1. **State the problem:** A motorist takes 2 hours to travel from one town to another and 1 hour 40 minutes to travel back. We need to calculate the percentage change in the speed of the motorist. 2. **Define variables:** Let the distance between the two towns be $d$ (same for both trips). 3. **Calculate speeds:** - Speed going to the town: $v_1 = \frac{d}{2}$ (distance over time) - Speed returning: Convert 1 hour 40 minutes to hours: $1 + \frac{40}{60} = \frac{5}{3}$ hours - Speed returning: $v_2 = \frac{d}{\frac{5}{3}} = \frac{3d}{5}$ 4. **Calculate percentage change in speed:** The formula for percentage change is: $$\text{Percentage change} = \frac{v_2 - v_1}{v_1} \times 100$$ Substitute values: $$= \frac{\frac{3d}{5} - \frac{d}{2}}{\frac{d}{2}} \times 100$$ Simplify numerator: $$\frac{3d}{5} - \frac{d}{2} = \frac{6d - 5d}{10} = \frac{d}{10}$$ Divide by $\frac{d}{2}$: $$\frac{\frac{d}{10}}{\frac{d}{2}} = \frac{d}{10} \times \frac{2}{d} = \frac{2}{10} = 0.2$$ Multiply by 100: $$0.2 \times 100 = 20\%$$ 5. **Interpretation:** The speed increased by 20% on the return trip compared to the trip to the town.