remainder Column
|0
|2
|0
|
2
4
Remainder Column
| 0
| 2
| 0
| 2
2
4
6
Be a Human Calculator
Exercise 7.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Divide the following numbers by using Vinculum method or by using
complements to find quotient and remainder:
1. Divide 9875 by 99
2. Divide 12345 by 979
3. Divide 56789 by 1099
4. Divide 6754 by 999
5. Divide 98765 by 1187
6. Divide 76234 by 199
7. Divide 98765 by 99
8. Divide 54634673 by 1999
9. Divide 686758 by 1089
10. Divide 998788 by 399
7.3 Modified Cancellation Technique:
Whenever we simplify terms like this 248 ÷ 48, we try to bring the fraction
into lowest form by dividing numerator and denominator both by suitable
numbers as follows:
(Cancelling Numerator and Denominator by 8)
2178 1089
= = 4.86
448
224
(Cancelling Numerator and Denominator by 2)
But when we have fractions like 445/48, 1297/33, 56789/21, we never
go for cancellation as numerator and denominator don’t have a common
factor or are not fully divisible by the same number. It is only a mental block
which stops us to cancel out the numbers in case only one number is fully
divisible. We may cancel out the numbers even if only the denominator is
fully divisible by some number as follows:
57
Be a Human Calculator
445 55.625 27.8125
=
=
= 9.27
48
6
3
(In the above example, we divide the numerator and denominator both by
8 and 2 successively even if the numerator was not fully divisible by these
numbers and that made our calculation easier).
129
=
33
432.3
= 39.3
11
(Divide numerator and denominator by 3)
Thus, by modifying our cancellation habits a bit, we may solve such
questions at a faster pace.
Exercise 7.2
Total Questions 10
Ideal Time 4 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 4 minutes)
1. Divide 123456 by 15
2. Divide 12345 by 24
3. Divide 56789 by 18
4. Divide 6754 by 21
5. Divide 98765 by 33
6. Divide 76234 by 38
7. Divide 98765 by 105
8. Divide 54634673 by 64
9. Divide 686758 by 63
10. Divide 998788 by 99
58
Chapter
8
Squaring Techniques
8.1 Squaring of Numbers ending with 5:
It is one of the easiest techniques used in mental calculations. It has two
simple steps:
Step-1: Square of any number ending with 5 will end with 25
Step-2: Multiply number prior to 5 with its successor and prefix the
answer before 25 to get the square root.
Examples
152
1 × 2 = 2
∴152 =
225
352
 3 × 4 = 12
 352 = 1225
552
 5 × 6 = 30
252
452
652
752
852
952
2 × 3 = 6
25
625
 4 × 5 = 20
 452 = 2025
 6 × 7 = 42
 652 = 4225
 8 × 9 = 72
 852 = 7225
 552 = 3025
 7 × 8 = 56
 752 = 5625
 9 × 10 = 90
 952 = 9025
1052
 10 × 11 = 110
105 2 = 11025
1252
 12 × 13 = 156
125 2 = 15625
 14 × 15 = 210
145 2 = 21025
1152
1352
1452
1552
 11 × 12 = 132
115 2 = 13225
 13 × 14 = 182
135 2 = 18225
 15 × 16 = 240
155 2 = 24025
Be a Human Calculator
Exercise 8.1
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Find squares of following numbers
1. 25
2. 35
3. 45
4. 55
5. 65
6. 75
7. 85
8. 95
9. 105
10. 115
8.2 Squaring of numbers from 51 to 59:
This is another great technique used in speed mathematics. It has two
simple steps:
Step-1: Answer will have two parts. The right most part will have two
digits and will be the square of number after 5.
Step-2: The leftmost part of the answer will be 25 + right digit of the
number.
Examples
512
522
532
542
552
562
572
582
592
25+1/12
25+2/22
25+3/32
25+4/42
25+5/52
25+6/62
25+7/72
25+8/82
25+9/92
60
2601
2704
2809
2916
3025
3136
3249
3364
3481
Be a Human Calculator
Exercise 8.2
Total Questions 9
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 9 questions in 2 minutes)
Find squares of following numbers
1. 51
2. 52
3. 53
4. 54
5. 55
6. 56
7. 57
8. 58
9. 59
8.3 General Method of Squaring a two digit number:
We may use the following algebraic identity to get square of any two digit
number irrespective of its format:
(a + b) 2 =a 2 + 2ab + b 2
The answer will have three parts. The Rightmost part will be given by b2,
the middle part of the answer will be given by 2ab and leftmost part of the
answer will be given by a2 . If the answer of rightmost and middle part so
obtained has more than 2 digits, unit’s digit will be noted as answer and ten’s
digit will be carried forward to the number on the immediate left side.
61
Be a Human Calculator
Examples
212
22/2 × 2 × 1/12
441
322
32/2 × 3 × 2/22 = 9124
1024
672
62/2 × 6 × 7/72
= 368449 = 36889
4489
472
732
42/2 × 4 × 7/72
= 165649 = 16609
72/2 × 7 × 3/32 = 49429
Exercise 8.3
Total Questions 10
2209
5329
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Find squares of following numbers
1. 78
2. 67
3. 43
4. 89
5. 83
6. 49
7. 94
8. 64
9. 74
10. 62
8.4 Squaring near a base number:
This technique is useful when given number is near to some base such as
100, 1000, 10000 etc. It has two steps:
Step-1: Answer will have two parts. The right most part will have digits
equal to number of zeroes in the base and will be equal to square of deficiency or excess of the number from the base.
62
Be a Human Calculator
Step-2: The leftmost part of the answer will be given number – deficiency
or given number + excess as the case may be.
Examples
98
Base = 100,
Deficiency = 2
1002 2
Base = 1000,
excess = 2
1002+02/(2)2
Base = 10000,
Deficiency =2
9998 - 2/(2)2
2
997 2
1009 2
9998 2
98-02/(2)2
Base = 1000,
Deficiency = 3
997-03/(3)2
Base = 1000,
excess = 9
1009+09/(9)2
9604
994009
1004004
1018081
99960004
Exercise 8.4
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 2 minutes)
Find squares of following numbers
1. 98
2. 998
3. 1002
4. 97
5. 96
6. 1009
7. 996
8. 1004
9. 99
10. 999
8.5 General Method of Squaring of any number:
This method may be used to find square root of any number irrespective
of its format or number of digits in it.
63
Be a Human Calculator
Before we learn this method, we have to learn concept of Index Number.
Index Number (IN): The concept of index number has been explained in
the following table:
Number Index Number Number
a
a
ab
2ab
abc
2ac + b
abcd
abcde
2ae + 2bd + c
32 = 9
23
2 × 2 × 3 = 12
123
2
2ad + 2bc
Index Number
3
2
1234
12345
2
2 × 1 × 3 + 22=10
2 × 1 × 4 + 2 × 2 × 3=20
2 × 1 × 5 + 2 × 2 × 4 + 32 = 35
Now we will learn this method through following example:
Example 1: 472 =?
Solution:
472
=
=
=
=
=
=
=
=
IN for 4/
42
/
16
/
16
/
16
/
16
/
16 + 6 /
22
/
Example 2: 2142 =?
IN for 47
2×4×7
56
56
56 + 4
0
6
0
0
/ IN for 7
/
72
/
49
/
9
4
/
9
/
9
/
9
/
9
∴472 = 2209
Solution:
2142
= IN for 2 / IN for 21 / IN for 214
/ IN for 14 / IN for 4
=
7
/
8
/ 16
7
/
9
/ 6
=
22
/ 2 × 2 × 1 / 2 × 2 × 4 + 12 / 2 × 1 × 4 / 4 2
=
4
/
=
4
4
/
4
/
/
5
/
4+1 /
∴ 2142 =
45796
1
7
64
/
8+1
/ 6
Be a Human Calculator
Example 3: 36472 =?
Solution:
36472 = IN for 3/IN for 36/IN for 364/IN for 3647/IN for
647/IN for 47/IN for7
=
=
32 / 2 × 3 × 6 / 2 × 3 × 4 + 6 2 / 2 × 3 × 7 + 2 × 6 × 4 /
2 × 6 × 7 + 42 / 2 × 4 × 7 / 72
9 /
3
6
/
6
0
/ 90
/
12 /
1
2
/ 9
/ 10
/ 5
0 / 56
/
4
/ 10
/
9
10
= 9+3 / 6+6 / 0+9 / 0+10/ 0+5 / 6+4 /
=
=
=
13 / 2
13 / 3
/ 10
/ 0
/ 0
/ 0
∴ 3647 2 =
13300609
/ 5+1 / 0
/ 6
/ 0
/
/
9
9
9
9
Note: Again this method looks very difficult and lengthy in the first look
as all the steps are explained and written in detail. In fact, once students
know, how to get IN for any number, they can find the answer orally in one
line as follows. Example 4: 12342 =?
Solution:
Thought process (not to be written on paper)
12342 = IN for 1/IN for 12/IN for 123/IN for 1234/IN for 234/
IN for 34/IN for4
Steps to be written actually
12342 = 1522756
1/4/10/20/25/24/16
65
Be a Human Calculator
Exercise 8.5
Total Questions 10
Ideal Time 4 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 4 minutes)
Find squares of following numbers
1. 214
2. 123
3. 3456
4. 128
5. 543
6. 1019
7. 778
8. 2134
9. 234
10. 456
66
Chapter
9
Cubing Techniques
In competitive examinations, students have to find the cube of a 2 digit
number often. Cubes of very large numbers are rarely used. To learn the
short cut tricks to find the cube of a two digit number, cubes of all the
single digits should be memorized.
The cubes of first ten natural numbers are given below-
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125,
63 = 216, 73 = 343, 83 = 512, 93 = 729, 103 = 1000
9.1 Cubing a 2 digit number
First Step: Write down the cube of the tens-digit followed by three more
numbers from left to right such that the last number will be the cube of
unit’s digit and the other two numbers in the first row should be in the
same ratio which is there between the digits of the given number.
Second Step: Put down, under the second and third numbers, double of
second and third number.
Third Step: Add up the two rows obtained in step one and two above.
While adding just write down the unit digit of the answer and carry forward the remaining number as we do in normal addition.
Example 1: Finding the cube of 12
Step I: Digit in ten’s place is 1, so write the first number as the cube of
1. The fourth number will be the cube of 2 i.e. 8. And also as the ratio
between 1 and 2 is 1:2, the next digits will be double the previous one. So,
the first row:
1248
Step II: In the above row 2nd and 3rd digits (from right) are 4 and 2 respectively. So, write down 8 and 4 below 4 and 2 respectively.
Be a Human Calculator
Step III: Then add up the two rows.
1
2
4
8
4
8
------------------------------------------
1
6
2
8 = 1728
1
Example 2: Finding the cube of 22
(As the ratio of digits in 22 is 1:1)
8
8
8
8
16
16
Carry Over
2
2
------------------------------------------
10
6
4
8 = 10648
2
2
Example 3: Finding the cube of 16
Solution:
(As the ratio of digits in 16 is 1:6)
1
6
36
216
12
72
Carry Over 3
12
21
-------------------------------------------
4
0
9
6 = 4096
3
12
21
Example 4: Finding the cube of 32
First row
27
18
12
8
27
18
12
8
(Note that, first number is cube of 3 and last number is cube of 2. 2nd number from left is 2/3 of 27 i.e. 18 and 3rd number from left is 2/3 of 18 i.e. 12).
Carry Over
36
24
5
3
-----------------------------------------32
7
6
8 = 32768
5
3
Example 5: Finding the cube of 79
First row
343
441
567
729
(Note that, first number is cube of 7 and last number is cube of 9. 2nd number from left is 9/7 of 343 i.e. 441 and 3rd number from left is 9/7 of 441
i.e. 567).
68
Be a Human Calculator
343 441
567
729
882 1134
Carry Over
150 177
72
-----------------------------------------------
493
0 1773
9 = 493039
150
72
Advantages of this Method
1. Best suited for numbers like 11, 22, 33, 44 etc.
2. Also good for numbers like 12, 21, 24, 42, 48, 84, 26, 62 where digits
are multiples of each other.
Limitations of this Method
1. Applicable only for two digit numbers.
2. The calculation is tedious for numbers like 79, 67, 85 etc., where digits
are not multiples of each other. In this case, it is better to square the
number and then multiply it again or to use method 9.2.
Exercise 9.1
Total Questions 89 Ideal Time 30 Min
(Students are advised to practice this exercise till they are able to
solve all 89 questions in 30 minutes)
Q 1 to 89: Find the cubes of all the numbers from 11 to 99.
9.2 Cubing near a base number
Step- I: Identify the nearest base to the given number and find the
excess or deficiency of the number from it.
For example:
For the number 97, base is 100 and deficiency is -3.
For 104, base is 100 and the excess is 4.
For 996, base is 1000 and deficiency is -4.
Step-II: The answer will have three parts. Left, Middle and Right. These
will be given by:
Left Part
Middle Part
Right Part
= Given number + 2 × Excess or Deficiency
= 3 × (Deficiency or Excess) 2
= (Deficiency or Excess) 3
69
Be a Human Calculator
Note 1: The number of digits in Middle and right part will be equal to the
number of zeroes in the base. The remaining part will be carried forward.
Note 2: In case of deficiency, the right part will come in negative or
vinculum number and it has to be converted to ordinary number by using
techniques explained in chapter-1.
Example 1: Finding the cube of 104
Base = 100, Excess = 4
Left Part
= 104 + 2 × 4 = 112
Middle Part
= 3 × (4)2 = 48
Right Part
= (4) 3=64
Therefore, answer will be 1124864.
Example 2: Finding the cube of 107
Base = 100, Excess = 7
Left Part
= 107 + 2 × 7 = 121
Middle Part
= 3 × (7)2 = 147
Right Part
= (7) 3= 343
Therefore, answer will be 121/147/343 or 1225043.
(See Note 1)
Example 3: Finding the cube of 97
Base = 100, Deficiency = -3
Left Part
= 97 + 2 × (-3) = 91
Middle Part
= 3 × (-3)2 = 27
Right Part
= (-3) 3= - 27 or 2 7
Therefore, answer will be 91/27/ 2 7 or 912626.
(See how to convert vinculum numbers into ordinary numbers –
Chapter-1).
Example 4: Finding the cube of 991
Base = 1000, Deficiency = -9
Left Part
= 991 + 2 × (-9) = 973
Middle Part
= 3 × (-9)2 = 243
Right Part
= (-9) 3= - 729 or 7 2 9
− − −
Therefore, answer will be 973/243/ 7 2 9 or 973242271.
70
Be a Human Calculator
Advantages of this Method
Best suited for numbers which are near to any base number like 10, 100,
1000 etc.
Limitations of this Method
The calculation is tedious if the given number is far away from the base.
Exercise 9.2
Total Questions 20
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 20 questions in 5 minutes)
Find the cube of following numbers:
Question
Number
Question
Number
1
91
11
992
4
94
14
995
2
3
5
6
7
8
9
10
92
12
93
13
95
15
96
16
97
17
98
18
99
19
991
20
71
993
994
996
997
998
999
9996
9997
Chapter
10
Square Root
10.1 Square Root of Perfect Square numbers consisting
of 3 or 4 digits (By Inspection)
For using this technique, students are advised to learn Squares of first 10
natural numbers and Square Roots of first 10 Perfect Square numbers by
heart, which are given below:
Table-1
Squares/Square Root
12 = 1 ⇒ 1 = 1
22 = 4 ⇒ 4 = 2
32 = 9 ⇒ 9 = 3
42 =
16 ⇒ 16 =
4
52 =
25 ⇒ 25 =
5
62 =
36 ⇒ 36 =
6
72 =
49 ⇒ 49 =
7
82 =
64 ⇒ 64 =
8
92 =
81 ⇒ 81 =
9
102 =
100 ⇒ 100 =
10
Observation
If a perfect Square ends with 1, its
Square root may end with 1
If a perfect Square ends with 4, its
Square root may end with 2
If a perfect Square ends with 9, its
Square root may end with 3
If a perfect Square ends with 6, its
Square root may end with 4
If a perfect Square ends with 5, its
Square root will end with 5
If a perfect Square ends with 6, its
Square root may end with 6
If a perfect Square ends with 9, its
Square root may end with 7
If a perfect Square ends with 4, its
Square root may end with 8
If a perfect Square ends with 1, its
Square root may end with 9
If a perfect Square ends with 0, its
Square root will end with 0
Be a Human Calculator
Summary of observations:
Perfect Square end with
Square root will end with
0
0
1
1 or 9
6
4 or 6
4
2 or 8
5
5
9
3 or 7
Perfect Square Numbers never end with 2,3,7 and 8
Step -1: Divide the number into two parts by making a partition after two
digits from right. So, 1936 should look like 19/36.
Step -2: Pick the left part of the number and by observation, ascertain as
to between squares of which of the two numbers, this part lie. The lower
number out of the two numbers observed, will be the ten’s digit of the
answer. In 19/36, left part 19 lies between 4 2 and 5 2 , so ten’s digit of
1936 will be 4.
Step -3: Now, see the unit digit of the given number and find unit digit of
the answer by observation. As the unit digit of 19/36, is 6, the square root
should end with 4 or 6 (see table 1). So, the answers may be 44 or 46.
Step -4: If there are two answers in step-3, eliminate one answer by
observation. As the likely answers are 44 or 46, and 452 = 2025 which is
more than 1936, so the answer will be 44.
Examples
Find the
Square Root
of following
numbers
Ten’s Digit of
Unit’s
the Answer Digit of the
answer
4489
(write it as
44/89 )
6 as 44 lies
between 6 2
and 7 2
3 or 7 as
4489ends
with 9.
74
Answer
63 or 67
(But,
2
65 = 4225), so
4489 > 65.
Answer is 67.
(Contd.)
Be a Human Calculator
8464
(write it as
84/64)
9 as 84 lies
between 92
and 102
2 or 8 as
8464 ends
with 4.
7225
(write it as
72/25)
8 as 72 lies
between 8 2
and 9 2
5 as 7225
ends with 5.
10816
(write it as
108/16)
92 or 98
(But,
952 = 9025),
8464
so
<95. Answer
is 92.
85
10 as 108 lies
4 or 6 as
104 or 106
2
between 10
10816 ends
(But,
2
and 11
with 6.
105 2 = 11025 ),
Exercise 10.1
Total Questions 50 so 10816
<105. Answer
is 104.
Ideal Time 10 Min
(Students are advised to practice this exercise till they are able to
solve all 50 questions in 10 minutes)
Find the square root of following perfect squares:
Question
Number
Question
Number
1
441
9
2401
12
2809
2
576
10
1521
13
3
1024
6
1681
4
5
7
8
11
1296
14
2209
15
2304
16
75
2601
2704
2916
3136
3249
3364
(Contd.)
Be a Human Calculator
Question
Number
Question
Number
17
3481
34
6241
20
3969
37
6889
18
19
21
22
23
24
25
26
27
28
29
30
31
32
33
3721
35
3844
36
4096
38
4356
39
4489
40
4624
41
4761
42
5041
43
5184
44
5329
45
5476
46
5625
47
5776
48
5929
49
6084
50
6561
6724
7056
7396
7569
7744
7921
8281
8649
8836
9025
9216
9409
9604
9801
10.2 Square Root of Perfect Square numbers consisting
of 5 digits (By Inspection)
For using this technique, students are advised to learn Squares of first 32
natural numbers, which are given below:
Number
1
2
3
4
5
6
7
8
9
Square
1
4
9
16
25
36
49
64
81
76
Number
10
11
12
13
14
15
16
17
18
Square
100
121
144
169
196
225
256
289
324
(Contd.)
Be a Human Calculator
Number
19
20
21
22
23
24
25
Square
361
400
441
484
529
576
625
Number
26
27
28
29
30
31
32
Square
676
729
784
841
900
961
1024
The method is exactly similar to method 11.1 and this fact will be clear to
students from the following examples:
Examples
Find the Square
Root of following
numbers
First two
Digits of the
Answer
Unit Digit of
the answer
Answer
13225
(write it as
132/25)
11 as 132 lies
between 112
and 122
5 as 13225
ends with 5.
115
54756
(write it as
547/54)
98596
(write it as
985/96)
47961
(write it as
479/61)
23 as 547 lies 4 or 6 as 54756
between 232
ends with 6
and 242
234 or 236
(But,
235 2 = 55225),
so 54756 <235.
Answer is 234
31 as 985 lies 4 or 6 as 98596
314 or 316
between 312
ends with 6
(But,
2
2
315 = 99225 ),
and 32
so 98596 <315.
Answer is 314
21 as 479 lies 1 or 9 as 47961
between 212
ends with 1
2
and 22
77
211 or 219
(But,
215 2 = 46225),
so 47961 >215.
Answer is 219
Be a Human Calculator
Exercise 10.2
Total Questions 50 Ideal Time 10 Min
(Students are advised to practice this exercise till they are able to
solve all 50 questions in 10 minutes)
Find the square root of following perfect squares:
Question
Number
Question
Number
1
10201
26
84681
4
11881
29
88804
2
3
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
10609
27
11236
28
14641
30
16129
31
17161
32
17956
33
18769
34
19321
35
19881
36
21609
37
22201
38
22801
39
24336
40
24964
41
25281
42
25921
43
28224
44
28561
45
30276
46
39204
44944
78
89401
90601
91204
91809
92416
93025
93636
94249
94864
95481
96100
96721
97344
97969
98596
99225
99856
100489
50
103041
49
77841
87616
47
48
65536
85849
101124
101761
(Contd.)
Be a Human Calculator
10.3 Approximate Square Root of non-perfect squares:
Step-1: Get Square root of nearest perfect square number by observation
to get our estimated answer.
For finding 7 , nearest number is
9 =3 (estimated answer)
Step-2: Find the difference between the given number and nearest perfect
square number and divide it by twice of our estimated answer. From,
difference of 9 and 7, divide twice of our estimated answer 3 to get:
9−7 2 1
= = = .33
6
6 3
Step-3: See if the required answer will be less than or more than our
estimated answer. As 9 =3, 7 < 3. So, approximate value of 7 will be
less than 3. So, subtract the value obtained in step 2 from our estimated
answer to get: 7 = 3- 0.33 = 2.67 (approx.)
Answer by using Calculator = 2.65 (So, method is perfect up-to first
place of decimal)
Note: If we see by observation that real answer is more than our estimated answer than we will add the value obtained in step-2 to our
estimated answer to get the approximate square root. This fact will
be evident from the following examples:
79
Be a Human Calculator
Examples
Find the
approx.
value of
Estimated
Answer
Approximate
deviation from
estimated answer
29
5 as
29 − 25 4
=
10
10
= .40
78
9 as
81 − 78 3
=
18
18
= .17
11 as
129 − 121 8
=
22
22
4
= = .36
11
129
4794
7987
25 =5
81 = 9
121 =11
70 as
4900 =70
90 as
8100 =90
4900 − 4794
140
106 53
= =
140 70
= .75
8100 − 7987
180
113
= = .63
180
80
Approx.
Answer
5 + 0.40 = 5.40
(Actual Answer
is 5.38)
9- 0.17 = 8.83
(Actual Answer is.
8.83)
11 + 0.40 = 11.40
(Actual Answer is
11.36)
70- 0.75 = 69.25
(Actual Answer is
69.24
90- 0.63 = 89.37
(Actual Answer is
89.37)
Be a Human Calculator
Exercise 10.3
Total Questions 50 Ideal Time 10 Min
(Students are advised to practice this exercise till they are able to
solve all 50 questions in 10 minutes)
Find the approximate square root of following numbers:
Question
Number
Question
Number
1
12
26
290
4
47
29
527
2
3
5
6
7
8
23
34
15
16
17
18
19
20
21
22
23
24
25
32
89
131
14
31
78
12
13
30
67
103
11
28
56
9
10
27
39
147
40
159
41
161
42
167
43
174
44
178
45
212
46
234
47
259
48
278
49
288
50
81
803
2056
38
141
731
36
37
138
629
1021
35
123
412
33
34
112
319
1259
1679
2890
3412
3819
4123
5123
5670
5918
6126
6789
7123
8234
9312
9789
9930
Chapter
11
Cube Root
Cube Root is just opposite process of cubing. For example, if cube
of 2 is 8 than cube root of 8 will be 2. Cube root of a number x is
represented as x1/3 or 3 x (commonly used).
∴ 23 =8 ⇒ 3 8 =2,
33 =
27 ⇒ 3 27 =
3
Perfect Cubes Numbers who have their cube roots as natural
numbers are called perfect cubes. For example: 27, 64, 1000 are
3
3
27 3,=
64 4, 3 =
1000 10
Perfect Cubes as=
Non-Perfect Cubes, on the other hand, the numbers who have their
cube roots in decimal numbers are called non-perfect cubes. For
example: 30, 68, 129 are Non-Perfect Cubes as their approximate
cube roots are
3
3
3
=
30 3.11,
=
68 4.08,
=
129 5.05 .
Traditional Method of finding Cube Roots of Perfect Cubes
Step 1: Factorize the given number into its prime factors;
Step 2: Divide the factors obtained in step 1 into group of three identical
factors;
Step 3: Pick one number each from each group and multiply them to get
the cube root.
Examples:
3
216 = 3 2 × 2 × 2 × 3 × 3 × 3 = 2 × 3 = 6
3
1728 = 3 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2 × 2 × 3 = 12
Be a Human Calculator
Traditional Method of finding Cube Roots of Non-Perfect Cubes
There is no traditional method available in elementary mathematics for
getting Cube Roots of Non-Perfect Cubes. Normally it is done by using
logarithm table or scientific calculator.
11.1 Cube roots of Perfect Cubes upto 6 digit numbers
(By Inspection)
For using this technique, students are advised to learn Cubes of first 10
natural numbers and Cube Roots of first 10 Perfect Cube numbers by
heart, which is given below:
Table-1
Cubes/Cube Root
13 =⇒
1 31=
1
23 =⇒
8 38=
2
33 =
27 ⇒ 3 27 =
3
43 =
64 ⇒ 3 64 =
4
53 =
125 ⇒ 3 125 =
5
63 =
216 ⇒ 3 216 =
6
73 =
343 ⇒ 3 343 =
7
83 =
512 ⇒ 3 512 =
8
93 =
729 ⇒ 3 729 =
9
Observation
If a perfect Cube ends with 1, its
Cube root also ends with 1
If a perfect Cube ends with 8, its
Cube root ends with 2
If a perfect Cube ends with 7, its
Cube root ends with 3
If a perfect Cube ends with 4, its
Cube root also ends with 4
If a perfect Cube ends with 5, its
Cube root also ends with 5
If a perfect Cube ends with 6, its
Cube root also ends with 6
If a perfect Cube ends with 3, its
Cube root ends with 7
If a perfect Cube ends with 2, its
Cube root ends with 8
If a perfect Cube ends with 9, its
Cube root also ends with 9
If a perfect Cube ends with 0,
103 =
1000 ⇒ 3 1000 =
10 its Cube root also ends with 0
84
Be a Human Calculator
Summary of observations:
(i) If a perfect cube ends with 0, 1, 4, 5, 6 and 9 than its cube root will also
end with the same number.
(ii) I f a perfect cube ends with 2, 3, 7 and 8 than its cube root will end
with its complement i.e. 8, 7, 3 and 2 respectively.
Step 1: Divide the given number into two 3-digit partitions starting
from the right. Answer will have two digits
For getting cube root of 2197, we will write it as 2/197,46656 as 46/656
and 857375 as 857/375.
Step 2:See the last (unit) digit of the given number and get the Unit’s
digit of the answer by observation as per table-1.
For example:
Unit’s digit of cube root of 2/197 will be 3.
Unit’s digit of cube root of 46/656 will be 6.
Unit’s digit of cube root of857/375 will be 5.
Step 3: Find the ten’s digit of the answer by inspection by seeing the
left partition.
For example:
Ten’s digit of cube root of 2/197 will be 1 as 2 lies between 13 and 23 .
Ten’s digit of cube root of 46/656 will be 3 as 46 lies between 33 and 43 .
Ten’s digit of cube root of 857/375 will be 9 as 857 lies between 93 and
103 .
Therefore,
Cube root of 2197
Cube root of 46656
Cube root of 857375
=
=
=
13
36
95
85
Be a Human Calculator
More Solved Examples:
Find the
Cube Root
of following
numbers
2197
(write it as
2/197)
46656
(write it as
46/656)
857375
(write it as
857/375)
Ten’s Digit of
the Answer
1 as 2 lies
between 13
and 23
3 as 46 lies
between 33
and 43
9 as 857 lies
between 93
and 103
Advantages of this Method
Unit’s
Digit of the
answer
Answer
3 as the
number ends
with 7
13
6 as the
number ends
with 6
5 as the
number ends
with 5
36
95
1. A
fter knowing this method, one can find cube root of any 4, 5 or 6 digit
perfect cube number within 4-5 seconds.
2. T
his method involves no factorization and no written calculations. One
can directly find the answer through inspection in seconds.
Limitations of this Method
1. A
pplicable only for perfect cubes.
2. A
pplicable up-to 6 digit numbers and not beyond.
3. O
ne should use this method only when he is sure that the given number
is a perfect cube.
4. O
ne should not use this method for questions involving approximate
cube root.
86
Be a Human Calculator
Finding Cube Root of a 4, 5, 6 digit Perfect Cube Number
(Summary of the Method)
Find the cube root of a b c d e f.
Step 1
Write or visualize a b c d e f as a b c/d e f.
Step 2
Find Ten’s digit of the answer
If abc is between x 3
3
and y , ten’s digit of
the answer will be x.
Step 3
Find Unit’s digit
of the answer
By seeing the last digit of the number
i.e. f, we may get the unit digit of the
answer by inspection.
If f is 0, 1, 4, 5, 6 & 9, the unit’s digit of
the answer will be the same.
If f is 2, 3, 7, 8 than the unit digit of
the answer will be it’s complement.
87
Be a Human Calculator
Exercise 11.1
Total Questions 82 Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 85 questions in 5 minutes)
Find the cube root of following perfect cubes:
Question
Number
Question
Number
1
1331
27
68921
4
2744
30
85184
2
3
5
6
7
8
9
10
1728
2197
5832
17
18
19
20
21
22
23
24
25
26
39
40
17576
41
19683
42
21952
43
24389
44
29791
45
32768
46
35937
47
39304
48
42875
49
46656
50
50653
54872
51
88
97336
132651
38
13824
91125
36
37
12167
79507
103823
35
9261
74088
33
34
6859
15625
16
32
4913
14
15
31
4096
10648
13
29
3375
11
12
28
52
110592
117649
140608
148877
157464
166375
175616
185193
195112
205379
226981
238328
250047
262144
274625
287496
300763
314432
(Contd.)
Be a Human Calculator
Question
Number
Question
Number
53
59319
68
614125
389017
71
658503
54
357911
57
405224
55
56
58
59
60
61
62
63
64
65
66
67
69
373248
70
72
421875
73
438976
74
456533
75
474552
76
493039
77
512000
78
531441
79
551368
80
571787
81
592704
82
328509
636056
681472
704969
753571
778688
804357
830584
857375
884736
912673
941192
970299
11.2 Cube roots of Perfect Cubes consisting 7, 8 and 9
digits (By Inspection)
This method is almost similar to the first method except for one additional
step that calculates the middle digit (ten’s digit) of the answer as in these
cases answers will be of three digits.
Step 1: Divide the given number into three 3-digit partitions starting
from the right. Answer will now have three digits
For example to get the cube root of 122,023,936, we will write or see it as
122/023/936
Step 2: See the last digit ( Unit digit) of the given number and get the
Unit digit of the answer by observation as per table-1. For example,
Unit digit of cube root of 122/023/936 will be 6
Step 3: Find the leftmost digit of the answer by observing the left partition.
For example: Leftmost digit of cube root of 122/023/936 will be 4
as 122 lies between 43 and 53.
89
Be a Human Calculator
Step 4: Find the ten’s (middle) digit of the answer by using the following method. Denote left, right and middle digits of the answer by L, R
and M respectively.
Now, L = 4, R = 6, M = ?
(i) Subtract R3 from the given number
122023936 – 216 = 122023720
(ii) Note down the penultimate digit (second last digit) from right. In
this case, it is 2.
(iii) Get M by using the result: 3 R2M will end with the penultimate
digit obtained in step 4 (ii).
In this case: 3 x 36 x M will end with 2 or 108 M will end with 2 =» M is 4
or 9. So our answer is 446 or 496.
But we know that, 4003= 64,000,000 and 5003= 125,000,000. So by
observation, given number is closer to 5003 and therefore our answer is 496.
Example2: Find Cube root of 72, 511, 713.
Step-I : 72/511/713
Step-II: Unit digit of the answer (R) will be 7
Step-III: Leftmost digit of the answer (L) will be 4 as 72 lies between 43
and 53
Step-IV: (i) Penultimate digit of 72511713 – 343 is 7.
(ii) 3 R2M should end with 7 =» 3 x 49 x M or 147 M should end
with 7 =» M =1
Therefore, answer is 417.
Example 3: Find Cube root of 10, 503, 459.
Step-I: 10/503/459
Step-II: Unit digit of the answer (c) will be 9
Step-III: Leftmost digit of the answer (a) will be 2.
Step-IV: (i) Penultimate digit of 10,503,459 – 729 is 3.
(ii) 3
R2M should end with 3 =» 3 x 81 x M or 243 M should end
with 3 =» M =1
Therefore, answer is 219.
90
Be a Human Calculator
Exercise 11.2
Total Questions 50
Ideal Time 10 Min
(Students are advised to practice this exercise till they are able to
solve all 50 questions in 10 minutes)
Find the cube root of following perfect cubes:
Question
Number
Question
Number
1
1030301
26
14706125
4
1295029
29
36926037
2
3
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1092727
27
1191016
28
1367631
30
1442897
31
1643032
32
1771561
2000376
2406104
39
2803221
40
2985984
41
3176523
42
3442951
43
3581577
44
3652264
45
3869893
46
4173281
47
5735339
48
7077888
49
9528128
50
91
94818816
254840104
38
2685619
69934528
36
37
2571353
49836032
134217728
35
2248091
30371328
33
34
2048383
21484952
194104539
229220928
360944128
395446904
465484375
469097433
535387328
580093704
700227072
758550528
814780504
846590536
904231063
935441352
988047936
994011992
Be a Human Calculator
11.3 Finding approximate Cube roots of non-Perfect Cubes
Case-I: T
he approximate cube root is more than the
approximate answer
To understand this method, let us go through an example. To find the cube
root of 250.
We know that 3 250 will be more than 6 and less than 7. But as the given
number is closer to 216, so our first approximation for the cube root of
250 is 6.
Now divide the given number by our first estimate twice to get second
approximation.
250 ÷ 6 = 41.67
41.67 ÷ 6 = 6.94 (second approximation)
Now divide the difference between second approximation and first approximation by 3, we get
0.94 ÷ 3 = 0.31
Adding this to our first approximation, we get 6 + .31 = 6.31. Round it
off little downward to get the approximate answer as 6.3. If we check it
through calculator, we will get the answer as 6.2996, which is very close
to our approximate answer.
Case-II : T
he approximate cube root is less than
the approximate answer
Let us find the cube root of 201.
We know that 3 201 will be more than 5 and less than 6. But as the given
number is closer to 216, so our first approximation for the cube root of
201 is 6.
Now divide the given number by our first estimate twice to get second
approximation.
201 ÷ 6 = 33.50
33.50 ÷ 6 = 5.58 (second approximation)
92
Be a Human Calculator
Now divide the difference between second approximation and first approximation by 3, we get
0.42 ÷ 3 = 0.14
Subtracting this from our first approximation, we get 6 - 0.14 = 5.86. If we
check it through calculator, we will get the answer as 5.8577, which is very
close to our approximate answer.
Exercise 11.3
Total Questions 20
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 20 questions in 5 minutes)
Find the approximate cube root of following numbers:
Question
Number
Question
Number
1
148
11
4758
4
3487
14
87868
7
3224
17
646
10
88
2
3
5
6
8
9
278
12
123
13
287
15
765
16
987
18
2134
19
20
93
8686
54
644
785
255
8866
612
Chapter
12
Fractions and Decimals
12.1 Addition of Mixed Fractions
Traditional Method:
1
1 13 29 13 × 7 + 29 × 4 91 + 116 207
11
3 +4 = +
=
=
=
=7
4
7 4 7
28
28
28
28
Alternative Method:
1
1
11
11
 1 1
3 + 4 =(3 + 4) +  +  =7 +
=7
 4 7
4
7
28
28
12.2 Subtraction of Mixed Fractions
Traditional Method:
1
1 21 29 21 × 7 − 29 × 4 147 − 116 31
3
5 −4 = −
=
=
=
=1
4
7 4
7
28
28
28
28
Alternative Method:
1
1
3
3
 1 1
5 − 4 =(5 − 4) +  −  =1 +
=1
 4 7
4
7
28
28
12.3 Multiplication/Square of Mixed Fractions:
(a) Square of Mixed Fractions with fractional part as ½:
(i) Fractional Part of the answer will always be ¼,
(ii) Whole number part of the answer will be whole number part of
the question multiplied by its successor.
Examples: (3 ½ )2 = 12 ¼, (4 ½ )2 = 20 ¼, (5 ½ )2 = 30 ¼, (6 ½ )2 = 42 ¼
(b) Multiplication of Mixed Fractions with sum of fractional parts as
1 and identical whole number part:
(i) Fractional Part of the answer will always be product of fractional parts
of the question,
Be a Human Calculator
(ii) Whole number part of the answer will be whole number part of
the question multiplied by its successor.
Examples:
2
(i) 4 1 × 4 2 =
20
3
3
9
6
(ii) 5 1 × 5 6 =
30
7
7
49
(c) Multiplication of Mixed Fractions with sum of fractional parts as
1/2 and identical whole number part:
(iii) F
ractional Part of the answer will always be product of fractional
parts of the question,
(iv) Whole number part of the answer will be whole number part of
the question multiplied by its successor added by ½ of whole
number part.
Examples:
3
(iii) 4 1 × 4 3 =
18
5
10
50
1
3
3
(iv) 6
×6
=
39
8
8
64
12.4 Converting Recurring decimals into Fractions
Type 1: When numbers are pure decimals and each number after
decimal is repeated i.e. numbers like 0.3333…, 0.353535…,
0.127127127… etc.
Traditional Approach
One line Approach
Let x = .333……
10x = 3.33…….
On subtraction, we get
9x = 3 or × = 3/9 = 1/3
−
.333….= .3 = 3/9 = 1/3
Let x = .353535……
100 x = 35.3535…….
On subtraction, we get
99 x = 35 or x = 35/99
Let x = .127127……
1000 x = 127.127127…….
On subtraction, we get
999 x = 127 or x = 127/999
− −
.353535……= .3 5 = 35/99
− − −
.127127127….= .1 2 7 =127/999
96
Be a Human Calculator
Step-I: Write the given number into bar form by showing bar over digits
which are repeated again and again.
Step-II: Write the answer with numerator as number obtained in first
step without decimal and denominator as a number with 9s only equal to
the number of digits in the numerator.
Type 2: When numbers are pure decimals and each number after
decimal is not repeated i.e. numbers like 0.43333…...,
0.1353535……, 0.21127127127… etc.
Traditional Approach
One line Approach
Let x = .4333……
10x = 4.3333…….
−
.4333….= .4 3 = 43-4/90 =
On subtraction, we get
9x = 3.9 or × = 3.9/9 = 39/90 39/90 = 13/30
=13/30
Let x = .1353535……
100x = 13.5353535…….
− −
.1353535……= .13 5 =135-1 /990
On subtraction, we get
99x = 13.4 or x = 13.4/99 = = 134/990 = 67/495
134/990 = 67/495
Let x =.21127127……
1000x = 211.27127127…….
.21127127127…. = 211 2 7
On subtraction, we get
= 21127-21/99900 =
999x = 211.06 or x = 211.06/999 21106/99900= 10553/49950
= 21106/99900 = 10553/49950
Step-I: Write the given number into bar form by showing bar over digits
which are repeated again and again.
Step-II: Write the answer with numerator as a number obtained in
first the step without decimal subtracted by number not repeated and
denominator as a number with 9s and 0s only. The number of 9s will be
equal to the number of digits with bar followed by number of zeroes equal
to number of digits without bar.
97
Be a Human Calculator
Type 3: When numbers are not pure decimals and each number
after the decimal is repeated i.e. numbers like 1.3333…...,
2.353535……, 3.127127127… etc.
Traditional Approach
One line Approach
Let x = 1.333……
10x = 13.33…….
On subtraction, we get
1
9x = 12 or x = 12/9 = 4/3 = 1
(From type 1-
3
.3 = 3/9 = 1/3)
Let x = 2.353535……
100x = 235.3535…….
On subtraction, we get
35
99x = 233 or x = 233/99 = 2
− −
1
3
35
2.353535……=2 .3 5 = 2
99
(From type 1.3 5= 35/99)
99
Let x = 3.127127……
1000x = 3127.127127…….
On subtraction, we get
999x = 3124 or x =
3124/999 = 3
3
9
1.333….= 1.3 = 1 = 1
− − −
3 . 1 2 7 1 2 7 1 2 7 … . = 3 .1 2 7 =
3
127
999
127
999
(From type 1− − −
.1 2 7 =127/999)
98
Be a Human Calculator
Type 4: When numbers are not pure decimals and each number
after the decimal is not repeated i.e. numbers like 1.43333…,
2.1353535…, 3.21127127127… etc.
Traditional Approach
Let x = 1.4333……
10x = 14.3333…….
On subtraction, we get
9x = 12.9 or x = 12.9/9 =
129/90 =43/30= 1
One line Approach
1.4333….= 1.4 3 = 1
−
13
30
(From Type-2, .4 3 = 13/30)
13
30
Let x = 2.1353535……
67
2.1353535……=2.1 3 5= 2
100x = 213.5353535…….
495
On subtraction, we get
(From Type-2, 1 3 5= 67/495)
99x = 211.4 or x = 211.4/99
= 2114/990 = 1057/495 =
2
67
495
Let x =3.21127127……
1000x = 3211 .27127127…….
On subtraction, we get
999x = 3208.06 or
x = 3208.06/999
= 320806/99900
= 160403/49950
3.21127127127…=3.21 1 2 7=
3
10553
49950
(From type -2, 21 1 2 7 =
10553/49950)
99
Be a Human Calculator
Exercise 12.1
Total Questions 8
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 8 questions in 2 minutes)
1
3
1.1.4 + 2
2.2.5 + 7
2
5
+7 =
5
6
3 2
2
+ +2 =
4 7
3
7
9
5 5
− =
4 12
1
8
2
=
13
3.3.11 − 12 +
4.4.5 + 27 +
5.5.19
6.6.2
7.7.
4 2
1
+ + 14 =
9 9
11
8
1
4
1 3
+2 +3 −4 − =
35
5
7
2 5
1
3
2
− 19 + 67 =
11
17
11
8.8.5
172
95 65 24
+ 10
+ +
=
110
100 52 32
Exercise 12.2
Total Questions 10
Ideal Time 2 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Covert the following decimal numbers into fraction:
Question
Number
Question
Number
1
2
3
4
5
.3535…..
.213213…
1.2929…..
.77777……
21.2353535..
6
7
8
9
10
0.9575757…
1.2222….
4.797979….
.234234……
12.57777…..
100
Chapter
13
LCM and HCF
Students are well aware of the traditional methods for finding HCF and LCM.
Here we will discuss some alternative methods for finding LCM and HCF.
13.1 Finding LCM and HCF of two numbers:
Method-1
Example 1: Find the HCF and LCM of 120 and 36
Solution: Write the numbers as follows:
36
3
= (In lowest form)
120 10
Now HCF = 36 ÷ 3 or 120 ÷ 10 = 12
and LCM = 36 × 10 or 120 × 3 = 360
Example 2: Find the HCF and LCM of 150 and 255
Solution: Write the numbers as follows:
150 10
=
(In lowest form)
255 17
Now HCF = 150 ÷ 10 or 255 ÷ 17 = 15
and LCM = 255 × 10 or 150 × 17 = 2550
Method-2
By using, Product of two numbers = L.C.M. × H.C.F.
Example 3: Find the HCF and LCM of 78 and 216.
Solution: Divide the given numbers by 2 to get 39 and 108 as quotients.
Again divide the 39 and 108 by 3 to get 13 and 36. Now, 13 and 36 are not
divisible by a single prime number simultaneously, so stop here.
2
3
| 78
| 39
13
216
108
36
|
|
So, HCF = 2 × 3 = 6 and LCM = (78 × 216) ÷ 6 = 2808
Be a Human Calculator
Example 4: Find the HCF and LCM of 129 and 219.
Solution:
3
| 129
| 43
219
73
|
|
So, HCF = 3 and LCM = (129 × 219) ÷ 3 = 9417
Method-3 If ratio of numbers is a : b and H is the HCF of the numbers,
then,
LCM of the numbers = H × a × b = HCF × Product of the ratios.
Example 5: Find the HCF and LCM of 16 and 40.
Solution:
By observation HCF = 8, Ratio of numbers is 2: 5,
Therefore, LCM = 8 × 2 × 5 =80
Exercise 13
Total Questions 10
Ideal Time 3 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 3 minutes)
Find the LCM and HCF of the following numbers:
1. 106, 54
2. 84, 212
3. 1208, 2112
4. 78, 98
5. 125, 650
6. 441, 9261
7. 180, 590
8. 2970, 8712
9. 252, 792
10. 192, 1056
102
Chapter
14
Checking Your Answer
We had learnt a powerful technique called ‘Digit Sum Method’ in chapter 1
of this book. Now we will learn, how to use this method to check or verify
your answer. This method normally works for addition, subtraction and
multiplication only but with little observation and common-sense, the same
may be made applicable to check your answers in case of squaring, square
root, cubing and cube roots as well. Students will be able to understand
the above facts in a better way after going through the following examples:
Checking Addition:
Example 1: Verify 98765 + 23875 = 122640
Solution: Digit Sum of 98765 = 8, Digit Sum of 23875 = 7
Digit Sum of Question (Left Hand Side) = 8 + 7 = 15 = 6
Digit Sum of Answer (Right Hand Side) = 122640 = 6
Therefore, Answer is verified.
Checking Subtraction:
Example 2: Verify 98765 - 23875 = 74890
Solution: Digit Sum of 98765 = 8, Digit Sum of 23875 = 7
Digit Sum of Question (Left Hand Side) = 8 - 7 = 1
Digit Sum of Answer (Right Hand Side) = 74890 = 19 = 1
Therefore, Answer is verified.
Note: While subtracting if digit sum of Left hand side comes out to
be negative, add nine to make it positive as we have already learnt
in Chapter 1 that nine means zero while calculating digit sum of a
number.
Example 3: Verify 9765 - 3875 = 4890
Solution: Digit Sum of 9765 = 0, Digit Sum of 3875 = 5
Digit Sum of Question (Left Hand Side) = 0 - 5 = -5 = -5 +9 = 4
Digit Sum of Answer (Right Hand Side) = 4890 = 3
Therefore, Answer is incorrect. The correct answer is 5890.
Be a Human Calculator
Checking Multiplication:
Example 4: Verify 12345 × 875 = 10801875
Solution: Digit Sum of 12345 = 6, Digit Sum of 875 = 2
Digit Sum of Question (Left Hand Side) = 6 × 2 = 12 = 3
Digit Sum of Answer (Right Hand Side) = 10801875 = 12 = 3
Therefore, Answer is verified.
Checking Division:
The process of verifying a division is slightly different from process of
checking addition, subtraction or multiplication. It may be checked as
follows:
Digit Sum (D.S.) of Dividend = D.S. of Divisor × D.S. of Quotient
Example 5: Verify 12625/25 = 505
Solution: Digit Sum of 12625 = 7, Digit Sum of 25 = 7
Digit Sum of 505 = 1
We may see that D.S. of 12625 = D.S. of 505 × D.S. of 25
Therefore, Answer is verified.
Checking Square of a number:
Example 5: Verify 532 = 3409
Solution: Digit Sum of 53 = 8, Digit Sum of 532 = 8 × 8 = 64 = 1
Digit Sum of 3409 = 7
Therefore, Answer is incorrect. The correct answer is 2809 with digit sum
as 1.
Checking Square Root of a number:
Example 6: Verify
7921 = 89
Solution: We have to check if 892 = 7921
Digit Sum of 89 = 8, Digit Sum of 892 = 8 × 8 = 64 = 1
Digit Sum of 7921 = 1
Therefore, Answer is correct.
104
Be a Human Calculator
Checking Cube of a Number:
Example 7: Verify 213 = 9261
Solution: Digit Sum of 21 = 3, Digit Sum of 213 = 3 × 3 × 3 = 27 = 9 = 0
Digit Sum of 9261 = 9 = 0
Therefore, Answer is correct.
Checking Cube Root of a Number:
Example 8: Verify 3 704969 = 89
Solution: We have to check if
Digit Sum of 89 = 8, Digit Sum of 893 = 8 × 8 × 8 = 512 = 8
Digit Sum of 704969 = 8
Therefore, Answer is correct.
Important Note:
This method will work for calculations involving decimals as well.
However, one has to check the position of decimal manually.
For example verify: 52.3 × 56.9 = 297.587
Digit Sum of LHS = 1 × 2 =2, Digit Sum of RHS =2
It seems that answer is right but if you see closely answer should be
2975.87
If the digit sum of Left side is not matching with digit sum of right
side, it is sure that your answer is wrong. Even if digit sum of both
sides are matching, still there is a possibility (though remote) that
your answer may be wrong.
For example verify: 123 × 456 = 56088 but suppose you have written the
answer as 56808. You will get Digit Sum of LHS = Digit Sum of RHS =0
105
Be a Human Calculator
Summary of Digit Sum Method
Operation
How to apply Digit Sum Method
Addition (a + b = c)
Check if D.S.(a) + D.S.(b)= D.S.(c)
Multiplication (a × b = c)
Check if D.S.(a) × D.S.(b)= D.S.(c)
Squaring (a2 = b)
Check if D.S.(a2) = D.S.(b)
Subtraction (a – b = c)
Division(a ÷ b = c)
Check if D.S.(a) - D.S.(b)= D.S.(c)
Check if D.S.(c) × D.S.(b)= D.S.(a)
Cubing (a3 = b)
Check if D.S.(a3) = D.S.(b)
Check if D.S.(a2) = D.S.(b)
Square Root ( b = a )
Cube Root ( 3 b = a)
Check if D.S.(a3) = D.S.(b)
Exercise 14
Check the answers of questions given in exercise 2, 3, 4, 8, 9, 10 and 11 by
using digit sum technique.
106
ALGEBRA
Chapter
15
Long Division or Synthetic
Division
15.1 Long Division
All the students must be aware of the process of long division of a
polynomial by another polynomial. An example of the same is given below:
Example1: Divide x 3 + 7 x 2 + 9 x + 11 by x − 2
x − 2 x 3 + 7 x 2 + 9 x + 11 x 2 + 9 x + 27 (Quotient)
x 3 − 2x 2
(-) (+)
---------------------9x2 + 9x
9x2 − 18x
(-) (+)
----------------------27x + 11
27x − 54
(-) (+)
--------------------------------65 (Remainder)
---------------------------------
15.2 Synthetic Division of a polynomial by a linear
polynomial of the type x ± a
A very powerful technique for simplifying the procedure of long division
is Synthetic Division. This technique, not only saves time but also saves
a lot of space, which is a constraint in all the competitive examinations
these days. In this technique, the quotient and the remainder are obtained
only with the help of coefficients.
Be a Human Calculator
Step-1: In place of dividend, put down all the coefficients of the original
dividend (including co-efficient of missing powers, if any) and in place of
the divisor put down the constant term of the divisor by changing its sign
as follows:
Let us take the same example of division ( as discussed on pre-page) of
x 3 + 7 x 2 + 9 x + 11 by x − 2
x 3 +x73 x+ 27 x+x239++x97x+x+211
+ 9 x + 11
11
Constant Term
|1
11
Coefficient of
2
7
9
Opposite of constant term (-2) in the Divisor x − 2
Step-2: Put 0 below the first coefficient of the dividend and draw a partition before the last coefficient of the dividend as follows:
2
|1
0
7
9
|
11
-------------------------------------------
Step-3: Add the first column to get 1. Multiply the result by divisor 2 and
put down the answer in the second row below 7. Add the second column
to get 9. Again multiply 9 by divisor 2 to get 18 and put it down in the third
column in the second row. Repeat the process till you get the sum in the
last column. Entries in the third column are the coefficients of our answer.
The power of first term in the quotient will be the difference in the highest
power of the term in the dividend and the divisor.
2
|1
7
9
|
11
0
2
18
|
54
x2
x2
Constant
-------------------------------------------------------1
9
27
|
65
Remainder
Therefore, the quotient is x 2 + 9 x + 27 and the remainder is 65.
110
Be a Human Calculator
Example 2: Divide 3 x 3 + 4 x 2 − 2 x − 1 by x + 4
-4
| 3
4
-2
|
-1
0
-12
32
|
-120
------------------------------------------------3
-8
30
|
-121
2
x
Constant Remainder
x
Therefore, the quotient is 3 x 2 − 8 x + 30 and the remainder is -121.
Example 3: Divide x 5 + 1 by x + 1 (observe the coefficients of the
dividend).
-1
| 1
0
0
0
0
| 1
0
-1
1
-1
1
| -1
--------------------------------------------------------------1
-1
1
-1
1
| 0
x4
x3
x Constant Remainder
x2
Therefore, the quotient is x 4 − x 3 + x 2 − x + 1 and the remainder is 0.
15.3 Synthetic division of a polynomial by a linear
polynomial of the type ax ± b
This method is almost similar to the first method except one step which
has been illustrated through the following example:
Example 4: Divide −4 x 3 + 9 x 2 + 9 x − 12 by 2 x − 4
3 9 2 9
3
2
3
2
−4 x + 9 x + 9 x −12 −4 x + 9 x + 9 x −12 −2 x + 2 x + 2 x − 6
= =
2x − 4
2( x − 2)
x−2
And after this step, the method of division is similar to method 15.2
2
|
-2
9/2
9/2
|
-6
0
-4
1
|
11
-------------------------------------------------2
1/2
11/2 |
5
x2
x
Constant
Remainder
111
Be a Human Calculator
Therefore, the quotient is −2 x 2 +
1
11
and the remainder is 5.
x+
2
2
15.4 Synthetic division by a polynomial of degree two
and above
This method has been illustrated through the following example:
Example 5: Divide x4 + 8x3+ 15x2 + 4x + 1 by x2 + 3x + 2
Steps involved:
1. I n place of dividend, put down the coefficients of the dividend (including co-efficient of missing powers, if any). The last two coefficients will
be used to find the remainder of the answer. In place of divisor, put
down the opposite of co-efficient of the linear term and the constant
term of the divisor as follows:
Divisor
-3
2 Row
nd
3rd Row
-2
Dividend/
Quotient
1
8
Remainder
15
4
1
4th Row
5th Row
2. P
ut 0 in the second, third and fourth rows under the first column of the
dividend to get sum as 1 as follows:
Divisor
-3
Dividend/
Quotient
-2
1
4 Row
0
2nd Row
3 Row
rd
th
5 Row
th
0
0
8
1
112
Remainder
15
4
1
Be a Human Calculator
3. Multiply the sum of first column (1) by both the dividends and put
the answer in the second and the third column of the second row as
follows:
Divisor
-3
Dividend/Quotient
-2
1
-3
4 Row
0
0
1
0
5
2 Row
nd
3rd Row
th
5 Row
th
8
0
0
15
-2
Remainder
4
1
0
0
4. Multiply the sum of the second column (5) by both the dividends and
put the answer in the third and the fourth column of the third row as
follows:
Divisor
-3
Dividend/Quotient
-2
1
4 Row
0
2 Row
nd
3 Row
rd
th
5 Row
th
8
15
0
-3
-15
1
5
-2
0
0
0
-2
0
Remainder
4
1
0
0
-10
0
5. Multiply the sum of the third column (-2) by both the dividends and
put the answer in the fourth and the fifth column of the fourth row as
follows:
Divisor
-3
Dividend/Quotient
-2
1
4 Row
0
2nd Row
3 Row
rd
th
5 Row
th
Answer
8
0
-3
0
0
1
0
5
x + 5x -2
2
15
-2
-15
0
-2
Remainder
4
1
6
4
0
5
0
0
-10
0
5
6. Since this is the division of a fourth-degree polynomial by a second degree polynomial, the answer will be of the second degree, and the last
two cells on the 5th row will represent coefficient of a linear remainder.
In this case the quotient is x2 + 5x -2 and the remainder is 0. x + 5 i.e. 5.
113
Be a Human Calculator
Note: I n first glance, it seems to be a lengthy process but after some
practice students will find the method easier, time saving and
space saving. This fact will be evident to students from the following example.
Example 6: Divide 2x4 + 4x3+ 5x2 + 1 by x2 + 4x + 1
Divisor
Dividend/Quotient
Remainder
-4
-1
2
5
0
1
4 Row
0
0
-76
-19
2 Row
nd
3 Row
rd
th
5 Row
th
Answer
4
0
-8
2
-4
0
-2
0
16
0
19
2x2 -4x +19
0
4
-72
0
0
-18
-72 x -18
Exercise 15
Total Questions 10
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Divide the following polynomials and find quotient and remainder:
3
2
1. x + 6 x + 8 x + 10 by x − 3
2. 2 x + 7 x + 9 x + 11 by x − 2
3
2
3. 3 x + 7 x + 10 x + 11 by 2 x − 3
3
2
2
3
2
4. x + 7 x + 9 x + 11 by x + x + 1
2
3
2
5. 2 x + 5 x + 9 x + 9 by x + 2 x + 1
6. 3 x 2 − 5 x − 10 by x + 1
7. x 2 − 3 x + 27 by 2 x + 1
8. 7 x 3 + 3 x + 2 by x + 1
9. x 4 + 3 x + 2 by x + 3
10. 3 x 2 + 5 x + 2 by 2 x + 1
114
Chapter
16
Factorization of Polynomials
16.1 Factorization of Quadratic Polynomials by inspection:
Type -1: Polynomial of the form x 2 + bx + c
Step 1: Split c into two parts such that their sum is b and product is c. Let
two parts are α and β
Step-2: Factorization of x 2 + bx + c will be ( x + α ).( x + β )
Example 1: Factorize x 2 + 7 x + 12
Step 1: Split 12 into the product of two numbers such that their sum is 7.
Such numbers are 4 and 3.
Step-2:∴ Factorization of x 2 + 7 x + 12 will be ( x + 4).( x + 3)
Example 2: Factorize x 2 − 6 x + 8
Step 1: Split 8 into the product of two numbers such that their sum is -6.
Such numbers are -4 and -2.
Step-2:∴ Factorization of x 2 − 6 x + 8 will be ( x − 4).( x − 2)
Example 3: Factorize x 2 − 3 x − 18
Step 1: Split -18 into the product of two numbers such that their sum is -3.
Such numbers are -6 and 3.
Step-2: ∴ Factorization of x 2 − 3 x − 18 will be ( x − 6).( x + 3)
Type -2: Polynomial of the form ax 2 + bx + c
Step 1: Split a × c into the product of two numbers such that their sum is
b. Let such numbers be α and β .
2
Step-2: Factorization of ax + bx + c will be
1
(ax + α ).(ax + β )
a
Be a Human Calculator
Example 4: Factorize 2 x 2 + 7 x + 6
Step 1: Split 2 × 6 = 12 into the product of two numbers such that their
sum is 7. Such numbers are 4 and 3.
Step-2:∴ Factorization of 2 x 2 + 7 x + 6 will be
1
(2 x + 4).(2 x + 3) = ( x + 2).(2 x + 3)
2
Example 5: Factorize 3 x 2 − 2 x − 8
Step 1: Split 3 × (-8) = -24 into the product of two numbers such that their
sum is -2. Such numbers are -6 and 4.
Step-2: ∴ Factorization of 3 x 2 − 2 x − 8 will be
1
(3 x − 6).(3 x + 4) = ( x − 2).(3 x + 4)
3
16.2 Factorization of Cubic Polynomials:
Type -1: Polynomial of the form x 3 + ax 2 + bx + c
Step-1: Let p ( x) = x 3 + ax 2 + bx + c .
Find all the possible factors of c including negative factors. Out of these
factors, let for x = α , we have p (x) =0, so α is a zero of x 3 + ax 2 + bx + c
and therefore, ( x − α ) will be a factor of x 3 + ax 2 + bx + c .
Step-2: Divide x 3 + ax 2 + bx + c by ( x − α ) using synthetic division
technique and get quotient which will be a quadratic polynomial.
Step-3: Factorize the quotient received in step-2 to get two factors. Write
down the answer by using factors obtained in step-1 and step-2.
Alternative Method (By inspection):
3
2
For polynomial p ( x) = x + ax + bx + c
(1) F
ind all possible factors of c including negative factors.
(2) I dentify three factors of c (let α , β , δ ) by inspection such that
2
+ bx
+ c+ δα =
α p+( βx) +=δx3=+−aax, αβ
+ βδ
b , and α.β.δ = −c
. x − β )(
. x −δ )
(3) T
herefore, x 3 + ax 2 + bx + c = ( x − α )(
116
Be a Human Calculator
Note: This method is very useful for those students whose observation
skills are strong. It is also useful for Multiple Choice Questions (MCQ)
where one can easily verify step-2 out of the available options.
Type -2: Polynomial of the form ax 3 + bx 2 + cx + d
Step-1: Let p(x) = ax3 + bx2 + cx + d
Find all possible factors of d including negative factors. The possible zeroes
of p(x) will be ±α, ±ß, ±γ, ±α/a, ±β/a and ±δ/a. Out of these numbers,
let for x = α , we have p(x) =0, so α is a zero of ax 3 + bx 2 + cx + d and
therefore, ( x − α ) will be a factor of ax 3 + bx 2 + cx + d .
Step 2: Divide ax 3 + bx 2 + cx + d by ( x − α ) using synthetic division
technique and get quotient which will be a quadratic polynomial. Let the
quotient be (ax 2 + px + q )
Write ax 3 + bx 2 + cx + d = ( x − α ).(ax 2 + px + q )
Factorize (ax 2 + px + q )
Step-3: Write down the answer by using factors obtained in step-1 and
step-3.
Example 1: Factorize x 3 − 4 x 2 − 7 x + 10
Solution:
Step-1: All possible factors of 10 are ±1, ±2, ±5, ±10
Step-2: Out of these factors, observe three numbers 5, 1 and -2 such that 5
× 1 × (-2) = -10, 5 + 1 + (-2) = 4 and 5 × 1+ 1 × (-2) + (-2) × 5 = -7.
Step-3: Therefore,
x 33 − 4 x 22 − 7 x + 10 = ( x 22 − 3 x − 10).( x − 1)
x − 4 x − 7 x + 10 = ( x − 3 x − 10).( x − 1)
=
( x + 2).( x − 5).( x − 1)
=
( x + 2).( x − 5).( x − 1)
117
Be a Human Calculator
Example 2: Factorize x3 - 2x2 - 5x + 6
Solution: Step – 1: Possible factors of 6 are ±1,±2,±3,±6
Step-2: By observation, 3, -2 and 1 are the numbers for which,
3 x (-2) x 1 = -6, 3 +(-2) + 1 = 2 and 3 x (-2) + (-2) x 1 + 1 x 3 = -5.
Step-3: Therefore, x3 - 2x2 - 5x + 6 = (x -3) (x +2) (x-1)
Example 3: Factorize 2x3 - 9x2 + 10x - 3
Solution: Step – 1: Possible zeroes of the polynomial are ±1,±3,±1/ 2,±3/ 2
Step-2: Putting x =1 in the given polynomial we have value of polynomial
as zero.
Therefore, (x-1) is a factor.
Step-3: Divide 2x3 - 9x2 + 10x – 3 by (x-1) using synthetic division, we
get quotient as = 2x2 - 7x + 3, which on factorization gives factors as
(x-3) (2x-1)
Step-4: Therefore, 2x3 - 9x2 + 10x – 3 = (x-1) (x-3) (2x-1)
Exercise 16
Total Questions 10
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Factorize the following polynomials into linear factors:
1. x 2 − 8 x + 15
2. x 2 + x − 42
3. 2 x 2 − 19 x + 24
4. x 2 + 4 x − 45
5. x 3 − 6 x 2 + 11x − 6
6. x 3 − 2 x 2 − 5 x + 6
7. 2 x 3 − 25 x 2 + 76 x − 32
118
Be a Human Calculator
8. x 3 − 11x 2 + 35 x − 25
9. x 3 + 7 x 2 − 6 x − 72
10. 2x4 + 15x3 + 5x2 - 15x -7
119
Chapter
17
Solving Equations
17.1 Linear Equations in two variables:
The standard form of linear equations in two variables is as follows:
a1 x + b1 y = c1
a2 x + b2 y = c2
Traditional Methods of Solving Linear Equations in two
variables:
(1) Elimination Method: Let us revisit the method through an example:
Example 1: Solve the following equations through elimination method:
2x + 3y = 7------(1)
3x + 7y = 11-----(2)
Step-I: To eliminate x, Multiplying equation (1) by 3 and equation (2) by
2, we have:
6x + 9y = 21------(3)
6x + 14y = 22----(4)
Step-II: Subtract equation (3) and (4) to get the value of y as follows:
-5y = -1 or y = 1/5
Step-III: Substitute the value of y in equation (1) to get the value of x as
follows:
2x + 3/5 = 7 or 2x = 7 – 3/5 =32/5 or x = 16/5
So, the solution of given equations is
x = 16/5 and y = 1/5
(2) Substitution Method: Let us revisit the method through an example:
Example 2: Solve the following equations through substitution method:
3x + 4 y = 10------(1)
5x + 6 y = 10------(2)
Be a Human Calculator
Step-I: Get the value of x from equation (1) in terms of y,
i.e. 3x = 10 – 4 y or x = (10-4 y )/3
Step-II: Substitute the value of x in equation (2) to get the value of y as
follows:
5(10 − 4 y )
5(10 − 4 y ) + 18 y
10 or
+ 6y =
= 10
3
3
50 − 20 y + 18 y
or 50 − 2 y =
30 or 2 y = 20 or y = 10
Step-III: Substitute the value of y in equation (1) to get the value of x as
follows:
3x + 40 = 10 or 3x = 10 – 40 =-30 or x = - 10
So, the solution of given equations is
x = - 10 and y = 10
(3) Cross Multiply Method
a1 x + b1 y = c1
a2 x + b2 y = c2
In this method, equations are solved using the following result:
−1
x
y
=
=
b1 c 2 − b2 c1 c1 a 2 − c 2 a1 a1 b2 − a 2 b1
Example 3: Solve the following equations through cross multiply method:
5x + 6 y = 7------(1)
8x + 9 y = 10----(2)
x
y
−1
= =
6 × 10 − 9 × 7 7 × 8 − 10 × 5 5 × 9 − 8 × 6
x
y −1
= =
⇒ x = −1, y = 2
−3 6 −3
122
Be a Human Calculator
(4) Graphical Method: As this method requires graph paper and so it is
not useful at all for the purpose of faster calculation.
Analysis of Traditional Methods: From the above discussion, it may
be seen that out of all the methods, cross multiply method is the shortest
and space saving method. Students, who are willing to solve the linear
equations in two variables traditionally, are advised to go through this
method only to save space and time.
Short cut methods of Solving Linear Equations in two variables: The
short-cut method of solving linear equation in two variables requires a close
observation of cross multiply method to make it further simpler and shorter:
a1 x + b1 y = c1
a2 x + b2 y = c2
(1) G
et D (by leaving coefficients of constant terms) by finding the
difference of cross product of coefficients of x and y as follows:
D = a1 b2 − a 2 b1
(2) G
et N x (by leaving the coefficients of x) by finding the difference
of cross product of coefficients of y and constant terms as
follows:
N x = b1 c 2 − b2 c1
(3) G
et N y (by leaving the coefficients of y) by finding the difference
of cross product of coefficients of x and constant terms as
follows:
N y = a1 c 2 − a 2 c1
(4) N
ow, found the answer using,
x=
Ny
−N x
,y=
D
D
Example 4: Solve the following equations:
8x + 3y = 11------(1)
5x + 7y = 9-------(2)
123
Be a Human Calculator
Solution: Here D = a1 b2 − a 2 b1 = 8.7 - 5.3 = 41
=
x
N x = b1 c 2 − b2 c1 = 3.9 - 7.11 = -50
N y = a1 c 2 − a 2 c1 = 8.9 - 5.11 = 17
N y 17
− N x − ( −50) 50
=
=
=
,y =
D
D
41
41
41
Note: Once students understand clearly how to get D, N x and N y ,
they may get their answer in only one step as follows:
Example 5: Solve the following equations:
3x - 5y = 7––––(1)
7x + 7y = -8––––(2)
Solution:
=
x
− N x − (( −5).( −8) − 7.7) 9
=
=
,
D
3.7 − 7.( −5)
56
y
=
N y 3.( −8) − 7.7 −73
=
=
D
56
56
17.2 Linear equations in three variables:
The standard form of linear equations in three variables is as follows:
a1 x + b1 y + c1 z = d 1
a 2 x + b2 y + c 2 z = d 2
a 3 x + b3 y + c 3 z = d 3
Traditional Methods of Solving Linear Equations in three variables
(Elimination Method): Let us revisit the method through an example.
Example 6: Solve the following equations:
5 x + 2 y + 3 z = 4 − − − −(1)
3 x + 4 y + 5 z = 6 − − − −(2)
2 x + 3 y + 4 z = 5 − − − −(3)
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Solution:
Step-I: Eliminate x from equations (1) and (2) by multiplying equation (1)
by 3 and equation (2) by 5 and subtracting them as follows:
15 x + 6 y + 9 z= 12 − − − −(1)
15 x + 20 y + 25 z= 30 − − (2)
On subtracting (1) and (2), we get:
−14 y − 16 z =
−18
or 7 y + 8 z = 9 − − − (4)
Step-II: Now, eliminate x from equations (2) and (3) by multiplying equation (2) by 2 and equation (3) by 3 and subtracting them as follows:
6 x + 8 y + 10 z= 12 − − − − (2)
6 x + 9 y + 12 z= 15 − − − − (3)
On subtracting (2) and (3), we get:
− y − 2 z =−3
or y + 2 z = 3 − − − −(5)
Step-III: Solve equation (4) and (5) to get y and z,
7 y + 8 z = 9 − − − −(4)
y + 2 z = 3 − − − −(5)
On solving, z = 2, y = -1
Step-IV: Substitute the values of y and z obtained in Step-III in equation
(1) to get x = 0.
Therefore, the solution is x = 0, y = -1 and z = 2.
Short cut method of Solving Linear Equations in three
variables:
a1 x + b1 y + c1 z = d 1
a 2 x + b2 y + c 2 z = d 2
a 3 x + b3 y + c 3 z = d 3
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Step-I: Observe the following four terms:
 a1 b1 c1 
 d 1 b1 c1 
 a1 d 1 c1 






D =  a 2 b2 c 2  , D x =  d 2 b2 c 2  , D y =  a 2 d 2 c 2 
a b c 
d b c 
a d c 
3
3
 3 3 3
 3 3 3
 3
 a1 b1 d 1 


D z =  a 2 b2 d 2 
a b d 
3
 3 3
Explanation: D contains coefficients of x, y, z retaining their position. Dx
is obtained by replacing the first column of D by coefficients of constant
terms, Dy is obtained by replacing the second column of D by coefficients
of constant terms and Dz is obtained by replacing the third column of D by
coefficients of constant terms
Step-II: The values of D, Dx, Dy and Dz may be obtained as follows:
Finding the value of D: Write first two columns of D again,
a1
b1
c1
a1
b1
a2
a3
b2
b3
c2
c3
a2
a3
b2
b3
D = ( a1 b2 c3 + b1 c 2 a3 + c1 a 2 b3)-(a3 b2 c1 + b3 c 2 a1 + c3 a 2 b1)
Finding the value of Dx:
d1
b1
c1
d1
b1
d2
d3
b2
b3
c2
c3
d2
d3
b2
b3
Dx: = (d 1 b2 c3 + b1 c 2 d 3 + c1 d 2 b3)-( d 3 b2 c1 + b3 c 2 d 1 + c3 d 2 b1)
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Be a Human Calculator
Finding the value of Dy:
a1
d1
c1
a1
d1
a2
a3
d2
d3
c2
c3
a2
a3
d2
d3
Dy = ( a1 d 2 c3 + d 1 c 2 a3 + c1 a 2 d 3 )-( a3 d 2 c1 + d 3 c 2 a1 + c3 a 2 d 1 )
Finding the value of Dz:
a1
b1
d1
a1
b1
a2
a3
b2
b3
d2
d3
a2
a3
b2
b3
Dz = ( a1 b2 d 3 + b1 d 2 a3 + d 1 a 2 b3 )-( a3 b2 d 1 + b3 d 2 a1 + d 3 a 2 b1 )
Step-III: Find the answer by using the following formulae:
Dy
x = Dx , y =
, z = Dz
D
D
D
Note: Initially students may find the process little difficult or lengthy but
after some practice, they will find that it is the fastest method to find the
solution of equations in three variables. For solving such questions fast,
students are advised to master the skills to get values of D, Dx, Dy and Dz.
Example 7: Solve the following equations:
x + 2 y + 3 z = 4 − − − −(1)
5 x + 6 y + z = 8 − − − −(2)
2 x + 3 y + 4 z = 5 − − − (3)
Solution:
1
5
2
2
6
3
3
1
4
1
5
2
2
6
3
D = (1.6.4 + 2.1.2 + 3.5.3)-(2.6.3 + 3.1.1 + 4.5.2) = 73 - 79 = -6
127
Be a Human Calculator
4
8
5
2
6
3
3
1
4
4
8
5
2
6
3
1
5
2
4
8
5
3
1
4
1
5
2
4
8
5
1
5
2
2
6
3
4
8
5
1
5
2
2
6
3
D x = (4.6.4+2.1.5+3.8.3)-(5.6.3+3.1.4+4.8.2) = 178-166=12
D y = (1.8.4+4.1.2+3.5.5)-(2.8.3+5.1.1+4.5.4) = 115-133=-18
D z = (1.6.5+2.8.2+4.5.3)-(2.6.4+3.8.1+5.5.2) = 122-122=0
Dx 12
Dy
−18
Dz
0
x ===
− 2, y = = =
3, z == =
0
D −6
D
−6
D
−16
17.3 Quadratic Equations:
The standard form of quadratic equation is as follows:
ax 2 + bx + c= 0, a ≠ 0
Traditional Methods of Solving Quadratic Equations:
1. Factorization Method:
Example 1: Solve x 2 − 7 x + 12 =
0 by factorization method.
x 2 − 7 x + 12 =
0
x 2 − 4 x − 3 x + 12 =
0
x( x − 4) − 3( x − 4) =
0
( x − 4).( x − 3) =
0
x = 3, 4
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Be a Human Calculator
Example 2: Solve 2 x 2 + 10 x + 8 =
0 by factorization method.
2 x 2 + 10 x + 8 =
0
2 x2 + 8x + 2 x + 8 =
0
2 x( x + 4) + 2( x + 4) =
0
( x + 4).(2 x + 2) =
0
x =−4, −1
2. Completion of Square Method:
00 by completion of square method.
Example 3: Solve x 2 −+76xx ++128 =
x2 + 6x + 8 = 0
( x) 2 + 2.3.x + (3) 2 − (3) 2 + 8 = 0
( x + 3) 2 − 1 =
0
( x + 3) 2 =
1
x + 3 =±1
x=
−2, − 4
3. Quadratic Formula Method:
For any quadratic equation:
ax 2 + bx + c= 0, a ≠ 0
x=
−b ± b 2 − 4ac
2a
Example 4: Solve x 2 + 3 x + 2 = 0 by quadratic formula method.
−b ± b 2 − 4ac −3 ± 32 − 4.1.2 −3 ± 1
=
=
2a
2.1
2
∴ x =−2, −1
=
x
Analysis of Traditional Methods: From the above discussion, it may
be seen that out of all the methods, Quadratic Formula method is the
shortest and space saving method and can be used to solve any quadratic
129
Be a Human Calculator
equation even with imaginary or irrational roots. Factorization method
may work only when roots are rational.
Short cut methods of Solving Quadratic Equations:
Type -1: Equations of the form x 2 + bx + c =
0
Step 1: Factorize c into two parts such that their sum is b. Let such factors
are α and β
Step-2: Solution is x =
−α , − β
Example 5: Solve x 2 + 11x + 24 =
0 by inspection.
Solution: Observe various factors of 24 in mind,
Ist
2nd
Sum
Factor Factor
1
24
8
11
4
6
10
2
25
12
3
14
We may see that, 24 = 8 × 3, also 11 = 8 + 3
So, the requisite factors are 8 and 3 and the solution is x = -8 and -3
Example 6: Solve x 2 − 7 x − 60 =
0 by inspection.
Solution: Observe various factors of -60 in mind,
Ist
2nd
Sum
Ist
2nd
Sum
Factor Factor
Factor Factor
1
-60
-59
-1
60
59
4
-15
-11
-4
15
11
2
3
-30
-20
5
-12
6
-10
-28
-2
-17
-3
-7
-5
-4
-6
130
30
20
12
10
28
17
7
4
Be a Human Calculator
By observation,
-12 × 5 = -60 and (-12) + 5 = -7
So, the requisite factors are -12 and 5 and solution of the given equation
is x = 12 and –5.
Type -2: Equations of the form ax 2 + bx + c =
0
Step 1: Factorize a.c into two parts such that their sum is b. Let such
factors are α and β .
Step-2: Solution is x = −α / a,− β / a
Example 7: Solve 3 x 2 − 2 x − 8 = 0
Solution: See various factors of 3 × (-8) = -24
Ist
2nd
Sum
Ist
2nd
Sum
Factor Factor
Factor Factor
1
-24
-23
-1
24
23
4
-6
-2
-4
6
2
2
3
-12
-8
-10
-2
-5
-3
12
8
It may be observed that, 4 × (-6) = -24, and 4 + (-6) = -2.
10
5
So, the requisite factors are 4 and -6 and solution of the given equation is
x = -4/3 and – (-6)/3 = 2
17.4 Solving Equations of degree more than 2:
The method for solving these equations will be same as of factorizing
them. The same has already been explained in chapter 15.
17.5 Some Special Equations:
(a) Equations of the form:
x+ y =a
x− y =b
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Method: Add both the equations to get 2x = a+b or x = (a+b)/2.
Subtract both the equations to get 2y = a-b or y = (a-b)/2
Thus, solution is x = (a+b)/2 and y = (a-b)/2.
Example 8: Solve following equations:
x+ y =
92
x− y =
42
Solution:=
x
92 + 42 134
92 − 42 50
=
= 67,=
y
= = 25
2
2
2
2
(b) Equations of the form:
ax + by = c1
bx + ay = c2
Method: Understand the method through the following example.
Example 9: Solve following equations:
37 x + 31 y =
105
31x + 37 y =
99
Solution: Add both the equations to get
68 (x +y) = 204 or x + y = 3-----(3)
Subtract both the equations to get
6 (x - y) = 6 or x - y = 1-----(4)
From (3) and (4) we get,
X = (3 + 1)/2 = 2 and y = (3-1)/2 = 1 (Method 17. 5 (a))
(c) Equations of the form:
( x + a ).( x + b) = ( x + c).( x + d ), where ab = cd
Traditional Method: Multiply and solve.
Short cut Method: Answer will always be x = 0.
Example 10: Solve ( x + 2).( x + 3) = ( x + 6).( x + 1)
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Solution:
Traditional Method:
On Multiplication, we get
x 2 + 5x + 6 = x 2 + 6 x + 6
⇒ 5x = 6 x ⇒ x = 0
Short cut Method:
As 2.3 = 6.1, we know that
x=0
(d) Equations of the form:
( x + a ).( x + b) = ( x + c).( x + d ), where ab ≠ cd
Traditional Method: Multiply and solve.
Short cut Method: Answer will always be
x=
cd − ab
a+b−c−d
if a + b # c + d and no solution if a + b = c +d
Example 11: Solve (x + 2).(x + 3) = (x + 4).(x + 2)
Solution:
Traditional Method:
Short cut Method:
On Multiplication, we get
x=
x2 + 5x + 6 = x2 + 6x + 8
cd − ab
a+b−c−d
x = −2
(e) Equations of the form:
1
1
+
=
0
ax + b cx + d
Traditional Method: Take LCM and solve.
Short cut Method: By observation, answer will be given by
(ax + b) + (cx + d ) =
0
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Example 12: Solve
1
1
+
=0
2 x + 3 3x + 9
Traditional Method:
Short cut Method:
1
1
0
+
=
2 x + 3 3x + 9
(3 x + 9) + (2 x + 3)
=0
(2 x + 3)(3 x + 9)
(3 x + 9) + (2 x + 3) =
0
5 x + 12 =0 ⇒ x =−12 / 5
By observation, we know that
(3 x + 9) + (2 x + 3) =
0
5 x + 12 =0 ⇒ x =−12 / 5
(f) Equations of the form:
ax + b ex + f
N1 N 2
=
⇒ =
, where N 1 + N 2 = D1 + D 2
cx + d
gx + h
D1 D 2
Traditional Method: Cross multiply and solve.
Short cut Method: Answer will always be given by N 1 + N 2 = D1 + D 2 = 0
and N 1 − D1 = N 2 − D 2 = 0
Example 13: Solve
2 x + 3 3x + 6
=
4x + 5 x + 4
Traditional Method:
2 x + 3 3x + 6
=
4x + 5 x + 4
(2 x + 3).( x + 4) = (4 x + 5).(3 x + 6)
2 x 2 + 8 x + 3 x + 12= 12 x 2 + 24 x + 15 x + 30
2 x 2 + 11x + 12= 12 x 2 + 39 x + 30
10 x 2 + 28 x + 18 = 0 ⇒ 5 x 2 + 14 x + 9 = 0
5x2 + 5x + 9 x + 9 =
0
5 x( x + 1) + 9( x + 1) = 0 ⇒ (5 x + 9).( x + 1) = 0
−9 / 5, −1
x=
134
Be a Human Calculator
Short cut Method:
Here, N 1 + N 2 = (2 x + 3) + (3 x + 6) = 5 x + 9
Also, D1 + D 2 = (4 x + 5) + (2 x + 4) = 5 x + 9 so the answer will be given by (2 x + 3) + (3 x + 6) = 0, ( x + 4) + (4 x + 5) = 0 ⇒ x = −9 / 5
(2 x + 3) − (4 x + 5) = 0, (3 x + 6) − ( x + 4) = 0 ⇒ x = −1
and
(g) Equations of the form:
1
1
1
1
+
=
+
x+a x+b x+c x+d
Where, (x+a) + (x+b) = (x+c) + (x+d)
Traditional Method: Take LCM of both sides and cross multiply to solve.
Short cut Method:The solution will be given by
(x+a) + (x+b) = (x+c) + (x+d) = 0
Example 14: Solve
1
1
1
1
+
=
+
x+3 x+9 x+5 x+7
Traditional Method:
1
1
1
1
+
=
+
x+3 x+9 x+5 x+7
( x + 9) + ( x + 3) ( x + 7) + ( x + 5)
=
( x + 3)( x + 9)
( x + 5)( x + 7)
(2 x + 12)( x + 5)( x + 7) =(2 x + 12)( x + 3)( x + 9)
(2 x + 12)( x 2 + 12 x + 35) = (2 x + 12)( x 2 + 12 x + 27)
(2 x + 12)( x 2 + 12 x + 35) − (2 x + 12)( x 2 + 12 x + 27) =
0
(2 x + 12) ( x 2 + 12 x + 35) − ( x 2 + 12 x + 27)  = 0
(2 x + 12).8 = 0 ⇒ (2 x + 12) = 0 ⇒ x = −6
Short cut Method:
Here ( x + 9) + ( x + 3) = ( x + 7) + ( x + 5) = 2 x + 12
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Be a Human Calculator
Therefore, the solution is given by,
(2 x + 12) =0 ⇒ x =−6
Example 15: Solve
1
1
1
1
−
=
−
x+2 x+3 x+5 x+6
Short cut Method: Transpose the given equation to get,
1
1
1
1
+
=
+
x+2 x+6 x+5 x+3
Here ( x + 2) + ( x + 6) = ( x + 5) + ( x + 3) = 2 x + 8
Therefore, the solution is given by,
(2 x + 8) = 0 ⇒ x = −4
Exercise 17
Total Questions 10
Ideal Time 5 Min
(Students are advised to practice this exercise till they are able to
solve all 10 questions in 5 minutes)
Solve the following equations:
1. 4 x − 5 y =
−1, 7 x − 3 y =
4
2. 3 x + 7 y =
17, 7 x − 5 y =
−3
3. 2 x 2 − 19 x + 24 =
0
4. 3 x + 7 y + 2 z = 12, 7 x − 5 y + 3 z = 5,3 x + 5 y + 2 z = 10
5. x 3 − 6 x 2 + 11x − 6 =
0
6. 3 x + 7 y + 2 z= 23, 7 x − 5 y + 3 z= 6,3 x + 5 y + 2 z= 19
7. 2 x 3 − 25 x 2 + 76 x − 32 =
0
8. x 3 − 11x 2 + 35 x − 25 =
0
9. x 3 + 7 x 2 − 6 x − 72 =
0
10. 2 x 4 + 15 x 3 + 5 x 2 − 15 x − 7 =
0
136
ANSWERS
Be a Human Calculator
Exercise 1.1
1.
4.
7.
10.
6
7
2
3
2.
5.
8.
1.
4.
7.
10.
766
89090
78025
010
2.
5.
8.
1.
24 3 10.
1 0 13
4.
7.
1.
4.
7.
10.
1.
2.
3.
4.
5.
6.
7.
6543
011
114326
2.
3.
6.
9.
476173
832676
321095
3 5 4 3
1 111 5.
1 11 4 3 3 4
8.
3.
6.
2.
5.
8.
4768 1929 19881 139
3.
6.
9.
524 2 2
2 3 3 3 24
9. 1 3 2 11 05
Exercises 1.3 (b)
Exercises 2
638625
180510
3884915
5813958
108645
767989
676129
0
0
8
Exercises 1.3 (a)
220 3 5
658
25215 81355
3.
6.
9.
Exercises 1.2
11 1 2 3 177
3
1
8
726
13799
998
Be a Human Calculator
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
132092.74723
61967.1043
13356.0603
4477.79935
5026.5825
78713.08903
4393.37424
38966.8665
64765.9686
11050.06484
243219.38018
689562.72344
132105.4743
5657278.803613
58851.91158
857210.2414
5352494.82665
3542066.9966
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
10361
72314
69889
2686989
31213
610998
498707
48760.10077
51227.8923
2929.0197
1.
2.
3.
4.
491779
71218
3869003
5329274
Exercises 3.1 and 3.2
Exercises 3.3
140
Be a Human Calculator
5.
6.
7.
8.
9.
10.
96867
616897
559865
131780.51723
51296.2243
6284.9897
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
984
4368
2457
6586
4104
5655
3268
2139
3486
4257
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
99292
438141
249249
662013
412783
568323
330812
218356
351513
429103
1.
2.
16892
43736
Exercises 4.1
Exercises 4.2
Exercises 4.3
3.
4.
42606
33558
141
Be a Human Calculator
Exercises 4.4
Question
Answer
1
69576
2
5489645
5
5332114
3
4
1055184
3015
6
670592745
9
353808
12
10488
7
8
6141
36342
10
384426
11
5313
13
1329615
16
4518486
19
263925
14
71799
15
132678
17
338688
18
26910
20
2652
142
Be a Human Calculator
Exercises 4.5
Question
Answer
1
10506
4
9024
2
11772
3
9702
5
990024
8
1003968
11
1006008
14
999919
6
2756
7
9984
9
3068
10
11663
12
5304
13
1001985
15
251502
16
248003
17
1003002
20
9579
18
3363
19
999964
143
Be a Human Calculator
Exercises 4.6
Question
Answer
1
720099
4
3452544
7
452133
2
9024
3
13574
5
1100099
6
560291
8
39865
9
10
6779322
166535
Exercises 4.7
Question
Multiply
1
1061106
2
1011038040
5
1009596
3
4
941094
994010994
6
1001983968
9
1003935744
7
8
10
124836
113122688
986050946
144
Be a Human Calculator
Exercises 5.1
Question
Number
2
4
8
16
32
1
998
Yes
No
No
No
No
1024
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
No
No
No
5
12345
123464
Yes
Yes
Yes
No
No
Yes
No
No
No
No
100400
Yes
Yes
Yes
Yes
No
8
9962
576775338
Yes
No
No
No
No
54676
Yes
Yes
No
No
No
99789
No
No
No
No
No
2
3
9967
4
6
7
9
10
Exercises 5.2
Question
Number
5
25
125
625
1
990
Yes
No
No
No
1025
Yes
Yes
No
No
Yes
No
No
No
12345
Yes
No
No
No
5
9960
123465
Yes
No
No
No
Yes
No
No
No
100400
Yes
Yes
No
No
8
9960
576775335
Yes
No
No
No
54675
Yes
Yes
No
No
91250
Yes
Yes
No
No
2
3
4
6
7
9
10
145
Be a Human Calculator
Exercises 5.3
Question
Number
3
9
1
990
Yes
Yes
No
No
Yes
No
Yes
No
123465
Yes
No
Yes
No
No
No
576775335
Yes
No
Yes
Yes
No
No
2
3
4
5
6
7
8
9
10
1025
9960
12345
9960
100400
54675
91250
Exercises 5.4
Question
Answer
1
Yes
4
Yes
2
No
3
Yes
5
Yes
6
Yes
7
No
8
No
9
No
10
No
146
Be a Human Calculator
Exercises 5.5
Question
Number
7
29
31
37
1
990
No
No
No
No
1025
No
No
No
No
No
No
No
No
No
No
No
No
5
10307
2244858
Yes
No
No
No
No
No
No
No
100400
No
No
No
No
8
9724
576775335
No
No
No
No
54675
No
No
No
No
91250
No
No
No
No
2
3
4
6
7
9
10
9966
Exercises 5.6
Question
Number
12
14
15
18
1
990
No
No
Yes
Yes
1025
No
No
No
No
No
No
No
No
10307
No
No
No
No
5
9966
2244858
No
Yes
No
No
No
No
No
No
100400
No
No
No
No
8
9724
576775335
No
No
Yes
No
54675
No
No
Yes
No
91250
No
No
No
No
2
3
4
6
7
9
10
147
Be a Human Calculator
Exercise 6
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
112.47
350
226.87
726.88
121
1350
2823.48
3234.18
15778.34
121.8
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Quotient : 99, remainder : 74
Quotient : 12, remainder : 597
Quotient : 51, remainder : 740
Quotient : 6, remainder : 760
Quotient : 83, remainder : 244
Quotient : 383, remainder : 17
Quotient : 997, remainder : 62
Quotient : 27331, remainder : 4
Quotient : 630, remainder : 688
Quotient : 2503, remainder : 91
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
8230.4
514.375
3154.94
321.619
2992.88
2006.16
940.62
853666.77
10900.92
10088.77
Exercise 7.1
Exercise 7.2
148
Be a Human Calculator
Exercise 8.1
1. 625
5. 4225
9. 11025
2. 1225
6. 5625
10. 13225
1. 2601
5. 3025
9. 3481
2. 2704
6. 3136
1. 6084
5. 6889
9. 5476
2. 4489
6. 2401
10. 3844
1. 9604
5. 9216
9. 9801
2. 996004
6. 1018081
10. 998001
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
3. 2025
7. 7225
4. 3025
8. 9025
Exercise 8.2
3. 2809
7. 3249
4. 2916
8. 3364
Exercise 8.3
3. 1849
7. 8836
4. 7921
8. 4096
Exercise 8.4
3. 1004004
7. 992016
Exercise 8.5
45796
15129
11943936
16384
294849
1038361
605284
4553956
54756
207936
149
4. 9409
8. 1008016
Be a Human Calculator
Exercise 9.1
Number
Cube
Number
Cube
11
1331
41
68921
14
2744
44
85184
12
13
15
16
17
18
19
20
21
1728
2197
5832
9261
28
29
30
31
32
33
34
35
36
37
38
39
40
53
54
55
17576
56
19683
57
21952
58
24389
59
27000
60
29791
61
32768
62
35937
63
39304
64
42875
65
46656
66
50653
67
54872
68
59319
64000
69
150
97336
125000
52
13824
91125
50
51
12167
79507
103823
49
8000
74088
47
48
6859
15625
27
46
4913
25
26
45
4096
10648
24
43
3375
22
23
42
70
110592
117649
132651
140608
148877
157464
166375
175616
185193
195112
205379
216000
226981
238328
250047
262144
274625
287496
300763
314432
328509
343000
(Contd.)
Be a Human Calculator
Number
Cube
Number
Cube
71
357911
86
636056
74
405224
89
704969
72
73
75
76
77
78
79
80
81
82
83
84
85
373248
87
389017
88
421875
90
438976
91
456533
92
474552
93
493039
94
512000
95
531441
96
551368
97
571787
98
592704
99
614125
658503
681472
729000
753571
778688
804357
830584
857375
884736
912673
941192
970299
Exercise 9.2
Question
Answer
Question
Answer
1
753571
778688
804357
830584
857375
884736
912673
941192
970299
11
976191488
14
985074875
2
3
4
5
6
7
8
9
10
12
13
15
16
17
18
19
973242271
20
151
979146657
982107784
988047936
991026973
994011992
997002999
998800479936
999100269973
Be a Human Calculator
Exercise 10.1
Question
Answer
Question
Answer
1
21
26
71
4
36
29
74
2
3
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
24
27
32
28
39
30
41
31
47
32
48
33
49
34
51
35
52
36
53
37
54
38
56
39
57
40
58
41
59
42
61
43
62
44
63
45
64
46
66
47
67
48
68
49
69
50
152
72
73
75
76
77
78
79
81
82
83
84
86
87
88
89
91
93
94
95
96
97
98
99
Be a Human Calculator
Exercise 10.2
Question
Answer
Question
Answer
1
101
26
291
4
109
29
298
2
3
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
103
27
106
28
121
30
127
31
131
32
134
33
137
34
139
35
141
36
147
37
149
38
151
39
156
40
158
41
159
42
161
43
168
44
169
45
174
46
198
47
212
48
256
49
279
50
153
293
296
299
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
321
Be a Human Calculator
Exercise 10.3
Question
Answer
Question
Answer
1
3.46
26
17.02
4
6.85
29
22.96
2
3
5
6
7
8
4.79
5.83
15
16
17
18
19
20
21
22
23
24
25
32
9.43
11.44
14
31
8.83
12
13
30
8.18
10.14
11
28
7.48
9
10
27
33
34
10.58
35
11.09
36
37
11.74
38
11.87
39
12.12
40
12.61
12.69
13.34
47
16.09
48
16.67
49
16.97
50
154
28.34
31.95
35.48
40.97
45.34
53.75
58.41
61.8
64.2
78.26
46
15.3
27.04
44
45
14.56
25.08
71.57
43
13.19
20.29
41
42
12.92
17.86
75.3
76.93
82.39
84.34
90.74
96.5
98.94
99.65
Be a Human Calculator
Exercise 11.1
Question
Answer
Question
Answer
1
11
31
45
4
14
34
48
2
3
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
12
32
13
33
15
35
16
36
17
37
18
38
19
39
21
40
22
41
23
42
24
43
25
44
26
45
27
46
28
47
29
48
31
49
32
50
33
51
34
52
35
53
36
54
37
55
38
56
41
57
42
58
43
44
59
155
60
46
47
49
51
52
53
54
55
56
57
58
59
61
62
63
64
65
66
67
68
39
71
72
73
74
75
76
77
(Contd.)
Be a Human Calculator
Question
Answer
Question
Answer
61
78
72
88
64
81
75
92
62
63
65
66
67
68
69
70
71
79
73
80
74
82
76
83
77
84
78
85
79
69
80
86
81
87
82
Exercise 11.2
90
91
93
94
95
96
97
98
99
Question
Answer
Question
Answer
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
101
103
106
109
111
113
118
121
126
127
131
134
137
139
141
144
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
147
151
153
154
157
161
179
192
212
245
278
312
333
368
412
456
156
(Contd.)
Be a Human Calculator
Question
Answer
Question
Answer
33
34
35
36
37
38
39
40
41
512
579
612
634
712
734
775
777
812
42
43
44
45
46
47
48
49
50
834
888
912
934
946
967
978
996
998
Question
Answer
Question
Answer
1
5.29
11
16.82
4
15.16
14
44.46
7
14.77
10
4.45
Exercise 11.3
2
6.53
3
4.97
5
5. 33
8.63
16
9.22
17
9.96
9
3.78
15
9.14
8
20.56
13
6.6
6
1. 14
12
8.64
18
12.87
6.34
19
20.7
20
8.49
Exercise 12.1
17
30
2. 15
59
84
3.
25
33
6. 2
11
18 9
10
7. 48
18
187
157
4. 32
8. 19
29
104
113
120
Be a Human Calculator
Exercise 12.2
Question
Answer
Question
Answer
1
35/99
6
474/495
2
213/999
7
2
9
3
1
29
99
8
4
4
7/9
9
234/999
5
21
10
233
990
Exercise 13
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1
LCM = 2862, HCF = 2
LCM = 4452, HCF =4
LCM = 318912, HCF =8
LCM = 3822, HCF =2
LCM = 3250, HCF =25
LCM = 9261, HCF =441
LCM = 10620, HCF =10
LCM = 130680, HCF =198
LCM = 5544, HCF =36
LCM =2112, HCF =96
158
79
99
12
26
45
Be a Human Calculator
Exercise 15
2
1.
Quotient: x + 9 x + 35
2.
Remainder: 115
Quotient: 2 x 2 + 11x + 31
Remainder: 73
3. Quotient: 3 x 2 + 23 x + 109
2
4
8
4.
Remainder: 415/8
Quotient: x + 6
Remainder: 2 x + 5
5.
Quotient: 2 x + 1
Remainder: 5 x + 8
6.
7.
8.
9.
10.
Quotient: 3x-8
Remainder: -2
Quotient: x/2 - 7/4
Remainder: 115/4
Quotient: 7x2-7x+10
Remainder: -8
Quotient: x3-3x2+9x-24
Remainder: 74
Quotient: 3x/2+7/4
Remainder: 1/4
Exercise 16
1. ( x − 3).( x − 5)
2. ( x − 6).( x + 7)
3. (2x- 3).(x-8)
4. ( x − 5).( x + 9)
5. ( x − 1).( x − 2).( x − 3)
6. ( x − 1).( x + 2).( x − 3)
7. (2x-1).(x-4).(x-8)
8. ( x − 1).( x − 5).( x − 5)
9. ( x − 3).( x + 4).( x + 6)
10. ( x − 1).( x + 1).( x + 7).(2 x + 1)
159
Be a Human Calculator
Exercise 17
1. x = 1, y = 1
2. x = 1, y = 2 3. x = 3 / 2, y = 8
4. x = 1, y = 1, z = 1
5. x = 1,2,3
6. x = 1, y = 2, z = 3
7. x = 1 / 2,4,8
8. x = 1,5,5
9. x = 3,−4,−6
10. x = 1,−1,−7,−1 / 2
160