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      Question

      The ratio of present ages of A and B is 5:8 respectively

      and the average of present ages of A and C is equal to 80% of the present age of B. 4 years ago from now, B was thrice as old as D. 4 years hence from now, age of D will increase by 25% as compared to his present age. Quantity I: Find the sum of present ages of A and B. Quantity II: Find the sum of present ages of C and D. In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity I and Quantity II and choose the correct option.
      A Quantity I > Quantity II Correct Answer Incorrect Answer
      B Quantity I < Quantity II Correct Answer Incorrect Answer
      C Quantity I ≥ Quantity II Correct Answer Incorrect Answer
      D Quantity I ≤ Quantity II Correct Answer Incorrect Answer
      E Quantity I = Quantity II or No relation Correct Answer Incorrect Answer

      Solution

      Quantity I: Let the present age of D be ‘x’ years. According to the question, => (x + 4) = 1.25x => x = 16 4 years ago from now, age of D = 16 – 4 = 12 years 4 years ago from now, age of B = 3 x 12 = 36 years So, the present age of B = 36 + 4 = 40 years Present age of A = 40 x (5/8) = 25 years 80% of the present age of B = 40 x 0.80 = 32 years Sum of the present ages of A and C = 32 x 2 = 64 years So, the present age of C = 64 – 25 = 39 years Quantity I: Sum of present ages of A and B = 25 + 40 = 65 years Quantity II: Sum of present ages of C and D = 39 + 16 = 55 years Hence, Quantity I > Quantity II

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