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# If vmt =/ 0, is v^2m^3t^-4 > 0 ? (1) m > v^2 (2) m

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Manager
Joined: 04 Jan 2008
Posts: 114
If vmt =/ 0, is v^2m^3t^-4 > 0 ? (1) m > v^2 (2) m  [#permalink]

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06 Sep 2008, 00:40
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If vmt =/ 0, is $$v^2m^3t^-4$$ > 0 ?

(1) m > $$v^2$$
(2) m > $$t^-4$$

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SVP
Joined: 29 Aug 2007
Posts: 2420

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06 Sep 2008, 01:00
dancinggeometry wrote:
If vmt =/ 0, is $$v^2m^3(1/t^4)$$ > 0 ?

(1) m > $$v^2$$
(2) m > $$t^-4$$

i guess it is : (v^2)(m^3)(1/t^4) >0

D. since v^2 and v^4 are positives, and m is > v^2 and v^4, (v^2)(m^3)(1/t^4) > 0.
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VP
Joined: 05 Jul 2008
Posts: 1329

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06 Sep 2008, 11:40
Assuming VMT is not equal to 0

and (v ^2 m ^3)/ t ^ 4 > 0 comes down to whether m >0 or not

From (1) and (2) individually we can say that m >0

Hence D
VP
Joined: 30 Jun 2008
Posts: 1004

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Updated on: 07 Sep 2008, 00:27
Statement 1 : helps us by telling us m is +ve ( if m is greater than the square of a number, which is always positive, then m is also positive). This shows the expression is positive. Stmt 1 is suff.

Statement 2: again shows m is +ve. Stmt 2 also suff.

Thus it is D
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Originally posted by amitdgr on 06 Sep 2008, 11:53.
Last edited by amitdgr on 07 Sep 2008, 00:27, edited 1 time in total.
Manager
Joined: 04 Jan 2008
Posts: 114

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06 Sep 2008, 23:47
OA is D. Bin 2 problem.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Zumit DS 013 &nbs [#permalink] 06 Sep 2008, 23:47
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