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Question

(B) Let the total distance be $d \, km$.The journey is divided into three parts:$1$. First part: Distance $= \frac{d}{3}$,Speed $= 25 \, km/hr$. Time

Vedclass · Accepted Answer

$33\frac{1}{3}$

Vedclass · Answer

(B) Let the total distance be $d \, km$.The journey is divided into three parts:$1$. First part: Distance $= \frac{d}{3}$,Speed $= 25 \, km/hr$. Time $t_1 = \frac{d/3}{25} = \frac{d}{75} \, hr$.$2$. Second part: Distance $= \frac{d}{4}$,Speed $= 30 \, km/hr$. Time $t_2 = \frac{d/4}{30} = \frac{d}{120} \, hr$.$3$. Third part: Distance $= d - (\frac{d}{3} + \frac{d}{4}) = d - \frac{7d}{12} = \frac{5d}{12}$,Speed $= 50 \, km/hr$. Time $t_3 = \frac{5d/12}{50} = \frac{d}{120} \, hr$.Total time $T = t_1 + t_2 + t_3 = \frac{d}{75} + \frac{d}{120} + \frac{d}{120} = \frac{d}{75} + \frac{2d}{120} = \frac{d}{75} + \frac{d}{60}$.Taking $LCM$ of $75$ and $60$ as $300$,$T = \frac{4d + 5d}{300} = \frac{9d}{300} = \frac{3d}{100} \, hr$.Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{d}{3d/100} = \frac{100}{3} = 33\frac{1}{3} \, km/hr$.

One third of a certain journey is covered at the rate of $25 \, km/hr$,one-fourth at the rate of $30 \, km/hr$,and the rest at $50 \, km/hr$. The average speed for the whole journey is (in $km/hr$):

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