Time for MATH MINUTE! (provide your favorite theme music
here).
Get out
your paper & pencil, because I have a new puzzle for you!!
Responses/Hints: PUZZLE #105 More sets
This is a
variation on Puzzle 104A.
Suppose
you have a group of 30 students which have the following characteristics:
10 have brown hair,
18 wear jeans,
16 wear glasses.
Also,
6 wear jeans and have glasses,
5 have brown hair and
wear jeans,
3 have brown hair and
glasses,
and
none of them have all 3 properties.
Can you determine how many
have brown hair but do not wear jeans?
How about how many wear
jeans and do not wear glasses?
Do all of the students have
brown hair, or wear jeans, or wear glasses?
If you made a
sketch like we did last time for #104A you should have something that looks
like this:
The
sets B, J, G correspond to the brown hair students, the jeans wearing students
and the glasses wearing students, and the subsets a, b, . . . are as before in
#104A.
Note
that there is no common intersection between all three sets.
From
the given we have the following equations, where the names of the subsets are
also used for the size of the subset.
B
= a+b+d = 10
J
= b+c+e = 18
G
= d+e+f = 16
e = 6
b = 5
d = 3
Working
back through the equations we have a = 2, c = 7, and f = 7.
This
means that all 30 students either have brown hair, or wear jeans, or wear
glasses.
I
should mention that this is not necessarily always the case for this type of
situation. It just happened that way for this example.
Have fun!
Send
your comments, ideas and solutions before Monday to the email below, and
in the subject line be sure to put
MM in the subject line
Visit
us here online at:
http://www2.potsdam.edu/parksjm/MM1.1.htm
to see the results every Friday.
See you next
time on MATH MINUTE! (theme music
fades out here).