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Problem 8 Advanced 800

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Problem 8 Advanced 800 [#permalink] New post 13 May 2012, 19:58
If -1/2<=x<=-1/3 and -1/4<=y<=-1/5, what is the minimum value of x(y^2)?

A. - 1/75

B. - 1/50

C. -1/48

D. -1/32

E. -1/16

[Reveal] Spoiler:
OA : D


The OA is calculated in the following manner

x-min = -1/2
y-min = -1/4 => y^2-min = 1/16
Hence, xy^2-min = -1/32

But shouldn't y^2-min be 1/25, making xy^2-min = -1/50?
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Re: Problem 8 Advanced 800 [#permalink] New post 14 May 2012, 21:22
Expert's post
aztec04 wrote:
If -1/2<=x<=-1/3 and -1/4<=y<=-1/5, what is the minimum value of x(y^2)?

A. - 1/75

B. - 1/50

C. -1/48

D. -1/32

E. -1/16

[Reveal] Spoiler:
OA : D


The OA is calculated in the following manner

x-min = -1/2
y-min = -1/4 => y^2-min = 1/16
Hence, xy^2-min = -1/32

But shouldn't y^2-min be 1/25, making xy^2-min = -1/50?
I don't have a copy of that book handy, but here is the the deal: x is negative, and Y^2 is positive. That means that we want the MAXIMUM value of Y^2 to get the minimum value of the product, not the minimum! The maximum positive y^2 (1/16) times the minimum value of X (-1/2) will get the correct minimum value.

Hope this helps!
_________________

Eli Meyer
Kaplan Teacher
http://www.kaptest.com/GMAT

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Problem 8 Advanced 800

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