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Difference between revisions of "1968 AHSME Problems"

(→‎Problem 6)
(→‎Problem 7)
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==Problem 7==
 
==Problem 7==
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Let <math>O</math> be the intersection point of medians <math>AP</math> and <math>CQ</math> of triangle <math>ABC.</math> if <math>OQ</math> is 3 inches, then <math>OP</math>, in inches, is:
 +
 +
<math>\text{(A) } 3\quad
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\text{(B) } \frac{9}{2}\quad
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\text{(C) } 6\quad
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\text{(D) } 6\quad
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\text{(E) } \text{undetermined}</math>
  
  
 
[[1968 AHSME Problems/Problem 7|Solution]]
 
[[1968 AHSME Problems/Problem 7|Solution]]
 +
 
==Problem 8==
 
==Problem 8==
  

Revision as of 22:27, 23 September 2014

Problem 1

Let $P$ units be the increase in circumference of a circle resulting from an increase in $\pi$ units in the diameter. Then $P$ equals:

$\text{(A) } \frac{1}{\pi}\quad\text{(B) } \pi\quad\text{(C) } \frac{\pi^2}{2}\quad\text{(D) } \pi^2\quad\text{(E) } 2\pi$


Solution

Problem 2

The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is:

$\text{(A) } -\frac{2}{3}\quad\text{(B) } -\frac{1}{3}\quad\text{(C) } 0\quad\text{(D) } \frac{1}{4}\quad\text{(E) } \frac{3}{8}$


Solution

Problem 3

A straight line passing through the point $(0,4)$ is perpendicular to the line $x-3y-7=0$. Its equation is:

$\text{(A) } y+3x-4=0\quad \text{(B) } y+3x+4=0\quad \text{(C) } y-3x-4=0\quad \\ \text{(D) } 3y+x-12=0\quad \text{(E) } 3y-x-12=0$


Solution

Problem 4

Define an operation $\star$ for positive real numbers as $a\star b=\frac{ab}{a+b}$. Then $4 \star (4 \star 4)$ equals:

$\text{(A) } \frac{3}{4}\quad \text{(B) } 1\quad \text{(C) } \frac{4}{3}\quad \text{(D) } 2\quad \text{(E )} \frac{16}{3}$


Solution

Problem 5

If $f(n)=\tfrac{1}{3} n(n+1)(n+2)$, then $f(r)-f(r-1)$ equals:

$\text{(A) } r(r+1)\quad \text{(B) } (r+1)(r+2)\quad \text{(C) } \tfrac{1}{3} r(r+1)\quad \\ \text{(D) } \tfrac{1}{3} (r+1)(r+2)\quad \text{(E )} \tfrac{1}{3} r(r+1)(2r+1)$

Solution

Problem 6

Let side $AD$ of convex quadrilateral $ABCD$ be extended through $D$, and let side $BC$ be extended through $C$, to meet in point $E.$ Let $S$ be the degree-sum of angles $CDE$ and $DCE$, and let $S'$ represent the degree-sum of angles $BAD$ and $ABC.$ If $r=S/S'$, then:

$\text{(A) } r=1 \text{ sometimes, } r>1 \text{ sometimes}\quad\\ \text{(B) } r=1 \text{ sometimes, } r<1 \text{ sometimes}\quad\\ \text{(C) } 0<r<1\quad \text{(D) } r>1\quad \text{(E) } r=1$

Solution

Problem 7

Let $O$ be the intersection point of medians $AP$ and $CQ$ of triangle $ABC.$ if $OQ$ is 3 inches, then $OP$, in inches, is:

$\text{(A) } 3\quad \text{(B) } \frac{9}{2}\quad \text{(C) } 6\quad \text{(D) } 6\quad \text{(E) } \text{undetermined}$


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

[asy] draw(circle((0,0),10),black+linewidth(.75)); fill((-11,0)--(11,0)--(11,-11)--(-11,-11)--cycle,white); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75)); draw((-8,6)--(8,6),black+linewidth(.75)); draw((0,0)--(0,10),black+linewidth(.75)); draw((-8,6)--(-8,2),black+linewidth(.75)); draw((8,6)--(8,2),black+linewidth(.75)); dot((0,0)); MP("O",(0,0),S);MP("a",(5,0),S); MP("J",(0,10),N);MP("D",(sqrt(96),2),E);MP("C",(-sqrt(96),2),W); MP("F",(8,6),E);MP("E",(-8,6),W);MP("G",(0,2),NE); MP("H",(0,6),NE);MP("L",(-8,2),S);MP("M",(8,2),S); [/asy]

In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD$. $O$,$G$,$H$,$J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF.$ Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ becomes arbitrarily close to:

$\text{(A)} 0\quad\text{(B)} 1\quad\text{(C)} \sqrt{2}\quad\text{(D)} \frac{1}{\sqrt{2}}+\frac{1}{2}\quad\text{(E)} \frac{1}{\sqrt{2}}+1$

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

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