Simple & Compound Interest – NDA Quantitative Aptitude Notes
Purpose – Quickly recall the definitions, formulas, and common tricks for solving interest‑related problems in the NDA exam.
1. Simple Explanation
| Simple Interest (SI) | Compound Interest (CI) | |
|---|---|---|
| Definition | Interest calculated only on the original principal. | Interest calculated on the principal and all previously accumulated interest. |
| Formula | (SI = P \times r \times t) | (CI = P \left[(1+r)^t - 1\right]) |
| Units | (P) – Principal, (r) – Rate per annum (decimal), (t) – Time in years. | Same, but (t) may be fractional for partial years. |
| Interest earned | Linear in time. | Exponential in time (growth factor ((1+r))). |
| Effect of compounding frequency | None (always linear). | Increases with more frequent compounding: (CI = P\left[(1+\tfrac{r}{n})^{nt} -1\right]) where (n) = compounding periods per year. |
Key Idea – Simple interest is a straight line on a time–interest graph, compound interest is a curve that opens upward.
2. Key Points & Quick Reference
| Topic | Note |
|---|---|
| Rate conversion | Percent to decimal: (r% = \dfrac{r}{100}). |
| Time conversion | 1 month = (\dfrac{1}{12}) year; 1 week = (\dfrac{1}{52}) year; 1 day = (\dfrac{1}{365}) year. |
| Compounding periods | (n = 1) (annually), (2) (semi‑annually), (4) (quarterly), (12) (monthly), (365) (daily). |
| Effective annual rate (EAR) | (EAR = (1+\tfrac{r}{n})^n - 1). |
| Rule of 72 | Years to double ≈ (\dfrac{72}{\text{interest rate %}}). |
| Compound interest with multiple rates | Treat each period separately: (P_1 = P(1+r_1)^{t_1}); then (P_2 = P_1(1+r_2)^{t_2}), etc. |
| Present value (PV) | (PV = \dfrac{FV}{(1+r)^t}). |
| Annuities | Future value of (k) equal payments: (FV = k \dfrac{(1+r)^t - 1}{r}). |
| Cumulative interest | (CI = FV - P). |
| Interest on interest | Only in CI; not in SI. |
| Typical NDA tricks | - “Simple + compound = double” → interpret as double the principal after one year at a rate (r). - “Compound for 2 yrs at 10% = Simple for 4 yrs at 5%” → compare growth factors. |
3. Worked Example – Step by Step
Example 1: Simple Interest
Problem – A sum of ₹50 000 is deposited at a rate of 8% per annum for 3.5 years. Find the simple interest earned.
Solution
- Convert rate: (r = 8% = 0.08).
- Convert time: (t = 3.5) years.
- Apply SI formula:
[ SI = P \times r \times t = 50,000 \times 0.08 \times 3.5 ] - Compute:
[ 50,000 \times 0.08 = 4,000 \ 4,000 \times 3.5 = 14,000 ] - Answer: ₹14 000 simple interest.
Quick check: 8% of ₹50 000 is ₹4 000 per year → 3.5 years → ₹14 000. ✅
Example 2: Compound Interest
Problem – ₹20 000 is invested at 6% per annum compounded quarterly for 4 years. Find the amount after 4 years.
Solution
- Rate per quarter: (r_q = \dfrac{6%}{4} = 1.5% = 0.015).
- Number of quarters: (n = 4 \times 4 = 16).
- Compound formula:
[ A = P (1+r_q)^n = 20,000 (1+0.015)^{16} ] - Compute the growth factor:
[ (1.015)^{16} \approx 1.268241 ] - Multiply:
[ A = 20,000 \times 1.268241 \approx 25,364.82 ] - Answer: ₹25 364.82 (rounded to ₹25 365).
Quick check: 6% per annum for 4 years → approximately 25% growth (rule of 72: 72/6 ≈ 12 years to double; 4 years ≈ 25% → close to 25%). ✅
Example 3: Comparing Simple & Compound (NDA‑style)
Problem – If ₹10 000 is invested at 10% per annum compounded annually, how many years will it take to reach ₹15 000? (Use compound interest).
Solution
We need (A =