Given:
- Number of students in class = 175
- Number of students enrolled in Mathematics = 100
- Number of students enrolled in Physics = 70
- Number of students enrolled in Chemistry = 46
- Number of students enrolled in Mathematics and Physics = 30
- Number of students enrolled in Physics and Chemistry = 23
- Number of students enrolled in Mathematics and Physics = 28
- Number of students enrolled in all three subjects = 18
To find:
(i) Number of students enrolled in Mathematics alone, Physics alone and Chemistry alone
Venn diagram:
Number of students enrolled in Mathematics = 100 = n(M)
Number of students enrolled in Physics = 70 = n(P)
Number of students enrolled in Chemistry = 46 = n(C)
Number of students enrolled in Mathematics and Physics
= 30 = n(M ∩ P)
Number of students enrolled in Mathematics and Chemistry
= 28 = n(M ∩ C)
Number of students enrolled in Physics and Chemistry
= 23 = n(P ∩ C)
Number of students enrolled in all the three subjects
= 18 = n(M ∩ P ∩ C) = g
We have,
n(M ∩ P) = e + g
30 = e + 18
e = 30 – 18 = 12
n(M ∩ C) = f + g
28 = f + 18
f = 28 – 18 = 10
n(P ∩ C) = d + g
23 = d + 18
d = 23 – 18 = 5
a = Number of students enrolled only in Mathematics
b = Number of students enrolled only in Physics c = Number of students enrolled only in Chemistry
We have,
M = a + e + f + g
100 = a + 12 + 10 + 18
a = 100 – 40
a = 60
Therefore,
Number of students enrolled only in Mathematics = 60
P = b + e + d + g
70 = b + 12 + 5 + 18
b = 70 – 35
b = 35
Therefore, Number of students enrolled only in Physics = 35
C = c + f + d + g
46 = c + 10 + 5 + 18
c = 46 – 33
c = 13
Therefore, Number of students enrolled only in Chemistry = 13
(ii) Number of students who have not offered any of these subjects
Number of students who have not offered any of these subjects
= 175 – {n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C)}
= 175 – (100 + 70 + 46 – 30 – 28 – 23 + 18)
= 175 – 153
= 22
Therefore,
Number of students who have not offered any of these subjects = 22