Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1986 AJHSME Problems/Problem 18: Difference between revisions

← Older editNewer edit →
Ryan 1235 (talk | contribs)
editor
6 edits
Forest3g (talk | contribs)
editor
196 edits
Line 14: Line 14:
==Solution==
==Solution==


The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.  
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.
 
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.
 
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 
 
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_18&oldid=176876"