In a class of 40 students, 32 offer mathematics, 24 offer physics and 4 offer neither mathematics nor physics. How many offer both mathematics and physics?

1. Analyze the question

We are given the total number of students in a class and the number of students who offer Mathematics, Physics, or neither. We need to find the number of students who offer both subjects. This is a classic Set Theory problem involving two sets.

2. Explain the concept

We use the principle of sets. Let:

• $n(U)$ = Total number of students = 40
• $n(M)$ = Students who offer Mathematics = 32
• $n(P)$ = Students who offer Physics = 24
• $n(M \cup P)'$ = Students who offer neither = 4
• $n(M \cap P)$ = Students who offer both (this is what we need to find)

The formula for the union of two sets is: $n(M \cup P) = n(M) + n(P) - n(M \cap P)$

Also, the total number of students is the sum of those who offer at least one subject and those who offer none: $n(U) = n(M \cup P) + n(M \cup P)'$

3. Show the solution

Step 1: Find the number of students who offer at least one subject. Since 4 students offer neither, the number of students who offer either Mathematics or Physics (or both) is: $n(M \cup P) = n(U) - n(M \cup P)'$ $n(M \cup P) = 40 - 4 = 36$

Step 2: Use the set formula to find the intersection. Substitute the values into the formula: $n(M \cup P) = n(M) + n(P) - n(M \cap P)$ $36 = 32 + 24 - n(M \cap P)$ $36 = 56 - n(M \cap P)$

Step 3: Solve for $n(M \cap P)$. $n(M \cap P) = 56 - 36$ $n(M \cap P) = 20$

4. Identify the answer

The number of students who offer both Mathematics and Physics is 20. This corresponds to option D.

5. Tip

When solving set problems, always subtract the "neither" group from the total first. A common mistake is to forget that the 4 students who offer neither must be excluded from the calculation of the union of the two sets. You can also visualize this using a Venn diagram to ensure your numbers add up correctly!