Compound Interest (கூட்டு வட்டி)

Compound Interest is the powerhouse of financial mathematics and a high-scoring topic in every competitive exam. Unlike simple interest, compound interest makes money grow exponentially - it's literally "interest earning interest!" Mastering compound interest is crucial for TNPSC, SSC, Banking, and UPSC exams because it appears in direct calculation problems, comparison questions, and real-world financial scenarios. Understanding this concept will not only boost your exam scores but also help you make smarter financial decisions in real life.

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🧠 Foundation: Understanding the Core Concept (அடிப்படை கருத்துகள்)

Compound Interest means calculating interest not just on the original principal amount, but also on the interest that has been added to the principal in previous periods. Think of it as a snowball effect 🏔️ - as it rolls down, it picks up more snow and grows bigger and faster!

Real-world examples:

• 🏦 Your savings account: Bank pays interest on your deposit + previous interest earned
• 💳 Credit card debt: You pay interest on the original debt + unpaid interest from previous months
• 📈 Investment growth: Your mutual fund earns returns on principal + previously earned returns

The Core Formula

English: Amount = Principal × (1 + Rate/100)^Time | CI = Amount - Principal

Tamil: தொகை = அசல் தொகை × (1 + வட்டி விகிதம்/100)^காலம் | கூட்டு வட்டி = தொகை - அசல் தொகை

ℹ️ Note on the ^ symbol (Power / Exponent)

• English: a^n means “a raised to the power n” (exponent). If n = 2, it’s square (²); if n = 3, it’s cube (³).
• தமிழ்: a^n என்பது a எண்ணை n முறை அதே எண்ணுடன் பெருக்குவது. n = 2 → வர்க்கம் (²), n = 3 → கன (³).
• Superscript view: a^2 ⇔ a², a^3 ⇔ a³ (e.g., (1.1)^2 = 1.1²).

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Level 0: Formula Playground

Compounding Period	English Formula	Tamil Term	Time Multiplier
Annual (Yearly)	A = P(1 + R/100)^n	ஆண்டுக்கு ஒருமுறை	n = years
Half-Yearly	A = P(1 + R/200)^(2n)	ஆண்டுக்கு இருமுறை	2n = twice per year
Quarterly	A = P(1 + R/400)^(4n)	காலாண்டுக்கு ஒருமுறை	4n = four times per year
Monthly	A = P(1 + R/1200)^(12n)	மாதந்தோறும்	12n = twelve times per year

Practice Drill 1: Find CI on ₹1000 at 10% for 2 years (annually) Solution: A = 1000(1 + 10/100)² = 1000(1.1)² = ₹1210, CI = 1210 - 1000 = ₹210

Practice Drill 2: Find Amount on ₹5000 at 8% for 1 year (half-yearly) Solution: A = 5000(1 + 8/200)² = 5000(1.04)² = 5000 × 1.0816 = ₹5408

Practice Drill 3: Find CI on ₹2000 at 12% for 1 year (quarterly) Solution: A = 2000(1 + 12/400)⁴ = 2000(1.03)⁴ = 2000 × 1.1255 = ₹2251, CI = ₹251

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⚙️ Unit Conversion Mastery (அலகு மாற்றம்)

Common conversion mistakes: Confusing the rate division and time multiplication for different compounding periods!

Key conversion factors:

• Annual: Rate stays same, Time = n years
• Half-Yearly: Rate ÷ 2, Time × 2
• Quarterly: Rate ÷ 4, Time × 4
• Monthly: Rate ÷ 12, Time × 12
• Percentage to decimal: 10% = 0.10, 8.5% = 0.085

Conversion Practice Problem 1: Convert "20% per annum for 3 years, compounded half-yearly" Solution: Rate = 20/2 = 10% per half-year, Time = 3 × 2 = 6 half-years

Conversion Practice Problem 2: Convert "15% per annum for 2 years, compounded quarterly"
Solution: Rate = 15/4 = 3.75% per quarter, Time = 2 × 4 = 8 quarters

Conversion Practice Problem 3: Convert "12% per annum for 18 months, compounded monthly" Solution: Rate = 12/12 = 1% per month, Time = 18 months

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Level 1: Problem Reading Strategy (கேள்வியை அணுகும் முறை)

G-F-V Framework:

• Given (கொடுக்கப்பட்டது): Principal, Rate, Time, Compounding frequency
• Find (கண்டுபிடி): Amount, Compound Interest, or comparison with Simple Interest
• Verify (சரிபார்): Check if CI > SI, reasonable growth rate

Example 1: Find the compound interest on ₹8000 at 15% per annum for 2 years compounded annually.

• Given: P = ₹8000, R = 15%, T = 2 years, compounded annually
• Find: Compound Interest
• Solution: A = 8000(1 + 15/100)² = 8000(1.15)² = 8000 × 1.3225 = ₹10,580, CI = 10,580 - 8000 = ₹2580
• Verify: SI would be 8000 × 15 × 2/100 = ₹2400, CI > SI ✓

Example 2: What will ₹12,000 amount to in 1½ years at 20% per annum compounded half-yearly?

• Given: P = ₹12,000, R = 20%, T = 1.5 years, compounded half-yearly
• Find: Amount after 1½ years
• Solution: Rate per half-year = 20/2 = 10%, Time = 1.5 × 2 = 3 half-years, A = 12000(1 + 10/100)³ = 12000(1.1)³ = 12000 × 1.331 = ₹15,972
• Verify: Amount increased by about 33% over 1.5 years ✓

Example 3: Find the difference between CI and SI on ₹5000 for 2 years at 10% per annum.

• Given: P = ₹5000, R = 10%, T = 2 years
• Find: CI - SI difference
• Solution: SI = 5000 × 10 × 2/100 = ₹1000, A = 5000(1.1)² = ₹6050, CI = 6050 - 5000 = ₹1050, Difference = 1050 - 1000 = ₹50
• Verify: For 2 years, difference = P × R²/100² = 5000 × 100/10000 = ₹50 ✓

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Level 2: Pattern Recognition (கணக்குகளின் வகையை அறிதல்)

Pattern	English Keywords	Core Concept	Tamil Term
Direct Calculation	"find amount", "calculate CI"	Basic formula application	நேரடி கணக்கீடு
Growth Multiples	"becomes x times", "doubles/triples"	Exponential growth patterns	பெருக்க வளர்ச்சி
Comparison Problems	"difference between CI & SI"	CI vs SI analysis	ஒப்பீட்டு கணக்குகள்
Mixed Compounding	"half-yearly", "quarterly"	Different compounding periods	கலப்பு கூட்டல் முறை

Pattern 1: Direct Calculation (நேரடி கணக்கீடு)

Keywords: "Find the compound interest", "What will be the amount", "Calculate CI" Core Insight: Straightforward application of the compound interest formula

Visual Diagram: Principal (₹P) → Year 1 → P(1+R/100) → Year 2 → P(1+R/100)² → Final Amount ↓ ↓ ↓ Interest More Interest Even More Interest on P on (P+Int1) on accumulated amount

Worked Example: Find CI on ₹4000 at 5% per annum for 3 years compounded annually. Solution:

• Apply formula: A = P(1 + R/100)^T
• A = 4000(1 + 5/100)³ = 4000(1.05)³ = 4000 × 1.157625 = ₹4630.50
• CI = A - P = 4630.50 - 4000 = ₹630.50

Pattern 2: Growth Multiples (பெருக்க வளர்ச்சி)

Keywords: "becomes 2 times", "doubles in x years", "amount to y times" Core Insight: Use logarithmic relationships and shortcut formulas for exponential growth

Visual Diagram: If money becomes x times in 'a' years and y times in 'b' years: (x)^(1/a) = (y)^(1/b) = (1 + r/100)

Principal → x times (a years) → y times (b years) P → xP → yP

Worked Example: A sum becomes 4 times in 8 years. In how many years will it become 8 times? Solution:

• Using shortcut: (x)^(1/a) = (y)^(1/b)
• (4)^(1/8) = (8)^(1/b)
• (2²)^(1/8) = (2³)^(1/b)
• 2^(2/8) = 2^(3/b)
• 2/8 = 3/b → b = 12 years

Pattern 3: Comparison Problems (ஒப்பீட்டு கணக்குகள்)

Keywords: "difference between CI and SI", "how much more than SI", "exceed SI by" Core Insight: For 2 years: Difference = P×R²/(100)², For 3 years: use direct calculation

Visual Diagram: Year 1: SI = CI (both calculate on principal only) Year 2: SI = P×R/100, CI = Interest on (P + Year1 interest) Year 3: Gap widens further as CI compounds on larger base

Worked Example: Find the difference between CI and SI on ₹6000 for 2 years at 8% per annum. Solution:

• Using 2-year shortcut: Difference = P × R²/(100)²
• Difference = 6000 × 8²/(100)² = 6000 × 64/10000 = ₹38.40
• Verification: SI = 6000×8×2/100 = ₹960, A = 6000(1.08)² = ₹6998.40, CI = ₹998.40, Difference = ₹38.40 ✓

Pattern 4: Mixed Compounding (கலப்பு கூட்டல் முறை)

Keywords: "compounded half-yearly", "quarterly", "monthly compounding" Core Insight: Adjust rate and time according to compounding frequency

Visual Diagram: Annual: 12% for 2 years → Rate=12%, Time=2 Half-yearly: 12% for 2 years → Rate=6%, Time=4 periods Quarterly: 12% for 2 years → Rate=3%, Time=8 periods Monthly: 12% for 2 years → Rate=1%, Time=24 periods

Worked Example: Find amount on ₹10,000 at 16% for 9 months compounded quarterly. Solution:

• Convert time: 9 months = 3 quarters
• Rate per quarter = 16/4 = 4%
• A = 10000(1 + 4/100)³ = 10000(1.04)³ = 10000 × 1.124864 = ₹11,248.64

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Level 3: Advanced Scenarios (சிக்கலான கணக்குகள்)

Advanced Pattern: Variable Rate Compounding When interest rates change over different years, calculate each year separately and compound the results.

Common Trap

Don't average the rates! Apply each rate to the accumulated amount from the previous period.

Example 1: ₹5000 is invested at 10% for the first year, 12% for the second year, and 8% for the third year, all compounded annually. Solution:

• After Year 1: A₁ = 5000(1.10) = ₹5500
• After Year 2: A₂ = 5500(1.12) = ₹6160
• After Year 3: A₃ = 6160(1.08) = ₹6652.80
• Total CI = 6652.80 - 5000 = ₹1652.80

Advanced Pattern: Population/Depreciation Growth Example 2: A machine worth ₹50,000 depreciates at 10% per annum. What will be its value after 3 years? Solution:

• Depreciation means value decreases, so rate becomes negative in effect
• Value = 50000(1 - 10/100)³ = 50000(0.9)³ = 50000 × 0.729 = ₹36,450

Advanced Strategy

For depreciation problems, use (1 - R/100) instead of (1 + R/100). For population growth, use the same CI formula with population as principal.

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⚠️ Common Beginner Mistakes (பொதுவான தவறுகள்)

Mistake 1: Wrong Rate Division for Compounding ❌ Wrong: For half-yearly compounding, using original rate and doubling time ✅ Correct: Divide rate by 2 and multiply time by 2

Mistake 2: Forgetting to Subtract Principal ❌ Wrong: Reporting Amount as Compound Interest ✅ Correct: CI = Amount - Principal (always subtract!)

Mistake 3: Using SI Formula for CI Problems ❌ Wrong: CI = P × R × T / 100 (this is SI formula!) ✅ Correct: CI = P(1 + R/100)^T - P

Mistake 4: Incorrect Difference Calculation ❌ Wrong: For 3 years, using the 2-year difference shortcut formula ✅ Correct: For 3+ years, calculate CI and SI separately, then find difference

Mistake 5: Unit Confusion in Time Period ❌ Wrong: Using months directly in years' place without conversion ✅ Correct: Convert months to years: 18 months = 1.5 years

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🚀 Key Takeaways (முக்கிய குறிப்புகள்)

1. Master the Base Formula: A = P(1 + R/100)^T is your foundation - memorize variations for different compounding
2. Compounding Frequency Rules: Divide rate by frequency, multiply time by frequency
3. Shortcut for 2-Year CI-SI Difference: P × R²/(100)² - saves precious exam time
4. Growth Multiple Formula: (x)^(1/a) = (y)^(1/b) for exponential growth problems
5. Always Verify: CI should always be ≥ SI for the same principal, rate, and time

Success Formula

Compound Interest Mastery = (Practice × Pattern Recognition)^(Consistent Effort) + Time Shortcuts² The magic of compounding works in exams too - small daily practice compounds into big scores! 🎯

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📝 Problems to Practice (பயிற்சிக் கேள்விகள்)

Pattern 1: Direct Calculation (நேரடி கணக்கீடு) - 3 Problems

Problem 1: Find the compound interest on ₹7500 at 12% per annum for 2 years compounded annually. Pattern Identification: Direct application of annual compounding formula Solution:

• Given: P = ₹7500, R = 12%, T = 2 years, compounded annually
• Step 1: Apply formula → A = P(1 + R/100)^T
• Step 2: A = 7500(1 + 12/100)² = 7500(1.12)²
• Step 3: A = 7500 × 1.2544 = ₹9408
• Step 4: CI = A - P = 9408 - 7500 = ₹1908 Answer: ₹1908

Problem 2: What will be the amount of ₹15,000 after 1.5 years at 20% per annum compounded half-yearly? Pattern Identification: Mixed compounding (half-yearly) with time conversion Solution:

• Given: P = ₹15,000, R = 20%, T = 1.5 years, compounded half-yearly
• Step 1: Convert for half-yearly → Rate = 20/2 = 10% per half-year, Time = 1.5 × 2 = 3 half-years
• Step 2: A = 15000(1 + 10/100)³ = 15000(1.1)³
• Step 3: A = 15000 × 1.331 = ₹19,965
• Step 4: Verification → Amount increased by ~33% over 1.5 years ✓ Answer: ₹19,965

Problem 3: Calculate the compound interest on ₹8000 at 15% per annum for 9 months compounded quarterly. Pattern Identification: Quarterly compounding with months-to-quarters conversion Solution:

• Given: P = ₹8000, R = 15%, T = 9 months, compounded quarterly
• Step 1: Convert time → 9 months = 3 quarters
• Step 2: Rate per quarter → 15/4 = 3.75%
• Step 3: A = 8000(1 + 3.75/100)³ = 8000(1.0375)³
• Step 4: A = 8000 × 1.1168 = ₹8934.40
• Step 5: CI = 8934.40 - 8000 = ₹934.40 Answer: ₹934.40

Pattern 2: Growth Multiples (பெருக்க வளர்ச்சி) - 3 Problems

Problem 4: A sum of money becomes 3 times itself in 6 years at compound interest. In how many years will it become 27 times? Pattern Identification: Exponential growth pattern using multiple relationship Solution:

• Given: Money becomes 3 times in 6 years, find time for 27 times
• Step 1: Apply shortcut formula → (x)^(1/a) = (y)^(1/b)
• Step 2: (3)^(1/6) = (27)^(1/b)
• Step 3: (3)^(1/6) = (3³)^(1/b) = (3)^(3/b)
• Step 4: 1/6 = 3/b → b = 18
• Step 5: Verification → 27 = 3³, so it makes sense that time triples Answer: 18 years

Problem 5: At compound interest, a certain sum doubles in 4 years. In how many years will it become 8 times the original amount? Pattern Identification: Doubling pattern with exponential growth Solution:

• Given: Sum doubles (2 times) in 4 years, find time for 8 times
• Step 1: (2)^(1/4) = (8)^(1/b)
• Step 2: (2)^(1/4) = (2³)^(1/b) = (2)^(3/b)
• Step 3: 1/4 = 3/b → b = 12
• Step 4: Logic check → 2 times in 4 years, 4 times in 8 years, 8 times in 12 years ✓ Answer: 12 years

Problem 6: A population grows from 50,000 to 1,25,000 in 10 years. In how many more years will it reach 3,12,500? Pattern Identification: Population growth with compound increase calculation Solution:

• Given: 50,000 → 1,25,000 (2.5 times) in 10 years, find time to reach 3,12,500
• Step 1: 3,12,500 ÷ 50,000 = 6.25 times the original
• Step 2: (2.5)^(1/10) = (6.25)^(1/b)
• Step 3: (2.5)^(1/10) = ((2.5)²)^(1/b) = (2.5)^(2/b)
• Step 4: 1/10 = 2/b → b = 20 years total
• Step 5: Additional years needed = 20 - 10 = 10 years Answer: 10 more years

Pattern 3: Comparison Problems (ஒப்பீட்டு கணக்குகள்) - 3 Problems

Problem 7: Find the difference between compound interest and simple interest on ₹12,000 for 2 years at 10% per annum. Pattern Identification: CI vs SI difference for 2 years using shortcut formula Solution:

• Given: P = ₹12,000, R = 10%, T = 2 years
• Step 1: Use 2-year shortcut → Difference = P × R²/(100)²
• Step 2: Difference = 12000 × (10)²/(100)² = 12000 × 100/10000
• Step 3: Difference = 12000 × 0.01 = ₹120
• Step 4: Verification → SI = 12000×10×2/100 = ₹2400, CI = 12000(1.1)² - 12000 = ₹2520, Difference = ₹120 ✓ Answer: ₹120

Problem 8: On what sum will the compound interest for 3 years at 5% per annum exceed the simple interest by ₹61? Pattern Identification: Reverse calculation from CI-SI difference to find principal Solution:

• Given: R = 5%, T = 3 years, CI - SI = ₹61, find P
• Step 1: Calculate SI → SI = P × 5 × 3/100 = 0.15P
• Step 2: Calculate CI → A = P(1.05)³ = P × 1.157625 = 1.157625P, CI = 0.157625P
• Step 3: Set up equation → CI - SI = 61
• Step 4: 0.157625P - 0.15P = 61 → 0.007625P = 61
• Step 5: P = 61/0.007625 = ₹8000 Answer: ₹8000

Problem 9: The compound interest on a sum for 2 years at 8% per annum is ₹166.40. Find the simple interest on the same sum for the same period. Pattern Identification: From CI to SI conversion using relationship formulas Solution:

• Given: CI = ₹166.40, R = 8%, T = 2 years, find SI
• Step 1: Use relationship → CI = SI + (SI × R × T/100) for 2 years becomes CI = SI + P×R²/(100)²
• Step 2: Let SI = x, then difference = P×R²/(100)² = P×64/10000
• Step 3: From A = P(1.08)² = 1.1664P, CI = 0.1664P = 166.40 → P = ₹1000
• Step 4: SI = P×R×T/100 = 1000×8×2/100 = ₹160
• Step 5: Check → Difference = 166.40 - 160 = ₹6.40 = 1000×64/10000 ✓ Answer: ₹160

Pattern 4: Mixed Compounding (கலப்பு கூட்டல் முறை) - 3 Problems

Problem 10: Find the amount on ₹18,000 at 16% per annum for 1 year if compounded quarterly. Pattern Identification: Quarterly compounding with rate and time adjustment Solution:

• Given: P = ₹18,000, R = 16%, T = 1 year, compounded quarterly
• Step 1: Adjust for quarterly → Rate = 16/4 = 4% per quarter, Time = 1×4 = 4 quarters
• Step 2: A = 18000(1 + 4/100)⁴ = 18000(1.04)⁴
• Step 3: A = 18000 × 1.16986 = ₹21,057.48
• Step 4: Verification → Annual would give 18000×1.16 = ₹20,880, quarterly > annual ✓ Answer: ₹21,057.48

Problem 11: What will ₹25,000 amount to in 18 months at 12% per annum compounded half-yearly? Pattern Identification: Half-yearly compounding with months conversion Solution:

• Given: P = ₹25,000, R = 12%, T = 18 months, compounded half-yearly
• Step 1: Convert time → 18 months = 3 half-years
• Step 2: Rate per half-year → 12/2 = 6%
• Step 3: A = 25000(1 + 6/100)³ = 25000(1.06)³
• Step 4: A = 25000 × 1.191016 = ₹29,775.40
• Step 5: CI = 29,775.40 - 25,000 = ₹4,775.40 Answer: ₹29,775.40

Problem 12: Find the compound interest on ₹20,000 for 8 months at 18% per annum compounded monthly. Pattern Identification: Monthly compounding with exact months calculation Solution:

• Given: P = ₹20,000, R = 18%, T = 8 months, compounded monthly
• Step 1: Rate per month → 18/12 = 1.5%
• Step 2: Time = 8 months
• Step 3: A = 20000(1 + 1.5/100)⁸ = 20000(1.015)⁸
• Step 4: A = 20000 × 1.12649 = ₹22,529.80
• Step 5: CI = 22,529.80 - 20,000 = ₹2,529.80 Answer: ₹2,529.80