Discussion
A bartender stole vodka from bottle A that contained 50% alcohol and he replaced what he had stolen with vodka from bottle B containing 20% alcohol. He used water to replenish bottle B, which is twice as large as bottle A. After the mixing, alcohol content of bottle A reduced to 30%. What is the alcohol content of bottle B now ?
Answer: C
Bottle B now holds 13.33% alcohol.
Let bottle A hold V and let x be the amount stolen. A keeps V - x at 50% and gains x at 20%, ending at 30%:
\[ \begin{aligned} 0.5(V - x) + 0.2x &= 0.3V \\ 0.2V &= 0.3x \\ x &= \tfrac{2}{3}V \end{aligned} \]
Bottle B, twice as large (2V, full), lost \(\tfrac{2}{3}V\) of its 20% vodka, replaced by water:
\[ \begin{aligned} \text{alcohol} &= 0.2 \times \tfrac{4}{3}V = \tfrac{4}{15}V \\ \text{strength} &= \tfrac{4}{15}V \div 2V = \tfrac{2}{15} \\ &\approx 13.33\% \end{aligned} \]
Forgetting that B is twice the size of A gives 6·67%. 10% and 15% do not follow from the 30% condition on bottle A.
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