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

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

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New post 06 Sep 2008, 00:40
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If vmt =/ 0, is \(v^2m^3t^-4\) > 0 ?

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

Kudos [?]: 97 [0], given: 0

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Re: Zumit DS 013 [#permalink]

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New post 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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Re: Zumit DS 013 [#permalink]

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New post 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

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Re: Zumit DS 013 [#permalink]

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New post 06 Sep 2008, 11:53
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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Last edited by amitdgr on 07 Sep 2008, 00:27, edited 1 time in total.

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Re: Zumit DS 013 [#permalink]

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

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Re: Zumit DS 013   [#permalink] 06 Sep 2008, 23:47
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If vmt =/ 0, is v^2m^3t^-4 > 0 ? (1) m > v^2 (2) m

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