Question

Solve the given worksheet:
In a survey of 25 students, it was found that 15 has taken Maths, 12 had taken Physics and
Chemistry, 5 has taken Maths and Chemistry, 9 had taken Maths and Physics, 4 had taken Physics
and 3 had taken all the three. Find the number of students that had taken,
aken Maths and Physics, 4 had taken Physics and Chemistry
(1) Only Chemistry
(ii) Only Maths (iii) Only Physics
(IV) Physics and Chemistry but not Maths (v) Maths and Physics but not Chemistry
(vi) Only one of the subjects (vii) At least one of the three subjects
(viii) None of the subjects
2 TF A
Pls solve and explain it I am so sad​

the one who answer I will do whatever they want

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

hope it helps