is x>y?
1) x^.5 > y
2) x^3 > y


From 1) x^0.5 > y
This means root x > y

we know that root of numbers are positive only

we have to consider two range of data here one between 0 and 1 and another >1
case a: if x is between 0 and 1 , example 0.25
root x would be 0.5 , means root(x) > x
point is the valus of x may have been less than y but making ti root(x) may make it more than Y
I mean x =0.25 and y = 0.3 , root(x)=0.5 > y, but you see x is not > y

INSUFFICIENT

case b: if x >1 , root(x) >x,
from this we can conclude that x> y for sure
SUFFICIENT


Since case a is INSUFFICIENT, statement1 by itself is not sufficient


From 2)
case a) x is positive: we know
There can be many cases here too,
When x is 2 and y is 3
X^3 > Y , we know x < Y
X could have been 4, and Y = 3
Still x^3 > y

As we observe we cannot conclude anything from the statement X^3 > Y
Hence INSUFFICIENT



Now combine both:
For x is between 0 and 1 , the root will increase the value and cube(x^3) will decrease the value

from case 1a) and 2) we can conclude that x >y

For x >1 , from case b itself we know x > y


Hence ANS: C

from 1
X^.5 >y
squaring both sides
X > Y^2
Y = 3 therefore let X be 10 ..satisfies----1
Y = 1/2 ...let X be 1/3 still satisfies-----2
from 1 X>Y from 2 Y>X
hence not suffiecient
from 2nd statement
X^3 >Y
Y = 3 therefore let X be 2 ..satisfies----3
Y = 2 therefore let X be 3 ..satisfies----4
from 3 X<Y from 4.. X>Y
hence not sufficent
now combine both .. we can validate eq 1 and with second statement
while eq 2 doesnt fit in
Hence C. Hop i am clear

(C)