An hourglass is formed from two identical cones. Initially, the upper cone is filled with sand and the lower one is empty. The sand flows at a constant rate from the upper to the lower cone. It takes exactly one hour to empty the upper cone. How long does it take for the depth of sand in the lower cone to be half the depth of sand in the upper cone? (Assume that the sand stays level in both cones at all times.)
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talkinaway
08.06.2017 07:19
If the top cone has radius $r$ and height $h$, then when the depth of sand is equal, the sand-filled part of the top cone will have height $\frac{h}{2}$, because the unfilled part will be a frustrum of height $\frac{h}{2}$ as well. (That frustrum will be the same frustrum of sand in the bottom cone.)
By similarity, the radius of the sand-filled cone of height $\frac{h}{2}$ will be $\frac{r}{2}$. Thus, it will be $\frac{1}{8}$ the volume of the remaining cone, with $\frac{7}{8}$ of the sand in the bottom. Intuitively, this should make some sense - when the hour glass starts, the sand must "spread out" along a wide circular base, and it takes a long time to fill. Thus, it should be more than $\frac{1}{2}$.
Since the bottom cone is $\frac{7}{8}$ full, it takes $\frac{7}{8}$ of an hour, or $\boxed{52.5}$ minutes.