From 50 Students taking examination in Mathematics, Physics and chemis

1.	From 50 Students taking examination in Mathematics, Physics and chemistry, each of the student has passed in at least one of the subject, 37 passes Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, atmost 29 Mathematics and Chemist and at most 20 Physics and Chemistry. What is the largest possible number that could have passes in all the three subjects?
Answer» Let M, P and C denote the students of Mathematics, Physics and Chemistry, respectively. Given, n(M ∪ P ∪ C) = 50 No. of students passed in Mathematics, n(M) = 37 No, of students passed in Physics, n(P) = 24 No. of students passed in Chemistry, n(C) = 43 No. of students passed in Mathematics and Physics, n(M ∩ P) = 19 No. of students passed in Mathematics and Chemistry, n(M ∩ C) = 29 No. of students passed in Physics and chemistry, n(P ∩ C) = 20 Using identity n(M ∪ P ∪ C) = n(M) + n(P) + n(C) − n(M ∩ P) − n(M ∩ C) − n(P ∩ C) + n(M ∩ P ∩ C) ∴ n(M ∪ P ∪ C) = n(M ∪ P ∪ C) − n(M) − n(P) − n(C) + n(M ∩ P) + n(M ∩ C) + n(P ∩ C) = 50 – 37 – 24 – 43 +19 + 29 + 20 = 50 – (37 + 24 + 43) + (19 + 29 + 20) = 50 – 104 + 68 = 50 – 36 = 14

From 50 Students taking examination in Mathematics, Physics and chemistry, each of the student has passed in at least one of the subject, 37 passes Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, atmost 29 Mathematics and Chemist and at most 20 Physics and Chemistry. What is the largest possible number that could have passes in all the three subjects?

Answer»

Let M, P and C denote the students of Mathematics, Physics and Chemistry, respectively.

Given,

n(M ∪ P ∪ C) = 50

No. of students passed in Mathematics, n(M) = 37

No, of students passed in Physics, n(P) = 24

No. of students passed in Chemistry, n(C) = 43

No. of students passed in Mathematics and Physics, n(M ∩ P) = 19

No. of students passed in Mathematics and Chemistry, n(M ∩ C) = 29

No. of students passed in Physics and chemistry, n(P ∩ C) = 20

Using identity

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

∴ n(M ∪ P ∪ C) = n(M ∪ P ∪ C) − n(M) − n(P) − n(C) + n(M ∩ P) + n(M ∩ C) + n(P ∩ C)

= 50 – 37 – 24 – 43 +19 + 29 + 20

= 50 – (37 + 24 + 43) + (19 + 29 + 20)

= 50 – 104 + 68

= 50 – 36

= 14