A container contains a mixture of wine and water in the ratio of 4:1. 150 litres of the mixture was taken out and 50 litres of water was added to it. If the water was 40% in the resultant mixture, what was the initial quantity of the mixture in the container?

Let the initial quantity of the mixture be 5x litres. Ratio of wine and water = 4 : 1 So, Quantity of wine = 4x litres Quantity of water = x litres Now, 150 litres of mixture is taken out. Since wine and water are in the ratio 4 : 1, quantity removed will also be in the same ratio. Wine removed = 150 × 4/5 = 120 litres Water removed = 150 × 1/5 = 30 litres So, remaining quantity of wine = 4x – 120 Remaining quantity of water = x – 30 Now, 50 litres of water is added. So, final quantity of water = x – 30 + 50 = x + 20 Final total quantity of mixture = 5x – 150 + 50 = 5x – 100 According to the question, Water is 40% of the resultant mixture. So, x + 20 = 40% of (5x – 100) x + 20 = 2x – 40 2x – x = 20 + 40 x = 60 Therefore, initial quantity of the mixture = 5x = 5 × 60 = 300 litres