Difference between revisions of "1965 AHSME"

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== Problem 1 ==
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'''1965 AHSME''' problems and solutions.  The first link contains the full set of test problems.  The rest contain each individual problem and its solution.
The number of real values of <math>x</math> satisfying the equation <math>2^{2x^2 - 7x + 5} = 1</math> is:
 
  
<math>\textbf{(A)}\ 0 \qquad
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* [[1965 AHSME Problems|Entire Exam]]
\textbf{(B) }\ 1 \qquad
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* [[1965 AHSME Problems|Answer Key]]
\textbf{(C) }\ 2 \qquad
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** [[1965 AHSME Problems/Problem 1|Problem 1]]  
\textbf{(D) }\ 3 \qquad
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** [[1965 AHSME Problems/Problem 2|Problem 2]]
\textbf{(E) }\ \text{more than 4}  </math> 
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** [[1965 AHSME Problems/Problem 3|Problem 3]]
 
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** [[1965 AHSME Problems/Problem 4|Problem 4]]
[[1965 AHSME Problems/Problem 1|Solution]]
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** [[1965 AHSME Problems/Problem 5|Problem 5]]
 
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** [[1965 AHSME Problems/Problem 6|Problem 6]]
== Problem 2==
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** [[1965 AHSME Problems/Problem 7|Problem 7]]
 
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** [[1965 AHSME Problems/Problem 8|Problem 8]]
A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the
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** [[1965 AHSME Problems/Problem 9|Problem 9]]
shorter of the arcs intercepted by the side, is:
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** [[1965 AHSME Problems/Problem 10|Problem 10]]
 
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** [[1965 AHSME Problems/Problem 11|Problem 11]]
<math>\textbf{(A)}\ 1: 1 \qquad
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** [[1965 AHSME Problems/Problem 12|Problem 12]]
\textbf{(B) }\ 1: 6 \qquad
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** [[1965 AHSME Problems/Problem 13|Problem 13]]
\textbf{(C) }\ 1: \pi \qquad
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** [[1965 AHSME Problems/Problem 14|Problem 14]]
\textbf{(D) }\ 3: \pi \qquad
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** [[1965 AHSME Problems/Problem 15|Problem 15]]
\textbf{(E) }\ 6:\pi  </math>   
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** [[1965 AHSME Problems/Problem 16|Problem 16]]
 
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** [[1965 AHSME Problems/Problem 17|Problem 17]]
[[1965 AHSME Problems/Problem 2|Solution]]
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** [[1965 AHSME Problems/Problem 18|Problem 18]]
 
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** [[1965 AHSME Problems/Problem 19|Problem 19]]
== Problem 3==
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** [[1965 AHSME Problems/Problem 20|Problem 20]]
 
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** [[1965 AHSME Problems/Problem 21|Problem 21]]
The expression <math>(81)^{ - 2^{ - 2}}</math> has the same value as:
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** [[1965 AHSME Problems/Problem 22|Problem 22]]
 
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** [[1965 AHSME Problems/Problem 23|Problem 23]]
<math>\textbf{(A)}\ \frac {1}{81} \qquad
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** [[1965 AHSME Problems/Problem 24|Problem 24]]
\textbf{(B) }\ \frac {1}{3} \qquad
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** [[1965 AHSME Problems/Problem 25|Problem 25]]
\textbf{(C) }\ 3 \qquad
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** [[1965 AHSME Problems/Problem 26|Problem 26]]
\textbf{(D) }\ 81\qquad
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** [[1965 AHSME Problems/Problem 27|Problem 27]]
\textbf{(E) }\ 81^4 </math> 
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** [[1965 AHSME Problems/Problem 28|Problem 28]]
 
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** [[1965 AHSME Problems/Problem 29|Problem 29]]
[[1965 AHSME Problems/Problem 3|Solution]]
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** [[1965 AHSME Problems/Problem 30|Problem 30]]
 
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** [[1965 AHSME Problems/Problem 31|Problem 31]]
== Problem 4==
+
** [[1965 AHSME Problems/Problem 32|Problem 32]]
 
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** [[1965 AHSME Problems/Problem 33|Problem 33]]
Line <math>\ell_2</math> intersects line <math>\ell_1</math> and line <math>\ell_3</math> is parallel to <math>\ell_1</math>. The three lines are distinct and lie in a plane.
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** [[1965 AHSME Problems/Problem 34|Problem 34]]
The number of points equidistant from all three lines is:
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** [[1965 AHSME Problems/Problem 35|Problem 35]]
 
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** [[1965 AHSME Problems/Problem 36|Problem 36]]
<math>\textbf{(A)}\ 0 \qquad
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** [[1965 AHSME Problems/Problem 37|Problem 37]]
\textbf{(B) }\ 1 \qquad
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** [[1965 AHSME Problems/Problem 38|Problem 38]]
\textbf{(C) }\ 2 \qquad
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** [[1965 AHSME Problems/Problem 39|Problem 39]]
\textbf{(D) }\ 4 \qquad
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** [[1965 AHSME Problems/Problem 40|Problem 40]]
\textbf{(E) }\ 8  </math> 
+
== See also ==
 
+
{{AHSME 40p box|year=1965|before=[[1964 AHSME]]|after=[[1966 AHSME]]}}
[[1965 AHSME Problems/Problem 4|Solution]]
 
 
 
== Problem 5==
 
 
 
When the repeating decimal <math>0.363636\ldots</math> is written in simplest fractional form, the sum of the numerator and denominator is:
 
 
 
<math>\textbf{(A)}\ 15 \qquad
 
\textbf{(B) }\ 45 \qquad
 
\textbf{(C) }\ 114 \qquad
 
\textbf{(D) }\ 135 \qquad
 
\textbf{(E) }\ 150    </math>
 
 
 
[[1965 AHSME Problems/Problem 5|Solution]]
 
 
 
== Problem 6==
 
 
 
If <math>10^{\log_{10}9} = 8x + 5</math> then <math>x</math> equals:
 
 
 
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B) }\ \frac {1}{2} \qquad
 
\textbf{(C) }\ \frac {5}{8} \qquad
 
\textbf{(D) }\ \frac{9}{8}\qquad
 
\textbf{(E) }\ \frac{2\log_{10}3-5}{8} </math>
 
 
 
[[1965 AHSME Problems/Problem 6|Solution]]
 
 
 
== Problem 7==
 
 
 
The sum of the reciprocals of the roots of the equation <math>ax^2 + bx + c = 0</math> is:
 
 
 
<math>\textbf{(A)}\ \frac {1}{a} + \frac {1}{b} \qquad
 
\textbf{(B) }\ - \frac {c}{b} \qquad
 
\textbf{(C) }\ \frac{b}{c}\qquad
 
\textbf{(D) }\ -\frac{a}{b}\qquad
 
\textbf{(E) }\ -\frac{b}{c} </math>
 
 
 
[[1965 AHSME Problems/Problem 7|Solution]]
 
 
 
== Problem 8==
 
 
 
One side of a given triangle is 18 inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid
 
whose area is one-third of that of the triangle. The length of this segment, in inches, is:
 
 
 
<math>\textbf{(A)}\ 6\sqrt {6} \qquad
 
\textbf{(B) }\ 9\sqrt {2} \qquad
 
\textbf{(C) }\ 12 \qquad
 
\textbf{(D) }\ 6\sqrt{3}\qquad
 
\textbf{(E) }\ 9 </math>
 
 
 
[[1965 AHSME Problems/Problem 8|Solution]]
 
 
 
== Problem 9==
 
 
 
The vertex of the parabola <math>y = x^2 - 8x + c</math> will be a point on the <math>x</math>-axis if the value of <math>c</math> is:
 
 
 
<math>\textbf{(A)}\ - 16 \qquad
 
\textbf{(B) }\ - 4 \qquad
 
\textbf{(C) }\ 4 \qquad
 
\textbf{(D) }\ 8 \qquad
 
\textbf{(E) }\ 16  </math> 
 
 
 
[[1965 AHSME Problems/Problem 9|Solution]]
 
 
 
== Problem 10==
 
 
 
The statement <math>x^2 - x - 6 < 0</math> is equivalent to the statement:
 
 
 
<math>\textbf{(A)}\ - 2 < x < 3 \qquad
 
\textbf{(B) }\ x > - 2 \qquad
 
\textbf{(C) }\ x < 3 \
 
\textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad
 
\textbf{(E) }\ x > 3 \text{ and }x < - 2  </math> 
 
 
 
[[1965 AHSME Problems/Problem 10|Solution]]
 
 
 
== Problem 11==
 
 
 
Consider the statements:
 
<cmath>I: (\sqrt { - 4})(\sqrt { - 16}) = \sqrt {( - 4)( - 16)}, \
 
II: \sqrt {( - 4)( - 16)} = \sqrt {64}, and \sqrt {64} = 8. </cmath>
 
Of these the following are incorrect.
 
 
 
<math>\textbf{(A)}\ \text{none} \qquad
 
\textbf{(B) }\ \text{I only} \qquad
 
\textbf{(C) }\ \text{II only} \qquad
 
\textbf{(D) }\ \text{III only}\qquad
 
\textbf{(E) }\ \text{I and III only} </math>
 
 
 
[[1965 AHSME Problems/Problem 11|Solution]]
 
 
 
== Problem 12==
 
 
 
A rhombus is inscribed in <math>\triangle ABC</math> in such a way that one of its vertices is <math>A</math> and two of its sides lie along <math>AB</math> and <math>AC</math>.
 
If <math>\overline{AC} = 6</math> inches, <math>\overline{AB} = 12</math> inches, and <math>\overline{BC} = 8</math> inches, the side of the rhombus, in inches, is:
 
 
 
<math>\textbf{(A)}\ 2 \qquad
 
\textbf{(B) }\ 3 \qquad
 
\textbf{(C) }\ 3 \frac {1}{2} \qquad
 
\textbf{(D) }\ 4 \qquad
 
\textbf{(E) }\ 5  </math> 
 
 
 
[[1965 AHSME Problems/Problem 12|Solution]]
 
 
 
== Problem 13==
 
 
 
Let <math>n</math> be the number of number-pairs <math>(x,y)</math> which satisfy <math>5y - 3x = 15</math> and <math>x^2 + y^2 \le 16</math>. Then <math>n</math> is:
 
 
 
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B) }\ 1 \qquad
 
\textbf{(C) }\ 2 \qquad
 
\textbf{(D) }\ \text{more than two, but finite}\qquad
 
\textbf{(E) }\ \text{greater than any finite number} </math>
 
 
 
[[1965 AHSME Problems/Problem 13|Solution]]
 
 
 
== Problem 14==
 
 
 
The sum of the numerical coefficients in the complete expansion of <math>(x^2 - 2xy + y^2)^7</math> is:
 
 
 
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B) }\ 7 \qquad
 
\textbf{(C) }\ 14 \qquad
 
\textbf{(D) }\ 128 \qquad
 
\textbf{(E) }\ 128^2  </math> 
 
 
 
[[1965 AHSME Problems/Problem 14|Solution]]
 
 
 
== Problem 15==
 
 
 
The symbol <math>25_b</math> represents a two-digit number in the base <math>b</math>. If the number <math>52_b</math> is double the number <math>25_b</math>, then <math>b</math> is:
 
 
 
<math>\textbf{(A)}\ 7 \qquad
 
\textbf{(B) }\ 8 \qquad
 
\textbf{(C) }\ 9 \qquad
 
\textbf{(D) }\ 11 \qquad
 
\textbf{(E) }\ 12 </math>   
 
 
 
[[1965 AHSME Problems/Problem 15|Solution]]
 
 
 
== Problem 16==
 
 
 
Let line <math>AC</math> be perpendicular to line <math>CE</math>. Connect <math>A</math> to <math>D</math>, the midpoint of <math>CE</math>, and connect <math>E</math> to <math>B</math>,
 
the midpoint of <math>AC</math>. If <math>AD</math> and <math>EB</math> intersect in point <math>F</math>, and <math>\overline{BC} = \overline{CD} = 15</math> inches,
 
then the area of triangle <math>DFE</math>, in square inches, is:
 
 
 
<math>\textbf{(A)}\ 50 \qquad
 
\textbf{(B) }\ 50\sqrt {2} \qquad
 
\textbf{(C) }\ 75 \qquad
 
\textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad
 
\textbf{(E) }\ 100 </math>
 
 
 
[[1965 AHSME Problems/Problem 16|Solution]]
 
 
 
== Problem 17==
 
 
 
Given the true statement: The picnic on Sunday will not be held only if the weather is not fair. We can then conclude that:
 
 
 
<math>\textbf{(A)}\ \text{If the picnic is held, Sunday's weather is undoubtedly fair.} \
 
\textbf{(B) }\ \text{If the picnic is not held, Sunday's weather is possibly unfair.} \
 
\textbf{(C) }\ \text{If it is not fair Sunday, the picnic will not be held.} \
 
\textbf{(D) }\ \text{If it is fair Sunday, the picnic may be held.} \
 
\textbf{(E) }\ \text{If it is fair Sunday, the picnic must be held.}    </math>
 
 
 
[[1965 AHSME Problems/Problem 17|Solution]]
 
 
 
== Problem 18==
 
 
 
If <math>1 - y</math> is used as an approximation to the value of <math>\frac {1}{1 + y}, |y| < 1</math>, the ratio of the error made to the correct value is:
 
 
 
<math>\textbf{(A)}\ y \qquad
 
\textbf{(B) }\ y^2 \qquad
 
\textbf{(C) }\ \frac {1}{1 + y} \qquad
 
\textbf{(D) }\ \frac{y}{1+y}\qquad
 
\textbf{(E) }\ \frac{y^2}{1+y}\qquad </math>
 
 
 
[[1965 AHSME Problems/Problem 18|Solution]]
 
 
 
== Problem 19==
 
 
 
If <math>x^4 + 4x^3 + 6px^2 + 4qx + r</math> is exactly divisible by <math>x^3 + 3x^2 + 9x + 3</math>, the value of <math>(p + q)r</math> is:
 
 
 
<math>\textbf{(A)}\ - 18 \qquad
 
\textbf{(B) }\ 12 \qquad
 
\textbf{(C) }\ 15 \qquad
 
\textbf{(D) }\ 27 \qquad
 
\textbf{(E) }\ 45 \qquad  </math> 
 
 
 
[[1965 AHSME Problems/Problem 19|Solution]]
 
 
 
== Problem 20==
 
 
 
For every <math>n</math> the sum of n terms of an arithmetic progression is <math>2n + 3n^2</math>. The <math>r</math>th term is:
 
 
 
<math>\textbf{(A)}\ 3r^2 \qquad
 
\textbf{(B) }\ 3r^2 + 2r \qquad
 
\textbf{(C) }\ 6r - 1 \qquad
 
\textbf{(D) }\ 5r + 5 \qquad
 
\textbf{(E) }\ 6r+2\qquad </math>
 
 
 
[[1965 AHSME Problems/Problem 20|Solution]]
 
 
 
== Problem 21==
 
 
 
It is possible to choose <math>x > \frac {2}{3}</math> in such a way that the value of <math>\log_{10}(x^2 + 3) - 2 \log_{10}x</math> is
 
 
 
<math>\textbf{(A)}\ \text{negative} \qquad
 
\textbf{(B) }\ \text{zero} \qquad
 
\textbf{(C) }\ \text{one} \
 
\textbf{(D) }\ \text{smaller than any positive number that might be specified} \
 
\textbf{(E) }\ \text{greater than any positive number that might be specified}    </math>
 
 
 
[[1965 AHSME Problems/Problem 21|Solution]]
 
 
 
== Problem 22==
 
 
 
If <math>a_2 \neq 0</math> and <math>r</math> and <math>s</math> are the roots of <math>a_0 + a_1x + a_2x^2 = 0</math>, then the equality
 
<math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds:
 
 
 
<math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0
 
\textbf{(B) }\ \text{for all values of }x \
 
\textbf{(C) }\ \text{only when }x = 0
 
\textbf{(D) }\ \text{only when }x = r \text{ or }x = s \
 
\textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0  </math> 
 
 
 
[[1965 AHSME Problems/Problem 22|Solution]]
 
 
 
== Problem 23==
 
 
 
If we write <math>|x^2 - 4| < N</math> for all <math>x</math> such that <math>|x - 2| < 0.01</math>, the smallest value we can use for <math>N</math> is:
 
 
 
<math>\textbf{(A)}\ .0301 \qquad
 
\textbf{(B) }\ .0349 \qquad
 
\textbf{(C) }\ .0399 \qquad
 
\textbf{(D) }\ .0401 \qquad
 
\textbf{(E) }\ .0499\qquad </math>
 
 
 
[[1965 AHSME Problems/Problem 23|Solution]]
 
 
 
== Problem 24==
 
 
 
Given the sequence <math>10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}</math>,
 
the smallest value of n such that the product of the first <math>n</math> members of this sequence exceeds <math>100000</math> is:
 
 
 
<math>\textbf{(A)}\ 7 \qquad
 
\textbf{(B) }\ 8 \qquad
 
\textbf{(C) }\ 9 \qquad
 
\textbf{(D) }\ 10 \qquad
 
\textbf{(E) }\ 11  </math> 
 
 
 
[[1965 AHSME Problems/Problem 24|Solution]]
 
 
 
== Problem 25==
 
 
 
Let <math>ABCD</math> be a quadrilateral with <math>AB</math> extended to <math>E</math> so that <math>\overline{AB} = \overline{BE}</math>.
 
Lines <math>AC</math> and <math>CE</math> are drawn to form <math>\angle{ACE}</math>. For this angle to be a right angle it is necessary that quadrilateral <math>ABCD</math> have:
 
 
 
<math>\textbf{(A)}\ \text{all angles equal}
 
\textbf{(B) }\ \text{all sides equal} \
 
\textbf{(C) }\ \text{two pairs of equal sides}
 
\textbf{(D) }\ \text{one pair of equal sides} \
 
\textbf{(E) }\ \text{one pair of equal angles} </math>   
 
 
 
[[1965 AHSME Problems/Problem 25|Solution]]
 
 
 
== Problem 26==
 
 
 
For the numbers <math>a, b, c, d, e</math> define <math>m</math> to be the arithmetic mean of all five numbers;
 
<math>k</math> to be the arithmetic mean of <math>a</math> and <math>b</math>; <math>l</math> to be the arithmetic mean of <math>c, d</math>, and <math>e</math>;
 
and <math>p</math> to be the arithmetic mean of <math>k</math> and <math>l</math>. Then, no matter how <math>a, b, c, d</math>, and <math>e</math> are chosen, we shall always have:
 
 
 
<math>\textbf{(A)}\ m = p \qquad
 
\textbf{(B) }\ m \ge p \qquad
 
\textbf{(C) }\ m > p \qquad \
 
\textbf{(D) }\ m < p\qquad
 
\textbf{(E) }\ \text{none of these} </math>
 
 
 
[[1965 AHSME Problems/Problem 26|Solution]]
 
 
 
== Problem 27==
 
 
 
When <math>y^2 + my + 2</math> is divided by <math>y - 1</math> the quotient is <math>f(y)</math> and the remainder is <math>R_1</math>.
 
When <math>y^2 + my + 2</math> is divided by <math>y + 1</math> the quotient is <math>g(y)</math> and the remainder is <math>R_2</math>. If <math>R_1 = R_2</math> then <math>m</math> is:
 
 
 
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B) }\ 1 \qquad
 
\textbf{(C) }\ 2 \qquad
 
\textbf{(D) }\ - 1 \qquad
 
\textbf{(E) }\ \text{an undetermined constant} </math>
 
 
 
[[1965 AHSME Problems/Problem 27|Solution]]
 
 
 
== Problem 28==
 
 
 
An escalator (moving staircase) of <math>n</math> uniform steps visible at all times descends at constant speed.
 
Two boys, <math>A</math> and <math>Z</math>, walk down the escalator steadily as it moves, A negotiating twice as many escalator
 
steps per minute as <math>Z</math>. <math>A</math> reaches the bottom after taking <math>27</math> steps while <math>Z</math> reaches the bottom after taking <math>18</math> steps. Then <math>n</math> is:
 
 
 
<math>\textbf{(A)}\ 63 \qquad
 
\textbf{(B) }\ 54 \qquad
 
\textbf{(C) }\ 45 \qquad
 
\textbf{(D) }\ 36 \qquad
 
\textbf{(E) }\ 30    </math>
 
 
 
[[1965 AHSME Problems/Problem 28|Solution]]
 
 
 
== Problem 29==
 
 
 
Of <math>28</math> students taking at least one subject the number taking Mathematics and English only equals the number
 
taking Mathematics only. No student takes English only or History only, and six students take Mathematics and
 
History, but not English. The number taking English and History only is five times the number taking all three subjects.
 
If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:
 
 
 
<math>\textbf{(A)}\ 5 \qquad
 
\textbf{(B) }\ 6 \qquad
 
\textbf{(C) }\ 7 \qquad
 
\textbf{(D) }\ 8 \qquad
 
\textbf{(E) }\ 9  </math> 
 
 
 
[[1965 AHSME Problems/Problem 29|Solution]]
 
 
 
== Problem 30==
 
 
 
Let <math>BC</math> of right triangle <math>ABC</math> be the diameter of a circle intersecting hypotenuse <math>AB</math> in <math>D</math>.
 
At <math>D</math> a tangent is drawn cutting leg <math>CA</math> in <math>F</math>. This information is not sufficient to prove that
 
 
 
<math>\textbf{(A)}\ DF \text{ bisects }CA \qquad
 
\textbf{(B) }\ DF \text{ bisects }\angle CDA \
 
\textbf{(C) }\ DF = FA \qquad
 
\textbf{(D) }\ \angle A = \angle BCD \qquad
 
\textbf{(E) }\ \angle CFD = 2\angle A    </math>
 
 
 
[[1965 AHSME Problems/Problem 30|Solution]]
 
 
 
== Problem 31==
 
 
 
The number of real values of <math>x</math> satisfying the equality <math>(\log_2x)(\log_bx) = \log_ab</math>, where <math>a > 0, b > 0, a \neq 1, b \neq 1</math>, is:
 
 
 
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B) }\ 1 \qquad
 
\textbf{(C) }\ 2 \qquad
 
\textbf{(D) }\ \text{a finite integer greater than 2}\qquad
 
\textbf{(E) }\ \text{not finite} </math>
 
 
 
[[1965 AHSME Problems/Problem 31|Solution]]
 
 
 
== Problem 32==
 
 
 
An article costing <math>C</math> dollars is sold for &#036;100 at a loss of <math>x</math> percent of the selling price.
 
It is then resold at a profit of <math>x</math> percent of the new selling price <math>S'</math>.
 
If the difference between <math>S'</math> and <math>C</math> is <math>1\frac {1}{9}</math> dollars, then x is:
 
 
 
<math>\textbf{(A)}\ \text{undetermined} \qquad
 
\textbf{(B) }\ \frac {80}{9} \qquad
 
\textbf{(C) }\ 10 \qquad
 
\textbf{(D) }\ \frac{95}{9}\qquad
 
\textbf{(E) }\ \frac{100}{9} </math> 
 
 
 
[[1965 AHSME Problems/Problem 32|Solution]]
 
 
 
== Problem 33==
 
 
 
If the number <math>15!</math>, that is, <math>15 \cdot 14 \cdot 13 \dots 1</math>, ends with <math>k</math> zeros when given to the base <math>12</math> and ends with <math>h</math> zeros
 
when given to the base <math>10</math>, then <math>k + h</math> equals:
 
 
 
<math>\textbf{(A)}\ 5 \qquad
 
\textbf{(B) }\ 6 \qquad
 
\textbf{(C) }\ 7 \qquad
 
\textbf{(D) }\ 8 \qquad
 
\textbf{(E) }\ 9  </math> 
 
 
 
[[1965 AHSME Problems/Problem 33|Solution]]
 
 
 
== Problem 34==
 
 
 
For <math>x \ge 0</math> the smallest value of <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> is:
 
 
 
<math>\textbf{(A)}\ 1 \qquad
 
\textbf{(B) }\ 2 \qquad
 
\textbf{(C) }\ \frac {25}{12} \qquad
 
\textbf{(D) }\ \frac{13}{6}\qquad
 
\textbf{(E) }\ \frac{34}{5} </math>
 
 
 
[[1965 AHSME Problems/Problem 34|Solution]]
 
 
 
== Problem 35==
 
 
 
The length of a rectangle is <math>5</math> inches and its width is less than <math>4</math> inches. The rectangle is folded so that two
 
diagonally opposite vertices coincide. If the length of the crease is <math>\sqrt {6}</math>, then the width is:
 
 
 
<math>\textbf{(A)}\ \sqrt {2} \qquad
 
\textbf{(B) }\ \sqrt {3} \qquad
 
\textbf{(C) }\ 2 \qquad
 
\textbf{(D) }\ \sqrt{5}\qquad
 
\textbf{(E) }\ \sqrt{\frac{11}{2}} </math>
 
 
 
[[1965 AHSME Problems/Problem 35|Solution]]
 
 
 
== Problem 36==
 
 
 
Given distinct straight lines <math>OA</math> and <math>OB</math>. From a point in <math>OA</math> a perpendicular is drawn to <math>OB</math>;
 
from the foot of this perpendicular a line is drawn perpendicular to <math>OA</math>.
 
From the foot of this second perpendicular a line is drawn perpendicular to <math>OB</math>;
 
and so on indefinitely. The lengths of the first and second perpendiculars are <math>a</math> and <math>b</math>, respectively.
 
Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:
 
 
 
<math>\textbf{(A)}\ \frac {b}{a - b} \qquad
 
\textbf{(B) }\ \frac {a}{a - b} \qquad
 
\textbf{(C) }\ \frac {ab}{a - b} \qquad
 
\textbf{(D) }\ \frac{b^2}{a-b}\qquad
 
\textbf{(E) }\ \frac{a^2}{a-b} </math>
 
 
 
[[1965 AHSME Problems/Problem 36|Solution]]
 
 
 
== Problem 37==
 
 
 
Point <math>E</math> is selected on side <math>AB</math> of <math>\triangle{ABC}</math> in such a way that <math>AE: EB = 1: 3</math> and point <math>D</math> is selected on side <math>BC</math>
 
such that <math>CD: DB = 1: 2</math>. The point of intersection of <math>AD</math> and <math>CE</math> is <math>F</math>. Then <math>\frac {EF}{FC} + \frac {AF}{FD}</math> is:
 
 
 
<math>\textbf{(A)}\ \frac {4}{5} \qquad
 
\textbf{(B) }\ \frac {5}{4} \qquad
 
\textbf{(C) }\ \frac {3}{2} \qquad
 
\textbf{(D) }\ 2\qquad
 
\textbf{(E) }\ \frac{5}{2} </math> 
 
 
 
[[1965 AHSME Problems/Problem 37|Solution]]
 
 
 
== Problem 38==
 
 
 
A takes <math>m</math> times as long to do a piece of work as <math>B</math> and <math>C</math> together; <math>B</math> takes <math>n</math> times as long as <math>C</math> and <math>A</math> together;
 
and <math>C</math> takes <math>x</math> times as long as <math>A</math> and <math>B</math> together. Then <math>x</math>, in terms of <math>m</math> and <math>n</math>, is:
 
 
 
<math>\textbf{(A)}\ \frac {2mn}{m + n} \qquad
 
\textbf{(B) }\ \frac {1}{2(m + n)} \qquad
 
\textbf{(C) }\ \frac{1}{m+n-mn}\qquad
 
\textbf{(D) }\ \frac{1-mn}{m+n+2mn}\qquad
 
\textbf{(E) }\ \frac{m+n+2}{mn-1} </math>
 
 
 
[[1965 AHSME Problems/Problem 38|Solution]]
 
 
 
== Problem 39==
 
 
 
A foreman noticed an inspector checking a <math>3</math>"-hole with a <math>2</math>"-plug and a <math>1</math>"-plug and suggested that two more gauges
 
be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, <math>d</math>, of each, to the nearest hundredth of an inch, is:
 
 
 
<math>\textbf{(A)}\ .87 \qquad
 
\textbf{(B) }\ .86 \qquad
 
\textbf{(C) }\ .83 \qquad
 
\textbf{(D) }\ .75 \qquad
 
\textbf{(E) }\ .71 </math>   
 
 
 
[[1965 AHSME Problems/Problem 39|Solution]]
 
 
 
== Problem 40==
 
 
 
Let <math>n</math> be the number of integer values of <math>x</math> such that <math>P = x^4 + 6x^3 + 11x^2 + 3x + 31</math> is the square of an integer. Then <math>n</math> is:
 
  
<math>\textbf{(A)}\ 4 \qquad
 
\textbf{(B) }\ 3 \qquad
 
\textbf{(C) }\ 2 \qquad
 
\textbf{(D) }\ 1 \qquad
 
\textbf{(E) }\ 0 </math>
 
 
[[1965 AHSME Problems/Problem 40|Solution]]
 
 
 
== See also ==
 
 
* [[AHSME]]
 
* [[AHSME]]
 
* [[AHSME Problems and Solutions]]
 
* [[AHSME Problems and Solutions]]
 
* [[AMC 12 Problems and Solutions]]
 
* [[AMC 12 Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
{{MAA Notice}}
 

Revision as of 13:40, 9 October 2014