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, 23:12

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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, 01:37, 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]

25 Sep 2012, 00:15

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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, 01:42

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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] 25 Sep 2012, 01:42

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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 ?

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