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 Post subject: A hiker walking at a constant rate of 4 miles per hour
 Post Posted: Fri Apr 17, 2009 12:21 pm 
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Course Students


Posts: 4
a hiker walking at a constant rate of 4 miles per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 20 miles per hour. The cyclist stops to wait for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up?

6 2/3
15
20
25
26 2/3

What is the Manhattan Gmat way to solving and approaching this?

Thank you in advance.


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 Post subject: Re: A hiker walking at a constant rate of 4 miles per hour
 Post Posted: Mon Apr 20, 2009 6:46 am 
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ManhattanGMAT Staff


Posts: 7146
probably the most efficient way to solve this one is with the concept of relative rates.

start at the time when the cyclist passes the hiker.
since they're both traveling in the same direction, their relative rate - i.e., the rate at which the distance between them is changing - is found by subtraction. so, as long as they're both moving the cyclist is gaining on the hiker at a rate of 20 - 4 = 16 mph.
after the cyclist stops, the hiker is making up that distance at a rate of 4 mph. that's only 1/4 as fast as the relative rate during the first 5 minutes, so it will take the hiker 4 times as long to catch back up. so, 4 x 5 = 20 minutes.

if you wanted to, you could calculate the actual distance between the cyclist and the hiker when the cyclist stops: (16 mi/h)(1/12 h) = d, where "1/12 h" is five minutes. then you could find the time taken by the hiker alone to travel the same distance "d".
you don't have to do this, though, if you realize the 1:4 ratio referenced above.


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 Post subject: Re: A hiker walking at a constant rate of 4 miles per hour
 Post Posted: Fri Apr 24, 2009 6:09 pm 
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Students


Posts: 11
i dont get ur explanation. the flaw is tht when u say tht the ratio is 1:4 and hence wht took cycler 5 min will take hiker 20 min is tht during the 5 min tht the cycler overtook and got away, hiker was travelling at 4kmph. hence, the ratio doesn't exlain it.

explanation: distance to be covered by hiker = distance by cycler - d by hiker
i.e. for cycler @ 20kmph for 5 min = 5/3 and for cycler @4kmph for 5 min = 1/3. now, the leftover distance to be covered by cycler = 5/3- 1/3 = 4/3.

time to cover 4/3km for cycler = 4/3 (distance) / 4 (speed) i.e. 1/3 hours i.e. 20 min.

guess this should help to understand the concept better.


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 Post subject: Re: A hiker walking at a constant rate of 4 miles per hour
 Post Posted: Mon Apr 27, 2009 4:12 am 
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ManhattanGMAT Staff


Posts: 7146
chintanjpatel wrote:
i dont get ur explanation. the flaw is tht when u say tht the ratio is 1:4 and hence wht took cycler 5 min will take hiker 20 min is tht during the 5 min tht the cycler overtook and got away, hiker was travelling at 4kmph. hence, the ratio doesn't exlain it.


no flaw.

check out the explanation again; the 1:4 ratio is the ratio of the hiker's catch-up speed to the original RELATIVE speed. all of the motion occurring during the original travel is subsumed under the relative speed, so we're good.

--

your explanation works too, of course.


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