Asked by Gaye
There are 47 in a club. 18 play chess and 13 play tennis. The number who play neither is 3 times the number who play both. How many play both?
Answers
Answered by
MathMate
Let x = number who play both
then 3x = number who play neither.
C = set of people who play chess
T = set of people who play tennis
U = total membership of the club
By the principle of inclusion-exclusion,
|C∪T| = |C| + |T| - |C∩T|
or
|U| - |C∪T| = |U| - (|C| + |T| - |C∩T|)
3x = 47 - (18+13-x)
2x = 16
x = 8 (number of members who play both)
3x = 24 (number of members who play neither)
Total number of players
= |C∪T|
= |C| + |T| - |C∩T|
= 18 + 13 - 8
= 23
Check:
|U|=|C∪T|+24
= 23+24
=47 OK
then 3x = number who play neither.
C = set of people who play chess
T = set of people who play tennis
U = total membership of the club
By the principle of inclusion-exclusion,
|C∪T| = |C| + |T| - |C∩T|
or
|U| - |C∪T| = |U| - (|C| + |T| - |C∩T|)
3x = 47 - (18+13-x)
2x = 16
x = 8 (number of members who play both)
3x = 24 (number of members who play neither)
Total number of players
= |C∪T|
= |C| + |T| - |C∩T|
= 18 + 13 - 8
= 23
Check:
|U|=|C∪T|+24
= 23+24
=47 OK
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.