Question 1145132
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At the start, the 10 quarts are 10 of water and 0 of alcohol.<br>
(1) First time:  When 1/10 (0.1) of the container is emptied, the amount of alcohol removed is (0.1)(0)=0, leaving 0; when 1 quart of alcohol is then added, the amount of alcohol is 0+1 = 1.<br>
(2) Second time:  When 1/10 (0.1) of the container is emptied, the amount of alcohol removed is (0.1)(1)=0.1, leaving 0.9; when 1 quart of alcohol is then added, the amount of alcohol is 0.9+1 = 1.9<br>
(3) Third time:  When 1/10 (0.1) of the container is emptied, the amount of alcohol removed is (0.1)(1.9)=0.19, leaving 1.71; when 1 quart of alcohol is then added, the amount of alcohol is 1.71+1 = 2.71<br>
(4) Fourth time:  When 1/10 (0.1) of the container is emptied, the amount of alcohol removed is (0.1)(2.71)=0.271, leaving 2.439; when 1 quart of alcohol is then added, the amount of alcohol is 2.439+1 = 3.439<br>
(5) Fifth time:  When 1/10 (0.1) of the container is emptied, the amount of alcohol removed is (0.1)(3.439)=0.3439, leaving 3.0951; when 1 quart of alcohol is then added, the amount of alcohol is 3.0951+1 = 4.0951<br>
After the 5th time, the percentage of alcohol in the container is 4.0951/10 converted to a percentage, which is 40.951%.<br>
ANSWER: 40.951%<br>
And now that we have obtained the answer by that method, there is a much easier path....<br>
Each time a quart of liquid is removed and replaced with a quart of alcohol, the fraction of water in the container gets multiplied by 0.9.
After 5 times, the fraction of the liquid in the container that is water is (0.9)^5 = 0.59049.
So after the 5th time the percentage of water in the container is 59.049%; so the percentage of alcohol is 100-59.049% = 40.951%.