Question 1206820: A school sports team contains 68 students. 33 do field events, 40 do track events, 23 do swimming, 14 do both field and track events, 8 do both swimming and field events. If 15 students do field events only and 10 do both swimming and track events, how many students do
a) swimming only,
b) track events only,
c) all three events?
Answer by ikleyn(52756)   (Show Source):
You can put this solution on YOUR website! .
A school sports team contains 68 students.
33 do field events, 40 do track events, 23 do swimming,
14 do both field and track events, 8 do both swimming and field events.
If 15 students do field events only and 10 do both swimming and track events, how many students do
a) swimming only,
b) track events only,
c) all three events?
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We are given a universal set of 68 students and 3 its basic subsets
F of 33 students
T of 40 students
S of 23 students.
We also are given three in-pair intersections
FT of 14 students
FS of 8 students
TS of 10 students.
We also are given that the set
F_only contains 15 students.
Let x be the number of students that do all 3 activities FTS.
Then, due to the Inclusion-Exclusion principle, we have this equation
68 - n(F) + n(T) + n(S) - n(FT) - n(FS) - n(TS) + x,
or
68 = 33 + 40 + 23 - 14 - 8 - 10 + x.
From it, we finf
x = 68 - 33 - 40 - 23 + 14 + 8 + 10 = 4.
So, the ANSWER to (c) is 4 students do all 3 activities.
Now for (a), "swimming only" has S_only = S - SF - TS + FTS = 23 - 8 - 10 + 4 = 9 students. ANSWER
Now for (b), "track only" has T_only = T - FT - TS + FTS = 40 - 14 - 10 + 4 = 20 students. ANSWER
At this point, the problem is solved in full.
All questions are answered.
The solution is complete.
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