13.194. The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $1 / 4$ of its length, and the circumference of the rear wheel is increased by $1 / 5$ of its length, then over the same distance, the front wheel will make 4 more revolutions than the rear wheel. Find the lengths of the circumferences of the front and rear wheels.
Solution.

The circumference of the wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $\mathrm{Cn}=s$. We will fill in the table of values for these quantities in the order indicated by the numbers (1), (2), ..., (12):

| Wheel | Before change |  |  | After change |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | $C$ | $s$ | $n$ | $C$ | $s$ | $n$ |
| Front | (1) $x$ | (3) 120 | (5) $\frac{120}{x}$ | (7) $\frac{5 x}{4}$ | (9) 120 | (11) $\frac{120 \cdot 4}{5 x}$ |
| Rear | (2) $y$ | (4) 120 | (6) $\frac{120}{y}$ | (8) $\frac{6 y}{5}$ | (10) 120 | (12) $\frac{120 \cdot 5}{6 y}$ |

According to the problem, we have the system $\left\{\begin{array}{l}\frac{120}{x}-\frac{120}{y}=6, \\ \frac{96}{x}-\frac{100}{y}=4,\end{array}\right.$ from which $x=4$ (m), $y=5$ (m).

Answer: 4 and $5 \mathrm{m}$.