## Task B-1.5.

A goldsmith has two alloys of silver and gold. In the first alloy, the ratio of silver to gold is $4: 5$, and in the second, it is $2: 5$. In what ratio should these two alloys be mixed to obtain a new alloy where the ratio of silver to gold is $7: 11$?
## Solution.

Let $x$ be the quantity of the first alloy and $y$ be the quantity of the second alloy. The amount of silver in the new alloy is $\frac{4}{9} x+\frac{2}{7} y$, and the amount of gold in the new alloy is $\frac{5}{9} x+\frac{5}{7} y$.

According to the condition of the problem,

$$
\left(\frac{4}{9} x+\frac{2}{7} y\right):\left(\frac{5}{9} x+\frac{5}{7} y\right)=7: 11
$$

Let's solve the obtained equation:

$$
\begin{gathered}
\frac{28 x+18 y}{63}: \frac{35 x+45 y}{63}=7: 11 \\
11(28 x+18 y)=7(35 x+45 y) \\
308 x+198 y=245 x+315 y, \quad 63 x=117 y \\
7 x=13 y
\end{gathered}
$$

The desired ratio is $x: y=13: 7$.

Note: Instead of the given equation, we can write the condition for the amount of 

silver $\frac{4}{9} x+\frac{2}{7} y=\frac{7}{18}(x+y) \quad$ or gold $\frac{5}{9} x+\frac{5}{7} y=\frac{11}{18}(x+y)$.

Setting up the equation is worth 3 points, and solving it is worth another 3 points.