A hard question on age problems for P3, with a step-by-step solution, diagrams and the answer. Age problems are common in the Hua Xia Cup's P1 to P4 first and qualifying rounds, and rarer in the P5 and P6 papers.
The question
1 year ago, Mum was 4 times as old as Ming.
In 21 years, Mum will be twice as old as Ming.
How old is Ming now?
Show the hints (3)
💡Hint 1: Does the gap between Mum and Ming change over time?
💡Hint 2: 1 year ago the gap was 3 times Ming's age then; in 21 years it will be 1 time Ming's age then.
💡Hint 3: Call Ming's age 1 year ago 1 part; in 21 years it is 3 parts. 1 part is 11.
Worked solution
Have a go first, then read the solution step by step.
1.The gap between them never changes (the two orange pieces are the same length).
2.1 year ago: gap = Ming's age then × 3. In 21 years: gap = Ming's age then × 1.
3.The gap is the same, so if Ming's earlier age is 1 part, the later one is 3 parts: 2 more.
4.The two moments are 1 + 21 = 22 years apart, which is 3 − 1 = 2 parts. 1 part = 22 ÷ 2 = 11.
5.1 year ago Ming was 1 × 11 = 11, so now 11 + 1 = 12.
6.Check: Mum is 45. 44 = 11 × 4 ✓; 66 = 33 × 2 ✓
⚠️Common mistake: stopping at the age 1 year ago (11) and forgetting to add the 1 back.
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