It is currently 19 Oct 2017, 10:06

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

Is (p/q) > ( /s) 1) (p/q) > (r/s) 2) r = 1

Author Message
Director
Joined: 01 May 2007
Posts: 795

Kudos [?]: 379 [0], given: 0

Is (p/q) > ( /s) 1) (p/q) > (r/s) 2) r = 1 [#permalink]

Show Tags

16 Jun 2008, 19:58
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.

Is (p/q) > ([r^2]/s)

1) (p/q) > (r/s)
2) r = 1

Kudos [?]: 379 [0], given: 0

Manager
Joined: 08 Jun 2008
Posts: 70

Kudos [?]: 9 [0], given: 0

Show Tags

16 Jun 2008, 20:15
1. It gives different (r^2)/s and r/s values for different numbers +ve, -ve and fractions. - Insufficient.
2. It simply says that p/q > 1/s - Insufficient.

Combine 1 & 2 it should be sufficient.

if r = 1 & p/q > r/s it means p/q > (r^2)/s

Kudos [?]: 9 [0], given: 0

CEO
Joined: 17 May 2007
Posts: 2947

Kudos [?]: 666 [0], given: 210

Show Tags

16 Jun 2008, 20:16
is it C ?

reason :
if $$r = 1$$ then $$\frac{r}{s}= \frac{r^2}{s}$$

Hence when we combine A and B
$$\frac{p}{q} > \frac{r}{s}$$ and hence$$\frac{p}{q} > \frac{r^2}{s}$$

Kudos [?]: 666 [0], given: 210

Director
Joined: 01 May 2007
Posts: 795

Kudos [?]: 379 [0], given: 0

Show Tags

16 Jun 2008, 20:26
It is C, but can you prove it by eliminating statement 1 and 2 first, before selecting C?

Kudos [?]: 379 [0], given: 0

CEO
Joined: 17 May 2007
Posts: 2947

Kudos [?]: 666 [0], given: 210

Show Tags

16 Jun 2008, 20:41
B is easy to eliminate - it provides no info what so ever on p and q.

A states that
$$\frac{p}{q}> \frac{r}{s}$$

consider $$p = 1, q = 1, r = -1, and, s = 1$$
in this case $$\frac{p}{q} = 1 > \frac{r}{s}$$ (which is -1)
however, $$\frac{r^2}{s}$$= 1 which is EQUAL to $$\frac{p}{q}$$

hence A is insufficient.

jimmyjamesdonkey wrote:
It is C, but can you prove it by eliminating statement 1 and 2 first, before selecting C?

Kudos [?]: 666 [0], given: 210

Senior Manager
Joined: 07 Jan 2008
Posts: 398

Kudos [?]: 290 [0], given: 0

Show Tags

17 Jun 2008, 05:12
C.

1. p/q > r/s. INSUF. (BSD has good explanation.)
2. r=1. INSUF

==> Both: SUF

Kudos [?]: 290 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 618

Kudos [?]: 198 [0], given: 1

Show Tags

17 Jun 2008, 06:27
jimmyjamesdonkey wrote:
Is (p/q) > ([r^2]/s)

1) (p/q) > (r/s)
2) r = 1

1: insuf
For all p,q,s > 0 and r < 0 condition 1 is satisfied but not necessarily (p/q) > (r^2/s)
2: insuf
(p,q,s) = (1,1,2) or (1,2,1)

1&2 suf (p/q) > (r/s) = (1*r)/s = r^2/s

Kudos [?]: 198 [0], given: 1

Re: DS Inequalities   [#permalink] 17 Jun 2008, 06:27
Display posts from previous: Sort by