Question

A car travels from place A to place B at $20$km/hour and returns at $30$km/hour. The average speed of the car the whole journey is

Answer 1

Hint: Concept of average speed is to be used. Average speed is the ratio of total distance covered to total time taken.
Average speed, ${v_{av}} = \dfrac{{{s_1} + {s_2}}}{{{t_1} + {t_2}}}$
Where in time ${t_1}$, speed is ${s_1}$
For time ${t_2}$, speed is ${s_2}$

Complete step by step answer:
When the car travels from place A to B,
speed $ = \;20$km/hour
Let the distance between place A and B be s km.
So, for journey from place A to place B,
distance, ${s_1} = s\;km$
speed, ${v_1} = 20$km/hour
time taken, ${t_1} = \dfrac{{dis\tan ce}}{{speed}}$
\[
   \Rightarrow \;\;time\;\;taken,\;\;{t_1}\; = \dfrac{{{s_1}}}{{{v_1}}} \\
   \Rightarrow \;\;{t_1} = \dfrac{s}{{20}}\;hours\;\;\;\;\;\;\;\;\;\;\;\;\;....(1) \\
 \]
Now, for the journey from place B to place A i.e. during return distance will be the same.
So, distance, ${s_2} = s\;km$
speed, ${v_2} = 30$ km/hr
time taken, ${t_2} = \dfrac{{{s_2}}}{{{v_2}}}\; \Rightarrow \;{t_2} = \dfrac{s}{{30}}$ hour
Now, average speed = $
   = \dfrac{{Total\;\;dis\tan ce}}{{Total\;\;time}} \\
  i.e.\;\;{v_{au}} = \dfrac{{{s_1} + {s_2}}}{{{t_1} + {t_2}}} \\
 $
Putting the values, we get
${v_{av}} = \dfrac{{s + s}}{{\dfrac{s}{{20}} + \dfrac{s}{{30}}}}$
$\Rightarrow {v_{av}} = \dfrac{{2s}}{{\dfrac{{30s + 20s}}{{20 \times 30}}}}$
$
  \Rightarrow {v_{av}} = 2s \times \dfrac{{600}}{{50s}} \\
  \Rightarrow {v_{av}} = 2 \times 12 \\
  \Rightarrow {v_{av}} = 24\;\;km/hour \\
 $
So, the average speed of the car through the whole journey is $24\;km/hour.$

Note:
Also when the distances are same, the average speed is given by
${v_{av}} = \dfrac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}$
as ${v_1} = 20\;\;km/hour\;\;\;\;\;\;\;{v_2} = 30\;km/hour$
So, \[
  {v_{av}} = \dfrac{{2 \times 20 \times 30}}{{20 + 30}} \\
   = \dfrac{{1200}}{{50}} \\
   = 24\;km/hr \\
 \]