|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 01 Mar 2009
Posts: 55
Followers: 1
Kudos [?]:
1
[0], given: 2
|
Is 1/p > r/(r^2+2)? 1) p=r 2) r>0 [#permalink]
17 Mar 2009, 22:55
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
Is 1/p > r/(r^2+2)?
1) p=r 2) r>0
Please explain
OA will be posted shortly.
Thanks,
|
|
|
|
|
|
|
Manager
Joined: 02 Mar 2009
Posts: 139
Followers: 1
Kudos [?]:
24
[0], given: 0
|
It is C
1. You can multiply both sides by r^2 + 2 because this quantity is always positive. Hence the inequality sign does not change. Then replacing p for r and simplifying, we get: Is p + 2/p >p? For p=1 yes For p=-1 no
So insufficient
2. Rearranging we get: Is (r^2 + 2)/p > r? We dont know anythinga bout p so Insufficient
Together.
the equation becomes: Is (r^2 + 2)/r > r? Since r>0, multiply both sides by r and the sign does not change. You get r^2 +2 > r^2
True!
|
|
|
|
|
|
SVP
Joined: 28 Dec 2005
Posts: 1612
Followers: 1
Kudos [?]:
53
[0], given: 2
|
i agree with C
From stat 1, you dont know the sign of p and r. Depending on whether p/r is positive or negative, the inequality either holds or doesnt. Insuff.
From stat 2, you know r is positive, but nothing about p. Insuff.
Together, you know that p and r are positive. Testing with fractions (i.e. btwn 0 and 1) and whole numbers, the inequality is always true. Suff
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|