1. Two ingots of gold and silver contain a total of $a \mathrm{~kg}$ of gold. If the percentage of gold in the first ingot is the same as the percentage of gold in the second ingot, the two ingots would have a total of $b \mathrm{~kg}$ of gold, and if the percentage of gold in the second ingot is the same as the percentage of gold in the first ingot, the two ingots would have a total of $c \mathrm{~kg}$ of gold. How many kilograms of gold are in the first, and how many in the second ingot?
Solution. Let the first piece contain $p$ percent gold and weigh $x \mathrm{~kg}$, and the second piece contain $q$ percent gold and weigh $y \mathrm{~kg}$. Then, according to the conditions, we have $p x+q)^{\prime}=100 a, \quad q x+q y=100 b$ and $p x+p y=100 c$. Therefore, $q y-p y=100(a-c)$ and $\quad p x-q x=100(a-b)$. We have $x=100 \frac{a-b}{p-q} \quad$ and
$y=100 \frac{c-a}{p-q}$. It also holds that $\frac{b}{q}=\frac{c}{p}$, so $p-q=p-p \frac{b}{c}=\frac{p(c-b)}{c}=\frac{q(c-b)}{b}$. Therefore,

$$
\begin{aligned}
& \frac{p}{100} x= \frac{p}{100} \cdot 100 \cdot \frac{a-b}{p-q}=p \frac{a-b}{\frac{p(c-b)}{c}}=\frac{c(a-b)}{c-b} \mathrm{~kg} \\
& \frac{q}{100} y=\frac{q}{100} \cdot 100 \cdot \frac{c-a}{p-q}=q \frac{c-a}{\frac{q(c-b)}{b}}=\frac{b(c-a)}{c-b} \mathrm{~kg}
\end{aligned}
$$