Difference between revisions of "2018 AMC 8 Problems/Problem 9"
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He will place <math>(12\cdot2)+(14\cdot2)=52</math> tiles around the border. For the inner part of the room, we have <math>10\cdot14=140</math> square feet. Each tile takes up <math>4</math> square feet, so he will use <math>\frac{140}{4}=35</math> tiles for the inner part of the room. Thus, the answer is <math>52+35=\boxed{\textbf{(B) }87}</math> | He will place <math>(12\cdot2)+(14\cdot2)=52</math> tiles around the border. For the inner part of the room, we have <math>10\cdot14=140</math> square feet. Each tile takes up <math>4</math> square feet, so he will use <math>\frac{140}{4}=35</math> tiles for the inner part of the room. Thus, the answer is <math>52+35=\boxed{\textbf{(B) }87}</math> | ||
| − | == | + | ==Solution 2== |
| + | The area around the border: (12 * 2) + (14 * 2) = 52. The area of tiles around the border: 1 * 1 = 1. Therefore, 52/1 = 52 is the amount of tiles around the border. | ||
| + | |||
| + | The inner part will have (12 - 2)(16 - 2) = 140. The area of those tiles are 2 * 2 = 4. 140/4 = 35 is the amount of tiles for the inner part. So, 52 + 35 = 87 | ||
==See Also== | ==See Also== | ||
Revision as of 12:59, 5 June 2019
Contents
Problem 9
Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?
Solution
He will place 
tiles around the border. For the inner part of the room, we have 
square feet. Each tile takes up 
square feet, so he will use 
tiles for the inner part of the room. Thus, the answer is 
Solution 2
The area around the border: (12 * 2) + (14 * 2) = 52. The area of tiles around the border: 1 * 1 = 1. Therefore, 52/1 = 52 is the amount of tiles around the border.
The inner part will have (12 - 2)(16 - 2) = 140. The area of those tiles are 2 * 2 = 4. 140/4 = 35 is the amount of tiles for the inner part. So, 52 + 35 = 87
See Also
| 2018 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
