Solved Problem 01 | Rectangular Parallelepiped

Volume of coal bin
$V = 19(6)(9)$

$V = 1026 \, \text{ ft}^3$
 

Weight 1026 ft³ coal
$W = \dfrac{1 \, \text{ ton}}{38 \, \text{ ft}^3}(1026 \, \text{ ft}^3)$

$W = 27 \, \text{ tons}$           answer