Adam, Borrek, and Cenda were comparing how many kilograms of chestnuts they had collected. They found that the arithmetic mean of what Adam and Borrek collected is 10 kg more than Cenda's contribution. And the arithmetic mean of what Adam and Cenda collected is 3 kg less than Borrek's contribution.

Determine the difference between the arithmetic mean of what Borrek and Cenda collected and Adam's contribution.

(M. Petrová)

Hint. Express the relationships from the problem using unknowns.
The amount of chestnuts ( $\mathrm{vg}$ ) collected by Adam, Borek, and Cenda will be denoted by $a, b$, and $c$ respectively. According to the problem statement, the following equations hold:

$$
\frac{a+b}{2}=c+10, \quad \frac{a+c}{2}=b-3 .
$$

We want to determine the difference between $\frac{b+c}{2}$ and $a$. To do this, it is sufficient to add the two equations and simplify:

$$
\begin{gathered}
a+\frac{b+c}{2}=b+c+7, \\
a-7=\frac{b+c}{2} .
\end{gathered}
$$

Adam's contribution is 7 kg greater than the arithmetic mean of what Borek and Cenda collected.

Notes. If we denote the desired difference by $x$ and add the equation $\frac{b+c}{2}=a-x$ to the initial two equations, then the sum of all three equations (and a simple rearrangement) gives $x=7$.

The initial pair of equations is equivalent to the pair

$$
a+b-2 c=20, \quad a-2 b+c=-6 \text {. }
$$

From these, two of the unknowns can be expressed in terms of the third, for example:

$$
b=\frac{26}{3}+c, \quad a=\frac{34}{3}+c
$$

The desired difference then becomes

$$
a-\frac{b+c}{2}=\frac{34}{3}+c-\frac{26}{6}-c=\frac{21}{3}=7 .
$$

Without additional information, it is not possible to determine how much each boy collected (fewer equations than unknowns). However, it is clear from the above process that the problem is feasible: regardless of what Cenda collected, as long as $c$ is non-negative, the contributions of the other boys, $a$ and $b$, are also non-negative according to the expressions (*).