In a vessel, the quantity of juice is 40% of the quantity of water. If 70 liters of the mixture are taken out from the vessel and 30 liters of juice are added to the remaining mixture, then water becomes 25% more than juice in the resultant mixture. Find the initial quantity of juice in the mixture?

Let the initial quantity of water in the vessel be $5x$ liters. So, the initial quantity of juice in the vessel = $40\%$ of $5x = 2x$ liters. Total initial quantity of the mixture = $2x + 5x = 7x$ liters. Ratio of juice to water in the mixture = $2 : 5$. When 70 liters of the mixture are taken out: Quantity of juice taken out = $70 \times \frac{2}{7} = 20$ liters. Quantity of water taken out = $70 \times \frac{5}{7} = 50$ liters. Remaining quantity of juice = $2x - 20$ liters. Remaining quantity of water = $5x - 50$ liters. After adding 30 liters of juice to the remaining mixture: New quantity of juice = $2x - 20 + 30 = 2x + 10$ liters. New quantity of water = $5x - 50$ liters. ATQ, New quantity of water is 25% more than the new quantity of juice. So, $5x - 50 = 1.25 \times (2x + 10)$ Or, $5x - 50 = 2.5x + 12.5$ Or, $2.5x = 62.5$ So, $x = 25$ Initial quantity of juice in the mixture = $2x = 2 \times 25 = 50$ liters. Hence, the correct answer is B.