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

If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \(\sqrt{x}>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(2) \(x>y\) --> if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is always YES: \(y<x^3\).

But:

For \(0<x<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\)

Answer: E.

Hope it's clear.