SOLUTION: can you please help me with this mind breaking question...and thank you for advance:). A traveler left a town A at 5:00 a.m. and traveled toward a town B. At 7:00 a.m. another tr

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: can you please help me with this mind breaking question...and thank you for advance:). A traveler left a town A at 5:00 a.m. and traveled toward a town B. At 7:00 a.m. another tr      Log On


   

Question 356064: can you please help me with this mind breaking question...and thank you for
advance:). A traveler left a town A at 5:00 a.m. and traveled toward a town B.
At 7:00 a.m. another traveler left town B and traveled toward A. They met at
9:05a.m. The first traveler arrived at B and the second arrived at A at the
same time. How long did each travel?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Let the 1st traveler's rate be x mph
Let the 2nd traveler's rate be y mph

Let's take a look at the moment when they met, which was 9:05.

At 9:05 the 1st traveler, who started at 5AM, had been traveling for
4 hours and 5 minutes, or 4%265%2F60 hours or 4%261%2F12 hours or 49%2F12 hours.  
At rate x mph, he had traveled expr%2849%2F12%29x or 49x%2F12 miles.

At 9:05 the 2nd traveler, who started at 7AM, had been traveling for
2 hours and 5 minutes, or 2%265%2F60 hours or 2%261%2F12 hours or 25%2F12 hours.  
At rate y mph, he had traveled expr%2825%2F12%29y or 25y%2F12 miles. 

When they met at 9:05, each had the same number of miles yet to go that 
the other had already been.  That is, the 1st traveler had 25y%2F12
miles yet to go and the 2nd traveler had 49x%2F12 miles yet to go.

We are told that they arrived at the same time.  So we will calculate each
traveler's time after 9:05 and set them equal. 

Since TIME=DISTANCE%2FRATE, the 1st traveler's time was his distance yet
to go, 25y%2F12, divided by his rate x, or %2825y%2F12%29%2Fx or %2825y%2F12%29%281%2Fx%29 or %2825y%29%2F%2812x%29 hours till he 
reached town B.

Also the 2nd traveler's time was his distance yet to go, 49x%2F12, divided
by his rate y, or %2849x%2F12%29%2Fy or %2849x%2F12%29%281%2Fy%29 or %2849x%29%2F%2812y%29 hours till he reached town A.

Since they arrived at their respective destinations at the same time, we set
these times equal:

     %2825y%29%2F%2812x%29%22%22=%22%22%2849x%29%2F%2812y%29

Multiplying both sides by 12 eliminates the 12's from the denominators:

     %2825y%29%2Fx%22%22=%22%22%2849x%29%2Fy

Cross multiplying:

     25y%5E2%22%22=%22%2249x%5E2

Taking positive square roots of both sides

            5y%22%22=%22%227x

Dividing both sides by 5x

            y%2Fx%22%22=%22%227%2F5


The 1st traveler's time to reach his destination after 9:05 was

%2825y%29%2F%2812x%29 which equals expr%2825%2F12%29%2Aexpr%28y%2Fx%29 which equals 
      5
expr%2825%2F12%29%2Aexpr%287%2F5%29%22%22=%22%22expr%28cross%2825%29%2F12%29%2Aexpr%287%2Fcross%285%29%29%22%22=%22%22expr%285%2F12%29%2Aexpr%287%2F1%29%22%22=%22%2235%2F12%22%22=%22%222%2611%2F12%22%22=%22%222%2655%2F60 or 2 hours and 55 minutes after 9:05.
So the 1st traveler reached his destination at 12 noon.

It isn't necessary, but, as a check, let's see if the 2nd traveler also reached
his destination at 12 noon.

The 2nd traveler's time to reach his destination after 9:05 was

%2849x%29%2F%2812y%29 which equals expr%2849%2F12%29%2Aexpr%28x%2Fy%29 which equals 
      7
expr%2849%2F12%29%2Aexpr%285%2F7%29%22%22=%22%22expr%28cross%2849%29%2F12%29%2Aexpr%285%2Fcross%287%29%29%22%22=%22%22expr%287%2F12%29%2Aexpr%285%2F1%29%22%22=%22%2235%2F12%22%22=%22%222%2611%2F12%22%22=%22%222%2655%2F60 or 2 hours and 55 minutes after 9:05.
So the 2nd traveler also reached his destination at 12 noon.

So the 1st traveler started at 5AM and arrived at noon, so his time was 7 hours.
The 2nd traveler started at 7AM and arrived at noon, so his time was 5 hours.

[Notice that this problem is independent of their speeds and the distance they
traveled.]

Edwin