Number System Questions for CAT With Solutions Test-2

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Number System For CAT (Test-2)
Number System For CAT (Test-2)

Number System is of the important topic for CAT. The number of questions asked from this topic is high in the exam.

Number System Questions for CAT:

You can download Number System questions or you can go through the details below.

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Question 1:

A number is formed by writing first 54 natural numbers next to each other as 12345678910111213 … Find the remainder when this number is divided by 8.

A. 1

B. 7

C. 2

D. 0

Question 2:

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

A. 
0

B.
 1

C. 
4

D. 
3

Question 3:

How many five digit numbers can be formed from 1, 2, 3, 4,  5, without repetition, when the digit at the unit’s place must be greater than that in the ten’s place?

A. 54

B. 60

C. 17

D. 2 x 4!

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Question 4:

If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is

A. 2

B. 5

C. 6

D. 12

Question 5:

Let k be a positive integer such that k+4 is divisible by 7.  Then the smallest positive integer n, greater than 2, such that k+2n is divisible by 7 equals

A. 
9

B. 7

C. 5

D. 3

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Solutions:

1) Answer (C)

For a number to be divisible by 8, last 3 digits must be
divisible by 8.
Last 3 digits of this number are 354.
354 mod 8 = 2
Hence, 2 is the remainder.

2) Answer (C)

As we know for a number to be divisible by 125, its last three digits should be divisible by 125 So for a five digit number, with digits 2,3,8,7,5 its last three  digits should be 875 and 375 Hence only 4 numbers are possible with its three digits as  875 and 375 I.e. 23875, 32875, 28375, 82375

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3) Answer (B)

Possible numbers with unit’s place as 5 = $$4 \times 3 \times 2 \times 1 = 24$$ Possible numbers with unit’s place as 4 and ten’s place 3,2,1 = $$3 \times 3 \times 2 \times 1 = 18$$
Possible numbers with unit’s place as 3 and ten’s place 2,1 = $$2 \times 3 \times 2 \times 1 = 12$$
Possible numbers with unit’s place as 3 and ten’s place 1 = $$1 \times 3 \times 2 \times 1 = 6$$
Total possible values = 24+18+12+6 = 60

4) Answer (C)

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

5) Answer (A)

let’s say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also  multiple of 7.
or 2n-4 should be a multiple of 7
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )

 

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