Difference between revisions of "1968 AHSME Problems"

Revision as of 00:02, 24 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

Let $P$ be the product of any three consecutive positive odd integers. The largest integer dividing all such $P$ is:

$\text{(A) } 15\quad \text{(B) } 6\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 1$

Solution

Problem 16

If $x$ is such that $\frac{1}{x}<2$ and $\frac{1}{x}>-3$ , then:

$\text{(A) } -\frac{1}{3}<x<\frac{1}{2}\quad \text{(B) } -\frac{1}{2}<x<3\quad \text{(C) } x>\frac{1}{2}\quad\\ \text{(D) } x>\frac{1}{2} \text{ or} -\frac{1}{3}<x<0\quad \text{(E) } x>\frac{1}{2} \text{ or } x<-\frac{1}{3}$

Solution

Problem 17

Let $f(n)=\frac{x_1+x_2+\cdots +x_n}{n}$ , where $n$ is a positive integer. If $x_k=(-1)^k, k=1,2,\cdots ,n$ , the set of possible values of $f(n)$ is:

$\text{(A) } \{0\}\quad \text{(B) } \{\frac{1}{n}\}\quad \text{(C) } \{0,-\frac{1}{n}\}\quad \text{(D) } \{0,\frac{1}{n}\}\quad \text{(E) } \{1,\frac{1}{n}\}$

Solution

Problem 18

Side $AB$ of triangle $ABC$ has length 8 inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$ , and $E$ is on segment $BC$ . Line $AE$ extended bisects angle $FEC$ . If $DE$ has length $5$ inches, then the length of $CE$ , in inches, is:

$\text{(A) } \frac{51}{4}\quad \text{(B) } 13\quad \text{(C) } \frac{53}{4}\quad \text{(D) } \frac{40}{3}\quad \text{(E) } \frac{27}{2}$

Solution

Problem 19

Let $n$ be the number of ways $10$ dollars can be changed into dimes and quarters, with at least one of each coin being used. Then $n$ equals:

$\text{(A) } 40\quad \text{(B) } 38\quad \text{(C) } 21\quad \text{(D) } 20\quad \text{(E) } 19$

Solution

Problem 20

The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:

$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$

Solution

Problem 21

If $S=1!+2!+3!+\cdots +99!$ , then the units' digit in the value of S is:

$\text{(A) } 9\quad \text{(B) } 8\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 0$

Solution

Problem 22

A segment of length $1$ is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:

$\text{(A) equal to } \frac{1}{4}\quad\\ \text{(B) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ \text{(C) greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ \text{(D) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{4}\quad\\ \text{(E) less than }\frac{1}{2}$

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 247: / Line 247: @@
 ==Problem 22==
+A segment of length <math>1</math> is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:
+<math>\text{(A) equal to } \frac{1}{4}\quad\\
+\text{(B) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\
+\text{(C) greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\
+\text{(D) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{4}\quad\\
+\text{(E) less than }\frac{1}{2}</math>
 [[1968 AHSME Problems/Problem 22|Solution]]
 ==Problem 23==

Page

Toolbox

Search

Difference between revisions of "1968 AHSME Problems"

Revision as of 00:02, 24 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