Solve Water Mixing Problem: Volume K Tank of Alcohol

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The problem involves a tank of K liters of alcohol where one liter is removed and replaced with water, repeated three times. After these iterations, the final ratio of water to alcohol should be 7:1. The solution approach involves calculating the alcohol and water components at each stage using the formula that accounts for the dilution effect of adding water. The initial state is defined with all alcohol and no water, and the calculations progress through each step to derive the final mixture. The goal is to determine the volume K of the tank based on the established ratio.
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I've struggled with this problem for a bit and I ran out of ideas.

"We have a tank of volume K liters of alcohol. We remove one liter from the tank and add one liter of water. From the mix, we remove one liter and add one liter of water. We do this one more time.

At the end of the process, there should be 7 times more water than alcohol in the tank.

What is the volume of the tank?"

My attempt at a solution is attached below. I hope it's legible.

The idea is that adding one liter of alcohol means multiplying the water and the alcohol components by (k-1)/k and adding 1 to the water component.

The alcohol component has been written down as a_i, the water component is b_i, where:
i={0,1,2,3} is the step
{a0, b0} is the initial state (a0=k, b0=0)
{a1,b1} is the state after adding one to b
{a3,b3} is the final state.
 

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Mixture(0) = alcohol.
After the first iteration, mixture(1) = (K-1)/K alcohol + 1/K water
After the second, mixture(2) = (K-1)/K mixture(1) + 1/K water
After the third, mixture(3) = (K-1)/K mixture(2) + 1/K water, and alcohol(3)/water(3) = 1/7.
 
Thanks. I mixed up stage one :)
 
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