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)

Question

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: \[ \boxed{267} \]