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DS: Is 1/p > r/(r^2+2) [#permalink ]

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05 Jul 2006, 21:38

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

1) p = r

2) r > 0

Please explain the answer

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Re: DS: Is 1/p > r/(r^2+2) [#permalink ]

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05 Jul 2006, 21:54

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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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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Yes 'C' it is.
We must be sure that r and p are both +ve or -ve.

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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 ]

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24 Jun 2007, 00:40

Pl exp.

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(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 ]

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17 Apr 2011, 10:42

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 ]

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17 Apr 2011, 11:02
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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 ]

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17 Apr 2011, 12:33

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 ]

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17 Apr 2011, 13:15
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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 ]

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17 Apr 2011, 19:15

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 ]

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18 Apr 2011, 12:58

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 ]

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18 Apr 2011, 13:06

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 ]

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22 Apr 2011, 19:09

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 ]

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04 May 2011, 09:33

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

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Re: DS - Inequalities [#permalink ]

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04 May 2011, 09:59

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 ]

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04 May 2011, 10:04

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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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 ]

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04 May 2011, 23:22

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)
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