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GmatPrep PS Question

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GmatPrep PS Question [#permalink]

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New post 02 Feb 2009, 11:46
00:00
A
B
C
D
E

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Is there a way to solve it less than 3 minutes?

OA will follow, please explain


** My apologies for not typing the question for future searches - don't you just hate all those signs? :wink: **
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Re: GmatPrep PS Question [#permalink]

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New post 02 Feb 2009, 12:36
GGUY wrote:
Is there a way to solve it less than 3 minutes?

OA will follow, please explain


** My apologies for not typing the question for future searches - don't you just hate all those signs? :wink: **



assume x=y=1 you can eliminate option I and III right away.

To find Option III

\(1/sqrt{x+y}=sqrt{x+y}/{x+y}\)
option III = \(sqrt{x}+sqrt{y}/{x+y}\)

Need to find out \(sqrt{x}+sqrt{y} > sqrt{x+y}\)

-- > square on both sides
--> \(x+y+2*sqrt{xy} > x+y\)
this is always true
so Square root of these also true

Fortunately I could solve within 3 minutes.. but will see if this answer match with OA or not.

C
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Re: GmatPrep PS Question [#permalink]

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New post 02 Feb 2009, 12:49
(1) rewrite \(\frac{1}{sqrt{x+y}}\) = \(\frac{1}{sqrt{x+y}}.\frac{sqrt{x+y}}{sqrt{x+y}}\)= \(\frac{sqrt{x+y}}{x+y}\)

(2) You do not know whether \(2x > x + y\) because \(y\) could be \(> x\) , so that leaves with II and III

(3) To test II and III check whether \(sqrt{x}+sqrt{y} > sqrt{x+y}\)

\(sqrt{x}+{sqrt{y} > {sqrt{x+y}\)
\((sqrt{x}+{sqrt{y})^2 > ({sqrt{x+y})^2\)
\(x+2{sqrt{xy} + y > x+y\)

So the left side is greater than the right side, which makes option II valid only (in option III the left side would be less than the right side)

I think doable in 2min, but hope it's right! :)

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Re: GmatPrep PS Question   [#permalink] 02 Feb 2009, 12:49
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