Question 433975: A long distance runner started on a course running at an average speed of 6 mph. one-half hour later, a second runner began the same course at an average speed of 7mph. How long after the second runner started did the second runner overtake the first runner?
Found 3 solutions by mananth, ikleyn, MathTherapy:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! Runner I 6 mph
Runner II 7 mph
Difference in time= 90 minutes
Train Awill have covered 9 miles before Runner II starts
catch up distance= 9 miles
catch up speed = 7-6 mph
catch up speed = 1 mph
Catchup time = catchup distance/catch up speed
catch up time= 9/1
catch up time= 9 hours
Answer by ikleyn(53427)  (Show Source):
You can put this solution on YOUR website! .
A long distance runner started on a course running at an average speed of 6 mph.
one-half hour later, a second runner began the same course at an average speed of 7mph.
How long after the second runner started did the second runner overtake the first runner?
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As I read the solution in the post by @mananth, it makes me laugh.
The runners are confused with trains, and the input numbers are confused with other, irrelevant numbers.
The solution and the answer in the post by @mananth both are incorrect.
So, I came to bring a correct normal solution.
Runner I 6 mph
Runner II 7 mph
Difference in time = 1/2 of an hour
Runner I will have covered 3 miles before Runner II starts
catch up distance= 3 miles
catch up speed = 7-6 mph
catch up speed = 1 mph
Catchup time = catchup distance/catch up speed
catch up time= 3/1
catch up time= 3 hours <<<---=== ANSWER
Solved correctly.
Answer by MathTherapy(10587)  (Show Source):
You can put this solution on YOUR website!
A long distance runner started on a course running at an average speed of 6 mph. one-half hour later, a second runner began the
same course at an average speed of 7mph. How long after the second runner started did the second runner overtake the first runner?
Let time taken by FASTER runner to get to meet up point, be T
With FASTER runner's speed being 7 mph, DISTANCE FASTER runner covers, to get to meet-up point = 7T
Since SLOWER runner started one-half ( )hour before FASTER runner, then time slower runner takes to get to meet-up point = T +
With SLOWER runner's speed being 6 mph, DISTANCE SLOWER runner covers, to get to meet-up point = 6(T + )
When both got to meet-up point, they'd covered the same distance
As such, we get the following DISTANCE equation: 7T = 6(T + )
7T = 6T + 3
7T - 6T = 3
Time FASTER runner takes to get to meet-up point/catch up with SLOWER runner, or T = 3 hours
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