Number System Questions for SSC CGL PDF

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Number System Questions for SSC CGL PDF
Number System Questions for SSC CGL PDF

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

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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.

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