Buses leave a bus terminal every 10 minutes, each travelling at a constant speed of 20 km/h along the same route. A man walks towards the terminal from the opposite direction and meets the buses at intervals of 8 minutes. What is the walking speed of the man (in km/h)?

Introduction / Context:
This problem tests the concept of relative speed and the idea of vehicles leaving a terminal at regular time intervals. The man is moving towards the terminal while the buses move away from it, and he meets them more frequently than they would pass a fixed point. Understanding the relationship between the spacing of buses and the relative speed between the man and the buses is the key to solving this question.

Given Data / Assumptions:
- Buses leave the terminal every 10 minutes, which is 10/60 = 1/6 hour.
- Speed of each bus = 20 km/h.
- The man walks towards the terminal from the opposite direction.
- The man meets successive buses every 8 minutes, which is 8/60 = 2/15 hour.
- All buses and the man move at constant speeds along the same straight route, with no stops.

Concept / Approach:
First, we compute the distance between two consecutive buses along the route in the ground frame. This distance depends on the speed of the buses and the time interval between their departures. Next, we consider the relative motion between the man and the buses. The time between successive meetings from the man's point of view equals the distance between buses divided by the sum of the speeds (because they are moving towards each other). We then solve for the man's speed using this relationship.

Step-by-Step Solution:
Step 1: Compute the distance between two consecutive buses.Time between departures = 1/6 hour. Bus speed = 20 km/h.Distance between buses = 20 * (1/6) = 20/6 = 10/3 km.Step 2: Let the man's speed be v km/h.Because they move towards each other, the relative speed = 20 + v km/h.Step 3: The man meets successive buses every 2/15 hour.So, time between meetings = distance between buses / relative speed.Therefore, 2/15 = (10/3) / (20 + v).Step 4: Solve for v: (10/3) / (20 + v) = 2/15.Cross multiply: (10/3) * 15 / 2 = 20 + v.This gives 10 * 5 / 2 = 20 + v, so 25 = 20 + v, hence v = 5 km/h.

Verification / Alternative check:
If the man walks at 5 km/h while the buses move at 20 km/h, their relative speed is 25 km/h. The separation between buses is 10/3 km. The time between meetings is then (10/3) / 25 hours = 10 / 75 hours = 2 / 15 hours, which is exactly 8 minutes. This matches the condition given in the question, confirming that the solution is correct.

Why Other Options Are Wrong:
- 6 km/h, 7.5 km/h, and 8 km/h give different relative speeds and hence different meeting intervals that do not match 8 minutes when calculated using the formula time = distance / relative speed.
- Only 5 km/h gives a meeting interval of exactly 8 minutes.

Common Pitfalls:
A typical mistake is to treat the problem as if the man were standing still and to assume that the time between buses remains 10 minutes. Another error is forgetting that the relative speed is the sum of speeds when objects move towards each other. Make sure to convert all time units to hours or minutes consistently and carefully apply the relationship between distance, speed, and time.

Final Answer:
The walking speed of the man is 5 km/h.