Problems on Ages Questions with Solutions
Problems on Ages are really just algebra in disguise — assign a variable to one age, express every other age in terms of it, and the given relationship becomes a solvable equation. Here’s the approach, four fully worked examples, and a few to try yourself.
The approach, quickly
- Assign a variable to one unknown age (usually the one appearing in the simplest ratio or relationship), then express every other age in the problem in terms of that same variable.
- "A is twice as old as B" means A = 2B — translate sentences in the same subject-first order they’re written, not reversed.
- For "after N years" or "N years ago" conditions, add or subtract N from EVERY person’s age in the problem, including the one already defined in terms of x — forgetting to adjust one person’s age is the most common mistake here.
- When a ratio is given directly (like "ages are in the ratio 3 : 5"), use a single multiplier — 3x and 5x — rather than two separate unknowns. This turns the problem into one variable, solvable with just one more equation from the rest of the problem.
Worked examples
1. 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.
Solution
- Let the son’s present age = x, so the father’s present age = 3x.
- After 12 years: father = 3x + 12, son = x + 12. Given: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12.
- Son = 12, Father = 36. Check: after 12 years, father = 48, son = 24, and 48 = 2 × 24.
Answer: Son = 12 years, Father = 36 years
2. 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.
Solution
- Let A = 4x and B = 5x.
- Eight years ago: A − 8 = 4x − 8, B − 8 = 5x − 8. Given: (4x − 8) / (5x − 8) = 3 / 4.
- Cross-multiply: 4(4x − 8) = 3(5x − 8) → 16x − 32 = 15x − 24 → x = 8.
- A = 4 × 8 = 32, B = 5 × 8 = 40. Check: 8 years ago, A = 24, B = 32, and 24 : 32 = 3 : 4.
Answer: A = 32 years, B = 40 years
3. 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.
Solution
- Let the daughter’s present age = x and Rohan’s present age = y.
- Five years ago: y − 5 = 3(x − 5) → y = 3x − 10 … (1)
- Ten years from now: y + 10 = 2(x + 10) → y = 2x + 10 … (2)
- Setting (1) = (2): 3x − 10 = 2x + 10 → x = 20. Then y = 2(20) + 10 = 50.
Answer: Daughter = 20 years, Rohan = 50 years
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.
Solution
- Let the daughter’s present age = x, so the mother’s present age = 40 − x.
- Five years ago: daughter = x − 5, mother = 35 − x. Given: 35 − x = 4(x − 5) → 35 − x = 4x − 20 → 55 = 5x → x = 11.
- Daughter = 11, Mother = 40 − 11 = 29. Check: 5 years ago, daughter = 6, mother = 24, and 24 = 4 × 6.
Answer: Daughter = 11 years, Mother = 29 years
Try these yourself
Work through these the same way, then check against the answer.
1. 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.
Answer: Son = 10 years, Mother = 40 years
2. 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.
Answer: P = 30 years, Q = 42 years
3. 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.
Answer: Son = 12 years, Man = 30 years
Where this comes up
Problems on Ages is tested directly in the Quantitative/Numerical Aptitude section of every exam Pariksha Saathi covers — SSC CGL, SSC MTS, SSC CHSL, IBPS PO, IBPS Clerk, SBI PO and SBI Clerk.
Practice more Problems on Ages questions
This page covers the approach and a handful of worked examples. Pariksha Saathi has full topic-wise practice sets for Problems on Ages (and every other Quant/Reasoning/English topic) with instant scoring and explanations — free, no account required.
Practice Problems on Ages free →