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

The number of students who play cricket is \(n(C)=50\)

The number of students who play hockey is \(n(H)=50\)

Number of students who play volley ball is \(n(V)=40\)

And \(n(C \cap H)=15\)

$$ \begin{aligned} &n(H \cap V)=20 \\ &n(C \cap V)=15 \\ &n(C \cap H \cap V)=10 \end{aligned} $$

Now, the total number of students is given by

$$ \begin{gathered} n(C \cup H \cup V)=n(C)+n(H)+n(V)-n(C \cap H)-n(H \cap V)-n(C \cap V) \\ +n(C \cap H \cap V) \\ =50+50+40-15-20-15+10 \\ =100 \end{gathered} $$

Hence, the total number of students is \(n(C \cup H \cup V)=100\)

Now, the number of students who played only cricket is given by

$$ \begin{aligned} n(C)-n(C \cap H)-n(C \cap V) &-n(C \cap H \cap V) \\ n(C)-n(C \cap H)-n(C \cap V) &-n(C \cap H \cap V)=50-15-15-10 \\ &=10 \end{aligned} $$

Hence, the number of students who played only cricket is \(10 .\)

Now, the number of students who played only hockey is given by

$$ \begin{aligned} &n(H)-n(C \cap H)-n(H \cap V)-n(C \cap H \cap V) \\ &n(H)-n(C \cap H)-n(H \cap V)-n(C \cap H \cap V)=50-15-20-10 \\ &=5 \end{aligned} $$

Hence, the number of students who played only hockey is \(5 .\)

Now, the number of students who played only Volley ball is given by

$$ n(V)-n(C \cap V)-n(H \cap V)-n(C \cap H \cap V) $$

$$ \begin{aligned} n(V)-n(C \cap V)-n(H \cap V) &-n(C \cap H \cap V)=40-15-20-10 \\ &=-5 \end{aligned} $$

Hence, no students who played only hockey.