It is currently 21 Oct 2017, 09:11

### 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

# Detailed solution will be appreciated. Thanks

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
VP
Joined: 18 May 2008
Posts: 1258

Kudos [?]: 527 [1], given: 0

Detailed solution will be appreciated. Thanks [#permalink]

### Show Tags

10 Jun 2008, 04:39
1
KUDOS
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.

Detailed solution will be appreciated.
Thanks
Attachments

matrix problem.jpg [ 93.94 KiB | Viewed 1150 times ]

Kudos [?]: 527 [1], given: 0

Manager
Joined: 24 Apr 2008
Posts: 161

Kudos [?]: 47 [0], given: 0

### Show Tags

10 Jun 2008, 06:42
Ok - will explain -

Firstly the students can be grouped into following:
LS for like sprout DS for dislike sprout similarly LL and DL
A student can be: LS-LL (P), LS-DL (Q), DS-LL (R) or DS-DL (S)
We need to find Q

Now: Q+S = 2/3N (N being total numner of students) and
S=3/5 X 2/3N = 2/5N

St1: N=120 hence from above equations : Q = 80-48 = 32 hence sufficient

St2: P + R = 40 doesnt help to find Q hence insufficient!!

Ans: A what is OA?

Edit: Yes: I now think D should be the answer
as statemt 2 implies P+R=N/3 thus can be solved for S. Sorry!! God save me!!!

Last edited by iamcartic on 11 Jun 2008, 06:34, edited 1 time in total.

Kudos [?]: 47 [0], given: 0

Intern
Joined: 10 Jun 2008
Posts: 7

Kudos [?]: 4 [1], given: 0

### Show Tags

10 Jun 2008, 06:43
1
KUDOS
let X = # of students

DISLIKE(lima beans) = (2/3 * X)
DISLIKE(lima beans) AND DISLIKE(brussels sprouts) = (2/3 * X) * (3/5)

let Y = DISLIKE(lima beans) AND LIKE(brussels sprouts) = (2/3 * X) * (1 - 3/5)

The question now becomes: do we have enough info to solve for Y?

Statement 1 tells us that X = 120; this alone is sufficient to solve for Y
Statement 2 tells us that...
LIKE(lima beans) = 40

Since
LIKE(lima beans) = 1 - DISLIKE(lima beans) = 1 - (2/3 * X) = 40
this lets us solve for X, so statement 2 alone is also sufficient

So the answer should be EACH is sufficient

Kudos [?]: 4 [1], given: 0

Intern
Joined: 29 Mar 2006
Posts: 26

Kudos [?]: 5 [1], given: 0

### Show Tags

10 Jun 2008, 14:00
1
KUDOS
i say D. I drew one of those boxes

Like Dislike
Lima 1/3 2/3
Brussel 2/3 *3/5

Kudos [?]: 5 [1], given: 0

Manager
Joined: 01 May 2008
Posts: 111

Kudos [?]: 7 [0], given: 0

Location: São Paulo

### Show Tags

10 Jun 2008, 19:10
Answer should be D!
Like Dislike
Lima 5/15 10/15
Brussels 6/15 9/12

Red- given
Blue- calculated

Statement (1) - 120 is the number of students, so 6/15, or 48 like Brussels Sprouts, and 5/15, or 40, like Lima Beans. 48-40 = 8
Statement (2) - 40 like Lima Beans, so 48 like Brussels Sprouts. 48-40 = 8

Kudos [?]: 7 [0], given: 0

Intern
Joined: 05 Mar 2008
Posts: 15

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

### Show Tags

11 Jun 2008, 07:44
Yes answer is D here.

Here is my calculations for each choice:

No. of students dislike Brussel sprouts who dislike Lima beans = 3/5 * 2/3N = 6/15N

1) Total # of students N = 120

No. of students dislike Brussel sprouts who dislike Lima beans = 2/5N = 2/5*120 = 48- Suff.

2) 40 of the students like lima beans

This is nothing but 1-2/3 N = 1/3N

So 1/3N = 40 --> N = 120
No. of students dislike Brussel sprouts who dislike Lima beans = 6/15N = 6/15*120 = 48 - Suff.

So answer is D.

Right ?

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

VP
Joined: 18 May 2008
Posts: 1258

Kudos [?]: 527 [0], given: 0

### Show Tags

11 Jun 2008, 22:23
try using matrix method. that is the bets way to do such kinda problems.yup answer is D
pbvmba wrote:
Yes answer is D here.

Here is my calculations for each choice:

No. of students dislike Brussel sprouts who dislike Lima beans = 3/5 * 2/3N = 6/15N

1) Total # of students N = 120

No. of students dislike Brussel sprouts who dislike Lima beans = 2/5N = 2/5*120 = 48- Suff.

2) 40 of the students like lima beans

This is nothing but 1-2/3 N = 1/3N

So 1/3N = 40 --> N = 120
No. of students dislike Brussel sprouts who dislike Lima beans = 6/15N = 6/15*120 = 48 - Suff.

So answer is D.

Right ?

Kudos [?]: 527 [0], given: 0

Re: GMATPREP-DS   [#permalink] 11 Jun 2008, 22:23
Display posts from previous: Sort by

# Detailed solution will be appreciated. Thanks

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.