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# If a > 0, b > 0 and c > 0, is a(b - c) = 0?

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Director
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If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]

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21 Aug 2009, 19:24
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If a > 0, b > 0 and c > 0, is a(b - c) = 0?

(1) b - c = c - b
(2) b/c = c/b
[Reveal] Spoiler: OA

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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Last edited by Bunuel on 30 Jun 2013, 08:03, edited 2 times in total.
Edited the question and added the OA.
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Re: if a>0, b>0, c>0, is a(b-c)=0? [#permalink]

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21 Aug 2009, 19:45
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a>0, b>0, c>0
for 1), b-c=c-b ==>> 2b=2c ==>>b=c
therefore a(b-c)=0, suff

for 2), b/c=c/b ==>> b^2=c^2
because b>0, c>0, ==>>b=c
therefore a(b-c)=0, suff
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Re: if a>0, b>0, c>0, is a(b-c)=0? [#permalink]

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25 Aug 2009, 10:32
[quote="tejal777"]if a>0, b>0, c>0, is a(b-c)=0?

I. b-c = c-b
II. b/c = c/b

A(B-C) = 0 and as a>0 thus it is POSSIBLE ONLY WHEN B = C

from 1

b-c-c+b = 0 thus 2b = 2c thus b=c...suff

from 2

b/c = c/b all +ve thus b^2 = c^2 thus b=c....suff

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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:33
For statement 2

we get $$b^2=c^2$$ then $$b^2 - C^2 = 0$$ >> $$(b-c) (b+c) = 0$$

Why is this manipulation incorrect?
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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:37
fozzzy wrote:
For statement 2

we get $$b^2=c^2$$ then $$b^2 - C^2 = 0$$ >> $$(b-c) (b+c) = 0$$

Why is this manipulation incorrect?

That is not incorrect.

You get $$(b-c) (b+c) = 0$$ so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0
Hence $$a(b-c)=0$$ => sufficient
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Director
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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:52
Zarrolou wrote:

That is not incorrect.

You get $$(b-c) (b+c) = 0$$ so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0
Hence $$a(b-c)=0$$ => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case?
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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:56
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fozzzy wrote:
Zarrolou wrote:

That is not incorrect.

You get $$(b-c) (b+c) = 0$$ so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0
Hence $$a(b-c)=0$$ => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case?

I do not know what you mean by "redundant information", but yes: we know that $$b+c$$ cannot be zero, so the other term ($$b-c$$) must be zero.

All we get from $$b^2-c^2=0$$ in this problem is that $$b-c=0$$
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Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]

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30 Jun 2013, 08:09
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If a > 0, b > 0 and c > 0, is a(b - c) = 0?

Is $$a(b - c) = 0$$? --> is $$a=0$$ or $$b-c=0$$? Since given that $$a > 0$$, then the questions basically asks whether $$b-c=0$$.

(1) b - c = c - b --> $$2b-2c=0$$ --> $$b-c=0$$. Sufficient.

(2) b/c = c/b --> $$b^2=c^2$$ --> $$(b-c)(b+c)=0$$ --> $$b+c=0$$ or $$b-c=0$$ but since $$b$$ and $$c$$ are positive, then $$b+c>0$$. Therefore $$b-c=0$$. Sufficient.

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Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]

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25 Sep 2014, 00:30
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Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0?   [#permalink] 25 Sep 2014, 00:30
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