Posted by Zach on Tuesday, May 4, 2010 at 11:57pm.
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%
Thank you
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