6th Class Mathematics Related to Competitive Exam Question Bank Series

  • question_answer
    3, 4, 5, 5, 12, 13, 7, 24, 25, 9, 41?

    A) 16                   

    B)          24

    C) 35                   

    D)          40

    Correct Answer: D

    Solution :

        Numbers are in sets of three such that each set forms a Pythagorean triple, i.e, \[{{3}^{2}}+{{4}^{2}}={{5}^{2}},\,{{12}^{2}}\,={{13}^{2}},\,{{7}^{2}}+{{24}^{2}}\,={{25}^{2}}\]  and so on. Quicker: 2nd number in each set is (1st number \[\times n+n\]) Method: Where, n = 1, 2, 3 ........ is set no. next number in the series is \[(9\times 4+4)=40\]


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