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Difference between revisions of "1986 AJHSME Problems/Problem 18"

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==Solution==
 
==Solution==
  
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The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.  
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On one side is 0, 12, 24,36, and the other side is the same. The top side is 12,24,36,48. There is no 0 or 60 because of the other two fences. 4*3=12. So B.
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Each of the sides of length 36 contributes <math>\frac{36}{12}=3</math> fence posts and the side of length 60 contributes <math>\frac{60}{12}=5</math> fence posts, so there are <math>2(5+3)=16</math> fence posts.
 
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Since we have the wall, we must cancel a side. This will be a 3 foot fenced side to limit loss. We have now a total of 9 
 
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However, the two corners where a 36-foot fence meets a 60-foot fence are not counted, so there are actually <math>9+2=11</math> fence posts.
 
  
 
<math>\boxed{\text{A}}</math>
 
<math>\boxed{\text{A}}</math>

Revision as of 22:05, 12 August 2022

Problem

A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60$ m?

[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution

On one side is 0, 12, 24,36, and the other side is the same. The top side is 12,24,36,48. There is no 0 or 60 because of the other two fences. 4*3=12. So B.

$\boxed{\text{A}}$

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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