Question 1142949
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The distance traveled either way is *[tex \Large d], the speed going is *[tex \Large r\ +\ 20], the time going is 2, the speed returning is *[tex \Large r] and the time returning is 3.


So the outbound trip is described by *[tex \Large d\ =\ 2(r\ +\ 20)]


And the return trip is described by *[tex \Large d\ =\ 3r]


Since the distance is the same both ways:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3r\ =\ 2(r\ +\ 20)]


Solve for *[tex \Large r] and then calculate *[tex \Large r\ +\ 20]


Of course, this solution will only be valid for Sarawang.  I have no idea about Sarawong's return speed.  Lesson:  Proofread your post BEFORE you click on Submit.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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