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

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Director
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A hiker walking at a constant rate of 4 miles per hour is [#permalink]

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16 Oct 2006, 18:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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16 Oct 2006, 19:11
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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16 Oct 2006, 21:13
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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17 Oct 2006, 03:44
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...

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17 Oct 2006, 04:52
OA is 20. Thanks guys.
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17 Oct 2006, 04:56
uphillclimb wrote:
OA is 20. Thanks guys.

Cannot believe that.... where am i wrong? somebody...
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17 Oct 2006, 05:35
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...

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, 05:40, edited 1 time in total.
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17 Oct 2006, 05:39
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...

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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17 Oct 2006, 05:43
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...

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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17 Oct 2006, 17:26
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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17 Oct 2006, 17:28
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...

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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18 Oct 2006, 06:42
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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18 Oct 2006, 07:44
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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18 Oct 2006, 11:19
I got tricked too earlier,..a very nice problem...
18 Oct 2006, 11:19
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A hiker walking at a constant rate of 4 miles per hour is

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