SSC CGL Exam > SSC CGL Notes > SSC CGL Mathematics Previous Year Paper (Topic-wise) > SSC CGL Previous Year Questions (2023 - 18): Simplification
SSC CGL Previous Year Questions (2023 - 18): Simplification | SSC CGL Mathematics Previous Year Paper (Topic-wise) PDF Download
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Page 1
Pinnacle Day: 8th - 10th Simplification
Simplification
KEY POINTS :-
1). (a + b)
2
= a
2
+ b
2
+ 2ab
2). (a b)
2
= a
2
+ b
2
2ab - -
3). a
2
b
2
= (a + b)(a b) - -
4). Square root of a number is a value
that, when multiplied by itself, gives the
given number.
Example :-
36 has two square roots 6 and -6;
6
2
= 36 and 6
2
= 36, hence we write -
36 = ±6
5). Square root by Factorization method
Find square root of 1089
First ?nd the factors of 1089.
1089 = 3 3 11 11 × × ×
1089 = 3 × 11 = 33
6). Square root by Division method
Find square root of 2304.
Note:- Make group of two digit from right
2304 = 48
7). If then the required number ? = ?? ,
will be (?) =
2
??
8). = a
2
b
2
c
2
??
4
× ??
4
× ??
4
9). = a b ??
??
× ??
??
??
2
×
??
2
10). =
??
??
??
??
11). ?? × ?? = ????
12). If any number ends with 1, its square
root will end with 1 or, 9, given that the
number is a perfect square.
Example :- , 81 = 9 121 = 11
13). If any number ends with 4, its square
root will end with 2 or 8, given that the
number is a perfect square.
Example: , 324 = 18 144 = 12
14). If any number ends with 5, its square
root will end with 5, given that the
number is a perfect square.
Example: , 225 = 15 625 = 25
15). If any number ends with 6, its square
root will end with 6, given that the
number is a perfect square.
Example: , 256 = 16 676 = 26
16). If any number ends with 9, its square
root will end with 3 or 7, given that the
number is a perfect square.
Example: , 49 = 7 169 = 13
17). The square of any number always
ends with 0, 1, 4, 5, 6 or, 9 but will never
end with 2, 3, 7 or, 8.
18). If in a given number, the total
number of digits are n and if n is even,
the square root of that number will have
digits and if is odd, then the number
??
2
??
of digits will be .
?? + 1
2
19). ?? ?? ?? … ?? = ??
2
??
- 1
2
??
Example
Find the value of 2 2 2 2 2
Here, n = 5, ? 2 2 2 2 2
= = 2 2
2
5
- 1
2
5
31
32
20). ; ?? ?? ?? …8 = ??
Example 7 7 7…8 = 7
21). To ?nd the value of
, ?nd the factors of ?? + ?? + ?? + …8
x, such that the difference between the
factors is 1, then the larger factor will be
the result.
22). To ?nd the value of
, ?nd the factors of ?? - ?? - ?? - …8
x, such that the difference between the
factors is 1, then the smaller factor will
be the result.
Example: Find the value of
12 + 12 + 12 + …8
The factors of 12 with a difference of 1
are 4, 3. Here, 4 is the larger number.
Value will be 4. ?
Conventional method
Let 12 + 12 + 12 + …8 = ??
? 12 + ?? = ??
? 12 + ?? = ??
2
? ??
2
- ?? - 12 = 0
? ??
2
- 4 ?? + 3 ?? - 12 = 0
? ?? - 4 ( ) ?? + 3 ( ) = 0
? ?? = 4 ?????? ?? = - 3
? ?? = 4
Example :- Find the value of
. 30 - 30 - 30 - …8
The factors of 30 with a difference of 1
are 5, 6. Here, 5 is the smaller number. ?
Value will be 5.
23). If any number is in form,
1
?? ± ??
then we multiply by its rationalization
factor in both numerator and ?? ± ??
denominator.
Example: Find the value of ;
1
9 - 8
=
1
9 - 8
1
9 - 8
×
9 + 8
9 + 8
= =
9 + 8
9
2
- 8
2
3 + 2 2
1
= 3 + 2 2
C ube and Cube Root
KEY POINTS
1) If a number is multiplied two times
with itself, then the result of this
multiplication is called the cube of that
number.
Example: 2 × 2 × 2 = 2
3
Cube root of = = 2
1
= 2 2
3
=
3
2
3
2
3×
1
3
2) Algebraic method to calculate cube:
( ?? + ?? )
3
= ??
3
+ 3 ???? ?? + ?? ( ) + ??
3
( ?? - ?? )
3
= ??
3
- 3 ???? ?? - ?? ( ) - ??
3
Indices and Surds
Key Points:
1) p
m
p
n
= p
m+n
2) × ( ??
??
)
??
= ??
????
3) = 4) =
??
??
??
??
??
?? - ??
(
??
??
)
??
??
??
??
??
5) 6) = ??
0
= 1 ??
- ?? 1
??
??
7) p = 8) = p
1
??
??
??
??
??
1
??
9) 10) =
??
???? =
??
?? ×
??
??
?? ??
??
??
??
??
??
11) = 12) = (
??
?? )
??
?? ??
??
??
??
??
??
13) =
??
??
??
?? × ??
??
KEY POINTS
1). Conversion of complex arithmetic
expressions into simple ones is called
Simpli?cation.
2). VBODMAS Rule: Vinculum > Brackets
> Of > division > multiplication > addition
> subtraction
* First solve vinculum i.e. bar.
Eg: 7 - 5 - 4 ( ) =?
First solve 5 – 4 i.e. 1 then 7 -1 = 6.
* For brackets open ?rst small brackets
www.ssccglpinnacle.com Download Pinnacle Exam Preparation App 38
Page 2
Pinnacle Day: 8th - 10th Simplification
Simplification
KEY POINTS :-
1). (a + b)
2
= a
2
+ b
2
+ 2ab
2). (a b)
2
= a
2
+ b
2
2ab - -
3). a
2
b
2
= (a + b)(a b) - -
4). Square root of a number is a value
that, when multiplied by itself, gives the
given number.
Example :-
36 has two square roots 6 and -6;
6
2
= 36 and 6
2
= 36, hence we write -
36 = ±6
5). Square root by Factorization method
Find square root of 1089
First ?nd the factors of 1089.
1089 = 3 3 11 11 × × ×
1089 = 3 × 11 = 33
6). Square root by Division method
Find square root of 2304.
Note:- Make group of two digit from right
2304 = 48
7). If then the required number ? = ?? ,
will be (?) =
2
??
8). = a
2
b
2
c
2
??
4
× ??
4
× ??
4
9). = a b ??
??
× ??
??
??
2
×
??
2
10). =
??
??
??
??
11). ?? × ?? = ????
12). If any number ends with 1, its square
root will end with 1 or, 9, given that the
number is a perfect square.
Example :- , 81 = 9 121 = 11
13). If any number ends with 4, its square
root will end with 2 or 8, given that the
number is a perfect square.
Example: , 324 = 18 144 = 12
14). If any number ends with 5, its square
root will end with 5, given that the
number is a perfect square.
Example: , 225 = 15 625 = 25
15). If any number ends with 6, its square
root will end with 6, given that the
number is a perfect square.
Example: , 256 = 16 676 = 26
16). If any number ends with 9, its square
root will end with 3 or 7, given that the
number is a perfect square.
Example: , 49 = 7 169 = 13
17). The square of any number always
ends with 0, 1, 4, 5, 6 or, 9 but will never
end with 2, 3, 7 or, 8.
18). If in a given number, the total
number of digits are n and if n is even,
the square root of that number will have
digits and if is odd, then the number
??
2
??
of digits will be .
?? + 1
2
19). ?? ?? ?? … ?? = ??
2
??
- 1
2
??
Example
Find the value of 2 2 2 2 2
Here, n = 5, ? 2 2 2 2 2
= = 2 2
2
5
- 1
2
5
31
32
20). ; ?? ?? ?? …8 = ??
Example 7 7 7…8 = 7
21). To ?nd the value of
, ?nd the factors of ?? + ?? + ?? + …8
x, such that the difference between the
factors is 1, then the larger factor will be
the result.
22). To ?nd the value of
, ?nd the factors of ?? - ?? - ?? - …8
x, such that the difference between the
factors is 1, then the smaller factor will
be the result.
Example: Find the value of
12 + 12 + 12 + …8
The factors of 12 with a difference of 1
are 4, 3. Here, 4 is the larger number.
Value will be 4. ?
Conventional method
Let 12 + 12 + 12 + …8 = ??
? 12 + ?? = ??
? 12 + ?? = ??
2
? ??
2
- ?? - 12 = 0
? ??
2
- 4 ?? + 3 ?? - 12 = 0
? ?? - 4 ( ) ?? + 3 ( ) = 0
? ?? = 4 ?????? ?? = - 3
? ?? = 4
Example :- Find the value of
. 30 - 30 - 30 - …8
The factors of 30 with a difference of 1
are 5, 6. Here, 5 is the smaller number. ?
Value will be 5.
23). If any number is in form,
1
?? ± ??
then we multiply by its rationalization
factor in both numerator and ?? ± ??
denominator.
Example: Find the value of ;
1
9 - 8
=
1
9 - 8
1
9 - 8
×
9 + 8
9 + 8
= =
9 + 8
9
2
- 8
2
3 + 2 2
1
= 3 + 2 2
C ube and Cube Root
KEY POINTS
1) If a number is multiplied two times
with itself, then the result of this
multiplication is called the cube of that
number.
Example: 2 × 2 × 2 = 2
3
Cube root of = = 2
1
= 2 2
3
=
3
2
3
2
3×
1
3
2) Algebraic method to calculate cube:
( ?? + ?? )
3
= ??
3
+ 3 ???? ?? + ?? ( ) + ??
3
( ?? - ?? )
3
= ??
3
- 3 ???? ?? - ?? ( ) - ??
3
Indices and Surds
Key Points:
1) p
m
p
n
= p
m+n
2) × ( ??
??
)
??
= ??
????
3) = 4) =
??
??
??
??
??
?? - ??
(
??
??
)
??
??
??
??
??
5) 6) = ??
0
= 1 ??
- ?? 1
??
??
7) p = 8) = p
1
??
??
??
??
??
1
??
9) 10) =
??
???? =
??
?? ×
??
??
?? ??
??
??
??
??
??
11) = 12) = (
??
?? )
??
?? ??
??
??
??
??
??
13) =
??
??
??
?? × ??
??
KEY POINTS
1). Conversion of complex arithmetic
expressions into simple ones is called
Simpli?cation.
2). VBODMAS Rule: Vinculum > Brackets
> Of > division > multiplication > addition
> subtraction
* First solve vinculum i.e. bar.
Eg: 7 - 5 - 4 ( ) =?
First solve 5 – 4 i.e. 1 then 7 -1 = 6.
* For brackets open ?rst small brackets
www.ssccglpinnacle.com Download Pinnacle Exam Preparation App 38
Pinnacle Day: 8th - 10th Simplification
(….), then {…} and then […].
* For modulus e.g. |-3| we write
magnitude only not sign i.e. 3 in this
case.
Important Formulas
1) . ( ?? + ?? )
2
= ??
2
+ 2 ???? + ??
2
2). ( ?? - ?? )
2
= ??
2
- 2 ???? + ??
2
3). ( ??
2
- ??
2
) = ( ?? + ?? )( ?? - ?? )
4). ( ?? + ?? )
3
= ??
3
+ ??
3
+ 3 ???? ( ?? + ?? )
5). ( ?? - ?? )
3
= ??
3
- ??
3
- 3 ???? ( ?? - ?? )
6). ( ??
3
+ ??
3
) = ( ?? + ?? )( ??
2
- ???? + ??
2
)
7). ( ??
3
- ??
3
) = ( ?? - ?? )( ??
2
+ ???? + ??
2
)
8). = ??
3
+ ??
3
+ ??
3
- 3 ??????
?? + ?? + ?? ( ) ??
2
+ ??
2
+ ??
2
- ???? - ???? - ????
( )
=
1
2
( ?? + ?? + ?? )[( ?? - ?? )
2
+ ( ?? - ?? )
2
+
( ?? - ?? )
2
]
Here, if ?? + ?? + ?? ( ) = 0 ,
then ??
3
+ ??
3
+ ??
3
- 3 ?????? = 0
? ??
3
+ ??
3
+ ??
3
= 3 ??????
9). ( ?? + ?? + ?? )
2
= ??
2
+ ??
2
+ ??
2
( )
+ 2 ( ????
+ ???? + ???? )
Variety Questions
Q.1. The simpli?ed form of the
expression
( ??
2
- 9 ) × ( 9 ??
2
- 1 )
( ?? -
1
3
) × ( 1 -
??
3
)
×
9
?? +
1
3
( )
1 +
??
3
( )
is _______.
SSC CHSL Tier II (10/01/2024)
(a) -729 (b) -81 (c) 729 (d) 81
Q.2. If , B = , ?? =
58
2
- 25
2
46
2
- 37
2
26
2
- 15
2
56
2
- 15
2
then the value of - is :
1
??
20
??
SSC CGL Tier II (26/10/2023)
(a) 2 (b) 1 (c) -1 (d) 0
Q.3. On comparing the following two
numeric expressions, we ?nd that ______.
[(2 ) ] and [(1 )
5
]
7
9
2
1
2
3
5
2
3
3
5
SSC MTS 14/09/2023 (2nd Shift)
(a) both the expressions are equal
(b) the ?rst expression is smaller than
the second
(c) the ?rst expression is larger than the
second
(d) the given two expressions cannot be
compared
Q.4. On simplifying the following
expression, we get ________.
24 45 - 16 20 + 245 - 47 5
3
1
5
+ 1
2
5
- 2
1
6
- 2
1
5
- 5
2
5
+ 3
1
6
SSC MTS 08/09/2023 (2nd Shift)
(a) 0 (b) 3 (c) 1 (d) 2
Q.5. Simplify :
( 3 . 321 )
3
+ ( 2 . 681 )
3
+ ( 1 . 245 )
3
- 3 × 3 . 321 × 2 . 681 × 1 . 245
( 3 . 321 )
2
+ ( 2 . 681 )
2
+ ( 1 . 245 )
2
- ( 3 . 321 × 2 . 681 ) - ( 2 . 681 × 1 . 245 ) - ( 1 . 245 × 3 . 321 )
= ?
SSC CHSL 04/08/2023 (1st Shift)
(a) 6.125 (b) 8.645 (c) 7.247 (d) 10.245
Q.6. Simplify the following expression.
0 . 01 + 0 . 68 × 0 . 68 + 0 . 22 × 0 . 22 +
0 . 044 + 0 . 136 + 0 . 44 × 0 . 68
SSC CHSL 03/08/2023 (1st Shift)
(a) 0.69 (b) 1 (c) 0.8 (d) 0.9
Q.7. Simplify the given expression .
0 . 09 × 0 . 09 + 0 . 04 × 0 . 04 + 0 . 16 × 0 . 16 + 2 × 0 . 09 × 0 . 04 + 2 × 0 . 04 × 0 . 16 + 2 × 0 . 09 × 0 . 16
0 . 3 × 0 . 3 + 0 . 2 × 0 . 2 + 0 . 4 × 0 . 4
SSC CHSL 02/08/2023 (4th Shift)
(a) 0.38 (b) 0.24 (c) 0.32 (d) 0.29
Q.8. What is the value of the given
Expression ?
4
?? + 4
- 5 × 4
?? + 2
15 × 4
??
- 2
2
× 4
??
SSC CGL 24/07/2023 (1st shift)
(a) 16 (b) 64 (c) 20 (d) 24
Q.9. What is the value of ( + 1)( + 1) 2
2
2
4
( + 1)( + 1) ....... ( + 1) ? 2
8
2
16
2
128
SSC CHSL 14/03/2023 (2nd Shift)
(a) (b)
2
256
- 1
2
2
256
- 1
3
(c) (d) 2
256
- 1
2
256
- 1
4
Q.10. Which of the following statement
is true ?
I. + + + ……
1
2
1
6
1
12
1
110
<
5
6
II. + + +...... >
1
3
1
15
1
35
1
143
7
13
SSC CHSL 13/03/2023 (4th Shift)
(a) Only I (b) Both I and II
(c) Only II (d) Neither I nor II
Q.11. Find the value of the following
expression.
[
5
8
- {
3
8
- (
5
8
-
3
8
)}] ???? 8 . 8 - 1 . 2
4
1
6
÷ 2 . 5 × 2 ÷
1
6
???? 60 + (
3
4
-
3
8
)
SSC CPO 11/11/2022 (Morning)
(a) 5 (b) 3 (c) 4 (d) 4
22
43
23
67
44
85
4
5
Q.12. Solve the following.
24 ÷
3
8
???? ( 8 + 2× 7 - 3 )+ [
2
11
÷
4
55
- {
5
8
+
6
16
}]
32÷ 15 - 7 + 75÷ ( 6 + 15÷3 + 4 )
SSC CPO 09/11/2022 (Morning)
(a) (b) (c) (d)
23
27
9
2
11
18
15
19
Q.13. Solve
( 6 . 9 )
3
+ ( 2 . 7 )
3
+ ( 5 . 8 )
3
- 3 × 6 . 9 × 2 . 7 ×5 . 8
( 6 . 9 )
2
+ ( 2 . 7 )
2
+ ( 5 . 8 )
2
-( 6 . 9 )( 2 . 7 )-( 2 . 7 )( 5 . 8 )-( 5 . 8 )( 6 . 9 )
SSC CHSL 07/06/2022(Afternoon)
(a) 15.1 (b) 15.3 (c) 15.2 (d) 15.4
Q.14. Simplify the following expression.
9 {(16 ) +
1
5
÷
1
5
÷ 12
2
3
- 9
2
3
17 of 3 }
1
2
1
21
SSC CHSL 26/05/2022(Evening)
(a) (b) (c) (d)
139
881
137
881
140
881
138
881
Q.15. The value of
+ is:
1
4
20 . 35 ( )
2
- 8 . 35 ( )
2
[ ]
× 0 . 0175
( 1 . 05 )
2
+ ( 1 . 05 )( 27 . 65 )
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
3
20
3
10
7
20
9
20
Q.16 The simpli?ed value of
119 -[ 48 ÷ 6 - 7 { 5 × 12 ÷ 3 - ( 15 - 3 - 8 )}]
24 ÷ 3 - 9 ???? 3 + ( 52 - 8 - 4 ) ÷ 6
is:
SSC MTS 11/10/2021(Morning)
(a) (b) (c) (d) -
111
11
-
127
27
127
43
111
11
Q.17. If -
1
4 - 8
+
3 + 2 2
3 - 2 2
3 - 2 2
3 + 2 2
, then what is the value of = ?? + ?? 2
(3a + 4b) ?
SSC CGL Tier II (15/11/2020)
(a) 99 (b) (c) 98 (d)
1
2
97
1
2
98
Q.18. If the following interchanges are
made in signs and numbers, which
equation would be correct ?
Interchanges: Signs: and - Numbers: ÷
15 and 5
SSC CHSL 13/10/2020 (Morning)
(a) 4 30 - 15 5 + 12 = 21 × ÷
(b) 12 30 - 15 5 + 4 = 37 × ÷
(c) 4 30 - 15 5 + 12 = 27 × ÷
(d) 4 30 - 5 15 + 14 = 21 × ÷
Q.19. If =
4
1 + 2 + 3
?? + ?? 2 + ?? 3
where a, b , c , d are natural - ?? 6 ,
numbers, then the value of a + b + c + d
is:
SSC CHSL 18/03/2020 (Morning)
(a) 0 (b) 2 (c) 4 (d) 1
Q.20.
If [{( )
3
}
(2x + 3)
] = [{( ) }
(3 x + 7)
]
2
3
- 3
4
2
3
2
3
- 6
5
then the value of is: 2 - 42 ??
SSC CHSL 18/03/2020(Morning)
(a) 5 (b) 6 (c) 3 (d) 4
Q.21. If ‘+’ means ‘-’, ‘-’ means ‘+’, ‘ ’ ×
means ‘ ’ and ‘ ’ means ‘ ’ then the ÷ ÷ ×
value of is:
42 - 12 × 3 + 8 ÷ 2 + 15
8 × 2 - 4 + 9 ÷ 3
www.ssccglpinnacle.com Download Pinnacle Exam Preparation App 39
Page 3
Pinnacle Day: 8th - 10th Simplification
Simplification
KEY POINTS :-
1). (a + b)
2
= a
2
+ b
2
+ 2ab
2). (a b)
2
= a
2
+ b
2
2ab - -
3). a
2
b
2
= (a + b)(a b) - -
4). Square root of a number is a value
that, when multiplied by itself, gives the
given number.
Example :-
36 has two square roots 6 and -6;
6
2
= 36 and 6
2
= 36, hence we write -
36 = ±6
5). Square root by Factorization method
Find square root of 1089
First ?nd the factors of 1089.
1089 = 3 3 11 11 × × ×
1089 = 3 × 11 = 33
6). Square root by Division method
Find square root of 2304.
Note:- Make group of two digit from right
2304 = 48
7). If then the required number ? = ?? ,
will be (?) =
2
??
8). = a
2
b
2
c
2
??
4
× ??
4
× ??
4
9). = a b ??
??
× ??
??
??
2
×
??
2
10). =
??
??
??
??
11). ?? × ?? = ????
12). If any number ends with 1, its square
root will end with 1 or, 9, given that the
number is a perfect square.
Example :- , 81 = 9 121 = 11
13). If any number ends with 4, its square
root will end with 2 or 8, given that the
number is a perfect square.
Example: , 324 = 18 144 = 12
14). If any number ends with 5, its square
root will end with 5, given that the
number is a perfect square.
Example: , 225 = 15 625 = 25
15). If any number ends with 6, its square
root will end with 6, given that the
number is a perfect square.
Example: , 256 = 16 676 = 26
16). If any number ends with 9, its square
root will end with 3 or 7, given that the
number is a perfect square.
Example: , 49 = 7 169 = 13
17). The square of any number always
ends with 0, 1, 4, 5, 6 or, 9 but will never
end with 2, 3, 7 or, 8.
18). If in a given number, the total
number of digits are n and if n is even,
the square root of that number will have
digits and if is odd, then the number
??
2
??
of digits will be .
?? + 1
2
19). ?? ?? ?? … ?? = ??
2
??
- 1
2
??
Example
Find the value of 2 2 2 2 2
Here, n = 5, ? 2 2 2 2 2
= = 2 2
2
5
- 1
2
5
31
32
20). ; ?? ?? ?? …8 = ??
Example 7 7 7…8 = 7
21). To ?nd the value of
, ?nd the factors of ?? + ?? + ?? + …8
x, such that the difference between the
factors is 1, then the larger factor will be
the result.
22). To ?nd the value of
, ?nd the factors of ?? - ?? - ?? - …8
x, such that the difference between the
factors is 1, then the smaller factor will
be the result.
Example: Find the value of
12 + 12 + 12 + …8
The factors of 12 with a difference of 1
are 4, 3. Here, 4 is the larger number.
Value will be 4. ?
Conventional method
Let 12 + 12 + 12 + …8 = ??
? 12 + ?? = ??
? 12 + ?? = ??
2
? ??
2
- ?? - 12 = 0
? ??
2
- 4 ?? + 3 ?? - 12 = 0
? ?? - 4 ( ) ?? + 3 ( ) = 0
? ?? = 4 ?????? ?? = - 3
? ?? = 4
Example :- Find the value of
. 30 - 30 - 30 - …8
The factors of 30 with a difference of 1
are 5, 6. Here, 5 is the smaller number. ?
Value will be 5.
23). If any number is in form,
1
?? ± ??
then we multiply by its rationalization
factor in both numerator and ?? ± ??
denominator.
Example: Find the value of ;
1
9 - 8
=
1
9 - 8
1
9 - 8
×
9 + 8
9 + 8
= =
9 + 8
9
2
- 8
2
3 + 2 2
1
= 3 + 2 2
C ube and Cube Root
KEY POINTS
1) If a number is multiplied two times
with itself, then the result of this
multiplication is called the cube of that
number.
Example: 2 × 2 × 2 = 2
3
Cube root of = = 2
1
= 2 2
3
=
3
2
3
2
3×
1
3
2) Algebraic method to calculate cube:
( ?? + ?? )
3
= ??
3
+ 3 ???? ?? + ?? ( ) + ??
3
( ?? - ?? )
3
= ??
3
- 3 ???? ?? - ?? ( ) - ??
3
Indices and Surds
Key Points:
1) p
m
p
n
= p
m+n
2) × ( ??
??
)
??
= ??
????
3) = 4) =
??
??
??
??
??
?? - ??
(
??
??
)
??
??
??
??
??
5) 6) = ??
0
= 1 ??
- ?? 1
??
??
7) p = 8) = p
1
??
??
??
??
??
1
??
9) 10) =
??
???? =
??
?? ×
??
??
?? ??
??
??
??
??
??
11) = 12) = (
??
?? )
??
?? ??
??
??
??
??
??
13) =
??
??
??
?? × ??
??
KEY POINTS
1). Conversion of complex arithmetic
expressions into simple ones is called
Simpli?cation.
2). VBODMAS Rule: Vinculum > Brackets
> Of > division > multiplication > addition
> subtraction
* First solve vinculum i.e. bar.
Eg: 7 - 5 - 4 ( ) =?
First solve 5 – 4 i.e. 1 then 7 -1 = 6.
* For brackets open ?rst small brackets
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Pinnacle Day: 8th - 10th Simplification
(….), then {…} and then […].
* For modulus e.g. |-3| we write
magnitude only not sign i.e. 3 in this
case.
Important Formulas
1) . ( ?? + ?? )
2
= ??
2
+ 2 ???? + ??
2
2). ( ?? - ?? )
2
= ??
2
- 2 ???? + ??
2
3). ( ??
2
- ??
2
) = ( ?? + ?? )( ?? - ?? )
4). ( ?? + ?? )
3
= ??
3
+ ??
3
+ 3 ???? ( ?? + ?? )
5). ( ?? - ?? )
3
= ??
3
- ??
3
- 3 ???? ( ?? - ?? )
6). ( ??
3
+ ??
3
) = ( ?? + ?? )( ??
2
- ???? + ??
2
)
7). ( ??
3
- ??
3
) = ( ?? - ?? )( ??
2
+ ???? + ??
2
)
8). = ??
3
+ ??
3
+ ??
3
- 3 ??????
?? + ?? + ?? ( ) ??
2
+ ??
2
+ ??
2
- ???? - ???? - ????
( )
=
1
2
( ?? + ?? + ?? )[( ?? - ?? )
2
+ ( ?? - ?? )
2
+
( ?? - ?? )
2
]
Here, if ?? + ?? + ?? ( ) = 0 ,
then ??
3
+ ??
3
+ ??
3
- 3 ?????? = 0
? ??
3
+ ??
3
+ ??
3
= 3 ??????
9). ( ?? + ?? + ?? )
2
= ??
2
+ ??
2
+ ??
2
( )
+ 2 ( ????
+ ???? + ???? )
Variety Questions
Q.1. The simpli?ed form of the
expression
( ??
2
- 9 ) × ( 9 ??
2
- 1 )
( ?? -
1
3
) × ( 1 -
??
3
)
×
9
?? +
1
3
( )
1 +
??
3
( )
is _______.
SSC CHSL Tier II (10/01/2024)
(a) -729 (b) -81 (c) 729 (d) 81
Q.2. If , B = , ?? =
58
2
- 25
2
46
2
- 37
2
26
2
- 15
2
56
2
- 15
2
then the value of - is :
1
??
20
??
SSC CGL Tier II (26/10/2023)
(a) 2 (b) 1 (c) -1 (d) 0
Q.3. On comparing the following two
numeric expressions, we ?nd that ______.
[(2 ) ] and [(1 )
5
]
7
9
2
1
2
3
5
2
3
3
5
SSC MTS 14/09/2023 (2nd Shift)
(a) both the expressions are equal
(b) the ?rst expression is smaller than
the second
(c) the ?rst expression is larger than the
second
(d) the given two expressions cannot be
compared
Q.4. On simplifying the following
expression, we get ________.
24 45 - 16 20 + 245 - 47 5
3
1
5
+ 1
2
5
- 2
1
6
- 2
1
5
- 5
2
5
+ 3
1
6
SSC MTS 08/09/2023 (2nd Shift)
(a) 0 (b) 3 (c) 1 (d) 2
Q.5. Simplify :
( 3 . 321 )
3
+ ( 2 . 681 )
3
+ ( 1 . 245 )
3
- 3 × 3 . 321 × 2 . 681 × 1 . 245
( 3 . 321 )
2
+ ( 2 . 681 )
2
+ ( 1 . 245 )
2
- ( 3 . 321 × 2 . 681 ) - ( 2 . 681 × 1 . 245 ) - ( 1 . 245 × 3 . 321 )
= ?
SSC CHSL 04/08/2023 (1st Shift)
(a) 6.125 (b) 8.645 (c) 7.247 (d) 10.245
Q.6. Simplify the following expression.
0 . 01 + 0 . 68 × 0 . 68 + 0 . 22 × 0 . 22 +
0 . 044 + 0 . 136 + 0 . 44 × 0 . 68
SSC CHSL 03/08/2023 (1st Shift)
(a) 0.69 (b) 1 (c) 0.8 (d) 0.9
Q.7. Simplify the given expression .
0 . 09 × 0 . 09 + 0 . 04 × 0 . 04 + 0 . 16 × 0 . 16 + 2 × 0 . 09 × 0 . 04 + 2 × 0 . 04 × 0 . 16 + 2 × 0 . 09 × 0 . 16
0 . 3 × 0 . 3 + 0 . 2 × 0 . 2 + 0 . 4 × 0 . 4
SSC CHSL 02/08/2023 (4th Shift)
(a) 0.38 (b) 0.24 (c) 0.32 (d) 0.29
Q.8. What is the value of the given
Expression ?
4
?? + 4
- 5 × 4
?? + 2
15 × 4
??
- 2
2
× 4
??
SSC CGL 24/07/2023 (1st shift)
(a) 16 (b) 64 (c) 20 (d) 24
Q.9. What is the value of ( + 1)( + 1) 2
2
2
4
( + 1)( + 1) ....... ( + 1) ? 2
8
2
16
2
128
SSC CHSL 14/03/2023 (2nd Shift)
(a) (b)
2
256
- 1
2
2
256
- 1
3
(c) (d) 2
256
- 1
2
256
- 1
4
Q.10. Which of the following statement
is true ?
I. + + + ……
1
2
1
6
1
12
1
110
<
5
6
II. + + +...... >
1
3
1
15
1
35
1
143
7
13
SSC CHSL 13/03/2023 (4th Shift)
(a) Only I (b) Both I and II
(c) Only II (d) Neither I nor II
Q.11. Find the value of the following
expression.
[
5
8
- {
3
8
- (
5
8
-
3
8
)}] ???? 8 . 8 - 1 . 2
4
1
6
÷ 2 . 5 × 2 ÷
1
6
???? 60 + (
3
4
-
3
8
)
SSC CPO 11/11/2022 (Morning)
(a) 5 (b) 3 (c) 4 (d) 4
22
43
23
67
44
85
4
5
Q.12. Solve the following.
24 ÷
3
8
???? ( 8 + 2× 7 - 3 )+ [
2
11
÷
4
55
- {
5
8
+
6
16
}]
32÷ 15 - 7 + 75÷ ( 6 + 15÷3 + 4 )
SSC CPO 09/11/2022 (Morning)
(a) (b) (c) (d)
23
27
9
2
11
18
15
19
Q.13. Solve
( 6 . 9 )
3
+ ( 2 . 7 )
3
+ ( 5 . 8 )
3
- 3 × 6 . 9 × 2 . 7 ×5 . 8
( 6 . 9 )
2
+ ( 2 . 7 )
2
+ ( 5 . 8 )
2
-( 6 . 9 )( 2 . 7 )-( 2 . 7 )( 5 . 8 )-( 5 . 8 )( 6 . 9 )
SSC CHSL 07/06/2022(Afternoon)
(a) 15.1 (b) 15.3 (c) 15.2 (d) 15.4
Q.14. Simplify the following expression.
9 {(16 ) +
1
5
÷
1
5
÷ 12
2
3
- 9
2
3
17 of 3 }
1
2
1
21
SSC CHSL 26/05/2022(Evening)
(a) (b) (c) (d)
139
881
137
881
140
881
138
881
Q.15. The value of
+ is:
1
4
20 . 35 ( )
2
- 8 . 35 ( )
2
[ ]
× 0 . 0175
( 1 . 05 )
2
+ ( 1 . 05 )( 27 . 65 )
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
3
20
3
10
7
20
9
20
Q.16 The simpli?ed value of
119 -[ 48 ÷ 6 - 7 { 5 × 12 ÷ 3 - ( 15 - 3 - 8 )}]
24 ÷ 3 - 9 ???? 3 + ( 52 - 8 - 4 ) ÷ 6
is:
SSC MTS 11/10/2021(Morning)
(a) (b) (c) (d) -
111
11
-
127
27
127
43
111
11
Q.17. If -
1
4 - 8
+
3 + 2 2
3 - 2 2
3 - 2 2
3 + 2 2
, then what is the value of = ?? + ?? 2
(3a + 4b) ?
SSC CGL Tier II (15/11/2020)
(a) 99 (b) (c) 98 (d)
1
2
97
1
2
98
Q.18. If the following interchanges are
made in signs and numbers, which
equation would be correct ?
Interchanges: Signs: and - Numbers: ÷
15 and 5
SSC CHSL 13/10/2020 (Morning)
(a) 4 30 - 15 5 + 12 = 21 × ÷
(b) 12 30 - 15 5 + 4 = 37 × ÷
(c) 4 30 - 15 5 + 12 = 27 × ÷
(d) 4 30 - 5 15 + 14 = 21 × ÷
Q.19. If =
4
1 + 2 + 3
?? + ?? 2 + ?? 3
where a, b , c , d are natural - ?? 6 ,
numbers, then the value of a + b + c + d
is:
SSC CHSL 18/03/2020 (Morning)
(a) 0 (b) 2 (c) 4 (d) 1
Q.20.
If [{( )
3
}
(2x + 3)
] = [{( ) }
(3 x + 7)
]
2
3
- 3
4
2
3
2
3
- 6
5
then the value of is: 2 - 42 ??
SSC CHSL 18/03/2020(Morning)
(a) 5 (b) 6 (c) 3 (d) 4
Q.21. If ‘+’ means ‘-’, ‘-’ means ‘+’, ‘ ’ ×
means ‘ ’ and ‘ ’ means ‘ ’ then the ÷ ÷ ×
value of is:
42 - 12 × 3 + 8 ÷ 2 + 15
8 × 2 - 4 + 9 ÷ 3
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Pinnacle Day: 8th - 10th Simplification
SSC CGL 03/03/2020 (Morning)
(a) - (b) - (c) (d)
5
3
15
19
15
19
5
3
Q.22. The value of
is:
( 4 . 6 )
4
+ ( 5 . 4 )
4
+ ( 24 . 84 )
2
( 4 . 6 )
2
+( 5 . 4 )
2
+ 24 . 84
SSC CGL Tier II (13/09/2019)
(a) 24.42 (b) 24.24 (c) 25.42 (d) 25.48
Q.23. The value of
is :
( 0 . 545 )( 0 . 081 )( 0 . 51 )( 5 . 2 )
( 0 . 324 )
3
+ ( 0 . 221 )
3
- ( 0 . 545 )
3
SSC CGL Tier II (12/09/2019)
(a) - 1 (b) 1 (c) 3 (d) - 3
Q.24. The value of 22. + 11.5 - 33.5 is 4 67 9
SSC CGL Tier II (11/09/2019)
(a) 0. (b) 0. (c) 0.3 (d) 0.4 32 412 1 12
Q.25. Find the value
{
( 0 . 7 )
2
÷ 0 . 14 + ( 0 . 6 )
2
÷ 0 . 18 + ( 0 . 5 )
2
÷ 0 . 05
4 ( 2 . 5 ???? 4 - 13 × 0 . 25 × 3 )
}
SSC MTS 07/08/2019 (Morning)
(a) (b) (c) (d)
25
2
19
2
23
2
21
2
Q.26. If x = +
1
12 . 13
1
13 . 14
+
1
14 . 15
….. + , y = +
1
23 . 24
1
36 . 37
1
37 . 38
+
….. + then is equal to :
1
38 . 39
1
71 . 72
??
??
SSC CHSL 10/07/2019 (Evening)
(a) (b) (c) (d) 3
1
3
1
24
1
72
Q.27. The simpli?ed value of
is:
0 . 01404
24
2
+ 6
2
- 144
SSC CHSL 10/07/2019 (Evening)
(a) 3 (b) × 10
- 5
6 × 10
- 5
(c) 2.4 (d) 3 × 10
- 4
× 10
- 4
Q.28. The simpli?ed value of
is:
1 . 0025 + 6 . 25 × 10
- 6
0 . 0025 + 0 . 95
SSC CHSL 09/06/2019 (Evening)
(a) 1.0025 (b) 1.0525
(c) 1.0005 (d) 1.0505
Q.29. The value of of ( of
9
15
2
3
÷
2
3
) ( of ) of (
3
2
÷
3
4
×
3
4
÷
3
4
4
3
5
4
÷
of ) is :
5
2
×
2
5
4
5
SSC CGL 06/06/2019 (Afternoon)
(a) (b) (c) (d)
20
9
4
25
18
125
40
9
Q.30. The value of (5 + 3 5 5) / (3 ÷ × ÷
3 of 6) of (4 4 4 of 4 + 4 4 4) is × ÷ ÷ ×
SSC CGL 06/06/2019 (Morning)
(a) 8 (b) 7 (c) 9 (d) 6
1
5
1
3
3
5
2
3
Q.31. The value of 2 ÷ (3 of
7
8
5
6
÷
2
7
2 ) [(2 of 4 ÷ ) ] is :
1
3
×
6
7
1
5
2
3
×
5
9
SSC CGL 04/06/2019 (Evening)
(a) (b) 4 (c) (d) 5
1
4
1
23
Q.32. The value of 2 3 2 of 3 2 × ÷ × ÷
(4 + 4 4 of 4 - 4 4 ) is: × 4 ÷ ÷ × 4
SSC CGL 04/06/2019 (Morning)
(a) 8 (b) 1 (c) 4 (d) 2
Q.33.
5 . 75 × 5 . 75 × 5 . 75 + 3 . 25 × 3 . 25 × 3 . 25
57 . 5 × 57 . 5 + 32 . 5 × 32 . 5 - 57 . 5 × 32 . 5
is equal to :
SSC CPO 12/03/2019 (Evening)
(a) 0.009 (b) 0.0009 (c) 0.9 (d) 0.09
Practice Questions
SSC CHSL 2023 Tier - 2
Q.34. If x = ( )
2
( )
-4
, ?nd the value
3
2
×
2
3
of ??
- 2
.
SSC CHSL Tier II (02/11/2023)
(a) ( )
6
(b) ( )
12
3
2
3
2
(c) ( )
12
(d) ( )
6
2
3
2
3
Q.35. If
.
3
0 . 08 × 0 . 8 × ?? = 0 . 008 × 0 . 8 ×
3
??
then ?nd the value of .
??
??
SSC CHSL Tier II (02/11/2023)
(a) (b) 8
3
× 10
- 8
8
3
× 10
- 9
(c) (d) 8
4
× 10
- 9
8
4
× 10
- 8
Q.36. Simplify :
.
3
3
343
( )
3
×
3
3
216
( )
3
SSC CHSL Tier II (10/01/2024)
(a) 56 (b) 42 (c) (d)
3
42 42
3
Q.37. Find the value of
+ + + 625
625
100
625
10000
625
1000000
SSC CHSL Tier II (10/01/2024)
(a) 27.075 (b) 27.775
(c) 27.75 (d) 27.025
SSC CGL 2023 Tier - 2
Q.38. What is the value of [ + { × (
1
8
1
6
÷ ) - ( × ÷ )} + ] ?
36
45
24
25
12
21
14
15
24
45
27
36
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
1
36
1
72
1
108
1
27
Q.39. Find the value of .
1 . 24 × 2 . 79
2 . 64 × 5 . 94
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
31
66
31
44
33
64
33
31
SSC CPO 2023 Tier - 1
Q.40. Evaluate
of
0 . 04
0 . 05
3
1
3
- 2
1
2
( )
÷
1
2
???? 1
1
4
1
3
+
1
5
????
1
9
SSC CPO 03/10/2023 (1st Shift)
(a) 5 (b) 0.3 (c) 3 (d) 0.03
Q.41. If (48 ÷ 72 × 3) - [15 ÷ 8 × (40 - 32) -
10] + 2P = 6 ÷ 2, then ?nd the value of P
SSC CPO 03/10/2023 (3rd Shift)
(a) 2 (b) 4 (c) 1 (d) 3
Q.42. Simplify .
1 -
3
7
÷
9
49
×
3
14
1 +
3
5
÷
1
25
×
1
30
SSC CPO 04/10/2023 (1st Shift)
(a) (b) (c) (d)
1
6
1
5
1
3
1
7
Q.43. The value of 1 - [3 ÷ {1 ÷
2
5
3
4
1
4
1
2
(1 × 3 ÷ 1 )}] is :
1
2
1
3
1
3
SSC CPO 04/10/2023 (1st Shift)
(a) 1 (b) 3 (c) 2 (d) 0
Q.44. The value of
is :
1
1
2
÷ 3
1
4
+
1
2
÷
13
14
+
1
5
1
5
× 3
1
2
-
1
3
÷ 1
3
4
× 3
1
2
SSC CPO 05/10/2023 (1st Shift)
(a) 38 (b) 33 (c) 36 (d) 40
Q.45. Simplify the given expression.
b + 2a - [(3b + a) - (2a + b) + 2a] - a
SSC CPO 05/10/2023 (3rd Shift)
(a) - a (b) - b (c) a (d) b
SSC MTS 2023 Tier - 1
Q.46. Simplify:
24 ÷ 6 × ( of + of ) = ?
3
4
1
3
5
3
3
4
SSC MTS 01/09/2023 (3rd Shift)
(a) 9 (b) 6 (c) 3 (d) 8
Q.47. Simplify the given expression
71 - ( - 3){ - 2 - } + 3{5 + ( - 2)(-1)} ( 8 - 3 )
SSC MTS 04/09/2023 (3rd Shift)
(a) 30 (b) 119 (c) 50 (d) 71
Q.48. Simplify the following. 15 ÷ 5 of 3 ×
4 + 8 ÷ 12 of 4 × 12 - 4 ÷ 5 × 20
SSC MTS 06/09/2023 (2nd Shift)
(a) (b) - 10 (c) 12 (d) 1
20
3
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Page 4
Pinnacle Day: 8th - 10th Simplification
Simplification
KEY POINTS :-
1). (a + b)
2
= a
2
+ b
2
+ 2ab
2). (a b)
2
= a
2
+ b
2
2ab - -
3). a
2
b
2
= (a + b)(a b) - -
4). Square root of a number is a value
that, when multiplied by itself, gives the
given number.
Example :-
36 has two square roots 6 and -6;
6
2
= 36 and 6
2
= 36, hence we write -
36 = ±6
5). Square root by Factorization method
Find square root of 1089
First ?nd the factors of 1089.
1089 = 3 3 11 11 × × ×
1089 = 3 × 11 = 33
6). Square root by Division method
Find square root of 2304.
Note:- Make group of two digit from right
2304 = 48
7). If then the required number ? = ?? ,
will be (?) =
2
??
8). = a
2
b
2
c
2
??
4
× ??
4
× ??
4
9). = a b ??
??
× ??
??
??
2
×
??
2
10). =
??
??
??
??
11). ?? × ?? = ????
12). If any number ends with 1, its square
root will end with 1 or, 9, given that the
number is a perfect square.
Example :- , 81 = 9 121 = 11
13). If any number ends with 4, its square
root will end with 2 or 8, given that the
number is a perfect square.
Example: , 324 = 18 144 = 12
14). If any number ends with 5, its square
root will end with 5, given that the
number is a perfect square.
Example: , 225 = 15 625 = 25
15). If any number ends with 6, its square
root will end with 6, given that the
number is a perfect square.
Example: , 256 = 16 676 = 26
16). If any number ends with 9, its square
root will end with 3 or 7, given that the
number is a perfect square.
Example: , 49 = 7 169 = 13
17). The square of any number always
ends with 0, 1, 4, 5, 6 or, 9 but will never
end with 2, 3, 7 or, 8.
18). If in a given number, the total
number of digits are n and if n is even,
the square root of that number will have
digits and if is odd, then the number
??
2
??
of digits will be .
?? + 1
2
19). ?? ?? ?? … ?? = ??
2
??
- 1
2
??
Example
Find the value of 2 2 2 2 2
Here, n = 5, ? 2 2 2 2 2
= = 2 2
2
5
- 1
2
5
31
32
20). ; ?? ?? ?? …8 = ??
Example 7 7 7…8 = 7
21). To ?nd the value of
, ?nd the factors of ?? + ?? + ?? + …8
x, such that the difference between the
factors is 1, then the larger factor will be
the result.
22). To ?nd the value of
, ?nd the factors of ?? - ?? - ?? - …8
x, such that the difference between the
factors is 1, then the smaller factor will
be the result.
Example: Find the value of
12 + 12 + 12 + …8
The factors of 12 with a difference of 1
are 4, 3. Here, 4 is the larger number.
Value will be 4. ?
Conventional method
Let 12 + 12 + 12 + …8 = ??
? 12 + ?? = ??
? 12 + ?? = ??
2
? ??
2
- ?? - 12 = 0
? ??
2
- 4 ?? + 3 ?? - 12 = 0
? ?? - 4 ( ) ?? + 3 ( ) = 0
? ?? = 4 ?????? ?? = - 3
? ?? = 4
Example :- Find the value of
. 30 - 30 - 30 - …8
The factors of 30 with a difference of 1
are 5, 6. Here, 5 is the smaller number. ?
Value will be 5.
23). If any number is in form,
1
?? ± ??
then we multiply by its rationalization
factor in both numerator and ?? ± ??
denominator.
Example: Find the value of ;
1
9 - 8
=
1
9 - 8
1
9 - 8
×
9 + 8
9 + 8
= =
9 + 8
9
2
- 8
2
3 + 2 2
1
= 3 + 2 2
C ube and Cube Root
KEY POINTS
1) If a number is multiplied two times
with itself, then the result of this
multiplication is called the cube of that
number.
Example: 2 × 2 × 2 = 2
3
Cube root of = = 2
1
= 2 2
3
=
3
2
3
2
3×
1
3
2) Algebraic method to calculate cube:
( ?? + ?? )
3
= ??
3
+ 3 ???? ?? + ?? ( ) + ??
3
( ?? - ?? )
3
= ??
3
- 3 ???? ?? - ?? ( ) - ??
3
Indices and Surds
Key Points:
1) p
m
p
n
= p
m+n
2) × ( ??
??
)
??
= ??
????
3) = 4) =
??
??
??
??
??
?? - ??
(
??
??
)
??
??
??
??
??
5) 6) = ??
0
= 1 ??
- ?? 1
??
??
7) p = 8) = p
1
??
??
??
??
??
1
??
9) 10) =
??
???? =
??
?? ×
??
??
?? ??
??
??
??
??
??
11) = 12) = (
??
?? )
??
?? ??
??
??
??
??
??
13) =
??
??
??
?? × ??
??
KEY POINTS
1). Conversion of complex arithmetic
expressions into simple ones is called
Simpli?cation.
2). VBODMAS Rule: Vinculum > Brackets
> Of > division > multiplication > addition
> subtraction
* First solve vinculum i.e. bar.
Eg: 7 - 5 - 4 ( ) =?
First solve 5 – 4 i.e. 1 then 7 -1 = 6.
* For brackets open ?rst small brackets
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Pinnacle Day: 8th - 10th Simplification
(….), then {…} and then […].
* For modulus e.g. |-3| we write
magnitude only not sign i.e. 3 in this
case.
Important Formulas
1) . ( ?? + ?? )
2
= ??
2
+ 2 ???? + ??
2
2). ( ?? - ?? )
2
= ??
2
- 2 ???? + ??
2
3). ( ??
2
- ??
2
) = ( ?? + ?? )( ?? - ?? )
4). ( ?? + ?? )
3
= ??
3
+ ??
3
+ 3 ???? ( ?? + ?? )
5). ( ?? - ?? )
3
= ??
3
- ??
3
- 3 ???? ( ?? - ?? )
6). ( ??
3
+ ??
3
) = ( ?? + ?? )( ??
2
- ???? + ??
2
)
7). ( ??
3
- ??
3
) = ( ?? - ?? )( ??
2
+ ???? + ??
2
)
8). = ??
3
+ ??
3
+ ??
3
- 3 ??????
?? + ?? + ?? ( ) ??
2
+ ??
2
+ ??
2
- ???? - ???? - ????
( )
=
1
2
( ?? + ?? + ?? )[( ?? - ?? )
2
+ ( ?? - ?? )
2
+
( ?? - ?? )
2
]
Here, if ?? + ?? + ?? ( ) = 0 ,
then ??
3
+ ??
3
+ ??
3
- 3 ?????? = 0
? ??
3
+ ??
3
+ ??
3
= 3 ??????
9). ( ?? + ?? + ?? )
2
= ??
2
+ ??
2
+ ??
2
( )
+ 2 ( ????
+ ???? + ???? )
Variety Questions
Q.1. The simpli?ed form of the
expression
( ??
2
- 9 ) × ( 9 ??
2
- 1 )
( ?? -
1
3
) × ( 1 -
??
3
)
×
9
?? +
1
3
( )
1 +
??
3
( )
is _______.
SSC CHSL Tier II (10/01/2024)
(a) -729 (b) -81 (c) 729 (d) 81
Q.2. If , B = , ?? =
58
2
- 25
2
46
2
- 37
2
26
2
- 15
2
56
2
- 15
2
then the value of - is :
1
??
20
??
SSC CGL Tier II (26/10/2023)
(a) 2 (b) 1 (c) -1 (d) 0
Q.3. On comparing the following two
numeric expressions, we ?nd that ______.
[(2 ) ] and [(1 )
5
]
7
9
2
1
2
3
5
2
3
3
5
SSC MTS 14/09/2023 (2nd Shift)
(a) both the expressions are equal
(b) the ?rst expression is smaller than
the second
(c) the ?rst expression is larger than the
second
(d) the given two expressions cannot be
compared
Q.4. On simplifying the following
expression, we get ________.
24 45 - 16 20 + 245 - 47 5
3
1
5
+ 1
2
5
- 2
1
6
- 2
1
5
- 5
2
5
+ 3
1
6
SSC MTS 08/09/2023 (2nd Shift)
(a) 0 (b) 3 (c) 1 (d) 2
Q.5. Simplify :
( 3 . 321 )
3
+ ( 2 . 681 )
3
+ ( 1 . 245 )
3
- 3 × 3 . 321 × 2 . 681 × 1 . 245
( 3 . 321 )
2
+ ( 2 . 681 )
2
+ ( 1 . 245 )
2
- ( 3 . 321 × 2 . 681 ) - ( 2 . 681 × 1 . 245 ) - ( 1 . 245 × 3 . 321 )
= ?
SSC CHSL 04/08/2023 (1st Shift)
(a) 6.125 (b) 8.645 (c) 7.247 (d) 10.245
Q.6. Simplify the following expression.
0 . 01 + 0 . 68 × 0 . 68 + 0 . 22 × 0 . 22 +
0 . 044 + 0 . 136 + 0 . 44 × 0 . 68
SSC CHSL 03/08/2023 (1st Shift)
(a) 0.69 (b) 1 (c) 0.8 (d) 0.9
Q.7. Simplify the given expression .
0 . 09 × 0 . 09 + 0 . 04 × 0 . 04 + 0 . 16 × 0 . 16 + 2 × 0 . 09 × 0 . 04 + 2 × 0 . 04 × 0 . 16 + 2 × 0 . 09 × 0 . 16
0 . 3 × 0 . 3 + 0 . 2 × 0 . 2 + 0 . 4 × 0 . 4
SSC CHSL 02/08/2023 (4th Shift)
(a) 0.38 (b) 0.24 (c) 0.32 (d) 0.29
Q.8. What is the value of the given
Expression ?
4
?? + 4
- 5 × 4
?? + 2
15 × 4
??
- 2
2
× 4
??
SSC CGL 24/07/2023 (1st shift)
(a) 16 (b) 64 (c) 20 (d) 24
Q.9. What is the value of ( + 1)( + 1) 2
2
2
4
( + 1)( + 1) ....... ( + 1) ? 2
8
2
16
2
128
SSC CHSL 14/03/2023 (2nd Shift)
(a) (b)
2
256
- 1
2
2
256
- 1
3
(c) (d) 2
256
- 1
2
256
- 1
4
Q.10. Which of the following statement
is true ?
I. + + + ……
1
2
1
6
1
12
1
110
<
5
6
II. + + +...... >
1
3
1
15
1
35
1
143
7
13
SSC CHSL 13/03/2023 (4th Shift)
(a) Only I (b) Both I and II
(c) Only II (d) Neither I nor II
Q.11. Find the value of the following
expression.
[
5
8
- {
3
8
- (
5
8
-
3
8
)}] ???? 8 . 8 - 1 . 2
4
1
6
÷ 2 . 5 × 2 ÷
1
6
???? 60 + (
3
4
-
3
8
)
SSC CPO 11/11/2022 (Morning)
(a) 5 (b) 3 (c) 4 (d) 4
22
43
23
67
44
85
4
5
Q.12. Solve the following.
24 ÷
3
8
???? ( 8 + 2× 7 - 3 )+ [
2
11
÷
4
55
- {
5
8
+
6
16
}]
32÷ 15 - 7 + 75÷ ( 6 + 15÷3 + 4 )
SSC CPO 09/11/2022 (Morning)
(a) (b) (c) (d)
23
27
9
2
11
18
15
19
Q.13. Solve
( 6 . 9 )
3
+ ( 2 . 7 )
3
+ ( 5 . 8 )
3
- 3 × 6 . 9 × 2 . 7 ×5 . 8
( 6 . 9 )
2
+ ( 2 . 7 )
2
+ ( 5 . 8 )
2
-( 6 . 9 )( 2 . 7 )-( 2 . 7 )( 5 . 8 )-( 5 . 8 )( 6 . 9 )
SSC CHSL 07/06/2022(Afternoon)
(a) 15.1 (b) 15.3 (c) 15.2 (d) 15.4
Q.14. Simplify the following expression.
9 {(16 ) +
1
5
÷
1
5
÷ 12
2
3
- 9
2
3
17 of 3 }
1
2
1
21
SSC CHSL 26/05/2022(Evening)
(a) (b) (c) (d)
139
881
137
881
140
881
138
881
Q.15. The value of
+ is:
1
4
20 . 35 ( )
2
- 8 . 35 ( )
2
[ ]
× 0 . 0175
( 1 . 05 )
2
+ ( 1 . 05 )( 27 . 65 )
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
3
20
3
10
7
20
9
20
Q.16 The simpli?ed value of
119 -[ 48 ÷ 6 - 7 { 5 × 12 ÷ 3 - ( 15 - 3 - 8 )}]
24 ÷ 3 - 9 ???? 3 + ( 52 - 8 - 4 ) ÷ 6
is:
SSC MTS 11/10/2021(Morning)
(a) (b) (c) (d) -
111
11
-
127
27
127
43
111
11
Q.17. If -
1
4 - 8
+
3 + 2 2
3 - 2 2
3 - 2 2
3 + 2 2
, then what is the value of = ?? + ?? 2
(3a + 4b) ?
SSC CGL Tier II (15/11/2020)
(a) 99 (b) (c) 98 (d)
1
2
97
1
2
98
Q.18. If the following interchanges are
made in signs and numbers, which
equation would be correct ?
Interchanges: Signs: and - Numbers: ÷
15 and 5
SSC CHSL 13/10/2020 (Morning)
(a) 4 30 - 15 5 + 12 = 21 × ÷
(b) 12 30 - 15 5 + 4 = 37 × ÷
(c) 4 30 - 15 5 + 12 = 27 × ÷
(d) 4 30 - 5 15 + 14 = 21 × ÷
Q.19. If =
4
1 + 2 + 3
?? + ?? 2 + ?? 3
where a, b , c , d are natural - ?? 6 ,
numbers, then the value of a + b + c + d
is:
SSC CHSL 18/03/2020 (Morning)
(a) 0 (b) 2 (c) 4 (d) 1
Q.20.
If [{( )
3
}
(2x + 3)
] = [{( ) }
(3 x + 7)
]
2
3
- 3
4
2
3
2
3
- 6
5
then the value of is: 2 - 42 ??
SSC CHSL 18/03/2020(Morning)
(a) 5 (b) 6 (c) 3 (d) 4
Q.21. If ‘+’ means ‘-’, ‘-’ means ‘+’, ‘ ’ ×
means ‘ ’ and ‘ ’ means ‘ ’ then the ÷ ÷ ×
value of is:
42 - 12 × 3 + 8 ÷ 2 + 15
8 × 2 - 4 + 9 ÷ 3
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Pinnacle Day: 8th - 10th Simplification
SSC CGL 03/03/2020 (Morning)
(a) - (b) - (c) (d)
5
3
15
19
15
19
5
3
Q.22. The value of
is:
( 4 . 6 )
4
+ ( 5 . 4 )
4
+ ( 24 . 84 )
2
( 4 . 6 )
2
+( 5 . 4 )
2
+ 24 . 84
SSC CGL Tier II (13/09/2019)
(a) 24.42 (b) 24.24 (c) 25.42 (d) 25.48
Q.23. The value of
is :
( 0 . 545 )( 0 . 081 )( 0 . 51 )( 5 . 2 )
( 0 . 324 )
3
+ ( 0 . 221 )
3
- ( 0 . 545 )
3
SSC CGL Tier II (12/09/2019)
(a) - 1 (b) 1 (c) 3 (d) - 3
Q.24. The value of 22. + 11.5 - 33.5 is 4 67 9
SSC CGL Tier II (11/09/2019)
(a) 0. (b) 0. (c) 0.3 (d) 0.4 32 412 1 12
Q.25. Find the value
{
( 0 . 7 )
2
÷ 0 . 14 + ( 0 . 6 )
2
÷ 0 . 18 + ( 0 . 5 )
2
÷ 0 . 05
4 ( 2 . 5 ???? 4 - 13 × 0 . 25 × 3 )
}
SSC MTS 07/08/2019 (Morning)
(a) (b) (c) (d)
25
2
19
2
23
2
21
2
Q.26. If x = +
1
12 . 13
1
13 . 14
+
1
14 . 15
….. + , y = +
1
23 . 24
1
36 . 37
1
37 . 38
+
….. + then is equal to :
1
38 . 39
1
71 . 72
??
??
SSC CHSL 10/07/2019 (Evening)
(a) (b) (c) (d) 3
1
3
1
24
1
72
Q.27. The simpli?ed value of
is:
0 . 01404
24
2
+ 6
2
- 144
SSC CHSL 10/07/2019 (Evening)
(a) 3 (b) × 10
- 5
6 × 10
- 5
(c) 2.4 (d) 3 × 10
- 4
× 10
- 4
Q.28. The simpli?ed value of
is:
1 . 0025 + 6 . 25 × 10
- 6
0 . 0025 + 0 . 95
SSC CHSL 09/06/2019 (Evening)
(a) 1.0025 (b) 1.0525
(c) 1.0005 (d) 1.0505
Q.29. The value of of ( of
9
15
2
3
÷
2
3
) ( of ) of (
3
2
÷
3
4
×
3
4
÷
3
4
4
3
5
4
÷
of ) is :
5
2
×
2
5
4
5
SSC CGL 06/06/2019 (Afternoon)
(a) (b) (c) (d)
20
9
4
25
18
125
40
9
Q.30. The value of (5 + 3 5 5) / (3 ÷ × ÷
3 of 6) of (4 4 4 of 4 + 4 4 4) is × ÷ ÷ ×
SSC CGL 06/06/2019 (Morning)
(a) 8 (b) 7 (c) 9 (d) 6
1
5
1
3
3
5
2
3
Q.31. The value of 2 ÷ (3 of
7
8
5
6
÷
2
7
2 ) [(2 of 4 ÷ ) ] is :
1
3
×
6
7
1
5
2
3
×
5
9
SSC CGL 04/06/2019 (Evening)
(a) (b) 4 (c) (d) 5
1
4
1
23
Q.32. The value of 2 3 2 of 3 2 × ÷ × ÷
(4 + 4 4 of 4 - 4 4 ) is: × 4 ÷ ÷ × 4
SSC CGL 04/06/2019 (Morning)
(a) 8 (b) 1 (c) 4 (d) 2
Q.33.
5 . 75 × 5 . 75 × 5 . 75 + 3 . 25 × 3 . 25 × 3 . 25
57 . 5 × 57 . 5 + 32 . 5 × 32 . 5 - 57 . 5 × 32 . 5
is equal to :
SSC CPO 12/03/2019 (Evening)
(a) 0.009 (b) 0.0009 (c) 0.9 (d) 0.09
Practice Questions
SSC CHSL 2023 Tier - 2
Q.34. If x = ( )
2
( )
-4
, ?nd the value
3
2
×
2
3
of ??
- 2
.
SSC CHSL Tier II (02/11/2023)
(a) ( )
6
(b) ( )
12
3
2
3
2
(c) ( )
12
(d) ( )
6
2
3
2
3
Q.35. If
.
3
0 . 08 × 0 . 8 × ?? = 0 . 008 × 0 . 8 ×
3
??
then ?nd the value of .
??
??
SSC CHSL Tier II (02/11/2023)
(a) (b) 8
3
× 10
- 8
8
3
× 10
- 9
(c) (d) 8
4
× 10
- 9
8
4
× 10
- 8
Q.36. Simplify :
.
3
3
343
( )
3
×
3
3
216
( )
3
SSC CHSL Tier II (10/01/2024)
(a) 56 (b) 42 (c) (d)
3
42 42
3
Q.37. Find the value of
+ + + 625
625
100
625
10000
625
1000000
SSC CHSL Tier II (10/01/2024)
(a) 27.075 (b) 27.775
(c) 27.75 (d) 27.025
SSC CGL 2023 Tier - 2
Q.38. What is the value of [ + { × (
1
8
1
6
÷ ) - ( × ÷ )} + ] ?
36
45
24
25
12
21
14
15
24
45
27
36
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
1
36
1
72
1
108
1
27
Q.39. Find the value of .
1 . 24 × 2 . 79
2 . 64 × 5 . 94
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
31
66
31
44
33
64
33
31
SSC CPO 2023 Tier - 1
Q.40. Evaluate
of
0 . 04
0 . 05
3
1
3
- 2
1
2
( )
÷
1
2
???? 1
1
4
1
3
+
1
5
????
1
9
SSC CPO 03/10/2023 (1st Shift)
(a) 5 (b) 0.3 (c) 3 (d) 0.03
Q.41. If (48 ÷ 72 × 3) - [15 ÷ 8 × (40 - 32) -
10] + 2P = 6 ÷ 2, then ?nd the value of P
SSC CPO 03/10/2023 (3rd Shift)
(a) 2 (b) 4 (c) 1 (d) 3
Q.42. Simplify .
1 -
3
7
÷
9
49
×
3
14
1 +
3
5
÷
1
25
×
1
30
SSC CPO 04/10/2023 (1st Shift)
(a) (b) (c) (d)
1
6
1
5
1
3
1
7
Q.43. The value of 1 - [3 ÷ {1 ÷
2
5
3
4
1
4
1
2
(1 × 3 ÷ 1 )}] is :
1
2
1
3
1
3
SSC CPO 04/10/2023 (1st Shift)
(a) 1 (b) 3 (c) 2 (d) 0
Q.44. The value of
is :
1
1
2
÷ 3
1
4
+
1
2
÷
13
14
+
1
5
1
5
× 3
1
2
-
1
3
÷ 1
3
4
× 3
1
2
SSC CPO 05/10/2023 (1st Shift)
(a) 38 (b) 33 (c) 36 (d) 40
Q.45. Simplify the given expression.
b + 2a - [(3b + a) - (2a + b) + 2a] - a
SSC CPO 05/10/2023 (3rd Shift)
(a) - a (b) - b (c) a (d) b
SSC MTS 2023 Tier - 1
Q.46. Simplify:
24 ÷ 6 × ( of + of ) = ?
3
4
1
3
5
3
3
4
SSC MTS 01/09/2023 (3rd Shift)
(a) 9 (b) 6 (c) 3 (d) 8
Q.47. Simplify the given expression
71 - ( - 3){ - 2 - } + 3{5 + ( - 2)(-1)} ( 8 - 3 )
SSC MTS 04/09/2023 (3rd Shift)
(a) 30 (b) 119 (c) 50 (d) 71
Q.48. Simplify the following. 15 ÷ 5 of 3 ×
4 + 8 ÷ 12 of 4 × 12 - 4 ÷ 5 × 20
SSC MTS 06/09/2023 (2nd Shift)
(a) (b) - 10 (c) 12 (d) 1
20
3
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Pinnacle Day: 8th - 10th Simplification
Q.49. The value of
is equal to:
3 ÷ 3 × 2 - 6 ÷ 3 + 7
5 ÷ 5 × 6 ÷ 6 + 3 + 3
SSC MTS 08/09/2023 (3rd Shift)
(a) 1 (b) (c) (d) 2
3
2
1
3
2
Q.50. Evaluate 3 - [2 ÷ {1 -
1
2
1
4
1
4
3
4
(1 - - 2 )}]
1
2
1
2
1
2
SSC MTS 11/09/2023 (1st Shift)
(a) 1 (b) 2 (c) 2 (d)
21
38
19
18
21
38
47
38
Q.51. What is the value of R, if,
10 (8 + 4R) ÷ (12 5) - 5 2
3
= 6 × ×
SSC MTS 11/09/2023 (2nd Shift)
(a) 86 (b) 68 (c) 76 (d) 67
Q.52. Identify the single rational number
that may replace the given expression.
[1² + 2² + 3² + 4²] × ( )
2
2
3 × 5
÷
1
4
SSC MTS 11/09/2023 (3rd Shift)
(a) 16 (b) 64 (c) (d)
1
64
1
16
Q.53. What is the value of
+ (1 - )
1
5
÷
1
5
×
1
5
?
?
?
?
×
1
5
1
9
+
1
9
×
1
9
?
?
?
?
×
1
10
6
25
SSC MTS 12/09/2023 (2nd Shift)
(a) 2 (b) 1 (c) 3 (d) 4
Q.54. Simplify the following expression.
( + ) of - ( ) of
5
3
6
5
÷
18
5
5
12
5
6
÷
5
12
+ of
5
12
2
3
3
2
SSC MTS 13/09/2023 (2nd Shift)
(a) 1 (b) (c) (d) 0
5
6
6
5
Q.55. The Value of [{2
4
(20 5 +10) ÷ 11} ×
- (2
3
15 ÷ 20) - (9 16)] is : × ×
SSC MTS 13/09/2023 (3rd Shift)
(a) 8 (b) 16 (c) 10 (d) 12
Q.56. Simplify: 6 ÷ [4 ÷ {5 × +
4
5
1
4
1
3
3
8
(3 + ) × }]
1
2
2
3
4
5
SSC MTS 14/09/2023 (1st Shift)
(a) 8 (b) 7 (c) 8 (d) 7
8
15
8
25
8
25
8
15
SSC CHSL 2023 Tier - 1
Q.57. Simplify the given expression .
4 17 ( )
3
- 7 ( )
3
[ ]
17 ( )
2
+ 7 ( )
2
+ 119
SSC CHSL 02/08/2023 (4th Shift)
(a) 40 (b) 50 (c) 60 (d) 30
Q.58. Simplify :
= ?
( 612 + 388 ) ² + ( 612 - 388 ) ²
( 612 × 612 ) + ( 388 × 388 )
SSC CHSL 04/08/2023 (4th Shift)
(a) 1 (b) 4 (c) 2 (d) 3
Q.59. If the mathematical symbols in the
expression given below are changed
from ‘+’ to ‘ ’ ‘-’ to ’, ’ to ‘-’ and ’ to ÷ ' × ÷ ×
‘+’. then what will be the value of the
expression ?
87 256 + 16 - 77 1359 × ÷
SSC CHSL 07/08/2023 (3rd Shift)
(a) 20 (b) - 40 (c) 40 (d) - 20
Q.60. Simplify the following expression
( 0 . 96 × 0 . 96 × 0 . 96 + 0 . 04 × 0 . 04 × 0 . 04 )
( 0 . 96 × 0 . 96 - 0 . 96 × 0 . 04 + 0 . 04 × 0 . 04 )
SSC CHSL 07/08/2023 (4th Shift)
(a) 0 (b) 0.82 (c) 2 (d) 1
Q.61. If , then is equal to : ?? = 3 ??
??
??
2
SSC CHSL 08/08/2023 (2nd Shift)
(a) (b) 6 (c) (d) 9
1
6
1
9
Q.62. Simplify the given expression.
( 0 . 25 )
4
+ 2 × ( 0 . 25 )
2
+ 1 - ( 0 . 25 )
2
( 0 . 25 )
2
+ 0 . 25 + 1
SSC CHSL 08/08/2023 (2nd Shift)
(a) 0.6755 (b) 0.9025
(c) 0.8125 (d) 0.7835
Q.63. The value of
is:
3 + 8 × 8 ÷ 8 ???? 8 + 7 ÷ 7 × 2
5 ÷ 5 ???? 4 + 3 × 3 ÷ 3 - 3 + 2
SSC CHSL 08/08/2023 (4th Shift)
(a) (b) (c) (d) 1
3
5
2
2
3
1
4
7
2
3
7
Q.64. Evaluate , given that
3 + 2
3 - 2
. 6 = 2 . 45
SSC CHSL 08/08/2023 (4th Shift)
(a) 7.7 (b) 9.9 (c) 8.8 (d) 6.6
Q.65. Simplify: [ of (35 5) + 4 ]
2
[9
1
5
- -
{ 7 + (2 3 )} + 5 ]
2
- -
SSC CHSL 09/08/2023 (3rd Shift)
(a) 100 (b) 36 (c) 25 (d) 64
Q.66. Simplify the given expression.
( 80×80×80 )+( 70×70×70 )+( 50×50×50 )- 840000
6400 + 4900 + 2500 - 5600 - 3500 - 4000
SSC CHSL 10/08/2023 (1st Shift)
(a) 100 (b) 200 (c) 400 (d) 300
Q.67. Find the value of the given
expression
( 4
1
3
+ 3
1
3
× 1
4
5
÷ 3
3
4
× ( 1
1
2
+ 1
1
3
))
2
3
÷
5
6
×
2
3
SSC CHSL 10/08/2023 (2nd Shift)
(a) 11 (b) 10 (c) 14 (d) 16
3
8
1
8
3
8
5
8
Q.68. Simplify :
( 379 + 276 )
2
+ ( 379 - 276 )
2
379 × 379 + 276 × 276
SSC CHSL 11/08/2023 (2nd Shift)
(a) 2 (b) 655 (c) 103 (d) 1
Q.69. What is the value of p, if, 25(4 + 3p)
÷ 5 - 3 × 8 = 26 ?
SSC CHSL 11/08/2023 (2nd Shift)
(a) 5 (b) 4 (c) 7 (d) 2
Q.70. Simplify the given expression.
( 381 + 119 )
2
+ ( 381 - 119 )
2
( 381 )
2
+ ( 119 )
2
SSC CHSL 14/08/2023 (1st Shift)
(a) - 1 (b) - 2 (c) 1 (d) 2
Q.71. Simplify the given expression
( 4 . 2 )
3
- 0 . 008
( 4 . 2 )
2
+ 0 . 84 + 0 . 04
SSC CHSL 14/08/2023 (2nd Shift)
(a) 4 (b) - 2 (c) 2 (d) - 4
Q.72. The value of
( ) is:
2 + 7 × 7 ÷ 9 ???? 9 + 6 ÷ 6 × 2
4 ÷ 4 ???? 5 + 7 × 7 ÷ 7 - 6 + 4
SSC CHSL 17/08/2023 (2nd Shift)
(a) (b) (c) (d)
1734
2407
1865
2106
1857
2507
1664
2305
Q.73. Simplify the expression:
143 × 143 + 143 × 139 + 139 × 139
143 × 143 × 143 - 139 × 139 × 139
SSC CHSL 17/08/2023 (3rd Shift)
(a) (b) 282 (c) (d) 4
1
2
1
4
Q.74. Simplify the given expression and
?nd the value for x = -1.
10 ??
2
+ 5? + 2?? + ?
5? + ??
SSC CHSL 17/08/2023 (3rd Shift)
(a) -1 (b) 0 (c) 1 (d) 2
SSC CGL 2023 Tier - 1
Q.75. Simplify the following expression
[ × - {( ÷ ) - ( × )}
85
34
1
18
46
69
27
135
86
129
91
14
of ]
112
36
SSC CGL 14/07/2023 (3rd shift)
(a) 3 (b) 3 (c) 3 (d) 3
1
5
3
4
1
2
1
4
Q.76. Simplify the given expression.
18 ÷ 3 of 2 × 5 +72 ÷ 18 of 2 ×3 - 4 ÷ 8 × 2
SSC CGL 14/07/2023 (4th shift)
(a) 20 (b) (c) (d) 25
20
3
25
3
Q.77. Simplify the given expression.
y + 2x - [y - (y - x + y) - (x + y) + y] - 2y.
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Page 5
Pinnacle Day: 8th - 10th Simplification
Simplification
KEY POINTS :-
1). (a + b)
2
= a
2
+ b
2
+ 2ab
2). (a b)
2
= a
2
+ b
2
2ab - -
3). a
2
b
2
= (a + b)(a b) - -
4). Square root of a number is a value
that, when multiplied by itself, gives the
given number.
Example :-
36 has two square roots 6 and -6;
6
2
= 36 and 6
2
= 36, hence we write -
36 = ±6
5). Square root by Factorization method
Find square root of 1089
First ?nd the factors of 1089.
1089 = 3 3 11 11 × × ×
1089 = 3 × 11 = 33
6). Square root by Division method
Find square root of 2304.
Note:- Make group of two digit from right
2304 = 48
7). If then the required number ? = ?? ,
will be (?) =
2
??
8). = a
2
b
2
c
2
??
4
× ??
4
× ??
4
9). = a b ??
??
× ??
??
??
2
×
??
2
10). =
??
??
??
??
11). ?? × ?? = ????
12). If any number ends with 1, its square
root will end with 1 or, 9, given that the
number is a perfect square.
Example :- , 81 = 9 121 = 11
13). If any number ends with 4, its square
root will end with 2 or 8, given that the
number is a perfect square.
Example: , 324 = 18 144 = 12
14). If any number ends with 5, its square
root will end with 5, given that the
number is a perfect square.
Example: , 225 = 15 625 = 25
15). If any number ends with 6, its square
root will end with 6, given that the
number is a perfect square.
Example: , 256 = 16 676 = 26
16). If any number ends with 9, its square
root will end with 3 or 7, given that the
number is a perfect square.
Example: , 49 = 7 169 = 13
17). The square of any number always
ends with 0, 1, 4, 5, 6 or, 9 but will never
end with 2, 3, 7 or, 8.
18). If in a given number, the total
number of digits are n and if n is even,
the square root of that number will have
digits and if is odd, then the number
??
2
??
of digits will be .
?? + 1
2
19). ?? ?? ?? … ?? = ??
2
??
- 1
2
??
Example
Find the value of 2 2 2 2 2
Here, n = 5, ? 2 2 2 2 2
= = 2 2
2
5
- 1
2
5
31
32
20). ; ?? ?? ?? …8 = ??
Example 7 7 7…8 = 7
21). To ?nd the value of
, ?nd the factors of ?? + ?? + ?? + …8
x, such that the difference between the
factors is 1, then the larger factor will be
the result.
22). To ?nd the value of
, ?nd the factors of ?? - ?? - ?? - …8
x, such that the difference between the
factors is 1, then the smaller factor will
be the result.
Example: Find the value of
12 + 12 + 12 + …8
The factors of 12 with a difference of 1
are 4, 3. Here, 4 is the larger number.
Value will be 4. ?
Conventional method
Let 12 + 12 + 12 + …8 = ??
? 12 + ?? = ??
? 12 + ?? = ??
2
? ??
2
- ?? - 12 = 0
? ??
2
- 4 ?? + 3 ?? - 12 = 0
? ?? - 4 ( ) ?? + 3 ( ) = 0
? ?? = 4 ?????? ?? = - 3
? ?? = 4
Example :- Find the value of
. 30 - 30 - 30 - …8
The factors of 30 with a difference of 1
are 5, 6. Here, 5 is the smaller number. ?
Value will be 5.
23). If any number is in form,
1
?? ± ??
then we multiply by its rationalization
factor in both numerator and ?? ± ??
denominator.
Example: Find the value of ;
1
9 - 8
=
1
9 - 8
1
9 - 8
×
9 + 8
9 + 8
= =
9 + 8
9
2
- 8
2
3 + 2 2
1
= 3 + 2 2
C ube and Cube Root
KEY POINTS
1) If a number is multiplied two times
with itself, then the result of this
multiplication is called the cube of that
number.
Example: 2 × 2 × 2 = 2
3
Cube root of = = 2
1
= 2 2
3
=
3
2
3
2
3×
1
3
2) Algebraic method to calculate cube:
( ?? + ?? )
3
= ??
3
+ 3 ???? ?? + ?? ( ) + ??
3
( ?? - ?? )
3
= ??
3
- 3 ???? ?? - ?? ( ) - ??
3
Indices and Surds
Key Points:
1) p
m
p
n
= p
m+n
2) × ( ??
??
)
??
= ??
????
3) = 4) =
??
??
??
??
??
?? - ??
(
??
??
)
??
??
??
??
??
5) 6) = ??
0
= 1 ??
- ?? 1
??
??
7) p = 8) = p
1
??
??
??
??
??
1
??
9) 10) =
??
???? =
??
?? ×
??
??
?? ??
??
??
??
??
??
11) = 12) = (
??
?? )
??
?? ??
??
??
??
??
??
13) =
??
??
??
?? × ??
??
KEY POINTS
1). Conversion of complex arithmetic
expressions into simple ones is called
Simpli?cation.
2). VBODMAS Rule: Vinculum > Brackets
> Of > division > multiplication > addition
> subtraction
* First solve vinculum i.e. bar.
Eg: 7 - 5 - 4 ( ) =?
First solve 5 – 4 i.e. 1 then 7 -1 = 6.
* For brackets open ?rst small brackets
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Pinnacle Day: 8th - 10th Simplification
(….), then {…} and then […].
* For modulus e.g. |-3| we write
magnitude only not sign i.e. 3 in this
case.
Important Formulas
1) . ( ?? + ?? )
2
= ??
2
+ 2 ???? + ??
2
2). ( ?? - ?? )
2
= ??
2
- 2 ???? + ??
2
3). ( ??
2
- ??
2
) = ( ?? + ?? )( ?? - ?? )
4). ( ?? + ?? )
3
= ??
3
+ ??
3
+ 3 ???? ( ?? + ?? )
5). ( ?? - ?? )
3
= ??
3
- ??
3
- 3 ???? ( ?? - ?? )
6). ( ??
3
+ ??
3
) = ( ?? + ?? )( ??
2
- ???? + ??
2
)
7). ( ??
3
- ??
3
) = ( ?? - ?? )( ??
2
+ ???? + ??
2
)
8). = ??
3
+ ??
3
+ ??
3
- 3 ??????
?? + ?? + ?? ( ) ??
2
+ ??
2
+ ??
2
- ???? - ???? - ????
( )
=
1
2
( ?? + ?? + ?? )[( ?? - ?? )
2
+ ( ?? - ?? )
2
+
( ?? - ?? )
2
]
Here, if ?? + ?? + ?? ( ) = 0 ,
then ??
3
+ ??
3
+ ??
3
- 3 ?????? = 0
? ??
3
+ ??
3
+ ??
3
= 3 ??????
9). ( ?? + ?? + ?? )
2
= ??
2
+ ??
2
+ ??
2
( )
+ 2 ( ????
+ ???? + ???? )
Variety Questions
Q.1. The simpli?ed form of the
expression
( ??
2
- 9 ) × ( 9 ??
2
- 1 )
( ?? -
1
3
) × ( 1 -
??
3
)
×
9
?? +
1
3
( )
1 +
??
3
( )
is _______.
SSC CHSL Tier II (10/01/2024)
(a) -729 (b) -81 (c) 729 (d) 81
Q.2. If , B = , ?? =
58
2
- 25
2
46
2
- 37
2
26
2
- 15
2
56
2
- 15
2
then the value of - is :
1
??
20
??
SSC CGL Tier II (26/10/2023)
(a) 2 (b) 1 (c) -1 (d) 0
Q.3. On comparing the following two
numeric expressions, we ?nd that ______.
[(2 ) ] and [(1 )
5
]
7
9
2
1
2
3
5
2
3
3
5
SSC MTS 14/09/2023 (2nd Shift)
(a) both the expressions are equal
(b) the ?rst expression is smaller than
the second
(c) the ?rst expression is larger than the
second
(d) the given two expressions cannot be
compared
Q.4. On simplifying the following
expression, we get ________.
24 45 - 16 20 + 245 - 47 5
3
1
5
+ 1
2
5
- 2
1
6
- 2
1
5
- 5
2
5
+ 3
1
6
SSC MTS 08/09/2023 (2nd Shift)
(a) 0 (b) 3 (c) 1 (d) 2
Q.5. Simplify :
( 3 . 321 )
3
+ ( 2 . 681 )
3
+ ( 1 . 245 )
3
- 3 × 3 . 321 × 2 . 681 × 1 . 245
( 3 . 321 )
2
+ ( 2 . 681 )
2
+ ( 1 . 245 )
2
- ( 3 . 321 × 2 . 681 ) - ( 2 . 681 × 1 . 245 ) - ( 1 . 245 × 3 . 321 )
= ?
SSC CHSL 04/08/2023 (1st Shift)
(a) 6.125 (b) 8.645 (c) 7.247 (d) 10.245
Q.6. Simplify the following expression.
0 . 01 + 0 . 68 × 0 . 68 + 0 . 22 × 0 . 22 +
0 . 044 + 0 . 136 + 0 . 44 × 0 . 68
SSC CHSL 03/08/2023 (1st Shift)
(a) 0.69 (b) 1 (c) 0.8 (d) 0.9
Q.7. Simplify the given expression .
0 . 09 × 0 . 09 + 0 . 04 × 0 . 04 + 0 . 16 × 0 . 16 + 2 × 0 . 09 × 0 . 04 + 2 × 0 . 04 × 0 . 16 + 2 × 0 . 09 × 0 . 16
0 . 3 × 0 . 3 + 0 . 2 × 0 . 2 + 0 . 4 × 0 . 4
SSC CHSL 02/08/2023 (4th Shift)
(a) 0.38 (b) 0.24 (c) 0.32 (d) 0.29
Q.8. What is the value of the given
Expression ?
4
?? + 4
- 5 × 4
?? + 2
15 × 4
??
- 2
2
× 4
??
SSC CGL 24/07/2023 (1st shift)
(a) 16 (b) 64 (c) 20 (d) 24
Q.9. What is the value of ( + 1)( + 1) 2
2
2
4
( + 1)( + 1) ....... ( + 1) ? 2
8
2
16
2
128
SSC CHSL 14/03/2023 (2nd Shift)
(a) (b)
2
256
- 1
2
2
256
- 1
3
(c) (d) 2
256
- 1
2
256
- 1
4
Q.10. Which of the following statement
is true ?
I. + + + ……
1
2
1
6
1
12
1
110
<
5
6
II. + + +...... >
1
3
1
15
1
35
1
143
7
13
SSC CHSL 13/03/2023 (4th Shift)
(a) Only I (b) Both I and II
(c) Only II (d) Neither I nor II
Q.11. Find the value of the following
expression.
[
5
8
- {
3
8
- (
5
8
-
3
8
)}] ???? 8 . 8 - 1 . 2
4
1
6
÷ 2 . 5 × 2 ÷
1
6
???? 60 + (
3
4
-
3
8
)
SSC CPO 11/11/2022 (Morning)
(a) 5 (b) 3 (c) 4 (d) 4
22
43
23
67
44
85
4
5
Q.12. Solve the following.
24 ÷
3
8
???? ( 8 + 2× 7 - 3 )+ [
2
11
÷
4
55
- {
5
8
+
6
16
}]
32÷ 15 - 7 + 75÷ ( 6 + 15÷3 + 4 )
SSC CPO 09/11/2022 (Morning)
(a) (b) (c) (d)
23
27
9
2
11
18
15
19
Q.13. Solve
( 6 . 9 )
3
+ ( 2 . 7 )
3
+ ( 5 . 8 )
3
- 3 × 6 . 9 × 2 . 7 ×5 . 8
( 6 . 9 )
2
+ ( 2 . 7 )
2
+ ( 5 . 8 )
2
-( 6 . 9 )( 2 . 7 )-( 2 . 7 )( 5 . 8 )-( 5 . 8 )( 6 . 9 )
SSC CHSL 07/06/2022(Afternoon)
(a) 15.1 (b) 15.3 (c) 15.2 (d) 15.4
Q.14. Simplify the following expression.
9 {(16 ) +
1
5
÷
1
5
÷ 12
2
3
- 9
2
3
17 of 3 }
1
2
1
21
SSC CHSL 26/05/2022(Evening)
(a) (b) (c) (d)
139
881
137
881
140
881
138
881
Q.15. The value of
+ is:
1
4
20 . 35 ( )
2
- 8 . 35 ( )
2
[ ]
× 0 . 0175
( 1 . 05 )
2
+ ( 1 . 05 )( 27 . 65 )
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
3
20
3
10
7
20
9
20
Q.16 The simpli?ed value of
119 -[ 48 ÷ 6 - 7 { 5 × 12 ÷ 3 - ( 15 - 3 - 8 )}]
24 ÷ 3 - 9 ???? 3 + ( 52 - 8 - 4 ) ÷ 6
is:
SSC MTS 11/10/2021(Morning)
(a) (b) (c) (d) -
111
11
-
127
27
127
43
111
11
Q.17. If -
1
4 - 8
+
3 + 2 2
3 - 2 2
3 - 2 2
3 + 2 2
, then what is the value of = ?? + ?? 2
(3a + 4b) ?
SSC CGL Tier II (15/11/2020)
(a) 99 (b) (c) 98 (d)
1
2
97
1
2
98
Q.18. If the following interchanges are
made in signs and numbers, which
equation would be correct ?
Interchanges: Signs: and - Numbers: ÷
15 and 5
SSC CHSL 13/10/2020 (Morning)
(a) 4 30 - 15 5 + 12 = 21 × ÷
(b) 12 30 - 15 5 + 4 = 37 × ÷
(c) 4 30 - 15 5 + 12 = 27 × ÷
(d) 4 30 - 5 15 + 14 = 21 × ÷
Q.19. If =
4
1 + 2 + 3
?? + ?? 2 + ?? 3
where a, b , c , d are natural - ?? 6 ,
numbers, then the value of a + b + c + d
is:
SSC CHSL 18/03/2020 (Morning)
(a) 0 (b) 2 (c) 4 (d) 1
Q.20.
If [{( )
3
}
(2x + 3)
] = [{( ) }
(3 x + 7)
]
2
3
- 3
4
2
3
2
3
- 6
5
then the value of is: 2 - 42 ??
SSC CHSL 18/03/2020(Morning)
(a) 5 (b) 6 (c) 3 (d) 4
Q.21. If ‘+’ means ‘-’, ‘-’ means ‘+’, ‘ ’ ×
means ‘ ’ and ‘ ’ means ‘ ’ then the ÷ ÷ ×
value of is:
42 - 12 × 3 + 8 ÷ 2 + 15
8 × 2 - 4 + 9 ÷ 3
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Pinnacle Day: 8th - 10th Simplification
SSC CGL 03/03/2020 (Morning)
(a) - (b) - (c) (d)
5
3
15
19
15
19
5
3
Q.22. The value of
is:
( 4 . 6 )
4
+ ( 5 . 4 )
4
+ ( 24 . 84 )
2
( 4 . 6 )
2
+( 5 . 4 )
2
+ 24 . 84
SSC CGL Tier II (13/09/2019)
(a) 24.42 (b) 24.24 (c) 25.42 (d) 25.48
Q.23. The value of
is :
( 0 . 545 )( 0 . 081 )( 0 . 51 )( 5 . 2 )
( 0 . 324 )
3
+ ( 0 . 221 )
3
- ( 0 . 545 )
3
SSC CGL Tier II (12/09/2019)
(a) - 1 (b) 1 (c) 3 (d) - 3
Q.24. The value of 22. + 11.5 - 33.5 is 4 67 9
SSC CGL Tier II (11/09/2019)
(a) 0. (b) 0. (c) 0.3 (d) 0.4 32 412 1 12
Q.25. Find the value
{
( 0 . 7 )
2
÷ 0 . 14 + ( 0 . 6 )
2
÷ 0 . 18 + ( 0 . 5 )
2
÷ 0 . 05
4 ( 2 . 5 ???? 4 - 13 × 0 . 25 × 3 )
}
SSC MTS 07/08/2019 (Morning)
(a) (b) (c) (d)
25
2
19
2
23
2
21
2
Q.26. If x = +
1
12 . 13
1
13 . 14
+
1
14 . 15
….. + , y = +
1
23 . 24
1
36 . 37
1
37 . 38
+
….. + then is equal to :
1
38 . 39
1
71 . 72
??
??
SSC CHSL 10/07/2019 (Evening)
(a) (b) (c) (d) 3
1
3
1
24
1
72
Q.27. The simpli?ed value of
is:
0 . 01404
24
2
+ 6
2
- 144
SSC CHSL 10/07/2019 (Evening)
(a) 3 (b) × 10
- 5
6 × 10
- 5
(c) 2.4 (d) 3 × 10
- 4
× 10
- 4
Q.28. The simpli?ed value of
is:
1 . 0025 + 6 . 25 × 10
- 6
0 . 0025 + 0 . 95
SSC CHSL 09/06/2019 (Evening)
(a) 1.0025 (b) 1.0525
(c) 1.0005 (d) 1.0505
Q.29. The value of of ( of
9
15
2
3
÷
2
3
) ( of ) of (
3
2
÷
3
4
×
3
4
÷
3
4
4
3
5
4
÷
of ) is :
5
2
×
2
5
4
5
SSC CGL 06/06/2019 (Afternoon)
(a) (b) (c) (d)
20
9
4
25
18
125
40
9
Q.30. The value of (5 + 3 5 5) / (3 ÷ × ÷
3 of 6) of (4 4 4 of 4 + 4 4 4) is × ÷ ÷ ×
SSC CGL 06/06/2019 (Morning)
(a) 8 (b) 7 (c) 9 (d) 6
1
5
1
3
3
5
2
3
Q.31. The value of 2 ÷ (3 of
7
8
5
6
÷
2
7
2 ) [(2 of 4 ÷ ) ] is :
1
3
×
6
7
1
5
2
3
×
5
9
SSC CGL 04/06/2019 (Evening)
(a) (b) 4 (c) (d) 5
1
4
1
23
Q.32. The value of 2 3 2 of 3 2 × ÷ × ÷
(4 + 4 4 of 4 - 4 4 ) is: × 4 ÷ ÷ × 4
SSC CGL 04/06/2019 (Morning)
(a) 8 (b) 1 (c) 4 (d) 2
Q.33.
5 . 75 × 5 . 75 × 5 . 75 + 3 . 25 × 3 . 25 × 3 . 25
57 . 5 × 57 . 5 + 32 . 5 × 32 . 5 - 57 . 5 × 32 . 5
is equal to :
SSC CPO 12/03/2019 (Evening)
(a) 0.009 (b) 0.0009 (c) 0.9 (d) 0.09
Practice Questions
SSC CHSL 2023 Tier - 2
Q.34. If x = ( )
2
( )
-4
, ?nd the value
3
2
×
2
3
of ??
- 2
.
SSC CHSL Tier II (02/11/2023)
(a) ( )
6
(b) ( )
12
3
2
3
2
(c) ( )
12
(d) ( )
6
2
3
2
3
Q.35. If
.
3
0 . 08 × 0 . 8 × ?? = 0 . 008 × 0 . 8 ×
3
??
then ?nd the value of .
??
??
SSC CHSL Tier II (02/11/2023)
(a) (b) 8
3
× 10
- 8
8
3
× 10
- 9
(c) (d) 8
4
× 10
- 9
8
4
× 10
- 8
Q.36. Simplify :
.
3
3
343
( )
3
×
3
3
216
( )
3
SSC CHSL Tier II (10/01/2024)
(a) 56 (b) 42 (c) (d)
3
42 42
3
Q.37. Find the value of
+ + + 625
625
100
625
10000
625
1000000
SSC CHSL Tier II (10/01/2024)
(a) 27.075 (b) 27.775
(c) 27.75 (d) 27.025
SSC CGL 2023 Tier - 2
Q.38. What is the value of [ + { × (
1
8
1
6
÷ ) - ( × ÷ )} + ] ?
36
45
24
25
12
21
14
15
24
45
27
36
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
1
36
1
72
1
108
1
27
Q.39. Find the value of .
1 . 24 × 2 . 79
2 . 64 × 5 . 94
SSC CGL Tier II (26/10/2023)
(a) (b) (c) (d)
31
66
31
44
33
64
33
31
SSC CPO 2023 Tier - 1
Q.40. Evaluate
of
0 . 04
0 . 05
3
1
3
- 2
1
2
( )
÷
1
2
???? 1
1
4
1
3
+
1
5
????
1
9
SSC CPO 03/10/2023 (1st Shift)
(a) 5 (b) 0.3 (c) 3 (d) 0.03
Q.41. If (48 ÷ 72 × 3) - [15 ÷ 8 × (40 - 32) -
10] + 2P = 6 ÷ 2, then ?nd the value of P
SSC CPO 03/10/2023 (3rd Shift)
(a) 2 (b) 4 (c) 1 (d) 3
Q.42. Simplify .
1 -
3
7
÷
9
49
×
3
14
1 +
3
5
÷
1
25
×
1
30
SSC CPO 04/10/2023 (1st Shift)
(a) (b) (c) (d)
1
6
1
5
1
3
1
7
Q.43. The value of 1 - [3 ÷ {1 ÷
2
5
3
4
1
4
1
2
(1 × 3 ÷ 1 )}] is :
1
2
1
3
1
3
SSC CPO 04/10/2023 (1st Shift)
(a) 1 (b) 3 (c) 2 (d) 0
Q.44. The value of
is :
1
1
2
÷ 3
1
4
+
1
2
÷
13
14
+
1
5
1
5
× 3
1
2
-
1
3
÷ 1
3
4
× 3
1
2
SSC CPO 05/10/2023 (1st Shift)
(a) 38 (b) 33 (c) 36 (d) 40
Q.45. Simplify the given expression.
b + 2a - [(3b + a) - (2a + b) + 2a] - a
SSC CPO 05/10/2023 (3rd Shift)
(a) - a (b) - b (c) a (d) b
SSC MTS 2023 Tier - 1
Q.46. Simplify:
24 ÷ 6 × ( of + of ) = ?
3
4
1
3
5
3
3
4
SSC MTS 01/09/2023 (3rd Shift)
(a) 9 (b) 6 (c) 3 (d) 8
Q.47. Simplify the given expression
71 - ( - 3){ - 2 - } + 3{5 + ( - 2)(-1)} ( 8 - 3 )
SSC MTS 04/09/2023 (3rd Shift)
(a) 30 (b) 119 (c) 50 (d) 71
Q.48. Simplify the following. 15 ÷ 5 of 3 ×
4 + 8 ÷ 12 of 4 × 12 - 4 ÷ 5 × 20
SSC MTS 06/09/2023 (2nd Shift)
(a) (b) - 10 (c) 12 (d) 1
20
3
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Pinnacle Day: 8th - 10th Simplification
Q.49. The value of
is equal to:
3 ÷ 3 × 2 - 6 ÷ 3 + 7
5 ÷ 5 × 6 ÷ 6 + 3 + 3
SSC MTS 08/09/2023 (3rd Shift)
(a) 1 (b) (c) (d) 2
3
2
1
3
2
Q.50. Evaluate 3 - [2 ÷ {1 -
1
2
1
4
1
4
3
4
(1 - - 2 )}]
1
2
1
2
1
2
SSC MTS 11/09/2023 (1st Shift)
(a) 1 (b) 2 (c) 2 (d)
21
38
19
18
21
38
47
38
Q.51. What is the value of R, if,
10 (8 + 4R) ÷ (12 5) - 5 2
3
= 6 × ×
SSC MTS 11/09/2023 (2nd Shift)
(a) 86 (b) 68 (c) 76 (d) 67
Q.52. Identify the single rational number
that may replace the given expression.
[1² + 2² + 3² + 4²] × ( )
2
2
3 × 5
÷
1
4
SSC MTS 11/09/2023 (3rd Shift)
(a) 16 (b) 64 (c) (d)
1
64
1
16
Q.53. What is the value of
+ (1 - )
1
5
÷
1
5
×
1
5
?
?
?
?
×
1
5
1
9
+
1
9
×
1
9
?
?
?
?
×
1
10
6
25
SSC MTS 12/09/2023 (2nd Shift)
(a) 2 (b) 1 (c) 3 (d) 4
Q.54. Simplify the following expression.
( + ) of - ( ) of
5
3
6
5
÷
18
5
5
12
5
6
÷
5
12
+ of
5
12
2
3
3
2
SSC MTS 13/09/2023 (2nd Shift)
(a) 1 (b) (c) (d) 0
5
6
6
5
Q.55. The Value of [{2
4
(20 5 +10) ÷ 11} ×
- (2
3
15 ÷ 20) - (9 16)] is : × ×
SSC MTS 13/09/2023 (3rd Shift)
(a) 8 (b) 16 (c) 10 (d) 12
Q.56. Simplify: 6 ÷ [4 ÷ {5 × +
4
5
1
4
1
3
3
8
(3 + ) × }]
1
2
2
3
4
5
SSC MTS 14/09/2023 (1st Shift)
(a) 8 (b) 7 (c) 8 (d) 7
8
15
8
25
8
25
8
15
SSC CHSL 2023 Tier - 1
Q.57. Simplify the given expression .
4 17 ( )
3
- 7 ( )
3
[ ]
17 ( )
2
+ 7 ( )
2
+ 119
SSC CHSL 02/08/2023 (4th Shift)
(a) 40 (b) 50 (c) 60 (d) 30
Q.58. Simplify :
= ?
( 612 + 388 ) ² + ( 612 - 388 ) ²
( 612 × 612 ) + ( 388 × 388 )
SSC CHSL 04/08/2023 (4th Shift)
(a) 1 (b) 4 (c) 2 (d) 3
Q.59. If the mathematical symbols in the
expression given below are changed
from ‘+’ to ‘ ’ ‘-’ to ’, ’ to ‘-’ and ’ to ÷ ' × ÷ ×
‘+’. then what will be the value of the
expression ?
87 256 + 16 - 77 1359 × ÷
SSC CHSL 07/08/2023 (3rd Shift)
(a) 20 (b) - 40 (c) 40 (d) - 20
Q.60. Simplify the following expression
( 0 . 96 × 0 . 96 × 0 . 96 + 0 . 04 × 0 . 04 × 0 . 04 )
( 0 . 96 × 0 . 96 - 0 . 96 × 0 . 04 + 0 . 04 × 0 . 04 )
SSC CHSL 07/08/2023 (4th Shift)
(a) 0 (b) 0.82 (c) 2 (d) 1
Q.61. If , then is equal to : ?? = 3 ??
??
??
2
SSC CHSL 08/08/2023 (2nd Shift)
(a) (b) 6 (c) (d) 9
1
6
1
9
Q.62. Simplify the given expression.
( 0 . 25 )
4
+ 2 × ( 0 . 25 )
2
+ 1 - ( 0 . 25 )
2
( 0 . 25 )
2
+ 0 . 25 + 1
SSC CHSL 08/08/2023 (2nd Shift)
(a) 0.6755 (b) 0.9025
(c) 0.8125 (d) 0.7835
Q.63. The value of
is:
3 + 8 × 8 ÷ 8 ???? 8 + 7 ÷ 7 × 2
5 ÷ 5 ???? 4 + 3 × 3 ÷ 3 - 3 + 2
SSC CHSL 08/08/2023 (4th Shift)
(a) (b) (c) (d) 1
3
5
2
2
3
1
4
7
2
3
7
Q.64. Evaluate , given that
3 + 2
3 - 2
. 6 = 2 . 45
SSC CHSL 08/08/2023 (4th Shift)
(a) 7.7 (b) 9.9 (c) 8.8 (d) 6.6
Q.65. Simplify: [ of (35 5) + 4 ]
2
[9
1
5
- -
{ 7 + (2 3 )} + 5 ]
2
- -
SSC CHSL 09/08/2023 (3rd Shift)
(a) 100 (b) 36 (c) 25 (d) 64
Q.66. Simplify the given expression.
( 80×80×80 )+( 70×70×70 )+( 50×50×50 )- 840000
6400 + 4900 + 2500 - 5600 - 3500 - 4000
SSC CHSL 10/08/2023 (1st Shift)
(a) 100 (b) 200 (c) 400 (d) 300
Q.67. Find the value of the given
expression
( 4
1
3
+ 3
1
3
× 1
4
5
÷ 3
3
4
× ( 1
1
2
+ 1
1
3
))
2
3
÷
5
6
×
2
3
SSC CHSL 10/08/2023 (2nd Shift)
(a) 11 (b) 10 (c) 14 (d) 16
3
8
1
8
3
8
5
8
Q.68. Simplify :
( 379 + 276 )
2
+ ( 379 - 276 )
2
379 × 379 + 276 × 276
SSC CHSL 11/08/2023 (2nd Shift)
(a) 2 (b) 655 (c) 103 (d) 1
Q.69. What is the value of p, if, 25(4 + 3p)
÷ 5 - 3 × 8 = 26 ?
SSC CHSL 11/08/2023 (2nd Shift)
(a) 5 (b) 4 (c) 7 (d) 2
Q.70. Simplify the given expression.
( 381 + 119 )
2
+ ( 381 - 119 )
2
( 381 )
2
+ ( 119 )
2
SSC CHSL 14/08/2023 (1st Shift)
(a) - 1 (b) - 2 (c) 1 (d) 2
Q.71. Simplify the given expression
( 4 . 2 )
3
- 0 . 008
( 4 . 2 )
2
+ 0 . 84 + 0 . 04
SSC CHSL 14/08/2023 (2nd Shift)
(a) 4 (b) - 2 (c) 2 (d) - 4
Q.72. The value of
( ) is:
2 + 7 × 7 ÷ 9 ???? 9 + 6 ÷ 6 × 2
4 ÷ 4 ???? 5 + 7 × 7 ÷ 7 - 6 + 4
SSC CHSL 17/08/2023 (2nd Shift)
(a) (b) (c) (d)
1734
2407
1865
2106
1857
2507
1664
2305
Q.73. Simplify the expression:
143 × 143 + 143 × 139 + 139 × 139
143 × 143 × 143 - 139 × 139 × 139
SSC CHSL 17/08/2023 (3rd Shift)
(a) (b) 282 (c) (d) 4
1
2
1
4
Q.74. Simplify the given expression and
?nd the value for x = -1.
10 ??
2
+ 5? + 2?? + ?
5? + ??
SSC CHSL 17/08/2023 (3rd Shift)
(a) -1 (b) 0 (c) 1 (d) 2
SSC CGL 2023 Tier - 1
Q.75. Simplify the following expression
[ × - {( ÷ ) - ( × )}
85
34
1
18
46
69
27
135
86
129
91
14
of ]
112
36
SSC CGL 14/07/2023 (3rd shift)
(a) 3 (b) 3 (c) 3 (d) 3
1
5
3
4
1
2
1
4
Q.76. Simplify the given expression.
18 ÷ 3 of 2 × 5 +72 ÷ 18 of 2 ×3 - 4 ÷ 8 × 2
SSC CGL 14/07/2023 (4th shift)
(a) 20 (b) (c) (d) 25
20
3
25
3
Q.77. Simplify the given expression.
y + 2x - [y - (y - x + y) - (x + y) + y] - 2y.
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Pinnacle Day: 8th - 10th Simplification
SSC CGL 17/07/2023 (1st shift)
(a) - y (b) - 2x (c) y (d) 2x
Q.78. Simplify the given expression.
( 326 + 222 )
2
- ( 326 - 222 )
2
( 326 × 222 )
SSC CGL 18/07/2023 (1st shift)
(a) 1 (b) 4 (c) 3 (d) 2
Q.79. Simplify the following expression.
[ + 12 ÷ 3 - {20 + (16 of 8 ÷ 16) - 25
(54 ÷ 18 of }]
1
2
SSC CGL 18/07/2023 (4th shift)
(a) -13 (b) 12 (c) 22 (d) 0
Q.80. Simplify the following
- + 2b
3 ?? + ??
2
?? - 3 ??
3
SSC CGL 19/07/2023 (1st shift)
(a) (b)
5 ( ?? + 3 ?? )
6
7 ?? + 3 ??
6
(c) (d)
7 ( ?? + 3 ?? )
6
?? + 3 ??
6
Q.81. Simplify the given expression.
(a - 2b) (b - 3a) + (a + b) (a - 3b) - (b - 3a)
(4a - 5b)
SSC CGL 21/07/2023 (1st shift)
(a) 8a
2
- 10ab (b) 8a
2
+ 10ab
(c) 10a
2
- 14ab (d) 10a
2
- 4ab
Q.82. The value of (3
2
)
2
+ 3 × 3 ÷ 3 - 3 is :
SSC CGL 21/07/2023 (2nd shift)
(a) 3
2
(b) 9
2
(c) 2
3
(d) 3
0
Q.83. If 720 ÷ 8 + 915 ÷ 15 - m + 32 × 5
= 1104 ÷ 16 × 111 ÷ 37, then the value of
m is:
SSC CGL 24/07/2023 (1st shift)
(a) 104 (b) 518 (c) 207 (d) 311
Q.84. Simplify the following expression.
(4x + 1)
2
- (4x + 3)(4x - 1)
SSC CGL 24/07/2023 (2nd shift)
(a) (4x + 1) (b) 4 (c) (4x - 3) (d) 4x
Q.85. Simplify the following.
0 . 01 ? 0 . 01 ? 0 . 01 + 0 . 003 ? 0 . 003 ? 0 . 003
0 . 05 ? 0 . 05 - 0 . 015 ? 0 . 05 + 0 . 015 ? 0 . 015
SSC CGL 24/07/2023 (3rd shift)
(a) (b)
13
25
× 10
3 13
15
× 10
- 3
(c) (d)
13
15
× 10
3 13
25
× 10
- 3
Q.86. What value of 'y' will satisfy the
following equation ?
= 2
3 ??
1 +
1
1 +
??
1 - ??
SSC CGL 25/07/2023 (1st shift)
(a) (b) (c) (d)
2
3
5
4
4
5
3
2
Q.87. Simplify the expression ,
?? - ??
?? + ??
where = 2 and = 3 . ?? ??
SSC CGL 25/07/2023 (4th shift)
(a) 2 - 6 (b) - 5 6 6
(c) 5 - 2 (d) 2 - 5 6 6
Q.88. Simplify the given expression
[ 120 × 120 × 120 - 100 × 100 × 100 ]
[ 120 × 120 + 120 × 100 + 100 × 100 ]
SSC CGL 27/07/2023 (2nd shift)
(a) 20 (b) 25 (c) 10 (d) 15
SSC Selection Post (Phase - XI)
Q.89. Simplify 52 52 48 48. × - ×
Higher Secondary 27/06/2023 (Shift - 3)
(a) 386 (b) 400 (c) 424 (d) 326
SSC CHSL 2022 Tier - 2
Q.90. What is the correct descending
order of the following ratios ?
, , ,
17
30
7
15
27
50
11
20
SSC CHSL Tier II (26/06/2023)
(a) > > >
11
20
17
30
27
50
7
15
(b) > > >
17
30
11
20
27
50
7
15
(c) > > >
17
30
27
50
11
20
7
15
(d) > > >
7
15
11
20
27
50
17
30
SSC MTS 2022 Tier - 1
Q.91. Find the value of
: [( 13² — 8² ) × 5 + 4 ]
SSC MTS 03/05/2023 (Afternoon)
(a) 22 (b) 24 (c) 26 (d) 23
Q.92. What is the value of
?
14 + 7 - 12 + 15 - 17 + 2 - 3 + 17
2 + 7 - 9 + 2 - 4 + 22
SSC MTS 03/05/2023 (Afternoon)
(a) (b) (c) (d)
23
20
17
20
19
20
33
20
Q.93. The value of
( ) + [1 - ] ( ) lies between :
1
5
1
4
1
2
×
2
5
SSC MTS 04/05/2023 (Morning)
(a) 2.1 and 2.4 (b) 0.2 and 0.6
(c) 1.5 and 1.8 (d) 0.8 and 1.2
Q.94. The value of ( + ) ( + )
1
2
3
4
÷
1
5
3
10
lies between:
SSC MTS 09/05/2023 (Afternoon)
(a) 2.7 and 3.0 (b) 2.0 and 2.2
(c) 1.5 and 1.8 (d) 2.3 and 2.6
Q.95. Find the value of 0.5 + 0.55 + 0.555
+ 0.5555 :
SSC MTS 09/05/2023 (Evening)
(a) 1.4405 (b) 1.0555
(c) 1.5555 (d) 2.1605
Q.96. Evaluate: 5 4 + (3 + 4)
²
÷ 7 4 : × ×
SSC MTS 12/05/2023 (Evening)
(a) 47 (b) 46 (c) 48 (d) 49
Q.97. Find the value of
[( 7² + 6³ + 9² ) ÷ 2 + 52 ]
SSC MTS 16/05/2023 (Morning)
(a) 14 (b) 15 (c) 16 (d) 13
Q.98. The value of 74 + 88 8 10 of 5 ÷ ×
90 15 + {44 (108 12)} is: - ÷ - ÷
SSC MTS 16/05/2023 (Evening)
(a) 700 (b) 653 (c) 770 (d) 605
Q.99. 48 ÷ 4 + A × 2 8 = 14, ?nd out the -
value of A.
SSC MTS 17/05/2023 (Evening)
(a) 10 (b) 5 (c) 14 (d) 8
Q.100. What is the value of x.
if ( ) of 20 + ( ) of 28 - ( ) of x = 10 ?
1
2
1
4
1
3
SSC MTS 19/05/2023 (Morning)
(a) 21 (b) 7 (c) 28 (d) 10
Q.101. What is the value of
?
2 . 5 × 2 . 5 + 0 . 3 × 0 . 3 + 5 × 0 . 3
1 . 6 × 1 . 6 + 0 . 6 × 0 . 6 - 1 . 2 × 1 . 6
SSC MTS 19/05/2023 (Evening)
(a) 7.64 (b) 8.64 (c) 2.8 (d) 7.84
Q.102. What is the value of
? 0 . 52 × 0 . 52 + 0 . 31 × 0 . 31 + 0 . 62 × 0 . 52
SSC MTS 13/06/2023 (Morning)
(a) 0.63 (b) 0.73 (c) 0.83 (d) 0.93
Q.103. What is the value of A .
if 2 + 4 (A + 6) × 2 = ? ÷
16
7
SSC MTS 14/06/2023 (Afternoon)
(a) 22 (b) 20 (c) 23 (d) 21
Q.104. What is the value of
?
1 . 2 × 1 . 2 + 0 . 2 × 0 . 2 + 1 . 2 × 0 . 4
1 . 8 × 1 . 8 + 0 . 4 × 0 . 4 - 1 . 8 × 0 . 8
SSC MTS 15/06/2023 (Morning)
(a) 0.1 (b) 0.25 (c) 0.2 (d) 1
Q.105. What is the value of
0 . 48 × 0 . 48 + 0 . 25 × 0 . 25 + 0 . 5 × 0 . 48
SSC MTS 16/06/2023 (Afternoon)
(a) 0.63 (b) 0.93 (c) 0.83 (d) 0.73
Q.106. Find the value of :
?
49
63
÷
21
9
-
85
510
+
88
132
÷
66
11
1
7
÷
33
21
+
26
143
-
1
209
÷
1
19
SSC MTS 19/06/2023 (Afternoon)
(a) (b) (c) (d)
55
36
55
39
13
46
44
13
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FAQs on SSC CGL Previous Year Questions (2023 - 18): Simplification - SSC CGL Mathematics Previous Year Paper (Topic-wise)
| 1. What is the importance of simplification in the SSC CGL exam? |  |
Ans. Simplification is crucial in the SSC CGL exam as it tests candidates' ability to solve mathematical problems quickly and accurately. It helps in saving time during the exam, allowing candidates to focus on other sections.
| 2. What types of questions related to simplification are commonly asked in SSC CGL? | |
Ans. Common types of questions include operations with fractions, decimals, percentages, square roots, and simplification of algebraic expressions. Candidates may also encounter questions that require them to simplify complex equations.
| 3. How can I improve my simplification skills for the SSC CGL exam? | |
Ans. To improve simplification skills, practice regularly using previous year papers and mock tests. Focus on understanding different mathematical operations and shortcuts, and work on time management to enhance speed and accuracy.
| 4. Are there any specific formulas or shortcuts for simplification that I should know for the SSC CGL? | |
Ans. Yes, some useful shortcuts include using the concept of BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction) for order of operations, and knowing common fraction conversions and percentages to quickly simplify calculations.
| 5. How much weight does the simplification section carry in the SSC CGL exam? | |
Ans. The simplification section is part of the quantitative aptitude segment of the SSC CGL exam. While it may not have a specific weight, mastering it can significantly impact overall performance, as it often leads to solving more complex problems efficiently.
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