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

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Senior Manager
Joined: 10 Mar 2008
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17 Jun 2008, 18:44
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Difficulty:

45% (medium)

Question Stats:

62% (02:15) correct 38% (01:10) wrong based on 103 sessions

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

(1) p=r
(2) r>0

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-1-p-r-r-2-2-1-p-r-2-r-86165.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Mar 2013, 05:47, edited 1 time in total.
Edited the question and added the OA.
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Re: DS: Inequalities [#permalink]

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17 Jun 2008, 20:27
C

statement 1: + and negative numbers have different results

statement 2: nothing about p

together suff
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Re: DS: Inequalities [#permalink]

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17 Jun 2008, 21:11
U r rite! C is the answer!

Detailed: 1/p > r/(r^2+2)

St1: p=r the inequality can be written as r^2 < r^2 + 2 when r>0 and as r^2 > r^2 + 2 for r<0

st2: r>0 is insufficient

Combining them: it is suffi.
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Re: Is 1/p>r/r^2+2? 1. p=r 2. r>0 [#permalink]

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14 Mar 2013, 05:38
How do you solve this question?
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Re: Is 1/p>r/r^2+2? 1. p=r 2. r>0 [#permalink]

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14 Mar 2013, 05:46
fozzzy wrote:
How do you solve this question?

Check here: is-1-p-r-r-2-2-1-p-r-2-r-86165.html

Hope it helps.
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Re: Is 1/p>r/r^2+2? 1. p=r 2. r>0   [#permalink] 14 Mar 2013, 05:46
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# Is 1/p>r/(r^2+2)?

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