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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]

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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, 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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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]

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New post 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

D
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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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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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New post 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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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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New post 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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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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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]

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

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

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New post 30 Nov 2016, 21:15
Whoo!!! another super tricky question which can take you off the hooks during exam.
The simplest way to look at this question is we have 3 conditions a>0, b>0 and c>0 so thy are positive numbers but can be integers or fraction. (always proceed with this thought process - even if irrelevant here, it helps in the long run).
Question asks if a(b-c)=0 now we know a>0 thus a not equal to 0 hence b-c=0 or B=C

Statement 1
B-c=0
B=c Sufficient

Statement 2
c^2=B^2
|C|=|B|
since they are both greater than 0 hence c=b
Suffcienet
Answer is D.

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

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New post 06 Feb 2017, 07:26
Bunuel wrote:
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.


Nice Explanation Bunuel. Thanks for sharing your approach.
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Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0?   [#permalink] 06 Feb 2017, 07:26
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