A cyclist overtakes a walker — Aptitude Question
A cyclist overtakes a walker: A walker leaves at 6 a.m. at 5 km/h. A cyclist leaves the same point at 8 a.m. Answer and worked solution on CodeKairo.
- Topic: Time, Speed & Distance
- Section: Quantitative Aptitude
- Difficulty: Hard
- Time target: 105 seconds
Question
A walker leaves at 6 a.m. at 5 km/h. A cyclist leaves the same point at 8 a.m. at 15 km/h along the same road. At what time does the cyclist catch up?
Options
- 8:30 a.m.
- 9:00 a.m.
- 9:30 a.m.
- 10:00 a.m.
Show the answer and the worked solution
Correct answer: B. 9:00 a.m.
Worked solution
- In two hours the walker covers 5 × 2 = 10 km.
- Closing speed = 15 − 5 = 10 km/h, so the gap closes in 10 ÷ 10 = 1 hour.
- The cyclist catches up at 9:00 a.m.
Answer: 9:00 a.m.
The idea
Compute the head start in distance, then divide by the relative speed. Add the result to the later departure time, not to the earlier one.
Shortcut. Catch-up time = head start × slower speed ÷ (faster − slower) = 2 h × 5 ÷ 10 = 1 h after 8 a.m., so 9:00 a.m.
Common trap. 8:30 a.m. — adding the speeds (10 km ÷ 20 km/h = 30 min) instead of subtracting them; the cyclist chases in the same direction.