A train traveling between $A$ and $B$ arrives 20 minutes earlier in $B$ if its speed exceeds the scheduled speed by $5 \mathrm{~km}$ per hour; however, it arrives 25 minutes late if its speed is $5 \mathrm{~km}$ per hour less than the scheduled speed. What is the scheduled speed?
I. Solution: Let the distance $AB$ be denoted by $a$. A train traveling at a constant speed will take as many hours to cover this distance as the number of kilometers it travels per hour is contained in $a$. If we denote the actual speed of the train by $v$ (and convert minutes to sixtieths of an hour), then the conditions of the problem can be written as:

$$
\frac{a}{v-5}-\frac{a}{v}=\frac{25}{60}=\frac{5}{12}, \quad \frac{a}{v}-\frac{a}{v+5}=\frac{20}{60}=\frac{1}{3}
$$

Dividing the equations by $a$ (which cannot be 0), and simplifying,

$$
\frac{5}{v(v-5)}=\frac{5}{12a}, \quad \frac{5}{v(v+5)}=\frac{1}{3a}=\frac{4}{12a}
$$

Multiplying the first equation by 4 and the second by 5, the right sides become equal, and thus the left sides and their reciprocals are also equal:

$$
\frac{1}{20} v(v-5)=\frac{1}{25} v(v+5)
$$

Multiplying the equation by 100, reducing to 0, and factoring out $v$, we get:

$$
v[5(v-5)-4(v+5)]=v(v-45)=0
$$

Since the train was not originally stationary, and thus $v=0$ is not a solution to the problem, the only possible solution is:

$$
v=45 \text{ km} / \text{hour.}
$$

Knowing this, the distance covered can be found from either of the equations in (1):

$$
a=150 \text{ km}
$$

Thus, the obtained values indeed provide a solution to the problem.

The above, somewhat mechanical calculation led to a seemingly quadratic equation, which, however, could be easily solved. If, however, we think through the relationships of the problem before writing down the formulas, we can obtain a simpler equation.

Essentially, this is the solution provided by József Békési (Nagykanizsa, D. Irányi).

II. Solution: Imagine that three trains depart from $A$ to $B$ at the same time on three parallel tracks. One travels at the actual speed $v$, the second at $v-5$, and the third at $v+5$, and this third train continues its journey without stopping at $B$. Since the second train covers less distance per hour than the first by the same amount that the third covers more, and the trains travel at a constant speed, the first train is always equidistant from the other two at any point in time.

When the first train arrives at $B$, the third has already left $B$ 20 minutes, or $\frac{1}{3}$ of an hour, ago, and is thus $\frac{1}{3}(v+5)$ away from $B$. The second train has 25 minutes, or $\frac{25}{60}=\frac{5}{12}$ of an hour, left to travel, so it still has to cover $\frac{5}{12}(v-5)$ km. Since the first train is always equidistant from the other two, we have:

$$
\frac{1}{3}(v+5)=\frac{5}{12}(v-5)
$$

Rearranging this,

$$
v=45 \text{ km} / \text{hour}
$$

is obtained.