1968 AHSME Problems

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: