Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?

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, $$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph.

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 = $$\frac{{20}}{{60}} \times 10$$   = $$\frac{{10}}{3}$$ 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.