Two cars start at the same time from city A heading towards city B. One travels at a constant speed of $60 \mathrm{~km} / \mathrm{h}$ and the other at a constant speed of $70 \mathrm{~km} / \mathrm{h}$. If the faster car completes the journey from A to B in 15 minutes less than the other car, what is the distance between these two cities?
We know that speed is the ratio of the distance traveled to the time spent.

Solution 1: Denoting by $t$ the time spent, in hours, by the slower car, which makes the trip at a speed of $60 \mathrm{~km} / \mathrm{h}$, we know that the time spent by the other car is $t-1 / 4$, since 15 minutes is a quarter of an hour. Since both cover the same distance, it follows that $60 \times t=70 \times(t-1 / 4)$, therefore, $t=7 / 4$ hours, or 1 hour and three quarters of an hour. Thus, the distance between the two cities is $60 \times 7 / 4=105 \mathrm{~km}$.

Solution 2: Let's represent by $d$ the distance, in kilometers, between cities A and B and by $T$ the time spent, in hours, by the faster car. Since the other car takes 15 minutes longer to cover the same distance, the time spent by the slower car is $T+0.25$ hours, as $15 \mathrm{~min}=0.25 \mathrm{~h}$. Since the faster car travels at $70 \mathrm{~km} / \mathrm{h}$, we have $70=d / T$ and, since the slower car travels at $60 \mathrm{~km} / \mathrm{h}$, we have $60=d /(T+0.25)$. Thus,

$$
d=70 \times T=60(T+0.25)
$$

that is, $T=1.5 \mathrm{~h}$, and the distance between cities A and B is equal to $d=70 \times 1.5=105 \mathrm{~km}$.