Hiker and cyclist :: Manhattan GMAT Forums

Posted: Fri Apr 04, 2008 7:15 pm

A hiker walking at a constant rate of 4 miles/hour is passed by a cyclist traveling in the same direction at a constant rate of 20 miles/hour. The cyclist stops to wait for the hiker 5 minutes after passing her, while the hiker continues to walk her constant rate. How many minutes must the cyclist wait until the runner catches up?
A) 6 and 2/3
B) 15
C) 20
D) 25
E) 26 and 2/3

Posted: Fri Apr 04, 2008 9:52 pm

tmmyc wrote:

For this problem, we use the d = r*t formula

Hiker distance after 5 minutes
r = 4 miles / 1 hour -> 4 miles / 60 minutes -> 1 mile / 15 minutes
t = 5 minutes
d = r*t -> 1/15 * 5 = 1/3 miles

Cyclist distance after 5 minutes
r = 20 miles / 1 hour -> 20 miles / 60 minutes -> 1 mile / 3 minutes
t = 5 minutes
d = r*t -> 1/3 * 5 = 5/3 miles

Distance between the two after 5 minutes
5/3 - 1/3 = 4/3 miles

Time it takes Hiker to travel 4/3 miles
4/3 = 1/15 * t
t = 20 minutes

Posted: Mon Apr 07, 2008 5:28 am

looks like the above poster's solution is valid. here's a shortcut, in case you like that sort of thing:

after the cyclist passes the pedestrian, their relative rate is 16 miles/hr (20 - 4, since they're traveling in the same direction): in other words, the cyclist is going to get 16 miles farther ahead of the pedestrian each hour. so, in five minutes, which is 1/12 hour, the cyclist will go (16 mi/hr)(1/12 hr) = 4/3 miles ahead of the pedestrian.

then, the cyclist must wait for the pedestrian to walk 4/3 mile. this takes t = d / r = (4/3 mi) / (4 mi/hr) = 1/3 hr = 20 minutes.