Container X contains a mixture of oil and water in the ratio 1 : 3, whereas container Y contains a mixture of oil and water in the ratio 2 : 3. If 2 litres of liquid from container X and 5 litres of liquid from container Y are mixed together in container Z that initially contains 1 liter of pure oil, then what is the percentage of oil in container Z?

Ratio of oil and water in container X = 1:3 Ratio of oil and water in container Y = 2:3 2 liter of mixture is taken out from X, Therefore, Oil = 2 × 1/4 = 1/2 liter Water = 2 × 3/4 = 3/2 liter Now, 5 liter of mixture is taken out from Y Therefore, Oil = 5 × 2/5 = 2 liter Water = 5 × 3/5 = 3 liter Now, In container Z, Quantity of oil = 1/2 + 2 + 1 = 7/2 liter Quantity of water = 3/2 + 3 = 9/2 liter Therefore, Required% = {(7/2)/(7/2 + 9/2)} × 100 = 7/2 × 2/16 × 100 = 175/4 = 43.75%