Difference between revisions of "2016 AMC 10B Problems/Problem 11"
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<math>\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512</math> | <math>\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512</math> | ||
− | ==Solution== | + | ==Solution 1== |
If the dimensions are <math>a\times b</math>, then one side will have <math>a+1</math> posts (including corners) and the other <math>b+1</math>. | If the dimensions are <math>a\times b</math>, then one side will have <math>a+1</math> posts (including corners) and the other <math>b+1</math>. | ||
Line 23: | Line 23: | ||
==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=336</math> |
− | < | + | <cmath>\boxed{\textbf{(B)}\ 336}</cmath>. |
+ | |||
+ | ==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 03:15, 2 November 2020
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 .
The total number of posts is .
Solve the system to get . Then the area is which is .
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 |
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