2018 AMC 8 Problems/Problem 9

Revision as of 13:44, 15 July 2019 by Olivera (talk | contribs) (Solution 2)

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?


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

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 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, $52 + 35 = 87$.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS