Age Word Problem Solver

Age word problems are the bread-and-butter of school algebra: a couple of sentences in plain English, two unknown ages, and one or two relationships that connect them. The Age Word Problem Solver translates those sentences into a small system of linear equations, solves the system step by step, and animates a timeline of past, present, and future ages so you can see why the answer makes sense. The five built-in patterns — sum-and-difference, multiple-and-difference, now-vs-future, now-vs-past, and three-person ratio — cover the vast majority of textbook puzzles.

How to use this solver

1. Pick the pattern that best matches your puzzle from the dropdown — for example, "X is N years older than Y; sum is S".
2. Type the names of the two (or three) people. The names appear inside the equations and in the timeline so the answer reads naturally.
3. Toggle the relation between "older than" and "younger than" — both work; the solver flips the sign of the difference automatically.
4. Fill in the numbers: age difference, sum, multiple, or years from now or ago, depending on the scenario.
5. Watch the live story preview at the top — if the sentence does not match your puzzle, adjust the inputs.
6. Click Solve. You will see both ages, the equations the solver set up, the algebraic steps, a verification, and an animated timeline showing the ages at every relevant moment.

The five canonical patterns at a glance

The trick that makes age problems easy

Everyone ages at the same rate. So if A is N years older than B today, A is still N years older than B in ten years, in twenty years, or ten years ago. That single invariant is what turns sentences like "in 5 years she will be twice as old as he is" into linear equations rather than a tangle of unknowns:

\[ \text{age difference} \;=\; \text{constant in time} \]

Once you write each person's age as "now" plus or minus the time shift, the equation becomes a single linear relation between two unknowns. With one more piece of information — a sum, a multiple, or a ratio — the system has a unique solution.

Worked example: now vs future

Anna is 8 years older than Ben. In 5 years, Anna will be twice as old as Ben. How old is each now?

1. Let Ben's current age be \( b \). Then Anna's current age is \( b + 8 \).
2. In 5 years, the ages are \( b + 5 \) and \( b + 13 \).
3. The condition "Anna will be twice as old as Ben" gives \( b + 13 = 2(b + 5) \).
4. Expand: \( b + 13 = 2b + 10 \), so \( b = 3 \).
5. Therefore Ben is 3 and Anna is 11.
6. Verify: in 5 years Ben is 8, Anna is 16, and \( 16 = 2 \cdot 8 \). ✓

Worked example: three-person ratio

The ages of Ava, Bea, and Cy are in the ratio 3 : 4 : 5, and the three together are 60 years old.

1. Let one ratio unit be \( x \). Then Ava is \( 3x \), Bea is \( 4x \), Cy is \( 5x \).
2. Their sum: \( 3x + 4x + 5x = 12x = 60 \).
3. Solve: \( x = 5 \). So Ava is 15, Bea is 20, Cy is 25.
4. Verify: \( 15 + 20 + 25 = 60 \). ✓

Common pitfalls and how to avoid them

• Forgetting that the difference is constant — students often write \( A + Y \) but forget that B has also aged Y years. Always shift both ages by the same amount.
• Confusing "K times" with "K times older than" — "twice as old" usually means \( A = 2B \). Some textbooks use "twice older" to mean \( A = 3B \). Pick the convention that matches your textbook and stick to it. The solver uses "K times" = \( A = K \cdot B \).
• K = 1 has no solution — that would mean A = B but you also said A is N years older than B, which contradicts a non-zero difference. The solver flags this case.
• Negative past ages — if a problem says "5 years ago A was 4 times as old as B" and the math gives B = 2 today, then 5 years ago B would be \( -3 \) — impossible. The solver checks this and warns you.
• Mixing "older" and "younger" — the relation toggle handles either direction. If A is younger, just swap names or toggle to "younger than"; the algebra is the same.

Quick translation table

Frequently asked questions