Age problems are among the oldest puzzles in algebra textbooks — some appear in collections of recreational math from the Middle Ages. They’re good practice because they force you to translate careful English (“in six years, the father will be twice as old as his son”) into equations and to remember that time moves for everyone at once. This generator makes a new problem each time you click and walks through the solution step by step.
How to use the age word problem generator
- Choose a problem type or leave it on Surprise me.
- Choose a difficulty: easy problems use small numbers and simple relationships; hard problems use larger multiples and “years ago” conditions.
- Click New problem, solve it on paper, then compare with the worked solution under the answer.
All generated ages are whole numbers, and parent–child gaps stay realistic.
A strategy that works every time
- Choose one unknown. Usually the younger person’s current age: let c = child’s age now.
- Write every other current age in terms of it: “four times as old” → 4c; “28 years older” → c + 28.
- Shift both ages for the past or future: in 5 years → add 5 to each; 3 years ago → subtract 3 from each.
- Translate the condition into an equation and solve.
- Answer the question asked (often both ages) and check the condition with real numbers.
A table keeps the setup straight:
| Now | In y years | y years ago | |
|---|---|---|---|
| Child | c | c + y | c − y |
| Parent | 3c | 3c + y | 3c − y |
Worked example
Problem: A father is three times as old as his son. In 12 years he will be twice as old as his son. How old are they now?
Let c = the son's age now; the father is 3c.
In 12 years: son c + 12, father 3c + 12
Equation: 3c + 12 = 2(c + 12)
Solve: 3c + 12 = 2c + 24 → c = 12
Answer: the son is 12 and the father is 36. In 12 years: 24 and 48, and 48 = 2 × 24 ✓
Problem types in this generator
- Sum and difference: “The sum of their ages is 73 and one is 9 years older.” Set the younger age as y and solve y + (y + 9) = 73.
- Ratio and sum: “Their ages are in the ratio 3 : 4 and add to 63.” Use 3k and 4k. The ratio calculator can split any total in a ratio.
- Future multiple: “Now 7 times as old; in 14 years, 3 times as old.”
- Past multiple: “Now 4 times as old; 12 years ago, 10 times as old.”
- Age gap and future multiple: “22 years older; in 4 years, 3 times as old.” Here the constant gap is the key fact.
Why the multiple shrinks over time
If a mother is 7 times her daughter’s age, the ratio can’t stay that way: their difference is fixed, so as both grow, the ratio drifts toward 1. A 49-year-old and a 7-year-old (7×) become 63 and 21 (3×) fourteen years later. Going backward, the ratio grows — which is why “years ago” problems use larger multiples. If your solution gives a younger person who is older than the parent, or a negative age, recheck the equation.
For a different flavor of algebra practice, try the coin word problems. To check a two-equation setup, use the system of linear equations calculator.
Frequently asked questions
How do you solve age word problems?
Let a variable stand for one person's current age and write the other person's age in terms of it. Then translate the condition about the past or future into an equation, remembering to add or subtract the same number of years from both ages, and solve.
What is the most common mistake in age problems?
Changing only one person's age. In five years, both people are five years older, so if the child is c now, the parent's future age is (parent's age now) + 5, not just the child's + 5.
How do I handle "years ago" problems?
Subtract the number of years from both current ages. If a mother is 4c now and the problem says 12 years ago, use 4c − 12 for her and c − 12 for the child.
Why does the age difference never change?
Both people get older at the same rate, so the gap between their ages stays fixed forever. A mother who is 28 years older than her son will always be 28 years older. That fact is often the quickest route to an answer.
Can age problems be solved with two variables?
Yes. Write one equation for the present relationship and one for the past or future relationship, then solve the system. Substituting the first into the second gives the one-variable equations shown in the solutions.