Question 221798
A 10-quart container is filled with water. 
One quart of water is drained out and replaced with alcohol. 
After mixing, a quart of the soluti0n is drained out and replaced with alcohol.
 This process continued until 5quarts of alcohol have been put into the container.
 The solution in the container is then what percent of alcohol?
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try it step by step
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After the 1st replacement, its a .1(10) mixture
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2nd replacement: 
.1(10-1) + 1 =
.1(9) + 1 = 
.9 + 1 = 1.9 = 1.9 qt of alcohol
that's a {{{1.9/10}}} = .19(10) mixture
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3rd replacement
.19(10-1) + 1 
.19(9) + 1
1.71 + 1 = 2.71 qt of alcohol
that's {{{2.71/10}}} = .27(10) mixture
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3rd replacement
.27(10-1) + 1 =
.27(9) + 1 = 
2.44 + 1 = 3.44 gt of alcohol
that's {{{3.44/10}}} = .344(10) mixture
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4th replacement
.344(10-1) + 1 =
.344(9) + 1 =
3.096 + 1 = 4.096 gt of alcohol
that's {{{4.096/10}}} = .4096(10) mixture
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5th replacement
.4096(10-1) + 1 =
.4096(9) + 1 =
3.6864 + 1 = 4.6864 qt of alcohol
that's {{{4.6864/10}}} = .46864(10) mixture
:
we can say 46.864% alcohol now