Compound Interest
Introduction
Compound interest is the interest calculated on both the principal and the accumulated interest from previous periods. Unlike simple interest, compound interest 'grows exponentially' because each period's interest is added to the principal for the next calculation.
Compound Interest Formula
Let:
- P = Principal (initial amount)
- R = Rate of interest per annum (in %)
- n = Number of years
- A = Amount after n years
Formula: A = P(1 + R/100)^n
Compound Interest (CI) = A — P
<ICSEExample title="Basic Compound Interest"> Find the amount and compound interest on Rs 10,000 at 8% per annum for 3 years. <Solution> P = 10000, R = 8%, n = 3 A = 10000(1 + 8/100)³ A = 10000(1.08)³ A = 10000 × 1.259712 A = Rs 12,597.12CI = A — P = 12597.12 — 10000 = Rs 2,597.12 </Solution> </ICSEExample>
Compounding Periods
Annually
Standard formula: A = P(1 + R/100)^n
Half-Yearly (Semi-Annually)
When interest is compounded half-yearly:
- Rate per half-year = R/2%
- Number of half-years = 2n
A = P(1 + R/200)^(2n)
<ICSEExample title="Half-Yearly Compounding"> Find compound interest on Rs 15000 at 10% per annum for 1.5 years, compounded half-yearly. <Solution> P = 15000, R = 10%, n = 1.5 Rate per half-year = 10/2 = 5% Number of half-years = 2 × 1.5 = 3 A = 15000(1 + 5/100)³ A = 15000(1.05)³ A = 15000 × 1.157625 A = Rs 17,364.38CI = 17364.38 — 15000 = Rs 2,364.38 </Solution> </ICSEExample>
Quarterly
When interest is compounded quarterly:
- Rate per quarter = R/4%
- Number of quarters = 4n
A = P(1 + R/400)^(4n)
Comparison Table
| Compounding | Formula | Effective Amount (Rs 1000, 12%, 1 year) |
|---|---|---|
| Annually | P(1 + R/100)ⁿ | Rs 1120.00 |
| Half-yearly | P(1 + R/200)^(2n) | Rs 1123.60 |
| Quarterly | P(1 + R/400)^(4n) | Rs 1125.51 |
More frequent compounding leads to a higher effective amount.
Finding Principal, Rate, and Time
Finding Principal
If A, R, and n are known: P = A / (1 + R/100)^n
Finding Rate
<ICSEExample title="Find the Rate"> At what rate per cent per annum will Rs 5000 amount to Rs 6655 in 3 years? <Solution> P = 5000, A = 6655, n = 3 A = P(1 + R/100)^n 6655 = 5000(1 + R/100)³ (1 + R/100)³ = 6655/5000 = 1.331 1 + R/100 = ³√1.331 = 1.10 R/100 = 0.10 R = 10%Thus, rate = 10% per annum. </Solution> </ICSEExample>
Finding Time
When A, P, and R are known, use logarithms: n = log(A/P) / log(1 + R/100)
Growth and Depreciation
Growth (Appreciation)
When the value of an item increases over time: A = P(1 + R/100)^n
Example: Population growth, property value appreciation.
Depreciation (Depreciation)
When the value of an item decreases over time: A = P(1 — R/100)^n
<ICSEExample title="Depreciation"> A machine worth Rs 50,000 depreciates at 10% per annum. Find its value after 3 years. <Solution> P = 50000, R = 10%, n = 3 Value after 3 years = 50000(1 — 10/100)³ = 50000(0.9)³ = 50000 × 0.729 = Rs 36,450 </Solution> </ICSEExample>Successive Growth
When growth rates differ each year: A = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)...
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Using simple interest formula for compound Interest | Always use A = P(1 + R/100)^n for CI |
| Forgetting to adjust rate for half-yearly/quarterly | Divide rate by number of periods per year |
| Confusing growth and depreciation | Use + for growth, - for depreciation |
| Rounding intermediate values incorrectly | Keep at least 4 decimal places during calculation |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Basic compound interest calculation | 3-4 marks | Very common |
| Half-yearly compounding | 3-4 marks | Common |
| Growth and depreciation | 3-4 marks | Common |
| Finding rate or principal | 3 marks | Occasionally asked |
Self-Test
Q1: Find the compound interest on Rs 8000 at 5% per annum for 2 years.
Q2: Calculate the amount on Rs 20000 for 1.5 years at 8% per annum compounded half-yearly.
Q3: The population of a town increases by 5% annually. If the present population is 1,85,220, what was it two years ago?
Q4: A sum of money becomes Rs 13310 in 3 years at 10% per annum compound interest. Find the sum.
Q5: A machine depreciates at 8% per annum. If its present value is Rs 40,000, what will be its value after 2 years?
