It is currently 19 Sep 2017, 14:05

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

GmatPrep PS Question

Author Message
Manager
Joined: 26 Jun 2007
Posts: 104

Kudos [?]: 78 [0], given: 0

Show Tags

02 Feb 2009, 11:46
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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

GP.JPG [ 49.53 KiB | Viewed 1207 times ]

Kudos [?]: 78 [0], given: 0

SVP
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1032 [0], given: 5

Location: New York

Show Tags

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

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

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
_________________

Smiling wins more friends than frowning

Kudos [?]: 1032 [0], given: 5

Intern
Joined: 25 Jan 2009
Posts: 4

Kudos [?]: [0], given: 0

Show Tags

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!

Kudos [?]: [0], given: 0

Re: GmatPrep PS Question   [#permalink] 02 Feb 2009, 12:49
Display posts from previous: Sort by