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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]  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.
Edited the question.
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Re: if m not equal to zero is m^3 > m^2 ? [#permalink]  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]  25 Sep 2012, 00:42
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Expert's post
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.

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Re: If m not equal to zero is m^3 > m^2 ?   [#permalink] 25 Sep 2012, 00:42
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