This is a compound interest problem where the principal amount needs to be calculated.

Given:

• Amount after 4 years (A₁) = ₹10,800
• Amount after 8 years (A₂) = ₹32,400
• Time difference = 8 - 4 = 4 years

Let the principal amount be $P$ and the annual compound interest rate be $r$ (in decimal).

Step 1: Compound Interest Formula

$A = P \cdot (1 + r)^t$ For two different times: $A_1 = P \cdot (1 + r)^4 \quad \text{and} \quad A_2 = P \cdot (1 + r)^8$

Step 2: Divide the Equations

$\frac{A_2}{A_1} = \frac{P \cdot (1 + r)^8}{P \cdot (1 + r)^4} = (1 + r)^4$ $\frac{32,400}{10,800} = (1 + r)^4$ $3 = (1 + r)^4$

Step 3: Solve for $1 + r$

$1 + r = \sqrt[4]{3}$ Using approximation: $1 + r \approx 1.316$ $r \approx 0.316 \, \text{or} \, 31.6\% \, \text{(annual rate of interest)}.$

Step 4: Calculate Principal $P$

Using $A_1 = P \cdot (1 + r)^4$: $10,800 = P \cdot (1.316)^4$ $10,800 = P \cdot 3$ $P = \frac{10,800}{3} = 3,600$

Final Answer:

The principal amount $P$ is ₹3,600.

Would you like to go through the explanation further or have any related questions?

5 Related Questions:

1. How would the principal change if the interest rate were higher?
2. Can you solve a similar problem with different amounts and time periods?
3. How does the compound interest formula differ from simple interest?
4. What is the significance of the time interval in compound interest calculations?
5. How can you approximate values like $\sqrt[4]{3}$ without a calculator?

Tip: Always use logarithms or exponential functions for precise interest rate calculations when the time period is extended!