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Difference between revisions of "1955 AHSME Problems/Problem 14"
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==Solution==
 
==Solution==
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Let each of the square's sides be <math>x</math>. The dimensions of the rectangle can be expressed as <math>1.1x</math> and <math>0.9x</math>. Therefore, the area of the rectangle is <math>0.99x^2</math>, while the square has an area of <math>x^2</math>. The ratio of <math>R : S</math> can be defined as <math>0.99x^2 : x^2</math>, which ultimately leads to <math>\textbf{(A)} 99 : 100</math>
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Let each of the square's sides be <math>x</math>. The dimensions of the rectangle can be expressed as <math>1.1x</math> and <math>0.9x</math>. Therefore, the area of the rectangle is <math>0.99x^2</math>, while the square has an area of <math>x^2</math>. The ratio of <math>R : S</math> can be defined as <math>0.99x^2 : x^2</math>, which ultimately leads to <math>\textbf{(A) } 99 : 100</math>
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1955|num-b=13|num-a=15}}
 
{{AHSME box|year=1955|num-b=13|num-a=15}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:17, 11 August 2021

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 (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
All AHSME Problems and Solutions


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