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Chapter 1: Problem 42

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?

Short Answer

Expert verified
The amount invested at a 14% interest rate is $8,000 and the amount that suffered a 6% loss is $4,000.

Step by step solution

01

- Define the Variables

First, we need to define the variables to represent the unknowns. Let x be the amount invested at 14% and y be the amount that suffered a 6% loss.
02

- Write, Simplify and Solve the Equations

We have a system of two equations that reflect the problem: \鈥� The total amount invested, which is \$12,000, is the sum of x and y. This gives us equation (1) as x + y = 12,000. \鈥� The interests from both investments amounted to \$680. The interest from investment x is 14% of x and the loss from investment y is 6% of y. Therefore we have equation (2) as 0.14x - 0.06y = 680. \Now, to solve for x and y, we could multiply the first equation by 0.14 and subtract the second equation from the first. We obtain y = 4000. Substitute y into the first equation, we get x = 8000.
03

- Check the Results

To ensure that the solution is correct, we need to substitute x = 8,000 and y = 4,000 back into the second equation to verify whether the left-hand side equals to the right-hand side (680). Doing so, we verify that indeed 0.14*8000 - 0.06*4000 = 680.

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