Question

In a class of 80 students numbered 1 to 80, all odd numbered students opt of 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

• A. 13
• B. 24
• C. 28
• D. 52

Answer 1

Answer: (C)

Numbers which are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80 they are 16 in numbers. Now, Numbers which are divisible by 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 they are 11 in numbers.
Also, total odd numbers = 40
Let C represents the students who opt. for cricket, F
for football and H for hockey.
$\therefore$ we have $n\left(C\right)=40,n\left(F\right)=16,n\left(H\right)=11$
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.
$n\left(C\cap F\right),8,n\left(C\cap H\right)=6$,
$n\left(F\cap H\right)=2,n\left(C\cap F\cap H\right)=1$
We Know
$n\left(C\cup F\cup H\right)=n\left(C\right)+n\left(F\right)+n\left(H\right)-n\left(C\cap F\right)-n\left(C\cap H\right)$
$-n\left(F\cap H\right)+n\left(C\cap H\cap F\right)$
$n\left(C\cup F\cup H\right)=67-16+1=52$
$\therefore n\left(C^{\prime }\cap F^{\prime }\cap H^{\prime }\right)$
= Total students $-n\left(C\cup F\cup H\right)$
$n\left(C^{\prime }\cap F^{\prime }\cap H^{\prime }\right)=80-52=28$