--- title: "Population 2090 by three methods, 110 lpcd" subject: "Water Supply Engineering" chapter: 3 key: "semester_5/water_supply_engineering:pop-b79" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-b79 site: BCT Engineering --- # Population 2090 by three methods, 110 lpcd Water Supply Engineering · Chapter 3: Quantity of Water - Asked in: 2079 Bhadra, 8 marks - 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-b79 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3 ## Question Determine the population of a city in the year 2090 by (i) Arithmetical increase method, (ii) Geometrical increase method, (iii) Decrease rate growth rate method. The census population of the city is as follows: | Year | 2030 | 2040 | 2050 | 2060 | 2070 | |---|---|---|---|---|---| | Population | 28,000 | 38,000 | 48,000 | 58,000 | 70,000 | Also calculate the total water demand for the city in 2090 using 110 lpcd. ## Short answer Arithmetical method gives 91,000; geometrical method 110,680; decreasing rate of growth method 89,620. Adopting the decreasing rate result, total water demand at 110 lpcd is 9,858 m³/day (114.1 L/s). ## Answer Census interval 10 years (B.S.), last census 2070; for 2090, $n = 2$ decades. **Arithmetical increase method** Average increase per decade: $\bar x = \dfrac{70{,}000-28{,}000}{4} = 10{,}500.0$ $$ P_{2090} = P_n + n\bar x = 70{,}000 + 2\times 10{,}500.0 = 91{,}000 $$ **Geometrical increase method** Average growth rate per decade (geometric mean): $r = \left(\dfrac{70{,}000}{28{,}000}\right)^{1/4}-1 = 25.743\%$ $$ P_{2090} = P_n(1+r)^n = 70{,}000\times(1+0.25743)^{2} = 110{,}680 $$ **Decreasing rate of growth method** | Decade | Increase | Growth rate (%) | Decrease in rate (%) | |---|---|---|---| | 2030-2040 | 10,000 | 35.71 | - | | 2040-2050 | 10,000 | 26.32 | 9.40 | | 2050-2060 | 10,000 | 20.83 | 5.48 | | 2060-2070 | 12,000 | 20.69 | 0.14 | Average decrease in rate $= 5.008\%$ per decade. | Year | Growth rate (%) | Population | |---|---|---| | 2080 | 15.68 | 80,977 | | 2090 | 10.67 | 89,620 | $P_{2090} = 89{,}620$ #### Summary | Method | Population in 2090 | |---|---| | Arithmetical increase | 91,000 | | Geometrical increase | 110,680 | | Decreasing rate of growth | 89,620 | #### Water demand in 2090 The decreasing rate of growth method follows the falling trend of the growth rate and is adopted for design. Per capita demand 110 lpcd. $$ Q = 89{,}620\times110 = 9{,}858{,}187\ \mathrm{l/day} = 9{,}858.2\ \mathrm{m^3/day} $$ **Answer: Population in 2090: arithmetical 91,000, geometrical 110,680, decreasing rate 89,620. Total water demand = 9,858 m$^3$/day (114.1 l/s) on the adopted population.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Average increase per decade 10,500 and n = 2. What is P2090? A. 91,000 B. 80,500 C. 110,680 D. 70,000 Answer: A. 70,000 + 2 x 10,500 = 91,000. (Key point: Arithmetical result) 2. Which expression gives the geometric growth rate? A. (70,000 - 28,000)/28,000 B. (70,000/28,000)^(1/4) - 1 C. (70,000/28,000)/4 D. 70,000/28,000 - 1 Answer: B. The geometric mean over 4 decades gives 25.743%. (Key point: Geometric rate) 3. Why is the decreasing rate method adopted? A. Gives the highest value B. It follows the falling trend of the growth rate C. It is easiest D. The arithmetic method fails Answer: B. Growth rates fall from 35.7% to 20.7%, which this method follows. (Key point: Adopted method) 4. How is total demand found from 89,620 people at 110 lpcd? A. Population x 110 l/day = 9,858 m3/day B. Population / 110 C. Population x 110 x 1.5 D. Population x 11 Answer: A. 89,620 x 110 = 9,858,200 l/day = 9,858 m3/day. 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