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)?

The correct option is (b) - 27.

[as per provisional answerkey]

Solution

This is a classic problem of repeated dilution. We need to find the final volume of chemical A and then subtract it from the total volume to find the volume of chemical B.

Given:
Initial volume of chemical A (V) = 10 litres
Volume removed and replaced each time (x) = 1 litre
Number of operations (n) = 3 (Initial replacement + second replacement + "once more")

Formula for remaining quantity of original liquid:
Final Quantity of A = Initial Quantity * $[1-(x/V){]}^n$
Final Quantity of A = $10\times [1-(1/10){]}^3$
Final Quantity of A = $10\times [9/10{]}^3$
Final Quantity of A = $10\times (729/1000)$
Final Quantity of A = 7.29 litres

Finding the volume of chemical B:
Since the total volume remains constant at 10 litres:
Volume of B = Total Volume - Final Volume of A
Volume of B = $10-7.29=2.71$ litres

Calculating the percentage of B:
Percentage of B = (Volume of B / Total Volume) * 100
Percentage of B = $(2.71/10)\times 100=27.1$

The approximate percentage is 27%.

Why the other options are incorrect

• Option (a) - 25: This value is too low; it assumes a linear reduction or fewer iterations than the three specified in the question.
• Option (c) - 29: This value is too high; it would result if more than 1 litre were replaced each time or if more than three iterations were performed.
• Option (d) - 31: This is a significant overestimation, likely resulting from calculating the addition of B without accounting for the fact that some of B is also removed during the second and third replacements.

Key Concept

The formula for repeated dilution: Final Quantity = Initial Quantity * (1 - Replacement Volume / Total Volume)^n.