very hard ds, pls, help

IMO D

given; x>0 and y>0
asked x^3>y

statement 1) sqrtx>y=>+-x>y^2;since x & y is positive

x>y^2

now x^3>x(for positive value of x)
hence X^3>y
suff

Statement 2)
x>y
since X^3>X therefore X^3>y
suff.

hence D

OA pls

the question, "if x and y are positive, is x^3>y ?", does it specify if it is an integer or not. Currently, the way the problem is stated, the answer would be A. If X =.2 and y=.1 then x^3 = .008 and if it x=2 and y =1 then x^3=8.

(1) sqrt(x) > y

let x = 4 and y = 1
then sqrt(x) > y

and x^3 = 64 is also > 1

but if x = 0.64 and y = 0.6
then sqrt (x) = 0.8 > 0.6

BUT x^3 = 0.26 is NOT greater than 0.6

Therefore statement 1 is not sufficient...

(2) x > y

if x = 2 and y = 1
then x > y and x^3 = 8 is greater than y = 1

but if x = 0.64 and y = 0.6
then x^3 = 0.26 is not greater than y = 0.6

therefore statement 2 is not sufficient


Statements 1 and 2 taken together

if x = 4 and y = 1
then x > y and sqrt(x) > y

and x^3 is also > y

now if x = 0.64 and y = 0.6

then x > y and sqrt (x) = 0.8 > y (0.6)

but x^3 = 0.26 < y (0.6)

Therefore options 1 and 2 taken together are also not suffient.

Hence, the answer is E