A train takes 3 hours less time for a journey of 360 km, if its speed is increased by 10 km/hr from its usual speed. What is its increased speed (in km/hr)?

Let usual speed = $$x$$ km/hr and new speed = $$(x+10)$$ km/hr

Usual time taken = $$\frac{360}{x}$$ and new time = $$\frac{360}{x+10}$$ -------------(i)

Also, new time = $$\frac{360}{x}-3$$ ------------(ii)

Comparing equations (i) and (ii), we get :

=> $$\frac{360}{x}-3=\frac{360}{x+10}$$

=> $$\frac{360-3x}{x}=\frac{360}{x+10}$$

=> $$360x+3600-3x^2-30x=360x$$

=> $$3x^2+30x-3600=0$$

=> $$x^2+10x-1200=0$$

=> $$x^2+40x-30x-1200=0$$

=> $$(x+40)(x-30)=0$$

=> $$x=-40,30$$

$$\because$$ Speed cannot be negative, => $$x\neq-40$$

$$\therefore$$ Increase speed = $$(x+10)=30+10=40$$ km/hr

=> Ans - (A)