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?
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Thanks and Regards,
Honneeey.
In former years,Used to run for "Likes", nowadays, craving for "Kudos". :D