Difference between revisions of "1969 AHSME Problems"

Revision as of 16:44, 1 October 2014

Problem 1

When $x$ is added to both the numerator and denominator of the fraction $\frac{a}{b},a \ne b,b \ne 0$ , the value of the fraction is changed to $\frac{c}{d}$ . Then $x$ equals:

$\text{(A) } \frac{1}{c-d}\quad \text{(B) } \frac{ad-bc}{c-d}\quad \text{(C) } \frac{ad-bc}{c+d}\quad \text{(D) }\frac{bc-ad}{c-d} \quad \text{(E) } \frac{bc+ad}{c-d}$

Solution

Problem 2

If an item is sold for $x$ dollars, there is a loss of $15\%$ based on the cost. If, however, the same item is sold for $y$ dollars, there is a profit of $15\%$ based on the cost. The ratio of $y:x$ is:

$\text{(A) } 23:17\quad \text{(B) } 17y:23\quad \text{(C) } 23x:17\quad \\ \text{(D) dependent upon the cost} \quad \text{(E) none of these.}$

Solution

Problem 3

If $N$ , written in base $2$ , is $11000$ , the integer immediately preceding $N$ , written in base $2$ , is:

$\text{(A) } 10001\quad \text{(B) } 10010\quad \text{(C) } 10011\quad \text{(D) } 10110\quad \text{(E) } 10111$

Solution

Problem 4

Let a binary operation $\star$ on ordered pairs of integers be defined by $(a,b)\star (c,d)=(a-c,b+d)$ . Then, if $(3,3)\star (0,0)$ and $(x,y)\star (3,2)$ represent identical pairs, $x$ equals:

$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 6$

Solution

Problem 5

If a number $N,N \ne 0$ , diminished by four times its reciprocal, equals a given real constant $R$ , then, for this given $R$ , the sum of all such possible values of $N$ is

$\text{(A) } \frac{1}{R}\quad \text{(B) } R\quad \text{(C) } 4\quad \text{(D) } \frac{1}{4}\quad \text{(E) } -R$

Solution

Problem 6

The area of the ring between two concentric circles is $12\tfrac{1}{2}\pi$ square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:

$\text{(A) } \frac{5}{\sqrt{2}}\quad \text{(B) } 5\quad \text{(C) } 5\sqrt{2}\quad \text{(D) } 10\quad \text{(E) } 10\sqrt{2}$

Solution

Problem 7

If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c$ , and $y_1-y_2=-6$ , then $b$ equals:

$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 3\quad \text{(D) } \sqrt{ac}\quad \text{(E) } \frac{a+c}{2}$

Solution

Problem 8

Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$ , $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$ . Then one interior angle of the triangle is:

$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$

Solution

Problem 9

The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:

$\text{(A) } 27\quad \text{(B) } 27\tfrac{1}{4}\quad \text{(C) } 27\tfrac{1}{2}\quad \text{(D) } 28\quad \text{(E) } 27\tfrac{1}{2}$

Solution

Problem 10

The number of points equidistant from a circle and two parallel tangents to the circle is:

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

Solution

Problem 11

Given points $P(-1,-2)$ and $Q(4,2)$ in the $xy$ -plane; point $R(1,m)$ is taken so that $PR+RQ$ is a minimum. Then $m$ equals:

$\text{(A) } -\tfrac{3}{5}\quad \text{(B) } -\tfrac{2}{5}\quad \text{(C) } -\tfrac{1}{5}\quad \text{(D) } \tfrac{1}{5}\quad \text{(E) either }-\tfrac{1}{5}\text{ or} \tfrac{1}{5}.$

Solution

Problem 12

Let $F=\frac{6x^2+16x+3m}{6}$ be the square of an expression which is linear in $x$ . Then $m$ has a particular value between:

$\text{(A) } 3 \text{ and } 4\quad \text{(B) } 4 \text{ and } 5\quad \text{(C) } 5 \text{ and } 6\quad \text{(D) } -4 \text{ and } -3\quad \text{(E) } -6 \text{ and } -5$

Solution

Problem 13

A circle with radius $r$ is contained within the region bounded by a circle with radius $R$ . The area bounded by the larger circle is $\frac{a}{b}$ times the area of the region outside the smaller circle and inside the larger circle. Then $R:r$ equals:

$\text{(A) }\sqrt{a}:\sqrt{b} \quad \text{(B) } \sqrt{a}:\sqrt{a-b}\quad \text{(C) } \sqrt{b}:\sqrt{a-b}\quad \text{(D) } a:\sqrt{a-b}\quad \text{(E) } b:\sqrt{a-b}$

Solution

Problem 14

The complete set of $x$ -values satisfying the inequality $\frac{x^2-4}{x^2-1}>0$ is the set of all $x$ such that:

$\text{(A) } x>2 \text{ or } x<-2 \text{ or} -1<x<1\quad \text{(B) } x>2 \text{ or } x<-2\quad \\ \text{(C) } x>1 \text{ or } x<-2\qquad\qquad\qquad\quad \text{(D) } x>1 \text{ or } x<-1\quad \\ \text{(E) } x \text{ is any real number except 1 or -1}$

Solution

Problem 15

In a circle with center $O$ and radius $r$ , chord $AB$ is drawn with length equal to $r$ (units). From $O$ , a perpendicular to $AB$ meets $AB$ at $M$ . From $M$ a perpendicular to $OA$ meets $OA$ at $D$ . In terms of $r$ the area of triangle $MDA$ , in appropriate square units, is:

$\text{(A) } \frac{3r^2}{16}\quad \text{(B) } \frac{\pi r^2}{16}\quad \text{(C) } \frac{\pi r^2\sqrt{2}}{8}\quad \text{(D) } \frac{r^2\sqrt{3}}{32}\quad \text{(E) } \frac{r^2\sqrt{6}}{48}$

Solution

Problem 16

When $(a-b)^n,n\ge2,ab\ne0$ , is expanded by the binomial theorem, it is found that when $a=kb$ , where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals:

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

Solution

Problem 17

The equation $2^{2x}-8\cdot 2^x+12=0$ is satisfied by:

$\text{(A) } log(3)\quad \text{(B) } \tfrac{1}{2}log(6)\quad \text{(C) } 1+log(\tfrac{3}{2})\quad \text{(D) } 1+\frac{log(3)}{log(2)}\quad \text{(E) none of these}$

Solution

Problem 18

The number of points common to the graphs of $(x-y+2)(3x+y-4)=0 \text{ and } (x+y-2)(2x-5y+7)=0$ is:

$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$

Solution

Problem 19

The number of distinct ordered pairs $(x,y)$ where $x$ and $y$ have positive integral values satisfying the equation $x^4y^4-10x^2y^2+9=0$ is:

$\text{(A) } 0\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 12\quad \text{(E) } \infty$

Solution

Problem 20

Let $P$ equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in $P$ is:

$\text{(A) } 36\quad \text{(B) } 35\quad \text{(C) } 34\quad \text{(D) } 33\quad \text{(E) } 32$

Solution

Problem 21

If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m}$ , then:

$\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ \text{(C) m must equal } \sqrt{2}\quad \text{(D) m must equal } 2\quad\\ \text{(E) m may be an non-negative real number}$

Solution

Problem 22

Let $K$ be the measure of the area bounded by the $x$ -axis, the line $x=8$ , and the curve defined by

$f={(x,y)\quad |\quad y=x \text{ when } 0 \le x \le 5, y=2x-5 \text{ when } 5 \le x \le 8}.$

Then $K$ is:

$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less than 44 but arbitrarily close to it}$

Solution

Problem 23

For any integer $n>1$ , the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:

$\text{(A) } 0\quad\qquad \text{(B) } 1\quad\\ \text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\ \text{(D) } n-1\quad \text{(E) } n$

Solution

Problem 24

When the natural numbers $P$ and $P'$ , with $P>P'$ , are divided by the natural number $D$ , the remainders are $R$ and $R'$ , respectively. When $PP'$ and $RR'$ are divided by $D$ , the remainders are $r$ and $r'$ , respectively. Then:

$\text{(A) } r>r' \text{ always}\quad \text{(B) } r<r' \text{ always}\quad\\ \text{(C) } r>r' \text{ sometimes and } r<r' \text{ sometimes}\quad\\ \text{(D) } r>r' \text{ sometimes and } r=r' \text{ sometimes}\quad\\ \text{(E) } r=r' \text{ always}$

Solution

Problem 25

If it is known that $log_2(a)+log_2(b) \ge 6$ , then the least value that can be taken on by $a+b$ is:

$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$

Solution

Problem 26

$[asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]$

A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:

$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

Solution

Problem 27

A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in $2$ hours, then the time, in hours, needed to traverse the $n$ th mile is:

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

Solution

Problem 28

Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$ , such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$ . Then $n$ is:

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

Solution

Problem 29

If $x=t^{1/(t-1)}$ and $y=t^{t/(t-1)},t>0,t \ne 1$ , a relation between $x$ and $y$ is:

$\text{(A) } y^x=x^{1/y}\quad \text{(B) } y^{1/x}=x^{y}\quad \text{(C) } y^x=x^y\quad \text{(D) } x^x=y^y\quad \text{(E) none of these}$

Solution

Problem 30

Let $P$ be a point of hypotenuse $AB$ (or its extension) of isosceles right triangle $ABC$ . Let $s=AP^2+PB^2$ . Then:

$\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\ \text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\ \text{(C) } s=2CP^2 \text{ only if P is the midpoint or an endpoint of AB}\quad\\ \text{(D) } s=2CP^2 \text{ always}\quad\\ \text{(E) } s>2CP^2 \text{ if P is a trisection point of AB}$

Solution

Problem 31

Let $OABC$ be a unit square in the $xy$ -plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$ . Let $u=x^2-y^2$ , and $v=xy$ be a transformation of the $xy$ -plane into the $uv$ -plane. The transform (or image) of the square is:

$[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot); MP("(A)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]$

$[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,234),black); draw(arc((-1.5,0),2.5,54,-54),black); MP("(B)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]$

$[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0),black); MP("(C)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); [/asy]$

$[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,180),black); draw(arc((-1.5,0),2.5,54,0),black); MP("(D)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]$

$[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot); MP("(E)",(-5,2),SW); MP("O",(.1,.1),SW); MP("(-1,0)",(-1,0),SW); MP("(0,1)",(0,1),NE); MP("(1,0)",(1,0),SE); MP("(0,-1)",(0,-1),SE); [/asy]$

Solution

Problem 32

Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relationship $u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.$ If $u_n$ is expressed as a polynomial in $n$ , the algebraic sum of its coefficients is:

$\text{(A) 3} \quad \text{(B) 4} \quad \text{(C) 5} \quad \text{(D) 6} \quad \text{(E) 11}$

Solution

Problem 33

Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n$ , the ratio of the eleventh term of the first series to the eleventh term of the second series is:

$\text{(A) } 4:3\quad \text{(B) } 3:2\quad \text{(C) } 7:4\quad \text{(D) } 78:71\quad \text{(E) undetermined}$

Solution

Problem 34

The remainder $R$ obtained by dividing $x^{100}$ by $x^2-3x+2$ is a polynomial of degree less than $2$ . Then $R$ may be written as:

$\text{(A) }2^{100}-1 \quad \text{(B) } 2^{100}(x-1)-(x-2)\quad \text{(C) } 2^{200}(x-3)\quad\\ \text{(D) } x(2^{100}-1)+2(2^{99}-1)\quad \text{(E) } 2^{100}(x+1)-(x+2)$

Solution

Problem 35

Let $L(m)$ be the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=m$ , where $-6<m<6$ . Let $r=[L(-m)-L(m)]/m$ . Then, as $m$ is made arbitrarily close to zero, the value of $r$ is:

$\text{(A) arbitrarily close to } 0\quad\\ \text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\quad \text{(C) arbitrarily close to }\frac{2}{\sqrt{6}} \quad \text{(D) arbitrarily large} \quad \text{(E) undetermined}$

Solution The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 254: / Line 254: @@
 ==Problem 21==
+If the graph of <math>x^2+y^2=m</math> is tangent to that of <math>x+y=\sqrt{2m}</math>, then:
+<math>\text{(A) m must equal } \tfrac{1}{2}\quad
+\text{(B) m must equal  } \frac{1}{\sqrt{2}}\quad\\
+\text{(C) m must equal } \sqrt{2}\quad
+\text{(D) m must equal } 2\quad\\
+\text{(E) m may be an non-negative real number} </math>
 [[1969 AHSME Problems/Problem 21|Solution]]
@@ Line 260: / Line 266: @@
 ==Problem 22==
+Let <math>K</math> be the measure of the area bounded by the <math>x</math>-axis, the line <math>x=8</math>, and the curve defined by
+<cmath>f={(x,y)\quad |\quad y=x \text{  when  } 0 \le x \le 5, y=2x-5 \text{  when  } 5 \le x \le 8}.</cmath>
+Then <math>K</math> is:
+<math>\text{(A) } 21.5\quad
+\text{(B) } 36.4\quad
+\text{(C) } 36.5\quad
+\text{(D) } 44\quad
+\text{(E) less than 44 but arbitrarily close to it} </math>
 [[1969 AHSME Problems/Problem 22|Solution]]
@@ Line 266: / Line 282: @@
 ==Problem 23==
+For any integer <math>n>1</math>, the number of prime numbers greater than <math>n!+1</math> and less than <math>n!+n</math> is:
+<math>\text{(A) } 0\quad\qquad
+\text{(B) } 1\quad\\
+\text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\
+\text{(D) } n-1\quad
+\text{(E) } n</math>
 [[1969 AHSME Problems/Problem 23|Solution]]
@@ Line 271: / Line 294: @@
 ==Problem 24==
+When the natural numbers <math>P</math> and <math>P'</math>, with <math>P>P'</math>, are divided by the natural number <math>D</math>, the remainders are <math>R</math> and <math>R'</math>, respectively. When <math>PP'</math> and <math>RR'</math> are divided by <math>D</math>, the remainders are <math>r</math> and <math>r'</math>, respectively. Then:
+<math>\text{(A) } r>r' \text{  always}\quad
+\text{(B) } r<r' \text{  always}\quad\\
+\text{(C) } r>r' \text{  sometimes and } r<r' \text{  sometimes}\quad\\
+\text{(D) } r>r' \text{  sometimes and } r=r' \text{  sometimes}\quad\\
+\text{(E) } r=r' \text{  always}</math>
 [[1969 AHSME Problems/Problem 24|Solution]]
@@ Line 277: / Line 306: @@
 ==Problem 25==
+If it is known that <math>log_2(a)+log_2(b) \ge 6</math>, then the least value that can be taken on by <math>a+b</math> is:
+<math>\text{(A) } 2\sqrt{6}\quad
+\text{(B) } 6\quad
+\text{(C) } 8\sqrt{2}\quad
+\text{(D) } 16\quad
+\text{(E) none of these} </math>
 [[1969 AHSME Problems/Problem 25|Solution]]
@@ Line 283: / Line 319: @@
+<asy>
+draw(arc((0,-1),2,30,150),dashed+linewidth(.75));
+draw((-1.7,0)--(0,0)--(1.7,0),dot);
+draw((0,0)--(0,.98),dot);
+MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N);
+</asy>
+A parabolic arch has a height of <math>16</math> inches and a span of <math>40</math> inches. The height, in inches, of the arch at the point <math>5</math> inches from the center <math>M</math> is:
+<math>\text{(A) } 1\quad
+\text{(B) } 15\quad
+\text{(C) } 15\tfrac{1}{3}\quad
+\text{(D) } 15\tfrac{1}{2}\quad
+\text{(E) } 15\tfrac{3}{4}</math>
 [[1969 AHSME Problems/Problem 26|Solution]]
@@ Line 288: / Line 338: @@
 ==Problem 27==
+A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in <math>2</math> hours, then the time, in hours, needed to traverse the <math>n</math>th mile is:
+<math>\text{(A) } \frac{2}{n-1}\quad
+\text{(B) } \frac{n-1}{2}\quad
+\text{(C) } \frac{2}{n}\quad
+\text{(D) } 2n\quad
+\text{(E) } 2(n-1)</math>
 [[1969 AHSME Problems/Problem 27|Solution]]
@@ Line 293: / Line 350: @@
 ==Problem 28==
+Let <math>n</math> be the number of points <math>P</math> interior to the region bounded by a circle with radius <math>1</math>, such that the sum of squares of the distances from <math>P</math> to the endpoints of a given diameter is <math>3</math>. Then <math>n</math> is:
+<math>\text{(A) } 0\quad
+\text{(B) } 1\quad
+\text{(C) } 2\quad
+\text{(D) } 4\quad
+\text{(E) } \infty</math>
 [[1969 AHSME Problems/Problem 28|Solution]]
@@ Line 298: / Line 362: @@
 ==Problem 29==
+If <math>x=t^{1/(t-1)}</math> and  <math>y=t^{t/(t-1)},t>0,t \ne 1</math>, a relation between <math>x</math> and <math>y</math> is:
+<math>\text{(A) } y^x=x^{1/y}\quad
+\text{(B) } y^{1/x}=x^{y}\quad
+\text{(C) } y^x=x^y\quad
+\text{(D) } x^x=y^y\quad
+\text{(E) none of these} </math>
 [[1969 AHSME Problems/Problem 29|Solution]]
@@ Line 303: / Line 374: @@
 ==Problem 30==
+Let <math>P</math> be a point of hypotenuse <math>AB</math> (or its extension) of isosceles right triangle <math>ABC</math>. Let <math>s=AP^2+PB^2</math>. Then:
+<math>\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\
+\text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\
+\text{(C) } s=2CP^2 \text{ only if P is the midpoint or an endpoint of AB}\quad\\
+\text{(D) } s=2CP^2 \text{ always}\quad\\
+\text{(E) } s>2CP^2 \text{ if P is a trisection point of AB}</math>
 [[1969 AHSME Problems/Problem 30|Solution]]
@@ Line 308: / Line 386: @@
 ==Problem 31==
+Let <math>OABC</math> be a unit square in the <math>xy</math>-plane with <math>O(0,0),A(1,0),B(1,1)</math> and <math>C(0,1)</math>. Let <math>u=x^2-y^2</math>, and <math>v=xy</math> be a transformation of the <math>xy</math>-plane into the <math>uv</math>-plane. The transform (or image) of the square is:
+<asy>
+draw((-3,0)--(3,0),EndArrow);
+draw((0,-4)--(0,4),EndArrow);
+draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot);
+MP("(A)",(-5,2),SW);
+MP("O",(0,0),SW);
+MP("(-1,0)",(-1,0),SW);
+MP("(0,2)",(0,2),NE);
+MP("(1,0)",(1,0),SE);
+MP("(0,-2)",(0,-2),SE);
+</asy>
+<asy>
+draw((-3,0)--(3,0),EndArrow);
+draw((0,-4)--(0,4),EndArrow);
+draw(arc((1.5,0),2.5,126,234),black);
+draw(arc((-1.5,0),2.5,54,-54),black);
+MP("(B)",(-5,2),SW);
+MP("O",(0,0),SW);
+MP("(-1,0)",(-1,0),SW);
+MP("(0,2)",(0,2),NE);
+MP("(1,0)",(1,0),SE);
+MP("(0,-2)",(0,-2),SE);
+</asy>
+<asy>
+draw((-3,0)--(3,0),EndArrow);
+draw((0,-4)--(0,4),EndArrow);
+draw((-1,0)--(0,2)--(1,0),black);
+MP("(C)",(-5,2),SW);
+MP("O",(0,0),SW);
+MP("(-1,0)",(-1,0),SW);
+MP("(0,2)",(0,2),NE);
+MP("(1,0)",(1,0),SE);
+</asy>
+<asy>
+draw((-3,0)--(3,0),EndArrow);
+draw((0,-4)--(0,4),EndArrow);
+draw(arc((1.5,0),2.5,126,180),black);
+draw(arc((-1.5,0),2.5,54,0),black);
+MP("(D)",(-5,2),SW);
+MP("O",(0,0),SW);
+MP("(-1,0)",(-1,0),SW);
+MP("(0,2)",(0,2),NE);
+MP("(1,0)",(1,0),SE);
+MP("(0,-2)",(0,-2),SE);
+</asy>
+<asy>
+draw((-3,0)--(3,0),EndArrow);
+draw((0,-4)--(0,4),EndArrow);
+draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot);
+MP("(E)",(-5,2),SW);
+MP("O",(.1,.1),SW);
+MP("(-1,0)",(-1,0),SW);
+MP("(0,1)",(0,1),NE);
+MP("(1,0)",(1,0),SE);
+MP("(0,-1)",(0,-1),SE);
+</asy>
 [[1969 AHSME Problems/Problem 31|Solution]]
@@ Line 313: / Line 456: @@
 ==Problem 32==
+Let a sequence <math>\{u_n\}</math> be defined by <math>u_1=5</math> and the relationship <math>u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.</math>If <math>u_n</math> is expressed as a polynomial in <math>n</math>, the algebraic sum of its coefficients is:
+<math>\text{(A) 3} \quad
+\text{(B) 4} \quad
+\text{(C) 5} \quad
+\text{(D) 6} \quad
+\text{(E) 11} </math>
 [[1969 AHSME Problems/Problem 32|Solution]]
@@ Line 318: / Line 468: @@
 ==Problem 33==
+Let <math>S_n</math> and <math>T_n</math> be the respective sums of the first <math>n</math> terms of two arithmetic series. If <math>S_n:T_n=(7n+1):(4n+27)</math> for all <math>n</math>, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
+<math>\text{(A) } 4:3\quad
+\text{(B) } 3:2\quad
+\text{(C) } 7:4\quad
+\text{(D) } 78:71\quad
+\text{(E) undetermined} </math>
 [[1969 AHSME Problems/Problem 33|Solution]]
 ==Problem 34==
+The remainder <math>R</math> obtained by dividing <math>x^{100}</math> by <math>x^2-3x+2</math> is a polynomial of degree less than <math>2</math>. Then <math>R</math> may be written as:
+<math>\text{(A) }2^{100}-1 \quad
+\text{(B) } 2^{100}(x-1)-(x-2)\quad
+\text{(C) } 2^{200}(x-3)\quad\\
+\text{(D) } x(2^{100}-1)+2(2^{99}-1)\quad
+\text{(E) } 2^{100}(x+1)-(x+2)</math>
 [[1969 AHSME Problems/Problem 34|Solution]]
@@ Line 328: / Line 494: @@
 ==Problem 35==
+Let <math>L(m)</math> be the <math>x</math> coordinate of the left end point of the intersection of the graphs of <math>y=x^2-6</math> and <math>y=m</math>, where <math>-6<m<6</math>. Let <math>r=[L(-m)-L(m)]/m</math>. Then, as <math>m</math> is made arbitrarily close to zero, the value of <math>r</math> is:
+<math>\text{(A) arbitrarily close to } 0\quad\\
+\text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\quad
+\text{(C) arbitrarily close to }\frac{2}{\sqrt{6}} \quad
+\text{(D) arbitrarily large} \quad
+\text{(E) undetermined} </math>
 [[1969 AHSME Problems/Problem 35|Solution]]
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Difference between revisions of "1969 AHSME Problems"

Revision as of 16:44, 1 October 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