You have an empty 3-liter container, a full 5-liter container with 50 degrees water, and a full 6-liter container with 90 degrees water.

Can you manage to get some 72 degrees water?

(In reply to Solution with good luck by nikki)

Sanity check:

I just wanted to check that my final solution did indeed have the 9 parts : 11 parts ratio. So I’m going to follow the same pouring, but this time I’m going to track 3(a, b) to mean "in the 3 liter jug we have a liters of 50 and b liters of 90).

3(0,0) 5(5,0) 6(0,6)
3(0,3) 5(5,0) 6(0,3)
3(0,3) 5(2,0) 6(3,3)
3(0,0) 5(2,3) 6(3,3)
3(1.5,1.5) 5(2,3) 6(1.5,1.5)
3(1.5,1.5) 5(0.8,1.2) 6(2.7,3.3)

The last step is the one with the most math. Basically when you pour 3 liters out of the 5 liter jug, that means you pour 3/5 of the jug out and keep 2/5 in the jug. Since in theory your mixture is uniform, that means that 2/5 of the 50 degree water stays in the jug, along with 2/5 of the 90 degree water. So 2*2/5 = 0.8 and 3*2/5 = 1.2

Next, you have to add the 2*3/5 @50 and 3*3/5 @90, that was poured out of the 5 liter jug, to the 1.5 and 1.5 of water that is already in the 6 liter jug. And that’s how we get 2.7 and 3.3 which do hold the 9 parts : 11 parts ratio that I was expecting.

Yay!

Posted by nikki on 2004-09-29 10:13:01