Author 
Message 
Manager
Joined: 01 Apr 2006
Posts: 180
Location: Toronto, Canada

1
This post received KUDOS
1
This post was BOOKMARKED
\(a\) , \(b\) , 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 \(c \ne 0\) ) Source: GMAT Club Tests  hardest GMAT questions REVISED VERSION OF THIS QUESTION IS HERE: m0370290.html#p1228282



CIO
Joined: 02 Oct 2007
Posts: 1218

Re: M03 DS [#permalink]
Show Tags
25 Sep 2008, 14:04
1
This post received KUDOS
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.
_________________
Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.
GMAT Club Premium Membership  big benefits and savings



Intern
Joined: 26 Apr 2010
Posts: 5

Re: M03 #19 [#permalink]
Show Tags
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
Joined: 15 Apr 2010
Posts: 192

Re: M03 #19 [#permalink]
Show Tags
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] = 11=0
since "c" is not = 0, "b" is zero to make the result "0". hence B is the answer!



Intern
Joined: 11 Jan 2010
Posts: 38

Re: M03 #19 [#permalink]
Show Tags
18 May 2010, 09:06
Question: Is abc = 0? Solution 1: a^2 = 2a => a . (a2) = 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!!!



Director
Joined: 21 Dec 2009
Posts: 583
Concentration: Entrepreneurship, Finance

Re: M03 #19 [#permalink]
Show Tags
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.
_________________
KUDOS me if you feel my contribution has helped you.



Manager
Joined: 27 May 2010
Posts: 199

Re: M03 #19 [#permalink]
Show Tags
22 Aug 2010, 05:47
A is insufficient as a=2 and B & C values are not given B is sufficient because b/c = 11 => 0
Since c cannot be zero, b = 0 ==> abc=0 Therefore answer is B



Director
Joined: 01 Feb 2011
Posts: 755

Re: M03 #19 [#permalink]
Show Tags
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 = 11 = 0. => b=0=> abc=0 Answer is B.
Posted from my mobile device



Manager
Joined: 15 Apr 2011
Posts: 64
Location: Bangalore India
Schools: LBS, HBS, ISB, Kelloggs, INSEAD

Re: M03 #19 [#permalink]
Show Tags
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 \(c \ne 0\) ) 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.
_________________
Thanks, AM



Manager
Joined: 11 Jul 2009
Posts: 167
WE: Design (Computer Software)

Re: M03 #19 [#permalink]
Show Tags
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
_________________
Kaustubh



Intern
Joined: 18 May 2011
Posts: 18

Re: M03 #19 [#permalink]
Show Tags
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



Intern
Joined: 22 May 2012
Posts: 1

Re: M03 #19 [#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 39615

Re: M03 #19 [#permalink]
Show Tags
23 May 2013, 05:08
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 \(c \ne 0\) ) 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) \(a^2 = 2a\) > \(a^22a=0\) > \(a(a2)=0\) > \(a=0\) 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) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) > \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\) > \(b=cc\) > \(b=0\). Sufficient. Answer: B.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 20 Nov 2012
Posts: 18
Location: United States
Concentration: Finance, International Business
GMAT Date: 06152013
GPA: 3.75
WE: Information Technology (Computer Software)

Re: M03 #19 [#permalink]
Show Tags
23 May 2013, 17:35
2
This post received KUDOS
(A)a(a2) = 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



Intern
Joined: 01 Mar 2013
Posts: 20

Re: M03 #19 [#permalink]
Show Tags
24 May 2013, 21:52
guptasulabh7 wrote: (A)a(a2) = 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?



Math Expert
Joined: 02 Sep 2009
Posts: 39615

Re: M03 #19 [#permalink]
Show Tags
25 May 2013, 03:13
crackgmat2013 wrote: guptasulabh7 wrote: (A)a(a2) = 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 01 Mar 2013
Posts: 20

Re: M03 #19 [#permalink]
Show Tags
25 May 2013, 11:39
Thank you Bunuel for clarifying d doubt Bunuel wrote: crackgmat2013 wrote: guptasulabh7 wrote: (A)a(a2) = 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.



Intern
Joined: 11 Aug 2013
Posts: 8
Concentration: Finance
Schools: Fuqua  Class of 0
GPA: 3.24

Re: M03 #19 [#permalink]
Show Tags
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 (stepbystep) and how do you come up with b/c being 11? Am I missing something basic here? Thanks! Posted from my mobile device



Math Expert
Joined: 02 Sep 2009
Posts: 39615

Re: M03 #19 [#permalink]
Show Tags
10 May 2014, 05:50











