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Director
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A hiker walking at a constant rate of 4 miles per hour is [#permalink]
 16 Oct 2006, 19:53
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 five 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?
6 2/3
15
20
25
26 2/3
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Manager
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1/3 + 4*(Time taken by the hiker to cover the remaining distance) = 5/3
So Time taken = 1/3 of an hour OR 20 mins ....ANS is C
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Director
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cyclist travelled 1.66 miles in 5 minutes (as per given speed data).
walker covered that 1.66 miles in 20 minutes( per his rate).
C is the answer.
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Director
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Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
Answer D....
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Director
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OA is 20. Thanks guys.
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Director
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uphillclimb wrote: OA is 20. Thanks guys.
Cannot believe that.... where am i wrong? somebody... 
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SVP
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SimaQ wrote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
Answer D....
After having calculated the cyclist distance, u have forgotten to remove the distance covered by the hiker during the 5 min
The distance separating the hiker and the cyclist when the cyclist stopped to ride is 1/12*20 - 1/12*4 = 4/3
Last edited by Fig on 17 Oct 2006, 06:40, edited 1 time in total.
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Director
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Fig wrote: SimaQ wrote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
Answer D.... After having calculated the cyclist distance, u have forgotten to remove the distance covered by the hiker during this 5 min The distance separating the hiker and the cyclist when the cyclist stopped to ride is 1/12*20 - 1/12*4 = 4/3 Now tell me, how to avoid such mistakes!!!! 
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SVP
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SimaQ wrote: Fig wrote: SimaQ wrote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
Answer D.... After having calculated the cyclist distance, u have forgotten to remove the distance covered by the hiker during this 5 min The distance separating the hiker and the cyclist when the cyclist stopped to ride is 1/12*20 - 1/12*4 = 4/3 Now tell me, how to avoid such mistakes!!!! To me, the only manner is to track all mistakes in an Errorlog or more appropriated a SillyMistakeslog 
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Director
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C
Relative speed = 16mph. In 5 minutes, cyclist had crossed = 16*5/60 miles
Time taken by hiker to cover this distance = 16*5/60 * 60/4 = 20 minutes
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Director
Joined: 23 Jun 2005
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SimaQ wrote: Fig wrote: SimaQ wrote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
Answer D.... After having calculated the cyclist distance, u have forgotten to remove the distance covered by the hiker during this 5 min The distance separating the hiker and the cyclist when the cyclist stopped to ride is 1/12*20 - 1/12*4 = 4/3 Now tell me, how to avoid such mistakes!!!! I try to use relative speeds.
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Director
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Quote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25...
I got this same answer. Will someone explain why this is wrong. I don't understand the above explanation.
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SVP
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ggarr wrote: Quote: Hiker is traveling at: 4miles/60minutes=1 mile/15 minutes
Cyclist: 20miles/60minutes= 1 mile/3 minutes
After 5 minutes the cyclist will cover: r*t=d....
(1/3)*5=d
d=5/3
The hiker will travel the mentioned above distance in r*t=d;
1/15*t=5/3
t=75/3
t=25... I got this same answer. Will someone explain why this is wrong. I don't understand the above explanation. While the cyclist is ridding during 5 min, the hiker does not stop : he goes ahead and covers a distance of 1/12*4.
This distance has to be considered in the calculation of the time required for the hicker to join the cyclist
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Director
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I got tricked too earlier,..a very nice problem...
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