We will use percent concentration to determine the answer.
At the beginning, one glass contains 500 ml of 100% wine and the second glass contains 500 ml of 100% vinegar.
After the first transfer, the first glass contains 400 ml of 100% wine and the second glass contains 600 ml of a wine-vinegar mixture. The concentration of wine and vinegar in the mixture can be determined by dividing the volume of each component by the...
We will use percent concentration to determine the answer.
At the beginning, one glass contains 500 ml of 100% wine and the second glass contains 500 ml of 100% vinegar.
After the first transfer, the first glass contains 400 ml of 100% wine and the second glass contains 600 ml of a wine-vinegar mixture. The concentration of wine and vinegar in the mixture can be determined by dividing the volume of each component by the total volume. So the mixture contains (100`-:` 600)`xx`100% = 16.67% wine and (500`-:` 600)`xx`100=83.33% vinegar.`<br> `
We will use a similar procedure to calculate the concentration in the first glass following the second transfer. The second transfer moved 100 ml of 16.67% wine and 83.33% vinegar into the first glass, which contained 400 ml of 100% wine.
Final concentration in the first glass is thus:
(83.33`-:` 500)`xx`100% = 16.67% vinegar
((400+16.67)`-:` 500)`xx`100 = 83.33% wine
Thus, if we consider wine and vinegar to be equally 'impure', neither of the glasses has more or less impurities than the other.
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