Age Problems: Method with Worked Examples

Age problems give information about people's ages at different times and ask for the ages now. The age problems below each reduce to one linear equation, under three stated assumptions: ages are whole years, every age in a condition is expressed at the time that condition states, and each person ages one year per year, so age differences stay fixed while ratios and multiples generally change.

The method for age problems

Pick a reference instant, usually now, and set one variable there. Write each age at the time its condition states, translate the condition into one equation, solve it, and check the answer against the original words at each stated time.

x

the unknown present age

x - k

that person's age k years ago

x + k

that person's age k years later

constant difference

if A is d years older than B now, A stays d years older at every time

Worked examples

Example 1: a multiple now, a different multiple later

A father is 3 times as old as his son. In 12 years he will be twice as old as his son. Find their present ages.

1. Let the son be x now; the father is 3x.
2. In 12 years: son x + 12, father 3x + 12, and the condition gives 3x + 12 = 2(x + 12).
3. 3x + 12 = 2x + 24, so x = 12.
4. Father now: 3 x 12 = 36. Check: in 12 years they are 24 and 48, and 48 = 2 x 24.

The son is 12 and the father is 36.

Example 2: a sum now, a ratio in the past

The present ages of A and B add to 50 years. Five years ago their ages were in the ratio 3 : 2. Find their present ages.

1. Five years ago the two ages added to 50 - 10 = 40.
2. Split 40 in the ratio 3 : 2: parts of 40/5 = 8, so the ages then were 24 and 16.
3. Now: 24 + 5 = 29 and 16 + 5 = 21.
4. Check: 29 + 21 = 50, and five years ago 24 : 16 = 3 : 2.

A is 29 and B is 21.

Example 3: an average changes when someone joins

The average age of 4 members of a group is 22 years. A fifth member joins and the average becomes 24 years. Find the new member's age.

1. Total age of the 4 members: 4 x 22 = 88 years.
2. Total age of the 5 members: 5 x 24 = 120 years.
3. New member's age: 120 - 88 = 32 years.
4. Check: (88 + 32)/5 = 120/5 = 24, the stated new average.

The new member is 32 years old.

Example 4: a constant difference with a future sum

A is 8 years older than B. In 10 years their ages will add to 60. Find their present ages.

1. Let B be x now; A is x + 8.
2. In 10 years: (x + 18) + (x + 10) = 60.
3. 2x + 28 = 60, so x = 16, and A is 16 + 8 = 24.
4. Check: in 10 years they are 34 and 26, and 34 + 26 = 60.

B is 16 and A is 24.

Example 5: two conditions at two different times

Six years ago Ravi was 4 times as old as his son. In 8 years he will be twice as old as his son. Find their present ages.

1. Let the son be s and Ravi r now, two unknowns reduced to one by substitution. Six years ago: r - 6 = 4(s - 6), so r = 4s - 18.
2. In 8 years: r + 8 = 2(s + 8). Substituting: 4s - 10 = 2s + 16.
3. 2s = 26, so s = 13, and r = 4 x 13 - 18 = 34.
4. Check: six years ago 28 and 7, and 28 = 4 x 7; in 8 years 42 and 21, and 42 = 2 x 21.

The son is 13 and Ravi is 34.

Summary

Stated time

write every age in a condition at the time that condition states

Ago and later

subtract k for the past and add k for the future, for every person in the condition

Difference versus ratio

differences stay fixed over time; ratios and multiples generally change

Check

substitute the answer into the original words at each stated time, not only into your equation

Frequently asked questions

Why must each age be written at its stated time?

Write every age in a condition at the time that condition states. Most conditions compare ages at one time, so both sides use it; a condition naming two times is valid only when each age is modeled at its own stated time.

Does the age difference stay constant?

Yes. Both people age one year per year, so the gap between their ages never changes. Multiples and ratios generally change as time passes; equal ages, for instance, keep the ratio 1.

What if the answer comes out negative or fractional?

In problems that expect whole-year ages, a negative or fractional answer means the equation or the copied numbers need rechecking. Verify that each condition was used at its stated time.

Can a problem have two unknowns?

Yes. Use one equation per condition, as in the two-times example above, or express both ages from one variable first when a sum or difference is given.

How do I check an age answer?

Put the ages back into the original words at each stated time: the past, present and future conditions must all hold exactly, not only the equation you built from them.

Sources and scope

Method checked against Purplemath, Age Word Problems, OnlineMathLearning, Algebra: Age Problems, and MathBitsNotebook, Linear Word Problems, all retrieved 15 November 2026. This article teaches arithmetic under stated assumptions only and makes no claim about any syllabus, board or class.

Related reading on this site: how to solve maths word problems, the Pair of Linear Equations practice page (age equations are linear), and maths tuition at EZ Tuitions.