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
$n\left(C\cup F\cup H\right) = n\left(C\right) + n\left(F\right) + n\left(H\right) - n\left(C \cap F\right) - n\left(C \cap H\right)$
$- n\left(F \cap H\right) + n\left(C \cap H \cap F\right)$
$n\left(C\cup F\cup H\right) = 67 - 16 + 1 = 52$
$\therefore n\left(C' \cap F' \cap H'\right)$
= Total students $- n\left(C \cup F \cup H\right)$
$n\left(C' \cap F'\cap H'\right)= 80 - 52 = 28$