Difference between revisions of "2018 AMC 8 Problems/Problem 9"

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The inner part will have <math>(12 - 2)(16 - 2) = 140</math>. The area of those tiles are <math>2 \cdot 2 = 4</math>. <math>\frac{140}{4} = 35</math> is the amount of tiles for the inner part. So, <math>52 + 35 = 87</math>.
 
The inner part will have <math>(12 - 2)(16 - 2) = 140</math>. The area of those tiles are <math>2 \cdot 2 = 4</math>. <math>\frac{140}{4} = 35</math> is the amount of tiles for the inner part. So, <math>52 + 35 = 87</math>.
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==Solution 3==
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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 \boxed{\textbf{(B)}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 17:32, 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?


$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$

Solution 1

He will place $(12\cdot2)+(14\cdot2)=52$ tiles around the border. For the inner part of the room, we have $10\cdot14=140$ square feet. Each tile takes up $4$ square feet, so he will use $\frac{140}{4}=35$ tiles for the inner part of the room. Thus, the answer is $52+35= \boxed{\textbf{(B)}87}$

Solution 2

The area around the border: $(12 \cdot 2) + (14 \cdot 2) = 52$. The area of tiles around the border: $1 \cdot 1 = 1$. Therefore, $\frac{52}{1} = 52$ is the number of tiles around the border.

The inner part will have $(12 - 2)(16 - 2) = 140$. The area of those tiles are $2 \cdot 2 = 4$. $\frac{140}{4} = 35$ is the amount of tiles for the inner part. So, $52 + 35 = 87$.

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 \boxed{\textbf{(B)}87}$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions