Question

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects

Answer 1

M = Mathematics ; P = Physics and C = Chemistry 

n(M) = 120 n(P) = 90 n (C) = 70 n ( M ∩ P) = 40 
n ( P ∩ C ) = 30 n ( C ∩ M ) = 50 n ( M ∪ P ∪ C )’ = 20 

Now n(M ∪ P ∪ C)’ = n(U) – n(M ∪ P ∪ C) 

20 = 200 – n (M ∪ P ∪ C) 

Therefore, n(M ∪ P ∪ C) = 200 – 20 = 180 

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

180 = 120 + 90 + 70 - 40 - 30 - 50 + n(M ∩ P ∩ C)

⇒ n(M ∩ P ∩ C) =180 - 120 - 90 - 70 + 40 + 30 + 50 

⇒ n(M ∩ P ∩ C) = 20.

Answer 2

Answer:

Step-by-step explanation:

M = Mathematics ; P = Physics and C = Chemistry  

n(M) = 120 n(P) = 90 n (C) = 70 n ( M ∩ P) = 40  

n ( P ∩ C ) = 30 n ( C ∩ M ) = 50 n ( M ∪ P ∪ C )’ = 20  

Now n(M ∪ P ∪ C)’ = n(U) – n(M ∪ P ∪ C)  

20 = 200 – n (M ∪ P ∪ C)  

Therefore, n(M ∪ P ∪ C) = 200 – 20 = 180  

n(M ∪ P ∪ C)  

= n(M) + n(P) + n(C) – n(M ∩ P) – n(P ∩ C) – n(C ∩ M) + n(M ∩ P ∩ C)

180 = 120 + 90 + 70 - 40 - 30 - 50 + n(M ∩ P ∩ C)

⇒ n(M ∩ P ∩ C) =180 - 120 - 90 - 70 + 40 + 30 + 50  

⇒ n(M ∩ P ∩ C) = 20.