Question

Financial Planning

Mark has $\$100,000$ to invest. His financial consultant advises him to diversify his investment in three types of bonds: short-term, intermediate-term, and long-term. The short-term bonds pay $4\%$, the intermediate term bonds pay $5\%$, and the long-term bonds pay $6\%$ simple interest per year. Mark wishes to realize a total annual income of $5.1\%$, with equal amounts invested in short- and intermediate-term bonds. How much should he invest in each type of bond?

Short answer

Thus the amount invested in a short-term bond is $\$30000$

Thus the amount invested in a long-term bond is $\$40000$

Step-by-step solution

Step 1. Given information

We have been given the three types of bonds along with their simple interest per year and we are asked to evaluate the investment in each bonds.

Step 2. Concept used

We will be assuming three types of given bonds as x be the short-term bond, be the short-term bond, z be the long-term bond.

Then, we will develop a system of three equations using the given three conditions.

By solving the system of the equation we can get values of $x,y,z$or of how much should he invest in each type of bond.

Step 3. Calculation

Let x be the short-term bond, y be the intermediate-term bond, z be the long-term bond.

The total amount to be invested is $100,000$

Let convert the conditions into the equations:

$\begin{array}{l}x+y+z=100,\mathrm{000....}\left(1\right)\\ 0.04x+0.05y+0.06z=0.051\times 100,\mathrm{000....}\left(2\right)\\ x=y\mathrm{....}\left(3\right)\end{array}$

Multiplying equation $1$ by $(-0.06)$and grouping with equation $2$we get

$\begin{array}{l}-0.06x-0.06y-0.06z=-\mathrm{6000....}\left(4\right)\\ 0.04x+0.05y+0.06z=\mathrm{5100....}\left(5\right)\end{array}$

Adding equations $4$and $5$, we get

$\begin{array}{l}(-0.06x+0.04x)+(-0.06y+0.05y)+(-0.06x+0.06y)=100000+5100\\ -0.02x-0.01y+0z=-900\\ 100(-0.02x-0.01y)=(-900)100\\ -2x-1y=-\mathrm{90000....}\left(6\right)\end{array}$

Again taking and grouping equations 6 and 3, we get

$\begin{array}{l}-2x-y=-90000\\ x-y=0\end{array}$

Subtracting the above equations we, get

$\begin{array}{l}(-2x-x)+(-y-(-y\left)\right)=(-90000-0)\\ -3x+0y=-90000\\ -3x=-90000\\ x=30000\end{array}$

Thus the amount x invested in a short-term bond is $\$30000$

Since $x=y$ then the amount invested in an intermediate-term bond is also $\$30000$

Substituting the values of x and $y=\$30000$

$\begin{array}{l}x+y+z=100000\\ 30000+300000+z=100000\\ 60000+z=100000\\ z=40000\end{array}$

Thus the amount z invested in a long-term bond is $\$40000$