Question 190059
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Let <i>r</i> be the rate of the slower plane and then <i>r</i> + 50 must be the rate of the faster plane.


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  d = rt]


we can also say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  t = \frac{d}{r}]


So for the slower plane's trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  t = \frac{600}{r} ]


and for the faster plane's trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  t = \frac{800}{r+50}]


Since both trips were completed in the same amount of time,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{600}{r} = \frac{800}{r+50}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 800r = 600(r + 50)]


Now solve for r to get the speed of the slower plane and add 50 to that to get the speed of the faster one.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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