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# M03-19

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:20
Expert's post
6
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BOOKMARKED
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Difficulty:

25% (medium)

Question Stats:

74% (00:55) correct 26% (01:35) wrong based on 189 sessions

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Is $$abc = 0$$?

(1) $$a^2 = 2a$$

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
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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.

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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.

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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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.

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.

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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.

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

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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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27 May 2015, 18:18
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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.

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

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

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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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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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09 Nov 2017, 21:12
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Expert's post
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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Re: M03-19   [#permalink] 09 Nov 2017, 21:12
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# M03-19

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