1955 AHSME Problems/Problem 14

Problem 14

The length of rectangle $R$ is $10$ % more than the side of square $S$ . The width of the rectangle is $10$ % less than the side of the square. The ratio of the areas, $R:S$ , is:

$\textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200$

Solution

Let each of the square's sides be $x$ . The dimensions of the rectangle can be expressed as $1.1x$ and $0.9x$ . Therefore, the area of the rectangle is $0.99x^2$ , while the square has an area of $x^2$ . The ratio of $R : S$ can be defined as $0.99x^2 : x^2$ , which ultimately leads to $\textbf{(A) } 99 : 100$

1955 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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 • 26 • 27 • 28 • 29 • 30
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1955 AHSME Problems/Problem 14

Problem 14

Solution

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