Difference between revisions of "1968 AHSME Problems"

(Created page with "==Problem 1== Solution ==Problem 2== Solution ==Problem 3== Solution...")
 
(Problem 31)
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==Problem 31==
 
==Problem 31==
  
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<asy>
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draw((0,0)--(10,20*sqrt(3)/2)--(20,0)--cycle,black+linewidth(.75));
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draw((20,0)--(20,12)--(32,12)--(32,0)--cycle,black+linewidth(.75));
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draw((32,0)--(37,10*sqrt(3)/2)--(42,0)--cycle,black+linewidth(.75));
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MP("I",(10,0),N);MP("II",(26,0),N);MP("III",(37,0),N);
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MP("A",(0,0),S);MP("B",(20,0),S);MP("C",(32,0),S);MP("D",(42,0),S);
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</asy>
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In this diagram, not drawn to scale, Figures <math>I</math> and <math>III</math> are equilateral triangular regions with respective areas of <math>32\sqrt{3}</math> and <math>8\sqrt{3}</math> square inches. Figure <math>II</math> is a square region with area <math>32</math> square inches. Let the length of segment <math>AD</math> be decreased by <math>12\tfrac{1}{2}</math> % of itself, while the lengths of <math>AB</math> and <math>CD</math> remain unchanged. The percent decrease in the area of the square is:
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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>
  
 
[[1968 AHSME Problems/Problem 31|Solution]]
 
[[1968 AHSME Problems/Problem 31|Solution]]
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==Problem 32==
 
==Problem 32==
  

Revision as of 21:22, 23 September 2014

Problem 1

Solution

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution

Problem 31

[asy] draw((0,0)--(10,20*sqrt(3)/2)--(20,0)--cycle,black+linewidth(.75)); draw((20,0)--(20,12)--(32,12)--(32,0)--cycle,black+linewidth(.75)); draw((32,0)--(37,10*sqrt(3)/2)--(42,0)--cycle,black+linewidth(.75)); MP("I",(10,0),N);MP("II",(26,0),N);MP("III",(37,0),N); MP("A",(0,0),S);MP("B",(20,0),S);MP("C",(32,0),S);MP("D",(42,0),S); [/asy]

In this diagram, not drawn to scale, Figures $I$ and $III$ are equilateral triangular regions with respective areas of $32\sqrt{3}$ and $8\sqrt{3}$ square inches. Figure $II$ is a square region with area $32$ square inches. Let the length of segment $AD$ be decreased by $12\tfrac{1}{2}$ % of itself, while the lengths of $AB$ and $CD$ remain unchanged. The percent decrease in the area of the square is:

$\text{(A)}\ 12\tfrac{1}{2}\qquad\text{(B)}\ 25\qquad\text{(C)}\ 50\qquad\text{(D)}\ 75\qquad\text{(E)}\ 87\tfrac{1}{2}$

Solution

Problem 32

Solution

Problem 33

Solution

Problem 34

Solution

Problem 35

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