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(D) Let the total distance of the journey be $D\, km$.Time taken to cover the first half of the distance $(D/2)$ at $50\, km/h$ is $t_1 = \frac{D/2}{5

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) Let the total distance of the journey be $D\, km$.Time taken to cover the first half of the distance $(D/2)$ at $50\, km/h$ is $t_1 = \frac{D/2}{50} = \frac{D}{100}\, hours$.Time taken to cover the remaining half of the distance $(D/2)$ at $70\, km/h$ is $t_2 = \frac{D/2}{70} = \frac{D}{140}\, hours$.The total time taken is $t_1 + t_2 = 6\, hours$.So,$\frac{D}{100} + \frac{D}{140} = 6$.Taking the least common multiple of $100$ and $140$,which is $700$:$\frac{7D + 5D}{700} = 6$$\frac{12D}{700} = 6$$12D = 4200$$D = \frac{4200}{12} = 350\, km$.Wait,re-evaluating the formula provided in the prompt: The average speed for equal distances is $v_{avg} = \frac{2s_1s_2}{s_1+s_2}$.Total distance $D = v_{avg} 	imes T = \left(\frac{2 	imes 50 	imes 70}{50+70}ight) 	imes 6 = \left(\frac{7000}{120}ight) 	imes 6 = \frac{7000}{20} = 350\, km$.Since $350$ is not among the options,the correct answer is $D$ (none of these).

$A$ car completes a journey in $6\, hours$. It covers half the distance at $50\, km/h$ and the rest at $70\, km/h$. The length of the journey is (in $km$):

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