Is 1/p > r/(r^2+2)? 1) p=r 2) r>0 : DS Archive
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# Is 1/p > r/(r^2+2)? 1) p=r 2) r>0

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Manager
Joined: 01 Mar 2009
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Is 1/p > r/(r^2+2)? 1) p=r 2) r>0 [#permalink]

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17 Mar 2009, 21:55
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Is 1/p > r/(r^2+2)?

1) p=r
2) r>0

OA will be posted shortly.

Thanks,
Manager
Joined: 02 Mar 2009
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Kudos [?]: 48 [0], given: 0

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17 Mar 2009, 22:08
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
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18 Mar 2009, 08:06
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
Re: DS: PT2   [#permalink] 18 Mar 2009, 08:06
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