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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Official Answer and Stats are available only to registered users. Register/Login.

_________________

Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690): http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

(1) a/(b+c) > b/(a+c). Suppose \(a\leq{b}\), then the numerator (n1) of LHS (a) is less than or equal to the numerator (n2) of RHS (b) AND the denominator (d1) of LHS (b+c) is more than or equal to the denominator (d2) of RHS (a+c). But if this is the case (if \(n_1\leq{n_2}\) and \(d_1\geq{d_2}\)), then \(\frac{n_1}{d_1}<\frac{n_2}{d_2}\). Therefore our assumption was wrong, which means that a>b. Sufficient.

(2) b + c < a. a is greater than b plus some positive number, thus a is greater than b. Sufficient.

Re: a, b, and c are positive, is a > b? [#permalink]

Show Tags

17 Dec 2012, 02:10

Bunuel wrote:

a, b, and c are positive, is a > b?

(1) a/(b+c) > b/(a+c). Suppose \(a\leq{b}\), then the numerator (n1) of LHS (a) is less than or equal to the numerator (n2) of RHS (b) AND the denominator (d1) of LHS (b+c) is more than or equal to the denominator (d2) of RHS (a+c). But if this is the case (if \(n_1\leq{n_2}\) and \(d_1\geq{d_2}\)), then \(\frac{n_1}{d_1}<\frac{n_2}{d_2}\). Therefore our assumption was wrong, which means that a>b. Sufficient.

(2) b + c < a. a is greater than b plus some positive number, thus a is greater than b. Sufficient.

Answer: D.

Hi Bunuel,

Could please explain why you have assumed \(a\leq{b}\) and not simply a<b.I guess same reason will apply for denominator as well. Is it because the Q asked whether a>b and hence we take it as \(a\leq{b}\).

Had the Question been is a>=b,perhaps we would have assumed a<b only.

Please confirm

If we try solving algebraically from st 1

a.a+a.c> b.b+b.c a2+ac-b2-bc >0

we end up with a condition

(a-b)(a+b-c)>0

We end up with a condition either a>b or a+b>c.This implies either both terms are negative or both positive.

What do we interpret from this

Thanks Mridul
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

(1) a/(b+c) > b/(a+c). Suppose \(a\leq{b}\), then the numerator (n1) of LHS (a) is less than or equal to the numerator (n2) of RHS (b) AND the denominator (d1) of LHS (b+c) is more than or equal to the denominator (d2) of RHS (a+c). But if this is the case (if \(n_1\leq{n_2}\) and \(d_1\geq{d_2}\)), then \(\frac{n_1}{d_1}<\frac{n_2}{d_2}\). Therefore our assumption was wrong, which means that a>b. Sufficient.

(2) b + c < a. a is greater than b plus some positive number, thus a is greater than b. Sufficient.

Answer: D.

Hi Bunuel,

Could please explain why you have assumed \(a\leq{b}\) and not simply a<b.I guess same reason will apply for denominator as well. Is it because the Q asked whether a>b and hence we take it as \(a\leq{b}\).

Had the Question been is a>=b,perhaps we would have assumed a<b only.

Please confirm

Yes, that's correct.

We are asked whether a>b. Assume that a>b is not true, so assume \(a\leq{b}\). Now, if after some reasoning based on a/(b+c) > b/(a+c) we'll get that \(a\leq{b}\) cannot hold true, then we'll get that our assumption (\(a\leq{b}\)) was wrong, thus it must be true that a>b.

mridulparashar1 wrote:

a.a+a.c> b.b+b.c a2+ac-b2-bc >0

we end up with a condition

(a-b)(a+b-c)>0

We end up with a condition either a>b or a+b>c.This implies either both terms are negative or both positive.

What do we interpret from this

Thanks Mridul

We can do this way too. The problem with your solution is that you factored a^2+ac-b^2-bc incorrectly: \(a^2+ac-b^2-bc=(a-b)(a+b+c)\).

So, we'd have: \(\frac{a}{b+c} > \frac{b}{a+c}\) --> \(a^2+ac>b^2+bc\) --> \((a-b)(a+b+c)>0\) Now, since we are give that a, b, and c are positive, then a+b+c>0, thus a-b>0 --> a>b. Sufficient.

Re: a, b, and c are positive, is a > b? [#permalink]

Show Tags

29 Apr 2013, 13:06

a, b, and c are positive, is a > b?

(1) a/(b+c) > b/(a+c) (2) b + c < a

from 1

The inequality boils down to a(a+c) > b(b+c) (a^2 +ac) - (b^2+bc) >0 . since the 3 unknown are given as +ve and the only difference between the values inside each of the 2 brackets is the values of a and b therefore a>b

from 2

a-b>c ,since c is +Ve therefore we can re write the ineq as a-b>0 therefore a>b

Re: a, b, and c are positive, is a > b? [#permalink]

Show Tags

18 Nov 2014, 04:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

This DS question can be dealt with in a variety of ways: Algebra, TESTing VALUES to discover patterns, or Number Properties.

We're told that A, B and C are POSITIVE. We're asked if A > B. This is a YES/NO question.

The "crux" of this question is the "C" and how it effects the relationship between A and B in the given inequalities. We can use Number Properties to get to the correct answer.

Fact 1: (A/B+C) > (B/A+C)

In these two fractions, notice that the only difference is that the "A" and the "B" switch positions. The "C" shows up in each denominator in the same capacity. Since the variables are all POSITIVE, they cannot be 0 or negative, so the C essentially has NO IMPACT on the inequality. Making the C "really small" or "really big" won't impact how A and B relate to one another.

For example... A = 2 B = 1 C = 1 2/2 > 1/3

and

A = 2 B = 1 C = 100 2/101 > 1/102

....have the same end results. We'll be left with...

A/B > B/A

This also means that A CANNOT equal B (otherwise the two fractions would equal one another). This only holds true when A > B The answer to the question is ALWAYS YES

Fact 2: B + C < A

Since B and C are both POSITIVE, A MUST be at least "C" greater than B. The answer to the question is ALWAYS YES. Fact 2 is SUFFICIENT