An hour-glass is formed by two identical cones.

Initially, the top cone is full of sand, and the bottom cone is empty.

The sand starts flowing down at a constant rate and the top cone is emptied in exactly 1:30 hours.

How long does it take for the height of the sand in the bottom cone be half of the height in the top cone?

As the top cone has its apex at its bottom and the bottom cone has its apex at its top, the empty portion of the bottom cone is congruent to the still-full portion of the top cone, whose sand will eventually fill in that remaining empty portion on the bottom.

So for the height of the bottom's sand to equal half the height of the sand in the top cone, it must be half the height of its own unfilled portion--that is, 1/3 the height of the cone.

So the sand that has not yet fallen fills a cone similar to the original full cone but 2/3 the size in each linear measure. As a result, it has 8/27 the volume of the original, so 8/27 of the 1.5 hours still remains, and that is 26 minutes, 40 seconds.

Subtracting 0:26:40 from 1:30:00 gives 1:03:20 as the length of time required to reach the state where the bottom sand level is half that in the top--that is, 1 hour, 3 minutes and 20 seconds.

Posted by Charlie on 2009-01-20 12:43:59