Growth and Decay
Growth and decay are real-world applications of the compound interest formula. When a quantity increases at a fixed rate over time, it is called growth. When it decreases, it is called decay (or depreciation).
Population of a city, value of a machine, growth of bacteria, and increase in prices all follow the compound growth or decay model.
In Class 8 NCERT Maths, growth and decay are studied as extensions of compound interest, using the same formula with slight modifications.
The key difference from simple interest is that in growth/decay, the rate applies not just to the original value but to the accumulated value after each time period. This leads to exponential increase or decrease.
What is Growth and Decay?
Definition:
- Growth occurs when a quantity increases at a fixed percentage rate per time period.
- Decay (Depreciation) occurs when a quantity decreases at a fixed percentage rate per time period.
Examples:
- Growth: Population increase, rise in property prices, bacterial growth, inflation.
- Decay: Depreciation of machines, vehicles, and electronics; radioactive decay; decrease in value of goods.
Growth and Decay Formula
Growth Formula:
A = P(1 + r/100)ⁿ
Decay (Depreciation) Formula:
A = P(1 − r/100)ⁿ
Where:
- A = final value (amount after n periods)
- P = initial value (principal/original value)
- r = rate of growth or decay per period (in %)
- n = number of time periods
Related formulas:
- Growth in value = A − P
- Depreciation in value = P − A
- Growth/Decay factor = (1 ± r/100)
Derivation and Proof
Derivation of the growth formula:
Step 1: Let the initial value be P and the growth rate be r% per year.
Step 2: After 1 year:
- Value = P + P × r/100 = P(1 + r/100)
Step 3: After 2 years:
- Value = P(1 + r/100) + P(1 + r/100) × r/100
- = P(1 + r/100)(1 + r/100)
- = P(1 + r/100)²
Step 4: After n years:
- Value = P(1 + r/100)ⁿ
Derivation of the decay formula:
The same logic applies, but the value decreases each year:
- After 1 year: P(1 − r/100)
- After n years: P(1 − r/100)ⁿ
Types and Properties
Problems on growth and decay can be classified as follows:
1. Population growth:
- Given the present population and growth rate, find the population after n years.
2. Depreciation of machinery/vehicles:
- Given the purchase price and depreciation rate, find the value after n years.
3. Finding the rate:
- Given initial value, final value, and time, find the rate of growth or decay.
4. Finding the time period:
- Given initial value, final value, and rate, find how many years.
5. Past value problems:
- Given the current value, find what the value was n years ago.
6. Changing rates:
- Growth rate changes from year to year. Apply different rates for each year.
Solved Examples
Example 1: Example 1: Population growth
Problem: The population of a town is 50,000. It increases at 4% per annum. Find the population after 2 years.
Solution:
Given:
- P = 50,000, r = 4%, n = 2
Using the formula:
- A = P(1 + r/100)ⁿ
- A = 50,000 × (1 + 4/100)²
- A = 50,000 × (1.04)²
- A = 50,000 × 1.0816
- A = 54,080
Answer: The population after 2 years is 54,080.
Example 2: Example 2: Depreciation of a car
Problem: A car worth ₹6,00,000 depreciates at 10% per year. Find its value after 3 years.
Solution:
- A = P(1 − r/100)ⁿ
- A = 6,00,000 × (1 − 10/100)³
- A = 6,00,000 × (0.9)³
- A = 6,00,000 × 0.729
- A = ₹4,37,400
Answer: The value after 3 years is ₹4,37,400.
Example 3: Example 3: Bacterial growth
Problem: A bacterial colony of 10,000 grows at 20% per hour. Find the number of bacteria after 3 hours.
Solution:
- A = 10,000 × (1 + 20/100)³
- A = 10,000 × (1.2)³
- A = 10,000 × 1.728
- A = 17,280
Answer: After 3 hours, there are 17,280 bacteria.
Example 4: Example 4: Depreciation amount
Problem: A machine bought for ₹2,00,000 depreciates at 15% per year. Find the depreciation in value after 2 years.
Solution:
- A = 2,00,000 × (1 − 15/100)²
- A = 2,00,000 × (0.85)²
- A = 2,00,000 × 0.7225
- A = ₹1,44,500
- Depreciation = P − A = 2,00,000 − 1,44,500 = ₹55,500
Answer: The depreciation is ₹55,500.
Example 5: Example 5: Finding past population
Problem: The present population of a village is 21,600. It has been growing at 20% per year. What was the population 2 years ago?
Solution:
- A = P(1 + r/100)ⁿ
- 21,600 = P × (1.20)²
- 21,600 = P × 1.44
- P = 21,600 / 1.44
- P = 15,000
Answer: The population 2 years ago was 15,000.
Example 6: Example 6: Different rates in different years
Problem: A city’s population is 1,00,000. It grows at 5% in the first year and 10% in the second year. Find the population after 2 years.
Solution:
- After Year 1: 1,00,000 × (1 + 5/100) = 1,00,000 × 1.05 = 1,05,000
- After Year 2: 1,05,000 × (1 + 10/100) = 1,05,000 × 1.10 = 1,15,500
Answer: The population after 2 years is 1,15,500.
Example 7: Example 7: Price inflation
Problem: The price of a commodity is ₹400. It increases at 8% per year. Find the price after 2 years.
Solution:
- A = 400 × (1.08)² = 400 × 1.1664 = ₹466.56
Answer: The price after 2 years is ₹466.56.
Example 8: Example 8: Value of a property
Problem: A property worth ₹20,00,000 appreciates at 12% per year. Find its value after 2 years.
Solution:
- A = 20,00,000 × (1.12)²
- A = 20,00,000 × 1.2544
- A = ₹25,08,800
Answer: The property value after 2 years is ₹25,08,800.
Example 9: Example 9: Depreciation with half-yearly rate
Problem: A computer worth ₹40,000 depreciates at 20% per annum. Find its value after 1.5 years if depreciation is applied half-yearly.
Solution:
- Half-yearly rate = 20/2 = 10%
- Number of half-years = 3
- A = 40,000 × (1 − 10/100)³
- A = 40,000 × (0.9)³ = 40,000 × 0.729
- A = ₹29,160
Answer: The value after 1.5 years is ₹29,160.
Example 10: Example 10: Comparison of growth and decay
Problem: Town A has population 80,000 growing at 5% per year. Town B has population 1,20,000 declining at 5% per year. Find both populations after 2 years.
Solution:
Town A (Growth):
- A = 80,000 × (1.05)² = 80,000 × 1.1025 = 88,200
Town B (Decay):
- A = 1,20,000 × (0.95)² = 1,20,000 × 0.9025 = 1,08,300
Answer: After 2 years, Town A = 88,200, Town B = 1,08,300.
Real-World Applications
Population Studies: Governments use the growth formula to project future population for urban planning, resource allocation, and policy making.
Finance — Depreciation: Companies calculate depreciation of assets (machinery, vehicles, equipment) for accounting, tax returns, and insurance claims.
Economics — Inflation: Economists use the growth model to predict price increases over time and to measure the purchasing power of money.
Biology — Bacterial Growth: Microbiologists model bacterial colony growth using the compound growth formula to predict population sizes.
Environmental Science: Decay models are used to study radioactive decay, pollution reduction over time, and deforestation rates.
Real Estate: Property values that appreciate or depreciate over time follow the growth/decay model.
Key Points to Remember
- Growth formula: A = P(1 + r/100)ⁿ.
- Decay formula: A = P(1 − r/100)ⁿ.
- These are extensions of the compound interest formula.
- Growth/decay is exponential, not linear.
- For different rates in different years, apply each rate separately year by year.
- To find the past value, rearrange: P = A / (1 ± r/100)ⁿ.
- Depreciation = P − A (for decay). Growth amount = A − P (for growth).
- Half-yearly: use rate/2 and n × 2 as the number of periods.
- Growth rate makes a quantity grow faster and faster over time (compound effect).
- Decay never reaches zero — the value keeps decreasing but never becomes exactly zero.
Practice Problems
- A city has a population of 2,00,000 growing at 6% per year. Find the population after 3 years.
- A scooter bought for ₹80,000 depreciates at 12% per year. Find its value after 2 years.
- Bacteria in a petri dish number 5,000. They increase at 25% per hour. How many bacteria are there after 4 hours?
- A painting worth ₹5,00,000 appreciates at 8% per year. Find its value after 3 years.
- The present population of a town is 1,32,300. It has been growing at 10% per year. What was the population 2 years ago?
- A laptop bought for ₹60,000 depreciates at 20% per year for the first year and 10% for the second year. Find its value after 2 years.
- The value of a machine decreases by 15% each year. If its present value is ₹1,00,000, what will it be worth in 3 years?
- A tree grows at 5% per year. If its height is 2 m now, what will it be after 3 years?
Frequently Asked Questions
Q1. What is the difference between growth and decay?
Growth means a quantity increases over time at a fixed rate. Decay means a quantity decreases over time at a fixed rate. Growth uses (1 + r/100); decay uses (1 − r/100).
Q2. How is this different from simple interest?
In simple interest, the increase is calculated on the original principal only. In growth/decay (compound model), the increase/decrease applies to the accumulated value after each period. This causes exponential change.
Q3. What is depreciation?
Depreciation is the decrease in value of an asset over time due to wear and tear, age, or obsolescence. It is calculated using the decay formula.
Q4. Can the growth rate change from year to year?
Yes. When rates differ, apply each rate separately: after Year 1 = P(1 + r₁/100), after Year 2 = result × (1 + r₂/100), and so on.
Q5. Does a decaying value ever reach zero?
Mathematically, no. The value keeps decreasing but never reaches exactly zero. However, in practice, a value may become negligible.
Q6. What is the formula for finding past value?
Rearrange: P = A / (1 + r/100)ⁿ for growth, or P = A / (1 − r/100)ⁿ for decay.
Q7. How is compound interest related to growth?
The growth formula A = P(1 + r/100)ⁿ is exactly the same as the compound interest formula. Compound interest is the financial application of the same mathematical model.
Q8. What does half-yearly depreciation mean?
The depreciation is applied every 6 months instead of every year. Use rate/2 as the period rate and 2n as the number of periods for n years.
Q9. Can growth be applied to things other than population?
Yes. Growth applies to prices, property values, bacterial colonies, investments, and any quantity that increases at a fixed percentage rate.
Q10. How do you find the depreciation amount?
Depreciation = Original value − Final value = P − A = P − P(1 − r/100)ⁿ = P[1 − (1 − r/100)ⁿ].
Related Topics
- Compound Interest
- Applications of Compound Interest
- Percentage Increase and Decrease
- Introduction to Exponents
- Introduction to Percentage
- Profit and Loss
- Discount Calculation
- Simple Interest
- Sales Tax and VAT
- Finding Percentage of a Number
- Converting Between %, Fraction and Decimal
- Word Problems on Comparing Quantities
- Word Problems on Profit and Loss
- Converting Percentage to Fraction
