Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 12 Apr 2006
Posts: 215
Location: India

DS: Is 1/p > r/(r^2+2) [#permalink]
Show Tags
05 Jul 2006, 21:38
3
This post was BOOKMARKED
Question Stats:
66% (02:22) correct
34% (00:59) wrong based on 90 sessions
HideShow timer Statistics
Is 1/p > r/(r^2+2)
1) p = r
2) r > 0
Please explain the answer
Official Answer and Stats are available only to registered users. Register/ Login.



VP
Joined: 25 Nov 2004
Posts: 1483

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



SVP
Joined: 30 Mar 2006
Posts: 1728

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)



Senior Manager
Joined: 08 Jun 2004
Posts: 495
Location: Europe

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



CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD  Class of 2008

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
_________________
SAID BUSINESS SCHOOL, OXFORD  MBA CLASS OF 2008



Director
Joined: 03 Sep 2006
Posts: 871

DSprep (1/p > r/(r^2+2) ?) [#permalink]
Show Tags
24 Jun 2007, 00:40
Pl exp.
Attachments
untitled.PNG [ 68.39 KiB  Viewed 3751 times ]



SVP
Joined: 01 May 2006
Posts: 1796

(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.



Intern
Joined: 23 Feb 2010
Posts: 19

Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
17 Apr 2011, 10:42
Is 1/p > r/(r^2 + 2)?
1. p = r 2. r > 0



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
17 Apr 2011, 11:02
1
This post received KUDOS
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"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Intern
Status: Life.. Be Kind to me
Joined: 12 Jan 2011
Posts: 20
Location: India
WE: Engineering (Telecommunications)

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
17 Apr 2011, 12:33
Please tell me what is wrong with my approach
\(1/p > r/(r^2+2)\)
\(r^2+2pr>0\) when p=r
\(r^2(r*r)+2=2 which is >0\)



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
17 Apr 2011, 13:15
2
This post received KUDOS
meprarth wrote: Please tell me what is wrong with my approach
\(1/p > r/(r^2+2)\)
\(r^2+2pr>0\) 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 \(r^2+2pr>0\) as the signs for p and r are unknown. You can cross multiply a "+ve" number because that doesn't hurt the inequality.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



SVP
Joined: 16 Nov 2010
Posts: 1663
Location: United States (IN)
Concentration: Strategy, Technology

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
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
_________________
Formula of Life > Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership  big benefits and savings



Manager
Status: ==GMAT Ninja==
Joined: 08 Jan 2011
Posts: 242
Schools: ISB, IIMA ,SP Jain , XLRI
WE 1: Aditya Birla Group (sales)
WE 2: Saint Gobain Group (sales)

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
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
_________________
WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/mytestexperience111610.html do not hesitate me giving kudos if you like my post.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
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.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 01 Feb 2011
Posts: 755

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



Senior Manager
Joined: 24 Mar 2011
Posts: 447
Location: Texas

DS  Inequalities [#permalink]
Show Tags
04 May 2011, 09:33
Is 1/p > r/(r^2 + 2)?
(1) p = r (2) r>0



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: DS  Inequalities [#permalink]
Show Tags
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. \(p=r\) \(\frac{1}{r} > \frac{r}{(r^2 + 2)}\) \(\frac{1}{r}  \frac{r}{(r^2 + 2)}>0\) \(\frac{r^2+2r^2}{r(r^2+2)}>0\) \(\frac{2}{r(r^2+2)}>0\) \(\frac{2}{r}>0\) 1But, we don't know anything about r's sign. Not Sufficient. 2. \(r>0\) 2p=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.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Intern
Status: ThinkTank
Joined: 07 Mar 2009
Posts: 28

Re: DS  Inequalities [#permalink]
Show Tags
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
_________________
http://www.hannibalprep.com



Senior Manager
Joined: 24 Mar 2011
Posts: 447
Location: Texas

Fig wrote: (C) for me 1/p > r/(r^2+2) ? <=> 1/p  r/(r^2+2) > 0 ? From 1p=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 2r > 0 and nothing about p. INSUFF. Both 1 & 2We have the condition r > 0 for the statment 1 to be concluded. SUFF. thank you. I like your explanation.



VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1326

Re: Is 1/p > r/(r^2 +2) [#permalink]
Show Tags
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
_________________
Visit  http://www.sustainablesphere.com/ Promote Green Business,Sustainable Living and Green Earth !!




Re: Is 1/p > r/(r^2 +2)
[#permalink]
04 May 2011, 23:22



Go to page
1 2
Next
[ 21 posts ]




