Rupees 10,000 is invested at 6% interest compounded annually. How long will it take to accumulate Rs. 20, 000 in the account?

Explanation

To solve the problem of how long it will take for an investment of Rs. 10,000 at an interest rate of 6% compounded annually to grow to Rs. 20,000, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Principal (P) = Rs. 10,000
- Rate of Interest (R) = 6% per annum
- Amount (A) = Rs. 20,000

2. **Use the Compound Interest Formula:**
The formula for compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^N \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( R \) is the annual interest rate (in percentage).
- \( N \) is the number of years the money is invested or borrowed.

3. **Substitute the Known Values into the Formula:**
\[ 20000 = 10000 \left(1 + \frac{6}{100}\right)^N \]

4. **Simplify the Equation:**
\[ 20000 = 10000 \left(1.06\right)^N \]
Dividing both sides by 10,000:
\[ 2 = (1.06)^N \]

5. **Take the Logarithm of Both Sides:**
\[ \log(2) = \log((1.06)^N) \]

6. **Apply the Logarithmic Property:**
Using the property \( \log(a^b) = b \cdot \log(a) \):
\[ \log(2) = N \cdot \log(1.06) \]

7. **Solve for N:**
\[ N = \frac{\log(2)}{\log(1.06)} \]

8. **Calculate the Logarithms:**
Using a calculator or logarithm table:
- \( \log(2) \approx 0.30103 \)
- \( \log(1.06) \approx 0.025306 \)

9. **Substitute the Logarithm Values:**
\[ N = \frac{0.30103}{0.025306} \]

10. **Perform the Division:**
\[ N \approx 11.89 \]

11. **Round to the Nearest Whole Number:**
Since time is typically expressed in whole years, we round 11.89 to approximately 12 years.

### Final Answer:
It will take approximately **12 years** for Rs. 10,000 to accumulate to Rs. 20,000 at an interest rate of 6% compounded annually.

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