A sum of money triples itself in 3 years at compound interest. In how many years will it become 9 times itself ?

Options

A. 4
B. 9
C. 6
D. 7

Explanation

To solve the problem step by step, we will use the concept of compound interest. Let's break it down:

### Step 1: Understand the Problem
We know that a sum of money triples itself in 3 years. We need to find out how many years it will take for the same sum to become 9 times itself.

### Step 2: Set Up the Equations
Let the principal amount be \( P \).
According to the problem, after 3 years, the amount becomes \( 3P \).

Using the formula for compound interest:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
where:
- \( A \) is the amount after time \( n \),
- \( P \) is the principal,
- \( r \) is the rate of interest,
- \( n \) is the time in years.

For the first scenario (tripling in 3 years):
\[ 3P = P \left(1 + \frac{r}{100}\right)^3 \]

### Step 3: Simplify the Equation
Dividing both sides by \( P \):
\[ 3 = \left(1 + \frac{r}{100}\right)^3 \]

### Step 4: Take the Cube Root
Taking the cube root of both sides:
\[ 1 + \frac{r}{100} = \sqrt[3]{3} \]

### Step 5: Set Up the Second Equation
Now we want to find out when the amount becomes \( 9P \):
\[ 9P = P \left(1 + \frac{r}{100}\right)^n \]

Dividing both sides by \( P \):
\[ 9 = \left(1 + \frac{r}{100}\right)^n \]

### Step 6: Substitute the Value
We already know that \( 1 + \frac{r}{100} = \sqrt[3]{3} \), so we substitute this into the equation:
\[ 9 = \left(\sqrt[3]{3}\right)^n \]

### Step 7: Rewrite 9 in Terms of Base 3
We can express 9 as \( 3^2 \):
\[ 3^2 = \left(3^{1/3}\right)^n \]

### Step 8: Apply the Power of a Power Rule
Using the property of exponents:
\[ 3^2 = 3^{n/3} \]

### Step 9: Set the Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 2 = \frac{n}{3} \]

### Step 10: Solve for \( n \)
Multiplying both sides by 3:
\[ n = 6 \]

### Conclusion
Thus, it will take **6 years** for the sum of money to become 9 times itself.

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