Question 390945: A vessel is filled with 20 liters of alcohol. A certain amount of alcohol is removed from the vessel and it is filled with water. than the same amount of the mixture is removed and the vessel is filled with water again. after this, vessel contains only 5 liters of pure alcohol. how many liters of the liquid were removed from the vessel each time?
The book says that the problem must be solved using quadratic equation. I got the answer logically, which is 10, but I could not write a proper equation.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
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
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