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# If vmt≠0, is v^2*m^3*t^(-4)>0?

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Joined: 09 Jul 2010
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If vmt≠0, is v^2*m^3*t^(-4)>0?  [#permalink]

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21 Jul 2010, 18:06
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25% (medium)

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72% (01:11) correct 28% (01:25) wrong based on 368 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: 02 Sep 2009
Posts: 59125
Re: Another DS problem.  [#permalink]

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21 Jul 2010, 18:16
1
1
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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17 Jun 2013, 05:02
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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17 Jun 2013, 20:36
1
sstudy wrote:
If vmt≠0, is v^2*m^3*t^(-4)>0?

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

This is a very easy question. I would say 500-600 level.

Question asks whether [(v^2)(m^3)]/(t^4) is greater than 0 or not. here, whether the value of expression is +ve or not depends only on sign of "m" as we know neither of the three is "0"

1. in first option "m" is given greater than square of something. Hence m is +ve.

2. same as first. m could be less then 1 but still positive.
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If vmt ≠ 0, is v^2m^3t^4 > 0?  [#permalink]

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Updated on: 29 Jul 2014, 22:23
If vmt ≠ 0, is v^2m^3t^4 > 0?
(1) m > $$v^2$$
(2) m >$$t^4$$

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Originally posted by kanusha on 29 Jul 2014, 22:19.
Last edited by Gnpth on 29 Jul 2014, 22:23, edited 1 time in total.
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Re: If vmt ≠ 0, is v^2m^3t^4 > 0?  [#permalink]

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29 Jul 2014, 23:11
1
kanusha wrote:
If vmt ≠ 0, is v^2m^3t^4 > 0?
(1) m > $$v^2$$
(2) m >$$t^4$$

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Based on the question stem, $$m^3$$ is the only element that needs to be evaluated to check if the entire equation is positive.

Statement 1: The only time that $$m>v^2$$ is when m is positive. Sufficient
Statement 2: The only time that $$m>t^4$$ is when m is positive. Sufficient

Note: Anything raised to an even number is always positive. (This means that regardless of sign, v & t will end up positive anyway)

Answer is D
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Re: If vmt ≠ 0, is v^2m^3t^4 > 0?  [#permalink]

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29 Jul 2014, 23:22
kanusha wrote:
Can any one solve this in simpliest Way,helpful for me

Thank you.

vmt ≠ 0, so we know that v, m, and t must all be either a positive or negative number, but cannot be 0.

(1) The value of m is greater than v^2. Whether we set v at a positive or negative number, it's square will be positive.
m has to be positive because it must be greater than v. With two positive numbers multiplied together, the only way that v^2m^3t^4 can be less than zero is if t^4 was negative. Since it is multiplied by an even exponent, t^4 will be a positive number regardless if t is negative or positive. Therefore, if (1) is valid, the solution can only be positive, for a "yes" response. Sufficient.

(2) If the value of m is greater than t^4, then both numbers must be positive, and again since v^2 is an even exponent, the square will be a positive number. Again, the only valid solution is a greater than zero, resulting in only a "yes" response. This is similar to (1). Sufficient.

Both statements are sufficient alone.
Answer: D
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Re: If vmt≠0, is v^2*m^3*t^(-4)>0?  [#permalink]

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29 Jul 2014, 23:43
kanusha wrote:
If vmt ≠ 0, is v^2m^3t^4 > 0?
(1) m > $$v^2$$
(2) m >$$t^4$$

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Merging topics. Please refer to the solution above.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1 and 3. Thank you.

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

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14 Nov 2018, 00:10
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If vmt≠0, is v^2*m^3*t^(-4)>0?   [#permalink] 14 Nov 2018, 00:10
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# If vmt≠0, is v^2*m^3*t^(-4)>0?

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