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# If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?

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Manager
Joined: 20 May 2008
Posts: 54
If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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Updated on: 17 Nov 2017, 02:21
3
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Difficulty:

25% (medium)

Question Stats:

72% (01:21) correct 28% (01:47) wrong based on 554 sessions

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If $$vmt ≠ 0$$, is $$v^2*m^3*t^{(-4)} > 0$$?

(1) $$m > v^2$$

(2) $$m > t^{(-4)}$$

Originally posted by LoyalWater on 02 Feb 2009, 22:05.
Last edited by Bunuel on 17 Nov 2017, 02:21, edited 4 times in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 53066
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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02 Oct 2012, 00:54
5
1
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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Manager
Joined: 04 Sep 2006
Posts: 112
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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04 Feb 2009, 00:11
1
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.
VP
Joined: 18 May 2008
Posts: 1093
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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04 Feb 2009, 03:28
1
1
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
Senior Manager
Joined: 30 Nov 2008
Posts: 467
Schools: Fuqua
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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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.
Intern
Joined: 28 Aug 2012
Posts: 7
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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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,
Weirdo
Director
Joined: 04 Jun 2016
Posts: 568
GMAT 1: 750 Q49 V43
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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19 Jul 2016, 12:19
LoyalWater wrote:
If vmt ≠ 0, is $$(v^2)*(m^3)*(t^(-4)) > 0$$

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

If vmt ≠ 0, is $$v^2*m^3*t^{-4} > 0$$
Lets analyse the question quickly
V*M*T is not zero ==> none of the digit is zero. But they can be -ve or +ve
IS $$v^2*m^3*t^{-4} > 0$$? ---------------> The expression $$>0$$ means positive; so the question is asking is the expression positive ?
ok...$$v^2$$ cannot be ever negative; squaring kills the polarity and always yield a positive number.
$$t^{-4}$$ cannot be ever negative . $$t^{-4}$$ is actually $$\frac{1}{t^4}$$ ; Diving 1 by a positive number will yield a positive number (fraction or otherwise but definitely positive)
$$m^3$$ can be positive or negative depending upon whether M is +ve or -ve
SO THE REAL QUESTION IS :- IS M POSITIVE?

(1)$$m > v^2$$
m = positive since squaring will give a positive number
SUFFICIENT

(2) $$m > t^{-4}$$
m will again be positive$$t^-4$$ means $$\frac{1}{t^4}$$; again positive fraction or integer
SUFFICIENT

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Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?  [#permalink]

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02 Dec 2017, 00:20
LoyalWater wrote:
If $$vmt ≠ 0$$, is $$v^2*m^3*t^{(-4)} > 0$$?

(1) $$m > v^2$$

(2) $$m > t^{(-4)}$$

WE are given that none of v or m or t is 0. So v^2 and t^4 will definitely be greater than 0 (even powers). All we need to know is whether m^3 > 0 or < 0 (or we need to know whether m is positive or negative).

(1) m > v^2.
Since v^2 is going to be positive, m > v^2.. this means m > 0. Sufficient.

(2) m > 1/t^4
Again t^4 or 1/t^4 are both going to be positive. So this also means m > 0. Sufficient.

Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?   [#permalink] 02 Dec 2017, 00:20
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