Question:
A 20 litre vessel is filled with alcohol. Some of the alcohol is poured out into another vessel of equal capacity, which is then completely filled by adding water. The mixture thus obtained is then poured into the first vessel to capacity. Then \(6\frac{2}{3}\) litres is poured from the first vessel into the second. Both vessels now contain an equal amount of alcohol. How much alcohol was originally poured from the first vessel into the second?
A 20 litre vessel is filled with alcohol. Some of the alcohol is poured out into another vessel of equal capacity, which is then completely filled by adding water. The mixture thus obtained is then poured into the first vessel to capacity. Then \(6\frac{2}{3}\) litres is poured from the first vessel into the second. Both vessels now contain an equal amount of alcohol. How much alcohol was originally poured from the first vessel into the second?
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When liquids are exchanged between vessels, track concentration changes step-by-step, not just total quantities.
Updated On: Jul 29, 2025
- 9 litres
- 10 litres
- 12 litres
- 12.5 litres
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The Correct Option is B
Solution and Explanation
Let \(x\) litres be poured from the first vessel (pure alcohol) into the second vessel.
- Second vessel: \(x\) litres alcohol + \((20 - x)\) litres water.
This mixture is poured back into the first vessel until it is full: the amount poured back = \(x\) litres of mixture (alcohol fraction in it = \(\frac{x}{20}\)).
After this exchange, when \(6\frac{2}{3}\) litres = \(20/3\) litres is transferred from the first to the second, they have equal alcohol.
Using concentration balancing equations, solving yields \(x = 10\) litres.
Thus the original transfer was \({10 \, \text{litres}}\).
This mixture is poured back into the first vessel until it is full: the amount poured back = \(x\) litres of mixture (alcohol fraction in it = \(\frac{x}{20}\)).
After this exchange, when \(6\frac{2}{3}\) litres = \(20/3\) litres is transferred from the first to the second, they have equal alcohol.
Using concentration balancing equations, solving yields \(x = 10\) litres.
Thus the original transfer was \({10 \, \text{litres}}\).
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