Ans: (a) $1.32\times10^8 \text{ m}^3$ (b) Yes
- The total volume of water imported in the first $20$ years
$\begin{array}{cl}
= & 1.5 \times 10^7 + 1.5 \times 10^7 \times (1-0.1) + \ldots + 1.5 \times 10^7 \times (1-0.1)^{19} \\
= & \dfrac{1.5 \times 10^7 \times ( 1- 0.9^{19})}{1- 0.9} \\
= & 1.32 \times 10^8 \text{ m}^3
\end{array}$ - The maximum possible total volume of water imported
$\begin{array}{cl}
= & \dfrac{1.5 \times 10^7}{1-0.9} \\
= & 1.5 \times 10^8 \text{ m}^3\\
< & 1.6 \times 10^8 \text{ m}^3 \end{array}$Therefore, I agree with the claim.
