SOLUTION: Wilma drove at an average speed of 35 mph from her home in City A to visit her sister in City B.
She stayed in City B 20 hours, and on the trip back averaged 55 mph. She returned
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-> SOLUTION: Wilma drove at an average speed of 35 mph from her home in City A to visit her sister in City B.
She stayed in City B 20 hours, and on the trip back averaged 55 mph. She returned
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Question 258710: Wilma drove at an average speed of 35 mph from her home in City A to visit her sister in City B.
She stayed in City B 20 hours, and on the trip back averaged 55 mph. She returned home 50 hours after leaving.
We have to determine how many miles x is it from City A to City B. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Wilma drove at an average speed of 35 mph from her home in City A to visit her sister in City B.
She stayed in City B 20 hours, and on the trip back averaged 55 mph. She returned home 50 hours after leaving.
We have to determine how many miles x is it from City A to City B
:
Obviously, we are only interested in her actual travel time
50 - 20 = 30 hrs total travel time
:
x = the miles between city A and B
:
Write a time equation; time = dist/speed
Time there + time back = 30 hrs + = 30
:
Multiply by a common denominator, 385 with do the job
385* + 385* = 385(30)
;
Cancel the denominators and you have:
11x + 7x = 11550
;
18x = 11550
x =
x = 641 miles between cities
;
:
Check this on calc; find the total time
641.67/35 + 641.67/55
18.33 + 11.67 = 30 hrs, confirms our solution