Question

What was the initial amount deposited in the account? 2. Kalleah deposited $45 into a bank account that accrued interest compounded monthly. If she emptied the account and withdrew $120 after 9 years, what was the interest rate on the account?

Answer 1

To find the interest rate, we'll use the formula for compound interest:

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

Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount deposited).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times interest is compounded per year.
- \(t\) is the time the money is invested for in years.

In this case:
- \(A = 120\) (the amount withdrawn after 9 years)
- \(P = 45\) (the amount deposited)
- \(n = 12\) (interest compounded monthly)
- \(t = 9\) years

Now substituting the known values into the formula, we have:

\[
120 = 45 \left(1 + \frac{r}{12}\right)^{12 \times 9}
\]

This simplifies to:

\[
120 = 45 \left(1 + \frac{r}{12}\right)^{108}
\]

Next, divide both sides by 45:

\[
\frac{120}{45} = \left(1 + \frac{r}{12}\right)^{108}
\]

Calculating \( \frac{120}{45} \):

\[
\frac{120}{45} = \frac{8}{3} \approx 2.6667
\]

Next, we have:

\[
2.6667 = \left(1 + \frac{r}{12}\right)^{108}
\]

To solve for \( r \), we first take the 108th root of both sides:

\[
1 + \frac{r}{12} = \left(2.6667\right)^{\frac{1}{108}}
\]

Calculating \( \left(2.6667\right)^{\frac{1}{108}} \):

Using a calculator, we find:

\[
\left(2.6667\right)^{\frac{1}{108}} \approx 1.009091
\]

Now we can solve for \( r \):

\[
1 + \frac{r}{12} \approx 1.009091
\]

Subtracting 1 from both sides gives:

\[
\frac{r}{12} \approx 0.009091
\]

Now, multiply both sides by 12 to isolate \( r \):

\[
r \approx 0.009091 \times 12 \approx 0.109092
\]

Converting to a percentage:

\[
r \approx 0.109092 \times 100 \approx 10.91\%
\]

Thus, the annual interest rate on the account was approximately **10.91%**.