Q.6 An automobile travels from city A to city B and returns to city A by the same route. The
speed of the vehicle during the onward and return journeys were constant at 60 km/h and
90 km/h, respectively. What is the average speed in km/h for the entire journey?
(A) 72 (B) 73 (C) 74 (D) 75

The average speed round trip from city A to B at 60 km/h outbound and 90 km/h return confuses many due to the common arithmetic mean trap. True average speed round trip uses total distance over total time, yielding 72 km/h for this classic problem.

Problem Solution

Let the one-way distance be d km. Total distance is 2d km.

Time outbound: d/60 hours
Time return: d/90 hours

Total time: d/60 + d/90 = d(3 + 2)/180 = 5d/180 = d/36 hours

Average speed: 2d ÷ (d/36) = 2 × 36 = 72 km/h

Harmonic Mean Method

For equal distances:

2 × 60 × 90 / (60 + 90) = 10800/150 = 72 km/h

Option Analysis

• (A) 72: Correct, matches all correct methods.
• (B) 73: Too high, arithmetic mean confusion.
• (C) 74: No basis, random midpoint.
• (D) 75: Arithmetic mean error—valid only if times were equal.

Why Harmonic Mean Rules

For equal distances, average speed = 2v₁v₂ / (v₁ + v₂).

Here, 2 × 60 × 90 / 150 = 72 km/h.

The slower outbound leg takes longer (d/60 > d/90), pulling the average below 75 km/h.

Common Mistake Exposed

Adding speeds (60 + 90) ÷ 2 = 75 fails because time durations are different.

At 72 km/h constant, total time equals the variable-speed trip.

MCQ Breakdown