Question

In a class of 80 students numbered a 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)

28

Answer 2

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 $- n\left(C \cup F \cup H\right)$ $n\left(C' \cap F'\cap H'\right)= 80 - 52 = 28$