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September 20, 2014

September 20, 2014

Posted by **Zach** on Tuesday, May 4, 2010 at 11:57pm.

thanks

~zach

- Math -
**Reiny**, Wednesday, May 5, 2010 at 7:20amAt 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%

- Math -
**Zach**, Wednesday, May 5, 2010 at 10:08pmThank you

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