SOLUTION: I need help! A person travels 15 miles in 1/3 of an hour, then returns by traveling that same 15 miles in 1/7 of an hour. What is that person's average rate for the trip?

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Question 529505: I need help!
A person travels 15 miles in 1/3 of an hour, then returns by traveling that same 15 miles in 1/7 of an hour. What is that person's average rate for the trip?

thank you
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A person travels 15 miles in 1/3 of an hour, then returns by traveling that same 15 miles in 1/7 of an hour.
What is that person's average rate for the trip?
:
Let a = the average speed of the 30 mi trip
:
Write a time equation, time = dist/speed
15%2F%281%2F3%29 + 15%2F%281%2F7%29 = 30%2Fa
invert the dividing fraction and multiply
3(15) + 7(15) = 30%2Fa
45 + 105 = 30%2Fa
150 = 30%2Fa
Multiply both sides by a
150a = 30
divide both sides by 150
a = 30%2F150
a = 1%2F5 mph hr is the average speed

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
I need help!
A person travels 15 miles in 1/3 of an hour, then returns by traveling that same 15 miles in 1/7 of an hour. What is that person's average rate for the trip?

Average rate = total distance, divided by total time

We then have: %2815+%2B+15%29%2F%28%281%2F3%29+%2B+%281%2F7%29%29

30%2F%2810%2F21%29, or 30%2821%2F10%29, or highlight_green%2863%29 mph

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