Simple and Compound Interest — Study Notes

Overview

Simple and Compound Interest is a core topic in SSC CGL Tier 1 Quantitative Aptitude, typically featuring 2–3 questions worth 4–6 marks. This topic tests your understanding of how money grows over time under different compounding scenarios. Mastery requires comfort with basic formulas, the ability to handle fractional time periods, and familiarity with non-annual compounding (half-yearly, quarterly). Questions on installments and population/depreciation analogies also appear regularly. The key to success is recognizing problem patterns quickly and applying the correct formula without calculation errors. Unlike more abstract topics, interest problems are highly formulaic—memorize the core equations and practice variations to build speed.

SI questions are generally straightforward plug-and-play, while CI problems demand careful attention to compounding frequency and the number of periods. Mixed problems (comparing SI vs. CI on the same principal) are common. The exam also tests reverse problems where you must find principal, rate, or time given the interest or amount. Understanding the relationship between these variables and practicing mental math shortcuts (like recognizing that CI for 2 years = P × r(200 + r)/10000) will save valuable exam time.

Key Concepts

• **Simple Interest grows linearly**: Each year's interest is calculated only on the original principal, so SI = (P × R × T)/100. The total amount A = P + SI.
• **Compound Interest grows exponentially**: Each period's interest is added to the principal before calculating the next period's interest. The amount formula is A = P(1 + R/100)^n, where n is the number of compounding periods.
• **Difference between SI and CI**: For small time periods (2–3 years), the difference is small but predictable. For 2 years: CI − SI = P(R/100)². This shortcut speeds up comparison problems.
• **Non-annual compounding**: When interest compounds half-yearly, divide the annual rate by 2 and double the time in years. For quarterly compounding, divide rate by 4 and multiply time by 4.
• **Rate and time adjustments**: Always match the rate period to the time period. If rate is annual but compounding is half-yearly, convert both consistently.
• **Installment problems**: When paying in installments, each installment's present value must sum to the principal. Use the formula for present value: PV = Amount/(1 + R/100)^n.
• **Depreciation and population**: These use the CI formula with a decreasing rate (depreciation) or increasing rate (population growth). The structure is identical to CI problems.
• **Fractional years**: For CI with fractional time (e.g., 2.5 years), compute CI for whole years first, then apply SI for the fractional part on the new amount.

Formulas / Key Facts

**Simple Interest:**

• SI = (P × R × T)/100
• Amount A = P + SI = P(1 + RT/100)
• Principal P = (SI × 100)/(R × T)
• Rate R = (SI × 100)/(P × T)
• Time T = (SI × 100)/(P × R)

**Compound Interest (Annual Compounding):**

• Amount A = P(1 + R/100)^T
• CI = A − P = P[(1 + R/100)^T − 1]
• If compounded half-yearly: A = P(1 + R/200)^(2T)
• If compounded quarterly: A = P(1 + R/400)^(4T)

**Difference between CI and SI:**

• For 2 years: CI − SI = P(R/100)²
• For 3 years: CI − SI = P(R/100)²(300 + R)/100

**Installment Problems:**

• For n equal installments x, each paid at the end of successive years: P = x/(1 + R/100) + x/(1 + R/100)² + ... + x/(1 + R/100)^n

**Depreciation:**

• Value after T years = P(1 − R/100)^T

**Population Growth:**

• Population after T years = P(1 + R/100)^T

Worked Examples

**Example 1: Basic SI Problem** Find the simple interest on Rs 8000 at 5% per annum for 3 years. **Solution:** SI = (P × R × T)/100 = (8000 × 5 × 3)/100 = 120000/100 = Rs 1200. Amount = 8000 + 1200 = Rs 9200.

**Example 2: Compound Interest with Annual Compounding** Find the compound interest on Rs 10000 at 10% per annum for 2 years. **Solution:** A = P(1 + R/100)^T = 10000(1 + 10/100)² = 10000(1.1)² = 10000 × 1.21 = Rs 12100. CI = 12100 − 10000 = Rs 2100. **Alternate (shortcut):** CI for 2 years = P(R/100)²(200 + R)/100 = 10000 × 0.1 × 0.1 × 210/100 = 2100. Actually, simpler: CI = P[2R/100 + (R/100)²] = 10000[0.2 + 0.01] = 2100.

**Example 3: Half-Yearly Compounding** Find CI on Rs 4000 at 20% per annum compounded half-yearly for 1 year. **Solution:** Half-yearly rate = 20/2 = 10%, time = 1 × 2 = 2 half-years. A = 4000(1 + 10/100)² = 4000(1.1)² = 4000 × 1.21 = Rs 4840. CI = 4840 − 4000 = Rs 840.

**Example 4: Difference between CI and SI** The difference between CI and SI on a sum for 2 years at 8% per annum is Rs 80. Find the principal. **Solution:** CI − SI = P(R/100)² 80 = P(8/100)² 80 = P × 64/10000 P = (80 × 10000)/64 = 12500. Principal = Rs 12500.

Common Mistakes

• **Confusing rate and time units**: Applying a 10% annual rate directly to half-yearly compounding without halving the rate. Always convert rate and time to match the compounding frequency.
• **Forgetting to subtract principal for CI**: Calculating amount A correctly but forgetting CI = A − P. Many students write the amount as the final answer when CI is asked.
• **Misapplying the difference formula**: Using CI − SI = P(R/100)² for 3 years instead of 2 years. The shortcut formula for difference changes with the number of years.
• **Rounding intermediate values prematurely**: Rounding 1.1² to 1.2 instead of 1.21 leads to wrong answers. Keep precision until the final step, or use exact fractions.
• **Mixing up installment formulas**: Treating all installments as paid at the start or end inconsistently. Read the problem carefully to determine if installments are at year-start or year-end, and discount accordingly.

Quick Reference

• SI formula: SI = PRT/100; grows linearly each year.
• CI formula (annual): A = P(1 + R/100)^T; exponential growth.
• Half-yearly CI: halve rate, double time periods.
• Quarterly CI: divide rate by 4, multiply time by 4.
• CI − SI for 2 years: P(R/100)².
• Installments: sum of present values = principal.
• Depreciation: same as CI with (1 − R/100)^T.