nsp007 wrote:

Sorry for the typo in the problem statement....

Here's the correct version !

Is \(\frac{1}{p} > \frac{r}{(r^2+2)}\) ?

1. p = r

2. r > 0

Can anyone pls. explain why the answer is C ?

stmnt1 - p=r = 1 we get 1/1 > 1/ (1^2+2) => 1>1/3 - suff

p=r = -3 => 1/-3 > -3> ([-3]^2 +2) => -1/3 > -3/11 insuff

stmnt2 - r>0 we have no info about p hence insuff

taking together have p=r > 0

p=r = 1 we get 1/1 > 1/ (1^2+2) => 1>1/3 - suff

p = r = 8 we get 1/8 > 8/ (8^2+2) => 1/8 > 8/66 suff

will go with C