Question

Исполнитель «Переливашка» имеет три сосуда: А(8л), Б(5л) и В(3л). Переливашка может переливать воду из одного сосуда в другой, пока он не станет полным. Изначально, сосуд А полный, Б и В - пустые. Команды для исполнителя записываются в виде: А-Б(перелить воду из сосуда А в сосуд Б) и т. д. Вода из каждого сосуда переливается в максимально возможном объёме. Сколько литров воды будет в каждом сосуде после выполнения алгоритма: А-Б, Б-В, В-А, Б-В, А-Б, Б-В, В-А?

Answer 1

А - 4 литра
Б - 4 литра
В- 0 литров

Answer 2

Problem Analysis

The problem describes a scenario where there are three containers labeled A, B, and C with capacities of 8 liters, 5 liters, and 3 liters respectively. Water can be transferred between the containers until one of them becomes full. Initially, container A is full, while containers B and C are empty. The commands for the performer are given in the format A-B (transfer water from container A to container B) and so on. The goal is to determine the amount of water in each container after executing the given algorithm.

Solution

To solve this problem, we can simulate the execution of the algorithm step by step and keep track of the water levels in each container.

1. Initially, container A is full (8 liters), and containers B and C are empty (0 liters). 2. The first command is A-B, which means transferring water from container A to container B. Since container B has a capacity of 5 liters, it can only hold 5 liters of water. Therefore, after this command, container A will have 8 - 5 = 3 liters of water, and container B will have 5 liters of water. 3. The second command is B-C, but container B only has 5 liters of water, so it cannot transfer any water to container C. Therefore, the water levels in containers A, B, and C remain the same: A = 3 liters, B = 5 liters, C = 0 liters. 4. The third command is C-A, but container C is empty, so it cannot transfer any water to container A. Therefore, the water levels in containers A, B, and C remain the same: A = 3 liters, B = 5 liters, C = 0 liters. 5. The fourth command is B-C, but container B only has 5 liters of water, so it cannot transfer any water to container C. Therefore, the water levels in containers A, B, and C remain the same: A = 3 liters, B = 5 liters, C = 0 liters. 6. The fifth command is A-B. Container A has 3 liters of water, and container B has 5 liters of water. Container B can only hold 5 liters, so it will be filled completely, and container A will have 3 - 5 = -2 liters of water. Since negative water levels are not possible, we assume that container A becomes empty. Therefore, the water levels in containers A, B, and C are: A = 0 liters, B = 5 liters, C = 0 liters. 7. The sixth command is B-C, but container B only has 5 liters of water, so it cannot transfer any water to container C. Therefore, the water levels in containers A, B, and C remain the same: A = 0 liters, B = 5 liters, C = 0 liters. 8. The seventh command is C-A, but container C is empty, so it cannot transfer any water to container A. Therefore, the water levels in containers A, B, and C remain the same: A = 0 liters, B = 5 liters, C = 0 liters.

Final Result

After executing the given algorithm, the amount of water in each container is as follows: - Container A: 0 liters - Container B: 5 liters - Container C: 0 liters

Please let me know if you have any further questions!