Question

Two Investments totaling $47,500 produce an annual income of $3000. One investment yields 9% per year, while the other yields 6% per year. How much is invested at each rate?

Answer 1

Let:

P1 = Investment #1

P2 = Investment #2

r1 = Rate #1

r2 = Rate #2

I1 = Income 1

I2 = Income 2

The 2 investment produce an annual income of $3000, so:

`I1+I2=3000`

One investment yields 9% per year, while the other yields 6% per year:

`\begin{gathered} I1=P1\cdot r1\cdot t_{\text{ }}(1) \\ I2=P2\cdot r2\cdot t_{\text{ }}(2) \\ where\colon \\ t=1 \\ r1=0.09 \\ r2=0.06 \\ I1=0.09P1_{\text{ }}(1) \\ I2=0.06P2_{\text{ }}(2) \end{gathered}`

Two Investments totaling $47,500 so:

`P1+P2=47500_{\text{ }}(3)`

Add (1) and (2):

`\begin{gathered} I1+I2=0.09P1+0.06P2 \\ so\colon \\ 0.09P1+0.06P2=3000_{\text{ }}(4) \end{gathered}`

From (4) solve for P1:

`P1=(3000-0.06P2)/(0.09)_{\text{ }}(5)`

Replace (5) into (3):