6 (1247). Three boxes are filled with nuts. In the second box, there are $10 \%$ more nuts than in the first, and $30 \%$ more than in the third. How many nuts are in each box if there are 80 more nuts in the first box than in the third?
Solution. Let there be $x$ nuts in the first box and $y$ in the third. Then in the second box, there were $x+0.1 x=1.1 x$ or $y+0.3 y=1.3 y$. Considering that there were 80 more nuts in the first box than in the third, we form the system of equations:

$$
\left\{\begin{array}{l}
1.1 x=1.3 y \\
x-y=80
\end{array}\right.
$$

from which $y=440, x=520, 1.1 x=572$.

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Remark. This problem can be solved without forming a system of equations. Let there be $x$ nuts in the first box, then in the third $-x-80$, in the second box $-1.1 x$, or $1.3(x-80)$. We have the equation: $1.1 x=1.3(x-80)$, from which $x=520$.

Answer: there were 520 nuts in the first box, 572 in the second, and 440 in the third.