Question

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

Answer 1

Solution


Let A be the set of students who play chess 


B be the set of students who play scrabble 


C be the set of students who play carrom 


Therefore, We are given 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 


We have - 


n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) 


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


40 = 69 – 19 - n(C ∩ A) 


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


n(C ∩ A) = 10 


Therefore, Number of students who play chess and carrom are 10. 


Also, number of students who play chess, carrom and not scrabble. 


= n(C ∩ A) - n(A ∩ B ∩ C) 


= 10 – 4 


= 6

Answer 2

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