Introduction / Context:
This problem involves average speed over a round trip with different speeds in the outbound and return journeys. The man cycles to a place at 16 km/h and returns at some lower speed. The overall average speed for the entire journey is 6.4 km/h. We must find the return speed. This is a classic harmonic mean style question in the time, speed and distance chapter, but here we solve directly using total distance and total time.
Given Data / Assumptions:
- Outbound speed = 16 km/h.
- Return speed = v km/h (unknown, but lower than 16 km/h).
- Average speed over the whole journey = 6.4 km/h.
- Distances outbound and return are equal.
Concept / Approach:
Let the one-way distance be D km. Then total distance = 2D km. Times for each leg are:
time_out = D / 16 hours.
time_back = D / v hours.
Average speed is defined as:
average speed = total distance / total time = 2D / (D / 16 + D / v).
We know the average speed is 6.4 km/h, so:
6.4 = 2D / (D / 16 + D / v).
Step-by-Step Solution:
Step 1: Simplify the expression for average speed.
6.4 = 2D / [D * (1 / 16 + 1 / v)].
Cancel D from numerator and denominator.
6.4 = 2 / (1 / 16 + 1 / v).
Step 2: Rearrange to solve for (1 / 16 + 1 / v).
1 / 16 + 1 / v = 2 / 6.4.
2 / 6.4 = 20 / 64 = 5 / 16.
So 1 / 16 + 1 / v = 5 / 16.
Step 3: Solve for 1 / v.
1 / v = 5 / 16 - 1 / 16 = 4 / 16 = 1 / 4.
Step 4: Invert to find v.
v = 4 km/h.
Therefore, his return speed is 4 km/h.
Verification / Alternative check:
Assume D = 16 km for simplicity. Then outbound time = 16 / 16 = 1 hour. Return speed = 4 km/h, so return time = 16 / 4 = 4 hours. Total distance = 32 km, total time = 5 hours. Average speed = 32 / 5 = 6.4 km/h, which matches the given value. Hence, v = 4 km/h is consistent.
Why Other Options Are Wrong:
6 km/h or 8 km/h would give different total times and hence different average speeds, not 6.4 km/h.
10 km/h would produce a relatively high average speed since both legs would be fairly fast.
5 km/h also fails to make the combined average exactly 6.4 km/h when checked with a sample distance.
Only 4 km/h yields the required 6.4 km/h average speed.
Common Pitfalls:
A common mistake is to average the speeds directly, for example taking (16 + v) / 2 = 6.4, which is incorrect because average speed is distance-based, not speed-based. Another error is to forget that distances are equal for a round trip and to assign different distances for the two legs. Algebra mistakes when handling fractions in 1 / 16 + 1 / v are also frequent. Handling the problem symbolically and checking with an assumed distance helps avoid these issues.
Final Answer:
The man's speed on the return journey is
4 km/h.
Discussion & Comments