A car's front wheels make 6 more revolutions than the rear wheels over a distance of $120 \mathrm{~m}$; if the circumference of the front wheels were a quarter larger, and that of the rear wheels a fifth larger, the excess in the number of revolutions would be 4 instead of 6. What is the circumference of the wheels?
Let the circumference of the front wheels be $x$, and that of the rear wheels be $y$. The number of revolutions of the front wheels over a 120-meter distance is $\frac{120}{x}$, and that of the rear wheels is $\frac{120}{y}$.

If the circumference of the front wheels is increased by a quarter, i.e., $\frac{5}{4} x$, then the number of revolutions is: $120: \frac{5}{4} x=\frac{4}{5} \cdot \frac{120}{x}$. If the circumference of the rear wheels is increased by a fifth, i.e., $\frac{6}{5} y$, then the number of revolutions is: $\frac{5}{6} \cdot \frac{120}{y}$.

Accordingly, the following two equations can be written:

$$
\begin{gathered}
\frac{120}{x}-\frac{120}{y}=6 \ldots \\
\frac{4}{5} \cdot \frac{120}{x}-\frac{5}{6} \cdot \frac{120}{y}=4 \ldots
\end{gathered}
$$

It is convenient to use the substitution $\frac{120}{x}=u, \frac{120}{y}=v$; thus,

$$
\begin{gathered}
u-v=6 \ldots \\
\frac{4}{5} u-\frac{5}{6} v=4 \ldots
\end{gathered}
$$

In 2a), we eliminate the fractions:

$$
24 u-25 v=120 \ldots
$$

We multiply the terms of 1a) by 24:

$$
24 u-24 v=144 \ldots
$$

By subtracting the corresponding terms of these two equations, we get

$$
\begin{gathered}
v=24 \text{ and thus } u=30 \\
\text{ therefore } \quad x=\frac{120}{u}=\frac{120}{30}=4 \text{ m} \text{ and } y=\frac{120}{v}=\frac{120}{24}=5 \text{ m} .
\end{gathered}
$$

Sándor Baka (Kemény Zsigmond r. V. in Bp. VI.).

Second Solution. From equations 1) and 2), by eliminating the fractions, we get:

$$
\begin{array}{r}
120 y-120 x=6 x y \ldots \\
96 y-100 x=4 x y .
\end{array}
$$

or

$$
\begin{aligned}
& 20 y-20 x=x y \ldots \\
& 24 y-25 x=x y \ldots
\end{aligned}
$$

From the last two equations: $20 y-20 x=24 y-25 x$ which simplifies to $4 y=5 x$. Substituting $y=\frac{5 x}{4}$ into, for example, 1d):

$$
25 x-20 x=\frac{5 x^{2}}{4} \quad \text { or } \quad x^{2}-4 x=x(x-4)=0
$$

From this: $x_{1}=0$ and $x_{2}=4$. According to the context of our problem, only $x=4$ can be considered. If $x=4, y=\frac{5 \cdot 4}{4}=5$.