Quick Concept Summary
Compound Interest (CI) is interest calculated on the principal plus the accumulated interest from previous periods. Unlike Simple Interest, the interest is reinvested each period.
Key difference from SI:
- In SI: interest is always calculated on original P
- In CI: interest is calculated on P + interest accumulated so far
Compounding frequency determines growth speed:
| Compounding | Formula Adjustment |
|---|---|
| Annually | A = P(1 + R/100)^n |
| Half-yearly | Rate → R/2, Periods → 2n |
| Quarterly | Rate → R/4, Periods → 4n |
| Monthly | Rate → R/12, Periods → 12n |
Practical applications in exams:
- Bank deposits and loans
- Population increase or decrease
- Machine/vehicle depreciation
- Value appreciation of assets
Key Formulas
| Scenario | Formula |
|---|---|
| CI (Annual compounding) | A = P(1 + R/100)^n, CI = A - P |
| CI (Half-yearly) | A = P(1 + R/200)^(2n) |
| CI (Quarterly) | A = P(1 + R/400)^(4n) |
| Depreciation | V = P(1 - R/100)^n |
| Population Growth | P_n = P_0 × (1 + R/100)^n |
| CI - SI (2 years) | Difference = P × (R/100)^2 |
| CI - SI (3 years) | Difference = P(R/100)^2 × (3 + R/100) |
| Effective Annual Rate | EAR = (1 + R/n)^n - 1 (n = compounding freq.) |
Quick values to memorize (for 10% annual rate):
- (1.1)^1 = 1.1
- (1.1)^2 = 1.21
- (1.1)^3 = 1.331
- (1.05)^2 = 1.1025
- (1.05)^4 = 1.2155 (approx)
Worked Examples
Example 1: Annual Compounding — Find CI and Amount
Example 2: Half-Yearly Compounding
Example 3: CI vs SI Difference Shortcut
Example 4: Population Growth Problem
Example 5: Depreciation Problem
Example 6: Find Rate When Amount and Time are Known
Shortcuts and Tricks
Trick 1: The Successive Percentage Method (for 2 years)
For 2 years at R%, effective CI percentage = R + R + R²/100
- At 10%: CI% = 10 + 10 + 100/100 = 21% → CI = 21% of P
- At 5%: CI% = 5 + 5 + 25/100 = 10.25% → CI = 10.25% of P
Example: P = Rs 2,000, R = 10%, n = 2 years CI = 21% of 2000 = Rs 420
Trick 2: Ratio Method for Finding Rate
If A₁ and A₂ are amounts at end of year 1 and year 2 respectively:
- A₂/A₁ = (1 + R/100)
- R = (A₂/A₁ - 1) × 100
Example: A₁ = Rs 11,000, A₂ = Rs 12,100 R = (12100/11000 - 1) × 100 = (1.1 - 1) × 100 = 10%
Trick 3: CI Difference for 2 Years (memorize this)
CI - SI = P × (R/100)²
This is always positive (CI > SI for n ≥ 2).
Example: P = Rs 10,000, R = 12% SI (2 years) = 10000 × 2 × 12/100 = Rs 2,400 CI - SI = 10000 × (12/100)² = 10000 × 0.0144 = Rs 144 CI = 2400 + 144 = Rs 2,544
Trick 4: Rule of 72 (Estimate Doubling Time)
Years to double ≈ 72 / R
Example: At 8% CI, money doubles in ≈ 72/8 = 9 years (Exact answer is 9.006 years — very accurate estimate)
Trick 5: Quarterly vs Annual Rate Comparison
For the same nominal annual rate R, quarterly compounding grows faster. Effective annual rate for quarterly = (1 + R/400)^4 - 1
At 8%: Effective = (1.02)^4 - 1 = 1.0824 - 1 = 8.24%
TNPSC Exam Pattern
How Compound Interest appears in TNPSC and Bank PO exams:
| Question Type | Exam Level | Frequency |
|---|---|---|
| Find CI for 2–3 years (annual) | TNPSC Group 2/4 | Very common |
| CI vs SI difference for 2 years | All exams | Common |
| Half-yearly compounding for 1–2 years | Bank PO | Common |
| Population growth in n years | TNPSC | Occasional |
| Depreciation value after n years | SSC, Railway | Occasional |
| Find rate when A and n are given | Bank PO | Occasional |
| Quarterly compounding | Bank PO Advanced | Less common |
Common TNPSC traps to avoid: 1. When "compounded half-yearly" — halve the rate AND double the periods 2. When "compounded quarterly" — divide rate by 4 AND multiply periods by 4 3. "Rate per annum" vs "rate per half-year" — always confirm what's given 4. Depreciation uses (1 - R/100), not (1 + R/100)
Practice Questions
Q1. Find the CI on Rs 5,000 at 8% per annum for 2 years, compounded annually. Answer: A = 5000 × (1.08)^2 = 5000 × 1.1664 = Rs 5,832. CI = 5832 - 5000 = Rs 832
Q2. Find the CI on Rs 16,000 at 10% per annum for 1.5 years, compounded half-yearly. Answer: Rate = 5%, periods = 3. A = 16000 × (1.05)^3 = 16000 × 1.157625 = Rs 18,522. CI = Rs 2,522
Q3. The SI on a sum for 2 years at 6% is Rs 720. Find the CI on the same sum for 2 years at 6%. Answer: P = (100 × 720)/(2 × 6) = Rs 6,000. CI - SI = 6000 × (0.06)^2 = 6000 × 0.0036 = Rs 21.60. CI = 720 + 21.60 = Rs 741.60
Q4. A city's population is 2,00,000 and grows at 4% per annum. What is the population after 2 years? Answer: P = 200000 × (1.04)^2 = 200000 × 1.0816 = 2,16,320
Q5. A car worth Rs 2,00,000 depreciates at 12% per annum. What is its value after 2 years? Answer: V = 200000 × (0.88)^2 = 200000 × 0.7744 = Rs 1,54,880
Q6. Rs 12,000 is invested at 10% per annum CI for 3 years. Find CI and compare with SI. Answer: CI: A = 12000 × (1.1)^3 = 12000 × 1.331 = Rs 15,972. CI = Rs 3,972. SI = (12000 × 3 × 10)/100 = Rs 3,600. CI exceeds SI by Rs 372.
Q7. A sum of money becomes Rs 6,050 in 2 years at 10% per annum CI. Find the principal. Answer: P = 6050 / (1.1)^2 = 6050 / 1.21 = Rs 5,000
Q8. Find the CI on Rs 1,00,000 at 5% per annum for 1 year, compounded quarterly. Answer: A = 100000 × (1 + 5/400)^4 = 100000 × (1.0125)^4 = 100000 × 1.05095 = Rs 1,05,095. CI = Rs 5,095
Q9. The difference between CI and SI on a sum for 2 years at 12% is Rs 72. Find the sum. Answer: CI - SI = P × (R/100)^2 → 72 = P × (0.12)^2 = P × 0.0144 → P = 72/0.0144 = Rs 5,000
Q10. A machine bought for Rs 80,000 has an annual depreciation of 15%. Find its value after 3 years. Answer: V = 80000 × (0.85)^3 = 80000 × 0.614125 = Rs 49,130