A 10-quart container is filled with water.

RedRapier

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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 solution is drained out and replaced with alcohol. This process is continued until 5 quarts of alcohol have been put into the container. The solution in the container is then what percent alcohol?

I understood the question as:
a1=100% water, a2=90% water
then 90 divided by 100 = 0.9
a1 = 100% (water percentage)
r = 0.9
n = 5+1

so I calculated for the 6th term = 59.049%
100%solution - 59.049%water = 40.951% alcohol

Am I correct or am I missing something?
 
Last edited:
Hi. Welcome to the boards. Did you miss the Read Before Posting message? Please take some time to familiarize yourself with the forum's guidelines. Thank you! :cool:

Then, show us what you've tried or thought about, so far. If you're stuck at the beginning, please explain why. For example, do you know how to calculate the percent of alcohol in the 10-quart solution, after replacing one quart of water with one quart of alcohol? If so, what percent did you get?
 
Hi. Welcome to the boards. Did you miss the Read Before Posting message? Please take some time to familiarize yourself with the forum's guidelines. Thank you! :cool:

Then, show us what you've tried or thought about, so far. If you're stuck at the beginning, please explain why. For example, do you know how to calculate the percent of alcohol in the 10-quart solution, after replacing one quart of water with one quart of alcohol? If so, what percent did you get?

Done, sorry for that.
 
It looks like what you did was compute 0.95, which seems to make sense to me.

After the first iteration you have a 90% water solution. In the second iteration, 90% of what you have is the 90% water solution, and 10% is pure alcohol. So that's 90% of 90%. Etc.
 
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