Q. 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 with 3 elements, let $\bigcup_{i = 1}^{30}{A_{i} =}\bigcup_{j = 1}^{n}{B_{j} =}S$ and each element of $S$ belongs to exactly 10 of the $A_{i}$'s and exactly 9 of the $B_{j}$'s. Then, $n$ is equal to
Explanation
(c)
Given, $A$'s are 30 sets with five elements each, so
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{i = 1}^{30}{n\left( A_{i} \right) = 5 \times 30 = 150}$ ...(i)
If the $m$ distinct elements in $S$ and each elements of $S$ belongs to exactly 10 of the $A_{i}$'s, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{i = 1}^{30}{n\left( A_{i} \right) = 10m}$ ...(ii)
From Eqs. (i) and (ii), $m = 15$
Similarly, $\sum_{j = 1}^{n}{n\left( B_{j} \right) = 3n}\ \$and$\ \ \sum_{j = 1}^{n}{n\left( B_{j} \right) = 9m}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3n = 9m$
$\Rightarrow \ \ \ \ \ n = \frac{9m}{3} = 3 \times 15 = 45$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.