1986 AJHSME Problems/Problem 18

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

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

Each of the sides of length 36 contribute $\frac{36}{12}+1=4$ fence posts and the side of length 60 contributes $\frac{60}{12}+1=6$ fence posts, so there are $4+4+6=14$ fence posts.

However, the two corners where a 36 foot fence meets an 60 foot fence are counted twice, so there are actually $14-2=12$ fence posts.

$\boxed{\text{B}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS