A person walks from his house to his office at a speed of \(x_1\) km/hr and returns by the same route at a speed of \(x_2\) km/hr. What is his average speed?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2x_1x_2}{x_1+x_2}\)
Average speed is total distance ÷ total time, not the plain average of the two speeds, because the slower trip takes longer.
Assuming the distance from house to office to be d km:
Time going \(\displaystyle =\frac{d}{x_1}\), time returning \(\displaystyle =\frac{d}{x_2}\)
Total time for both trips \(\displaystyle =\frac{d}{x_1}+\frac{d}{x_2}=\frac{d(x_1+x_2)}{x_1x_2}\)
Total distance there and back \(=2d\)
Average speed \(\displaystyle =2d\div\frac{d(x_1+x_2)}{x_1x_2}=\frac{2x_1x_2}{x_1+x_2}\) km/hr
Tip: For equal distances at two speeds, the average speed is always \(\displaystyle\frac{2x_1x_2}{x_1+x_2}\) (the harmonic mean); the distance d cancels, so it is never needed.