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    Question

    Present average age of 4 friends: β€˜A’, β€˜B’,

    β€˜C’ and β€˜D’ is 25 years. If their ages are in the arithmetic progression in the same order and β€˜D’ is 4 years older than β€˜B’, then find the sum of present ages of β€˜A’ and β€˜C’.
    A 42 years Correct Answer Incorrect Answer
    B 45 years Correct Answer Incorrect Answer
    C 48 years Correct Answer Incorrect Answer
    D 41 years Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    Common difference between ages = (4/2) = 2 years. So, let present ages of β€˜A’, β€˜B’, β€˜C’ and β€˜D’ be β€˜x’, (x + 2) years, (x + 4) years and (x + 6) years, respectively According to question: x + x + 2 + x + 4 + x + 6 = 100 Or, 4x = 88 Or, x = 22 years (Age of A) Present age of β€˜C’ = x + 4 = 22 + 4 = 26 years Required sum = 22 + 26 = 48 years

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