7.2. A car, moving at a constant speed, traveled from point A to point B in 3 hours. To reduce the travel time on the return trip, the driver left point B at a speed 25% higher, and upon reaching the midpoint of the journey between A and B, increased the speed by another 20%. How long will the return trip take?
Answer: 2 hours 12 minutes. Solution. Let the distance from A to B be $a$ (km), and the speed of movement from A to B be $v$ (km/h). Then $\frac{a}{v}=3$. Let $\mathrm{C}$ be the midpoint of the path between A and B. Then the time of movement on the return trip from B to C is $\frac{a / 2}{v \cdot 1.25}=\frac{2}{5} \frac{a}{v}=\frac{6}{5}$ (hour), and the time of movement from C to A is $\frac{a / 2}{v \cdot 1.25 \cdot 1.2}=\frac{1}{3} \frac{a}{v}=1$ (hour). Therefore, the time of the return trip $\frac{6}{5}+1=\frac{11}{5}$ (hour).