Of the students who eat in a certain cafeteria, each student

[amahajan01]

(1)
120 students total
--> 40 YES LIMA, 80 NO LIMA
split up the NO LIMA crowd:
--> 3/5 of 80 = 48 NO LIMA + NO BRUSSELS
--> 2/5 of 80 = 36 NO LIMA + YES BRUSSELS
sufficient

Hi Ron,

Am re-opening the thread as I am still not a 100% sure about the solution. Shouldn't the above read as 3/5 of 80 = 48 no lima + no brussels and 2/5 of 80 = 32 no lima + yes brussels. Let me know if I am missing something here.

[RonPurewal]

you are correct; i also went back and fixed that old post. thanks.

[FarukhA694]

Hello Ron and Team,

First of all - Happy Thanksgiving to everyone!

I believe there is something which is still missing. This is how I did it:

https://drive.google.com/file/d/0B_mCLX ... sp=sharing

The top table is how I did it and the numbers come out to be 120 for the whole group. However, if we do it while considering it a mutually exclusive set then the numbers don't come out to be 120 for the LBs and BSs. Could you please check.

Kindest Regards,
FA

[RonPurewal]

"liking lima beans" and "liking brussels sprouts" are not mutually exclusive.
neither are "disliking lima beans" and "disliking brussels sprouts".

this is just basic common sense -- these are 2 totally different foods -- but, also, the problem statement explicitly says there are people who dislike both.

no, there are not "multiple interpretations".