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Of the students who eat in a certain cafeteria, each student

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rschunti
 Post subject: Of the students who eat in a certain cafeteria, each student
 Post Posted: Wed Dec 19, 2007 7:35 pm 
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 sprout but dislike lima beans?

1). 120 students eat in the cafeteria.

2). 40 of the students like lima beans.

This is a GMATPREP question. What is the best way to solve it?
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RonPurewal
 Post subject:
 Post Posted: Fri Dec 21, 2007 5:17 am 
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ManhattanGMAT Staff


Posts: 7146
work backwards from the information they give you.


(1)
120 students total
--> 40 YES LIMA, 80 NO LIMA
split up the NO LIMA crowd:
--> 3/5 of 80 = 48 NO LIMA + YES BRUSSELS
--> 2/5 of 80 = 36 NO LIMA + YES BRUSSELS
sufficient

(2)
the fraction 2/3 tells you that the other 1/3 like lima beans (so that the ratio YES LIMA : NO LIMA is 1 : 2)
therefore, x (the unknown multiplier) = 40, and 2x is therefore 80
--> 40 YES LIMA, 80 NO LIMA
rest of the problem proceeds as above
sufficient

answer = d
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tarek99
 Post subject:
 Post Posted: Thu Jan 24, 2008 5:11 am 
but ron, in statement 1, you're assuming that only from the remaining 80 of whose who dislike lima beans would also like brussel beans? what if there is a portion from those who like lima beans would also like brussel beans?
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Guest
 Post subject:
 Post Posted: Thu Jan 24, 2008 5:18 am 
oh it's ok! i got it now after carefully reading the question again. :)
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StaceyKoprince
 Post subject:
 Post Posted: Fri Jan 25, 2008 2:24 am 
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ManhattanGMAT Staff


Posts: 6077
Location: San Francisco
Great! Love it when you figure it out for yourself!

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Stacey Koprince
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lalitkc
 Post subject: Re: Of the students who eat in a certain cafeteria, each student
 Post Posted: Mon Nov 16, 2009 1:55 pm 
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Students


Posts: 11
Let L = students who like Lima beans ; L’ = students who dont like Lima beans
B = students who like Brussel beans B’ = students who don’t like Brussel beans
S = total number of students who eat in the cafeteria

Given: L’ = 2/3 S ==> L = 1/3 S -- eqn 1
Students who dislike both L & B = B’L’= 3/5 L’ ==> students who like B but dislike L = BL’ = 2/5 L’ –- eqn 2
Question is: Find BL’
St 1 ==> S = 120 ==> From eqn 1 & 2, we can find BL’. So Stmt 1 is sufficient
St 2 ==> L = 40 ==> L’ = 80 ==> From eqn 2 we can find BL’. So Stmt 2 is sufficient
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suhpra
 Post subject: Re: Of the students who eat in a certain cafeteria, each student
 Post Posted: Wed Dec 01, 2010 3:37 pm 
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Posts: 2
I want to re-open the thread as my understanding is different from the others.

I thought this way X-Y = those who like BS but dislike LB
where X= all those ppl who like BS and Y= all those who dislike LB

So we know that X= 32 [Like BS but dislike LB] + ? [Like BS and like LB]

So, I would expect the answer to be X-80. And since, we do not know how many among those who Like LM, Like BS, X can not be known.

Can anyone please help me out!

Thanks in advance.
Suhas
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RonPurewal
 Post subject: Re: Of the students who eat in a certain cafeteria, each student
 Post Posted: Thu Dec 02, 2010 9:17 am 
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ManhattanGMAT Staff


Posts: 7146
suhpra wrote:
I want to re-open the thread as my understanding is different from the others.

I thought this way X-Y = those who like BS but dislike LB
where X= all those ppl who like BS and Y= all those who dislike LB

So we know that X= 32 [Like BS but dislike LB] + ? [Like BS and like LB]

So, I would expect the answer to be X-80. And since, we do not know how many among those who Like LM, Like BS, X can not be known.

Can anyone please help me out!

Thanks in advance.
Suhas


whoa, no, you can't do that.

here's an analogy for why not:
let's say that, in an auditorium, there are 10 people from fresno, california.
now let's say that the same auditorium contains 40 people who don't play football.
according to your reasoning above, then, the number of people from fresno who do play football would be ... negative 30.
so you can see why that wouldn't work.

the easiest way for you to see which quantities legitimately do add and subtract is to construct a double set matrix (see our word translations strategy guide, if you don't know what that is), and then just look carefully at the headings on the rows and columns, all of which are additive.
if you do that, you will discover that the correct subtraction is
(like B but dislike L)
= (ALL who like B) - (like B and like L)
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