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

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If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink] New post 21 Aug 2009, 18:24
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A
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C
D
E

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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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Last edited by Bunuel on 30 Jun 2013, 07:03, 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] New post 21 Aug 2009, 18:45
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Answer is D.

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] New post 25 Aug 2009, 09:32
[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] New post 30 Jun 2013, 06:33
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] New post 30 Jun 2013, 06:37
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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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink] New post 30 Jun 2013, 06:52
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] New post 30 Jun 2013, 06:56
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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] New post 30 Jun 2013, 07:09
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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.

Answer: D.
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Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0?   [#permalink] 30 Jun 2013, 07:09
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