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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? [#permalink] New post 14 Dec 2012, 08:28
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a, b, and c are positive, is a > b?

(1) a/(b+c) > b/(a+c)
(2) b + c < a
[Reveal] Spoiler: OA

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Re: a, b, and c are positive, is a > b? [#permalink] New post 14 Dec 2012, 09:10
daviesj wrote:
a, b, and c are positive, is a > b?

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


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}
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Re: a, b, and c are positive, is a > b? [#permalink] New post 14 Dec 2012, 09:18
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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.

Answer: D.
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Re: a, b, and c are positive, is a > b? [#permalink] New post 17 Dec 2012, 01:10
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] New post 17 Dec 2012, 03:48
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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.

Hope it's clear.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: a, b, and c are positive, is a > b? [#permalink] New post 29 Apr 2013, 12:06
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

D
Re: a, b, and c are positive, is a > b?   [#permalink] 29 Apr 2013, 12:06
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