Answer:
The volume of the first acid solution (containing 20% acid) in the mixture is 14 liters
The volume of the second acid solution (containing 40% acid) in the mixture is 14 liters
The volume of the third acid solution (containing 70% acid) in the mixture is 28 liters
Step-by-step explanation:
The percentage of acid in the first acid solution = 20%
The percentage of acid in the second acid solution = 40%
The percentage of acid in the third acid solution = 70%
The volume they want to use the three solutions to obtain a mixture of = 56 liters
The concentration of the mixture they want to obtain = 50%
The amount of the 70% solution to be used = 2 × The volume of the 40% solution to be included
Let 'x', 'y', and 'z', represent the volumes of the first second and third acid solutions in the mixture
Therefore, from the question parameters, we get;
x + y + z = 56...(1)
z = 2·y...(2)
0.2·x + 0.4·y + 0.7·z = 0.5 × 56...(3)
Multiplying equation (3) by 10 and then subtracting equation (1) multiplied by 2 from the result gives;
10 × (0.2·x + 0.4·y + 0.7·z = 0.5 × 56) = 2·x + 4·y + 7·z = 5 × 56 = 280
2·x - 2·x + 4·y - 2·y + 7·z - 2·y = 280 - 2 × 56 = 168
2·y + 5·z = 168
From equation (2) and (4), we get;
2·y + 5·z = 2·y + 5×(2·y) = 168
12·y = 168
y = 168/12 = 14
The volume of the second acid solution in the mixture, y = 14 liters
From equation (2), we have;
z = 2·y
∴ z = 2 × 14 = 28
The volume of the third acid solution in the mixture, z = 28 liters
From equation (1), we get;
x + y + z = 56
x = 56 - (y + z)
∴ x = 56 - (14 + 28) = 14
The volume of the first acid solution in the mixture, x = 14 liters.
