First slide

Operations on Sets

Question

Suppose $A_{1}, A_{2}, ......... A_{30}$ are 30 sets each having 5 elements and $B_{1}, B_{2}, ......... B_{n}$ are n sets each With 3 elements. Let $\cup_{i = 1}^{30} A_{i} = \cup_{j = 1}^{n} B_{j} = S$ and each element of S belongs to exactly 10 of the $A_{i}' s$ and exactly 9 of the $B_{j}' s$ , then n =

Moderate

A

15
B

3
C

45
D

50

Solution

Since $s = \cup_{i = 1}^{30} A_{i}$ Now element in the union S belongs to exactly 10 of the $A^{1}_{i} s$ so $n (s) = \frac{1}{10}$ ( 5 x 30 ) = 15
Again $s = \cup_{j = 1}^{n} B_{j}$

$∴$ $n (s) = \frac{1}{9} (3 x n) = \frac{n}{3}$

Thus $\frac{n}{3} = 15$ $\Rightarrow n = 45$

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