A laboratory container holds a solution composed solely of alcohol, acid, and water. Initi… — GRE General Test

A laboratory container holds a solution composed solely of alcohol, acid, and water. Initially, alcohol accounts for $\frac{3}{8}$ of the solution's total volume, and acid accounts for $\frac{1}{4}$ of the total volume. In a two-step process, a chemist first removes $\frac{1}{3}$ of the alcohol present and $\frac{1}{2}$ of the acid present, with no water removed. Next, the chemist adds pure water until water accounts for $\frac{3}{5}$ of the solution's new total volume. What fraction of the final solution's total volume is alcohol?

$\frac{4}{15}$ Answer
B
$\frac{1}{4}$
C
$\frac{4}{9}$
D
$\frac{2}{5}$
E
$\frac{2}{3}$

Answer

The fraction of the final solution's total volume that is alcohol is

\frac{4}{15}

The correct fraction

\frac{4}{15}

is derived by calculating the remaining alcohol volume as

\frac{1}{4}

of the initial total volume

V

, and determining the final total volume

V_{\text{final}} = \frac{15}{16}V

using the constant non-water volume of

\frac{3}{8}V

. Dividing

\frac{1}{4}V

\frac{15}{16}V

yields

\frac{4}{15}

Step-by-Step Solution

Determine initial volume fractions for all three components.

Let

V

be the initial total volume. Alcohol is

\frac{3}{8}V

, acid is

\frac{1}{4}V = \frac{2}{8}V

, and water is

1 - (\frac{3}{8} + \frac{2}{8}) = \frac{3}{8}V

Establishing the initial baseline volumes allows accurate tracking through the multi-step changes.

Calculate component volumes after the removal step.

Remaining alcohol =

\frac{3}{8}V \times (1 - \frac{1}{3}) = \frac{1}{4}V

. Remaining acid =

\frac{1}{4}V \times (1 - \frac{1}{2}) = \frac{1}{8}V

. Water remains

\frac{3}{8}V

Removing specified fractions of individual components changes their absolute volumes.

Calculate non-water volume and final total volume after water is added.

Total non-water volume =

\frac{1}{4}V + \frac{1}{8}V = \frac{3}{8}V

. Since water becomes

\frac{3}{5}

of the final volume

V_{\text{final}}

, non-water is

1 - \frac{3}{5} = \frac{2}{5}

V_{\text{final}}

. Thus,

\frac{2}{5} V_{\text{final}} = \frac{3}{8}V \implies V_{\text{final}} = \frac{5}{2} \times \frac{3}{8}V = \frac{15}{16}V

Adding only water keeps the non-water volume constant, providing a fixed reference point to find the new total volume.

Compute the final fraction of alcohol in the solution.

\text{Alcohol Fraction} = \frac{\text{Alcohol Volume}}{V_{\text{final}}} = \frac{\frac{1}{4}V}{\frac{15}{16}V} = \frac{1}{4} \times \frac{16}{15} = \frac{4}{15}.

The required quantity is the part-to-whole ratio of remaining alcohol to the final total volume.

Key Concept

Multi-step fraction operations involving component-wise removals and fixed non-water volumes in liquid mixtures.

Estimated Time:2m 30s

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