## Task 4 - 180734

In a container, there are exactly $25 \mathrm{~kg}$ of a $4 \%$ aqueous solution, i.e., $4 \%$ of this solution consists of the dissolved substance, the rest consists of water.

How much percent of the water must be removed from this solution so that a new solution is created whose water content is only $90 \%$?
In the initial solution, there are exactly $4 \%$ of the dissolved substance, which is exactly $1 \mathrm{~kg}$ in $25 \mathrm{~kg}$ of solution. This amount represents exactly $10 \%$ of the new solution after a certain amount of water has been removed, if the new solution totals $10 \mathrm{~kg}$.

Thus, if $15 \mathrm{~kg}$ of water is removed from the initial solution, the proportion will be $90\%$, as required. Therefore, it needs to be determined what percentage of $24 \mathrm{~kg}$ of water $15 \mathrm{~kg}$ of water is. For this sought percentage $x$, the relationship is

$$
x: 100 \% = 15: 24 \quad \Rightarrow \quad x = \frac{1500}{24} \% = 62.5 \%
$$

Accordingly, $62.5 \%$ of the water contained in the initial solution must be removed to obtain a new solution with a $90 \%$ water content.