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

Hiker's speed : Cyclist's speed = 4 : 20 = 1 : 5
To cover the same distance, Time taken by Hiker : Time taken by Cyclist = 5 : 1

(If distance is same, speed is inversely proportional to time)

If cyclist took 5 mins, Hiker will take 25 mins. So she will need another 20 mins. (When cyclist was covering the distance in 5 mins, the Hiker was also walking for those 5 mins)
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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

In 1/12 hours (5 minutes) after the hiker is passed by the cyclist the distance between them will comprise \((20-4)*\frac{1}{12}=\frac{4}{3}\) miles (note that during these 5 minute hiker walks too, so their relative rate is 20-4 miles per hour). The hiker thus will need \(\frac{\frac{4}{3}}{4}=\frac{1}{3}\) hours, or 20 minutes to catch up.

Answer: C.
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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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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

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



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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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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When cyclist stops, the distance between he and the hiker is 20*5-4*580
So, the cyclist will wait 80/420 minutes.

Relative Speed Ratio = 5x-x = 4x

Relative Time Taken by Cyclist = 5 mins
The time hiker will take = 5x4 = 20 mins