(C) Let the distance between the two places be $x$.Total distance covered $= x + x = 2x$.Time taken to go $= \frac{x}{u}$.Time taken to return $= \fra

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$\frac{1}{\frac{1}{2}(\frac{1}{u}+\frac{1}{v})}$

(C) Let the distance between the two places be $x$.Total distance covered $= x + x = 2x$.Time taken to go $= \frac{x}{u}$.Time taken to return $= \frac{x}{v}$.Total time taken $= \frac{x}{u} + \frac{x}{v} = x(\frac{1}{u} + \frac{1}{v})$.Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x(\frac{1}{u} + \frac{1}{v})} = \frac{2}{\frac{1}{u} + \frac{1}{v}}$.This can be rewritten as $\frac{1}{\frac{1}{2}(\frac{1}{u} + \frac{1}{v})}$.

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$A$ cyclist goes from one place to another and returns by the same route. He pedals his way uniformly with speed $u$ while going and with speed $v$ while returning. The average speed of his journey is

A
$\frac{u+v}{2}$
B
$\frac{u-v}{2}$
C
$\frac{1}{\frac{1}{2}(\frac{1}{u}+\frac{1}{v})}$
D
$\frac{1}{\frac{1}{2}(\frac{1}{u}-\frac{1}{v})}$

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$A$ cyclist goes from one place to another and returns by the same route. He pedals his way uniformly with speed $u$ while going and with speed $v$ while returning. The average speed of his journey is

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The average age of a committee of $8$ members is $40 \text{ years}$. $A$ member,aged $55 \text{ years}$,retired and was replaced by a member aged $39 \text{ years}$. The average age of the present committee is (in $\text{years}$):

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