1999 CEMC Gauss (Grade 7) Problems/Problem 11

Problem

The floor of a rectangular room is covered with square tiles. The room is 10 tiles long and 5 tiles wide. The number of tiles that touch the walls of the room is

$\text{(A)}\ 26 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 34 \qquad \text{(D)}\ 46 \qquad \text{(E)}\ 50$

Solution 1

We can construct a rectangle to represent the floor of the room, and shade in the area of tiles that touch the walls of the room.

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The area of a rectangle is its width times its height.

Therefore, the total number of tiles in the rectangular room would be $10 \times 5 = 50$ .

The tiles that touch the walls of the room would be on the edges of the rectangle. This means that we can subtract the number of tiles in the middle of the rectangle from the total number of tiles in the room to get the shaded area.

Looking at the diagram, the number of tiles in the middle of the rectangle forms a smaller rectangle with side lengths that are $2$ tiles shorter than the width and height of the large rectangle. This means that the smaller rectangle is 8 tiles long and 3 tiles wide.

The area of this rectangle would be $8 \times 3$ = $24$ .

$50 - 24 = \boxed {\textbf {(A)} 26}$

Solution 2

There would be $10$ tiles on the top and bottom of the sides of the rectangle each. There would then be $3$ tiles on the left and right sides of the rectangle that haven't been counted for. This means we have:

$10 \times 2 + 3 \times 2 = \boxed {\textbf {(A)} 26}$

Solution 3 (slow)

We can count each of the tiles on the edges of the rectangle to get $\boxed {\textbf {(A)} 26}$ .

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1999 CEMC Gauss (Grade 7) Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Solution 3 (slow)