AGE PROBLEM
Age problems, one invariant solves them all.
The key to age problems is a single invariant: the age gap between two people never changes — only the multiples do. This page covers four types (sum-and-difference, difference-multiple, sum-multiple and shifted-multiple); enter the conditions to get both ages now with step-by-step reasoning and verification, plus random practice questions.
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How to use
Pick a type, fill the conditions, press Solve. Sum & difference takes the age sum and gap; difference & multiple takes the multiple and the gap; sum & multiple takes the sum and the multiple; shifted multiple takes the multiple from x years ago, the gap and x. Every step leans on the same invariant: the gap = older − younger, past, present or future. Results include a verification you can trace back to the question.
FAQ
Why does the age gap never change?
Each passing year adds one year to both ages, so the difference is constant. That is the starting point of every age problem: find the invariant gap first, then combine it with the multiple or sum at the moment in question.
Why can’t I just divide the gap by the multiple?
The multiple holds only at that moment. “A was 3 times B x years ago” means the gap equalled 2 units of B’s age then (the multiple minus one). First get B’s age at that time as gap ÷ (multiple − 1), then shift to today. Dividing today’s gap by the multiple is the classic mistake.
The ages came out non-integer — what now?
Check the conditions first: parity mismatch between sum and gap, or a multiple that cannot pair with the gap, are common causes. Textbook problems usually have integer answers, so a decimal means the conditions contradict each other.
Is any data uploaded?
No. All solving and question generation run locally in the browser with no network requests — fine for classroom or home practice.