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If m not equal to zero is m^3 > m^2 ?

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If m not equal to zero is m^3 > m^2 ? [#permalink] New post 24 Sep 2012, 22:12
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If m not equal to zero is m^3 > m^2 ?

(1) m > 0

(2) m^2 > m
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2012, 00:37, edited 1 time in total.
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Re: if m not equal to zero is m^3 > m^2 ? [#permalink] New post 24 Sep 2012, 23:15
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harikris wrote:

if m not equal to zero is m^3 > m^2 ?

1) m>0

2) m^2 > m


The question asks if m^3 > m^2. We can safely divide by m^2 on both sides without worrying about whether to reverse the inequality, because m^2 can never be negative. So the question is just asking "Is m > 1?"

Statement 1 is now clearly not sufficient. Statement 2 is true for every negative number, but is also true when m > 1, so is not sufficient. When we combine the statements, we know m is positive from Statement 1, so we can safely divide both sides by m in Statement 2, and we find m > 1, which is what we wanted to prove. So the answer is C.
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Re: If m not equal to zero is m^3 > m^2 ? [#permalink] New post 25 Sep 2012, 00:42
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If m not equal to zero is m^3 > m^2 ?

Since \(m\neq{0}\), then \(m^2>0\) and we can safely divide \(m^3 > m^2\) by it. Thus, the question becomes: is \(m>1\)?

(1) m > 0. Not sufficient.

(2) m^2 > m --> \(m(m-1)>0\) --> \(m<0\) or \(m>1\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(m>1\). Sufficient.

Answer: C.
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Re: If m not equal to zero is m^3 > m^2 ? [#permalink] New post 24 Jun 2015, 02:35
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Re: If m not equal to zero is m^3 > m^2 ?   [#permalink] 24 Jun 2015, 02:35
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