Difference between revisions of "1958 AHSME Problems"
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+ | {{AHSC 50 Problems | ||
+ | |year=1958 | ||
+ | }} | ||
== Problem 1 == | == Problem 1 == | ||
+ | The value of <math> [2 - 3(2 - 3)^{-1}]^{-1}</math> is: | ||
+ | <math> \textbf{(A)}\ 5\qquad | ||
+ | \textbf{(B)}\ -5\qquad | ||
+ | \textbf{(C)}\ \frac{1}{5}\qquad | ||
+ | \textbf{(D)}\ -\frac{1}{5}\qquad | ||
+ | \textbf{(E)}\ \frac{5}{3}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 1|Solution]] |
== Problem 2 == | == Problem 2 == | ||
+ | If <math> \frac {1}{x} - \frac {1}{y} = \frac {1}{z}</math>, then <math> z</math> equals: | ||
+ | <math> \textbf{(A)}\ y - x\qquad \textbf{(B)}\ x - y\qquad \textbf{(C)}\ \frac {y - x}{xy}\qquad \textbf{(D)}\ \frac {xy}{y - x}\qquad \textbf{(E)}\ \frac {xy}{x - y}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 2|Solution]] |
== Problem 3 == | == Problem 3 == | ||
+ | Of the following expressions the one equal to <math> \frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}}</math> is: | ||
+ | <math> \textbf{(A)}\ \frac{a^2b^2}{b^2 - a^2}\qquad | ||
+ | \textbf{(B)}\ \frac{a^2b^2}{b^3 - a^3}\qquad | ||
+ | \textbf{(C)}\ \frac{ab}{b^3 - a^3}\qquad | ||
+ | \textbf{(D)}\ \frac{a^3 - b^3}{ab}\qquad | ||
+ | \textbf{(E)}\ \frac{a^2b^2}{a - b}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 3|Solution]] |
== Problem 4 == | == Problem 4 == | ||
+ | In the expression <math> \frac{x + 1}{x - 1}</math> each <math> x</math> is replaced by <math> \frac{x + 1}{x - 1}</math>. The resulting expression, evaluated for <math> x = \frac{1}{2}</math>, equals: | ||
+ | <math> \textbf{(A)}\ 3\qquad | ||
+ | \textbf{(B)}\ -3\qquad | ||
+ | \textbf{(C)}\ 1\qquad | ||
+ | \textbf{(D)}\ -1\qquad | ||
+ | \textbf{(E)}\ \text{none of these}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 4|Solution]] |
== Problem 5 == | == Problem 5 == | ||
+ | The expression <math> 2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}</math> equals: | ||
+ | <math> \textbf{(A)}\ 2\qquad | ||
+ | \textbf{(B)}\ 2 - \sqrt{2}\qquad | ||
+ | \textbf{(C)}\ 2 + \sqrt{2}\qquad | ||
+ | \textbf{(D)}\ 2\sqrt{2}\qquad | ||
+ | \textbf{(E)}\ \frac{\sqrt{2}}{2}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 5|Solution]] |
== Problem 6 == | == Problem 6 == | ||
+ | The arithmetic mean between <math> \frac {x + a}{x}</math> and <math> \frac {x - a}{x}</math>, when <math> x \neq 0</math>, is: | ||
+ | <math> \textbf{(A)}\ {2}\text{, if }{a \neq 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 6|Solution]] |
== Problem 7 == | == Problem 7 == | ||
+ | A straight line joins the points <math> (-1,1)</math> and <math> (3,9)</math>. Its <math> x</math>-intercept is: | ||
+ | <math> \textbf{(A)}\ -\frac{3}{2}\qquad | ||
+ | \textbf{(B)}\ -\frac{2}{3}\qquad | ||
+ | \textbf{(C)}\ \frac{2}{5}\qquad | ||
+ | \textbf{(D)}\ 2\qquad | ||
+ | \textbf{(E)}\ 3</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 7|Solution]] |
== Problem 8 == | == Problem 8 == | ||
+ | Which of these four numbers <math> \sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}</math>, is (are) rational: | ||
+ | <math> \textbf{(A)}\ \text{none}\qquad | ||
+ | \textbf{(B)}\ \text{all}\qquad | ||
+ | \textbf{(C)}\ \text{the first and fourth}\qquad | ||
+ | \textbf{(D)}\ \text{only the fourth}\qquad | ||
+ | \textbf{(E)}\ \text{only the first}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 8|Solution]] |
== Problem 9 == | == Problem 9 == | ||
+ | A value of <math> x</math> satisfying the equation <math> x^2 + b^2 = (a - x)^2</math> is: | ||
+ | <math> \textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad | ||
+ | \textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad | ||
+ | \textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad | ||
+ | \textbf{(D)}\ \frac{a - b}{2}\qquad | ||
+ | \textbf{(E)}\ \frac{a^2 - b^2}{2}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 9|Solution]] |
== Problem 10 == | == Problem 10 == | ||
+ | For what real values of <math> k</math>, other than <math> k = 0</math>, does the equation <math> x^2 + kx + k^2 = 0</math> have real roots? | ||
+ | <math> \textbf{(A)}\ {k < 0}\qquad | ||
+ | \textbf{(B)}\ {k > 0} \qquad | ||
+ | \textbf{(C)}\ {k \ge 1} \qquad | ||
+ | \textbf{(D)}\ \text{all values of }{k}\qquad | ||
+ | \textbf{(E)}\ \text{no values of }{k}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 10|Solution]] |
== Problem 11 == | == Problem 11 == | ||
+ | The number of roots satisfying the equation <math> \sqrt{5 - x} = x\sqrt{5 - x}</math> is: | ||
+ | <math> \textbf{(A)}\ \text{unlimited}\qquad | ||
+ | \textbf{(B)}\ 3\qquad | ||
+ | \textbf{(C)}\ 2\qquad | ||
+ | \textbf{(D)}\ 1\qquad | ||
+ | \textbf{(E)}\ 0</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 11|Solution]] |
== Problem 12 == | == Problem 12 == | ||
+ | If <math> P = \frac{s}{(1 + k)^n}</math> then <math> n</math> equals: | ||
+ | <math> \textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad | ||
+ | \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad | ||
+ | \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \ | ||
+ | \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad | ||
+ | \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 12|Solution]] |
== Problem 13 == | == Problem 13 == | ||
+ | The sum of two numbers is <math> 10</math>; their product is <math> 20</math>. The sum of their reciprocals is: | ||
+ | <math> \textbf{(A)}\ \frac{1}{10}\qquad | ||
+ | \textbf{(B)}\ \frac{1}{2}\qquad | ||
+ | \textbf{(C)}\ 1\qquad | ||
+ | \textbf{(D)}\ 2\qquad | ||
+ | \textbf{(E)}\ 4</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 13|Solution]] |
== Problem 14 == | == Problem 14 == | ||
+ | At a dance party a group of boys and girls exchange dances as follows: one boy dances with <math> 5</math> girls, a second boy dances with <math> 6</math> girls, and so on, the last boy dancing with all the girls. If <math> b</math> represents the number of boys and <math> g</math> the number of girls, then: | ||
+ | <math> \textbf{(A)}\ b = g\qquad | ||
+ | \textbf{(B)}\ b = \frac{g}{5}\qquad | ||
+ | \textbf{(C)}\ b = g - 4\qquad | ||
+ | \textbf{(D)}\ b = g - 5\qquad \ | ||
+ | \textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 14|Solution]] |
== Problem 15 == | == Problem 15 == | ||
+ | A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is: | ||
+ | <math> \textbf{(A)}\ 1080\qquad | ||
+ | \textbf{(B)}\ 900\qquad | ||
+ | \textbf{(C)}\ 720\qquad | ||
+ | \textbf{(D)}\ 540\qquad | ||
+ | \textbf{(E)}\ 360</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 15|Solution]] |
== Problem 16 == | == Problem 16 == | ||
+ | The area of a circle inscribed in a regular hexagon is <math> 100\pi</math>. The area of hexagon is: | ||
+ | <math> \textbf{(A)}\ 600\qquad | ||
+ | \textbf{(B)}\ 300\qquad | ||
+ | \textbf{(C)}\ 200\sqrt{2}\qquad | ||
+ | \textbf{(D)}\ 200\sqrt{3}\qquad | ||
+ | \textbf{(E)}\ 120\sqrt{5}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 16|Solution]] |
== Problem 17 == | == Problem 17 == | ||
+ | If <math> x</math> is positive and <math> \log{x} \ge \log{2} + \frac{1}{2}\log{x}</math>, then: | ||
+ | <math> \textbf{(A)}\ {x}\text{ has no minimum or maximum value}\qquad \ | ||
+ | \textbf{(B)}\ \text{the maximum value of }{x}\text{ is }{1}\qquad \ | ||
+ | \textbf{(C)}\ \text{the minimum value of }{x}\text{ is }{1}\qquad \ | ||
+ | \textbf{(D)}\ \text{the maximum value of }{x}\text{ is }{4}\qquad \ | ||
+ | \textbf{(E)}\ \text{the minimum value of }{x}\text{ is }{4}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 17|Solution]] |
== Problem 18 == | == Problem 18 == | ||
+ | The area of a circle is doubled when its radius <math> r</math> is increased by <math> n</math>. Then <math> r</math> equals: | ||
+ | <math> \textbf{(A)}\ n(\sqrt{2} + 1)\qquad | ||
+ | \textbf{(B)}\ n(\sqrt{2} - 1)\qquad | ||
+ | \textbf{(C)}\ n\qquad | ||
+ | \textbf{(D)}\ n(2 - \sqrt{2})\qquad | ||
+ | \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 18|Solution]] |
== Problem 19 == | == Problem 19 == | ||
+ | The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b = 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is: | ||
+ | <math> \textbf{(A)}\ 1 : 3\qquad | ||
+ | \textbf{(B)}\ 1 : 9\qquad | ||
+ | \textbf{(C)}\ 1 : 10\qquad | ||
+ | \textbf{(D)}\ 3 : 10\qquad | ||
+ | \textbf{(E)}\ 1 : \sqrt{10}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 19|Solution]] |
== Problem 20 == | == Problem 20 == | ||
+ | If <math> 4^x - 4^{x - 1} = 24</math>, then <math> (2x)^x</math> equals: | ||
+ | <math> \textbf{(A)}\ 5\sqrt{5}\qquad | ||
+ | \textbf{(B)}\ \sqrt{5}\qquad | ||
+ | \textbf{(C)}\ 25\sqrt{5}\qquad | ||
+ | \textbf{(D)}\ 125\qquad | ||
+ | \textbf{(E)}\ 25</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 20|Solution]] |
== Problem 21 == | == Problem 21 == | ||
+ | In the accompanying figure <math> \overline{CE}</math> and <math> \overline{DE}</math> are equal chords of a circle with center <math> O</math>. Arc <math> AB</math> is a quarter-circle. Then the ratio of the area of triangle <math> CED</math> to the area of triangle <math> AOB</math> is: | ||
+ | <asy> | ||
+ | draw(circle((0,0),10),black+linewidth(.75)); | ||
+ | draw((-10,0)--(0,0)--(10,0)--(0,10)--cycle,dot); | ||
+ | MP("O",(0,0),N);MP("C",(-10,0),W);MP("D",(10,0),E);;MP("E",(0,10),N); | ||
+ | draw((-sqrt(70),-sqrt(30))--(sqrt(30),-sqrt(70))--(0,0)--cycle,dot); | ||
+ | MP("A",(-sqrt(70),-sqrt(30)),SW);MP("B",(sqrt(30),-sqrt(70)),SE); | ||
+ | </asy> | ||
− | [[ | + | <math> \textbf{(A)}\ \sqrt {2} : 1\qquad \textbf{(B)}\ \sqrt {3} : 1\qquad \textbf{(C)}\ 4 : 1\qquad \textbf{(D)}\ 3 : 1\qquad \textbf{(E)}\ 2 : 1</math> |
+ | |||
+ | [[1958 AHSME Problems/Problem 21|Solution]] | ||
== Problem 22 == | == Problem 22 == | ||
+ | A particle is placed on the parabola <math> y = x^2 - x - 6</math> at a point <math> P</math> whose <math> y</math>-coordinate is <math> 6</math>. It is allowed to roll along the parabola until it reaches the nearest point <math> Q</math> whose <math> y</math>-coordinate is <math> -6</math>. The horizontal distance traveled by the particle (the numerical value of the difference in the <math> x</math>-coordinates of <math> P</math> and <math> Q</math>) is: | ||
− | [[ | + | <math> \textbf{(A)}\ 5\qquad |
+ | \textbf{(B)}\ 4\qquad | ||
+ | \textbf{(C)}\ 3\qquad | ||
+ | \textbf{(D)}\ 2\qquad | ||
+ | \textbf{(E)}\ 1</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 22|Solution]] | ||
== Problem 23 == | == Problem 23 == | ||
+ | If, in the expression <math> x^2 - 3</math>, <math> x</math> increases or decreases by a positive amount of <math> a</math>, the expression changes by an amount: | ||
+ | <math> \textbf{(A)}\ {\pm 2ax + a^2}\qquad | ||
+ | \textbf{(B)}\ {2ax \pm a^2}\qquad | ||
+ | \textbf{(C)}\ {\pm a^2 - 3} \qquad | ||
+ | \textbf{(D)}\ {(x + a)^2 - 3}\qquad\ | ||
+ | \textbf{(E)}\ {(x - a)^2 - 3}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 23|Solution]] |
== Problem 24 == | == Problem 24 == | ||
+ | A man travels <math> m</math> feet due north at <math> 2</math> minutes per mile. He returns due south to his starting point at <math> 2</math> miles per minute. The average rate in miles per hour for the entire trip is: | ||
+ | <math> \textbf{(A)}\ 75\qquad | ||
+ | \textbf{(B)}\ 48\qquad | ||
+ | \textbf{(C)}\ 45\qquad | ||
+ | \textbf{(D)}\ 24\qquad\ | ||
+ | \textbf{(E)}\ \text{impossible to determine without knowing the value of }{m}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 24|Solution]] |
== Problem 25 == | == Problem 25 == | ||
+ | If <math> \log_{k}{x}\cdot \log_{5}{k} = 3</math>, then <math> x</math> equals: | ||
+ | <math> \textbf{(A)}\ k^6\qquad | ||
+ | \textbf{(B)}\ 5k^3\qquad | ||
+ | \textbf{(C)}\ k^3\qquad | ||
+ | \textbf{(D)}\ 243\qquad | ||
+ | \textbf{(E)}\ 125</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 25|Solution]] |
== Problem 26 == | == Problem 26 == | ||
+ | A set of <math> n</math> numbers has the sum <math> s</math>. Each number of the set is increased by <math> 20</math>, then multiplied by <math> 5</math>, and then decreased by <math> 20</math>. The sum of the numbers in the new set thus obtained is: | ||
+ | <math> \textbf{(A)}\ s + 20n\qquad | ||
+ | \textbf{(B)}\ 5s + 80n\qquad | ||
+ | \textbf{(C)}\ s\qquad | ||
+ | \textbf{(D)}\ 5s\qquad | ||
+ | \textbf{(E)}\ 5s + 4n</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 26|Solution]] |
== Problem 27 == | == Problem 27 == | ||
+ | The points <math> (2,-3)</math>, <math> (4,3)</math>, and <math> (5, k/2)</math> are on the same straight line. The value(s) of <math> k</math> is (are): | ||
+ | <math> \textbf{(A)}\ 12\qquad | ||
+ | \textbf{(B)}\ -12\qquad | ||
+ | \textbf{(C)}\ \pm 12\qquad | ||
+ | \textbf{(D)}\ {12}\text{ or }{6}\qquad | ||
+ | \textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 27|Solution]] |
== Problem 28 == | == Problem 28 == | ||
+ | A <math> 16</math>-quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is: | ||
+ | <math> \textbf{(A)}\ \frac{1}{4}\qquad | ||
+ | \textbf{(B)}\ \frac{81}{256}\qquad | ||
+ | \textbf{(C)}\ \frac{27}{64}\qquad | ||
+ | \textbf{(D)}\ \frac{37}{64}\qquad | ||
+ | \textbf{(E)}\ \frac{175}{256}</math> | ||
− | [[ | + | [[1958 AHSME Problems/Problem 28|Solution]] |
== Problem 29 == | == Problem 29 == | ||
+ | In a general triangle <math> ADE</math> (as shown) lines <math> \overline{EB}</math> and <math> \overline{EC}</math> are drawn. Which of the following angle relations is true? | ||
+ | <asy> | ||
+ | draw((-8,0)--(-2,0)--(4,0)--(10,0)--(0,10)--cycle,dot); | ||
+ | draw((-2,0)--(0,10),dot);draw((4,0)--(0,10),dot); | ||
+ | MP("A",(-8,0),S);MP("B",(-2,0),S);MP("C",(4,0),S);MP("D",(10,0),S);MP("E",(0,10),N); | ||
+ | MP("x",(-7.9,.4),E);MP("z",(-2,.4),W);MP("m",(-2,.4),E);MP("n",(4,.4),W);MP("c",(4,.4),E);MP("a",(9.9,.4),W); | ||
+ | MP("y",(-.2,8.8),SW);MP("w",(.1,8.8),S);MP("b",(.7,9),SE); | ||
+ | </asy> | ||
− | [[ | + | <math> \textbf{(A)}\ x + z = a + b\qquad \textbf{(B)}\ y + z = a + b\qquad \textbf{(C)}\ m + x = w + n\qquad \ |
+ | \textbf{(D)}\ x + z + n = w + c + m\qquad \textbf{(E)}\ x + y + n = a + b + m</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 29|Solution]] | ||
== Problem 30 == | == Problem 30 == | ||
+ | If <math> xy = b</math> and <math> \frac{1}{x^2} + \frac{1}{y^2} = a</math>, then <math> (x + y)^2</math> equals: | ||
+ | |||
+ | <math> \textbf{(A)}\ (a + 2b)^2\qquad | ||
+ | \textbf{(B)}\ a^2 + b^2\qquad | ||
+ | \textbf{(C)}\ b(ab + 2)\qquad | ||
+ | \textbf{(D)}\ ab(b + 2)\qquad | ||
+ | \textbf{(E)}\ \frac{1}{a} + 2b</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 30|Solution]] | ||
+ | |||
+ | == Problem 31 == | ||
+ | The altitude drawn to the base of an isosceles triangle is <math> 8</math>, and the perimeter <math> 32</math>. The area of the triangle is: | ||
+ | |||
+ | <math> \textbf{(A)}\ 56\qquad | ||
+ | \textbf{(B)}\ 48\qquad | ||
+ | \textbf{(C)}\ 40\qquad | ||
+ | \textbf{(D)}\ 32\qquad | ||
+ | \textbf{(E)}\ 24</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 31|Solution]] | ||
+ | |||
+ | == Problem 32 == | ||
+ | With $ <math> 1000</math> a rancher is to buy steers at $ <math>25</math> each and cows at $ <math>26</math> each. If the number of steers <math> s</math> and the number of cows <math> c</math> are both positive integers, then: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{this problem has no solution}\qquad\ | ||
+ | \textbf{(B)}\ \text{there are two solutions with }{s}\text{ exceeding }{c}\qquad \ | ||
+ | \textbf{(C)}\ \text{there are two solutions with }{c}\text{ exceeding }{s}\qquad \ | ||
+ | \textbf{(D)}\ \text{there is one solution with }{s}\text{ exceeding }{c}\qquad \ | ||
+ | \textbf{(E)}\ \text{there is one solution with }{c}\text{ exceeding }{s}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 32|Solution]] | ||
+ | |||
+ | == Problem 33 == | ||
+ | For one root of <math> ax^2 + bx + c = 0</math> to be double the other, the coefficients <math> a,\,b,\,c</math> must be related as follows: | ||
+ | |||
+ | <math> \textbf{(A)}\ 4b^2 = 9c\qquad | ||
+ | \textbf{(B)}\ 2b^2 = 9ac\qquad | ||
+ | \textbf{(C)}\ 2b^2 = 9a\qquad \ | ||
+ | \textbf{(D)}\ b^2 - 8ac = 0\qquad | ||
+ | \textbf{(E)}\ 9b^2 = 2ac</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 33|Solution]] | ||
+ | |||
+ | == Problem 34 == | ||
+ | The numerator of a fraction is <math> 6x + 1</math>, then denominator is <math> 7 - 4x</math>, and <math> x</math> can have any value between <math> -2</math> and <math> 2</math>, both included. The values of <math> x</math> for which the numerator is greater than the denominator are: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{3}{5} < x \le 2\qquad | ||
+ | \textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad | ||
+ | \textbf{(C)}\ 0 < x \le 2\qquad \ | ||
+ | \textbf{(D)}\ 0 \le x \le 2\qquad | ||
+ | \textbf{(E)}\ -2 \le x \le 2</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 34|Solution]] | ||
+ | |||
+ | == Problem 35 == | ||
+ | A triangle is formed by joining three points whose coordinates are integers. If the <math> x</math>-coordinate and the <math> y</math>-coordinate each have a value of <math> 1</math>, then the area of the triangle, in square units: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{must be an integer}\qquad | ||
+ | \textbf{(B)}\ \text{may be irrational}\qquad | ||
+ | \textbf{(C)}\ \text{must be irrational}\qquad | ||
+ | \textbf{(D)}\ \text{must be rational}\qquad \ | ||
+ | \textbf{(E)}\ \text{will be an integer only if the triangle is equilateral.}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 35|Solution]] | ||
+ | |||
+ | == Problem 36 == | ||
+ | The sides of a triangle are <math> 30</math>, <math> 70</math>, and <math> 80</math> units. If an altitude is dropped upon the side of length <math> 80</math>, the larger segment cut off on this side is: | ||
+ | |||
+ | <math> \textbf{(A)}\ 62\qquad | ||
+ | \textbf{(B)}\ 63\qquad | ||
+ | \textbf{(C)}\ 64\qquad | ||
+ | \textbf{(D)}\ 65\qquad | ||
+ | \textbf{(E)}\ 66</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 36|Solution]] | ||
+ | |||
+ | == Problem 37 == | ||
+ | The first term of an arithmetic series of consecutive integers is <math> k^2 + 1</math>. The sum of <math> 2k + 1</math> terms of this series may be expressed as: | ||
+ | |||
+ | <math> \textbf{(A)}\ k^3 + (k + 1)^3\qquad | ||
+ | \textbf{(B)}\ (k - 1)^3 + k^3\qquad | ||
+ | \textbf{(C)}\ (k + 1)^3\qquad \ | ||
+ | \textbf{(D)}\ (k + 1)^2\qquad | ||
+ | \textbf{(E)}\ (2k + 1)(k + 1)^2</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 37|Solution]] | ||
+ | |||
+ | == Problem 38 == | ||
+ | Let <math> r</math> be the distance from the origin to a point <math> P</math> with coordinates <math> x</math> and <math> y</math>. Designate the ratio <math> \frac{y}{r}</math> by <math> s</math> and the ratio <math> \frac{x}{r}</math> by <math> c</math>. Then the values of <math> s^2 - c^2</math> are limited to the numbers: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both excluded}\qquad\ | ||
+ | \textbf{(B)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both included}\qquad \ | ||
+ | \textbf{(C)}\ \text{between }{-1}\text{ and }{+1}\text{, both excluded}\qquad \ | ||
+ | \textbf{(D)}\ \text{between }{-1}\text{ and }{+1}\text{, both included}\qquad \ | ||
+ | \textbf{(E)}\ {-1}\text{ and }{+1}\text{ only}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 38|Solution]] | ||
+ | |||
+ | == Problem 39 == | ||
+ | We may say concerning the solution of <math>|x|^2 + |x| - 6 =0</math> that: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{there is only one root}\qquad | ||
+ | \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad | ||
+ | \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \ | ||
+ | \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad | ||
+ | \textbf{(E)}\ \text{the product of the roots is }{-6}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 39|Solution]] | ||
+ | |||
+ | == Problem 40 == | ||
+ | Given <math> a_0 = 1</math>, <math> a_1 = 3</math>, and the general relation <math> a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n</math> for <math> n \ge 1</math>. Then <math> a_3</math> equals: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{13}{27}\qquad | ||
+ | \textbf{(B)}\ 33\qquad | ||
+ | \textbf{(C)}\ 21\qquad | ||
+ | \textbf{(D)}\ 10\qquad | ||
+ | \textbf{(E)}\ -17</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 40|Solution]] | ||
+ | |||
+ | == Problem 41 == | ||
+ | The roots of <math> Ax^2 + Bx + C = 0</math> are <math> r</math> and <math> s</math>. For the roots of | ||
+ | <math> x^2+px +q =0</math> | ||
+ | |||
+ | to be <math> r^2</math> and <math> s^2</math>, <math> p</math> must equal: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad | ||
+ | \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad | ||
+ | \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \ | ||
+ | \textbf{(D)}\ B^2 - 2C\qquad | ||
+ | \textbf{(E)}\ 2C - B^2</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 41|Solution]] | ||
+ | |||
+ | == Problem 42 == | ||
+ | In a circle with center <math> O</math>, chord <math> \overline{AB}</math> equals chord <math> \overline{AC}</math>. Chord <math> \overline{AD}</math> cuts <math> \overline{BC}</math> in <math> E</math>. If <math> AC = 12</math> and <math> AE = 8</math>, then <math> AD</math> equals: | ||
+ | |||
+ | <math> \textbf{(A)}\ 27\qquad | ||
+ | \textbf{(B)}\ 24\qquad | ||
+ | \textbf{(C)}\ 21\qquad | ||
+ | \textbf{(D)}\ 20\qquad | ||
+ | \textbf{(E)}\ 18</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 42|Solution]] | ||
+ | |||
+ | == Problem 43 == | ||
+ | <math> \overline{AB}</math> is the hypotenuse of a right triangle <math> ABC</math>. Median <math> \overline{AD}</math> has length <math> 7</math> and median <math> \overline{BE}</math> has length <math> 4</math>. The length of <math> \overline{AB}</math> is: | ||
+ | |||
+ | <math> \textbf{(A)}\ 10\qquad | ||
+ | \textbf{(B)}\ 5\sqrt{3}\qquad | ||
+ | \textbf{(C)}\ 5\sqrt{2}\qquad | ||
+ | \textbf{(D)}\ 2\sqrt{13}\qquad | ||
+ | \textbf{(E)}\ 2\sqrt{15}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 43|Solution]] | ||
+ | == Problem 44 == | ||
+ | Given the true statements: (1) If <math> a</math> is greater than <math> b</math>, then <math> c</math> is greater than <math> d</math> (2) If <math> c</math> is less than <math> d</math>, then <math> e</math> is greater than <math> f</math>. A valid conclusion is: | ||
− | [[ | + | <math> \textbf{(A)}\ \text{If }{a}\text{ is less than }{b}\text{, then }{e}\text{ is greater than }{f}\qquad \ |
+ | \textbf{(B)}\ \text{If }{e}\text{ is greater than }{f}\text{, then }{a}\text{ is less than }{b}\qquad \ | ||
+ | \textbf{(C)}\ \text{If }{e}\text{ is less than }{f}\text{, then }{a}\text{ is greater than }{b}\qquad \ | ||
+ | \textbf{(D)}\ \text{If }{a}\text{ is greater than }{b}\text{, then }{e}\text{ is less than }{f}\qquad \ | ||
+ | \textbf{(E)}\ \text{none of these}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 44|Solution]] | ||
+ | |||
+ | == Problem 45 == | ||
+ | A check is written for <math> x</math> dollars and <math> y</math> cents, <math> x</math> and <math> y</math> both two-digit numbers. In error it is cashed for <math> y</math> dollars and <math> x</math> cents, the incorrect amount exceeding the correct amount by <math> $17.82</math>. Then: | ||
+ | |||
+ | <math> \textbf{(A)}\ {x}\text{ cannot exceed }{70}\qquad \ | ||
+ | \textbf{(B)}\ {y}\text{ can equal }{2x}\qquad\ | ||
+ | \textbf{(C)}\ \text{the amount of the check cannot be a multiple of }{5}\qquad \ | ||
+ | \textbf{(D)}\ \text{the incorrect amount can equal twice the correct amount}\qquad \ | ||
+ | \textbf{(E)}\ \text{the sum of the digits of the correct amount is divisible by }{9}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 45|Solution]] | ||
+ | |||
+ | == Problem 46 == | ||
+ | For values of <math> x</math> less than <math> 1</math> but greater than <math> -4</math>, the expression | ||
+ | <math>\frac{x^2 - 2x + 2}{2x - 2}</math> | ||
+ | has: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{no maximum or minimum value}\qquad \ | ||
+ | \textbf{(B)}\ \text{a minimum value of }{+1}\qquad \ | ||
+ | \textbf{(C)}\ \text{a maximum value of }{+1}\qquad \ | ||
+ | \textbf{(D)}\ \text{a minimum value of }{-1}\qquad \ | ||
+ | \textbf{(E)}\ \text{a maximum value of }{-1}</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 46|Solution]] | ||
+ | |||
+ | == Problem 47 == | ||
+ | <math> ABCD</math> is a rectangle (see the accompanying diagram) with <math> P</math> any point on <math> \overline{AB}</math>. <math> \overline{PS} \perp \overline{BD}</math> and <math> \overline{PR} \perp \overline{AC}</math>. <math> \overline{AF} \perp \overline{BD}</math> and <math> \overline{PQ} \perp \overline{AF}</math>. Then <math> PR + PS</math> is equal to: | ||
+ | |||
+ | <asy> | ||
+ | draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--cycle,dot); | ||
+ | draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); | ||
+ | draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); | ||
+ | draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); | ||
+ | draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); | ||
+ | MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); | ||
+ | MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); | ||
+ | MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); | ||
+ | </asy> | ||
+ | |||
+ | <math> \textbf{(A)}\ PQ\qquad | ||
+ | \textbf{(B)}\ AE\qquad | ||
+ | \textbf{(C)}\ PT + AT\qquad | ||
+ | \textbf{(D)}\ AF\qquad | ||
+ | \textbf{(E)}\ EF</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 47|Solution]] | ||
+ | |||
+ | == Problem 48 == | ||
+ | Diameter <math> \overline{AB}</math> of a circle with center <math> O</math> is <math> 10</math> units. <math> C</math> is a point <math> 4</math> units from <math> A</math>, and on <math> \overline{AB}</math>. <math> D</math> is a point <math> 4</math> units from <math> B</math>, and on <math> \overline{AB}</math>. <math> P</math> is any point on the circle. Then the broken-line path from <math> C</math> to <math> P</math> to <math> D</math>: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\ | ||
+ | \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \ | ||
+ | \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \ | ||
+ | \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \ | ||
+ | \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 48|Solution]] | ||
+ | |||
+ | == Problem 49 == | ||
+ | In the expansion of <math> (a + b)^n</math> there are <math> n + 1</math> dissimilar terms. The number of dissimilar terms in the expansion of <math> (a + b + c)^{10}</math> is: | ||
+ | |||
+ | <math> \textbf{(A)}\ 11\qquad | ||
+ | \textbf{(B)}\ 33\qquad | ||
+ | \textbf{(C)}\ 55\qquad | ||
+ | \textbf{(D)}\ 66\qquad | ||
+ | \textbf{(E)}\ 132</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 49|Solution]] | ||
+ | |||
+ | == Problem 50 == | ||
+ | In this diagram a scheme is indicated for associating all the points of segment <math> \overline{AB}</math> with those of segment <math> \overline{A'B'}</math>, and reciprocally. To described this association scheme analytically, let <math> x</math> be the distance from a point <math> P</math> on <math> \overline{AB}</math> to <math> D</math> and let <math> y</math> be the distance from the associated point <math> P'</math> of <math> \overline{A'B'}</math> to <math> D'</math>. Then for any pair of associated points, if <math> x = a,\, x + y</math> equals: | ||
+ | |||
+ | <asy> | ||
+ | draw((0,-3)--(0,3),black+linewidth(.75)); | ||
+ | draw((1,-2.5)--(5,-2.5),black+linewidth(.75)); | ||
+ | draw((3,2.5)--(4,2.5),black+linewidth(.75)); | ||
+ | draw((1,-2.5)--(4,2.5),black+linewidth(.75)); | ||
+ | draw((5,-2.5)--(3,2.5),black+linewidth(.75)); | ||
+ | draw((2.6,-2.5)--(3.6,2.5),black+linewidth(.75)); | ||
+ | dot((0,2.5));dot((1,2.5));dot((2,2.5));dot((3,2.5));dot((4,2.5));dot((5,2.5)); | ||
+ | dot((0,-2.5));dot((1,-2.5));dot((2,-2.5));dot((3,-2.5));dot((4,-2.5));dot((5,-2.5)); | ||
+ | MP("D",(0,2.5),NW);MP("A",(3,2.5),N);MP("P",(3.5,2.5),N);MP("B",(4,2.5),N); | ||
+ | MP("D'",(0,-2.5),NW);MP("B'",(1,-2.5),NW);MP("P'",(2.25,-2.5),N);MP("A'",(5,-2.5),NE); | ||
+ | MP("0",(0,2.5),SE);MP("1",(1,2.5),SE);MP("2",(2,2.5),SE);MP("3",(3,2.5),SE);MP("4",(4,2.5),SE);MP("5",(5,2.5),SE); | ||
+ | MP("0",(0,-2.5),SE);MP("1",(1,-2.5),SE);MP("2",(2,-2.5),SE);MP("3",(3,-2.5),SE);MP("4",(4,-2.5),SE);MP("5",(5,-2.5),SE); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | <math> \textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a - 51\qquad \textbf{(C)}\ 17 - 3a\qquad \textbf{(D)}\ \frac {17 - 3a}{4}\qquad \textbf{(E)}\ 12a - 34</math> | ||
+ | |||
+ | [[1958 AHSME Problems/Problem 50|Solution]] | ||
== See also == | == See also == | ||
− | + | ||
− | * [[ | + | * [[AMC 12 Problems and Solutions]] |
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 50p box|year=1958|before=[[1957 AHSME|1957 AHSC]]|after=[[1959 AHSME|1959 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 13:21, 20 February 2020
1958 AHSC (Answer Key) Printable version: | AoPS Resources • PDF | ||
Instructions
| ||
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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 Problem 41
- 42 Problem 42
- 43 Problem 43
- 44 Problem 44
- 45 Problem 45
- 46 Problem 46
- 47 Problem 47
- 48 Problem 48
- 49 Problem 49
- 50 Problem 50
- 51 See also
Problem 1
The value of is:
Problem 2
If , then equals:
Problem 3
Of the following expressions the one equal to is:
Problem 4
In the expression each is replaced by . The resulting expression, evaluated for , equals:
Problem 5
The expression equals:
Problem 6
The arithmetic mean between and , when , is:
Problem 7
A straight line joins the points and . Its -intercept is:
Problem 8
Which of these four numbers , is (are) rational:
Problem 9
A value of satisfying the equation is:
Problem 10
For what real values of , other than , does the equation have real roots?
Problem 11
The number of roots satisfying the equation is:
Problem 12
If then equals:
Problem 13
The sum of two numbers is ; their product is . The sum of their reciprocals is:
Problem 14
At a dance party a group of boys and girls exchange dances as follows: one boy dances with girls, a second boy dances with girls, and so on, the last boy dancing with all the girls. If represents the number of boys and the number of girls, then:
Problem 15
A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is:
Problem 16
The area of a circle inscribed in a regular hexagon is . The area of hexagon is:
Problem 17
If is positive and , then:
Problem 18
The area of a circle is doubled when its radius is increased by . Then equals:
Problem 19
The sides of a right triangle are and and the hypotenuse is . A perpendicular from the vertex divides into segments and , adjacent respectively to and . If , then the ratio of to is:
Problem 20
If , then equals:
Problem 21
In the accompanying figure and are equal chords of a circle with center . Arc is a quarter-circle. Then the ratio of the area of triangle to the area of triangle is:
Problem 22
A particle is placed on the parabola at a point whose -coordinate is . It is allowed to roll along the parabola until it reaches the nearest point whose -coordinate is . The horizontal distance traveled by the particle (the numerical value of the difference in the -coordinates of and ) is:
Problem 23
If, in the expression , increases or decreases by a positive amount of , the expression changes by an amount:
Problem 24
A man travels feet due north at minutes per mile. He returns due south to his starting point at miles per minute. The average rate in miles per hour for the entire trip is:
Problem 25
If , then equals:
Problem 26
A set of numbers has the sum . Each number of the set is increased by , then multiplied by , and then decreased by . The sum of the numbers in the new set thus obtained is:
Problem 27
The points , , and are on the same straight line. The value(s) of is (are):
Problem 28
A -quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is:
Problem 29
In a general triangle (as shown) lines and are drawn. Which of the following angle relations is true?
Problem 30
If and , then equals:
Problem 31
The altitude drawn to the base of an isosceles triangle is , and the perimeter . The area of the triangle is:
Problem 32
With $ a rancher is to buy steers at $ each and cows at $ each. If the number of steers and the number of cows are both positive integers, then:
Problem 33
For one root of to be double the other, the coefficients must be related as follows:
Problem 34
The numerator of a fraction is , then denominator is , and can have any value between and , both included. The values of for which the numerator is greater than the denominator are:
Problem 35
A triangle is formed by joining three points whose coordinates are integers. If the -coordinate and the -coordinate each have a value of , then the area of the triangle, in square units:
Problem 36
The sides of a triangle are , , and units. If an altitude is dropped upon the side of length , the larger segment cut off on this side is:
Problem 37
The first term of an arithmetic series of consecutive integers is . The sum of terms of this series may be expressed as:
Problem 38
Let be the distance from the origin to a point with coordinates and . Designate the ratio by and the ratio by . Then the values of are limited to the numbers:
Problem 39
We may say concerning the solution of that:
Problem 40
Given , , and the general relation for . Then equals:
Problem 41
The roots of are and . For the roots of
to be and , must equal:
Problem 42
In a circle with center , chord equals chord . Chord cuts in . If and , then equals:
Problem 43
is the hypotenuse of a right triangle . Median has length and median has length . The length of is:
Problem 44
Given the true statements: (1) If is greater than , then is greater than (2) If is less than , then is greater than . A valid conclusion is:
Problem 45
A check is written for dollars and cents, and both two-digit numbers. In error it is cashed for dollars and cents, the incorrect amount exceeding the correct amount by . Then:
Problem 46
For values of less than but greater than , the expression has:
Problem 47
is a rectangle (see the accompanying diagram) with any point on . and . and . Then is equal to:
Problem 48
Diameter of a circle with center is units. is a point units from , and on . is a point units from , and on . is any point on the circle. Then the broken-line path from to to :
Problem 49
In the expansion of there are dissimilar terms. The number of dissimilar terms in the expansion of is:
Problem 50
In this diagram a scheme is indicated for associating all the points of segment with those of segment , and reciprocally. To described this association scheme analytically, let be the distance from a point on to and let be the distance from the associated point of to . Then for any pair of associated points, if equals:
See also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1957 AHSC |
Followed by 1959 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.