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M03 #19 [#permalink] New post 15 Sep 2008, 12:36
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Question Stats:

75% (01:43) correct 25% (00:18) wrong based on 1 sessions
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 )

[Reveal] Spoiler: OA
B

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

Thanks,
John.
[Reveal] Spoiler: OA
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Re: M03 DS [#permalink] New post 22 Sep 2008, 08:34
Can anyone help? :)
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Re: M03 DS [#permalink] New post 25 Sep 2008, 14:04
Hi.
From S1 we see that a is either 0 or 2. Insufficient.
S2. You have to notice that we need to simplify the fraction in S2:
\frac{b}{c} = \frac{(a+b)^2}{a^2+2ab+b^2} - 1
\frac{b}{c} = \frac{(a+b)^2}{(a+b)^2} - 1
\frac{b}{c} = 1 - 1 = 0
b = 0

Hope this helps.
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Re: M03 DS [#permalink] New post 25 Sep 2008, 14:38
dzyubam wrote:
Hi.
From S1 we see that a is either 0 or 2. Insufficient.
S2. You have to notice that we need to simplify the fraction in S2:
\frac{b}{c} = \frac{(a+b)^2}{a^2+2ab+b^2} - 1
\frac{b}{c} = \frac{(a+b)^2}{(a+b)^2} - 1
\frac{b}{c} = 1 - 1 = 0
b = 0

Hope this helps.


he is missing -1 in S2.

otherwise good one dzyubam
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Re: M03 #19 [#permalink] New post 17 May 2010, 05:28
The answer is B right?
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Re: M03 #19 [#permalink] New post 17 May 2010, 06:35
Can some one explain Statement 1? Thanks!
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Re: M03 #19 [#permalink] New post 17 May 2010, 07:24
nshah12 wrote:
Can some one explain Statement 1? Thanks!


Statement one is saying that the square of A is equal to 2 x A. So this can be 2 or 0.

0 squared = 0 * 0 = 0
2 * 0 = 0

AND

2 squared = 2 * 2 = 4
2 * 2 = 4

So statement implies that A is either 0 or 2, therefore we can't tell if ABC = 0.

Hope that helps.
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Re: M03 #19 [#permalink] New post 17 May 2010, 08:10
Yes you're right. You can see the OA in the spoiler after clicking "Reveal".
tiruraju wrote:
The answer is B right?

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Re: M03 #19 [#permalink] New post 17 May 2010, 16:36
B for the reasons already stated.

b^2 + 2bc + c^2 = (b + c)^2
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Re: M03 #19 [#permalink] New post 17 May 2010, 18:26
this is how I approached it:

1) First condition gives us 2 possible values of a => a=0 or 2

2) Second condition comes as [b][/c] = 1-1=0

since "c" is not = 0, "b" is zero to make the result "0". hence B is the answer!
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Re: M03 #19 [#permalink] New post 17 May 2010, 23:53
I will go with option B
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Re: M03 #19 [#permalink] New post 18 May 2010, 06:35
At first I forgot the value 0 of a in (1). Luckily it's correct to choose B. But should be careful next time.
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Re: M03 #19 [#permalink] New post 18 May 2010, 09:06
Question: Is abc = 0?

Solution 1:
a^2 = 2a => a . (a-2) = 0 => a = 0 or a = 2

When a = 0 => abc = 0
When a = 2 => abc may or may not be zeros depending on the values of b and c.

Thus, solution 1 does not give a clear yes or no answer and is not sufficient.

Solution 2:
Solving the given equation: a^2 + 2ab + b^2 = (a + b)^2
Thus, b/c = 1 - 1 = 0
=> b = 0

When b = 0 => abc = 0
Thus, Solution 2 gives a clear answer that, yes, abc = 0.

Therefore, the correct answer choice is B.

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Re: M03 #19 [#permalink] New post 18 May 2010, 10:24
Enough solution has been given from the previous posts.
(1) a= 0 or 2...insufficient
(2) b=0...sufficient to answer whether abc = 0 (yes).
B, off course is the answer.
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Re: M03 #19 [#permalink] New post 18 May 2010, 20:10
S1 ,a=2 or 0 not sufficieant
S2,B=0 so sufficieant
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Re: M03 #19 [#permalink] New post 20 Aug 2010, 04:22
Its B.

Only B gives a distinct value Zero.
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Re: M03 #19 [#permalink] New post 22 Aug 2010, 05:47
A is insufficient as a=2 and B & C values are not given
B is sufficient because b/c = 1-1 => 0

Since c cannot be zero, b = 0 ==> abc=0
Therefore answer is B
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Re: M03 #19 [#permalink] New post 19 May 2011, 05:57
1. Not sufficient
As a can be 0 or 2,abc may or may not be 0.
2.Sufficient
b/c = 1-1 = 0.
=> b=0=> abc=0
Answer is B.

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Re: M03 #19 [#permalink] New post 19 May 2011, 06:39
Yey! Correct! I answered B.
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Re: M03 #19 [#permalink] New post 19 May 2011, 10:43
It's B. Took me a minute to reread statement 2. Once I realized it stated C couldn't be 0 it became much clearer.
Re: M03 #19   [#permalink] 19 May 2011, 10:43
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