Difference between revisions of "2018 AMC 8 Problems/Problem 9"
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==Problem 9== | ==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? | 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? | ||
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| + | =Solution== | ||
He will place <math>(12\cdot2)+(14\cdot2)=52</math> tiles around the border. Now, we have <math>10\cdot14</math> inner rectangle, so <math>\frac{10\cdot14}{4}=35</math> tiles. Thus, the answer is <math>52+35=\boxed{87}, \textbf{(B)}</math> | He will place <math>(12\cdot2)+(14\cdot2)=52</math> tiles around the border. Now, we have <math>10\cdot14</math> inner rectangle, so <math>\frac{10\cdot14}{4}=35</math> tiles. Thus, the answer is <math>52+35=\boxed{87}, \textbf{(B)}</math> | ||
Revision as of 12:57, 21 November 2018
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. Now, we have 
inner rectangle, so 
tiles. Thus, the answer is 
| 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 | ||
