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GMATPREP is a(b-c)=0?

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Source: — Data Sufficiency |
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by cramya » Mon Dec 15, 2008 12:42 pm
Another way to look at stm II

b^2 = c^2

b^2-c^2 = 0

(b+c) (b-c) = 0

since b>0 and c>0 b+c cannot be 0 so b-c=0

SUFF

or

In general b^2=c^2 could mean b = - c or b = c but since its given b>0 and c>0 b=c i.e. b-c=0

SUFF
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by pbanavara » Mon Dec 15, 2008 6:50 pm
cramya wrote:Another way to look at stm II

b^2 = c^2

b^2-c^2 = 0

(b+c) (b-c) = 0

since b>0 and c>0 b+c cannot be 0 so b-c=0

SUFF

or

In general b^2=c^2 could mean b = - c or b = c but since its given b>0 and c>0 b=c i.e. b-c=0

SUFF
Totally - It's very easy to get stuck into the thought that b2=c2 => +/-b=+/-c - One glance at the question stem will reveal that all are positive so b=c.
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by vittalgmat » Mon Dec 15, 2008 11:36 pm
so is the OA D ??

I did the following for stmt 1:
b -c = c -b
=> 2(b-c) = 0
=> (b-c) = 0
multiplying by a both sides
a(b-c) = 0

Sufficient.

Is this correct? esp the * by a

thanks
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by cramya » Mon Dec 15, 2008 11:42 pm
=> (b-c) = 0
multiplying by a both sides
a(b-c) = 0
OA should be D) imo

Vittal,
I would say yes! Once u proved b-c = 0 u can skip the multiplying by a part since a(b-c) will always be 0 if either a or b-c is 0
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by ddo » Wed Dec 17, 2008 3:22 am
Answer is D.

Thanks to all! I missed the part that they are all positive
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