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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

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

"A" it is . From statment 1 we have x> y^2.....and for all x and y > 1 , x has to be > y...Suff

From statement 2 we have y < x^2...for x = 2 and y = 3, it is true, so x<y in this case , but for x = 4 and y = 3, above exp is still true but x > y...thus insuff.