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Difference between revisions of "1969 AHSME Problems"

(Created page with "==Problem 1== Solution ==Problem 2== Solution ==Problem 3== [[1969 AHSME Problems/Problem 3|Solution]...")
 
m
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==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>
  
  
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==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>
  
  
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==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]]
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==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]]
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==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 92:
 
==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 104:
 
==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) } 27\tfrac{1}{2}</math>
  
 
[[1969 AHSME Problems/Problem 9|Solution]]
 
[[1969 AHSME Problems/Problem 9|Solution]]
Line 46: Line 116:
 
==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]]

Revision as of 16:35, 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

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Let $n$ be the number of ways $10$ dollars can be changed into dimes and quarters, with at least one of each coin being used. Then $n$ equals:


Solution

Problem 20

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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