Question

Let $A=\{1,2,3,4,...,40\}$ be the set of registration number of students of class $12$. Students having registration number as multiple of $4$ play cricket but do not play football, students having even registration number play either cricket or football or both and students having registration number as multiple of $10$ play both cricket and football. Find the number of students who play football.

• A. $6$
• B. $14$
• C. $16$
• D. $10$

Answer 1

The correct option is D $10$

Let $X$ be the set of students who play cricket and set of students who play football be $Y$.
$\text{Then}X-Y=\{4,8,12,16,20,24,28,32,36,40\}$
$\Rightarrow n(X-Y)$ $=10$
and $X\cup Y$ $=\{2,4,6,...,40\}$
$\Rightarrow n(X\cup Y)$ $=20$
also $X\cap Y$ $=\{10,20,30,40\}$
$\Rightarrow n(X\cap Y)$ $=4$

Using, $n(X\cup Y)=n(X-Y)+n(Y-X)+n(X\cap Y)$
$\Rightarrow n(Y-X)=20-10-4$
$\Rightarrow n(Y-X)=6$

Here we have
$n\left(Y\right)=n(Y-X)+n(X\cap Y)$
$\Rightarrow n\left(Y\right)=6+4=10$