Problem
There are two sections with different heights, and the areas of the upper and lower sections are 155 $m^2$ and 175 $m^2$ respectively. The height difference between the upper and lower sections is 1.5 $m$. The soil obtained by digging up 3 $m$ in an area of 20 $m^2$ is brought from the outside and filled in the lower section. After that, how much the upper section should be dug up in order to fill in the lower section and make the upper and lower sections one leveled field?
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Solution
Assuming that the height of the lower section is 0 m, the volume of soil that makes up the upper section is
$$155×1.5=232.5 m^3. —[1]$$
The volume of soil brought from outside is
$$20×3=60m^3. —[2]$$
The sum of [1] and [2] is
$$232.5+60=292.5 m^3. —[3]$$
The total area of one field created by leveling the upper and lower section is
$$155+175=330 m^2. —[4]$$
When the height of one leveled field is $x m$, from [3] and [4], we see that
$$330x=292.5,$$
$$∴ x≒0.886364 m.$$
Therefore, the height at which the upper section should be dug down is $$1.5-0.886364=0.613636 m.$$
(Answer) approximately 0.6136 $m$.
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), pp.49-50; pp.312-313.