1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and an equal volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again topped up with the same volume of argon. In the new mixture, $11.76\%$ oxygen was found. How many liters of the mixture were released each time from the vessel? Give your answer as a number without specifying the unit.

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Solution. Let $x$ be the volume of the mixture released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x$. The percentage of oxygen in the mixture after adding argon in this case is $10(2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x)$. According to the condition, $10(2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x) = 11.76$. Solving the equation, we arrive at the quadratic equation $x^2 - 20x + 51 = 0, x_1 = 3, x_2 = 17$. The second root does not fit the condition.

Answer: 3.