Correct Answer - Option 2 : 65
Concept:
Number of elements present in exactly only one set
n(A) + n(B) + n(C) – 2 × [n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] + 3 × n(A ⋂ B ⋂ C)]
Calculations:
Given, U : Total number of students.
n(U) = 100
M : Student who liked mathematics
n(M) = 32
B : Student who liked Business
n(B) = 38
L : Student who liked Literature
n(B) = 30
Moreover, students liked both Mathematics and Literature = n(\(\rm M\cap L\)) = 7
students liked both Mathematics and Business = n(\(\rm M\cap B\)) = 10
students liked both Business and Literature = n(\(\rm B\cap L\)) = 8
students liked all three subjects = n(\(\rm M\cap B\cap L\)) = 5
Now, number of people who liked exactly one subject
= n(M) + n(B) + n(L) – 2 × [n(M ⋂ B) + n(B ⋂ L) + n(M ⋂ L)] + 3 × n(M ⋂ B ⋂ L)]
= 32 + 38 + 30 - 2 (10 + 8 + 7) + 3(5)
= 65
Hence, option (2) is correct.