Four-leg round trip — average over mixed speeds: A driver travels from City 1 to City 2 at 60 km/h and returns at 40 km/h. He then repeats the journey: going again at 120 km/h (twice the original onward speed) and returning at 20 km/h (half the original return speed). What is his overall average speed for the entire four-leg journey?

Difficulty: Medium

Correct Answer: 40 km/h

Explanation:


Introduction / Context:
Average speed over multiple legs depends on total distance and total time, not the average of speeds. With equal leg distances D each way, the computation simplifies.


Given Data / Assumptions:

  • Leg speeds: 60, 40, 120, 20 km/h over equal distances D.
  • Total distance = 4D.


Concept / Approach:
Compute total time as the sum of per-leg times, then divide total distance by total time.


Step-by-Step Solution:

Time = D/60 + D/40 + D/120 + D/20= D*(1/60 + 1/40 + 1/120 + 1/20)= D*(2/120 + 3/120 + 1/120 + 6/120) = D*(12/120) = D/10Total distance = 4D ⇒ V_avg = 4D / (D/10) = 40 km/h


Verification / Alternative check:
Pick D = 120 km; compute exact times; the ratio remains 40 km/h.


Why Other Options Are Wrong:
50–55 km/h arise from naïve averaging of speeds rather than distance/time aggregation.


Common Pitfalls:
Using arithmetic mean of speeds for round trips or mixed legs with unequal times.


Final Answer:
40 km/h

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