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(D) Let the total distance be $D = 1 \, km$.Time taken for the first part $(t_1)$ $= \frac{1/5}{8} = \frac{1}{40} \, h$.Time taken for the second part

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$15$

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(D) Let the total distance be $D = 1 \, km$.Time taken for the first part $(t_1)$ $= \frac{1/5}{8} = \frac{1}{40} \, h$.Time taken for the second part $(t_2)$ $= \frac{1/10}{25} = \frac{1}{250} \, h$.Remaining distance $= 1 - (\frac{1}{5} + \frac{1}{10}) = 1 - \frac{3}{10} = \frac{7}{10} \, km$.Time taken for the remaining part $(t_3)$ $= \frac{7/10}{20} = \frac{7}{200} \, h$.Total time taken $(T)$ $= t_1 + t_2 + t_3 = \frac{1}{40} + \frac{1}{250} + \frac{7}{200}$.Finding a common denominator $(1000)$: $T = \frac{25}{1000} + \frac{4}{1000} + \frac{35}{1000} = \frac{64}{1000} = \frac{8}{125} \, h$.Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{1}{8/125} = \frac{125}{8} = 15.625 \, km/h$.

$A$ car covers $\frac{1}{5}$ of the distance from $A$ to $B$ at the speed of $8 \, km/h$,$\frac{1}{10}$ of the distance at $25 \, km/h$,and the remaining at the speed of $20 \, km/h$. Find the average speed of the whole journey in $km/h$. (in $.625$)

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