Applications of Compound Interest
Compound interest is not limited to bank deposits and loans. The same formula A = P(1 + R/100)ⁿ applies to many real-life situations where a quantity increases or decreases at a fixed percentage over time.
The three most common applications are population growth, depreciation of machinery or vehicles, and bacterial growth. In each case, you use the compound interest formula with slight modifications depending on whether the quantity is increasing or decreasing.
When a quantity increases at a fixed rate, use: A = P(1 + R/100)ⁿ. When a quantity decreases (depreciates) at a fixed rate, use: A = P(1 − R/100)ⁿ.
What is Applications of Compound Interest?
Definition: Applications of compound interest refer to the use of the CI formula in situations beyond banking — such as population growth, depreciation, and inflation.
Growth: A = P(1 + R/100)ⁿ
Depreciation: A = P(1 − R/100)ⁿ
Where:
- P = initial value (population, price, etc.)
- R = rate of growth or depreciation per annum (%)
- n = number of years
- A = final value after n years
Types and Properties
Type 1: Population Growth
- A city or town grows at a fixed percentage per year.
- Formula: Population after n years = P(1 + R/100)ⁿ
- P = present population, R = growth rate %, n = number of years.
Type 2: Depreciation
- Value of a machine, car, or electronic item decreases each year.
- Formula: Value after n years = P(1 − R/100)ⁿ
- P = present value, R = rate of depreciation %, n = number of years.
Type 3: Bacterial/Cell Growth
- Bacteria or cells multiply at a fixed rate per hour or per day.
- Formula: Count after n periods = P(1 + R/100)ⁿ
Type 4: Inflation
- Price of goods increases at a fixed rate per year.
- Formula: Future price = Current price × (1 + R/100)ⁿ
Type 5: Mixed rates for different years
- When the growth or depreciation rate changes each year:
- A = P × (1 + R₁/100) × (1 + R₂/100) × ...
Solved Examples
Example 1: Example 1: Population growth
Problem: The population of a town is 1,25,000. It increases at 4% per annum. Find the population after 3 years.
Solution:
Given:
- P = 1,25,000
- R = 4%
- n = 3
Using growth formula:
- A = P(1 + R/100)ⁿ
- A = 1,25,000 × (1 + 4/100)³
- A = 1,25,000 × (1.04)³
- A = 1,25,000 × 1.124864
- A = 1,40,608
Answer: Population after 3 years = 1,40,608 (approx.).
Example 2: Example 2: Depreciation of a car
Problem: A car is purchased for Rs 6,00,000. Its value depreciates at 10% per annum. Find its value after 2 years.
Solution:
Given:
- P = Rs 6,00,000
- R = 10%
- n = 2
Using depreciation formula:
- A = P(1 − R/100)ⁿ
- A = 6,00,000 × (1 − 10/100)²
- A = 6,00,000 × (0.9)²
- A = 6,00,000 × 0.81
- A = Rs 4,86,000
Answer: Value after 2 years = Rs 4,86,000.
Example 3: Example 3: Bacterial growth
Problem: A colony of bacteria has 5,000 bacteria. It doubles every hour. How many bacteria will there be after 4 hours?
Solution:
Given:
- P = 5,000
- R = 100% (doubles means 100% growth)
- n = 4 hours
Using growth formula:
- A = P(1 + R/100)ⁿ
- A = 5,000 × (1 + 100/100)⁴
- A = 5,000 × 2⁴
- A = 5,000 × 16
- A = 80,000
Answer: After 4 hours, there will be 80,000 bacteria.
Example 4: Example 4: Depreciation of machinery
Problem: A machine worth Rs 2,50,000 depreciates at 8% per annum. What is its value after 3 years?
Solution:
Given:
- P = Rs 2,50,000
- R = 8%
- n = 3
Using depreciation formula:
- A = 2,50,000 × (1 − 8/100)³
- A = 2,50,000 × (0.92)³
- A = 2,50,000 × 0.778688
- A = Rs 1,94,672
Answer: Value after 3 years = Rs 1,94,672.
Example 5: Example 5: Population with different rates
Problem: A city has a population of 2,00,000. It grew by 5% in the first year and 3% in the second year. Find the population after 2 years.
Solution:
Given:
- P = 2,00,000
- R₁ = 5%, R₂ = 3%
Using different-rate formula:
- A = P × (1 + R₁/100) × (1 + R₂/100)
- A = 2,00,000 × 1.05 × 1.03
- A = 2,00,000 × 1.0815
- A = 2,16,300
Answer: Population after 2 years = 2,16,300.
Example 6: Example 6: Inflation
Problem: A kg of rice costs Rs 60 today. If prices increase at 5% per annum, what will it cost after 2 years?
Solution:
- P = Rs 60, R = 5%, n = 2
- A = 60 × (1.05)²
- A = 60 × 1.1025
- A = Rs 66.15
Answer: Cost after 2 years = Rs 66.15.
Example 7: Example 7: Population decrease
Problem: Due to migration, a village's population of 50,000 decreases by 2% per year. Find the population after 3 years.
Solution:
- P = 50,000, R = 2%, n = 3
- A = 50,000 × (1 − 2/100)³
- A = 50,000 × (0.98)³
- A = 50,000 × 0.941192
- A = 47,059 (approx.)
Answer: Population after 3 years ≈ 47,059.
Example 8: Example 8: Finding initial population
Problem: The present population of a city is 2,16,000. If it increased at 20% per year, what was the population 2 years ago?
Solution:
- A = 2,16,000, R = 20%, n = 2
- A = P(1 + R/100)ⁿ
- 2,16,000 = P × (1.2)²
- 2,16,000 = P × 1.44
- P = 2,16,000 ÷ 1.44
- P = 1,50,000
Answer: Population 2 years ago was 1,50,000.
Example 9: Example 9: Depreciation — finding the rate
Problem: A machine bought for Rs 1,00,000 is valued at Rs 81,000 after 2 years. Find the rate of depreciation.
Solution:
- P = 1,00,000, A = 81,000, n = 2
- 81,000 = 1,00,000 × (1 − R/100)²
- (1 − R/100)² = 81,000/1,00,000 = 0.81
- 1 − R/100 = √0.81 = 0.9
- R/100 = 0.1
- R = 10%
Answer: Rate of depreciation = 10% per annum.
Example 10: Example 10: Growth then depreciation
Problem: The value of a property is Rs 10,00,000. It appreciates by 10% in the first year and depreciates by 5% in the second year. Find its value after 2 years.
Solution:
- After Year 1 (growth): 10,00,000 × (1 + 10/100) = 10,00,000 × 1.10 = Rs 11,00,000
- After Year 2 (depreciation): 11,00,000 × (1 − 5/100) = 11,00,000 × 0.95 = Rs 10,45,000
Answer: Value after 2 years = Rs 10,45,000.
Real-World Applications
Where CI applications are used:
- Census and urban planning: Government uses population growth formulas to predict future population and plan infrastructure.
- Insurance: Depreciation formulas determine the current value of insured items (cars, machinery).
- Biology: Growth of bacteria, viruses, and cell cultures uses the compound growth formula.
- Economics: Inflation rates determine future cost of goods and services.
- Real estate: Property appreciation over years follows compound growth patterns.
- Automobile industry: Resale value of vehicles is calculated using depreciation formula.
Key Points to Remember
- For growth (population, price increase): A = P(1 + R/100)ⁿ
- For depreciation (value decrease): A = P(1 − R/100)ⁿ
- For different rates in different years: multiply separate growth factors.
- Population growth and bacterial growth use the same formula as CI.
- Depreciation means value decreases — use minus in the formula.
- To find initial value, rearrange: P = A ÷ (1 ± R/100)ⁿ.
- If a quantity grows and then shrinks, apply each factor in sequence.
- All values should be rounded to the nearest whole number for population problems.
Practice Problems
- The population of a town is 80,000 and grows at 5% per annum. Find the population after 2 years.
- A computer worth Rs 50,000 depreciates at 20% per annum. Find its value after 3 years.
- Bacteria in a culture increase at 10% per hour. If there are 2,000 initially, how many after 3 hours?
- A painting is bought for Rs 3,00,000. It appreciates at 12% in the 1st year and 8% in the 2nd year. Find its value after 2 years.
- The present population of a city is 1,21,000. If it grew at 10% per year, what was the population 2 years ago?
- A machine depreciates at 15% per annum. After 2 years, its value is Rs 72,250. Find the original price.
- The price of petrol is Rs 100/litre. If it increases at 6% per year, what will be the price after 3 years?
- A village population decreases by 4% per year from 25,000. Find the population after 2 years.
Frequently Asked Questions
Q1. What is the formula for population growth?
Population after n years = P(1 + R/100)ⁿ, where P is the current population and R is the annual growth rate in %.
Q2. What is depreciation?
Depreciation is the decrease in value of an asset over time. The formula is: Value after n years = P(1 − R/100)ⁿ.
Q3. How is the CI formula used for population growth?
Replace P (principal) with current population, R (rate) with growth rate %, n (time) with number of years, and A (amount) with future population. The formula is the same: A = P(1 + R/100)ⁿ.
Q4. What if the rate changes each year?
Use: A = P × (1 + R₁/100) × (1 + R₂/100) × ... where R₁, R₂ are the rates for each year.
Q5. How do you find the population 'n years ago'?
Use P = A ÷ (1 + R/100)ⁿ, where A is the present population.
Q6. Can both growth and depreciation happen in the same problem?
Yes. Apply growth factor for years with increase and depreciation factor for years with decrease, in sequence.
Q7. What is the difference between appreciation and depreciation?
Appreciation means value increases over time (use + in formula). Depreciation means value decreases (use − in formula).
Q8. Is compound interest used in inflation calculations?
Yes. If inflation is R% per year, the future price of an item = Current price × (1 + R/100)ⁿ.
Related Topics
- Compound Interest
- Percentage Increase and Decrease
- Simple Interest
- Growth and Decay
- Introduction to Percentage
- Profit and Loss
- Discount Calculation
- 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
- Compound Interest (Half-Yearly & Quarterly)
