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if p = r,
(r^2 + 2 -r^2) / [r(r^2 + 2)] >0
2/ [r(r^2 + 2)] >0
if r or p >0, then true.
from ii, we know only r is +ve. nothing about p.
so from i and ii, it is sufficient.

Re: DS: inequality [#permalink]
05 May 2006, 07:42

jodeci wrote:

macca wrote:

Is 1/p > r/(r^2 + 2) ?

1. p=r

2.r>0

A is my answer:

questions stem gives us r^2 + 2 >pr

statement 1. p=r,,which means that pr = r^2, then we have r^2+2>r^2 which is true.

statement 2. Does not help.

Without knowing the sign of p/r, you can't multiply both sides and say
if 1/p = r/(r^2 +2) means (r^2 + 2) > pr. It will hold true only if p & r both are +ve.

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