Bunuel wrote:

dimitri92 wrote:

What is the best approach to tackle questions like these ?

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?I'd advise to make a table:

Note that:

"2/3 dislike lima beans" means 2/3 of total dislike lima;

"

of those who dislike lima beans, 3/5 also dislike brussels sprouts" means

of those who dislike lima \(1-\frac{3}{5}=\frac{2}{5}\) like sprout, or \(\frac{2}{3}*\frac{2}{5}=\frac{4}{15}\) of total

dislike lima but like sprouts. So to calculate # of students who

dislike lima but like sprouts we should now total # of students (t).

(1) 120 students eat in the cafeteria --> \(t=120\) --> \(x=\frac{4}{15}t=32\). Sufficient.

(2) 40 of the students like lima beans --> total students who like lima + total students who dislike lima = total --> \(40+\frac{2}{3}t=t\) --> \(t=120\) --> \(x=\frac{4}{15}t=32\). Sufficient.

Answer: D.

Attachment:

Lima-Sprouts.JPG

hi Bunuel,

I reached the same answer but my values were different than yours.

Lima Beans Brussels Total

Like X/3 X-(2X/5)=3X/5

Dislike (2/3 )X 2/3*3/5=(2/5)X

Total X X X

Now as we are told that EVery student either likes or dislikes each bean. So the total of Beans like + dislike=X

and total of Brussels Like + dislike=X

Statement 1 says X=120 therefore sufficient

Statement 2 directly gives X/3=40 therefore X=120.

Is this approach correct?

Cos the answer that we are looking for is "Students who like Brussels but dislike Lima"

Will this approach furnish an answer?

_________________

Thanks and Regards,

Honneeey.

In former years,Used to run for "Likes", nowadays, craving for "Kudos". :D