NIMCET 2021 Question Paper with solutions

It is given that each A has 5 elements
$\sum _{\mathit{i}=1}^{30}\mathit{n}\left({\mathit{A}}_{\mathit{i}}\right)=5\times 30=150$
Suppose s has $\mathit{m}$ distinct elements of $\mathit{A}$ 's.
Since each element of $\mathit{S}$ belongs to exactly 10 of ${\mathit{A}}_{\mathit{i}}$ 's
$\therefore \sum _{\mathit{i}=1}^{30}\mathit{n}\left({\mathit{A}}_{\mathit{i}}\right)=10\mathit{m}$
$\Rightarrow 10\mathit{m}=150$
$\Rightarrow \mathit{m}=15$
Since each ${\mathit{B}}_{\mathit{j}}$ has 3 elements and each element of $\mathit{S}$ belongs to exactly 9 of the $\mathit{B}$ 's
$\sum _{\mathit{j}=1}^{\mathit{n}}\mathit{n}\left({\mathit{B}}_{\mathit{j}}\right)=3\mathit{n}$ and $\sum _{\mathit{j}=1}^{\mathit{n}}\mathit{n}\left({\mathit{B}}_{\mathit{j}}\right)=9\mathit{m}$
$\Rightarrow 3\mathit{n}=9\mathit{m}$
$\Rightarrow \mathit{n}=3\mathit{m}$
$\Rightarrow \mathit{m}=15$ so $\mathit{n}=3\times 15$
$\Rightarrow \mathit{n}=45$