Difference between revisions of "1995 AJHSME Problems/Problem 11"
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:15, 4 July 2013
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]](http://latex.artofproblemsolving.com/1/d/7/1d722c9b7817a874ae13c9c5f3bee0f7bc2fa033.png)
Solution
Counting around, when Jane walks 
steps, she will be at 
. When Hector walks 
steps, he will also be at . Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is 
.
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
