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      Question

      A container X has (20 + x) litres of pure milk. Another

      container Y has 40 litres of a solution with 30% milk and the rest water. All the liquid from both containers is mixed together. Then 10 litres of this mixture is taken out and replaced by pure water. If after mixing (but before taking anything out) the percentage of milk in the mixture is 60%, what is the quantity of milk (in litres) in the final mixture after the replacement?
      A 24 liters Correct Answer Incorrect Answer
      B 36 liters Correct Answer Incorrect Answer
      C 48 liters Correct Answer Incorrect Answer
      D 60 liters Correct Answer Incorrect Answer
      E None of these Correct Answer Incorrect Answer

      Solution

      ATQ, Container X: (20 + x) litres milk (pure). Container Y: 40 litres of 30% milk ⇒ Milk in Y = 0.30 × 40 = 12 litres After mixing: Total volume = (20 + x) + 40 = 60 + x Total milk = (20 + x) + 12 = x + 32 Given that after mixing, milk is 60%: (x + 32) / (60 + x) = 0.60 x + 32 = 0.60x + 36 x − 0.60x = 36 − 32 0.40x = 4 x = 10 So after mixing: Total volume = 60 + 10 = 70 litres Total milk = x + 32 = 10 + 32 = 42 litres Now 10 litres of mixture is taken out: Milk removed = 42 × (10 / 70) = 42 × (1/7) = 6 litres So milk left = 42 − 6 = 36 litres Then 10 litres of pure water is added (no milk added). Final volume = 70 litres Final milk = 36 litres

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