Questions: What annual interest rate is required for a debt of 11,339 to grow into 13,880 in 6 years if interest compounds monthly? Round your answer to the nearest tenth of a percent.

Solution Steps

To find the annual interest rate required for a debt to grow to a certain amount with monthly compounding, we can use the compound interest formula:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

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 (decimal).
• \( n \) is the number of times that interest is compounded per year.
• \( t \) is the time the money is invested for in years.

We need to solve for \( r \) given:

• \( A = 13,880 \)
• \( P = 11,339 \)
• \( n = 12 \) (since interest compounds monthly)
• \( t = 6 \)

Rearrange the formula to solve for \( r \):

\[ \left(1 + \frac{r}{12}\right)^{12 \cdot 6} = \frac{13880}{11339} \]

Then, solve for \( r \).

We are given the following values:

• Principal amount \( P = 11339 \)
• Amount after 6 years \( A = 13880 \)
• Compounding frequency \( n = 12 \) (monthly)
• Time period \( t = 6 \) years

Using the compound interest formula:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

we can rearrange it to solve for the annual interest rate \( r \):

\[ \left(1 + \frac{r}{12}\right)^{12 \cdot 6} = \frac{13880}{11339} \]

Calculating the right-hand side:

\[ \frac{13880}{11339} \approx 1.222 \]

Now we need to find \( r \) such that:

\[ \left(1 + \frac{r}{12}\right)^{72} \approx 1.222 \]

After solving for \( r \), we find:

\[ r \approx 0.0337 \]

To express \( r \) as a percentage:

\[ \text{Annual Rate} = r \times 100 \approx 3.374750582208063 \approx 3.4\% \]

Final Answer