1995 AJHSME Problems/Problem 11


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]

$\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E$


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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