Question
A 320 litre mixture of water and acid contains 80 litres more water than acid. If 25% of the mixture is removed and then ‘k’ litres of acid is added, then the quantity of water in the resultant mixture would be 25% more than that of acid in it. Find the value of ‘k’.
Solution
Let the original quantity of acid in the mixture = ‘y’ litres Then, original quantity of water in the mixture = (y + 80) litres We have, y + y + 80 = 320 Or, 2y + 80 = 320 So, y = (320 - 80) ÷ 2 = 120 litres And so, original quantities of acid and water in the mixture are 120 litres and 200 litres, respectively After removing 25% of the mixture, Remaining quantity of water = 200 × 0.75 = 150 litres Remaining quantity of acid = 120 × 0.75 = 90 litres Final quantity of acid = 150 ÷ 1.25 = 120 litres So, k = 120 - 90 = 30
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