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Difference between revisions of "1969 AHSME Problems"

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==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 138: Line 154:
 
==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 143: Line 167:
 
==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]]
Line 149: Line 179:
 
==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 167: Line 229:
 
==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>
  
  
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==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]]

Revision as of 16:39, 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

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution

Problem 31

Solution

Problem 32

Solution

Problem 33

Solution

Problem 34

Solution

Problem 35

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

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