You have 3 unmarked pitchers that hold 6 cups, 10 cups, and 15 cups. The 10 cup pitcher is filled with lemonade, but you have to dilute it so people at the party don't complain that it's too sour. Unfortunately, I have, in the name of logic, broken your only measuring cup.

How can you dilute the lemonade to 20 cups such that all of the lemonade is of equal concentration? You have a sink with unlimited water, but you can't waste any lemonade. Try using as few moves as possible.

1. Pour the 10 cups of lemonade into the 15-cup pitcher, top off with water. You now have a 2:1 dilution (10:5) until further notice. ;)
2. Fill the 6-c pitcher from the 15-c, leaving 9 cups.
3. Empty the 9 cups from the 15-c pitcher into the 10-c.
4. Transfer the contents of 6-c back into the 15-c.
5. Fill the 6-c from the 10-c, leaving 3 cups behind.
6. Add this 6-c to the 15-c again, filling it to 12 cups total.
7. Pour the 3 cups from the 10-c pitcher into the 6-c pitcher.

At this point, you have:
6-c pitcher with 2 cups lemonade and 1 cup water
10-c pitcher empty
15-c pitcher with 8 cups lemonade and 4 cups water

8. Add water to the 6-c pitcher, (2 cups lemonade & 4 cups water)
9. Move this liquid to the 10-c pitcher.
10. Fill the 6-c pitcher from the 15-c, resulting in both of them containing 4 cups lemonade and 2 cups water.
11. Add the contents of the 10-c pitcher to the 15-c to get (2:4) and (4:2) = (6:6) Yaaayyyyyy! 12 cups of drinkable lemonade!
12. Fill the 10-c pitcher with the drinkable lemonade, wait until the guests drink it empty, then:
13: Add all the remaining partially-diluted lemonade to the 10-c pitcher, top it off with water to produce the final 10 cups of the desired 20 cups of beverage.

Sheesh, my arms are tired and the counter is sticky! I like Paul's approach much better.

Posted by gail on 2005-07-26 06:25:22