Difference between revisions of "1965 AHSME"

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1965 AHSME problems. Community discussions on each problem can be found by clicking on the problem number.

Problem 1

The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:

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

Solution

Problem 2

A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:

$\textbf{(A)}\ 1: 1 \qquad \textbf{(B) }\ 1: 6 \qquad \textbf{(C) }\ 1: \pi \qquad \textbf{(D) }\ 3: \pi \qquad \textbf{(E) }\ 6:\pi$

Solution

Problem 3

The expression $(81)^{ - 2^{ - 2}}$ has the same value as:

$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$

Solution

Problem 4

Line $\ell_2$ intersects line $\ell_1$ and line $\ell_3$ is parallel to $\ell_1$ . The three lines are distinct and lie in a plane. The number of points equidistant from all three lines is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 8$

Solution

Problem 5

When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:

$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$

Solution

Problem 6

If $10^{\log_{10}9} = 8x + 5$ then $x$ equals:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ \frac {1}{2} \qquad \textbf{(C) }\ \frac {5}{8} \qquad \textbf{(D) }\ \frac{9}{8}\qquad \textbf{(E) }\ \frac{2\log_{10}3-5}{8}$

Solution

Problem 7

The sum of the reciprocals of the roots of the equation $ax^2 + bx + c = 0$ is:

$\textbf{(A)}\ \frac {1}{a} + \frac {1}{b} \qquad \textbf{(B) }\ - \frac {c}{b} \qquad \textbf{(C) }\ \frac{b}{c}\qquad \textbf{(D) }\ -\frac{a}{b}\qquad \textbf{(E) }\ -\frac{b}{c}$

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:

$\textbf{(A)}\ 6\sqrt {6} \qquad \textbf{(B) }\ 9\sqrt {2} \qquad \textbf{(C) }\ 12 \qquad \textbf{(D) }\ 6\sqrt{3}\qquad \textbf{(E) }\ 9$

Solution

Problem 9

The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:

$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$

Solution

Problem 10

The statement $x^2 - x - 6 < 0$ is equivalent to the statement:

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

Solution

Problem 11

Consider the statements: $I: (\sqrt { - 4})(\sqrt { - 16}) = \sqrt {( - 4)( - 16)}, \\ II: \sqrt {( - 4)( - 16)} = \sqrt {64}, and \sqrt {64} = 8.$ Of these the following are incorrect.

$\textbf{(A)}\ \text{none} \qquad \textbf{(B) }\ \text{I only} \qquad \textbf{(C) }\ \text{II only} \qquad \textbf{(D) }\ \text{III only}\qquad \textbf{(E) }\ \text{I and III only}$

Solution

Problem 12

A rhombus is inscribed in $\triangle ABC$ in such a way that one of its vertices is $A$ and two of its sides lie along $AB$ and $AC$ . If $\overline{AC} = 6$ inches, $\overline{AB} = 12$ inches, and $\overline{BC} = 8$ inches, the side of the rhombus, in inches, is:

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

Solution

Problem 13

Let $n$ be the number of number-pairs $(x,y)$ which satisfy $5y - 3x = 15$ and $x^2 + y^2 \le 16$ . Then $n$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ \text{more than two, but finite}\qquad \textbf{(E) }\ \text{greater than any finite number}$

Solution

Problem 14

The sum of the numerical coefficients in the complete expansion of $(x^2 - 2xy + y^2)^7$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 7 \qquad \textbf{(C) }\ 14 \qquad \textbf{(D) }\ 128 \qquad \textbf{(E) }\ 128^2$

Solution

Problem 15

The symbol $25_b$ represents a two-digit number in the base $b$ . If the number $52_b$ is double the number $25_b$ , then $b$ is:

$\textbf{(A)}\ 7 \qquad \textbf{(B) }\ 8 \qquad \textbf{(C) }\ 9 \qquad \textbf{(D) }\ 11 \qquad \textbf{(E) }\ 12$

Solution

Problem 16

Let line $AC$ be perpendicular to line $CE$ . Connect $A$ to $D$ , the midpoint of $CE$ , and connect $E$ to $B$ , the midpoint of $AC$ . If $AD$ and $EB$ intersect in point $F$ , and $\overline{BC} = \overline{CD} = 15$ inches, then the area of triangle $DFE$ , in square inches, is:

$\textbf{(A)}\ 50 \qquad \textbf{(B) }\ 50\sqrt {2} \qquad \textbf{(C) }\ 75 \qquad \textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad \textbf{(E) }\ 100$

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:

$\textbf{(A)}\ \text{If the picnic is held, Sunday's weather is undoubtedly fair.} \\ \textbf{(B) }\ \text{If the picnic is not held, Sunday's weather is possibly unfair.} \\ \textbf{(C) }\ \text{If it is not fair Sunday, the picnic will not be held.} \\ \textbf{(D) }\ \text{If it is fair Sunday, the picnic may be held.} \\ \textbf{(E) }\ \text{If it is fair Sunday, the picnic must be held.}$

Solution

Problem 18

If $1 - y$ is used as an approximation to the value of $\frac {1}{1 + y}, |y| < 1$ , the ratio of the error made to the correct value is:

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

Solution

Problem 19

If $x^4 + 4x^3 + 6px^2 + 4qx + r$ is exactly divisible by $x^3 + 3x^2 + 9x + 3$ , the value of $(p + q)r$ is:

$\textbf{(A)}\ - 18 \qquad \textbf{(B) }\ 12 \qquad \textbf{(C) }\ 15 \qquad \textbf{(D) }\ 27 \qquad \textbf{(E) }\ 45 \qquad$

Solution

Problem 20

For every $n$ the sum of n terms of an arithmetic progression is $2n + 3n^2$ . The $r$ th term is:

$\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad$

Solution

Problem 21

It is possible to choose $x > \frac {2}{3}$ in such a way that the value of $\log_{10}(x^2 + 3) - 2 \log_{10}x$ is

$\textbf{(A)}\ \text{negative} \qquad \textbf{(B) }\ \text{zero} \qquad \textbf{(C) }\ \text{one} \\ \textbf{(D) }\ \text{smaller than any positive number that might be specified} \\ \textbf{(E) }\ \text{greater than any positive number that might be specified}$

Solution

Problem 22

If $a_2 \neq 0$ and $r$ and $s$ are the roots of $a_0 + a_1x + a_2x^2 = 0$ , then the equality $a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )$ holds:

$\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 \textbf{(B) }\ \text{for all values of }x \\ \textbf{(C) }\ \text{only when }x = 0 \textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0$

Solution

Problem 23

If we write $|x^2 - 4| < N$ for all $x$ such that $|x - 2| < 0.01$ , the smallest value we can use for $N$ is:

$\textbf{(A)}\ .0301 \qquad \textbf{(B) }\ .0349 \qquad \textbf{(C) }\ .0399 \qquad \textbf{(D) }\ .0401 \qquad \textbf{(E) }\ .0499\qquad$

Solution

Problem 24

Given the sequence $10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}$ , the smallest value of n such that the product of the first $n$ members of this sequence exceeds $100000$ is:

$\textbf{(A)}\ 7 \qquad \textbf{(B) }\ 8 \qquad \textbf{(C) }\ 9 \qquad \textbf{(D) }\ 10 \qquad \textbf{(E) }\ 11$

Solution

Problem 25

Let $ABCD$ be a quadrilateral with $AB$ extended to $E$ so that $\overline{AB} = \overline{BE}$ . Lines $AC$ and $CE$ are drawn to form $\angle{ACE}$ . For this angle to be a right angle it is necessary that quadrilateral $ABCD$ have:

$\textbf{(A)}\ \text{all angles equal} \textbf{(B) }\ \text{all sides equal} // \textbf{(C) }\ \text{two pairs of equal sides} \textbf{(D) }\ \text{one pair of equal sides} // \textbf{(E) }\ \text{one pair of equal angles}$

Solution

Problem 26

For the numbers $a, b, c, d, e$ define $m$ to be the arithmetic mean of all five numbers; $k$ to be the arithmetic mean of $a$ and $b$ ; $l$ to be the arithmetic mean of $c, d$ , and $e$ ; and $p$ to be the arithmetic mean of $k$ and $l$ . Then, no matter how $a, b, c, d$ , and $e$ are chosen, we shall always have:

$\textbf{(A)}\ m = p \qquad \textbf{(B) }\ m \ge p \qquad \textbf{(C) }\ m > p \qquad \textbf{(D) }\ m < p\qquad \textbf{(E) }\ \text{none of these}$

Solution

Problem 27

When $y^2 + my + 2$ is divided by $y - 1$ the quotient is $f(y)$ and the remainder is $R_1$ . When $y^2 + my + 2$ is divided by $y + 1$ the quotient is $g(y)$ and the remainder is $R_2$ . If $R_1 = R_2$ then $m$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ - 1 \qquad \textbf{(E) }\ \text{an undetermined constant}$

Solution

Problem 28

An escalator (moving staircase) of $n$ uniform steps visible at all times descends at constant speed. Two boys, $A$ and $Z$ , walk down the escalator steadily as it moves, A negotiating twice as many escalator steps per minute as $Z$ . $A$ reaches the bottom after taking $27$ steps while $Z$ reaches the bottom after taking $18$ steps. Then $n$ is:

$\textbf{(A)}\ 63 \qquad \textbf{(B) }\ 54 \qquad \textbf{(C) }\ 45 \qquad \textbf{(D) }\ 36 \qquad \textbf{(E) }\ 30$

Solution

Problem 29

Of $28$ 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:

$\textbf{(A)}\ 5 \qquad \textbf{(B) }\ 6 \qquad \textbf{(C) }\ 7 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 9$

Solution

Problem 30

Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ in $D$ . At $D$ a tangent is drawn cutting leg $CA$ in $F$ . This information is not sufficient to prove that

$\textbf{(A)}\ DF \text{ bisects }CA \qquad \textbf{(B) }\ DF \text{ bisects }\angle CDA \\ \textbf{(C) }\ DF = FA \qquad \textbf{(D) }\ \angle A = \angle BCD \qquad \textbf{(E) }\ \angle CFD = 2\angle A$

Solution

Problem 31

The number of real values of $x$ satisfying the equality $(\log_2x)(\log_bx) = \log_ab$ , where $a > 0, b > 0, a \neq 1, b \neq 1$ , is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ \text{a finite integer greater than 2}\qquad \textbf{(E) }\ \text{not finite}$

Solution

Problem 32

An article costing $C$ dollars is sold for $100 at a loss of $x$ percent of the selling price. It is then resold at a profit of $x$ percent of the new selling price $S'$ . If the difference between $S'$ and $C$ is $1\frac {1}{9}$ dollars, then x is:

$\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B) }\ \frac {80}{9} \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ \frac{95}{9}\qquad \textbf{(E) }\ \frac{100}{9}$

Solution

Problem 33

If the number $15!$ , that is, $15 \cdot 14 \cdot 13 \dots 1$ , ends with $k$ zeros when given to the base $12$ and ends with $h$ zeros when given to the base $10$ , then $k + h$ equals:

$\textbf{(A)}\ 5 \qquad \textbf{(B) }\ 6 \qquad \textbf{(C) }\ 7 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 9$

Solution

Problem 34

For $x \ge 0$ the smallest value of $\frac {4x^2 + 8x + 13}{6(1 + x)}$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ \frac {25}{12} \qquad \textbf{(D) }\ \frac{13}{6}\qquad \textbf{(E) }\ \frac{34}{5}$

Solution

Problem 35

The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is $\sqrt {6}$ , then the width is:

$\textbf{(A)}\ \sqrt {2} \qquad \textbf{(B) }\ \sqrt {3} \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ \sqrt{5}\qquad \textbf{(E) }\ \sqrt{\frac{11}{2}}$

Solution

Problem 36

Given distinct straight lines $OA$ and $OB$ . From a point in $OA$ a perpendicular is drawn to $OB$ ; from the foot of this perpendicular a line is drawn perpendicular to $OA$ . From the foot of this second perpendicular a line is drawn perpendicular to $OB$ ; and so on indefinitely. The lengths of the first and second perpendiculars are $a$ and $b$ , 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:

$\textbf{(A)}\ \frac {b}{a - b} \qquad \textbf{(B) }\ \frac {a}{a - b} \qquad \textbf{(C) }\ \frac {ab}{a - b} \qquad \textbf{(D) }\ \frac{b^2}{a-b}\qquad \textbf{(E) }\ \frac{a^2}{a-b}$

Solution

Problem 37

Point $E$ is selected on side $AB$ of $\triangle{ABC}$ in such a way that $AE: EB = 1: 3$ and point $D$ is selected on side $BC$ such that $CD: DB = 1: 2$ . The point of intersection of $AD$ and $CE$ is $F$ . Then $\frac {EF}{FC} + \frac {AF}{FD}$ is:

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

Solution

Problem 38

A takes $m$ times as long to do a piece of work as $B$ and $C$ together; $B$ takes $n$ times as long as $C$ and $A$ together; and $C$ takes $x$ times as long as $A$ and $B$ together. Then $x$ , in terms of $m$ and $n$ , is:

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

Solution

Problem 39

A foreman noticed an inspector checking a $3$ "-hole with a $2$ "-plug and a $1$ "-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, $d$ , of each, to the nearest hundredth of an inch, is:

$\textbf{(A)}\ .87 \qquad \textbf{(B) }\ .86 \qquad \textbf{(C) }\ .83 \qquad \textbf{(D) }\ .75 \qquad \textbf{(E) }\ .71$

Solution

Problem 40

Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:

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

Solution

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

1965 AHSC (Problems • Answer Key • Resources)
Preceded by 1964 AHSME	Followed by 1966 AHSME
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
 '''1965 AHSME''' problems. Community discussions on each problem can be found by clicking on the problem number.
-* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=1965&sid=e43c943ffbc35a0dc367ec4e52f1708b 1965 AHSME Problems] (PDF:  [http://www.artofproblemsolving.com/Forum/resources/files/usa/44/USA-AMC_12/AHSME-1965-44.pdf])
+== Problem 1 ==
+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
+\textbf{(B) }\ 1 \qquad
+\textbf{(C) }\ 2 \qquad
+\textbf{(D) }\ 3 \qquad
+\textbf{(E) }\ \text{more than 4}   </math>
+[[1965 AHSME Problems/Problem 1|Solution]]
+== Problem 2==
+A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the
+shorter of the arcs intercepted by the side, is:
+<math>\textbf{(A)}\ 1: 1 \qquad
+\textbf{(B) }\ 1: 6 \qquad
+\textbf{(C) }\ 1: \pi \qquad
+\textbf{(D) }\ 3: \pi \qquad
+\textbf{(E) }\ 6:\pi  </math>
+[[1965 AHSME Problems/Problem 2|Solution]]
+== Problem 3==
+The expression <math>(81)^{ - 2^{ - 2}}</math> has the same value as:
+<math>\textbf{(A)}\ \frac {1}{81} \qquad
+\textbf{(B) }\ \frac {1}{3} \qquad
+\textbf{(C) }\ 3 \qquad
+\textbf{(D) }\ 81\qquad
+\textbf{(E) }\ 81^4 </math>
+[[1965 AHSME Problems/Problem 3|Solution]]
+== Problem 4==
+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.
+The number of points equidistant from all three lines is:
+<math>\textbf{(A)}\ 0 \qquad
+\textbf{(B) }\ 1 \qquad
+\textbf{(C) }\ 2 \qquad
+\textbf{(D) }\ 4 \qquad
+\textbf{(E) }\ 8  </math>
+[[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]]
 {{MAA Notice}}
@@ Line 11: / Line 526: @@
 * [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=1965&sid=e43c943ffbc35a0dc367ec4e52f1708b 1965 AHSME Problems] (PDF:  [http://www.artofproblemsolving.com/Forum/resources/files/usa/44/USA-AMC_12/AHSME-1965-44.pdf])

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Difference between revisions of "1965 AHSME"

Revision as of 12:16, 9 October 2014

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

See also