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

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)

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

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?

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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