Question

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?

Answers

GPT-4o mini
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.