JeetGulia wrote:Of Students who eats in cafe, each student either like or dislikes L and M beans. of these students 2/3 dislikes L beans. and of those who dislikes L beans, 3/5 also dislikes M beans. How many of the students like M beans and but dislikes L beans.

a) 120 Students eat in cafe
b) 40 of the students like L beans.

And D

Given,
Let total students be = x
# Dislike L beans = 2x/3
# Dislike both L & M beans = (2x/3) * 3/5 = 2x/5

a) Total students = 120
So, x = 120
# Dislike L beans = 120 * (2x/3) = 48
# Dislike both L and M beans = 120 * (2x/5) = 80
# Students who like M but dislike L beans = (80 - 48) = 32
Sufficient.

b) # students who like L beans = 40
This means, 40 = (1 - 2x/3) = x/3
Solving for x, you get x = 120.
Sufficient.

Hence, D.

is this an overlapping sets problem or just fractions?

Gurpinder wrote:is this an overlapping sets problem or just fractions?

Little bit of both

P.S. I got this problem in my GMAT Prep Test too

beatthegmatinsept wrote:

Gurpinder wrote:is this an overlapping sets problem or just fractions?

Little bit of both

P.S. I got this problem in my GMAT Prep Test too

The reason I am asking this is because, if I do this one using the chart method that Mgmat outlined in its books for overlapping sets, the answer i seem to get is (E).

both statements are saying the exact same thing.

beatthegmatinsept -- can you try this one using the table method?

---------------like l --------- dislike l--
like m ------------------------------------
dislike m --
--------------------------------------------120

we are looking for the part in red, but all i get are the totals.

Gurpinder wrote:

beatthegmatinsept wrote:

Gurpinder wrote:is this an overlapping sets problem or just fractions?

Little bit of both

P.S. I got this problem in my GMAT Prep Test too

The reason I am asking this is because, if I do this one using the chart method that Mgmat outlined in its books for overlapping sets, the answer i seem to get is (E).

both statements are saying the exact same thing.

beatthegmatinsept -- can you try this one using the table method?

---------------like l --------- dislike l--
like m ------------------------------------
dislike m --
--------------------------------------------120

we are looking for the part in red, but all i get are the totals.

i didnt try solving it using the chart method. i have noticed some sets problem are not easy to solve using the chart method.

Hello beatthegmatinsept,

In stmt 1 you have taken

120- (dislike l)- dislike(L&M)= (Like M) & dislike (L) which I believe is wrong as its not taking into account of Stundents who like (L)

deepaks04 wrote:Hello beatthegmatinsept,

In stmt 1 you have taken

120- (dislike l)- dislike(L&M)= (Like M) & dislike (L) which I believe is wrong as its not taking into account of Stundents who like (L)

If you read the question again, it says each student either likes or dislikes L and/or M beans. If out of 120 total students, 48 dislike L beans, that must mean that the remaining 72 students like L beans.
The question asks students like L beans, but dislike M beans. We know there are a total of 80 students who dislike BOTH beans, but there are 48 that we know dislike L beans only. So the difference, 32 must be students who don't dislike L beans, but dislike M beans.
Where is the problem?