Difference between revisions of "2016 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
+ | By Albert471 | ||
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To start, use algebra to determine the number of posts on each side. You have (the long sides count for <math>2</math> because there are twice as many) <math>6x = 20 + 4</math> (each corner is double counted so you must add <math>4</math>) Making the shorter end have <math>4</math>, and the longer end have <math>8</math>. <math>((8-1)*4)*((4-1)*4) = 28*12 = 336</math>. Therefore, the answer is <math>\boxed{\textbf{(B)}\ 336}</math> | To start, use algebra to determine the number of posts on each side. You have (the long sides count for <math>2</math> because there are twice as many) <math>6x = 20 + 4</math> (each corner is double counted so you must add <math>4</math>) Making the shorter end have <math>4</math>, and the longer end have <math>8</math>. <math>((8-1)*4)*((4-1)*4) = 28*12 = 336</math>. Therefore, the answer is <math>\boxed{\textbf{(B)}\ 336}</math> | ||
Revision as of 12:02, 22 February 2016
Problem
Carl decided to 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
By Albert471
To start, use algebra to determine the number of posts on each side. You have (the long sides count for because there are twice as many) (each corner is double counted so you must add ) Making the shorter end have , and the longer end have . . Therefore, the answer is
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.