A self-employed contractor nearing retirement made two investments totaling $15,000. In one year, these investments yielded $1023 in simple interest. Part of the money was invested at 6% and the rest at 7.5%. How much was invested at each rate?
amount invested at 6% .... x
amount invested at 7.5% ----- 15000 - x
.06x + .075(15000-x) = 1023
solve for x, etc
notice this equation follows the same thinking as the equation I gave you in your previous post with the mixtures.
this is just another mixture problem. How did you set it up?
To solve this problem, we need to set up a system of equations based on the given information. Let's say the amount invested at 6% is x, and the amount invested at 7.5% is y.
According to the problem, the total amount invested is $15,000, so we can write the equation:
x + y = 15000
We also know that the total interest earned in one year is $1023. The interest earned on x is given by the formula:
Interest on x = x * 0.06
Similarly, the interest earned on y is given by:
Interest on y = y * 0.075
So, we can write another equation based on the total interest earned:
Interest on x + Interest on y = 1023
(x * 0.06) + (y * 0.075) = 1023
Now, we have a system of two equations that we can solve simultaneously to find the values of x and y.
To solve this system of equations, we can use substitution or elimination method. I will use the elimination method to demonstrate:
Multiply the first equation by 0.06 to match the coefficients of x in both equations:
0.06x + 0.06y = 900
Now, subtract the second equation from this:
(0.06x + 0.06y) - (0.06x + 0.075y) = 900 - 1023
0.06x - 0.06x + 0.06y - 0.075y = -123
-0.015y = -123
y = (-123) / (-0.015)
y = 8200
Now, substitute the value of y back into the first equation to solve for x:
x + 8200 = 15000
x = 15000 - 8200
x = 6800
Therefore, $6800 was invested at 6% and $8200 was invested at 7.5%.