DS problem

[sanjay]

GMAT prep problem. Test-1.
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?

A. 120 students eat in the cafeteria.

B. 40 of the students like lima beans.

OA is C.

I would appreciate if someone can explain me how to solve it.
Sanjay

[alok.sarsidharan]

Are you sure the answer is C ??? I was getting D as the answer, each statement alone is sufficient.

[sanjay]

I am so so sorry!!!

Yes, the answer is "D".

I'd appreciate your effort, it you please explain me how to solve it.

Sanjay

[lalitkc]

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?

A. 120 students eat in the cafeteria.

B. 40 of the students like lima beans.

Let the total students who eat in the cafeteria be 'x'.
therefore we can say that 2x/3 dislike lima beans
and 3/5 (2x/3) = 2x/5 dislike both lima beans and brussels sprouts.
So students who dislike only lima beans (and like brussels sprouts) = 2x/3 - 2x/5 = 4x/15 ---- (i)

Statement 1 => x = 120
From (i) => Students who like brussel sprouts but dislike Lima beans = 4x/15 = 32

Hence Statement 1 is sufficient.

Statement 2 => 40 students like Lima beans = (x-40) students dislike Lima Beans = 2x/3 ==> x = 120 which is statement 1. Hence Statement 2 is sufficient

Answer : D

[sanjay]

Makes sense now! Seems so easy once you get to know it.

Thanks for helping me out, Lalit.
Sanjay

[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

[Ben Ku]

lalitkc offers two good solutions. I will only add that setting up a double-set matrix as described in our classes and Strategy Guides is a helpful approach to this problem. Let me know if there are additional questions. Thanks!

[me.parashar]

can you please tell me how can we solve this question using matrix. i tried but failed.

[jlucero]

Here's the matrix for this problem. Note that I chose 15 as the common denominator of 3 and 5 so that we can calculate the fraction of people in each category. Once we have a specific number for any of the parts that we have filled out, we can set up a ratio to solve- i.e. 5/15 = 40/x