Question

in a group of 84 persons each plays at least one game out of three viz. tennis, badminton and cricket. 28 of them play cricket ,40 play tennis and 48 play badminton. if 6 play both cricket and badminton and 4 play tennis and badminton and no one plays all the three games. find the number of persons who play cricket but not tennis.

Answer 1

Answer:

Students who plays all three games = 5

Step-by-step explanation:

Let C, T, B denote the set of students who play cricket, tennis and badminton respectively

Then we have, n(U) = n(C ∪ T ∪ B) = 84,

n(C) = 28, n(T) = 40, n(B) = 48,

n(C ∩ B) = 6, n(T ∩ B) = 4, n(C ∩ T ∩ B) = 0

Now, n(C ∪ F ∪ V) = n(C) + n(T) + n(B) - n(C ∩ B) - n(T ∩ B) - n(C ∩ T) + n(C ∩ T ∩ B)

⇒ 84 = 28 + 40 + 48 - 6 - 4 -  n(C ∩ T)

⇒ n(C ∩ T)  = 116 - 10 - 84

⇒ n(C ∩ T) = 22

Hence, No. of Students who plays cricket but tennis = n(C) - n(C ∩ T)

                                                                                           = 28 - 22

                                                                                           = 6

Answer 2

Is this Question answer if 6