brstorewala wrote:

from statement 1....

square root of any number x (where x>1), is less than the number itself.......

so if the square root of x is greater than y, then obviously x >y always.....so statement 1 is sufficient....

from statement 2.......

Assume y = 9 and x = 4........in this case

sqrt 9 < 4 ( i.e sqrt y < x) and y > x

Assume y = 9 and x = 12.....in this case

sqrt 9 < 12 ( i.e sqrt y < x) and x > y

so statement 2 is insufficient

so answer A

Sorry for taking up your time. I was doing this one late at night and just did not understand the question. It is a one line explanation; sorry about that

bb