Let bigcup_{i=1}^{50} X_{i} = bigcup_{i=1}^{n | vedclass

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(D) Let $n(T)$ be the number of elements in set $T$. Given that $\bigcup_{i=1}^{50} X_{i} = T$ and each $X_{i}$ has $10$ elements,the sum of the numbe

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$30$

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(D) Let $n(T)$ be the number of elements in set $T$. Given that $\bigcup_{i=1}^{50} X_{i} = T$ and each $X_{i}$ has $10$ elements,the sum of the number of elements in all $X_{i}$ is $50 	imes 10 = 500$. Since each element of $T$ belongs to exactly $20$ sets $X_{i}$,we have $20 	imes n(T) = 500$,which gives $n(T) = \frac{500}{20} = 25$. Similarly,for the sets $Y_{i}$,we have $\bigcup_{i=1}^{n} Y_{i} = T$ and each $Y_{i}$ has $5$ elements,so the sum of the number of elements in all $Y_{i}$ is $n 	imes 5 = 5n$. Since each element of $T$ belongs to exactly $6$ sets $Y_{i}$,we have $6 	imes n(T) = 5n$. Substituting $n(T) = 25$,we get $6 	imes 25 = 5n$,which simplifies to $150 = 5n$,so $n = 30$.

Let $\bigcup_{i=1}^{50} X_{i} = \bigcup_{i=1}^{n} Y_{i} = T$,where each $X_{i}$ contains $10$ elements and each $Y_{i}$ contains $5$ elements. If each element of the set $T$ is an element of exactly $20$ of the sets $X_{i}$ and exactly $6$ of the sets $Y_{i}$,then $n$ is equal to:

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