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
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.