Study Notes – Simple & Compound Interest
Subject: Quantitative Aptitude
Exam: SSC CPO
1. Quick Overview
| Concept | Definition | Formula | Notes |
|---|---|---|---|
| Simple Interest (SI) | Interest calculated only on the principal amount. | (\displaystyle SI = \frac{P \times R \times T}{100}) | (P) = Principal, (R) = Rate (%), (T) = Time (years). |
| Compound Interest (CI) | Interest calculated on the principal and on accumulated interest. | (\displaystyle CI = P\left(1+\frac{R}{n}\right)^{nT} - P) | (n) = compounding frequency per year (e.g., 4 for quarterly). |
| Total Amount (A) | Principal + Interest | (A = P + SI) for simple; (A = P\left(1+\frac{R}{n}\right)^{nT}) for compound |
Key takeaway:
Simple interest is linear in time; compound interest grows exponentially.
2. Simple Interest (SI)
2.1 How to Compute
- Identify (P), (R), and (T).
- Apply the SI formula.
- Total Amount: (A = P + SI).
2.2 Common Variations
| Variation | Formula |
|---|---|
| Find (P) when (A) & (SI) known | (P = \dfrac{A}{1+\frac{R T}{100}}) |
| Find (R) | (R = \dfrac{SI \times 100}{P \times T}) |
| Find (T) | (T = \dfrac{SI \times 100}{P \times R}) |
2.3 Practice Question
A sum of ₹5,000 is invested at 6% per annum for 3 years. What is the simple interest?
Solution
[ SI = \frac{5000 \times 6 \times 3}{100} = \frac{90,000}{100} = ₹900 ]
Answer: ₹900
3. Compound Interest (CI)
3.1 How to Compute
- Identify (P), (R), (T), and (n).
- Compute the factor: (\displaystyle \left(1+\frac{R}{n}\right)^{nT}).
- Multiply by (P) to get the final amount (A).
- Interest earned: (CI = A - P).
3.2 Common Variations
| Variation | Formula |
|---|---|
| Find (P) when (A) & CI known | (P = \dfrac{A}{\left(1+\frac{R}{n}\right)^{nT}}) |
| Find (R) | (\displaystyle R = n\left[\left(\frac{A}{P}\right)^{1/(nT)} - 1\right] \times 100) |
| Find (T) | (\displaystyle T = \frac{\ln(A/P)}{n \ln(1+R/n)}) |
Tip: For quick mental calculations, if (n=1) (annual compounding) use (A=P(1+R/100)^T).
For quarterly, (n=4). For monthly, (n=12).
3.3 Worked Example
Problem:
₹12,000 is deposited in a bank at an annual rate of 8% compounded quarterly. Find the amount after 4 years.
Step‑by‑Step
- (P = 12{,}000)
- (R = 8%) → (R/n = 8/4 = 2%)
- (nT = 4 \times 4 = 16) compounding periods
- Compute factor:
[ \left(1+\frac{R}{n}\right)^{nT} = \left(1+0.02\right)^{16} = 1.02^{16} ] - (1.02^{16}) ≈ 1.3686 (use a calculator or log table).
- Final amount:
[ A = 12{,}000 \times 1.3686 \approx ₹16{,}423 ] - Compound interest earned:
[ CI = A - P \approx 16{,}423 - 12{,}000 = ₹4{,}423 ]
Answer: Approx. ₹16,423 after 4 years (CI ≈ ₹4,423).
4. Quick Revision Cheat Sheet
| Concept | Formula | Quick Tips |
|---|---|---|
| Simple Interest | (SI = \dfrac{PRT}{100}) | Use a single multiplication & division by 100. |
| Total Amount (SI) | (A = P + SI) | “Add” the computed SI to the principal. |
| Compound Interest | (CI = P\left(1+\dfrac{R}{n}\right)^{nT} - P) | When (n=1), (CI = P(1+R/100)^T - P). |
| Total Amount (CI) | (A = P\left(1+\dfrac{R}{n}\right)^{nT}) | Use exponent rules; calculator helps. |
| Finding Unknowns | Invert formulas accordingly (e.g., (P = A/(1+RT/100)) for SI). | Remember to keep units consistent |