is x^3>y>

[alex.gellatly]

If x and y are positive, is x^3>y?
1. sqroot(x)>y
2. x>y

The OA is E, but I got D. Could someone please explain. Thanks

[Birottam Dutta]

Hey, x and y are positive but the factor here is that they are not integers. They can be fractions and that is what the question is asking.

(1) sqrt (x)>y.

Let x=1/4. and y=1/3.

sqrt (1/4) = 1/2 >1/3 or y.

but x =1/4 is less than 1/3. so x^3 = 1/64 is also < y. So, this is not sufficient.

(2) x>y.

Using same logic, let x=1/2 and y = 1/3. Again here, x^3 is less than y. So this is not sufficient.

Combining both, sqrt (x)>y and x>y.

Let x = 1/4 and y = 1/5.

here, both x and sqrt (x) are greater than y.

but x^3 = 1/64 is less than 1/5.

So, even this is not sufficient.

All of these conditions are sufficient if x and y are integers but not sufficient for fractions.

So, E is the correct answer.

[tutorphd]

The two statements together are not sufficient because of the weird behavior of numbers between 0 and 1:

If 0 < x < 1, raising it to higher power brings the result closer to zero, so sqrt(x)>y and x>y do not guarantee that x^3 (which is brought closer to zero) will remain larger than y.

If we knew that x>1, there is no weirdness. The fact that x>y quarantees that x^3=x*x*x>y because the factors x are larger than 1 and they scale up/increase the result with each multiplication.