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If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

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5va
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If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   26 Sep 2010, 02:02
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If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

A. 2
B. 4
C. 8
D. 16
E. 32

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PROPER VERSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html

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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   26 Sep 2010, 03:55
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If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

A. 2
B. 4
C. 8
D. 16
E. 32

\(\frac{2(x+y)^2}{2(x-y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{x^2+2+y^2}{x^2-2+y^2}\) --> we cannot find the value of this fraction without some additional info.

For example if \(x=2\) and \(y=\frac{1}{2}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{25}{9}\);

But if if \(x=4\) and \(y=\frac{1}{4}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{17^2}{15^2}\).

So you are right there is something wrong with this question.

I think the question should be: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16\).

If so, then the answer would be D.

Hope it helps.
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5va
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   26 Sep 2010, 08:41
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thanks a ton Bunuel :)
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   03 Oct 2010, 19:20
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I had this problem today, but it's actually this:

2^[(x+y)^2] / 2^[(x-y)^2]

It won't show up correctly using LATEX.

You simplify it to 2^[(x+y)^2 - (x-y)^2], and:

\((x+y)^2 - (x-y)^2 = x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)\)
\(= 2xy + 2xy = 4xy\)

And, since xy = 1, you're left with 2^4 = 16.
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   14 May 2015, 22:35
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TehJay wrote:
I had this problem today, but it's actually this:

It won't show up correctly using LATEX.

You simplify it to 2^[(x+y)^2 - (x-y)^2], and:

\((x+y)^2 - (x-y)^2 = x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)\)
\(= 2xy + 2xy = 4xy\)

And, since xy = 1, you're left with 2^4 = 16.


Assuming the question is as you mentioned:

\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)

Given that xy = 1, if you are asked a unique value of the given expression, it means it will have the same value for all x and y under the constraint xy = 1. So just put x = 1, y = 1, you get

\(\frac{2^{(1+1)^2}}{2^{(1-1)^2}}\)

\(2^4\)

16
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   25 Nov 2018, 04:07
VeritasKarishma wrote:
TehJay wrote:
I had this problem today, but it's actually this:

It won't show up correctly using LATEX.

You simplify it to 2^[(x+y)^2 - (x-y)^2], and:

\((x+y)^2 - (x-y)^2 = x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)\)
\(= 2xy + 2xy = 4xy\)

And, since xy = 1, you're left with 2^4 = 16.


Assuming the question is as you mentioned:

\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)

Given that xy = 1, if you are asked a unique value of the given expression, it means it will have the same value for all x and y under the constraint xy = 1. So just put x = 1, y = 1, you get

\(\frac{2^{(1+1)^2}}{2^{(1-1)^2}}\)

\(2^4\)

16



VeritasKarishma , so in official GMATprep question we are dealing with exponents you mean ? :?
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   12 Jan 2019, 14:38
5va just wasted 20 minutes, trying to solve the issue. Can you please edit the question for other people?
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink] New post   23 Jul 2019, 08:28
Expert Reply
5va wrote:
If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

A. 2
B. 4
C. 8
D. 16
E. 32


PROPER VERSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html

This Question is Locked Due to Poor Quality
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The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
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Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? [#permalink]
23 Jul 2019, 08:28
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