119(988). In a flask, there is a solution of table salt. From the flask, $\frac{1}{5}$ of the solution is poured into a test tube and evaporated until the percentage of salt in the test tube doubles. After this, the evaporated solution is poured back into the flask. As a result, the salt content in the flask increases by $3 \%$. Determine the initial percentage of salt.
Solution. Before solving the problem, it is reasonable to clarify how the expression "percentage content of salt" should be understood (this is the ratio of the mass of salt to the mass of the solution, expressed as a percentage).

Let the initial mass of the solution in the flask be $m$ grams, the mass of salt in it be $m_{\text {s }}$ grams, and the initial percentage content of salt be $x \%$, i.e.,

$$
x=\frac{m_{s}}{m} \cdot 100
$$

According to the problem, $\frac{1}{5}$ of the solution was poured out from the flask, i.e., $\frac{m}{5}$ grams, so $\frac{4}{5} m$ grams remained. Since the amount of salt does not change during evaporation, to double the percentage content of salt in the test tube, the solution after evaporation should be $\frac{m}{10}$ grams, and a total of $\frac{9}{10}$ grams $\left(\frac{4}{5} m+\frac{m}{10}=\frac{9}{10} m\right)$.

Since the percentage content of salt increased by $3 \%$, we have:

$$
x+3=\frac{m_{s} \cdot 100}{\frac{9}{10} m}
$$

From equations (1) and (2), we get the equation

$$
x+3=\frac{10}{9} x, \text { from which } x=27
$$

Answer: $27 \%$.