Difference between revisions of "1968 AHSME Problems"

Latest revision as of 13:13, 20 February 2020

1968 AHSC (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 35-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test.
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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:

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

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

If all the logarithms are real numbers, the equality $log(x+3)+log(x-1)=log(x^2-2x-3)$ is satisfied for:

$\text{(A) all real values of }x \quad\\ \text{(B) no real values of } x\quad\\ \text{(C) all real values of } x \text{ except } x=0\quad\\ \text{(D) no real values of } x \text{ except } x=0\quad\\ \text{(E) all real values of } x \text{ except } x=1$

Solution

Problem 24

A painting $18$ " X $24$ " is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:

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

Solution

Problem 25

Ace runs with constant speed and Flash runs $x$ times as fast, $x>1$ . Flash gives Ace a head start of $y$ yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

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

Solution

Problem 26

Let $S=2+4+6+\cdots +2N$ , where $N$ is the smallest positive integer such that $S>1,000,000$ . Then the sum of the digits of $N$ is:

$\text{(A) } 27\quad \text{(B) } 12\quad \text{(C) } 6\quad \text{(D) } 2\quad \text{(E) } 1$

Solution

Problem 27

Let $S_n=1-2+3-4+\cdots +(-1)^{n-1}n$ , where $n=1,2,\cdots$ . Then $S_{17}+S_{33}+S_{50}$ equals:

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

Solution

Problem 28

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$ , then a possible value for the ratio $a/b$ , to the nearest integer, is:

$\text{(A) } 5\quad \text{(B) } 8\quad \text{(C) } 11\quad \text{(D) } 14\quad \text{(E) none of these}$

Solution

Problem 29

Given the three numbers $x,y=x^x,z=x^{x^x}$ with $.9<x<1.0$ . Arranged in order of increasing magnitude, they are:

$\text{(A) } x,z,y\quad \text{(B) } x,y,z\quad \text{(C) } y,x,z\quad \text{(D) } y,z,x\quad \text{(E) } z,x,y$

Solution

Problem 30

Convex polygons $P_1$ and $P_2$ are drawn in the same plane with $n_1$ and $n_2$ sides, respectively, $n_1\le n_2$ . If $P_1$ and $P_2$ do not have any line segment in common, then the maximum number of intersections of $P_1$ and $P_2$ is:

$\text{(A) } 2n_1\quad \text{(B) } 2n_2\quad \text{(C) } n_1n_2\quad \text{(D) } n_1+n_2\quad \text{(E) } \text{none of these}$

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

$A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O$ . When $A$ is at $O$ , $B$ is $500$ yards short of $O$ . In two minutes they are equidistant from $O$ , and in $8$ minutes more they are again equidistant from $O$ . Then the ratio of $A$ 's speed to $B$ 's speed is:

$\text{(A) } 4:5\quad \text{(B) } 5:6\quad \text{(C) } 2:3\quad \text{(D) } 5:8\quad \text{(E) } 1:2$

Solution

Problem 33

A number $N$ has three digits when expressed in base $7$ . When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

Problem 34

With $400$ members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was $\frac{12}{11}$ of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?

$\text{(A) } 75\quad \text{(B) } 60\quad \text{(C) } 50\quad \text{(D) } 45\quad \text{(E) } 20$

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

1968 AHSC (Problems • Answer Key • Resources)
Preceded by 1967 AHSC	Followed by 1969 AHSC
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+{{AHSC 35 Problems
+|year = 1968
+}}
 ==Problem 1==
+Let <math>P</math> units be the increase in circumference of a circle resulting from an increase in <math>\pi</math> units in the diameter. Then <math>P</math> equals:
+<math>\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</math>
 [[1968 AHSME Problems/Problem 1|Solution]]
 ==Problem 2==
+The real value of <math>x</math> such that <math>64^{x-1}</math> divided by <math>4^{x-1}</math> equals <math>256^{2x}</math> is:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 2|Solution]]
 ==Problem 3==
+A straight line passing through the point <math>(0,4)</math> is perpendicular to the line <math>x-3y-7=0</math>. Its equation is:
+<math>\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</math>
 [[1968 AHSME Problems/Problem 3|Solution]]
 ==Problem 4==
+Define an operation <math>\star</math> for positive real numbers as <math>a\star b=\frac{ab}{a+b}</math>. Then <math>4 \star (4 \star 4) </math> equals:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 4|Solution]]
 ==Problem 5==
+If <math>f(n)=\tfrac{1}{3} n(n+1)(n+2)</math>, then <math>f(r)-f(r-1)</math> equals:
+<math>\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)</math>
 [[1968 AHSME Problems/Problem 5|Solution]]
 ==Problem 6==
+Let side <math>AD</math> of convex quadrilateral <math>ABCD</math> be extended through <math>D</math>, and let side <math>BC</math> be extended through <math>C</math>, to meet in point <math>E.</math> Let <math>S</math> be the degree-sum of angles <math>CDE</math> and <math>DCE</math>, and let <math>S'</math> represent the degree-sum of angles <math>BAD</math> and <math>ABC.</math> If <math>r=S/S'</math>, then:
+<math>\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</math>
 [[1968 AHSME Problems/Problem 6|Solution]]
 ==Problem 7==
+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
+\text{(B) } \frac{9}{2}\quad
+\text{(C) } 6\quad
+\text{(D) } 9\quad
+\text{(E) } \text{undetermined}</math>
 [[1968 AHSME Problems/Problem 7|Solution]]
 ==Problem 8==
+A positive number is mistakenly divided by <math>6</math> instead of being multiplied by <math>6.</math> Based on the correct answer, the error thus committed, to the nearest percent, is :
+<math>\text{(A) } 100\quad
+\text{(B) } 97\quad
+\text{(C) } 83\quad
+\text{(D) } 17\quad
+\text{(E) } 3</math>
 [[1968 AHSME Problems/Problem 8|Solution]]
 ==Problem 9==
+The sum of the real values of <math>x</math> satisfying the equality <math>|x+2|=2|x-2|</math> is:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 9|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:
+<math>\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.}</math>
 [[1968 AHSME Problems/Problem 10|Solution]]
 ==Problem 11==
+If an arc of <math>60^{\circ}</math> on circle <math>I</math> has the same length as an arc of <math>45^{\circ}</math> on circle <math>II</math>, the ratio of the area of circle <math>I</math> to that of circle <math>II</math> is:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 11|Solution]]
 ==Problem 12==
+A circle passes through the vertices of a triangle with side-lengths <math>7\tfrac{1}{2},10,12\tfrac{1}{2}.</math> The radius of the circle is:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 12|Solution]]
 ==Problem 13==
+If <math>m</math> and <math>n</math> are the roots of <math>x^2+mx+n=0 ,m \ne 0,n \ne 0</math>, then the sum of the roots is:
+<math>\text{(A) } -\frac{1}{2}\quad
+\text{(B) } -1\quad
+\text{(C) } \frac{1}{2}\quad
+\text{(D) } 1\quad
+\text{(E) } \text{undetermined}</math>
 [[1968 AHSME Problems/Problem 13|Solution]]
 ==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==
+Let <math>P</math> be the product of any three consecutive positive odd integers. The largest integer dividing all such <math>P</math> is:
+<math>\text{(A) } 15\quad
+\text{(B) } 6\quad
+\text{(C) } 5\quad
+\text{(D) } 3\quad
+\text{(E) } 1</math>
 [[1968 AHSME Problems/Problem 15|Solution]]
 ==Problem 16==
+If <math>x</math> is such that <math>\frac{1}{x}<2</math> and <math>\frac{1}{x}>-3</math>, then:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 16|Solution]]
 ==Problem 17==
+Let <math>f(n)=\frac{x_1+x_2+\cdots +x_n}{n}</math>, where <math>n</math> is a positive integer. If <math>x_k=(-1)^k, k=1,2,\cdots ,n</math>, the set of possible values of <math>f(n)</math> is:
+<math>\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}\}</math>
 [[1968 AHSME Problems/Problem 17|Solution]]
 ==Problem 18==
+Side <math>AB</math> of triangle <math>ABC</math> has length 8 inches. Line <math>DEF</math> is drawn parallel to <math>AB</math> so that <math>D</math> is on segment <math>AC</math>, and <math>E</math> is on segment <math>BC</math>. Line <math>AE</math> extended bisects angle <math>FEC</math>. If <math>DE</math> has length <math>5</math> inches, then the length of <math>CE</math>, in inches, is:
+<math>\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}</math>
 [[1968 AHSME Problems/Problem 18|Solution]]
 ==Problem 19==
+Let <math>n</math> be the number of ways <math>10</math> dollars can be changed into dimes and quarters, with at least one of each coin being used. Then <math>n</math> equals:
+<math>\text{(A) } 40\quad
+\text{(B) } 38\quad
+\text{(C) } 21\quad
+\text{(D) } 20\quad
+\text{(E) } 19</math>
 [[1968 AHSME Problems/Problem 19|Solution]]
 ==Problem 20==
+The measures of the interior angles of a convex polygon of <math>n</math> sides are in arithmetic progression. If the common difference is <math>5^{\circ}</math> and the largest angle is <math>160^{\circ}</math>, then <math>n</math> equals:
+<math>\text{(A) } 9\quad
+\text{(B) } 10\quad
+\text{(C) } 12\quad
+\text{(D) } 16\quad
+\text{(E) } 32</math>
 [[1968 AHSME Problems/Problem 20|Solution]]
 ==Problem 21==
+If <math>S=1!+2!+3!+\cdots +99!</math>, then the units' digit in the value of S is:
+<math>\text{(A) } 9\quad
+\text{(B) } 8\quad
+\text{(C) } 5\quad
+\text{(D) } 3\quad
+\text{(E) } 0</math>
 [[1968 AHSME Problems/Problem 21|Solution]]
 ==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==
+If all the logarithms are real numbers, the equality
+<math>log(x+3)+log(x-1)=log(x^2-2x-3)</math>
+is satisfied for:
+<math>\text{(A) all real values of }x \quad\\
+\text{(B) no real values of } x\quad\\
+\text{(C) all real values of } x \text{ except } x=0\quad\\
+\text{(D) no real values of } x \text{ except } x=0\quad\\
+\text{(E) all real values of } x \text{ except } x=1</math>
 [[1968 AHSME Problems/Problem 23|Solution]]
 ==Problem 24==
+A painting <math>18</math>" X <math>24</math>" is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:
+<math>\text{(A) } 1:3\quad
+\text{(B) } 1:2\quad
+\text{(C) } 2:3\quad
+\text{(D) } 3:4\quad
+\text{(E) } 1:1</math>
 [[1968 AHSME Problems/Problem 24|Solution]]
 ==Problem 25==
+Ace runs with constant speed and Flash runs <math>x</math> times as fast, <math>x>1</math>. Flash gives Ace a head start of <math>y</math> yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:
+<math>\text{(A) } xy\quad
+\text{(B) } \frac{y}{x+y}\quad
+\text{(C) } \frac{xy}{x-1}\quad
+\text{(D) } \frac{x+y}{x+1}\quad
+\text{(E) } \frac{x+y}{x-1}</math>
 [[1968 AHSME Problems/Problem 25|Solution]]
 ==Problem 26==
+Let <math>S=2+4+6+\cdots +2N</math>, where <math>N</math> is the smallest positive integer such that <math>S>1,000,000</math>. Then the sum of the digits of <math>N</math> is:
+<math>\text{(A) } 27\quad
+\text{(B) } 12\quad
+\text{(C) } 6\quad
+\text{(D) } 2\quad
+\text{(E) } 1</math>
 [[1968 AHSME Problems/Problem 26|Solution]]
 ==Problem 27==
+Let <math>S_n=1-2+3-4+\cdots +(-1)^{n-1}n</math>, where <math>n=1,2,\cdots</math>. Then <math>S_{17}+S_{33}+S_{50}</math> equals:
+<math>\text{(A) } 0\quad
+\text{(B) } 1\quad
+\text{(C) } 2\quad
+\text{(D) } -1\quad
+\text{(E) } -2</math>
 [[1968 AHSME Problems/Problem 27|Solution]]
 ==Problem 28==
+If the arithmetic mean of <math>a</math> and <math>b</math> is double their geometric mean, with <math>a>b>0</math>, then a possible value for the ratio <math>a/b</math>, to the nearest integer, is:
+<math>\text{(A) } 5\quad
+\text{(B) } 8\quad
+\text{(C) } 11\quad
+\text{(D) } 14\quad
+\text{(E) none of these}</math>
 [[1968 AHSME Problems/Problem 28|Solution]]
 ==Problem 29==
+Given the three numbers <math>x,y=x^x,z=x^{x^x}</math> with <math>.9<x<1.0</math>. Arranged in order of increasing magnitude, they are:
+<math>\text{(A) } x,z,y\quad
+\text{(B) } x,y,z\quad
+\text{(C) } y,x,z\quad
+\text{(D) } y,z,x\quad
+\text{(E) } z,x,y</math>
 [[1968 AHSME Problems/Problem 29|Solution]]
 ==Problem 30==
+Convex polygons <math>P_1</math> and <math>P_2</math> are drawn in the same plane with <math>n_1</math> and <math>n_2</math> sides, respectively, <math>n_1\le n_2</math>. If <math>P_1</math> and <math>P_2</math> do not have any line segment in common, then the maximum number of intersections of <math>P_1</math> and <math>P_2</math> is:
+<math>\text{(A) } 2n_1\quad
+\text{(B) } 2n_2\quad
+\text{(C) } n_1n_2\quad
+\text{(D) } n_1+n_2\quad
+\text{(E) } \text{none of these}</math>
 [[1968 AHSME Problems/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 <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:
+<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]]
 ==Problem 32==
+<math>A</math> and <math>B</math> move uniformly along two straight paths intersecting at right angles in point <math>O</math>. When <math>A</math> is at <math>O</math>, <math>B</math> is <math>500</math> yards short of <math>O</math>. In two minutes they are equidistant from <math>O</math>, and in <math>8</math> minutes more they are again equidistant from <math>O</math>. Then the ratio of <math>A</math>'s speed to <math>B</math>'s speed is:
+<math>\text{(A) } 4:5\quad
+\text{(B) } 5:6\quad
+\text{(C) } 2:3\quad
+\text{(D) } 5:8\quad
+\text{(E) } 1:2</math>
 [[1968 AHSME Problems/Problem 32|Solution]]
 ==Problem 33==
+A number <math>N</math> has three digits when expressed in base <math>7</math>. When <math>N</math> is expressed in base <math>9</math> the digits are reversed. Then the middle digit is:
+<math>\text{(A) } 0\quad
+\text{(B) } 1\quad
+\text{(C) } 3\quad
+\text{(D) } 4\quad
+\text{(E) } 5</math>
 [[1968 AHSME Problems/Problem 33|Solution]]
 ==Problem 34==
+With <math>400</math> members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was <math>\frac{12}{11}</math> of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?
+<math>\text{(A) } 75\quad
+\text{(B) } 60\quad
+\text{(C) } 50\quad
+\text{(D) } 45\quad
+\text{(E) } 20</math>
 [[1968 AHSME Problems/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 <math>O</math>, the radius is <math>a</math> inches, chord <math>EF</math> is parallel to chord <math>CD</math>. <math>O</math>,<math>G</math>,<math>H</math>,<math>J</math> are collinear, and <math>G</math> is the midpoint of <math>CD</math>. Let <math>K</math> (sq. in.) represent the area of trapezoid <math>CDFE</math> and let <math>R</math> (sq. in.) represent the area of rectangle <math>ELMF.</math> Then, as <math>CD</math> and <math>EF</math> are translated upward so that <math>OG</math> increases toward the value <math>a</math>, while <math>JH</math> always equals <math>HG</math>, the ratio <math>K:R</math> becomes arbitrarily close to:
+<math>\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</math>
 [[1968 AHSME Problems/Problem 35|Solution]]
+== See also ==
+* [[AMC 12 Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{AHSME 35p box|year=1968|before=[[1967 AHSME|1967 AHSC]]|after=[[1969 AHSME|1969 AHSC]]}}
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Difference between revisions of "1968 AHSME Problems"

Latest revision as of 13:13, 20 February 2020

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

See also