Replacement lowers concentration: A jar of whisky is 40% alcohol. A portion is removed and replaced with another whisky that is 19% alcohol, resulting in a final alcohol percentage of 26%. What fraction of the original whisky was replaced?

Introduction / Context: In replacement problems, concentrations combine linearly. If a fraction r of volume is replaced, the final concentration equals the remaining part of the old concentration plus the added part of the new concentration. Volumes cancel, leaving an equation in r only.

Given Data / Assumptions:

• Initial alcohol fraction = 0.40.
• Replacing fraction = r (of the total volume).
• Incoming alcohol fraction = 0.19.
• Final fraction = 0.26.

Concept / Approach: Final fraction = (1 − r)*0.40 + r*0.19. Solve for r to find the portion replaced.

Step-by-Step Solution:

Verification / Alternative check: Substitute r = 2/3: final = (1/3)*0.40 + (2/3)*0.19 = 0.1333… + 0.1266… ≈ 0.26, matching the target.

Why Other Options Are Wrong: Other fractions produce final concentrations different from 26%; only 2/3 balances the convex combination between 40% and 19%.

Common Pitfalls: Averaging 40% and 19% without weighting by r and (1 − r); ignoring that only a fraction of the original remains.

Final Answer: 2/3