Question

A class has 175 students. The following is the description showing the number of students studying one or more of following subjects in this class. Mathematics 100; Physics 70; Chemistry 46; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 22 ; Mathematics ,Physics and chemistry 18 :- find - (i) the number of students who are enrolled in Mathematics, Physics alone and Chemistry alone; (ii) the number of students who have not offered any of these three subjects.

Answer 1

Solution:-
Let M, P and C represents the sets of the students who studied mathematics, physics and chemistry respectively.
Given : n(M) = 100 ; n(P) = 70 and n(C) = 46
n(M∩P) = 30 ; n(M∩C) = 28 and n(P∩C) = 22 and n(M∩P∩C) = 18
n(M∪P∪C) = n(M) + n(P) + n(C) - n(M∩P) - n(M∩C) - n(P∩C) + n(M∩P∩C)
= 100 + 70 + 46 - 30 - 28 - 22 + 18
= 234 - 80
= 154
So, the number of students who any of the subjects = 154
Therefore, the number of students who have not any of these three subjects = 175 - 154
= 21 students.
Thus, 21 students have not offered any of these three subjects.
Now,
Number of students who are enrolled in mathematics alone = n(M) - n(M∩P) - n(M∩C) + n(M∩P∩C)
= 100 - 30 - 28 +18
= 118 - 58
= 60 students enrolled in mathematics alone.

Number of students who are enrolled in physics alone =
n(P) - n(P∩C) - n(M∩P) + n(M∩P∩C)
= 70 - 22 - 30 + 18
= 88 - 52
= 36 students are enrolled in physics alone.

Number of students enrolled in chemistry alone =
n(C) - n(P∩C) - n(M∩C) + n(M∩P∩C)
46 - 22 - 28 + 18
= 64 - 50
= 14 students enrolled in chemistry alone.
Answer.

Answer 2

Solution:-

Let M, P and C represents the sets of the students who studied mathematics, physics and chemistry respectively.

Given : n(M) = 100 ; n(P) = 70 and n(C) = 46

n(M∩P) = 30 ; n(M∩C) = 28 and n(P∩C) = 22 and n(M∩P∩C) = 18

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

= 100 + 70 + 46 - 30 - 28 - 22 + 18

= 234 - 80

= 154

So, the number of students who any of the subjects = 154

Therefore, the number of students who have not any of these three subjects = 175 - 154

= 21 students.

Thus, 21 students have not offered any of these three subjects.

Now,

Number of students who are enrolled in mathematics alone = n(M) - n(M∩P) - n(M∩C) + n(M∩P∩C)

= 100 - 30 - 28 +18

= 118 - 58

= 60 students enrolled in mathematics alone.

Number of students who are enrolled in physics alone =

n(P) - n(P∩C) - n(M∩P) + n(M∩P∩C)

= 70 - 22 - 30 + 18

= 88 - 52

= 36 students are enrolled in physics alone.

Number of students enrolled in chemistry alone =

n(C) - n(P∩C) - n(M∩C) + n(M∩P∩C)

46 - 22 - 28 + 18

= 64 - 50

= 14 students enrolled in chemistry alone.

Answer.