You have three cups which differ in volume and shape. You know that the largest cup is less than three times the volume of the smallest cup. Each cup has a mark showing how much one-third is. You also have a drain.

The largest cup is already full of water. Pour water back and forth to fill all three cups to their one-third mark, pouring any excess down the drain.

Note: If a cup is less than one third filled, it may be filled up to the one third line BUT if a cup is over one third full it cannot be poured out to reduce it to exactly one third filled because when the cup is tipped, you have no indicator for one third full when tipped.

The key to this problem is to reach this situation: the largest cup has 1/3 G + 1/3 M + 1/3 S water where G, M and S are the sizes of the largest cup, the middle size cup and the smallest cup, respectively. Or the largest cup has 1/3G and the middle cup has 1/3 M and 1/3 S water. (There could be some other similar cases).

My solution is slightly different from others.
1. Fill up the middle size cup and the smallest cup
2. Empty the greatest cup's water to the drain
3. Fill the greatest cup to 1/3 with the water from the smallest cup.
4. Empty the smallest cup's water to the drain
5. Fill smallest cup to 1/3 with water from the middle size cup
6. Pour the water in the smallest cup to the greatest cup
7. Pour all water in the middle size cup to smallest
8. Fill the middle size cup to 1/3.
9. Empty the smallest cup's water to drain
10. Fill the smallest cup to 1/3 with the water from the greatest cup.

Posted by Jeffery on 2004-12-01 15:54:15