The difference between the compound interest and the simple interest acrued on an amount of 18,000 in 2 years is 405. Find the rate of interest per annum

Answer

To solve the problem, we need to find the rate of interest per annum given the difference between compound interest (CI) and simple interest (SI) for a principal amount of ₹18,000 over 2 years is ₹405.

### Step-by-Step Solution:

1. **Identify the given values**:
- Principal (P) = ₹18,000
- Time (T) = 2 years
- Difference between CI and SI = ₹405

2. **Write the formulas for CI and SI**:
- The formula for Simple Interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
- The formula for Compound Interest (CI) is:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]

3. **Express the difference between CI and SI**:
\[ CI - SI = 405 \]
Substituting the formulas:
\[ P \left(1 + \frac{R}{100}\right)^T - P - \frac{P \times R \times T}{100} = 405 \]

4. **Factor out P**:
\[ P \left(\left(1 + \frac{R}{100}\right)^T - 1 - \frac{R \times T}{100}\right) = 405 \]
Substitute \(P = 18000\):
\[ 18000 \left(\left(1 + \frac{R}{100}\right)^2 - 1 - \frac{R \times 2}{100}\right) = 405 \]

5. **Simplify the equation**:
Divide both sides by 18000:
\[ \left(1 + \frac{R}{100}\right)^2 - 1 - \frac{2R}{100} = \frac{405}{18000} \]
Simplifying \(\frac{405}{18000}\) gives:
\[ \frac{405}{18000} = \frac{9}{400} \]
So, we have:
\[ \left(1 + \frac{R}{100}\right)^2 - 1 - \frac{2R}{100} = \frac{9}{400} \]

6. **Expand the left-hand side**:
\[ \left(1 + \frac{R}{100}\right)^2 = 1 + 2\frac{R}{100} + \left(\frac{R}{100}\right)^2 \]
Therefore:
\[ 1 + 2\frac{R}{100} + \left(\frac{R}{100}\right)^2 - 1 - \frac{2R}{100} = \frac{9}{400} \]
This simplifies to:
\[ \left(\frac{R}{100}\right)^2 = \frac{9}{400} \]

7. **Solve for R**:
Multiply both sides by \(100^2\):
\[ R^2 = 9 \times 100^2 / 400 \]
Simplifying gives:
\[ R^2 = 9 \times 25 = 225 \]
Therefore:
\[ R = \sqrt{225} = 15 \]

8. **Conclusion**:
The rate of interest per annum is:
\[ R = 15\% \]