A car and a minivan drive from Alphaville to Betatown. The car travels at a constant speed of $40 \mathrm{~km} / \mathrm{h}$ and the minivan travels at a constant speed of $50 \mathrm{~km} / \mathrm{h}$. The minivan passes the car 10 minutes before the car arrives at Betatown. How many minutes pass between the time at which the minivan arrives in Betatown and the time at which the car arrives in Betatown?
The car takes 10 minutes to travel from the point at which the minivan passes it until it arrives in Betatown.

Since the car drives at $40 \mathrm{~km} / \mathrm{h}$ and since 10 minutes equals $\frac{1}{6}$ hour, then the car travels $40 \mathrm{~km} / \mathrm{h} \cdot \frac{1}{6} \mathrm{~h}=\frac{20}{3} \mathrm{~km}$ in these 10 minutes.

Thus, the distance between the point where the vehicles pass and Betatown is $\frac{20}{3} \mathrm{~km}$.

Since the minivan travels at $50 \mathrm{~km} / \mathrm{h}$, it covers this distance in $\frac{203 \mathrm{~km}}{50 \mathrm{~km} / \mathrm{h}}=\frac{2}{15} \mathrm{~h}$.

Now $\frac{2}{15} \mathrm{~h}=\frac{8}{60} \mathrm{~h}$ which equals 8 minutes, and so the minivan arrives in Betatown $10-8=2$ minutes before the car.

ANSWER: 2 minutes