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      Question

      Product of two consecutive positive even numbers is 440.

      Find the sum of the digits of the two numbers. 
      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.  ATQ,  (2n * (2n + 2) = 440 )  Or, (n * (2n + 2) = 220 )  Or, ( 2n² + 2n = 220 )  Or, ( n² + n = 110 )  Or, ( n² + n - 110 = 0 )  Or, ( n² + 11n - 10n - 110 = 0 )  Or, ( n(n + 11) - 10(n + 11) = 0 )  Or, ( (n + 11)(n - 10) = 0 )  So, ( n = -11 ) or ( 10 )  So, ( n = 10 )  So, numbers are ( 20 ) and ( 22 ).  Therefore, required sum = ( 2 + 0 + 2 + 2 = 6 ) 

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