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

Question

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 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

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Official Answer

C

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KarishmaB · Accepted Answer

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)

Bunuel · Answer

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

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

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

Answer: C.

krushna · Answer

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

SVaidyaraman · Answer

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.

JeffTargetTestPrep · Answer

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

fskilnik · Answer

?\,\,\,:\,\,\,{	ext{minutes}}\,\,\,\left( {{	ext{last}}\,\,{	ext{diagram}}} ight)

https://preview.ibb.co/jsiRTe/04_Oct18_8t.gif

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

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

} } ight)\,\,\, = {5 \over 3}\,\,{m{miles}}

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

} } ight)\,\,\, = 25\,\,{m{minutes}}

Obs.: arrows indicate  licit converters .

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

\left( * ight)\,\,{m{used}}\,\,{m{while}}\,\,\left( {{m{also}}} ight)\,\,{m{cyclist}}\,\,{m{was}}\,\,{m{moving}}!

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

Regards, 
Fabio.

AlN · Answer

Hi Sir,

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

fskilnik · Answer

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.

DanTheGMATMan · Answer

Sort through the sequence of events and remember that the final catch-up episode means subtracting rates to get the rate of gain:

https://www.youtube.com/watch?v=yJPsYqc8J7Q

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=88088 --- Harsha GMAT-Club-Forum-x6dx3m12.png

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

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