A car covers the first-half of the distance between two places at $$40\,km/h$$   and other half at $$60\,km/h.$$   The average speed of the car is

Solution :
Let the distance between two places be $$d$$ and $${t_1}$$ is time taken by car to travel first-half length, $${t_2}$$ is time taken by car to travel second-half length. Time taken by car to travel first-half length,
$${t_1} = \frac{{\left( {\frac{d}{2}} \right)}}{{40}} = \frac{d}{{80}}$$
Time taken by car to travel second-half length,
$$\eqalign{ & {t_2} = \frac{{\left( {\frac{d}{2}} \right)}}{{60}} = \frac{d}{{120}} \cr & \therefore {\text{Total time}} = {t_1} + {t_2} \cr & = \frac{d}{{80}} + \frac{d}{{120}} \cr & = d\left( {\frac{1}{{80}} + \frac{1}{{120}}} \right) \cr & = \frac{d}{{48}} \cr & \therefore {\text{Average}}\,{\text{speed}} = \frac{d}{{{t_1} + {t_2}}} \cr & = \frac{d}{{\left( {\frac{d}{{48}}} \right)}} \cr & = 48\,km/h \cr} $$
Alternative
$$\eqalign{ & {v_{av}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}} \cr & = \frac{{2 \times 40 \times 60}}{{40 + 60}} \cr & = 48\,km/h \cr} $$