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(1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\);

(1)+(2) We have \(y=positive\) and \(x=negative\), thus \(y>x\) (this can be rewritten as \(y-x>0\) or \(0>x-y\)). Now: \(LHS=\frac{1}{x-y}=\frac{1}{negative}=negative\), and \(RHS=y-x=positive\) thus \(\frac{1}{x-y}=negative<y-x=positive\). Sufficient.

In order to separate your x's and y's from each other, you need to know which term is larger. This is because you need to know whether x-y or y-x is negative (one of them will be, unless they are equal). Taking both statements together, you know that the two variables are not equal, and you can manipulate the inequality, keeping in mind that you need to flip the sign when multiplying or dividing by a negative number.
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I explained it in kind of a backwards way. The first thing you need to know in that problem is whether x and y are equal. No single statement tells you that.

Sorry for the double post, but it seems I can't edit my previous post while viewing the forum with Tapatalk.

Sent from my HTC Glacier using Tapatalk 2
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(1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\);

(1)+(2) We have \(y=positive\) and \(x=negative\), thus \(y>x\) (this can be rewritten as \(y-x>0\) or \(0>x-y\)). Now: \(LHS=\frac{1}{x-y}=\frac{1}{negative}=negative\), and \(RHS=y-x=positive\) thus \(\frac{1}{x-y}=negative<y-x=positive\). Sufficient.

Answer: C.

Why can't we do the following rephrase?

1/(x-y) < (y-x)

-1/(y-x) < (y-x)

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.

(1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\);

(1)+(2) We have \(y=positive\) and \(x=negative\), thus \(y>x\) (this can be rewritten as \(y-x>0\) or \(0>x-y\)). Now: \(LHS=\frac{1}{x-y}=\frac{1}{negative}=negative\), and \(RHS=y-x=positive\) thus \(\frac{1}{x-y}=negative<y-x=positive\). Sufficient.

Answer: C.

Why can't we do the following rephrase?

1/(x-y) < (y-x)

-1/(y-x) < (y-x)

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.

The point is that we don't know whether y-x is positive or negative. If it's positive, then yes we'd have -1 < (y-x)^2 but if it's negative, then we'd have -1 > (y-x)^2 (flip the sign when multiplying by negative value).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. _________________

With the way you put it, this question seemed like a below 500 question :\

I don't know why I took so long to decipher it was option C .

It happens to the best of people especially if the question is done during exam time.... One more thing..I think you need to look at these links for books for your preparation

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Excellent Explanation dabral Also i did it a bit differently.. I made two cases => one for Y>x and the other for Y<x hence i marked C
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It's not clear what you mean at all. The solutions above does not distinguish between rational and irrational numbers. Why/how should they? The solutions are fine and are correct generally for all real numbers.
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