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If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]
 02 Feb 2009, 23:05
Question Stats:
56% (02:09) correct
43% (01:16) wrong based on 2 sessions
If vmt ≠ 0, is v^2m^3t^(-4) > 0? (1) m > v^2 (2) m > t^(-4)
Last edited by Bunuel on 02 Oct 2012, 01:55, edited 2 times in total.
Renamed the topic and edited the question.
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LoyalWater wrote: 1.If vmt ≠ 0, is v2m3t-4 > 0?
(1) m > v2
(2) m > t-4 v^2*m^2*m*t-4 > 0 1) m>v^2 t can be +Ve or -ve v^2*m^2*m*t-4 > 0 can be true or false not sufficient 2) m > t-4 m can be +ve or -ve can be true or false not sufficient when combined. m is positive --> t must be +Ve v^2*m^2*m*t > 4 or <4 eventhough all v,m,t are positive. v can be fraction value which can lead to v^2*m^2*m*t to <4 not sufficient E
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LoyalWater wrote: 1.If vmt ≠ 0, is v2m3t-4 > 0?
(1) m > v2
(2) m > t-4 please post the question properly.. I believe v2 is power 2 of V .. then use exponent symbol "^" (v^2)
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LoyalWater wrote: 1.If vmt ≠ 0, is v2m3t-4 > 0?
(1) m > v2
(2) m > t-4 v^2 *m^3*t -4 >0 v^2 * m^2*m*t>4 1) m>v^2 ok..insuff..all we know is that m is positive 2) m>t-4 insuff we dont know..if m<0 or positive..insuff together.. all we know is that M>0 i.e however we dont know if T<0 or not v^2 *M^3 >0 we dont know if T>0 or not..Insuff E it is..
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sorry guys..
i have put the exponent sign..
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I think it is D .
m is +ve .
The powers of v & t are even . SO the combined product is +ve only.
from stat 2 : m is again +ve . so Suff.
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It shld be D
Is v^2 *m^3*t^-4>0?
Since v^2 and 1/t^4are positive, so it depends upon whether m is positive or not
(1) m>v^2 >0 hence sufficient
(2) m> 1/t^4 >0 hence sufficient
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I agree with D.
Since the question stem converts to (v^2)(m^3) / (t^4) - IS this > 0
We know V^2 is alwats positive, t^4 is always positve. Hence the value of m decides whether the expression is > 0 or not.
From stmt 1, it is given m > v2 ==> m is positive. Hence the expression > 0. Sufficient.
From Stmt 2, it is given, m > (1/(t ^4) ==> m positive. Hence the expression > 0. Sufficient.
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mrsmarthi wrote: I agree with D.
Since the question stem converts to (v^2)(m^3) / (t^4) - IS this > 0
We know V^2 is alwats positive, t^4 is always positve. Hence the value of m decides whether the expression is > 0 or not.
From stmt 1, it is given m > v2 ==> m is positive. Hence the expression > 0. Sufficient.
From Stmt 2, it is given, m > (1/(t ^4) ==> m positive. Hence the expression > 0. Sufficient. ------------------------------------------------------------------------------------------------------- Can anyone please explain in Stmt2, how m> (1/(t^4))? Thanks, Weirdo
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Weirdo2989 wrote: mrsmarthi wrote: I agree with D.
Since the question stem converts to (v^2)(m^3) / (t^4) - IS this > 0
We know V^2 is alwats positive, t^4 is always positve. Hence the value of m decides whether the expression is > 0 or not.
From stmt 1, it is given m > v2 ==> m is positive. Hence the expression > 0. Sufficient.
From Stmt 2, it is given, m > (1/(t ^4) ==> m positive. Hence the expression > 0. Sufficient. ------------------------------------------------------------------------------------------------------- Can anyone please explain in Stmt2, how m> (1/(t^4))? Thanks, Weirdo means that none of the unknowns equals to zero. Is v^2*m^3*t^{-4}>0? --> is \frac{v^2*m^3}{t^4}> 0? As v^2 and t^4 are positive (remember none of the unknowns equals to zero) this inequality will hold true if and only m^3>0, or, which is the same, when m>0. (1) m>v^2 --> m is more than some positive number ( v^2), hence m is positive. Sufficient. (2) m>t^{-4} --> m>\frac{1}{t^4} --> Again m is more than some positive number ( \frac{1}{t^4} ), hence m is positive. Sufficient. Answer: D. Hope it's clear.
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