Fundamentals of Quantitative Aptitude

QUANTITATIVE
APTITUDE
Pawan Mishra

Important Concepts and
Formulas - Numbers
 Number Sets and Properties of Numbers

 Counting Numbers (Natural numbers) : 1, 2, 3 ...

 Whole Numbers : 0, 1, 2, 3 ...

 Integers : -3, -2, -1, 0, 1, 2, 3 ...

 Rational Numbers
 Rational numbers can be expressed as ab where a and b are
integers and b≠0
Examples: 112, 42, 0, −811 etc.
 All integers, fractions and terminating or recurring decimals are
rational numbers.

Irrational Numbers
Any number which is not a rational number is an irrational
number. In other words, an irrational number is a number
which cannot be expressed as ab where a and b are
integers.
For instance, numbers whose decimals do not terminate and do
not repeat cannot be written as a fraction and hence they are
irrational numbers.
Example : π, 2√, (3+5√), 43√ (meaning 4×3√), 6√3 etc
Please note that the value of π = 3.14159 26535 89793 23846
26433 83279 50288 41971 69399 37510 58209 74944 59230
78164 06286 20899 86280 34825 34211 70679...
We cannot π as a simple fraction (The fraction 22/7 = 3.14.... is
just an approximate value of π)

Real Numbers
Real numbers include counting numbers, whole numbers,
integers, rational numbers and irrational numbers.
Surds
Let a be any rational number and n be any positive integer
such that a√n is irrational. Then a√n is a surd.
Example : 3√, 10−−√6, 43√ etc
Please note that numbers like 9√, 27−−√3 etc are not surds
because they are not irrational numbers
Every surd is an irrational number. But every irrational number
is not a surd. (eg : π , e etc are not surds though they are
irrational numbers.)

ADDITION, SUBTRACTIONAND
MULTIPLICATION RULES FOR EVENAND
ODD NUMBERS

Addition Rules for Even and Odd
Numbers
The sum of any number of even numbers is
always even
The sum of even number of odd numbers is
always even
The sum of odd number of odd numbers is
always odd

Subtraction Rules for Even and
Odd Numbers
The difference of two even numbers is always
even
The difference of two odd numbers is always
even

Multiplication Rules for Even
and Odd Numbers
The product of even numbers is always even
The product of odd numbers is always odd
If there is at least one even number multiplied
by any number of odd numbers, the product is
always even

3STEPMULTIPLICATIONTRICK-A
SHORTCUTMETHOD

Step 1
First step is same as conventional method, here
we multiply 2 with 2.

Step 2
This is an interesting step. Now multiply last digit
first value and first digit of second value and
vice-versa. Then we add outcomes. But we
need the last number that is 8 here.

Step 3
This is the last step, in this step we do
multiplication ten's digit of both value and add
the remainder from previous calculation. That's
it, we completed the calculation in 3 steps
instead of six steps.

We can use this method for multiplication of
three or even four digit numbers but time
management is really important in IBPS exam
and other recruitment exams so for longer
calculations, estimation is the best trick. I will
post an article about how to do long
calculations using estimation and result is 95%
accurate which is enough to arrive at answer.
As two of the readers namely Rahul and Ansh
have requested me to use this technique in
longer calculations multiplications. I am
updating this article.

Multiplication of 3 digit
numbers
In this example I will multiply 432 with 346. Now
the 3 step multiplication method will become 5
step. This method can be used for 4 and even
5 digit numbers but as in bank exams there is
lack of time available for calculations I
recommend you to use approximation for long
calculations.

Step 1
Step 2
Step 3
Step 4
Step 5
In case you find any difficulty to
understand the above
multiplication method then ask
your question in the comments. I
will try to answer every query
asap

Divisible By
One whole number is divisible by another if
the remainder we get after the division is
zero.
Examples
36 is divisible by 4 because 36 ÷ 4 = 9 with a
remainder of 0.
36 is divisible by 6 because 36 ÷ 6 = 6 with a
remainder of 0.
36 is not divisible by 5 because 36 ÷ 5 = 7 with
a remainder of 1.

Divisibility Rules
By using divisibility rules we can easily find out
whether a given number is divisible by another
number without actually performing the
division. This helps to save time especially
when working with numbers.

DivisibilityRule Description Examples
Divisibility
by 2
A number is divisible by
2 if the last digit is even.
i.e., if the last digit is 0 or
2 or 4 or 6 or 8
Example1: Check if 64 is
divisible by 2.
The last digit of 64 is 4
(even). Hence 64 is divisible
by 2
divisible by 2.
The last digit of 69 is 9 (not
even). Hence 69 is not
divisible by 2

Divisibility
by 3
A number is divisible
by 3 if the sum of the
digits is divisible by 3
divisible by 3.
3 + 8 + 7 = 18.
18 is divisible by 3. Hence
387 is divisible by 3
divisible by 3.
4 + 2 + 1 = 7.
7 is not divisible by 3.
Hence 421 is not divisible
by 3

Divisibility by
4
4 if the number formed
by the last two digits is
divisible by 4.
divisible by 4.
Number formed by the last
two digits = 16.
16 is divisible by 4. Hence
416 is divisible by 4
divisible by 4.
Number formed by the last
two digits = 81.
81 is not divisible by 4.
Hence 481 is not divisible
by 4

Divisibility by
5
5 if the last digit is
either 0 or 5.
divisible by 5.
Last digit is 5. Hence 305 is
divisible by 5.
divisible by 5.
divisible by 5.
divisible by 5.
not divisible by 5.

Divisibility by
6
6 if it is divisible by both
2 and 3.
Example1: Check if 546 is divisible by 6.
546 is divisible by 2. 546 is also divisible by 3. (Check
the divisibility rule of 2 and 3 to find out this)
Hence 546 is divisible by 6
633 is not divisible by 2 though 633 is divisible by 3.
(Check the divisibility rule of 2 and 3 to find out this)
Hence 633 is not divisible by 6
635 is not divisible by 2. 635 is also not divisible by 3.
(Check the divisibility rule of 2 and 3 to find out this)
428 is divisible by 2 but 428 is not divisible by
3.(Check the divisibility rule of 2 and 3 to find out
this)

Divisibility by
7
To find out if a number
is divisible by 7, double
the last digit and
subtact it from the
number formed by the
remaining digits.
Repeat this process until
we get at a smaller
number whose
divisibility we know.
If this smaller number is
0 or divisible by 7, the
original number is also
divisible by 7.
Example1: Check if 349 is divisible by
7.
Given number = 349
34 - (9 × 2) = 34 - 18 = 16
16 is not divisible by 7. Hence 349 is
not divisible by 7
7.
Given number = 364
36 - (4 × 2) = 36 - 8 = 28
28 is divisible by 7. Hence 364 is also
divisible by 7
Example3: Check if 3374 is divisible
by 7.
Given number = 3374
337 - (4 × 2) = 337 - 8 = 329
32 - (9 × 2) = 32 - 18 = 14
divisible by 7.
Hence 3374 is also divisible by 7.

Divisibility by 8
A number is divisible by 8 if
the number formed by the last
three digits is divisible by 8.
by 8.
The number formed by the last three
digits of 7624 = 624.
divisible by 8.
divisible by 8.
digits of 129437464 = 464.
464 is divisible by 8. Hence 129437464
is also divisible by 8.
by 8.
digits of 737460 = 460.
460 is not divisible by 8. Hence 737460
is also not divisible by 8.

Divisibility by 9
the sum of its digits is divisible
by 9.
(Please note that we can apply
this rule to the answer again
and again if we need)
3 + 6 + 7 + 8 + 2 + 1 = 27
divisible by 9.
4 + 7 + 1 + 2 + 8 = 22
22 is not divisible by 9. Hence 47128 is not
divisible by 9.
9.
4 + 9+ 7 + 5 + 2 + 9 + 1 + 9 + 8 + 9= 63
Since 63 is big, we can use the same method to
see if it is divisible by 9.
6 + 3 = 9
9 is divisible by 9. Hence 63 is also divisible by
9.
Hence 4975291989 is also divisible by 9.

Divisibility by 10
the last digit is 0.
divisible by 10.
divisible by 10.
divisible by 10.
Last digit is not 0. Hence 5462 is
not divisible by 10

Divisibility by 11
To find out if a number is divisible by 11, find the
sum of the odd numbered digits and the sum of the
even numbered digits.
Now substract the lower number obtained from the
bigger number obtained.
If the number we get is 0 or divisible by 11, the
original number is also divisible by 11.
8 + 1 + 6 = 15
5 + 3 = 8
15 - 8 = 7
7 is not divisible by 11. Hence 85136 is not divisible by 11.
2 + 3 + 1 + 2 = 8
7 + 7 + 5 = 19
19 - 8 = 11
11 is divisible by 11. Hence 2737152 is also divisible by 11.
9 + 7 = 16
5 = 5
16 - 5 = 11
9 + 4 = 13
5 + 8 = 13
13 - 13 = 0
We got the difference as 0. Hence 9548 is divisible by 11.

Divisibility by
12
12 if the number is
divisible by both 3 and 4
720 is divisible by 3 and 720 is also divisible by 4. (Check
the divisibility rules of 3 and 4 to find out this)
Hence 720 is also divisible by 12
916 is not divisible by 3 , though 916 is divisible by
4.(Check the divisibility rules of 3 and 4 to find out this)
921 is divisible by 3. But 921 is not divisible by 4.(Check
the divisibility rules of 3 and 4 to find out this)
827 is not divisible by 3. 827 is also not divisible by
4.(Check the divisibility rules of 3 and 4 to find out this)

Divisibility by
13
To find out if a number
is divisible by 13,
multiply the last digit by
4 and add it to the
remaining digits.
Repeat this process until
we get at a smaller
number whose
divisibility we know.
divisible by 13, the
original number is also
divisible by 13.
Given number = 349
34 + (9 × 4) = 34 + 36 = 70
70 is not divisible by 13. Hence 349 is not divisible by 349
Given number = 572
57 + (2 × 4) = 57 + 8 = 65
65 is divisible by 13. Hence 572 is also divisible by 13
Given number = 68172
6817 + (2 × 4) = 6817 + 8 = 6825
682 + (5 × 4) = 682 + 20 = 702
70 + (2 × 4) = 70 + 8 = 78
Given number = 651
65 + (1 × 4) = 65 + 4 = 69
69 is not divisible by 13. Hence 651 is not divisible by 13

Divisibility by
14
14 if it is divisible by
both 2 and 7.
Example1: Check if 238 is divisible by 14
238 is divisible by 2 . 238 is also divisible by 7. (Please
check the divisibility rule of 2 and 7 to find out this)
342 is divisible by 2 , but 342 is not divisible by 7.(Please
175 is not divisible by 2 , though it is divisible by 7.(Please
337 is not divisible by 2 and also by 7 (Please check the
divisibility rule of 2 and 7 to find out this)

Divisibility by 15
A number is divisible by 15 If
it is divisible by both 3 and 5.
483 is divisible by 3 , but 483 is not divisible by 5. (Please
485 is not divisible by 3 , though it is divisible by 5. (Please
487 is not divisible by 3 . It is also not divisible by 5
(Please check the divisibility rule of 3 and 5 to find out
this)

Divisibility by 16
A number is divisible by 16 if the number formed
by the last four digits is divisible by 16.
The number formed by the last four digits of 5696512 =
6512
16.
The number formed by the last four digits of 3326976 =
6976
16.
The number formed by the last three digits of 732374360
= 4360
4360 is not divisible by 16. Hence 732374360 is also not
divisible by 16.

Divisibility by 17
To find out if a number is
divisible by 17, multiply the
last digit by 5 and subtract it
from the number formed by
the remaining digits.
Repeat this process until you
arrive at a smaller number
whose divisibility you know.
divisible by 17, the original
number is also divisible by 17.
Given Number = 500327
50032 - (7 × 5 )= 50032 - 35 = 49997
4999 - (7 × 5 ) = 4999 - 35 = 4964
496 - (4 × 5 ) = 496 - 20 = 476
47 - (6 × 5 ) = 47 - 30 = 17
17 is divisible by 17. Hence 500327 is also divisible by 17
52146 - (1 × 5 )= 52146 -5 = 52141
5214 - (1 × 5 ) = 5214 - 5 = 5209
520 - (9 × 5 ) = 520 - 45 = 475
47 - (5 × 5 ) = 47 - 25 = 22
22 is not divisible by 17. Hence 521461 is not divisible by
17

Divisibility by 18 A number is divisible by 18 if it is divisible by both 2 and 9.
31104 is divisible by 2. 31104 is also divisible by 9. (Please
1170 is divisible by 2. 1170 is also divisible by 9. (Please
1182 is divisible by 2 , but 1182 is not divisible by 9.
(Please check the divisibility rule of 2 and 9 to find out
this)
1287 is not divisible by 2 though it is divisible by 9. (Please

Divisibility by 19
To find out if a number is
divisible by 19, multiply the
last digit by 2 and add it to the
remaining digits.
Repeat this process until you
arrive at a smaller number
whose divisibility you know.
divisible by 19, the original
number is also divisible by 19.
divisible by 19.
7468 + (9 × 2 )= 7468 + 18 = 7486
748 + (6 × 2 ) = 748 + 12 = 760
76 + (0 × 2 ) = 76 + 0 = 76
76 is divisible by 19. Hence 74689
is also divisible by 19
divisible by 19.
7123 + (4 × 2 )= 7123 + 8 = 7131
713 + (1 × 2 )= 713 + 2 = 715
71 + (5 × 2 )= 71 + 10 = 81
81 is not divisible by 19. Hence
71234 is not divisible by 19

Divisibility by 20
it is divisible by 10 and the
tens digit is even.
(There is one more rule to see
if a number is divisible by 20
which is given below.
the number is divisible by
both 4 and 5)
720 is divisible by 10. (Please check the divisibility rule of
10 to find out this).
The tens digit = 2 = even digit.
1340 is divisible by 10. (Please check the divisibility rule
of 10 to find out this).
The tens digit = 2 = even digit.
1350 is divisible by 10. (Please check the divisibility rule
of 10 to find out this).
But the tens digit = 5 = not an even digit.
1325 is not divisible by 10 (Please check the divisibility
rule of 10 to find out this) though the tens digit = 2 = even
digit.

WHAT ARE FACTORS OF A
NUMBER AND HOW TO
FIND IT OUT?

Factors of a number
If one number is divisible by a second number,
the second number is a factor of the first
number
The lowest factor of any positive number = 1
The highest factor of any positive number = the
number itself
Example
The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36
because each of these numbers divides 36
with a remainder of 0

How to find out factors of a
number
Write down 1 and the number itself (lowest and
highest factors).
Check if the given number is divisible by 2
(Reference: Divisibility by 2 rule)
If the number is divisible by 2, write down 2
as the second lowest factor and divide the
given number by 2 to get the second
highest factor
Check for divisibility by 3, 4,5, and so on. till the
beginning of the list reaches the end

Example1: Find out the factors
of 72
Write down 1 and the number itself (72) as lowest and
highest factors.
1 . . . 72
72 is divisible by 2 (Reference: Divisibility by 2 Rule).
72 ÷ 2 = 36. Hence 2nd lowest factor = 2 and 2nd highest
factor = 36. So we can write as
1, 2 . . . 36, 72
72 ÷ 3 = 24 . Hence 3rd lowest factor = 3 and 3rd highest
factor = 24. So we can write as
1, 2, 3, . . . 24, 36, 72

72 ÷ 4 = 18. Hence 4th lowest factor = 4 and 4th
highest factor = 18. So we can write as
1, 2, 3, 4, . . . 18, 24, 36, 72
72 is not divisible by 5 (Reference: Divisibility by 5
Rule)
highest factor = 12. So we can write as

72 is not divisible by 7 (Reference: Divisibility by
7 Rule)
72 is divisible by 8 (Reference: Divisibility by 8
Rule).
highest factor = 9.
Now our list is complete and the factors of
72 are
1, 2, 3, 4, 6, 8, 9 12, 18, 24, 36, 72

Example2: Find out the factors
of 22
Write down 1 and the number itself (22) as lowest and highest factors
1 . . . 22
22 ÷ 2 = 11. Hence 2nd lowest factor = 2 and 2nd highest factor = 11. So
we can write as
1, 2 . . . 11, 22
22 is not divisible by 3 (Reference: Divisibility by 3 Rule).
Now our list is complete and the factors of 22 are
1, 2, 11, 22

Important Properties of
Factors
If a number is divisible by another number, then it is
also divisible by all the factors of that number.
Example : 108 is divisible by 36 because 106 ÷ 38
= 3 with remainder of 0.
The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36
because each of these numbers divides 36 with a
remainder of 0.
Hence, 108 is also divisible by each of the
numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.

WHAT ARE PRIME
NUMBERS AND
COMPOSITE NUMBERS?

Prime Numbers
A prime number is a positve integer that is divisible
by itself and 1 only. Prime numbers will have
exactly two integer factors.
Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc.
Please note the following facts
Zero is not a prime number because zero is divisible
by more than two factors. Zero can be divided by
1, 2, 3 etc.
(0 ÷ 1 = 0, 0÷ 2 = 0 ...)
One is not a prime number because it does not have
two factors. It is divisible by only 1

Composite Numbers
Composite numbers are numbers that have
more than two factors. A composite number is
divisible by at least one number other than 1
and itself.
Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20,
etc.
Please note that zero and 1 are neither prime
numbers nor composite numbers.
Every whole number is either prime or
composite, with two exceptions 0 and 1 which
are neither prime nor composite

WHAT ARE PRIME
FACTORIZATION AND
PRIME FACTORS ?

Prime factor Prime
factorization
he factors which are prime numbers are called prime
factors
Prime factorization of a number is the expression of the
number as the product of its prime factors
Example 1:
Prime factorization of 280 can be written as 280 = 2 ×
2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280
are 2, 5 and 7
Example 2:
Prime factorization of 72 can be written as 72 = 2 × 2
× 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2
and 3

How to find out prime
factorization and prime factors
of a number
Repeated Division Method : In order to find out
the prime factorization of a number, repeatedly
divide the number by the smallest prime
number possible(2,3,5,7,11, ...) until the
quotient is 1.

Example 1: Find out Prime
factorization of 280
2*140=280
2*2*70=280
2*2*2*35=280
2*2*2*5*7=280
Hence, prime factorization of 280 can be written
as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the
prime factors of 280 are 2, 5 and 7

Example 2: Find out Prime
factorization of 72
2*36=72
2*2*18=72
2*2*2*9=72
2*2*2*3*3=72
Hence, prime factorization of 72 can be written
as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the
prime factors of 72 are 2 and 3

Important Properties
Every whole number greater than 1 can be
uniquely expressed as the product of its prime
factors. For example, 700 = 22 × 52 × 7

Multiples
Multiples of a whole number are the products of
that number with 1, 2, 3, 4, and so on
Example : Multiples of 3 are 3, 6, 9, 12, 15, ...
If a number x divides another number y exactly
with a remainder of 0, we can say that x is a
factor of y and y is a multiple of x
For instance, 4 divides 36 exactly with a
remainder of 0. Hence 4 is a factor of 36 and
36 is a multiple of 4

WHAT IS LEAST COMMON
MULTIPLE (LCM) AND HOW
TO FIND LCM

Least Common Multiple
(LCM)
Least Common Multiple (LCM) of two or more
numbers is the smallest number that is a
multiple of all the numbers
Example: LCM of 3 and 4 = 12 because 12 is the
smallest multiple which is common to 3 and 4
(In other words, 12 is the smallest number
which is divisible by both 3 and 4)
We can find out LCM using prime factorization
method or division method

How to find out LCM using prime
factorization method
Step1 : Express each number as a product of
prime factors.
Step2 : LCM = The product of highest powers of
all prime factors

Example 1 : Find out LCM of 8
and 14
Step1 : Express each number as a product of prime
factors.
8 = 23
14 = 2 × 7
Step2 : LCM = The product of highest powers of all
prime factors
Here the prime factors are 2 and 7
The highest power of 2 here = 23
Hence LCM = 23 × 7 = 56

Example 2 : Find out LCM of 18,
24, 9, 36 and 90
Step1 : Express each number as a product of prime factors
18 = 2 × 32
24 = 23 × 3
9 = 32
36 = 23 × 32
90 = 2 × 5 × 32
Step2 : LCM = The product of highest powers of all prime
factors
Here the prime factors are 2, 3 and 5
Hence LCM = 23 × 32 × 5 = 360

How to find out LCM using
Division Method (shortcut)
Step 1 : Write the given numbers in a horizontal line
separated by commas.
Step 2 : Divide the given numbers by the smallest
prime number which can exactly divide at least
two of the given numbers.
Step 3 : Write the quotients and undivided numbers
in a line below the first.
Step 4 : Repeat the process until we reach a stage
where no prime factor is common to any two
numbers in the row.
Step 5 : LCM = The product of all the divisors and
the numbers in the last line.

Example 1 : Find out LCM of 8
and 14
Hence Least common multiple (L.C.M) of 8 and
14 = 2 × 4 × 7 = 56

Example 2 : Find out LCM of 18,
24, 9, 36 and 90
 Hence Least common multiple (L.C.M) of 18,
24, 9, 36 and 90 = 2 × 2 × 3 × 3 × 2 × 5 = 360

WHAT IS HIGHEST COMMON
FACTOR (HCF) OR GREATEST
COMMON MEASURE (GCM)
OR GREATEST COMMON
DIVISOR (GCD) AND HOW TO
FIND IT OUT ?

Highest Common Factor(H.C.F) or Greatest
Common Measure(G.C.M) or Greatest Common
Divisor (G.C.D)
Highest Common Factor(H.C.F) or Greatest
Common Measure(G.C.M) or Greatest
Common Divisor (G.C.D) of two or more
numbers is the greatest number which divides
each of them exactly.
Example : HCF or GCM or GCD of 60 and 75 =
15 because 15 is the highest number which
divides both 60 and 75 exactly.
We can find out HCF using prime factorization
method or division method

How to find out HCF using prime
factorization method
Step1 : Express each number as a product of
prime factors. (Reference: Prime Factorization
and how to find out Prime Factorization)
Step2 : HCF is the product of all common prime
factors using the least power of each common
prime factor.

Example 1 : Find out HCF of 60 and 75 (Reference:
Prime Factorization and how to find out Prime
Factorization)
factors.
60 = 22 × 3 × 5
75 = 3 × 52
Step2 : HCF is the product of all common prime
factors using the least power of each common
prime factor.
Here, common prime factors are 3 and 5
The least power of 3 here = 3
Hence, HCF = 3 × 5 = 15

Example 2 : Find out HCF of 36,
24 and 12
factors. (Reference: Prime Factorization and how to
find out Prime Factorization)
36 = 22 × 32
24 = 23 × 3
12 = 22 × 3
Step2 : HCF is the product of all common prime factors
using the least power of each common prime factor.
Here 2 and 3 are common prime factors.
Hence, HCF = 22 × 3 = 12

27 and 80
factors. (Reference: Prime Factorization and how
to find out Prime Factorization)
36 = 22 × 32
27 = 33
80 = 24 × 5
Step2 : HCF = HCF is the product of all common
prime factors using the least power of each
common prime factor.
Here you can see that there are no common prime
factors.
Hence, HCF = 1

How to find out HCF using prime factorization
method - By dividing the numbers (shortcut)
Step 1 : Write the given numbers in a horizontal line
separated by commas.
Step 2 : Divide the given numbers by the smallest
prime number which can exactly divide all of the
given numbers.
Step 3 : Write the quotients in a line below the first.
Step 4 : Repeat the process until we reach a stage
where no common prime factor exists for all of the
numbers.
Step 5 :We can see that the factors mentioned in
the left side clearly divides all the numbers exactly
and they are common prime factors. Their product
is the HCF

Example 1 : Find out HCF of 60
and 75
We can see that the prime factors mentioned in the left side clearly divides all the
numbers exactly and they are common prime factors. no common prime factor is exists
for the numbers came at the bottom.
Hence HCF = 3 × 5 =15

24 and 12
We can see that the prime factors mentioned in the
left side clearly divides all the numbers exactly
and they are common prime factors. no common
prime factor is exists for the numbers came at the
bottom.
Hence HCF = 2 × 2 × 3 = 12.

24 and 48
We can see that the prime factors mentioned in
the left side clearly divides all the numbers
exactly and they are common prime factors. no
common prime factor is exists for the numbers
came at the bottom.
Hence HCF = 2 × 2 × 3 = 12.

HOW TO FIND OUT HCF
USING DIVISION METHOD
(SHORTCUT)

To find out HCF of two given numbers using division method
Step 1: Divide the larger number by the smaller
number
Step 2: Divisor of step 1 is divided by its
remainder
Step 3: Divisor of step 2 is divided by its
remainder. Continue this process till we get 0
as remainder.
Step 4: Divisior of the last step is the HCF.

To find out HCF of three given numbers using division method
Step 1: Find out HCF of any two numbers.
Step 2: Find out the HCF of the third number
and the HCF obtained in step 1
Step 3: HCF obtained in step 2 will be the HCF
of the three numbers

To find out HCF of more than three numbers using division method
In a similar way as explained for three numbers,
we can find out HCF of more than three
numbers also

Example 1 : Find out HCF of 60
and 75

Hence HCF of 3556 and 3224 = 4

How to calculate LCM and HCF for
fractions
Least Common Multiple (L.C.M.) for fractions
LCM for fractions = LCM of Numerators
HCF of Denominators
Example 1: Find out LCM of 1/2, 3/8, 3/4
LCM = LCM (1, 3, 3) / HCF (2, 8, 4)=32
Example 2: Find out LCM of 2/5, 3/10
LCM = LCM (2, 3) / HCF (5, 10)=65

Highest Common Multiple (H.C.F) for fractions
HCF for fractions = HCF of Numerators / LCM of
Denominators
Example 1: Find out HCF of 3/5, 6/11, 9/20
HCF = HCF (3, 6, 9) / LCM (5, 11, 20)=3220
Example 2: Find out HCF of 4/5, 2/3
HCF = HCF (4, 2) / LCM (5, 3)=215

How to calculate LCM and HCF for
Decimals
Step 1 : Make the same number of decimal
places in all the given numbers by suffixing
zero(s) in required numbers as needed.
Step 2 : Now find the LCM/HCF of these
numbers without decimal.
Step 3 : Put the decimal point in the result
obtained in step 2 leaving as many digits on its
right as there are in each of the numbers.

Example1 : Find the LCM and
HCF of .63, 1.05, 2.1
Step 1 : Make the same number of decimal places in all the
given numbers by suffixing zero(s) in required numbers as
needed.
i.e., the numbers can be written as .63, 1.05, 2.10
Step 2 : Now find the LCM/HCF of these numbers without
decimal.
Without decimal, the numbers can be written as 63, 105 and
210 .
LCM (63, 105 and 210) = 630
HCF (63, 105 and 210) = 21
Step 3 : Put the decimal point in the result obtained in step 2
leaving as many digits on its right as there are in each of the
numbers.
i.e., here, we need to put decimal point in the result obtained
in step 2 leaving two digits on its right.
i.e., the LCM (.63, 1.05, 2.1) = 6.30
HCF (.63, 1.05, 2.1) = .21

Type 1 : Fractions with same
denominators.
Compare 3/5 and 1/5
These fractions have same denominator. So just
compare the numerators. Bigger the numerator,
bigger the number.
3 > 1. Hence 3/5>1/5
Example 2: Compare 2/7 and 3/7 and 8/7
These fractions have same denominator. So just
compare the numerators. Bigger the numerator,
bigger the number.
8 > 3 > 2. Hence 8/7>3/7>2/7

Type 2 : Fractions with same
numerators.
Example 1: Compare 3/5 and 3/8
These fractions have same numerator. So just
compare the denominators. Bigger the
denominator, smaller the number.
8 > 5. Hence 3/8<3/5
Example 2: Compare 7/8 and 7/2 and 7/5
These fractions have same numerator. So just
compare the denominators. Bigger the
denominator, smaller the number.
8 > 5 > 2. Hence 7/8<7/5<7/2

Type 3 : Fractions with different
numerators and denominators.
Example 1: Compare 3/5 and 4/7
To compare such fractions, find out LCM of the denominators. Here, LCM(5, 7) = 35
Now , convert each of the given fractions into an equivalent fraction with 35 (LCM) as
the denominator.
The denominator of 3/5 is 5. 5 needs to be multiplied with 7 to get 35. Hence,
3/5=3×7/5×7=2135
The denominator of 4/7 is 7. 7 needs to be multiplied with 5 to get 35. Hence,
4/7=4×5/7×5=2035
21/35>20/35
Hence, 3/5>4/7
Or
Convert the fractions to decimals
3/5=.6
4/7=.5... (Need not find out the complete decimal value; just find out up to what is
required for comparison. In this case the first digit itself is sufficient to do the
comparison)
.6 > .5...
Hence, 3/5>4/7

CO-PRIME NUMBERS OR
RELATIVELY PRIME
NUMBERS

Two numbers are said to be co-prime (also spelled coprime) or
relatively prime if they do not have a common factor other than 1. i.e.,
if their HCF is 1.
Example1: 3, 5 are co-prime numbers (Because
HCF of 3 and 5 = 1)
Example2: 14, 15 are co-prime numbers
(Because HCF of 14 and 15 = 1)
A set of numbers is said to be pair wise co-prime
(or pair wise relatively prime) if every two
distinct numbers in the set are co-prime

Example1 : The numbers 10, 7, 33, 13 are pair wise co-prime, because HCF of any pair
of the numbers in this is 1.
HCF (10, 7) = HCF (10, 33) = HCF (10, 13) = HCF (7, 33) = HCF (7, 13) = HCF (33,
13) = 1.
Example2 : The numbers 10, 7, 33, 14 are not pair wise co-prime because HCF(10, 14)
= 2 ≠ 1 and HCF(7, 14) = 7 ≠ 1.
If a number is divisible by two co-prime numbers, then the number is divisible by their
product also.
Example
3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1)
14325 is divisible by 3 and 5.
3 × 5 = 15
Hence 14325 is divisible by 15 also
If a number is divisible by more than two pair wise co-prime numbers, then the number
is divisible by their product also.

Example1 : The numbers 3, 4, 5 are pair wise co-prime because
HCF of any pair of numbers in this is 1
1440 is divisible by 3, 4 and 5.
3 × 4 × 5 = 60. Hence 1440 is also divisible by 60
Example2
The numbers 3, 4, 9 are not pair wise co-prime because HCF
(3, 9 ) = 3 ≠ 1
1440 is divisible by 3, 4 and 9.
3 X 4 X 9 = 108. However 1440 is not divisible by 108 as 3, 4,
9 are not pair wise co-prime

Important Points to Note on LCM
and HCF
Product of two numbers = Product of their HCF and
LCM.
Example
LCM (8, 14) = 56
HCF (8, 14) = 2
LCM (8, 14) × HCF (8, 14) = 56 × 2 = 112
8 × 14 = 112
Hence LCM (8, 14) × HCF (8, 14) = 8 × 14

GENERAL SHORTCUT
METHOD FOR FINDING
SQUARES

You can find square of any number in the world
with this method.
Let’s say the number is two digit number. i.e. AB.
So B is units digit and A is tens digit.
Step 1: Find Square of B
Step 2: Find 2×A×B
Step 3: Find Square of A
Let’s take an example.

We want to find square of 37.
Step 1: Find square of 7.
Square of 7 = 49.
So write 9 in the answer and 4 as carry to the second
step.
Step 2: Find 2×(3×7)
2 × (3 × 7) = 42.
42 + 4(Carry) = 46.
Write 6 in the answer and 4 as a carry to the third
step.
Step 3: Find square of 3
Square of 3 = 9
9 + 4(Carry) = 13.
Write 13 in the answer.
So the answer is 1369.

Now, If the number is of three digit i.e. ABC
Here C is unit’s digit, B is ten’s digit and A is hundredth digit.
Step 1: Find Square of C
Step 2: Find 2 × (B × C)
Step 3: Find 2 × (A × C) + B2
(NOTE: You may observe that in odd number of digit case, we are
multiplying end two digits with 2 (here: A and C) and squaring single
digit (here B).
Step 4: Find 2 × (A × B)
(NOTE: You may observe that whenever there are double digits, we
are multiplying it with 2. And whenever there is single digit, we are
squaring it.)
Step 5: Find square of A
(NOTE: Here is single digit, so we are squaring it.)

Find square of 456.
Step 1: Find square of 6.
Square of 6 = 36.
So write 6 in the answer and 3 as a carry to the second step.
Step 2: Find 2 × (5 × 6)
2 × (5 × 6) = 60
60 + 3(Carry) = 63
Write 3 in the answer and 6 as a carry to the third step.
Step 3: Find 2 × (4 × 6) + 52
2 × (4 × 6) + 52 = 73
73 + 6(Carry) = 79
Write 9 in the answer and 7 as a carry to the fourth step.
Step 4: Find 2 × (4 × 5)
2 × (4 × 5) = 40
40 + 7 = 47
Write 7 in the answer and 4 as a carry to the fifth step.
Step 5: Find square of 4
Square of 4 = 16
16 + 4(Carry) = 20
Write 20 in the answer.
So 4562 = 207936.

Now, If the number is of four digit i.e. ABCD
Here D is unit’s digit, C is ten’s digit, B is hundredth digit and A is thousands
digit.
Step 1: Find Square of D
Step 2: Find 2 × (C × D)
Step 3: Find 2 × (B × D) + C2
(NOTE: You may observe that in odd number of digit case, we are multiplying
end two digits with 2 (here: B and D) and squaring single remaining digit
(here C).
Step 4: Find 2 × (A × D) + 2 × (B × C)
(NOTE: You may observe that where ever there is even digits, we are
multiplying end two digits with 2 + remaining two digits with 2.)
Step 5: Find 2 × (A × C) + B2
(NOTE: You may observe that in odd number of digit case, we are multiplying
end two digits with 2 (here: A and C) and squaring single remaining digit
(here B).
Step 6: Find 2 × (A × B)
(NOTE: You may observe that here even digits so we are multiplying them with
2, and no remaining digits so we are not adding anything.)
Step 7: Find square of A

Square Of The Numbers With
Unit Digit As 5
This method is to find square of the numbers which
has unit’s digit as 5. i.e.: 25, 45, 65, etc.
You can find square of these numbers by three
easy steps.
Step 1: Multiply ten’s digit with its next number.
Step 2: Find square of unit’s digit. i.e.: Square of 5.
Step 3: Write answers of step 1 and step 2 to
together or side by side.
Let’s take examples.

Find square of 35
3 × ( 3 + 1 ) = 3 × 4 = 12
Step 2: Find square of unit’s digit. i.e.: Square of
5.
Square of 5 = 25
together.
Answer = 1225

Find square of 65
6 × ( 6 + 1 ) = 6 × 7 = 42
5.
Square of 5 = 25
together.
Answer = 4225

Find square of 95
9 × ( 9 + 1 ) = 9 × 10 = 90
5.
Square of 5 = 25
together.
Answer = 9025

Find square of 115
11 × ( 11 + 1 ) = 11 × 12 = 132
(Note: We are taking whole 11 as a ten’s
digit.)
5.
Square of 5 = 25
together.
Answer = 13225

Find square of 215
21 × ( 21 + 1 ) = 21 × 22 = 462
(Note: We are taking whole 21 as a ten’s
digit.)
5.
Square of 5 = 25
together.
Answer = 46225

Square Of The Numbers In
50s
This method is used to find square of the numbers
in 50s i.e. numbers from 51 to 59.
You can find square of these numbers in three
simple steps.
Step 1: Add 25 to the unit’s digit.
Step 2: Square the unit’s digit.
Step 3: Write the answers of step 1 and step 2
together or side by side.

Square of 56
Step 1: Add 25 to the unit’s digit
6 + 25 = 31
Step 2: Square the unit’s digit
62 = 36
together.
Answer = 3136

Square of 59
9 + 25 = 34
92 = 81
together.
Answer = 3481

Square of 53.
Step 1: Add 25 to the unit’s digit.
3 + 25 = 28
Step 2: Square the unit’s digit.
32 = 9
Step 3: Write answers of step 1 and step 2
together.
Answer: 2809
(NOTE: Whenever square of unit’s
digit is on only single digit then we are adding
0 before it.)

Square of 52
2 + 25 = 27
22 = 4
together.
Answer = 2704
(NOTE: Whenever square of unit’s digit
is on only single digit then we are adding 0 before
it.)

Square The Number If You Know
Square Of Previous Number
This method is to find square of the number if you
know square of the previous number.
You can find answers in three simple steps.
Step 1: Find square of the previous number which is
known.
Step 2: Multiply the number being squared by 2 and
subtract 1.
Step 3: Add Step 1 and Step 2
Let’s take some examples.

Find square of 31.
Step 1: Find square of previous number (30)
which is known.
302 = 900
Step 2: Multiply the number being squared (31)
by 2 and subtract 1.
(31 × 2) – 1 = 62 – 1 = 61
900 + 61 = 961

Find square of 26.
which is known.
252 = 625
(26 × 2) – 1 = 52 – 1 = 51
625 + 51 = 676

Find square of 81.
which is known.
802 = 6400
(81 × 2) – 1 =162 – 1 = 161
6400 + 161 = 6561

SHORTCUT METHOD FOR
FINDING CUBE

If a2 = b,
b√=a (i.e., square root of b is a.)
· Square Root
Examples
22=4and4√=2
52=25and25−−√=5
If a3 = b,
b√3=a (i.e., cube root of b is a.)
Cube Root
Examples
23=8and8√3=2
53=125and125−−−√3=5
If a2 = b,
b√=a (i.e., square root of b is
a.)
If a3 = b,
b√3=a (i.e., cube root of b is a.)

BASICS CONCEPTS AND
FORMULAS - SIMPLIFICATION

Order of Operations - BODMAS
Rule
What is BODMAS Rule?
BODMAS rule defines the correct sequence in
which operations are to be performed in a
given mathematical expression to find its
value.
In BODMAS,
B = Bracket,
O = Order (Powers, Square Roots, etc.)
DM = Division and Multiplication (left-to-right)
AS = Addition and Subtraction (left-to-right)

This means, to simplify an expression, the following order must
be followed.
Do operations in Brackets first, strictly in the order (), {}
and []
Evaluate exponents (Powers, Roots, etc.)
Perform division and multiplication, working from left to
right. (division and multiplication rank equally and
done left to right).
Perform addition and subtraction, working from left to
right. (addition and subtraction rank equally and done
left to right).

In India, the term BODMAS is used whereas in
USA the acronym PEMDAS is used. The full
form of PEMDAS is "Parentheses, Exponents,
Multiplication, Division, Addition, and
Subtraction". Both are correct and only
difference is that these are used in different
parts of the world. Some other variations which
are used across the world to represent the
same concept are BIDMAS, ERDMAS,
PERDMAS and BPODMAS.

Examples
12 + 22 ÷ 11 × (18 ÷ 3)2 - 10
= 12 + 22 ÷ 11 × 62 - 10 (∵ Brackets first)
= 12 + 22 ÷ 11 × 36 - 10 (∵ exponents)
= 12 + 2 × 36 - 10 = 12 + 72 - 10 (∵ division and
multiplication, left to right)
= 84 - 10 = 74 (∵ Addition and Subtraction, left to
right)
4 + 10 - 3 × 6 / 3 + 4
= 4 + 10 - 18/3 + 4 = 4 + 10 - 6 + 4 (∵ division and
multiplication, left to right)
= 14 - 6 + 4 = 8 + 4 = 12 (∵ Addition and Subtraction,
left to right)

Modulus of a Real Number
For any real number x, the absolute value or
modulus of x is denoted by |x| and is defined
as
|x|={x−xif x≥0if x<0
Hence, the Modulus(absolute value) of x is
always either positive or zero, but never
negative
Example
|8| = |-8| = 8

A fraction is an expression that indicates the quotient of
two quantities.
Examples of fractions: etc.
A fraction has two parts, Numerator and Denominator.
Numerator is the number at the top of the fraction and
denominator is the number at the bottom of the
fraction.
For example, in the fraction ¼ , 1 is numerator and 4
is denominator
The denominator of a fraction cannot be zero

Type of Fractions
i. Common Fraction
A common fraction (also known as Vulgar fraction and simple fraction) is a fraction in which both
numerator and denominator are integers (As with other fractions, the denominator cannot be zero)
Examples: 2/3 , 7/12 , −6/5
i. Decimal Fraction
A decimal fraction is a fraction in which denominator is an integer power of ten. (The term decimals are commonly used to refer
decimal fractions).
Generally, a decimal fraction is expressed using decimal notation and its denominator is not mentioned explicitly
Examples:
1/10 = .1
1/100 = .01
17/100 = .17
4/10000 = .0004
121/100 = 1.21
Decimal fractions can also be expressed using scientific notation with negative exponents
Example: 4.193 × 10−7 is a decimal fraction which represents .0000004193

 Generally, a zero is added to the left of the decimal point if a whole number is absent to draw the attention to the fact that the number is a
decimal fraction.
For example, .1 and 0.1 represents the same
 Annexing Zeros to the extreme right of a decimal fraction
Annexing zeros to the extreme right of a decimal fraction does not change its value.
Examples
0.4 = 0.40 = 0.400 = 0.4000, etc.
1.21 = 1.210 = 1.2100 = 1.21000, etc
 If the numerator and denominator contain the same number of decimal places, we can remove the decimal sign
Example
1.2/4.8=12/48=1/4
0.03/0.24=3/24=1/8
 Conversion of a Decimal into Common Fraction
Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove
the decimal point and reduce the fraction to its lowest terms.
Examples
0.5 = 5/10=1/2
.75 = 75/100=3/4
1.25 = 125/100=5/4

IMPORTANT CONCEPTS AND
FORMULAS - ALGEBRA
(INCLUSIVE OF SURDS AND
INDICES )

 a(b+c)=ab+ac(Distributive Law)
 (a+b)2=a2+2ab+b2
 (a−b)2=a2−2ab+b2
 (a+b)2+(a−b)2=2(a2+b2)
 (a+b)3=a3+3a2b+3ab2+b3=a3+3ab(a+b)+b3
 (a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3
 a2−b2=(a−b)(a+b)
 a3+b3=(a+b)(a2−ab+b2)
 a3−b3=(a−b)(a2+ab+b2)
 an−bn=(a−b)(an−1+an−2b+an−3b2+ ... +bn−1)
 (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
 (a+b)(a+c)=a2+(b+c)a+bc

 an=a.a.a ... (n times)
 am.an...ap=am+n+...+p
 am.an=am+n
 am/an=am−n
 (am)n=amn=(an)m
 (ab)n=anbn
 (a/b)n=an / bn
 a−n=1/an
 an=1/a−n
 a0=1 where a∈ R, a ≠0
 ap/q=ap−−√q
 if am=an where a≠0 and a≠±1, then m=n
 if an=bn where n≠0, then a=±b

 (x+a)2=x2+2ax+a2
 (x−a)2=x2−2ax+a2
 (x+a)(x+b)=x2+(a+b)x+ab
 x2−a2=(x−a)(x+a)
 (xn+1) is completely divisible by (x + 1) when n is odd
 (xn+an) is completely divisible by (x + a) when n is odd
 (xn−an) is completely divisible by (x−a) for every natural number n
 (xn−an) is completely divisible by (x+a) when n is even

SURDS AND IMPORTANT
PROPERTIES

Surds
Let a be a rational number and n be a positive integer
such that a√n is irrational. Then a√n is called a surd of
order n.
Examples :
3√ is a surd of order 2
7√3 is a surd of order 3
43√ is a surd of order 2
Please note that numbers like 9√, 27−−√3 etc are not
surds because they are not irrational numbers
Every surd is an irrational number. But every irrational
number is not a surd. (eg : π , e etc are not surds
though they are irrational numbers.)

Quadratic Surds
A surd of order 2 is called a quadratic surd
Examples : 2√,3√,(3+5√), etc.

Laws of Indices:
i. am * an = am+n
ii. am/an = am-n
iii. (am)n =amn
iv. (ab)n = anbn
v. (a/b)n = an/bn
vi. a0= 1

QUADRATIC EQUATIONS AND
HOW TO SOLVE QUADRATIC
EQUATIONS

A. Quadratic Equations
A quadratic equation is a second degree univariate
polynomial equation.
A quadratic equation can be written as (general form
or standard form of quadratic equation)
ax2+bx+c=0 where x is a variable, a, b and c are
constants and a≠0
Example : x2+5x+6=0
(Please note that if a=0, equation becomes a linear
equation)
The solutions of a quadratic equation are called its
roots.

B. How to Solve Quadratic
Equations
There are many methods to solve Quadratic
equations. Quadratic equations can be solved
by factoring, completing the square, graphing,
Newton's method, and using the quadratic
formula. We can go through some of the
popular methods for solving quadratic
equations here.

How to Solve Quadratic
Equations By Factoring
Step 1: Write the quadratic equation in standard
form.
i.e. , bring all terms on the left side of the equal
sign and 0 on the right side of the equal sign.
Step 2: Factor the left side of the equation
Step 3: Equate each factor to 0 and solve the
equations
Factoring is one of the fastest ways for solving
quadratic equations. The concept will be clear
from the following examples.

Example1 : Solve the Quadratic
Equation x2+4x−11=x−1
Step 1 : Write the quadratic equation in standard form.
x2+4x−11=x−1
⇒x2+4x−11−x+1=0
⇒x2+3x−10=0
Step 2 : Factor the left side of the equation.
We know that x2+(a+b)x+ab=(x+a)(x+b). We will use the same concept for factoring.
Assume x2+3x−10=x2+(a+b)x+ab
We need to find out a and b such that a + b = 3 and ab = -10
a = +5 and b = -2 satisfies the above condition.
Hence x2+3x−10=(x+5)(x−2)
x2+3x−10=0
⇒(x+5)(x−2)=0
Step 3 :Equate each factor to 0 and solve the equations
(x + 5)(x - 2) = 0
=> (x + 5) = 0 or (x - 2) = 0
=> x = -5 or 2
Hence, the solutions of the quadratic equation x2+4x−11=x−1 are x = -5 and x = 2.
(In other words, x = -5 and x = 2 are the roots of the quadratic equation x2+4x−11=x−1

Equation x2−7x+10=0
This equation is already in the standard form. Hence let's go to step 2
Here we need to find out a and b such that a + b = -7 and ab = +10
a = -5 and b = -2 satisfies the above condition.
Hence x2−7x+10=(x−5)(x−2)
Hence, x2+3x−10=0
⇒(x−5)(x−2)=0
(x - 5)(x - 2) = 0
=> (x - 5) = 0 or (x - 2) = 0
=> x = 5 or 2
Hence, the solutions of the quadratic equation x2−7x+10=0 are x = 5 and x = 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation x2−7x+10=0)

Equation 3x2−14x+8=0
Here we need to follow a slightly different approach for factoring because here the coefficient of x2 ≠ 1 (whereas In
example 1 and example 2, the coefficient of x2 was 1)
Hence, to factor 3x2−14x+8, we need to do the followings
1. Product of the second degree term and the constant. i.e., 3x2×8=24x2
2. We got product as 24x2 and have middle term as -14x. From this , it is clear that we can take -12x and -2x such
that their sum is -14x and product is 24x2
Hence, 3x2−14x+8=0 = 3x2−12x−2x+8=3x(x−4)−2(x−4)=(x−4)(3x−2)
Hence, 3x2−14x+8=0
⇒(x−4)(3x−2)=0
(x - 5)(x - 2) = 0
=> (x - 4) = 0 or (3x - 2) = 0
=> x = 4 or 23.
Hence, the solutions of the quadratic equation 3x2−14x+8=0 are x = 4 and x = 23
(In other words, x = 4 and x = 23 are the roots of the quadratic equation 3x2−14x+8=0

Equation 6x2−17x+12=0
coefficient of x2 ≠ 1
Hence, to factor 6x2−17x+12, we need to do the followings
1. Product of the second degree term and the constant. i.e., 6x2×12=72x2
2. We got product as 72x2 and have middle term as -17x. From this , it is clear that we can take -8x and -9x such
that their sum is -17x and product is 72x2
Hence, 6x2−17x+12=6x2−8x−9x+12=2x(3x−4)−3(3x−4)=(3x−4)(2x−3)
Hence, 6x2−17x+12=0
⇒(3x−4)(2x−3)=0
(3x - 4)(2x - 3) = 0
=> (3x - 4) = 0 or (2x - 3) = 0
=> x = 43 or 32.
Hence, the solutions of the quadratic equation 6x2−17x+12=0 are x = 43 and x = 32.
(In other words, x = 43 and x = 32 are the roots of the quadratic equation 6x2−17x+12=0

(SPECIAL FACTORIZATION
EXAMPLES)

Equation x2−9=0
Here,x2−9 is in the form a2−b2 where a = x and b = 3
We know that a2−b2=(a−b)(a+b)We will use the same concept for factoring this.
x2−9=x2−32=(x−3)(x+3)
Hence, x2−9=0
⇒(x−3)(x+3)=0
(x - 3)(x + 3) = 0
=> x = 3 or -3
Hence, the solutions of the quadratic equation x2−9=0 are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation x2−9=0

Example6: Solve the Quadratic
Equation x2+6x+9=0
Here,x2+6x+9 is in the form a2+2ab+b2 where a = x and b = 3
We know that a2+2ab+b2=(a+b)2 We will use the same concept for factoring this.
x2+6x+9=x2+2×x×3+32=(x+3)2
Hence, x2+6x+9=0
⇒(x+3)2=0
Step 3 :
(x+3)2=0
=> x + 3 = 0
=> x = -3
Hence, the solution of the quadratic equation x2+6x+9=0 is x = -3
(In other words, x = -3 is the root of the quadratic equation x2+6x+9=0

Equation x2−6x+9=0
Here,x2−6x+9 is in the form a2−2ab+b2 where a = x and b = 3
We know that a2−2ab+b2=(a−b)2 We will use the same concept for factoring this.
x2−6x+9=x2−2×x×3+32=(x−3)2
Hence, x2+6x+9=0
⇒(x−3)2=0
Step 3 :
(x−3)2=0
=> x - 3 = 0
=> x = 3
Hence, the solution of the quadratic equation x2−6x+9=0 is x = 3
(In other words, x = 3 is the root of the quadratic equation x2−6x+9=0

Types of questions based on
cyclicity of numbers
There are mainly 3 categories of questions
which fall under cyclicity of numbers, that
include
How to find units digit of ab
How to find units digit of ab * cd * ef
How to find units digit of abc
Let's see how to solve these questions using the
concept of cyclicity of numbers, with examples.

Find units digit of ab
Given ab, units place digit of the result depends on units place digit of a and
the divisibility of power b.
Consider powers of 2
As we know,
21 =2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128.. and so on
What do you observe here? We can see that the units place digit for powers of
2 repeat in an order: 2, 4, 8, 6. So the "cyclicity" of number 2 is 4 (that
means the pattern repeats after 4 occurrences) and the cycle pattern is 2,
4, 8, 6.
From this you can see that to find the units place digit of powers of 2, you
have to divide the exponent by 4.
Let's check the validity of above formula with an example.

Find the units place digit of 299?
Using the above observation of cyclicity of
powers of 2, divide the exponent by 4
99/4 gives reminder as 3. That means, units
place digit of 299 is the 3rd item in the cycle
which is 8.

Shortcuts to solve problems
related to units place digit of ab
Case 1: If b is a multiple of 4
If a is an even number, ie: 2, 4, 6 or 8 then the units
place digit is 6
If a is an odd number, ie: 1, 3, 7 or 9 then the units
place digit is 1
Case 2: If b is not a multiple of 4
Let r be the reminder when b is divided by 4, then
units place of ab will be equal to units place of ar
Let's call it the general rule of cyclicity. Using
these rules you can solve all the problems
related to cyclicity of numbers.

Number ^1 ^2 ^3 ^4 Cyclicity
2 2 4 8 6 4
3 3 9 7 1 4
4 4 6 4 6 2
5 5 5 5 5 1
6 6 6 6 6 1
7 7 9 3 1 4
8 8 4 2 6 4
9 9 1 9 1 2

Find units digit for numbers of
the form ab * cd
Question of this type is similar to first category.
First find the unit digit of ab and cd separately.
Let the answers be x and y
Then unit digit of ab * cd = units digit of x * y

Find units digit of abc
Questions of this type have to be approached on a case by case basis.
Case 1: If cyclicity of units place digit of a is 4 then we have to divide
the exponent of a by 4 and find out the remainder. Depending on
the value of remainder we can apply the general rule of cyclicity
given above and reach the solution.
Case 2: If cyclicity of units place digit of a is 2, only extra information
we need to find is if the exponent will be even or odd. Then we can
apply the general rule of cyclicity given above and reach the
solution.
Let's see application of these rules with help of examples

Find the units place digit of 24344
Here cyclicity of units place digit is 4 (Units place digit is 2, from the above table we can
see the cyclicity of 2 is 4). Hence case 1 is applicable.
Now we have to find the remainder when exponent of 2 is divided by 4, that is the
remainder when 4344 is divided by 4.
Remainder of 4344/4 = Remainder of (44 – 1)44/4
Using the binomial theorem, (as explained in number system tutorial) we can see
that there is only one term in the expansion of (44 – 1)44 which is not divisible by 4.
The term is 144/4
Remainder of 144/4 = 1
Now we can apply the general rules of cyclicity, (since reminder is 1, case 2 of
general rule of cyclicity is applicable) which says, units place of 24344 = units place of
21 = 2.

Find the units place digit of
293945
Here cyclicity of units place digit of a, that is cyclicity of 9 is 2. Hence
case 2 is applicable
Next step is we need to find if 3945 is even or odd
3945 will always result in an odd number, because we are
multiplying an odd number 39, odd number of times (45).
Now we can apply the general rule of cyclicity described in the first
section. General rule says, first check if exponent is divisible by 4.
Since the exponent here is odd, it's not divisible by 4.
Again according to general rule, if exponent is not divisible by 4, find
the reminder when exponent is divided by 4.
Since the exponent is odd here, the possible reminders when it's
divided by 4 are 1 and 3.
This means, the units place digit of number will be 1st or 3rd
element in the cyclicity of units place digit of a.
In this case 1st and 3rd elements of 9 (units place digit of 29) is 9.
Hence units place digit of 2293945 is 9.

Fundamentals of Quantitative Aptitude

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Andere mochten auch

Andere mochten auch (10)

Ähnlich wie Fundamentals of Quantitative Aptitude

Ähnlich wie Fundamentals of Quantitative Aptitude (20)

Mehr von Pawan Mishra

Mehr von Pawan Mishra (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Fundamentals of Quantitative Aptitude