1. In the "Clumsy Hands" club, unbalanced lever scales with arms of different lengths and pans of different weights were made in a slapdash manner. As a result of four weighings on these scales, the following "equilibria" were obtained:

$$
\begin{array}{ll}
{[\text { left } 3 \text { kg }=\text { right melon ]; }} & \text { [left melon }=\text { right } 5.5 \text { kg]; } \\
\text { [left } 5 \text { kg = right watermelon]; } & \text { [left watermelon }=\text { right } 10 \text { kg]. }
\end{array}
$$

What are the true weights (masses) of the melon and the watermelon? (12 points)
Solution. The mass ratio [left $x=$ right $y$] is linear: $y=k x+b$, where $k>0$ and $b$ are constants. If [left $x=$ right $y$] and [left $y=$ right $z$], then $z=K x+B$, where $K=k^{2}, B=(k+1) b$. We have

$$
\left\{\begin{array} { l } 
{ K \cdot 3 + B = 5 . 5 } \\
{ K \cdot 5 + B = 1 0 }
\end{array} \Longrightarrow \left\{\begin{array} { c } 
{ K = 9 / 4 } \\
{ B = - 5 / 4 }
\end{array} \Longrightarrow \left\{\begin{array}{c}
k=3 / 2 \\
b=-1 / 2
\end{array}\right.\right.\right.
$$

from which we find the masses: melon $=k \cdot 3+b$, watermelon $=k \cdot 5+b$.

Answer: melon $=4$ kg, watermelon $=7$ kg.