Difference between revisions of "1968 AHSME Problems"

Revision as of 23:04, 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

A positive number is mistakenly divided by $6$ instead of being multiplied by $6.$ Based on the correct answer, the error thus committed, to the nearest percent, is :

$\text{(A) } 100\quad \text{(B) } 97\quad \text{(C) } 83\quad \text{(D) } 17\quad \text{(E) } 3$

Solution

Problem 9

The sum of the real values of $x$ satisfying the equality $|x+2|=2|x-2|$ is:

$\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{3}\quad \text{(C) } 6\quad \text{(D) } 6\tfrac{1}{3}\quad \text{(E) } 6\tfrac{2}{3}$

Solution

Problem 10

Assume that, for a certain school, it is true that

I: Some students are not honest. II: All fraternity members are honest.

A necessary conclusion is:

$\text{(A) Some students are fraternity members.} \quad\\ \text{(B) Some fraternity member are not students.} \quad\\ \text{(C) Some students are not fraternity members.} \quad\\ \text{(D) No fraternity member is a student.} \quad\\ \text{(E) No student is a fraternity member.}$

Solution

Problem 11

If an arc of $60^{\circ}$ on circle $I$ has the same length as an arc of $45^{\circ}$ on circle $II$ , the ratio of the area of circle $I$ to that of circle $II$ is:

$\text{(A) } 16:9\quad \text{(B) } 9:16\quad \text{(C) } 4:3\quad \text{(D) } 3:4\quad \text{(E) } \text{none of these}$

Solution

Problem 12

A circle passes through the vertices of a triangle with side-lengths $7\tfrac{1}{2},10,12\tfrac{1}{2}.$ The radius of the circle is:

$\text{(A) } \frac{15}{4}\quad \text{(B) } 5\quad \text{(C) } \frac{25}{4}\quad \text{(D) } \frac{35}{4}\quad \text{(E) } \frac{15\sqrt{2}}{2}$

Solution

Problem 13

If $m$ and $n$ are the roots of $x^2+mx+n=0 ,m \ne 0,n \ne 0$ , then the sum of the roots is:

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

Solution

Problem 14

If $x$ and $y$ are non-zero numbers such that $x=1+\frac{1}{y}$ and $y=1+\frac{1}{x}$ , then $y$ equals

$\text{(A) } x-1\quad \text{(B) } 1-x\quad \text{(C) } 1+x\quad \text{(D) } -x\quad \text{(E) } x$

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.

@@ Line 152: / Line 152: @@
 ==Problem 14==
+If <math>x</math> and <math>y</math> are non-zero numbers such that <math>x=1+\frac{1}{y}</math> and <math>y=1+\frac{1}{x}</math>, then <math>y</math> equals
+<math>\text{(A) } x-1\quad
+\text{(B) } 1-x\quad
+\text{(C) } 1+x\quad
+\text{(D) } -x\quad
+\text{(E) } x</math>
 [[1968 AHSME Problems/Problem 14|Solution]]
 ==Problem 15==

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Difference between revisions of "1968 AHSME Problems"

Revision as of 23:04, 23 September 2014

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35