Simple & compound interest

Simple & Compound Interest – Study Notes

Himachal Pradesh Police Sub‑Inspector (SI) – Quantitative Aptitude

These notes are crafted to help you grasp the concepts, solve typical exam problems, and revise quickly for the HP Police SI exam.

1. Quick Conceptual Overview

Key:
P = Principal (initial amount)
R = Annual rate of interest (in %)
T = Time (in years)
I = Simple interest earned
A = Total amount after compounding
n = Number of compounding periods per year (1 = annual, 2 = half‑annual, 4 = quarterly, 12 = monthly, 365 = daily)

2. Simple Interest (SI)

2.1 Formula

[ I = \frac{P \times R \times T}{100} ]

2.2 How to Find:

• Interest (I): Use the above formula directly.
• Principal (P): (P = \dfrac{I \times 100}{R \times T}).
• Rate (R): (R = \dfrac{I \times 100}{P \times T}).
• Time (T): (T = \dfrac{I \times 100}{P \times R}).

2.3 Worked Example (SI)

Problem:
A person borrows ₹12,000 at a simple interest rate of 7 % per annum for 3 years. Find the total interest payable and the total amount to be repaid.

Solution:

1. Use SI formula
   (I = \dfrac{12{,}000 \times 7 \times 3}{100} = \dfrac{252{,}000}{100} = ₹2{,}520.)

2. Total amount
   (A = P + I = 12{,}000 + 2{,}520 = ₹14{,}520.)

Answer:
Interest = ₹2,520; Total payable = ₹14,520.

3. Compound Interest (CI)

3.1 Formula

[ A = P \left(1 + \frac{R}{n}\right)^{nT} ]

• If only the interest is needed:
  [ CI = A - P ]

3.2 Common Compounding Frequencies

3.3 How to Find:

• Amount (A): Directly from the formula.
• Compound Interest (CI): (CI = A - P).
• Principal (P): (P = \dfrac{A}{\left(1 + \frac{R}{n}\right)^{nT}}).
• Rate (R): Solve for (R) using logarithms, e.g.
  [ R = n \left[\left(\frac{A}{P}\right)^{\frac{1}{nT}} - 1\right] \times 100 ]
• Time (T): (T = \dfrac{\ln\left(\frac{A}{P}\right)}{n \ln\left(1 + \frac{R}{n}\right)}).

3.4 Worked Example (CI)

Problem:
₹5,000 is invested at an annual interest rate of 8 % compounded quarterly. Find the amount after 2 years.

Solution:

1. Identify values
   (P = 5{,}000), (R = 8%), (n = 4), (T = 2).

2. Compute (A)
   [ A = 5{,}000 \left(1 + \frac{8}{4 \times 100}\right)^{4\times 2} = 5{,}000 \left(1 + 0.02\right)^{8} = 5{,}000 \times 1.02^{8} ]

3. Calculate (1.02^{8})
   (1.02^{8} \approx 1.171661).

4. Final amount
   (A \approx 5{,}000 \times 1.171661 = ₹5{,}858.31).

Answer:
After 2 years, the investment grows to approximately ₹5,858.31 (CI ≈ ₹858.31).

4. Key Points & Tricks for the Exam