A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is

Let the distance between the two cities = $$d$$ km and the speed be $$s$$ km/hr

=> Time taken to travel = $$\frac{d}{s}$$ hr

Now, in condition 1, the car takes 1 hr lesser when speed is increased by 10 km/h

=> $$\frac{d}{s+10} = \frac{d}{s} - 1$$

=> $$\frac{d}{s} - \frac{d}{s+10} = 1$$

=> $$10d = s(s+10)$$ ---------Eqn(1)

and according to condition 2,

=> $$\frac{d}{s+20} = \frac{d}{s} - 1 - \frac{45}{60}$$

=> $$\frac{d}{s} - \frac{d}{s+20} = \frac{7}{4}$$

=> $$20d = \frac{7}{4} s(s+20)$$ ---------------Eqn(2)

Solving eqns (1) & (2), we get

$$s$$ = 60 km/hr and $$d$$ = 420 km