Question 718810
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Let's say that it takes *[tex \LARGE x] hours for the new machine to do the job by itself.  Then it must take *[tex \LARGE 4x] hours for the old machine by itself.  Then we can say that the new machine can do *[tex \LARGE \frac{1}{x}] of the job in one hour, and the old machine can do *[tex \LARGE \frac{1}{4x}] of the job in one hour. Since we know that it takes 12 hours for both machines working together, we know that the two machines working together can do *[tex \LARGE \frac{1}{12}] of the job in one hour.  All of that leads us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{4x}\ =\ \frac{1}{12}]


Solve for *[tex \LARGE x], then calculate *[tex \LARGE 4x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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