A person goes from point N to point P and comes back. His average speed for the whole journey is 60 km/h. If his speed while going from N to P is 40 km/h, what is his speed (in km/h) while coming back from P to N?

Introduction / Context:
This problem is a classic example of average speed over a round trip when the speeds in the two directions are different. Many students incorrectly assume that the average speed is the simple mean of the two speeds, but that is only true when the times are equal, not when the distances are equal. Here, the person travels the same distance from N to P and from P back to N, with different speeds in each direction, and we are told the average for the entire journey. We must determine the return speed.

Given Data / Assumptions:
- One-way distance between N and P = d km (same both ways).
- Speed from N to P (going) = 40 km/h.
- Speed from P to N (return) = v km/h (unknown).
- Average speed for the entire to-and-fro trip = 60 km/h.
- Motion is along a straight path with constant speeds and no additional stops.

Concept / Approach:
Average speed for a round trip with equal distances is defined as total distance divided by total time. The total distance is 2d. The total time is the sum of the time taken for the outward and return journeys: d / 40 + d / v. We are told that the average speed is 60 km/h, so 60 must equal 2d divided by this total time. This provides an equation that we can solve for v. The distance d cancels out, so the actual numerical value of d is not needed.

Step-by-Step Solution:
Step 1: Express total distance and total time.Total distance = 2d km.Time going = d / 40 hours; time returning = d / v hours.Total time = d / 40 + d / v.Step 2: Use the formula for average speed.Average speed = total distance / total time = 60 km/h.So 60 = 2d / (d / 40 + d / v).Step 3: Simplify by cancelling d (since d > 0):60 = 2 / (1 / 40 + 1 / v).Step 4: Invert both sides after rearranging:1 / 60 = (1 / 2) * (1 / 40 + 1 / v).So 1 / 40 + 1 / v = 2 / 60 = 1 / 30.Step 5: Solve for 1 / v:1 / v = 1 / 30 - 1 / 40 = (4 / 120 - 3 / 120) = 1 / 120.Therefore v = 120 km/h.

Verification / Alternative check:
Assume for convenience that d = 120 km. Time going at 40 km/h = 120 / 40 = 3 hours. Time returning at 120 km/h = 120 / 120 = 1 hour. Total distance = 240 km, total time = 4 hours. Average speed = 240 / 4 = 60 km/h, which matches the given average. This confirms that 120 km/h is correct for the return speed.

Why Other Options Are Wrong:
- 80 km/h or 100 km/h, when used as return speeds, give total times that produce average speeds different from 60 km/h.
- 140 km/h is larger than necessary and would yield an average speed greater than 60 km/h for any fixed distance d.

Common Pitfalls:
A very common error is to compute the average speed as (40 + v) / 2 = 60, which leads to v = 80 km/h. This misapplies the arithmetic mean, which is not appropriate when distances, not times, are fixed. Always return to the fundamental definition of average speed as total distance divided by total time, especially for round-trip problems.

Final Answer:
The person’s speed while coming back from P to N is 120 km/h.