1180.84 – Mixing Wine and Water


Tags:Problem Set 10Algebra/TrigonometryVisual

Two identical barrels stand next to each other. One is half full of water; the other is half full of wine. Take a dipper and pour one dipperful of wine into the barrel of water. Stir. Now take a dipperful of this mixture and pour it into the barrel of wine. Both barrels now contain a mixture of wine and water. Which is larger: the percentage of wine in the water barrel or the percentage of water in the wine barrel? Explain your reasoning.


Solution

The percentages are the same. Here’s why. Let D be the number of dippers in a half-barrel. The water barrel, after the first dipper of wine is poured into it, has D + 1 dippers but only one of them is wine. The percent of wine in the water barrel is therefore, 1/(D + 1) and this does not change as one dipper of the stirred-up mixture is removed and poured into the wine barrel. The wine barrel now is restored to D dippers but how much of this is water? The added dipper is 1/(D + 1) percent wine, hence D/(D + 1) percent water. Thus the percentage of water in the wine barrel is

DD+11D=1D+1\frac{\frac{D}{D+1}*1}{D} = \frac{1}{D+1}

The percentages are the same.