A person goes from point N to P and comes back. His average speed for the whole journey is 60 km/hr. If his speed while going from N to P is 40 km/hr, then what will be the speed of the person (in km/hr) while coming back from P to N?

Speed of the person from point N to P = 40 km/hr

Let speed on return journey = $$s$$ km/hr

Since, equal distances are covered, average speed = harmonic mean of both speeds = $$\frac{2xy}{x+y}$$

=> $$\frac{2\times s\times 40}{s+40}=60$$

=> $$\frac{80s}{s+40}=60$$

=> $$\frac{4s}{s+40}=3$$

=> $$4s=3s+120$$

=> $$4s-3s=s=120$$

$$\therefore$$ Speed of the person (in km/hr) while coming back from P to N = 120 km/hr

=> Ans - (C)