Question

Among a group of students, 50 played cricket, 50 played hockey and 40 played volley ball. 15 played both cricket and hockey, 20 played both hockey and volley ball, 15 played cricket and volley ball and 10 played all three. If every student played at least one game, find the number of students and how many played only cricket, only hockey and only volley ball?

Answer 1

Given:

Students playing cricket = n(C) = 50

Students playing hockey = n(H) = 50

Students playing volleyball = n(V) = 40

Students playing both hockey and cricket = n(C∩H) = 15

Students playing both hockey and volleyball = n(H∩V) = 20

Students playing both volleyball and cricket = n(C∩V) = 15

Students playing all three = n(C∩H∩V) = 10

To Find:

Students who played only cricket, only hockey and only volley ball?

Solution:

Students who played at least one game -

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

= 100.

Let number of students who played cricket and volleyball = a

Let number of students who played cricket and hockey = b

Let number of students who played hockey and volleyball = c

Let the number of students who played all three = d

Thus, d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Therefore, a = 15 – 10 = 5 [cricket and volleyball]

b = 15 – 10 = 5 [cricket and hockey]

c = 20 – 10 = 10 [hockey and volleyball]

Students who played only cricket

= n(C) – [a + b + d] = 50 – (5 + 5 + 10)

= 30

Students who played only hockey

= n(H) – [b + c + d]

= 50 – ( 5 + 10 + 10) = 25

Students who played only volley ball

= n(V) – [a + c + d] = 40 – (10 + 5 + 10)

= 15