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# If vmt ≠ 0, is v^2m^3t^-4 > 0?

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Manager
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If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]

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02 Feb 2009, 23:05
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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, 04:30, edited 3 times in total.
Renamed the topic and edited the question.
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03 Feb 2009, 05:31
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

E
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03 Feb 2009, 05:33
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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03 Feb 2009, 10:16
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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03 Feb 2009, 23:12
sorry guys..
i have put the exponent sign..
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04 Feb 2009, 01:11
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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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04 Feb 2009, 04:28
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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04 Feb 2009, 12:35
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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01 Oct 2012, 23:57
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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02 Oct 2012, 01:54
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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.

Hope it's clear.
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]

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30 Dec 2013, 22:27
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]

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03 Jan 2014, 01:13
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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]

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03 Jan 2014, 04:31
Expert's post
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]

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29 Apr 2015, 09:38
Hello from the GMAT Club BumpBot!

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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0?   [#permalink] 29 Apr 2015, 09:38
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# If vmt ≠ 0, is v^2m^3t^-4 > 0?

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