A hard question on surplus and shortage for P4–P5, with a step-by-step solution, diagrams and the answer. Surplus-and-shortage questions appear in the Hua Xia Cup's P1 to P5 papers; harder ones become questions about distance and time.
The question
Ming walks from home to school, always setting off at the same time.
At 65 m a minute Ming arrives 4 minutes after school starts; at 95 m a minute, 8 minutes before.
How far is it from Ming's home to school, in metres?
Show the hints (3)
💡Hint 1: How many minutes apart are the two walking times?
💡Hint 2: In the time the walk takes at 95 m a minute, how much less ground would 65 m a minute cover?
💡Hint 3: At 95 m a minute the walk takes 26 minutes.
Worked solution
Have a go first, then read the solution step by step.
1.The fast walk (95 m a minute) is 8 early and the slow one (65 m a minute) 4 late: 4 + 8 = 12 minutes apart.
2.When the fast walk arrives, the slow one still has 12 minutes to go: 65 × 12 = 780 m behind.
3.Setting off together, the fast walk gains 95 − 65 = 30 m a minute. To get 780 m ahead takes 780 ÷ 30 = 26 minutes, so the fast walk takes 26 minutes.
4.Distance = 95 × 26 = 2470 m.
5.Check: at 65 m a minute it takes 2470 ÷ 65 = 38 minutes, 12 more than 26 ✓
⚠️Common mistake: using only the 4 late or the 8 early. The two times differ by 12 minutes.
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