Respuesta :
Given:
The total investment = $36000.
The interest rate of the first investment = 8%.
The interest rate of the second investment = 6%.
The interest rate of the third investment = 9%.
The total interest = $ 2820.
The interest from the first investment was 3 times the interest from the second.
Required:
We need to find the amounts of the three parts of the investment.
Explanation:
Let x, y, and z be the amounts of the three parts of the investment.
The total investment = $36000.
[tex]x+y+z=36000[/tex]The total interest = $ 2820.
[tex]8\text{\% of x +6 \% of y + 9 \% of z = 2820}[/tex][tex]\frac{8}{100}x+\frac{6}{100}y+\frac{9}{100}z=2820[/tex][tex]0.08x+0.06y+0.09z=2820[/tex]The interest from the first investment was 3 times the interest from the second.
[tex]0.08x=3\times0.06y[/tex][tex]0.08x=0.18y[/tex]Divide both sides by 0.08.
[tex]\frac{0.08x}{0.08}=\frac{0.18y}{0.08}[/tex][tex]x=2.25y[/tex][tex]Substitute\text{ x=2.25y in the equation }0.08x+0.06y+0.09z=2820.[/tex][tex]0.08(2.25y)+0.06y+0.09z=2820[/tex][tex]0.18y+0.06y+0.09z=2820[/tex][tex]0.24y+0.09z=2820[/tex][tex]0.09z=2820-0.24y[/tex]Divide both sides by 0.09.
[tex]\frac{0.09z}{0.09}=\frac{2820}{0.09}-\frac{0.24y}{0.09}[/tex][tex]z=31333.33-2.67y[/tex][tex]Substitute\text{ x =2.25y and z=31333.33y-2.67y in the equation }x+y+z=36000.[/tex][tex]2.25y+y+31333.33-2.67y=36000[/tex][tex]2.25y+y-2.67y=36000-31333.33[/tex][tex]0.58y=4666.67[/tex]Divide both sides by 0.58.
[tex]\frac{0.58y}{0.58}=\frac{4666.67}{0.58}[/tex][tex]y=8045.98[/tex][tex]Substitute\text{ y =8045.98 in the equation }x=2.25y.[/tex][tex]x=2.25\times8045.98[/tex][tex]x=18103.46[/tex][tex]Substitute\text{ }x=18103.46\text{, and y =8045.98 in the equation x+y+z=36000.}[/tex][tex]18103.46+8045.98+z=36000[/tex][tex]26149.44+z=36000[/tex][tex]z=36000-26149.44[/tex][tex]z=9850.56[/tex]Final answer:
The first part of the investment was $ 18103.46.
The second part of the investment was $ 8045.98.
The third part of the investment was $ 9850.56
