Let $ABCDA'B'C'D'$ be a rectangular parallelipiped, where $ABCD$ is the lower face and $A, B, C$ and $D$ are below $A', B', C'$ and $D'$, respectively. The parallelipiped is divided into eight parts by three planes parallel to its faces. For each vertex $P$, let $V_P$ denote the volume of the part containing $P$. Given that $V_A= 40$, $V_C = 300$ , $V_B' = 360$ and $V_C'= 90$, find the volume of $ABCDA'B'C'D'$.

By considering the plane separating $A, A'$ we have $V_A / V_{A'} = V_B / V_{B'} = V_C / V_{C'} = 10/3 = V_D / V_{D'}.$ By considering the plane separating $C, D$ we have $V_C / V_D = V_{C'} / V_{D'} = V_B / V_A = 30/1 = V_{B'} / V_{A'}.$ We deduce from these relations all eight of the $V_P$s and add them together to get the total volume $\boxed{2015}.$