Population forecast to 2040, geometric method

The last census is 2021; for 2040, $n = (2040-2021)/10 = 1.9$ decades.

Population by geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{42{,}000}{15{,}000}\right)^{1/4}-1 = 29.357\%$

$$P_{2040} = P_n(1+r)^n = 42{,}000\times(1+0.29357)^{1.9} = 68{,}493$$

Water demand in 2040

Domestic demand $= 68{,}493\times120/1000 = 8{,}219.2\ \mathrm{m^3/day}$.

Industrial demand (20%) and losses (15%) are fractions of the total demand $Q$, so the domestic demand is 65% of $Q$.

$$\begin{aligned} Q &= \frac{8{,}219.2}{1-0.20-0.15} = \frac{8{,}219.2}{0.65} = 12{,}645.0\ \mathrm{m^3/day}\\ \text{Industrial} &= 0.20Q = 2{,}529.0\ \mathrm{m^3/day},\quad \text{Losses} = 0.15Q = 1{,}896.7\ \mathrm{m^3/day} \end{aligned}$$

Answer: Population in 2040 = 68,493; total water demand = 12,645 m$^3$/day (146.4 l/s).