A hiker walking at a constatnt rate of 4 miles/hr is passed

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A hiker walking at a constatnt rate of 4 miles/hr is passed [#permalink]

14 Aug 2007, 11:12

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A hiker walking at a constatnt rate of 4 miles/hr is passed by a cyclist travelling in the same direction along the same path at a constant rate of 20miles/hr. The cyclist stops to wait for the hiker 5 mins after passing her, while the hiker continues to walk at the constant rate. How many mins must the cyclist wait until the hiker catches up.

A 6 2/3
B 15
C 20
D 25
E 26 2/3

pls xplain y oa is C here...i thought it was D

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Re: gmat prp cd parctice 2 PS [#permalink]

14 Aug 2007, 11:33

crazy123 wrote:

In 5 mins, both traveled:
Hiker: (4/60)*5 = 1/3 mile
Biker: (20/60)*5 = 5/3 mile
Therefore, Biker has to waited: 5/3 - 1/3 = 4/3 miles.
Time for Hiker to travel 4/3 mile: (60/4)*(4/3) = 20mins.

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14 Aug 2007, 12:54

Good call on this one. The key is that the cyclist keeps going after passing the hiker for 5 min, but this does not mean that the hiker stops walking. If you assume the hiker stops walking, then you get 5/3 mile as the distance between the two, which the hiker will make up in 25 minutes.

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14 Aug 2007, 13:28

Hi,
I also go with the answer C.

My approach:

Relative speed of Cycle wrt hiker (20 - 4) = 16 Miles/ Hr
in 5 mins the Difference between both 16/60 * 5 = 4/3 Miles

Hiker will travel 4/3 miles in (4/3)/4 * 60 = 20 Mins :lol:

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14 Aug 2007, 22:31

C too...

the hiker must have travelled 20 miles in 1 hour before the cyclist will catch up. After that, the cyclist will cover 5/3miles in 5 minutes, while the hiker will cover 1/3 mile in 5 minutes. This is a difference of 4/3 miles. The hiker will take 20 minutes to catch up.

[#permalink] 14 Aug 2007, 22:31

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A hiker walking at a constatnt rate of 4 miles/hr is passed

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A hiker walking at a constatnt rate of 4 miles/hr is passed

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