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# If x and y are positive, which of the following must be

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Intern
Joined: 07 May 2006
Posts: 16

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If x and y are positive, which of the following must be [#permalink]

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17 Jun 2006, 15:15
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Difficulty:

55% (hard)

Question Stats:

60% (01:12) correct 40% (01:26) wrong based on 258 sessions

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If x and y are positive, which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$?

1. $$\frac{\sqrt{x+y}}{2x}$$

2. $$\frac{\sqrt{x}+\sqrt{y}}{x+y}$$

3. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-which-of-the-following-must-be-85276.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Aug 2012, 02:40, edited 5 times in total.
Edited the question and added the OA.

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SVP
Joined: 24 Sep 2005
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18 Jun 2006, 01:18
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For this question, one way is plugging in number. Another way is making use of some rules of inequalities of fractions:
for a, b, c, and d >0
for two fraction: a/b and c/d
if ad> bc, a/b> c/d and vice versa.

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Director
Joined: 07 Jun 2006
Posts: 513

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02 Jul 2006, 07:10
ns wrote:
Please explain how do we solve this question. Also are there any number properties that helps solve this question quicker?

Thanks

Note: Edited: Answer is (C) i.e. II. I had 2 windows open so mistook C to B (though I identified II as the answer).

1/sqrt(x+y)) is same as sqrt(x+y)/(x+y)

from properties, we know that if x,y > 0 then sqrt(x+y) < sqrt(x)+sqrt(y) (try plugging - rt(16)+rt(9) = 7 is more than rt(16+9) = 5)

then, II is definitely 'in' => answer can be C, E (A is ruled out because II is B is possible)

Now beginning elimination, E is ruled out as we are subtracting therefore reducing the numerator, and if Y is > X this would become -ve and we cannot say with certainty.

The answer can only be (B) i.e. II.

(Though the one angle that worries me a bit of the sign of the resultant operations...)

What is the OA?

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Intern
Joined: 03 Jun 2012
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17 Aug 2012, 14:56
Not sure about alternative methods, but plugging in numbers works relatively quickly. If you use 1 for x & y (only restriction is that they must be positive), none of the 3 options is greater than the query. While one is equal to the query, we're only asked to demonstrate that the option(s) must (ie always) be greater. So, none.

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Intern
Joined: 01 Jan 2011
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Location: Kansas, USA

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17 Aug 2012, 17:48
Plug in x= y= 1. With this, you will be able to eliminate all the answer choices except for A and C. Now, test for case A. Could it be true that for [1][/(x+y)^1/2]> [x^1/2+ y^1/2][(x+y)^1/2]? For this to be true 2(xy)^1/2 will have to be less than zero and that is not possible since x and y are both >0. Hence, answer choice C remains.

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Intern
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17 Aug 2012, 19:46
Yes. I should have included more details in my post & I would have seen my error. Sorry for any confusion I've caused. Definitely have to make sure I don't make careless errors on the test. I apologize again.

Posted from my mobile device

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Re: number property   [#permalink] 17 Aug 2012, 19:46
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