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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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cyclist is relatively 4times faster
So in 5 minutes, he has taken the lead for 20 minutes relative to the Hiker.

So he has to wait for 20 minutes
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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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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

1. Let us start at the point when the cyclist and the hiker are together
2. In 5 min, cyclist travels 20/12 miles, whereas the hiker travels 4/12 miles.
3. The hiker has to make up 20/12 - 4/12 miles which is 4/3 miles.
4. Time to travel 4/3 miles by the hiker is 4/3 /(4) = 4/12 hrs = 20 min.
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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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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

We are given that a cyclist travels at a rate of 20 mph, passes a hiker, and then stops to wait for the hiker after traveling for 5 minutes. Since 5 minutes = 1/12 hours, the cyclist travels a distance of 20/12 = 5/3 miles.

We let the extra time, in hours, of the hiker = t, and then the hiker’s total time is t + 1/12; thus, the distance in miles of the hiker is 4(t + 1/12) = 4t + 1/3.

Since the hiker catches the cyclist, we set their distances equal and determine t:

4t + ⅓ = 5/3

4t = 4/3

t = (4/3)/4 = 4/12 = 1/3 hours = 20 minutes

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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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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

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

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

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

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]
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]
When cyclist stops, the distance between he and the hiker is 20*5-4*580
So, the cyclist will wait 80/420 minutes.
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a [#permalink]
Relative Speed Ratio = 5x-x = 4x

Relative Time Taken by Cyclist = 5 mins
The time hiker will take = 5x4 = 20 mins
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a [#permalink]
JeffTargetTestPrep wrote:
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

We are given that a cyclist travels at a rate of 20 mph, passes a hiker, and then stops to wait for the hiker after traveling for 5 minutes. Since 5 minutes = 1/12 hours, the cyclist travels a distance of 20/12 = 5/3 miles.

We let the extra time, in hours, of the hiker = t, and then the hiker’s total time is t + 1/12; thus, the distance in miles of the hiker is 4(t + 1/12) = 4t + 1/3.

Since the hiker catches the cyclist, we set their distances equal and determine t:

4t + ⅓ = 5/3

4t = 4/3

t = (4/3)/4 = 4/12 = 1/3 hours = 20 minutes

­Why haven't we equted the total distance for cyclists in the above equation? 5/3 miles is the distance for only 5 mins and not the entire distance?
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a [#permalink]
nikitathegreat wrote:
JeffTargetTestPrep wrote:
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

We are given that a cyclist travels at a rate of 20 mph, passes a hiker, and then stops to wait for the hiker after traveling for 5 minutes. Since 5 minutes = 1/12 hours, the cyclist travels a distance of 20/12 = 5/3 miles.

We let the extra time, in hours, of the hiker = t, and then the hiker’s total time is t + 1/12; thus, the distance in miles of the hiker is 4(t + 1/12) = 4t + 1/3.

Since the hiker catches the cyclist, we set their distances equal and determine t:

4t + ⅓ = 5/3

4t = 4/3

t = (4/3)/4 = 4/12 = 1/3 hours = 20 minutes

­Why haven't we equted the total distance for cyclists in the above equation? 5/3 miles is the distance for only 5 mins and not the entire distance?

The question is looking only for the time from when the cyclist first passes the walker to when the walker catches up and because of this, information about what happens before they first meet doesn't matter.

Posted from my mobile device
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Re: A hiker walking at a constant rate of 4 miles per hour is passed by a [#permalink]
­Sort through the sequence of events and remember that the final catch-up episode means subtracting rates to get the rate of gain:

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