Difference between revisions of "2018 AMC 8 Problems/Problem 9"
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==Solution 3== | ==Solution 3== | ||
− | Inside the room, there is a rectangle of 1 times 1 squares. There are 2(12+16) - 4 small tiles, getting 52 1 times 1 tiles. The space in the room for 2 times 2 tiles is 14*10, because subtract 2 from each dimension due to the 1 times 1 squares. The area that is left is 140. The area of each 2 times 2 square is 4, so 140/4 is 35, and 52+35 is 87, so | + | Inside the room, there is a rectangle of 1 times 1 squares. There are 2(12+16) - 4 small tiles, getting 52 1 times 1 tiles. The space in the room for 2 times 2 tiles is 14*10, because subtract 2 from each dimension due to the 1 times 1 squares. The area that is left is 140. The area of each 2 times 2 square is 4, so 140/4 is 35, and 52+35 is 87, so 87$ |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2018|num-b=8|num-a=10}} | {{AMC8 box|year=2018|num-b=8|num-a=10}} |
Revision as of 17:33, 20 December 2021
Problem
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 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, .
Solution 3
Inside the room, there is a rectangle of 1 times 1 squares. There are 2(12+16) - 4 small tiles, getting 52 1 times 1 tiles. The space in the room for 2 times 2 tiles is 14*10, because subtract 2 from each dimension due to the 1 times 1 squares. The area that is left is 140. The area of each 2 times 2 square is 4, so 140/4 is 35, and 52+35 is 87, so 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 |