Q. 153 - Intermediate Algebra

Question

Arnold invested \($64,000$, some at $5.5$% interest and the rest at $9$%. How much did he invest at each rate if he received \)$4500$in interest in one year?

Step-by-Step Solution

Answer

Arnold should invest $\$36,000$ at $5.5\%$ rate and $\$28,000$ at $9\%$.

Step 1. Given Information

We are given that Arnold invested $\$64,000$ in two account with interest $5.5\%$ and $9\%$. 

Step 2. Assumptions and Formation of equations.

Let $x$ and $y$ be the amount invested in first and second account respectively.

Total amount invested is equal to $\$64,000$, so

$x+y=64000$

He received $\$4500$ every year, so the other equation is given by

$$\begin{gathered} 5.5\%x+9\%y=4500 \\ 0.055x+0.09y=4500 -\left(2\right) \end{gathered}$$

Step 3. Solving the equations.

The first equation can be written as,

$x=64000-y$

Now, putting the value of $x$ in second equation, we get

$$\begin{gathered} 0.055x+0.09y=4500 \\ 0.055(64000-y)+0.09y=4500 \\ 3520-0.055y+0.09y=4500 \\ 0.35y=980 \end{gathered}$$

Dividing by $0.35$, we get

$y=28000$

Now, putting the value of $x$ in third equation, we get

$$\begin{gathered} x=64000-28000 \\ x=36,000 \end{gathered}$$

Hence he should invest $\$36,000$ in first account with $5.5\%$ and $\$28,000$ in second account with $9\%$ interest.

Step 4. Checking the solution

Checking the solution by putting the value of $x, y$ in the equations, we get

$$\begin{gathered} x+y=64000 \\ 36000+28000=64000 \\ 64000=64000 \\ 0.055x+0.09y=4500 \\ 0.055\left(36000\right)+0.09\left(28000\right)=4500 \\ 1980+2520=4500 \\ 4500=4500 \end{gathered}$$

This is true, hence the solution is correct.