During a gallery tour, a group of boys and girls met. During the tour, no one disturbed or left. After the tour, 15 girls left, and thus there were twice as many boys as girls left in the gallery. Subsequently, 45 boys left, and five times as many girls as boys remained in the gallery.

How many girls were in the gallery during the tour?

(L. Hozová)
Let $d$ and $ch$ denote the number of girls and boys during the tour. The relationships from the problem can be written as

$$
ch=2(d-15), \quad d-15=5(ch-45),
$$

The number of girls after the tour, expressed from the first equation, is $d-15=\frac{1}{2} ch$. Together with the second equation, we get an equation with the unknown $ch$,

$$
\frac{1}{2} ch=5(ch-45)
$$

which we solve as follows:

$$
\begin{aligned}
ch & =10 ch-450, \\
9 ch & =450, \\
ch & =50 .
\end{aligned}
$$

From this, we get $\frac{1}{2} ch=5(ch-45)=25$, which corresponds to $d-15$. Thus, $d=25+15=40$. During the tour, there were 40 girls in the gallery.

Notes. The system of equations (*) can be solved in various ways. For example, substituting the first equation into the second gives an equation with the unknown $d$, which we solve as follows:

$$
\begin{aligned}
d-15 & =5(2(d-15)-45), \\
d-15 & =10 d-375, \\
9 d & =360, \\
d & =40 .
\end{aligned}
$$

The number of girls after some left was divisible by five. Since 15 girls left, the number of girls during the tour was also divisible by five. Therefore, it is possible to substitute multiples of five greater than 15 for $d$, compute $ch$ from the first equation in (*), and verify whether the second equation holds. The only valid option is $d=40$.

Evaluation. 2 points for formulating the relationships (*); 2 points for solving the system of equations; 2 points for the quality of the commentary. When testing the options, consider the completeness of the commentary; a randomly discovered but unexplained solution should be rated 2 points.