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M03-19

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New post 16 Sep 2014, 00:20
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New post 16 Sep 2014, 00:20
Official Solution:


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

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) 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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New post 27 May 2015, 07:23
Bunuel wrote:
Official Solution:


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

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) 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


I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=-b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.).

so it should be E.

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New post 27 May 2015, 07:33
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suhasreddy wrote:
Bunuel wrote:
Official Solution:


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

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) 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


I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=-b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.).

so it should be E.


That's not true. If a+b were 0, then \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) would not be correct for any real a, b, and c.

When having something like x^2/x=y you can ALWAYS reduce by x and write x = y.

The answer is B.
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New post 27 May 2015, 11:35
Bunuel wrote:
Official Solution:

(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


Can someone explain the second step in this simplification process. It looks like factoring out the \sqrt{c}, but somehow \sqrt{c} became c. Is there some rule that allows this?

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New post 27 May 2015, 17:24
The square root of c needs to be squared too

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New post 27 May 2015, 17:48
I guess what I'm saying is that I've never seen this operation before. I believe it, because I basically get the same answer the long way:

(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\))
...and then the foil method

If it's a valid shortcut then I wonder if there is a proof or something for it so I know when I can use it in other situations.

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New post 27 May 2015, 18:18
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I guess what I'm saying is that I've never seen this operation before. I believe it, because I basically get the same answer the long way:

(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\))
...and then the foil method

If it's a valid shortcut then I wonder if there is a proof or something for it so I know when I can use it in other situations.


This is like the example below:

(2*5+3*5)^2 = 5^2 * (2+3) = 25*5 = 125

The way you are mentioning the equation is also correct but instead of the traditional foil method , try to see it this way:

(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\))

Take sqroot(c) common from both the terms on RHS of the equation above. sqroot (c)*sqroot (c) = c as sqroot (any number)= number ^0.5 (and x^p * x^q = x^ (p+q)).

So after doing this you get, sqroot (c) * sqroot (c) * (a+b)^2 = c * (a+b)^2. This is what Bunuel has done.

Last edited by ENGRTOMBA2018 on 10 Jun 2015, 19:02, edited 1 time in total.

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New post 27 May 2015, 21:11
Thanks, that clears it up!

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New post 31 May 2015, 01:12
B
(1) Gives a= 2,0 Insuff
(2) Gives b = 0 or a=-b but a cannot be = -b as then equation's value will become undefined
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New post 09 Nov 2017, 15:04
Bunuel wrote:
Is \(abc = 0\)?


(1) \(a^2 = 2a\)

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


Why in (1) \(a^2 = 2a\), I cannot divide both sides by 'a' and get \(a = 2\)?
\(a^2 = 2a\) -> \((1/a) * a^2 = 2a * (1/a)\) -> \(a=2\)

Is it wrong to do this way?

Edit: Answering my own question with a passage from Manhattan's Guide Book #2:
"The GMAT will often attempt to disguise quadratic equations by putting them in forms that do not quite look like the traditional form o f \(ax^2+ bx - c = 0\)."

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New post 09 Nov 2017, 21:12
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guireif wrote:
Bunuel wrote:
Is \(abc = 0\)?


(1) \(a^2 = 2a\)

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


Why in (1) \(a^2 = 2a\), I cannot divide both sides by 'a' and get \(a = 2\)?
\(a^2 = 2a\) -> \((1/a) * a^2 = 2a * (1/a)\) -> \(a=2\)

Is it wrong to do this way?

Edit: Answering my own question with a passage from Manhattan's Guide Book #2:
"The GMAT will often attempt to disguise quadratic equations by putting them in forms that do not quite look like the traditional form o f \(ax^2+ bx - c = 0\)."


Yes, it's wrong. You cannot reduce this by a, because you'll loose a possible root a = 0. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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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] 09 Nov 2017, 21:12
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