Difference between revisions of "1960 AHSME Problems"
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+ | {{AHSC 40 Problems | ||
+ | |year = 1960 | ||
+ | }} | ||
== Problem 1== | == Problem 1== | ||
If <math>2</math> is a solution (root) of <math>x^3+hx+10=0</math>, then <math>h</math> equals: | If <math>2</math> is a solution (root) of <math>x^3+hx+10=0</math>, then <math>h</math> equals: | ||
<math>\textbf{(A)}10\qquad | <math>\textbf{(A)}10\qquad | ||
− | \textbf{(B )}9 \qquad | + | \textbf{(B)}9 \qquad |
− | \textbf{(C )}2\qquad | + | \textbf{(C)}2\qquad |
− | \textbf{(D )}-2\qquad | + | \textbf{(D)}-2\qquad |
− | \textbf{(E )}-9 </math> | + | \textbf{(E)}-9 </math> |
[[1960 AHSME Problems/Problem 1|Solution]] | [[1960 AHSME Problems/Problem 1|Solution]] | ||
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expressed in dollars, is: | expressed in dollars, is: | ||
− | <math>\textbf{(A)}0\qquad | + | <math>\textbf{(A) }0\qquad |
− | \textbf{(B )}144\qquad | + | \textbf{(B) }144\qquad |
− | \textbf{(C )}256\qquad | + | \textbf{(C) }256\qquad |
− | \textbf{(D )}400\qquad | + | \textbf{(D) }400\qquad |
− | \textbf{(E )}416 </math> | + | \textbf{(E) }416 </math> |
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Each of two angles of a triangle is <math>60^{\circ}</math> and the included side is <math>4</math> inches. The area of the triangle, in square inches, is: | Each of two angles of a triangle is <math>60^{\circ}</math> and the included side is <math>4</math> inches. The area of the triangle, in square inches, is: | ||
− | <math>\textbf{(A)}8\sqrt{3}\qquad | + | <math>\textbf{(A) }8\sqrt{3}\qquad |
− | \textbf{(B )}8\qquad | + | \textbf{(B) }8\qquad |
− | \textbf{(C )}4\sqrt{3}\qquad | + | \textbf{(C) }4\sqrt{3}\qquad |
− | \textbf{(D )}4\qquad | + | \textbf{(D) }4\qquad |
− | \textbf{(E )}2\sqrt{3} </math> | + | \textbf{(E) }2\sqrt{3} </math> |
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The number of distinct points common to the graphs of <math>x^2+y^2=9</math> and <math>y^2=9</math> is: | The number of distinct points common to the graphs of <math>x^2+y^2=9</math> and <math>y^2=9</math> is: | ||
− | <math>\textbf{(A)}\text{infinitely many}\qquad | + | <math>\textbf{(A) }\text{infinitely many}\qquad |
− | \textbf{(B )}\text{four}\qquad | + | \textbf{(B) }\text{four}\qquad |
− | \textbf{(C )}\text{two}\qquad | + | \textbf{(C) }\text{two}\qquad |
− | \textbf{(D )}\text{one}\qquad | + | \textbf{(D) }\text{one}\qquad |
− | \textbf{(E )}\text{none} </math> | + | \textbf{(E) }\text{none} </math> |
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The circumference of a circle is <math>100</math> inches. The side of a square inscribed in this circle, expressed in inches, is: | The circumference of a circle is <math>100</math> inches. The side of a square inscribed in this circle, expressed in inches, is: | ||
− | <math>\textbf{(A)}\frac{25\sqrt{2}}{\pi}\qquad | + | <math>\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad |
− | \textbf{(B )}\frac{50\sqrt{2}}{\pi}\qquad | + | \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad |
− | \textbf{(C )}\frac{100}{\pi}\qquad | + | \textbf{(C) }\frac{100}{\pi}\qquad |
− | \textbf{(D )}\frac{100\sqrt{2}}{\pi}\qquad | + | \textbf{(D) }\frac{100\sqrt{2}}{\pi}\qquad |
− | \textbf{(E )}50\sqrt{2} </math> | + | \textbf{(E) }50\sqrt{2} </math> |
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Then the area of circle <math>II</math>, in square inches, is: | Then the area of circle <math>II</math>, in square inches, is: | ||
− | <math>\textbf{(A)}8\qquad | + | <math>\textbf{(A) }8\qquad |
− | \textbf{(B )}8\sqrt{2}\qquad | + | \textbf{(B) }8\sqrt{2}\qquad |
− | \textbf{(C )}8\sqrt{\pi}\qquad | + | \textbf{(C) }8\sqrt{\pi}\qquad |
− | \textbf{(D )}16\qquad | + | \textbf{(D) }16\qquad |
− | \textbf{(E )}16\sqrt{2} </math> | + | \textbf{(E) }16\sqrt{2} </math> |
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When reduced to lowest terms the sum of the numerator and denominator of this fraction is: | When reduced to lowest terms the sum of the numerator and denominator of this fraction is: | ||
− | <math>\textbf{(A)}7\qquad | + | <math>\textbf{(A) }7\qquad |
− | \textbf{(B)} 29\qquad | + | \textbf{(B) } 29\qquad |
− | \textbf{(C )}141\qquad | + | \textbf{(C) }141\qquad |
− | \textbf{(D )}349\qquad | + | \textbf{(D) }349\qquad |
− | \textbf{(E )}\text{none of these} </math> | + | \textbf{(E) }\text{none of these} </math> |
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== Problem 9== | == Problem 9== | ||
− | The fraction <math>\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}</math> is (with suitable restrictions of the values of | + | The fraction <math>\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}</math> is (with suitable restrictions of the values of a, b, and c): |
− | <math>\text{(A) irreducible}\qquad | + | <math>\text{(A) irreducible}\qquad</math> |
− | \text{(B) reducible to negative 1}\qquad | + | |
− | \text{(C) reducible to a polynomial of three terms}\qquad | + | <math>\text{(B) reducible to negative 1}\qquad</math> |
− | \text{(D) reducible to}\frac{a-b+c}{a+b-c}\qquad | + | |
− | \text{(E) reducible to}\frac{a+b-c}{a-b+c} </math> | + | <math>\text{(C) reducible to a polynomial of three terms}\qquad</math> |
− | + | ||
+ | <math>\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad</math> | ||
+ | |||
+ | <math>\text{(E) reducible to} \frac{a+b-c}{a-b+c}</math> | ||
[[1960 AHSME Problems/Problem 9|Solution]] | [[1960 AHSME Problems/Problem 9|Solution]] | ||
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Given the following six statements: | Given the following six statements: | ||
− | < | + | <cmath>\text{(1) All women are good drivers}</cmath> |
− | \text{(2) Some women are good drivers} | + | <cmath>\text{(2) Some women are good drivers}</cmath> |
− | \text{(3) No men are good drivers} | + | <cmath>\text{(3) No men are good drivers}</cmath> |
− | \text{(4) All men are bad drivers} | + | <cmath>\text{(4) All men are bad drivers}</cmath> |
− | \text{(5) At least one man is a bad driver} | + | <cmath>\text{(5) At least one man is a bad driver}</cmath> |
− | \text{(6) All men are good drivers.}</math><math> | + | <cmath>\text{(6) All men are good drivers.}</cmath> |
+ | |||
+ | |||
+ | The statement that negates statement <math>(6)</math> is: | ||
− | |||
− | < | + | <math>\textbf{(A) }(1)\qquad |
− | \textbf{(B )}(2)\qquad | + | \textbf{(B) }(2)\qquad |
− | \textbf{(C )}(3)\qquad | + | \textbf{(C) }(3)\qquad |
− | \textbf{(D )}(4)\qquad | + | \textbf{(D) }(4)\qquad |
− | \textbf{(E )}(5) | + | \textbf{(E) }(5)</math> |
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== Problem 11== | == Problem 11== | ||
− | For a given value of < | + | For a given value of <math>k</math> the product of the roots of <math>x^2-3kx+2k^2-1=0</math> |
− | is < | + | is <math>7</math>. The roots may be characterized as: |
− | < | + | <math>\textbf{(A) }\text{integral and positive} \qquad\textbf{(B) }\text{integral and negative} \qquad \ |
− | \textbf{(B )}\text{integral and negative} \qquad | + | \textbf{(C) }\text{rational, but not integral} \qquad\textbf{(D) }\text{irrational} \qquad\textbf{(E) } \text{imaginary} </math> |
− | \textbf{(C )}\text{rational, but not integral} \qquad | ||
− | \textbf{(D )}\text{irrational} \qquad | ||
− | \textbf{(E )} \text{imaginary} | ||
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== Problem 12== | == Problem 12== | ||
− | The locus of the centers of all circles of given radius < | + | The locus of the centers of all circles of given radius <math>a</math>, in the same plane, passing through a fixed point, is: |
− | < | + | <math>\textbf{(A) }\text{a point}\qquad |
− | \textbf{(B )}\text{ a straight line}\qquad | + | \textbf{(B) }\text{ a straight line}\qquad |
− | \textbf{(C )}\text{two straight lines}\qquad | + | \textbf{(C) }\text{two straight lines}\qquad |
− | \textbf{(D )}\text{a circle}\qquad | + | \textbf{(D) }\text{a circle}\qquad |
− | \textbf{(E )}\text{two circles} <math> | + | \textbf{(E) }\text{two circles} </math> |
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== Problem 13== | == Problem 13== | ||
− | The polygon(s) formed by < | + | The polygon(s) formed by <math>y=3x+2, y=-3x+2</math>, and <math>y=-2</math>, is (are): |
− | < | + | <math>\textbf{(A) }\text{an equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \ |
− | \textbf{(B )}\text{an isosceles triangle}\qquad | + | \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral} </math> |
− | \textbf{(C )}\text{a right triangle}\qquad | ||
− | \textbf{(D )}\text{a triangle and a trapezoid}\qquad | ||
− | \textbf{(E )}\text{a quadrilateral} | ||
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== Problem 14== | == Problem 14== | ||
− | If < | + | If <math>a</math> and <math>b</math> are real numbers, the equation <math>3x-5+a=bx+1</math> has a unique solution <math>x</math> [The symbol <math>a \neq 0</math> means that <math>a</math> is different from zero]: |
− | < | + | <math>\text{(A) for all a and b} \qquad |
− | \ | + | \text{(B) if a }\neq\text{2b}\qquad |
− | \ | + | \text{(C) if a }\neq 6\qquad \ |
− | \ | + | \text{(D) if b }\neq 0\qquad |
− | \ | + | \text{(E) if b }\neq 3 </math> |
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== Problem 15== | == Problem 15== | ||
− | Triangle < | + | Triangle <math>I</math> is equilateral with side <math>A</math>, perimeter <math>P</math>, area <math>K</math>, and circumradius <math>R</math> (radius of the circumscribed circle). |
− | Triangle < | + | Triangle <math>II</math> is equilateral with side <math>a</math>, perimeter <math>p</math>, area <math>k</math>, and circumradius <math>r</math>. If <math>A</math> is different from <math>a</math>, then: |
− | < | + | <math>\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad |
− | \textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad | + | \textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad \ |
\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad | \textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad | ||
\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad | \textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad | ||
− | \textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes} <math> | + | \textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes} </math> |
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== Problem 16== | == Problem 16== | ||
− | In the numeration system with base < | + | In the numeration system with base <math>5</math>, counting is as follows: <math>1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots</math>. |
− | The number whose description in the decimal system is < | + | The number whose description in the decimal system is <math>69</math>, when described in the base <math>5</math> system, is a number with: |
− | < | + | <math>\textbf{(A)}\ \text{two consecutive digits} \qquad\textbf{(B)}\ \text{two non-consecutive digits} \qquad \ |
− | \textbf{(B)}\ \text{two non-consecutive digits} \qquad | + | \textbf{(C)}\ \text{three consecutive digits} \qquad\textbf{(D)}\ \text{three non-consecutive digits} \qquad \ |
− | \textbf{(C)}\ \text{three consecutive digits} \qquad | + | \textbf{(E)}\ \text{four digits} </math> |
− | \textbf{(D)}\ \text{three non-consecutive digits} \qquad | ||
− | \textbf{(E)}\ \text{four digits} | ||
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== Problem 17== | == Problem 17== | ||
− | The formula < | + | The formula <math>N=8 \times 10^{8} \times x^{-3/2}</math> gives, for a certain group, the number of individuals whose income exceeds <math>x</math> dollars. |
− | The lowest income, in dollars, of the wealthiest < | + | The lowest income, in dollars, of the wealthiest <math>800</math> individuals is at least: |
− | < | + | <math>\textbf{(A)}\ 10^4\qquad |
\textbf{(B)}\ 10^6\qquad | \textbf{(B)}\ 10^6\qquad | ||
\textbf{(C)}\ 10^8\qquad | \textbf{(C)}\ 10^8\qquad | ||
\textbf{(D)}\ 10^{12} \qquad | \textbf{(D)}\ 10^{12} \qquad | ||
− | \textbf{(E)}\ 10^{16} <math> | + | \textbf{(E)}\ 10^{16} </math> |
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== Problem 18== | == Problem 18== | ||
− | The pair of equations < | + | The pair of equations <math>3^{x+y}=81</math> and <math>81^{x-y}=3</math> has: |
− | < | + | <math>\textbf{(A)}\ \text{no common solution} \qquad \ |
− | \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad | + | \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad \ |
− | \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad | + | \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad \ |
− | \textbf{(D)}\text{ a common solution in positive and negative integers} \qquad | + | \textbf{(D)}\text{ a common solution in positive and negative integers} \qquad \ |
− | \textbf{(E)}\ \text{none of these} <math> | + | \textbf{(E)}\ \text{none of these} </math> |
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== Problem 19== | == Problem 19== | ||
− | Consider equation < | + | Consider equation <math>I: x+y+z=46</math> where <math>x, y</math>, and <math>z</math> are positive integers, and equation <math>II: x+y+z+w=46</math>, |
− | where < | + | where <math>x, y, z</math>, and <math>w</math> are positive integers. Then |
− | < | + | <math>\textbf{(A)}\ \text{I can be solved in consecutive integers} \qquad \ |
− | \ | + | \textbf{(B)}\ \text{I can be solved in consecutive even integers} \qquad \ |
− | \ | + | \textbf{(C)}\ \text{II can be solved in consecutive integers} \qquad \ |
− | \ | + | \textbf{(D)}\ \text{II can be solved in consecutive even integers} \qquad \ |
− | \ | + | \textbf{(E)}\ \text{II can be solved in consecutive odd integers} </math> |
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== Problem 20== | == Problem 20== | ||
− | The coefficient of < | + | The coefficient of <math>x^7</math> in the expansion of <math>(\frac{x^2}{2}-\frac{2}{x})^8</math> is: |
− | < | + | <math>\textbf{(A)}\ 56\qquad |
\textbf{(B)}\ -56\qquad | \textbf{(B)}\ -56\qquad | ||
\textbf{(C)}\ 14\qquad | \textbf{(C)}\ 14\qquad | ||
\textbf{(D)}\ -14\qquad | \textbf{(D)}\ -14\qquad | ||
− | \textbf{(E)}\ 0 <math> | + | \textbf{(E)}\ 0 </math> |
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== Problem 21== | == Problem 21== | ||
− | The diagonal of square < | + | The diagonal of square <math>I</math> is <math>a+b</math>. The perimeter of square <math>II</math> with twice the area of <math>I</math> is: |
− | < | + | <math>\textbf{(A)}\ (a+b)^2\qquad |
\textbf{(B)}\ \sqrt{2}(a+b)^2\qquad | \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad | ||
\textbf{(C)}\ 2(a+b)\qquad | \textbf{(C)}\ 2(a+b)\qquad | ||
\textbf{(D)}\ \sqrt{8}(a+b) \qquad | \textbf{(D)}\ \sqrt{8}(a+b) \qquad | ||
− | \textbf{(E)}\ 4(a+b) <math> | + | \textbf{(E)}\ 4(a+b) </math> |
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== Problem 22== | == Problem 22== | ||
− | The equality < | + | The equality <math>(x+m)^2-(x+n)^2=(m-n)^2</math>, where <math>m</math> and <math>n</math> are unequal non-zero constants, is satisfied by <math>x=am+bn</math>, where: |
− | < | + | <math>\textbf{(A)}\ a = 0, b \text{ } \text{has a unique non-zero value}\qquad \ |
− | \textbf{(B)}\ a = 0, b \text{ } \text{has two non-zero values}\qquad | + | \textbf{(B)}\ a = 0, b \text{ } \text{has two non-zero values}\qquad \ |
− | \textbf{(C)}\ b = 0, a \text{ } \text{has a unique non-zero value}\qquad | + | \textbf{(C)}\ b = 0, a \text{ } \text{has a unique non-zero value}\qquad \ |
− | \textbf{(D)}\ b = 0, a \text{ } \text{has two non-zero values}\qquad | + | \textbf{(D)}\ b = 0, a \text{ } \text{has two non-zero values}\qquad \ |
− | \textbf{(E)}\ a \text{ } \text{and} \text{ } b \text{ } \text{each have a unique non-zero value} <math> | + | \textbf{(E)}\ a \text{ } \text{and} \text{ } b \text{ } \text{each have a unique non-zero value} </math> |
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== Problem 23== | == Problem 23== | ||
− | The radius < | + | The radius <math>R</math> of a cylindrical box is <math>8</math> inches, the height <math>H</math> is <math>3</math> inches. |
− | The volume < | + | The volume <math>V = \pi R^2H</math> is to be increased by the same fixed positive amount when <math>R</math> |
− | is increased by < | + | is increased by <math>x</math> inches as when <math>H</math> is increased by <math>x</math> inches. This condition is satisfied by: |
− | < | + | <math>\textbf{(A)}\ \text{no real value of x} \qquad \\ |
− | \textbf{(B)}\ \text{one integral value of} \ | + | \textbf{(B)}\ \text{one integral value of x} \qquad \\ |
− | \textbf{(C)}\ \text{one rational, but not integral, value of} \ | + | \textbf{(C)}\ \text{one rational, but not integral, value of x} \qquad \\ |
− | \textbf{(D)}\ \text{one irrational value of} \ | + | \textbf{(D)}\ \text{one irrational value of x}\qquad \\ |
− | \textbf{(E)}\ \text{two real values of} | + | \textbf{(E)}\ \text{two real values of x} </math> |
− | |||
[[1960 AHSME Problems/Problem 23|Solution]] | [[1960 AHSME Problems/Problem 23|Solution]] | ||
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== Problem 24== | == Problem 24== | ||
− | If < | + | If <math>\log_{2x}216 = x</math>, where <math>x</math> is real, then <math>x</math> is: |
− | < | + | <math> \textbf{(A)}\ \text{A non-square, non-cube integer}\qquad </math> |
− | \textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number} \qquad | + | <math> \textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad </math> |
− | \textbf{(C)}\ \text{An irrational number} \qquad | + | <math> \textbf{(C)}\ \text{An irrational number}\qquad </math> |
− | \textbf{(D)}\ \text{A perfect square}\qquad | + | <math> \textbf{(D)}\ \text{A perfect square}\qquad </math> |
− | \textbf{(E)}\ \text{A perfect cube} | + | <math> \textbf{(E)}\ \text{A perfect cube} </math> |
− | |||
[[1960 AHSME Problems/Problem 24|Solution]] | [[1960 AHSME Problems/Problem 24|Solution]] | ||
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== Problem 25== | == Problem 25== | ||
− | Let < | + | Let <math>m</math> and <math>n</math> be any two odd numbers, with <math>n</math> less than <math>m</math>. |
− | The largest integer which divides all possible numbers of the form < | + | The largest integer which divides all possible numbers of the form <math>m^2-n^2</math> is: |
− | < | + | <math>\textbf{(A)}\ 2\qquad |
\textbf{(B)}\ 4\qquad | \textbf{(B)}\ 4\qquad | ||
\textbf{(C)}\ 6\qquad | \textbf{(C)}\ 6\qquad | ||
\textbf{(D)}\ 8\qquad | \textbf{(D)}\ 8\qquad | ||
− | \textbf{(E)}\ 16 <math> | + | \textbf{(E)}\ 16 </math> |
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== Problem 26== | == Problem 26== | ||
− | Find the set of < | + | Find the set of <math>x</math>-values satisfying the inequality <math>|\frac{5-x}{3}|<2</math>. [The symbol <math>|a|</math> means <math>+a</math> if <math>a</math> is positive, |
− | < | + | <math>-a</math> if <math>a</math> is negative,<math>0</math> if <math>a</math> is zero. The notation <math>1<a<2</math> means that a can have any value between <math>1</math> and <math>2</math>, excluding <math>1</math> and <math>2</math>. ] |
− | < | + | <math>\textbf{(A)}\ 1 < x < 11\qquad |
\textbf{(B)}\ -1 < x < 11\qquad | \textbf{(B)}\ -1 < x < 11\qquad | ||
\textbf{(C)}\ x< 11\qquad | \textbf{(C)}\ x< 11\qquad | ||
\textbf{(D)}\ x>11\qquad | \textbf{(D)}\ x>11\qquad | ||
− | \textbf{(E)}\ |x| < 6 <math> | + | \textbf{(E)}\ |x| < 6 </math> |
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== Problem 27== | == Problem 27== | ||
− | Let < | + | Let <math>S</math> be the sum of the interior angles of a polygon <math>P</math> for which each interior angle is <math>7\frac{1}{2}</math> times the |
exterior angle at the same vertex. Then | exterior angle at the same vertex. Then | ||
− | < | + | <math>\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \ |
− | \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad | + | \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \ |
− | \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad | + | \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \ |
− | \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad | + | \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \ |
− | \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular} <math> | + | \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular} </math> |
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== Problem 28== | == Problem 28== | ||
− | The equation < | + | The equation <math>x-\frac{7}{x-3}=3-\frac{7}{x-3}</math> has: |
− | < | + | <math> \textbf{(A)}\ \text{infinitely many integral roots}\qquad\textbf{(B)}\ \text{no root}\qquad\textbf{(C)}\ \text{one integral root}\qquad </math> |
− | \textbf{(B)}\ \text{no root}\qquad | + | <math> \textbf{(D)}\ \text{two equal integral roots}\qquad\textbf{(E)}\ \text{two equal non-integral roots} </math> |
− | \textbf{(C)}\ \text{one integral root}\qquad | ||
− | \textbf{(D)}\ \text{two equal integral roots} \qquad | ||
− | \textbf{(E)}\ \text{two equal non-integral roots} | ||
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== Problem 29== | == Problem 29== | ||
− | Five times < | + | Five times <math>A</math>'s money added to <math>B</math>'s money is more than <math>\$51.00</math>. Three times <math>A</math>'s money minus <math>B</math>'s money is <math>\$21.00</math>. |
− | If < | + | If <math>a</math> represents <math>A</math>'s money in dollars and <math>b</math> represents <math>B</math>'s money in dollars, then: |
− | < | + | <math>\textbf{(A)}\ a>9, b>6 \qquad |
\textbf{(B)}\ a>9, b<6 \qquad | \textbf{(B)}\ a>9, b<6 \qquad | ||
\textbf{(C)}\ a>9, b=6\qquad | \textbf{(C)}\ a>9, b=6\qquad | ||
\textbf{(D)}\ a>9, \text{but we can put no bounds on} \text{ } b\qquad | \textbf{(D)}\ a>9, \text{but we can put no bounds on} \text{ } b\qquad | ||
− | \textbf{(E)}\ 2a=3b <math> | + | \textbf{(E)}\ 2a=3b </math> |
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== Problem 30== | == Problem 30== | ||
− | Given the line < | + | Given the line <math>3x+5y=15</math> and a point on this line equidistant from the coordinate axes. Such a point exists in: |
− | < | + | <math> \textbf{(A)}\ \text{none of the quadrants}\qquad\textbf{(B)}\ \text{quadrant I only}\qquad\textbf{(C)}\ \text{quadrants I, II only}\qquad </math> |
− | \textbf{(B)}\ \text{quadrant I only}\qquad | + | <math> \textbf{(D)}\ \text{quadrants I, II, III only}\qquad\textbf{(E)}\ \text{each of the quadrants} </math> |
− | \textbf{(C)}\ \text{quadrants I, II only}\qquad | ||
− | \textbf{(D)}\ \text{quadrants I, II, III only} \qquad | ||
− | \textbf{(E)}\ \text{each of the quadrants} | ||
− | |||
[[1960 AHSME Problems/Problem 30|Solution]] | [[1960 AHSME Problems/Problem 30|Solution]] | ||
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== Problem 31== | == Problem 31== | ||
− | For < | + | For <math>x^2+2x+5</math> to be a factor of <math>x^4+px^2+q</math>, the values of <math>p</math> and <math>q</math> must be, respectively: |
− | < | + | <math>\textbf{(A)}\ -2, 5\qquad |
\textbf{(B)}\ 5, 25\qquad | \textbf{(B)}\ 5, 25\qquad | ||
\textbf{(C)}\ 10, 20\qquad | \textbf{(C)}\ 10, 20\qquad | ||
\textbf{(D)}\ 6, 25\qquad | \textbf{(D)}\ 6, 25\qquad | ||
− | \textbf{(E)}\ 14, 25 <math> | + | \textbf{(E)}\ 14, 25 </math> |
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== Problem 32== | == Problem 32== | ||
− | In this figure the center of the circle is < | + | In this figure the center of the circle is <math>O</math>. <math>AB \perp BC</math>, <math>ADOE</math> is a straight line, <math>AP = AD</math>, and <math>AB</math> has a length twice the radius. Then: |
<asy> | <asy> | ||
Line 444: | Line 435: | ||
label("$P$",P,SW);</asy> | label("$P$",P,SW);</asy> | ||
− | <math>\textbf{(A)} AP^2 = PB \times AB\qquad | + | <math>\textbf{(A)} AP^2 = PB \times AB\qquad \ |
− | \textbf{(B)}\ AP \times DO = PB \times AD\qquad | + | \textbf{(B)}\ AP \times DO = PB \times AD\qquad \ |
− | \textbf{(C)}\ AB^2 = AD \times DE\qquad | + | \textbf{(C)}\ AB^2 = AD \times DE\qquad \ |
− | \textbf{(D)}\ AB \times AD = OB \times AO\qquad | + | \textbf{(D)}\ AB \times AD = OB \times AO\qquad \ |
\textbf{(E)}\ \text{none of these} </math> | \textbf{(E)}\ \text{none of these} </math> | ||
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== Problem 37== | == Problem 37== | ||
− | The base of a triangle is of length <math>b</math>, and the | + | The base of a triangle is of length <math>b</math>, and the altitude is of length <math>h</math>. |
A rectangle of height <math>x</math> is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is: | A rectangle of height <math>x</math> is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is: | ||
Line 561: | Line 552: | ||
To satisfy the equation <math>\frac{a+b}{a}=\frac{b}{a+b}</math>, <math>a</math> and <math>b</math> must be: | To satisfy the equation <math>\frac{a+b}{a}=\frac{b}{a+b}</math>, <math>a</math> and <math>b</math> must be: | ||
− | <math>\textbf{(A)}\ \text{both rational}\qquad | + | <math> \textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad </math> |
− | \textbf{(B)}\ \text{both real but not rational}\qquad | + | <math> \textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real} </math> |
− | \textbf{(C)}\ \text{both not real}\qquad | + | |
− | \textbf{(D)}\ \text{one real, one not real}\qquad | ||
− | \textbf{(E)}\ \text{one real, one not real or both not real} | ||
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Given right <math>\triangle ABC</math> with legs <math>BC=3, AC=4</math>. Find the length of the shorter angle trisector from <math>C</math> to the hypotenuse: | Given right <math>\triangle ABC</math> with legs <math>BC=3, AC=4</math>. Find the length of the shorter angle trisector from <math>C</math> to the hypotenuse: | ||
− | <math>\textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad | + | <math> \textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad</math> |
− | \textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad | ||
− | \textbf{(C)}\ 6\sqrt{3}-8\qquad | ||
− | \textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad | ||
− | \textbf{(E)}\ \frac{25}{12} | ||
[[1960 AHSME Problems/Problem 40|Solution]] | [[1960 AHSME Problems/Problem 40|Solution]] | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | * [[AMC 12 Problems and Solutions]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 40p box|year=1960|before=[[1959 AHSME|1959 AHSC]]|after=[[1961 AHSME|1961 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 20:27, 30 December 2020
1960 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 • 36 • 37 • 38 • 39 • 40 |
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
If is a solution (root) of , then equals:
Problem 2
It takes seconds for a clock to strike o'clock beginning at o'clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike o'clock?
Problem 3
Applied to a bill for the difference between a discount of % and two successive discounts of % and %, expressed in dollars, is:
Problem 4
Each of two angles of a triangle is and the included side is inches. The area of the triangle, in square inches, is:
Problem 5
The number of distinct points common to the graphs of and is:
Problem 6
The circumference of a circle is inches. The side of a square inscribed in this circle, expressed in inches, is:
Problem 7
Circle passes through the center of, and is tangent to, circle . The area of circle is square inches. Then the area of circle , in square inches, is:
Problem 8
The number can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:
Problem 9
The fraction is (with suitable restrictions of the values of a, b, and c):
Problem 10
Given the following six statements:
The statement that negates statement is:
Problem 11
For a given value of the product of the roots of is . The roots may be characterized as:
Problem 12
The locus of the centers of all circles of given radius , in the same plane, passing through a fixed point, is:
Problem 13
The polygon(s) formed by , and , is (are):
Problem 14
If and are real numbers, the equation has a unique solution [The symbol means that is different from zero]:
Problem 15
Triangle is equilateral with side , perimeter , area , and circumradius (radius of the circumscribed circle). Triangle is equilateral with side , perimeter , area , and circumradius . If is different from , then:
Problem 16
In the numeration system with base , counting is as follows: . The number whose description in the decimal system is , when described in the base system, is a number with:
Problem 17
The formula gives, for a certain group, the number of individuals whose income exceeds dollars. The lowest income, in dollars, of the wealthiest individuals is at least:
Problem 18
The pair of equations and has:
Problem 19
Consider equation where , and are positive integers, and equation , where , and are positive integers. Then
Problem 20
The coefficient of in the expansion of is:
Problem 21
The diagonal of square is . The perimeter of square with twice the area of is:
Problem 22
The equality , where and are unequal non-zero constants, is satisfied by , where:
Problem 23
The radius of a cylindrical box is inches, the height is inches. The volume is to be increased by the same fixed positive amount when is increased by inches as when is increased by inches. This condition is satisfied by:
Problem 24
If , where is real, then is:
Problem 25
Let and be any two odd numbers, with less than . The largest integer which divides all possible numbers of the form is:
Problem 26
Find the set of -values satisfying the inequality . [The symbol means if is positive, if is negative, if is zero. The notation means that a can have any value between and , excluding and . ]
Problem 27
Let be the sum of the interior angles of a polygon for which each interior angle is times the exterior angle at the same vertex. Then
Problem 28
The equation has:
Problem 29
Five times 's money added to 's money is more than . Three times 's money minus 's money is . If represents 's money in dollars and represents 's money in dollars, then:
Problem 30
Given the line and a point on this line equidistant from the coordinate axes. Such a point exists in:
Problem 31
For to be a factor of , the values of and must be, respectively:
Problem 32
In this figure the center of the circle is . , is a straight line, , and has a length twice the radius. Then:
Problem 33
You are given a sequence of terms; each term has the form where stands for the product of all prime numbers less than or equal to , and takes, successively, the values . Let be the number of primes appearing in this sequence. Then is:
Problem 34
Two swimmers, at opposite ends of a -foot pool, start to swim the length of the pool, one at the rate of feet per second, the other at feet per second. They swim back and forth for minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
Problem 35
From point outside a circle, with a circumference of units, a tangent is drawn. Also from a secant is drawn dividing the circle into unequal arcs with lengths and . It is found that , the length of the tangent, is the mean proportional between and . If and are integers, then may have the following number of values:
Problem 36
Let be the respective sums of terms of the same arithmetic progression with as the first term and as the common difference. Let . Then is dependent on:
Problem 37
The base of a triangle is of length , and the altitude is of length . A rectangle of height is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
Problem 38
In this diagram and are the equal sides of an isosceles , in which is inscribed equilateral . Designate by , by , and by . Then:
Problem 39
To satisfy the equation , and must be:
Problem 40
Given right with legs . Find the length of the shorter angle trisector from to the hypotenuse:
See also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1959 AHSC |
Followed by 1961 AHSC | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.