SOLUTION: Cheryl left home on a business trip in her car and averaged 50mph. The next day she started to return at an average rate of 38 mph. After driving 2 hours longer than she spent on h

Question 475494: Cheryl left home on a business trip in her car and averaged 50mph. The next day she started to return at an average rate of 38 mph. After driving 2 hours longer than she spent on her outward trip. She discovered that she was still 14 miles from home. How far from home had she traveled?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Cheryl left home on a business trip in her car and averaged 50mph.
The next day she started to return at an average rate of 38 mph.
After driving 2 hours longer than she spent on her outward trip, she discovered that she was still 14 miles from home.
How far from home had she traveled?
:
Let t = travel time on the 1st day
then
(t+2) = travel time on the 2nd day to get 14 mi from home
:
Write a distance equation, dist = speed * time
:
1st day dist = 2nd day dist + 14 mi
50t = 38(t+2) + 14
50t = 38t + 76 + 14
50t - 38t = 90
12t = 90
t =
t = 7.5 hrs
:
Find the distance:
50 * 7.5 = 375 mi
:
Check the dist
38(7.5+2) + 14
361 + 14 = 375
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