From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the student has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. What is the largest possible number that could have passed all three examination?
We are given: \[\] \( n(M \cup P \cup C) = 50 \) \[\] \( n(M) = 37, \, n(P) = 24, \, n(C) = 43 \) \[\] \( n(M \cap P) = 19, \, n(M \cap C) = 29, \, n(P \cap C) = 20 \) \[\] We need to find \( n(M \cap P \cap C) \) \[\] The formula for the union of three sets is: \[ n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C) \] Substituting the given values: \[ 50 = 37 + 24 + 43 - 19 - 20 - 29 + n(M \cap P \cap C) \] Simplify: \[ 50 = 104 - 68 + n(M \cap P \cap C) \] \[ 50 = 36 + n(M \cap P \cap C) \] Solve for \( n(M \cap P \cap C) \): \[ n(M \cap P \cap C) = 50 - 36 = 14 \] Final Answer: \[ n(M \cap P \cap C) = 14 \]
Reinforce your conceptual grasp with related questions from NIMCET 2024:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →