Two cars start at the same time from city A heading towards city B. One of them 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 makes this trip in 15 minutes less than the other car, what is the distance between the two cities?
Solution 1: We know that distance $=$ speed $\times$ time. Let $t$ be the time spent by the slower car (the one that travels at a speed of $60 \mathrm{~km} / \mathrm{h}$). Therefore, the time spent by the other car was $t-15$. Since both cover the same distance, converting hours to minutes, we have:

$$
\frac{60}{60} \times t=\frac{70}{60} \times(t-15) \Rightarrow t=105 \min =1 \frac{3}{4} \mathrm{~h}
$$

Thus, the distance between the two cities is:

$$
60 \times 1 \frac{3}{4}=60 \times \frac{7}{4}=105 \mathrm{~km}
$$

Solution 2: Let $d$ represent the distance between cities A and B, and $T$ 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 it is $T+0.25$ hours, as $15 \mathrm{~min}=0.25 \mathrm{~h}$.

Since speed is the ratio of the distance traveled to the time spent, we conclude that $70=\frac{d}{T}$ and $60=\frac{d}{T+0.25}$. Therefore, $d=70 T=60(T+0.25)$, which means $T=1.5 \mathrm{~h}$. Thus, $d=70 \times 1.5=105 \mathrm{~km}$.