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

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Wed Oct 03, 2018 2:27 pm

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Source: GMAT Prep

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 & 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

The OA is C.

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Source: Beat The GMAT — Problem Solving | Wed Oct 03, 2018 2:27 pm

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by [email protected] » Wed Oct 03, 2018 4:39 pm

Hi All,

We're told that 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 & waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate. We're asked for the number of MINUTES the cyclist must wait until the hiker catches up. This question requires the use of the Distance Formula - and you might find it easiest to approach the math in small 'steps.'

Since the cyclist continues riding for 5 minutes AFTER passing the hiker, we can determine how far the cyclist went during that time. It's worth noting that BOTH the cyclist and the hiker were moving during that time, so the cyclist was moving away from the hiker at a speed of 20 - 4 = 16 miles/hour. Five minutes = 5/60 = 1/12 of an hour.

Distance = (Rate)(Time)
Distance = (16 miles/hour)(1/12 hour)
D = 16/12 = 4/3 of a mile

The cyclist stops and waits for the hiker to 'catch up.' Now that we know that distance that the hiker must travel, we can determine how long that would take.

Distance = (Rate)(Time)
4/3 miles = (4 miles/hour)(T hour)
(4/3)/4 = T
4/12 hour = T
1/3 hour = T

Since there are 60 minutes in 1 hour, 1/3 of an hour = 20 minutes

Final Answer: C

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Rich

Contact Rich at [email protected]

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by GMATGuruNY » Wed Oct 03, 2018 4:49 pm

BTGmoderatorLU wrote:Source: GMAT Prep

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 & 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

The OA is C.

Rate and time have a RECIPROCAL RELATIONSHIP.
The RATE RATIO for the hiker and cyclist = 4 mph : 20 mph = 1:5.
Thus:
The TIME RATIO for the hiker and cyclist = the reciprocal of the rate ratio = 5:1.
Implication:
The hiker will take 5 TIMES AS LONG as the cyclist to travel the same distance.
Since the cyclist travels for 5 minutes, the hiker will take 5 times as long -- 25 minutes -- to travel the same distance.
Thus, the cyclist will have to wait 20 minutes for the hiker to catch up.

The correct answer is C.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

a hiker walking at a constant rate of 4 miles per hour is pa

by fskilnik@GMATH » Thu Oct 04, 2018 2:13 pm

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 hikerÂ´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

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

LetÂ´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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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by Scott@TargetTestPrep » Thu Oct 04, 2018 6:34 pm

BTGmoderatorLU wrote:Source: GMAT Prep

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 & 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.

During these 5 minutes, the hiker walks 4/12 = 1/3 miles. So the difference between the cyclist and the hiker is 5/3 - 1/3 = 4/3 miles when the cyclist stops and waits for the hiker. Therefore, it will take the hiker (4/3)/4 = 1/3 hour = 20 minutes to catch up with the cyclist.

Alternate Solution:

Notice that the cyclist travels precisely 5 times as fast as the hiker. Therefore, the distance traveled by the cyclist in 5 minutes will be traveled by the hiker in 25 minutes. Since the hiker had already been walking for 5 minutes when the cyclist stopped to wait for her, the hiker must walk 25 - 5 = 20 more minutes to catch up with the cyclist.

Answer: C

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Founder and CEO
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