Beans (GMAT Prep)

[alex.gellatly]

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.

[Anurag@Gurome]

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.

Say, number of students who eats in the cafeteria = n
Hence, 2n/3 students dislikes Lima beans.
Hence, (1 - 3/5)*(2n/3) = (2/5)*(2n/3) = 4n/15 students dislikes Lima beans but likes Brussels sprouts.

Statement 1: n = 120 --> Sufficient

Statement 1: (1 - 2/3)*n = 40 --> n = 3*40 = 120 --> Sufficient

The correct answer is D.

[alex.gellatly]

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.

Say, number of students who eats in the cafeteria = n
Hence, 2n/3 students dislikes Lima beans.
Hence, (1 - 3/5)*(2n/3) = (2/5)*(2n/3) = 4n/15 students dislikes Lima beans but likes Brussels sprouts.

Statement 1: n = 120 --> Sufficient

Statement 1: (1 - 2/3)*n = 40 --> n = 3*40 = 120 --> Sufficient

The correct answer is D.

I'm sorry, but I still don't understand why statement 2 is sufficient. Could you please explain? Thanks

[Anurag@Gurome]

I'm sorry, but I still don't understand why statement 2 is sufficient. Could you please explain? Thanks

I am guessing that you have no problem with statement 1.

Now, we know that each student either either likes or dislikes Lima beans. Now, 2/3 of the students dislikes Lima beans.
Hence, (1 - 2/3) = 1/3 of the students likes Lima beans.

From statement 2, 1/3 of the students = 40
Hence, number of students who eat in the cafeteria = 40*3 = 120
Therefore, statement 2 is equivalent to statement 1.

[niketdoshi123]

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.

This is a good overlapping sets question



Here we need to find the # of students who like brussels sprouts but dislike lima beans => 4x/15
We can find this if we know the value of x.
Hence the question can be re-framed as : How many students eat in cafeteria?Ans = x

Statement 1: SUFFICIENT
Ans : there are 120 students who eat in the cafeteria.

Statement 2: SUFFICIENT
From the overlapping sets diagram we know that x/3 students like lima beans.
so x/3=40 =>x=120.

[Gaurav 2013-fall]

thanks alex for the question! Good one.

[GMATGuruNY]

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.

This is an EITHER/OR group question.
Each student EITHER likes limas OR dislikes limas.
Each student EITHER likes brussels sprouts OR dislikes brussels sprouts.
To organize the information, use a GROUP GRID:



In the grid above, every row must add up to the total, as must every column.
In the top row, (like brussels and like limas) + (like brussels but dislike limas) = total who like brussels.
In the leftmost column, (like limas and like brussels) + (like limas but dislike brussels) = total who like limas.

Let T = total students. Thus, the bottom right corner = T.

2/3 dislike lima beans:
Thus, the bottom of the middle column = (2/3)T.
Since the sum of the bottom row is T, the bottom left corner (the total who like limas) = (1/3)T.

Of those who dislike lima beans, 3/5 also dislike brussels sprouts.
Thus, the middle box = (3/5)(2/3)T.
Since the sum of the middle column = (2/3)T, the top box of the middle column (the total who dislike limas but like brussels sprouts) = (2/5)(2/3)T.

Thus, to determine the number who like brussels sprouts but dislike lima beans, we need to know the value of T.

Statement 1: 120 students eat in the cafeteria
T = 120.
SUFFICIENT.

Statement 2: 40 students like lima beans
The bottom left corner shows that the number who like limas = (1/3)T.
Thus, (1/3)T = 40.
T = 120.
SUFFICIENT.

The correct answer is D.