$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 HELP
if x is at 12%, then the rest (9500-x) is at 7%.
so, add up the interest for each part. The sum must be 960.00
To solve this problem, you can use a system of equations.
Assume that the amount invested at 12% is "x" dollars, and the amount invested at 7% is "9500 - x" dollars.
Now, let's calculate the yield from each investment:
Yield from the investment at 12% = (12/100) * x
Yield from the investment at 7% = (7/100) * (9500 - x)
According to the problem, the total yield is $960:
(12/100) * x + (7/100) * (9500 - x) = 960
Simplifying the equation:
0.12x + 0.07(9500 - x) = 960
Now, let's solve for x:
0.12x + 0.07 * 9500 - 0.07x = 960
0.12x - 0.07x = 960 - 0.07 * 9500
0.05x = 960 - 665
0.05x = 295
x = 295 / 0.05
x = 5900
So, $5900 was invested at a 12% interest rate, and $9500 - $5900 = $3600 was invested at a 7% interest rate.