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A hiker walking at a constant rate of 4 miles per hour was

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A hiker walking at a constant rate of 4 miles per hour was [#permalink] New post 28 Sep 2007, 15:07
A hiker walking at a constant rate of 4 miles per hour was passed by a cyclist traveling at a rate of 20 miles per hour on the same path and in the same direction. The cyclist stops to wait for the hiker 5 minutes after passing him, while the hiker continues to walk at his constant rate. How many minutes must the cyclist wait until the hiker catches up?

1) 6 2/3
2) 15
3) 20
4) 25
5) 26 2/3
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Re: Rate Problem [#permalink] New post 28 Sep 2007, 15:26
Witchiegrlie wrote:
A hiker walking at a constant rate of 4 miles per hour was passed by a cyclist traveling at a rate of 20 miles per hour on the same path and in the same direction. The cyclist stops to wait for the hiker 5 minutes after passing him, while the hiker continues to walk at his constant rate. How many minutes must the cyclist wait until the hiker catches up?

1) 6 2/3
2) 15
3) 20
4) 25
5) 26 2/3


D. 25 mins.-----CHANGED BELOW NEXT POST

Vh=4
Vb=20
t=5 mins after passing the hiker

D=20*5=100 mile passed by biker so hiker should walk 100 miles at Vb=4

t=100/4=25

Last edited by Ravshonbek on 28 Sep 2007, 15:38, edited 2 times in total.
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 [#permalink] New post 28 Sep 2007, 15:27
I will go with C.

The diff in speeds is 20-4=16mph
Every hour, cyclist would be 16 miles ahead of the hiker
In 5 mins, the cyclist would be (5/60)*16=4/3 miles ahead

Hiker will take = (4/3)/(4/60) = 20 mins to catch-up.
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 [#permalink] New post 28 Sep 2007, 15:35
subhen wrote:
I will go with C.

The diff in speeds is 20-4=16mph
Every hour, cyclist would be 16 miles ahead of the hiker
In 5 mins, the cyclist would be (5/60)*16=4/3 miles ahead

Hiker will take = (4/3)/(4/60) = 20 mins to catch-up.


wow. right
the hiker did not stop. sorry.
then Sh=5*4=20 so he as well travels 20miles
then Sb=5*20=100 biker travels 100 miles

then 100-20=80 miles has to be passed while biker waits
thus t=80/4=20

noch changed C .20
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Re: Rate Problem [#permalink] New post 28 Sep 2007, 15:39
Witchiegrlie wrote:
A hiker walking at a constant rate of 4 miles per hour was passed by a cyclist traveling at a rate of 20 miles per hour on the same path and in the same direction. The cyclist stops to wait for the hiker 5 minutes after passing him, while the hiker continues to walk at his constant rate. How many minutes must the cyclist wait until the hiker catches up?

1) 6 2/3
2) 15
3) 20
4) 25
5) 26 2/3


Speed of Hiker = 4 miles per hour = (1/15) miles per min
Speed of Cyclist = 20 miles per hour = (1/3) miles per min

Now Cyclist stops after 5 minutes ride.
That means he travelled for 5 minutes before he stopped.
So Distance covered by the cyclist = Speed * Time
= 1/3 * 5 = (5/3) miles

But Hiker has also walked for those 5 minutes so he must have covered the distance = (1/15) * 5 = (1/3)

Therefore, remaining distance to be covered by Hiker = 5/3 - 1/3 = 4/3

We need to find, how much time Hiker will take to cover 4/3 miles

Time = D/S = (4/3)/(1/15) = 20 minutes

And time taken by Hiker to travel (4/3) miles is equal to the time for which Cyclist should wait.

Hence C.

Good question.....

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 [#permalink] New post 28 Sep 2007, 16:43
Biker's relative speed = 16 m/h
Biker travelled in 5 min: 16 * 5/60 = 4/3 miles.
Time taken by hiker to cover 4/3 miles: 4/(4/3) = 1/3 hrs = 20 mins

Ans C.
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Re: Rate Problem [#permalink] New post 28 Sep 2007, 17:29
Witchiegrlie wrote:
A hiker walking at a constant rate of 4 miles per hour was passed by a cyclist traveling at a rate of 20 miles per hour on the same path and in the same direction. The cyclist stops to wait for the hiker 5 minutes after passing him, while the hiker continues to walk at his constant rate. How many minutes must the cyclist wait until the hiker catches up?

1) 6 2/3
2) 15
3) 20
4) 25
5) 26 2/3


C.

This question has been posted multiple times. Essentially the way to solve it is to remember that the hiker continues to run after being passed. You must account for this by subtracting the hikers distance traveled by the total distance traveled by the biker in 5min.

Then just figure out how long it takes the hiker to travel the difference.
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 [#permalink] New post 28 Sep 2007, 20:32
I had to re-read the question a few times but I think I got it.

The cyclist passed the hiker, continued to cycle for another 5 mintues and stopped to wait for the hiker. So, in that 5 minutes, the cyclist travelled 5/3 miles (see calc below) and the hiker travelled 1/3 miles (see calc below). So, at the end of that 5 minutes, the hiker has 4/3 miles remaining before reaching the cyclist.

So, in 5 minutes, the cyclist travelled c miles:
20 mi : 60 min = c : 5 minutes
c = 5/3 miles

And, in 5 minutes the hiker travelled c miles:
4 mi : 60 min = h : 5 minutes
h = 1/3 miles

The distance between the hiker and the cyclist:
4/3 mi = 5/3 - 1/3

Speed = Distance / Time
Time = Distance / Speed
Time = (4/3) / 4
Time = 1/3 hour or 20 minutes
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 [#permalink] New post 05 Oct 2007, 08:05
The problem is I had to read the question multiple times during the test and I couldnt get it bcos of time pressure. Any tips for that.
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 [#permalink] New post 05 Oct 2007, 09:53
lnaik wrote:
The problem is I had to read the question multiple times during the test and I couldnt get it bcos of time pressure. Any tips for that.


I have the same problem. My suggestion: Do a search on these forums and look for rate problems. Read them. Solve them. READ AND STUDY alternative methods of solving the problem. This is broaden your knowledge of different rate problems. The problem is that there are many ways of asking essentially the same question so going through several rate problems helps to understand and grasp what is being asked. Your memory of past problems will help to "understand" new problems faster.

On a side note, if you have the Kaplan 800 book I would recommend looking through the rate word problems in that book. It contains several different ways the GMAT can ask a rate question.
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 [#permalink] New post 25 Dec 2007, 20:28
This was the very last question on my practice test, I guessed it right at random, since i only had 3 seconds left.

my math score was 48, so this should be a 44+ difficulty.
  [#permalink] 25 Dec 2007, 20:28
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