# Plep's Puzzles

Name:

## Saturday, October 23, 2004

### 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?

Anonymous said...

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

October 23, 2004 at 11:08 PM
mick said...

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!

October 25, 2004 at 12:30 PM
Anonymous said...

You're both right!

Steven

October 27, 2004 at 5:36 AM
ME-L said...

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.

November 8, 2004 at 12:50 PM