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Is 1/p > r/(r^2+2) ? 1. p = r 2. r > 0 Can

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Is 1/p > r/(r^2+2) ? 1. p = r 2. r > 0 Can  [#permalink]

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New post Updated on: 21 Mar 2010, 00:17
2
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

80% (01:00) correct 20% (00:39) wrong based on 59 sessions

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

1. p = r
2. r > 0


Can anyone pls. explain why the answer is C ?

Originally posted by nsp007 on 20 Mar 2010, 22:17.
Last edited by nsp007 on 21 Mar 2010, 00:17, edited 3 times in total.
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Re: Gmat Prep DS question !  [#permalink]

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New post 20 Mar 2010, 23:49
nsp007 wrote:
Is \(\frac{1}{p} > \frac{r}{(r+2)}\) ?

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) => 1>1/3 - suff

p=r = -3 => 1/-3 > -3> (-3 +2) => -1/3 > 3 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) => 1>1/3 - suff
p = r = 8 we get 1/8 > 8/ (8+2) => 1/8 > 8/10 insuff

will go with E
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Re: Gmat Prep DS question !  [#permalink]

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New post 20 Mar 2010, 23:57
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 ?
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Re: Gmat Prep DS question !  [#permalink]

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New post 21 Mar 2010, 00:02
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
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Re: Gmat Prep DS question !  [#permalink]

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New post 21 Mar 2010, 02:58
1/p > r/r^2 +2
this condition satisfy only when p=r and both are positive

will go with C
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Re: Gmat Prep DS question !  [#permalink]

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New post 21 Mar 2010, 03:36
nsp007 wrote:
Is \(\frac{1}{p} > \frac{r}{(r^2+2)}\) ?

1. p = r
2. r > 0


Can anyone pls. explain why the answer is C ?


Algebraic approach:

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

(1) \(p=r\) --> is \(\frac{1}{r}>\frac{r}{r^2+2}?\) --> as \(r^2+2\) is always positive, multiplying inequality by this expression we'll get: is \(\frac{r^2+2}{r}>r?\) --> is \(r+\frac{2}{r}>r?\) --> is \(\frac{2}{r}>0?\) This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.

(2) \(r>0\). Not sufficient by itself.

(1)+(2) As from (2) \(r>0\), then \(\frac{2}{r}>0\) is true. Sufficient.

Answer: C.
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Re: Gmat Prep DS question !  [#permalink]

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New post 22 Mar 2010, 16:14
thanks for the explanation !
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Re: Is 1/p > r/(r^2+2) ? 1. p = r 2. r > 0 Can  [#permalink]

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Re: Is 1/p > r/(r^2+2) ? 1. p = r 2. r > 0 Can &nbs [#permalink] 02 Jul 2018, 21:33
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