Solve problems about ages and their relationships over time.
Free, timed SHSAT practice sets on age problems— every question includes a full worked explanation, and there's no account required.
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Two ideas solve almost every age problem.
1. Everyone gets older by the same number of years. In \(n\) years, a 40-year-old is \(40 + n\) and a 15-year-old is \(15 + n\). For "in how many years will the first be twice as old as the second?" write: \(40 + n = 2(15 + n)\), so \(40 + n = 30 + 2n\), and \(n = 10\). Check: in 10 years they are 50 and 25.
2. Sum and difference. If two ages add to \(S\) and differ by \(D\), the older person is \(\dfrac{S + D}{2}\) and the younger is \(\dfrac{S - D}{2}\). Ages that add to 40 and differ by 6 are 23 and 17.
Traps:
Set up \(39 + n = 2(14 + n)\). Solving: \(n = \dfrac{39 - 2 \times 14}{2 - 1} = 11\) years. (Reporting Zoe's future age instead of the number of years gives 50 years. Reporting Liam's future age instead of the number of years gives 25 years. Reporting the sum of their ages at that time gives 75 years.)
\(\text{Nina} = \dfrac{21 + 3}{2} = 12\). (Using \(\dfrac{21 - 3}{2}\), which is Zoe's age, gives 9 years old. Forgetting to divide \(21 + 3\) by 2 gives 24 years old. Forgetting to divide \(21 - 3\) by 2 gives 18 years old.)