GMatPrep-DS : DS Archive
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 08 Dec 2016, 22:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# GMatPrep-DS

Author Message
Intern
Joined: 05 Mar 2007
Posts: 24
Followers: 0

Kudos [?]: 1 [0], given: 0

### Show Tags

09 Sep 2007, 07:31
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

Attachments

DS7.JPG [ 77.42 KiB | Viewed 657 times ]

Senior Manager
Joined: 27 Aug 2007
Posts: 253
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

09 Sep 2007, 07:39
Just plug in the numbers once negative than positive, you will see that they are siffices together

Ans: C
Director
Joined: 17 Sep 2005
Posts: 924
Followers: 4

Kudos [?]: 72 [0], given: 0

### Show Tags

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
Director
Joined: 08 Jun 2007
Posts: 583
Followers: 2

Kudos [?]: 97 [0], given: 0

### Show Tags

16 Sep 2007, 11:20
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.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 7

Kudos [?]: 187 [0], given: 0

### Show Tags

16 Sep 2007, 12:19
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

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30

Kudos [?]: 348 [0], given: 0

### Show Tags

16 Sep 2007, 22:15
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
Display posts from previous: Sort by