A container has 3 liters of pure wine. 1 liter from the container is

The question can be solved in under a minute if you understand the concept of concentration and volume.

Removal and addition happen 19 times so:

\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\)
All terms get canceled (4 in num with 4 in den, 5 in num with 5 in den etc) and you are left with \(C_f = \frac{1}{77}\)
Since Volume now is 22 lt, Volume of wine = \(22*(\frac{1}{77}) = \frac{2}{7}\)

Theory:
1. When a fraction of a solution is removed, the percentage of either part does not change. If milk:water = 1:1 in initial solution, it remains 1:1 in final solution.

2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)

3.
Amount of A = Concentration of A * Volume of mixture
Amount = C*V
( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt)
When you add water to this solution, the amount of milk does not change.
So Initial Conc * Initial Volume = Final Conc * Final Volume
\(C_i * V_i = C_f * V_f\)
\(C_f = C_i * (V_i/V_f)\)
In the question above, we find the final concentration of wine. Initial concentration \(C_i\) = 1 (because it is pure wine)
When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt.
So \(C_f = 1 * 2/4\)
Since it is done 19 times, \(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\)

The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has \(V_f\) as 22, you get amount of wine = 1/77 * 22 = 2/7 lt