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

Question

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

sahibr · Accepted Answer

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 4537 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

meghdutgorai · Answer

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 =  / 
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 =  /

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

Hence answer is A

gracie · Answer

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)

effatara · Answer

Assumptions:
Length of tunnel=80 meters {so the railway worker (RW) is 20m from the near end of the tunnel (NE) and 60m from the far end (FE)}
Speed of the train (T) = 'x' m/s
Speed of RW = 'y' m/s
Distance of T from NE = 'd' meters

(a) RW and T take the same amount of time to reach NE (RW running at twice his usual speed):
x/2y = d/20.....> x/y=d/10..........(i)
(b) RW and T take the same time to reach FE (RW running at 3/4 his usual speed):
x/y*(3/4)=(d+80)/60.....> 4x/3y = (d+80)/60.....> x/y=3(d+80)/4*60........(ii)

From Eqns. (i) and (ii), we get: x/y=d/10=3(d+80)/240.....> d=80/7
Substituting this value in (i), we get x/y=(80/7)/10=8/7. ANS: A

sujoykrdatta · Answer

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

Answer: 
Since we are dealing with a numeric 25%, for both the tunnel length and speed of the worker, let's assume:

the tunnel as 4 units long => thus, he is 1 unit away from the near end

the worker's speed as 4n units (note: we could have simply taken 4 as the speed as well since we only want the speed ratio)

Let the train be at a distance of D units from the near end

# When the man runs to near end with double the speed:
The man covers distance 1 at double his speed in the same time when the train covers distance D ...(i)

# When the man runs to far end with 75% speed:
The man covers distance 3 at 75% his speed in the same time when the train covers distance (D+4) ... (ii)

From (i):

The man covers distance  1  at double his speed in the same time when the train covers distance  D

=> The man covers distance  3  at double his speed in the same time when the train covers distance  3D

=> The man covers distance  3  at his  normal speed  in the same time when the train covers distance  3D*2  ... since the man will now take double the time -- note: time is inversely proportional to speed

=> The man covers distance  3  at  75% his speed  in the same time when the train covers distance  3D*2*4/3 = 8D

Thus, from (ii):

=> D+4 = 8D
=> D = 4/7

Thus, again from (i):

The man covers distance 1 at double his speed in the same time when the train covers distance D = 4/7

=> The man covers distance 1 at his normal speed in the same time when the train covers distance 2D = 8/7

=> Speed ratio of train and worker
= Ratio of distances they cover in unit time (since D is directly proportional to speed)
= 8/7 : 1
 = 8 : 7

Answer A

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

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