Q. 60 - Precalculus Enhanced with Graphing Utilities

Question

Financial Planning 

A recently retired couple needs $\backslash (12,000$ per year to supplement their Social Security. They have $\backslash )150,000$ to invest to obtain this income. They have decided on two investment options: AA bonds yielding $10\%$ per annum and a Bank Certificate yielding $5\%$.

(a) How much should be invested in each to realize exactly $\backslash (12,000$?

(b) If, after $2$ years, the couple requires $\backslash )14,000$ per year in income, how should they reallocate their investment to achieve the new amount? 

Step-by-Step Solution

Answer

Part (a) Investments in AA bonding and bank are $\$90,000$ and $\$60,000$ respectively.

Part (b) Investments in AA bonding and bank are $\$130,000$ and $\$20,000$ respectively.

Part (a) Step 1. Given Information

Total investment amount $=\$150,000$

Interest on AA bonds $=10\%$

Interest on Bank Certificate $=5\%$

Amount required by couple per year $=\$12,000$

Part (a) Step 2. Explanation

Let $x$ and $y$ be the investment in AA Banks and Bank Certificate respectively.

Required equations are:

$$\begin{gathered} 0.1x+0.05y=12000......\left(i\right) \\ x+y=150000......\left(ii\right) \end{gathered}$$

We use the elimination method. Multiplying equation $\left(ii\right)$ by $0.1$ we get,

$$\begin{gathered} -0.1x-0.1y=-0.1\times 150000 \\ \Rightarrow -0.1x-0.1y=-15000......\left(iii\right) \end{gathered}$$

Adding equations $\left(i\right)$ and $\left(iii\right)$ we get,

$$\begin{gathered} -0.05y=-3000 \\ \Rightarrow y=\frac{3000}{0.05} \\ \Rightarrow y=60000 \end{gathered}$$

Substituting the value of $y$ in equation $\left(ii\right)$ we get,

$$\begin{gathered} x+60000=150000 \\ \Rightarrow x=150000-60000 \\ \Rightarrow x=90000 \end{gathered}$$

Part (b) Step 1. Explanation

Let $x$ and $y$ be the investment in AA Banks and Bank Certificate respectively when the return is $\$14000$.

Required equations are:

$$\begin{gathered} 0.1x+0.05y=14000......\left(iv\right) \\ x+y=150000......\left(v\right) \end{gathered}$$

We use the elimination method. Multiplying equation $\left(v\right)$ by $-0.1$ we get,

$$\begin{gathered} -0.1x-0.1y=-0.1\times 150000 \\ \Rightarrow -0.1x-0.1y=-15000......\left(vi\right) \end{gathered}$$

Adding equations $\left(iv\right)$ and $\left(vi\right)$ we get,

$$\begin{gathered} -0.05y=-1000 \\ \Rightarrow y=\frac{1000}{0.05} \\ \Rightarrow y=20000 \end{gathered}$$

Substituting the value of $y$ in equation $\left(v\right)$ we get,

$$\begin{gathered} x+20000=150000 \\ \Rightarrow x=150000-20000 \\ \Rightarrow x=130000 \end{gathered}$$