Puducherry TET – Quantitative Aptitude
Topic: Simple & Compound Interest
1. Simple Explanation
| Concept | What it means | When it is used |
|---|---|---|
| Simple Interest (SI) | Interest calculated only on the original principal amount. | Short‑term loans, savings where interest is credited once a year. |
| Compound Interest (CI) | Interest calculated on the original principal and on the interest that has already been accumulated. | Long‑term deposits, loans with monthly/quarterly compounding. |
Both are expressed as a percentage of the principal.
2. Key Points & Formulas
| Item | Formula | Variables | Notes |
|---|---|---|---|
| Simple Interest | SI = P × R × T |
P = Principal, R = Rate per annum (decimal), T = Time in years |
SI is linear with time. |
| Total Amount (A) | A = P + SI |
Equivalent to A = P (1 + R T) |
|
| Compound Interest | A = P (1 + R/n)^(nT) |
P = Principal, R = Rate per annum (decimal), n = compounding periods per year, T = Time in years |
If n = 1 → annual compounding. |
| Compound Interest Earned | CI = A – P |
||
| Effective Annual Rate (EAR) | EAR = (1 + R/n)^n – 1 |
Useful when comparing different compounding periods. | |
| Rule of 72 | Years to double ≈ 72 / (annual rate in %) |
Rough estimate for continuous growth. | |
| Conversion to Monthly Rate | r_monthly = (1 + R)^(1/12) – 1 |
For monthly compounding. | |
| Time required to reach a target amount | T = ln(A/P) / (n ln(1 + R/n)) |
Works for compound interest. |
Common Misconceptions
- Rate vs. Rate per period – Always convert the annual rate to the rate per compounding period before plugging into the CI formula.
- Time units – Keep all time units consistent (years, months, days).
- Adding interest vs. Adding principal – In SI, interest is added only once at the end; in CI, it is added each compounding period.
3. Worked Examples
Example 1 – Simple Interest
Problem
A student borrows ₹12,000 at a simple interest rate of 8% per annum for 3 years. What is the total amount payable?
Solution
- Convert rate to decimal:
R = 8% = 0.08 - Apply SI formula:
SI = P × R × T = 12,000 × 0.08 × 3 = 2,880 - Total amount:
A = P + SI = 12,000 + 2,880 = 14,880
Answer – ₹14,880
Example 2 – Compound Interest (Quarterly Compounding)
Problem
₹5,000 is deposited in a bank at 12% per annum, compounded quarterly, for 2 years. Find the maturity value and the interest earned.
Solution
- Annual rate → decimal:
R = 12% = 0.12 - Compounding periods per year:
n = 4(quarterly) - Time in years:
T = 2 - Compute
A:
A = P (1 + R/n)^(nT)
= 5,000 (1 + 0.12/4)^(4×2)
= 5,000 (1 + 0.03)^(8)
= 5,000 (1.03)^8
-
Calculate
(1.03)^8 ≈ 1.26677 -
A ≈ 5,000 × 1.26677 = 6,333.85 -
Compound interest earned:
CI = A – P = 6,333.85 – 5,000 = 1,333.85
Answer – Maturity value ≈ ₹6,333.85; Interest earned ≈ ₹1,333.85
Example 3 – Time Required (Compound Interest)
Problem
How many years will it take ₹8,000 to grow to ₹12,000 at 10% per annum compounded annually?
Solution
P = 8,000,A = 12,000,R = 0.10,n = 1- Use
T = ln(A/P) / (n ln(1 + R/n))
T = ln(12,000/8,000) / (1 × ln(1 + 0.10))
= ln(1.5) / ln(1.10)
≈ 0.405465 / 0.095310
≈ 4.25 years
Answer – Approximately 4.25 years.
4. Quick Revision Summary
| Topic | Formula | Quick Tip |
|---|---|---|
| Simple Interest | SI = PRT |
Think “Principal × Rate × Time” (no compounding). |
| Total Amount (SI) | A = P(1 + RT) |
Add the interest to the principal. |
| Compound Interest | A = P(1 + R/n)^(nT) |
Convert annual rate to period rate first |