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GPREP2 Question
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parallel_chase
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k cannot be 1, 0, or -1
Statement I
1/(k-1) >0
Therefore k could be anything greater than 1 and I/k will always be greater than 0.
Sufficient.
Statement II
1/(k+1) > 0
k could be -0.5 in which case 1/k < 0
k could be +0.5 in which case 1/k > 0
Therefore Insufficient.
Hence A is the answer.
Let me know if you still have any doubts.
Statement I
1/(k-1) >0
Therefore k could be anything greater than 1 and I/k will always be greater than 0.
Sufficient.
Statement II
1/(k+1) > 0
k could be -0.5 in which case 1/k < 0
k could be +0.5 in which case 1/k > 0
Therefore Insufficient.
Hence A is the answer.
Let me know if you still have any doubts.
IMO Amedea66 wrote:Please help.........
ST 1. In order to be (1/(k-1))>0 k-1 also must be >0. => k>1 In every number that k>1 1/k always >0 SUFF
ST 2. Like the first statement k+1>0 => k>-1. Here we couln't say anything. If k=-0.5 then 1/k<0 but if k=0.5 then 1/k>0 INSUFF
