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(C) Let the initial positions of $X$ and $Y$ be $P$ and $Q$ respectively,with $PQ = 200 \, m$. $Y$ walks $60 \, m$ along the path,then turns left $(20

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let the initial positions of $X$ and $Y$ be $P$ and $Q$ respectively,with $PQ = 200 \, m$. $Y$ walks $60 \, m$ along the path,then turns left $(20 \, m)$,then right $(40 \, m)$,and then right again $(20 \, m)$ to return to the original road. The total horizontal distance covered by $Y$ is $60 + 40 = 100 \, m$. The total distance covered by $Y$ is $60 + 20 + 40 + 20 = 140 \, m$. Since $X$ and $Y$ walk at the same speed,$X$ also covers a total distance of $140 \, m$ in the same time. $Y$ is now at a point $100 \, m$ away from $Q$ towards $P$. $X$ is at a point $140 \, m$ away from $P$ towards $Q$. The distance between $X$ and $Y$ is $200 - (100 + 140)$ is not possible as they cross each other. Actually,$X$ is at $140 \, m$ from $P$ and $Y$ is at $100 \, m$ from $Q$ (which is $200 - 100 = 100 \, m$ from $P$). Distance between $X$ and $Y = |140 - 100| = 40 \, m$.

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