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

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

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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, 23:05.
Last edited by Bunuel on 17 Nov 2017, 03:21, edited 4 times in total.
Renamed the topic and edited the question.
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Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0? [#permalink]

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New post 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.

Answer: D.

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

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

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New post 04 Feb 2009, 04:28
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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: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0? [#permalink]

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

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

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New post 19 Jul 2016, 13:19
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

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

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New post 02 Dec 2017, 01:20
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.

Hence D answer
Re: If vmt ≠ 0, is (v^2)*(m^3)*(t^(-4)) > 0?   [#permalink] 02 Dec 2017, 01:20
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