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1.Set Theory
hard
Suppose ${A_1},\,{A_2},\,{A_3},........,{A_{30}}$ are thirty sets each having $5$ elements and ${B_1},\,{B_2}, ......., B_n$ are $n$ sets each with $3$ elements. Let $\bigcup\limits_{i = 1}^{30} {{A_i}} = \bigcup\limits_{j = 1}^n {{B_j}} = S$ and each elements of $S$ belongs to exactly $10$ of the $A_i's$ and exactly $9$ of the $B_j's$. Then $n$ is equal to
A
$15$
B
$3$
C
$45$
D
None of these
Solution
(c) $O(S)$ = $O\left( {\bigcup\limits_{i = 1}^{30} {{A_i}} } \right) = \frac{1}{{10}}(5 \times 30) = 15$
Since, element in the union $S$ belongs to $10$ of $Ai$' s
Also, $O(S)$ = $O\left( {\bigcup\limits_{j\, = 1}^n {{B_j}} } \right) = \frac{{3n}}{9} = \frac{n}{3}$,
$\therefore$ $\frac{n}{3} = 15 \Rightarrow n = 45$.
Standard 11
Mathematics
