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by ruplun » Fri Jul 15, 2011 8:04 am
Of the students who eat in a certain cafeteria , each student either likes or dislikes lime beans and each student either likes or dislikes brussel sprouts. Of these students,2/3 dislike lima beans ; and of those who dislike lima beans , 3/5 also dislike brussel sprouts. How many of the students like brussel sprouts but dislike lima beans?

a. 120 students eat in the cafeteria
b. 40 of the students like lima beans
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by krishnasty » Fri Jul 15, 2011 8:53 am
ruplun wrote:Of the students who eat in a certain cafeteria , each student either likes or dislikes lime beans and each student either likes or dislikes brussel sprouts. Of these students,2/3 dislike lima beans ; and of those who dislike lima beans , 3/5 also dislike brussel sprouts. How many of the students like brussel sprouts but dislike lima beans?

a. 120 students eat in the cafeteria
b. 40 of the students like lima beans
I guess this question should have been posted in DS and not in PS.
nevertheless, IMO E .

Reason:
we are required to get the number of students who like brussel sprouts but dislike lima beans.
let x = number of students
number of students who dislike lima beans = (2/3)x
...and...
(3/5) of (2/3)x dislike brussel sprouts.

(1) - jst gives the value of x and hence, we can calculate the people who dislike lima beans and out of those, who like brussel sprouts. but wat about people who like brussel sprouts and same out of those, dislike brussel sprouts.
INSUFFICIENT

(2) - saying the same thing as said in (1).
INSUFFICIENT

combining both , INSUFFICIENT

hence, E
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by Frankenstein » Fri Jul 15, 2011 9:13 am
ruplun wrote:Of the students who eat in a certain cafeteria , each student either likes or dislikes lime beans and each student either likes or dislikes brussel sprouts. Of these students,2/3 dislike lima beans ; and of those who dislike lima beans , 3/5 also dislike brussel sprouts. How many of the students like brussel sprouts but dislike lima beans?

a. 120 students eat in the cafeteria
b. 40 of the students like lima beans
Hi,
Let n be the total number of students
2n/3 dislike Lima
Of these (2n/3), 3/5 also dislike brussel. So, the remaining 2/5 are the ones who like brussel.
So, essentially these are the number of students who dislike Lima but like brussel and this number will be (2n/3)*(2/5) = 4n/15
So, we need to calculate 4n/15
From(1):
n = 120
So, 4n/15 can be calculated and it is (4/15)*120 = 32
Sufficient
From(2):
40 like Lima
2n/3 dislike Lima So, n/3 like Lima.
So, n/3 = 40 =>n =120
Again we can calculate 4n/15 = 32
Sufficient

Hence, D
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by nafiul9090 » Sat Jul 16, 2011 6:51 am
its D
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by tzohrabyan » Sat Jul 16, 2011 11:31 am
I agree with Frankenstein that with the first statement it is sufficient to solve the problem. However, i want to ask - the second info says 40 of the students - but which students? the first one specifically mentions 120 students who eat in cafeteria. What if these 40 people are people who do not eat or is the total of those who eat and not eat. So, unless i am too pick about words of gmat, it should be A because only the first one specifically mentions that the total of the students who EAT in CAFETERIA is 120.
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by Frankenstein » Sat Jul 16, 2011 9:22 pm
tzohrabyan wrote:I agree with Frankenstein that with the first statement it is sufficient to solve the problem. However, i want to ask - the second info says 40 of the students - but which students? the first one specifically mentions 120 students who eat in cafeteria. What if these 40 people are people who do not eat or is the total of those who eat and not eat. So, unless i am too pick about words of gmat, it should be A because only the first one specifically mentions that the total of the students who EAT in CAFETERIA is 120.
Hi,
Even the question stem says :
"How many of the students like brussel sprouts but dislike lima beans? "
It doesn't explicitly state students who eat in the cafeteria. So, I guess we can take it for granted. Anyway, I think questions will be clear on the real GMAT.
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