Quantitative Aptitude

Problems on Ages: Practice Questions with Answers

15 practice questions on Problems on Ages, with answers; most also have an explanation or a worked solution. Try each one before you open its answer.

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1. Q.1

   Five years ago, the ratio of A's and B's ages was 3 : 5. Ten years from now, the ratio will be 3 : 4. Find their present ages.

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   Answer: A = 20 years, B = 30 years

   Solution:

   1. Let present ages be A and B. Five years ago: (A−5)/(B−5) = 3/5 → 5A − 25 = 3B − 15 → 5A − 3B = 10 … (1)
   2. Ten years from now: (A+10)/(B+10) = 3/4 → 4A + 40 = 3B + 30 → 4A − 3B = −10 … (2)
   3. Subtract (2) from (1): A = 20. Substitute into (2): 4(20) − 3B = −10 → 80 − 3B = −10 → B = 30.
   4. Check: 5 years ago, 15 : 25 = 3 : 5. 10 years from now, 30 : 40 = 3 : 4. Both match.

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2. Q.2

   A grandfather is 4 times as old as his grandson. In 8 years, the grandfather will be 3 times as old as the grandson will be then. Find their present ages.

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   Answer: Grandson = 16 years, Grandfather = 64 years

   Solution:

   1. Let the grandson's present age = x, so the grandfather's present age = 4x.
   2. In 8 years: 4x + 8 = 3(x + 8) → 4x + 8 = 3x + 24 → x = 16.
   3. Grandson = 16, Grandfather = 64. Check: in 8 years, grandson = 24, grandfather = 72, and 72 = 3 × 24.

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3. Q.3

   The present ages of A and B are in the ratio 4 : 5. Eight years ago, their ages were in the ratio 3 : 4. Find their present ages.

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   Answer: A = 32 years, B = 40 years

   Solution:

   1. Let A = 4x and B = 5x.
   2. Eight years ago: A − 8 = 4x − 8, B − 8 = 5x − 8. Given: (4x − 8) / (5x − 8) = 3 / 4.
   3. Cross-multiply: 4(4x − 8) = 3(5x − 8) → 16x − 32 = 15x − 24 → x = 8.
   4. A = 4 × 8 = 32, B = 5 × 8 = 40. Check: 8 years ago, A = 24, B = 32, and 24 : 32 = 3 : 4.

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4. Q.4

   The sum of the present ages of a mother and her daughter is 40 years. Five years ago, the mother’s age was 4 times the daughter’s age. Find their present ages.

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   Answer: Daughter = 11 years, Mother = 29 years

   Solution:

   1. Let the daughter’s present age = x, so the mother’s present age = 40 − x.
   2. Five years ago: daughter = x − 5, mother = 35 − x. Given: 35 − x = 4(x − 5) → 35 − x = 4x − 20 → 55 = 5x → x = 11.
   3. Daughter = 11, Mother = 40 − 11 = 29. Check: 5 years ago, daughter = 6, mother = 24, and 24 = 4 × 6.

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5. Q.5

   Five years ago, Rohan’s age was three times his daughter’s age. Ten years from now, Rohan’s age will be twice his daughter’s age. Find their present ages.

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   Answer: Daughter = 20 years, Rohan = 50 years

   Solution:

   1. Let the daughter’s present age = x and Rohan’s present age = y.
   2. Five years ago: y − 5 = 3(x − 5) → y = 3x − 10 … (1)
   3. Ten years from now: y + 10 = 2(x + 10) → y = 2x + 10 … (2)
   4. Setting (1) = (2): 3x − 10 = 2x + 10 → x = 20. Then y = 2(20) + 10 = 50.

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6. Q.6

   A father is 3 times as old as his son. After 12 years, the father will be twice as old as his son. Find their present ages.

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   Answer: Son = 12 years, Father = 36 years

   Solution:

   1. Let the son’s present age = x, so the father’s present age = 3x.
   2. After 12 years: father = 3x + 12, son = x + 12. Given: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12.
   3. Son = 12, Father = 36. Check: after 12 years, father = 48, son = 24, and 48 = 2 × 24.

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7. Q.7

   The difference between the ages of a father and his son is 24 years. In 4 years, the father will be 4 times as old as the son will be then. Find their present ages.

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   Answer: Son = 4 years, Father = 28 years

   Solution:

   1. Let the son's present age = x, so the father's present age = x + 24 (the age difference stays 24 years at every point in time).
   2. In 4 years: (x + 24) + 4 = 4(x + 4) → x + 28 = 4x + 16 → 3x = 12 → x = 4.
   3. Son = 4, Father = 28. Check the difference: 28 − 4 = 24. Check the given condition: in 4 years, father = 32, son = 8, and 32 = 4 × 8.

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8. Q.8

   The difference between the ages of two friends is 15 years. Eight years ago, the elder was twice as old as the younger. Find their present ages.

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   Answer: Younger = 23 years, Elder = 38 years. Elder = younger+15; 8 years ago, (younger+15−8) = 2(younger−8).

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9. Q.9

   Six years ago, a man was 4 times as old as his son. Six years from now, he will be twice as old as his son. Find their present ages.

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   Answer: Son = 12 years, Man = 30 years

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10. Q.10

    The present ages of P and Q are in the ratio 5 : 7. Six years ago, their ages were in the ratio 2 : 3. Find their present ages.

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    Answer: P = 30 years, Q = 42 years

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11. Q.11

    Five years ago, the ratio of X's and Y's ages was 1 : 2. Ten years from now, the ratio will be 2 : 3. Find their present ages.

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    Answer: X = 20 years, Y = 35 years. Set up (X−5)/(Y−5)=1/2 and (X+10)/(Y+10)=2/3, then solve together.

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12. Q.12

    The sum of the present ages of a man and his wife is 72 years. Four years ago, the ratio of their ages was 5 : 3. Find their present ages.

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    Answer: Man = 44 years, Wife = 28 years. (M−4)/(W−4)=5/3 with M+W=72.

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13. Q.13

    A mother is 5 times as old as her daughter. In 6 years, she will be 3 times as old as the daughter will be then. Find their present ages.

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    Answer: Daughter = 6 years, Mother = 30 years. 5x+6 = 3(x+6) gives x = 6.

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14. Q.14

    A mother is 4 times as old as her son. After 20 years, she will be twice as old as her son. Find their present ages.

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    Answer: Son = 10 years, Mother = 40 years

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15. Q.15

    A mother is 5 times as old as his daughter now. After 10 years, the mother will be 3 times as old as the daughter. Find mother's present age.

    1. A 50 years
    2. B 10 years
    3. C 60 years
    4. D 40 years
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    Correct answer: (A) 50 years

    Let the daughter's present age be s. mother's age = 5s. After 10 years: 5s+10 = 3(s+10) → s = 10. So mother = 50, daughter = 10, sum = 60, difference = 40.

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