Question

In a survey of 25 students, it was found that 15 had taken mathematics, 12 had taken physics and 11 had taken chemistry. 5 had taken mathematics and chemistry, 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all subjects. find number of students who had taken only chemistry? i) only maths ii) only physics iii) only chemistry iv) only one of three v) only physics and chemistry vi) only mathematics and physics vii) only mathematics and chemistry viii) only two of the subjects ix) at least one subject x) none of the subject ​

Answer 1

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

5. Number of students taking Maths and Physics but not Chemistry = n((M ∩ P) - C)

= n(M ∩ P) - n(P ∩ M ∩ C)

= 9 - 3

= 6

6. Number of students taking only one subject = n((only M) + (only P) + (only C))

= n(only M) + n(only P) + n(only C)

= 4 + 2 + 5

= 11

7. Number of students taking at least one subject = n(M ∪ P ∪ C)

= n M) + n(P) + n(C) - n(M ∩ P) - n(P ∩ C) - n(M ∩ C) + n(M ∩ P ∩ C)

= 15 + 12 + 11 - 9 - 4 - 5 + 3

= 41 - 18

= 23

8. Number of students taking none of three subject = 25 - n(M ∪ P ∪ C)

= 25 - 23

= 2