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:

$\frac{2145}{0.055}=P_{8.5}$

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