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A hiker walking at a constant rate of 4 miles per hour is passed by a

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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 16 Dec 2018, 05:21
Speed of hiker is 4 miles per hour = 1/15 per minute
Speed of cyclist is 20 miles per hour =1/3 per minute

1/3*5 = 5/3 (Distance = Speed* Time)
1/15 *X = 5/3 (Since eventually both have to meet)

Thus X=25 and when 5 minutes are already covered by cyclist, thus 25-5 = 20

Ans : 20
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 26 Jun 2019, 23:51
fskilnik wrote:
Quote:
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. Five minutes after the passing the cyclist stops, while the hiker continues to walk at the hiker´s rate. How many minutes must the cyclist wait until the hiker catches up?

A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3

\(?\,\,\,:\,\,\,{\text{minutes}}\,\,\,\left( {{\text{last}}\,\,{\text{diagram}}} \right)\)

Image

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

\({\rm{cyclist}}:\,\,\,5\min \,\,\left( {{{20\,\,{\rm{miles}}} \over {60\,\,\min }}\,\matrix{
\nearrow \cr
\nearrow \cr

} } \right)\,\,\, = {5 \over 3}\,\,{\rm{miles}}\)

\({\rm{hiker}}:\,\,\,{5 \over 3}{\rm{miles}}\,\,\left( {{{60\,\,{\rm{min}}} \over {4\,\,{\rm{miles}}}}\,\matrix{
\nearrow \cr
\nearrow \cr

} } \right)\,\,\, = 25\,\,{\rm{minutes}}\)

Obs.: arrows indicate licit converters.

\(? = 25 - 5\left( * \right) = 20\min\)

\(\left( * \right)\,\,{\rm{used}}\,\,{\rm{while}}\,\,\left( {{\rm{also}}} \right)\,\,{\rm{cyclist}}\,\,{\rm{was}}\,\,{\rm{moving}}!\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.



Hi Sir,

I gt 25 minutes but I amnot able to understand why are we subtracting 5 from it.
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 27 Jun 2019, 03:47
AlN wrote:
Hi Sir,

I gt 25 minutes but I amnot able to understand why are we subtracting 5 from it.

Hi AIN,

Thank you for your interest in our solution!

During 5min the cyclist is able to run 5/3 miles (1st line), and the hiker needs 25min to run this distance (2nd line).

The fact is that the hiker "uses" 5min of "this" 25min while the cyclist runs 5/3 miles, therefore from the 25min needed by the hiker, he "saves" 5min using the interval of time before the cyclist stops!

Regards and success in your studies,
Fabio.

P.S.: more details related to the relative velocity (speed) are presented in our course.
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 24 Sep 2019, 10:48
anilnandyala wrote:
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 & waits for the hiker 5 min 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

A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3


Given:
1. 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.
2. the cyclist stops & waits for the hiker 5 min after passing her while the hiker continues to walk at her constant rate.

Asked: how many minutes must the cyclist wait until the hiker catches up?

In 5 minutes cyclist travels more than hiker = 16*5/60 = 4/3 miles
Hiker will take time to catch up = 4/3 / 4 = 1/3 hour = 60/3 = 20 minutes

IMO C
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A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 13 Nov 2019, 07:31
\(\frac{(16*\frac{5}{60})}{4}\)
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a  [#permalink]

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New post 24 Nov 2019, 01:46
In 5 minutes (1/12 hrs) the cyclist travels 20*(1/12) = 5/3 miles. Let's assume he has to wait for 't' hrs for the cyclist to catch up.
Therefore, the total time that the cyclist takes to cover 5/3 miles is the sum of the time the cyclist traveled (1/12) hrs and the time he waited ('t' hrs).
Thus, [(1/12) + t] hrs*4 mph = 5/3 miles.... > t=(1/3) hrs=20 minutes. ANS: C
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a   [#permalink] 24 Nov 2019, 01:46

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