Question 1172721
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If *[tex \Large \ \frac{4}{7}\ \ ] are red, and the rest are blue, then *[tex \Large \ \frac{3}{7}\ \ ] are blue.  Assuming that the total number of chips, both red and blue, is an integer, then the total number of chips must be an integer multiple of 7, which is to say that the total number of chips is *[tex \Large 7n] such that *[tex \Large n] is an integer.  Therefore the original number of red chips is *[tex \Large 4n] and the number of blue chips is *[tex \Large 3n].


Half of the red chips are *[tex \Large 2n], and double the blue chips is *[tex \Large 6n].  Now there are a total of *[tex \Large 2n\ +\ 6n\ =\ 8n] chips in the urn. *[tex \Large \frac{2n}{8n}\ =\ \frac{1}{4}] of the chips are red, and therefore *[tex \Large \frac{3}{4}] of the chips are blue.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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