If 8100 becomes 10000 at the rate of `11(1)/(9)%` compound interest in certain time period then find the Time Period?

Options

A. 2.5 year
B. 2 year
C. 1.5 year
D. 3 year

Answer

To solve the problem of finding the time period for which an amount of ₹8100 grows to ₹10000 at a compound interest rate of \(11 \frac{1}{9}\%\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Principal (P) = ₹8100
- Amount (A) = ₹10000
- Rate of Interest (R) = \(11 \frac{1}{9}\%\) = \(\frac{100}{9}\%\)

2. **Convert the Rate of Interest to Decimal:**
- To convert \(11 \frac{1}{9}\%\) to a decimal, we divide by 100:
\[ R = \frac{100}{9} \times \frac{1}{100} = \frac{1}{9} \]

3. **Calculate the Compound Interest (CI):**
- The formula for compound interest is:
\[ CI = A - P \]
- Substituting the values:
\[ CI = 10000 - 8100 = ₹1900 \]

4. **Set Up the Compound Interest Formula:**
- The formula for the amount in compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
- Rearranging for \(n\):
\[ 10000 = 8100 \left(1 + \frac{1}{9}\right)^n \]

5. **Simplify the Equation:**
- Calculate \(1 + \frac{1}{9} = \frac{10}{9}\):
\[ 10000 = 8100 \left(\frac{10}{9}\right)^n \]

6. **Divide Both Sides by 8100:**
\[ \frac{10000}{8100} = \left(\frac{10}{9}\right)^n \]
- Simplifying \(\frac{10000}{8100} = \frac{100}{81}\):
\[ \frac{100}{81} = \left(\frac{10}{9}\right)^n \]

7. **Take Logarithms:**
- Taking logarithm on both sides:
\[ \log\left(\frac{100}{81}\right) = n \cdot \log\left(\frac{10}{9}\right) \]

8. **Calculate \(n\):**
- Rearranging gives:
\[ n = \frac{\log\left(\frac{100}{81}\right)}{\log\left(\frac{10}{9}\right)} \]
- Using logarithm values, we can calculate \(n\).

9. **Final Calculation:**
- After calculating, we find that \(n = 2\) years.

### Conclusion:
The time period for which ₹8100 becomes ₹10000 at a compound interest rate of \(11 \frac{1}{9}\%\) is **2 years**.