Macedonian Mathematical Competitions

The area of the land where the dug soil is allotted is $(12^2 - 4^2\pi)\text{ m}^2 = (144 - 16\pi)\text{ m}^2$. Let $x$ be the height of the hole before the dug soil is allotted. Then the thickness of the layer of allotted and then pressed soil is $(3-x)\text{ m}$. Because the dug soil is pressed as much as it was before the digging it, the volume is the same, i.e.

$4^2\pi \cdot x = (144 - 16\pi) \cdot (3 - x)$.

If we solve the last equation we get that $x = \left(3 - \frac{\pi}{3}\right)\text{ m}$. Because $\pi \approx 3.14$ we get that $x \approx 1.95\text{ m}$ should be the dug.