### 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?

A. 3 km/h B. 4 km/h C. 5 km/h D. 7 km/h Answer: Option C
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. $\begin{array}{l} \text { Or, } 20 \times \frac{10}{60}=\frac{8}{60} \times(20+x) \\ \text { Or, } 200=160+8 x \\ \text { Or, } 8 x=40 \\ \text { Hence, } x=5 \mathrm{kmph} \end{array}$
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 $=\frac{20}{60} \times 10=\frac{10}{3} \mathrm{~km}$ 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.