$\dfrac{a}{150 ~ \text{ft.}} = \dfrac{\frac{1}{4} ~ \text{in.}}{1 ~ \text{ft.}}$
$a = 37.5 ~ \text{in.} = 3.125 ~ \text{ft.}$
$r = 55 - a = 55 - 3.125$
$r = 51.875 ~ \text{ft.}$
Total volume (water + concrete)
$V_t = \text{Frustum of a cone}$
$V_t = \frac{1}{3}\pi h(R^2 + r^2 + Rr)$
$V_t = \frac{1}{3}\pi (150)[ \, 55^2 + 51.875^2 + 55(51.875) \, ]$
$V_t = 1,346,037.462 ~ \text{ft.}^3$
Volume of water
$V_w = \text{Cylinder}$
$V_w = \pi (50^2)(150)$
$V_w = 1,178,097.245 ~ \text{ft.}^3$
Volume of concrete
$V_c = V_t - V_w$
$V_c = 1,346,037.462 - 1,178,097.245$
$V_c = 167,940.217 ~ \text{ft.}^3$
Weight of water
$W_w = \gamma_w V_w = 62.4(1,178,097.245)$
$W_w = 73,513,268.09 ~ \text{lb}$
Weight of concrete
$W_c = \gamma_c V_c = 150(167,940.217)$
$W_c = 25,191,032.55 ~ \text{lb}$
Total weight when full of water
$W_t = W_w + W_c = 73,513,268.09 + 25,191,032.55$
$W_t = 98,704,300.64 ~ \text{lb} \times \dfrac{1 ~ \text{short ton}}{2000 ~ \text{lb}}$
$W_t = 49,352.15 ~ \text{short tons}$ answer
