Difference between revisions of "1995 AJHSME Problems/Problem 11"
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<math>\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E</math> | <math>\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Counting around, when Jane walks <math>12</math> steps, she will be at <math>D</math>. When Hector walks <math>6</math> steps, he will also be at <math>D</math>. Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is <math>\boxed{D}</math>. | Counting around, when Jane walks <math>12</math> steps, she will be at <math>D</math>. When Hector walks <math>6</math> steps, he will also be at <math>D</math>. Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We count to find that the perimeter of the rectangle is 18. We divide by 3 to see how far Hector has walked when he meets Jane, this distance is 6. We t(en move 6 steps counterclockwise to find the point they met or <math>\boxed{D}</math>. | ||
==See Also== | ==See Also== | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:33, 12 January 2025
Contents
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 1
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 
.
Solution 2
We count to find that the perimeter of the rectangle is 18. We divide by 3 to see how far Hector has walked when he meets Jane, this distance is 6. We t(en move 6 steps counterclockwise to find the point they met or .
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
