Difference between revisions of "1962 AHSME Problems"

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==Problem 1==
+
== 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 ==
==Problem 2==
 
  
 
The expression <math>\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}</math> is equal to:  
 
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>
+
<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:
 +
 +
\textbf{(A)}\ 5 \qquad
 +
\textbf{(B)}\ 4 \frac{1}{2} \qquad
 +
\textbf{(C)}\ 4 \qquad \textbf{(D)}\ 2 \qquad
 +
\textbf{(E)}\ 1   
 +
 +
[[1962 AHSME Problems/Problem 10|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   
 +
 +
[[1962 AHSME Problems/Problem 11|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   
 +
 +
[[1962 AHSME Problems/Problem 12|Solution]]
 +
 +
== Problem 13 ==
 +
 
 +
R varies directly as S and inverse 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   
 +
 +
[[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 121_b, written in the integral base b, 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 in 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 short 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)}\ \text{larger than any finite quantity} </math>
  
  
==See Also==
+
== See also ==
[[Category:Introductory Algebra Problems]]
+
* [[AHSME]]
 +
* [[AHSME Problems and Solutions]]
 +
* [[AMC 12 Problems and Solutions]]
 +
* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:04, 3 October 2014

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 $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}$ (Error compiling LaTeX. Unknown error_msg) 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 inverse 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:

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

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 short 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:

$\textbf{(A)}\ \frac{1}9\qquad \textbf{(B)}\ \frac{10}{81}\qquad \textbf{(C)}\ \frac{1}8\qquad \textbf{(D)}\ \frac{17}{72}\qquad \textbf{(E)}\ \text{larger than any finite quantity}$


See also

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