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

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==Problem==
 
==Problem==
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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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Jenica is tiling the floor of her 12 foot by 16 foot living room. She 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 she use?
  
  

Revision as of 11:56, 2 January 2023

Problem

Jenica is tiling the floor of her 12 foot by 16 foot living room. She 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 she 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, $52 + 35 = \boxed{87}$.

Video Solution

https://youtu.be/V5h00z5MDEI

~savannahsolver

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
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