## Task A-1.1.

To purchase a school bus that will transport children from four locations $A, B, C, D$, a total of $1050000 \mathrm{kn}$ is required. The locations will bear a portion of the costs proportional to their population. The number of residents in location $D$ is equal to the sum of the residents in locations $A$ and $C$, the number of residents in location $A$ is $25 \%$ less than in $B$, and $20 \%$ more than in $C$. Determine how much each location will pay.
## Solution.

Let's denote the number of inhabitants in places $A, B, C, D$ as $a, b, c, d$ respectively.

According to the conditions of the problem, we have:

$$
d=a+c, \quad a=\frac{3}{4} b, \quad a=\frac{6}{5} c
$$

so we can express these numbers in terms of $a$:

$$
b=\frac{4}{3} a, \quad c=\frac{5}{6} a, \quad d=a+\frac{5}{6} a=\frac{11}{6} a
$$

Thus,

$$
a: b: c: d=1: \frac{4}{3}: \frac{5}{6}: \frac{11}{6}=6: 8: 5: 11
$$

The amount of $1050000 \mathrm{kn}$ needs to be divided according to the obtained ratio. Let's denote the amounts to be paid to each place as $A, B, C, D$.

By substituting $A=6 k, B=8 k, C=5 k, D=11 k$

into the expression $A+B+C+D=1050000$, we get:

$$
\begin{gathered}
6 k+8 k+5 k+11 k=1050000 \\
k=35000
\end{gathered}
$$

The cost amounts for each place are:

$$
A=210000 \mathrm{kn}, \quad B=280000 \mathrm{kn}, \quad C=175000 \mathrm{kn}, \quad D=385000 \mathrm{kn}
$$