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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]
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 _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]
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? _________________
Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]
30 Jun 2013, 06:56
1
This post received KUDOS
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 \(b-c=0\) _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]
30 Jun 2013, 07:09
Expert's post
3
This post was BOOKMARKED
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.
Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]
24 Sep 2014, 23:30
Hello from the GMAT Club BumpBot!
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