A) 30
B) 40
C) 45
D) 50
Correct Answer: C
Solution :
Since, \[S=\,\bigcup\limits_{i=1}^{30}{{{A}_{i}}}\] and each element of S is in 9 \[10\,{{A}_{i}}'s\]we have \[n(S)=\frac{1}{10}\,\sum\limits_{i\,=\,1}^{30}{n({{A}_{i}})=}({{A}_{i}})=\frac{1}{10}\,(30\times 5)=15\] Also, \[S=\bigcup\limits_{j=1}^{n}{{{B}_{j}}}\] and each element of S is in \[9\,{{B}_{j}}'s\] we have \[n(S)=\frac{1}{9}\sum\limits_{j=1}^{n}{n({{B}_{j}})}\] \[\Rightarrow \] \[15=\frac{1}{9}(n\times 3)\Rightarrow \,n=45\]You need to login to perform this action.
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