A car travels 50% faster than a bike. Both start at the same time from A to B. The car reaches 25 minutes earlier than the bike. If the distance from A to B is 100 km, find the speed of the bike.

Explanation:-

P __________100km__________Q.
Let car takes time T hours to reach destination. So,
Bike will take (T+(²⁵/₆₀)).
Let speed of the bike = S kmph.
Speed of Car = S + 50% of S = ^3S/₂ kmph
For the both the case distance is constant. And when distance remain constant then time is inversely propositional to speed (As ST + D).
So, $$\eqalign{ & \frac{{\left( {\frac{{3S}}{2}} \right)}}{{\left( S \right)}} = \frac{{\left\{ {T + \left( {\frac{5}{{12}}} \right)} \right\}}}{T} \cr & 3T = 2T + \frac{{10}}{{12}} \cr & T = \frac{{10}}{{12}}{\text{hours}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}} \cr & \frac{{3S}}{2} = \frac{{100}}{{\left( {\frac{{10}}{{12}}} \right)}} \cr & \frac{{3S}}{2} = 120 \cr & S = 80\,kmph \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{bike}} \cr & = 80\,kmph \cr}$$