Population Growth & Compoun Ch 3 Notes

Complete bilingual study notes for Chapter 3: Growth and Depreciation — every concept explained step by step, with definitions, formulas, and worked examples.

3.0 Review

In Nepal, the census began in 1968 B.S. (1911 A.D.) and since then it has been conducted almost every 10 years. The bar chart shows total population (in lakhs) by census year: 1968: 56, 1977: 56, 1987: 55, 1998: 63, 2009/11: 82, 2018: 94, 2028: 116, 2038: 150, 2048: 185, 2058: 232, 2068: 265, 2078 (preliminary): 292.

a) Population of Nepal in 1968 B.S. $= 56$ lakhs. b) The initial report of the 2078 B.S. census showed the population reached $292$ lakhs. c) Comparing successive census years: population stayed the same from 1968 to 1977 ( $56 \to 56$ ), decreased from 1977 to 1987 ( $56 \to 55$ ), and increased in every census from 1987 onward. d) Percentage increase of 2078 B.S. relative to 2068 B.S.: $\frac{292 - 265}{265} \times 100% \approx 10.19%$ . e) Percentage decrease of 1987 B.S. relative to 1977 B.S.: $\frac{56 - 55}{56} \times 100% \approx 1.79%$ . f) Change in population compared to the previous census (in lakhs): $1977 - 1968 = 0$ ; $1987 - 1977 = - 1$ ; $1998 - 1987 = 8$ ; $2009/11 - 1998 = 19$ ; $2018 - 2009/11 = 12$ ; $2028 - 2018 = 22$ ; $2038 - 2028 = 34$ ; $2048 - 2038 = 35$ ; $2058 - 2048 = 47$ ; $2068 - 2058 = 33$ ; $2078 - 2068 = 27$ . g) Other everyday examples of such increment include bank savings growing with interest, prices rising with inflation, the height of a growing plant, and the number of users of an app or service over time.

3.1.1 Growth

Activity 1: According to the census 2078, the population of a municipality was $25, 000$ . If the population increases by $2%$ every year: a) The population in 2079 B.S. $= 25, 000 + 2%$ of $25, 000 = 25, 000 + 25, 000 \times \frac{2}{100} = 25, 500$ . b) The population in 2080 B.S. $= 25, 500 + 2%$ of $25, 500 = 25, 500 + 25, 500 \times \frac{2}{100} = 26, 010$ . c) Yes — this population growth can be computed the same way as compound interest: with $P = 25, 000$ , growth rate $R = 2%$ , time $T = 2$ years, $P_{T} = P (1 + \frac{R}{100})^{T} = 25000 (1 + \frac{2}{100})^{2} = 25000 (\frac{51}{50})^{2} = 26, 010$ , matching the year-by-year calculation.

If external setup does not affect it, the population of a place increases at a certain rate. This is called population growth rate, and the increased population is called growth population. Problems related to growth population can be solved by comparing with the compound interest system. Population after $T$ years: $P_{T} = P (1 + \frac{R}{100})^{T}$ . Increased population after $T$ years $= P [(1 + \frac{R}{100})^{T} - 1]$ . Here, $P$ = population of the initial year, $R$ = population growth rate, $T$ = time. (Compare: in compound interest, $P$ denotes principal, $R$ denotes rate, $T$ denotes time.) Population growth is affected by migration and death. If the population growth rate differs year by year, then the population after $T$ years: $P_{T} = P (1 + \frac{R _{1}}{100}) (1 + \frac{R _{2}}{100}) (1 + \frac{R _{3}}{100}) \dots (1 + \frac{R _{T}}{100})$ , where $R_{1}, R_{2}, R_{3}, \dots, R_{T}$ respectively represent the population growth rates of the first, second, third, ..., $T^{th}$ year.

The population of a city in 2078 B.S. was 50,000. If the annual population growth rate was 2%, what will be the population in 2080 B.S.? Calculate it.

Solution: Initial population of the city $P = 50, 000$ . Population growth rate $R = 2%$ p.a. Time $T = 2080 - 2078 = 2$ years. $P_{T} = P (1 + \frac{R}{100})^{T} = 50, 000 (1 + \frac{2}{100})^{2} = 50, 000 (\frac{102}{100})^{2} = 50, 000 \times 1.0404 = 52, 020$ . Thus, the population in 2080 B.S. will be $52, 020$ .

Two years ago, the price of a sack of 25 kg jeera masino rice was Rs. 1,300. If the inflation rate was 5% p.a., by how much is the price increased now? Find it.

Solution: Initial price of jeera masino rice $P = Rs. 1, 300$ . Increased rate of price $R = 5%$ p.a. Time $T = 2$ years. Increased price $= P [(1 + \frac{R}{100})^{T} - 1] = 1, 300 [(1 + \frac{5}{100})^{2} - 1] = 1, 300 [(\frac{105}{100})^{2} - 1] = 1, 300 \times 0.1025 = Rs. 133.25$ . Therefore, the increased price of jeera masino rice is Rs. $133.25$ .

If the price of a photocopy machine increases from Rs. 1,00,000 to Rs. 1,21,000 in 2 years, find the rate of yearly increment.

Solution: Initial price of the photocopy machine $P = Rs. 1, 00, 000$ . Time $T = 2$ years. Present price $P_{T} = Rs. 1, 21, 000$ . $P_{T} = P (1 + \frac{R}{100})^{T}$ : $1, 21, 000 = 1, 00, 000 (1 + \frac{R}{100})^{2}$ , so $\frac{1 , 21 , 000}{1 , 00 , 000} = (1 + \frac{R}{100})^{2}$ , giving $(\frac{11}{10})^{2} = (1 + \frac{R}{100})^{2}$ , so $\frac{11}{10} = 1 + \frac{R}{100}$ , giving $\frac{11}{10} - 1 = \frac{R}{100}$ , i.e. $\frac{1}{10} = \frac{R}{100}$ , so $R = \frac{1}{10} \times 100 = 10%$ .

The number of students of a basic school is 500 now. In how many years will the number be 720 if the number of students increases by 20% p.a. every year?

Solution: Initial number of students $P = 500$ . Increase rate $R = 20%$ p.a. Time $T = T$ years. Number of students after $T$ years $P_{T} = 720$ . $P_{T} = P (1 + \frac{R}{100})^{T}$ : $720 = 500 (1 + \frac{20}{100})^{T}$ , so $\frac{720}{500} = (1 + \frac{20}{100})^{T}$ , giving $\frac{720}{500} = (\frac{120}{100})^{T}$ , i.e. $\frac{36}{25} = (\frac{6}{5})^{T}$ , so $(\frac{6}{5})^{2} = (\frac{6}{5})^{T}$ . Since the bases are the same, the exponents must be equal: $T = 2$ . Therefore, after 2 years, the number of students will be 720.

The price of a plot of land is Rs. 15,97,200 per aana. If the rate of increase in price is 10% p.a., what was the price of the land per ropani before 3 years? Find it. (1 ropani = 16 aana)

Solution: Present price of the plot of land per aana $P_{T} = Rs. 15, 97, 200$ . Time $T = 3$ years. Rate of increase in price $R = 10%$ p.a. $P_{T} = P (1 + \frac{R}{100})^{T}$ : $15, 97, 200 = P (1 + \frac{10}{100})^{3} = P (\frac{110}{100})^{3} = P \times 1.331$ , so $P = \frac{15 , 97 , 200}{1.331} = Rs. 12, 00, 000$ . Thus, the price of land per aana before 3 years $P = Rs. 12, 00, 000$ . Since $16$ aana $= 1$ ropani, the price of land per ropani $= 16 \times Rs. 12, 00, 000 = Rs. 1, 92, 00, 000$ .

The number of SEE appeared students from a district in 2076 B.S. was 50,000. If in the coming 3 years, the number increased by 5%, 6% and 4% respectively, find how many students will appear in the SEE in 2079 B.S.

Solution: Number of SEE-appeared students 3 years before $P = 50, 000$ . Time $T = 3$ years. Rate of increase in the first year $R_{1} = 5%$ p.a., second year $R_{2} = 6%$ p.a., third year $R_{3} = 4%$ p.a. $P_{3} = P (1 + \frac{R _{1}}{100}) (1 + \frac{R _{2}}{100}) (1 + \frac{R _{3}}{100}) = 50, 000 (1 + \frac{5}{100}) (1 + \frac{6}{100}) (1 + \frac{4}{100}) = 50, 000 (\frac{105}{100}) (\frac{106}{100}) (\frac{104}{100}) = 57, 876$ . Therefore, $57, 876$ students will appear in the SEE in 2079 B.S.

The population of a municipality in 2078 B.S. was 1,00,000. In 2079 B.S., 8000 migrated there from other places and 500 died due to several circumstances. If the population increase rate is 2% p.a. every year, what will be the population in 2081 B.S.? Find it.

Solution: Case I: Population in 2078 B.S. $P = 1, 00, 000$ . Population increase rate $R = 2%$ p.a. Time $T = 1$ year (2078 to 2079). In-migration population $M_{in} = 8, 000$ . Death population $D = 500$ . Since the population increases by $2%$ every year, $P_{T} = P (1 + \frac{R}{100})^{T} = 1, 00, 000 (1 + \frac{2}{100})^{1} = 1, 00, 000 (\frac{102}{100})^{1} = 1, 02, 000$ . The final population in 2079 B.S., accounting for migration in and deaths, $= 1, 02, 000 + 8, 000 - 500 = 1, 09, 500$ . Case II: Population in 2079 B.S. $P = 1, 09, 500$ . Time $T = 2$ years (2079 to 2081). Population growth rate $R = 2%$ p.a. $P_{T} = P (1 + \frac{R}{100})^{T} = 1, 09, 500 (1 + \frac{2}{100})^{2} = 1, 09, 500 (\frac{102}{100})^{2} = 1, 13, 924$ . Therefore, the population of the municipality in 2081 B.S. will be $1, 13, 924$ .

3.2 Depreciation

Activity 1: a) A farmer sells a tractor for Rs. $3, 60, 000$ which was purchased for Rs. $5, 50, 000$ before 2 years — the price has decreased because the tractor has been used and has lost some of its efficiency over time. b) A cupboard can be bought for Rs. $15, 000$ less than the price of a new one from a second-hand shop, because a used cupboard is worth less than a new one. c) A photocopy machine bought for Rs. $2, 10, 000$ some years ago now costs only Rs. $1, 00, 000$ , because after using machinery for some time its price depreciates. From these situations: the price of the tractor diminishes by Rs. $1, 90, 000$ in (a); in (b), second-hand goods are cheaper than new ones; in (c), after using machinery goods, the price depreciates.

A product is prepared for a certain period. Its efficiency decreases in accordance with its increased use. Thus, after using some machinery items for a certain period of time, their cost diminishes at a certain rate. This is called depreciation. The price that diminishes at a certain rate during a certain period of time is called compound depreciation.

Activity 2 (general derivation): Let the initial price of a good $= V_{0}$ , yearly depreciation rate $= R %$ , time duration $= T$ , and price of the good after $T$ years $= V_{T}$ . Price after 1 year: $V_{1} = V_{0} - V_{0}$ of $R % = V_{0} - V_{0} \times \frac{R}{100} = V_{0} (1 - \frac{R}{100})$ . Price after 2 years: $V_{2} = V_{0} (1 - \frac{R}{100}) - R %$ of $V_{0} (1 - \frac{R}{100}) = V_{0} (1 - \frac{R}{100}) - V_{0} (1 - \frac{R}{100}) \times \frac{R}{100} = V_{0} (1 - \frac{R}{100}) (1 - \frac{R}{100}) = V_{0} (1 - \frac{R}{100})^{2}$ . Similarly, price after 3 years $V_{3} = V_{0} (1 - \frac{R}{100})^{3}$ . In general: a) Price after $T$ years, $V_{T} = V_{0} (1 - \frac{R}{100})^{T}$ . b) The depreciated price $= V_{0} - V_{T} = V_{0} - V_{0} (1 - \frac{R}{100})^{T} = V_{0} [1 - (1 - \frac{R}{100})^{T}]$ . c) The price of a good need not depreciate at the same rate every year. If $R_{1} %, R_{2} %, R_{3} %, \dots, R_{T} %$ are respectively the depreciating rates for the first, second, third, ..., $T^{th}$ year, then the price after $T$ years, $V_{T} = V_{0} (1 - \frac{R _{1}}{100}) (1 - \frac{R _{2}}{100}) (1 - \frac{R _{3}}{100}) \dots (1 - \frac{R _{T}}{100})$ .

In short: growth increases a quantity, while depreciation decreases it — both follow the same compound-style formula, with $+$ for growth and $-$ for depreciation.

3 years ago a book was published costing Rs. 200, and is being sold in an exhibition at 5% rate of yearly depreciation. What is the price of the book this year?

Solution: Initial price of the book $V_{0} = Rs. 200$ . Rate of depreciation $R = 5%$ p.a. Time $T = 3$ years. $V_{T} = V_{0} (1 - \frac{R}{100})^{T} = 200 (1 - \frac{5}{100})^{3} = 200 (\frac{95}{100})^{3} = 200 \times 0.857375 = Rs. 171.47$ . Therefore, the book costs Rs. $171.47$ this year.

Seema admitted in BBA. She purchased a computer for Rs. 40,000 for her study. After using it for 2 years, if the price of the computer depreciates by Rs. 7,600, then find the rate of depreciation.

Solution: Initial price of the computer $V_{0} = Rs. 40, 000$ . Time $T = 2$ years. Depreciated price (amount lost) $= Rs. 7, 600$ . So the price after 2 years $V_{T} = V_{0} - 7, 600 = 40, 000 - 7, 600 = Rs. 32, 400$ . $V_{T} = V_{0} (1 - \frac{R}{100})^{T}$ : $32, 400 = 40, 000 (1 - \frac{R}{100})^{2}$ , so $\frac{32 , 400}{40 , 000} = (1 - \frac{R}{100})^{2}$ , giving $(\frac{18}{20})^{2} = (1 - \frac{R}{100})^{2}$ , so $\frac{18}{20} = 1 - \frac{R}{100}$ , giving $\frac{R}{100} = 1 - \frac{18}{20} = \frac{2}{20}$ , so $R = \frac{2}{20} \times 100 = 10%$ . Therefore, the rate of depreciation of the computer is $10%$ p.a.

The price of a house is Rs. 20,00,000 now. If its price decreases by 10% every year, then in how many years will its price be Rs. 14,58,000?

Solution: Present price of the house $V_{0} = Rs. 20, 00, 000$ . Rate of depreciation $R = 10%$ . Price after $T$ years $V_{T} = Rs. 14, 58, 000$ . $V_{T} = V_{0} (1 - \frac{R}{100})^{T}$ : $14, 58, 000 = 20, 00, 000 (1 - \frac{10}{100})^{T}$ , so $\frac{14 , 58 , 000}{20 , 00 , 000} = (\frac{90}{100})^{T}$ , giving $0.729 = (0.9)^{T}$ . Since $(0.9)^{3} = 0.729$ , we get $(0.9)^{3} = (0.9)^{T}$ , so $T = 3$ years. Thus, 3 years later, the price of the house will be Rs. $14, 58, 000$ .

A factory established with the capital of Rs. 4 crore gained Rs. 75 lakhs in 3 years but its cost depreciated by 2.5% p.a. Then the company was sold after 3 years. Now, calculate whether the factory is in profit or loss.

Solution: Initial price of investment $V_{0} = Rs. 4, 00, 00, 000$ . Rate of depreciation $R = 2.5%$ . Time $T = 3$ years. Price after 3 years $V_{3} = V_{0} (1 - \frac{R}{100})^{T} = 4, 00, 00, 000 (1 - \frac{2.5}{100})^{3} = 4, 00, 00, 000 (\frac{97.5}{100})^{3} = Rs. 3, 70, 74, 375$ . The profit gained by the factory (from operations) after 3 years $= Rs. 75, 00, 000$ . The total amount obtained from the factory $= Rs. 3, 70, 74, 375 + Rs. 75, 00, 000 = Rs. 4, 45, 74, 375$ . Total investment $= Rs. 4, 00, 00, 000$ . Thus, the profit while selling it after 3 years $= Rs. 4, 45, 74, 375 - Rs. 4, 00, 00, 000 = Rs. 45, 74, 375$ .

When the price of a share of a finance company depreciates continuously for 2 years by 10% p.a. and it is sold for Rs. 25,920, then how many shares of Rs. 100 were sold? Find it.

Solution: Present value of the share of the finance company $V_{T} = Rs. 25, 920$ . Depreciation rate $R = 10%$ . Time $T = 2$ years. $V_{T} = V_{0} (1 - \frac{R}{100})^{T}$ : $25, 920 = V_{0} (1 - \frac{10}{100})^{2} = V_{0} (\frac{90}{100})^{2} = V_{0} \times \frac{81}{100}$ , so $V_{0} = \frac{25 , 920 \times 100}{81} = Rs. 32, 000$ . Hence, the value of the share before 2 years $V_{0} = Rs. 32, 000$ . Two years ago, the price of a share was Rs. $100$ , so the total number of shares $= \frac{32 , 000}{100} = 320$ . Therefore, two years ago, the finance company sold $320$ shares of Rs. $100$ .