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

Hence answer is A

The question isn't very difficult. Visualisation is key here.

Take a moment to the understand the figure below.


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)


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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