📚 SBI PO – Quantitative Aptitude
Topic: Simple & Compound Interest
Goal: Master the concepts, formulas, and quick‑solve techniques for Simple Interest (SI) and Compound Interest (CI) – the most frequent topics in the SBI PO exam.
1. Conceptual Overview
| Simple Interest | Compound Interest | |
|---|---|---|
| Definition | Interest calculated only on the principal amount. | Interest earned on the principal and on the interest accumulated over time. |
| Growth nature | Linear | Exponential (or geometric) |
| Formula | SI = P × R × T / 100 |
CI = P × (1 + R/100)^T – P |
| When to use | Short‑term loans, intra‑day transactions, simple savings. | Long‑term investments, mutual funds, savings accounts with periodic compounding. |
| Key Parameter | Principal (P) is the only amount that matters. | Principal (P) plus compounding frequency (n). |
Remember:
If the interest is compounded annually, the formula above is used.
If compounding frequency differs (monthly, quarterly, daily, etc.), adjust the rate and time accordingly.
2. Simple Interest (SI)
2.1 Formula
[ \text{SI} = \frac{P \times R \times T}{100} ]
- P – Principal amount
- R – Rate of interest per annum (in %).
- T – Time in years (if time is in months, divide by 12).
2.2 Key Points
| Item | Detail |
|---|---|
| Time units | Convert months to years: T (years) = months / 12. |
| Principal | The initial amount invested or borrowed. |
| Rate | Expressed as a percentage; convert to decimal for calculations. |
| Interest amount | Add to principal to get total amount: A = P + SI. |
| Backward problems | Often given SI, P, and R, ask for T. Rearrange: T = (SI × 100) / (P × R). |
2.3 Worked Example – 1
Problem:
A bank pays 12% simple interest per annum on a deposit of ₹10,000. How much interest will be earned in 9 months?
Solution:
- Convert time to years:
[ T = \frac{9}{12} = 0.75 \text{ years} ] - Apply SI formula:
[ \text{SI} = \frac{10,000 \times 12 \times 0.75}{100} = \frac{10,000 \times 9}{100} = 900 ] - Total amount:
[ A = P + \text{SI} = 10,000 + 900 = 10,900 ]
Answer: ₹900 interest; total ₹10,900.
3. Compound Interest (CI)
3.1 General Formula
[ \text{CI} = P \left(1 + \frac{R}{100}\right)^T - P ]
- T – Number of compounding periods (years if annual compounding).
- If compounding n times per year, use:
[ \text{CI} = P \left(1 + \frac{R}{100n}\right)^{nT} - P ]
3.2 Key Points
| Item | Detail |
|---|---|
| Compounding frequency (n) | 1 = annual, 2 = semi‑annual, 4 = quarterly, 12 = monthly, 365 = daily. |
| Time in periods | Convert years to number of periods: periods = T × n. |
| Effective annual rate (EAR) | EAR = (1 + R/(100n))^n – 1. Useful for comparing rates. |
| Backward problems | Given CI, P, R, n, solve for T: |
| [ | |
| T = \frac{\ln\left(1 + \frac{\text{CI}}{P}\right)}{n \ln\left(1 + \frac{R}{100n}\right)} | |
| ] | |
| Quick mental trick | For small R and n, use the approximation: |
| [ | |
| \text{CI} \approx P \left(1 + \frac{R}{100} \times T\right) + \frac{P \times R^2 \times T}{2 \times 100^2} | |
| ] | |
| (Use only when memory/time is precious.) |
3.3 Worked Example – 2
Problem:
₹5,000 is invested at 8% per annum compounded semi‑annually. Find the amount after 3 years.
Solution:
-
Parameters
- (P = 5,000)
- (R = 8%)
- (n = 2) (semi‑annual)
- (T = 3) years
-
Compute compounding periods
[ \text{periods} = n \times T = 2 \times 3 = 6 ] -
Rate per period
[ \frac{R}{100n} = \frac{8}{200} = 0.04 ] -
Apply CI formula
[ A = P \left(1 + 0.04\right)^6 = 5,000 \times 1.04^6 ]Compute (1.04^6):
(1.04^2