A man makes four trips of equal distances. His speed on the first trip is 60 km/h, and on each subsequent trip his speed is half of the previous trip. What is his average speed (in km/h) over all four trips?

Introduction / Context:
This question deals with average speed over several trips of equal distance but different speeds. The man makes four trips, and on each subsequent trip his speed is half of the previous one. We must determine the overall average speed across all four trips. Because the distances are equal for each leg, average speed is not the simple mean of the speeds; instead, it is the total distance divided by the total time across all legs.

Given Data / Assumptions:
- Distance of each trip = d km (same for all four trips).
- Speed on trip 1 = 60 km/h.
- Speed on trip 2 = 30 km/h (half of 60).
- Speed on trip 3 = 15 km/h (half of 30).
- Speed on trip 4 = 7.5 km/h (half of 15).
- No extra waiting time between trips; only travel times are considered.

Concept / Approach:
Let the distance of each trip be d. The total distance travelled is 4d. The time taken for each trip is distance / speed, so the total time is the sum of all four individual times. The overall average speed is then computed as total distance / total time. Since d will appear in every term, it will cancel out, and we can find the average speed purely from speed values. This is a typical harmonic-mean style situation for equal distances.

Step-by-Step Solution:
Step 1: Express time for each trip.Trip 1 time = d / 60 hours.Trip 2 time = d / 30 hours.Trip 3 time = d / 15 hours.Trip 4 time = d / 7.5 hours.Step 2: Simplify the fourth time term.7.5 km/h = 15/2 km/h, so d / 7.5 = d / (15/2) = (2d) / 15.Step 3: Sum all times.Total time T = d / 60 + d / 30 + d / 15 + 2d / 15.Convert to a common denominator of 60:d / 60 = d/60.d / 30 = 2d / 60.d / 15 = 4d / 60.2d / 15 = 8d / 60.So T = (1 + 2 + 4 + 8)d / 60 = 15d / 60 = d / 4 hours.Step 4: Total distance = 4d.Average speed V_avg = total distance / total time = 4d / (d / 4) = 4d * 4 / d = 16 km/h.

Verification / Alternative check:
If we assume a specific distance, say d = 60 km for simplicity, then trip distances are 60 km each, total distance = 240 km. Times: 60/60 = 1 hour; 60/30 = 2 hours; 60/15 = 4 hours; 60/7.5 = 8 hours. Total time = 1 + 2 + 4 + 8 = 15 hours. Average speed = 240 / 15 = 16 km/h, confirming our result.

Why Other Options Are Wrong:
- 30 km/h is close to the first two speeds but ignores the very slow final trips which heavily increase total time and reduce average speed.
- 28.125 km/h and 27.5 km/h are too high because they do not properly account for how the extremely slow speeds on later trips dominate the total time when distances are equal.

Common Pitfalls:
Many students incorrectly compute the arithmetic mean of the speeds: (60 + 30 + 15 + 7.5) / 4 = 28.125 km/h, which is not the correct overall average speed. This ignores the fact that equal distances at very different speeds do not contribute equally to the total time. Always use total distance divided by total time for problems involving varying speeds over equal distances.

Final Answer:
The man’s average speed over the four trips is 16 km/h.