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    Question

    Product of two consecutive positive even numbers is 440.

    Find the sum of the digits of the two numbers.┬а
    A 6 Correct Answer Incorrect Answer
    B 8 Correct Answer Incorrect Answer
    C 10 Correct Answer Incorrect Answer
    D 12 Correct Answer Incorrect Answer

    Solution

    Let the numbers be ( 2n ) and ( (2n + 2) ) respectively.┬а ATQ,┬а (2n * (2n + 2) = 440 )┬а Or, (n * (2n + 2) = 220 )┬а Or, ( 2n┬▓ + 2n = 220 )┬а Or, ( n┬▓ + n = 110 )┬а Or, ( n┬▓ + n - 110 = 0 )┬а Or, ( n┬▓ + 11n - 10n - 110 = 0 )┬а Or, ( n(n + 11) - 10(n + 11) = 0 )┬а Or, ( (n + 11)(n - 10) = 0 )┬а So, ( n = -11 ) or ( 10 )┬а So, ( n = 10 )┬а So, numbers are ( 20 ) and ( 22 ).┬а Therefore, required sum = ( 2 + 0 + 2 + 2 = 6 )┬а

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