Difference between revisions of "1968 AHSME Problems/Problem 31"
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<math>\text{(A)}\ 12\tfrac{1}{2}\qquad\text{(B)}\ 25\qquad\text{(C)}\ 50\qquad\text{(D)}\ 75\qquad\text{(E)}\ 87\tfrac{1}{2}</math> | <math>\text{(A)}\ 12\tfrac{1}{2}\qquad\text{(B)}\ 25\qquad\text{(C)}\ 50\qquad\text{(D)}\ 75\qquad\text{(E)}\ 87\tfrac{1}{2}</math> | ||
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+ | == Solution == | ||
+ | The side of the square after the length decrease is <math>2\sqrt{2}</math>, which gives an area of <math>8</math> square inches. <math>32</math> to <math>8</math> is a decrease of <math>75</math> %. Therefore, answer is <math>\boxed{\textbf{(D)}}</math>. | ||
== Solution == | == Solution == |
Revision as of 09:45, 23 February 2024
Contents
[hide]Problem
In this diagram, not drawn to scale, Figures and are equilateral triangular regions with respective areas of and square inches. Figure is a square region with area square inches. Let the length of segment be decreased by % of itself, while the lengths of and remain unchanged. The percent decrease in the area of the square is:
Solution
The side of the square after the length decrease is , which gives an area of square inches. to is a decrease of %. Therefore, answer is .
Solution
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.