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A man goes to market with a speed of $ 30 \mathrm{~km} / \mathrm{h} $ and returns back with thespeed of $ 40 \mathrm{~km} / \mathrm{h} $. Find his average speed.
Given:
Speed, $S_1$ = 30 km/h (while going to the market)
Speed, $S_2$ = 40 km/h (while returning from the market)
To find: Average speed, $S$.
Solution:
Let $t_1$ is the time taken by the man while going to the market, and it is given as-
$t_1=\frac {d}{30}$ $[\because t=\frac {D}{S}]$
and $t_2$ is the time taken by the man while returning from the market, and it is given as-
$t_2=\frac {d}{40}$ $[\because t=\frac {D}{S}]$
Let $d$ is the distance from the initial position of the man and the market.
Thus,
The total distance travelled, $D=d+d=2d$ $(\because the\ distance\ is\ same\ while\ going\ to\ the\ market\ and\ returning\ from\ the\ market)$
Now, we know that average speed is the ratio of total distance travelled and the total time taken.
It is given as-
$Average\ Speed\ (S)=\frac{Total\ distance\ \ travelled\ (d)}{Total\ time\ \ taken\ (t)}$
$S=\frac {2d}{\frac {d}{30}+\frac {d}{40}}$
$S=\frac {2d}{\frac {4d+3d}{120}}$
$S=\frac {2d}{\frac {7d}{120}}$
$S=\frac {2d\times {120}}{7d}$
$S=\frac {240d}{7d}$
$S=34.28km/h$
Therefore, the average speed, $S$ is 34.28km/h.