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== | ||
Latest revision as of 19:33, 12 January 2025
Contents
[hide]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 | ||
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