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Difference between revisions of "1969 AHSME Problems"

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 +
{{AHSC 35 Problems
 +
|year = 1969
 +
}}
 
==Problem 1==
 
==Problem 1==
 +
When <math>x</math> is added to both the numerator and denominator of the fraction
 +
<math>\frac{a}{b},a \ne b,b \ne 0</math>, the value of the fraction is changed to <math>\frac{c}{d}</math>.
 +
Then <math>x</math> equals:
  
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 1|Solution]]
 
[[1969 AHSME Problems/Problem 1|Solution]]
  
 
==Problem 2==
 
==Problem 2==
 +
 +
 +
If an item is sold for <math>x</math> dollars, there is a loss of <math>15\%</math> based on the cost. If, however, the same item is sold for <math>y</math> dollars, there is a profit of <math>15\%</math> based on the cost. The ratio of <math>y:x</math> is:
 +
 +
<math>\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.} </math>
  
  
Line 11: Line 31:
  
 
==Problem 3==
 
==Problem 3==
 +
 +
 +
If <math>N</math>, written in base <math>2</math>, is <math>11000</math>, the integer immediately preceding <math>N</math>, written in base <math>2</math>, is:
 +
 +
<math>\text{(A) } 10001\quad
 +
\text{(B) } 10010\quad
 +
\text{(C) } 10011\quad
 +
\text{(D) } 10110\quad
 +
\text{(E) } 10111</math>
  
  
Line 17: Line 46:
 
==Problem 4==
 
==Problem 4==
  
 +
Let a binary operation <math>\star</math> on ordered pairs of integers be defined by <math>(a,b)\star (c,d)=(a-c,b+d)</math>. Then, if <math>(3,3)\star (0,0)</math> and <math>(x,y)\star (3,2)</math> represent identical pairs, <math>x</math> equals:
 +
 +
<math>\text{(A) } -3\quad
 +
\text{(B) } 0\quad
 +
\text{(C) } 2\quad
 +
\text{(D) } 3\quad
 +
\text{(E) } 6</math>
  
 
[[1969 AHSME Problems/Problem 4|Solution]]
 
[[1969 AHSME Problems/Problem 4|Solution]]
  
 
==Problem 5==
 
==Problem 5==
 +
 +
If a number <math>N,N \ne 0</math>, diminished by four times its reciprocal, equals a given real constant <math>R</math>, then, for this given <math>R</math>, the sum of all such possible values of <math>N</math> is
 +
 +
<math>\text{(A) } \frac{1}{R}\quad
 +
\text{(B) } R\quad
 +
\text{(C) } 4\quad
 +
\text{(D) } \frac{1}{4}\quad
 +
\text{(E) } -R</math>
 +
  
 
[[1969 AHSME Problems/Problem 5|Solution]]
 
[[1969 AHSME Problems/Problem 5|Solution]]
Line 26: Line 71:
 
==Problem 6==
 
==Problem 6==
  
 +
The area of the ring between two concentric circles is <math>12\tfrac{1}{2}\pi</math> square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:
 +
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 6|Solution]]
 
[[1969 AHSME Problems/Problem 6|Solution]]
Line 31: Line 83:
 
==Problem 7==
 
==Problem 7==
  
 +
If the points <math>(1,y_1)</math> and <math>(-1,y_2)</math> lie on the graph of <math>y=ax^2+bx+c</math>, and <math>y_1-y_2=-6</math>, then <math>b</math> equals:
 +
 +
<math>\text{(A) } -3\quad
 +
\text{(B) } 0\quad
 +
\text{(C) } 3\quad
 +
\text{(D) } \sqrt{ac}\quad
 +
\text{(E) } \frac{a+c}{2}</math>
  
 
[[1969 AHSME Problems/Problem 7|Solution]]
 
[[1969 AHSME Problems/Problem 7|Solution]]
Line 36: Line 95:
 
==Problem 8==
 
==Problem 8==
  
 +
Triangle <math>ABC</math> is inscribed in a circle. The measure of the non-overlapping minor arcs <math>AB</math>, <math>BC</math> and <math>CA</math> are, respectively, <math>x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}</math>. Then one interior angle of the triangle is:
 +
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 8|Solution]]
 
[[1969 AHSME Problems/Problem 8|Solution]]
Line 41: Line 107:
 
==Problem 9==
 
==Problem 9==
  
 +
The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:
 +
 +
<math>\text{(A) } 27\quad
 +
\text{(B) } 27\tfrac{1}{4}\quad
 +
\text{(C) } 27\tfrac{1}{2}\quad
 +
\text{(D) } 28\quad
 +
\text{(E) } 28\tfrac{1}{2}</math>
  
 
[[1969 AHSME Problems/Problem 9|Solution]]
 
[[1969 AHSME Problems/Problem 9|Solution]]
Line 46: Line 119:
 
==Problem 10==
 
==Problem 10==
  
 +
The number of points equidistant from a circle and two parallel tangents to the circle is:
  
 +
<math>\text{(A) } 0\quad
 +
\text{(B) } 2\quad
 +
\text{(C) } 3\quad
 +
\text{(D) } 4\quad
 +
\text{(E) } \infty</math>
  
 
[[1969 AHSME Problems/Problem 10|Solution]]
 
[[1969 AHSME Problems/Problem 10|Solution]]
Line 52: Line 131:
 
==Problem 11==
 
==Problem 11==
  
 +
 +
Given points <math>P(-1,-2)</math> and <math>Q(4,2)</math> in the <math>xy</math>-plane; point <math>R(1,m)</math> is taken so that <math>PR+RQ</math> is a minimum. Then <math>m</math> equals:
 +
 +
<math>\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}. </math>
  
 
[[1969 AHSME Problems/Problem 11|Solution]]
 
[[1969 AHSME Problems/Problem 11|Solution]]
  
 
==Problem 12==
 
==Problem 12==
 +
 +
Let <math>F=\frac{6x^2+16x+3m}{6}</math> be the square of an expression which is linear in <math>x</math>. Then <math>m</math> has a particular value between:
 +
 +
<math>\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</math>
  
  
Line 62: Line 157:
 
==Problem 13==
 
==Problem 13==
  
 +
 +
A circle with radius <math>r</math> is contained within the region bounded by a circle with radius <math>R</math>. The area bounded by the larger circle is <math>\frac{a}{b}</math> times the area of the region outside the smaller circle and inside the larger circle. Then <math>R:r</math> equals:
 +
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 13|Solution]]
 
[[1969 AHSME Problems/Problem 13|Solution]]
Line 67: Line 170:
 
==Problem 14==
 
==Problem 14==
  
 +
The complete set of <math>x</math>-values satisfying the inequality <math>\frac{x^2-4}{x^2-1}>0</math> is the set of all <math>x</math> such that:
  
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 14|Solution]]
 
[[1969 AHSME Problems/Problem 14|Solution]]
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==Problem 15==
 
==Problem 15==
  
 +
In a circle with center <math>O</math> and radius <math>r</math>, chord <math>AB</math> is drawn with length equal to <math>r</math> (units). From <math>O</math>, a perpendicular to <math>AB</math> meets <math>AB</math> at <math>M</math>. From <math>M</math> a perpendicular to <math>OA</math> meets <math>OA</math> at <math>D</math>. In terms of <math>r</math> the area of triangle <math>MDA</math>, in appropriate square units, is:
  
 +
<math>\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}</math>
  
 
[[1969 AHSME Problems/Problem 15|Solution]]
 
[[1969 AHSME Problems/Problem 15|Solution]]
  
 
==Problem 16==
 
==Problem 16==
 +
 +
When <math>(a-b)^n,n\ge2,ab\ne0</math>, is expanded by the binomial theorem, it is found that when <math>a=kb</math>, where <math>k</math> is a positive integer, the sum of the second and third terms is zero. Then <math>n</math> equals:
 +
 +
<math>\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</math>
  
 
[[1969 AHSME Problems/Problem 16|Solution]]
 
[[1969 AHSME Problems/Problem 16|Solution]]
  
 
==Problem 17==
 
==Problem 17==
 +
 +
The equation <math>2^{2x}-8\cdot 2^x+12=0</math> is satisfied by:
 +
 +
<math>\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} </math>
  
 
[[1969 AHSME Problems/Problem 17|Solution]]
 
[[1969 AHSME Problems/Problem 17|Solution]]
  
 
==Problem 18==
 
==Problem 18==
 +
 +
The number of points common to the graphs of
 +
<math>(x-y+2)(3x+y-4)=0 \text{    and    } (x+y-2)(2x-5y+7)=0</math>
 +
is:
 +
 +
<math>\text{(A) } 2\quad
 +
\text{(B) } 4\quad
 +
\text{(C) } 6\quad
 +
\text{(D) } 16\quad
 +
\text{(E) } \infty</math>
  
 
[[1969 AHSME Problems/Problem 18|Solution]]
 
[[1969 AHSME Problems/Problem 18|Solution]]
Line 91: Line 232:
 
==Problem 19==
 
==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:
+
The number of distinct ordered pairs <math>(x,y)</math> where <math>x</math> and <math>y</math> have positive integral values satisfying the equation <math>x^4y^4-10x^2y^2+9=0</math> is:
 +
 
 +
<math>\text{(A) } 0\quad
 +
\text{(B) } 3\quad
 +
\text{(C) } 4\quad
 +
\text{(D) } 12\quad
 +
\text{(E) } \infty</math>
  
  
Line 98: Line 245:
 
==Problem 20==
 
==Problem 20==
  
 +
Let <math>P</math> equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in <math>P</math> is:
  
 +
<math>\text{(A) } 36\quad
 +
\text{(B) } 35\quad
 +
\text{(C) } 34\quad
 +
\text{(D) } 33\quad
 +
\text{(E) } 32</math>
  
 
[[1969 AHSME Problems/Problem 20|Solution]]
 
[[1969 AHSME Problems/Problem 20|Solution]]
Line 104: Line 257:
 
==Problem 21==
 
==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]]
 
[[1969 AHSME Problems/Problem 21|Solution]]
Line 110: Line 269:
 
==Problem 22==
 
==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]]
 
[[1969 AHSME Problems/Problem 22|Solution]]
Line 116: Line 285:
 
==Problem 23==
 
==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]]
 
[[1969 AHSME Problems/Problem 23|Solution]]
Line 121: Line 297:
 
==Problem 24==
 
==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]]
 
[[1969 AHSME Problems/Problem 24|Solution]]
Line 127: Line 309:
 
==Problem 25==
 
==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]]
 
[[1969 AHSME Problems/Problem 25|Solution]]
Line 133: Line 322:
  
  
 +
<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]]
 
[[1969 AHSME Problems/Problem 26|Solution]]
Line 138: Line 341:
 
==Problem 27==
 
==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]]
 
[[1969 AHSME Problems/Problem 27|Solution]]
Line 143: Line 353:
 
==Problem 28==
 
==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]]
 
[[1969 AHSME Problems/Problem 28|Solution]]
Line 148: Line 365:
 
==Problem 29==
 
==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]]
 
[[1969 AHSME Problems/Problem 29|Solution]]
Line 153: Line 377:
 
==Problem 30==
 
==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]]
 
[[1969 AHSME Problems/Problem 30|Solution]]
Line 158: Line 389:
 
==Problem 31==
 
==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=2xy</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]]
 
[[1969 AHSME Problems/Problem 31|Solution]]
Line 163: Line 459:
 
==Problem 32==
 
==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]]
 
[[1969 AHSME Problems/Problem 32|Solution]]
Line 168: Line 471:
 
==Problem 33==
 
==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]]
 
[[1969 AHSME Problems/Problem 33|Solution]]
Line 173: Line 484:
 
==Problem 34==
 
==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]]
 
[[1969 AHSME Problems/Problem 34|Solution]]
Line 178: Line 497:
 
==Problem 35==
 
==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]]
 
[[1969 AHSME Problems/Problem 35|Solution]]
 +
 +
== See also ==
 +
 +
* [[AMC 12 Problems and Solutions]]
 +
* [[Mathematics competition resources]]
 +
 +
{{AHSME 35p box|year=1969|before=[[1968 AHSME|1968 AHSC]]|after=[[1970 AHSME|1970 AHSC]]}} 
 +
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:41, 13 November 2024

1969 AHSC (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF

Instructions

  1. 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.
  2. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers.
  4. Figures are not necessarily drawn to scale.
  5. 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

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:

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

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 $v=2xy$ 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

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
1968 AHSC 		Followed by
1970 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


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