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

[rschunti]

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?

[RonPurewal]

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 + NO 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

[tarek99]

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?

[Guest]

oh it's ok! i got it now after carefully reading the question again. :)

[StaceyKoprince]

Great! Love it when you figure it out for yourself!

[lalitkc]

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

[suhpra]

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

[RonPurewal]

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)

[gmatwork]

Hi,

I tried solving this question using double set matrix as discussed in Strategy guides but was not able to do so. I am sure I am doing something wrong. After filling out the table, when I look at what is unknown it just doesn't make sense.

I used B (Beans), S(sprouts) - on the top row of the table and L (like), D (Dislike) on the vertical row of the table.

What is wrong with this? I am missing out something here?

Will double set matrix approach always work for an overlapping sets question involving two sets? Under what circumstance will this approach not work?

[RonPurewal]

erpriyankabishnoi, ya, you are setting up the matrix incorrectly.

the sine qua non of the double-set matrix is that the first two rows must represent mutually exclusive situations, and the first two columns must also represent mutually exclusive situations.

so, you've actually got two issues here: not only are "lima beans" and "brussels sprouts" not mutually exclusive, but they are in fact not situations at all.

the proper way to set up the double-set matrix here is to have "like lima beans" and "dislike lima beans" (mutually exclusive) on the rows, and "like brussels sprouts" and "dislike brussels sprouts" (also mutually exclusive) on the columns. or, of course, vice versa.

try it that way; you should be able to get it to work.

[gmatwork]

Thank you, Ron. As matter of fact I did the same mistake today on one another problem and realized this issue. Thanks for pointing it out!

Also for some reason, while doing D/S Overlapping sets questions I tend to get slowed down by drawing three separate matrices- for each statement and given info.

The complex language (way the percentage figures are presented some with respect to a subgroup and others with respect to the whole population are so mind boggling some times. Is there something ( apart from practice) I can do speed up overlapping sets' problems?

[tim]

practice! :)

[michael.lattari]

Keep in mind the statement says "each either likes or dislikes lima and each likes or dislikes brussels" I believe this means that none of the students can like both.

-mike

[RonPurewal]

Keep in mind the statement says "each either likes or dislikes lima and each likes or dislikes brussels" I believe this means that none of the students can like both.

-mike

No. The point of this wording is that the blue thing and the purple thing are separate issues.

That's why the wording is so bulky and un-smooth in the first place"”to emphasize the fact that absolutely any combination of these is possible.
If "not both" were the intended interpretation, the words would say "... but not both", or "exactly one", or some other equivalent.

[RonPurewal]

To see that they're separate issues, note that you're perfectly able to replace either the blue or the purple thing with a completely dissimilar issue.

E.g.,
each student either likes or dislikes lima beans and each student either can or can't drive a stick shift.

If the conditions were actually related to each other"”e.g., "Joe likes X or likes Y, but not both""”then this substitution wouldn't be possible.
In this problem, the fact that both conditions deal with "liking things" is completely a coincidence.