Problems Based on Ages
Competitive Mathematics · Arithmetic
You are 20 and your uncle is 35. In 10 years you will be 30 and he 45 — but the gap of 15 years stays the same. That is the key to every age problem: the difference between two people’s ages never changes while their ratio changes every year. So most problems reduce to one linear equation; you only have to keep “today”, “ years ago” and “ years hence” straight.
Age ratios
Finding ages from the present ratio and the ratio years ago or hence.
The method
Take today’s ages as and and get from the condition.
- Present ratio ⟹ ages .
- “ years hence” ⟹ add to both; “ years ago” ⟹ subtract from both.
- Equate the new ratio to the stated one, cross-multiply and find .
- Check: is the difference between the two ages unchanged?
- 1Ages : ⟹ .
- 2 ⟹ ; ages 15 and 21 years.
- 1After 5 years the sum is ; split 3 : 4 gives and .
- 2Today both are 5 years less: and . The younger is .
- 1After 10 years the ages are and ⟹ ratio .
- 2The gap is — just as before.
Multiple-type questions
Today one is times the other; years hence / ago times.
One unknown, one equation
Call the younger and the elder ; shift the time and form the equation.
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Average age and timelines
Family or class average ages, a three-person ratio, the middle person and event timelines.
Averages, ratios and timelines
Solve average-age questions through the total age; in timelines add the durations.
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Traps and fast tricks
Where marks go in age questions.
The commonest mistakes
The slips that cost marks.
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Fast tricks
Habits for speed.
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The practice ladder
New questions of every type, in three levels.
Level 1 — Basic
One-rule questions.
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Level 2 — Standard
Exam-level questions.
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Level 3 — Advanced
Questions at the hard end of competitive exams.
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Revision — formula cards
Read the front, think, then flip.
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