Question

A combined total of $55,000 is invested in two bonds that pay 3% and 8.5% simple interest. The annual interest is $3,795.00. How much is invested in each bond?

Answer 1

We are given that a combined total of $55000 is invested in two bonds that pay 3% and 8% respectively. The interest gained is given by the following formula:

`i=Prt`

Where:

`\begin{gathered} i=\text{ interest} \\ P=\text{ amount invested} \\ r=\text{ interest rate in decimal form} \\ t=\text{ time} \end{gathered}`

Since the given interest is annual this means that the value of time is 1:

`t=1`

Also, since the interest is the combination of the interest of both bonds this means that the total interest is the sum of the interest of each bond:

`i_T=i_3+i_(8.5)`

Where:

`i_T=\text{ total interest}`

Now, we will use the formula for the interest_

`i_T=P_3r_3t_3+P_(8.5)r_(8.5)t_(8.5)`

Now, we substitute the value of time:

`i_T=P_3r_3+P_(8.5)r_(8.5)`

Now, we substitute the interest rates:

`i_T=0.03P_3+0.085P_(8.5)`

The total interest is $3795:

`3795=0.03P_3+0.085P_(8.5)`

Since the total investment is $55000 this means that the sum of the amounts invested in the 3% and 8.5% bonds is:

`P_3+P_(8.5)=55000`

Now, we solve for P3:

`P_3=55000-P_(8.5)`

Now, we substitute in the equation of the total interest:

`3795=0.03(55000-P_(8.5))+0.085P_(8.5)`

Now, we solve for P3. First, we use the distributive law in the first parenthesis:

`3795=1650-0.03P_(8.5)+0.085P_(8.5)`

Now, we add like terms;

`3795=1650+0.055P_(8.5)`

Now, we subtract 1650 from both sides:

`\begin{gathered} 3795-1650=0.055P_(8.5) \\ 2145=0.055P_(8.5) \end{gathered}`

Now, we divide both sides by 0.055:

`(2145)/(0.055)=P_(8.5)`

solving the operations:

`39000=P_(8.5)`

Now, we substitute this value in the formula for P3:

`P_3=55,000-P_(8.5)`

Now, we substitute the value of P8.5:

`\begin{gathered} P_3=55000-39000 \\ P_3=16000 \end{gathered}`

Therefore, the amount invested in the 3% bond is $16000, and the amount invested in the 8.5% bond is $39000