SOLUTION: Mark made a business trip of 247.5 miles. He averaged 51 mph for the first part of the trip and 60 mph for the second part. If the trip took 4.5 hours, how long did he tr

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Mark made a business trip of 247.5 miles. He averaged 51 mph for the first part of the trip and 60 mph for the second part. If the trip took 4.5 hours, how long did he tr      Log On


   

Question 996027: Mark made a business trip of
247.5 miles. He averaged
51 mph for the first part of the trip and
60 mph for the second part. If the trip took
4.5 hours, how long did he travel at each rate?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Mark made a business trip of 247.5 miles.
He averaged 51 mph for the first part of the trip and
60 mph for the second part.
If the trip took 4.5 hours, how long did he travel at each rate?
:
let d = distance traveled at 51 mph
the total distance is 247.5 mi therefore:
(247.5-d) = distance traveled at 60 mph
:
Write a time equation; time = dist/speed
:
51mph time + 60 mph time = 4.5 hrs
d%2F51 + %28247.5-d%29%2F60 = 4.5
Least common factor of 51 and 60 is 1020, multiply equation by 1020
1020*d%2F51 + 1020*%28247.5-d%29%2F60 = 1020(4.5)
Cancel the denominators and we have
20d + 17(247.5-d) = 4590
20d + 4207.5 - 17d = 4590
20d - 17d = 4590 - 4207.5
3d = 382.5
d = 382.5/3
d = 127.5 mi at 51 mph
then
247.5 - 127.5 = 120 mi at 60 mph
:
:
Confirm this answer, find the actual time at each speed
127.5/51 = 2.5 hrs
120/60 = 2 hrs
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total time 4.5 hrs