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

In every meet their combined distance travelled is 100 m. Let there be "a" meets. =>[(100a) * 5/(5+3)]=300Which gives a=4.8 Though it cant be decimal. But What is wrong with this approach???? As there is a similar question in gyan room-arithmetic shortcuts...and i hv applied the same approach here..

@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