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

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==Solution==
 
==Solution==
  
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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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Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a <math>36</math>m by <math>60</math>m rectangle, so the <math>60</math>m side should be parallel to the wall. That means the two fences perpendicular to the wall are <math>36</math>m. We can start by counting <math>60\div12+1</math> on the <math>60</math>m fence (since we also count the <math>0</math>m post). Next, we have the two <math>36</math>m fences. There are a total of <math>36\div12+1-1</math> fences on that side since the <math>0</math>m and <math>60</math>m fence posts are also part of the <math>36</math>m fences. So we have <math>6+2\cdot3=12</math> minimum fence posts needed to box a <math>36</math>m by <math>60</math>m grazing area, or answer <math>\boxed{\text{B}}</math>.
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<math>\boxed{\text{B}}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 22:24, 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]

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

Solution

Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$m by $60$m rectangle, so the $60$m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$m. We can start by counting $60\div12+1$ on the $60$m fence (since we also count the $0$m post). Next, we have the two $36$m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$m and $60$m fence posts are also part of the $36$m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$m by $60$m grazing area, or answer $\boxed{\text{B}}$.

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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