SOLUTION: In the morning, Maria drove to a business appointment at 60mph. Her average speed on the return trip in the afternoon was 50mph. The return trip took 1/5 hr longer because of heavy

Question 1017095: In the morning, Maria drove to a business appointment at 60mph. Her average speed on the return trip in the afternoon was 50mph. The return trip took 1/5 hr longer because of heavy traffic. How far did she travel to the appointment?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
In the morning, Maria drove to a business appointment at 60mph.
Her average speed on the return trip in the afternoon was 50mph.
The return trip took 1/5 hr longer because of heavy traffic.
How far did she travel to the appointment?
:
let d = one-way distance
:
Write a time equation; time = dist/speed
:
"The return trip took 1/5 hr longer because of heavy traffic."
1/5 = .2 hr
return time = "to" time + .2 hrs
= () + .2
multiply equation by 300, cancel the denominators and you have
6d = 5d + 300(.2)
6d - 5d = 60
d = 60 mi one-way. Total travel dist = 120 mi
:
:
Confirm this, find the actual time each way
60/50 = 1.2 hrs
60/60 = 1 hr

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