Answer :
The controller puts money in 3 places.
First,
He puts money at 2% per year
Let it be "x"
The interest accumulated is
[tex]0.02x[/tex]Second, he puts 1/3rd of previous at 3% per year.
That will be investing (1/3)x at 3%
The interest accumulated is
[tex]\begin{gathered} 0.03(\frac{1}{3}x) \\ =0.01x \end{gathered}[/tex]Thirdly, he puts another balance, Let it be "y", at 9% CD.
The interest accumulated is:
[tex]0.09y[/tex]The sum of all the 3 account interest is $603, thus we can write:
[tex]\begin{gathered} 0.02x+0.01x+0.09y=603 \\ 0.03x+0.09y=603 \end{gathered}[/tex]This is equation 1.
Also, we are told that the total amount of investments is 10,000, thus we can write:
[tex]\begin{gathered} x+\frac{1}{3}x+y=10000 \\ \frac{4}{3}x+y=10000 \end{gathered}[/tex]We have to solve these equations simultaneously to find x and y.
Lets solve for y of the 2nd equation and put it into the first equation and solve for x. Shown below:
[tex]\begin{gathered} \frac{4}{3}x+y=10000 \\ y=10000-\frac{4}{3}x \\ \text{ We have} \\ 0.03x+0.09y=603 \\ \text{Substituting, we have:} \\ 0.03x+0.09(10,000-\frac{4}{3}x)=603 \\ \text{Solving for x:} \\ 0.03x+900-0.12x=603 \\ 900-603=0.12x-0.03x \\ 297=0.09x \\ x=\frac{297}{0.09} \\ x=3300 \end{gathered}[/tex]So, y is:
[tex]\begin{gathered} y=10000-\frac{4}{3}x \\ y=10000-\frac{4}{3}(3300) \\ y=5600 \end{gathered}[/tex]So,
Answer
$3300 invested in first account$1100 invested in second account$5600 invested in third account