# 1995 AJHSME Problems/Problem 11

## Problem

Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label("A",(-3,0),SW); dot((-3,3)); label("B",(-3,3),NW); dot((0,3)); label("C",(0,3),N); dot((3,3)); label("D",(3,3),NE); dot((3,0)); label("E",(3,0),SE); dot((0,0)); label("start",(0,0),S); label("\longrightarrow",(0,-0.75),E); label("\longleftarrow",(0,-0.75),W); label("\textbf{Jane}",(0,-1.25),W); label("\textbf{Hector}",(0,-1.25),E); [/asy]$ $\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E$

## Solution

Counting around, when Jane walks $12$ steps, she will be at $D$. When Hector walks $6$ steps, he will also be at $D$. Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is $\boxed{D}$.

## See Also

 1995 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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