Henry Ernest Dudeney/Puzzles and Curious Problems/294 - The Keg of Wine/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $294$

The Keg of Wine
A man had a $10$-gallon keg of wine and a jug.
One day he drew off a jugful of wine and filled up the keg with water.
Later on, when the wine and water had got thoroughly mixed, he drew off another jugful, and again filled up the keg with water.
The keg then contained equal quantities of wine and water.
What was the capacity of the jug?


Solution

Approximately $2.93$ gallons.


Proof

Let $J$ gallons be the volume of the jug.

At the first drawing there will be $10 - J$ gallons of wine in the keg.

After mixing it with water, the concentration of wine per unit volume is $\dfrac {10 - J} {10}$.

The second drawing takes $J \dfrac {10 - J} {10}$ gallons of wine out of the keg.

Hence there remains $10 - J - J \dfrac {10 - J} {10}$ gallons of wine in the keg.

Then the keg is refilled.

Because there are equal quantities of wine and water in the keg, that means there must be $5$ gallons of each.

That is:

\(\ds 10 - J - J \dfrac {10 - J} {10}\) \(=\) \(\ds 5\)
\(\ds \leadsto \ \ \) \(\ds 100 - 10 J - 10 J + J^2\) \(=\) \(\ds 50\) multiplying through by $10$ to clear fractions and rearranging
\(\ds \leadsto \ \ \) \(\ds J^2 - 20 J + 50\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds J\) \(=\) \(\ds \dfrac {20 \pm \sqrt {20^2 - 4 \times 50} } 2\) Quadratic Formula
\(\ds \) \(=\) \(\ds 10 \pm \sqrt {50}\) simplifying
\(\ds \) \(=\) \(\ds 10 - \sqrt {50}\) as $10 + \sqrt {50}$ is greater than $10$ and hence absurd

This evaluates to approximately $10 - 7.07$, or about $2.93$ gallons.

$\blacksquare$


Sources

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