It is currently 22 Nov 2017, 01:01

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

There was a prob with my calc the first time around. I've

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Director
Director
avatar
Joined: 12 Jun 2006
Posts: 531

Kudos [?]: 167 [0], given: 1

There was a prob with my calc the first time around. I've [#permalink]

Show Tags

New post 02 Mar 2007, 00:14
00:00
A
B
C
D
E

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.

There was a prob with my calc the first time around. I've since corrected this prob.

if m>0 and n>0 is m+x/n+x > m/n?

stmnt 1: m is greater than n
stmnt 2: x is greater than 0

m+x/n+x > m/n =

m+x/n+x - m/n > 0 =

n(m+x)/n(n+x) - m(n+x)/n(n+x) > 0 = nm + nx - mn - mx/n^2 + nx =

nx - mx/n^2 + nx = -mx/n^2

Rephrased, the question reads: is -mx/n^2 > 0?

I picked b. The stem states that m and n are greater than 0. Statement 2 says that x is greater than 0. If m, n and x are all greater than 0 the equation must be greater than 0 too. The answer is C. Where did I go wrong?

Last edited by ggarr on 02 Mar 2007, 04:40, edited 5 times in total.

Kudos [?]: 167 [0], given: 1

Manager
Manager
avatar
Joined: 12 May 2006
Posts: 175

Kudos [?]: 72 [0], given: 0

 [#permalink]

Show Tags

New post 02 Mar 2007, 03:27
I guess there is a small problem in the calculation below. Following is my explaination:

m+x/n+x > m/n =

m+x/n+x - m/n > 0 =

[n(m+x) - m(n+x)]/n(n+x) > 0

[n(m+x) - m(n+x)] > 0

mn + nx - mn - mx > 0

x(n-m) > 0

=> x > 0 & n>m OR x< 0 & n<m

However both of these does not match with the given statements hence the answer should be E.

Kudos [?]: 72 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 29 Jan 2007
Posts: 439

Kudos [?]: 69 [0], given: 0

Location: Earth
 [#permalink]

Show Tags

New post 02 Mar 2007, 11:38
I agree answer has to be C.

I did it this way.

We know that a/b = c/d
then

(a-b)/b = (c-d)/d

If you apply that
Question becomes
Is (m+x-n-x)/(n+x) > (m-n)/n

so

Is (m-n)/(n+x) > (m-n)/n

Stmt 1
m > n ...so numerators are positive...but X is not known to be -ve or +ve...

so NOT SUFF

Stmt 2
x >0..so Denominator on left is greater than that on right...
but Is m-n negative or positive? Not known so we can NOT tell if left > right ( example here is 2 < 5 but -2 > -5... so we need to know the sign)

so NOT SUFF.

If you combine 1 and 2...its SUFF so answer C.

Kudos [?]: 69 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 29 Jan 2007
Posts: 439

Kudos [?]: 69 [0], given: 0

Location: Earth
 [#permalink]

Show Tags

New post 02 Mar 2007, 11:59
grepro
you had it all correct except your last reasoning.

you simplified it to

Is x(n-m) > 0 ?

You can answer this question when you know
Sign of X and Sign of (n-m)

Stmt 1 and 2 combined give you that information. So answer is C.

Kudos [?]: 69 [0], given: 0

  [#permalink] 02 Mar 2007, 11:59
Display posts from previous: Sort by

There was a prob with my calc the first time around. I've

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.