Question

Out of $800$ boys in a school, $224$ played circket, $240$ played hockey and $336$ played basketball. Of the total, $64$ played both basketball and hockey; $80$ played circket and basketball and $40$ played cricket and hockey; $24$ played all the three games. The number of boys who did not play any game is?

• A. $128$
• B. $216$
• C. $240$
• D. $160$

Answer 1

Answer: (D)

$160$

Given, $n\left(U\right)=800,$
Also $n\left(C\right)=224,n\left(H\right)=240,n\left(B\right)=336$
$n(H\cap B)=64,n(B\cap C)=80$
$n(H\cap C)=40,n(C\cap H\cap B)=24$
Therefore, $n(C^c\cap H^c\cap B^c)=n\left[\right(C\cup H\cup B{)}^c]$
$n\left(U\right)-n(C\cup H\cup B)$
$=800-\left[n\right(C)+n(H)+n(B)-n(H\cap C)-n(H\cap B)-n(C\cap B)+n(C\cap H\cap B\left)\right]$
$=800-[224+240+336-40-64-80+24]$
$=800-640=160$