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      Question

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

      β€˜C’ and β€˜D’ is 50 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 95 years Correct Answer Incorrect Answer
      B 92 years Correct Answer Incorrect Answer
      C 98 years Correct Answer Incorrect Answer
      D 93 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 = 200 Or, 4x = 188 Or, x = 47 years (Age of A) Present age of β€˜C’ = x + 4 = 47 + 4 = 51 years Required sum = 47 + 51 = 98 years

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