Question 340403
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Let *[tex \Large x] represent the amount of water in full glass units contained in glass B.


After the first pouring,


Glass A contains *[tex \Large \frac{x}{2}]
Glass B contains *[tex \Large \frac{x}{2}]


After the second pouring


Glass A contains *[tex \Large \frac{x}{2}\ +\ \frac{x}{4}\ =\ \frac{3x}{4}]
Glass B contains *[tex \Large \frac{x}{4}]


After the third pouring


Glass A contains *[tex \Large \frac{3x}{4}\ +\ \frac{x}{8}\ =\ \frac{7x}{8}]
Glass B contains *[tex \Large \frac{x}{8}]


But since Glass A is now one-half full, we know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{7x}{8}\ =\ \frac{1}{2}]


Solving


*[tex \LARGE \ \ \ \ \ \ \ \ \ x\ =\ \frac{1}{2}\,\cdot\,\frac{8}{7}\ =\ \frac{4}{7}]


of a glass full was the beginning amount in Glass B.  Since Glass B now contains *[tex \Large \frac{1}{8}] of the original amount, Glass B must now contain *[tex \Large \frac{1}{8}\,\cdot\,\frac{4}{7}\ =\ \frac{1}{14}] of a full glass.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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