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1.Set Theory
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Out of $800$ boys in a school, $224$ played cricket, $240$ played hockey and $336$ played basketball. Of the total, $64$ played both basketball and hockey; $80$ played cricket 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
$160$
Solution
(d) $n\,(C) = 224,\,n\,(H) = 240,n\,(B) = 336$
$n\,(H \cap B) = 64,\,\,n(B \cap C) = 80$
$n(H \cap C) = 40$, $n(C \cap H \cap B) = 24$
$n\,({C^c} \cap {H^c} \cap {B^C}) = n\,[{(C \cup H \cup B)^c}]$
$ = n( \cup ) – n(C \cup H \cup B)$
$ = 800 – [n(C) + n(H) + n(B) – n(H \cap C)$
$ – n(H \cap B) – n(C \cap B) + n(C \cap H \cap B)]$
$ = 800 – [224 + 240 + 336 – 64 – 80 – 40 + 24]$
$ = 800 – 640 = 160$.
Standard 11
Mathematics
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