Question

Each student in a class of 40 plays at least one indoor game badminton, table tennis and
scrabble. 18 play badminton, 20 play scrabble and 27 play table tennis. 7 play badminton
and scrabble, 12 play scrabble and table tennis and 4 play badminton, table tennis and
scrabble. Find the number of students who play
(i) badminton and table tennis.
(ii) badminton, table tennis but not scrabble

Answer 1

$\huge \tt \pink{ \underline{ \underline{Answer : - }}}$

Let,

 A --> set of students who play chess. 

B --> set of students who play scrabble. 

C --> set of students who play carrom 

$∴ n(A ∪ B ∪ C) = 40,$

$n(A) = 18,     \\ n(B) = 20     \\  n(C) = 27,$

$n(A ∩ B) = 7,   \\   n(C ∩ B) = 12 \\ n(A ∩ B ∩ C) = 4$

i) We know that,

$n(A ∪ B ∪ C) =$

$n(A) + n(B) + n(C) - n(A ∩ B) - n(B∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)$

$∴ 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4$

$40 = 50 - n(C ∩ A)$

$40 = 50 - n(C ∩ A) n(C ∩ A) = 50 -40$

$⇒n(C ∩ A) = 10$

$∴ Number \: of \: students \: who \: play \\ chess \: and \: carrom = 10.$

ii) number of students who play chess, carrom but not scrabble. 

$= n(C ∩ A) - n(A ∩ B ∩ C)$

$= 10 – 4$

$= 6$