M03 #19 : GMAT Club Tests

M03 #19

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Re: M03 #19 [#permalink]

19 May 2011, 21:22

jjhko wrote:

a , b , and c are integers. Is abc = 0 ?

1. a^2 = 2a
2. \frac{b}{c} = \frac{(a+b)^2}{a^2+2ab+b^2} - 1 ( a \ne -b and c \ne 0 )

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Can't seem to figure out the explanation that I saw on the test.

Thanks,
John.

I think the answer is B as st 2 gives us b=0 because C \ne 0. so abc = 0.
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Re: M03 #19 [#permalink]

20 May 2011, 05:47

\frac{b}{c} = \frac{(a+b)^2}{a^2+2ab+b^2} - 1 ( a \ne -b and c \ne 0 )
(a+b)^2=a^2+2ab+b^2
therefore frac{(a+b)^2}{a^2+2ab+b^2}=1
\frac{(a+b)^2}{a^2+2ab+b^2} - 1 =0
so 2 is sufficient
so Answer=B
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Re: M03 #19 [#permalink]

21 May 2011, 12:34

1) A = 0 or 2 therefore insuff
2) B is sufficient since b/c = 1-1 => 0 and C != 0, therefore B = 0
B
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Re: M03 #19 [#permalink]

22 May 2011, 11:49

Interpretation:
Statement: a=0 or b=0 or c=0
all three a,b,c =0?

option 1: insufficient since a=0 or a=2
option 2: sufficient since b=0

Ans: B

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Re: M03 #19 [#permalink]

22 May 2011, 23:16

I missed the - 1 in the question ...

It gets mixed with conditions given at the side of question...

Anyways B is correct
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Re: M03 #19 [#permalink]

23 May 2012, 22:13

As we know a^2 + 2ab +b^2 is(a+b)^2
Therefore the expression (a+b)^2/(a^2 +2ab + b^2) can be reduced to 1
So substituting this in our equation we get
b/c=1 - 1 =0
which implies b=0, which satisfies our condition

Re: M03 #19 [#permalink] 23 May 2012, 22:13

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