Question
Leo invests $2,000 at an interest rate of 4%, compounded quarterly, and another $2,000 at an interest rate of 3.75%, compounded annually. How much are the investments worth in total at the end of 3 years?
A. $4,513.74
B. $4,487.19
C. $3,296.58
D. $4,155.00
A. $4,513.74
B. $4,487.19
C. $3,296.58
D. $4,155.00
Answers
GPT-4o mini
To find the total value of Leo's investments at the end of 3 years, we'll calculate the amount for each investment separately and then sum them up.
### Investment 1: $2,000 at 4% 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.
- \(P\) = principal amount (the initial amount of money).
- \(r\) = annual interest rate (decimal).
- \(n\) = number of times that interest is compounded per year.
- \(t\) = the number of years the money is invested or borrowed for.
For the first investment:
- \(P = 2000\)
- \(r = 0.04\) (4% as a decimal)
- \(n = 4\) (quarterly)
- \(t = 3\)
Plugging the values into the formula:
\[
A_1 = 2000 \left(1 + \frac{0.04}{4}\right)^{4 \times 3}
\]
\[
A_1 = 2000 \left(1 + 0.01\right)^{12}
\]
\[
A_1 = 2000 \left(1.01\right)^{12}
\]
Calculating \(1.01^{12}\):
\[
1.01^{12} \approx 1.126825
\]
Now calculate \(A_1\):
\[
A_1 = 2000 \times 1.126825 \approx 2253.65
\]
### Investment 2: $2,000 at 3.75% compounded annually
Using the same formula for the second investment:
- \(P = 2000\)
- \(r = 0.0375\) (3.75% as a decimal)
- \(n = 1\) (annually)
- \(t = 3\)
\[
A_2 = 2000 \left(1 + \frac{0.0375}{1}\right)^{1 \times 3}
\]
\[
A_2 = 2000 \left(1 + 0.0375\right)^{3}
\]
\[
A_2 = 2000 \left(1.0375\right)^{3}
\]
Calculating \(1.0375^{3}\):
\[
1.0375^{3} \approx 1.115978
\]
Now calculate \(A_2\):
\[
A_2 = 2000 \times 1.115978 \approx 2231.96
\]
### Total Investment Value
Now, we sum up \(A_1\) and \(A_2\):
\[
A_{total} = A_1 + A_2 = 2253.65 + 2231.96 \approx 4485.61
\]
Checking back with the answer choices provided, it looks like I made a slight miscalculation or rounding difference. However, our answer \(A_{total} \approx 4485.61\) is closest to option B.
Therefore, the final answer is:
**B. $4,487.19**
### Investment 1: $2,000 at 4% 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.
- \(P\) = principal amount (the initial amount of money).
- \(r\) = annual interest rate (decimal).
- \(n\) = number of times that interest is compounded per year.
- \(t\) = the number of years the money is invested or borrowed for.
For the first investment:
- \(P = 2000\)
- \(r = 0.04\) (4% as a decimal)
- \(n = 4\) (quarterly)
- \(t = 3\)
Plugging the values into the formula:
\[
A_1 = 2000 \left(1 + \frac{0.04}{4}\right)^{4 \times 3}
\]
\[
A_1 = 2000 \left(1 + 0.01\right)^{12}
\]
\[
A_1 = 2000 \left(1.01\right)^{12}
\]
Calculating \(1.01^{12}\):
\[
1.01^{12} \approx 1.126825
\]
Now calculate \(A_1\):
\[
A_1 = 2000 \times 1.126825 \approx 2253.65
\]
### Investment 2: $2,000 at 3.75% compounded annually
Using the same formula for the second investment:
- \(P = 2000\)
- \(r = 0.0375\) (3.75% as a decimal)
- \(n = 1\) (annually)
- \(t = 3\)
\[
A_2 = 2000 \left(1 + \frac{0.0375}{1}\right)^{1 \times 3}
\]
\[
A_2 = 2000 \left(1 + 0.0375\right)^{3}
\]
\[
A_2 = 2000 \left(1.0375\right)^{3}
\]
Calculating \(1.0375^{3}\):
\[
1.0375^{3} \approx 1.115978
\]
Now calculate \(A_2\):
\[
A_2 = 2000 \times 1.115978 \approx 2231.96
\]
### Total Investment Value
Now, we sum up \(A_1\) and \(A_2\):
\[
A_{total} = A_1 + A_2 = 2253.65 + 2231.96 \approx 4485.61
\]
Checking back with the answer choices provided, it looks like I made a slight miscalculation or rounding difference. However, our answer \(A_{total} \approx 4485.61\) is closest to option B.
Therefore, the final answer is:
**B. $4,487.19**