The roots are a and b: a + b = p and ab = 12 (a + b)^2 = p^2 (a - b)^2 = (a + b)^2 - 4ab => (a - b)^2 = p^2 - 12 * 4 = p^2 - 48 If |a - b| ≥ 12 { Difference between the roots is at least 12} then, (a - b)^2 ≥ 144 p^2 - 48 ≥ 144 p^2 ≥ 192 P ≥ 8√3 or P ≤ -8√3

@rajesh_balasubramanian Hello. I was just looking for an online CAT coaching and stumbled upon this. The answer seems to be 7. And i don't think 10 can be a value for B. Eg: Take the Track length as 300m. A will complete it in 50secs. And B will take 75 secs to overlap A for the first time(300/4). I can email you my solution if you'd like to give it a look. Thank you.

@rajesh_balasubramanian Sir, if we take x=0 and y = 1050, then HCF(0,1050) = 1050. Or am I missing something?

Exactly one of ab, bc and ca is odd => Two are odd and one is even abc is a multiple of 4 => the even number is a multiple of 4 The arithmetic mean of a and b is an integer => a and b are odd and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3 c can be 4 or 8. c = 4; a, b can be 3, 5 or 5, 9 c = 8; a, b can be 3, 7 or 7, 9 Four triplets are possible.

No two boys sit next to each other => Boys and girls must alternate. As they are seated around a circular table, there is no other possibility. Now, 4 boys and 4 girls need to be seated around a circular table such that they alternate. Again, let us do this in two steps. Step I: Let 4 boys occupy seats around a circle. This can be done in 3! ways. Step II: Let 4 girls take the 4 seats between the boys. This can be done in 4! ways. Note that when the girls go to occupy seats around the table, the idea of the circular arrangement is gone. Girls occupy seats between the boys. The seats are defined as seat between B1 & B2, B2 & B3, B3 & B4 or B4 & B1. So there are 4! ways of doing this. Total number of ways = 3! × 4! = 6 × 24 = 144

Really enjoyed it.Thanks for such a deep insight