How many liters each of a 55% acid solution and a 80% acid solution must be used to produce 50 liters of a 60% acid solution? (Round to two decimal places if necessary.)
.55 V1 + .80 V2 = .6(50)
and
V1 + V2 = 50 so V2 = 50 - V1
.55 V1 + .8(50 -V1) = 30 etc
To solve this problem, we need to set up a system of equations based on the information given.
Let's represent the amount of the 55% acid solution as "x" liters and the amount of the 80% acid solution as "y" liters.
We know that the total volume of the acid solution produced is 50 liters. Therefore, we have the equation:
x + y = 50 -- Equation 1
We also know that the resulting acid solution is 60% acid. This means that the amount of acid in each solution, when combined, should be 60% of the total volume. We can calculate the amount of acid as follows:
For the 55% acid solution:
Amount of acid = 0.55x
For the 80% acid solution:
Amount of acid = 0.80y
Since we want a total of 50 liters of the resulting 60% acid solution, the amount of acid in this solution would be 0.60 * 50 = 30 liters.
So, we have the equation:
0.55x + 0.80y = 30 -- Equation 2
Now, we need to solve this system of equations to find the values of x and y.
To eliminate one variable, let's multiply Equation 1 by 0.55:
0.55x + 0.55y = 27.50 -- Equation 3
Now, subtract Equation 3 from Equation 2:
0.55x + 0.80y - 0.55x - 0.55y = 30 - 27.50
0.25y = 2.50
Divide both sides by 0.25:
y = 2.50 / 0.25
y = 10
Now that we have the value of y, we can substitute it back into Equation 1 to find x:
x + 10 = 50
x = 50 - 10
x = 40
Therefore, to produce 50 liters of a 60% acid solution, you will need 40 liters of the 55% acid solution and 10 liters of the 80% acid solution.