Q59·CSAT · Prelims 2026

Quantitative — Successive Fluid Dilution

NumericalMixtures & AlligationArithmetic Problem● Hard

Question

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)? [cite: 4879, 4880, 4881, 4882, 4903, 4904, 4905, 4906]

Options

Answer

Explanation

To solve for the final concentration, apply the standard successive formula for fluid replacement: V₍final₎ = V₍initial₎ \left(1 - x/V\right)^n.

Parameters: Total volume V = 10 liters, replacement volume x = 1 liter, number of operational cycles n = 3. [cite: 4879, 4880, 4882, 4904, 4906]

Calculate the remaining volume of Chemical A:

V_A = 10 × \left(1 - 1/10\right)³ = 10 × \left(9/10\right)³ = 10 × 729/1000 = 7.29 litres

Calculate the final volume of Chemical B: Since the total fluid volume remains constant at 10 liters, the remaining balance belongs to chemical B:

V_B = 10 - 7.29 = 2.71 litres

Convert to Percentage:

Percentage of B = 2.71/10 × 100\% = 27.1\% ≈ 27\% [cite: 4882, 4906]

For multi-cycle fluid dilution problems, use the standard formula to find the remaining volume of the initial fluid first, then subtract from the total volume to find the new fluid's share.

Answer: (b).

Question details

Year

2026

Paper

CSAT

Question

Q59

Section

Quantitative Aptitude

Sub-topic

Mixtures & Alligation

Type

Arithmetic Problem

Difficulty

Hard

Source hint

Successive replacement — fractional dilution tracking formula

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