A) 15
B) 135
C) 45
D) 90
Correct Answer: C
Solution :
Since, each A; has 5 elements. \[\therefore \]\[\sum\limits_{i=1}^{30}{n({{A}_{i}})=5\times 30=150}\] ?(i) If m distinct elements in S and each element of S belongs to exactly 10 of \[A{{'}_{i}}s,\]so we have \[\sum\limits_{i=1}^{30}{n({{A}_{i}})=10m}\] ?(ii) From Eqs. (i) and (ii), we get 10m =150 \[\Rightarrow \] m = 15 ?(iii) Similarly, \[\sum\limits_{j=1}^{n}{n({{B}_{j}})=3n}\]and \[\sum\limits_{j=1}^{n}{n({{B}_{j}})=9m}\] \[\therefore \] \[3n=9m\] \[\Rightarrow \]\[n=\frac{9m}{3}=3m=3\times 15=45\][from Eq. (iii)] Hence, \[n=45\]You need to login to perform this action.
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