Answer
The answer is: 3.2 liters of pure water and 0.8 liters of 50% acid solution.
Work Step by Step
We assign the variables:
x = initial liters of acid
y = pure water volume
Step 1:
Find the equations that represent the problem.
We are mixing both solutions to create a new one, but it has to be in the same terms. We will represent it in acid percentages. Pure (100%) water means 0% acid.
50% acid solution + 0% acid solution = 10% acid solution -> Eq. 1
We can get the second equation form the wording in the problem.
From "a solution that is 50% acid",
50% acid solution $=0.5x$ -> Eq. 2
We can get the third equation form the wording in the problem.
From "a solution that is 100% water",
0% acid solution $=0$ -> Eq. 3
We can get the fourth equation form the wording in the problem.
From "to make 4 L of a solution that is 10% acid",
10% acid solution $=0.1(4)=0.4$ -> Eq. 4
Remember that the volume from the new solution has to equal the sum of both volumes of the initial solutions, so
$x+y=4$ -> Eq. 5
Step 2:
Solve the system of equations using the substitution method,
-> Substitute Eq. 2 and Eq. 4 into Eq. 1
$0.5x+0=0.4$
$0.5x=0.4$
$x=\frac{0.4}{0.5}$
$x=0.8L$
->Substitute the value for $x$ into Eq. 5
$0.8+y=4$
$y=4-0.8$
$y=3.2$
Step 3:
The answer is: 3.2 liters of pure water and 0.8 liters of 50% acid solution.
