SOLUTION: Please help me solve this problem...I dont know how to start solving this.. A 10-quart container is filled with water. One quart of water is drained out and replaced with alcoho

Question 221798: Please help me solve this problem...I dont know how to start solving this..
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?
Thank you in advance..
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
try it step by step
:
After the 1st replacement, its a .1(10) mixture
:
2nd replacement:
.1(10-1) + 1 =
.1(9) + 1 =
.9 + 1 = 1.9 = 1.9 qt of alcohol
that's a = .19(10) mixture
:
3rd replacement
.19(10-1) + 1
.19(9) + 1
1.71 + 1 = 2.71 qt of alcohol
that's = .27(10) mixture
:
3rd replacement
.27(10-1) + 1 =
.27(9) + 1 =
2.44 + 1 = 3.44 gt of alcohol
that's = .344(10) mixture
:
4th replacement
.344(10-1) + 1 =
.344(9) + 1 =
3.096 + 1 = 4.096 gt of alcohol
that's = .4096(10) mixture
:
5th replacement
.4096(10-1) + 1 =
.4096(9) + 1 =
3.6864 + 1 = 4.6864 qt of alcohol
that's = .46864(10) mixture
:
we can say 46.864% alcohol now
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