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The controller (money manager) for a small company puts some money in a bank account paying 2% per year. He uses some additional money, amounting to 1/3 the amount placed in the bank to buy bonds paying 3% per year. With the balance of the funds he buys a 9% certificate of deposit. The first year the investments bring a return of $603. If the total of the investments is $10,000 how much is invested at each rate?

Sagot :

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

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