Question

A combined total of $21,000 is invested in two bonds that pay 4% and 8.5% simple interest. The annual interest is $1,515.00. How much is invested in each bond?The amount invested at 4% is $ .The amount invested at 8.5% is $ .

Answer 1

Given:

Total investment is, T (P) = $21,000.

First type of interest is, r(1) = 4%=0.04.

Second type of interest is, r(2) = 8.5%=0.085.

Annual interest amount is, T (I) = $1,515.00.

The objective is to find the amount invested in each type of interest.

Consider the amount invested in first type as x and the amount invested in second type as y.

Then the equation of total investment can be written as,

`x+y=21000\ldots\ldots..(1)`

Now, the equation of total interest amount can be written as,

`\begin{gathered} r(1)x+r(2)y=T(I) \\ 0.04x+0.085y=1515\ldots\ldots\ldots\ldots.(2) \end{gathered}`

Multiply equation (1) by 0.04.

`\begin{gathered} 0.04x+0.04y=0.04(21000) \\ 0.04x+0.04y=840\ldots\ldots\ldots\ldots(3) \end{gathered}`

On solving equation (2) and (3),

`\begin{gathered} 0.04x+0.085y=1515 \\ (0.04x+0.04y=840)/(0.045y=675) \\ y=(675)/(0.045) \\ y=15000 \end{gathered}`

Substitute the value of y in equation (1) to find x.

`\begin{gathered} x+y=21000 \\ x+15000=21000 \\ x=21000-15000 \\ x=6000 \end{gathered}`

Hence, the amount invested at 4% is $6000 and the amount invested at 8.5% is $15000.