$9500 is invested part of it at 12% and part of it at 7%. For a certain year the total yield is $960.00. How much was invested at each rate? Please show work
Yolo
0.15x+0.05(50000-x)=5000
To solve this problem, we can use the concept of mixtures and the formula for the interest earned.
Let's say the amount invested at 12% is x, and the amount invested at 7% is (9500 - x) since the total amount invested is $9500.
We know that the total yield from both investments is $960.00.
Now, we can calculate the interest earned from each investment:
Interest from the investment at 12% = x * 0.12
Interest from the investment at 7% = (9500 - x) * 0.07
According to the given information, the total interest earned is $960.00:
x * 0.12 + (9500 - x) * 0.07 = 960
Now, let's solve this equation to find the value of x:
0.12x + 0.07(9500 - x) = 960
0.12x + 665 - 0.07x = 960
0.12x - 0.07x = 960 - 665
0.05x = 295
x = 295 / 0.05
x = 5900
So, $5900 was invested at 12% and the remaining amount, which is $9500 - $5900 = $3600, was invested at 7%.
Therefore, $5900 was invested at 12% and $3600 was invested at 7%.