According to the question,
n(C) = 50, n(H) = 50, n(V) = 40
n(C∩H) = 15
n(H∩V) = 20
n(C∩V) = 15
n(C∩H∩V) = 10
No. of students who played at least one game
n(CUHUV) = 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
Total number of students = 100.
Let a denote the number of people who played cricket and volleyball only.
Let b denote the number of people who played cricket and hockey only.
Let c denote the number of people who played hockey and volleyball only.
Let d denote the number of people who played all three games.
Accordingly, d = n (C∩H∩V) = 10
Now, n(C∩V) = a + d = 15
n(C∩H) = b + d = 15
n(H∩V) = c + d = 20
Therefore, a = 15 – 10 = 5 [cricket and volleyball only]
b = 15 – 10 = 5 [cricket and hockey only]
c = 20 – 10 = 10 [hockey and volleyball only]
No. of students who played only cricket = n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30
No. of students who played only hockey = n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25
No. of students who played only volley ball = n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Alternatively, we can solve it faster with the help of a Venn diagram.
The Venn diagram for the given information looks like this.

Subtracting the values in the intersections from the individual values gives us the number of students who played only one game.