1. Consider a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, where $A B C D$ is the bottom face with the letters assigned in a clockwise direction, and $A, B, C$, and $D$ are directly below $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$ respectively. The parallelepiped is divided into eight pieces by three planes that are orthogonal to each other and parallel to the faces of the parallelepiped. For each vertex $P$ of the parallelepiped, let $V_{P}$ denote the volume of the piece of the parallelepiped that contains $P$. Given that $V_{A}=40, V_{C}=300, V_{B^{\prime}}=360$ and $V_{C^{\prime}}=90$, what is the volume of the parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$?
Solution: Let $x, y, z$ be the distances from $A$ to the three planes of the cuts, and $x^{\prime}, y^{\prime}, z^{\prime}$ the distances from $C^{\prime}$ to the same planes, so that $x+x^{\prime}=A B, y+y^{\prime}=A D, z+z^{\prime}=A A^{\prime}$.

When two pieces share a face, their volumes are proportional to their respective heights, and thus we obtain the following equalities:

$$
\frac{y}{y^{\prime}}=4=\frac{V_{B^{\prime}}}{V_{C^{\prime}}}=\frac{V_{B}}{V_{C}}, \quad \frac{z}{z^{\prime}}=\frac{10}{3}=\frac{V_{C}}{V_{C^{\prime}}}, \quad \frac{x}{x^{\prime}}=\frac{V_{A}}{V_{B}}
$$

From the first equality, we find that $V_{B}=4 V_{C}=1200$, and substituting into the last one, we get $\frac{x}{x^{\prime}}=\frac{1}{30}$. If we call $V$ the volume of the parallelepiped, we have:

$$
\frac{V}{V_{C^{\prime}}}=\frac{\left(x+x^{\prime}\right)\left(y+y^{\prime}\right)\left(z+z^{\prime}\right)}{x^{\prime} y^{\prime} z^{\prime}}=\left(\frac{x}{x^{\prime}}+1\right)\left(\frac{y}{y^{\prime}}+1\right)\left(\frac{z}{z^{\prime}}+1\right)=\frac{403}{18}
$$

from which $V=\frac{403}{18} V_{C^{\prime}}=\frac{403}{18} \cdot 90=2015$.