Statement 1:
if x is 2, y=5, then 4<5, here 2<5, that is x<y
if x is 0.5, y=0.3, then 0.25<0.3, here 0.5>0.3, that is x>y
So we cant conclude that if x>y of x<y

Statement 2:
Here x>=0, if square both sides of the inequality,
we have x>y^2, then x>y

Thus, statement 2 only is sufficient.

Please let me know if this is correct

piyush2694 wrote:Is x > y?

(1) x^2 < y

(2) x^(1/2) >y

$$x\,\,\mathop > \limits^? \,\,y$$
$$\left( {1 + 2} \right)\,\,\,\,{x^2} < y < \sqrt x \,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 5}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.