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

The total number of students = 100

Number of students that play cricket only = 30

Number of students that play hockey only = 25

Number of students that play volleyball only = 15

Step-by-step explanation:

Let,

Total number of students be n(T)

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

A.T.Q., It is given that:

n(C) = 50

n(H) = 50

n(V) = 40

Number of students played both cricket and hockey both = n(C&H) = 15

Number of students played both hockey and  volley ball = n(H&V) = 20

Number of students played both cricket and volley ball  = n(C&V) = 15

Number of students played all three = n(C&H&V) = 10

Now,

The number of students playing at least one game n(T) = 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

Now,

Let p = number of students playing cricket and volleyball only.

Let q = number of students playing cricket and hockey only.

Let r = number of students playing hockey and volleyball only.

Let s = number of students playing all three games.

This shows that:

s = n (C&H&V) = 10

n(C&V) = p + s = 15

n(C&H) = q + s = 15

n(H&V) = r + s = 20

Thus,

Hence,

p = 15 – 10

= 5

q = 15 – 10

= 5

r = 20 – 10

= 10

So,

The no. of students playing cricket only

=   n(C) – [p + q + s]

= 50 – (5 + 5 + 10)

= 30

The no. of students playing hockey only

= n(H) – [q + r + s]

= 50 – ( 5 + 10 + 10)

= 25

The no. of students playing volleyball only

= n(V) – [p + r + s]

= 40 – (10 + 5 + 10)

= 15

Learn more: Find the number of students

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