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If m≠0, is m^3>m^2?

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If m≠0, is m^3>m^2?  [#permalink]

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New post 11 Sep 2018, 19:09
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If \(m≠0\), is \(m^3>m^2\)?

1) \(m>0\)
2) \(m^2>m\)

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Re: If m≠0, is m^3>m^2?  [#permalink]

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

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New post 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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New post 11 Sep 2018, 19:49
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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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New post 11 Sep 2018, 21:41
ammuseeru wrote:
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On gmatclub app/quant question, question formatting does appear correct. Please find below snapshot.

Attachment:
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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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If m≠0, is m^3>m^2?  [#permalink]

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New post 11 Sep 2018, 22:11
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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.

Answer: C.

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

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