There are two chemicals which do not react with each other. A container contains 10 litres of the chemical A. One litre of this chemical is removed from it and one litre of the chemical B is poured. Then one litre of the mixture is removed from the container and one litre of B is poured. If this process of replacing one litre of the mixture by one litre of B is performed once more, then what is the volume of B that is present in the container approximately (in percentage)?

Correct Option: B (27)

This problem evaluates the mathematical concept of successive replacement in mixtures, a standard topic in quantitative aptitude examinations such as the UPSC Civil Services Aptitude Test (CSAT), introduced in 2011.

To find the final volume of the original chemical (A) after multiple identical replacements, we apply the successive replacement formula: Final Volume = Initial Volume × (1 - (Replacement Volume / Total Volume))ⁿ, where 'n' is the number of replacement operations.

Here, the total volume of the mixture remains constant at 10 litres.

• Initial Volume of A = 10 L
• Replacement Volume = 1 L
• Number of operations (n) = 3 (the initial replacement, the second replacement, and the process 'performed once more').

Calculation: Final Volume of A = 10 × (1 - 1/10)³ Final Volume of A = 10 × (0.9)³ = 10 × 0.729 = 7.29 L.

Since the total container volume is 10 L, the volume of chemical B is the remainder: Volume of B = 10 L - 7.29 L = 2.71 L. To find the approximate percentage of B in the container: (2.71 / 10) × 100 = 27.1%, which rounds to approximately 27%. Thus, Option B is mathematically correct as established by foundational mixture principles (e.g., referenced in authoritative textbooks like R.S. Aggarwal's Quantitative Aptitude).

Why other options are incorrect:

• Option A (25): Incorrect. Choosing 25% assumes an erroneous calculation where the dilution effect is misjudged, yielding exactly 2.5 L of B.
• Option C (29): Incorrect. This results from arithmetic errors in exponentiation or misjudging the fractional remainders across the three iterative steps.
• Option D (31): Incorrect. This implies a final volume of B (3.1 L) that exceeds the absolute total amount of B poured into the container across all three operations (which is exactly 3 L). This is mathematically and physically impossible.

Takeaway: For successive replacement problems, always track the remaining volume of the original substance using the formula Final = Initial × (1 - x/V)ⁿ. Calculating the introduced substance directly is complex because a fraction of it is removed with every subsequent step.