The <em><u>correct answer</u></em> is:
He needs 20 liters of the 20% solution and 40 liters of the 50% solution.
Explanation:
Let x represent the amount of the 20% solution and y represent the amount of the 50% solution.
The total amount of acid in the 20% solution would then be 0.2x; the total amount of acid in the 50% solution would be 0.5y.
We know that together, these make 60 liters of a 40% solution; this gives us the equation
0.2x+0.5y = 0.4(60)
Simplifying, we get
0.2x+0.5y = 2.4
We also know that the amounts of the 20% solution and 50% solution together give us 60 liters; this gives us the equation
x+y = 60
We now have a system of equations:
![\left \{ {{0.2x+0.5y=24} \atop {x+y=60}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B0.2x%2B0.5y%3D24%7D%20%5Catop%20%7Bx%2By%3D60%7D%7D%20%5Cright.)
We will use elimination to solve this. First we will make the coefficients of y equal; to do this, we will multiply the top equation by 2:
![\left \{ {{2(0.2x+0.5y=24)} \atop {x+y=60}} \right. \\\\\left \{ {{0.4x+y=48} \atop {x+y=60}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B2%280.2x%2B0.5y%3D24%29%7D%20%5Catop%20%7Bx%2By%3D60%7D%7D%20%5Cright.%20%3C%2Fp%3E%3Cp%3E%5C%5C%3C%2Fp%3E%3Cp%3E%5C%5C%5Cleft%20%5C%7B%20%7B%7B0.4x%2By%3D48%7D%20%5Catop%20%7Bx%2By%3D60%7D%7D%20%5Cright.)
We will now cancel the y variables by subtracting the bottom equation:
![\left \{ {{0.4x+y=48} \atop {-(x+y=60)}} \right. \\\\-0.6x = -12](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B0.4x%2By%3D48%7D%20%5Catop%20%7B-%28x%2By%3D60%29%7D%7D%20%5Cright.%20%3C%2Fp%3E%3Cp%3E%5C%5C%3C%2Fp%3E%3Cp%3E%5C%5C-0.6x%20%3D%20-12)
Divide both sides by -0.6:
![\frac{-0.6x}{-0.6}=\frac{-12}{-0.6}\\\\x=20](https://tex.z-dn.net/?f=%5Cfrac%7B-0.6x%7D%7B-0.6%7D%3D%5Cfrac%7B-12%7D%7B-0.6%7D%3C%2Fp%3E%3Cp%3E%5C%5C%3C%2Fp%3E%3Cp%3E%5C%5Cx%3D20)
Substituting this into the second equation, we have
x+y = 60
20+y = 60
Subtract 20 from each side:
20+y-20 = 60-20
y = 40