Difference between revisions of "2016 AMC 10B Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
| − | If the dimensions are <math> | + | If the dimensions are <math>4a\times 4b</math>, then one side will have <math>a+1</math> posts (including corners) and the other <math>b+1</math> (including corners). |
The total number of posts is <math>2(a+b)=20</math>. | The total number of posts is <math>2(a+b)=20</math>. | ||
| + | |||
| + | This diagram represents the number of posts around the garden. | ||
<asy>size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); | <asy>size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); | ||
for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); | for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); | ||
| Line 15: | Line 17: | ||
draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); | draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); | ||
draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); | draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); | ||
| − | label("$a$",(0,-1)--(5,-1),S);label("$b$",(6,0)--(6,7),E); | + | label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E); |
label("$a$",(1,1));label("$b$",(4,5)); | label("$a$",(1,1));label("$b$",(4,5)); | ||
</asy> | </asy> | ||
| − | + | We solve the system <math>\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}</math> to get <math>a=3,\ b=7</math>. Then the area is <math>4a\cdot 4b=336</math> which is <math>\boxed{\textbf{(B) } 336}</math>. | |
| + | |||
| + | ~Edits by BakedPotato66 | ||
==Solution 2== | ==Solution 2== | ||
| − | To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only <math>16</math> posts. Now there are twice as many posts on the longer side then the shorter side. From this we can see that we can put <math>8</math> posts on the longer side and <math>4</math> posts on the shorter side. On the shorter side, we have <math>3</math> spaces between the <math>4</math> posts. On the longer side, we have 7 spaces between the 8 fence posts. There are <math>4</math> yards between each post. Therefore, the answer is <math>12 | + | To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only <math>16</math> posts. Now there are twice as many posts on the longer side then the shorter side. From this, we can see that we can put <math>8</math> posts on the longer side and <math>4</math> posts on the shorter side. On the shorter side, we have <math>3</math> spaces between the <math>4</math> posts. On the longer side, we have <math>7</math> spaces between the <math>8</math> fence posts. There are <math>4</math> yards between each post. Therefore, the answer is <math>12\times28=\boxed{\textbf{(B)}\ 336}</math>. |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/yVAQ-bt7s-U | ||
| − | + | ~savannahsolver | |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2016|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:34, 15 April 2022
Problem
Carl decided to fence in his rectangular garden. He bought 
fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 
yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
Solution 1
If the dimensions are 
, then one side will have 
posts (including corners) and the other 
(including corners).
The total number of posts is 
.
This diagram represents the number of posts around the garden.
![[asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E); label("$a$",(1,1));label("$b$",(4,5)); [/asy]](http://latex.artofproblemsolving.com/f/b/3/fb3330e0e1124978ddeb7b2f002fd64388a1cc5a.png)
We solve the system 
to get 
. Then the area is 
which is 
.
~Edits by BakedPotato66
Solution 2
To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only 
posts. Now there are twice as many posts on the longer side then the shorter side. From this, we can see that we can put 
posts on the longer side and posts on the shorter side. On the shorter side, we have 
spaces between the posts. On the longer side, we have 
spaces between the fence posts. There are yards between each post. Therefore, the answer is 
.
Video Solution
~savannahsolver
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
