**Number System Questions for SSC CGL PDF:**

Number System Questions for SSC CGL PDF

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**Question 1:** The product of two 2-digit numbers is 2160 and their H.C.F. is 12. The numbers are

a) (12, 60)

b) (72, 30)

c) (36, 60)

d) (60, 72)

**Question 2:** The least number to be added to 13851 to get a number which is divisible by 87 is

a) 18

b) 43

c) 54

d) 69

**Question 3:** The least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case, but when divided by 9 leaves no remainder, is:

a) 1677

b) 1683

c) 2523

d) 3363

**Question 4:** When a number is divided by 56, the remainder will be 29. If the same number i divided by 8, then the remainder will be

a) 6

b) 7

c) 5

d) 3

**Question 5:** The sum of three consecutive natural numbers each divisible by 5, is 225. The

largest among them is

a) 85

b) 75

c) 70

d) 80

**Question 6:** What is the average of all numbers between 8 and 74 which are divisible by 7?

a) 40

b) 41

c) 42

d) 43

**Question 7:** What is the remainder when 2468 is divided by 37?

a) 26

b) 36

c) 18

d) 14

**Question 8:** What is the remainder when 2468 is divided by 37?

a) 26

b) 36

c) 18

d) 14

**Question 9:** What least number must be added to 329, so that the sum is completely divisible by

7?

a) 1

b) 0

c) 2

d) 3

**Question 10:** What is the largest 4 digit number that is exactly divisible by 93?

a) 9961

b) 9971

c) 9981

d) 9951

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**Answer & Solutions for Number System Questions for SSC CGL PDF:**

**1) Answer (C)**

Let the two numbers be 12x and 12y respectively where x and y are co-primes

Product of numbers = (12x) x (12Y) = 2160

=> xy = 2160 / 144 = 15

Possible pairs of x and y whose H.C.F. is 1 = (3,5)

∴ Required numbers = (12×3) , (12×5)

= 36 , 60

=> Ans – (C)

**2) Answer (D)**

Factorizing 13851 = 87 \times× 159 + 18

Thus, on dividing 13851 by 87, the remainder is 18

∴ The number that must be added to 13851 so that the sum obtained is completely divisible by 87

= 87 – 18 = 69

=> Ans – (D)

**3) Answer (B)**

L.C.M. of 5, 6, 7 and 8 = 840

Thus, required number is of the form = 840k + 3

Least value of k for which (840k+3) is divisible by 9 is k = 2

∴ Required number = 840 × 2 + 3 = 1683

=> Ans – (B)

**4) Answer (C)**

When the number is divided by 56, remainder is 29

=> Let the number = 56+29 = 85

Now, if 85 is divided by 8, => 85 = 8 × 10 +5

Thus, remainder = 5

=> Ans – (C)

**5) Answer (D)**

Let the three consecutive natural numbers each divisible by 5 be (5x – 5),(5x),(5x + 5)

Sum = (5x – 5) + (5x) + (5x + 5) = 225

=> 15x = 225

=> x = 225 / 15

∴ Largest number = 5(15) + 5 = 75 + 5 = 80

=> Ans – (D)

**6) Answer (C)**

The numbers between 8 and 74 which are divisible by 7 are 14, 21, 28, 35, 42, 49, 56, 63, 70.

sum = 378

average = 378/9 = 42.

SHORTCUT:

average = 7*(average of 2, 3, 4, 5, 6, 7, 8, 9, 10) = 7*(6) = 42.

so the answer is option C.

**7) Answer (A)**

37*66 = 2442 is the least nearest multiple of 37.

The remainder when 2468 is divided by 37 = 2468 – 2442 = 26

so the answer is option A.

**8) Answer (B)**

6345 = 3*3*3*5*47

2160 = 3*3*3*5*16

HCF = product of common prime factors = 3*3*3*5 = 135

So the answer is option B.

**9) Answer (B)**

329 / 7 = 47

no need to add any number as 329 is divisible by 7.

so the answer is option B.

**10) Answer (D)**

Largest 4 digit number = 9999

9999/93 = 107.52

So largest 4 digit multiple of 93 = 107*93 = 9951

So the answer is option D.