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⚡ AZUCATION SHORTCUT:
After the two-way exchange of volume $x$, the volume in A returns to 60L.
Final Alcohol in A = $\frac{15}{19} \times 60$.
Using the exchange relation: $\frac{60^2}{60+x} = \frac{15 \times 60}{19}$
$\implies \frac{60}{60+x} = \frac{15}{19} \implies \frac{4}{60+x} = \frac{1}{19} \implies 60+x = 76 \implies \mathbf{x = 16}$.

1. First Transfer: $x$ L alcohol moves A → B. Vessel B total volume becomes $60+x$ L.

2. Second Transfer: $x$ L mixture moves B → A. The alcohol concentration in B was $\frac{x}{60+x}$.
Alcohol moved back to A $= x \times \left( \frac{x}{60+x} \right) = \frac{x^2}{60+x}$.

3. Final Balance in A:
Final Alcohol in A $= (60 - x) + \frac{x^2}{60+x} = \frac{15}{19} \times 60$
$\implies \frac{60(60+x) - x(60+x) + x^2}{60+x} = \frac{900}{19}$
$\implies \frac{3600 + 60x - 60x - x^2 + x^2}{60+x} = \frac{900}{19}$
$\implies \frac{3600}{60+x} = \frac{900}{19} \implies 4 = \frac{60+x}{19}$
$\implies 60+x = 76 \implies \mathbf{x = 16}$.

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