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# GMatPrep-DS

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GMatPrep-DS [#permalink]  09 Sep 2007, 07:31
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions

From 1) the inequality becomes is r^2 + 2 > r^2, regardless of r being postive or negative or 0, the ineqiality holds good.

From 2) it does not say anything about p.

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Senior Manager
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Just plug in the numbers once negative than positive, you will see that they are siffices together

Ans: C
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Re: GMatPrep-DS [#permalink]  16 Sep 2007, 10:06
zion wrote:

From 1) the inequality becomes is r^2 + 2 > r^2, regardless of r being postive or negative or 0, the ineqiality holds good.

From 2) it does not say anything about p.

Ans can not be A.

Let's take r = 0 then as per ST1 p=r
So (1/p) = (1/0) which is not defined.

If we combine ST1 and ST2,
Then "r" should be greater than 0 and in this case, 1/P will always be greater than r/(r^2+2).
You can verify this by putting any value of r >0.

Hence C.

- Brajesh
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Ferihere wrote:
Just plug in the numbers once negative than positive, you will see that they are siffices together

Ans: C

I am not sure if thats the correct way. Even if you substiute once -ve and one +ve number , you should still get A .
I think brajesh is correct that we need to consider the number 0 here.
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1/p > r/(r^2+2) ---> true ?

statement 1

p = r

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

1 > p^2/(p^2+2) ---> if p>0

p^2+2 > p^2

2 > 0 ---> true

1 < p^2/(p^2+2) ---> if p<0

Note ! if you double by p to get this outcome you have doubled by a negative number ! to get p^2 you need to flip signs !!

p^2+2 < P^2

2 < 0 ---> false

insufficent

statement 2

r > 0

insufficent

both statements

since r = p

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

1 > p^2/(p^2+2) ---> if p>0 ---> given

p^2+2 > p^2

2 > 0 ---> true

sufficient

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St1:
If p = r = 2,
1/2 > 2/6 ? --> yes.

If p = r = -2
-1/2 > -2/6 --> No.

Insufficient.

St2:

St1 & St2:
If p = r = 2,
1/2 > 2/6? --> yes.

If p = r = 1/2
2 > 2/9? --> yes.

Sufficient.

Ans C
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