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JEE Main & Advanced Mathematics Sets Question Bank Self Evaluation Test - Sets

  • question_answer
    Let A, B, C be finite sets. Suppose that \[n(A)=10,\]\[n(B)=15,\] \[n(C)=20,\] \[n(A\cap B)=8\] and \[n(B\cap C)=9.\] Then the possible value of \[n(A\cup B\cup C)\] is

    A) 26

    B) 27

    C) 28

    D) Any of the three values 26, 27, 28 is possible

    Correct Answer: D

    Solution :

    [d]We have
    \[n(A\cup B\cup C)=n(A)+n(B)+n(C)-\]
    \[n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\]
    \[=10+15+20-8-9-n(C\cap A)+n(A\cap B\cap C)\]
    \[=28-\{n(C\cap A)-n(A\cap B\cap C)\}\]               (i)
    Since \[n(C\cap A)\ge n(A\cap B\cap C)\]
    We have \[n(C\cap A)\ge n(A\cap B\cap C)\ge 0\]   (ii)
    From (i) and (ii): \[n(A\cup B\cup C)\le 28\]            (iii)
    Now, \[n(A\cup B)=n(A)+n(B)-n(A\cap B)=10+15-8=17\]and \[n(B\cup C)=n(B)+n(C)-n(B\cap C)=15+20-9=26\]since, \[n(A\cup B\cup C)\ge n(A\cup C)\] and \[n(A\cup B\cup C)\ge n(B\cup C)\] we have
    \[n(A\cup B\cup C)\ge 17\] and \[n(A\cup B\cup C)\ge 26\]
    Hence \[n(A\cup B\cup C)\ge 26\]             (iv)
    From (iii) and (iv) we obtain
    \[26\le n(A\cup B\cup C)\le 28\]
    Also \[n(A\cup B\cup C)\] is a positive integer \[\therefore \,\,\,n(A\cup B\cup C)=26\,\,or\,\,27\,\,or\,\,28\]


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