Question 237015
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First let's test your theory.


We know that distance equals rate times time, or *[tex \Large d\ =\ rt],


So if the express train went 50 miles at 40 miles per hour, it would have taken *[tex \Large \frac{50}{40}\ =\ 1.25] hours.  And it would have taken the local train @ 20mph *[tex \Large \frac{50}{20}\ =\ 2.5] hours.  That means that the express train would have arrived an hour and 15 minutes ahead of the local, not 1 hour like the problem states.  [BUZZZ!!!] Thank you for playing, we have some lovely parting gifts...


Let *[tex \Large r] represent the rate of the local train.  Then *[tex \Large 2r] must be the rate of the express train.  Let *[tex \Large t] be the time it took the local train, then *[tex \Large t\ -\ 1] is the time it took the express train.  The distance is the same 50 miles for both trains, so:


For the local train:  *[tex \LARGE 50\ =\ rt]


For the express:  *[tex \LARGE \ \ \,50\ =\ 2r(t\ -\ 1)]


Multiply both sides of the first equation by *[tex \LARGE \frac{1}{r}]


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


Multiply both sides of the second equation by *[tex \LARGE \frac{1}{2r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 1\ =\ \frac{50}{2r}\ =\ \frac{25}{r}]


Add 1 to both sides of the second equation and combine using the LCD of *[tex \LARGE r]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{25\ +\ r}{r}]


Now we have two expressions for *[tex \LARGE t] in terms of *[tex \LARGE r].  Since *[tex \LARGE t\ =\ t]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{50}{r}\ =\ \frac{25\ +\ r}{r}]


Multiply both sides by *[tex \LARGE \frac{1}{r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25\ +\ r\ =\ 50]


I'll let you take it from there.


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