Difference between revisions of "1966 AHSME Problems"
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== Problem 2 == | == 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]] | [[1966 AHSME Problems/Problem 2|Solution]] | ||
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== Problem 12 == | == 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]] | [[1966 AHSME Problems/Problem 12|Solution]] | ||
== Problem 13 == | == 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]] | [[1966 AHSME Problems/Problem 13|Solution]] | ||
== Problem 14 == | == 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]] | [[1966 AHSME Problems/Problem 14|Solution]] | ||
== Problem 15 == | == 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]] | [[1966 AHSME Problems/Problem 15|Solution]] | ||
== Problem 16 == | == 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]] | [[1966 AHSME Problems/Problem 16|Solution]] | ||
== Problem 17 == | == 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]] | [[1966 AHSME Problems/Problem 17|Solution]] | ||
== Problem 18 == | == 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]] | [[1966 AHSME Problems/Problem 18|Solution]] | ||
== Problem 19 == | == 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]] | [[1966 AHSME Problems/Problem 19|Solution]] | ||
== Problem 20 == | == 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]] | [[1966 AHSME Problems/Problem 20|Solution]] | ||
== Problem 21 == | == 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]] | [[1966 AHSME Problems/Problem 21|Solution]] | ||
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[[1966 AHSME Problems/Problem 30|Solution]] | [[1966 AHSME Problems/Problem 30|Solution]] | ||
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+ | == Problem 31 == | ||
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+ | [[1966 AHSME Problems/Problem 31|Solution]] | ||
+ | == Problem 32 == | ||
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+ | [[1966 AHSME Problems/Problem 32|Solution]] | ||
+ | == Problem 33 == | ||
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+ | [[1966 AHSME Problems/Problem 33|Solution]] | ||
+ | == Problem 34 == | ||
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+ | [[1966 AHSME Problems/Problem 34|Solution]] | ||
+ | == Problem 35 == | ||
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+ | [[1966 AHSME Problems/Problem 35|Solution]] | ||
+ | == Problem 36 == | ||
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+ | [[1966 AHSME Problems/Problem 36|Solution]] | ||
+ | == Problem 37 == | ||
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+ | [[1966 AHSME Problems/Problem 37|Solution]] | ||
+ | == Problem 38 == | ||
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+ | [[1966 AHSME Problems/Problem 38|Solution]] | ||
+ | == Problem 39 == | ||
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+ | [[1966 AHSME Problems/Problem39|Solution]] | ||
+ | == Problem 40 == | ||
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+ | [[1966 AHSME Problems/Problem40|Solution]] | ||
== See also == | == See also == |
Revision as of 22:46, 14 September 2014
Contents
[hide]- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 See also
Problem 1
Given that the ratio of to is constant, and when , then, when , equals:
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
Problem 3
If the arithmetic mean of two numbers is and their geometric mean is , then an equation with the given two numbers as roots is:
Problem 4
Circle I is circumscribed about a given square and circle II is inscribed in the given square. If is the ratio of the area of circle to that of circle , then equals:
Problem 5
The number of values of satisfying the equation
is:
Problem 6
is the diameter of a circle centered at . is a point on the circle such that angle is . If the diameter of the circle is inches, the length of chord , expressed in inches, is:
Problem 7
Let be an identity in . The numerical value of is:
Problem 8
The length of the common chord of two intersecting circles is feet. If the radii are feet and feet, a possible value for the distance between the centers of teh circles, expressed in feet, is:
Problem 9
If , then equals:
Problem 10
If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:
Problem 11
The sides of triangle are in the ratio . is the angle-bisector drawn to the shortest side , dividing it into segments and . If the length of is , then the length of the longer segment of is:
Problem 12
The number of real values of that satisfy the equation is:
Problem 13
The number of points with positive rational coordinates selected from the set of points in the -plane such that , is:
Problem 14
The length of rectangle is 5 inches and its width is 3 inches. Diagonal is divided into three equal segments by points and . The area of triangle , expressed in square inches, is:
Problem 15
If and , then
Problem 16
If and , and real numbers, then equals:
Problem 17
The number of distinct points common to the curves and is:
Problem 18
In a given arithmetic sequence the first term is , the last term is , and the sum of all the terms is . The common difference is:
Problem 19
Let be the sum of the first terms of the arithmetic sequence and let be the sum of the first terms of the arithmetic sequence . Assume . Then for:
Problem 20
The negation of the proposition "For all pairs of real numbers , if , then " is: There are real numbers such that
Problem 21
An "-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively ; for all values of , sides and are non-parallel, sides and being respectively identical with sides and ; prolong the pairs of sides numbered and until they meet. (A figure is shown for the case ).
Let be the degree-sum of the interior angles at the points of the star; then equals:
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Five points are taken in order on a straight line with distances , , , and . is a point on the line between and and such that . Then equals:
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.