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Detailed solution will be appreciated. Thanks

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Detailed solution will be appreciated. Thanks [#permalink] New post 10 Jun 2008, 03:39
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A
B
C
D
E

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Re: GMATPREP-DS [#permalink] New post 10 Jun 2008, 05:42
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, 05:34, edited 1 time in total.
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Re: GMATPREP-DS [#permalink] New post 10 Jun 2008, 05:43
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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
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Re: GMATPREP-DS [#permalink] New post 10 Jun 2008, 13:00
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i say D. I drew one of those boxes

Like Dislike
Lima 1/3 2/3
Brussel 2/3 *3/5
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Re: GMATPREP-DS [#permalink] New post 10 Jun 2008, 18:10
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
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Re: GMATPREP-DS [#permalink] New post 11 Jun 2008, 06:44
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 ?
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Re: GMATPREP-DS [#permalink] New post 11 Jun 2008, 21:23
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 ?
Re: GMATPREP-DS   [#permalink] 11 Jun 2008, 21:23
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