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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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If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 19:09
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Difficulty:

45% (medium)

Question Stats:

63% (00:47) correct 37% (00:38) wrong based on 82 sessions

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If $$m≠0$$, is $$m^3>m^2$$?

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

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Intern
Joined: 03 Apr 2018
Posts: 37
Concentration: Strategy, Leadership
GMAT 1: 750 Q50 V41
Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 19:15
This is definitely not 700 level..
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Joined: 29 Aug 2018
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If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 19:38
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
C is the answer
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Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 19:49
bb
On gmatclub app/quant question, question formatting does appear correct. Please find below snapshot.

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Re: If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 21:41
ammuseeru wrote:
bb
On gmatclub app/quant question, question formatting does appear correct. Please find below snapshot.

Attachment:
IMG_4617.JPG

Sent from my iPhone using GMAT Club Forum mobile app

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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If m≠0, is m^3>m^2?  [#permalink]

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11 Sep 2018, 22:11
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1
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.

Answer: C.

Hope it's clear.
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If m≠0, is m^3>m^2?  [#permalink]

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04 Nov 2018, 06:44
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, 06:52
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, 22:11
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.

C is the answer.
Re: If m≠0, is m^3>m^2? &nbs [#permalink] 10 Nov 2018, 22:11
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# If m≠0, is m^3>m^2?

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