From 50 students taking examination in mathematics, physics and chemistry , 37 passed maths ,24 physics and 43 chemistry .Atmost 19 passed maths and physics ,atmost 29 passed maths and chemistry and atmost 20 passed physics and chemistry .The largest possible no. that could have passed all three examination is
a -11. b-12. c-13. d-14.
Answers
Answer:
The answer is 14.
Steps:
Given,
n(M∪P∪C) = 50
n(M) = 57
n(P) = 24
n(C) = 43
n(M∩P) ≤ 19
n(M∩C) ≤ 29
n(P∩C) ≤ 20
Let n(M∩P∩C) = x
We know that
n(M∪P∪C) = n(M) + n(P) + n(C) - n(M∩P) - n(P∩C) - n(M∩C) - n(M∩P∩C)
⇒ n(M∪P∪C) ≤ n(M) + n(P) + n(C) - n(M∩P) - n(P∩C) - n(M∩C) - n(M∩P∩C)
⇒ 50 ≤ 37 + 24 + 43 - 19 - 20 - 29 - x
⇒ 50 ≤ 104 - 68 + x
⇒ 50 ≤ 36 + x
⇒ 50 - 36 ≤ x
⇒ 14 ≤ x
⇒ x ≥ 14
Regarding to that, we can say that the largest possible number that that could have passed in 3 subjects is 14.
PS: Please look at the attachment to understand better.
❍ Given : From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the student has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. What is the largest possible number that could have passed all three examination?
❍ To Find : The largest possible no. that could have passed all three examination is .
✬ Solution :-
❒ Let,
- ☯ M be the set of students passing in Mathematics
- ☯ P be the set of students passing in Physics.
- ☯ C be the set of students passing in Chemistry
❒ Now,
➯ n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24, n(C) = 43
➯ n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, n(P ∩ C) ≤ 20 (Given)
➯ n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C) ≤ 50
➯ 37 + 24 + 43 – 19 – 29 – 20 + n(M ∩ P ∩ C) ≤ 50
➯ n(M ∩ P ∩ C) ≤ 50 – 36
➯ n(M ∩ P ∩ C) ≤ 14
✬ Thus, the largest possible number that could have passed all the three examinations is 14.
Answer :-
D) 14