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is 1/p > r/(r^2+2)

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Manager
Joined: 20 Mar 2005
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04 Aug 2007, 07:25
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is 1/p > r/(r^2+2)

1) p=r

2 r> 0

Kudos [?]: 104 [0], given: 0

VP
Joined: 28 Mar 2006
Posts: 1367

Kudos [?]: 38 [0], given: 0

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04 Aug 2007, 07:31
Balvinder wrote:
is 1/p > r/(r^2+2)

1) p=r

2 r> 0

Should be C

if p=r it doesnt mean that r^2+2>r^2 since we do not know the sign of r

if it is ve the ">" stays the same but if r is -ve then the sign flips

So we need a condition that r>0 which is exactly what B gives

Hence C stands

Kudos [?]: 38 [0], given: 0

Director
Joined: 03 May 2007
Posts: 867

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Schools: University of Chicago, Wharton School

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04 Aug 2007, 10:19
Balvinder wrote:
is 1/p > r/(r^2+2)

1) p=r
2 r> 0

C is it.

using both information, it is true that 1/p > r/(r^2+2).

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Manager
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05 Aug 2007, 00:13
trivikram wrote:
Balvinder wrote:
is 1/p > r/(r^2+2)

1) p=r

2 r> 0

Should be C

if p=r it doesnt mean that r^2+2>r^2 since we do not know the sign of r

if it is ve the ">" stays the same but if r is -ve then the sign flips

So we need a condition that r>0 which is exactly what B gives

Hence C stands

trivikram can you clear my doubts over this ; both side are square, which make it positive irrespective of the sign of r.

so r^2 +2 will definetly will be great than r^2

OA is C but still i want to clear my doubts

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VP
Joined: 28 Mar 2006
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05 Aug 2007, 08:46
Balvinder wrote:
trivikram wrote:
Balvinder wrote:
is 1/p > r/(r^2+2)

1) p=r

2 r> 0

Should be C

if p=r it doesnt mean that r^2+2>r^2 since we do not know the sign of r

if it is ve the ">" stays the same but if r is -ve then the sign flips

So we need a condition that r>0 which is exactly what B gives

Hence C stands

trivikram can you clear my doubts over this ; both side are square, which make it positive irrespective of the sign of r.

so r^2 +2 will definetly will be great than r^2

OA is C but still i want to clear my doubts

When are they squares? Only after cross-multiplying.

But if we have a situation like the one in thsi problem dont ever think it is right to do the way you are thinking.

First ask yourself do I know the sign of this variable. If you aint sure then you better account for it.

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Manager
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05 Aug 2007, 11:50

1/p>r/r^2+2
Thus if p>o, r^2+2>pr Case 1
If P<0, r^2+2<pr>r^2. Case 1.
Case2 does not hold true as r^2+2 can not be less than -r^2.
Thus 1 is sufficient.

Maybe I am very tired and talking completely nonsense. But please explain.

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Manager
Joined: 27 May 2007
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05 Aug 2007, 12:51
I didn't do any algebraic manipulation on this one. I just plugged in the number 3, then the number -3, for p and r, to check the first statement. With 3 it was true:
1/p>r/r^2+2 becomes 1/3>3/11, true

With -3 it was false:
1/p>r/r^2+2 becomes -1/3>-3/11, false
so insufficient.

Statement 2 gives no info about the value of p, so insufficient.

With both of them, we know that p and r are equal and positive. Any positive value makes the statement true.

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05 Aug 2007, 12:51
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