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when they both negative.. |x|>|y| (because x/y>1) in this case stat1 fails 2x-2y=1 --> this is not possible at all 2x-2y ---> always negative.. (when both x and y are -ve)
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I don't agree with this solution with reference to statement 2, One cannot conveniently multiply both sides of the equation with y assuming y is positive. If y is negative, the inequality sign changes.

I don't agree with this solution with reference to statement 2, One cannot conveniently multiply both sides of the equation with y assuming y is positive. If y is negative, the inequality sign changes.

There is no multiplication by \(y\): it should be \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> as \(x-y=\frac{1}{2}\) --> \(\frac{1}{2y}>0\) --> \(y>0\) --> as \(\frac{x}{y}>1\), they both have the same sign, thus \(x>0\).

Answer: C.

Complete solution:

Are x and y both positive?

(1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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