In one container, there is $80 \mathrm{~cm}^{3}$ of a salt solution with a density of $1.2 \mathrm{g} / \mathrm{cm}^{3}$, and in another container, there is $100 \mathrm{~cm}^{3}$ of an ammonia solution with a density of $0.8 \mathrm{g} / \mathrm{cm}^{3}$. From the first container, $x \mathrm{~cm}^{3}$ is poured into the second one; after thorough mixing, $x \mathrm{~cm}^{3}$ is poured back into the first one. What is $x$, if after the pouring back, the liquid in both containers has the same weight?
I. Solution: In one of the containers, there is $80 \cdot 1.2 = 96$ g of salt solution, and in the other, there is $100 \cdot 0.8 = 80$ g of ammonia. After pouring $\mathrm{x} \mathrm{~cm}^{3}$ of salt solution and mixing, the specific gravity of the solution in the second container can be expressed using $x$ as the ratio of the weight of the liquid to its volume: $\frac{80 + 1.2 x}{100 + x}$. However, this specific gravity can also be numerically calculated, as after pouring back, there is $100 \mathrm{~cm}^{3}$ of salt solution-ammonia mixture in the second container, which, according to the problem, weighs $\frac{96 + 80}{2} = 88$ g. Therefore, the specific gravity is $\frac{88}{100} = 0.88$. Considering our previous result,

$$
\frac{80 + 1.2 x}{100 + x} = 0.88
$$

From this,

$$
0.32 x = 8, \quad \text{and thus} \quad x = 25 \mathrm{~cm}^{3} \text{.}
$$

Gáspár Ágnes (Szeged, Hunyadi Mátyás g. I. o. t.)

II. Solution: The problem can also be solved by reasoning. The 96 g of liquid in the first container, after being poured and poured back, will be 88 g. The 8 g weight loss is due to the fact that $x \mathrm{~cm}^{3}$ of salt solution with a specific gravity of $1.2 \mathrm{g} / \mathrm{cm}^{3}$ is replaced by the same amount of a salt solution-ammonia mixture with a smaller specific gravity $\left(0.88 \mathrm{g} / \mathrm{cm}^{3}\right)$. The change in specific gravity is $1.2 - 0.88 = 0.32 \mathrm{g} / \mathrm{cm}^{3}$, which means that the replacement of every $\mathrm{cm}^{3}$ of salt solution results in a decrease of 0.32 g. The 8 g decrease is therefore the result of the replacement of $\frac{8}{0.32} = 25 \mathrm{~cm}^{3}$ of salt solution.

Deák Miklós (Bp. XI., József A. g. II. o. t.)