### Another Pouring Puzzle

Suppose that you have access to an 11-gallon container, a 6-gallon container, and a source of running water (such as a water tap). You do not have access to any measuring devices, but you are allowed to pour water down a drain.

How would you set about measuring out 8 gallons of water?

Answers in the comments please. Another puzzle in about a week.

## 4 Comments:

Ok, there could be a shorter way, but here goes...

11...6 --> 11...0 --> 5...6

5...0 --> 0...5 --> 11...5

10...6 --> 10...0 --> 4...6

4...0 --> 0...4 --> 11...4

9...6 --> 9...0 --> 3...6

3...0 --> 0...3 --> 11...3

8...3 --> 8...0

prodigaltom

I'm not sure I understand the previous notation - here's how I would do it (which may or may not be the same).

Fill the 6 gal, pour into the 11 gal.

Fill the 6 gal, pour into the 11 gal; you have 1 gal left left over.

Empty the 11 gal; pour the 1 gal from the 6 gal into the 11 gal.

Fill the 6 gal, pour into the 11 gal.

Fill the 6 gal, pour into the 11 gal; you have 2 gal left over

Empty the 11 gal; pour the 2 gal from the 6 gal into the 11 gal.

Fill the 6 gal, pour into the 11 gal. 6+2 =8!

You're both right!

Steven

Ah, that would be the mathematician's method. Here's the physics method:

1) Fill the 11-gallon jug and the 6-gallon jug.

2) Wait for the 11-gallon jug to evaporate entirely. Measure the number of hours this takes under fixed circumstances.

3) Repeat step 2 one hundred times so that you can normalize for weather conditions, etc. While you are waiting, drink from the 6-gallon jug.

4) Calculate the mean evaporation rate. Use this rate to determine how long it will take for 3 gallons to evaporate.

5) Fill the 11-gallon jug again. Wait for 3 gallons to evaporate.

6) While you are waiting, drink from the 6-gallon jug.

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