Difference between revisions of "1966 AHSME Problems"

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1966 AHSC (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 40-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.
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 • 36 • 37 • 38 • 39 • 40

Problem 1

Given that the ratio of $3x - 4$ to $y + 15$ is constant, and $y = 3$ when $x = 2$ , then, when $y = 12$ , $x$ equals:

$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$

Solution

Problem 2

When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is

$\text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease}$

Solution

Problem 3

If the arithmetic mean of two numbers is $6$ and their geometric mean is $10$ , then an equation with the given two numbers as roots is:

$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$ $\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$

Solution

Problem 4

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle $I$ to that of circle $II$ , then $r$ equals:

$\text{(A)} \sqrt 2 \qquad \text{(B)} 2 \qquad \text{(C)} \sqrt 3 \qquad \text{(D)} 2\sqrt 2 \qquad \text{(E)} 2\sqrt 3$

Solution

Problem 5

The number of values of $x$ satisfying the equation

$\frac {2x^2 - 10x}{x^2 - 5x} = x - 3$

is:

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$

Solution

Problem 6

$AB$ is the diameter of a circle centered at $O$ . $C$ is a point on the circle such that angle $BOC$ is $60^\circ$ . If the diameter of the circle is $5$ inches, the length of chord $AC$ , expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

Solution

Problem 7

Let $\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}$ be an identity in $x$ . The numerical value of $N_1N_2$ is:

$\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$

Solution

Problem 8

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

Problem 9

If $x = (\log_82)^{(\log_28)}$ , then $\log_3x$ equals:

$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$

Solution

Problem 10

If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:

$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$

Solution

Problem 11

The sides of triangle $BAC$ are in the ratio $2: 3: 4$ . $BD$ is the angle-bisector drawn to the shortest side $AC$ , dividing it into segments $AD$ and $CD$ . If the length of $AC$ is $10$ , then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

Solution

Problem 12

The number of real values of $x$ that satisfy the equation $(2^{6x+3})(4^{3x+6})=8^{4x+5}$ is:

$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$

Solution

Problem 13

The number of points with positive rational coordinates selected from the set of points in the $xy$ -plane such that $x+y \le 5$ , is:

$\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}$

Solution

Problem 14

The length of rectangle $ABCD$ is 5 inches and its width is 3 inches. Diagonal $AC$ is divided into three equal segments by points $E$ and $F$ . The area of triangle $BEF$ , expressed in square inches, is:

$\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}$

Solution

Problem 15

If $x-y>x$ and $x+y<y$ , then

$\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0$

Solution

Problem 16

If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$ , $x$ and $y$ real numbers, then $xy$ equals:

$\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4$

Solution

Problem 17

The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=4$ is:

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

Solution

Problem 18

In a given arithmetic sequence the first term is $2$ , the last term is $29$ , and the sum of all the terms is $155$ . The common difference is:

$\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}$

Solution

Problem 19

Let $s_1$ be the sum of the first $n$ terms of the arithmetic sequence $8,12,\cdots$ and let $s_2$ be the sum of the first $n$ terms of the arithmetic sequence $17,19,\cdots$ . Assume $n \ne 0$ . Then $s_1=s_2$ for:

$\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n$

Solution

Problem 20

The negation of the proposition "For all pairs of real numbers $a,b$ , if $a=0$ , then $ab=0$ " is: There are real numbers $a,b$ such that

$\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0 \qquad \text{(C) } a=0 \text{ and } ab\ne 0$

$\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0$

Solution

Problem 21

$[asy] draw((0,-5)--(-6,10),black+dashed+linewidth(1)); draw((-6,10)--(10,0),black+dashed+linewidth(1)); draw((10,0)--(-10,0),black+dashed+linewidth(1)); draw((-10,-0)--(10,10),black+dashed+linewidth(1)); draw((10,10)--(0,-5),black+dashed+linewidth(1)); draw((-2,0)--(-10/3,10/3),black+linewidth(2)); draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2)); draw((10/9,50/9)--(90/17,50/17),black+linewidth(2)); draw((90/17,50/17)--(10/3,0),black+linewidth(2)); draw((10/3,0)--(-2,0),black+linewidth(2)); MP("1", (1,0), N); MP("2", (-2.5,2), E); MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N); [/asy]$

An " $n$ -pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively $1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5$ ; for all $n$ values of $k$ , sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2$ ; prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5$ ).

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals:

$\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)$

Solution

Problem 22

Consider the statements: (I) $\sqrt{a^2+b^2}=0$ , (II) $\sqrt{a^2+b^2}=ab$ , (III) $\sqrt{a^2+b^2}=a+b$ , (IV) $\sqrt{a^2+b^2}=a - b$ , where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions other than $a=0$ and $b=0$ , are:

$\text{(A) (I),(II),(III),(IV)}\quad \text{(B) (II),(III),(IV) only} \quad \text{(C) (I),(III),(IV) only}\quad \text{(D) (III),(IV) only} \quad \text{(E) (I) only}$

Solution

Problem 23

If $x$ is real and $4y^2+4xy+x+6=0$ , then the complete set of values of $x$ for which $y$ is real, is:

$\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) } x\le2 \text{ or } x\ge3 \quad \text{(C) } x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3$

Solution

Problem 24

If $Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1$ , then $MN$ equals:

$\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}$

Solution

Problem 25

If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$ , then $F(101)$ equals:

$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$

Solution

Problem 26

Let $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then m can be:

$\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}$

Solution

Problem 27

At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:

$\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3 \quad \text{(D) } \frac{7}{2} \quad \text{(E) } 4$

Solution

Problem 28

Five points $O,A,B,C,D$ are taken in order on a straight line with distances $OA = a$ , $OB = b$ , $OC = c$ , and $OD = d$ . $P$ is a point on the line between $B$ and $C$ and such that $AP: PD = BP: PC$ . Then $OP$ equals:

$\textbf{(A)} \frac {b^2 - bc}{a - b + c - d} \qquad \textbf{(B)} \frac {ac - bd}{a - b + c - d} \\ \textbf{(C)} - \frac {bd + ac}{a - b + c - d} \qquad \textbf{(D)} \frac {bc + ad}{a + b + c + d} \qquad \textbf{(E)} \frac {ac - bd}{a + b + c + d}$

Solution

Problem 29

The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:

$\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630$

Solution

Problem 30

If three of the roots of $x^4+ax^2+bx+c=0$ are $1$ , $2$ , and $3$ then the value of $a+c$ is:

$\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63$

Solution

Problem 31

$[asy] draw(circle((0,0),10),black+linewidth(1)); draw(circle((-1.25,2.5),4.5),black+linewidth(1)); dot((0,0)); dot((-1.25,2.5)); draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5)); MP("O'", (0,0), W); MP("O", (-2,2), W); MP("A", (-10,-2), W); MP("B", (10,-2), E); MP("C", (-2,sqrt(96)), N); MP("D", (sqrt(96)-2.5,7), NE); [/asy]$ Triangle $ABC$ is inscribed in a circle with center $O'$ . A circle with center $O$ is inscribed in triangle $ABC$ . $AO$ is drawn, and extended to intersect the larger circle in $D$ . Then we must have:

$\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD$

Solution

Problem 32

Let $M$ be the midpoint of side $AB$ of triangle $ABC$ . Let $P$ be a point on $AB$ between $A$ and $M$ , and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D$ . If the ratio of the area of triangle $BPD$ to that of triangle $ABC$ is denoted by $r$ , then

$\text{(A) } \frac{1}{2}<r<1 \text{, depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{, independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{, depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{, depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{, independent of the position of P}$

Solution

Problem 33

If $ab \ne 0$ and $|a| \ne |b|$ , the number of distinct values of $x$ satisfying the equation

$\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},$

is:

$\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four}$

Solution

Problem 34

Let $r$ be the speed in miles per hour at which a wheel, $11$ feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by $\frac{1}{4}$ of a second, the speed $r$ is increased by $5$ miles per hour. Then $r$ is:

$\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \quad \text{(D) } 11 \quad \text{(E) } 12$

Solution

Problem 35

Let $O$ be an interior point of triangle $ABC$ , and let $s_1=OA+OB+OC$ . If $s_2=AB+BC+CA$ , then

$\text{(A) for every triangle } s_2>2s_1,s_1 \le s_2 \\ \text{(B) for every triangle } s_2>2s_1,s_1 < s_2 \\ \text{(C) for every triangle } s_1> \tfrac{1}{2}s_2,s_1 < s_2 \\ \text{(D) for every triangle } s_2\ge 2s_1,s_1 \le s_2 \\ \text{(E) neither (A) nor (B) nor (C) nor (D) applies to every triangle}$

Solution

Problem 36

Let $(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}$ be an identity in $x$ . If we let $s=a_0+a_2+a_4+\cdots +a_{2n}$ , then $s$ equals:

$\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}$

Solution

Problem 37

Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:

$\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}$

Solution

Problem 38

In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB$ , respectively, intersect in point $O$ . $P$ is the midpoint of side $AC$ , and $MP$ intersects $CN$ in $Q$ . If the area of triangle $OMQ$ is $n$ , then the area of triangle $ABC$ is:

$\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n$

Solution

Problem 39

In base $R_1$ the expanded fraction $F_1$ becomes $.373737\cdots$ , and the expanded fraction $F_2$ becomes $.737373\cdots$ . In base $R_2$ fraction $F_1$ , when expanded, becomes $.252525\cdots$ , while the fraction $F_2$ becomes $.525252\cdots$ . The sum of $R_1$ and $R_2$ , each written in the base ten, is:

$\text{(A) } 24 \quad \text{(B) } 22 \quad \text{(C) } 21 \quad \text{(D) } 20 \quad \text{(E) } 19$

Solution

Problem 40

$[asy] draw(circle((0,0),10),black+linewidth(1)); MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N); MP("a", (-5,0), S);MP("E", (-6,3), N); dot((0,0));dot((-6,2)); draw((-10,0)--(10,0),black+linewidth(1)); draw((-10,0)--(10,10),black+linewidth(1)); draw((-10,-12)--(-10,12),black+linewidth(1)); draw((10,-12)--(10,12),black+linewidth(1)); [/asy]$ In this figure $AB$ is a diameter of a circle, centered at $O$ , with radius $a$ . A chord $AD$ is drawn and extended to meet the tangent to the circle at $B$ in point $C$ . Point $E$ is taken on $AC$ so the $AE=DC$ . Denoting the distances of $E$ from the tangent through $A$ and from the diameter $AB$ by $x$ and $y$ , respectively, we can deduce the relation:

$\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}$

Solution

1966 AHSC (Problems • Answer Key • Resources)
Preceded by 1965 AHSC	Followed by 1967 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 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+{{AHSC 40 Problems
+|year = 1966
+}}
 == Problem 1 ==
 Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, when <math>y = 12</math>, <math>x</math> equals:
@@ Line 7: / Line 10: @@
 == Problem 2 ==
+When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is
+<math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math>
 [[1966 AHSME Problems/Problem 2|Solution]]
 == Problem 3 ==
-If the arithmetic mean of two numbers is <math>6</math> and thier geometric mean is <math>10</math>, then an equation with the given two numbers as roots is:
+If the arithmetic mean of two numbers is <math>6</math> and their geometric mean is <math>10</math>, then an equation with the given two numbers as roots is:
 <math>\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0</math>
@@ Line 22: / Line 27: @@
 Circle I is circumscribed about a given square and circle II is inscribed in the given square. If <math>r</math> is the ratio of the area of circle <math>I</math> to that of circle <math>II</math>, then <math>r</math> equals:
-<math>\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt 3 \qquad \text{(D)} \ 2\sqrt 2 \qquad \text{(E)} \ 2\sqrt 3</math>
+<math>\text{(A)}  \sqrt 2 \qquad \text{(B)}  2 \qquad \text{(C)}  \sqrt 3 \qquad \text{(D)}  2\sqrt 2 \qquad \text{(E)}  2\sqrt 3</math>
 [[1966 AHSME Problems/Problem 4|Solution]]
@@ Line 28: / Line 33: @@
 == Problem 5 ==
 The number of values of <math>x</math> satisfying the equation
-\[
-\frac {2x^2 - 10x}{x^2 - 5x} = x - 3
+<math>\frac {2x^2 - 10x}{x^2 - 5x} = x - 3</math>
-\]
 is:
@@ Line 52: / Line 57: @@
 == Problem 8 ==
-The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the distance between the centers of teh circles, expressed in feet, is:
+The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the distance between the centers of the circles, expressed in feet, is:
 <math>\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math>
@@ Line 73: / Line 78: @@
 == Problem 11 ==
+The sides of triangle <math>BAC</math> are in the ratio <math>2: 3: 4</math>. <math>BD</math> is the angle-bisector drawn to the shortest side <math>AC</math>, dividing it into segments <math>AD</math> and <math>CD</math>. If the length of <math>AC</math> is <math>10</math>, then the length of the longer segment of <math>AC</math> is:
+<math>\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12</math>
 [[1966 AHSME Problems/Problem 11|Solution]]
 == Problem 12 ==
+The number of real values of <math>x</math> that satisfy the equation <cmath>(2^{6x+3})(4^{3x+6})=8^{4x+5}</cmath> is:
+<math>\text{(A)  zero} \qquad \text{(B)  one} \qquad \text{(C)  two} \qquad \text{(D)  three} \qquad \text{(E)  greater than 3}</math>
 [[1966 AHSME Problems/Problem 12|Solution]]
 == Problem 13 ==
+The number of points with positive rational coordinates selected from the set of points in the <math>xy</math>-plane such that <math>x+y \le 5</math>, is:
+<math>\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}</math>
 [[1966 AHSME Problems/Problem 13|Solution]]
 == Problem 14 ==
+The length of rectangle <math>ABCD</math> is 5 inches and its width is 3 inches. Diagonal <math>AC</math> is divided into three equal segments by points <math>E</math> and <math>F</math>. The area of triangle <math>BEF</math>, expressed in square inches, is:
+<math>\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}</math>
 [[1966 AHSME Problems/Problem 14|Solution]]
 == Problem 15 ==
+If <math>x-y>x</math> and <math>x+y<y</math>, then
+<math>\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0</math>
 [[1966 AHSME Problems/Problem 15|Solution]]
 == Problem 16 ==
+If <math>\frac{4^x}{2^{x+y}}=8</math> and <math>\frac{9^{x+y}}{3^{5y}}=243</math>, <math>x</math> and <math>y</math> real numbers, then <math>xy</math> equals:
+<math>\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4</math>
 [[1966 AHSME Problems/Problem 16|Solution]]
 == Problem 17 ==
+The number of distinct points common to the curves <math>x^2+4y^2=1</math> and <math>4x^2+y^2=4</math> is:
+<math>\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4</math>
 [[1966 AHSME Problems/Problem 17|Solution]]
 == Problem 18 ==
+In a given arithmetic sequence the first term is <math>2</math>, the last term is <math>29</math>, and the sum of all the terms is <math>155</math>. The common difference is:
+<math>\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}</math>
 [[1966 AHSME Problems/Problem 18|Solution]]
 == Problem 19 ==
+Let <math>s_1</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>8,12,\cdots</math> and let <math>s_2</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>17,19,\cdots</math>. Assume <math>n \ne 0</math>. Then <math>s_1=s_2</math> for:
+<math>\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n</math>
 [[1966 AHSME Problems/Problem 19|Solution]]
 == Problem 20 ==
+The negation of the proposition "For all pairs of real numbers <math>a,b</math>, if <math>a=0</math>, then <math>ab=0</math>" is: There are real numbers <math>a,b</math> such that
+<math>\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0  \qquad \text{(C) } a=0 \text{ and } ab\ne 0</math>
+<math>\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0</math>
 [[1966 AHSME Problems/Problem 20|Solution]]
@@ Line 124: / Line 151: @@
 == Problem 21 ==
+<asy>
+draw((0,-5)--(-6,10),black+dashed+linewidth(1));
+draw((-6,10)--(10,0),black+dashed+linewidth(1));
+draw((10,0)--(-10,0),black+dashed+linewidth(1));
+draw((-10,-0)--(10,10),black+dashed+linewidth(1));
+draw((10,10)--(0,-5),black+dashed+linewidth(1));
+draw((-2,0)--(-10/3,10/3),black+linewidth(2));
+draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2));
+draw((10/9,50/9)--(90/17,50/17),black+linewidth(2));
+draw((90/17,50/17)--(10/3,0),black+linewidth(2));
+draw((10/3,0)--(-2,0),black+linewidth(2));
+MP("1", (1,0), N);  MP("2", (-2.5,2), E);  MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N);
+</asy>
+An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>).
+Let <math>S</math> be the degree-sum of the interior angles at the <math>n</math> points of the star; then <math>S</math> equals:
+<math>\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)</math>
 [[1966 AHSME Problems/Problem 21|Solution]]
 == Problem 22 ==
+Consider the statements: (I)<math>\sqrt{a^2+b^2}=0</math>,   (II) <math>\sqrt{a^2+b^2}=ab</math>,   (III) <math>\sqrt{a^2+b^2}=a+b</math>,   (IV) <math>\sqrt{a^2+b^2}=a - b</math>, where we allow <math>a</math> and <math>b</math> to be real or complex numbers. Those statements for which there exist solutions other than <math>a=0</math> and <math>b=0</math>, are:
+<math>\text{(A)  (I),(II),(III),(IV)}\quad \text{(B)  (II),(III),(IV) only} \quad \text{(C)  (I),(III),(IV) only}\quad \text{(D)  (III),(IV) only} \quad \text{(E)  (I) only}</math>
 [[1966 AHSME Problems/Problem 22|Solution]]
 == Problem 23 ==
+If <math>x</math> is real and <math>4y^2+4xy+x+6=0</math>, then the complete set of values of <math>x</math> for which <math>y</math> is real, is:
+<math>\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) }  x\le2 \text{ or } x\ge3 \quad \text{(C) }  x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3</math>
 [[1966 AHSME Problems/Problem 23|Solution]]
 == Problem 24 ==
+If <math>Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1</math>, then <math>MN</math> equals:
+<math>\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}</math>
 [[1966 AHSME Problems/Problem 24|Solution]]
 == Problem 25 ==
+If <math>F(n+1)=\frac{2F(n)+1}{2}</math> for <math>n=1,2,\cdots</math> and <math>F(1)=2</math>, then <math>F(101)</math> equals:
+<math>\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53</math>
 [[1966 AHSME Problems/Problem 25|Solution]]
 == Problem 26 ==
+Let <math>m</math> be a positive integer and let the lines <math>13x+11y=700</math> and <math>y=mx-1</math> intersect in a point whose coordinates are integers. Then m can be:
+<math>\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}</math>
 [[1966 AHSME Problems/Problem 26|Solution]]
 == Problem 27 ==
+At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:
+<math>\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3 \quad \text{(D) } \frac{7}{2} \quad \text{(E) } 4</math>
 [[1966 AHSME Problems/Problem 27|Solution]]
 == Problem 28 ==
+Five points <math>O,A,B,C,D</math> are taken in order on a straight line with distances <math>OA = a</math>, <math>OB = b</math>, <math>OC = c</math>, and <math>OD = d</math>.  <math>P</math> is a point on the line between <math>B</math> and <math>C</math> and such that <math>AP: PD = BP: PC</math>.  Then <math>OP</math> equals:
+<math>\textbf{(A)} \frac {b^2 - bc}{a - b + c - d} \qquad \textbf{(B)} \frac {ac - bd}{a - b + c - d} \\
+ \textbf{(C)} - \frac {bd + ac}{a - b + c - d} \qquad \textbf{(D)} \frac {bc + ad}{a + b + c + d} \qquad \textbf{(E)} \frac {ac - bd}{a + b + c + d}</math>
 [[1966 AHSME Problems/Problem 28|Solution]]
 == Problem 29 ==
+The number of positive integers less than <math>1000</math> divisible by neither <math>5</math> nor <math>7</math> is:
+<math>\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630</math>
 [[1966 AHSME Problems/Problem 29|Solution]]
 == Problem 30 ==
+If three of the roots of <math>x^4+ax^2+bx+c=0</math> are <math>1</math>, <math>2</math>, and <math>3</math> then the value of <math>a+c</math> is:
+<math>\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63</math>
 [[1966 AHSME Problems/Problem 30|Solution]]
+== Problem 31 ==
+<asy>
+draw(circle((0,0),10),black+linewidth(1));
+draw(circle((-1.25,2.5),4.5),black+linewidth(1));
+dot((0,0));
+dot((-1.25,2.5));
+draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5));
+draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5));
+draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5));
+draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5));
+MP("O'", (0,0), W);
+MP("O", (-2,2), W);
+MP("A", (-10,-2), W);
+MP("B", (10,-2), E);
+MP("C", (-2,sqrt(96)), N);
+MP("D", (sqrt(96)-2.5,7), NE);
+</asy>
+Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have:
+<math>\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD</math>
+[[1966 AHSME Problems/Problem 31|Solution]]
+== Problem 32 ==
+Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Let <math>P</math> be a point on <math>AB</math> between <math>A</math> and <math>M</math>, and let <math>MD</math> be drawn parallel to <math>PC</math> and intersecting <math>BC</math> at <math>D</math>. If the ratio of the area of triangle <math>BPD</math> to that of triangle <math>ABC</math> is denoted by <math>r</math>, then
+<math>\text{(A) } \frac{1}{2}<r<1 \text{,  depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{,  independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{,  depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{,  depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{,  independent of the position of P}</math>
+[[1966 AHSME Problems/Problem 32|Solution]]
+== Problem 33 ==
+If <math>ab \ne 0</math> and <math>|a| \ne |b|</math>, the number of distinct values of <math>x</math> satisfying the equation
+<cmath>\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},</cmath>
+is:
+<math>\text{(A) zero}  \quad \text{(B) one}  \quad \text{(C) two}  \quad \text{(D) three}  \quad \text{(E) four} </math>
+[[1966 AHSME Problems/Problem 33|Solution]]
+== Problem 34 ==
+Let <math>r</math> be the speed in miles per hour at which a wheel, <math>11</math> feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by <math>\frac{1}{4}</math> of a second, the speed <math>r</math> is increased by <math>5</math> miles per hour. Then <math>r</math> is:
+<math>\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \quad \text{(D) } 11 \quad \text{(E) } 12</math>
+[[1966 AHSME Problems/Problem 34|Solution]]
+== Problem 35 ==
+Let <math>O</math> be an interior point of triangle <math>ABC</math>, and let <math>s_1=OA+OB+OC</math>. If <math>s_2=AB+BC+CA</math>, then
+<math>\text{(A) for every triangle } s_2>2s_1,s_1 \le s_2 \\
+\text{(B) for every triangle } s_2>2s_1,s_1 < s_2 \\
+\text{(C) for every triangle } s_1> \tfrac{1}{2}s_2,s_1 < s_2 \\
+\text{(D) for every triangle } s_2\ge 2s_1,s_1 \le s_2 \\
+\text{(E) neither (A) nor (B) nor (C) nor (D) applies to every triangle}</math>
+[[1966 AHSME Problems/Problem 35|Solution]]
+== Problem 36 ==
+Let <math>(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}</math> be an identity in <math>x</math>. If we let <math>s=a_0+a_2+a_4+\cdots +a_{2n}</math>, then <math>s</math> equals:
+<math>\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}</math>
+[[1966 AHSME Problems/Problem 36|Solution]]
+== Problem 37 ==
+Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let <math>h</math> be the number of hours needed by Alpha and Beta, working together, to do the job. Then <math>h</math> equals:
+<math>\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}</math>
+[[1966 AHSME Problems/Problem 37|Solution]]
+== Problem 38 ==
+In triangle <math>ABC</math> the medians <math>AM</math> and <math>CN</math> to sides <math>BC</math> and <math>AB</math>, respectively, intersect in point <math>O</math>. <math>P</math> is the midpoint of side <math>AC</math>, and <math>MP</math> intersects <math>CN</math> in <math>Q</math>. If the area of triangle <math>OMQ</math> is <math>n</math>, then the area of triangle <math>ABC</math> is:
+<math>\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n</math>
+[[1966 AHSME Problems/Problem 38|Solution]]
+== Problem 39 ==
+In base <math>R_1</math> the expanded fraction <math>F_1</math> becomes <math>.373737\cdots</math>, and the expanded fraction <math>F_2</math> becomes <math>.737373\cdots</math>. In base <math>R_2</math> fraction <math>F_1</math>, when expanded, becomes <math>.252525\cdots</math>, while the fraction <math>F_2</math> becomes <math>.525252\cdots</math>. The sum of <math>R_1</math> and <math>R_2</math>, each written in the base ten, is:
+<math>\text{(A) } 24 \quad \text{(B) } 22 \quad \text{(C) } 21 \quad \text{(D) } 20 \quad \text{(E) } 19</math>
+[[1966 AHSME Problems/Problem 39|Solution]]
+== Problem 40 ==
+<asy>
+draw(circle((0,0),10),black+linewidth(1));
+MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N);
+MP("a", (-5,0), S);MP("E", (-6,3), N);
+dot((0,0));dot((-6,2));
+draw((-10,0)--(10,0),black+linewidth(1));
+draw((-10,0)--(10,10),black+linewidth(1));
+draw((-10,-12)--(-10,12),black+linewidth(1));
+draw((10,-12)--(10,12),black+linewidth(1));
+</asy>
+In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to meet the tangent to the circle at <math>B</math> in point <math>C</math>. Point <math>E</math> is taken on <math>AC</math> so the <math>AE=DC</math>. Denoting the distances of <math>E</math> from the tangent through <math>A</math> and from the diameter <math>AB</math> by <math>x</math> and <math>y</math>, respectively, we can deduce the relation:
+<math>\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}</math>
+[[1966 AHSME Problems/Problem 40|Solution]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
-* [[1966 AHSME]]
 * [[Mathematics competition resources]]
+{{AHSME 40p box|year=1966|before=[[1965 AHSME|1965 AHSC]]|after=[[1967 AHSME|1967 AHSC]]}}
+{{MAA Notice}}

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Difference between revisions of "1966 AHSME Problems"

Revision as of 21:10, 11 March 2024

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

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

See also