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# M03 #19

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Manager
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15 Sep 2008, 12:36
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$$a$$ , $$b$$ , 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

Source: GMAT Club Tests - hardest GMAT questions

REVISED VERSION OF THIS QUESTION IS HERE: m03-70290.html#p1228282
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25 Sep 2008, 14:04
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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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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.
Manager
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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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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.

Award me with Kudos+1, if this helps!! Cheers!!!
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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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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
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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

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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 $$c \ne 0$$ )

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

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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Thanks,
AM

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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
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Kaustubh

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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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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
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23 May 2013, 05:08
1
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Expert's post
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 $$c \ne 0$$ )

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

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

Thanks,
John.

BELOW IS REVISED VERSION OF THIS QUESTION:

Is $$abc = 0$$ ?

In order $$abc = 0$$ to be true at least one of the unknowns must be zero.

(1) $$a^2 = 2a$$ --> $$a^2-2a=0$$ --> $$a(a-2)=0$$ --> $$a=0$$ or $$a=2$$. If $$a=0$$ then the answer is YES but if $$a=2$$ then $$abc$$ may not be equal to zero (for example consider: $$a=2$$, $$b=3$$ and $$c=4$$). Not sufficient.

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$ --> $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$ --> $$b=c-c$$ --> $$b=0$$. Sufficient.

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23 May 2013, 17:35
2
KUDOS
(A)a(a-2) = 0

a = 0 or a =2

In sufficient
b/c = ((a+b)^2 - (a+b)^2)/(a+b)^2

b/c=0

b=0

Sufficient

Ans:B
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24 May 2013, 21:52
guptasulabh7 wrote:
(A)a(a-2) = 0

a = 0 or a =2

In sufficient
b/c = ((a+b)^2 - (a+b)^2)/(a+b)^2

b/c=0

b=0

Sufficient

Ans:B

Is this the right simplification for statement 1:-

I got only 1 value for a i.e. a=2

a square = 2a
a * a = 2a
a = 2

if its not correct simplification please explain, why?
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25 May 2013, 03:13
crackgmat2013 wrote:
guptasulabh7 wrote:
(A)a(a-2) = 0

a = 0 or a =2

In sufficient
b/c = ((a+b)^2 - (a+b)^2)/(a+b)^2

b/c=0

b=0

Sufficient

Ans:B

Is this the right simplification for statement 1:-

I got only 1 value for a i.e. a=2

a square = 2a
a * a = 2a
a = 2

if its not correct simplification please explain, why?

Yes, that's not correct.

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) $$a^2=a$$ by $$a$$ you assume, with no ground for it, that $$a$$ does not equal to zero thus exclude a possible solution (notice that both $$a=2$$ AND $$a=0$$ satisfy the equation).

Hope it's clear.
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25 May 2013, 11:39
Thank you Bunuel for clarifying d doubt

Bunuel wrote:
crackgmat2013 wrote:
guptasulabh7 wrote:
(A)a(a-2) = 0

a = 0 or a =2

In sufficient
b/c = ((a+b)^2 - (a+b)^2)/(a+b)^2

b/c=0

b=0

Sufficient

Ans:B

Is this the right simplification for statement 1:-

I got only 1 value for a i.e. a=2

a square = 2a
a * a = 2a
a = 2

if its not correct simplification please explain, why?

Yes, that's not correct.

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) $$a^2=a$$ by $$a$$ you assume, with no ground for it, that $$a$$ does not equal to zero thus exclude a possible solution (notice that both $$a=2$$ AND $$a=0$$ satisfy the equation).

Hope it's clear.
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09 May 2014, 22:14
I have a couple of questions. I get what is done with condition 1, but with condition 2 I don't properly follow your thinking.

Could someone show me how the statement 2 is simplified (step-by-step) and how do you come up with b/c being 1-1? Am I missing something basic here?

Thanks!

Posted from my mobile device
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10 May 2014, 05:50
financebunny wrote:
I have a couple of questions. I get what is done with condition 1, but with condition 2 I don't properly follow your thinking.

Could someone show me how the statement 2 is simplified (step-by-step) and how do you come up with b/c being 1-1? Am I missing something basic here?

Thanks! :)

Posted from my mobile device

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$.

Factor out $$\sqrt{c}$$ from the numerator and apply $$a^2+2ab+b^2=(a+b)^2$$ to the denominator: $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$

Reduce by $$(a+b)^2$$: $$b=c-c$$ --> $$b=0$$. Sufficient.

Does this make sense?
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Re: M03 #19   [#permalink] 10 May 2014, 05:50
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# M03 #19

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