It is currently 12 Dec 2017, 16:01

### 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

# m02 #20

Author Message
Senior Manager
Joined: 08 Jun 2010
Posts: 386

Kudos [?]: 102 [0], given: 13

Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32

### Show Tags

08 Dec 2011, 06:34

Kudos [?]: 102 [0], given: 13

Manager
Joined: 05 Sep 2012
Posts: 78

Kudos [?]: 17 [0], given: 17

### Show Tags

07 Dec 2012, 05:32
Yep. Got it.

I guess the funda here is hidden in the Q stem.
Does not matter if the answer is a negative or positive as long as it is an integer.

Kudos [?]: 17 [0], given: 17

Current Student
Joined: 24 Apr 2012
Posts: 38

Kudos [?]: 18 [0], given: 35

GMAT 1: 660 Q48 V33
WE: Information Technology (Health Care)

### Show Tags

20 Dec 2012, 14:19
the question boils down to P*sqrt(q) is an integer only when p is an integer and q is a perfect sq

Statement 1 - both are satisfied so sufficient
statemnt 2 not sufficient

Kudos [?]: 18 [0], given: 35

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135392 [0], given: 12691

### Show Tags

16 Oct 2013, 10:54
nikhil007 wrote:
If we had to add a constraint that is $$p\frac{q}{\sqrt{q}}$$ a positive integer ? (or negative integer for that matter), will answer still be the same?
coz Sqroot Q = +-P as per statement 1. isn't it?

Yes and in this case the answer would be C.
_________________

Kudos [?]: 135392 [0], given: 12691

Current Student
Joined: 20 Jul 2013
Posts: 14

Kudos [?]: 24 [0], given: 309

Concentration: General Management
GMAT 1: 710 Q49 V37

### Show Tags

05 Dec 2013, 06:02
Bunuel wrote:

(1) $$q = p^2$$ --> $$p=\sqrt{q}$$ --> $$p\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}}=q=integer$$. Sufficient.

Bunuel, Isn't $$p\frac{q}{\sqrt{q}}$$ a mixed number. Equivalent to $$p + \frac{q}{\sqrt{q}}$$

Kudos [?]: 24 [0], given: 309

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135392 [0], given: 12691

### Show Tags

05 Dec 2013, 06:19
sathishkumar434 wrote:
Bunuel wrote:

(1) $$q = p^2$$ --> $$p=\sqrt{q}$$ --> $$p\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}}=q=integer$$. Sufficient.

Bunuel, Isn't $$p\frac{q}{\sqrt{q}}$$ a mixed number. Equivalent to $$p + \frac{q}{\sqrt{q}}$$

Nope. It's $$p*\frac{q}{\sqrt{q}}$$.
_________________

Kudos [?]: 135392 [0], given: 12691

Manager
Status: Work hard in silence, let success make the noise
Joined: 25 Nov 2013
Posts: 157

Kudos [?]: 84 [0], given: 84

Location: India
Concentration: Finance, General Management
GMAT 1: 540 Q50 V15
GMAT 2: 640 Q50 V27
GPA: 3.11
WE: Consulting (Computer Software)

### Show Tags

06 Dec 2013, 06:49
IMO A...

Statement 1 is sufficient since q is +ve (given) and p can be +ve or -ve, the expression would always be an integer.
Statement 2 is not sufficient since we do not know whether (sqrt)q is integer or non-integer. So the output of the expression could be integer or non-integer.

So, A.
_________________

Sahil Chaudhary
If you find this post helpful, please take a moment to click on the "+1 KUDOS" icon.
My IELTS 7.5 Experience
From 540 to 640...Done with GMAT!!!
http://www.sahilchaudhary007.blogspot.com

Kudos [?]: 84 [0], given: 84

Re: m02 #20   [#permalink] 06 Dec 2013, 06:49

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by

# m02 #20

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.