Suppose the fraction of the original mixture that is acid is p, and we start with T litres in total. Then we have pT litres of acid, and the rest, T - pT litres, must be water.
When we add one litre of water, we still have pT litres of acid, and now we have T - pT + 1 litres of water. We know this is now 20% acid (and so is 80% water), so acid and water are now in a 1 to 4 ratio, and T - pT + 1 = 4pT.
When we next add one litre of acid, we still have T - pT + 1 litres of water, but now we have pT + 1 litres of acid. If we now have a solution which is 1/3 acid, then acid and water are in a 1 to 2 ratio, and T - pT + 1 = 2(pT + 1).
Since we now know T - pT + 1 is equal both to 4pT and to 2(pT + 1), those must equal each other, so
4pT = 2pT + 2
pT = 1
and we started out with 1 litre of acid. We know after adding 1 litre of water, the solution is 20% acid, and since we have 1 litre of acid, we must then have 4 litres of water. So in the initial situation, before adding the 1 litre of water, we had 3 litres of water and 1 litre of acid, and a 25% acid solution.
The OA currently listed in the first post is "C", which is not correct.
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