SOLUTION: A high school student in Seattle, Washington, attends the University of Central Florida. On the way to UCF he took a southern route. After graduation he returned to Seattle via a
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Question 630689: A high school student in Seattle, Washington, attends the University of Central Florida. On the way to UCF he took a southern route. After graduation he returned to Seattle via a northern trip. On both trips he had the same average speed. If the southern trek took 47 hours, the northern trek took 52 hours, the northern trek was 310 miles longer, how long was each trip?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A high school student in Seattle, Washington, attends the University of Central Florida.
On the way to UCF he took a southern route.
After graduation he returned to Seattle via a northern trip.
On both trips he had the same average speed.
If the southern trek took 47 hours, the northern trek took 52 hours,
the northern trek was 310 miles longer, how long was each trip?
:
Let d = the southern distance
then
(d+310) = the northern distance
:
Write a speed equation. Speed = dist/time
:
Northern speed = southern speed
=
Cross multiply
52d = 47(d+310)
52d = 47d + 14750
52d - 47d = 14750
d = 14750/5
d = 2,914 mi is the southern route
then
2,914 + 310 = 3,224 mi is the northern route
;
:
We can confirm this by finding the speeds, they should be equal
2914/47 = 62 mph
3224/52 = 62 mph
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