Question

An investor invested a total of 57000 into two bonds. One bond paid 3% simple interest, and the other paid 2 7/8% interest. The investor earned a total of $1676.25 in annual interest. How much was originally invested in each account?

Answer 1

$30,000 (at 3%) and $27,000 (at 2 7/8%)

Step-by-step explanation:

The investor invested a total of 57000

• Let the amount invested at 3% simple interest = x

,

• Then, the amount invested at 2 7/8% simple interest = 57000-x

Recall the formula for simple interest:

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

Interest at 3%

`\begin{gathered} S\mathrm{}I\mathrm{}=(x*3*1)/(100) \\ =(3x)/(100) \\ =0.03x \end{gathered}`

Interest at 2 7/8%

`\begin{gathered} SI=((57000-x)*2(7)/(8)*1)/(100) \\ =(2.875(57000-x))/(100) \\ =0.02875(57000-x) \end{gathered}`

Total Interest

The investor earned a total of $1676.25 in annual interest. Therefore:

`\begin{gathered} 0.03x+0.02875(57000-x)=1676.25 \\ 0.03x+1638.75-0.02875x=1676.25 \\ 0.03x-0.02875x=1676.25-1638.75 \\ 0.00125x=37.5 \\ x=(37.5)/(0.00125) \\ x=30,000 \end{gathered}`

Thus, the amount invested at 3% simple interest = $30,000

The amount invested at 2 7/8% simple interest = 57,000-30,000 = $27,000