Mixture word problem

Question

A tank contains 20 / of a solution that is 60% acid. How much water must be added to reduce it to a 20% solution

Answer 1

I am not sure what 20/ means, but I'll assume that you mean 

20 liters.  

 

Let x =  number  of liters of water to be added

     20 = number of liters of 60% acid solution

    x + 20 = number of liters of 20% acid solution

 

(Acid content of the 60% solution) + (Acid content of water) = (Acid content of the mixture)

 

0.60(20) + 0.00x = 0.20(x + 20)

 

                      12 = 0.20x + 4

                        8 = 0.20x

                      40 = x

 

Add 40 liters of water to 20 liters of 60% acid solution to obtain 60 liters of 20% acid solution.