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| − | == Problem 1 ==
| + | '''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
| + | * [[1965 AHSME Problems|Entire Exam]] |
| − | \textbf{(B) }\ 1 \qquad
| + | * [[1965 AHSME Problems|Answer Key]] |
| − | \textbf{(C) }\ 2 \qquad
| + | ** [[1965 AHSME Problems/Problem 1|Problem 1]] |
| − | \textbf{(D) }\ 3 \qquad
| + | ** [[1965 AHSME Problems/Problem 2|Problem 2]] |
| − | \textbf{(E) }\ \text{more than 4} </math>
| + | ** [[1965 AHSME Problems/Problem 3|Problem 3]] |
| − | | + | ** [[1965 AHSME Problems/Problem 4|Problem 4]] |
| − | [[1965 AHSME Problems/Problem 1|Solution]] | + | ** [[1965 AHSME Problems/Problem 5|Problem 5]] |
| − | | + | ** [[1965 AHSME Problems/Problem 6|Problem 6]] |
| − | == Problem 2==
| + | ** [[1965 AHSME Problems/Problem 7|Problem 7]] |
| − | | + | ** [[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
| + | ** [[1965 AHSME Problems/Problem 9|Problem 9]] |
| − | shorter of the arcs intercepted by the side, is:
| + | ** [[1965 AHSME Problems/Problem 10|Problem 10]] |
| − | | + | ** [[1965 AHSME Problems/Problem 11|Problem 11]] |
| − | <math>\textbf{(A)}\ 1: 1 \qquad
| + | ** [[1965 AHSME Problems/Problem 12|Problem 12]] |
| − | \textbf{(B) }\ 1: 6 \qquad
| + | ** [[1965 AHSME Problems/Problem 13|Problem 13]] |
| − | \textbf{(C) }\ 1: \pi \qquad
| + | ** [[1965 AHSME Problems/Problem 14|Problem 14]] |
| − | \textbf{(D) }\ 3: \pi \qquad
| + | ** [[1965 AHSME Problems/Problem 15|Problem 15]] |
| − | \textbf{(E) }\ 6:\pi </math>
| + | ** [[1965 AHSME Problems/Problem 16|Problem 16]] |
| − | | + | ** [[1965 AHSME Problems/Problem 17|Problem 17]] |
| − | [[1965 AHSME Problems/Problem 2|Solution]] | + | ** [[1965 AHSME Problems/Problem 18|Problem 18]] |
| − | | + | ** [[1965 AHSME Problems/Problem 19|Problem 19]] |
| − | == Problem 3==
| + | ** [[1965 AHSME Problems/Problem 20|Problem 20]] |
| − | | + | ** [[1965 AHSME Problems/Problem 21|Problem 21]] |
| − | The expression <math>(81)^{ - 2^{ - 2}}</math> has the same value as:
| + | ** [[1965 AHSME Problems/Problem 22|Problem 22]] |
| − | | + | ** [[1965 AHSME Problems/Problem 23|Problem 23]] |
| − | <math>\textbf{(A)}\ \frac {1}{81} \qquad
| + | ** [[1965 AHSME Problems/Problem 24|Problem 24]] |
| − | \textbf{(B) }\ \frac {1}{3} \qquad
| + | ** [[1965 AHSME Problems/Problem 25|Problem 25]] |
| − | \textbf{(C) }\ 3 \qquad
| + | ** [[1965 AHSME Problems/Problem 26|Problem 26]] |
| − | \textbf{(D) }\ 81\qquad
| + | ** [[1965 AHSME Problems/Problem 27|Problem 27]] |
| − | \textbf{(E) }\ 81^4 </math>
| + | ** [[1965 AHSME Problems/Problem 28|Problem 28]] |
| − | | + | ** [[1965 AHSME Problems/Problem 29|Problem 29]] |
| − | [[1965 AHSME Problems/Problem 3|Solution]] | + | ** [[1965 AHSME Problems/Problem 30|Problem 30]] |
| − | | + | ** [[1965 AHSME Problems/Problem 31|Problem 31]] |
| − | == Problem 4==
| + | ** [[1965 AHSME Problems/Problem 32|Problem 32]] |
| − | | + | ** [[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.
| + | ** [[1965 AHSME Problems/Problem 34|Problem 34]] |
| − | The number of points equidistant from all three lines is:
| + | ** [[1965 AHSME Problems/Problem 35|Problem 35]] |
| − | | + | ** [[1965 AHSME Problems/Problem 36|Problem 36]] |
| − | <math>\textbf{(A)}\ 0 \qquad
| + | ** [[1965 AHSME Problems/Problem 37|Problem 37]] |
| − | \textbf{(B) }\ 1 \qquad
| + | ** [[1965 AHSME Problems/Problem 38|Problem 38]] |
| − | \textbf{(C) }\ 2 \qquad
| + | ** [[1965 AHSME Problems/Problem 39|Problem 39]] |
| − | \textbf{(D) }\ 4 \qquad
| + | ** [[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==
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| − | | |
| − | When the repeating decimal <math>0.363636\ldots</math> is written in simplest fractional form, the sum of the numerator and denominator is:
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| − | | |
| − | <math>\textbf{(A)}\ 15 \qquad
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| − | \textbf{(B) }\ 45 \qquad
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| − | \textbf{(C) }\ 114 \qquad
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| − | \textbf{(D) }\ 135 \qquad
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| − | \textbf{(E) }\ 150 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 5|Solution]] | |
| − | | |
| − | == Problem 6==
| |
| − | | |
| − | If <math>10^{\log_{10}9} = 8x + 5</math> then <math>x</math> equals:
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| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad
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| − | \textbf{(B) }\ \frac {1}{2} \qquad
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| − | \textbf{(C) }\ \frac {5}{8} \qquad
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| − | \textbf{(D) }\ \frac{9}{8}\qquad
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| − | \textbf{(E) }\ \frac{2\log_{10}3-5}{8} </math>
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| − | | |
| − | [[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:
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| − | | |
| − | <math>\textbf{(A)}\ \frac {1}{a} + \frac {1}{b} \qquad
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| − | \textbf{(B) }\ - \frac {c}{b} \qquad
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| − | \textbf{(C) }\ \frac{b}{c}\qquad
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| − | \textbf{(D) }\ -\frac{a}{b}\qquad
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| − | \textbf{(E) }\ -\frac{b}{c} </math>
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| − | | |
| − | [[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:
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| − | | |
| − | <math>\textbf{(A)}\ 6\sqrt {6} \qquad
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| − | \textbf{(B) }\ 9\sqrt {2} \qquad
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| − | \textbf{(C) }\ 12 \qquad
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| − | \textbf{(D) }\ 6\sqrt{3}\qquad
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| − | \textbf{(E) }\ 9 </math>
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| − | | |
| − | [[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
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| − | \textbf{(B) }\ - 4 \qquad
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| − | \textbf{(C) }\ 4 \qquad
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| − | \textbf{(D) }\ 8 \qquad
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| − | \textbf{(E) }\ 16 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 9|Solution]] | |
| − | | |
| − | == Problem 10==
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| − | | |
| − | The statement <math>x^2 - x - 6 < 0</math> is equivalent to the statement:
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| − | | |
| − | <math>\textbf{(A)}\ - 2 < x < 3 \qquad
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| − | \textbf{(B) }\ x > - 2 \qquad
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| − | \textbf{(C) }\ x < 3 \\
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| − | \textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad
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| − | \textbf{(E) }\ x > 3 \text{ and }x < - 2 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 10|Solution]] | |
| − | | |
| − | == Problem 11==
| |
| − | | |
| − | Consider the statements:
| |
| − | <cmath>I: (\sqrt { - 4})(\sqrt { - 16}) = \sqrt {( - 4)( - 16)}, \\
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| − | II: \sqrt {( - 4)( - 16)} = \sqrt {64}, and \sqrt {64} = 8. </cmath>
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| − | Of these the following are incorrect.
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| − | | |
| − | <math>\textbf{(A)}\ \text{none} \qquad
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| − | \textbf{(B) }\ \text{I only} \qquad
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| − | \textbf{(C) }\ \text{II only} \qquad
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| − | \textbf{(D) }\ \text{III only}\qquad
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| − | \textbf{(E) }\ \text{I and III only} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 11|Solution]] | |
| − | | |
| − | == Problem 12==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 2 \qquad
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| − | \textbf{(B) }\ 3 \qquad
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| − | \textbf{(C) }\ 3 \frac {1}{2} \qquad
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| − | \textbf{(D) }\ 4 \qquad
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| − | \textbf{(E) }\ 5 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 12|Solution]] | |
| − | | |
| − | == Problem 13==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad
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| − | \textbf{(B) }\ 1 \qquad
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| − | \textbf{(C) }\ 2 \qquad
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| − | \textbf{(D) }\ \text{more than two, but finite}\qquad
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| − | \textbf{(E) }\ \text{greater than any finite number} </math>
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| − | | |
| − | [[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:
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| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad
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| − | \textbf{(B) }\ 7 \qquad
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| − | \textbf{(C) }\ 14 \qquad
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| − | \textbf{(D) }\ 128 \qquad
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| − | \textbf{(E) }\ 128^2 </math>
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| − | | |
| − | [[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
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| − | \textbf{(B) }\ 8 \qquad
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| − | \textbf{(C) }\ 9 \qquad
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| − | \textbf{(D) }\ 11 \qquad
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| − | \textbf{(E) }\ 12 </math>
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| − | | |
| − | [[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:
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| − | | |
| − | <math>\textbf{(A)}\ 50 \qquad
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| − | \textbf{(B) }\ 50\sqrt {2} \qquad
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| − | \textbf{(C) }\ 75 \qquad
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| − | \textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad
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| − | \textbf{(E) }\ 100 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 16|Solution]] | |
| − | | |
| − | == Problem 17==
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| − | | |
| − | Given the true statement: The picnic on Sunday will not be held only if the weather is not fair. We can then conclude that:
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| − | | |
| − | <math>\textbf{(A)}\ \text{If the picnic is held, Sunday's weather is undoubtedly fair.} \\
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| − | \textbf{(B) }\ \text{If the picnic is not held, Sunday's weather is possibly unfair.} \\
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| − | \textbf{(C) }\ \text{If it is not fair Sunday, the picnic will not be held.} \\
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| − | \textbf{(D) }\ \text{If it is fair Sunday, the picnic may be held.} \\
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| − | \textbf{(E) }\ \text{If it is fair Sunday, the picnic must be held.} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 17|Solution]] | |
| − | | |
| − | == Problem 18==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ y \qquad
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| − | \textbf{(B) }\ y^2 \qquad
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| − | \textbf{(C) }\ \frac {1}{1 + y} \qquad
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| − | \textbf{(D) }\ \frac{y}{1+y}\qquad
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| − | \textbf{(E) }\ \frac{y^2}{1+y}\qquad </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 18|Solution]] | |
| − | | |
| − | == Problem 19==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ - 18 \qquad
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| − | \textbf{(B) }\ 12 \qquad
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| − | \textbf{(C) }\ 15 \qquad
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| − | \textbf{(D) }\ 27 \qquad
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| − | \textbf{(E) }\ 45 \qquad </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 19|Solution]] | |
| − | | |
| − | == Problem 20==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 3r^2 \qquad
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| − | \textbf{(B) }\ 3r^2 + 2r \qquad
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| − | \textbf{(C) }\ 6r - 1 \qquad
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| − | \textbf{(D) }\ 5r + 5 \qquad
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| − | \textbf{(E) }\ 6r+2\qquad </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 20|Solution]] | |
| − | | |
| − | == Problem 21==
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| − | | |
| − | 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
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| − | | |
| − | <math>\textbf{(A)}\ \text{negative} \qquad
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| − | \textbf{(B) }\ \text{zero} \qquad
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| − | \textbf{(C) }\ \text{one} \\
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| − | \textbf{(D) }\ \text{smaller than any positive number that might be specified} \\
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| − | \textbf{(E) }\ \text{greater than any positive number that might be specified} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 21|Solution]] | |
| − | | |
| − | == Problem 22==
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| − | | |
| − | 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
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| − | <math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds:
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| − | | |
| − | <math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0
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| − | \textbf{(B) }\ \text{for all values of }x \\
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| − | \textbf{(C) }\ \text{only when }x = 0
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| − | \textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\
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| − | \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 22|Solution]] | |
| − | | |
| − | == Problem 23==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ .0301 \qquad
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| − | \textbf{(B) }\ .0349 \qquad
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| − | \textbf{(C) }\ .0399 \qquad
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| − | \textbf{(D) }\ .0401 \qquad
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| − | \textbf{(E) }\ .0499\qquad </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 23|Solution]] | |
| − | | |
| − | == Problem 24==
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| − | | |
| − | Given the sequence <math>10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}</math>,
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| − | the smallest value of n such that the product of the first <math>n</math> members of this sequence exceeds <math>100000</math> is:
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| − | | |
| − | <math>\textbf{(A)}\ 7 \qquad
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| − | \textbf{(B) }\ 8 \qquad
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| − | \textbf{(C) }\ 9 \qquad
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| − | \textbf{(D) }\ 10 \qquad
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| − | \textbf{(E) }\ 11 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 24|Solution]] | |
| − | | |
| − | == Problem 25==
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| − | | |
| − | Let <math>ABCD</math> be a quadrilateral with <math>AB</math> extended to <math>E</math> so that <math>\overline{AB} = \overline{BE}</math>.
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ \text{all angles equal}
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| − | \textbf{(B) }\ \text{all sides equal} \\
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| − | \textbf{(C) }\ \text{two pairs of equal sides}
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| − | \textbf{(D) }\ \text{one pair of equal sides} \\
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| − | \textbf{(E) }\ \text{one pair of equal angles} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 25|Solution]] | |
| − | | |
| − | == Problem 26==
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| − | | |
| − | For the numbers <math>a, b, c, d, e</math> define <math>m</math> to be the arithmetic mean of all five numbers;
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| − | <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>;
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ m = p \qquad
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| − | \textbf{(B) }\ m \ge p \qquad
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| − | \textbf{(C) }\ m > p \qquad \\
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| − | \textbf{(D) }\ m < p\qquad
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| − | \textbf{(E) }\ \text{none of these} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 26|Solution]] | |
| − | | |
| − | == Problem 27==
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| − | | |
| − | 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>.
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad
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| − | \textbf{(B) }\ 1 \qquad
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| − | \textbf{(C) }\ 2 \qquad
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| − | \textbf{(D) }\ - 1 \qquad
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| − | \textbf{(E) }\ \text{an undetermined constant} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 27|Solution]] | |
| − | | |
| − | == Problem 28==
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| − | | |
| − | An escalator (moving staircase) of <math>n</math> uniform steps visible at all times descends at constant speed.
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| − | Two boys, <math>A</math> and <math>Z</math>, walk down the escalator steadily as it moves, A negotiating twice as many escalator
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 63 \qquad
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| − | \textbf{(B) }\ 54 \qquad
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| − | \textbf{(C) }\ 45 \qquad
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| − | \textbf{(D) }\ 36 \qquad
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| − | \textbf{(E) }\ 30 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 28|Solution]] | |
| − | | |
| − | == Problem 29==
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| − | | |
| − | Of <math>28</math> students taking at least one subject the number taking Mathematics and English only equals the number
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| − | taking Mathematics only. No student takes English only or History only, and six students take Mathematics and
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| − | History, but not English. The number taking English and History only is five times the number taking all three subjects.
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| − | If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:
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| − | | |
| − | <math>\textbf{(A)}\ 5 \qquad
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| − | \textbf{(B) }\ 6 \qquad
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| − | \textbf{(C) }\ 7 \qquad
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| − | \textbf{(D) }\ 8 \qquad
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| − | \textbf{(E) }\ 9 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 29|Solution]] | |
| − | | |
| − | == Problem 30==
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| − | | |
| − | 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>.
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| − | 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
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| − | | |
| − | <math>\textbf{(A)}\ DF \text{ bisects }CA \qquad
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| − | \textbf{(B) }\ DF \text{ bisects }\angle CDA \\
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| − | \textbf{(C) }\ DF = FA \qquad
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| − | \textbf{(D) }\ \angle A = \angle BCD \qquad
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| − | \textbf{(E) }\ \angle CFD = 2\angle A </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 30|Solution]] | |
| − | | |
| − | == Problem 31==
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| − | | |
| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad
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| − | \textbf{(B) }\ 1 \qquad
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| − | \textbf{(C) }\ 2 \qquad
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| − | \textbf{(D) }\ \text{a finite integer greater than 2}\qquad
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| − | \textbf{(E) }\ \text{not finite} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 31|Solution]] | |
| − | | |
| − | == Problem 32==
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| − | | |
| − | An article costing <math>C</math> dollars is sold for $100 at a loss of <math>x</math> percent of the selling price.
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| − | It is then resold at a profit of <math>x</math> percent of the new selling price <math>S'</math>.
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| − | If the difference between <math>S'</math> and <math>C</math> is <math>1\frac {1}{9}</math> dollars, then x is:
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| − | | |
| − | <math>\textbf{(A)}\ \text{undetermined} \qquad
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| − | \textbf{(B) }\ \frac {80}{9} \qquad
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| − | \textbf{(C) }\ 10 \qquad
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| − | \textbf{(D) }\ \frac{95}{9}\qquad
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| − | \textbf{(E) }\ \frac{100}{9} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 32|Solution]] | |
| − | | |
| − | == Problem 33==
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| − | | |
| − | 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
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| − | when given to the base <math>10</math>, then <math>k + h</math> equals:
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| − | | |
| − | <math>\textbf{(A)}\ 5 \qquad
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| − | \textbf{(B) }\ 6 \qquad
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| − | \textbf{(C) }\ 7 \qquad
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| − | \textbf{(D) }\ 8 \qquad
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| − | \textbf{(E) }\ 9 </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 33|Solution]] | |
| − | | |
| − | == Problem 34==
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| − | | |
| − | For <math>x \ge 0</math> the smallest value of <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> is:
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| − | | |
| − | <math>\textbf{(A)}\ 1 \qquad
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| − | \textbf{(B) }\ 2 \qquad
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| − | \textbf{(C) }\ \frac {25}{12} \qquad
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| − | \textbf{(D) }\ \frac{13}{6}\qquad
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| − | \textbf{(E) }\ \frac{34}{5} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 34|Solution]] | |
| − | | |
| − | == Problem 35==
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| − | | |
| − | 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
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| − | diagonally opposite vertices coincide. If the length of the crease is <math>\sqrt {6}</math>, then the width is:
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| − | | |
| − | <math>\textbf{(A)}\ \sqrt {2} \qquad
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| − | \textbf{(B) }\ \sqrt {3} \qquad
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| − | \textbf{(C) }\ 2 \qquad
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| − | \textbf{(D) }\ \sqrt{5}\qquad
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| − | \textbf{(E) }\ \sqrt{\frac{11}{2}} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 35|Solution]] | |
| − | | |
| − | == Problem 36==
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| − | | |
| − | 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>;
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| − | from the foot of this perpendicular a line is drawn perpendicular to <math>OA</math>.
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| − | From the foot of this second perpendicular a line is drawn perpendicular to <math>OB</math>;
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| − | and so on indefinitely. The lengths of the first and second perpendiculars are <math>a</math> and <math>b</math>, respectively.
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| − | Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:
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| − | | |
| − | <math>\textbf{(A)}\ \frac {b}{a - b} \qquad
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| − | \textbf{(B) }\ \frac {a}{a - b} \qquad
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| − | \textbf{(C) }\ \frac {ab}{a - b} \qquad
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| − | \textbf{(D) }\ \frac{b^2}{a-b}\qquad
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| − | \textbf{(E) }\ \frac{a^2}{a-b} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 36|Solution]] | |
| − | | |
| − | == Problem 37==
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| − | | |
| − | 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>
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ \frac {4}{5} \qquad
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| − | \textbf{(B) }\ \frac {5}{4} \qquad
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| − | \textbf{(C) }\ \frac {3}{2} \qquad
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| − | \textbf{(D) }\ 2\qquad
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| − | \textbf{(E) }\ \frac{5}{2} </math>
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| − | | |
| − | [[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;
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ \frac {2mn}{m + n} \qquad
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| − | \textbf{(B) }\ \frac {1}{2(m + n)} \qquad
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| − | \textbf{(C) }\ \frac{1}{m+n-mn}\qquad
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| − | \textbf{(D) }\ \frac{1-mn}{m+n+2mn}\qquad
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| − | \textbf{(E) }\ \frac{m+n+2}{mn-1} </math>
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| − | | |
| − | [[1965 AHSME Problems/Problem 38|Solution]] | |
| − | | |
| − | == Problem 39==
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| − | | |
| − | 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
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| − | 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:
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| − | | |
| − | <math>\textbf{(A)}\ .87 \qquad
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| − | \textbf{(B) }\ .86 \qquad
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| − | \textbf{(C) }\ .83 \qquad
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| − | \textbf{(D) }\ .75 \qquad
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| − | \textbf{(E) }\ .71 </math>
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| − | | |
| − | [[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:
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| | | | |
| − | <math>\textbf{(A)}\ 4 \qquad
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| − | \textbf{(B) }\ 3 \qquad
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| − | \textbf{(C) }\ 2 \qquad
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| − | \textbf{(D) }\ 1 \qquad
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| − | \textbf{(E) }\ 0 </math>
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| − |
| |
| − | [[1965 AHSME Problems/Problem 40|Solution]]
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| − |
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| − |
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| − | == 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}}
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