1. One car covers a distance of 120 km 18 minutes faster than the other. If the first car reduced its speed by 12 km/h, and the second car increased its speed by $10 \%$, they would spend the same amount of time on the same distance. Find the speeds of the cars.
Solution: Let $x$ be the speed of the first car, $y$ be the speed of the second car $\frac{120}{x}+\frac{18}{60}=\frac{120}{y}+5, \quad x-12=1.1 y, \quad \frac{40}{x}+\frac{1}{10}=\frac{44}{x-12}, \quad x^{2}-52 x-4800=0, \quad x=100, \quad y=80$ Answer: $100 \mathrm{km} / \mathrm{h}, 80 \mathrm{km} / \mathrm{h}$