Question 909803: A traveler left a town A at 5:00 am, and traveled toward a town B. At 7:00 o’clock another traveler left town B and traveled toward A. They met at 9:05 am. The first traveler arrived at B and the second at A at the same time. How long did each make the journey?
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! A traveler left a town A at 5:00 am, and traveled toward a town B. At 7:00
o’clock another traveler left town B and traveled toward A. They met at 9:05 am.
The first traveler arrived at B and the second at A at the same time. How long
did each make the journey?
Let traveler 1's rate be r
Let traveler 2's rate be s
Suppose they each arrive at their destinations at t hours after
meeting at M
From 5:00AM to 7:00AM is 2 hours
From 7:00AM to 9:05AM is 2 hours 5 minute = 1 1/12 hours = 25/12 hours
A------P---------M-------B
Suppose traveler 1 is at P at 7:00 when traveler 2 starts.
Suppose they meet at M.
Using distance = rate*time for traveler 1:
Distance AP = 2r
Distance PM = (25/12)r
Adding, distance AM = AP+PM = 2r + (25/12)r = (49/12)r
Distance MB = tr
For traveler 2
Distance BM = (25/12)s
Distance MA = ts
--------------------------------
BM = MB, so (25/12)s = tr
or
(1) 25s = 12tr
MA = AM, so ts = (49/12)r
or
(2) 12ts = 49r
Dividing equals by equals, from (1) and (2) we get
25/(12t) = (12t)/49
144t^2 = 1225
t^2 = 1225/144
t = 35/12 = 2 11/12 hours = 2 hours and 55 minutes.
After meeting at 9:05AM they arrived at their respective towns at 12 noon.
So traveler 1's trip took 7 hours and traveler 2's trip took 5 hours.
Edwin
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