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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • 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 set each with three elements. Let \[\bigcup\limits_{i\,=\,1}^{30}{{{A}_{i}}=}\,\bigcup\limits_{j\,=\,1}^{n}{{{B}_{j}}=S}\]. Assume that each element of S belongs to exactly 10 of the \[{{A}_{i}}'s\] and exactly 9 of \[{{B}_{j}}'s\]. The value of n must be

    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 \[n(S)=\frac{1}{9}\sum\limits_{j=1}^{n}{n({{B}_{j}})}\] \[\Rightarrow \]            \[15=\frac{1}{9}(n\times 3)\Rightarrow \,n=45\]


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