Three years ago , sum of the ages of a, b and c was 52 years. Four years ago…

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Three years ago , sum of the ages of a, b and c was 52 years. Four years ago B, who is a year younger than C, was 1(1/2) times as old as A. The sum of the ages of B and C two years hence will be -

  1. A.

    43 years

  2. B.

    51 years

  3. C.

    49 years

  4. D.

    42 years

Attempted by 10 students.

Show answer & explanation

Correct answer: C

Answer: 49 years

  1. Let the present ages be A = a, B = b, C = c.

  2. Three years ago the sum was 52, so (a - 3) + (b - 3) + (c - 3) = 52, which gives a + b + c = 61.

  3. B is a year younger than C, so c = b + 1.

  4. Four years ago B was 3/2 times A: b - 4 = (3/2)(a - 4). Multiply both sides: 2b - 8 = 3a - 12, so 2b - 3a = -4.

  5. Substitute c = b + 1 into a + b + c = 61 to get a + 2b = 60, so a = 60 - 2b.

  6. Plug a into 2b - 3a = -4: 2b - 3(60 - 2b) = -4 => 8b - 180 = -4 => 8b = 176 => b = 22.

  7. Then a = 60 - 2(22) = 16, and c = b + 1 = 23.

  8. Two years hence, sum of B and C = (b + 2) + (c + 2) = (22 + 2) + (23 + 2) = 24 + 25 = 49.

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