Problem 1. Five boys are weighed in pairs in all possible combinations. The scale showed the following weights: $72 \mathrm{~kg}, 75 \mathrm{~kg}, 76 \mathrm{~kg}, 77 \mathrm{~kg}, 78$ $\mathrm{kg}, 80 \mathrm{~kg}, 81 \mathrm{~kg}, 82 \mathrm{~kg}, 85 \mathrm{~kg}$ and $86 \mathrm{~kg}$. What is the total weight of all five boys?
Solution. First method. Each boy was weighed once with each of the other four boys, which means that in the given ten weighings, the weight of each boy appears four times. If we add up the given weighings, then in the sum, the weights of the five boys will participate four times. According to this, the weight of the five boys is a quarter of the calculated sum. So, if $x$ is the weight of the five boys, then

$$
x=(72+75+76+77+78+80+81+82+85+86): 4=792: 4=198 \text{ kg}
$$

Second method. Let $x, y, z, u, v$ denote the weights of the first, second, third, fourth, and fifth boy, respectively. Then, from the condition of the problem we have

$$
\begin{aligned}
& x+y=72, y+z=75, z+u=76, u+v=77, v+x=78 \\
& x+z=80, y+u=81, z+v=82, u+x=85, v+y=86
\end{aligned}
$$

If we add these 10 equations, we will get $4(x+y+z+u+v)=792$. So, the five boys together weigh $x+y+z+u+v=198 \text{ kg}$.