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
May is meeting a friend at the MTR station.
She walks from home, setting off at a fixed time.
At 60 m a minute she would arrive 4 minutes late; at 80 m a minute, 7 minutes early.
To arrive exactly on time, how many metres a minute must May walk?
Show the hints (3)
💡Hint 1: How many minutes apart are the two walking times? Find the distance first.
💡Hint 2: With the distance, use either walk to find how many minutes there are from setting off to the meeting time.
💡Hint 3: The distance is 2640 m, and there are 40 minutes from setting off to the meeting time.
Worked solution
Have a go first, then read the solution step by step.
1.The fast walk (80 m a minute) is 7 early and the slow one (60 m a minute) 4 late: 4 + 7 = 11 minutes apart.
2.When the fast walk arrives, the slow one is 11 minutes short: 60 × 11 = 660 m.
3.The gap grows 80 − 60 = 20 m a minute, so the fast walk takes 660 ÷ 20 = 33 minutes.
4.Distance = 80 × 33 = 2640 m.
5.The fast walk is 7 early, so there are 33 + 7 = 40 minutes to the meeting time.
6.On time: 2640 ÷ 40 = 66 m a minute.
7.Check: at 60 m a minute it takes 44 minutes, 4 late ✓; at 66 it takes exactly 40 ✓
⚠️Common mistake: averaging the two speeds. Find the distance and the time first, then divide.
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