Difference between revisions of "2018 AMC 8 Problems/Problem 9"
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==Solution 1== | ==Solution 1== | ||
− | 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= -AyUsH AgRaWaL \boxed{\textbf{(B)}87}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 12:18, 13 November 2019
Contents
Problem 9
Ayush 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, but he is too weak to lift any tiles. How many tiles will he use?
Solution 1
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: . The area of tiles around the border: . Therefore, is the number of tiles around the border.
The inner part will have . The area of those tiles are . is the amount of tiles for the inner part. So, .
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 |