Answer:
Margaret traveled 210 miles and then back.
Step-by-step explanation:
Question:
Margaret drove to a business appointment at 70 mph
.
Her average speed on the return trip was 60 mph. The return trip took 1/2 hour longer because of heavy traffic . How far did she travel?
Given:
Average speed during trip = 70 mph
Average speed during return trip = 60 mph
Time taken for return trip was [tex]\frac{1}{2}[/tex] hour longer than the actual trip.
To find the length of the trip.
Solution:
Let time taken for trip in hours be = [tex]x[/tex]
Average speed for the onward trip =70 mph
Distance covered in miles = [tex]Speed\times time = 70x[/tex]
The time taken for return trip in hours will be = [tex]x+\frac{1}{2}[/tex]
Average speed for the return trip =60 mph
Distance covered in miles = [tex]Speed\times time = 60(x+\frac{1}{2})[/tex]
As the distance for onward trip and return trip will be same, so we have :
[tex]70x=60(x+\frac{1}{2})[/tex]
Using distribution.
[tex]70x=60x+30[/tex]
subtracting both sides by [tex]60x[/tex]
[tex]70x-60x=60x-60x+30[/tex]
[tex]10x=30[/tex]
Dividing both sides by 10.
[tex]\frac{10x}{10}=\frac{30}{10}[/tex]
∴ [tex]x=3[/tex]
Thus, it took Margaret 3 hours to complete the trip onward.
Distance traveled = [tex]Speed\times time = 70(3)\ miles = 210\ miles[/tex]
Thus, Margaret traveled 210 miles and then back.
