Simple & Compound Interest

Simple & Compound Interest

🟢 Lite

Key Formula/Rule

SI = (P × R × T) ÷ 100. Amount (SI) = P + SI = P(1 + RT/100). CI: Amount = P(1 + R/100)^T.

Memory Trick

SI = “P × R × T over 100” — three ingredients, one division. For CI, you’re multiplying the growth factor (1 + R/100) by itself T times because each period’s amount becomes the next period’s base.

1-Sentence Summary

Tests whether you understand the difference between interest calculated only on the original principal (SI) vs. interest calculated on the accumulated amount each period (CI).

30-Second Example

Q: SI on ₹5,000 at 6% per annum for 2 years? A: ₹600 — (5000 × 6 × 2) ÷ 100 = 600

🟡 Standard

Concept

Interest is basically the “rent” you pay for using someone else’s money, or the “bonus” you earn for lending yours. Simple Interest (SI) is the straightforward version: you borrow ₹10,000 at 8% per year, you pay 8% of ₹10,000 = ₹800 every single year, flat. The interest never changes because it’s always calculated on the original ₹10,000, never on the accumulated amount. So over 3 years, SI = ₹800 × 3 = ₹2,400.

Compound Interest (CI) is where it gets interesting. Instead of calculating interest on just the original principal, you calculate it on the accumulated amount each period. In year one, interest is 8% of ₹10,000 = ₹800, so you now owe ₹10,800. In year two, interest is 8% of ₹10,800 = ₹864. You’re now paying interest on interest — which is both good when you’re earning and painful when you’re borrowing.

The formula for CI is Amount = P(1 + R/100)^T. That exponent T is doing heavy lifting — it means you’re multiplying (1 + R/100) by itself T times. For 3 years at 8%, that’s (1.08)^3 = 1.2597, so ₹10,000 becomes ₹12,597. The extra ₹197 compared to SI (₹12,400) is the “interest on interest” effect.

When interest is compounded half-yearly (twice a year), the rate per half-year becomes R/2 and the number of periods becomes 2T. So 8% per annum compounded half-yearly means 4% every 6 months for 2 years (4 periods).

Key Formulas

Formula	Use
SI = (P × R × T) ÷ 100	Simple interest for one year
Amount (SI) = P + SI = P(1 + RT/100)	Total amount with SI
CI: A = P(1 + R/100)^T	Amount with CI (annual compounding)
CI − SI difference	Extra amount earned with CI vs. SI
Rate per period = R ÷ n, Periods = n × T	For n-times-per-year compounding

Worked Example

Q: Find the compound interest on ₹20,000 at 10% per annum for 2 years, compounded annually.

Step 1: Year 1 interest = 10% of 20,000 = ₹2,000 Year 1 amount = 20,000 + 2,000 = ₹22,000

Step 2: Year 2 interest = 10% of 22,000 = ₹2,200 Year 2 amount = 22,000 + 2,200 = ₹24,200

Step 3: CI = Final Amount − Principal = 24,200 − 20,000 = ₹4,200

Or use formula: A = 20000 × (1 + 10/100)^2 = 20000 × 1.1 × 1.1 = ₹24,200 → CI = 24,200 − 20,000 = ₹4,200

Answer: ₹4,200

Common Errors

• Using time T directly for half-yearly compounding → Convert properly: 2 years half-yearly means 4 periods at (R/2)% each
• Confusing SI and CI formulas → SI uses P × R × T (linear); CI uses P raised to power T (exponential)
• Forgetting to subtract the principal from CI → CI is not the final amount; it’s the interest portion only: CI = A − P

🔴 Extended

Full Concept

Why CI Grows Exponentially

This is the key insight. With SI, after T years your money grows linearly: P + P×R×T/100. With CI, it grows multiplicatively: P × (1 + R/100)^T. The difference is enormous over long periods because (1 + R/100)^T grows exponentially while P×R×T grows only linearly.

Compare ₹10,000 at 12% per year for 30 years:

• SI: 10,000 + (10,000 × 12 × 30)/100 = ₹46,000
• CI: 10,000 × (1.12)^30 = ₹29,95,992 ≈ ₹30 lakhs

Same principal, same rate, same time — but CI gives you roughly 65× more. That’s the power of compounding, and it’s why starting to invest early matters so much.

More Frequent Compounding

When CI is compounded more than once a year, the formula adjusts: A = P(1 + R/(100×n))^(n×T) where n = number of compounding periods per year. For monthly compounding at 12% per annum: n = 12, so rate per month = 1%, periods = 12T.

For quarterly: A = P(1 + R/400)^(4T) For monthly: A = P(1 + R/1200)^(12T)

The more frequent the compounding, the higher the effective amount — because you’re earning “interest on interest” more often. But notice the rate per period shrinks proportionally, so the benefit is real but not dramatic.

Effective Rate vs. Nominal Rate

The nominal rate is the stated annual rate. The effective rate is what you actually earn or pay when compounding is factored in. If the nominal rate is 12% compounded monthly, the effective annual rate is (1 + 0.12/12)^12 − 1 = (1.01)^12 − 1 ≈ 0.1268 = 12.68%. So the effective rate is slightly higher than the nominal rate because of more frequent compounding.

The Rule of 72

This is a beautiful shortcut for doubling time. Divide 72 by the annual rate (r%) and you get the approximate number of years to double your money. At 6% per year: 72 ÷ 6 = 12 years. At 9%: 72 ÷ 9 = 8 years. It’s not exact, but for CUET it’s close enough and saves precious exam time.

For more precision: Rule of 69.3 (better for continuous compounding) or Rule of 70 (close enough for most rates).

Depreciation

This is just CI in reverse — a constant percentage decrease each period. If a machine worth ₹1,00,000 depreciates at 20% per year: after year 1 = 1,00,000 × 0.80 = ₹80,000. After year 2 = 80,000 × 0.80 = ₹64,000. Formula: Value after T years = P × (1 − r/100)^T. Notice the minus sign instead of plus.

Multiple Approaches

Standard — Formula Method: CI = P[(1 + R/100)^T − 1] This is cleaner when you just need the interest portion, not the full amount.

Shortcut — Rule of 72: Years to double ≈ 72 ÷ r. Works best for rates between 4% and 20%. Outside this range, error increases.

Shortcut — Net Change Approximation: For two successive rates r₁% and r₂%: Net factor = (1 + r₁/100)(1 + r₂/100). Equivalent to net% = r₁ + r₂ + (r₁×r₂)/100.

CUET-Level Problems

Q1: The difference between CI and SI on a sum at 10% per annum for 2 years is ₹31. Find the sum. Working: SI for 2 years = (P × 10 × 2)/100 = 0.20P CI Year 1 = 0.10P, Amount after Y1 = 1.10P CI Year 2 = 0.10 × 1.10P = 0.11P, Total CI = 0.10P + 0.11P = 0.21P Difference = 0.21P − 0.20P = 0.01P = 31 P = ₹3,100 Answer: ₹3,100

Q2: A car worth ₹8,00,000 depreciates at 10% per year. What will it be worth after 3 years? Working: After Y1: 8,00,000 × 0.90 = 7,20,000 After Y2: 7,20,000 × 0.90 = 6,48,000 After Y3: 6,48,000 × 0.90 = 5,83,200 Or: 8,00,000 × (0.9)^3 = 8,00,000 × 0.729 = ₹5,83,200 Answer: ₹5,83,200

Tricky Cases

• CI when rate differs each year: Don’t use the single formula. Apply each year’s rate sequentially: A = P × (1 + r₁/100) × (1 + r₂/100) × (1 + r₃/100).
• SI and CI being equal for specific combinations: This happens at certain T and R values. For 2 years, CI − SI = P × (R/100)^2. Setting this equal to some value lets you solve backwards.
• Decimal years: 6 months = 0.5 years, 18 months = 1.5 years. Always convert time to the same unit as the rate before applying formulas.

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