Chapter 1 - Measurement and Problem Solving - Learning Path Questions and Exercises - Exercises - Page 30: 39

a). $149618m^{3}$ b). $149618 \times 10^{3} kg.$ c). $329159 \times 10^{3} lb. $

a). Volume of the Coliseum is $= area \times depth = \frac{22}{7} \times r^{2} \times depth$ $= \frac{22}{7} \times 125^{2}m^{2} \times 3.048m$ $= 149618 m^{3}$ So, amount of water required is $149618m^{3}$ b). $1 m^{3} = 10^{3}L$ Hence, mass = $149618 \times 10^{3} kg$ c). 1 kg=2.2lb Hence, mass = $149618 \times 10^{3} \times 2.2 lb$ $= 329159 \times 10^{3}lb.$