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Question

(B) Let the distance between $A$ and $B$ be $d 	ext{ km}$.Time taken for the onward journey from $A$ to $B$ is $t_1 = \frac{d}{10} 	ext{ hours}$.Tim

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let the distance between $A$ and $B$ be $d 	ext{ km}$.Time taken for the onward journey from $A$ to $B$ is $t_1 = \frac{d}{10} 	ext{ hours}$.Time taken for the return journey from $B$ to $A$ is $t_2 = \frac{d}{8} 	ext{ hours}$.The total time taken is $t_1 + t_2 = 4 \frac{1}{2} = \frac{9}{2} 	ext{ hours}$.So,$\frac{d}{10} + \frac{d}{8} = \frac{9}{2}$.Taking the least common multiple of $10$ and $8$,which is $40$,we get $\frac{4d + 5d}{40} = \frac{9}{2}$.$\frac{9d}{40} = \frac{9}{2}$.$d = \frac{9}{2} 	imes \frac{40}{9} = 20 	ext{ km}$.The total distance covered during the entire journey is the sum of the onward and return distances,which is $d + d = 2d$.Total distance $= 2 	imes 20 = 40 	ext{ km}$.

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