In a parking lot, there are motorcycles, cars, buses, and trucks, totaling 80 vehicles and 540 wheels. Each motorcycle has 2 wheels, each car has 4, each bus has 6, and each truck has 8. The number of cars is the sum of the number of motorcycles and the number of buses. How many trucks are there in this parking lot, if this number is less than 50?

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Solution

Let the quantities of motorcycles, cars, buses, and trucks be $m, k, o$, and $c$ respectively. Organizing the information, we have:

$$
\left\{\begin{array}{cccc}
m+k+o+c & = & 80 & (I) \\
2 m+4 k+6 o+8 c & = & 540 & (II) \\
k & = & m+o & (III)
\end{array}\right.
$$

Substituting (III) into (II) and (I), we arrive at:

$$
\left\{\begin{array}{ccc}
2 m+2 o+c & =80 & (IV) \\
3 m+5 o+4 c & =270 & (V)
\end{array}\right.
$$

Substituting (IV) into (V), we get:

$$
\begin{aligned}
3 m+5 o+4(80-2 m-2 o) & =270 \\
3 m+5 o+320-8 m-8 o & =270 \\
5 m+3 o & =50 \\
o & =\frac{50-5 m}{3}
\end{aligned}
$$

Returning to (IV), we have:

$$
\begin{aligned}
2 m+2\left(\frac{50-5 m}{3}\right)+c & =80 \\
c & =80-2 m+\frac{10 m-100}{3} \\
c & =\frac{4 m+140}{3}
\end{aligned}
$$

Substituting the results found into (I), we arrive at:

$$
\begin{aligned}
m+k+\frac{50-5 m}{3}+\frac{4 m+140}{3} & =80 \\
k & =\frac{50-2 m}{3}
\end{aligned}
$$

Thus, we arrive at the solution of the system

$$
\{(m, k, o, c)\}=\left\{\left(m, \frac{50-2 m}{3}, \frac{50-5 m}{3}, \frac{4 m+140}{3}\right)\right\}
$$

From $o=\frac{50-5 m}{3} \in \mathbb{N}$, we conclude that $m \in\{1,4,7,10\}$. Let's analyze each of the cases:

| case | $m$ | $k$ | $o$ | $c$ |
| :---: | :---: | :---: | :---: | :---: |
| $1^{\circ}$ | 1 | 16 | 15 | 48 |
| $2^{\circ}$ | 4 | 14 | 10 | 52 |


| case | $m$ | $k$ | $o$ | $c$ |
| :---: | :---: | :---: | :---: | :---: |
| $3^{\circ}$ | 7 | 12 | 5 | 56 |
| $4^{\circ}$ | 10 | 10 | 0 | 60 |

Since the number of trucks must be less than 50, there are 48 trucks in the parking lot.