When one litre water is added to a mixture containing acid and water, the new mixture has 20% acid. If one litre of acid is added to the new mixture, the resulting mixture has $$33\frac{1}{3}\%$$ acid. The percentage of water in the original mixture is

Let the mixture initially has "a" litres acid and "b" liters water. Now you add 1-litre water, hence the total water is "b+1" litres.

According to question    a=0.2(a+b+1) or 0.8a=0.2b +0.2  or 4a=b+1

Now one litre of acid is added to the mixture. Hence acid content is "a+1"  and water content is "b+1"

Now a+1= 1/3(a+b+2) or 3a+3 = a+b+2 or 2a+1=b

Substituting above 4a= 2a+1+1 or 2a=2 or a=1 and b=3 

Hence percentage of water in the original mixture is 3/(3+1) =3/4 or 75%