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Help on Rate Problem

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by DCJ » Thu Jul 23, 2009 10:26 pm
A hiker walking at a constant rate of 4mph is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20mph. 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 bike catches up?

A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3
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by Rajani » Thu Jul 23, 2009 10:51 pm
Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins
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by Rajani » Thu Jul 23, 2009 10:58 pm
Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins
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by cammijc » Fri Jul 24, 2009 6:25 am
Cyclist:

60 min = 20 m
5 min = x
x = 5/3

Hiker:

60 min = 4m
5 min = y
y = 1/3

So, during the five min after the cyclist passes the hiker, the cyclist traveled 5/3 mile while the hiker traveled 1/3 mile (because the hiker keeps going after the cyclist passes her.

This means that when the cyclist stops, the hiker has to go 4/3 mile to catch up (5/3 - 1/3).

If she goes 1/3 m in 5 min, then it will take her 20 min to go 4/3 mile (5 min*4).

IMO C.
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by cata1yst » Fri Jul 24, 2009 6:44 am
cammijc wrote:Cyclist:

60 min = 20 m
5 min = x
x = 5/3

Hiker:

60 min = 4m
5 min = y
y = 1/3

So, during the five min after the cyclist passes the hiker, the cyclist traveled 5/3 mile while the hiker traveled 1/3 mile (because the hiker keeps going after the cyclist passes her.

This means that when the cyclist stops, the hiker has to go 4/3 mile to catch up (5/3 - 1/3).

If she goes 1/3 m in 5 min, then it will take her 20 min to go 4/3 mile (5 min*4).

IMO C.
I got the same answer the same way. You need to take into account the distance traveled by the hiker during the 5 minutes the biker uses after he passes the hiker.

C.
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by kaulnikhil » Fri Jul 24, 2009 6:51 am
it should be C
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Help on Rate Problem

by DCJ » Fri Jul 24, 2009 9:42 am
Thanks everyone. The correct answer is C. 20. I also kept getting 25 until I realized I needed to take into account that the hiker was still walking while the cyclist waited.
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by jjk » Fri Jul 24, 2009 11:39 am
Rajani wrote:Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins
What you should have done is see that the relative speed of the cyclist is 20 - 14 = 16 mph. 5 minutes elapse after the cyclist passes the hiker, and 5 minutes equals 1/12 of an hour. So 16 X 1/12 = 16/12 = 4/3 miles.

4/3 miles is the distance between the cyclist and the hiker when the cyclist decides to stop and wait for the hiker to catch up.

From there, it's a simple d = rt problem. 4/3 = 4t, so t = 1/3 of an hour, 20 minutes.
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