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

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Intern
Joined: 10 Jul 2010
Posts: 34
If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?  [#permalink]

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26 Sep 2010, 01:02
2
4
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Question Stats:

88% (01:08) correct 13% (02:09) wrong based on 364 sessions

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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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File comment: This questions seems to be wrong, altough it is from GMAT PreP!!

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Math Expert
Joined: 02 Sep 2009
Posts: 51122
If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?  [#permalink]

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26 Sep 2010, 02:55
3
1
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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26 Sep 2010, 07:41
thanks a ton Bunuel
Manager
Joined: 06 Aug 2010
Posts: 173
Location: Boston

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03 Oct 2010, 18:20
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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8665
Location: Pune, India
If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?  [#permalink]

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14 May 2015, 21:35
1
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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Karishma
Veritas Prep GMAT Instructor

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Joined: 09 Mar 2016
Posts: 1215
Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?  [#permalink]

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25 Nov 2018, 03:07
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 ?
Re: If xy=1, what is the value of 2(x+y)^2/2(x-y)^2? &nbs [#permalink] 25 Nov 2018, 03:07
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