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Question: In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student ...

In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student like to play at least one of the two games. How many students like to play both cricket and football?

Explanation

Solution

Hint: First, as per the given data we should draw the diagram for better and clear understanding. Then we know that union of set cricket and set football is given i.e. 35 and value of set football and value of set cricket is also given. Just we need to put the data in the formula n(cf)=n(c)+n(f)n(cf)n\left( c\cup f \right)=n\left( c \right)+n\left( f \right)-n\left( c\cap f \right) and we will get the answer.

Complete step-by-step answer:

Here, we are given that total students are 35 out of which 24 like to play cricket which we can denote as n(c)=24n\left( c \right)=24 and 16 like to play football which is denoted as n(f)=16n\left( f \right)=16 . So, total students who like to play cricket or football are denoted as union i.e. n(cf)=35n\left( c\cup f \right)=35 . This, can be shown as Venn diagram as below:

Cricket set is denoted as c and football set is denoted as f. So, we have to find black -colored region which is the intersection region, so we will come to know students who like to play both cricket and football.

So, we will use the union of 2 sets formula, given as:

n(cf)=n(c)+n(f)n(cf)n\left( c\cup f \right)=n\left( c \right)+n\left( f \right)-n\left( c\cap f \right)

Substituting all the values in the above formula, we get

35=24+16n(cf)\Rightarrow 35=24+16-n\left( c\cap f \right)

35=40n(cf)\Rightarrow 35=40-n\left( c\cap f \right)

Taking constant on right-hand side and intersection term on left-hand side, we get

n(cf)=4035=5\Rightarrow n\left( c\cap f \right)=40-35=5

Thus, 5 students like to play both i.e. cricket and football.

Note: Sometimes, students get confused in the data whether it is given as a class of 35 students should be taken as a universal set or not. If it is taken as a universal set i.e. n(U)=35n\left( U \right)=35 and then see that the value of set football and set cricket should be equal to universal set. But here it will not match with the universal set. So, by seeing this student sometimes they don't understand which formula to apply here. So, don’t make this mistake if the universal word is given in question then only take that otherwise consider it as union of set football and cricket.