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?
If x was invested at 12%, then 8000-x suffered a 5% loss.
0.12x -0.05(8000-x) = 620
0.12x - 400 + 0.05x = 620
0.17x = 1020
Solve for x
6000
Let's assume you invested x amount of money in the stock that paid 12% annual interest.
The remaining amount you invested, (8000 - x), suffered a 5% loss.
The income from the investment in the stock is given by:
x * 0.12
The income from the investment that suffered a 5% loss is given by:
(8000 - x) * (-0.05)
According to the given information, the total annual income from both investments was $620.
So, we can set up the following equation:
x * 0.12 + (8000 - x) * (-0.05) = 620
Now, let's solve this equation to find the value of x.
To solve this problem, let's break it down step by step.
Let's assume you invested x dollars in the stock that paid a 12% annual interest rate. Consequently, the remaining amount invested would be (8000 - x) dollars.
The income from the investment in the stock can be calculated by multiplying the invested amount (x) by the interest rate (12%), giving us an annual income of 0.12x.
On the other hand, the remaining amount invested (8000 - x) suffered a 5% loss, resulting in a -5% annual income, which can be calculated as -0.05 * (8000 - x).
The total annual income is given as $620. Therefore, we can set up the equation:
0.12x + (-0.05)(8000 - x) = 620
Simplifying the equation further:
0.12x - 0.05 * 8000 + 0.05x = 620
0.17x - 400 = 620
0.17x = 620 + 400
0.17x = 1020
x = 1020 / 0.17
x ≈ 6000
Therefore, you invested $6000 in the stock that paid a 12% annual interest rate, and the remaining amount ($8000 - $6000 = $2000) was invested at a 5% loss.