Question

Имеются два сосуда. В первом содержится 1 литр 10-ти процентного раствора кислоты, во втором – 2 литра 60-ти процентного. Проделали следующее действие, состоящее из двух этапов: на первом этапе из второго сосуда перелили в первый 1 литр раствора, на втором из первого перелили обратно во второй 1 литр полученной смеси. Укажите в ответе, какое минимальное количество раз нужно проделать такое действие, чтобы концентрация растворов в сосудах отличалась менее чем на 0,1%?

Answer 1

Решение

Пусть и -- концентрации 1-го и 2-го растворов соответственно. После переливания концентрация 1-го станет а после второго переливания концентрация 2-го станет . Тогда Следовательно, через действий разность концентраций станет равна Отсюда наименьшим решением неравенства является  

наверное так!!!

Answer 2

Problem Analysis

To solve this problem, we need to determine the minimum number of times the described action needs to be repeated so that the concentration of the solutions in the vessels differs by less than 0.1%.

Solution

Let's start by calculating the concentration of the solutions in the vessels after each step of the described action.

1. Initial Concentrations: - Vessel 1: 1 liter of 10% acid solution - Vessel 2: 2 liters of 60% acid solution

2. Step 1: Transfer 1 liter from Vessel 2 to Vessel 1: - After the transfer, Vessel 1 contains 2 liters of a solution. - The concentration of the solution in Vessel 1 after the transfer can be calculated using the formula: \[C_1V_1 + C_2V_2 = C_fV_f\] where \(C_1\) and \(C_2\) are the initial concentrations, \(V_1\) and \(V_2\) are the volumes, and \(C_f\) is the final concentration. - Substituting the given values: \[10% \times 1L + 60% \times 1L = C_f \times 2L\] \[10% + 60% = 2C_f\] \[70% = 2C_f\] \[C_f = 35%\]

3. Step 2: Transfer 1 liter from Vessel 1 back to Vessel 2: - After the transfer, Vessel 1 contains 1 liter of a solution, and Vessel 2 contains 2 liters of a solution. - The concentration of the solution in Vessel 2 after the transfer can be calculated using the same formula: \[C_1V_1 + C_2V_2 = C_fV_f\] Substituting the given values: \[35% \times 1L + 60% \times 2L = C_f \times 2L\] \[35% + 120% = 2C_f\] \[95% = 2C_f\] \[C_f = 47.5%\]

4. Analysis: - After each complete cycle of the described action, the concentration of the solutions in the vessels changes. - To minimize the difference in concentration to less than 0.1%, we need to continue the action until the difference is less than 0.1%.

Conclusion

The minimum number of times the described action needs to be repeated to achieve a difference in concentration of less than 0.1% is infinite. Due to the nature of the concentrations and the action described, the concentrations will never be exactly the same, and the difference will never be less than 0.1%.