Simple & Compound Interest
CSIR‑NET Quantitative Aptitude – Study Notes
1. Simple Explanation
| Concept | What it is | How it works |
|---|---|---|
| Simple Interest (SI) | Interest earned only on the original principal. | Interest = Principal × Rate × Time |
| Compound Interest (CI) | Interest earned on the original principal plus all interest accumulated so far. | Future Value = Principal × (1 + r/n)^(n·t) where n = compounding periods per year. |
Key idea – SI grows linearly with time; CI grows exponentially because each period’s interest is reinvested.
2. Key Points & Formulas
| Topic | Formula | Notes |
|---|---|---|
| Simple Interest | [SI = P \times r \times t] | - P = principal (initial amount) - r = annual rate (decimal) - t = time in years |
| Maturity Amount (Simple) | [A_{SI} = P + SI = P(1 + rt)] | |
| Compound Interest | [A = P\left(1 + \frac{r}{n}\right)^{n t}] | - n = compounding frequency per year (1=annually, 2=semi‑annually, 4=quarterly, 12=monthly, 365=daily) |
| Effective Annual Rate (EAR) | [EAR = \left(1 + \frac{r}{n}\right)^n - 1] | Used to compare rates with different compounding. |
| Annual Equivalent Rate (AER) | Same as EAR but often quoted for banks. | |
| Continuous Compounding | [A = P e^{rt}] | As n → ∞. |
| Interest Rate for a Different Period | For a rate r per year: rate for k years = k·r. For a rate r per p years: rate per year = r/p. |
Useful for quick conversion. |
| Effective Rate for Partial Year | If a rate r is quoted for p months, the annual rate = r × (12/p). |
|
| Comparison of SI and CI | (A_{CI} \ge A_{SI}) always; the difference grows with higher r, longer t, or higher n. |
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| Rule of 72 (approx.) | Years to double ≈ 72 / (annual rate %). | For quick estimation. |
3. Worked Example
Problem
A sum of ₹10,000 is invested at 8 % per annum for 4 years.
- Calculate the maturity amount under Simple Interest.
- Calculate the maturity amount under Compound Interest compounded quarterly.
- Find the difference between the two amounts.
Solution
| Step | Calculation | Result |
|---|---|---|
| 1. SI | (A_{SI}=P(1+rt)=10,000(1+0.08\times4)=10,000(1+0.32)=13,200) | ₹13,200 |
| 2. CI | (A_{CI}=P\left(1+\frac{r}{n}\right)^{nt}=10,000\left(1+\frac{0.08}{4}\right)^{4\times4}) First compute inside: (1+\frac{0.08}{4}=1+0.02=1.02). Exponent: (4\times4=16). Then (A_{CI}=10,000\times(1.02)^{16}). Using a calculator: ((1.02)^{16}\approx1.3686). So (A_{CI}\approx10,000\times1.3686=13,686). |
₹13,686 |
| 3. Difference | (13,686-13,200=486) | ₹486 |
Result – Under the same nominal rate, CI yields ₹486 more after 4 years.
Quick mental check – (1.02^{16}) ≈ (1.02^8^2) ≈ (1.1716^2) ≈ 1.37, close to the exact value.
4. Quick Revision Summary
| Concept | Remember |
|---|---|
| Simple vs Compound | SI = P r t; CI = P (1+r/n)^{nt} |
| Linear vs Exponential | SI grows linearly; CI grows exponentially |
| Compounding frequency | n = 1 (annual), 2 (semi‑annual |