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speed and time

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by mkhanna » Thu Sep 24, 2009 8:35 am
A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the path at a constant rate of 20 miles per hour. 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 hiker catches up?
a) 6 2/3
b) 15
c) 20
d) 25
e) 26 2/3
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Source: — Problem Solving |

by sreak1089 » Thu Sep 24, 2009 8:42 am
IMO D
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by glorydefined » Thu Sep 24, 2009 8:46 am
OA : c, please let me know if iam correct
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by m&m » Thu Sep 24, 2009 10:10 am
the biker goes 20-4 = 16mi/hr faster, so he is gaining 16mi on the biker per hour...

16 mi/hr*1 hr/60mins*5 mins = 16/12 mi to catch up = 4/3 mi

4/3 mi / 4 mi/hr = 1/3 hr = 20 mins

hope that helps
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by akhilkathuria » Thu Sep 24, 2009 10:14 am
answer is C.

A(Hiker)4m/h
--------------------------------
B(Cyclist)20m/h

For 5 minutes A would travel D=S*T= 1/3 (4/60*5)
For 5 minutes B would travel D=S*T= 5/3 (5/60*20)

Now after 5 miniutes Biker Stops

<-----1/3---->A<----4/3-->
--------------------------
<----------5/3----------->B

now A has to travel 4/3 inorder to reach B.
hence D=S*T = 4/3=4*T = T =1/3 which is 1/3*60=20 mins
:)
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by cbenk121 » Thu Sep 24, 2009 1:26 pm
Ahhh tricky! Forgot to account for the fact that the biker was only gaining 16 mph instead of 20. So my original answer was 25.

My method was exactly as same as m&m, no point in reposting.
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