13.234. There were two copper alloys with different percentages of copper in each. The number expressing the percentage of copper in the first alloy is 40 less than the number expressing the percentage of copper in the second alloy. Then, both these alloys were melted together, after which the copper content was $36 \%$. Determine the percentage of copper in each alloy, if the first alloy contained 6 kg of copper, and the second - 12 kg.
## Solution.

Let in the first alloy 6 kg of copper constitute $x \%$, in the second 12 kg $-y \%$; $m_{1}$ - the mass of the first alloy, $m_{2}$ - the mass of the second. According to the condition

$$
\left\{\begin{array}{l}
0.01 x \cdot m_{1}=6 \\
0.01 y \cdot m_{2}=12 \\
0.36\left(m_{1}+m_{2}\right)=18 \\
x+40=y
\end{array}\right.
$$

from which $x=20 \%, y=60 \%$.

Answer: 20 and $60 \%$.