Problem 43 - College Algebra Essentials

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

The amount of money invested at 14% interest was $5,000 and at a 6% loss was $7,000.

Step 1 - Formulate the equations based on the problem

Create two equations based on the given conditions, which are the sum of the investments and the income from the investments. The first equation would be: \(x + (12,000 - x) = 12,000\), representing the sum of the investments. The second equation is: \(0.14x - 0.06(12,000 - x) = 680\), representing the total income from the investments.

Step 2 - Solve the first equation

Solve the first equation for simplicity. Since \(x + (12,000 - x) = 12,000\), this simplifies to \(x = 12,000\).

Step 3 - Substitute the result from the first equation in the second equation

Substituting the result from the first equation into the second, gives us: \(0.14(12,000) - 0.06(12,000 - 12,000) = 680\). Solving for \(x\) in the above equation gives \(x = 5,000\). Meaning, $5,000 was invested at 14%.

Step 4 - Calculate the amount invested at -6%

Now, subtract the amount invested at 14% from the total investment to find the amount invested at -6%. So, it's \(12,000 - 5,000 = 7,000\). Hence, $7,000 was invested with a 6% loss.