Chapter-3 (Growth and Depreciation) Notes

Growth and Depreciation: Notes

Mathematics • Grade 10 • Complete notes, formulas and solved examples

In this post

1. 3.0 Review: percentage change
2. 3.1 Growth
3. Solved examples on growth
4. 3.2 Depreciation
5. Solved examples on depreciation

3.0 Review: Percentage Change

Nepal’s census has been held almost every 10 years since 1968 B.S. Comparing two census figures tells us how much the population has grown or fallen.

• Change = new value − old value
• Percentage change = (change ÷ old value) × 100% (always measured against the earlier figure)
• A positive change is an increase (growth); a negative change is a decrease.

Everyday examples of this kind of growth: population, number of students, price of goods (inflation), bank deposits, number of bacteria, number of vehicles.

3.1 Growth

Activity: population growing 2% every year

A municipality had 25,000 people in 2078 B.S., growing 2% a year.

Year	Working	Population
2078	given	25,000
2079	25,000 + 2% of 25,000	25,500
2080	25,500 + 2% of 25,500	26,010

Each year’s increase is calculated on the previous year’s population, exactly like compound interest. So we can use the compound interest formula.

Growth formulas

1. Population after T years: P_T = P(1 + R/100)^T
2. Increase after T years: P_T − P = P[(1 + R/100)^T − 1]
3. If the rate differs each year: P_T = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)…(1 + R_T/100)

Here P = population (or number, or price) in the initial year, R = growth rate per year, T = time in years.

Compound interest	Growth
Principal P	Initial population / number / price P
Rate R% p.a.	Growth rate R% per year (or per hour, per month…)
Compound amount CA	Population after T years, PT
Compound interest CI	Increase in population

Migration and deaths: population growth is also affected by people moving in (in-migration), moving out (out-migration) and deaths. The formula gives only the natural growth. Add in-migrants and subtract people who move out or die at the time they happen, then continue growing from the adjusted figure.

Quick results worth remembering

• To find the rate: calculate P_T ÷ P and write it as a power of (1 + R/100).
• To find the time: write P_T ÷ P as a power of the same base and compare the powers.
• To find the earlier value: P = P_T ÷ (1 + R/100)^T.
• If the values after n and n + 1 years are given, divide the later by the earlier to get (1 + R/100).
• “Every 5 people add 1 more” means growth of 1/5 = 20%.
• 1 ropani = 16 aana.

Solved Examples on Growth

Ex 1 A city had 50,000 people in 2078 B.S. with 2% annual growth. Find the population in 2080 B.S.

T = 2080 − 2078 = 2 years.

P₂ = 50,000(1.02)² = 50,000 × 1.0404

52,020 people

Ex 2 Two years ago a 25 kg sack of rice cost Rs. 1,300. With inflation of 5% p.a., by how much has the price increased?

Increase = 1,300[(1.05)² − 1] = 1,300 × 0.1025

Rs. 133.25

Ex 3 The price of a photocopy machine rose from Rs. 1,00,000 to Rs. 1,21,000 in 2 years. Find the yearly rate of increase.

1,21,000/1,00,000 = (1 + R/100)² ⇒ (11/10)² = (1 + R/100)²

So 1 + R/100 = 11/10 ⇒ R/100 = 1/10 ⇒ R = 10%

Ex 4 A school has 500 students. In how many years will it have 720 if the number rises 20% a year?

720/500 = (1.2)^T ⇒ 36/25 = (6/5)^T ⇒ (6/5)² = (6/5)^T

Same base, so the powers are equal: T = 2 years

Ex 5 Land costs Rs. 15,97,200 per aana now. If prices rise 10% p.a., what was the price per ropani 3 years ago? (1 ropani = 16 aana)

15,97,200 = P(1.1)³ = 1.331P ⇒ P = Rs. 12,00,000 per aana

Per ropani = 16 × 12,00,000 = Rs. 1,92,00,000

Ex 6 50,000 students appeared in SEE in 2076 B.S. If the number rose 5%, 6% and 4% in the next three years, how many will appear in 2079 B.S.?

P₃ = 50,000 × 1.05 × 1.06 × 1.04 = 57,876

57,876 students

Ex 7 A municipality had 1,00,000 people in 2078 B.S. In 2079 B.S., 8,000 people migrated in and 500 died. If the growth rate is 2% p.a., find the population in 2081 B.S.

Case I (2078 → 2079): 1,00,000 × 1.02 = 1,02,000.

Adjust: 1,02,000 + 8,000 − 500 = 1,09,500.

Case II (2079 → 2081, 2 years): 1,09,500 × (1.02)² = 1,09,500 × 1.0404 = 1,13,923.8

≈ 1,13,924 people

Your textbook prints “+ 8,000 + 500”, but the figure it uses (1,09,500) is obtained by subtracting the 500 deaths. Deaths reduce the population.

3.2 Depreciation

A tractor bought for Rs. 5,50,000 sells for Rs. 3,60,000 two years later, and a second-hand cupboard is cheaper than a new one. Goods lose efficiency with use and age, so their price falls. This is depreciation. When the price falls by a fixed percentage each year, on the reduced price of the previous year, it is called compound depreciation.

Deriving the formula

Let the initial price be V₀ and the yearly depreciation rate be R%.

V₁ = V₀ − V₀·R/100 = V₀(1 − R/100) V₂ = V₁(1 − R/100) = V₀(1 − R/100)² V₃ = V₀(1 − R/100)³  ⇒  V_T = V₀(1 − R/100)^T

Depreciation formulas

1. Value after T years: V_T = V₀(1 − R/100)^T
2. Depreciated amount: V₀ − V_T = V₀[1 − (1 − R/100)^T]
3. If the rate differs each year: V_T = V₀(1 − R₁/100)(1 − R₂/100)…(1 − R_T/100)

	Growth	Depreciation
Factor each year	(1 + R/100)	(1 − R/100)
Value after T years	P(1 + R/100)T	V₀(1 − R/100)T
Effect	Value increases	Value decreases
Change	PT − P	V₀ − VT

Growth and depreciation use the same method. The only difference is “+” in growth and “−” in depreciation. For unknown rate or time, divide V_T by V₀ and write the result as a power of (1 − R/100), then compare.

Solved Examples on Depreciation

Ex 1 A book published 3 years ago at Rs. 200 is sold at an exhibition. If it depreciates 5% a year, find its price this year.

V₃ = 200(0.95)³ = 200 × 0.857375 = 171.475

≈ Rs. 171.47

Ex 2 A computer bought for Rs. 40,000 depreciates by Rs. 7,600 in 2 years. Find the yearly rate of depreciation.

V₂ = 40,000 − 7,600 = 32,400.

32,400/40,000 = (1 − R/100)² ⇒ (18/20)² = (1 − R/100)²

So 1 − R/100 = 18/20 ⇒ R/100 = 2/20 ⇒ R = 10%

Ex 3 A house costing Rs. 20,00,000 loses 10% of its price every year. In how many years will it be worth Rs. 14,58,000?

14,58,000/20,00,000 = 0.729 = (0.9)^T = (0.9)³

T = 3 years

Ex 4 A factory set up with Rs. 4 crore earned Rs. 75 lakh in 3 years, while its value depreciated 2.5% a year. It is sold after 3 years. Is there a profit or loss?

V₃ = 4,00,00,000(0.975)³ = Rs. 3,70,74,375

Total received = 3,70,74,375 + 75,00,000 = Rs. 4,45,74,375.

Profit = 4,45,74,375 − 4,00,00,000 = Rs. 45,74,375 profit

Ex 5 A finance company’s share price fell 10% a year for 2 years, and the shares were sold for Rs. 25,920. How many shares of Rs. 100 each were sold?

25,920 = V₀(0.9)² = 0.81V₀ ⇒ V₀ = 25,920 ÷ 0.81 = 32,000

Number of shares = 32,000 ÷ 100 = 320 shares

Growth and Depreciation • Grade 10 Mathematics • Notes