Question

Jamar invested $3,700 in an account paying an interest rate of 3.375% compounded continuously. Jordan invested $3,700 in an account paying an interest rate of 3.75% compounded quarterly. After 20 years, how much more money would Jordan have in his account than Jamar, to the nearest dollar?(1 point)
Responses

$998
$998

$539
$539

$627
$627

$267

Answer 1

To determine how much more money Jordan would have than Jamar after 20 years, we can use the formulas for continuous and compound interest.

1. **For Jamar's investment** (compounded continuously):
\[
A = Pe^{rt}
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money).
- \( r \) = the annual interest rate (decimal).
- \( t \) = the time the money is invested for in years.
- \( e \) = Euler's number (approximately equal to 2.71828).

Given:
- \( P = 3700 \)
- \( r = 0.03375 \) (3.375% as a decimal)
- \( t = 20 \)

Plugging in the values:
\[
A = 3700 \cdot e^{0.03375 \cdot 20}
\]
\[
A \approx 3700 \cdot e^{0.675}
\]
\[
A \approx 3700 \cdot 1.966
\]
\[
A \approx 7264.2
\]

2. **For Jordan's investment** (compounded quarterly):
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( n \) = the number of times that interest is compounded per year.

Given:
- \( P = 3700 \)
- \( r = 0.0375 \) (3.75% as a decimal)
- \( n = 4 \) (quarterly)
- \( t = 20 \)

Plugging in the values:
\[
A = 3700 \left(1 + \frac{0.0375}{4}\right)^{4 \cdot 20}
\]
\[
= 3700 \left(1 + 0.009375\right)^{80}
\]
\[
= 3700 \left(1.009375\right)^{80}
\]
\[
A \approx 3700 \cdot 2.035
\]
\[
A \approx 7529.3
\]

3. **Calculating the difference**:
\[
\text{Difference} = A_{\text{Jordan}} - A_{\text{Jamar}} \approx 7529.3 - 7264.2 \approx 265.1
\]

Rounding to the nearest dollar, Jordan would have approximately **$267** more than Jamar.

Thus, the answer is:
\[
\boxed{267}
\]