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?
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?
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