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Difference between revisions of "1962 AHSME Problems"
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==Problem 1==
+
{{AHSC 40 Problems
−
 
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|year = 1962
 +
}}
 +
== Problem 1 ==
 
The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to:  
 
The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to:  
  
−
<math> \textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8} </math>
+
<math>\textbf{(A)}\ \frac{4y-1}{8} \qquad  
−
 
+
\textbf{(B)}\ 8 \qquad  
 +
\textbf{(C)}\ \frac{15}{2} \qquad  
 +
\textbf{(D)}\ \frac{15}{8}\qquad
 +
\textbf{(E)}\ \frac{1}{8}</math>    
 +
 
 
[[1962 AHSME Problems/Problem 1|Solution]]
 
[[1962 AHSME Problems/Problem 1|Solution]]
−
==Problem 2==
 
  
−
The expression <math>\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}</math> is equal to:
+
== Problem 2 ==
  
−
<math> \textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1 </math>
+
The expression <math>\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}</math> is equal to:
  
 +
<math>\textbf{(A)}\ \frac{\sqrt{3}}{6} \qquad
 +
\textbf{(B)}\ \frac{-\sqrt{3}}{6} \qquad
 +
\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad
 +
\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad
 +
\textbf{(E)}\ 1</math>
 +
 
[[1962 AHSME Problems/Problem 2|Solution]]
 
[[1962 AHSME Problems/Problem 2|Solution]]
−
==Problem 3==
 
  
 +
== Problem 3 ==
 +
 
The first three terms of an arithmetic progression are <math>x - 1, x + 1, 2x + 3</math>, in the order shown. The value of <math>x</math> is:  
 
The first three terms of an arithmetic progression are <math>x - 1, x + 1, 2x + 3</math>, in the order shown. The value of <math>x</math> is:  
  
−
<math> \textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined} </math>
+
<math>\textbf{(A)}\ - 2 \qquad  
−
 
+
\textbf{(B)}\ 0 \qquad  
 +
\textbf{(C)}\ 2 \qquad  
 +
\textbf{(D)}\ 4 \qquad  
 +
\textbf{(E)}\ \text{undetermined}</math>  
 +
 
[[1962 AHSME Problems/Problem 3|Solution]]
 
[[1962 AHSME Problems/Problem 3|Solution]]
−
==Problem 4==
 
  
−
If <math>8^x = 32</math>, then x equals:  
+
== Problem 4 ==
 +
 +
If <math>8^x = 32</math>, then <math>x</math> equals:  
  
−
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4} </math>
+
<math>\textbf{(A)}\ 4 \qquad  
 +
\textbf{(B)}\ \frac{5}{3} \qquad  
 +
\textbf{(C)}\ \frac{3}{2} \qquad  
 +
\textbf{(D)}\ \frac{3}{5} \qquad  
 +
\textbf{(E)}\ \frac{1}{4} </math>  
  
 
[[1962 AHSME Problems/Problem 4|Solution]]
 
[[1962 AHSME Problems/Problem 4|Solution]]
  
 +
== Problem 5 ==
 +
 +
If the radius of a circle is increased by <math>1</math> unit, the ratio of the new circumference to the new diameter is:
 +
 +
<math>\textbf{(A)}\ \pi + 2 \qquad
 +
\textbf{(B)}\ \frac{2 \pi + 1}{2} \qquad
 +
\textbf{(C)}\ \pi \qquad
 +
\textbf{(D)}\ \frac{2\pi-1}{2}\qquad
 +
\textbf{(E)}\ \pi-2    </math>
 +
 +
[[1962 AHSME Problems/Problem 5|Solution]]
 +
 +
== Problem 6 ==
 +
 +
A square and an equilateral triangle have equal perimeters. The area of the triangle is <math>9 \sqrt{3}</math> square inches.
 +
Expressed in inches the diagonal of the square is:
 +
 +
<math>\textbf{(A)}\ \frac{9}{2} \qquad
 +
\textbf{(B)}\ 2 \sqrt{5} \qquad
 +
\textbf{(C)}\ 4 \sqrt{2} \qquad
 +
\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad
 +
\textbf{(E)}\ \text{none of these}  </math>
 +
 +
[[1962 AHSME Problems/Problem 6|Solution]]
 +
 +
== Problem 7 ==
 +
 
 +
Let the bisectors of the exterior angles at <math>B</math> and <math>C</math> of <math>\triangle ABC</math> meet at <math>D</math>. Then, if all measurements are in degrees, <math>\angle BDC</math> equals:
 +
 +
<math>\textbf{(A)}\ \frac {1}{2} (90 - A) \qquad
 +
\textbf{(B)}\ 90 - A \qquad
 +
\textbf{(C)}\ \frac {1}{2} (180 - A) \qquad \\
 +
\textbf{(D)}\ 180-A\qquad
 +
\textbf{(E)}\ 180-2A </math>   
 +
 +
[[1962 AHSME Problems/Problem 7|Solution]]
 +
 +
== Problem 8 ==
 +
 +
Given the set of <math>n</math> numbers; <math>n > 1</math>, of which one is <math>1 - \frac {1}{n}</math> and all the others are <math>1</math>. The arithmetic mean of the <math>n</math> numbers is:
 +
 +
<math>\textbf{(A)}\ 1 \qquad
 +
\textbf{(B)}\ n - \frac {1}{n} \qquad
 +
\textbf{(C)}\ n - \frac {1}{n^2} \qquad
 +
\textbf{(D)}\ 1-\frac{1}{n^2}\qquad
 +
\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}  </math>
 +
 +
[[1962 AHSME Problems/Problem 8|Solution]]
 +
 +
== Problem 9 ==
 +
 +
When <math>x^9-x</math> is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
 +
 +
<math>\textbf{(A)}\ \text{more than 5} \qquad
 +
\textbf{(B)}\ 5 \qquad
 +
\textbf{(C)}\ 4 \qquad
 +
\textbf{(D)}\ 3 \qquad
 +
\textbf{(E)}\ 2    </math>
 +
 +
[[1962 AHSME Problems/Problem 9|Solution]]
 +
 +
== Problem 10 ==
 +
 +
A man drives <math>150</math> miles to the seashore in <math>3</math> hours and <math>20</math> minutes. He returns from the shore to the starting point in <math>4</math> hours and <math>10</math> minutes.
 +
Let <math>r</math> be the average rate for the entire trip. Then the average rate for the trip going exceeds <math>r</math> in miles per hour, by:
 +
 +
<math>\textbf{(A)}\ 5 \qquad
 +
\textbf{(B)}\ 4 \frac{1}{2} \qquad
 +
\textbf{(C)}\ 4 \qquad \textbf{(D)}\ 2 \qquad
 +
\textbf{(E)}\ 1    </math>
 +
 +
[[1962 AHSME Problems/Problem 10|Solution]]
 +
 +
== Problem 11 ==
 +
 +
The difference between the larger root and the smaller root of <math>x^2 - px + (p^2 - 1)/4 = 0</math> is:
 +
 +
<math>\textbf{(A)}\ 0 \qquad
 +
\textbf{(B)}\ 1 \qquad
 +
\textbf{(C)}\ 2 \qquad
 +
\textbf{(D)}\ p \qquad
 +
\textbf{(E)}\ p+1    </math>
 +
 +
[[1962 AHSME Problems/Problem 11|Solution]]
 +
 +
== Problem 12 ==
 +
 +
When <math>\left ( 1 - \frac{1}{a} \right ) ^6</math> is expanded the sum of the last three coefficients is:
 +
 +
<math>\textbf{(A)}\ 22 \qquad
 +
\textbf{(B)}\ 11 \qquad
 +
\textbf{(C)}\ 10 \qquad
 +
\textbf{(D)}\ -10 \qquad
 +
\textbf{(E)}\ -11    </math>
 +
 +
[[1962 AHSME Problems/Problem 12|Solution]]
 +
 +
== Problem 13 ==
 +
 
 +
<math>R</math> varies directly as <math>S</math> and inversely as <math>T</math>. When <math>R = \frac43</math> and <math>T = \frac {9}{14}, S = \frac37</math>. Find <math>S</math> when <math>R = \sqrt {48}</math> and <math>T = \sqrt {75}</math>.
 +
 +
<math>\textbf{(A)}\ 28 \qquad
 +
\textbf{(B)}\ 30 \qquad
 +
\textbf{(C)}\ 40 \qquad
 +
\textbf{(D)}\ 42 \qquad
 +
\textbf{(E)}\ 60  </math> 
 +
 +
[[1962 AHSME Problems/Problem 13|Solution]]
 +
 +
== Problem 14 ==
 +
 +
Let <math>s</math> be the limiting sum of the geometric series <math>4- \frac{8}{3} + \frac{16}{9} - \dots</math>, as the number of terms increases without bound. Then <math>s</math> equals:
 +
 +
<math>\textbf{(A)}\ \text{a number between 0 and 1} \qquad
 +
\textbf{(B)}\ 2.4 \qquad
 +
\textbf{(C)}\ 2.5 \qquad
 +
\textbf{(D)}\ 3.6\qquad
 +
\textbf{(E)}\ 12  </math>
 +
 +
[[1962 AHSME Problems/Problem 14|Solution]]
 +
 +
== Problem 15 ==
 +
 +
Given <math>\triangle ABC</math> with base <math>AB</math> fixed in length and position. As the vertex <math>C</math> moves on a straight line,
 +
the intersection point of the three medians moves on:
 +
 +
<math>\textbf{(A)}\ \text{a circle} \qquad
 +
\textbf{(B)}\ \text{a parabola} \qquad
 +
\textbf{(C)}\ \text{an ellipse} \qquad
 +
\textbf{(D)}\ \text{a straight line}\qquad
 +
\textbf{(E)}\ \text{a curve here not listed}  </math>
 +
 +
[[1962 AHSME Problems/Problem 15|Solution]]
 +
 +
== Problem 16 ==
 +
 +
Given rectangle <math>R_1</math> with one side <math>2</math> inches and area <math>12</math> square inches.
 +
Rectangle <math>R_2</math> with diagonal <math>15</math> inches is similar to <math>R_1</math>. Expressed in square inches the area of <math>R_2</math> is:
 +
 +
<math>\textbf{(A)}\ \frac{9}{2} \qquad
 +
\textbf{(B)}\ 36 \qquad
 +
\textbf{(C)}\ \frac{135}{2} \qquad
 +
\textbf{(D)}\ 9\sqrt{10}\qquad
 +
\textbf{(E)}\ \frac{27\sqrt{10}}{4}  </math>
 +
 +
[[1962 AHSME Problems/Problem 16|Solution]]
 +
 +
== Problem 17 ==
 +
 
 +
If <math>a = \log_8 225</math> and <math>b = \log_2 15</math>, then <math>a</math>, in terms of <math>b</math>, is:
 +
 +
<math>\textbf{(A)}\ \frac{b}{2} \qquad
 +
\textbf{(B)}\ \frac{2b}{3}\qquad
 +
\textbf{(C)}\ b \qquad
 +
\textbf{(D)}\ \frac{3b}{2}\qquad
 +
\textbf{(E)}\ 2b</math>
 +
 +
[[1962 AHSME Problems/Problem 17|Solution]]
 +
 +
== Problem 18 ==
 +
 +
A regular dodecagon (<math>12</math> sides) is inscribed in a circle with radius <math>r</math> inches. The area of the dodecagon, in square inches, is:
 +
 +
<math>\textbf{(A)}\ 3r^2 \qquad
 +
\textbf{(B)}\ 2r^2 \qquad
 +
\textbf{(C)}\ \frac{3r^2 \sqrt{3}}{4} \qquad
 +
\textbf{(D)}\ r^2\sqrt{3}\qquad
 +
\textbf{(E)}\ 3r^2\sqrt{3} </math> 
 +
 +
[[1962 AHSME Problems/Problem 18|Solution]]
 +
 +
== Problem 19 ==
 +
 
 +
If the parabola <math>y = ax^2 + bx + c</math> passes through the points <math>( - 1, 12), (0, 5)</math>, and <math>(2, - 3)</math>, the value of <math>a + b + c</math> is:
 +
 +
<math>\textbf{(A)}\ - 4 \qquad
 +
\textbf{(B)}\ - 2 \qquad
 +
\textbf{(C)}\ 0 \qquad
 +
\textbf{(D)}\ 1 \qquad
 +
\textbf{(E)}\ 2    </math>
 +
 +
[[1962 AHSME Problems/Problem 19|Solution]]
 +
 +
== Problem 20 ==
 +
 
 +
The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
 +
 +
<math>\textbf{(A)}\ 108 \qquad
 +
\textbf{(B)}\ 90 \qquad
 +
\textbf{(C)}\ 72 \qquad
 +
\textbf{(D)}\ 54 \qquad
 +
\textbf{(E)}\ 36    </math>
 +
 +
[[1962 AHSME Problems/Problem 20|Solution]]
 +
 +
== Problem 21 ==
 +
 
 +
It is given that one root of <math>2x^2 + rx + s = 0</math>, with <math>r</math> and <math>s</math> real numbers, is <math>3+2i (i = \sqrt{-1})</math>. The value of <math>s</math> is:
 +
 +
<math>\textbf{(A)}\ \text{undetermined} \qquad
 +
\textbf{(B)}\ 5 \qquad
 +
\textbf{(C)}\ 6 \qquad
 +
\textbf{(D)}\ -13\qquad
 +
\textbf{(E)}\ 26  </math>
 +
 +
[[1962 AHSME Problems/Problem 21|Solution]]
 +
 +
== Problem 22 ==
 +
 
 +
The number <math>121_b</math>, written in the integral base <math>b</math>, is the square of an integer, for
 +
 +
<math>\textbf{(A)}\ b = 10,\text{ only} \qquad
 +
\textbf{(B)}\ b = 10 \text{ and } b = 5, \text{ only} \qquad \\
 +
\textbf{(C)}\ 2\leq b\leq 10\qquad
 +
\textbf{(D)}\ b > 2\qquad
 +
\textbf{(E)}\ \text{no value of }b </math> 
 +
 +
[[1962 AHSME Problems/Problem 22|Solution]]
 +
 +
== Problem 23 ==
 +
 +
In <math>\triangle ABC</math>, <math>CD</math> is the altitude to <math>AB</math> and <math>AE</math> is the altitude to <math>BC</math>.
 +
If the lengths of <math>AB, CD</math>, and <math>AE</math> are known, the length of <math>DB</math> is:
 +
 +
<math>\textbf{(A)}\ \text{not determined by the information given} \qquad \\
 +
\textbf{(B)}\ \text{determined only if A is an acute angle} \qquad \\
 +
\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad \\
 +
\textbf{(D)}\ \text{determined only if ABC is an acute triangle} \qquad \\
 +
\textbf{(E)}\ \text{none of these is correct} </math>   
 +
 +
[[1962 AHSME Problems/Problem 23|Solution]]
 +
 +
== Problem 24 ==
 +
 +
Three machines <math>\text{P, Q, and R,}</math> working together, can do a job in <math>x</math> hours.
 +
When working alone, <math>\text{P}</math> needs an additional <math>6</math> hours to do the job; <math>\text{Q}</math>, one additional hour; and <math>R</math>, <math>x</math> additional hours. The value of <math>x</math> is:
 +
 +
<math>\textbf{(A)}\ \frac23 \qquad
 +
\textbf{(B)}\ \frac{11}{12} \qquad
 +
\textbf{(C)}\ \frac32 \qquad
 +
\textbf{(D)}\ 2\qquad
 +
\textbf{(E)}\ 3  </math>
 +
 +
[[1962 AHSME Problems/Problem 24|Solution]]
 +
 +
== Problem 25 ==
 +
 +
Given square <math>ABCD</math> with side <math>8</math> feet. A circle is drawn through vertices <math>A</math> and <math>D</math> and tangent to side <math>BC</math>. The radius of the circle, in feet, is:
 +
 +
<math>\textbf{(A)}\ 4 \qquad
 +
\textbf{(B)}\ 4 \sqrt{2} \qquad
 +
\textbf{(C)}\ 5 \qquad
 +
\textbf{(D)}\ 5 \sqrt{2} \qquad
 +
\textbf{(E)}\ 6  </math> 
 +
 +
[[1962 AHSME Problems/Problem 25|Solution]]
 +
 +
== Problem 26 ==
 +
 +
For any real value of <math>x</math> the maximum value of <math>8x - 3x^2</math> is:
 +
 +
<math>\textbf{(A)}\ 0 \qquad
 +
\textbf{(B)}\ \frac83 \qquad
 +
\textbf{(C)}\ 4 \qquad
 +
\textbf{(D)}\ 5 \qquad
 +
\textbf{(E)}\ \frac{16}{3}  </math> 
 +
 +
[[1962 AHSME Problems/Problem 26|Solution]]
 +
 +
== Problem 27 ==
 +
 +
Let <math>a @ b</math> represent the operation on two numbers, <math>a</math> and <math>b</math>, which selects the larger of the two numbers,
 +
with <math>a@a = a</math>. Let <math>a ! b</math> represent the operator which selects the smaller of the two numbers, with <math>a ! a = a</math>.
 +
Which of the following three rules is (are) correct?
 +
 +
<math>\textbf{(1)}\ a@b = b@a \qquad \\
 +
\textbf{(2)}\ a@(b@c) = (a@b)@c \qquad \\
 +
\textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c) </math>
 +
 +
 +
<math>\textbf{(A)}\ (1)\text{ only} \qquad
 +
\textbf{(B)}\ (2) \text{ only} \qquad
 +
\textbf{(C)}\ \text{(1) and (2) only}\qquad
 +
\textbf{(D)}\ \text{(1) and (3) only}\qquad
 +
\textbf{(E)}\ \text{all three} </math> 
 +
 +
[[1962 AHSME Problems/Problem 27|Solution]]
 +
 +
== Problem 28 ==
 +
 +
The set of <math>x</math>-values satisfying the equation <math>x^{\log_{10} x} = \frac{x^3}{100}</math> consists of:
 +
 +
<math>\textbf{(A)}\ \frac{1}{10} \qquad
 +
\textbf{(B)}\ \text{10, only} \qquad
 +
\textbf{(C)}\ \text{100, only} \qquad
 +
\textbf{(D)}\ \text{10 or 100, only}\qquad
 +
\textbf{(E)}\ \text{more than two real numbers.}  </math>
 +
 +
[[1962 AHSME Problems/Problem 28|Solution]]
 +
 +
== Problem 29 ==
 +
 +
Which of the following sets of <math>x</math>-values satisfy the inequality <math>2x^2 + x < 6</math>?
 +
 +
<math>\textbf{(A)}\ - 2 < x < \frac{3}{2} \qquad
 +
\textbf{(B)}\ x > \frac32 \text{ or }x < - 2 \qquad
 +
\textbf{(C)}\ x <\frac{3}2\qquad\\
 +
\textbf{(D)}\ \frac{3}2 < x < 2\qquad
 +
\textbf{(E)}\ x <-2 </math> 
 +
 +
[[1962 AHSME Problems/Problem 29|Solution]]
 +
 +
== Problem 30 ==
 +
 
 +
Consider the statements:
 +
 +
<math> \textbf{(1)}\ \text{p and q are both true}\qquad\\
 +
\textbf{(2)}\ \text{p is true and q is false}\qquad\\
 +
\textbf{(3)}\ \text{p is false and q is true}\qquad\\
 +
\textbf{(4)}\ \text{p is false and q is false.} </math>
 +
 +
How many of these imply the negative of the statement "<math>p</math> and <math>q</math> are both true?"
 +
 +
<math>\textbf{(A)}\ 0 \qquad
 +
\textbf{(B)}\ 1 \qquad
 +
\textbf{(C)}\ 2 \qquad
 +
\textbf{(D)}\ 3 \qquad
 +
\textbf{(E)}\ 4  </math> 
 +
 +
[[1962 AHSME Problems/Problem 30|Solution]]
 +
 +
== Problem 31 ==
 +
 +
The ratio of the interior angles of two regular polygons with sides of unit length is <math>3: 2</math>. How many such pairs are there?
 +
 +
<math>\textbf{(A)}\ 1 \qquad
 +
\textbf{(B)}\ 2 \qquad
 +
\textbf{(C)}\ 3 \qquad
 +
\textbf{(D)}\ 4 \qquad
 +
\textbf{(E)}\ \infty</math>   
 +
 +
[[1962 AHSME Problems/Problem 31|Solution]]
 +
 +
== Problem 32 ==
 +
 +
If <math>x_{k+1} = x_k + \frac12</math> for <math>k=1, 2, \dots, n-1</math> and <math>x_1=1</math>, find <math>x_1 + x_2 + \dots + x_n</math>.
 +
 +
<math>\textbf{(A)}\ \frac{n+1}{2} \qquad
 +
\textbf{(B)}\ \frac{n+3}{2} \qquad
 +
\textbf{(C)}\ \frac{n^2-1}{2} \qquad
 +
\textbf{(D)}\ \frac{n^2+n}{4}\qquad
 +
\textbf{(E)}\ \frac{n^2+3n}{4} </math> 
 +
 +
[[1962 AHSME Problems/Problem 32|Solution]]
 +
 +
== Problem 33 ==
 +
 +
The set of <math>x</math>-values satisfying the inequality <math>2 \leq |x-1| \leq 5</math> is:
 +
 +
<math> \textbf{(A)}\ -4\leq x\leq-1\text{ or }3\leq x\leq 6\qquad
 +
\textbf{(B)}\ 3\leq x\leq 6\text{ or }-6\leq x\leq-3\qquad\\
 +
\textbf{(C)}\ x\leq-1\text{ or }x\geq 3\qquad
 +
\textbf{(D)}\ -1\leq x\leq 3\qquad
 +
\textbf{(E)}\ -4\leq x\leq 6 </math>
 +
 +
[[1962 AHSME Problems/Problem 33|Solution]]
 +
 +
== Problem 34 ==
 +
 +
For what real values of <math>K</math> does <math>x = K^2 (x-1)(x-2)</math> have real roots?
 +
 +
<math> \textbf{(A)}\ \text{none}\qquad
 +
\textbf{(B)}\ -2<K<1\qquad
 +
\textbf{(C)}\ -2\sqrt{2}< K < 2\sqrt{2}\qquad\\
 +
\textbf{(D)}\ K>1\text{ or }K<-2\qquad
 +
\textbf{(E)}\ \text{all} </math>
 +
 +
[[1962 AHSME Problems/Problem 34|Solution]]
 +
 +
== Problem 35 ==
 +
 +
A man on his way to dinner shortly after 6: 00 p.m. observes that the hands of his watch form an angle of <math>110^{\circ}</math>.
 +
Returning before 7: 00 p.m. he notices that again the hands of his watch form an angle of <math>110^{\circ}</math>.
 +
The number of minutes that he has been away is:
 +
 +
<math>\textbf{(A)}\ 36 \frac23 \qquad
 +
\textbf{(B)}\ 40 \qquad
 +
\textbf{(C)}\ 42 \qquad
 +
\textbf{(D)}\ 42.4 \qquad
 +
\textbf{(E)}\ 45  </math> 
 +
 +
[[1962 AHSME Problems/Problem 35|Solution]]
 +
 +
== Problem 36 ==
 +
 
 +
If both <math>x</math> and <math>y</math> are both integers, how many pairs of solutions are there of the equation <math>(x-8)(x-10) = 2^y</math>?
 +
 +
<math>\textbf{(A)}\ 0 \qquad
 +
\textbf{(B)}\ 1 \qquad
 +
\textbf{(C)}\ 2 \qquad
 +
\textbf{(D)}\ 3 \qquad
 +
\textbf{(E)}\ \text{more than 3}  </math> 
 +
 +
[[1962 AHSME Problems/Problem 36|Solution]]
 +
 +
== Problem 37 ==
 +
 
 +
<math>ABCD</math> is a square with side of unit length. Points <math>E</math> and <math>F</math> are taken respectively on sides <math>AB</math> and <math>AD</math>
 +
so that <math>AE = AF</math> and the quadrilateral <math>CDFE</math> has maximum area. In square units this maximum area is:
 +
 +
<math>\textbf{(A)}\ \frac12 \qquad
 +
\textbf{(B)}\ \frac {9}{16} \qquad
 +
\textbf{(C)}\ \frac{19}{32} \qquad
 +
\textbf{(D)}\ \frac{5}{8}\qquad
 +
\textbf{(E)}\ \frac{2}3 </math>   
 +
 +
[[1962 AHSME Problems/Problem 37|Solution]]
  
 +
== Problem 38 ==
 +
 +
The population of Nosuch Junction at one time was a perfect square.
 +
Later, with an increase of <math>100</math>, the population was one more than a perfect square.
 +
Now, with an additional increase of <math>100</math>, the population is again a perfect square.
  
 +
The original population is a multiple of:
  
 +
<math>\textbf{(A)}\ 3 \qquad
 +
\textbf{(B)}\ 7 \qquad
 +
\textbf{(C)}\ 9 \qquad
 +
\textbf{(D)}\ 11 \qquad
 +
\textbf{(E)}\ 17  </math> 
 +
 +
[[1962 AHSME Problems/Problem 38|Solution]]
  
 +
== Problem 39 ==
 +
 +
Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches.
 +
Its area is <math>3 \sqrt{15}</math> square inches. The length of the third median in inches, is:
  
 +
<math>\textbf{(A)}\ 4 \qquad
 +
\textbf{(B)}\ 3 \sqrt{3} \qquad
 +
\textbf{(C)}\ 3 \sqrt{6} \qquad
 +
\textbf{(D)}\ 6\sqrt{3}\qquad
 +
\textbf{(E)}\ 6\sqrt{6}  </math> 
 +
 +
[[1962 AHSME Problems/Problem 39|Solution]]
  
 +
== Problem 40 ==
 +
 +
The limiting sum of the infinite series, <math>\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots</math> whose <math>n</math>th term is <math>\frac{n}{10^n}</math> is:
  
 +
<math> \textbf{(A)}\ \frac{1}9\qquad
 +
\textbf{(B)}\ \frac{10}{81}\qquad
 +
\textbf{(C)}\ \frac{1}8\qquad
 +
\textbf{(D)}\ \frac{17}{72}\qquad
 +
\textbf{(E)}\ \infty </math>
  
 +
[[1962 AHSME Problems/Problem 40|Solution]]
  
 +
== See also ==
  
 +
* [[AMC 12 Problems and Solutions]]
 +
* [[Mathematics competition resources]]
  
 +
{{AHSME 40p box|year=1962|before=[[1961 AHSME|1961 AHSC]]|after=[[1963 AHSME|1963 AHSC]]}} 
  
−
==See Also==
 
−
[[Category:Introductory Algebra Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:40, 10 April 2023

1962 AHSC (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF

Instructions

  1. 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.
  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 • 36 • 37 • 38 • 39 • 40

Problem 1

The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:

$\textbf{(A)}\ \frac{4y-1}{8} \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ \frac{15}{2} \qquad \textbf{(D)}\ \frac{15}{8}\qquad \textbf{(E)}\ \frac{1}{8}$

Solution

Problem 2

The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ is equal to:

$\textbf{(A)}\ \frac{\sqrt{3}}{6} \qquad \textbf{(B)}\ \frac{-\sqrt{3}}{6} \qquad \textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad \textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad \textbf{(E)}\ 1$

Solution

Problem 3

The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:

$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \text{undetermined}$

Solution

Problem 4

If $8^x = 32$, then $x$ equals:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ \frac{5}{3} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \frac{3}{5} \qquad \textbf{(E)}\ \frac{1}{4}$

Solution

Problem 5

If the radius of a circle is increased by $1$ unit, the ratio of the new circumference to the new diameter is:

$\textbf{(A)}\ \pi + 2 \qquad \textbf{(B)}\ \frac{2 \pi + 1}{2} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{2\pi-1}{2}\qquad \textbf{(E)}\ \pi-2$

Solution

Problem 6

A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:

$\textbf{(A)}\ \frac{9}{2} \qquad \textbf{(B)}\ 2 \sqrt{5} \qquad \textbf{(C)}\ 4 \sqrt{2} \qquad \textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 7

Let the bisectors of the exterior angles at $B$ and $C$ of $\triangle ABC$ meet at $D$. Then, if all measurements are in degrees, $\angle BDC$ equals:

$\textbf{(A)}\ \frac {1}{2} (90 - A) \qquad \textbf{(B)}\ 90 - A \qquad \textbf{(C)}\ \frac {1}{2} (180 - A) \qquad \\ \textbf{(D)}\ 180-A\qquad \textbf{(E)}\ 180-2A$

Solution

Problem 8

Given the set of $n$ numbers; $n > 1$, of which one is $1 - \frac {1}{n}$ and all the others are $1$. The arithmetic mean of the $n$ numbers is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ n - \frac {1}{n} \qquad \textbf{(C)}\ n - \frac {1}{n^2} \qquad \textbf{(D)}\ 1-\frac{1}{n^2}\qquad \textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}$

Solution

Problem 9

When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:

$\textbf{(A)}\ \text{more than 5} \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 2$

Solution

Problem 10

A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \frac{1}{2} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$

Solution

Problem 11

The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ p \qquad \textbf{(E)}\ p+1$

Solution

Problem 12

When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ -10 \qquad \textbf{(E)}\ -11$

Solution

Problem 13

$R$ varies directly as $S$ and inversely as $T$. When $R = \frac43$ and $T = \frac {9}{14}, S = \frac37$. Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$.

$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$

Solution

Problem 14

Let $s$ be the limiting sum of the geometric series $4- \frac{8}{3} + \frac{16}{9} - \dots$, as the number of terms increases without bound. Then $s$ equals:

$\textbf{(A)}\ \text{a number between 0 and 1} \qquad \textbf{(B)}\ 2.4 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 3.6\qquad \textbf{(E)}\ 12$

Solution

Problem 15

Given $\triangle ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:

$\textbf{(A)}\ \text{a circle} \qquad \textbf{(B)}\ \text{a parabola} \qquad \textbf{(C)}\ \text{an ellipse} \qquad \textbf{(D)}\ \text{a straight line}\qquad \textbf{(E)}\ \text{a curve here not listed}$

Solution

Problem 16

Given rectangle $R_1$ with one side $2$ inches and area $12$ square inches. Rectangle $R_2$ with diagonal $15$ inches is similar to $R_1$. Expressed in square inches the area of $R_2$ is:

$\textbf{(A)}\ \frac{9}{2} \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ \frac{135}{2} \qquad \textbf{(D)}\ 9\sqrt{10}\qquad \textbf{(E)}\ \frac{27\sqrt{10}}{4}$

Solution

Problem 17

If $a = \log_8 225$ and $b = \log_2 15$, then $a$, in terms of $b$, is:

$\textbf{(A)}\ \frac{b}{2} \qquad \textbf{(B)}\ \frac{2b}{3}\qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ \frac{3b}{2}\qquad \textbf{(E)}\ 2b$

Solution

Problem 18

A regular dodecagon ($12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:

$\textbf{(A)}\ 3r^2 \qquad \textbf{(B)}\ 2r^2 \qquad \textbf{(C)}\ \frac{3r^2 \sqrt{3}}{4} \qquad \textbf{(D)}\ r^2\sqrt{3}\qquad \textbf{(E)}\ 3r^2\sqrt{3}$

Solution

Problem 19

If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12), (0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:

$\textbf{(A)}\ - 4 \qquad \textbf{(B)}\ - 2 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

Solution

Problem 20

The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:

$\textbf{(A)}\ 108 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 72 \qquad \textbf{(D)}\ 54 \qquad \textbf{(E)}\ 36$

Solution

Problem 21

It is given that one root of $2x^2 + rx + s = 0$, with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$. The value of $s$ is:

$\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ -13\qquad \textbf{(E)}\ 26$

Solution

Problem 22

The number $121_b$, written in the integral base $b$, is the square of an integer, for

$\textbf{(A)}\ b = 10,\text{ only} \qquad \textbf{(B)}\ b = 10 \text{ and } b = 5, \text{ only} \qquad \\ \textbf{(C)}\ 2\leq b\leq 10\qquad \textbf{(D)}\ b > 2\qquad \textbf{(E)}\ \text{no value of }b$

Solution

Problem 23

In $\triangle ABC$, $CD$ is the altitude to $AB$ and $AE$ is the altitude to $BC$. If the lengths of $AB, CD$, and $AE$ are known, the length of $DB$ is:

$\textbf{(A)}\ \text{not determined by the information given} \qquad \\ \textbf{(B)}\ \text{determined only if A is an acute angle} \qquad \\ \textbf{(C)}\ \text{determined only if B is an acute angle} \qquad \\ \textbf{(D)}\ \text{determined only if ABC is an acute triangle} \qquad \\ \textbf{(E)}\ \text{none of these is correct}$

Solution

Problem 24

Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$, one additional hour; and $R$, $x$ additional hours. The value of $x$ is:

$\textbf{(A)}\ \frac23 \qquad \textbf{(B)}\ \frac{11}{12} \qquad \textbf{(C)}\ \frac32 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 3$

Solution

Problem 25

Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$. The radius of the circle, in feet, is:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4 \sqrt{2} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 5 \sqrt{2} \qquad \textbf{(E)}\ 6$

Solution

Problem 26

For any real value of $x$ the maximum value of $8x - 3x^2$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac83 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \frac{16}{3}$

Solution

Problem 27

Let $a @ b$ represent the operation on two numbers, $a$ and $b$, which selects the larger of the two numbers, with $a@a = a$. Let $a ! b$ represent the operator which selects the smaller of the two numbers, with $a ! a = a$. Which of the following three rules is (are) correct?

$\textbf{(1)}\ a@b = b@a \qquad \\ \textbf{(2)}\ a@(b@c) = (a@b)@c \qquad \\ \textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c)$


$\textbf{(A)}\ (1)\text{ only} \qquad \textbf{(B)}\ (2) \text{ only} \qquad \textbf{(C)}\ \text{(1) and (2) only}\qquad \textbf{(D)}\ \text{(1) and (3) only}\qquad \textbf{(E)}\ \text{all three}$

Solution

Problem 28

The set of $x$-values satisfying the equation $x^{\log_{10} x} = \frac{x^3}{100}$ consists of:

$\textbf{(A)}\ \frac{1}{10} \qquad \textbf{(B)}\ \text{10, only} \qquad \textbf{(C)}\ \text{100, only} \qquad \textbf{(D)}\ \text{10 or 100, only}\qquad \textbf{(E)}\ \text{more than two real numbers.}$

Solution

Problem 29

Which of the following sets of $x$-values satisfy the inequality $2x^2 + x < 6$?

$\textbf{(A)}\ - 2 < x < \frac{3}{2} \qquad \textbf{(B)}\ x > \frac32 \text{ or }x < - 2 \qquad \textbf{(C)}\ x <\frac{3}2\qquad\\ \textbf{(D)}\ \frac{3}2 < x < 2\qquad \textbf{(E)}\ x <-2$

Solution

Problem 30

Consider the statements:

$\textbf{(1)}\ \text{p and q are both true}\qquad\\ \textbf{(2)}\ \text{p is true and q is false}\qquad\\ \textbf{(3)}\ \text{p is false and q is true}\qquad\\ \textbf{(4)}\ \text{p is false and q is false.}$

How many of these imply the negative of the statement "$p$ and $q$ are both true?"

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

Solution

Problem 31

The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$. How many such pairs are there?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$

Solution

Problem 32

If $x_{k+1} = x_k + \frac12$ for $k=1, 2, \dots, n-1$ and $x_1=1$, find $x_1 + x_2 + \dots + x_n$.

$\textbf{(A)}\ \frac{n+1}{2} \qquad \textbf{(B)}\ \frac{n+3}{2} \qquad \textbf{(C)}\ \frac{n^2-1}{2} \qquad \textbf{(D)}\ \frac{n^2+n}{4}\qquad \textbf{(E)}\ \frac{n^2+3n}{4}$

Solution

Problem 33

The set of $x$-values satisfying the inequality $2 \leq |x-1| \leq 5$ is:

$\textbf{(A)}\ -4\leq x\leq-1\text{ or }3\leq x\leq 6\qquad \textbf{(B)}\ 3\leq x\leq 6\text{ or }-6\leq x\leq-3\qquad\\ \textbf{(C)}\ x\leq-1\text{ or }x\geq 3\qquad \textbf{(D)}\ -1\leq x\leq 3\qquad \textbf{(E)}\ -4\leq x\leq 6$

Solution

Problem 34

For what real values of $K$ does $x = K^2 (x-1)(x-2)$ have real roots?

$\textbf{(A)}\ \text{none}\qquad \textbf{(B)}\ -2<K<1\qquad \textbf{(C)}\ -2\sqrt{2}< K < 2\sqrt{2}\qquad\\ \textbf{(D)}\ K>1\text{ or }K<-2\qquad \textbf{(E)}\ \text{all}$

Solution

Problem 35

A man on his way to dinner shortly after 6: 00 p.m. observes that the hands of his watch form an angle of $110^{\circ}$. Returning before 7: 00 p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$. The number of minutes that he has been away is:

$\textbf{(A)}\ 36 \frac23 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 42.4 \qquad \textbf{(E)}\ 45$

Solution

Problem 36

If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{more than 3}$

Solution

Problem 37

$ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE = AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:

$\textbf{(A)}\ \frac12 \qquad \textbf{(B)}\ \frac {9}{16} \qquad \textbf{(C)}\ \frac{19}{32} \qquad \textbf{(D)}\ \frac{5}{8}\qquad \textbf{(E)}\ \frac{2}3$

Solution

Problem 38

The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$, the population was one more than a perfect square. Now, with an additional increase of $100$, the population is again a perfect square.

The original population is a multiple of:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 17$

Solution

Problem 39

Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt{3} \qquad \textbf{(C)}\ 3 \sqrt{6} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ 6\sqrt{6}$

Solution

Problem 40

The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is:

$\textbf{(A)}\ \frac{1}9\qquad \textbf{(B)}\ \frac{10}{81}\qquad \textbf{(C)}\ \frac{1}8\qquad \textbf{(D)}\ \frac{17}{72}\qquad \textbf{(E)}\ \infty$

Solution

See also

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
1961 AHSC 		Followed by
1963 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


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