In the dairy technology lab at a university, a technician fills a holding tank of capacity 50 litres by using 3 different pipes. The rate of flow through each pipe is equal to 1 litre/min. The 1st pipe pours milk only, the 2nd pipe pours water only and the 3rd pipe pours a mixture of milk and water, in which the ratio of milk to water is 2 : 3. At any given instance, only 1 of the 3 pipes is opened. Each of the pipes is opened to fill its contents into the holding tank for exactly 5 minutes after which it is replaced by one of the other 2 pipes. The ratio of milk to water in the tank after 50 minutes cannot be

The total process runs for 50 minutes, consisting of exactly ten 5-minute intervals. The contents poured are 5 litres of milk from pipe 1, 5 litres of water from pipe 2, and 5 litres of a 2:3 milk-to-water mixture, providing 2 litres of milk and 3 litres of water from pipe 3. Depending on how many times each pipe is used, the total milk will be 5x + 2y and the total water will be 5z + 3y, where x, y and z are the number of times pipes 1, 3 and 2 are used, respectively. The sum x + y + z must equal 10, meaning the total liquid volume is always 50 litres. For the ratio 6:19, the required milk is 50 multiplied by 6 divided by 25, which equals 12 litres; this implies 5x + 2y equals 12. Testing small values shows the only non-negative solutions for this equation are x=2 and y=1, which would require z to equal 7; however, z cannot be greater than 5 because pipes must be swapped after each 5-minute interval, meaning a single pipe cannot be used more than 5 times. Therefore, a ratio of 6:19 is impossible to achieve.

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