Thats a too tough to calculated.

Got (2/7)L in A - After a lengthy calculation.
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GT

Let's go step by step:

First operation: 3L-1L=2=6/3L of wine left, total 4L;
#2: 6/3L-(6/3)/4=6/3-6/12=18/12=6/4L of wine left, total 5L;
#3: 6/4L-(6/4)/5=6/4-6/20=24/20=6/5L, total 6L;
#4: 6/5L-(6/5)/6=6/5-6/30=30/30=6/6L, total 7L;
....

At this point it's already possible to see the pattern: x=6/(n+2)

n=19 --> x=6/(19+2)=6/21=2/7L

Answer: A.
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I think it's almost impossible within 2-3 minutes to figure out and build the pattern of 6/3, 6/4, 6/5,.....
Just wonder how did you notice that 3L-1L=2=6/3L and why not 4/2L or 10/5L?

Bunuel could you please explain the calculation for the 2nd op... Thanks

Folks, have we seen any task like this, but easier? It can be useful to apply this concept on a gmat-like example.

After the first operation there will be 2L of wine left. 2 can be expressed as a fraction and be written as 6/3. Then as you can see in the solution, for any operation the amount of wine can be expressed as 6/(# of operations+2) L.

Anyway: this is not a GMAT question, so don't worry about it.
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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
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First time, C_f = 1 * 2/4
Second time when you remove 1 litre from 4 litres, Initial Volume becomes 3 lts. When you add 2 lts of water back, final volume becomes 5 lts.

Now for second step, C_f = C_i * 3/5
C_i for the second step is the final concentration obtained from step 1 i.e. C_i for second step = 1 * 2/4
So, C_f = 1 * 2/4 * 3/5
and so on for every subsequent step... It doesn't matter whether we have 19 or 119 steps, we still get our answer in 2 steps.
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First of all the question has only three answer choices instead of five and also its difficulty level is higher than that of GMAT's.
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Thanks Bunuel- im not going to spend my time on this after seeing your repeated posts about this not being gmat qn!