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If x does not equal y, and xy does not equal 0, then when x [#permalink]

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02 Nov 2010, 01:48

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69% (02:01) correct
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\(\frac{x+y}{x-y}\)

If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to:

If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to:

A) \(-\frac{x+y}{x-y}\)

B) \(\frac{x-y}{x+y}\)

C) \(\frac{x+y}{x-y}\)

D) \(\frac{x^2-y^2}{xy}\)

E) \(\frac{y-x}{x+y}\)

Please explain your answers if you can. OA will be provided in a few hours.

When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}\).

If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to:

A) \(-\frac{x+y}{x-y}\)

B) \(\frac{x-y}{x+y}\)

C) \(\frac{x+y}{x-y}\)

D) \(\frac{x^2-y^2}{xy}\)

E) \(\frac{y-x}{x+y}\)

Please explain your answers if you can. OA will be provided in a few hours.

When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}\).

Answer: A.

Yes, that is correct. But can you explain how the denominator ends up as y-x instead of x-y? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A".

If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to:

A) \(-\frac{x+y}{x-y}\)

B) \(\frac{x-y}{x+y}\)

C) \(\frac{x+y}{x-y}\)

D) \(\frac{x^2-y^2}{xy}\)

E) \(\frac{y-x}{x+y}\)

Please explain your answers if you can. OA will be provided in a few hours.

When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}\).

Answer: A.

Yes, that is correct. But can you explain how the denominator ends up as y-x instead of x-y? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A".

If x does not equal y, and xy does not equal 0, then when x is replaced by 1/x and y is replaced by 1/y everywhere in the expression above, the resulting expression is equivalent to...

I think the formatting of your solution (inside the spoiler tags) became illegible, but I may be able to guess at the mistake. When we replace 'x' with '1/x', and 'y' with '1/y', we have the following:

I think you then rewrote this as \(\left( \frac{1}{x} + \frac{1}{y} \right) (x - y)\), and if so, that isn't right. If that's what you did, you were probably trying to apply the rule "when you divide by a fraction, multiply by its reciprocal". There are two problems with trying to do that right away: we aren't dividing by a simple fraction here, so we can't use that rule (in the denominator, we're subtracting one fraction from another), and regardless, x-y is not the reciprocal of (1/x) - (1/y). If those terms were reciprocals of each other, their product would be 1, and if you multiply them out, you'll see their product is not 1.

As a first step, since we have fractions in both our numerator and denominator, I'd just multiply the top and bottom by xy. We then need to multiply by -1 on top and bottom only because of how the question designer chose to write the correct answer choice:

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: If x does not equal y, and xy does not equal 0, then when x [#permalink]

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29 Oct 2014, 23:44

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Re: If x does not equal y, and xy does not equal 0, then when x [#permalink]

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14 Aug 2016, 18:10

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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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