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ds - set [#permalink] New post 25 Sep 2005, 12:54
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A
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D
E

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Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans, and of those who dislike lima beans, 3/5 also dislike Brussels sprouts. How many students like brussels sprouts but dislike lima beans ?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans

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 [#permalink] New post 25 Sep 2005, 13:48
I would pick D...

X is equal to total students.

X/3 like lima beans...2X/3 dislike lima beans

we are told 3/5 of 2X/3 like brussel sprouts...

we need to know number of students

(1) is suff

(2) is suff, we can derive total number of students from info given...
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 [#permalink] New post 25 Sep 2005, 16:06
D here as well. Same reasoning as above.
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 [#permalink] New post 26 Sep 2005, 21:11
According to 1,
If we have 120 students, 80 students dislike lima beans and out of these 80, 48 dislike brussel sprouts...

But we dont know that, 48 dislike both, and 32 dislike lima beans but we still dont know the exact number that dislike just brussel sprouts...
So insufficient


Same with 2,
We know the number of people that like lima beans, but not the exact number that like brussel sprouts...

Even with 1 and 2, insufficient. I will go for E.
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Don't Like OA [#permalink] New post 27 Sep 2005, 13:50
I don't know about this OA... the answer only addresses the small group of students who dislike limabeans, and not the group that likes lima beans. Could there be brussel sprout lovers in that group? Is this a real GMAT question... and if so how are we to handle this?

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 [#permalink] New post 28 Sep 2005, 04:11
The answer is indeed D.

(1) Suppose there are 120 students.
Then 2/3*120 = 80 dislike Lima beans.
And 3/5*80 = 48 dislike both Lima beans and Brussels sprouts.
We are told that all students have an opinion.
So 80-48 = 32 must dislike Lima beans and like Brussels sprouts.

(2) Suppose there are 40 students who like Lima beans.
We know that 2/3 of the total dislike Lima beans.
We are told that all students have an opinion.
So 1/3 of the total like Lima beans.
Hence there must be 40/(1/3) = 120 students.
Argument follows from above.

We do not know how the 40 are split, but we are not asked for that.
  [#permalink] 28 Sep 2005, 04:11
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