gmatusa2010 wrote:Is X>y?

1) sqrt (X) >y

2) X^3>y

To answer this question we must have a clear idea about the relation of x with x³ and √x. Note the followings:

• 1. √x ≥ x for (0 ≤ x ≤ 1)
  2. √x ≤ x for (1 ≤ x)
  3. x³ ≤ x for (x ≤ -1) and (0 ≤ x ≤ 1)
  4. x³ ≥ x for (-1 ≤ x ≤ 0) and (1 ≤ x)

Now analyze the problem:

Statement 1: √x > y
For 0 ≤ x ≤ 1, √x ≥ x : In this case x may or may not be greater than y.
For x ≥ 1, √x ≤ x : In this case x is definitely greater than y.

Not sufficient.

Statement 2: x³ > y
For x ≤ -1 and 0 ≤ x ≤ 1, x³ ≤ x : In this case x is definitely greater than y.
For 1 ≤ x and -1 ≤ x ≤ 0, x³ ≥ x : In this case x may or may not be greater than y.

Not sufficient.

1 & 2 Together: Negative values of x are discarded due to √x. Only two regions of concern: 0 ≤ x ≤ 1 and x ≥ 1

• 1. For 0 ≤ x ≤ 1 : √x ≥ x ≥ x³ > y => x > y
  2. For x ≥ 1 : x³ ≥ x ≥ √x > y => x > y

Sufficient.

The correct answer is C.