Out of 25 students in a class, 15 play basketball and 11 play chess. What is the maximum number of students who play both basketball and chess?

1. Problem Description

Out of 25 students in a class, 15 play basketball and 11 play chess. What is the maximum number of students who play both basketball and chess?

2. Solution Steps

Let $B$ be the set of students who play basketball, and $C$ be the set of students who play chess.

We are given that the total number of students in the class is

2

5. The number of students who play basketball is $|B| = 15$.

The number of students who play chess is $|C| = 11$.

We want to find the maximum number of students who play both basketball and chess, which is the maximum value of $|B \cap C|$.

We know that

$|B \cup C| = |B| + |C| - |B \cap C|$.

We also know that the number of students who play either basketball or chess or both cannot exceed the total number of students in the class. Therefore, $|B \cup C| \leq 25$.

Substituting the given values, we have:

$|B \cup C| = 15 + 11 - |B \cap C| = 26 - |B \cap C|$.

Since $|B \cup C| \leq 25$, we have:

$26 - |B \cap C| \leq 25$.

$26 - 25 \leq |B \cap C|$.

$1 \leq |B \cap C|$.

However, $|B \cap C|$ can be at most the smaller of $|B|$ and $|C|$. So, $|B \cap C| \leq \min(|B|, |C|)$.

In our case, $|B \cap C| \leq \min(15, 11) = 11$.

We want to maximize $|B \cap C|$. If all 11 chess players also play basketball, then $|B \cap C| = 11$.

In this case, $|B \cup C| = 15 + 11 - 11 = 15$, which is less than or equal to

2

5.

Let's say $|B \cap C| = x$. Then $|B \cup C| = 15 + 11 - x = 26 - x$. We require that $26 - x \le 25$, so $x \ge 1$. Also, $x$ cannot exceed the number of students playing chess, so $x \le 11$. Thus the maximal value is $x = 11$.

3. Final Answer

The maximum number of students who play both basketball and chess is 11.