Simple and Compound Interest
1. Simple Interest (SI)
Interest calculated ONLY on the original principal for the entire period.
Formula: SI = (P × R × T) / 100
Where: P = Principal (initial amount) R = Rate of interest per annum (%) T = Time (in years)
Amount (A) = P + SI = P + PRT/100 = P(1 + RT/100)
Worked Example: Find the simple interest on Rs 6000 at 8% per annum for 3 years.
SI = (6000 × 8 × 3)/100 = 1440 Amount = 6000 + 1440 = Rs 7440
Worked Example: At what rate per cent per annum will Rs 5000 amount to Rs 6200 in 4 years?
SI = 6200 — 5000 = 1200 1200 = (5000 × R × 4)/100 = 200R R = 6% per annum
2. Compound Interest (CI)
Interest is calculated on the PRINCIPAL as well as the ACCUMULATED interest from previous periods.
'Compound interest is 'interest on interest.' The amount grows FASTER than simple interest.'
Formula for Annual Compounding
A = P(1 + R/100)ⁿ CI = A — P = P[(1 + R/100)ⁿ — 1]
Where n = number of years.
Worked Example: Find the compound interest on Rs 8000 at 10% per annum for 3 years.
A = 8000(1 + 10/100)³ = 8000(1.1)³ = 8000 × 1.331 = Rs 10,648 CI = 10648 — 8000 = Rs 2648
Comparison: SI vs CI
| Year | SI (Rs) | CI (Rs) |
|---|---|---|
| 1 | 800 | 800 |
| 2 | 800 | 880 |
| 3 | 800 | 968 |
| Total | 2400 | 2648 |
CI is greater than SI for the same principal, rate, and time (when time > 1 year).
3. CI Compounded Half-Yearly
When interest is compounded HALF-YEARLY:
- Rate becomes R/2 per half-year
- Number of periods becomes 2n
A = P(1 + R/(2×100))²ⁿ = P(1 + R/200)²ⁿ
Worked Example: Find the CI on Rs 10,000 at 8% per annum compounded half-yearly for 1.5 years.
R per half-year = 8/2 = 4% Number of half-years = 1.5 × 2 = 3 A = 10000(1 + 4/100)³ = 10000(1.04)³ = 10000 × 1.124864 = Rs 11,248.64 CI = 11248.64 — 10000 = Rs 1248.64
4. CI Compounded Quarterly
When interest is compounded QUARTERLY:
- Rate becomes R/4 per quarter
- Number of periods becomes 4n
A = P(1 + R/(4×100))⁴ⁿ = P(1 + R/400)⁴ⁿ
5. Growth and Depreciation
Population Growth
Population after n years = P₀(1 + r/100)ⁿ
This follows the SAME formula as compound interest.
Depreciation
Value after n years = P₀(1 — r/100)ⁿ
Worked Example: The population of a city is 2,00,000. It increases at 5% per annum. Find the population after 3 years.
Population = 200000(1 + 5/100)³ = 200000(1.05)³ = 200000 × 1.157625 = 231525
Worked Example: The value of a machine depreciates at 10% per annum. If its present value is Rs 1,00,000, what was its value 2 years ago?
Let value 2 years ago = V. V(1 — 10/100)² = 100000 V(0.9)² = 100000 V × 0.81 = 100000 V = 100000/0.81 = Rs 1,23,456.79
6. Difference Between CI and SI
For the SAME principal, rate, and time:
- For 2 years: CI — SI = P(R/100)²
- For 3 years: CI — SI = P(R/100)²(300 + R)/100
Worked Example: Find the difference between CI and SI on Rs 10,000 at 10% for 2 years.
CI — SI = P(R/100)² = 10000(10/100)² = 10000 × 0.01 = Rs 100
Verification: SI = (10000 × 10 × 2)/100 = Rs 2000 CI = 10000(1.1)² — 10000 = 12100 — 10000 = Rs 2100 Difference = Rs 100 ✓
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| 'SI and CI are always equal for 1 year' | TRUE — they ARE equal for 1 year |
| 'Population growth rate r% means r% of ORIGINAL each year' | It means r% of the CURRENT population each year (compound) |
| 'For half-yearly, use R/2 and n remains same' | Use R/2 AND double the time (n → 2n) |
| 'CI formula gives the interest, not amount' | CI formula A = P(1+R/100)ⁿ gives AMOUNT. Subtract P for CI |
ICSE Exam Focus (5–8 marks)
- 2-mark questions: SI calculations (find SI, amount, rate, or time)
- 3-mark questions: CI for 2-3 years, annually compounded
- 4-mark questions: CI with half-yearly or quarterly compounding
- 5-mark questions: Population growth or depreciation problems
- 6-mark questions: Difference between CI and SI
- 8-mark questions: Multi-step problems comparing different compounding periods
Self-Test
Q1. Find SI on Rs 5000 at 6% per annum for 2½ years. A1. SI = (5000 × 6 × 2.5)/100 = 5000 × 15/100 = Rs 750.
Q2. Find the amount and CI on Rs 15,000 at 10% for 2 years compounded annually. A2. A = 15000(1.1)² = 15000 × 1.21 = Rs 18,150. CI = 18150 — 15000 = Rs 3150.
Q3. Find CI on Rs 8000 at 12% per annum for 1 year compounded half-yearly. A3. R per half-year = 6%, n = 2. A = 8000(1.06)² = 8000 × 1.1236 = Rs 8988.80. CI = Rs 988.80.
Q4. The population of a town increases 4% annually. If it is 1,56,000 now, what was it 3 years ago? A4. Let past population = P. P(1.04)³ = 156000. P = 156000/1.124864 ≈ Rs 1,38,672.
Q5. A sum amounts to Rs 13,310 in 3 years at 10% compounded annually. Find the sum. A5. P(1.1)³ = 13310. P × 1.331 = 13310. P = 13310/1.331 = Rs 10,000.
Q6. Find the difference between CI and SI on Rs 20,000 at 5% for 3 years. A6. CI — SI = P(R/100)²(300 + R)/100 = 20000(0.05)²(305)/100 = 20000 × 0.0025 × 3.05 = Rs 152.50.
