overlapping sets in gmat prep

[nageshb25]

Of the students who eat in a certain cafeteria, each student either likes or dislikes Lima beans and each studende either likes or dislikes Bussels sprouts. Of these students 2/3 dislike Lima beans; and of these who dislike Lima beans, 3/5 also dislike Burssel sprouts. How many of the students like brussel sprouts but dislike Lima beans

1) the total number of students are 120

2) students who dislike Lima beans are 40

[poonammarwah23]

Like L Dislike L Total
Like B 4/15t
Dislike B 2/5 t
Total 2/3 t t

Say total number of students who eat in a cafeteria = t

Out of t, 2/3 dislike Lima beans = 2/3t

Out of 2/3t, 3/5 also dislike Burssel sprouts= 2/3x3/5= 2/5t dislike both L and B
That means Out of 2/3t ,2/5(1-3/5) Dislike L and Like B : 2/3x2/5= 4/15t

if we know t we can find out the answer.

Statement 1: Total number of students are given but out of these how many eat in cafeteria is not known -- Not Sufficient.

Statement 2: 2/3t = 40 ; t =60 Sufficient.

Ans: B

OA please

[luc2r4]

From the Stem : Dislike Lima = ( 2/3 )* total of students

I made a kind of binomial tree and found that :

Dislike ( Brussels and Lima) = (3/5)*( people who dislike LImA)

Then Dislike ( Brussels and Lima) = (3/5)*( ( 2/3 )* total of students)

Given

1) the total number of students are 120

2) students who dislike Lima beans are 40

I BELIEVE that answer = D each alone

[tejkumar.m]

The OA must be Either of the ones is sufficient.

Explanation:

It can be easily explained using a figure.

let

a = ppl who like L only
b = ppl who like both L and S
c = ppl who like S only
d = ppl who dont like either

it is given that

2/3 (a+b+c+d) = c+d ----- equation 1

and (3/5)*(2/3)* (a+b+c+d) = d ===>>> 2/5 * (a+b+c+d)

For knowing how many ppl like S only i.e. "c" we need to have either the total number of students i.e. (a+b+c+d) OR value of "c+d" i.e. students who dont like L using which one can find (a+b+c+d) value from equation 1..

1) a+b+c+d value is given = 120 ... Sufficient
2) c+d is given = 40 ... can find a+b+c+d and there by value "c"

[mschwrtz]

nageshb25, please be sure to post OA with question. Should be D.

poonammarwah23, nice matrix. However, the "students" in the statements are the very same "students who eat in a certain cafeteria" we met in the question, so S1 is sufficient. You seem to have recognized who the "students" were when you turned to statement 2, since you correctly inferred that S2 is sufficient.

[kanchana.ramar]

It is easier to visualize this with a Venn diagram. We get 4 circles. One each for L, L',B and B'. The intersection of L and L' and B and B' are null (can't have it both ways and sit on the fence!).

The interesting thing is that is that the students who dislike like lima beans, which is 2/3 of the population, have to belong to either one of the groups-> L'B or L'B'. i.e. L'B + L'B=2/3 of the population.(This is true for each of the 4 circles of the Venn diagram).
We know L'B' (=2/5 of the population) and therefore L'B= 4/15 of the population, so, we could work out the number if we knew the population.
Statement 1 gives us the population straight away!
Statement 2 also gives a clue to the population (40 like lima beans and hence constitute 1/3 of the population).
Hope this helps!

[RonPurewal]

Of the students who eat in a certain cafeteria, each student either likes or dislikes Lima beans and each studende either likes or dislikes Bussels sprouts. Of these students 2/3 dislike Lima beans; and of these who dislike Lima beans, 3/5 also dislike Burssel sprouts. How many of the students like brussel sprouts but dislike Lima beans

1) the total number of students are 120

2) students who dislike Lima beans are 40

well, it's rather pointless to argue about which of two equivalent diagrams is "easier"; that's basically a personal issue. (note that the previous poster's idea of 4 intersecting venn diagrams is extremely similar to the 4 squares at the top left of the double-set matrix, which correspond exactly to what the poster is calling BL, B'L, BL', and B'L'.)

in any case, if x is the total population, it's fairly easy to see that 2x/3 is the number of people who don't like lima beans; therefore, the crux of the problem is the realization that 3/5 of that result, or (3/5)(2x/3) = 2x/5, represents the number of people who dislike lima beans *and* dislike brussels sprouts, and that the other 2/5 of that result, (2/5)(2/5) = 4x/15 (which can also be gotten by subtracting the 2x/5 from the 2x/3), represents people who dislike lima beans but *like* brussels sprouts.

so, basically, whichever diagram helps you to reach that conclusion is the diagram that you should use.

[sourabh.ag]

Aren't the statements contradictory

Of these students 2/3 dislike Lima beans

1) the total number of students are 120

2) students who dislike Lima beans are 40

If we take the statement 1, the total number of student who dislike lima beans is 2/3 * 120 = 80; statement 2 suggests that the same is 40.

[RonPurewal]

Aren't the statements contradictory

Of these students 2/3 dislike Lima beans

1) the total number of students are 120

2) students who dislike Lima beans are 40

If we take the statement 1, the total number of student who dislike lima beans is 2/3 * 120 = 80; statement 2 suggests that the same is 40.

this is a good point; yes, this problem is transcribed incorrectly. (it has clearly been transcribed ... um ... not carefully; the sentences have been written in very poor english.)

here's a quote from another forum, which is definitely the real version:

Lima Beans and Brussels Sprouts:
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.