To enter a park, a group with two men, four women, and two children paid 226 reais, while a group with three men, three women, and one child paid 207 reais.

a) How much would a group with 8 men, 10 women, and 4 children pay to enter the park?

b) If the ticket prices are all natural numbers, how many possible prices are there for the tickets?

#
Solution

a) Let $h$ be the ticket price for men, $m$ be the ticket price for women, and $c$ be the ticket price for children. Organizing the information, we have:

$$
\left\{\begin{aligned}
2 h+4 m+2 c & =226 \\
3 h+3 m+c & =207
\end{aligned}\right.
$$

A group with 8 men, 10 women, and 4 children is the same as a group of type (I) plus two groups of type (II), meaning they would pay $226 + 2 \cdot 207 = 640$ reais.

b) By performing 2(II) - (I), we obtain the equation $4 h + 2 m = 188$, from which $m = 94 - 2 h$. Substituting this into (II), we get $c = 3 h - 75$. The solution to the system can be written as $\{(h, m, c)\} = \{h, 94 - 2 h, 3 h - 75\}$. Since $m = 94 - 2 h > 0$, it follows that $h < 47$. Also, since $c = 3 h - 75 > 0$, it follows that $h > 25$. Therefore, $25 < h < 47$, which results in $46 - 25 = 21$ possible solutions.