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Bunuel
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Bunuel
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can we rewrite xy > x/y => xy^2 >x => y^2 > x/x => y^2 >1 or y>1?

No, that's wrong. We cannot multiply xy > x/y by y because we don't know its sign. If y<0, then we'd get xy^2 < x (flip the sign when multiplying by negative value) but if y > 0, then we'd have xy^2 > x (keep the sign when multiplying by positive value). For the same reason we cannot reduce by x.
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I think this is a high-quality question and I agree with explanation. When i solved this problem i used exactly the same thought process - absolutely no deviation- as given in the explanation! Wow kudos to me :-)
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Bunuel

(1) \(0 \lt y \lt 1\). From this statement it follows that \((y-\frac{1}{y}) \lt 0\).

I don't understand this rule and the logic behind it. Where can I learn more information about it.

Thank you
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Bunuel

(1) \(0 \lt y \lt 1\). From this statement it follows that \((y-\frac{1}{y}) \lt 0\).

I don't understand this rule and the logic behind it. Where can I learn more information about it.

Thank you

Since y is from 0 to 1, then 1/y is more than 1. Thus y - 1/y = (positive number less than 1) - (number more than 1) < 0.

Hope it's clear.
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Bunuel
Is \(xy \gt \frac{x}{y}\)?


(1) \(0 \lt y \lt 1\)

(2) \(xy \gt 1\)

Statement 1: implies that \(y\) is positive but nothing mentioned about \(x\). Insufficient

Statement 2: we need values of \(x\) & \(y\) to determine the value of \(\frac{x}{y}\). this statement implies that both \(x\) & \(y\) are of same sign but their values cannot be deduced. Insufficient

Combining 1 & 2, we know that \(y\) is positive so \(x\) is also positive and as \(y<1\) so it is reciprocal of any positive integer

so if \(x\) is integer for e.g \(x=2\) and \(y=\frac{1}{5}\), then \(xy=2*\frac{1}{5}=0.4\) but \(\frac{x}{y}=2/\frac{1}{5}=10\). Hence \(xy<\frac{x}{y}\)

and if \(x\) is not integer for e.g \(x=\frac{2}{5}\) and \(y=\frac{1}{2}\), then \(xy=\frac{2}{5}*\frac{1}{2}=0.2\) but \(\frac{x}{y}=\frac{2}{5}/\frac{1}{2}=0.8\). Hence \(xy<\frac{x}{y}\)

So we have a NO for our question stem. Sufficient

Option C
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I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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