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# If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? 1) m>v^2 2)

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If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? 1) m>v^2 2) [#permalink]  22 Jan 2010, 04:21
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If v*m*t not = 0, is v^2*m^3*t^{-4} > 0?

1) m>v^2
2) m>t^{-4}
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Re: Another DS [#permalink]  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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Re: Another DS [#permalink]  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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Re: Another DS [#permalink]  22 Jan 2010, 09:31
1
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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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Re: Another DS [#permalink]  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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Re: Another DS [#permalink]  22 Jan 2010, 09:40
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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}

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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Re: Another DS [#permalink]  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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Re: Another DS [#permalink]  22 Jan 2010, 10:13
D it is!

thank you folks
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Re: Another DS [#permalink]  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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Need Solution for some DS problems from SET1 [#permalink]  23 Jun 2010, 12:30
I have problem in solving following DS questions from SET1 (I don't know the source of this). Can anyone help me in resolvign them?

DS Questions: Need help

Q5:
If vmt ≠ 0, is v2m3t-4 > 0?
(1) m > v2
(2) m > t-4
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Re: Another DS [#permalink]  24 Jun 2010, 02:35
D for Sure
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Re: Another DS   [#permalink] 24 Jun 2010, 02:35
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