$6300 is invested, part of it at 11% and part of it at 8%. For a certain year, the total yield is $597.00. How much was invested at each rate?
amount invested at 11% --- x
amount invested at 8% ---- 6300-x
solve for x .....
.11x + .08(6300-x) = 597
To solve this problem, you can set up a system of equations based on the given information. Let's use variables to represent the amounts invested at each rate.
Let's say x represents the amount invested at 11% and y represents the amount invested at 8%.
From the problem, we know that the total amount invested is $6300, so we have the equation:
x + y = $6300 -- Equation 1
We also know that the total yield from the investments for the year was $597. To calculate the yield, we need to multiply the amount invested at each rate by the corresponding rate and add them together. The yield from the amount invested at 11% is 11% of x, and the yield from the amount invested at 8% is 8% of y. So we have another equation:
0.11x + 0.08y = $597 -- Equation 2
Now we have a system of equations.
To solve this system, we can either use substitution or elimination.
Let's use elimination. Multiply Equation 2 by 100 to remove the decimals:
11x + 8y = $59700 -- Equation 3
Now we can eliminate y by multiplying Equation 1 by 8:
8x + 8y = $50400 -- Equation 4
Subtract Equation 4 from Equation 3 to eliminate y:
11x + 8y - (8x + 8y) = $59700 - $50400
3x = $9300
Divide both sides of the equation by 3:
x = $3100
Now substitute the value of x into Equation 1 to solve for y:
$3100 + y = $6300
y = $3200
Therefore, $3100 was invested at 11% and $3200 was invested at 8%.