Author
Message
TAGS:
Director

Joined: 11 Sep 2006

Posts: 514

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

A hiker walking at a constant rate of 4 miles per hour is [#permalink ]
16 Oct 2006, 18:53

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

_________________

...there ain't no such thing as a free lunch...

Manager

Joined: 07 Jun 2006

Posts: 113

Followers: 2

Kudos [? ]:
2
[0 ] , given: 0

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

Director

Joined: 17 Jul 2006

Posts: 716

Followers: 1

Kudos [? ]:
8
[0 ] , given: 0

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.

Director

Joined: 05 Feb 2006

Posts: 904

Followers: 1

Kudos [? ]:
33
[0 ] , given: 0

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....

Director

Joined: 11 Sep 2006

Posts: 514

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

OA is 20. Thanks guys.

_________________

...there ain't no such thing as a free lunch...

Director

Joined: 05 Feb 2006

Posts: 904

Followers: 1

Kudos [? ]:
33
[0 ] , given: 0

uphillclimb wrote:

OA is 20. Thanks guys.

Cannot believe that.... where am i wrong? somebody...

SVP

Joined: 01 May 2006

Posts: 1814

Followers: 8

Kudos [? ]:
89
[0 ] , given: 0

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, 05:40, edited 1 time in total.

Director

Joined: 05 Feb 2006

Posts: 904

Followers: 1

Kudos [? ]:
33
[0 ] , given: 0

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!!!!

SVP

Joined: 01 May 2006

Posts: 1814

Followers: 8

Kudos [? ]:
89
[0 ] , given: 0

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

Director

Joined: 23 Jun 2005

Posts: 847

GMAT 1 : 740 Q48 V42

Followers: 5

Kudos [? ]:
24
[0 ] , given: 1

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

Director

Joined: 23 Jun 2005

Posts: 847

GMAT 1 : 740 Q48 V42

Followers: 5

Kudos [? ]:
24
[0 ] , given: 1

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.

Director

Joined: 12 Jun 2006

Posts: 539

Followers: 1

Kudos [? ]:
14
[0 ] , given: 1

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.

SVP

Joined: 01 May 2006

Posts: 1814

Followers: 8

Kudos [? ]:
89
[0 ] , given: 0

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

Director

Joined: 01 Oct 2006

Posts: 500

Followers: 1

Kudos [? ]:
16
[0 ] , given: 0

I got tricked too earlier,..a very nice problem...