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# m02 #13

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Manager
Joined: 10 Jul 2010
Posts: 196
Followers: 1

Kudos [?]: 20 [0], given: 12

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02 Feb 2012, 05:02
x and y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of x?
A. 2
B. 4
C. 8
D. 12
E. 16

X and Y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of X ?

OA is
[Reveal] Spoiler:
A

Question: How did the explanation get -3 in the numerator??

http://gmatclub.com/tests/m02#expl13

Last edited by Bunuel on 02 Feb 2012, 12:21, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 36597
Followers: 7093

Kudos [?]: 93448 [0], given: 10563

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02 Feb 2012, 12:20
menacel wrote:
X and Y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of X ?

OA is
[Reveal] Spoiler:
A

Question: How did the explanation get -3 in the numerator??

http://gmatclub.com/tests/m02#expl13

x and y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of x?
A. 2
B. 4
C. 8
D. 12
E. 16

$$y= \sqrt{64}=8$$;

$$\frac{-4*8^3+64*8}{96}$$ --> reduce by 8*4: $$\frac{-8^2+16}{3}=\frac{-48}{3}=-16$$ (the way they got 3 in nominator: $$-4*8^3+8^3=1*8^3-4*8^3=-3*8^3$$);

Next, $$x^2-10x=-16$$ --> $$x=2$$ or $$x=8$$. Hence, the minimum possible value of x is 2.

Hope it's clear.
_________________
Manager
Joined: 10 Jul 2010
Posts: 196
Followers: 1

Kudos [?]: 20 [0], given: 12

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03 Feb 2012, 06:28
Bunuel wrote:
menacel wrote:
X and Y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of X ?

OA is
[Reveal] Spoiler:
A

Question: How did the explanation get -3 in the numerator??

http://gmatclub.com/tests/m02#expl13

x and y are positive integers. If $$y= sqrt(64)$$ and $$x^2-10x=\frac{-4y^3+64y}{96}$$ , what is the minimum possible value of x?
A. 2
B. 4
C. 8
D. 12
E. 16

$$y= \sqrt{64}=8$$;

$$\frac{-4*8^3+64*8}{96}$$ --> reduce by 8*4: $$\frac{-8^2+16}{3}=\frac{-48}{3}=-16$$ (the way they got 3 in nominator: $$-4*8^3+8^3=1*8^3-4*8^3=-3*8^3$$);

Next, $$x^2-10x=-16$$ --> $$x=2$$ or $$x=8$$. Hence, the minimum possible value of x is 2.

Hope it's clear.

much clearer thanks a lot!
Re: m02 #13   [#permalink] 03 Feb 2012, 06:28
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# m02 #13

Moderator: Bunuel

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