[Algebra]
Problem: Linh is driving at $60 \mathrm{~km} / \mathrm{h}$ on a long straight highway parallel to a train track. Every 10 minutes, she is passed by a train travelling in the same direction as she is. These trains depart from the station behind her every 3 minutes and all travel at the same constant speed. What is the constant speed of the trains, in $\mathrm{km} / \mathrm{h}$ ?

Solution: Suppose that the trains are travelling at $v \mathrm{~km} / \mathrm{h}$.

Consider two consecutive points in time at which the car is passed by a train.

Since these points are 10 minutes apart, and 10 minutes equals $\frac{1}{6}$ hour, and the car travels at $60 \mathrm{~km} / \mathrm{h}$, then the car travels $(60 \mathrm{~km} / \mathrm{h}) \cdot\left(\frac{1}{6} \mathrm{~h}\right)=10 \mathrm{~km}$.

During these 10 minutes, each train travels $\frac{1}{6} v \mathrm{~km}$, since its speed is $v \mathrm{~km} / \mathrm{h}$.

At the first instance, Train A and the car are next to each other.

At this time, Train B is " 3 minutes" behind Train A.

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Since 3 minutes is $\frac{1}{20}$ hour, then Train B is $\frac{1}{20} v \mathrm{~km}$ behind Train A and the car.

Therefore, the distance from the location of Train B at the first instance to the location where it passes the car is $\left(\frac{1}{20} v+10\right) \mathrm{km}$.

But this distance also equals $\frac{1}{6} v \mathrm{~km}$, since Train B travels for 10 minutes.

Thus, $\frac{1}{6} v=\frac{1}{20} v+10$ or $\frac{10}{60} v-\frac{3}{60} v=10$ and so $\frac{7}{60} v=10$ or $v=\frac{600}{7}$.

Therefore, the trains are travelling at $\frac{600}{7} \mathrm{~km} / \mathrm{h}$.
Suppose that the trains are travelling at $v \mathrm{~km} / \mathrm{h}$.

Consider the following three points in time: the instant when the car and Train A are next to each other, the instant when Train B is at the same location that the car and Train A were at in the previous instant, and the instant when the car and Train B are next to each other.

<img_3611>

From the first instant to the second, Train B "catches up" to where Train A was, so this must take a total of 3 minutes, because the trains leave the station 3 minutes apart.

Since 3 minutes equals $\frac{3}{60}$ hour and the car travels at $60 \mathrm{~km} / \mathrm{h}$, then the car travels $(60 \mathrm{~km} / \mathrm{h}) \cdot\left(\frac{3}{60} \mathrm{~h}\right)=3 \mathrm{~km}$ between these two instants.

From the first instant to the third, 10 minutes passes, since these are consecutive points at which the car is passed by trains. In 10 minutes, the car travels $10 \mathrm{~km}$.

Therefore, between the second and third instants, $10-3=7$ minutes pass. During these 7 minutes, Train B travels $10 \mathrm{~km}$.

Since 7 minutes equals $\frac{7}{60}$ hour, then $v \mathrm{~km} / \mathrm{h}=\frac{10 \mathrm{~km}}{7 / 60 \mathrm{~h}}=\frac{600}{7} \mathrm{~km} / \mathrm{h}$, and so the trains are travelling at $\frac{600}{7} \mathrm{~km} / \mathrm{h}$.

Final answer: $\frac{600}{7}$ km/h