Two containers of the same volume are 40% and 50% full of water, respectively. They are then filled completely with milk. If the contents of both containers are mixed in a larger vessel, what is the percentage of milk in the final mixture?

Correct — C. Take each container's volume as 100 units.

Container 1: 40 water, so milk = 100 − 40 = 60
Container 2: 50 water, so milk = 100 − 50 = 50

Total milk: 60 + 50 = 110 out of 200 units
Percentage: 110⁄200 × 100 = 55% → option (c)

Why the others are wrong

• (a)
  45% — 45% is the water share, (40 + 50) out of 200 units. The question asks for milk, which fills the rest: 100% − 45% = 55%.
• (b)
  50% — 50% is the milk share of the second container alone. Mixing it with the first container's 60% milk lifts the combined share to 55%.
• (d)
  60% — 60% is the milk share of the first container alone. The second container holds 50% milk, so the equal-volume mix averages to 55%.

When equal volumes are mixed, the concentration of the mixture is the simple average of the two concentrations: (60% + 50%) ÷ 2 = 55%.

When volumes differ, weight each concentration by its volume instead. Taking each container as 100 units makes either case plain arithmetic.

Read which liquid is asked for: the containers are described by water, but the question asks for milk.

Key facts

• A container that is x% water, topped up with milk, is (100 − x)% milk.
• Mixing equal volumes averages the concentrations: (60 + 50) ÷ 2 = 55.
• The water and milk shares of the final mix add to 100%: 45% + 55%.

Common traps

• Answering with the water share, 45%, because the stem gives water percentages.
• Averaging two percentages when the containers differ in volume: the simple average needs equal volumes.

No directly related past PYQ was found.