A beaker holds a mixture of wine and water in which 50% is wine. A second…
2026
A beaker holds a mixture of wine and water in which 50% is wine. A second beaker holds twice as much mixture as the first, in which 25% is wine and the rest is water. Both beakers are poured completely into a third (empty) beaker. What fraction of wine does the third beaker contain?
- A.
1/2
- B.
2/3
- C.
1/4
- D.
1/3
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Correct answer: D
CONCEPT: When two solutions with different concentrations are mixed, the total quantity of the solute (here, wine) in the combined mixture is the sum of the solute contributed by each source, and the resulting concentration is the total solute divided by the total volume -- not a plain average of the two percentages.
Let the mixture in the first beaker be 1 unit; since the second beaker holds twice as much mixture as the first, it has 2 units.
Wine in the first beaker = 50% of 1 unit = 0.5 unit.
Wine in the second beaker = 25% of 2 units = 0.5 unit.
Total wine poured into the third beaker = 0.5 + 0.5 = 1 unit.
Total mixture volume in the third beaker = 1 + 2 = 3 units.
Fraction of wine in the third beaker = total wine divided by total volume = 1 / 3.
Cross-check: water contributed = 50% of 1 unit + 75% of 2 units = 0.5 + 1.5 = 2 units, so wine + water = 1 + 2 = 3 units, matching the total mixture volume -- and the wine and water fractions (1/3 and 2/3) add to 1, confirming the result.