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

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==Solution==
 
==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>
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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 4 square feet, so he will use <math>\frac{140}{4}=35</math> tiles for in inner part of the room. Thus, the answer is <math>52+35=\boxed{\textbf{(B) }87}</math>
  
  
 
<math>\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120</math>
 
<math>\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120</math>
 
{{AMC8 box|year=2018|num-b=8|num-a=10}}
 
{{AMC8 box|year=2018|num-b=8|num-a=10}}

Revision as of 16:35, 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 $(12\cdot2)+(14\cdot2)=52$ 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 in inner part of the room. Thus, the answer is $52+35=\boxed{\textbf{(B) }87}$


$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120$

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