Simple Interest
Definition: Simple interest is calculated only on the principal amount (original sum) for the entire period.
Formula:
Simple Interest = (Principal × Rate × Time) / 100
Amount = Principal + Simple Interest
Where:
- Principal (P) = Initial amount of money
- Rate (R) = Annual interest rate (as a percentage)
- Time (T) = Time period (in years)
Simple Interest Examples
Example 1: Basic Calculation
Problem: Find the simple interest on ₹5,000 at 8% per annum for 3 years.
Solution:
- Principal (P) = ₹5,000
- Rate (R) = 8% per annum
- Time (T) = 3 years
Simple Interest = (5,000 × 8 × 3) / 100 = ₹1,200 Amount = ₹5,000 + ₹1,200 = ₹6,200
Example 2: Finding Principal
Problem: What principal will amount to ₹8,800 in 4 years at 10% simple interest?
Solution: Let Principal = P Amount = P + SI = P + (P × 10 × 4)/100 = P + 0.4P = 1.4P
Given: 1.4P = ₹8,800 Therefore: P = ₹8,800 ÷ 1.4 = ₹6,286 (approximately)
Example 3: Finding Rate
Problem: At what rate percent will ₹2,500 amount to ₹3,500 in 5 years?
Solution: Simple Interest = ₹3,500 – ₹2,500 = ₹1,000 Using SI = (P × R × T) / 100 1,000 = (2,500 × R × 5) / 100 1,000 = 125R R = 8% per annum
Example 4: Finding Time
Problem: In how many years will ₹6,000 amount to ₹7,800 at 6% simple interest?
Solution: Simple Interest = ₹7,800 – ₹6,000 = ₹1,800 Using SI = (P × R × T) / 100 1,800 = (6,000 × 6 × T) / 100 1,800 = 360T T = 5 years
Example 5: Monthly Interest
Problem: Find simple interest on ₹12,000 at 9% per annum for 8 months.
Solution: Time = 8 months = 8/12 years = 2/3 years SI = (12,000 × 9 × 2/3) / 100 = ₹720
Example 6: Days Calculation
Problem: Calculate SI on ₹15,000 at 10% per annum for 73 days.
Solution: Time = 73 days = 73/365 years = 1/5 years SI = (15,000 × 10 × 1/5) / 100 = ₹300
Example 7: Mixed Time Units
Problem: Find SI on ₹8,000 at 12% per annum for 2 years 4 months 15 days.
Solution: Time = 2 + 4/12 + 15/365 = 2 + 1/3 + 3/73 ≈ 2.374 years SI = (8,000 × 12 × 2.374) / 100 = ₹2,279.04
Example 8: Borrowing Money
Problem: Ravi borrows ₹25,000 at 15% simple interest. How much does he pay back after 18 months?
Solution: Time = 18 months = 18/12 = 1.5 years SI = (25,000 × 15 × 1.5) / 100 = ₹5,625 Amount to pay back = ₹25,000 + ₹5,625 = ₹30,625
Example 9: Investment Growth
Problem: Priya invests ₹40,000 at 7% simple interest. After how many years will her investment become ₹54,400?
Solution: SI needed = ₹54,400 – ₹40,000 = ₹14,400 Using SI = (P × R × T) / 100 14,400 = (40,000 × 7 × T) / 100 14,400 = 2,800T T = 14,400 ÷ 2,800 = 5.14 years (approximately 5 years 2 months)
Example 10: Double Investment
Problem: How long will it take for ₹20,000 to double at 8% simple interest?
Solution: Final amount needed = ₹40,000 SI needed = ₹40,000 – ₹20,000 = ₹20,000 Using SI = (P × R × T) / 100 20,000 = (20,000 × 8 × T) / 100 20,000 = 1,600T T = 12.5 years
Example 11: Comparing Rates
Problem: Compare investments: ₹30,000 at 9% for 4 years vs ₹30,000 at 11% for 3 years.
Solution: Option 1: SI = (30,000 × 9 × 4) / 100 = ₹10,800, Amount = ₹40,800 Option 2: SI = (30,000 × 11 × 3) / 100 = ₹9,900, Amount = ₹39,900 Option 1 is better by ₹900
Example 12: Partial Payment
Problem: Amit borrows ₹50,000 at 12% SI. After 2 years, he pays ₹30,000. What balance remains after 1 more year?
Solution: SI for 2 years = (50,000 × 12 × 2) / 100 = ₹12,000 Amount after 2 years = ₹62,000 Balance after paying ₹30,000 = ₹32,000 SI on ₹32,000 for 1 year = (32,000 × 12 × 1) / 100 = ₹3,840 Final balance = ₹32,000 + ₹3,840 = ₹35,840
Compound Interest
Definition: Compound interest is calculated on the principal amount plus any previously earned interest. Interest earns interest!
Formula:
Amount = P(1 + R/100)^T
Compound Interest = Amount - Principal
Alternative Formula (when compounded annually):
CI = P[(1 + R/100)^T - 1]
Compound Interest Examples
Example 1: Annual Compounding
Problem: Find compound interest on ₹10,000 at 12% per annum for 3 years, compounded annually.
Solution: Amount = 10,000(1 + 12/100)³ = 10,000(1.12)³ = 10,000 × 1.404928 = ₹14,049.28 CI = ₹14,049.28 – ₹10,000 = ₹4,049.28
Year-by-year breakdown:
- Year 1: ₹10,000 + ₹1,200 = ₹11,200
- Year 2: ₹11,200 + ₹1,344 = ₹12,544
- Year 3: ₹12,544 + ₹1,505.28 = ₹14,049.28
Example 2: Half-yearly Compounding
Problem: Calculate CI on ₹8,000 at 10% per annum for 1.5 years, compounded half-yearly.
Solution: When compounded half-yearly:
- Rate per half-year = 10%/2 = 5%
- Number of half-years = 1.5 × 2 = 3
Amount = 8,000(1 + 5/100)³ = 8,000(1.05)³ = 8,000 × 1.157625 = ₹9,261 CI = ₹9,261 – ₹8,000 = ₹1,261
Example 3: Quarterly Compounding
Problem: Find CI on ₹5,000 at 8% per annum for 2 years, compounded quarterly.
Solution: When compounded quarterly:
- Rate per quarter = 8%/4 = 2%
- Number of quarters = 2 × 4 = 8
Amount = 5,000(1 + 2/100)⁸ = 5,000(1.02)⁸ = 5,000 × 1.171659 = ₹5,858.30 CI = ₹5,858.30 – ₹5,000 = ₹858.30
Example 4: Different Rates for Different Years
Problem: Find CI on ₹6,000 at 10% for 1st year, 12% for 2nd year, and 15% for 3rd year.
Solution:
- After 1st year: 6,000 × 1.10 = ₹6,600
- After 2nd year: 6,600 × 1.12 = ₹7,392
- After 3rd year: 7,392 × 1.15 = ₹8,500.80
CI = ₹8,500.80 – ₹6,000 = ₹2,500.80
Example 5: Finding Principal with CI
Problem: What sum will amount to ₹13,310 in 3 years at 10% compound interest?
Solution: Let Principal = P P(1 + 10/100)³ = 13,310 P(1.1)³ = 13,310 P × 1.331 = 13,310 P = ₹10,000
Example 6: Monthly Compounding
Problem: Calculate CI on ₹25,000 at 12% per annum for 2 years, compounded monthly.
Solution: Monthly rate = 12%/12 = 1% Number of months = 2 × 12 = 24 Amount = 25,000(1 + 1/100)²⁴ = 25,000(1.01)²⁴ = 25,000 × 1.2697 = ₹31,742.50 CI = ₹31,742.50 – ₹25,000 = ₹6,742.50
Example 7: Daily Compounding
Problem: Find CI on ₹10,000 at 6% per annum for 1 year, compounded daily.
Solution: Daily rate = 6%/365 ≈ 0.0164% Number of days = 365 Amount = 10,000(1 + 6/36500)³⁶⁵ = 10,000(1.000164)³⁶⁵ ≈ ₹10,618.31 CI = ₹618.31
Example 8: Finding Rate with CI
Problem: At what rate will ₹8,000 amount to ₹9,261 in 3 years with compound interest?
Solution: 8,000(1 + R/100)³ = 9,261 (1 + R/100)³ = 9,261/8,000 = 1.1576 1 + R/100 = ∛1.1576 = 1.05 R/100 = 0.05 R = 5%
Example 9: Finding Time with CI
Problem: In how many years will ₹12,000 become ₹19,487 at 10% compound interest?
Solution: 12,000(1.1)ᵀ = 19,487 (1.1)ᵀ = 19,487/12,000 = 1.624 Taking log: T log(1.1) = log(1.624) T = log(1.624)/log(1.1) = 5 years
Example 10: Depreciation (Negative Growth)
Problem: A car worth ₹8,00,000 depreciates at 15% per annum. What’s its value after 3 years?
Solution: Value = 8,00,000(1 – 15/100)³ = 8,00,000(0.85)³ = 8,00,000 × 0.614125 = ₹4,91,300
Example 11: Population Growth
Problem: A city’s population is 2,00,000. If it grows at 8% annually, what will be the population after 4 years?
Solution: Population = 2,00,000(1 + 8/100)⁴ = 2,00,000(1.08)⁴ = 2,00,000 × 1.3605 = 2,72,098
Example 12: Effective Annual Rate
Problem: Compare 12% compounded quarterly vs 11.8% compounded monthly.
Solution: Option 1: Effective rate = (1 + 12/400)⁴ – 1 = (1.03)⁴ – 1 = 12.55% Option 2: Effective rate = (1 + 11.8/1200)¹² – 1 = (1.009833)¹² – 1 = 12.47% Option 1 (12% quarterly) is better
Example 13: Doubling Time
Problem: How long will it take for ₹50,000 to double at 9% compound interest?
Solution: 50,000(1.09)ᵀ = 1,00,000 (1.09)ᵀ = 2 Taking log: T = log(2)/log(1.09) = 8.04 years
Example 14: Investment Split
Problem: Divide ₹1,00,000 between two schemes: one at 8% SI and other at 10% CI, both for 3 years, such that both give equal returns.
Solution: Let amount in SI = x, amount in CI = (1,00,000 – x) SI = (x × 8 × 3)/100 = 0.24x CI = (1,00,000 – x)[(1.1)³ – 1] = (1,00,000 – x)(0.331) Setting equal: 0.24x = (1,00,000 – x)(0.331) 0.24x = 33,100 – 0.331x 0.571x = 33,100 x = ₹57,970 (in SI), ₹42,030 (in CI)
Example 15: Loan EMI Calculation
Problem: Calculate monthly EMI for ₹5,00,000 loan at 12% annual interest for 5 years.
Solution: Monthly rate = 12%/12 = 1% = 0.01 Number of months = 60 EMI = P × [r(1+r)ⁿ]/[(1+r)ⁿ-1] EMI = 5,00,000 × [0.01(1.01)⁶⁰]/[(1.01)⁶⁰-1] EMI = 5,00,000 × 0.0222/0.8167 = ₹11,122
Example 16: Future Value Planning
Problem: Meera wants ₹10,00,000 in 8 years. How much should she invest today at 11% compound interest?
Solution: P(1.11)⁸ = 10,00,000 P × 2.3045 = 10,00,000 P = ₹4,33,926
Example 17: Inflation Impact
Problem: If inflation is 6% annually, what will be the real value of ₹1,00,000 after 5 years?
Solution: Real value = 1,00,000/(1.06)⁵ = 1,00,000/1.3382 = ₹74,726
Example 18: Mixed Compounding
Problem: ₹20,000 at 8% for first 2 years (annual), then 10% for next 3 years (half-yearly).
Solution: After 2 years: 20,000(1.08)² = ₹23,328 For next 3 years at 10% half-yearly: 23,328(1.05)⁶ = ₹31,250
Comparison: Simple vs Compound Interest
Example: ₹1,000 at 10% for 3 years
Simple Interest:
- Year 1: ₹1,000 + ₹100 = ₹1,100
- Year 2: ₹1,100 + ₹100 = ₹1,200
- Year 3: ₹1,200 + ₹100 = ₹1,300
- Total SI = ₹300
Compound Interest:
- Year 1: ₹1,000 × 1.10 = ₹1,100
- Year 2: ₹1,100 × 1.10 = ₹1,210
- Year 3: ₹1,210 × 1.10 = ₹1,331
- Total CI = ₹331
Difference = ₹331 – ₹300 = ₹31
Advanced Examples & Applications
Banking & Finance Examples
Example 1: Fixed Deposit Ladder
Problem: You have ₹3,00,000. Compare investing all at once vs splitting into 3 FDs of ₹1,00,000 each at different rates.
Solution: Option 1: ₹3,00,000 at 7% for 3 years = 3,00,000(1.07)³ = ₹3,67,507 Option 2:
- ₹1,00,000 at 6.5% for 1 year = ₹1,06,500
- ₹1,00,000 at 7% for 2 years = ₹1,14,490
- ₹1,00,000 at 7.5% for 3 years = ₹1,24,230
- Total = ₹3,45,220 Option 1 is better by ₹22,287
Example 2: Credit Card Minimum Payment
Problem: Credit card balance ₹25,000 at 36% annual (3% monthly). Paying only minimum 2% of balance each month.
Solution: Month 1: Balance ₹25,000, Interest ₹750, Payment ₹500, New balance ₹25,250 Month 2: Balance ₹25,250, Interest ₹757.50, Payment ₹505, New balance ₹25,502.50 This debt keeps growing! Minimum payments don’t cover interest.
Example 3: Home Loan with Prepayment
Problem: ₹50,00,000 home loan at 8.5% for 20 years. What if you prepay ₹1,00,000 in 5th year?
Solution: Regular EMI = ₹43,391 After 4 years, outstanding ≈ ₹42,50,000 After ₹1,00,000 prepayment: ₹41,50,000 Savings: About ₹2,50,000 in total interest and 2 years reduction
Example 4: Recurring Deposit
Problem: Monthly RD of ₹5,000 for 5 years at 7% compound interest.
Solution: This involves compound interest on multiple deposits: Amount = 5,000 × [((1.07)⁵ – 1) / 0.07] × (1.07) Amount = 5,000 × 5.7507 × 1.07 = ₹3,07,662 Total invested = ₹3,00,000, Interest earned = ₹7,662
Business Applications
Example 5: Business Loan Impact
Problem: A business needs ₹10,00,000. Compare bank loan at 12% CI vs private lender at 18% SI, both for 3 years.
Solution: Bank loan: 10,00,000(1.12)³ = ₹14,04,928, Interest = ₹4,04,928 Private lender: SI = (10,00,000 × 18 × 3)/100 = ₹5,40,000 Bank loan saves ₹1,35,072
Example 6: Equipment Purchase vs Lease
Problem: Buy equipment for ₹5,00,000 (depreciates 20% annually) vs lease at ₹1,50,000 per year for 3 years.
Solution: Purchase: Value after 3 years = 5,00,000(0.8)³ = ₹2,56,000 Net cost = ₹5,00,000 – ₹2,56,000 = ₹2,44,000 Lease: Total cost = ₹4,50,000 Purchase is better by ₹2,06,000
Example 7: Revenue Growth Projection
Problem: Company revenue is ₹50 lakhs. If it grows 15% annually, what will it be in 7 years?
Solution: Revenue = 50,00,000(1.15)⁷ = 50,00,000 × 2.6600 = ₹1,33,00,000
Investment Portfolio Examples
Example 8: SIP (Systematic Investment Plan)
Problem: Monthly SIP of ₹10,000 for 15 years at 12% annual return.
Solution: Using compound interest formula for annuities: Amount = 10,000 × [((1.01)¹⁸⁰ – 1) / 0.01] Amount = 10,000 × 499.58 = ₹49,95,800 Total invested = ₹18,00,000, Gains = ₹31,95,800
Example 9: Dollar Cost Averaging
Problem: Invest ₹1,00,000 monthly in mutual fund with variable returns: 15%, -5%, 20%, 8%, 12% over 5 years.
Solution: Year 1: 1,00,000 × 1.15 = 1,15,000 Year 2: (1,15,000 + 1,00,000) × 0.95 = 2,04,250 Year 3: (2,04,250 + 1,00,000) × 1.20 = 3,65,100 Year 4: (3,65,100 + 1,00,000) × 1.08 = 5,02,308 Year 5: (5,02,308 + 1,00,000) × 1.12 = 6,82,585
Example 10: Risk vs Return
Problem: Compare three investments over 10 years:
- Safe: ₹2,00,000 at 6% CI
- Moderate: ₹2,00,000 at 10% CI
- Risky: ₹2,00,000 at 15% CI
Solution: Safe: 2,00,000(1.06)¹⁰ = ₹3,58,169 Moderate: 2,00,000(1.10)¹⁰ = ₹5,18,748 Risky: 2,00,000(1.15)¹⁰ = ₹8,09,374
Real Estate Examples
Example 11: Property Appreciation
Problem: House bought for ₹50 lakhs appreciates 8% annually. What’s its value in 12 years?
Solution: Value = 50,00,000(1.08)¹² = 50,00,000 × 2.5182 = ₹1,25,91,000
Example 12: Rental Yield Calculation
Problem: Property worth ₹80 lakhs gives ₹25,000 monthly rent, increasing 5% annually.
Solution: Annual rent = ₹3,00,000 Rental yield = (3,00,000/80,00,000) × 100 = 3.75% After 5 years: Rent = 3,00,000(1.05)⁵ = ₹3,82,884
Retirement Planning Examples
Example 13: Pension Fund Growth
Problem: Starting at age 25, invest ₹20,000 annually for 35 years at 11% return.
Solution: Amount = 20,000 × [((1.11)³⁵ – 1) / 0.11] Amount = 20,000 × 271.024 = ₹54,20,480
Example 14: Late Start Impact
Problem: Compare starting retirement savings at 25 vs 35, both saving ₹50,000 annually at 10%.
Solution: Starting at 25 (40 years): 50,000 × [((1.10)⁴⁰ – 1) / 0.10] = ₹2,21,25,000 Starting at 35 (30 years): 50,000 × [((1.10)³⁰ – 1) / 0.10] = ₹82,17,000 Difference: ₹1,39,08,000 (starting 10 years earlier)
Education Planning Examples
Example 15: Child’s Education Fund
Problem: Child is 5 years old. College fees will be ₹25 lakhs when child turns 18. Invest now at 9% CI.
Solution: Time = 13 years Required investment = 25,00,000/(1.09)¹³ = 25,00,000/3.0658 = ₹8,15,180
Example 16: Education Loan
Problem: Education loan of ₹15 lakhs at 10% CI. 4-year course + 1 year grace period. EMI for 10 years.
Solution: Amount after 5 years = 15,00,000(1.10)⁵ = ₹24,15,765 Monthly rate = 10%/12 = 0.833% EMI = 24,15,765 × [0.00833(1.00833)¹²⁰]/[(1.00833)¹²⁰-1] = ₹31,924
Tax Planning Examples
Example 17: Tax-Free vs Taxable Bonds
Problem: ₹5 lakhs investment choice: Tax-free bonds at 6% vs Corporate bonds at 9% (30% tax bracket).
Solution: Tax-free: Annual return = 5,00,000 × 0.06 = ₹30,000 Taxable: Pre-tax return = 5,00,000 × 0.09 = ₹45,000 Post-tax return = ₹45,000 × 0.70 = ₹31,500 Taxable bonds better by ₹1,500 annually
Example 18: ELSS vs PPF
Problem: ₹1.5 lakhs annual investment: ELSS (15% return, 3-year lock) vs PPF (7.5% return, 15-year lock).
Solution: ELSS (over 15 years): Assuming average 15% return Amount ≈ 1,50,000 × [((1.15)¹⁵ – 1) / 0.15] = ₹1,22,37,000
PPF (15 years): Amount = 1,50,000 × [((1.075)¹⁵ – 1) / 0.075] = ₹39,69,000 ELSS potentially better but higher risk
Special Cases & Complex Scenarios
Example 19: Inflation-Adjusted Returns
Problem: Investment of ₹1 lakh at 12% nominal return with 6% inflation for 10 years.
Solution: Nominal value = 1,00,000(1.12)¹⁰ = ₹3,10,585 Real rate = (1.12/1.06) – 1 = 5.66% Real value = 1,00,000(1.0566)¹⁰ = ₹1,73,168
Example 20: Currency Depreciation Impact
Problem: NRI invests $10,000 in India when $1 = ₹75. After 3 years, investment grows to ₹10 lakhs but $1 = ₹85.
Solution: Initial investment = $10,000 × 75 = ₹7,50,000 Final value in dollars = ₹10,00,000 ÷ 85 = $11,765 Dollar return = (11,765 – 10,000)/10,000 = 17.65% over 3 years
Example 21: Staggered Investment
Problem: Invest ₹1 lakh each at the beginning of years 1, 3, and 5. Total value after 8 years at 11% CI.
Solution: First investment: 1,00,000(1.11)⁸ = ₹2,14,359 Second investment: 1,00,000(1.11)⁶ = ₹1,68,506
Third investment: 1,00,000(1.11)⁴ = ₹1,35,131 Total value = ₹5,17,996
Example 22: Break-Even Analysis
Problem: Business invests ₹20 lakhs in equipment. Annual cash flow ₹4 lakhs. At what discount rate does NPV = 0?
Solution: For NPV = 0: 20,00,000 = 4,00,000 × [1 – (1+r)⁻ⁿ]/r This requires iterative calculation, but IRR ≈ 15% for 7-8 years payback.
1. Bank Savings Account
You deposit ₹50,000 in a savings account at 4% annual interest, compounded monthly.
After 5 years: Amount = 50,000(1 + 4/1200)⁶⁰ = 50,000(1.003333)⁶⁰ = ₹61,051
2. Fixed Deposit
₹2,00,000 FD at 7% for 3 years, compounded annually: Amount = 2,00,000(1.07)³ = ₹2,45,007
3. Loan Interest (Simple)
Car loan of ₹5,00,000 at 12% simple interest for 5 years: Interest = (5,00,000 × 12 × 5) / 100 = ₹3,00,000 Total repayment = ₹8,00,000
4. Credit Card Debt (Compound)
Outstanding balance of ₹25,000 at 36% annual interest (3% monthly): After 1 year: 25,000(1.03)¹² = ₹35,694
More Practice Problems
Basic Level Problems
Problem 1: Find SI on ₹8,500 at 6% for 2 years 3 months. Answer: Time = 2.25 years, SI = (8,500 × 6 × 2.25)/100 = ₹1,147.50
Problem 2: What principal amounts to ₹9,680 in 4 years at 8% SI? Answer: Let P be principal. P + (P × 8 × 4)/100 = 9,680, 1.32P = 9,680, P = ₹7,333.33
Problem 3: Find CI on ₹12,000 at 5% for 3 years, compounded annually. Answer: Amount = 12,000(1.05)³ = ₹13,891.50, CI = ₹1,891.50
Problem 4: At what rate will ₹6,000 double in 8 years with SI? Answer: For doubling, SI = ₹6,000. Rate = (6,000 × 100)/(6,000 × 8) = 12.5%
Problem 5: Compare SI and CI on ₹10,000 at 10% for 2 years. Answer: SI = ₹2,000, CI = 10,000(1.1)² – 10,000 = ₹2,100, Difference = ₹100
Intermediate Level Problems
Problem 6: A sum becomes ₹4,800 in 2 years and ₹5,400 in 3 years at SI. Find principal and rate. Answer: SI for 1 year = ₹600, SI for 2 years = ₹1,200, Principal = ₹3,600, Rate = 16.67%
Problem 7: ₹15,000 is invested partly at 8% SI and partly at 12% SI for 3 years. Total interest = ₹4,200. Find the amounts. Answer: Let x be amount at 8%. (x × 8 × 3)/100 + ((15,000-x) × 12 × 3)/100 = 4,200 Solving: x = ₹7,500 at 8%, ₹7,500 at 12%
Problem 8: Find CI on ₹20,000 at 15% for 2 years 4 months, compounded annually. Answer: For 2 years: 20,000(1.15)² = ₹26,450 For 4 months (1/3 year): SI = (26,450 × 15 × 1/3)/100 = ₹1,322.50 Total amount = ₹27,772.50, CI = ₹7,772.50
Problem 9: What should be the CI on ₹8,000 for 1.5 years at 10% half-yearly? Answer: Amount = 8,000(1.05)³ = ₹9,261, CI = ₹1,261
Problem 10: A man invests ₹1,000 at the beginning of each year for 5 years at 8% CI. Find total amount. Answer: Sum of geometric series with different starting points: Total = 1,000[(1.08)⁵ + (1.08)⁴ + (1.08)³ + (1.08)² + (1.08)] = ₹6,336.60
Advanced Level Problems
Problem 11: The difference between CI and SI on a sum for 3 years at 10% is ₹310. Find the sum. Answer: Difference = P × R² × (300 + R)/100³ 310 = P × 10² × (300 + 10)/100³ = P × 0.031 P = ₹10,000
Problem 12: A loan of ₹10,000 at 20% CI becomes ₹17,280 in n years. Find n. Answer: 10,000(1.2)ⁿ = 17,280, (1.2)ⁿ = 1.728 Taking log: n × log(1.2) = log(1.728), n = 3 years
Problem 13: If the difference between CI and SI on ₹5,000 for 2 years is ₹50, find the rate. Answer: Difference = P × R²/100² = 50 5,000 × R²/10,000 = 50, R² = 100, R = 10%
Problem 14: A sum triples in 10 years at CI. In how many years will it become 27 times? Answer: If P becomes 3P in 10 years: (1+r)¹⁰ = 3 For 27P: (1+r)ⁿ = 27 = 3³ = ((1+r)¹⁰)³ = (1+r)³⁰ Therefore, n = 30 years
Problem 15: ₹10,000 is invested at CI. After 2 years it becomes ₹11,664. After 2 more years, what will it become? Answer: Rate can be found: (1+r)² = 11,664/10,000 = 1.1664 1+r = 1.08, so r = 8% After 4 years total: 10,000(1.08)⁴ = ₹13,604.89
Real-World Application Problems
Problem 16: A company’s profit grows from ₹2 lakhs to ₹2.88 lakhs in 2 years. Assuming compound growth, what will be the profit in year 5? Answer: (1+r)² = 2.88/2 = 1.44, so r = 20% Year 5 profit = 2,00,000(1.2)⁵ = ₹4,97,664
Problem 17: A machine costs ₹5 lakhs and depreciates at 12% annually. After how many years will its value be ₹2.5 lakhs? Answer: 5,00,000(0.88)ⁿ = 2,50,000, (0.88)ⁿ = 0.5 Taking log: n = log(0.5)/log(0.88) = 5.43 years
Problem 18: You want to accumulate ₹50 lakhs in 15 years. How much should you invest monthly at 12% CI? Answer: Using annuity formula: PMT = FV × r/[(1+r)ⁿ – 1] Monthly rate = 1%, n = 180 months PMT = 50,00,000 × 0.01/[(1.01)¹⁸⁰ – 1] = ₹6,735
Problem 19: Compare two insurance policies: Policy A: Pay ₹10,000 annually for 20 years, get ₹5 lakhs at maturity Policy B: Pay ₹15,000 annually for 15 years, get ₹5 lakhs at maturity Which is better at 8% discount rate?
Answer: Policy A NPV = -10,000 × [(1-(1.08)⁻²⁰)/0.08] + 5,00,000/(1.08)²⁰ = ₹9,515 Policy B NPV = -15,000 × [(1-(1.08)⁻¹⁵)/0.08] + 5,00,000/(1.08)¹⁵ = ₹29,465 Policy B is better
Problem 20: A startup’s valuation grows from ₹10 crores to ₹810 crores in 6 years. What’s the compound annual growth rate? Answer: (1+r)⁶ = 810/10 = 81, 1+r = ⁶√81 = 3, r = 200%
Challenge Problems
Problem 21: Three friends invest ₹20,000 each in different schemes:
- Friend A: 8% SI for 5 years
- Friend B: 6% CI for 5 years
- Friend C: 10% CI for first 2 years, then 4% CI for next 3 years Who gets the maximum return?
Answer: A: SI = (20,000 × 8 × 5)/100 = ₹8,000, Total = ₹28,000 B: Amount = 20,000(1.06)⁵ = ₹26,764 C: After 2 years: 20,000(1.1)² = ₹24,200 After 5 years: 24,200(1.04)³ = ₹27,217 Friend A gets maximum return
Problem 22: A person borrows ₹1 lakh at 10% CI and invests it at 12% CI. What’s the net gain after 4 years? Answer: Amount to repay = 1,00,000(1.1)⁴ = ₹1,46,410 Amount received = 1,00,000(1.12)⁴ = ₹1,57,352 Net gain = ₹10,942
Problem 23: If ₹1 invested today becomes ₹16 after n years at CI, and the interest rate doubles, in how many years will ₹1 become ₹16? Answer: Original: (1+r)ⁿ = 16 New rate: (1+2r)ᵐ = 16 If original rate was such that (1+r) = 2, then n = 4 New: (1+2r) = (1+2×1) = 3, so 3ᵐ = 16 m = log(16)/log(3) = 2.52 years
Problem 24: The population of a city decreases from 8 lakhs to 6.48 lakhs in 2 years due to migration. If this trend continues, what will be the population after 5 more years? Answer: Decrease rate: (6.48/8)^(1/2) = 0.9, so 10% annual decrease After 5 more years: 6,48,000(0.9)⁵ = ₹3,82,206
Problem 25: An article is sold at 20% profit. If both cost price and selling price increase by ₹100, the profit becomes 16%. Find the original cost price. Answer: Let CP = x, SP = 1.2x New: (x+100) and (1.2x+100) Profit = 16%: (1.2x+100) = 1.16(x+100) 1.2x + 100 = 1.16x + 116 0.04x = 16, x = ₹400
Problem 1
Find both SI and CI on ₹15,000 at 8% per annum for 2 years.
Simple Interest: (15,000 × 8 × 2) / 100 = ₹2,400 Compound Interest: 15,000(1.08)² – 15,000 = ₹2,496
Problem 2
At what rate will ₹4,000 amount to ₹5,324 in 3 years with compound interest?
Let rate = R% 4,000(1 + R/100)³ = 5,324 (1 + R/100)³ = 1.331 1 + R/100 = 1.1 R = 10%
Problem 3
How much more does ₹20,000 earn in 4 years at 6% compound interest compared to simple interest?
SI: (20,000 × 6 × 4) / 100 = ₹4,800 CI: 20,000(1.06)⁴ – 20,000 = ₹5,249.79 Difference: ₹5,249.79 – ₹4,800 = ₹449.79
Key Takeaways
- Simple Interest: Linear growth, interest calculated only on principal
- Compound Interest: Exponential growth, interest calculated on principal + accumulated interest
- Time Factor: The longer the time period, the greater the difference between SI and CI
- Frequency: More frequent compounding periods result in higher returns
- Real Applications: Most real-world financial products use compound interest
Remember: Compound interest is more powerful for investments, but more expensive for loans!