Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or,

Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5 kmph.

Detailed Explanation:
A _____________M_______________B
A = Bus Terminal.

B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well.

Distance Covered by Bus in 10 min = AB = 2060×102060×10

= 103103 km.

Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes.
Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph.

Relative speed = 20 + x

To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph

Speed of Express A = x/4kmph
Speed of Express B = 2x/7kmph =>(t=3.30 = 3(1/2) = 7/2)
Trains will meet T hours after 7 a.m
(x/4)(y+2) + (2x/7)y = x
(y+2)/4 + (2y/7) = 1
(7y+14+8y)/28 = 1
15y = 28-14
Y =(14/15)×60 = 56 min

The second car overtake the first car in x hours
Distance covered by the first car in x hours = Distance covered by the second car in x hours
10x = x/2[2a + (x-1)d]
10x = x/2[2*8 + (x-1)1/2]
x = 40 -31 = 9

First person speed = 25 m/s * 18/5 = 90 kmph
Second person speed = 35 m/s * 18/5 = 126 kmph
First person covers 90 * 10 = 900km
900/450 = 2

d = 200 m, a = 16kmph = 40/9 m/s, b = 20kmph = 50/9 m/s
Required Distance D = d*(a/b-a)b= 200*(40/9/10/9) = 800m