Questions: Harry deposited 3,000.00 into a new savings account that earns interest compounded monthly. After 2 years, the balance in the account was 3,078.00. What was the interest rate on the account? Round your answer to the nearest tenth of a percent.

Solution Steps

We are given the following values:

• Final amount \( A = 3078.00 \)
• Principal amount \( P = 3000.00 \)
• Number of compounding periods per year \( n = 12 \)
• Time in years \( t = 2 \)

We apply the compound interest formula:

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

Rearranging this formula to solve for the annual interest rate \( r \):

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

Substituting the known values into the rearranged formula:

\[ r = 12 \left( \left( \frac{3078.00}{3000.00} \right)^{\frac{1}{12 \cdot 2}} - 1 \right) \]

Calculating this gives us:

\[ r \approx 0.01284073866759794 \]

To express \( r \) as a percentage, we multiply by 100:

\[ \text{Interest Rate} \approx 1.284073866759794 \% \]

Rounding this to the nearest tenth of a percent results in:

\[ \text{Interest Rate} \approx 1.3 \% \]

Final Answer