**Question:**

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

**Solution:**

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 = 2*n*

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.

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