Quant Boosters  Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta  Set 1

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

This post is deleted!

@soumil_jain 0.04

Exactly 3 factors means the number is a perfect square.
We need to get this by multiplying two numbers, means only two ways
Case 1 : a = 1 and b is prime such that b^2 < 100. b can be 2^2 , 3^2, 5^2 or 7^2  4 cases
Case 2 : a is a prime such that a^2 < 100 and b = 1. a can be 2^2 , 3^2, 5^2 or 7^2  again, 4 cases.
So total favourable cases = 4 + 4 = 8
Total cases = 100 * 99 (a can be picked in 100 ways and b can can be picked in 99 ways)
So required probability = 8/9900 = 2/2475

with repetition or without repetition?
seems with repetition for the sake easy calculation: 2x5/100x100

ccb = 441
ab = 21
b = 1

@MaliNaveenReddy it is mentioned as distinct integers.
Also, how repetition is significant here in identifying the favourable cases ?
as the only 8 cases possible with a/b = 1 and b/a = prime^2 < 100
Can you please share how you got 10 favourable cases ?