Question
90 members of a sport club played at least one of the games tennis, football and volleyball. 10 people play tennis and football, 19 people play football and volleyball, 29 plays tennis and volleyball, n people play the three game games, 2n people each play only one game. Find those that play at least 2 games
Answers
GPT 3.5
To find the number of people who play at least two games, we can use the principle of inclusion-exclusion.
Let's denote the number of people who play tennis as $T$, the number of people who play football as $F$, and the number of people who play volleyball as $V$. We are given:
$T + F + V = 90$
$T \cap F = 10$
$F \cap V = 19$
$T \cap V = 29$
$2n$ people play only one game.
To find the number of people who play at least two games, we need to find $|T \cup F \cup V|$ (the number of people who play at least one of the three games). We can calculate this as:
$|T \cup F \cup V| = |T| + |F| + |V| - |T \cap F| - |F \cap V| - |T \cap V| + |T \cap F \cap V|$
$|T \cup F \cup V| = T + F + V - (T \cap F) - (F \cap V) - (T \cap V) + n$
Since $2n$ people play only one game, we have $|T \cup F \cup V| = 90 - 2n$. Substituting the given values, we have:
$90 - 2n = T + F + V - (10 + 19 + 29) + n$
Simplifying, we get:
$2n + n = T + F + V - 58$
$3n = T + F + V - 58$
Now, let's look at the total number of people who play only one game. This is equal to:
$|T \cap F| + |F \cap V| + |T \cap V|$
Since each of these intersections has $n$ people, we have:
$|T \cap F| + |F \cap V| + |T \cap V| = 3n$
Substituting this back into the previous equation, we get:
$3n = 3n - 58$
Simplifying, we find:
$0 = -58$
This implies that there is no solution to the problem as stated. Please check the given values and provide any missing information if you need further assistance.
Let's denote the number of people who play tennis as $T$, the number of people who play football as $F$, and the number of people who play volleyball as $V$. We are given:
$T + F + V = 90$
$T \cap F = 10$
$F \cap V = 19$
$T \cap V = 29$
$2n$ people play only one game.
To find the number of people who play at least two games, we need to find $|T \cup F \cup V|$ (the number of people who play at least one of the three games). We can calculate this as:
$|T \cup F \cup V| = |T| + |F| + |V| - |T \cap F| - |F \cap V| - |T \cap V| + |T \cap F \cap V|$
$|T \cup F \cup V| = T + F + V - (T \cap F) - (F \cap V) - (T \cap V) + n$
Since $2n$ people play only one game, we have $|T \cup F \cup V| = 90 - 2n$. Substituting the given values, we have:
$90 - 2n = T + F + V - (10 + 19 + 29) + n$
Simplifying, we get:
$2n + n = T + F + V - 58$
$3n = T + F + V - 58$
Now, let's look at the total number of people who play only one game. This is equal to:
$|T \cap F| + |F \cap V| + |T \cap V|$
Since each of these intersections has $n$ people, we have:
$|T \cap F| + |F \cap V| + |T \cap V| = 3n$
Substituting this back into the previous equation, we get:
$3n = 3n - 58$
Simplifying, we find:
$0 = -58$
This implies that there is no solution to the problem as stated. Please check the given values and provide any missing information if you need further assistance.