Question

A total of $9500 is invested, part at 10% simple interest and part at 8%. If the total annual return from the two investments is $ 878.00, how much is invested at each rate?

Answer 1

• The amount invested at 10% = $5,900

,

• The amount invested at 8% =$3600

Step-by-step explanation:

`Simple\: Interest=(Principal* Rate* Time)/(100)`

Let the amount invested at 10%=y

`\begin{gathered} \text{Simple Interest}=(y*10*1)/(100) \\ =0.1y \end{gathered}`

Then, the amount invested at 8% =$(9500-y)

`\begin{gathered} \text{Simple Interest}=((9500-y)*8*1)/(100) \\ =0.08(9500-y) \\ =760-0.08y \end{gathered}`

The return from the two investments is $ 878.00, thus:

`\begin{gathered} 0.1y+(760-0.08y)=878 \\ 0.1y-0.08y=878-760 \\ 0.02y=118 \\ y=(118)/(0.02) \\ y=5,900 \end{gathered}`

We conclude that:

• The amount invested at 10% = $5,900

,

• The amount invested at 8% =9500-5900=$3600