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1) 1/M>1; i.e. 0<M<1; i.e. m is positive and is less than 1, therefore MM will be less than 1 for all values of M that lies between 0 and 1, hence 1 alone is sufficient.sanju09 wrote:Is 1 > MM?
(1) 1/M > 1.
(2) M > 0.
Made up
From 1 you wrote 0<M<1 ; how do you know M>0manpsingh87 wrote:1) 1/M>1; i.e. 0<M<1; i.e. m is positive and is less than 1, therefore MM will be less than 1 for all values of M that lies between 0 and 1, hence 1 alone is sufficient.sanju09 wrote:Is 1 > MM?
(1) 1/M > 1.
(2) M > 0.
Made up
2) M>0, now when m is less than 1, then we have MM<1, when m is greater than 1 we have MM>1, as here two results are possible for two different cases, hence 2 alone is not sufficient to answer the question..!!
A
if we consider M to be negative then the condition 1/M >1 will get violated..!!!HSPA wrote:From 1 you wrote 0<M<1 ; how do you know M>0manpsingh87 wrote:1) 1/M>1; i.e. 0<M<1; i.e. m is positive and is less than 1, therefore MM will be less than 1 for all values of M that lies between 0 and 1, hence 1 alone is sufficient.sanju09 wrote:Is 1 > MM?
(1) 1/M > 1.
(2) M > 0.
Made up
2) M>0, now when m is less than 1, then we have MM<1, when m is greater than 1 we have MM>1, as here two results are possible for two different cases, hence 2 alone is not sufficient to answer the question..!!
A
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