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Both runners and walkers are participating in a marathon on exactly th

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New post 19 Oct 2016, 21:15
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Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?

a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles

Please assist with above problem.
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Both runners and walkers are participating in a marathon on exactly th [#permalink]

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New post 19 Oct 2016, 21:23
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Distance walked by Bill until 8 am = 4*1.25=5miles

Relative velocity between walker and runner = 9-4 = 5miles/hr

So fastest runner catches Bill 1hour after 8am

Total distance covered by Bill = 5 + 4 = 9miles

D


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Re: Both runners and walkers are participating in a marathon on exactly th [#permalink]

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New post 19 Oct 2016, 21:55
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Bill's speed = 4 mi/hr Runner's speed = 9 mi/hr

Relative speed = 9 - 4 = 5 mi/hr

Distance covered by Bill before Runner started running = 1.25*4 = 5

Time taken by runner to overtake Bill = 5/5 = 1 hr

Distance covered by Bill in 1 hr = 4*1 = 4 mi

Total Distance covered by Bill = 5 + 4 = 9 mi

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Re: Both runners and walkers are participating in a marathon on exactly th [#permalink]

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New post 09 May 2017, 11:35
How is Time taken by runner to overtake Bill = 1?
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New post 10 May 2017, 19:31
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ashikshetty wrote:
How is Time taken by runner to overtake Bill = 1?


Short version: Distance between them is 5 miles. Bill covered 5 miles before runner started. "Combined" or relative rate, derived from subtracting Bill's rate of 4 from runner's rate of 9, is 9-4=5 mph. 5/5 = 1 hour

Longer version: Bill has already traveled 1.25 hrs * 4 mph, 5 miles. That's the distance / gap between them. D/r=t. To get time, that 5 miles is divided by their "relative rate," which is runner(9)- Bill(4) = 5 mph==>

Runner runs at 9mph but Bill is still walking too, at 4mph, so you find their "combined" or relative rate by subtracting: 9-4 = 5 mph. That's actually the rate at which the gap is closing.

Distance (gap) = 5 miles. Rate that gap is closing is 5 miles/hr. At a rate of 5 miles per hour, it takes one hour to go 5 miles.

(And although the gap between them has closed when the runner overtakes Bill at 9 a.m., Bob has still walked 4 miles MORE miles in that one hour, for a total of 9 miles. Maybe that's the confusing part?)

Hope it helps.
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Re: Both runners and walkers are participating in a marathon on exactly th [#permalink]

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New post 15 May 2017, 17:31
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alanforde800Maximus wrote:
Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?

a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles


We can classify this problem as a “catch-up” rate problem, in which we use the formula:

distance of Bill = distance of fastest runner

We are given that Bill starts at 6:45 AM and walks at a constant rate of 4 mph and that the fastest runner starts at 8 AM and runs at a constant rate of 9 mph.

Since Bill started an hour and 15 minutes, or 5/4 hours, before the fastest runner, we can let the fastest runner’s time = t hours and Bill’s time = t + 5/4 hours. Since distance = rate x time, we can calculate each person’s distance in terms of t.

Bill’s distance = 4 x (t + 5/4) = 4t + 5

fastest runner’s distance = 9t

We can equate the two distances and determine t.

4t + 5 = 9t

5 = 5t

t = 1 hour

Now we can determine how far Bill will have walked by the time he is passed by the fastest runner.

Bill’s distance = 4(1) + 5 = 9 miles

Answer: D
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Re: Both runners and walkers are participating in a marathon on exactly th [#permalink]

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New post 24 Jan 2018, 21:27
Hi All,

This is an example of a 'chase down' question - the first entity gets out ahead and the second entity chases the first down.

Based on the information in the prompt, the walkers have a 5/4 hour head start on the runners. Since the walkers average 4 miles/hour, the 'head start' of a walker will be...

D = (R)(T)
D = (4 miles/hour)(5/4 hours)
D = 5 miles

So the walker is 5 miles ahead when the runner gets started. Since both walker and runner are now moving, the walker will continue to move at 4 miles/hour wile the runner will move at 9 miles/hour. Thus, the runner will 'catch up' 5 miles every hour on the walker. Since the walker is currently 5 miles ahead of the runner, the runner will need 1 hour to catch up to the walker. In that 1 hour, the runner will travel 9 miles.

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Re: Both runners and walkers are participating in a marathon on exactly th   [#permalink] 24 Jan 2018, 21:27
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