Difference between revisions of "1955 AHSME Problems/Problem 14"
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==Solution== | ==Solution== | ||
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> | 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 == | ||
+ | {{AHSME box|year=1955|num-b=13|num-a=15}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 16:26, 2 August 2020
Problem 14
The length of rectangle is % more than the side of square . The width of the rectangle is % less than the side of the square. The ratio of the areas, , is:
Solution
Let each of the square's sides be . The dimensions of the rectangle can be expressed as and . Therefore, the area of the rectangle is , while the square has an area of . The ratio of can be defined as , which ultimately leads to
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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