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?

Analyze the question: This is a two-set Venn diagram problem. We're given the total number of students, the number in each subject, and the number in neither. We need the number in both (the intersection).

Concept: For two sets M and P:

n(M ∪ P) = n(M) + n(P) − n(M ∩ P)

Also, n(M ∪ P) = Total − Neither

Solution:

1. Students who offer at least one subject: n(M ∪ P) = 40 − 4 = 36

2. Substitute into the formula: 36 = 32 + 24 − n(M ∩ P) 36 = 56 − n(M ∩ P)

3. Solve: n(M ∩ P) = 56 − 36 = 20

Check:

• Only Mathematics = 32 − 20 = 12
• Only Physics = 24 − 20 = 4
• Both = 20
• Neither = 4
• Total = 12 + 4 + 20 + 4 = 40 ✓

Answer: 20 students offer both Mathematics and Physics, which is option D.

Exam tip: For two-set problems, always subtract "neither" from the total first to get the union. Then use n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Always verify by adding all the regions of your Venn diagram to see if they give the total.