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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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