Doc has available to have a 20% alcohol solution and a 70% alcohol solution how many liters of each solution should be mixed to make 30 L of a 50% alcohol solution?

12 L of the 20% solution should be used with 18 L of the 70% alcohol solution

Step-by-step explanation:

assuming that the concentration given is in terms of mass/volume, then doing an mass balance

C₁*V₁ + C₂*V₂ = C₃*V₃

where C = concentration, V = volume, 1,2,3 denotes first, second and final solution respectively.

and since we are mixing the same substance, the total volume V₃ will be the sum of the solutions mixed

V₁ + V₂ = V₃ → V₂=V₃-V₁

thus

C₁*V₁ + C₂ * (V₃-V₁) = C₃*V₃

(C₁-C₂) * V₁ + C₂*V₃ = C₃*V₃

(C₁-C₂) * V₁ = (C₃ - C₂) * V₃

V₁ = V₃ * (C₃ - C₂) / (C₁-C₂)

replacing values

V₁ = V₃ * (C₃ - C₂) / (C₁-C₂) = 30 L * (50%-70%) / (20% - 70%) = 12 L

then

V₂=V₃-V₁=30 L - 12 L = 18 L

therefore 12 L of the 20% solution should be used with 18 L of the 70% alcohol solution