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Bunuel
If x and y are integers and xy ≠ 0, what is the value of \(\frac{x^{-2y}}{y^{2x}}\)?

(1) x + y = 0

(2) xy = y/x

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MAGOOSH OFFICIAL SOLUTION:
Attachment:
powersinafraction_text.png
powersinafraction_text.png [ 14.03 KiB | Viewed 13062 times ]
FAQ: How do we choose what numbers to plug in to test statement 2?

The second statement tells us that xy = y/x, which means we can't just pick any numbers. If, for example, x=2, then the statement wouldn't be possible (and remember that's not how data sufficiency works—all statements must be mathematically possible, sufficient or not). That is, any 2y≠y/2, unless y=0.

But we know that y≠0 by the statement xy≠0 at the beginning of the question, so that's not allowed.

So let's rephrase this: we need a value of x that gives us the same result whether we multiply or divide by x. The only possibilities are 1 and -1.

So we plug in x=1, then try a couple different values of y, and find that the results are insufficient, as shown in the video above.
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[quote="Bunuel"][quote="Bunuel"]If x and y are integers and xy ≠ 0, what is the value of \(\frac{x^{-2y}}{y^{2x}}\)?

So let's rephrase this: we need a value of x that gives us the same result whether we multiply or divide by x. The only possibilities are 1 and -1.



I believe the above text should say - we need a value of y (and not x) that gives us the same result whether we multiply or divide by x. The only possibilities are 1 and -1.
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If \(x\) and \(y\) are integers and \(xy\neq{0}\), what is the value of \(\frac{x^{-2y}}{y^{2x}}\)?

[1] \(x+y =0\)
[2] \(xy=\frac{y}{x}\)
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St1)
x+y = 0
or, x = -y
x^(-2y)/y^(2x) = (-y)^(-2y)/y^(-2y) = y^(-2y)/y^(-2y) =1
Sufficient.

St2) xy = y/x
or, \(x^2\)*y-y =0
or, y(\(x^2\)-1) = 0
or, y =0, or x = +/- 1
Now as xy is not equal to 0, y is not equal to 0.
x = +/- 1, value of y is not known.
hence we can not find the value of given expression.
Not sufficient.

Answer A
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Princ
If \(x\) and \(y\) are integers and \(xy\neq{0}\), what is the value of \(\frac{x^{-2y}}{y^{2x}}\)?

[1] \(x+y =0\)
[2] \(xy=\frac{y}{x}\)

Merging topics. Please check the discussion above.
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