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

Question

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

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

knightofdelta · Answer

1. a/(b+c) > b/(a+c) =a(a+c) > b(b+c) This means a > b {sufficient} 2. b+c < a = b < a-c This means a is still > b despite subtracting c {sufficient}

Bunuel · Answer

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

D

WoundedTiger · Answer

Hi Bunuel, Could please explain why you have assumed a\leq{b} and not simply ab and hence we take it as a\leq{b} . Had the Question been is a>=b,perhaps we would have assumed a 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

Bunuel · Answer

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.

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.

Hope it's clear.

yezz · Answer

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

D

Cadaver · Answer

Statement 1 :- a/(b+c)>b/(a+c)

Add 1 both the sides 
(a+b+c)/(b+c)>(a+b+c)/(a+c)
1/(b+c)>1/(a+c)
a+c>b+c

As all are positive so its safe to presume a>b

Statement 2 :- b+c<a

As all are positive so its safe to presume a>b

Hence D

EMPOWERgmatRichC · Answer

Hi All,

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

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

anairamitch1804 · Answer

Posting official solution of this problem.

bumpbot · Answer

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

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

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