A certain sum becomes 3 times itself in 4 years at compound interest. In how many years does it become 27 times itself ?

Options

A. 15 years
B. 12 years
C. 36 years
D. 21 years

Explanation

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

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that a certain sum becomes 3 times itself in 4 years at compound interest. We need to find out how many years it will take for the same sum to become 27 times itself.

2. **Setting Up the Equation**:
Let's denote the principal amount as \( P \).
- After 4 years, the amount becomes \( 3P \).
- The formula for compound interest is given by:
\[ A = P \left(1 + \frac{r}{100}\right)^t \]
where \( A \) is the amount, \( P \) is the principal, \( r \) is the rate of interest, and \( t \) is the time in years.

3. **First Condition**:
From the first condition:
\[ 3P = P \left(1 + \frac{r}{100}\right)^4 \]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ 3 = \left(1 + \frac{r}{100}\right)^4 \]

4. **Second Condition**:
For the second condition where the amount becomes \( 27P \):
\[ 27P = P \left(1 + \frac{r}{100}\right)^n \]
Again, dividing both sides by \( P \):
\[ 27 = \left(1 + \frac{r}{100}\right)^n \]

5. **Relating the Two Conditions**:
We can express \( 27 \) in terms of \( 3 \):
\[ 27 = 3^3 \]
Therefore, we can write:
\[ 27 = \left(3\right)^3 = \left(1 + \frac{r}{100}\right)^{3 \cdot 4} \]
This implies:
\[ \left(1 + \frac{r}{100}\right)^n = \left(1 + \frac{r}{100}\right)^{12} \]

6. **Equating the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ n = 12 \]

### Conclusion:
Thus, it will take **12 years** for the sum to become 27 times itself at compound interest.