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8th Class Mathematics Factorisation Question Bank Factors & Multiples

  • question_answer
    The sum of two numbers is 462 and their highest common factor is 22. What is the minimum number of pair that satisfy these conditions?

    A)  5                                

    B)  6            

    C)  7                                

    D)  8

    Correct Answer: B

    Solution :

    Let the required number be 22a and 22b. Then, \[22a+22b=462\,\] \[22(a+b)=462\] \[a+b=21.\] Now, Co - Primes with sum 21 are \[(1,\,\,20),\,\,(2,\,\,19),\,\,(4,17),\,\,(5,16),\,\,(8,\,\,13)\]and \[(10,\,\,11)\] \[\therefore \] Required numbers are \[(22\times 1,\,\,22\times 20),\]\[(22\times 2,\,\,22\times 19),\]\[(22\times 4,\,\,22\times 17),\]\[(22\times 5,\,\,22\times 16),\]\[(22\times 8,\,\,22\times 13),\]and \[(22\times 10,\,\,22\times 11).\] Clearly, the number of such pairs is 6.


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