Difference between revisions of "1969 AHSME Problems"
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+ | {{AHSC 35 Problems | ||
+ | |year = 1969 | ||
+ | }} | ||
==Problem 1== | ==Problem 1== | ||
When <math>x</math> is added to both the numerator and denominator of the fraction | When <math>x</math> is added to both the numerator and denominator of the fraction | ||
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\text{(C) } 27\tfrac{1}{2}\quad | \text{(C) } 27\tfrac{1}{2}\quad | ||
\text{(D) } 28\quad | \text{(D) } 28\quad | ||
− | \text{(E) } | + | \text{(E) } 28\tfrac{1}{2}</math> |
[[1969 AHSME Problems/Problem 9|Solution]] | [[1969 AHSME Problems/Problem 9|Solution]] | ||
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==Problem 21== | ==Problem 21== | ||
+ | If the graph of <math>x^2+y^2=m</math> is tangent to that of <math>x+y=\sqrt{2m}</math>, then: | ||
+ | <math>\text{(A) m must equal } \tfrac{1}{2}\quad | ||
+ | \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ | ||
+ | \text{(C) m must equal } \sqrt{2}\quad | ||
+ | \text{(D) m must equal } 2\quad\\ | ||
+ | \text{(E) m may be an non-negative real number} </math> | ||
[[1969 AHSME Problems/Problem 21|Solution]] | [[1969 AHSME Problems/Problem 21|Solution]] | ||
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==Problem 22== | ==Problem 22== | ||
+ | Let <math>K</math> be the measure of the area bounded by the <math>x</math>-axis, the line <math>x=8</math>, and the curve defined by | ||
+ | <cmath>f={(x,y)\quad |\quad y=x \text{ when } 0 \le x \le 5, y=2x-5 \text{ when } 5 \le x \le 8}.</cmath> | ||
+ | |||
+ | Then <math>K</math> is: | ||
+ | |||
+ | <math>\text{(A) } 21.5\quad | ||
+ | \text{(B) } 36.4\quad | ||
+ | \text{(C) } 36.5\quad | ||
+ | \text{(D) } 44\quad | ||
+ | \text{(E) less than 44 but arbitrarily close to it} </math> | ||
[[1969 AHSME Problems/Problem 22|Solution]] | [[1969 AHSME Problems/Problem 22|Solution]] | ||
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==Problem 23== | ==Problem 23== | ||
+ | For any integer <math>n>1</math>, the number of prime numbers greater than <math>n!+1</math> and less than <math>n!+n</math> is: | ||
+ | |||
+ | <math>\text{(A) } 0\quad\qquad | ||
+ | \text{(B) } 1\quad\\ | ||
+ | \text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\ | ||
+ | \text{(D) } n-1\quad | ||
+ | \text{(E) } n</math> | ||
[[1969 AHSME Problems/Problem 23|Solution]] | [[1969 AHSME Problems/Problem 23|Solution]] | ||
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==Problem 24== | ==Problem 24== | ||
+ | When the natural numbers <math>P</math> and <math>P'</math>, with <math>P>P'</math>, are divided by the natural number <math>D</math>, the remainders are <math>R</math> and <math>R'</math>, respectively. When <math>PP'</math> and <math>RR'</math> are divided by <math>D</math>, the remainders are <math>r</math> and <math>r'</math>, respectively. Then: | ||
+ | <math>\text{(A) } r>r' \text{ always}\quad | ||
+ | \text{(B) } r<r' \text{ always}\quad\\ | ||
+ | \text{(C) } r>r' \text{ sometimes and } r<r' \text{ sometimes}\quad\\ | ||
+ | \text{(D) } r>r' \text{ sometimes and } r=r' \text{ sometimes}\quad\\ | ||
+ | \text{(E) } r=r' \text{ always}</math> | ||
[[1969 AHSME Problems/Problem 24|Solution]] | [[1969 AHSME Problems/Problem 24|Solution]] | ||
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==Problem 25== | ==Problem 25== | ||
+ | If it is known that <math>\log_2(a)+\log_2(b) \ge 6</math>, then the least value that can be taken on by <math>a+b</math> is: | ||
+ | |||
+ | <math>\text{(A) } 2\sqrt{6}\quad | ||
+ | \text{(B) } 6\quad | ||
+ | \text{(C) } 8\sqrt{2}\quad | ||
+ | \text{(D) } 16\quad | ||
+ | \text{(E) none of these} </math> | ||
[[1969 AHSME Problems/Problem 25|Solution]] | [[1969 AHSME Problems/Problem 25|Solution]] | ||
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+ | <asy> | ||
+ | draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); | ||
+ | draw((-1.7,0)--(0,0)--(1.7,0),dot); | ||
+ | draw((0,0)--(0,.98),dot); | ||
+ | MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); | ||
+ | </asy> | ||
+ | |||
+ | A parabolic arch has a height of <math>16</math> inches and a span of <math>40</math> inches. The height, in inches, of the arch at the point <math>5</math> inches from the center <math>M</math> is: | ||
+ | |||
+ | <math>\text{(A) } 1\quad | ||
+ | \text{(B) } 15\quad | ||
+ | \text{(C) } 15\tfrac{1}{3}\quad | ||
+ | \text{(D) } 15\tfrac{1}{2}\quad | ||
+ | \text{(E) } 15\tfrac{3}{4}</math> | ||
[[1969 AHSME Problems/Problem 26|Solution]] | [[1969 AHSME Problems/Problem 26|Solution]] | ||
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==Problem 27== | ==Problem 27== | ||
+ | A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in <math>2</math> hours, then the time, in hours, needed to traverse the <math>n</math>th mile is: | ||
+ | |||
+ | <math>\text{(A) } \frac{2}{n-1}\quad | ||
+ | \text{(B) } \frac{n-1}{2}\quad | ||
+ | \text{(C) } \frac{2}{n}\quad | ||
+ | \text{(D) } 2n\quad | ||
+ | \text{(E) } 2(n-1)</math> | ||
[[1969 AHSME Problems/Problem 27|Solution]] | [[1969 AHSME Problems/Problem 27|Solution]] | ||
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==Problem 28== | ==Problem 28== | ||
+ | Let <math>n</math> be the number of points <math>P</math> interior to the region bounded by a circle with radius <math>1</math>, such that the sum of squares of the distances from <math>P</math> to the endpoints of a given diameter is <math>3</math>. Then <math>n</math> is: | ||
+ | |||
+ | <math>\text{(A) } 0\quad | ||
+ | \text{(B) } 1\quad | ||
+ | \text{(C) } 2\quad | ||
+ | \text{(D) } 4\quad | ||
+ | \text{(E) } \infty</math> | ||
[[1969 AHSME Problems/Problem 28|Solution]] | [[1969 AHSME Problems/Problem 28|Solution]] | ||
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==Problem 29== | ==Problem 29== | ||
+ | If <math>x=t^{1/(t-1)}</math> and <math>y=t^{t/(t-1)},t>0,t \ne 1</math>, a relation between <math>x</math> and <math>y</math> is: | ||
+ | |||
+ | <math>\text{(A) } y^x=x^{1/y}\quad | ||
+ | \text{(B) } y^{1/x}=x^{y}\quad | ||
+ | \text{(C) } y^x=x^y\quad | ||
+ | \text{(D) } x^x=y^y\quad | ||
+ | \text{(E) none of these} </math> | ||
[[1969 AHSME Problems/Problem 29|Solution]] | [[1969 AHSME Problems/Problem 29|Solution]] | ||
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==Problem 30== | ==Problem 30== | ||
+ | Let <math>P</math> be a point of hypotenuse <math>AB</math> (or its extension) of isosceles right triangle <math>ABC</math>. Let <math>s=AP^2+PB^2</math>. Then: | ||
+ | |||
+ | <math>\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\ | ||
+ | \text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\ | ||
+ | \text{(C) } s=2CP^2 \text{ only if P is the midpoint or an endpoint of AB}\quad\\ | ||
+ | \text{(D) } s=2CP^2 \text{ always}\quad\\ | ||
+ | \text{(E) } s>2CP^2 \text{ if P is a trisection point of AB}</math> | ||
[[1969 AHSME Problems/Problem 30|Solution]] | [[1969 AHSME Problems/Problem 30|Solution]] | ||
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==Problem 31== | ==Problem 31== | ||
+ | Let <math>OABC</math> be a unit square in the <math>xy</math>-plane with <math>O(0,0),A(1,0),B(1,1)</math> and <math>C(0,1)</math>. Let <math>u=x^2-y^2</math>, and <math>v=2xy</math> be a transformation of the <math>xy</math>-plane into the <math>uv</math>-plane. The transform (or image) of the square is: | ||
+ | |||
+ | <asy> | ||
+ | draw((-3,0)--(3,0),EndArrow); | ||
+ | draw((0,-4)--(0,4),EndArrow); | ||
+ | draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot); | ||
+ | MP("(A)",(-5,2),SW); | ||
+ | MP("O",(0,0),SW); | ||
+ | MP("(-1,0)",(-1,0),SW); | ||
+ | MP("(0,2)",(0,2),NE); | ||
+ | MP("(1,0)",(1,0),SE); | ||
+ | MP("(0,-2)",(0,-2),SE); | ||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | draw((-3,0)--(3,0),EndArrow); | ||
+ | draw((0,-4)--(0,4),EndArrow); | ||
+ | draw(arc((1.5,0),2.5,126,234),black); | ||
+ | draw(arc((-1.5,0),2.5,54,-54),black); | ||
+ | MP("(B)",(-5,2),SW); | ||
+ | MP("O",(0,0),SW); | ||
+ | MP("(-1,0)",(-1,0),SW); | ||
+ | MP("(0,2)",(0,2),NE); | ||
+ | MP("(1,0)",(1,0),SE); | ||
+ | MP("(0,-2)",(0,-2),SE); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | draw((-3,0)--(3,0),EndArrow); | ||
+ | draw((0,-4)--(0,4),EndArrow); | ||
+ | draw((-1,0)--(0,2)--(1,0),black); | ||
+ | MP("(C)",(-5,2),SW); | ||
+ | MP("O",(0,0),SW); | ||
+ | MP("(-1,0)",(-1,0),SW); | ||
+ | MP("(0,2)",(0,2),NE); | ||
+ | MP("(1,0)",(1,0),SE); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | draw((-3,0)--(3,0),EndArrow); | ||
+ | draw((0,-4)--(0,4),EndArrow); | ||
+ | draw(arc((1.5,0),2.5,126,180),black); | ||
+ | draw(arc((-1.5,0),2.5,54,0),black); | ||
+ | MP("(D)",(-5,2),SW); | ||
+ | MP("O",(0,0),SW); | ||
+ | MP("(-1,0)",(-1,0),SW); | ||
+ | MP("(0,2)",(0,2),NE); | ||
+ | MP("(1,0)",(1,0),SE); | ||
+ | MP("(0,-2)",(0,-2),SE); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | draw((-3,0)--(3,0),EndArrow); | ||
+ | draw((0,-4)--(0,4),EndArrow); | ||
+ | draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot); | ||
+ | MP("(E)",(-5,2),SW); | ||
+ | MP("O",(.1,.1),SW); | ||
+ | MP("(-1,0)",(-1,0),SW); | ||
+ | MP("(0,1)",(0,1),NE); | ||
+ | MP("(1,0)",(1,0),SE); | ||
+ | MP("(0,-1)",(0,-1),SE); | ||
+ | </asy> | ||
[[1969 AHSME Problems/Problem 31|Solution]] | [[1969 AHSME Problems/Problem 31|Solution]] | ||
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==Problem 32== | ==Problem 32== | ||
+ | Let a sequence <math>\{u_n\}</math> be defined by <math>u_1=5</math> and the relationship <math>u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.</math>If <math>u_n</math> is expressed as a polynomial in <math>n</math>, the algebraic sum of its coefficients is: | ||
+ | |||
+ | <math>\text{(A) 3} \quad | ||
+ | \text{(B) 4} \quad | ||
+ | \text{(C) 5} \quad | ||
+ | \text{(D) 6} \quad | ||
+ | \text{(E) 11} </math> | ||
[[1969 AHSME Problems/Problem 32|Solution]] | [[1969 AHSME Problems/Problem 32|Solution]] | ||
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==Problem 33== | ==Problem 33== | ||
+ | |||
+ | Let <math>S_n</math> and <math>T_n</math> be the respective sums of the first <math>n</math> terms of two arithmetic series. If <math>S_n:T_n=(7n+1):(4n+27)</math> for all <math>n</math>, the ratio of the eleventh term of the first series to the eleventh term of the second series is: | ||
+ | |||
+ | <math>\text{(A) } 4:3\quad | ||
+ | \text{(B) } 3:2\quad | ||
+ | \text{(C) } 7:4\quad | ||
+ | \text{(D) } 78:71\quad | ||
+ | \text{(E) undetermined} </math> | ||
[[1969 AHSME Problems/Problem 33|Solution]] | [[1969 AHSME Problems/Problem 33|Solution]] | ||
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==Problem 34== | ==Problem 34== | ||
+ | The remainder <math>R</math> obtained by dividing <math>x^{100}</math> by <math>x^2-3x+2</math> is a polynomial of degree less than <math>2</math>. Then <math>R</math> may be written as: | ||
+ | |||
+ | |||
+ | <math>\text{(A) }2^{100}-1 \quad | ||
+ | \text{(B) } 2^{100}(x-1)-(x-2)\quad | ||
+ | \text{(C) } 2^{200}(x-3)\quad\\ | ||
+ | \text{(D) } x(2^{100}-1)+2(2^{99}-1)\quad | ||
+ | \text{(E) } 2^{100}(x+1)-(x+2)</math> | ||
[[1969 AHSME Problems/Problem 34|Solution]] | [[1969 AHSME Problems/Problem 34|Solution]] | ||
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==Problem 35== | ==Problem 35== | ||
+ | Let <math>L(m)</math> be the <math>x</math> coordinate of the left end point of the intersection of the graphs of <math>y=x^2-6</math> and <math>y=m</math>, where <math>-6<m<6</math>. Let <math>r=[L(-m)-L(m)]/m</math>. Then, as <math>m</math> is made arbitrarily close to zero, the value of <math>r</math> is: | ||
+ | |||
+ | <math>\text{(A) arbitrarily close to } 0\quad\\ | ||
+ | \text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\quad | ||
+ | \text{(C) arbitrarily close to }\frac{2}{\sqrt{6}} \quad | ||
+ | \text{(D) arbitrarily large} \quad | ||
+ | \text{(E) undetermined} </math> | ||
[[1969 AHSME Problems/Problem 35|Solution]] | [[1969 AHSME Problems/Problem 35|Solution]] | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | * [[AMC 12 Problems and Solutions]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 35p box|year=1969|before=[[1968 AHSME|1968 AHSC]]|after=[[1970 AHSME|1970 AHSC]]}} | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:41, 13 November 2024
1969 AHSC (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
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 |
Contents
- 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 See also
Problem 1
When is added to both the numerator and denominator of the fraction , the value of the fraction is changed to . Then equals:
Problem 2
If an item is sold for dollars, there is a loss of based on the cost. If, however, the same item is sold for dollars, there is a profit of based on the cost. The ratio of is:
Problem 3
If , written in base , is , the integer immediately preceding , written in base , is:
Problem 4
Let a binary operation on ordered pairs of integers be defined by . Then, if and represent identical pairs, equals:
Problem 5
If a number , diminished by four times its reciprocal, equals a given real constant , then, for this given , the sum of all such possible values of is
Problem 6
The area of the ring between two concentric circles is square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:
Problem 7
If the points and lie on the graph of , and , then equals:
Problem 8
Triangle is inscribed in a circle. The measure of the non-overlapping minor arcs , and are, respectively, . Then one interior angle of the triangle is:
Problem 9
The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:
Problem 10
The number of points equidistant from a circle and two parallel tangents to the circle is:
Problem 11
Given points and in the -plane; point is taken so that is a minimum. Then equals:
Problem 12
Let be the square of an expression which is linear in . Then has a particular value between:
Problem 13
A circle with radius is contained within the region bounded by a circle with radius . The area bounded by the larger circle is times the area of the region outside the smaller circle and inside the larger circle. Then equals:
Problem 14
The complete set of -values satisfying the inequality is the set of all such that:
Problem 15
In a circle with center and radius , chord is drawn with length equal to (units). From , a perpendicular to meets at . From a perpendicular to meets at . In terms of the area of triangle , in appropriate square units, is:
Problem 16
When , is expanded by the binomial theorem, it is found that when , where is a positive integer, the sum of the second and third terms is zero. Then equals:
Problem 17
The equation is satisfied by:
Problem 18
The number of points common to the graphs of is:
Problem 19
The number of distinct ordered pairs where and have positive integral values satisfying the equation is:
Problem 20
Let equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in is:
Problem 21
If the graph of is tangent to that of , then:
Problem 22
Let be the measure of the area bounded by the -axis, the line , and the curve defined by
Then is:
Problem 23
For any integer , the number of prime numbers greater than and less than is:
Problem 24
When the natural numbers and , with , are divided by the natural number , the remainders are and , respectively. When and are divided by , the remainders are and , respectively. Then:
Problem 25
If it is known that , then the least value that can be taken on by is:
Problem 26
A parabolic arch has a height of inches and a span of inches. The height, in inches, of the arch at the point inches from the center is:
Problem 27
A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in hours, then the time, in hours, needed to traverse the th mile is:
Problem 28
Let be the number of points interior to the region bounded by a circle with radius , such that the sum of squares of the distances from to the endpoints of a given diameter is . Then is:
Problem 29
If and , a relation between and is:
Problem 30
Let be a point of hypotenuse (or its extension) of isosceles right triangle . Let . Then:
Problem 31
Let be a unit square in the -plane with and . Let , and be a transformation of the -plane into the -plane. The transform (or image) of the square is:
Problem 32
Let a sequence be defined by and the relationship If is expressed as a polynomial in , the algebraic sum of its coefficients is:
Problem 33
Let and be the respective sums of the first terms of two arithmetic series. If for all , the ratio of the eleventh term of the first series to the eleventh term of the second series is:
Problem 34
The remainder obtained by dividing by is a polynomial of degree less than . Then may be written as:
Problem 35
Let be the coordinate of the left end point of the intersection of the graphs of and , where . Let . Then, as is made arbitrarily close to zero, the value of is:
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1968 AHSC |
Followed by 1970 AHSC | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.