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m02 #13 [#permalink] New post 02 Feb 2012, 06:02
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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, 13:21, edited 1 time in total.
Added the answer choices
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Re: m02 #13 [#permalink] New post 02 Feb 2012, 13:20
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


Always post answer choices for PS questions. Original question should read:

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.

Answer: A.

Hope it's clear.
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Re: m02 #13 [#permalink] New post 03 Feb 2012, 07:28
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


Always post answer choices for PS questions. Original question should read:

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.

Answer: A.

Hope it's clear.



much clearer thanks a lot!
Re: m02 #13   [#permalink] 03 Feb 2012, 07:28
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