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A hiker began walking at a constant rate of 4mph and is

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A hiker began walking at a constant rate of 4mph and is [#permalink] New post 17 Jun 2006, 15:45
A hiker began walking at a constant rate of 4mph and is passed by a cyclist traveling in the same direction along the path at a constant rate of 20mph. The cyclist stops to wait for the hiker 5 minutes after passing her. How many minutes must the cyclist wait for the hiker to catch up?

Please explain your work.
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 [#permalink] New post 17 Jun 2006, 16:14
Answer is 25 minutes.

1. cyclist travels= (20m/hr)(1h/60min)(5min)= 5/3 miles
2. hiker's rate= (4m/hr)(1hr/60min)= 1m/15min
3. Plug in the values from 1 & 2 in T=D/R
4. T=(5/3)/(1/15) = 25 minutes
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 [#permalink] New post 17 Jun 2006, 21:01
20 min
cyclist related speed ( 20-4=16 )
distance in 5 minutes 16/12 =4/3
hiker does 1/3 in 5 minutes, so he need 5*4=20 minutes

Last edited by deowl on 18 Jun 2006, 01:53, edited 1 time in total.
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 [#permalink] New post 17 Jun 2006, 21:12
relative speed = 20 - 4 mph = 16 mph
distance travelled relative to hiker = 5*16/60=4/3miles

time taken by hiker = (4/3)*(1/4)*60 = 20 minutes
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 [#permalink] New post 17 Jun 2006, 23:30
I overlooked the relative speed but how did you come up with the hiker's 1/4 rate?

Thanks in advance for the feedback.

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 [#permalink] New post 18 Jun 2006, 01:54
brcinsf wrote:
I overlooked the relative speed but how did you come up with the hiker's 1/4 rate?

Thanks in advance for the feedback.

-Tru


Made a typo while posting :)
Obviousy it is 1/3, otherwise the answer wouldn't be 20 :lol:
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 [#permalink] New post 18 Jun 2006, 06:38
I totally forgot relative speed, so..

cyclist = 20-4=16
conversion = 16/60 = 4/15
Hiker = 4/60 = 1/15

5*4/15 = 4/3 miles

4/3 = 1/15t
t = 20

THANKS!
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 [#permalink] New post 18 Jun 2006, 08:38
My thinking is:

a) Find out how far the cyclist went in 5 minutes,

d = st
d = (u-v)*(5/60) = (20 -4 )*(5/60) 16*5/60 = 4/3

b) next, how long will it take for A to go that distance
d = st
(4/3) = (4)*t/60 => 5 = (4*60)/(3*4) = 20

I think as long as we remember these:

d = (u - v)* t; if both parties are travelling in the same direction
d = (u + v)*t; if both in opposite directions

It should be simple after a couple of tries. I used to find these a little hard before, now I'm getting used to them.
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  [#permalink] 18 Jun 2006, 08:38
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