From container A containing 54 liter of mixture of milk and water in ratio of 8 : 1 , 18 liter of the mixture is taken out and poured into container B in which ratio of milk to water is 3 : 1. If difference between total milk and total water in container B is 30 liter then find the quantity of initial mixture in container B.

Step 1: Analyze Container A

• Total mixture in A = 54 liters
• Milk : Water = 8 : 1 → Total parts = 9
• Milk in A = (8/9) × 54 = 48 liters
• Water in A = (1/9) × 54 = 6 liters

Step 2: Take out 18 liters from A and move to B

• Since ratio remains the same, taken portion will also have milk and water in 8:1
• Milk transferred = (8/9) × 18 = 16 liters
• Water transferred = (1/9) × 18 = 2 liters

Step 3: Let the initial quantity in Container B be x liters

• Given Milk : Water in B = 3 : 1 → Total parts = 4
• Milk in B = (3/4) × x
• Water in B = (1/4) × x

Step 4: Add the transferred 16 L milk and 2 L water to B

Total milk in B = (3/4)x + 16  
Total water in B = (1/4)x + 2

Step 5: Given that the difference between total milk and water in B is 30 liters

[(3/4)x + 16] - [(1/4)x + 2] = 30
=> (3x - x)/4 + 14 = 30
=> (2x)/4 + 14 = 30
=> x/2 + 14 = 30
=> x/2 = 16
=> x = 32

Answer: 32 liters

The initial quantity of mixture in container B was 32 liters.

This problem demonstrates how mixture problems can combine ratios, algebra, and logical analysis. It’s a classic type of question seen in quantitative aptitude exams and useful in real-life scenarios involving blending and resource allocation.