Solve Water Mixing Problem: Volume K Tank of Alcohol

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SUMMARY

The water mixing problem involves a tank of volume K liters of alcohol, where a series of operations are performed: removing one liter of alcohol, adding one liter of water, and repeating this process three times. The final ratio of water to alcohol in the tank must be 7:1. The solution requires calculating the alcohol and water components at each stage, represented as a_i for alcohol and b_i for water, with initial conditions set at a0 = K and b0 = 0. The iterative process leads to the conclusion that after three iterations, the ratio of alcohol to water is established as 1:7.

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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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