A person has to cover a distance of 60 km. If he increases his speed by 3 km/hr he reaches 1 hr early. Find his initial speed (in km/hr).

16 15 12 10

12

Let the initial speed = x km/hr ATQ, \(\frac{60}{x}-\frac{60}{x+3}=1\) let x = 12 (by using option) \(\frac{60}{12}-\frac{60 }{150}=1\) 5 – 4 = 1 (satisfied)