A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself ?

Answer

To solve the problem step by step, we will use the properties of compound interest.

### Step-by-Step Solution:

1. **Understanding the Problem**: We know that a sum of money doubles in 4 years. We need to find out how many years it will take for the same sum to become eight times itself.

2. **Using the Compound Interest Formula**: The formula for compound interest is given by:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]
where:
- \( A \) is the amount after time \( T \),
- \( P \) is the principal amount (initial sum),
- \( R \) is the rate of interest,
- \( T \) is the time in years.

3. **Setting Up the First Case**: Since the amount doubles in 4 years, we can set up the equation:
\[ 2P = P \left(1 + \frac{R}{100}\right)^4 \]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ 2 = \left(1 + \frac{R}{100}\right)^4 \]

4. **Setting Up the Second Case**: Now, we want to find out when the amount will be eight times the principal:
\[ 8P = P \left(1 + \frac{R}{100}\right)^T \]
Again, dividing both sides by \( P \):
\[ 8 = \left(1 + \frac{R}{100}\right)^T \]

5. **Relating the Two Cases**: From the first case, we have:
\[ 2 = \left(1 + \frac{R}{100}\right)^4 \]
We can express \( 8 \) as \( 2^3 \):
\[ 8 = \left(2\right)^3 = \left(1 + \frac{R}{100}\right)^{3 \cdot 4} \]
This means:
\[ 8 = \left(1 + \frac{R}{100}\right)^{12} \]

6. **Comparing the Powers**: Now we have two equations:
- \( 2 = \left(1 + \frac{R}{100}\right)^4 \)
- \( 8 = \left(1 + \frac{R}{100}\right)^{12} \)

Since the bases are the same, we can compare the exponents:
\[ T = 12 \text{ years} \]

7. **Conclusion**: It will take 12 years for the amount to become eight times itself.

### Final Answer:
The time required for the amount to become eight times itself is **12 years**.