Number System Questions for SSC CGL PDF

0
2295
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

100 SSC CGL (latest pattern mocks) – Rs. 399

Download All SSC CGL Questions & Tricks PDF

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

SSC CGL 2018 Syllabus PDF

Take SSC CGL Free Mock Test

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

Top Rated APP for SSC CGL Preparation

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.

SSC CGL Free Previous Papers

LEAVE A REPLY

Please enter your comment!
Please enter your name here