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 $\textbf{(A) } 99 : 100$

See Also

1955 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions


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