Problem 41 - College Algebra Essentials

Question

Things did not go quite as planned. You invested \(\$ 8000\), part of it in stock that paid \(12 \%\) annual interest. However, the rest of the money suffered a \(5 \%\) loss. If the total annual income from both investments was \(\$ 620,\) how much was invested at each rate?

Step-by-Step Solution

Answer

So, about $5000 was invested at a 12% profit rate and the remaining $3000 was the portion that suffered a 5% loss.

Step 1: Define the Variables

Let's say that \(x\) is the amount invested at a 12% profit, and \(y\) is the portion that had a 5% loss. Therefore, from the total value of the investment, we have the equation \(x + y = 8000\).

Step 2: Set up Second Equation

We know that the 12% profitable portion gave 12% of \(x\) as a profit and the 5% loss portion gave a decrease of 5% of \(y\). As combined, the total gain was $620, we can set up the equation \(0.12x - 0.05y = 620\).

Step 3: Solve the System of Equations

We now have a system of two equations to determine the values of \(x\) and \(y\). They can be solved using substitution or elimination method. Here, we'll use substitution for simplicity since we can easily express \(y\) as \(8000 - x\) from the first equation, and substitute this in the second equation to solve for \(x\). Then, the value of \(x\) can be substituted back to get the value of \(y\).

Step 4: Find the Values of x and y

Upon substituting and solving, \(x\) becomes approximately \(5000\) and \(y\) thus becomes approximately \(3000\).