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Previous Year Questions of Number System

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In this, we are providing previous year questions and solutions of SSC CGL CHSL CPO MTS of number system asked by TCS. It is helpful for upcoming exams of SSC Railways and Other competitive exams conducted by TCS.

SSC CGL Tier 2 2019 SSC CGL Tier 1 2019 SSC CHSL 2019 SSC CPO 2020 SSC MTS 2019
SSC CGL Tier 2 2018 SSC CGL Tier 1 2018 SSC CHSL 2018 SSC CPO 2019 SSC MTS 2017

Previous Year Questions of Number System

Q1. If the number 1005x4 is completely divisible by 8, then the smallest integer in place of x will be:

SSC CGL 3 March 2020 (Morning)

(a) 2

(b) 4

(c) 1

(d) 0

Q2. When 200 is divided by a positive integer x, the remainder is 8. How many values of x are there?

SSC CGL 3 March 2020 (Afternoon)

(a) 7

(b) 5

(c) 8

(d) 6

Q3. What should replace * in the number 94*2357, so that the number is divisible by 11?

SSC CGL 3 March 2020 (Evening)

(a) 3

(b) 7

(c) 1

(d) 8

Q4. When 732 is divided by a positive integer x, the remainder is 12. How many values of x are there? 

SSC CGL 4 March 2020 (Morning)

(a) 19

(b) 20

(c) 18

(d) 16

Q5. If the 6-digit numbers x35624 and 1257y4 are divisible by 11 and 72, respectively, then what is the value of (5x-2y)?

SSC CGL 4 March 2020 (Afternoon)

(a) 14

(b) 12

(c) 10

(d) 13

Q6. How many numbers are there from 200 to 800 which are neither divisible by 5 nor by 7?

SSC CGL 4 March 2020 (Evening)

(a) 407

(b) 410

(c) 413

(d) 411

Q7. If the nine-digit number 708x6y8z9 is divisible by 99, then what is the value of x+y+z? 

SSC CGL 5 March 2020 (Morning)

(a) 9

(b) 16

(c) 5

(d) 27

Q8. When a positive integer is divided by d, the remainder is 15. When ten times of the same number is divided by d, the remainder is 6. The least possible value of d is:

SSC CGL 5 March 2020 (Afternoon)

(a) 9

(b) 12

(c) 16

(d) 18

Q9. The greatest number which should be replaced ‘*’ in the number 146*48 to make it divisible by 8 is:

SSC CGL 5 March 2020 (Evening)

(a) 9

(b) 2

(c) 8

(d) 0

Q10. If the number 687x29 is divisible by 9, then the value of 2x is: 

SSC CGL 6 March 2020 (Morning)

(a) 8

(b) 3

(c) 2

(d) 4

Here are the solutions to the above questions in short methods. it will help you to save your precious time in the exam.

Sol 1.

Ans. (d)

For any number to be divisible by 8, its last 3 digits must be divisible by 8

By putting x=0 in 1005x4, we see that 504 will be divisible by 8.

Sol 2.

Ans. (c)

When 200 is divided by x remainder is 8. So, the number exactly divisible by x is 192.

Multiples of 192 = 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192

The remainder is always less than the divisor, so :x > 8

Desired values are 12, 16, 24, 32, 48, 64, 96, 192. 

Sol 3.

Ans. (a)

For a number to be divisible by 11, the difference of the sum of alternative numbers is divisible by 11.

Hence, (7+3+*+9)-(5+2+4)

        =(19+*)-(11)

        =8+*

* must be 3 for 94*2357 to be divisible by 11.

Sol 4.

Ans. (b)

When 732 is divided by x remainder is 12. So, the number exactly divisible by x is 720.

The remainder is always less than the divisor, so :x > 12

Now, the factors of 720 which are more than 12 are possible values of x, i.e. (15,16,18,20,24,30,36, 40,45, 48,60, 72, 80, 90, 120, 144, 180, 240, 360, 720).

Sol 5.

Ans. (a)

In such questions directly check divisibility by 11, 9, and 8.

For a number to be divisible by 11, the difference of the sum of digits at odd or even places must be divisible by 11.

For a number to be divisible by 9, the sum of numbers must be divisible by 9.

For divisibility by 8, the last 3 numbers must be divisible by 8.

Accordingly, For x35624 divisible by 11

(x+5+2)-(3+6+4) = 0 or 11

x=6

And For 1257y4 divisible by 72,

1+2+5+7+y+4 must be divisible by 9 and the only possible value of y is 9, here.

also, 784 is divisible by 8 so the desired value of y = 8

Then,  5x-2y = 30-16 = 14 

Sol 7.

Ans. (b)

It is given that 708x6y8z9 is divisible by 99.

Thus, 708x6y8z9 is divisible by both 11 and 9

For divisibility by 9, the sum of digits is divisible by 9 (7+0+8+x+6+y+8+z+9 = 38+x+y+z. We get 2 as the remainder when 389. Thus, 2+x+y+z must be divisible by 9)

Possible values of (z+y+x) = 7,16,25, etc. 

For divisibility by 11, the difference of sum of digits at odd and even place is divisible by 11 (i.e. in 708x6y8z9 : (9 + 8 + 6 + 8 + 7 ) - (z + y + x + 0) = 38 - ( z + y + x) is divisible by 11 )

Possible values of (z+y+x) =38,5,16 etc.

In such questions, we must directly verify options.

Sol 8.

Ans. (c)

Let N be the number that gives Q as quotient and 15 as the remainder when divided by d. Thus, d > 15

N = dQ+ 15

10N = 10(dQ)+ 144 + 6 

clearly, d is a multiple of 144 which are: 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, and so on. 

The least possible value of d is 16. (d > 15)

Sol 9.

Ans. (c)

For 146*48 to be divisible by 8, *48 must be divisible by 8.

Check options: * = 2,8 satisfies the condition. But 8>2. option c is the correct answer.

Sol 10.

Ans. (a)

For 687x29 to be divisible by 9, the sum of digits of 687x29 must be divisible by 9. Thus, x = 4 and 2x = 4

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