Question

В бассейне было 250 кубических метров воды. Через первую трубу вода выливается из бассейна, причём только порциями по 150 кубических метров, а через вторую трубу наливается, причём порциями по 99 кубических метров. Манипулируя этими трубами, какое наименьшее количество воды можно оставить в бассейне?

Answer 1

Ответ 97 так как
Решение :
1) 250-150+99=199
2) 199-150+99=148
3)148-150+99=97
4) 97-150+99=-53+99=46
Дальше вода начнёт уходить в минус и мы не сможем из 46 вылить 150 а потом 99 налить

Answer 2

Calculating the Minimum Remaining Water in the Pool

To find the minimum amount of water that can be left in the pool, we can calculate the amount of water that can be drained and added through the two pipes.

1. Water Drained through the First Pipe: - The first pipe drains water in portions of 150 cubic meters. - To drain the entire pool, the number of portions needed can be calculated as: ``` 250 / 150 = 1.67 ``` - Since the portions must be whole numbers, the first pipe can drain a maximum of 150 cubic meters of water.

2. Water Added through the Second Pipe: - The second pipe adds water in portions of 99 cubic meters. - To minimize the remaining water, the number of portions added should be maximized. - The maximum number of portions that can be added can be calculated as: ``` 250 / 99 = 2.52 ``` - Since the portions must be whole numbers, the second pipe can add a maximum of 198 cubic meters of water.

3. Calculating the Minimum Remaining Water: - To minimize the remaining water, the first pipe should drain 150 cubic meters, and the second pipe should add 198 cubic meters. - The minimum remaining water can be calculated as: ``` 250 - 150 + 198 = 298 - 150 = 100 cubic meters ```

Therefore, the minimum amount of water that can be left in the pool is 100 cubic meters.

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