SOLUTION: Wyatt left his house and rode his bike into town at 10 mph. Along the way he got a flat so he had to turn around and walk his bike back to his house traveling 5 mph. If the trip do
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Question 1144630: Wyatt left his house and rode his bike into town at 10 mph. Along the way he got a flat so he had to turn around and walk his bike back to his house traveling 5 mph. If the trip down and back took 9 hours, how far did he get before his tire went flat? Found 3 solutions by Alan3354, josgarithmetic, ikleyn:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Wyatt left his house and rode his bike into town at 10 mph. Along the way he got a flat so he had to turn around and walk his bike back to his house traveling 5 mph. If the trip down and back took 9 hours, how far did he get before his tire went flat?
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Avg speed of the round trip = 2*10*5/(10+5) = 20/3 mi/hr
RT distance = (20/3)*9 = 60 miles
Let the unknown distance under the question be " d " miles.
Then the time biking is the distance " d " divided by the average rate biking
= hours.
The time walking is the same distance " d " divided by the average rate walking
= hours.
The total time + is 9 hours.
It gives you an equation
+ = 9 hours.
This equation is called "time" equation, since each its term is time.
To solve it, multiply both sides by 10. You will get
d + 2d = 90,
3d = 90,
d = = 30.
ANSWER. One way distance is 30 miles.
Solved.
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Using "time" equation is the STANDARD method of solving such problems.
From my post, learn on how to write, how to use and how to solve a "time" equation.