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If vmt ≠ 0, is v^2m^3t^-4 > 0?

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If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink] New post 02 Feb 2009, 22:05
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If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?

(1) m > v^2
(2) m > t^(-4)
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Jan 2014, 03:30, edited 3 times in total.
Renamed the topic and edited the question.
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Re: vmt [#permalink] New post 03 Feb 2009, 04:31
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

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Re: vmt [#permalink] New post 03 Feb 2009, 04:33
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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Re: vmt [#permalink] New post 03 Feb 2009, 09:16
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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Re: vmt [#permalink] New post 03 Feb 2009, 22:12
sorry guys..
i have put the exponent sign..
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Re: vmt [#permalink] New post 04 Feb 2009, 00:11
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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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Re: vmt [#permalink] New post 04 Feb 2009, 03:28
It shld be D
Is \(v^2 *m^3*t^-4>0?\)
Since \(v^2\) and \(1/t^4\)are 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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Re: vmt [#permalink] New post 04 Feb 2009, 11:35
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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Re: vmt [#permalink] New post 01 Oct 2012, 22:57
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,
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Re: vmt [#permalink] New post 02 Oct 2012, 00:54
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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


If \(v*m*t\) not \(= 0\), is \(v^2*m^3*t^{-4} > 0\)?

\(v*m*t\neq{0}\) 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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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink] New post 30 Dec 2013, 21:27
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink] New post 03 Jan 2014, 00:13
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The original question is confusing because there is no * sign.

One could interpret it as V raised to a long exponent (2m^3t^(-4)) since there are no * in between each product.
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink] New post 03 Jan 2014, 03:31
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catalysis wrote:
The original question is confusing because there is no * sign.

One could interpret it as V raised to a long exponent (2m^3t^(-4)) since there are no * in between each product.


Edited the stem as suggested. Thank you.
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink] New post 29 Apr 2015, 08:38
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0?   [#permalink] 29 Apr 2015, 08:38
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