A man makes his onward journey at 16 km/hr and backward journey at 28 km/hr. What is his average speed during the entire journey (correct to 2 decimal places)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20.36 km/hr
Average speed is total distance ÷ total time. The onward and backward journeys cover the same distance \(d\), but more time is spent at the slower speed, so the average comes out below the simple average of the two speeds.
Time going \(\displaystyle =\frac{d}{16}\), time coming back \(\displaystyle =\frac{d}{28}\)
Average speed \(\displaystyle =\frac{2d}{\frac{d}{16}+\frac{d}{28}}=\frac{2\times16\times28}{16+28}\)
\(\displaystyle =\frac{896}{44}=20.36\) km/hr (to 2 decimal places)
Tip: For the same distance at two speeds \(x\) and \(y\), the average speed is \(\frac{2xy}{x+y}\). The simple average \(\frac{16+28}{2}=22\) km/hr is wrong, because the time at each speed is not the same.