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?

Well, you are going to have x% of 50 degree water, and (100-x)% of 90 degree water to get 100% of 72 degree water.

So 50x + 90(1-x) = 72*1.  So x = 0.45. So you have 45% 50 degree water and 55% 90 degree water. Also, 0.45 = 9/20 and 0.55 = 11/20. Or you can think of it as 9 parts 50 degree with 11 parts 90 degree water = 20 parts.

Don't ask me how I got it… I just started playing around and got two quantities whose average was 72, and then I was able to pour even parts together. I didn't even use any of the 9/20 or 11/20 stuff up there.

So I'm going to use the following notation where 3(x, y) mean "in the 3 liter jug we have x liters of y degree water.

Start
3(0, --) 5(5, 50) 6(6, 90)
Fill the 3 liter from the 6 liter
3(3, 90) 5(5, 50) 6(3, 90)
Fill the 6 liter from the 5 liter ([3*50 + 3*90]/6 = 70)
3(3, 90) 5(2, 50) 6(6, 70)
Fill the 5 liter from the 3 liter ([2*50 + 3*90]/5 = 74)
3(0, --) 5(5, 74) 6(6, 70)
Fill the 3 liter from the 6 liter
3(3, 70) 5(5, 74) 6(3, 70)
Fill the 6 liter from the 5 liter ([3*74 + 3*70]/6 = 72)
3(3, 70) 5(2, 74) 6(6, 72)

I think I just basically thought that my first two moves couldn't be right, because then you might have to dump the 90 water that is in the 3 liter jug, and then you could never get water warmer than 70. In an attempt to show myself that I shouldn't make those first two moves, I went through it to see if I really would have to dump the 90 degree water out of the 3-liter jug.

Then I saw that I didn't HAVE to dump it – I could pour it into the 5 liter jug but I was sure that wouldn't get me anywhere. But when I saw that the result was 74 degrees, I knew right away that I could finish the problem because I could add even amounts of the 70 and 74 water to get 72.

Edited on September 29, 2004, 9:57 am

Posted by nikki on 2004-09-29 09:54:36