(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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After replacing an old member with a new member,it was found that the average age of five members of a club is the same as it was $3 \, \text{years}$ ago. What is the difference between the ages of the replaced member and the new member (in $\text{years}$)?

The average of $8$ numbers is $20$. The average of the first two numbers is $15 \frac{1}{2}$ and that of the next three is $21 \frac{1}{3}$. If the sixth number is less than the seventh and eighth numbers by $4$ and $7$ respectively,then the eighth number is:

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