How many liters of a 20% alcohol solution must be mixed with 10 liters of an 80% solution to get a 40% solution?

Let the number of liters of 20% alcohol solution =L

 

0.20L  +  0.80*(10) =0.40*[L + 10], solve for L

0.20L  +  8                 =0.40L + 4

8 - 4                            =0.40L - 0.20L

4                                  =0.20L

L = 4 / 0.20

L = 20 Liters of 20% alcohol solution will be required.