Simple & Compound Interest – Study Notes
For the Rajasthan Forest Guard Quantitative Aptitude module
1. Introduction
Interest is the cost of borrowing money or the return on money invested.
Two primary types are used in exams:
| Type | When it is paid | Typical use in tests |
|---|---|---|
| Simple Interest (SI) | At the end of the whole period | Basic word‑problem, “interest on a deposit” |
| Compound Interest (CI) | At the end of each period (year, month, etc.) | “Interest on a loan after 3 years” |
Knowing the formulas, how to apply them, and common pitfalls is key.
2. Simple Interest
2.1 Quick Definition
The interest earned or paid is proportional only to the principal and the time.
SI = P × R × T
P = principal (₹), R = annual rate (%) , T = time (years)
2.2 Core Relationships
| Quantity | Symbol | Formula |
|---|---|---|
| Interest | I | (I = PRT) |
| Maturity amount | A | (A = P + I = P(1+RT)) |
| Rate | R | (R = \dfrac{I}{PT}) |
| Time | T | (T = \dfrac{I}{PR}) |
| Principal | P | (P = \dfrac{I}{RT}) |
2.3 Common Mistakes
| Mistake | Why it fails | Fix |
|---|---|---|
| Using R/100 in the formula | Confusing rate units | Remember R is already in percent; formula uses R directly |
| Mixing units of time | Seconds vs years | Convert all time units to years (or months consistently) |
| Adding interest to principal before using it in a new formula | Creates double counting | Use only the original P for each calculation |
3. Compound Interest
3.1 Quick Definition
Interest is calculated on the accumulated amount at the end of each period.
CI = (A - P)
(A = P\left(1+\frac{R}{100}\right)^{n})
- n = number of compounding periods (years, months, etc.)
- If compounding m times per year:
(A = P\left(1+\frac{R}{100m}\right)^{mn})
3.2 Core Relationships
| Quantity | Symbol | Formula |
|---|---|---|
| Accumulated amount | A | (A = P(1+r)^n) where (r=\frac{R}{100}) |
| Interest | CI | (CI = A - P) |
| Rate | R | (R = \left[\left(\frac{A}{P}\right)^{1/n} - 1\right] \times 100) |
| Time | T | (T = \dfrac{\ln(A/P)}{\ln(1+r)}) |
| Principal | P | (P = A/(1+r)^n) |
3.3 Common Mistakes
| Mistake | Why it fails | Fix |
|---|---|---|
| Forgetting to convert rate to decimal | Incorrect growth factor | Use (r = R/100) |
| Using years for n when compounding monthly | Wrong exponent | Set (n = \text{months}), (r = R/(12×100)) |
| Ignoring “compounding frequency” | Underestimates amount | Apply (m) compounding periods per year |
4. Quick Comparison
| Feature | Simple Interest | Compound Interest |
|---|---|---|
| Interest per period | Same amount | Increases each period |
| Formula for amount | (P(1+RT)) | (P(1+r)^n) |
| Use when | Time < 1 year, or exam remains simple | Long‑term loan/ investment |
| Derivative of amount w.r.t. time | Linear | Exponential |
5. Worked Examples
Example 1 – Simple Interest
A ₹12,000 deposit is kept for 4 years at 5% per annum. Find the total amount.
Solution
[ I = PRT = 12,000 \times 0.05 \times 4 = 2,400 ]
[ A = P + I = 12,000 + 2,400 = 14,400 ]
Answer: ₹14,400
Example 2 – Compound Interest (Annual)
₹5,000 is invested for 3 years at 6% per annum compounded annually. Find the amount.
Solution
[ A = 5,000 \left(1+\frac{6}{100}\right)^{3} = 5,000 (1.06)^3 ]
[ (1.06)^3 \approx 1.191016 ]
[ A \approx 5,000 \times 1.191016 = 5,955.08 ]
Answer: ₹5,955.08
Example 3 – Compound Interest (Quarterly)
A ₹10,000 loan is to be repaid in 2 years, with interest charged quarterly at 8% per annum. Find the total amount payable.
Parameters
- m = 4 (quarterly)
- n = 2 × 4 = 8 quarters
- r = R/(100m) = 8/(100×4) = 0.02
Solution
[ A = 10,000 \left(1+0.02\right)^{8} ]
[ (1.02)^8 \approx 1.171659 ]
[ A \approx 10,000 \times 1.171659 = 11,716.59 ]
Answer: ₹11,716.59