Let the sets $A=\left\{2,\mathrm{}4,\mathrm{}\mathrm{6,8}\dots \right\}$ and $B=\left\{3,\mathrm{}6,\mathrm{}9,\mathrm{}12\dots \right\}$ such that $n\left(A\right)=200$ and  $n\left(B\right)=250.$ If $n\left(A\cup B\right)=k$, then $\frac{k}{100}$ is equal to

Let, the last element of set $A$ is, $t_{200}=2+\left(200-1\right)2=400$

and the last element of set $B$ is $t_{250}=3+\left(250-1\right)3=750$

Common elements are $\to 6,\mathrm{}12,\mathrm{}18\dots$

We have to find the common elements upto $400$
$t_n=6+\left(\mathrm{}n-1\right)6\Rightarrow 400=6+6n-6$ $\Rightarrow n=\frac{400}{6}=66.6\dots$
$t_{66}=6+\left(66-1\right)6=396$
Number of common elements $=66$

$n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$
$=200+250-66=384\Rightarrow k=384$
$\Rightarrow \frac{k}{100}=3.84$