Lincoln invested $6,500 in an account paying an interest rate of 3, start fraction, 3, divided by, 4, end fraction3

4
3

% compounded quarterly. Savannah invested $6,500 in an account paying an interest rate of 4, one eigth4
8
1

% compounded continuously. After 20 years, how much more money would Savannah have in her account than Lincoln, to the nearest dollar?

1 answer

To solve this problem, we will calculate the final amounts in both Lincoln's and Savannah's accounts after 20 years and then find the difference.

Lincoln's Investment

Lincoln's investment earns interest compounded quarterly. The formula for compound interest is:

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

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).
  • \(n\) = the number of times that interest is compounded per year.
  • \(t\) = the number of years the money is invested for.

For Lincoln:

  • \(P = 6500\)
  • \(r = \frac{3.75}{100} = 0.0375\) (3.75% as decimal)
  • \(n = 4\) (compounded quarterly)
  • \(t = 20\)

Now we can calculate \(A\) for Lincoln:

\[ A_L = 6500 \left(1 + \frac{0.0375}{4}\right)^{4 \times 20} \]

Calculating the individual parts:

  • \(\frac{0.0375}{4} = 0.009375\)
  • \(4 \times 20 = 80\)

So the equation becomes:

\[ A_L = 6500 \left(1 + 0.009375\right)^{80} \] \[ A_L = 6500 \left(1.009375\right)^{80} \]

Now calculate \(1.009375^{80}\):

\[ 1.009375^{80} \approx 2.030856 \]

Now, plug this back into the equation for \(A_L\):

\[ A_L = 6500 \times 2.030856 \approx 13200.57 \]

Savannah's Investment

Savannah’s investment earns interest compounded continuously. The formula for continuous compounding is:

\[ A = Pe^{rt} \]

Where:

  • \(A\) = the amount of money accumulated after an amount of time, including interest.
  • \(P\) = the principal amount.
  • \(e\) = the base of the natural logarithm (approximately equal to 2.71828).
  • \(r\) = the annual interest rate (decimal).
  • \(t\) = the number of years the money is invested for.

For Savannah:

  • \(P = 6500\)
  • \(r = \frac{4.125}{100} = 0.04125\) (4.125% as decimal)
  • \(t = 20\)

Calculating \(A\) for Savannah:

\[ A_S = 6500 e^{0.04125 \times 20} \]

Calculating \(0.04125 \times 20 = 0.825\):

Now calculate \(e^{0.825}\):

\[ e^{0.825} \approx 2.283 \]

Now plug this into the equation for \(A_S\):

\[ A_S \approx 6500 \times 2.283 \approx 14839.5 \]

Find the Difference

Now, we can find how much more money Savannah has compared to Lincoln:

\[ \text{Difference} = A_S - A_L = 14839.5 - 13200.57 \approx 1638.93 \]

Rounding to the nearest dollar, the difference is:

\[ \text{Difference} \approx 1639 \]

Conclusion

Savannah would have approximately $1,639 more than Lincoln after 20 years.