Problem 42 - College Algebra

Question

Things did not go quite as planned. You invested \(\$ 12,000\), part of it in stock that paid \(14 \%\) annual interest. However, the rest of the money suffered a \(6 \%\) loss. If the total annual income from both investments was \(\$ 680,\) how much was invested at each rate?

Step-by-Step Solution

Answer

You invested \$7000 at a \(14\%\) annual interest rate and \$5000 in a venture that suffered a \(6 \%\) loss.

Step 1: Set Up The Equations

Here’s what we know: there are two amounts of money invested, let's call them \(x\) and \(y\), with \(x\) being the amount that made a \(14 \%\) profit and \(y\) being the amount that lost \(6 \%\) of its value. In total there was \$12,000 invested which gives us our first equation: \(x + y = 12000\). For the return on investment, \(x\) provided a \(14 \%\) profit and \(y\) resulted in a \(6 \%\) loss, for a total profit of \$680. This gives a second equation: \(0.14x - 0.06y = 680\).

Step 2: Solve The Equations

These are now simple simultaneous equations that can be solved. First, multiply the first equation by 0.14: \(0.14x + 0.14y = 1680\). Then subtract the second equation from our newly formed equation: \(0.14x - 0.06y = 680\), which gives \(0.2y = 1000\). Solving for \(y\) gives \(y = \$5000\). Substituting \(y\) into the equation \(x + y = 12000\) gives \(x = \$7000\).

Step 3: Interpret the Results

These results simply need to be interpreted within the context of the problem. This means that, of the initial \$12000, \$7000 was invested in a scheme that attracted a \(14 \%\) annual interest rate, while \$5000 was invested in a scheme that suffered a \(6 \%\) loss.