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

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DS: Is 1/p > r/(r^2+2) [#permalink] New post 05 Jul 2006, 20:38
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Is 1/p > r/(r^2+2)

1) p = r
2) r > 0

Please explain the answer
[Reveal] Spoiler: OA
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Re: DS: Is 1/p > r/(r^2+2) [#permalink] New post 05 Jul 2006, 20:54
humans wrote:
Is 1/p > r/(r^2+2)

1) p = r
2) r > 0

Please explain the answer


from 1, if p=r> -ve, 1/p < r / (r^2+2)
if 0<p=r, 1/p > r / (r^2+2). so insuffcient..

from 2, r>0 is also insuffcient.

togather 1/p > r / (r^2+2).

so C.
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 [#permalink] New post 05 Jul 2006, 21:39
Will go with C

1) p = r

When p=r=+ve, it satisfies the equation
But when p=r=-ve , the equation fails hence Insuff

2) r>0
P can be anything hence Insuff

Together

p=r=+ve
Hence

1/p > r/(r^2+2)
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 [#permalink] New post 05 Jul 2006, 22:09
Yes 'C' it is.
We must be sure that r and p are both +ve or -ve.
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 [#permalink] New post 06 Jul 2006, 07:48
C

St1: Fails for -ve fraction values. Pass for all other values.: INSUFF

St2: we don't know about p: INSUFF

Combined: From st1, it was failing for only -ve fraction values of p (or r) but st2 removes that condition. : SUFF
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DS-prep (1/p > r/(r^2+2) ?) [#permalink] New post 23 Jun 2007, 23:40
Pl exp.
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 [#permalink] New post 24 Jun 2007, 04:08
(C) for me :)

1/p > r/(r^2+2) ?
<=> 1/p - r/(r^2+2) > 0 ?

From 1
p=r

So,
1/p - r/(r^2+2)
= 1/r - r/(r^2+2)
= [(r^2+2) - r^2] / [r*(r^2+2)]
= 2 / [r*(r^2+2)]
= 1/r * 2/(r^2+2)

As, r^2 >=0, we know that r^2 + 2>= 2 > 0

So, 1/p - r/(r^2+2) > 0 if and only if r > 0. But we do not know if r > 0.

INSUFF.

From 2
r > 0 and nothing about p.

INSUFF.

Both 1 & 2
We have the condition r > 0 for the statment 1 to be concluded.

SUFF.
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Is 1/p > r/(r^2 +2) [#permalink] New post 17 Apr 2011, 09:42
Is 1/p > r/(r^2 + 2)?

1. p = r
2. r > 0
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 17 Apr 2011, 10:02
junior wrote:
Is 1/p > r/(r^2 + 2)?

1. p = r
2. r > 0


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

1. p=r

\frac{1}{r} > \frac{r}{r^2+2}
\frac{r^2+2}{r} > r
r+\frac{2}{r} > r
Is \frac{2}{r} > 0

We don't know sign of r.

Not Sufficient.

2. r > 0
We don't know anything about p.
Not Sufficient.

Combining both;
We know \frac{2}{r} > 0 as r>0
Sufficient.

Ans: "C"
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 17 Apr 2011, 11:33
Please tell me what is wrong with my approach

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

r^2+2-pr>0
when p=r

r^2-(r*r)+2=2 which is >0
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 17 Apr 2011, 12:15
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meprarth wrote:
Please tell me what is wrong with my approach

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

r^2+2-pr>0
when p=r

r^2-(r*r)+2=2 which is >0


You can't cross multiply "-ve" numbers. We don't know what's "p", a "-ve" or "+ve"
If p=-ve, then cross multiplying will reverse the inequality sign.

\frac{1}{-2}>\frac{-2}{1}

Cross multiply:
1>4, which is WRONG.

Thus,
1/p > r/(r^2+2)
cannot be written as
r^2+2-pr>0 as the signs for p and r are unknown.

You can cross multiply a "+ve" number because that doesn't hurt the inequality.
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 17 Apr 2011, 18:15
The question asks :

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

(1) says p = r, so

1/r - r/(r^2 + 2) > 0

(r^2 + 2 - r^2)/r(r^2 + 2) > 0

or, 2/r(r^2 + 2) > 0

r^2 + 2 is positive, but the sign r is not known, so (1) is insufficient.


(2) says r > 0


but we still don't know for sure, e.g.,

p = 1, r = 1

r^2 + 2 = 3

1/p > 1/3


p = 10, r = 3

1/10 < 1/3

So (2) is insufficient

(1) and (2) says 2/r(r^2 + 2) > 0 and r > 0

So sufficient

Answer - C
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 18 Apr 2011, 11:58
fluke wrote:
junior wrote:
Is 1/p > r/(r^2 + 2)?

1. p = r
2. r > 0


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

1. p=r

\frac{1}{r} > \frac{r}{r^2+2}
\frac{r^2+2}{r} > r
r+\frac{2}{r} > r
Is \frac{2}{r} > 0

We don't know sign of r.

Not Sufficient.

2. r > 0
We don't know anything about p.
Not Sufficient.

Combining both;
We know \frac{2}{r} > 0 as r>0
Sufficient.

Ans: "C"

Dear Fluke
as far as i know inequilities
you cant cross multiply untill the signs are positive
you yourself said following first statement that we dont know the signs
and you even cross multiplied

rather it should have been
\frac{2}{r(r^2+2)} > 0 not 2/r

please correct me if i am wrong
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 18 Apr 2011, 12:06
Warlock007 wrote:
fluke wrote:
junior wrote:
Is 1/p > r/(r^2 + 2)?

1. p = r
2. r > 0


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

1. p=r

\frac{1}{r} > \frac{r}{r^2+2}
\frac{r^2+2}{r} > r
r+\frac{2}{r} > r
Is \frac{2}{r} > 0

We don't know sign of r.

Not Sufficient.

2. r > 0
We don't know anything about p.
Not Sufficient.

Combining both;
We know \frac{2}{r} > 0 as r>0
Sufficient.

Ans: "C"

Dear Fluke
as far as i know inequilities
you cant cross multiply untill the signs are positive
you yourself said following first statement that we dont know the signs
and you even cross multiplied

rather it should have been
\frac{2}{r(r^2+2)} > 0 not 2/r

please correct me if i am wrong


r^2+2 \ge 2, which is +ve. Thus, I multiplied.
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 22 Apr 2011, 18:09
1. Not sufficient.

we need to find out the following is true or not

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

=> 2/(r* (r^2+2)) >0 ?

=> r*(r^2+2)>0 ?

=> r >0 , (r^2+2)>0 (case 1 )
or r <0 , (r^2+2)<0 (case 2) case 2 cant be true as r^2 cannot be negative.

=> r>0?

2. Not sufficient as we dont know anything about p.

together we know p=r and r>0. Sufficient.

Answer is C.
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DS - Inequalities [#permalink] New post 04 May 2011, 08:33
Is 1/p > r/(r^2 + 2)?

(1) p = r
(2) r>0
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Re: DS - Inequalities [#permalink] New post 04 May 2011, 08:59
agdimple333 wrote:
Is 1/p > r/(r^2 + 2)?

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


Sol:

Is \hspace{3} \frac{1}{p} > \frac{r}{(r^2 + 2)}?

1. p=r

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

\frac{1}{r} - \frac{r}{(r^2 + 2)}>0

\frac{r^2+2-r^2}{r(r^2+2)}>0

\frac{2}{r(r^2+2)}>0

\frac{2}{r}>0-------------------------1

But, we don't know anything about r's sign.
Not Sufficient.

2. r>0------------------------2
p=0.000000000000000000000001 and r=1; answer will be YES.
p=-1 and r=1; answer will be NO.
Not Sufficient.

Combined, using eq1 and eq2:
\frac{2}{r}>0 as r>0
Sufficient.

Ans: "C"

P.S.: I remember answering this before. I am just not able to locate that thread.
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Re: DS - Inequalities [#permalink] New post 04 May 2011, 09:04
The concept tested here is inequalities

I dont see a rephrasing here because it depends on the sign of p. We can not cross multiply.

1) r > 0 but we have no info on p so Insuff - AD out

2) p =r so the question is 1/r > r/ r^2 +2 ?. If we plug r = 1 we get yes answer. However, if we plug -2 we get no. Insuff - B out

1+2) we can infer p^2 +2 > p^2 ? => 2 > 0? always yes for any p. Suff

The answer is C
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Re: [#permalink] New post 04 May 2011, 18:43
Fig wrote:
(C) for me :)

1/p > r/(r^2+2) ?
<=> 1/p - r/(r^2+2) > 0 ?

From 1
p=r

So,
1/p - r/(r^2+2)
= 1/r - r/(r^2+2)
= [(r^2+2) - r^2] / [r*(r^2+2)]
= 2 / [r*(r^2+2)]
= 1/r * 2/(r^2+2)

As, r^2 >=0, we know that r^2 + 2>= 2 > 0

So, 1/p - r/(r^2+2) > 0 if and only if r > 0. But we do not know if r > 0.

INSUFF.

From 2
r > 0 and nothing about p.

INSUFF.

Both 1 & 2
We have the condition r > 0 for the statment 1 to be concluded.

SUFF.


thank you. I like your explanation.
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Re: Is 1/p > r/(r^2 +2) [#permalink] New post 04 May 2011, 22:22
a p=r= positive LHS > RHS
p=r= negative LHS <RHS not sufficient.

b p<0 r>0 and p,r>0 give different values. not sufficient.

a+b p,r>0 hence LHS > RHS. hence C
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Re: Is 1/p > r/(r^2 +2)   [#permalink] 04 May 2011, 22:22
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