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by maolivie » Fri Jul 13, 2007 3:20 pm
A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist travelling in the same direction at t he 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?

1- 6 2/3
2- 15
3- 20
4- 25
5 - 26 2/3

I keep getting 2 even after converting the rates to minues. OA is C.
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Re: Gmatprep rates

by jayhawk2001 » Fri Jul 13, 2007 3:30 pm
The speed of the cyclist relative to the hiker = 20-4 = 16 mph

Since the cyclist travels for 5 min after passing the hiker, the
distance traveled by the cyclist = 16 * 5 / 60 miles

This distance can be traveled by the hiker in 16 * 5/60 * 1/4
= 20/60 hrs = 20 min
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by maolivie » Fri Jul 13, 2007 3:37 pm
where did the 60 in the denominator to 5/60 come from?
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by jayhawk2001 » Fri Jul 13, 2007 3:50 pm
maolivie wrote:where did the 60 in the denominator to 5/60 come from?
All 'rate' values are in miles per "hour". So, you either have to convert
rates to miles per minute OR convert time from minutes to hour for
proper calculation

We are given that the cyclist travels 5 minutes i.e. 5/60 hours.
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by perpetual_stud » Mon Jul 16, 2007 9:40 am
Hello Guys,

I am new to the forum.

The biker passes the hiker and bikes for another 5 mins. So the distance traveled by biker is 5/3 miles in 5 mins.

The time that the biker is biking for 5 mins, the hiker is walking to so you have to take into acct the distance the hiker travels in 5 mins.

In 5 mins the hiker travels 1/3 miles. So the distance traveled by the hiker after the biker stops is 5/3 – 1/3 = 4/3.

The time the biker has to wait is therefore 4/3/4 = 4/12 = 1/3.
So the biker has to wait 1/3 hour which is equal to 1/3 X 60 = 20 mins.
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