an 8% solution, a 10% solution, and a 20% solution of sulfuric acid are to be mixed to get 100 milliliters of a 12% solution. How many milliliters of each must be mixed if the volume of the 20% solution must be 25 milliliters less than the volume of the 10% solution
To solve this problem, we can use the method of solving systems of equations. Let's first define our variables:
Let x be the volume (in milliliters) of the 8% solution.
Let y be the volume (in milliliters) of the 10% solution.
Let z be the volume (in milliliters) of the 20% solution.
Now, we can set up the equations based on the given information:
Equation 1: The total volume of the solution is 100 milliliters.
x + y + z = 100
Equation 2: The concentration of the final solution is 12%.
(0.08x + 0.1y + 0.2z) / 100 = 0.12
Equation 3: The volume of the 20% solution is 25 milliliters less than the volume of the 10% solution.
z = y - 25
Now, we have a system of three equations. Let's solve it step-by-step:
Step 1: Substituting equation 3 into equation 1, we get:
x + y + (y - 25) = 100
2y - 25 = 100
2y = 125
y = 62.5
Step 2: Substituting the value of y in equation 3, we find:
z = 62.5 - 25
z = 37.5
Step 3: Substituting the values of x, y, and z into equation 1, we can solve for x:
x + 62.5 + 37.5 = 100
x + 100 = 100
x = 0
Thus, we have found that we need 0 milliliters of the 8% solution, 62.5 milliliters of the 10% solution, and 37.5 milliliters of the 20% solution to obtain 100 milliliters of a 12% solution, with the volume of the 20% solution being 25 milliliters less than the volume of the 10% solution.