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Source: — Data Sufficiency |
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by neelgandham » Sun Jan 29, 2012 2:05 pm
If m!=1, Is m^2/(1-m) > m ?
m^2/(1-m) > m
m^2/(1-m)-(m)>0
m((m/(1-m))-1)>0
m*(2m-1)/(1-m)>0?

So the question can be rephrased to
If m!=1, m*(2m-1)/(1-m)>0?
1) m>0

We shall consider plugging in numbers in three intervals (0,1/2), (1/2,1), and >1
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 0
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
m = 2, m*(2m-1)/(1-m) = +ve*(+ve)/(-ve) < 0.
Insufficient to answer the question
2) m is not an integer
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 0
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
Insufficient to answer the question
From 1 and 2
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 0
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
Insufficient to answer the question

IMO E
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by pemdas » Sun Jan 29, 2012 2:58 pm
st(1) m>0 and we safely continue m/(1-m)>1. Since m is +ve, (1-m) must be +ve too.
m>1-m, m>1/2 Insuff as we consider all +ve values (m>0)
st(2) Insuff
combining st(1&2): Insuff

e
sud21 wrote:If m not equal to 1, Is m2/(1-m) > m?
1). m>0
2). m is not an integer
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