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(C) Total distance $= 6 \, km$.Total time $= 45 \, minutes = \frac{45}{60} \, hr = 0.75 \, hr$.First half distance $= 3 \, km$.Time taken for first ha

Vedclass · Accepted Answer

$12$

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(C) Total distance $= 6 \, km$.Total time $= 45 \, minutes = \frac{45}{60} \, hr = 0.75 \, hr$.First half distance $= 3 \, km$.Time taken for first half $= \frac{2}{3} 	imes 45 \, minutes = 30 \, minutes = 0.5 \, hr$.Remaining distance $= 6 \, km - 3 \, km = 3 \, km$.Remaining time $= 45 \, minutes - 30 \, minutes = 15 \, minutes = \frac{15}{60} \, hr = 0.25 \, hr$.Required speed for remaining distance $= \frac{	ext{Remaining distance}}{	ext{Remaining time}} = \frac{3 \, km}{0.25 \, hr} = 12 \, km/hr$.

$A$ person has to cover a distance of $6 \, km$ in $45 \, minutes$. If he covers one-half of the distance in two-third of the total time,to cover the remaining distance in the remaining time,his speed (in $km/hr$) must be:

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