saikiranjadav wrote:

Is p > q ?

(1) p^2 > q^2

(2) p^3 > q^3

Condition 1) \(p^2 - q^2 > 0\)

\((p+q)(p-q) > 0\)

\(p + q > 0\) implies \(p - q > 0\) or \(p > q\).

\(p - q < 0\) implies \(p - q < 0\) or \(p < q\).

Thus we have \(p = 2\), \(q = 1\) and \(p = -2\), \(q = -1\).

Thus this condition is not sufficient.

Condition 2) \(p^3 > q^3\)

\(p^3 - q^3 = (p-q)(p^2+pq+q^2) > 0\)

Without loss of generality, we can assume \(q \ne 0\).

\(p^2 + pq + q^2\)

\(= q^2 \{ \frac{p^2}{q^2} + \frac{p}{q} + 1 \}\)

\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + 1 \}\)

\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + \frac{1}{4} + \frac{3}{4} \}\)

\(= q^2 \{ ( \frac{p}{q} + \frac{1}{2} )^2 + \frac{3}{4} \} \ge \frac{3}{4} q^2 > 0\)

Thus we have \(p^2 + pq + q^2 > 0\).

Since \(p^3 - q^3 > 0\), \(p^2 + pq + q^2 > 0\) and \(p^3 - q^3 = (p-q)(p^2 + pq + q^2) > 0\), we have \(p - q > 0\) or \(p > q\).

This is sufficient.

Answer: B

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare

The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.

"Only $99 for 3 month Online Course"

"Free Resources-30 day online access & Diagnostic Test"

"Unlimited Access to over 120 free video lessons - try it yourself"