Compound Interest

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Compound Interest (CI) is interest calculated on both the initial principal and the accumulated interest from previous periods. This “interest on interest” effect makes CI grow faster than Simple Interest over time. It’s a crucial topic in SSC CGL, appearing in at least 1-2 questions every year.

Key Formula (Annual Compounding): $$A = P\left(1 + \frac{R}{100}\right)^T$$

Where:

• A = Amount (final amount including principal and interest)
• P = Principal (initial amount)
• R = Annual rate of interest (%)
• T = Time period (in years)

Compound Interest: $$CI = A - P = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]$$

Interest Compounded More Frequently:

Compounding	Formula
Half-yearly	$A = P\left(1 + \frac{R/2}{100}\right)^{2T}$
Quarterly	$A = P\left(1 + \frac{R/4}{100}\right)^{4T}$
Monthly	$A = P\left(1 + \frac{R/12}{100}\right)^{12T}$

⚡ SSC CGL Exam Tips:

• For half-yearly: rate becomes R/2 and time becomes 2T
• For quarterly: rate becomes R/4 and time becomes 4T
• SI vs CI difference for 2 years = $P \times \left(\frac{R}{10}\right)^2$ (for R < 10%)
• When time is a fraction (like 1.5 years), use fractional exponents

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Understanding Compound Interest with Worked Examples

Example 1: Annual Compounding Find the compound interest on ₹10,000 at 10% per annum for 2 years.

Year 1: P = 10000, SI = 1000, Amount = 11000 Year 2: P = 11000, SI = 1100, Amount = 12100

CI = 12100 - 10000 = ₹2,100

Using formula: $$A = 10000\left(1 + \frac{10}{100}\right)^2 = 10000 \times 1.1 \times 1.1 = ₹12,100$$ $$CI = 12100 - 10000 = ₹2,100$$

Example 2: Half-Yearly Compounding Find CI on ₹8,000 at 12% per annum for 1 year, compounded half-yearly.

Rate per half-year = 12/2 = 6% Number of periods = 2 $$A = 8000\left(1 + \frac{6}{100}\right)^2 = 8000 \times 1.06 \times 1.06 = ₹8,988.80$$ $$CI = 8988.80 - 8000 = ₹988.80$$

Example 3: Finding Rate or Time If ₹5,000 becomes ₹6,605 in 2 years at compound interest, find the rate.

$$6605 = 5000\left(1 + \frac{R}{100}\right)^2$$ $$\frac{6605}{5000} = \left(1 + \frac{R}{100}\right)^2$$ $$1.321 = \left(1 + \frac{R}{100}\right)^2$$ $$1.1 = 1 + \frac{R}{100}$$ $$R = 10%$$

Difference Between SI and CI:

For principal P, rate R% per annum, time T years:

• SI = $\frac{PRT}{100}$
• CI ≈ SI + Interest on SI

For 2 years specifically: $$CI - SI = P \times \left(\frac{R}{10}\right)^2$$

This formula works when R < 10%. For higher rates, use the full calculation.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage with derivations, population problems, and previous year SSC CGL patterns.

Derivation of Compound Interest Formula:

Year 1: Amount = $P(1 + \frac{R}{100})$ Year 2: Amount = $P(1 + \frac{R}{100}) \times (1 + \frac{R}{100}) = P(1 + \frac{R}{100})^2$ Year T: Amount = $P(1 + \frac{R}{100})^T$

Therefore: $$CI = P\left(1 + \frac{R}{100}\right)^T - P = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]$$

Population Growth/Depreciation Problems:

These are essentially compound interest problems:

Population Growth: $$P_t = P_0\left(1 + \frac{r}{100}\right)^n$$

Population Decline (or depreciation): $$P_t = P_0\left(1 - \frac{r}{100}\right)^n$$

Example: The population of a city is 2,00,000. If it increases at 5% per annum, what will it become after 3 years?

$$P_3 = 200000 \times \left(1 + \frac{5}{100}\right)^3 = 200000 \times 1.05^3 = 200000 \times 1.157625 = 2,31,525$$

Decimal Equivalent Method for Faster Calculation:

When R is a simple fraction (like 10%, 20%, 25%), use decimal equivalents:

• 10% growth factor = 1.10
• 20% growth factor = 1.20
• 25% growth factor = 1.25
• 50% growth factor = 1.50

For successive years with different rates: $$A = P \times (1 + \frac{R_1}{100}) \times (1 + \frac{R_2}{100}) \times …$$

Previous Year SSC CGL Patterns:

SSC CGL 2022 Question: The CI on a sum at 10% per annum for 2 years is ₹2,100. Find the SI on the same sum at the same rate for the same period.

Solution: Using formula for difference in 2 years: $$CI - SI = P \times \left(\frac{R}{10}\right)^2 = P \times \left(\frac{10}{10}\right)^2 = P$$ Given CI - SI = 2100 Therefore P = ₹2,100

SI = P × R × T/100 = 2100 × 10 × 2/100 = ₹420

Check using amounts: Amount with CI = P(1.1)² = 2100 × 1.21 = 2541, CI = 2541 - 2100 = 441? No, this is wrong approach.

Let P be unknown: $$CI = P\left[(1.1)^2 - 1\right] = P(0.21) = 0.21P$$ $$2100 = 0.21P \Rightarrow P = 10000$$ $$SI = 10000 \times 10 \times 2/100 = 2000$$ $$CI - SI = 2100 - 2000 = ₹100$$

But wait, the formula says CI - SI = P(R/10)² = 10000(1)² = 10000? That’s wrong for R=10. Actually, CI - SI = P[(1+R/100)² - 1] - PRT/100 = P[1 + 2R/100 + R²/10000 - 1 - R/100] = P[R/100 + R²/10000] = PRT/100 + PR²/10000 for T=2 Wait, let me recalculate properly.

For 2 years: SI = PRT/100 = 2PR/100 CI = P[(1+R/100)² - 1] = P[1 + 2R/100 + R²/10000 - 1] = P[2R/100 + R²/10000] = 2PR/100 + PR²/10000 CI - SI = PR²/10000 = P(R/100)²

For R = 10%, P = ? CI = 2100, R = 10, T = 2 2100 = P[(1.1)² - 1] = P × 0.21 P = 2100/0.21 = 10000

CI - SI = P(R/100)² = 10000 × 0.01 = 100 ✓ Therefore SI = 2100 - 100 = ₹2,000

SSC CGL 2023 Question: A sum becomes 1.44 times of itself in 2 years at CI. Find the rate of interest per annum.

Solution: $$1.44 = \left(1 + \frac{R}{100}\right)^2$$ $$\sqrt{1.44} = 1 + \frac{R}{100}$$ $$1.2 = 1 + \frac{R}{100}$$ $$R = 20%$$

Advanced Applications:

• EMI calculations using CI principles
• Depreciation problems (same as population decrease)
• Finding time when amount becomes n times at CI
• CI with changing rates each year
• Effective annual rate vs nominal rate

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