Divisibility Test - Practice Test


5 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Divisibility Test - Practice Test


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Attempt Divisibility Test - Practice Test | 5 questions in 10 minutes | Mock test for UPSC preparation | Free important questions MCQ to study UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making for UPSC Exam | Download free PDF with solutions
QUESTION: 1

What is the smallest number that should be added to 27452 to make it exactly divisible by 9? 

Solution:

If we divide 27452 by 9 the remainder will be 2. Since the question asks for the number to added, we will find out the difference between next number divisible.
9-2= 7
We will add 7 to 27452 = 27459 which is divisible by 9. 

QUESTION: 2

If 522x is a three digit number with as a  digit x . If the number is divisible by 6, What is the value of the  digit  x is?

Solution:

The correct option is Option E.

Given, 

522x is a three digit number but actually, it's a four digit number

522x is divisible by 6 if it's divisible by 2 and 3 both

522x would be divisible by 2

If 5 + 2 + 2 + x is divisible by 3

If 9 + x is divisible by 3

If x = 0, 3, 6 or 9

Taking intersection of 0, 2, 4, 6 or 8 and 0, 3, 6 or 9

x = 0 or 6

5220 and 5226is divisible by 6

If we take 522x is three digit number then 52y should be divisible by 6 where y = 2x

52y is divisible by 3

If 5 + 2 + y is divisible by 3 

If y = 2, 5, 8

Taking common y = 2 or 8

     y = 2x 

=> x = 1 or 4

522 and 528 are divisible by 6

QUESTION: 3

What is the least value of x such that 7648x is divisible by 11?

Solution:

A number is divisible by 11 when the difference between the sum of digits at even places and at odd places is 0 or multiple of 11

The given number is 7648x.  

(Sum of digits at EVEN places) – (sum of digits at ODD places) = 0

⇒ (6 + 8 ) - ( X + 7 + 4 ) = 0

⇒ 14 - ( X + 11 ) = 0   

Here the value of x must be 3
So, the least value of x is 3.

 

 

QUESTION: 4

 If M183 is divisible by 11, find the value of the smallest natural number M ? 

Solution:

A number is divisible by 11, when the difference between the sum of digits at even places and at odd places is 0 or multiple of 11. The given number is M183.    

(Sum of digits at EVEN places) – (sum of digits at ODD places) = 0

(8 + M )- (3+1)= 0                        

(8 + M ) -  4  =  0           

Here the value of M must be 7 .

QUESTION: 5

What least whole number should be added to 532869 to make it divisible by 9? 

Solution:

Answer : C

Explanation:If a number is divisible by 9, the sum of its digits must be a multiple of 9.     Here, 5+3+2+8+6+9=33, the next multiple of 9 is 36.3 must be added to 532869 to make it divisible by 9.

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