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# A railroad worker finds himself trapped inside a tunnel at a point

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Intern
Joined: 01 Jan 2016
Posts: 26
A railroad worker finds himself trapped inside a tunnel at a point  [#permalink]

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Updated on: 12 Sep 2017, 19:16
2
17
00:00

Difficulty:

95% (hard)

Question Stats:

30% (03:12) correct 70% (03:11) wrong based on 90 sessions

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A railroad worker finds himself trapped inside a tunnel at a point that is 25% of the length of the tunnel with a train approaching towards the near end with constant speed. He calculates that if he runs towards the near end doubling his speed then he would just manage to save himself. If on the other hand, he reduced his speed by 25% and ran towards the far end of the tunnel, then he would also just manage to avert danger. What is the the ratio of the speed of the train to the speed of the worker?
(1) 8 : 7
(2) 7 : 8
(3) 3 : 1
(4) 2 : 1
(5) None of these

Originally posted by cptholt on 12 Sep 2017, 10:51.
Last edited by cptholt on 12 Sep 2017, 19:16, edited 1 time in total.
Intern
Joined: 29 Dec 2016
Posts: 1
Re: A railroad worker finds himself trapped inside a tunnel at a point  [#permalink]

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12 Sep 2017, 11:31
2
Let length of tunnel be - x
Let distance of train from nearest end be - y
Let speed of man be - m
Let speed of train be - t

Condition 1 : when the man runs to the nearest end of tunnel
To be just in time we can write the equation as

y/t = [25x]/[100*(2m)]
or y = xt/8m ..........eq (1)

Condition 2 : When the man runs to the further end of the tunnel
To be just in time we can write the equation as

(y+x)/t = [75x/100]/[75m/100]

On simplifing we get

y = (xt/m) - x .........(eq 2)

equating eq 1 and eq 2 we get

xt/8m = (xt/m) - x
simpliying we get

t/m = 8/7

VP
Joined: 07 Dec 2014
Posts: 1222
Re: A railroad worker finds himself trapped inside a tunnel at a point  [#permalink]

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12 Sep 2017, 11:33
theperfectgentleman wrote:
A railroad worker finds himself trapped inside a tunnel at a point that is 25% of the length of the tunnel with a train approaching towards the near end with constant speed. He calculates that if he runs towards the near end doubling his speed then he would just manage to save himself. If on the other hand, he reduced his speed by 25% and ran
towards the far end of the tunnel, then he would also just manage to avert danger. What is the the ratio of the speed of the train to the speed of the worker?
(1) 8 : 7
(2) 7 : 8
(3) 3 : 1
(4) 2 : 1
(5) None of these

rt=rate of train
rw=rate of worker
d=distance from engine to tunnel
t=distance of tunnel
equation1: d/rt=(t/4)/2rw➡
rt/rw=8d/t
equation2: (d+t)/rt=(3t/4)/(3rw/4)➡
rt/rw=t+d/t
8d/t=(t+d)/t➡
8d=t+d➡
t=7d
rt/rw=t+d/t=8d/7d=8:7
(1)
Intern
Joined: 24 Jun 2018
Posts: 35
Re: A railroad worker finds himself trapped inside a tunnel at a point  [#permalink]

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21 Feb 2019, 17:29
3
1
The question isn't very difficult. Visualisation is key here.

Take a moment to the understand the figure below.
Attachment:

Screenshot 2019-02-22 at 05.37.45.png [ 24.41 KiB | Viewed 775 times ]

Now we know p is stranded at 25% of the near end.
Let the distance of the entire tunnel be x.
We may divide the tunnel into two parts; x/4 and 3x/4

Also, the distance between the train and tunnel is unknown. Lets call it d.

Let the speed of the person be s
Let the speed of the train be a.

Question is: Find the ratio a/s

Case I
The person runs towards the near end with twice the speed ie. 2s
The train is also approaching the near end.
Both reach at exact same time.
Hence time is common and can be equated.

Time taken by train = d/a
Time taken by person = (x/4)/2s= x/8s

d/a = x/8s

a/s=8d/x -----> Eqn (1)

Case II

The person runs towards the farther end
He needs to cover the distance 3x/4
He also runs at 75% speed ie. 3s/4
Time taken by person = (3x/4)/(3s/4) = x/s {time=distance/speed}

The train is also approaching the farther end.
The train needs to cover the distance d and then the distance x ie. d + x
The speed of the train is again a.
Time taken by train = (d+x)/a

We can again equate the two time periods since they are equal.

(d+x)/a=x/s

a/s=(d+x)/x

a/s = d/x + 1 -----> Eqn (2)

-----------------------------------------------------

Substituting 1 in 2 we get

8d/x=d/x +1

7d/x=1

x=7d

Using the value of x found in eqn (1)
a/s=8d/7d

a/s=8/7
Re: A railroad worker finds himself trapped inside a tunnel at a point   [#permalink] 21 Feb 2019, 17:29
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