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(D) Let the total distance between $P$ and $Q$ be $D \, km$.The time taken to travel the first $\frac{1}{4}$ of the distance at $10 \, km/h$ is $t_1 =

Vedclass · Accepted Answer

$80$

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(D) Let the total distance between $P$ and $Q$ be $D \, km$.The time taken to travel the first $\frac{1}{4}$ of the distance at $10 \, km/h$ is $t_1 = \frac{D/4}{10} = \frac{D}{40} \, hours$.The remaining distance is $D - \frac{D}{4} = \frac{3D}{4} \, km$.The time taken to travel the remaining distance at $12 \, km/h$ is $t_2 = \frac{3D/4}{12} = \frac{3D}{48} = \frac{D}{16} \, hours$.The total time taken is $t_1 + t_2 = 7 \, hours$.$\frac{D}{40} + \frac{D}{16} = 7$.Taking the least common multiple of $40$ and $16$,which is $80$:$\frac{2D + 5D}{80} = 7$.$\frac{7D}{80} = 7$.$D = 80 \, km$.

$A$ man starts from a place $P$ and reaches the place $Q$ in $7 \, hours$. He travels $\frac{1}{4}^{th}$ of the distance at $10 \, km/h$ and the remaining distance at $12 \, km/h$. The distance,in $km$,between $P$ and $Q$ is

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