Sets-Relations-And-Functions Question 126
Question: In a class of 80 students numbered a to 80, all odd numbered students opt if Cricket, students whose numbers are divisible by 5 opt for football and those whose numbers are divisible by 7 opt for Hockey. The number of students who do not opt any of the three games, is
Options:
A) 13
B) 24
C) 28
D) 52
Correct Answer: C $ \therefore $ we have $ n(C)=40,n(F)=16,n(H)=11 $ now, $ C\cap F= $ odd numbers which are divisible by 5.
$ C\cap H $ = Odd numbers which are divisible by 7.
$ F\cap H= $ Numbers which are divisible by both 5 and 7.
$ n(C\cap F),8,n(C\cap H)=6, $ $ \therefore n(C’’\cap F’\cap H’) $ Show Answer
Answer:
Solution:
$ n(F\cap H)=2,n(C\cap F\cap H)=1 $
We know
$ n(C\cup F\cup H)=n(C)+n(F)+n(H)-n(C\cap F) $
$ -n(C\cap H)-n(F\cap H)+n(C\cap H\cap F) $
$ n(C\cup F\cup H)=67-16+1=52 $
= Total students $ -n(C\cup F\cup H) $
$ n(C’\cap F’\cap H’)=80-52=28 $
BETA
