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(B) Let the initial distance between $A$ and $B$ be $200 \,m$.$B$ walks $60 \,m$ along the road,then turns left and walks $20 \,m$,then turns right an

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let the initial distance between $A$ and $B$ be $200 \,m$.$B$ walks $60 \,m$ along the road,then turns left and walks $20 \,m$,then turns right and walks $40 \,m$,and finally turns right to return to the original road.Since $B$ turns right after walking $40 \,m$ and returns to the road,the distance covered by $B$ along the original road is $60 \,m + 40 \,m = 100 \,m$.Since $A$ and $B$ walk at the same speed,$A$ also covers a total distance of $140 \,m$ (the total path length of $B$ is $60 + 20 + 40 + 20 = 140 \,m$).$A$ moves $140 \,m$ towards $B$ along the original road.Initially,$A$ and $B$ were $200 \,m$ apart. After $A$ moves $140 \,m$ and $B$ effectively moves $100 \,m$ along the road,the remaining distance between them is $200 - (140 + 100) = -40 \,m$. This implies they have crossed each other.The distance between them is $|200 - (140 + 100)| = 40 \,m$.

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