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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