Growth and Depreciation: Grade 10 Math

Nepal counted 2,64,94,504 people at the 2068 census and 2,91,64,578 at the 2078 one: 10.08% more in ten years. So did the population grow by about 1% a year? Not quite, and the reason is the whole of this unit.

Things also lose value. Nepal’s tax law lets a business count its cars, buses and minibuses as losing 20% of their value every year. Five years at 20% sounds like 100%: is a five-year-old car worth nothing? It isn’t, and by the end of this unit you will be able to say exactly what it is worth, and what rate really turns 10.08% in ten years into a yearly figure. One idea does both: repeating a percentage means multiplying, again and again.
Figures: National Statistics Office, 2021 census (Onlinekhabar, March 2023); Income Tax Act 2058, Schedule 2.

For the SEE

The question: Arithmetic is 13 marks in three questions: compound interest, growth and depreciation, and currency. In CDC’s model paper the growth-and-depreciation question is 4 marks: a. a formula (1 mark), b. a value after some years (1), c. a twist such as an unknown rate, two things at once or migration (2, higher ability).
The formulas it needs:

Growth: $P_T = P\left(1 + \frac{R}{100}\right)^T$ , increase $= P_T - P$
Depreciation: $V_T = V_0\left(1 - \frac{R}{100}\right)^T$ , depreciated amount $= V_0 - V_T$
A different rate each year: one factor per year, P(1 + R₁/100)(1 + R₂/100)… or V₀(1 − R₁/100)(1 − R₂/100)…
Migration or deaths: grow to the year they happen, add or take them away, then grow on.

How the SEE asks it has a full question marked part by part; Exam practice has every exercise question to mark yourself.

3.0 Review: Nepal’s censuses

Nepal’s first census was taken in 1968 BS (1911 AD), and there has been one about every ten years since. Figure 3.1 shows what each counted.

Figure 3.1 Nepal’s population at its twelve censuses, in lakhs (1 lakh = 1,00,000), years in BS. The two red bars are the censuses whose count was lower than the one before. Final counts of the National Statistics Office.

Look at (f) again. From 2038 to 2048 the population grew by about the same number of people as in the ten years before, 34.7 lakh, yet the percentage fell, from 30% to 23%, because the same number was added to a bigger population. In this unit a growth rate is always a percentage of what there already is.

3.1 Growth

Activity 1: a municipality grows

a. 2079: 25000 + 2% of 25000 = 25000 + 500 = 25500.

b. 2080: the 2% is now of 25,500, not of 25,000: 25500 + 510 = 26010.

c. Yes, exactly. The second year’s growth (510) is bigger than the first (500), because the 500 people added in the first year have children too, just as the first year’s interest earns interest. With P = 25000, R = 2 and T = 2, the compound-amount formula gives the same answer:

$P_2 = 25000\left(1 + \frac{2}{100}\right)^2 = 25000 \times \left(\frac{51}{50}\right)^2 = 26010$

Figure 3.2 Growth and depreciation as multiplication. Adding 2% is multiplying by 1.02; taking 5% off is multiplying by 0.95. Every year is one more multiplication.

Adding R% is multiplying by (1 + R/100). So a population that grows by R% every year is multiplied by the same number every year, and after T years by that number T times (Figure 3.2).

When a population (or a price, or anything) increases by the same percentage of itself every year, that percentage is its growth rate (questions often write a yearly rate as “2% p.a.”, per annum, Latin for “each year”), and what it has grown to is the grown population (or grown price). With P the population at the start, R% the yearly growth rate and T the number of years:

$\text{Population after } T \text{ years: } P_T = P\left(1 + \frac{R}{100}\right)^T$

$\text{Increase in } T \text{ years: } P_T - P = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]$

These are Unit 2’s compound amount and compound interest with new names: P is the population at the start instead of the principal, R the growth rate instead of the interest rate, T the time in years in both.

If the rate is different each year, R₁% in the first, R₂% in the second and so on to $R_T\%$ , each year multiplies by its own factor:

$P_T = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)\cdots\left(1 + \frac{R_T}{100}\right)$

A real population doesn’t follow one rate for ever. Births and deaths set its natural growth, and people moving in (in-migration) and moving out (out-migration) change it more. In a question, people who move in, move away or die in a given year are added or taken away at that point, and the population grows on from the new number.

A story from maths

Malthus and the race between people and food

In 1798 an English clergyman, Thomas Malthus, published An Essay on the Principle of Population. He argued that a population left alone grows by multiplying, like this unit’s formula (doubling about every 25 years), while food can only grow by adding, the same amount each time. Multiplying always wins in the end, so, he warned, hunger would keep populations in check. He was wrong about food: farming has grown far faster than he imagined. But the difference between adding and multiplying is the one that matters, and it gave Charles Darwin a key idea. Reading Malthus in 1838, Darwin saw that if more are born than can survive, “favourable variations would tend to be preserved, and unfavourable ones to be destroyed”: natural selection.

A town of 40,000 grows by 5% a year. Which gives its population after 3 years?

Pick an answer to see why.

Back to the census

A steady 0.965% a year, not 1%, takes Nepal from its 2068 count to its 2078 count.

Ten years of 1.008% (10.08% ÷ 10) would give $26494504 \times 1.01008^{10} \approx 29289605$ , or 2,92,89,605: about 1.25 lakh too many, because each year’s 1% is of a bigger population than the last. Find the rate yourself:

Move the rate until the last dot lands on the 2078 census. Is it more or less than 1%?

Growth rate: 1.20% a year

Grown to 2078: 298.5 lakh
Census 2078: 291.6 lakh
Ten years’ growth: 12.67%

Too slow: the last dot falls short of the 2078 count. Raise the rate.

Too fast: the last dot overshoots the 2078 count. Lower the rate.

That’s it: a steady 1.20% a year lands within 687 thousand people of the 2078 count. That is less than 1.008% a year (10.08% ÷ 10), because each year grows a bigger population.

Answer

The National Statistics Office gives the rate as 0.92% a year. It is lower than 0.965% for two reasons. The 2078 count was held five months later than planned, because of COVID-19, so a little more than ten years separate the two censuses. And the office’s formula assumes a population that grows a little every moment, not in one jump a year, which reaches the same total at a slightly lower rate. Both ways of counting agree on the big fact: the growth was the slowest of any ten years since the 1987 census.

Worked examples: growth

In each one, write down P, R and T first, decide whether the question wants the population, the increase, the rate or the time, then put the numbers in.

The population, and the increase

Finding the rate

Finding the time

Working backwards

A different rate each year

People moving in, and deaths

A Royal Bengal tiger in tall grass in Chitwan National Park, looking at the camera — **Figure 3.3** A Royal Bengal tiger in Chitwan National Park. Photo: Supradai, CC BY 4.0

Exercise 3.1

Question 1

A place has population P now, and $P_T$ after T years, growing at R% a year. Write the formula for $P_T$ .

Answer

Chitwan National Park had x tigers in 2079 BS, and their number grows by R% a year. How many tigers will it have after N years?

Answer

The number of Nepalis going abroad for work grows by R₁%, R₂% and R₃% in the first, second and third years. If P went in the year before, write the formula for the number after 3 years.

Answer

Question 2

If a country’s death rate is lower than its birth rate, does its population increase or decrease?

Answer

A year ago a 25 kg sack of rice cost Rs 1,500. Its price has risen by 10% a year. What does the sack cost now?

Answer

Question 3

At the 2021 census a city had 5,18,452 people. If it grows by 4.5% a year, what will its population be after 3 years?

Answer

A landlord agrees with a businessman to raise the rent by 15% a year. If a shutter’s rent is Rs 10,000 now, what will it be after 3 years?

Answer

The bacteria that set curd grow by 10% an hour. If there are $4 \times 10^{11}$ at 6 a.m., how many are there 2 hours later?

Answer

Question 4

Two years ago a rural municipality had 28,500 people. If it grows by 2% a year, by how much has its population increased in 2 years?

Answer

Three years ago a school’s Grade 10 monthly fee was Rs 6,500. Its rules raise fees by 10% every year. By how much has the fee risen in 3 years?

Answer

A plot of land costs Rs 6,00,000 now. If its price rises by 10% a year, by how much will it rise in 2 years?

Answer

Question 5

A university has 21,632 students now; 2 years ago it had 20,000. What is the yearly rate of increase?

Answer

A rural municipality had 40,000 people at the end of 2018 AD and 44,100 at the end of 2020 AD. Find its yearly growth rate.

Answer

Three years ago a liter of cooking oil cost Rs 125; now it costs Rs 216. What is the yearly rate of inflation?

Answer

Question 6

A village has 13,310 people, and its population grows by 10% a year. How many years ago did it have 10,000?

Answer

At the start of Baisakh a papaya plant was 4 m tall. If it grows by 4% a month, after how many months will it be 4.3264 m tall?

Answer

Question 7

Land now costs Rs 19,96,500 an aana, and its price has risen by 10% a year. What was its price per ropani 3 years ago? (1 ropani = 16 aana)

Answer

Land now costs Rs 2,66,200 an aana, and its price has risen by 10% a year. What was its price per ropani 3 years ago?

Answer

Question 8

A patient’s urine sample had $1 \times 10^5$ bacteria at 6 a.m. Tests at 7, 8 and 9 a.m. showed they grew by 3%, 4% and 5% in those hours. How many bacteria were there at 9 a.m.?

Answer

Three years ago a city had 1,50,000 people. Its population then grew by 2% in the first year, 4% in the second and 5% in the third. What is it now?

Answer

Question 9

An insurance company started 3 years ago with 1,000 trained agents. From the start, its rule has been that every group of 5 agents adds 1 new agent each year. How many agents does it have now?

Answer

A financial institution started at the beginning of 2078 with 200 marketers. Its rule is that every group of 5 marketers adds 1 new marketer each year. How many marketers will it have at the end of 2079?

Answer

Question 10

10.

A rural municipality’s population grows by 10% a year. At the end of 2 years, 5,800 people moved in, and the population then was 30,000. What was the population at the start?

Answer

Question 11

11.

Over two years the US dollar rose against the Nepali rupee by 5% a year, and now $\$1 =$ Rs 120. How many rupees was $\$1$ worth 2 years ago?

Answer

Question 12

12.

Two years ago a village had 31,250 people, and it grows by 6% a year. One year ago, 625 people moved away. What is its population now?

Answer

Question 13

13.

Three years ago a district had 3,75,000 people, growing by 2% a year. At the end of the second year 1,480 people moved in and 2,750 died in a natural disaster. What is its population now?

Answer

Question 14

At the beginning of 2075 BS a metropolitan city had 5,00,000 people; at the end of 2077 BS it had 6,65,500.

What was its yearly growth rate?

Answer

If it keeps growing the same way, what will its population be at the end of 2079?

Answer

3.2 Depreciation

Activity 1: why things lose value

Most things people use, machines, vehicles, phones, furniture, wear out and go out of date, so their value falls as time passes. This fall in value is depreciation. When the value falls by the same percentage of what is left every year, it is compound depreciation, and that percentage is the rate of depreciation.

Activity 2: the formula

Let a thing be worth V₀ at the start, and lose R% of its value every year. Let $V_T$ be its value after T years.

After 1 year it has lost R% of V₀:

$V_1 = V_0 - V_0 \times \frac{R}{100} = V_0\left(1 - \frac{R}{100}\right)$

Taking R% off is multiplying by (1 − R/100). In the second year it loses R% of what is left, V₁, not of V₀:

$\begin{aligned} V_2 &= V_1 - V_1 \times \frac{R}{100} \\ &= V_1\left(1 - \frac{R}{100}\right) \\ &= V_0\left(1 - \frac{R}{100}\right)\left(1 - \frac{R}{100}\right) = V_0\left(1 - \frac{R}{100}\right)^2 \end{aligned}$

The same way, $V_3 = V_0\left(1 - \frac{R}{100}\right)^3$ , and after T years:

$V_T = V_0\left(1 - \frac{R}{100}\right)^T$

The value lost in that time, the depreciated amount, is

$V_0 - V_T = V_0\left[1 - \left(1 - \frac{R}{100}\right)^T\right]$

Does a thing lose the same percentage every year? Not always: a rate may change from year to year. With R₁%, R₂%, …, $R_T\%$ :

$V_T = V_0\left(1 - \frac{R_1}{100}\right)\left(1 - \frac{R_2}{100}\right)\left(1 - \frac{R_3}{100}\right)\cdots\left(1 - \frac{R_T}{100}\right)$

It is growth’s formula with one sign changed. Growth multiplies by a number more than 1 every year, so the value increases; depreciation multiplies by a number less than 1, so it decreases.

Rs 1,00,000, ten years. Try Depreciation at 20%: is each year’s loss bigger or smaller than the last? Then switch to Growth at the same rate.

Rate: 20% a year Year: 3

Value after year 3: Rs 51,200
Lost in year 3: Rs 12,800
Multiplied by: ×0.80 a year

Each year adds 20% of a bigger value, so the green pieces get bigger every year and the value runs away upwards.

Each year takes 20% of a smaller value, so the red pieces get smaller every year, and the value never reaches zero.

Back to the car

A car losing 20% a year is still worth about a third of its price after 5 years, not nothing.

Each year takes 20% of what is left, so it keeps 80%: $0.8^5 \approx 0.328$ . A Rs 40,00,000 car is worth $4000000 \times 0.8^5 = 1310720$ , Rs 13,10,720. “5 × 20% = 100%” would be true only if every year took 20% of the first price. And because each year keeps 80% of something more than zero, the value never reaches zero, however many years pass. Half of it is gone a little after 3 years ( $0.8^3 = 0.512$ ).

A laptop worth Rs 80,000 depreciates by 10% a year. Which is its value after 2 years?

Pick an answer to see why.

A machine depreciates by 20% a year. In which year does it lose the most rupees?

Pick an answer to see why.

Worked examples: depreciation

The value after some years

Finding the rate

Finding the time

Profit or loss

Working backwards

How the SEE asks it

In the SEE’s Compulsory Mathematics paper, Arithmetic carries 13 marks in three questions (compound interest; growth and depreciation; currency and exchange), with knowledge, understanding, application and higher-ability parts spread among them (CDC’s specification grid). In CDC’s model paper the growth-and-depreciation question looks like this:

Part	Level	Marks	What it asks
a.	knowledge	1	a formula: grown value, depreciated value or amount
b.	understanding	1	a value after a few years
c.	higher ability	2	a twist: an unknown rate from a total, two things at once, migration

Here is a question in that shape, and a longer one to practice with.

Exercise 3.2

Question 1

A thing costs Rs P at first and depreciates by R% a year. Write the formula for its value after T years.

Answer

In $V_T = V_0\left(1 - \frac{R}{100}\right)^T$ , the value after T years, what does R stand for?

Answer

Question 2

Ram sells a watch that cost Rs 5,000 after 1 year, at 7% depreciation. How much value has it lost?

Answer

After 1 year, at 5% depreciation a year, a motorcycle sells for Rs 57,000. What did it cost?

Answer

Question 3

What will a cupboard costing Rs 16,800 be worth after 2 years at 15% depreciation a year?

Answer

A new 125 cc motorcycle made in India costs Rs 2,50,000 in Nepal. If its value falls by 4% a year, what will it be worth after 3 years of use?

Answer

Question 4

Sameer bought a mobile phone for Rs 30,000. Because of a problem at home he has to sell it after 2 years, at 30% depreciation a year. Find the depreciated amount.

Answer

A maths teacher bought a laptop for Rs 96,000 for online classes. It depreciates by 15% a year. If he sells it after 3 years, how much value has it lost?

Answer

Question 5

Three years ago, 4 ropani of land in the hills was bought for Rs 12,50,000. Now it can be sold for only Rs 1,60,000 a ropani. By what percentage a year has its price depreciated?

Answer

A man bought a watch for Rs 5,000 and sold it after 3 years for only Rs 625. Find the yearly rate of depreciation.

Answer

Question 6

A printing press bought for Rs 4,00,000 some years ago has lost 30% of its value every year and is now worth Rs 1,96,000. How many years ago was it bought?

Answer

A car costs Rs 8,00,000. If its value falls by 10% a year, after how many years will it be worth Rs 5,83,200?

Answer

Question 7

Over 2 years the US dollar lost 5% of its value against the rupee each year, and now $\$1 =$ Rs 125. How many rupees was $\$1$ worth 2 years ago?

Answer

Three years ago a group of young people back from work abroad started a cow farm with Rs 2,80,000. Because of political instability, the farm’s value has fallen by 5% a year. What is it worth now?

Answer

Question 8

A photocopy machine costs Rs 5,00,000 now. It depreciates by 15% in the first year, 10% in the second and 5% in the third. What will it be worth after 3 years?

Answer

A lamination machine was sold for Rs 24,168 after depreciating by 4% in the first year and 5% in the second. What did it cost 2 years ago?

Answer

Question 9

A businessman bought a heavy truck for Rs 48,00,000 and earned Rs 6,80,000 with it in 2 years. It depreciates by 10% a year, and he sells it after 2 years. Find his profit or loss.

Answer

Question 10

10.

A bus owner bought a bus for Rs 16,00,000 and ran it on the Kathmandu–Baglung route for 3 years, earning only Rs 5,10,000. Its value depreciates by 5% a year, and he sells it after 3 years. Find his profit or loss.

Answer

Question 11

11.

A finance company’s shares on the Nepal Stock Exchange fell in price by 10% a year for 2 years. Your neighbor sold all his shares for Rs 28,350. If he had bought them 2 years ago at Rs 100 a share, how many shares had he bought?

Answer

Question 12

A hydropower company’s share price is falling by 10% a year. Sashi sells all his shares now for Rs 7,10,775.

What were his shares worth 2 years ago?

Answer

He bought them in the IPO, 2 years ago, at Rs 100 a share. How many shares did he buy?

Answer

Question 13

At a fixed yearly rate of depreciation, a thing will be worth Rs 10,240 after 2 years and Rs 8,192 after 3 years.

Find the rate of depreciation.

Answer

What was it worth at first?

Answer

Question 14

At a fixed yearly rate of depreciation, a thing will be worth Rs 5,41,500 after 2 years and Rs 5,14,425 after 3 years.

What is the rate of depreciation?

Answer

What was it worth at first?

Answer

Worksheet to print (PDF, 4 pages) Answer key for teachers

Harder worksheet (PDF, 3 pages) Answer key for teachers

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

Nepal’s 2021 census, the final count of 29,164,578 and the growth rate of 0.92% a year (the opening, 3.0, Back to the census): Onlinekhabar, “29,164,578 is the official size of Nepal’s population”, 24 March 2023, reporting the National Statistics Office.
The populations at every census, 1911–2021 (3.0, Figure 3.1): Wikipedia, “Census in Nepal”, read October 2026, giving the National Statistics Office’s table.
The 2078 census postponed by five months because of COVID-19 (Back to the census): Onlinekhabar, “Deferred national population census from November 11 to 25”, September 2021.
Consumer price inflation of 7.74%, 5.44% and 4.06% (3.1, “Prices grow the same way”): Nepal Rastra Bank, “Current Macroeconomic and Financial Situation, based on annual data of 2024/25”, Table 2, August 2025.
Malthus’s Essay on the Principle of Population, 1798 (3.1): Wikipedia, “An Essay on the Principle of Population”, read October 2026.
Darwin reading Malthus in 1838, and the words quoted from his Autobiography (3.1): University of California Museum of Paleontology, “Thomas Malthus (1766–1834)”, read October 2026.
Nepal’s tigers, 121 in 2010, 355 in 2022 and 429 in 2026, Chitwan’s 128 in 2022 and 145 in 2026 (3.1): Onlinekhabar, “Tiger census 2026: tiger population reaches 429 in Nepal”, 29 July 2026.
The depreciation pools and their rates, 5% (buildings), 25% (computers, furniture, office equipment), 20% (automobiles, buses and minibuses) and 15% (construction machines and other depreciable assets) of the pool’s value each year (the opening, 3.2): Income Tax Act 2058, Schedule 2, “Assessment of depreciation deduction”, read October 2026.

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