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dimmak
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11 Sep 2018, 20:09
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
65% (01:11) correct
35%
(01:03)
wrong
based on 317
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11 Sep 2018, 20:38
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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
--------------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
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.
Answer: C.
Hope it's clear.
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.
