how much pure alcohol should be added to 6 gallons of 10% acid solution to obtain a solution that is 40%

3

Step-by-step explanation:

Assume that the total amount of alcohol at the end is x, and that you are adding y gallons to the mixture. We have 6+y=x, and x*0.4 (40% of x) = y+current amount of alcohol. The current amount of alcohol is 10%*6, or 0.6 gallons, so x*0.4 = y+0.6. Solving for y, we have

x-6=y (from 6+y=x),

and putting that into our second equation, we have

x*0.4 = (x-6) + 0.6, so

x*0.4 = x-5.4, and subtracting x from both sides, we get

-0.6*x = - 5.4, and dividing both sides by - 0.6, we get that

x=9

Thus, y=x-6=3, our answer