A container initially holds 50 litres of pure milk. First, 8 litres of milk are removed from the container and replaced with 8 litres of water. This process of removing 8 litres of the current mixture and replacing it with 8 litres of water is repeated two more times. After these three operations in total, how many litres of milk are now contained in the container (approximately)?

Aptitude Alligation or Mixture Difficulty: Medium
Choose an option
  • A
    24.52 litres
  • B
    29.63 litres
  • C
    28.21 litres
  • D
    25.14 litres
  • E
    30.00 litres

Answer

Correct Answer: 29.63 litres

Explanation

Introduction: This question is a classic repeated replacement problem where pure milk in a container is gradually replaced by water in several steps. Each time a fixed amount of the current mixture is removed and replaced with water. We must find the quantity of milk remaining after three such operations.

Given Data / Assumptions:

  • Initial volume in the container = 50 litres, all milk.
  • In each step, 8 litres of the current mixture are removed.
  • The 8 litres removed are replaced with 8 litres of water.
  • This process is repeated 3 times in total.
  • The total volume in the container remains 50 litres at all times.
  • We must compute the approximate volume of milk left after these 3 steps.

Concept / Approach: If V is the total volume and x litres are removed and replaced each time, the fraction of the original liquid remaining after one operation is (1 - x / V). After n identical operations, the fraction remaining is (1 - x / V)^n. Here V = 50, x = 8, and n = 3. Multiplying this fraction by 50 litres gives the milk left after three operations.

Step-by-Step Solution: Step 1: Total volume V = 50 litres, removed each time x = 8 litres. Step 2: Fraction of milk remaining after one operation = 1 - x / V = 1 - 8 / 50. Step 3: 8 / 50 = 0.16, so fraction remaining each time = 1 - 0.16 = 0.84. Step 4: After 3 operations, fraction of milk remaining = (0.84)^3. Step 5: Compute (0.84)^3 = 0.84 * 0.84 * 0.84 ≈ 0.5927. Step 6: Milk remaining = 50 * 0.5927 ≈ 29.635 litres. Step 7: Rounding to two decimal places, milk remaining ≈ 29.63 litres.

Verification / Alternative check: We can also approximate stepwise. After first operation, milk = 50 * 0.84 = 42 litres. After second operation, milk = 42 * 0.84 ≈ 35.28 litres. After third operation, milk = 35.28 * 0.84 ≈ 29.63 litres. This matches the value obtained using the direct exponential formula and confirms the result is correct.

Why Other Options Are Wrong: 24.52, 28.21, and 25.14 litres are too low compared to the calculated value and do not match the repeated 16% removal pattern correctly. 30.00 litres is close but slightly higher than the precise result; it is an overestimate and does not match the exact calculation of 29.63 litres.

Common Pitfalls: Students sometimes subtract 8 litres of milk each time, ignoring that after the first step the liquid is a mixture, not pure milk. Another mistake is to forget that the removal each time is of the current mixture in the same proportion, so a constant fraction, not a constant amount, of milk is removed. Using linear subtraction instead of an exponential fraction is a frequent source of error.

Final Answer: The container now contains approximately 29.63 litres of milk.

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