A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have 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/hr. The distance between two cities is :

Answer: Option B

Let distance = x km and usual rate = y kmph. Then,
$$\begin{array}{rl}& \frac{x}{y}-\frac{x}{y+10}=1\\ & \text{or,}y\left(y+10\right)=10x........\left(1\right)\end{array}$$
Now, in the 2^nd scenario with a further increase in speed the driver could have saved another 45 min = $$\frac{3}{4}$$ hrs.
Therefore, total time saved
$$=1+\frac{3}{4}=\frac{7}{4}\text{hrs}\text{.}$$
Putting it in equation, we get
$$\begin{array}{rl}& \frac{x}{y}-\frac{x}{y+20}=\frac{7}{4}\\ & \text{or,}y\left(y+20\right)=\frac{80x}{7}.........\left(2\right)\end{array}$$
On dividing (1) by (2), we get y = 60
Substituting y = 60 in (1), we get :
x = 420 km.