A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 {\mathrm{kmh}}^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 {\mathrm{kmh}}^{-1}$. The average speed of the man over the interval of time 0 to 50 min is equal to

Man walks from his home to market with a speed of $5 {\mathrm{kmh}}^{-1}$. Distance = 2.5 km and time

                              $=\frac{d}{v}=\frac{2.5}{5}=\frac{1}{2}=30 \min$

and he returns back with speed of $7.5 {\mathrm{kmh}}^{-1}$ in rest a time of 20 min.

Distance $=7.5\times \frac{20}{60}=2.5 \mathrm{km}$

So, average speed $=\frac{\text{ total distance }}{\text{ total time }}=\frac{(2.5+2.5)}{(50/60)\mathrm{h}}=\frac{30}{5}\mathrm{km}/\mathrm{h}$