Question

Suppose you invest $10,000 in a savings account with 2.5% compound interest. Which account has the greatest value after 10 years?
Responses

A an account compounded annuallyan account compounded annually

B an account compounded semiannuallyan account compounded semiannually

C an account compounded quarterlyan account compounded quarterly

D an account compounded monthly

Answer 1

To determine which account will have the greatest value after 10 years with a 2.5% compound interest rate, 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 unit year.
- \(t\) is the time the money is invested for in years.

Let's calculate for each compounding frequency:

1. **Annually (n = 1)**:
\[
A = 10000 \left(1 + \frac{0.025}{1}\right)^{1 \times 10} = 10000 \left(1.025\right)^{10}
\]
\[
A \approx 10000 \times 1.28008 \approx 12800.80
\]

2. **Semiannually (n = 2)**:
\[
A = 10000 \left(1 + \frac{0.025}{2}\right)^{2 \times 10} = 10000 \left(1.0125\right)^{20}
\]
\[
A \approx 10000 \times 1.282037 \approx 12820.37
\]

3. **Quarterly (n = 4)**:
\[
A = 10000 \left(1 + \frac{0.025}{4}\right)^{4 \times 10} = 10000 \left(1.00625\right)^{40}
\]
\[
A \approx 10000 \times 1.28368 \approx 12836.76
\]

4. **Monthly (n = 12)**:
\[
A = 10000 \left(1 + \frac{0.025}{12}\right)^{12 \times 10} = 10000 \left(1.00208333\right)^{120}
\]
\[
A \approx 10000 \times 1.284024 \approx 12840.24
\]

Now, let's summarize the results:
- Annually: \(\$12800.80\)
- Semiannually: \(\$12820.37\)
- Quarterly: \(\$12836.76\)
- Monthly: \(\$12840.24\)

From these calculations, we can see that the account compounded monthly has the greatest value of approximately \$12,840.24 after 10 years.

Thus, the answer is **D: an account compounded monthly**.