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# If r*s*t>0; is r^2*s^3*t^-4>0 1)s>r^2 2)s>t^-4

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Director
Joined: 17 Oct 2005
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If r*s*t>0; is r^2*s^3*t^-4>0 1)s>r^2 2)s>t^-4 [#permalink]

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07 Jan 2006, 17:20
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If r*s*t>0; is r^2*s^3*t^-4>0

1)s>r^2

2)s>t^-4

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Intern
Joined: 13 Nov 2005
Posts: 29

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07 Jan 2006, 18:37
I'd say D.

r^2 is always > 0
Similarly t^4 is always > 0
Hence if s is positive then, r^2*s^3*t^-4>0

From (1) - s is greater than square of a number - so s > 0
From (2) - s > 1/t^4 - since t^4 > 0, s > 0

Hence either statement gives enough information to say that the expression will be > 0

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SVP
Joined: 28 Dec 2005
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07 Jan 2006, 20:40
Agree with D

From statement 1: s>r^2. r can be positive or negative, but r^2 will always be positive. Since s>r^2, s will also always be positive. And t^4 will be positive as well. So three positive terms will have a positive product.

From statement 2: s*t^4 > 1 . t^4 will always be positive, and to meet this condition, s will also have to be positive. And r^2 will always be positive. Again, enough to answer the question

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Director
Joined: 17 Oct 2005
Posts: 921

Kudos [?]: 280 [0], given: 0

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07 Jan 2006, 22:57
great work guys

OA is D

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07 Jan 2006, 22:57
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