In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?

Analysis

This is a classic set theory problem involving two overlapping sets. We need to find the intersection of two groups (students offering French and students offering Arabic) within a universal set (the total number of students).

Concept

We use the principle of inclusion-exclusion for two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Where:

• $n(A \cup B)$ is the number of students who offer at least one subject.
• $n(A)$ is the number of students who offer French.
• $n(B)$ is the number of students who offer Arabic.
• $n(A \cap B)$ is the number of students who offer both (our unknown).

Step-by-Step Solution

1. Identify the Universal Set ($U$): Total students = 150

2. Find the number of students who offer at least one subject ($n(A \cup B)$): Since 20 students offer neither, we subtract them from the total: $n(A \cup B) = 150 - 20 = 130$

3. Apply the formula: Let $x$ be the number of students who offer both subjects ($n(A \cap B)$). $130 = 80 + 60 - x$

4. Solve for $x$: $130 = 140 - x$ $x = 140 - 130$ $x = 10$

Conclusion

The number of students who offer both subjects is 10.

Exam Tip

When solving set theory problems, always remember to subtract the "neither" group from the total population first. If you forget this step, you will likely get an incorrect answer. Also, drawing a quick Venn Diagram can help you visualize the overlap and ensure your numbers add up correctly!