Each set $X_{r}$ contains 5 elements and each set $Y_{r}$ contains 2 elements and $\bigcup_{r=1}^{20} X_{r}=S=\bigcup_{r=1}^{n} Y_{r}$. If each element of $S$ belongs to exactly 10 of the $X_{r}^{\prime s}$ and to exactly 4 of the $Y_{r}^{\prime} s$, then $n$ is
(a) 10
(b) 20
(c) 100
(d) 50
Let us suppose
Each $x_{r}$ contains 5 elements and each $y_{r}$ contains 2 elements such that $\bigcup_{r=1}^{20} X_{r}=S=\bigcup_{r=1}^{n} Y_{r}$
$\therefore n(S)=20 \times 5 \quad\left(\because\right.$ each $x_{r}$ has 5 elements $)$
n(S) = 100
It is given that each element of 5 belong to exactly 10 of the xr's.
$\therefore$ Number of distinct elements in $S=\frac{100}{10}=10$
Since each $y_{r}$ has 2 elements and $\bigcup_{r=1}^{u} Y_{r}=S$
∴ n(S) = n × 2 = 2n
And each element of S belong to exactly 4 of yr's
$\Rightarrow$ number of distinct elements in $S=\frac{2 n}{4}$ ...(2)
from (1) and (2)
$10=\frac{2 n}{4}$
i. e $n=20$
Hence, the correct answer is option B.