Two solutions of acid were mixed to obtain 10 liters of new solution. ...
Let's assume the volume of the first solution (with higher concentration) is x liters.
According to the problem, the first solution contains 0.8 liters of acid, so its concentration is 0.8/x (liters of acid per liter of solution).
The second solution contains 0.6 liters of acid, and the percentage of acid in the first solution is twice that in the second. This means that the percentage of acid in the second solution is half that in the first solution.
Let's calculate the percentage of acid in the second solution:
Percentage of acid in the second solution = (0.6 liters of acid / x liters of solution) * 100
Since the percentage of acid in the second solution is half that in the first solution, we can write:
0.5 * (0.8 liters of acid / x liters of solution) * 100 = (0.6 liters of acid / x liters of solution) * 100
Now we can simplify and solve this equation:
0.4/x = 0.6/x
0.4 = 0.6
This equation is not possible, so our initial assumption that the volume of the first solution is x liters is incorrect.
Let's try another assumption:
Let the volume of the second solution be y liters.
So the volume of the first solution, which contains 0.8 liters of acid, must be (10 - y) liters (since the total volume of the new solution is 10 liters).
According to the problem, the percentage of acid in the first solution is twice that in the second:
(0.8 liters of acid / (10 - y) liters of solution) = 2 * (0.6 liters of acid / y liters of solution)
Now we can solve this equation:
0.8/y = 2 * 0.6/(10 - y)
Simplifying further:
0.8/y = 1.2/(10 - y)
Cross-multiplying:
0.8 * (10 - y) = 1.2 * y
8 - 0.8y = 1.2y
Combining like terms:
8 = 2y
y = 8/2
y = 4
Therefore, the volume of the second solution is 4 liters. Since the total volume of the new solution is 10 liters, the volume of the first solution is (10 - 4) = 6 liters.
So, the correct answer is 6 liters, which corresponds to option D.