Respuesta :
Using simple interest, it is found that Joe invested $1,200 at 8% interest and $2,800 at 11% interest.
Simple Interest
Simple interest is used when there is a single compounding per time period.
The interest after t years in is modeled by:
[tex]I = Prt[/tex]
In which:
- P is the initial amount.
- r is the interest rate, as a decimal.
In this problem:
- Joe invests a total of $4000 in two plans, hence [tex]P_1 + P_2 = 4000[/tex].
- Part of the money is invested at 8% per year and the rest at 11% per year, hence [tex]r_1 = 0.08, r_2 = 0.11[/tex].
- The interest paid after 1 year on the 11% investment is $212 more than the interest paid on the 8% investment, hence [tex]I_2 = 212 + I_1[/tex].
The equations are:
[tex]I_1 = 0.08P_1[/tex]
[tex]I_2 = 0.11P_2[/tex]
Considering that [tex]P_1 + P_2 = 4000 \rightarrow P_2 = 4000 - P_1[/tex]:
[tex]I_2 = 0.11(4000 - P_1)[/tex]
Considering that [tex]I_2 = 212 + I_1[/tex]:
[tex]212 + I_1 = 0.11(4000 - P_1)[/tex]
Since [tex]I_1 = 0.08P_1[/tex]:
[tex]212 + 0.08P_1 = 0.11(4000 - P_1)[/tex][/tex]
[tex]0.19P_1 = 228[/tex]
[tex]P_1 = \frac{228}{0.19}[/tex]
[tex]P_1 = 1200[/tex]
[tex]P_2 = 4000 - P_1 = 4000 - 1200 = 2800[/tex]
Hence, Joe invested $1,200 at 8% interest and $2,800 at 11% interest.
To learn more about simple interest, you can take a look at https://brainly.com/question/26473004
