is x > y?

1) sqrt(x) > y

2) cube(x) > y


Here is what i think:-

1) sqrt(x) > y

Case 1: 0<x<1

Let x = 0.36 and y = 0.50

sqrt(0.36) > 0.50

=> 0.6 > 0.5 But x=0.36 and y = 0.50 --- And 0.36 > 0.50 is false

Case 2: Let x = 16 and y=3

=> and sqrt(16) > 3 -- which is true

So Case 1 is "False" and Case 2 is "true"

Hence Statement 1 is insufficient.


Statement 2:-

cube(x) > y

Case 1:

cube(2) > 7 but 2 > 7 is "false"

Case 2:

I am NOT able to find a value of x such that 0<x<1

Probably because if cube(x) > y then x>1

Is my line of reasoning correct to decide that

Answer is (B)

Is there a way to think about this problem without substituting numnbers?

Thanks a lot.