Difference between revisions of "1966 AHSME Problems"

Revision as of 23:46, 14 September 2014

Problem 1

Given that the ratio of $3x - 4$ to $y + 15$ is constant, and $y = 3$ when $x = 2$ , then, when $y = 12$ , $x$ equals:

$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$

Solution

Problem 2

When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is

$\text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease}$

Solution

Problem 3

If the arithmetic mean of two numbers is $6$ and their geometric mean is $10$ , then an equation with the given two numbers as roots is:

$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$ $\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$

Solution

Problem 4

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle $I$ to that of circle $II$ , then $r$ equals:

$\text{(A)} \sqrt 2 \qquad \text{(B)} 2 \qquad \text{(C)} \sqrt 3 \qquad \text{(D)} 2\sqrt 2 \qquad \text{(E)} 2\sqrt 3$

Solution

Problem 5

The number of values of $x$ satisfying the equation

$\frac {2x^2 - 10x}{x^2 - 5x} = x - 3$

is:

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$

Solution

Problem 6

$AB$ is the diameter of a circle centered at $O$ . $C$ is a point on the circle such that angle $BOC$ is $60^\circ$ . If the diameter of the circle is $5$ inches, the length of chord $AC$ , expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

Solution

Problem 7

Let $\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}$ be an identity in $x$ . The numerical value of $N_1N_2$ is:

$\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$

Solution

Problem 8

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of teh circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

Problem 9

If $x = (\log_82)^{(\log_28)}$ , then $\log_3x$ equals:

$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$

Solution

Problem 10

If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:

$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$

Solution

Problem 11

The sides of triangle $BAC$ are in the ratio $2: 3: 4$ . $BD$ is the angle-bisector drawn to the shortest side $AC$ , dividing it into segments $AD$ and $CD$ . If the length of $AC$ is $10$ , then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

Solution

Problem 12

The number of real values of $x$ that satisfy the equation $(2^{6x+3})(4^{3x+6})=8^{4x+5}$ is:

$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$

Solution

Problem 13

The number of points with positive rational coordinates selected from the set of points in the $xy$ -plane such that $x+y \le 5$ , is:

$\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}$

Solution

Problem 14

The length of rectangle $ABCD$ is 5 inches and its width is 3 inches. Diagonal $AC$ is divided into three equal segments by points $E$ and $F$ . The area of triangle $BEF$ , expressed in square inches, is:

$\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}$

Solution

Problem 15

If $x-y>x$ and $x+y<y$ , then

$\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0$

Solution

Problem 16

If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$ , $x$ and $y$ real numbers, then $xy$ equals:

$\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4$

Solution

Problem 17

The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=4$ is:

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

Solution

Problem 18

In a given arithmetic sequence the first term is $2$ , the last term is $29$ , and the sum of all the terms is $155$ . The common difference is:

$\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}$

Solution

Problem 19

Let $s_1$ be the sum of the first $n$ terms of the arithmetic sequence $8,12,\cdots$ and let $s_2$ be the sum of the first $n$ terms of the arithmetic sequence $17,19,\cdots$ . Assume $n \ne 0$ . Then $s_1=s_2$ for:

$\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n$

Solution

Problem 20

The negation of the proposition "For all pairs of real numbers $a,b$ , if $a=0$ , then $ab=0$ " is: There are real numbers $a,b$ such that

$\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0 \qquad \text{(C) } a=0 \text{ and } ab\ne 0$

$\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0$

Solution

Problem 21

An " $n$ -pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively $1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5$ ; for all $n$ values of $k$ , sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2$ ; prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5$ ).

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals:

$\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)$

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Five points $O,A,B,C,D$ are taken in order on a straight line with distances $OA = a$ , $OB = b$ , $OC = c$ , and $OD = d$ . $P$ is a point on the line between $B$ and $C$ and such that $AP: PD = BP: PC$ . Then $OP$ equals:

$\textbf{(A)} \frac {b^2 - bc}{a - b + c - d} \qquad \textbf{(B)} \frac {ac - bd}{a - b + c - d} \\ \textbf{(C)} - \frac {bd + ac}{a - b + c - d} \qquad \textbf{(D)} \frac {bc + ad}{a + b + c + d} \qquad \textbf{(E)} \frac {ac - bd}{a + b + c + d}$

Solution

Problem 29

Solution

Problem 30

Solution

Problem 31

Solution

Problem 32

Solution

Problem 33

Solution

Problem 34

Solution

Problem 35

Solution

Problem 36

Solution

Problem 37

Solution

Problem 38

Solution

Problem 39

Solution

Problem 40

Solution

@@ Line 7: / Line 7: @@
 == Problem 2 ==
+When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is
+<math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math>
 [[1966 AHSME Problems/Problem 2|Solution]]
@@ Line 80: / Line 82: @@
 == Problem 12 ==
+The number of real values of <math>x</math> that satisfy the equation <cmath>(2^{6x+3})(4^{3x+6})=8^{4x+5}</cmath> is:
+<math>\text{(A)  zero} \qquad \text{(B)  one} \qquad \text{(C)  two} \qquad \text{(D)  three} \qquad \text{(E)  greater than 3}</math>
 [[1966 AHSME Problems/Problem 12|Solution]]
 == Problem 13 ==
+The number of points with positive rational coordinates selected from the set of points in the <math>xy</math>-plane such that <math>x+y \le 5</math>, is:
+<math>\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}</math>
 [[1966 AHSME Problems/Problem 13|Solution]]
 == Problem 14 ==
+The length of rectangle <math>ABCD</math> is 5 inches and its width is 3 inches. Diagonal <math>AC</math> is divided into three equal segments by points <math>E</math> and <math>F</math>. The area of triangle <math>BEF</math>, expressed in square inches, is:
+<math>\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}</math>
 [[1966 AHSME Problems/Problem 14|Solution]]
 == Problem 15 ==
+If <math>x-y>x</math> and <math>x+y<y</math>, then
+<math>\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0</math>
 [[1966 AHSME Problems/Problem 15|Solution]]
 == Problem 16 ==
+If <math>\frac{4^x}{2^{x+y}}=8</math> and <math>\frac{9^{x+y}}{3^{5y}}=243</math>, <math>x</math> and <math>y</math> real numbers, then <math>xy</math> equals:
+<math>\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4</math>
 [[1966 AHSME Problems/Problem 16|Solution]]
 == Problem 17 ==
+The number of distinct points common to the curves <math>x^2+4y^2=1</math> and <math>4x^2+y^2=4</math> is:
+<math>\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4</math>
 [[1966 AHSME Problems/Problem 17|Solution]]
 == Problem 18 ==
+In a given arithmetic sequence the first term is <math>2</math>, the last term is <math>29</math>, and the sum of all the terms is <math>155</math>. The common difference is:
+<math>\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}</math>
 [[1966 AHSME Problems/Problem 18|Solution]]
 == Problem 19 ==
+Let <math>s_1</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>8,12,\cdots</math> and let <math>s_2</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>17,19,\cdots</math>. Assume <math>n \ne 0</math>. Then <math>s_1=s_2</math> for:
+<math>\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n</math>
 [[1966 AHSME Problems/Problem 19|Solution]]
 == Problem 20 ==
+The negation of the proposition "For all pairs of real numbers <math>a,b</math>, if <math>a=0</math>, then <math>ab=0</math>" is: There are real numbers <math>a,b</math> such that
+<math>\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0  \qquad \text{(C) } a=0 \text{ and } ab\ne 0</math>
+<math>\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0</math>
 [[1966 AHSME Problems/Problem 20|Solution]]
 == Problem 21 ==
+An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>).
+Let <math>S</math> be the degree-sum of the interior angles at the <math>n</math> points of the star; then <math>S</math> equals:
+<math>\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)</math>
 [[1966 AHSME Problems/Problem 21|Solution]]
@@ Line 176: / Line 202: @@
 [[1966 AHSME Problems/Problem 30|Solution]]
+== Problem 31 ==
+[[1966 AHSME Problems/Problem 31|Solution]]
+== Problem 32 ==
+[[1966 AHSME Problems/Problem 32|Solution]]
+== Problem 33 ==
+[[1966 AHSME Problems/Problem 33|Solution]]
+== Problem 34 ==
+[[1966 AHSME Problems/Problem 34|Solution]]
+== Problem 35 ==
+[[1966 AHSME Problems/Problem 35|Solution]]
+== Problem 36 ==
+[[1966 AHSME Problems/Problem 36|Solution]]
+== Problem 37 ==
+[[1966 AHSME Problems/Problem 37|Solution]]
+== Problem 38 ==
+[[1966 AHSME Problems/Problem 38|Solution]]
+== Problem 39 ==
+[[1966 AHSME Problems/Problem39|Solution]]
+== Problem 40 ==
+[[1966 AHSME Problems/Problem40|Solution]]
 == See also ==

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Difference between revisions of "1966 AHSME Problems"

Revision as of 23:46, 14 September 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