A laboratory vessel contains a liquid solution consisting of water and ethanol, where wate… — GRE General Test

A laboratory vessel contains a liquid solution consisting of water and ethanol, where water makes up $\frac{3}{8}$ of the total volume. First, $\frac{1}{5}$ of the total volume of the solution is drained and replaced with an equal volume of pure ethanol. Next, $\frac{1}{4}$ of the resulting mixture is evaporated, removing water and ethanol in proportion to their presence. Finally, pure water is added to fill the vessel back to its original total volume. What fraction of the final solution is water? Express your answer as a decimal.

Cevap: 0.475

Cevap

The fraction of the final solution that is water is 0.475 (or 19/40).

By following the multi-step fractional changes to the liquid volume, the remaining water prior to refilling is

\frac{9}{40}

of the original capacity. Refilling the missing

\frac{1}{4}

(or

\frac{10}{40}

) volume with pure water yields

\frac{19}{40} = 0.475

of the total solution as water.

Adım Adım Çözüm

Track water content after the initial replacement.

Water fraction becomes

\frac{3}{10}

of the original volume.

Let the original total volume be

V

. Initially, water volume is

\frac{3}{8}V

. Removing

\frac{1}{5}

of the solution leaves

\frac{4}{5}

of the original solution, so the water volume becomes

\frac{4}{5} \times \frac{3}{8}V = \frac{3}{10}V

. Replacing the removed volume with pure ethanol brings total volume back to

V

, with water occupying

\frac{3}{10}V

Track water content after evaporation.

Water volume becomes

\frac{9}{40}V

and total solution volume becomes

\frac{3}{4}V

Evaporating

\frac{1}{4}

of the solution leaves

\frac{3}{4}

of the mixture intact. The remaining water volume is

\frac{3}{4} \times \frac{3}{10}V = \frac{9}{40}V

Calculate the final water fraction after refilling with pure water.

Final water volume is

\frac{19}{40}V = 0.475V

To restore the total volume from

\frac{3}{4}V

back to

V

, an amount equal to

V - \frac{3}{4}V = \frac{1}{4}V

of pure water is added. Adding this to the existing water gives

\frac{9}{40}V + \frac{1}{4}V = \frac{9}{40}V + \frac{10}{40}V = \frac{19}{40}V

. Dividing by total volume

V

yields

\frac{19}{40} = 0.475

Anahtar Kavram

Sequential fractional reduction and component tracking

Tahmini Süre:2m 30s

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