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Sajjad1994
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Official Explanation

We can reword the question as either “Is \(x^2y > \frac{x^2}{y}\)?” or “Is \(x^2y < \frac{x^2}{y}\)?” A valid assumption is \(x^2y ≠ \frac{x^2}{y}\) since the original question asks for the greater one of the two. Let's work on “Is \(x^2y > \frac{x^2}{y}\)?” Dividing this inequality by the positive value \(x^2\) changes the question to “Is \(y > \frac{1}{y}\)?” Since the inequality in the question no longer has \(x^2\), any further information about \(x\) is irrelevant. So, Statement (1) is not required.

From Statement (2) alone, we have that \(y < –1\). Hence, \(y\) is negative. Dividing both sides of this inequality by \(–y\), a positive number, yields \(–1 < \frac{1}{y}\). Combining the inequalities \(y < –1\) and \(–1 < \frac{1}{y}\) yields \(y < –1 < \frac{1}{y}\) from which we have \(y < \frac{1}{y}\). Hence, Statement (2) alone answers the question. Hence, the answer is (B). The answer to the question is “No. \(y\) is not greater than \(\frac{1}{y}\).” So, the answer to the original question is “No. \(x^2y\) is not greater than \(\frac{x^2}{y}\),” or “\(\frac{x^2}{y}\) is the greater one.”

Answer: B

stne
SajjadAhmad
Which one of the expressions \(x^2y\) and \(\frac{x^2}{y}\) is greater?

(1) \(y > x\)

(2) \(y < –1\)

Source: Nova GMAT

Hi SajjadAhmad,

Already proved B is also Insuff. Please check OA and provide OE
For statement two
2)Y=-2 and X= 0 both are same
y=-2 and x= -3 second exp. is greater.
Hence B is infuff.

Correct answer C
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