📚 PTE – Quantitative Aptitude
Topic: Simple & Compound Interest
Goal: Master the fundamentals, formulae, and problem‑solving techniques for Simple Interest (SI) and Compound Interest (CI)—the two most common questions in the PTE Quant section.
1️⃣ Simple Explanation
| Concept | What it means | When it applies |
|---|---|---|
| Simple Interest (SI) | Interest calculated only on the original principal for each time period. | Short‑term loans, fixed‑rate savings, PTE practice questions that ask for “simple interest” or “interest earned after X years”. |
| Compound Interest (CI) | Interest calculated on the principal plus all previously earned interest. The compounding frequency (annually, semi‑annually, quarterly, monthly, daily) determines how often the interest is added. | Long‑term investments, savings accounts, loans with periodic compounding, PTE questions that mention “compounded” or give a compounding frequency. |
Key Idea:
- SI = Linear growth (straight line).
- CI = Exponential growth (curved line).
2️⃣ Simple Interest (SI)
🔢 Formula
[ \text{SI} = \frac{P \times R \times T}{100} ]
- P = Principal (initial amount)
- R = Annual Rate of Interest (in %)
- T = Time (in years)
Total amount after T years
[ A = P + \text{SI} = P \left( 1 + \frac{R T}{100} \right) ]
📌 Key Points
| Point | Explanation |
|---|---|
| Rate & Time Units | Both must be in the same time unit. If rate is per annum and time is in months, convert months → years. |
| No compounding | Interest never earns interest. |
| Linear growth | Doubling the time doubles the interest. |
| Simple interest on a loan | The borrower pays the same percentage of the original loan each year. |
| SI vs CI | For the same P, R, T, CI is always ≥ SI. |
📚 Worked Example
Problem:
A student invests ₹12,000 at a simple interest rate of 5 % per annum for 4 years. What is the total amount at the end of the period?
Solution:
-
Identify variables:
- (P = ₹12,000)
- (R = 5%)
- (T = 4) years
-
Compute SI:
[ \text{SI} = \frac{12,000 \times 5 \times 4}{100} = \frac{240,000}{100} = ₹2,400 ]
- Compute total amount:
[ A = P + \text{SI} = 12,000 + 2,400 = ₹14,400 ]
Answer: ₹14,400 after 4 years.
3️⃣ Compound Interest (CI)
🔢 Formula
[ A = P \left(1 + \frac{R}{n}\right)^{nT} ]
- P = Principal
- R = Annual Rate of Interest (in %)
- n = Number of compounding periods per year
- T = Time (in years)
- CI = (A - P)
Special Cases
| Scenario | Formula |
|---|---|
| Annual compounding (n = 1) | (A = P(1 + R/100)^T) |
| Semi‑annual compounding (n = 2) | (A = P(1 + R/(2 \times 100))^{2T}) |
| Quarterly (n = 4) | (A = P(1 + R/(4 \times 100))^{4T}) |
| Monthly (n = 12) | (A = P(1 + R/(12 \times 100))^{12T}) |
| Continuous compounding | (A = P e^{(R/100)T}) (use Excel or calculator to approximate e) |
📌 Key Points
| Point | Explanation |
|---|---|
| Compounding frequency | More frequent compounding → higher maturity amount. |
| Effective annual rate (EAR) | (\text{EAR} = \left(1 + \frac{R}{n}\right)^n - 1) – useful to compare instruments with different compounding frequencies. |
| Time unit | Must match compounding periods. If T = 3 years and n = 4 (quarterly), exponent = 12. |
| Exponential growth | Small differences in R or n can lead to large differences in A over long periods. |
| CI > SI | For the same P, R, T, CI is always higher. |
| Continuous compounding | Gives the theoretical maximum for a given R and T. |
📚 Worked Example
Problem:
You invest ₹25,000 at an annual rate of 8 % compounded quarterly for 5 years. Find the maturity amount and the compound interest earned.
Solution:
-
Identify variables:
- (P = ₹25,000)
- (R = 8%)
- (n = 4) (quarterly)
- (T = 5) years
-
Compute (A) using the formula:
[ A = 25,000 \times \left(1 +