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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
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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
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 Updated on: 10 Jun 2015, 19:02
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meshackb wrote:
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.

Originally posted by ENGRTOMBA2018 on 27 May 2015, 18:18.
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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New post 13 Mar 2018, 08:05
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


Bunuel,

In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared.

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New post 13 Mar 2018, 08:24
wildhorn 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


Bunuel,

In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared.

Posted from my mobile device


Not sure I can follow you...

Anyway, we are factoring \(\sqrt{c}\) from the numerator in \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\):

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

\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\).
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New post 13 Mar 2018, 08:57
Bunuel wrote:
wildhorn 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


Bunuel,

In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared.

Posted from my mobile device


Not sure I can follow you...

Anyway, we are factoring \(\sqrt{c}\) from the numerator in \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\):

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

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


Bunuel,

thank you for the explanation. I see my mistake now. I have a follow up question. Would there be a simple way to reduce the numerator if there wasn't a constant multiple being used? For example, if the numerator was (A*C^1/2+B*D^1/2)^2 rather than (A*C^1/2+B*C^1/2)^2
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New post 24 Jun 2019, 05:39
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Okay from my personal experience, whenever you get a DS question, solve the Second Statement first.

In the above question . solving the Second statment first -

Taking \sqrt{c} common from the RHS , we get
\(b =( \sqrt{c}^2 *(a+b)^2)/(a+b)^2 - c\)
=> b = \(c - c\)
=> b = 0.

Hence we see that b = 0. Therefore the product abc will be 0. We can conclude from this statement that we dont even require the 1st Statement anymore.

Therefore Option B (Statement 2 ALONE is sufficient to solve the question).


Kindly Give Kudos if you liked my explanation.
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New post 24 Jun 2019, 08:32
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'm getting confused on a particular step, when \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) is being simplified to \(b= \frac{c*(a+b)^2}{(a+b)^2} - c\).
When solving it I'm getting \(b= \frac{c*(a^2+b^2)}{(a+b)^2} - c\). And I'm not sure how \((a^2+b^2)\) is being written as \((a+b)^2\), as \((a+b)^2 = {a^2+2ab+b^2}\)
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New post 24 Jun 2019, 09:04
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banerjr1 wrote:

I'm getting confused on a particular step, when \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) is being simplified to \(b= \frac{c*(a+b)^2}{(a+b)^2} - c\).
When solving it I'm getting \(b= \frac{c*(a^2+b^2)}{(a+b)^2} - c\). And I'm not sure how \((a^2+b^2)\) is being written as \((a+b)^2\), as \((a+b)^2 = {a^2+2ab+b^2}\)


Let's do this with some intermediate steps so the math is more clear:

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

Factor out \(\sqrt{c}\) from \(a*\sqrt{c}+b*\sqrt{c}\)

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

Distribute the square to \(\sqrt{c}\) and \(a+b\) (This is where you made your mistake — we can distribute a square to terms that are being multiplied together, but not terms that are being added together. This is why we get \((a+b)^2\) and not \(a^2 + b^2\).)

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

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

Either expand \((a+b)^2\) in numerator to \(a^2+2ab+b^2\) OR factor \(a^2+2ab+b^2\) in denominator to \((a+b)^2\)

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

Cancel like terms from numerator and denominator (either \(a^2+2ab+b^2\) or \((a+b)^2\))

\(b= c - c\)

\(b= 0\)

So you're totally correct that \(a^2 + b^2\) doesn't equal \((a+b)^2\), but we should never get \(a^2 + b^2\) based on exponent rules for distributing exponents.
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New post 12 Jul 2019, 20:10
i am super confused,

Bunuel suppose a = (b* c)/b -c here as per this example we can write as a = c-c = 0

if a =(b* c)/b can we write as a = c by cancelling b rather if ab = bc can we cancel the b ???
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New post 12 Jul 2019, 22:27
(1)
\(a^2 = 2a\)
\(a = 2\)

Don't know about any other value.
Not sufficient.


(2)
\(b = (a\sqrt{c} + b\sqrt{c})^2\)/\((a + b)^2\)
\(b = {[a^2*c + b^2*c + 2abc]/[(a + b)^2]} - c\)
Taking C common from the numerator
\(b = {c*[a^2 + b^2 + 2ab]/[(a + b)^2]} - c\)
\(b = {c*[(a + b)^2]/[(a + b)^2]} - c\)
\(b = {c*[(a + b)^2]/[(a + b)^2]} - c\)
\(b = c - c\)
\(b = 0\)

if \(b = 0\), then \(abc = 0\). Hence, (2) is sufficient.

Answer B
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