Q. 418 - 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 \)4,500 in

interest in one year?

Step-by-Step Solution

Answer

He invested $\$36000$ at $5.5\%$ interest rate and for rest $9\%$ interest rate he invested $\$28000$.

Step 1. Given information.

Total amount invested $=\$64000$

Interest at which investment will be done $=5.5\% and 9\%$

Return after one year $=\$28000$

Step 2. Write the concept.

Let $x\$$be the amount invested with $5.5\%$ and $y\$$ be the amount invested with $9\%$.

Therefore equation will be:

$55x+90y=4500000$

Step 3. Calculation.

$$$$\begin{gathered} 55x+90y=4500\times 1000 ........\left(1\right) \\ x+y=64000 ..........\left(2\right) \end{gathered}$$

Multiplying equation (2) by $90$ and subtracting with equation (1).

$$\begin{gathered} Solve for x: \\ 35x=1260000 \\ x=36000 \end{gathered}$$

Put the value of $x$ in equation (1) :

$$\begin{gathered} Solve for y: \\ 55\left(36000\right)+90y=4500000 \\ 90y=2520000 \\ y=28000 \end{gathered}$$

Step 4. Check the values

Substituting the values of $x and y$ in equation (2):

$$\begin{gathered} x+y=64000 \\ 36000+28000 \\ 64000=64000 \end{gathered}$$

which holds true.