A medium question on meeting for P4–P5, with a step-by-step solution, diagrams and the answer. The basis of journey problems: meeting and catching up appear in the Hua Xia Cup's P4 to P6 qualifying rounds and finals.
The question
The loop path in the park is 300 m a lap.
Ming runs faster than May, and both keep steady speeds.
If they start together from the same point and run in opposite directions, they meet every 50 seconds.
If they start together from the same point and run the same way, Ming catches May every 150 seconds.
How many metres a second does Ming run?
Show the hints (3)
💡Hint 1: Running opposite ways, how far have they run between them when they meet? Running the same way, how much further has Ming run than May when he catches her?
💡Hint 2: Opposite ways: sum of speeds = 300 ÷ 50. Same way: difference of speeds = 300 ÷ 150.
💡Hint 3: With the sum and the difference of the speeds: the faster is half of (sum + difference), the slower half of (sum − difference).
Worked solution
Have a go first, then read the solution step by step.
1.Opposite ways, they cover one lap between them per meeting: sum = 300 ÷ 50 = 6 m/s.
2.Same way, each catch means Ming has gained one lap: difference = 300 ÷ 150 = 2 m/s.
3.Give May the missing 2 and they are equal: Ming = (6 + 2) ÷ 2 = 4 m/s.
⚠️Common mistake: thinking they cover a lap 'between them' when running the same way. That is a chase: what matters is the difference of their speeds.
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