Abhijit K wrote:If x and y are positive, is x³ > y?
(1) √x > y
(2) x > y

First, since we know x and y are both positive, we don't have to worry about 0 or negative numbers.

Statement 1 is tricky. If √x > y, then it seems that x³ would also be greater than y, but things are not so simple, because fractions can work differently from whole numbers.

If x = 1/4 and y = 1/3, then √x = 1/2 and 1/2 > 1/3.

However (1/4)³ is way less than 1/3.

So Statement 1 is insufficient.

Statement 2 also seems to indicate that x³ > y. For instance, if x = 2 and y = 1, then x³ > y.

But once again given fractional values of x and y, x could be greater than y and x³ could be less than y. For instance if x = 1/2 and y = 1/3, then x³ = (1/2)³ = 1/8 which is less than 1/3.

So Statement 2 is insufficient.

Combining the statements we can still find examples that make x³ > y and fractions that fit both statements and result in x³ < y.

For example if x = 1/2 and y = 1/3, then x > y and √x > y, but (1/2)³ = 1/8 which is less than 1/3.

So the statements in combination are still insufficient.

Choose E.

This is a classic number-picking question.

S1: square both sides to get x > y^2

x = 2, y^2 = 1
x^3 = 8 y = 1, so YES, x^3 is bigger

x = 1/2, y^2 = 1/4
x^3 = 1/8, y= 1/2, so NO, x^3 is not bigger

S1: Not Sufficient


S2: Reuse x = 2, y = 1
x^3 = 8, y = 1, YES, x^3 is bigger

x = 1/2, y = 1/3
x^3 = 1/8, y = 1/3, NO, x^3 is not bigger


Together: We know: x > y^2 and x > y
x = 2, y = 1 satisfies both
x^3 = 8, y = 1, YES, x^3 is bigger

x = 1/2, y = 1/3 satisfies both (y^2 = 1/9, and 1/2 > 1/9)
x^3 = 1/8, y = 1/3, NO, x^3 is not bigger

Still not sufficient.

Answer is E

Abhijit K wrote:If x and y are positive, is x³ > y?
(1) √x > y
(2) x > y

Question stem: Is y < x³?
This problem can also be solved GRAPHICALLY.


The red region in the figure above consists of all points such that y < x³.
Question stem, rephrased:
Is y a positive value in the red region?

Statement 1: y < √x
Statement 2: y < x

The dark purple region in the figure above consists of all points such that y < √x and y < x.
We need to determine whether this dark purple region lies completely below the graph of y = x³.

Statements combined:
Putting together y < √x, y < x and y = x³, we get:


The dark green region in the figure above consists of all points such that y < √x and y < x.
Some of the dark green region lies ABOVE y = x³ (implying that y can be GREATER than x³), while some of the dark green region lies BELOW y = x³ (implying that y can be LESS than x³).
Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.