A boy goes to school from his home at a speed | vedclass

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(C) Let the distance between the home and the school be $d$ km.The time taken to go to school is $t_1 = \frac{d}{x}$ hours.The time taken to return ho

Vedclass · Accepted Answer

$H.M.$

Vedclass · Answer

(C) Let the distance between the home and the school be $d$ km.The time taken to go to school is $t_1 = \frac{d}{x}$ hours.The time taken to return home is $t_2 = \frac{d}{y}$ hours.Average speed is defined as the total distance divided by the total time taken.$	ext{Average speed} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{d + d}{t_1 + t_2} = \frac{2d}{\frac{d}{x} + \frac{d}{y}}$.Simplifying the expression: $\frac{2d}{d(\frac{1}{x} + \frac{1}{y})} = \frac{2}{\frac{x+y}{xy}} = \frac{2xy}{x+y}$.This expression $\frac{2xy}{x+y}$ is the Harmonic Mean $(H.M.)$ of $x$ and $y$.

$A$ boy goes to school from his home at a speed of $x$ km/hour and comes back at a speed of $y$ km/hour,then the average speed is given by

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