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rate distance

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by raunekk » Sun Jun 29, 2008 12:12 am
15. A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling 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 continue to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up?

can anyone explanin me how to do this sum..
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Re: rate distance

by durgesh79 » Sun Jun 29, 2008 2:22 am
raunekk wrote:15. A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling 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 continue to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up?

can anyone explanin me how to do this sum..
First we need to find out what is the distance between the cyclist and hiker when the cyclist stops. this happened aftet 5 minutes they passed.

time = 5 minutes = 5/60 hrs
speed = relative speed between cyclist and hiker = 20-4 = 16 miles/hr
distance = 16 * 5/60

Now hiker has to travel this distance with a speed of 4 miles/hr

time = (16 * 5/60)/4 hr
=20/60 hr
=20 minutes
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by Carol » Mon Jun 30, 2008 5:42 am
This is my reasoning, although I am not quite sure...

we know the cyclist rides 20 miles/hour and the hiker walks 4 miles/hour.

the formula: D (distance) = R (rate) x T (time)

after passing the hiker the cyclist stoped, and we want to know for how long does he need to wait until the hiker reaches him. with the data we have, we must know what distance has the cyclist reached, after the 5 minutes he had crossed the hiker?

d = 20 x 5

D = 100 miles

the hiker:

D = 4 x 5

D = 20 miles

the difference between the cyclist and the hiker is 80 miles.

the hiker, in order to reach the cyclest, has 80 more miles to walk.

80 = 4 x t

T = 20 minutes . 20 minutes is the time the cyclist has to wait.
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by rocket » Sat Feb 27, 2010 10:54 am
Carol,

You have combined hour and minutes. Cycle speed is 20 miles in 1 hr not in 1 min
there fore, D is not eq to 5x20.

I hope u agree.
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by harshavardhanc » Sat Feb 27, 2010 1:24 pm
rocket wrote:Carol,

You have combined hour and minutes. Cycle speed is 20 miles in 1 hr not in 1 min
there fore, D is not eq to 5x20.

I hope u agree.
ha ha! perhaps, this can be termed as an "useful oblivion" :)

the fact that Carol forgot an inessential ( for this problem ) conversion, did not affect the answer. Her approach is absolutely correct, and not doing the "hour->minute conversion" makes the calculation easier .

@roket: observe that durgesh has not touched the "1/60 hr " factor anywhere in his calculation and this is what you are referring to in your query.
Regards,
Harsha
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