Out of a group of $50$ students taking examinations in Mathematics, Physics and Chemistry, $37$ passed Mathematics, $24$ passed Physics and $43$ passed Chemistry. Given that no more than $19$ passed both Mathematics and Physics, no more than $29$ passed both Mathematics and Chemistry and no more than $20$ passed both Physics and Chemistry, what is the maximum number of students who could have passed all three examinations?
- A. 12
- B. 9
- C. 14
- D. 10
✅ Correct Answer: 14
Explanation
Let $M$, $P$ and $C$ denote the sets of students passing Mathematics, Physics and Chemistry Given $n(M)=37$, $n(P)=24$, $n(C)=43$ $n(M \cap P) \le 19$, $n(M \cap C) \le 29$, $n(P \cap C) \le 20$ Using $n(M \cup P \cup C) = n(M)+n(P)+n(C) - n(M \cap P) - n(M \cap C) - n(P \cap C) + n(M \cap P \cap C)$ $50 \ge 37+24+43-19-29-20 + n(M \cap P \cap C)$ $50 \ge 36 + n(M \cap P \cap C)$ $n(M \cap P \cap C) \le 14$ Hence the maximum possible number is $14$
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