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# Problem 8 Advanced 800

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Intern
Joined: 09 Apr 2012
Posts: 10

Kudos [?]: [0], given: 3

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13 May 2012, 20:58
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?

Kudos [?]: [0], given: 3

Kaplan GMAT Instructor
Joined: 25 Aug 2009
Posts: 644

Kudos [?]: 306 [0], given: 2

Location: Cambridge, MA

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14 May 2012, 22:22
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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Kudos [?]: 306 [0], given: 2

Re: Problem 8 Advanced 800   [#permalink] 14 May 2012, 22:22
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# Problem 8 Advanced 800

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