Question

In a class of $50$ students, $20$ play cricket, $15$ play hockey and $5$ play both the games. Then, the number of students who play neither is
A)$5$
B)$10$
C)$15$
D)$20$

Answer 1

Hint: Apply the formula $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ , the value of $n\left( {A \cup B} \right)$ gives the number of students who play at least one game. Hence, students who play neither game could be found by taking the difference of total and students and $n\left( {A \cup B} \right)$

Complete step-by-step answer: Total number of students $ = 50$
Total number of students $ = 50$
Number of students who play cricket; $n\left( C \right) = 20$ (1)
Number of students who play hockey; $n\left( H \right) = 15$ (2)
Number of students who play both the games; $n\left( {C \cap H} \right) = 5$ (3)
We know that $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Here, it becomes,
$n\left( {C \cup H} \right) = n\left( C \right) + n\left( H \right) - n\left( {C \cap H} \right)$ (4)
Where, $n\left( {C \cup H} \right)$ provides the value of the number of students who play at least one game.
Substituting the values of $n\left( C \right){\text{ , }}n\left( H \right){\text{ , }}n\left( {C \cap H} \right)$ from equation (1), (2), (3) in equation (4)
We get,
$
  n\left( {C \cup H} \right) = 20 + 15 - 5 \\
  n\left( {C \cup H} \right) = 30 \\
 $
Students who play neither
$
   = {\text{ total students }} - {\text{ students playing at least one game}} \\
  {\text{ = }}50 - 30 \\
   = 20 \\
 $

So, the correct answer is “Option D”.


Note: These types of problems could also be solved using different techniques , i.e., by venn diagram. This approach to the problem involves drawing two circles for cricket and hockey and they must overlap since the overlap part indicates the students who play both the games. The outside of the venn diagram must be equal to the total number of students in the class, i.e., $50$