Physics

No need for calculations, to find the average of two numbers $((a*b)/(a+b))*2$ $((20*30)/(20+30))*2$ =24km/hr

$We\quad know\quad that\quad t\quad =\quad \frac { d }{ v } \\ let's\quad call{ { \quad t }_{ 1 } }\quad the\quad time\quad that\quad the\quad car\quad takes\quad \quad to\quad go\quad from\quad A\quad to\quad B\quad \\ then\quad \\ { _{ }{ t }_{ 1\quad } }=\quad \frac { d }{ 20 } \quad .\quad Similarly,\quad { _{ }{ t }_{ 2 } }\quad =\quad \frac { d }{ 30 } .\quad \\ let's\quad call\quad v\quad the\quad averge\quad velocity\quad .\quad Since\quad the\quad total\quad time\quad needs\quad to\\ be\quad the\quad same\quad then\quad :\\ \frac { d }{ 20 } \quad +\quad \frac { d }{ 30 } \quad =\quad \frac { 2d }{ v } \\ \\ So\quad :\quad \frac { 50d }{ 600 } =\quad \frac { 2d }{ v } \\ \\ v\quad =\quad \frac { 1200d }{ 50d } \\ finally\quad v\quad =\quad 24km/h\\ \\ \therefore \quad \boxed { v\quad =\quad 24\quad km/h }$

$V=\dfrac{d}{t}$ where $V=speed, d=distance, t=time$

$ave.~speed=\dfrac{total~distance}{total~time}$

$d_1=20t_1$ and $d_2=30t_2$

but $d_1=d_2$ ,

$20t_1=30t_2$ $\implies$ $t_1=\dfrac{3}{2}t_2$

Substituting, we have

$ave.~speed=\dfrac{20t_1+30t_2}{t_1+t_2}=\dfrac{20\left(\dfrac{3}{2}\right)t_2+30t_2}{\dfrac{3}{2}t_2+t_2}=24~kph$

Let: Distance $=x$

$1^{st}$ case:

Distance $=x$

Speed $=20km/hr$

Time $=\frac{x}{20}$

$2^{nd}$ case:

Distance $=x$

Speed $=30km/hr$

Time $=\frac{x}{30}$

Total Distance $=x+x=2x$

Total Time $=\frac{x}{20}+\frac{x}{30}=\frac{5x}{60}=\frac{x}{12}$

Average Speed $= \frac{2x}{\frac{x}{12}}$

$=2x\times\frac{12}{x}=2\times12=24km/hr$

Thus, the answer is: Average Speed $= \boxed{24km/hr}$

Formula ( 2 xy) / ( x + y)