# Problem Involving Sets

Printable View

• Nov 9th 2013, 04:11 PM
Fratricide
Problem Involving Sets
In a class of 100 students, 55 are girls, 45 have blue eyes, 40 are blond, 25 are blond girls, 15 are blue-eyed blonds, 20 are blue-eyed girls, and 5 are blue-eyed blond girls. Find
a) The number of blond boys
I know that if there are 55 girls there must be 45 boys. If there are 40 blond people, and 25 of them are girls, then there must be 15 blond boys.

b) The number of boys who are not blond or blue-eyed
If there are 15 blonde boys, there are 30 boys who are not blond. If there are 45 people that have blue eyes, and there are 20 blue-eyed girls, then there must be 25 blue-eyed boys. If 25 boys have blue eyes, then there must be 20 boys who are not blue-eyed. If there are 15 people that are blue-eyed blonds, and 5 of them are girls, then there are 10 blue-eyed blond boys. So far, we have 30 boys who are not blond and 20 boys who are not blue-eyed. Where do I go from here?
• Nov 9th 2013, 05:15 PM
chiro
Re: Problem Involving Sets
Hey Fratricide.

Hint: You should try using de-morgans theorems where not (A OR B) = not(A) AND not(B) to find not(Blond) OR not(Blue-Eyed).
• Nov 9th 2013, 09:06 PM
Fratricide
Re: Problem Involving Sets
Could you elaborate, please?
• Nov 9th 2013, 11:27 PM
chiro
Re: Problem Involving Sets
You are basically trying to find not(Blond) OR not(Blue-eyed). Using de-morgans laws this is equivalent to not(Blond AND Blue-Eyed).

Note that the last thing that was mentioned was 5 are blue-eyed-blonde-girls or Blue-eyed AND Blonde. (You need to use this piece of information).

Remember that
(Blonde AND Blue-Eyed) OR (Note Blonde AND Blue-Eyed) = Blue-Eyed and
(Blond and Non-Blue-Eyed) OR (Blonde and Blue-Eyed) = Blonde.