Simple & Compound Interest – Study Notes (Quantitative Aptitude)
Target Exam – Assam Patwari / Lekhpal
Subject – Quantitative Aptitude
Topic – Simple & Compound Interest
1. Simple Explanation
| Concept | Simple Interest (SI) | Compound Interest (CI) |
|---|---|---|
| What is it? | Interest that is earned only on the original principal. | Interest that is earned on the principal and on all previously earned interest. |
| Formula | SI = P × R × T / 100 |
A = P (1 + r/n)^(nT) (if compounding n times per year) |
Amount after T years |
A = P + SI = P (1 + R T / 100) |
A = P (1 + r/n)^(nT) |
Examples of n |
1 (annual), 4 (quarterly), 12 (monthly), 365 (daily) | Same as above |
P = Principal (initial amount)
R = Annual rate of interest (percentage)
r = Annual rate in decimal (i.e. R/100)
T = Time period in years
n = Number of compounding periods per year
2. Key Points (Important for Exam)
| # | Point | Why it matters |
|---|---|---|
| 1 | SI depends only on the initial principal | Easy to compute; used for short‑term loans or simple savings. |
| 2 | CI grows faster because interest is added to the principal | Key to understanding long‑term investment growth. |
| 3 | Compounding frequency n matters |
More frequent compounding → higher effective yield. |
| 4 | Effective Annual Rate (EAR) | EAR = (1 + r/n)^n – 1 – useful to compare different rates or compounding schedules. |
| 5 | Continuous compounding | A = P e^(rt) – rarely used in exams but good to know for theory. |
| 6 | Fractional periods | Use proportional time: T = (days elapsed) / (days in year) (usually 365). |
| 7 | Reverse calculations | Given A, P, T, find R or t using the formulas. |
| 8 | Common pitfalls | • Mixing up R and r. • Forgetting to divide by 100. • Using wrong compounding frequency. |
3. Worked Example
Problem
A person deposits ₹10,000 in a savings account.
(i) The bank offers simple interest at 12 % per annum. Find the amount after 3 years.
(ii) The same bank offers compound interest at 12 % per annum compounded quarterly. Find the amount after 3 years.
(iii) Which scheme yields a higher return? By how much?
Solution
| Scheme | Formula | Calculation | Result |
|---|---|---|---|
| (i) Simple Interest | A = P (1 + R T / 100) |
A = 10,000 × (1 + 12 × 3 / 100) = 10,000 × (1 + 0.36) = 10,000 × 1.36 |
₹13,600 |
| (ii) Compound Interest (quarterly) | A = P (1 + r/n)^(nT) where r = 0.12, n = 4, T = 3 |
A = 10,000 × (1 + 0.12/4)^(4×3) = 10,000 × (1 + 0.03)^12 = 10,000 × (1.03)^12 ≈ 10,000 × 1.42576 ≈ ₹14,257.60 |
₹14,257.60 |
| (iii) Difference | ₹14,257.60 – ₹13,600 = ₹657.60 |
₹657.60 more with compound interest |
Result: Compound interest yields a higher return by ₹657.60.
4. Quick Revision Summary
| Topic | Formula | Use‑case | Quick Tip |
|---|---|---|---|
| Simple Interest | SI = P × R × T / 100 |
Short‑term loans, simple savings | Remember R is in percent, divide by 100 |
| Amount (SI) | A = P + SI |
Final payout | A = P (1 + R T / 100) |
| Compound Interest | A = P (1 + r/n)^(nT) |
Long‑term investments, loans with |