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M03 #19

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M03 #19 [#permalink] New post 15 Sep 2008, 11:36
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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 )

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B

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REVISED VERSION OF THIS QUESTION IS HERE: m03-70290.html#p1228282
[Reveal] Spoiler: OA
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Re: M03 DS [#permalink] New post 25 Sep 2008, 13:04
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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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Re: M03 #19 [#permalink] New post 17 May 2010, 06:24
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, 17:26
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 18 May 2010, 08:06
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, 09:24
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 22 Aug 2010, 04:47
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, 04:57
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, 20:22
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

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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] New post 20 May 2011, 04:47
\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] New post 22 May 2011, 10:49
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] New post 23 May 2012, 21:13
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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Re: M03 #19 [#permalink] New post 23 May 2013, 04:08
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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.

Answer: B.
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Re: M03 #19 [#permalink] New post 23 May 2013, 16:35
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(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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Re: M03 #19 [#permalink] New post 24 May 2013, 20:52
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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Re: M03 #19 [#permalink] New post 25 May 2013, 02:13
Expert's post
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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Re: M03 #19 [#permalink] New post 25 May 2013, 10:39
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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Re: M03 #19 [#permalink] New post 09 May 2014, 21:14
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! :)

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Re: M03 #19 [#permalink] New post 10 May 2014, 04:50
Expert's post
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! :)

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(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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M03 #19   [#permalink] 10 May 2014, 04:50
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