https://file.aiquickdraw.com/mathbot/file/17304700814240l2gnxz4.jpg

To solve this problem, we will use the compound interest formula:

\[
A = P \left(1 + \frac{r}{100}\right)^n
\]

where:
- \( A \) is the total amount after the specified time,
- \( P \) is the principal amount (initial amount),
- \( r \) is the interest rate per period,
- \( n \) is the number of periods.

Given values:
- Principal, \( P = 5000 \) rupees,
- Interest rate, \( r = 9\% \),
- Time period, \( n = 3 \) years (compounded annually).

### Step-by-Step Calculation:
1. Substitute the values into the formula:

\[
A = 5000 \left(1 + \frac{9}{100}\right)^3
\]

2. Simplify inside the parentheses:

\[
A = 5000 \left(1 + 0.09\right)^3 = 5000 \times (1.09)^3
\]

3. Calculate \( (1.09)^3 \):

\[
(1.09)^3 = 1.295029
\]

4. Multiply by the principal amount:

\[
A = 5000 \times 1.295029 = 6475.145
\]

### Final Answer:
The total amount after 3 years is approximately **Rs. 6475.14**.

The correct answer is:
- **Rs. 6475.14** (highlighted option).

Would you like further details on this calculation?

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**Related Questions:**
1. What would the total amount be if the interest rate was 10%?
2. How would the total amount change if the compounding was semi-annual instead of annual?
3. What would be the interest earned if the principal amount was Rs. 6000 instead of Rs. 5000?
4. How does compound interest differ from simple interest in terms of final amount?
5. How is the formula modified for monthly compounding?

**Tip:** When solving compound interest problems, remember that the interest compounds on the new balance each period, leading to exponential growth in the total amount.

If the principal amount is Rs.5000, the interest rate is 9% per year compounded annually, and the time period is 3 years, what is the total amount after 3 years?

Find the total amount after 3 years for a principal of Rs. 5000 at a 9% annual compound interest rate. This problem demonstrates compound interest calculations, suitable for Grades 8-10, and includes a step-by-step solution.

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