NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 18:29 It is currently 26 Apr 2024, 18:29
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

m12#05 - DS

User avatar
ykaiim
Director
Director
Joined: 25 Aug 2007
Posts: 520
Own Kudos [?]: 5425 [0]
Given Kudos: 40
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Send PM
m12#05 - DS [#permalink] New post   12 Jun 2010, 05:40
If \(a,b,c,d\) are positive numbers, is \(\frac{a}{b}<\frac{(a+c)}{(b+d)}\)?

1. \(\frac{a}{b}<\frac{c}{d}\)
2. \(a+c<b+d\)

The OA is A, but, I have one doubt:
From S1, we can say \(\frac{a}{c}<\frac{b}{d}\)

On adding 1 on both sides, the equation becomes:
\(\frac{(a+c)}{c}<\frac{(b+d)}{d}\)
\(\frac{(a+c)}{(b+d)}<\frac{c}{d}\)

But, we are given \(\frac{a}{b}<\frac{c}{d}\)

So, S1 is insufficient. Please explain where I am missing the point.

Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests for the latest questions.
Still interested? Check out the "Best Topics" block below for better discussion and related questions.
Thank you for understanding, and happy exploring!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92948
Own Kudos [?]: 619246 [0]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Re: m12#05 - DS [#permalink] New post   12 Jun 2010, 08:47
Expert Reply
ykaiim wrote:
If \(a,b,c,d\) are positive numbers, is \(\frac{a}{b}<\frac{(a+c)}{(b+d)}\)?

1. \(\frac{a}{b}<\frac{c}{d}\)
2. \(a+c<b+d\)

The OA is A, but, I have one doubt:
From S1, we can say \(\frac{a}{c}<\frac{b}{d}\)

On adding 1 on both sides, the equation becomes:
\(\frac{(a+c)}{c}<\frac{(b+d)}{d}\)
\(\frac{(a+c)}{(b+d)}<\frac{c}{d}\)

But, we are given \(\frac{a}{b}<\frac{c}{d}\)

So, S1 is insufficient. Please explain where I am missing the point.


Given: \(a\), \(b\), \(c\), and \(d\) are positive numbers. Question: is \(\frac{a}{b}<\frac{(a+c)}{(b+d)}\)? --> as all numbers are positive we can safely crossmultiply --> is \(ab+ad<ab+bc\)? --> is \(ad<bc\)? is \(\frac{a}{b}<\frac{c}{d}\)?

(1) Directly gives the answer. Sufficient.
(2) Not sufficient.

Answer: A.

We can get the answer the way you were doing too: \(\frac{a}{b}<\frac{c}{d}\) --> \(\frac{d}{b}<\frac{c}{a}\) --> add 1 to both sides --> \(\frac{d}{b}+1<\frac{c}{a}+1\) --> \(\frac{d+b}{b}<\frac{c+a}{a}\) --> \(\frac{a}{b}<\frac{a+c}{b+d}\).

Hope it helps.
User avatar
ykaiim
Director
Director
Joined: 25 Aug 2007
Posts: 520
Own Kudos [?]: 5425 [0]
Given Kudos: 40
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Send PM
Re: m12#05 - DS [#permalink] New post   12 Jun 2010, 08:56
Bunuel,

Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92948
Own Kudos [?]: 619246 [0]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Re: m12#05 - DS [#permalink] New post   12 Jun 2010, 09:21
Expert Reply
ykaiim wrote:
Bunuel,

Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?


You are not missing anything.

If we are given that \(a\), \(b\), \(c\), and \(d\) are positive numbers and \(\frac{a}{b}<\frac{c}{d}\), then \(\frac{(a+c)}{(b+d)}<\frac{c}{d}\) (the inequality you derived) is true. We are just asked to derive another inequality (the one I derived) not this one.

Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests for the latest questions.
Still interested? Check out the "Best Topics" block above for better discussion and related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: m12#05 - DS [#permalink]
12 Jun 2010, 09:21
Moderator:
Bunuel
Math Expert
92948 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions