Last year, there were 30 more boys than girls in our scout troop. This year, the number of children in the troop increased by $10 \%$, while the number of boys increased by $5 \%$ and the number of girls increased by $20 \%$.

How many children do we have in the troop this year?

(L. Hozová)
Both the number of children and their current year's increments are expressed by natural numbers. Given that $10 \% = 10 / 100 = 1 / 10$, the original total number of children must have been a multiple of 10. Similarly, the original number of boys was a multiple of 20, since $5 \% = 5 / 100 = 1 / 20$, and the original number of girls was a multiple of 5, since $20 \% = 20 / 100 = 1 / 5$.

Since there were 30 more boys than girls, the smallest possible original numbers were as follows ( $ch$, resp. $d$ denote the original number of boys, resp. girls):

| $ch$ | 40 | 60 | 80 | 100 | 120 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $d$ | 10 | 30 | 50 | 70 | 90 | $\ldots$ |
| $ch+d$ | 50 | 90 | 130 | 170 | 210 | $\ldots$ |

After accounting for this year's increments, we get the following overview:

| $1.05 ch$ | 42 | 63 | 84 | 105 | 126 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $1.2 d$ | 12 | 36 | 60 | 84 | 108 | $\ldots$ |
| $1.05 ch + 1.2 d$ | 54 | 99 | 144 | 189 | 234 | $\ldots$ |
| $1.1 (ch + d)$ | 55 | 99 | 143 | 187 | 231 | $\ldots$ |

The sum of the new numbers of boys and girls equals the original sum increased by $10 \%$ precisely in the second case; with increasing numbers, the difference between these two values only increases. This year, there are 99 children in the scout troop.

Note. The problem can be expressed using the equation

$$
1.05 ch + 1.2 d = 1.1 (ch + d)
$$

where $ch = d + 30$. After substitution and simplification, we get:

$$
\begin{aligned}
2.25 d + 31.5 & = 2.2 d + 33, \\
0.05 d & = 1.5, \\
d & = 30,
\end{aligned}
$$

which corresponds to the second case mentioned above.