Question:
In a survey of 25 students it was found that 15 had taken Maths, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Maths and chemistry, 9 had taken Maths and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects. Find the number of students that had taken: i) Only Chemistry ii) Only Maths iii) Only Physics iv) Physics and Chemistry but not Maths.
Answer:
Let M: Set of students who have taken Maths
P: Set of students who have taken Physics
C: Set of students who have taken Chemistry
Given,
Total students n(U) = 25
n(M) = 15, n(P) = 12, n(C) = 11
n(M ∩ C) = 5, n(P ∩ C) = 4, n(M ∩ P) = 9,
n(M ∩ P ∩ C) = 3
1. Number of students taking only Chemistry = n(C - (M ∪ P))
= n(C) - n(C ∩ (M ∪ P))
= n(C) - [n(C ∩ M) + n(C ∩ P) - n((C ∩ M) ∩ (C ∩ P)) ]
= n(C) - n(C ∩ M) - n(C ∩ P) + n(C ∩ M ∩ P)
= 11 - 5 - 4 + 3
= 14 - 9
= 5
2. Number of students taking only Maths = n(M - (P ∪ C))
= n(M) - n(M ∩ (P ∪ C))
= n(M) - [n(M ∩ P) + n(M ∩ C) - n((M ∩ P) ∩ (M ∩ C)) ]
= n(M) - n(M ∩ P) - n(M ∩ C) + n(M ∩ P ∩ C)
= 15 - 9 - 5 + 3
= 18 - 14
= 4
3. Number of students taking only Physics = n(P - (M ∪ C))
= n(P) - n(P ∩ (M ∪ C))
= n(P) - [n(P ∩ M) + n(P ∩ C) - n((P ∩ M) ∩ (P ∩ C)) ]
= n(P) - n(P ∩ M) - n(P ∩ C) + n(P ∩ M ∩ C)
= 12 - 9 - 4 + 3
= 15 - 13
= 2
4. Number of students taking Physics and Chemistry but not Maths = n((P ∩ C) - M)
= n(P ∩ C) - n(P ∩ M ∩ C)
= 4 - 3
= 1
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