A 360 litre mixture of water and acid contains 72 litres more water than acid. If one-third of the mixture is removed and then ‘k’ litres of acid is added, then quantity of water in the resultant mixture would be 20% more than that of acid in it. Find the value of ‘k’.

Let the original quantity of acid in the mixture = 'y' litres Then, original quantity of water in the mixture = (y + 72) litres We have, y + y + 72 = 360 Or, 2y + 72 = 360 So, y = (360 - 72) ÷ 2 = 144 litres And so, original quantities of acid and water in the mixture are 144 litres and 216 litres, respectively After removing one-third of the mixture, Remaining quantity of water = 216 X 2/3 = 144 litres Remaining quantity of acid = 144 X 2/3 = 96 litres Final quantity of acid = 144 ÷ 1.2 = 120 litres So, k = 120 - 96 = 24