posted by Zach .
A large tank contains 400-kg mixture of water and alcohol. The mixture is 64% alcohol by weight. At each step, 100 kg of the mixture will be drained from the tank, replaced with 100 kg of water and then stirred. After three steps, what percent of the final solution will be alchohol?
At the start we had >64(400) kg of alcohol
100 kg were removed,
so we had .64(300) left, adding 100 kg of water does not change that, so
after 1st step, alcohol left is .64(300) of the 400 kg
in the 2nd step the same argument can be applied
so the amount of alcohol left is .64(.64(300))
after the 3rd step, amount left is .64(.64(.64(300)))
= 300(.64)^3 = 78.64 kg
so the percentage of the tank that is alcohol is 78.64/400 = 19.66%