# Filling water jugs

You have three jugs with volumes of 12, 8, and 5 liters. The 12L jug is full of water, while the other two jugs are empty. You have no other source of water.

Using these jugs, how can you obtain exactly 6 liters of water in the 12L jug?

• I've tried a lot of things but can't seem to know what to do Sep 1, 2017 at 9:35
• Hello and welcome to Puzzling. I've edited your question slightly for clarity. Please feel free to edit it further or to roll back my changes. Sep 1, 2017 at 11:08
• Possible duplicate of A Set of Water Jug Challenges
– Oray
Sep 1, 2017 at 11:14
• This is confusingly written. You mix units (inches or liters? Pick one!) and it seems likely that your intention is that the 12 jug is full and there is no other source of water, but it doesn't say that - if that's what you intend, please edit the puzzle so that is clear.
– Rubio
Sep 1, 2017 at 11:28
• Possible duplicate of A general solution to the decanting problem? (aka jug-pouring, water-pouring) Sep 1, 2017 at 12:04

1. 12j-12, 8j-0, 5j-0

2. 12j-4, 8j-8, 5j-0

3. 12j-4, 8j-3, 5j-5

4. 12j-9, 8j-3, 5j-0

5. 12j-9, 8j-0, 5j-3

6. 12j-1, 8j-8, 5j-3

7. 12j-1, 8j-6, 5j-5

8. 12j-6, 8j-6, 5j-0 (done!)

Ends up with 6 inches of water in the 12-inch jug with no water wasted!

• After step 6, you could fill 5 from 8, then pour 5 into 12 to not waste anything.
– Apep
Sep 1, 2017 at 11:52
• Right, that's even better. I'll update my answer, thank you! Sep 1, 2017 at 11:53
• This is the only way to perform the task in seven transfers, and it is not possible with less. Sep 6, 2017 at 19:53

Assume all the jugs have the same cross-section, so the volume is proportional to the height of water in the jugs.

Fill the 8-inch jug from the 12-inch jug, then fill the 5-inch jug from the 8-inch jug. This leaves 3 inches in the 8-inch jug. Discard that 3 inches of water. Empty the 5-inch jug into the 12-inch jug. Repeat.

This discards 6 inches of water and leaves 6 inches in the 12-inch jug.

here is another way of doing this:

12j-12, 8j-0, 5j-0
12j-0, 8j-8, 5j-4
12j-8, 8j-4, 5j-0
12j-3, 8j-4, 5j-5
12j-3, 8j-8, 5j-1
12j-11, 8j-1, 5j-0
12j-6, 8j-1, 5j-5(Done!)

• I don't know how to hide the text, but these steps are correct to get the 6L in 12L jug Sep 6, 2017 at 19:22
• It's another way yes, but do note that this takes nine transfers rather than the (minimal) seven (even though it was not asked of us to be optimal in transfers). Sep 6, 2017 at 19:52