We have three types of alloys - made of gold, silver, and copper - with the ratios of these metals being $1: 3: 5$, $3: 5: 1$, and $5: 1: 3$. Create an alloy from these alloys weighing $351 \mathrm{~g}$, in which the ratio of gold, silver, and copper is $7: 9: 11$.
A $351 \mathrm{~g}$ alloy contains metals in the ratio $7: 9: 11$, so by proportional division, the alloy to be produced should contain $91 \mathrm{~g}$ of gold, $117 \mathrm{~g}$ of silver, and $143 \mathrm{~g}$ of copper.

Suppose that from the three available alloys, we use $x, y, z$ grams respectively to produce the new alloy, then according to the problem,

$$
\begin{aligned}
& \frac{x}{9}+\frac{3 y}{9}+\frac{5 z}{9}=91 \\
& \frac{3 x}{9}+\frac{5 y}{9}+\frac{z}{9}=117 \\
& \frac{5 x}{9}+\frac{y}{9}+\frac{3 z}{9}=143
\end{aligned}
$$

which simplifies to

$$
\begin{aligned}
x+3 y+5 z & =819 \\
3 x+5 y+z & =1053 \\
5 x+y+3 z & =1287
\end{aligned}
$$

Subtract 3 times (1) from (2), and 5 times (1) from (3). Dividing the two resulting equations by 2, we get

$$
\begin{aligned}
2 y+7 z & =702 \\
7 y+11 z & =1404
\end{aligned}
$$

Subtracting 2 times (5) from 7 times (4), we get

$$
27 z=2106, \quad \text { from which } \quad z=78 \text {. }
$$

Substituting this value of $z$ into (4),

$$
y=\frac{702-7 \cdot 78}{2}=351-7 \cdot 39=351-273=78
$$

and thus from (1),

$$
x=819-3 \cdot 78-5 \cdot 78=819-8 \cdot 78=819-624=195
$$

Eva Cinkotai (Bp. XIII. Court of Conciliation I. Dept.)