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Can someone confirm the OA, it is B according to me as well;

For the following question the answer is B.

If x>1 and y>1, is x<y?

(1) \(\frac{x^2}{xy+x}<1\) --> reduce by \(x\): \(\frac{x}{y+1}<1\) --> cross multiply, notice that we can safely do that since \(y+1>0\): \(x<y+1\) --> if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient.

(2) \(\frac{xy}{y^2-y}<1\) --> reduce by \(y\): \(\frac{x}{y-1}<1\) --> cross multiply, notice that we can safely do that since \(y-1>0\): \(x<y-1\) --> \(x+1<y\) (\(y\) is more than \(x\) plus 1) --> \(y>x\). Sufficient.

Can someone confirm the OA, it is B according to me as well;

If (2) is \(\frac{xy}{y^2}-y<1\), then the answer is indeed E.

If x>1 and y>1, is x<y?

(1) \(\frac{x^2}{xy+x}<1\) --> reduce by \(x\): \(\frac{x}{y+1}<1\) --> cross multiply, notice that we can safely do that since \(y+1>0\): \(x<y+1\) --> if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient.

(2) \(\frac{xy}{y^2}-y<1\) --> reduce by \(y\): \(\frac{x}{y}-y<1\) --> \(x<y^2\) --> if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient.

(1)+(2) If \(x=2\) and \(y=3\), then the answer is YES but If \(x=2\) and \(y=\sqrt{3}\approx{1.7}\) (notice that this set of numbers satisfy both statements), then the answer is NO. Not Sufficient.

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