--- title: "Population 2011 and 2016 by three methods" subject: "Water Supply Engineering" chapter: 3 key: "semester_5/water_supply_engineering:pop-j66" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-j66 site: BCT Engineering --- # Population 2011 and 2016 by three methods Water Supply Engineering · Chapter 3: Quantity of Water - Asked in: 2066 Jestha (old course) - Video: [population forecasting, Incremental increase method with example, water supply engineering numerical](https://www.youtube.com/watch?v=oAKAJPr_-Ug) — Civil Engineering Exam - 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-j66 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3 ## Question Population of a town as obtained from the census report is as follows: | Year A.D. | 1951 | 1961 | 1971 | 1981 | 1991 | 2001 | |---|---|---|---|---|---|---| | Population | 18000 | 27000 | 38000 | 51000 | 66000 | 83000 | Determine the population of the town in the year 2011 and 2016 by (i) Arithmetical increase method, (ii) Geometrical increase method and (iii) Decreased rate of growth method. ## Short answer For 2011: arithmetic 96,000, geometric 112,678, decreased rate 99,348. For 2016: arithmetic 102,500, geometric 131,286, decreased rate 106,122. ## Answer Census interval 10 years; last census 2001. For 2011, $n = 1$ decade; for 2016, $n = 1.5$ decades. ### Population in 2011 (n = 1) **Arithmetical increase method** Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$ $$ P_{2011} = P_n + n\bar x = 83{,}000 + 1\times 13{,}000.0 = 96{,}000 $$ **Geometrical increase method** Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$ $$ P_{2011} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1} = 112{,}678 $$ **Decreasing rate of growth method** | Decade | Increase | Growth rate (%) | Decrease in rate (%) | |---|---|---|---| | 1951-1961 | 9,000 | 50.00 | - | | 1961-1971 | 11,000 | 40.74 | 9.26 | | 1971-1981 | 13,000 | 34.21 | 6.53 | | 1981-1991 | 15,000 | 29.41 | 4.80 | | 1991-2001 | 17,000 | 25.76 | 3.65 | Average decrease in rate $= 6.061\%$ per decade. | Year | Growth rate (%) | Population | |---|---|---| | 2011 | 19.70 | 99,348 | $P_{2011} = 99{,}348$ ### Population in 2016 (n = 1.5) **Arithmetical increase method** Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$ $$ P_{2016} = P_n + n\bar x = 83{,}000 + 1.5\times 13{,}000.0 = 102{,}500 $$ **Geometrical increase method** Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$ $$ P_{2016} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1.5} = 131{,}286 $$ **Decreasing rate of growth method** | Decade | Increase | Growth rate (%) | Decrease in rate (%) | |---|---|---|---| | 1951-1961 | 9,000 | 50.00 | - | | 1961-1971 | 11,000 | 40.74 | 9.26 | | 1971-1981 | 13,000 | 34.21 | 6.53 | | 1981-1991 | 15,000 | 29.41 | 4.80 | | 1991-2001 | 17,000 | 25.76 | 3.65 | Average decrease in rate $= 6.061\%$ per decade. | Year | Growth rate (%) | Population | |---|---|---| | 2011 | 19.70 | 99,348 | | 2021 | 13.64 | 112,896 | Population at 2016 is obtained by linear interpolation between the decade values: $P_{2016} = 106{,}122$. #### Summary | Method | 2011 | 2016 | |---|---|---| | (i) Arithmetical increase | 96,000 | 102,500 | | (ii) Geometrical increase | 112,678 | 131,286 | | (iii) Decreased rate of growth | 99,348 | 106,122 | **Answer: 2011: arithmetical 96,000, geometrical 112,678, decreased rate 99,348. 2016: arithmetical 102,500, geometrical 131,286, decreased rate 106,122.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 2016 with last census 2001, n is A. 15 decades B. 1.0 decade C. 1.5 decades D. 0.5 decades Answer: C. (2016-2001)/10 = 1.5. (Key point: Value of n) 2. The geometric growth rate is found as A. (83,000/18,000)^(1/5) - 1 = 35.757% B. (83,000-18,000)/18,000 C. (83,000/18,000)^(1/6) - 1 D. (83,000/18,000)^5 - 1 Answer: A. Five decade intervals between 1951 and 2001. (Key point: Geometric rate) 3. How is the 2016 population by decreasing rate method found? A. Using n = 1.5 in the arithmetic formula B. Linear interpolation between 2011 and 2021 values C. Taking the 2021 value D. Applying 19.70% for 1.5 decades Answer: B. 2016 lies between the 2011 (99,348) and 2021 (112,896) values: 106,122. (Key point: 2016 decreasing rate) ## Close variants set in other exams - [Population 2021 and 2026 by three methods](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-c68) — Water Supply Engineering, chapter 3, asked 2× ## Related in Water Supply Engineering - [Population 2021 and 2026, arithmetic and decreasing rate](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-as69) — Water Supply Engineering, chapter 3 - [Population forecast 2021 census, 2024 base year](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-b81) — Water Supply Engineering, chapter 3 - [Water demand 2030, fully plumbed town](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-a74) — Water Supply Engineering, chapter 3 - [Population forecast to 2040, geometric method](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-bk81) — Water Supply Engineering, chapter 3 - [Population 2088 by three methods, design year demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-k72) — Water Supply Engineering, chapter 3 - [Population 2021 and 2028, incremental method](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-b80) — Water Supply Engineering, chapter 3