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