A 180 litre mixture of water and acid contains 36 litres

Solution

Let the original quantity of acid in the mixture = ‘y’ litres Then, original quantity of water in the mixture = (y + 36) litres We have, y + y + 36 = 180 Or, 2y + 36 = 180 So, y = (180 - 36) ÷ 2 = 72 litres And so, original quantities of acid and water in the mixture are 72 litres and 108 litres, respectively After removing one-third of the mixture, Remaining quantity of water = 108 × 2/3 = 72 litres Remaining quantity of acid = 72 × 2/3 = 48 litres Final quantity of acid = 72 ÷ 1.2 = 60 litres So, k = 60 - 48 = 12