x > y

[j_shreyans]

Is x > y?

(1) under root x > y

(2) x^3 > y

[Brent@GMATPrepNow]

Is x > y?

(1) under root x > y

(2) x^3 > y

"under root"?

Here are a few symbols to help your post: √x ∛x ∜x

Cheers,
Brent

[j_shreyans]

Hi Brent ,

Thanks for your help!!

Is x > y?

(1) √x > y

(2) x^3 > y

[GMATGuruNY]

Is x > y?

(1) √x > y

(2) x³ > y

Statement 1: √x > y
It's possible that x=4 and y=1.
In this case, x>y.
It's possible that x=1/4 and y=1/4.
In this case, x=y.
INSUFFICIENT.

Statement 2: x³ > y
It's possible that x=4 and y=1.
In this case, x>y.
It's possible that x=4 and y=4.
In this case, x=y.
INSUFFICIENT.

Statements combined:
Statement 1: If x≤y and y<√x, then x≤y<√x, implying that x<√x.
Statement 2: If x≤y and y<x³, then x≤y<x³, implying that x<x³.
Not possible.
There is no value of x that is BOTH less than √x AND less than x³.
Since it is not possible that x≤y, it must be true that x>y.
SUFFICIENT.

The correct answer is C.

[GMATGuruNY]

Is x > y?

(1) √x > y

(2) x³ > y

Alternate way to combine the two statements:

Since we can't take the square root of a negative, statement 1 implies that x≥0.
Thus, when we combine the two statements, if y<0, we know that y<x.
Our concern is what happens when y≥0.
One approach is to memorize the shapes of some basic graphs:



Only in the yellow region is y<√x and y<x³.
The entire yellow region is below the graph of y=x, implying that y<x throughout the entire region.
Thus, combining the two statements, we know that y<x.
SUFFICIENT.

The correct answer is C.

[Matt@VeritasPrep]

Just to illustrate algebraically why some of those inequalities are not possible ...

If √x > x, then x > x², or x - x² > 0, or x(1-x) > 0.

x(1-x) > 0 implies that either x and (1-x) are both positive or that x and (1-x) are both negative.

If x and (1-x) are both positive, then 1 > x > 0. There's no way for them to both be negative, however (if x is negative, then 1-x = 1-(neg) = 1+(pos)). So we know x is positive.

If x³ > x and x is positive, we can divide both sides by x, which gives us x² > 1. Since x is positive, this implies x > 1.

So we see that √x > x implies 1 > x > 0 and x³ > x implies that x > 1. It's impossible for x to be both greater than 1 and less than 1, so there are NO values of x that satisfy BOTH inequalities.