Let's analyze this problem step-by-step using a Venn diagram.

The problem states:

• 64 students took a mathematics course.
• 94 students took a chemistry course.
• 58 students took a physics course.
• 28 students took both mathematics and chemistry.
• 26 students took both mathematics and physics.
• 22 students took both chemistry and physics.
• The goal is to find how many students took exactly one course only.

Let's denote:

• $M$: the set of students who took mathematics
• $C$: the set of students who took chemistry
• $P$: the set of students who took physics

Using the inclusion-exclusion principle for three sets, we know:

$|M \cup C \cup P| = |M| + |C| + |P| - |M \cap C| - |M \cap P| - |C \cap P| + |M \cap C \cap P|$

Step 1: Define the Variables

Let:

• $x$ be the number of students who took all three courses (mathematics, chemistry, and physics).

From the problem, we know:

1. $|M| = 64$
2. $|C| = 94$
3. $|P| = 58$
4. $|M \cap C| = 28$
5. $|M \cap P| = 26$
6. $|C \cap P| = 22$

Step 2: Calculate Total Students

We don't have the total number of students yet, but we can determine it by solving for $x$ if we set up an equation using the inclusion-exclusion principle. Let's substitute these values into the equation:

$|M \cup C \cup P| = 64 + 94 + 58 - 28 - 26 - 22 + x$ $|M \cup C \cup P| = 140 + x$

The total number of students will give us $|M \cup C \cup P|$, which we need to find how many took exactly one course.

Step 3: Calculate Students Taking Only One Course

Using the principle of inclusion-exclusion, the number of students who took only one course is calculated as:

• Only Mathematics: $|M| - |M \cap C| - |M \cap P| + |M \cap C \cap P|$
• Only Chemistry: $|C| - |C \cap M| - |C \cap P| + |M \cap C \cap P|$
• Only Physics: $|P| - |P \cap M| - |P \cap C| + |M \cap C \cap P|$

Let's substitute the values