Arithmetic Questions for SSC CHSL Exam

Download Arithmetic Questions for SSC CHSL exam 2020. Most important arithmetic practice questions based on asked questions in previous exam papers for SSC CHSL. Start practice and download the questions along with solutions. It will be helpful to crack the SSC CHSL exam 2020.

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Question 1: Which two signs should be interchanged in the following equation to make it correct?
$5 \times 15 \div 7 – 20 + 4 = 77$

a) $(+  and  -)$

b) $(\div  and  +)$

c) $(-  and  \times)$

d) $(+  and  \times)$

Question 2: Which two signs should be interchanged in the following equation to make it correct?
$405 + 27 \times 40 – 308 \div 22 = 314$

a) $+  and  \div$

b) $\times  and  +$

c) $\div  and  \times$

d) $+  and  -$

Question 3: Three persons A, B and C have different amounts of rupees with them. If C takes ₹11 from A, C will have equal amount as B has. B and C together have total ₹111. How many rupees does C have?

a) 60

b) 65

c) 50

d) 70

Question 4: Which two signs should be interchanged in the following equation to make it correct?
$211 \times 14 + 627 \div 33 – 17 = 520$

a) $+  and  -$

b) $+  and  \div$

c) $\div  and  \times$

d) $\times  and  -$

Question 5: By interchanging the given two signs which of the following equation will be correct?
$+  and  -$

a) $13 – 14 \times 3 + 7 = 48$

b) $9 + 16 – 11 \times 2 = 3$

c) $11 \times 3 + 5 – 2 = 36$

d) $16 \times 4 + 11 – 10 = 65$

Question 6: By interchanging which two signs the equation will be correct?
$11 + 9 – 4 \times 12 \div 6 = 32$

a) $\div  and  -$

b) $-  and  +$

c) $\times  and  +$

d) $\times  and  \div$

Question 7: After interchanging the given two signs, what will be the value of $9 + 143 \div 13 \times 11 – 160 = ?$
$\times  and  \div$

a) 18

b) -32

c) -17

d) 28

Question 8: If the given two numbers are interchanged, which of the following equations will become correct?
3 and 5

a) $6 + 5 \times 2 – 3 = 7$

b) $3 + 4 – 1 \times 5 = 2$

c) $3 + 5 – 4 \times 2 = -1$

d) $1 + 9 \div 3 + 5 = 9$

Question 9: If the two signs, ‘$\times$ and $\div$’ are interchanged, which of the following equations will be correct?

a) $14 + 62 \times 13 \div 31 = 38$

b) $14 + 9 \div 24 \times 3 = 86$

c) $63 + 59 \div 4 \times 8 = 171$

d) $11 \times 12 \div 48 + 6 = 40$

Question 10: After interchanging the given two signs, what will be the value of $11 \div 9 – 63 + 7 \times 2$ ?
$\div  and  +$

a) 2

b) 5

c) -4

d) -2

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Question 11: By interchanging the given two signs which of the following equation will be correct?
$\times$ and $+$

a) $4 + 9 \times 11 \div 3 = 37$

b) $11 \times 6 \div 18 + 54 = 29$

c) $13 – 12 \times 5 + 11 = 46$

d) $6 \times 11 + 13 \div 11 = 17$

Question 12: After interchanging the given signs, what will be the value of $16 + 19 \times 51 \div 153 – 12 ?$
$\times  and  \div$

a) 63

b) 51

c) 61

d) 57

Question 13: By interchanging which two numbers the equation will be correct?
$9 – 2 \times 4 \div 6 = -3$

a) 6 and 9

b) 6 and 2

c) 4 and 9

d) 2 and 4

Question 14: By interchanging which two signs the equation will be correct?
$19 + 63 \div 7 – 8 \times 12 = 79$

a) $-  and  \times$

b) $\times  and  +$

c) $\div  and  -$

d) $\times  and  \div$

Question 15: By interchanging which two signs the equation will be correct?
$6 + 9 \times 2 – 3 \div 4 = 8$

a) $-  and  \div$

b) $-  and  +$

c) $+  and  \div$

d) $\div  and  \times$

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Question 16: By interchanging which two numbers, the value obtained after solving the given equation will be ’13’?
$7 + 8 \div 4 \times 3$

a) 8 and 3

b) 8 and 4

c) 7 and 3

d) 3 and 4

Question 17: By interchanging which two signs the equation will be correct?
$11 + 16 \times 12 \div 4 – 2 = 21$

a) $\times  and  -$

b) $-  and  +$

c) $\div  and  +$

d) $\div  and  \times$

Question 18: By interchanging which two signs the equation will be correct?
$9 \times 11 \div 31 + 62 – 13 = 18$

a) $\times  and  \div$

b) $-  and  \times$

c) $\div  and  +$

d) $+  and  \times$

Question 19: After interchanging the given two numbers, what will be the value of $3 + 2 \div 1 \times 4 – 7$?
4 and 7

a) 10

b) 12

c) 13

d) 11

Question 20: After interchanging which two numbers, the value of given equation will be ‘4’?
6 + 3 $\div$ 9 $\times$ 7 – 5

a) 7 and 6

b) 3 and 5

c) 5 and 6

d) 9 and 5

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Answers & Solutions:

1) Answer (B)

Trial and error method.

2) Answer (A)

3) Answer (C)

4) Answer (D)

if we interchange $+  and  -$  answer would be 2952

if we interchange $+  and  \div$ answer would be 20.71

if we interchange  $\div  and  \times$ answer would be 20689.07

if we interchange  $\times  and  -$ answer would be  520

5) Answer (A)

By interchanging ‘$+$’ and ‘$-$’

(A) : $13 – 14 \times 3 + 7 = 48$

L.H.S. $\equiv13+14\times3-7$

= $13+42-7=48=$ R.H.S.

=> Ans – (A)

6) Answer (A)

7) Answer (A)

8) Answer (A)

if we interchange 3 and 5 then-

6 + 3 $\times$ 2 – 5

= 6+6-5

=7

9) Answer (B)

If we interchange ‘$\times$ and $\div$’ then-

14 + 9 $ \times $ 24 $ \div $ 3

= 14+72

= 86

10) Answer (A)

$11 \div 9 – 63 + 7 \times 2$

= $11 + 9 – 63 \div 7 \times 2$

=  $11 + 9 – 9 \times 2$

= 11 – 9 =2

So , the answer would be option a )2.

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11) Answer (B)

On interchanging the two signs $ +$ and $\times $

we get the equation as

$ 11+6 \div 18\times 54 $

or $11+1\div 3\times 54 =29$

$ 29=29 $ which is true

12) Answer (C)

On interchanging the signs $\times and \div$

we get the equation as

$ 16+19\div 51 \times 153 -12 $

or $ 16+19 \times 3-12$

16+57-12=61

13) Answer (B)

$9 – 2 \times 4 \div 6 = -3$

By option (2) 6 and 2, on interchanging we get, $9 – 6 \times 4 \div 2 = -3$

Using BODMAS rule, we get, $9 – 6 \times 2 = -3$

9 – 12 = -3

-3 = -3

Hence LHS = RHS.

Hence our answer.

14) Answer (A)

For this question we have to go by option, so when we replace $-  and  \times$ as per option A,

$19 + 63 \div 7 \times 8 – 12 = 79$

$19 + {\frac{63}{7}}\times 8 – 12 = 79$

$19 + {9\times8} – 12 = 79$

$19 + 72 – 12 = 79$

$79 = 79$

Therefore option A is the answer.

15) Answer (A)

$6 + 9 \times 2 – 3 \div 4 = 8$

this results in a fraction so in order to get a whole number which is 8 it is important to place divide symbol such that a whole number is obtained

now if  $-  and  \div$ is interchanged 9 gets divided by 3 thus resulting in a whole number

$6 + 9 \times 2\div 3-  4 $ (using BODMAS rule)

$ 6 + \frac{ 9 \times 2}{ 3} – 4 $

= 6+ 6 – 4 = 8

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16) Answer (A)

going by the options

A) 8 and 3

$7 + 3 \div 4 \times 8$=  $7 + \frac{8}{4}\times 3$ = 7 + 6 = 13 (Option A is correct )

B) 8 and 4

$7 + 4 \div 8 \times 3$ =  $7 + \frac{4}{8}\times 3$ = 7 + $\frac{3}{2}$= $\frac{17}{2}$   (option B is incorrect)

c) 7 and 3

$3 + 8 \div 4 \times 7$ = 3+$ \frac{8}{4}\times 7$ = 3 +14 = 17 (option C is incorrect)

d) 3 and 4

$7 + 8 \div 3 \times 4$ = 7+ $ \frac{8}{3}\times 4$ = 7 +$ \frac{32}{3}$ = $ \frac{53}{3}$  (option D is incorrect)

17) Answer (A)

Given, 11+16 $\times$ 12 $ $ $\div$ 4 $ $ – 2 = 21

These type of questions can be quickly solved by checking options.

So,from the options we get option A would be correct.

On solving,11+16-12 $\div $ 4 $ $ $\times$2 $ $

$\Rightarrow$  27 – 6 = 21.

18) Answer (D)

19) Answer (C)

given  3 + 2 ÷ 1 × 4 – 7

on interchanging 4 and 7

BODMAS rule{ brackets > of > division > multiplication > addition > subtraction }

we get  3 + 2 ÷ 1 × 7 – 4 ( using BODMAS rule)

3+$\frac{2}{1} \times$ 7 -4 = 3 + 2 $\times$ 7  – 4 = 3 + 14 – 4 = 13

20) Answer (A)

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