Question

Each student in a class of 35 plays at least one
game among chess, carrom and table tennis. 22
play chess, 21 play carrom, 15 play table tennis,
10 play chess and table tennis, 8 play carrom and
table tennis and 6 play all the three games. Find the number of students who
play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom
(Hint: Use Venn diagram)

Answer 1

Answer

4.7/5

69

Let Number of Students Playing Chess be : n(Ch) = 22

Let Number of Students Playing Carrom be : n(Ca) = 21

Let Number of Students Playing Table Tennis be : n(TT) = 15

Given Number of Students Playing both Chess and Table Tennis = 10

⇒ n(Ch ∩ TT) = 10

Given Number of Students Playing both Carrom and Table Tennis = 8

⇒ n(Ca ∩ TT) = 8

Given Number of Students Playing all the three Games = 6

⇒ n(Ch ∩ Ca ∩ TT) = 6

Given that Each Student in the Class of 35 Plays atleast one Game among Chess, Carrom and Table Tennis

⇒ Total number of Students Playing atleast one Game = 35

⇒ n(Ch ∪ Ca ∪ TT) = 35

We know that :

n(Ch ∪ Ca ∪ TT) = n(Ch) + n(TT) + n(Ca) - n(Ch ∩ TT) - n(Ca ∩ TT) - n(Ch ∩ Ca) + n(Ch ∩ Ca ∩ TT)

Substituting all the Respective values we get :

⇒ 35 = 22 + 15 + 21 - 10 - 8 - n(Ch ∩ Ca) + 6

⇒ 35 = 46 - n(Ch ∩ Ca)

⇒ n(Ch ∩ Ca) = 11

⇒ Number of Students Playing both Carrom and Chess = 11

From the Venn Diagram :

We can Notice that Number of Students Playing both Carrom and Chess also includes the students who are playing all the Three Games.

⇒ Number of Students who are playing both Carrom and Chess but not Table Tennis can be found by Subtracting Number of students playing all the Three games from Number of students playing both Carrom and Chess.

(i) Number of Students Playing both Carrom and Chess but not Table Tennis = n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) = 11 - 6 = 5

(ii) Number of Students playing only Chess will be :

n(Ch) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ch ∩ TT) - n(Ch ∩ Ca ∩ TT) }

⇒ 22 - {11 - 6 + 6 + 10 - 6}

⇒ 22 - 15 = 7

Number of Students only playing Chess will be : 7

(iii) Number of Students playing only Carrom will be :

n(Ca) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ca ∩ TT) - n(Ch ∩ Ca ∩ TT) }

⇒ 21 - {11 - 6 + 6 + 8 - 6}

⇒ 21 - 13 = 8

Number of Students only playing Carrom will be : 8

Answer 2

A – Chess

B – Carrom

C – Table Tennis

n(A) = 22

n(B) = 21

n(C) = 15

n(A ∩ C) = 10

n(B ∩ C) = 8

n(A ∩ B ∩ C) = 6 y = 22 – (x + 6 + 4) = 22 – (x + 10) = 22 – x – 10 = 12 – x z = 21 – (x + 6 + 2) = 21 – (8 + x) 21 – 8 – x = 13 – x y + z + 3 + x + 2 + 4 + 6 = 35 12 – x + 13 – x + 15 + x = 35 40 – x = 35 x = 40 – 35 = 5 (i) Number of students who play only chess and Carrom but not table tennis = 5 (ii) Number of students who play only chess = 12 – x = 12 – 5 = 7 (iii) Number of students who play only carrom = 13 – x = 13 – 5 = 8

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