A. 76
B. 32
C. 45
D. 23
B = Set of students who fair in mathematics
Then , n(A) = 50, n(B) = 30 and n(A∩ B) = 12.
n(AUB) = n(A) + n(B)- n(A∩B) = (50 + 30 -12) = 68
Number of students who fail in one or both the students = 68
Number of those who pass in both = (100 – 68) = 32