1. The pond is filled with 100% alcohol. We remove 2 l of alcohol and add the same amount of distilled water. This procedure is repeated once more, i.e., we remove 2 l of the mixture and add 2 l of distilled water. In this way, the pond contains a 36% alcohol solution. How many liters of solution does this pond hold?
Solution. When we pour out 2 liters from $x$ liters of alcohol and add 2 liters of water to the solution, there will be $x-2$ liters of alcohol, i.e., we will get a solution with $\frac{100(x-2)}{x}$ percent alcohol. The total amount of pure alcohol after the second transfer will be $x-2-2 \frac{x-2}{x}$ liters, and according to the condition, this is $0.36 x$. Therefore, $x-2-2 \frac{x-2}{x}=0.36 x$. The last equation is equivalent to the equation $x(x-2)-2(x-2)=0.36 x^{2}$, i.e., to the equation $(x-2)^{2}=(0.6 x)^{2}$. Since $x-2$ and $0.6 x$ are positive numbers, we get $x-2=0.6 x$, i.e., $x=5$ liters.