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      Question

      If x + y + xy = 118, such that x < y and both 'x' and

      'y' are positive integers, then find minimum value of (x + y) .
      A 24 Correct Answer Incorrect Answer
      B 34 Correct Answer Incorrect Answer
      C 22 Correct Answer Incorrect Answer
      D 32 Correct Answer Incorrect Answer

      Solution

      x + y + xy = 118

      Add 1 on both sides of the equation, we get,

      1 + x + y + xy = 118 + 1

      Or, (1 + x) + y X (1 + x) = 119

      So, (1 + x) X (1 + y) = (7 X 17) or (1 X 119)

      When, (1 + x) X (1 + y) = 7 X 17

      We can say that, 1 + x = 17 or 1 + y = 17

      So, 1 + x = 17 or, 1 + y = 17

      So, x = 16 or y = 16

      When, x = 16, y = 6 and when y = 16, x = 6

      So, for both cases, x + y = 16 + 6 = 22

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