Sand deposit reserve in tons

Principle: for a horizontal deposit the thickness is the true thickness, so reserve (volume) = area $\times$ thickness; mass = volume $\times$ density.

Given

Thickness $t = 16$ m; area $= 150 \times 150 = 22{,}500\ \text{m}^2$; density $\rho = 2870\ \text{kg/m}^3$.

Calculation

$$\begin{aligned} V &= 22{,}500 \times 16 = 360{,}000\ \text{m}^3 \\ M &= V\rho = 360{,}000 \times 2870 = 1.0332\times 10^{9}\ \text{kg} \\ &= 1{,}033{,}200\ \text{tonnes} \end{aligned}$$

(1 tonne = 1000 kg.)

Answer: reserve of sand = 3.6 × 10⁵ m³, i.e. about 1.03 × 10⁶ tonnes (1,033,200 t).