Suppose A1, A2,....., A30 are thirty sets each having 5 elements and B1, B2,..., Bn are n sets each having 3 elements. Let $\cup_{i=1}^{30^{\prime}} A_i=\bigcup_{j=1}^n B_j=S$ and each element of $S$ belongs to exactly 10 of $A_i$ 's and exactly 9 of $B_j$ 's

Question

Suppose $A_1, A_2, \ldots, A_{30}$ are thirty sets each having 5 elements and $B_1, B_2, \ldots, B_n$ are $n$ sets each having 3 elements. Let $\cup_{i=1}^{30^{\prime}} A_i=\bigcup_{j=1}^n B_j=S$ and each element of $S$ belongs to exactly 10 of $A_i$ 's and exactly 9 of $B_j$ 's. The value of $n$ is equal to

(a) 30

(b) 38

(c) 45

(d) 50

Answer 1

SOLUTION : If elements are not repeated, then number of elements in $A_1 \cup A_2 \cup \ldots \cup A_n$ is $30 \times 5$. But each element is used 10 times so, $S=\frac{30 \times 5}{10}=15$.

Similarly, if elements in $B_1, B_2, \ldots, B_n$ are not repeated, then total number of elements is $3 n$ but each element is repeated 9 times so,

$S=\frac{3 n}{9}=15$

$\Rightarrow \quad n=45$

So, The correct option of this question will be (c).