Difference between revisions of "1986 AJHSME Problems/Problem 18"
(→Problem) |
(→Solution) |
||
| Line 16: | Line 16: | ||
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. | 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. | ||
| − | <math>\boxed{\text{ | + | <math>\boxed{\text{B}}</math> |
==See Also== | ==See Also== | ||
Revision as of 23:08, 12 August 2022
Problem
A rectangular grazing area is to be fenced off on three sides using part of a 
meter rock wall as the fourth side. Fence posts are to be placed every 
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 
m by 
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]](http://latex.artofproblemsolving.com/0/3/c/03c15a06354898452ca990f3cafbdccb11317cbe.png)
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.
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. 
