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Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 07:31

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D

E

Difficulty:

55% (hard)

Question Stats:

59% (02:45) correct
41% (01:44) wrong based on 69 sessions

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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 of the students like brussels sprouts but dislike lima beans?

(1) 120 students eat in the cafeteria (2) 40 of the students like lima beans

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 09:17

'D' it is.

1) 120 students total.
2/3(120) = 80 students dislike lima beans. So 40 like lima beans.
3/5(80) = 48 dislike brussel sprouts, So 32 students like brussel sprouts but dislike lima beans.

Sufficient.

2) 40 students like lima beans.
From the question stem 2/3 dislikes lima beans, so 1/3 likes lima beans.
1/3 = 40, 2/3 = 80.
Solve further the same way with 1).

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 09:18

We know each student either (likes OR dislikes) - A. beans; B. sprouts.

Out of all the students, 2/3 dislikes lima beans.... (I) Therefore 1/3 likes lima beans ..... (II)

..and of those who dislike lima beans, 3/5 also dislikes brussel sprouts:
That means 3/5th of 2/3rd students dislike sprouts. So:
2/3 of 3/5 = 2/3 x 3/5 = 2/5 th of all students dislike sprouts .... (III)

From (III), if 2/5 of all students dislike sprouts, then the remaining 3/5 like sprouts ... as they all either like or dislike sprouts....(IV)

Also, of those who dislike lima beans, 3/5 also dislikes brussel sprouts. That means 2/5 (of those whole dislike lima beans) likes sprouts. So:
2/5 of 2/3 = 2/5 x 2/3 = 4/15 of all students like sprouts but dislike means ..... (V)

NOW, (1) says 120 students eat at cafeteria. Then our answer is 4/15 of 120.

(2) says 40 students like lima beans.
From (II) above, 40 students = 1/3 of total # of students. From here we can find what 4/15 of students will be.

Hence D.

Last edited by gmatmba on 22 May 2006, 21:55, edited 1 time in total.

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 11:43

M8 wrote:

While I was solving the qustion you guys gave the answer. But I hope that my explanation is clear for you.

Ur explanation is absolutely valid!! _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 12:33

remgeo wrote:

We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.

1) Out of these 80, some like/dislike sprouts. 2)Out of these 40 also some like/dislike sprouts.

We are given only the 1st part.

We do not know anything about no of people who like beans but like/dislike.

But we are also given in the stem: "Of these students, 2/3 dislikes lima beans; and of those who dislike lima beans, 3/5 also dislikes brussel sprouts".
So from here we know it all

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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22 May 2006, 12:42

gmatmba wrote:

remgeo wrote:

We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.

1) Out of these 80, some like/dislike sprouts. 2)Out of these 40 also some like/dislike sprouts.

We are given only the 1st part.

We do not know anything about no of people who like beans but like/dislike.

But we are also given in the stem: "Of these students, 2/3 dislikes lima beans; and of those who dislike lima beans, 3/5 also dislikes brussel sprouts". So from here we know it all

I think I'm thinking too much and getting confused
Time for a good sleep I guess.

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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23 May 2006, 10:51

remgeo wrote:

We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.

1) Out of these 80, some like/dislike sprouts. 2)Out of these 40 also some like/dislike sprouts.

We are given only the 1st part.

We do not know anything about no of people who like beans but like/dislike.

Hi Remgeo.
I see your problem. But we know needed info about those people.

Will try to explain.

From the question stem: each student either likes or dislikes lime beans and each students either likes or dislikes brussel sprouts.

Of these students, 2/3 dislikes lima beans 1) There are 120 people in the cafeteria.
2/3 (80) dislikes lima beans, so the rest (40) of the 120 - likes lime beans.
and of those who dislike lima beans, 3/5 also dislikes brussel sprouts. 3/5 (of 80 who dislikes lima beans) dislikes brussel sprouts = 48, So the rest of 80-48=32 students like brussel sprouts but at the same time dislike lima beans.

2) 40 students like lima beans.
From the question stem we have 2/3 who dislikes lima beans. So 1/3 who likes lima beans. And those 1/3 = 40, so 2/3 = 80, TOTAL = 120 students, then solve by analogy with 1).

Re: Of the students who eat in a certain cafeteria, each student either [#permalink]

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24 May 2006, 04:42

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D

1) total students 120
From the question,
Number of people who dislike lima = 2/3*120 = 80
We have to find number of people who like brussel sprouts and dislike lima beans. One part we hae found number of people who dislike lima. Lets find the second part

Also from the question ,
Number of people who dislike bussel sprout as well as dislike lima = 3/5*80 = 48
Now number of people who like brussel sprout but dislike lima = 80-48= 32

2) 40 donot like lima beans. hence no of people = 3/2*40 = 120
Now use the same reasoning as 1

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