Sol:

Given

 Subject Number of students Mathematics 100 Physics 70 Chemistry 46 Mathematics and Physics 30 Mathematics and Chemistry 28 Physics and Chemistry 23 Mathematics, Physics and Chemistry 18

As given in the table:

Total number of students, n(U) = 175

Total number of students enrolled in Mathematics alone, n(M) = 100

Total number of students enrolled in Physics alone, n(M) = 70

Total number of students enrolled in Chemistry alone, n(M) = 46

Therefore, n(U) = 175, n(M)=100, n(P)=70 and n(C) =46

$$n(M\cap P) = 30, n(M\cap C) = 28, n (P\cap C) = 26 and n(M\cap P\cap C) = 18$$ $$m(M\cup P\cup C) = n(M)+ n(P)+ n(C)- n(M\cap P)- n(M\cap C)- n(P\cap C)+ n(M\cap P\cap C)$$ $$\Rightarrow$$100 + 70 + 46 – 30 – 28 – 22 + 18

$$\Rightarrow$$ 234 – 80

$$\Rightarrow$$ 154

We have to find:

(i) Total number of students enrolled in Mathematics alone are

$$n(M)- n(M\cap P) – n(M\cap C) + n(M\cap P\cap C)$$ $$\Rightarrow$$100 – 30 – 28 + 18

$$\Rightarrow$$ 118 – 58

$$\Rightarrow$$60

(ii) Total number of students enrolled in Physics alone are

$$n(P)- n(P\cap M) – n(P\cap C) + n(P\cap M\cap C)$$ $$\Rightarrow$$70 – 30 – 23 + 18

$$\Rightarrow$$ 118 – 58

$$\Rightarrow$$ 35

(iii) Total number of students enrolled in Chemistry alone are

$$n(C)- n(C\cap M) – n(C\cap P) + n(C\cap M\cap P)$$ $$\Rightarrow$$46 – 28 – 23 + 18

$$\Rightarrow$$-5 + 18

$$\Rightarrow$$ 13

(iv) There are no students who have not offered any of these three subjects.

The number of students who any of the subjects = 154

$$\Rightarrow$$175-154

$$\Rightarrow 21$$ students.

Therefore,

The total number of students enrolled in Mathematics alone are = 60 students.

The total number of students enrolled in Physics alone are = 35 students.

The total number of students enrolled in Chemistry alone is = 13 students.

There are no students who have not offered any of these three subjects = 21students.

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