Free Numbers Set II - 02 Practice Test - CAT 

Question 1

How many numbers between 1 and 500 can be expressed as a difference of squares in atleast one way?

A. 125
B. 250
C. 375
D. none of these


Solution : C

option (c)

Numbers which cannot be expressed as difference of squares are all even non-multiples of 4. There are 250 even non-multiples of 4 from 1 to 1000 and 125 such numbers between 1 and 500 (2,6,10......). 125 cannot be expressed hence 375 can be expressed. Hence option c

Question 2

Multiplying prime numbers >1 to <100 consecutively, how many zeroes will result at the end?

A. 0
B. 2
C. 3
D. None of these


Solution : D

option (d)

First prime number =2 .To need a zero, we need a 2 and a 5 as 2×5=10. this occurs only once,

hence there will be only one zero. Answer = option d

Question 3

A number N leaves a remainder 3 when it is divided by 15. Which of the following when divided by 15 will leave a remainder 6?
I) 6N
II) 7N
IV) 5N+6

A. I and II
B. II,III and IV
C. III and IV
D. II and IV


Solution : B


take N=3 as 315 gives a remainder 3.

6N=18 1815 does not give a remainder 6

7N=2115 remainder =6

2N=615. remainder =6

5N+6=2115. Remainder = 6

Question 4

A number is formed by writing the same digit 8 times. That number is always divisible by?

A. 7
B. 13
C. 19
D. none of these


Solution : D

Take an example of 11111111. This can be any of the 8 digits.

It would be divisible by 13 and 7 as if we could divide it into equal even groups of 3.

For 19, we need to divide it into even groups of 8. This is not true

This number is divisible by 11, where the rule of divisibility is sum of digits at odd places- sum of digits at even places.

Hence, Answer is none of these.

Question 5

Find the ten’s digit of 74288

A. 3
B. 6
C. 7
D. 8


Solution : C

Ans. (c)

74288=(37×2)288 (refer demo tutorial for last 2 digits technique)






(__21)×(__56)=__76. answer is 7

Question 6

 What is the remainder when 1091 is divided by 13? 

A. 0
B. 1
C. 2
D. none of these


Solution : D

option d

Since, 10613R=1=103×10313R=(1)×(1)=1

Hence we get 1090R=1, we are left with 101 which will give a remainder of 10, hence the remainder will be 10 or option d.

Question 7

How many factors of 24000 are odd numbers?

A. 32
B. 7
C. 16
D. none of these


Solution : D

Answer=option d

Prime factorize 24000=26×3×53

Take out 26 and then find the number of factors. This will give all the odd factors

In this case, number of factors =(3+1)×(1+1)=4×2=8

Question 8

Given that p,q,r,s,t,u are integers and p+q+r+s+t+u=2005, what is the minimum value of (1)p+(1)q+(1)r+(1)s+(1)t+(1)u?

A. 0
B. -2
C. -3
D. none of these


Solution : D

p+q+r+s+t+u=2005 of the 6 integers, the number of integers that can be odd is 5,3 or 1. In the expression (1)p+(1)q+(1)r+(1)s+(1)t+(1)u

1)When one integer is even and the others are odd, then only one among is +1 and others are -1 each. Hence the sum of the terms is -5 + 1 = -4.

2)When three of the integers are even, then three of the terms are +1 each and the remaining three are -1 each. Hence the sum is 0.

3)When five of the integers are even, then five terms will be +1 each and the other term is -1. Hence the sum is 4. The minimum value of the sum is -4.

Question 9

Findthe remainder when 10347 is divided by 3.

A. 1
B. -1
C. 0
D. 2


Solution : C

option (c)

Any power of 10 when divided by 3 gives a remainder of 1. Also,7 divided by 3 remainder is 1. Therefore individual remainders are 1 and 1.

Thus, 1-1=0. Answer is 0

Question 10

 When (629)24 is divided by 21, find the remainder.

A. 1
B. 2
C. 5
D. 11


Solution : A

629 divided by 21 remainder is -1. Thus the question changes to 12421 Remainder =+1

Question 11

Given that p is a prime number. p2[p271p] is always divisible by?

A. 25
B. 23
C. 27
D. none of these


Solution : D

According to fermat's theorem " If P is a prime number and N is prime to P, then NpN is divisible by P."

Open the brackets, p2[p271p]=p29p will always be divisible by 29, as it is prime.

Question 12

A lies between -4 and 10. B lies between 20 and 50. BA lies between?

A. 2, 25
B. 0,50
C. 5,12.5
D. none of these


Solution : D

Find the least value and the highest value of BA

Least BA=50(1)=50

Highest BA=500=


Question 13

The 66th term of the series is

A. odd
B. even
C. cannot be determined
D. none of these


Solution : B


Observe the pattern


Every 3rd term is even. Hence a term which is a multiple of 3 will be even. 66 is a multiple of 3. Hence, the 66th term is even

Question 14

If the 12th term is 144 and the 14th term is 377. What is the 15th term?

A. 502
B. 401
C. 610
D. cannot be determined


Solution : C

option (c)

Tn+1=Tn1+Tn. 12th term=144

14th term =12th term +13th term

144+13th term=377


15th term =13th term +14th term =233+377=610

Question 15

Find the remainder when 9400+9401+9402.......9410 is divided by 6?

A. 0
B. 1
C. 3
D. 2


Solution : C

Any power of 9 when divided by 6, gives a remainder 3. hence answer =11×36((As there are 11 terms)= 336 = Remainder =3