---
title: "Population forecast to 2040, geometric method"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:pop-bk81"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-bk81
site: BCT Engineering
---

# Population forecast to 2040, geometric method

Water Supply Engineering · Chapter 3: Quantity of Water

- Asked in: 2081 Baisakh, 8 marks
- Video: [Population Forecasting Numericals Using Arithmetic| Geometric and Incremental Increase Method](https://www.youtube.com/watch?v=dpVlwFlbUpw) — Master Civil Engineering Easily
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-bk81
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Population of a town in Nepal as obtained from the census report is as follows:

| Year (AD) | 1981 | 1991 | 2001 | 2011 | 2021 |
|---|---|---|---|---|---|
| Population | 15000 | 21000 | 27000 | 34000 | 42000 |

Forecast the population of the town in the year 2040 by geometrical increase method and calculate the design year total water demand assuming per capita demand of 120 lpcd. Take industrial demand as 20% of total demand and losses wastage as 15% of total demand. Neglect demands other than domestic, industrial and losses.

## Short answer

By geometrical increase method (r = 29.36% per decade), population in 2040 is 68,493. Domestic demand plus 20% industrial and 15% losses gives total water demand 12,645 m³/day (146.4 L/s).

## Answer

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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Last census 2021, forecast 2040. What is n?
   A. 1.9 decades
   B. 2.0 decades
   C. 19 decades
   D. 0.19 decades
   Answer: A. (2040-2021)/10 = 1.9 decades. (Key point: n for 2040)

2. How is the average geometric growth rate per decade found?
   A. Arithmetic mean of all increases
   B. (Pn/P0)^(1/4) - 1 using the first and last census
   C. (Pn - P0)/P0
   D. Pn/P0 - 1 divided by 4
   Answer: B. Geometric mean over 4 decades gives r = 29.357%. (Key point: Growth rate)

3. With industrial 20% and losses 15%, domestic is what share of total demand?
   A. 80%
   B. 85%
   C. 35%
   D. 65%
   Answer: D. 100 - 20 - 15 = 65%, so Q = 8,219.2/0.65 = 12,645 m3/day. (Key point: Domestic share)

4. What is the 2040 population?
   A. 68,493
   B. 42,000
   C. 58,000
   D. 75,918
   Answer: A. 42,000 x 1.29357^1.9 = 68,493. (Key point: Result)

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