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

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Senior Manager
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Is 1/p > r/(r^2 + 2) ? 1. p=r 2.r>0 [#permalink] New post 08 Apr 2006, 10:54
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A
B
C
D
E

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

1. p=r

2.r>0
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Re: DS: inequality [#permalink] New post 08 Apr 2006, 11:02
macca wrote:
Is 1/p > r/(r^2 + 2) ?
1. p=r
2.r>0


1/p > r/(r^2 + 2)
1/p - r/(r^2 + 2) >0
[(r^2 + 2)-pr]/[p(r^2 + 2)]>0
(r^2 + 2 -pr) / [p(r^2 + 2)] >0

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.
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 [#permalink] New post 08 Apr 2006, 13:10
IMO its C

from 1 substituting p = r

we get [2/r*(r^2 +1)] ----> the value of this expression depends on r since r^2 + 1 is always +ve.
so for r>0, its +ve.
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 [#permalink] New post 08 Apr 2006, 15:33
Answer is C

1) Depending whether R is greater/less than 1 the answer is yes/no => Insufficient

2) Does not tell us anything about P => Insufficient

Taken together they are sufficent since the left side is always greater than the right if R is positive.
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 [#permalink] New post 08 Apr 2006, 15:34
[quote]1) Depending whether R is greater/less than 1 the answer is yes/no => Insufficient[/code]

Meant 0 instead of 1
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Re: DS: inequality [#permalink] New post 08 Apr 2006, 15:41
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.
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 [#permalink] New post 08 Apr 2006, 15:49
Quote:
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.


Try plugging -1 for R and then try 2.
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 [#permalink] New post 08 Apr 2006, 16:05
Charlie45 wrote:
Quote:
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.


Try plugging -1 for R and then try 2.


I thought it works in all instances: for -1, you have -1^2 +2> -1*-1 which is true, for 2 you have 2^2 +2 > 2*2 which is also true.

Explain what you meant
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Re: DS: inequality [#permalink] New post 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.

hence the answer shd be C
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 [#permalink] New post 07 May 2006, 13:11
Agree Answer is C :wink:

Statement 1 will fail for negative values of R
  [#permalink] 07 May 2006, 13:11
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Is 1/p > r/(r^2 + 2) ? 1. p=r 2.r>0

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