Number system is important topic in quantitative aptitude section of SSC CHSL and CGL exams. You can download the SSC CHSL number system questions and answers PDF.

**Number System Questions for SSC CHSL Exam:**

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**Question 1:** What smallest number should be added to 2957 so that the sum is completely divisible by 17 ?

a) 9

b) 2

c) 3

d) 1

**Question 2:** Which of the following numbers is not a prime number?

a) 731

b) 227

c) 347

d) 461

**Question 3:** The largest 5 digit number exactly divisible by 95 is

a) 99936

b) 99935

c) 99940

d) 99933

**Question 4:** If the number 583_437 is completely divisible by 9, then the smallest whole number in the place of the blank digit will be

a) 4

b) 5

c) 3

d) 6

**Question 5:** The sum of all prime numbers between 30 and 42 is

a) 103

b) 109

c) 105

d) 104

**Question 6:** What smallest number should be added to 2401 so that the sum is completely divisible by 14 ?

a) 8

b) 7

c) 4

d) 5

**Question 7:** Which of the following numbers is completely divisible by 99?

a) 51579

b) 51557

c) 55036

d) 49984

**Question 8:** If in a two digit number, the digit at unit place is z and the digit at tens place is 8, then the number is

a) 80z + z

b) 80 + z

c) 8z + 8

d) 80z + 1

**Question 9:** What least number must be added to 1039, so that the sum obtained is completely divisible by 29?

a) 4

b) 5

c) 8

d) 6

**Question 10:** Which of the following numbers is not a prime number?

a) 197

b) 313

c) 439

d) 391

**Solutions for Number System Questions for SSC CHSL:**

**Solutions:**

**1) Answer (D)**

If 2957 is divisible 17, => 2957 = 17 $×$ 173 + 16

Quotient is 173 and remainder is 16

Thus, smallest number that should be added to 2957 so that the sum is completely divisible by 17 = (17 – 16) = 1

=> Ans – (D)

**2) Answer (A)**

Prime Factorization of 731 = 17 $×$ 43

For all the remaining numbers, the only factors are 1 and the number themselves.

Thus, only 731 is not a prime number.

=> Ans – (A)

**3) Answer (C)**

Largest 5 digit number = 99999

If 99999 is divided by 95, => 99999 = 95 $×$ 1052 + 59

Quotient is 1052 and remainder is 59

Thus, largest 5 digit number which is divisible by 95 = 99999 – 59

= 99940

=> Ans – (C)

**4) Answer (D)**

Number : 583_437

Let the missing digit = $$x$$

=> Sum of digits = 5 + 8 + 3 + $$x$$ + 4 + 3 + 7 = 30 + $$x$$

Clearly, sum of digits is between 30 and 39. The only number divisible by 9 and which lies between 30 and 39 is 36

=> $$30 + x = 36$$

=> $$x = 36 – 30 = 6$$

=> Ans – (D)

**5) Answer (B)**

Sum of prime numbers between 30 and 42 :

= 31 + 37 + 41

= 109

=> Ans – (B)

**6) Answer (B)**

Factorizing 2401 = 14 $×$ 171 + 7

Thus, on dividing 2401 by 14, the remainder is 7

$∴$ The number that must be added to 2401 so that the sum obtained is completely divisible by 14

= 14 – 7 = 7

=> Ans – (B)

**7) Answer (A)**

For a number to be divisible by 99, it must be divisible by 9 and 11

(A) : 51579 = 5 + 1 + 5 + 7 + 9 = 27 which is divisible by 9 and also by 11

(B) : 51557 = 5 + 1 + 5 + 5 + 7 = 23 which is not divisible by 9

(C) : 55036 = 5 + 5 + 0 + 3 + 6 = 19 which is not divisible by 9

(D) : 49984 = 4 + 9 + 9 + 8 + 4 = 34 which is not divisible by 9

Thus, only option (A) is divisible by 99

**8) Answer (B)**

Digit at unit’s place = z

Digit at ten’s place = 8

=> 2-digit number = $$(10 \times 8) + (1 \times z)$$

= 80 + z

=> Ans – (B)

**9) Answer (B)**

Factorizing 1039 = 29 $×$ 35 + 24

Thus, on dividing 1039 by 29, the remainder is 24

$∴$ The number that must be added to 1039 so that the sum obtained is completely divisible by 29

= 29 – 24 = 5

=> Ans – (B)

**10) Answer (D)**

Among the given numbers, there are no factors of 197, 313, 439 other than 1 and the numbers themselves.

391 = 23 $×$ 17

=> 391 is not a prime number.

=> Ans – (D)

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