$6300 is invested, part of it at 11% and part of it at 77%. For a certain year, the total yield is $565.00. How much was invested at each rate?
how much was invested at 11% and 7%?
thank you
If $x at 11%, then the rest $(6300-x) is at 7%. So, just add up the interest:
.11x + .07(6300-x) = 565
To find out how much was invested at each rate, we can set up a system of equations based on the given information.
Let's assume that the amount invested at 11% is A, and the amount invested at 77% is B.
From the given information, we know that the total amount invested is $6300, so we can write the first equation as:
A + B = 6300 ----(1)
We also know that the total yield from the investments is $565. We can calculate the yield from each amount invested and add them up to get the total yield. The yield from A invested at 11% is 0.11A, and the yield from B invested at 77% is 0.77B. So the second equation is:
0.11A + 0.77B = 565 ----(2)
Now we can solve the system of equations (1) and (2) to find the values of A and B.
Here's one way to solve the system of equations:
1. Multiply equation (1) by 0.11 to get 0.11A + 0.11B = 0.11 * 6300 = 693
2. Subtract the equation (2) from the equation obtained in step 1:
0.11A + 0.11B - 0.11A - 0.77B = 693 - 565
Simplifying, we get:
0.66B = 128
3. Divide both sides of the equation by 0.66 to solve for B:
B = 128 / 0.66 = 194.24
4. Substitute the value of B back into equation (1) to solve for A:
A + 194.24 = 6300
A = 6300 - 194.24 = 6105.76
So, approximately $6105.76 was invested at 11%, and approximately $194.24 was invested at 77%.