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
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-> 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
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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?
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let d = one-way distance
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Write a time equation; time = dist/speed
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"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
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Confirm this, find the actual time each way
60/50 = 1.2 hrs
60/60 = 1 hr