1. First, solve the arithmetic expression given: 3 \times 16 \div 3 = ? and 24 \times 8 + ? \div 2. Step 1: Calculate 3 \times 16 = 48. Step 2: Divide 48 by 3 to get 16. So, the value replacing ? in the expression is 16. 2. Given the class of 40 students and sets for Mathematics (M=18), English (E=20), Chemistry (C=15), with intersections: E \cap C = 11, M \cap E = 6, M \cap C = 4, and all three M \cap E \cap C = 2. 3. Calculate the number of students taking only one subject. Only M = M - (M \cap E + M \cap C - M \cap E \cap C) = 18 - (6 + 4 - 2) = 18 - 8 = 10. Only E = E - (M \cap E + E \cap C - M \cap E \cap C) = 20 - (6 + 11 - 2) = 20 - 15 = 5. Only C = C - (M \cap C + E \cap C - M \cap E \cap C) = 15 - (4 + 11 - 2) = 15 - 13 = 2. Total only one subject = 10 + 5 + 2 = 17. 4. Calculate the number taking exactly two subjects only. Two subjects only = (M \cap E + E \cap C + M \cap C) - 3 \times (M \cap E \cap C) = (6 + 11 + 4) - 3 \times 2 = 21 - 6 = 15. 5. Calculate students not taking any subject. Using formula for union: |M \cup E \cup C| = |M| + |E| + |C| - |M \cap E| - |E \cap C| - |M \cap C| + |M \cap E \cap C| = 18 + 20 + 15 - 6 - 11 - 4 + 2 = 53 - 21 + 2 = 34. So, not taking any = Total - |M \cup E \cup C| = 40 - 34 = 6. 6. Students taking Mathematics but not English: = Only M + (M \cap C - M \cap E \cap C) = 10 + (4 - 2) = 10 + 2 = 12. 7. Students taking English and Chemistry but not Mathematics: = (E \cap C - M \cap E \cap C) = 11 - 2 = 9. Final answers: (a) 17 (b) 15 (c) 6 (d) 12 (e) 9