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

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Intern
Joined: 17 Jan 2010
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If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? [#permalink]

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22 Jan 2010, 04:21
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25% (medium)

Question Stats:

71% (01:51) correct 29% (01:11) wrong based on 109 sessions

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If vmt≠0, is v^2*m^3*t^(-4)>0?

(1) m > v^2
(2) m > t^(-4)
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Joined: 20 Aug 2009
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22 Jan 2010, 04:30
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Sign of $$v^2m^3t^{-4}$$ depends on m, because both x^2 and t^(-4) are always positive.

Each statement alone is sufficient t determine that m>0. So (D)

Last edited by shalva on 22 Jan 2010, 12:03, edited 1 time in total.
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22 Jan 2010, 08:11
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shalva wrote:
Sign of $$v^2m^3t^4$$ depends on m, because both x^2 and t^(-4) are always positive.

Each statement alone is sufficient t determine that m>0. So (D)

Statement 1

$$m > v^2$$

So, m is positive as it's greater than $$v^2$$
but, what about t? what if t's negative?
couldn't $$t^{-4}$$ be a negative number as it's $$^{-4}$$?

Statement 2

$$m > t^{-4}$$

again, t could be negative or positive.

Combining the two statements

we do not know whether t is positive

So, IMO it's E
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22 Jan 2010, 09:31
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rahulms wrote:
If $$v*m*t$$ not $$= 0$$, is $$v^2*m^3*t^{-4} > 0$$?

1) $$m>v^2$$
2) $$m>t^{-4}$$

$$V^2$$ will always be positive so $$V^2>0$$
Same for $$T^{-4}$$ so $$T^{-4} >0$$

Question stem asks whether m is positive or negative (m>0 or m<0)

Yes/No question

1. $$m>v^2$$

Given $$V^2>0$$ so m is positive or $$m >0$$
Ans Yes STATEMENT SUFFICIENT

2. $$m>t^{-4}$$

given $$T^{-4}$$so $$T^{-4} >0$$
so m is positive or $$m >0$$
Ans Yes STATEMENT SUFFICIENT

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22 Jan 2010, 09:34
rahulms wrote:
shalva wrote:
Sign of $$v^2m^3t^4$$ depends on m, because both x^2 and t^(-4) are always positive.

Each statement alone is sufficient t determine that m>0. So (D)

Statement 1

$$m > v^2$$

So, m is positive as it's greater than $$v^2$$
but, what about t? what if t's negative?
couldn't $$t^{-4}$$ be a negative number as it's $$^{-4}$$?

Statement 2

$$m > t^{-4}$$

again, t could be negative or positive.

Combining the two statements

we do not know whether t is positive

So, IMO it's E

Even power of any number is always positive... i.e. greater than zero.

$$t^{-4}$$ can be written as $$\frac{1}{t^{4}}$$

result will be less than one but greater than zero.
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22 Jan 2010, 09:40
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Expert's post
rahulms wrote:
If $$v*m*t$$ not $$= 0$$, is $$v^2*m^3*t^{-4} > 0$$?

1) $$m>v^2$$
2) $$m>t^{-4}$$

Shalva's solution is correct.

First of all: knowing that $$v*m*t\neq{0}$$ implies that none of the variables equals to zero.

Second of all: $$v^2*m^3*t^{-4}>0$$? --> $$\frac{v^2*m^3}{t^4}> 0$$? Now when this inequality holds true? Obviously as $$v^2$$ and $$t^4$$ are positive 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.

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22 Jan 2010, 09:43
agree with D

in both conditions the only variable that can change the sign of the expression is M. Since in both conditions we are told that M most be greater than some always positive number than in both condition we can answer the question if M is positive and thus the expression is positive.
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22 Jan 2010, 10:38
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A and B alone are sufficient to answer.

If m > v^2 => m has to be positive.

thus m * v^2 = +ve and t^-4 is going to be positive irrespective of value of t.

Hence A sufficient.

Similarly if m > t^-4 => m should be +ve.

Hence m * t^-4 will be +ve.

v^2 will be +ve irrespective of the value of v.

Hence B alone is also sufficient.

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Re: If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? [#permalink]

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18 Dec 2013, 05:30
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Re: If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ?   [#permalink] 18 Dec 2013, 05:30
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