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Problems on Ages — Questions and Answers

Age problems are linear equations wearing a disguise. Assign a variable to the present age, then add or subtract the same number of years from everyone in the family — the classic mistake is applying the shift to one person only. When a ratio of present ages is given, use the k-method and let the future or past condition determine k.

22 solved questionsArithmetic AptitudeFree · no signup

Problems on Ages— Concepts, Formulas & Shortcuts

  • Let the present age be x; n years ago it was x − n and n years hence it will be x + n.
  • A time shift applies to every person in the problem simultaneously.
  • For ratio questions, write present ages as ax and bx and translate the second condition into an equation.
  • The difference between two people's ages is constant forever — often the fastest route to the answer.
  • Average age of a group × number of members = total age; use totals when members join or leave.
  • "x% of what it was n years ago" converts directly into x = (percentage/100)(x − n).

Problems on Ages Practice Questions with Answers

Attempt each question first, then open the explanation. All 22 questions below are free to read and require no signup.

  1. Q1.A father is three times as old as his son. After 12 years he will be twice as old. Their present ages are:

    Moderate
    • A30 and 10
    • B36 and 12
    • C42 and 14
    • D45 and 15
    +Show Answer & Explanation

    Answer: B. 36 and 12

    Explanation: Let son = x, father = 3x. 3x + 12 = 2(x + 12) → x = 12, so father = 36 and son = 12.

    Problems on Ages question 1 of 22

  2. Q2.The sum of the present ages of A and B is 50 years. Ten years ago A was twice as old as B. A's present age is:

    Moderate
    • A25 years
    • B28 years
    • C30 years
    • D32 years
    +Show Answer & Explanation

    Answer: C. 30 years

    Explanation: A + B = 50 and A − 10 = 2(B − 10) → (50 − B) − 10 = 2B − 20 → 3B = 60 → B = 20, A = 30.

    Problems on Ages question 2 of 22

  3. Q3.The present ages of a father and son are in the ratio 7 : 3. After 6 years the ratio becomes 2 : 1. Their present ages are:

    Moderate
    • A35 and 15
    • B42 and 18
    • C49 and 21
    • D56 and 24
    +Show Answer & Explanation

    Answer: B. 42 and 18

    Explanation: (7x + 6)/(3x + 6) = 2 → 7x + 6 = 6x + 12 → x = 6 → 42 and 18 years.

    Problems on Ages question 3 of 22

  4. Q4.A is two years older than B, who is twice as old as C. If the total of their ages is 27, then B's age is:

    Moderate
    • A7 years
    • B8 years
    • C9 years
    • D10 years
    +Show Answer & Explanation

    Answer: D. 10 years

    Explanation: C = x, B = 2x, A = 2x + 2 → 5x + 2 = 27 → x = 5 → B = 10 years.

    Problems on Ages question 4 of 22

  5. Q5.A man is 4 times as old as his son. Five years ago he was 9 times as old. Their present ages are:

    Moderate
    • A28 and 7
    • B32 and 8
    • C36 and 9
    • D40 and 10
    +Show Answer & Explanation

    Answer: B. 32 and 8

    Explanation: 4x − 5 = 9(x − 5) → 4x − 5 = 9x − 45 → 5x = 40 → x = 8 → man 32, son 8.

    Problems on Ages question 5 of 22

  6. Q6.Three years ago the average age of a husband, wife and child was 27 years. Five years ago the average age of the wife and child was 20 years. The husband's present age is:

    Difficult
    • A35 years
    • B38 years
    • C40 years
    • D42 years
    +Show Answer & Explanation

    Answer: C. 40 years

    Explanation: Present total of all three = 3 × 27 + 3 × 3 = 90. Present total of wife and child = 2 × 20 + 2 × 5 = 50. Husband = 90 − 50 = 40 years.

    Problems on Ages question 6 of 22

  7. Q7.Ten years ago P was half of Q's age. If the ratio of their present ages is 3 : 4, the sum of their present ages is:

    Difficult
    • A30 years
    • B35 years
    • C40 years
    • D45 years
    +Show Answer & Explanation

    Answer: B. 35 years

    Explanation: P = 3k, Q = 4k and 3k − 10 = (4k − 10)/2 → 6k − 20 = 4k − 10 → k = 5 → P = 15, Q = 20 → sum 35.

    Problems on Ages question 7 of 22

  8. Q8.A man's present age is 125% of what it was 10 years ago. His present age is:

    Moderate
    • A40 years
    • B45 years
    • C50 years
    • D55 years
    +Show Answer & Explanation

    Answer: C. 50 years

    Explanation: x = 1.25(x − 10) → x = 1.25x − 12.5 → 0.25x = 12.5 → x = 50 years.

    Problems on Ages question 8 of 22

  9. Q9.The present age of a father is twice that of his son. Twenty years ago the father was four times as old. The son's present age is:

    Moderate
    • A20 years
    • B25 years
    • C30 years
    • D35 years
    +Show Answer & Explanation

    Answer: C. 30 years

    Explanation: 2x − 20 = 4(x − 20) → 2x − 20 = 4x − 80 → 2x = 60 → x = 30 years.

    Problems on Ages question 9 of 22

  10. Q10.The ratio of the ages of two friends is 5 : 7. After 6 years it becomes 3 : 4. Their present ages are:

    Moderate
    • A25 and 35
    • B30 and 42
    • C35 and 49
    • D40 and 56
    +Show Answer & Explanation

    Answer: B. 30 and 42

    Explanation: (5x + 6)/(7x + 6) = 3/4 → 20x + 24 = 21x + 18 → x = 6 → 30 and 42 years.

    Problems on Ages question 10 of 22

  11. Q11.A mother is 30 years older than her daughter. In 12 years she will be three times as old. The daughter's present age is:

    Moderate
    • A3 years
    • B5 years
    • C6 years
    • D8 years
    +Show Answer & Explanation

    Answer: A. 3 years

    Explanation: (x + 30) + 12 = 3(x + 12) → x + 42 = 3x + 36 → 2x = 6 → x = 3 years.

    Problems on Ages question 11 of 22

  12. Q12.The sum of the ages of a father and son is 60 years. Six years ago the father's age was five times the son's. The son's present age is:

    Difficult
    • A12 years
    • B14 years
    • C16 years
    • D20 years
    +Show Answer & Explanation

    Answer: B. 14 years

    Explanation: (60 − x) − 6 = 5(x − 6) → 54 − x = 5x − 30 → 6x = 84 → x = 14 years.

    Problems on Ages question 12 of 22

  13. Q13.The average age of a class of 20 students is 14 years. If the teacher's age is included, the average becomes 15. The teacher's age is:

    Moderate
    • A30 years
    • B32 years
    • C35 years
    • D40 years
    +Show Answer & Explanation

    Answer: C. 35 years

    Explanation: New total = 21 × 15 = 315; old total = 280 → teacher = 35 years.

    Problems on Ages question 13 of 22

  14. Q14.A is 5 years older than B, and B is 3 years older than C. If the sum of their ages is 41, then B's age is:

    Moderate
    • A11 years
    • B12 years
    • C13 years
    • D14 years
    +Show Answer & Explanation

    Answer: C. 13 years

    Explanation: Write everything in terms of B: A = B + 5 and C = B − 3. So (B + 5) + B + (B − 3) = 41 → 3B + 2 = 41 → B = 13 years.

    Problems on Ages question 14 of 22

  15. Q15.Six years ago a man was three times as old as his son. In six years he will be twice as old. The man's present age is:

    Difficult
    • A30 years
    • B36 years
    • C42 years
    • D48 years
    +Show Answer & Explanation

    Answer: C. 42 years

    Explanation: M − 6 = 3(S − 6) and M + 6 = 2(S + 6). Subtracting: 12 = 2S + 12 − 3S + 18 → S = 18, M = 42 years.

    Problems on Ages question 15 of 22

  16. Q16.The present age of a woman is four times her daughter's. In 8 years it will be twice. The daughter's present age is:

    Moderate
    • A4 years
    • B5 years
    • C6 years
    • D8 years
    +Show Answer & Explanation

    Answer: A. 4 years

    Explanation: 4x + 8 = 2(x + 8) → 4x + 8 = 2x + 16 → 2x = 8 → x = 4 years.

    Problems on Ages question 16 of 22

  17. Q17.A father is 30 years older than his son. In 10 years he will be twice as old. The son's present age is:

    Moderate
    • A15 years
    • B20 years
    • C25 years
    • D30 years
    +Show Answer & Explanation

    Answer: B. 20 years

    Explanation: (x + 30) + 10 = 2(x + 10) → x + 40 = 2x + 20 → x = 20 years.

    Problems on Ages question 17 of 22

  18. Q18.The ratio of the present ages of A and B is 2 : 3 and the sum is 50 years. B's age is:

    Easy
    • A20 years
    • B25 years
    • C30 years
    • D35 years
    +Show Answer & Explanation

    Answer: C. 30 years

    Explanation: Total 5 parts → 1 part = 10 → B = 3 × 10 = 30 years.

    Problems on Ages question 18 of 22

  19. Q19.Five years ago a man was 25 years old. His age after 10 years will be:

    Easy
    • A35 years
    • B40 years
    • C45 years
    • D50 years
    +Show Answer & Explanation

    Answer: B. 40 years

    Explanation: Present age = 30, so after 10 years he is 40.

    Problems on Ages question 19 of 22

  20. Q20.A is twice as old as B. Ten years ago A was three times as old. A's present age is:

    Moderate
    • A20 years
    • B30 years
    • C40 years
    • D50 years
    +Show Answer & Explanation

    Answer: C. 40 years

    Explanation: 2x − 10 = 3(x − 10) → 2x − 10 = 3x − 30 → x = 20 (B), so A = 40 years.

    Problems on Ages question 20 of 22

  21. Q21.The average age of a family of 5 is 28 years. The total of their ages is:

    Easy
    • A120 years
    • B130 years
    • C140 years
    • D150 years
    +Show Answer & Explanation

    Answer: C. 140 years

    Explanation: 5 × 28 = 140 years.

    Problems on Ages question 21 of 22

  22. Q22.A mother is three times as old as her son. In 12 years she will be twice as old. The son's present age is:

    Moderate
    • A10 years
    • B12 years
    • C14 years
    • D16 years
    +Show Answer & Explanation

    Answer: B. 12 years

    Explanation: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12 years.

    Problems on Ages question 22 of 22

Problems on Ages — Frequently Asked Questions

What is the most common mistake in age problems?+

Applying a time shift to only one person. If the question says "after 6 years", you must add 6 to every age in the equation, not just to the one you are solving for.

Why is the age difference useful?+

Because it never changes. If a father is 24 years older than his son today, he will still be 24 years older in any past or future year — this often replaces a second equation entirely.

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