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 :

Solution (By Examveda Team)

Let distance = x km and usual rate = y kmph. Then,
$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr & {\text{or,}}\,y\left( {y + 10} \right) = 10x\,.\,.\,.\,.\,.\,.\,.\,.\,\,\left( 1 \right) \cr} $$
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
$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr & {\text{or,}}\,y\left( {y + 20} \right) = \frac{{80x}}{7}\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
On dividing (1) by (2), we get y = 60
Substituting y = 60 in (1), we get :
x = 420 km.