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

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14 Dec 2012, 10:18

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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.

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

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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 _________________

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

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17 Dec 2012, 04:48

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Expert's post

mridulparashar1 wrote:

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

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]

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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]

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18 Nov 2014, 04:47

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

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14 Jan 2015, 14:37

Expert's post

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

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