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JEE Main & Advanced Sample Paper JEE Main Sample Paper-41

  • question_answer
    Suppose \[{{A}_{1}},{{A}_{2}},...,{{A}_{30}}\]are thirty sets each with five elements and \[{{B}_{1}},{{B}_{2}},...,{{B}_{n}}\]are n sets each with three elements such that \[\underset{i=1}{\overset{30}{\mathop{\cup }}}\,{{A}_{i}}=\underset{i=1}{\overset{n}{\mathop{\cup }}}\,{{B}_{i}}=S\]. If each element of S belongs to exactly ten of \[{{A}_{i}}'s\] and exactly 9 of the \[{{B}_{j}}'s\], then the value of n is

    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, \[n=45\]


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