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# If m≠0, is m^3>m^2?

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Manager
Joined: 08 Sep 2017
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Location: Colombia
GMAT 1: 710 Q49 V39

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11 Sep 2018, 20:09
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45% (medium)

Question Stats:

66% (01:14) correct 34% (00:57) wrong based on 109 sessions

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If $$m≠0$$, is $$m^3>m^2$$?

1) $$m>0$$
2) $$m^2>m$$

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Joined: 03 Apr 2018
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Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 20:15
This is definitely not 700 level..
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Joined: 29 Aug 2018
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11 Sep 2018, 20:38
m^3<m^2 m^3<m^2<m m^3>m^2>m
--------------0------------------------1---------------- Number line
m<0 0<m<1 m>1
1)Two possibilities for m>0 so insufficient
2)m(m-1)>0
m<0 or m>1
Two possibilities so insufficient.
1)+2)
m>1 Sufficient
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Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 20:49
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Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 22:41
ammuseeru wrote:
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I'm no expert to answer this. But, it takes some time to process math functions. After that it'll be fine.
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11 Sep 2018, 23:11
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If $$m≠0$$, is $$m^3>m^2$$?

The square of a number is always non-negative, so 0 or positive (m^2 in the problem is positive only since given that m is not 0). Thus we can safely reduce $$m^3>m^2$$ by m^2 to get the simplified question: is m > 1?

(1) $$m>0$$. Not sufficient.

(2) $$m^2>m$$. This is true for all negative numbers (because in this case (m^2 = positive) > (m = negative)) as well as for numbers greater than 1. Not sufficient.

(1)+(2) Since from (1) m > 0, then by reducing (2) by m we'll get m > 1. Sufficient.

Hope it's clear.
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04 Nov 2018, 07:44
Bunuel

2) $$m^2>m$$. This is true for all negative numbers (because in this case (m^2 = positive) > (m = negative)) as well as for numbers greater than 1. Not sufficient.

But m^2>m will be m>1 which is equal to the question stem. So can you explain statement 2? Or else even the question can be neg^3 > neg^2?
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Re: If m≠0, is m^3>m^2?  [#permalink]

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04 Nov 2018, 07:52
topper97 wrote:
Bunuel

2) $$m^2>m$$. This is true for all negative numbers (because in this case (m^2 = positive) > (m = negative)) as well as for numbers greater than 1. Not sufficient.

But m^2>m will be m>1 which is equal to the question stem. So can you explain statement 2? Or else even the question can be neg^3 > neg^2?

As given in the solution m^2 > m is true if m < 0 AND m > 1, so from (2) we cannot be sure that m > 1. Here you cannot reduce m^2 > m by m because we don't know its sign. If m > 0, then by reducing we'd get m > 1 but if m < 0, then by reducing we'd get m < 1 (recall that we should flip the sign when reducing/multiplying by a negative value).
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Re: If m≠0, is m^3>m^2?  [#permalink]

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10 Nov 2018, 23:11
dimmak wrote:
If $$m≠0$$, is $$m^3>m^2$$?

1) $$m>0$$
2) $$m^2>m$$

1 alone is of course not sufficient because each provides little information.
2 alone will lead to "yes" for positive values such as m=3 and no for negative values such m=-3. Hence. 2 is also insufficient.

It's between C and E.

Combining, if m is positive and m^2 > m which means that m is greater than 1. For all m>1, m^3 > m^2. Hence, yes, m^3 > m^2.

Re: If m≠0, is m^3>m^2?   [#permalink] 10 Nov 2018, 23:11
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