Bunuel wrote:

If x and y are positive numbers, is x > y?

(1) \(\frac{x^3}{y} < 1\)

(2) \(\frac{\sqrt[3]{x}}{y} < 1\)

Statement 1\(\frac{x^3}{y} < 1\)

x and y are positive numbers so we can cross multiply.

\(x^3 < y\)

x = 2, y = 9

\(x^3\) = 8. This is less than y. So statement (1) is satisfied.

Since x=2, and y =9,

x is not more than y.

x = (1/2), y = (1/7)

\(x^3\) = (1/8). This is less than y. So statement (1) is satisfied.

Since x=(1/2), and y =(1/7),

x is more than y.

Statement (1) is not sufficientStatement 2\(\frac{\sqrt[3]{x}}{y} < 1\)

\({\sqrt[3]{x}} < y\)

x and y are positive numbers so we can raise each of them to cubic powers

so x < \(y^3\)

x = 7, y = 2

\(y^3\) = 8. x =7. x is less than \(y^3\). So statement (2) is satisfied.

Since x=7, and y =9,

x is more than y.

x = 1, y = 3

\(y^3\) = 27. x =1. x is less than \(y^3\). So statement (2) is satisfied.

Since x=1, and y =3,

x is not more than y.

Statement (2) is not sufficient

Combining Statement 1 and 2\(x^3 < y\)

x < \(y^3\)

Since x is positive we can cube it. so \(x^3 < y^6\)

Now \(x^3 < y\) and \(x^3 < y^6\)

This means that we can take \(x^3 < y\)

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

E is the answer.