rates

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rates

by divya23 » Thu Jun 16, 2011 12:17 am
a hiker walking at a constant rate of 4 miles per hr is passed by a cyclist traveling in the same direction along the same path at a const rate of 20 miles/hr.the cyclist stops to wait for the hiker 5 minutes after passing her. while the hiker continues to walk at the same rate how many minutes do the cyclist wait until the hiker catcher up
15
20
25
30
35

[spoiler]oa = 20[/spoiler]

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by Roy@MasterGmat » Thu Jun 16, 2011 1:13 am
First, find the gap created between the hiker and the cyclist in the first phase of the question. Then, find how long it takes the hiker to close the gap.


Hiker: 4 Mph

Cyclist: 20 Mph



the gap between the cyclist and the hiker is the distance created between them during the first 5 minutes after the cyclist passed the biker. To find it, first find the combined rate by subtracting the hiker's speed for the cyclist's:



Combined rate = 20 - 4 = 16 Mph



Calculate the gap with the [Speed * Time = Distance] formula. The speed is the combined rate (16) and the time is 5 minutes (which is 5/60 hours):



16 * 5/60 = d

d = 80/60 = 4/3 miles



Finally, calculate how long it would take the hiker to complete 4/3 miles:



4 * T = 4/3

T = 1/3



1/3 of an hour is 20 minutes, so the correct answer is B
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by rohangupta83 » Thu Jun 16, 2011 1:54 am
Alternatively:

Speed of walker = 4 m/hr or 4/60 = 1/15 m/min
Speed of Cyclist= 20 m/hr or 20/60 =1/3 m/min

Distance travelled by cyclist in 5 mins = 1/3*5 = 5/3

Therefore the walker needs to travel 5/3 miles

Hence,
5/3 = 1/15(speed of the walker) * time
or
time = 5/3 * 15
time = 25 mins

As the cyclist waits after traveling for 5 mins.

He waits for 25 - 5 = 20 mins.