a bottle contains 1000ml of pure ethanol. 300 ml is removed and the bottle is topped up with pure water. the mixture is stirred.

Question

what is the volume of ethanol in the bottle if the process is repeated five times?
how many times should the process be repeated for there to be less than 1 ml of ethanol in the bottle?

Answer 1

first procedure:
amount alcohol removed = 300 ml
amount of alcohol left = 700 ml or .7(1000)
topped up with 300 ml of water
alcohol left in the 1000 ml = 700 ml
percentage of alcohol in the mixture = 70% = .7

2nd procedure:
amount of alcohol removed = .7(300) ml
amount of alcohol left = 700 - .7(300) = 100(10 - .7(3))
=490
topped up with 300 ml of water
alcohol left in the 1000 ml = 490 ml
percentage of alcohol in the mixture = 490/1000 = .49 = .7^2

3rd procedure:
amount of alcohol removed = .49(300)
amount of alcohol left = 490 = .49(300) = 343
topped up with 300 of water
alcohol left in the 1000 ml mixture = 343
percentage of alcohol in the mixture = 343/1000 = .343 = .7^3

.....

percentage of alcohol in the mixture after the 5th procedure = .7^5 or .16807
volume left = .16807(1000) = 168.07 ml

we want .7^n (1000) < 1
.7^n ≤ .001

let's look at .7^n = .001
take logs of both sides and use log rules:
n log.7 = log .001
n = log .001/log .7
n = 19.3

so to be less than 1 ml you would have to repeat 20 times.

Answer 2

Thank you!