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Official Solution:

Question asks: is \(xy \gt \frac{x}{y}\)? Or: is \(x(y-\frac{1}{y}) \gt 0\)?

(1) \(0 \lt y \lt 1\). From this statement it follows that \((y-\frac{1}{y}) \lt 0\). Now, if \(x \gt 0\) then \(x(y-\frac{1}{y}) \lt 0\) but if \(x \lt 0\) then \(x(y-\frac{1}{y}) \gt 0\). Not sufficient.

(2) \(xy \gt 1\). If \(x=y=2\) then the answer is YES but if \(x=4\) and \(y=\frac{1}{2}\) then the answer is NO. Not sufficient.

(1)+(2) Since from (1) \(y \gt 0\) and from (2) \(xy \gt 1\) then \(x \gt 0\). So, we have that \(x \gt 0\) and \((y-\frac{1}{y}) \lt 0\) (from 1), which means that their product \(x(y-\frac{1}{y}) \lt 0\). Sufficient.

Answer: C

Explanation for C :- is \(x(y-\frac{1}{y}) \gt 0\)? or is \(x\frac{(y^2-1)}{y} \gt 0\)?

According to statement (1) y is positive fraction less than 1 but greater than 0 and according to statement (2) xy is positive. which means when we combine statement (1) and (2) we get x must be postive.

To answer this question; is \(x\frac{(y^2-1)}{y} \gt 0\)? we can rewrite this as is \(x\frac{(y-1)(y+1)}{y} \gt 0\)?

Note that everything on the lefthand side will be positive except for (y-1) hence the answer to this question is \(x\frac{(y^2-1)}{y} \gt 0\)? will be NO hence together the statements are sufficient to answer this question. Correct answer is Option C

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Manish

"Only I can change my life. No one can do it for me"