Question

How much pure acid should be mixed with 3 gallons of a 20% acid solution in order to get a 40% acid solution?

Answer 1

100 y+(20*3)=40(y+3)
100y +60 = 40y +120
100y-40y =120-60
60y = 60
y=1
need 1 gallons

Answer 2

3(.20) + b(1) = 3(.40) + b(.40)
b(1-.40) = 3(.40-.20) i get 1 as well