A liter of $a \%$ alcohol is diluted with water to $b \%$. A part of the mixture is set aside, and the remainder is further diluted until it becomes $c \%$. How much water did we use in total, if we know that the amount of the resulting $c \%$ mixture is equal to the amount of the $b \%$ mixture that was set aside?
First, let's add $x$ liters of water to the alcohol. In this case, the amount of alcohol in the $a\%$ mixture is equal to the amount of alcohol in the $b\%$ mixture, that is,

$$
\frac{a}{100}=\frac{(1+x) b}{100}, \quad \text { hence } \quad x=\frac{a-b}{b}
$$

Let the amount of the set-aside mixture and the amount of the $c\%$ mixture both be $y$. The latter was created by adding $y-\left(\frac{a}{b}-y\right)=2 y-\frac{a}{b}$ liters of water, so

$$
\left(\frac{a}{b}-y\right) \frac{b}{100}=y \frac{c}{100}, \quad \text { hence } \quad y=\frac{a}{b+c}
$$

Therefore, the amount of water used is

$$
x+\left(2 y-\frac{a}{b}\right)=\frac{a-b}{b}+2 \frac{a}{b+c}-\frac{a}{b}=\frac{2 a}{b+c}-1
$$