Question 1011003: Mr. Walker walks at 6mph and Mr. Jogger jogs at 10mph. Mr Walker walks to a nearby town and back in 10 hours, not stopping at all until his return. Mr. Jogger starts jogging from the same place at the same time to the same town and back, also not stopping at all until his return. For what fraction of the Mr. Walkers trip is he within 4 miles of Mr. Jogger?
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! I like this problem. Gonna take some explanation, so check it out...
Okay here is how we go...remember rate times time equals distance, or rt=d...
Mr. Walker walks at 6 mph and the whole trip, there and back, takes him 10 hours. Thus the distance, there and back, is 60 miles. (30 miles each way.)
Mr. Jogger takes but 60 miles/10 mph = 6 hours for the round trip.
If they leave at the same time, they will be within four miles of each other for the first hour only...after that Jogger will have gone 10 miles and Walker only 6 miles...
When Jogger finishes one-way, he is at mile 30, while Walker is at mile 18, twelve miles apart.
Jogger turns around. They have to close their distance by eight miles to get within four miles of each other. That takes thirty minutes (their combined speed is 16 miles per hour).
At that point Jogger will be at mile 25 (going back) and Walker will be at mile 21 (still going forward). For the next half-hour they will be within four miles of each other, until Jogger is at mile 20 and Walker is at mile 24.
After that, they will be spread apart until Walker is four miles from home, coming back. Jogger will have been waiting there. The last four miles takes Walker 2/3 hour or 40 minutes.
So adding those times together, we have 1 hour + 30 minutes + 40 minutes = 2 1/6 hours.
Thus the fraction Mr. Walker will be within four miles of Mr. Jogger will be 2 1/6 divided by 10 or 13/60 of the time.
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