Question 390945

Let x=amount that is removed and replaced each time

After initial action (remove alcohol and fill with water )we have 20 liters that contains (20-x)/20 parts of pure alcohol or (20-x) liters of pure alcohol

After 2nd removal, we remove ((20-x)/20)*x parts of pure alcohol leaving
(20-x)-((20-x)/20)*x of pure alcohol and we are told that this equals 5 liters, sooooo:
(20-x)-(20x-x^2)/20 =5 multiply each term by 20
400-20x-20x+x^2=100 simplify
x^2-40x+300=0----quadratic in standard form and it can be factored
(x-10)(x-30)=0
x=10 liters ------------amount that is removed and replaced
x=30  NO GOOD --This amount is more than 20 liters 

After first action, we remove 10 liters of pure alcohol leaving 20 liters that is 50% alcohol
After 2nd action, we remove 10 liters that is 50% alcohol or 5 liters of pure alcohol
In total we remove 10+5 or 15 liters of pure alcohol leaving 5 liters

Hope this helps---ptaylor