SOLUTION: On the first part of a 317-mile trip, a salesman averaged 58 miles per hour. He averaged only 52 miles per hour on the last part of the trip because of an increased volume of traff

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Question 1001336: On the first part of a 317-mile trip, a salesman averaged 58 miles per hour. He averaged only 52 miles per hour on the last part of the trip because of an increased volume of traffic. Find the amount of time at EACH of the speeds if the total time was 5 hours and 45 minutes. ---I'm confused on how to do this, if the 52 miles per hour happened because of traffic, are we supposed to assume 58 miles per hour again?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
On the first part of a 317-mile trip, a salesman averaged 58 miles per hour.
He averaged only 52 miles per hour on the last part of the trip because of an increased volume of traffic.
Find the amount of time at EACH of the speeds if the total time was 5 hours and 45 minutes.
:
Change 5 hr 45 min to 5.75 hrs
:
let d = distance driven at 58 mph
Total dist is 317, therefore
(317-d) = distance driven at 52 mph
:
Write a time equation; time = dist/speed
time at 58mph + time at 52mph = 5.75 hrs
d%2F58 + %28%28317-d%29%29%2F52 = 5.75
multiply equation by least common multiple of 58 & 52, that is 1508
1508*d%2F58 + 1508*%28%28317-d%29%29%2F52 = 1508*5.75
cancel the denominators
26d + 29(317-d) = 8671
26d + 9193 - 29d = 8671
26d - 29d = 8671 - 9193
-3d = -522
d = -522/-3
d = +174 mi at 58 mph
then
317 - 174 = 143 mi at 52 mph
"Find the amount of time at EACH of the speeds"
174/58 = 3hrs at 58 mph
and
143/52 = 2.75 hrs at 52 mph
;
:
We can check this by noting the two times do add up 5.75 which is 5 hrs 45 min