Difference between revisions of "1958 AHSME Problems"

Latest revision as of 13:21, 20 February 2020

1958 AHSC (Answer Key) Printable version: Wiki \| AoPS Resources • PDF
Instructions This is a 50-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test.
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Problem 1

The value of $[2 - 3(2 - 3)^{-1}]^{-1}$ is:

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ -5\qquad \textbf{(C)}\ \frac{1}{5}\qquad \textbf{(D)}\ -\frac{1}{5}\qquad \textbf{(E)}\ \frac{5}{3}$

Solution

Problem 2

If $\frac {1}{x} - \frac {1}{y} = \frac {1}{z}$ , then $z$ equals:

$\textbf{(A)}\ y - x\qquad \textbf{(B)}\ x - y\qquad \textbf{(C)}\ \frac {y - x}{xy}\qquad \textbf{(D)}\ \frac {xy}{y - x}\qquad \textbf{(E)}\ \frac {xy}{x - y}$

Solution

Problem 3

Of the following expressions the one equal to $\frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}}$ is:

$\textbf{(A)}\ \frac{a^2b^2}{b^2 - a^2}\qquad \textbf{(B)}\ \frac{a^2b^2}{b^3 - a^3}\qquad \textbf{(C)}\ \frac{ab}{b^3 - a^3}\qquad \textbf{(D)}\ \frac{a^3 - b^3}{ab}\qquad \textbf{(E)}\ \frac{a^2b^2}{a - b}$

Solution

Problem 4

In the expression $\frac{x + 1}{x - 1}$ each $x$ is replaced by $\frac{x + 1}{x - 1}$ . The resulting expression, evaluated for $x = \frac{1}{2}$ , equals:

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 5

The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$ equals:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2 - \sqrt{2}\qquad \textbf{(C)}\ 2 + \sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{2}\qquad \textbf{(E)}\ \frac{\sqrt{2}}{2}$

Solution

Problem 6

The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \neq 0$ , is:

$\textbf{(A)}\ {2}\text{, if }{a \neq 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$

Solution

Problem 7

A straight line joins the points $(-1,1)$ and $(3,9)$ . Its $x$ -intercept is:

$\textbf{(A)}\ -\frac{3}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ \frac{2}{5}\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 3$

Solution

Problem 8

Which of these four numbers $\sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}$ , is (are) rational:

$\textbf{(A)}\ \text{none}\qquad \textbf{(B)}\ \text{all}\qquad \textbf{(C)}\ \text{the first and fourth}\qquad \textbf{(D)}\ \text{only the fourth}\qquad \textbf{(E)}\ \text{only the first}$

Solution

Problem 9

A value of $x$ satisfying the equation $x^2 + b^2 = (a - x)^2$ is:

$\textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad \textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad \textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad \textbf{(D)}\ \frac{a - b}{2}\qquad \textbf{(E)}\ \frac{a^2 - b^2}{2}$

Solution

Problem 10

For what real values of $k$ , other than $k = 0$ , does the equation $x^2 + kx + k^2 = 0$ have real roots?

$\textbf{(A)}\ {k < 0}\qquad \textbf{(B)}\ {k > 0} \qquad \textbf{(C)}\ {k \ge 1} \qquad \textbf{(D)}\ \text{all values of }{k}\qquad \textbf{(E)}\ \text{no values of }{k}$

Solution

Problem 11

The number of roots satisfying the equation $\sqrt{5 - x} = x\sqrt{5 - x}$ is:

$\textbf{(A)}\ \text{unlimited}\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solution

Problem 12

If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:

$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$

Solution

Problem 13

The sum of two numbers is $10$ ; their product is $20$ . The sum of their reciprocals is:

$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$

Solution

Problem 14

At a dance party a group of boys and girls exchange dances as follows: one boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all the girls. If $b$ represents the number of boys and $g$ the number of girls, then:

$\textbf{(A)}\ b = g\qquad \textbf{(B)}\ b = \frac{g}{5}\qquad \textbf{(C)}\ b = g - 4\qquad \textbf{(D)}\ b = g - 5\qquad \\ \textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}$

Solution

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:

$\textbf{(A)}\ 1080\qquad \textbf{(B)}\ 900\qquad \textbf{(C)}\ 720\qquad \textbf{(D)}\ 540\qquad \textbf{(E)}\ 360$

Solution

Problem 16

The area of a circle inscribed in a regular hexagon is $100\pi$ . The area of hexagon is:

$\textbf{(A)}\ 600\qquad \textbf{(B)}\ 300\qquad \textbf{(C)}\ 200\sqrt{2}\qquad \textbf{(D)}\ 200\sqrt{3}\qquad \textbf{(E)}\ 120\sqrt{5}$

Solution

Problem 17

If $x$ is positive and $\log{x} \ge \log{2} + \frac{1}{2}\log{x}$ , then:

$\textbf{(A)}\ {x}\text{ has no minimum or maximum value}\qquad \\ \textbf{(B)}\ \text{the maximum value of }{x}\text{ is }{1}\qquad \\ \textbf{(C)}\ \text{the minimum value of }{x}\text{ is }{1}\qquad \\ \textbf{(D)}\ \text{the maximum value of }{x}\text{ is }{4}\qquad \\ \textbf{(E)}\ \text{the minimum value of }{x}\text{ is }{4}$

Solution

Problem 18

The area of a circle is doubled when its radius $r$ is increased by $n$ . Then $r$ equals:

$\textbf{(A)}\ n(\sqrt{2} + 1)\qquad \textbf{(B)}\ n(\sqrt{2} - 1)\qquad \textbf{(C)}\ n\qquad \textbf{(D)}\ n(2 - \sqrt{2})\qquad \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}$

Solution

Problem 19

The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c$ . A perpendicular from the vertex divides $c$ into segments $r$ and $s$ , adjacent respectively to $a$ and $b$ . If $a : b = 1 : 3$ , then the ratio of $r$ to $s$ is:

$\textbf{(A)}\ 1 : 3\qquad \textbf{(B)}\ 1 : 9\qquad \textbf{(C)}\ 1 : 10\qquad \textbf{(D)}\ 3 : 10\qquad \textbf{(E)}\ 1 : \sqrt{10}$

Solution

Problem 20

If $4^x - 4^{x - 1} = 24$ , then $(2x)^x$ equals:

$\textbf{(A)}\ 5\sqrt{5}\qquad \textbf{(B)}\ \sqrt{5}\qquad \textbf{(C)}\ 25\sqrt{5}\qquad \textbf{(D)}\ 125\qquad \textbf{(E)}\ 25$

Solution

Problem 21

In the accompanying figure $\overline{CE}$ and $\overline{DE}$ are equal chords of a circle with center $O$ . Arc $AB$ is a quarter-circle. Then the ratio of the area of triangle $CED$ to the area of triangle $AOB$ is:

$[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(0,0)--(10,0)--(0,10)--cycle,dot); MP("O",(0,0),N);MP("C",(-10,0),W);MP("D",(10,0),E);;MP("E",(0,10),N); draw((-sqrt(70),-sqrt(30))--(sqrt(30),-sqrt(70))--(0,0)--cycle,dot); MP("A",(-sqrt(70),-sqrt(30)),SW);MP("B",(sqrt(30),-sqrt(70)),SE); [/asy]$

$\textbf{(A)}\ \sqrt {2} : 1\qquad \textbf{(B)}\ \sqrt {3} : 1\qquad \textbf{(C)}\ 4 : 1\qquad \textbf{(D)}\ 3 : 1\qquad \textbf{(E)}\ 2 : 1$

Solution

Problem 22

A particle is placed on the parabola $y = x^2 - x - 6$ at a point $P$ whose $y$ -coordinate is $6$ . It is allowed to roll along the parabola until it reaches the nearest point $Q$ whose $y$ -coordinate is $-6$ . The horizontal distance traveled by the particle (the numerical value of the difference in the $x$ -coordinates of $P$ and $Q$ ) is:

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$

Solution

Problem 23

If, in the expression $x^2 - 3$ , $x$ increases or decreases by a positive amount of $a$ , the expression changes by an amount:

$\textbf{(A)}\ {\pm 2ax + a^2}\qquad \textbf{(B)}\ {2ax \pm a^2}\qquad \textbf{(C)}\ {\pm a^2 - 3} \qquad \textbf{(D)}\ {(x + a)^2 - 3}\qquad\\ \textbf{(E)}\ {(x - a)^2 - 3}$

Solution

Problem 24

A man travels $m$ feet due north at $2$ minutes per mile. He returns due south to his starting point at $2$ miles per minute. The average rate in miles per hour for the entire trip is:

$\textbf{(A)}\ 75\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 24\qquad\\ \textbf{(E)}\ \text{impossible to determine without knowing the value of }{m}$

Solution

Problem 25

If $\log_{k}{x}\cdot \log_{5}{k} = 3$ , then $x$ equals:

$\textbf{(A)}\ k^6\qquad \textbf{(B)}\ 5k^3\qquad \textbf{(C)}\ k^3\qquad \textbf{(D)}\ 243\qquad \textbf{(E)}\ 125$

Solution

Problem 26

A set of $n$ numbers has the sum $s$ . Each number of the set is increased by $20$ , then multiplied by $5$ , and then decreased by $20$ . The sum of the numbers in the new set thus obtained is:

$\textbf{(A)}\ s + 20n\qquad \textbf{(B)}\ 5s + 80n\qquad \textbf{(C)}\ s\qquad \textbf{(D)}\ 5s\qquad \textbf{(E)}\ 5s + 4n$

Solution

Problem 27

The points $(2,-3)$ , $(4,3)$ , and $(5, k/2)$ are on the same straight line. The value(s) of $k$ is (are):

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ -12\qquad \textbf{(C)}\ \pm 12\qquad \textbf{(D)}\ {12}\text{ or }{6}\qquad \textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}$

Solution

Problem 28

A $16$ -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:

$\textbf{(A)}\ \frac{1}{4}\qquad \textbf{(B)}\ \frac{81}{256}\qquad \textbf{(C)}\ \frac{27}{64}\qquad \textbf{(D)}\ \frac{37}{64}\qquad \textbf{(E)}\ \frac{175}{256}$

Solution

Problem 29

In a general triangle $ADE$ (as shown) lines $\overline{EB}$ and $\overline{EC}$ are drawn. Which of the following angle relations is true?

$[asy] draw((-8,0)--(-2,0)--(4,0)--(10,0)--(0,10)--cycle,dot); draw((-2,0)--(0,10),dot);draw((4,0)--(0,10),dot); MP("A",(-8,0),S);MP("B",(-2,0),S);MP("C",(4,0),S);MP("D",(10,0),S);MP("E",(0,10),N); MP("x",(-7.9,.4),E);MP("z",(-2,.4),W);MP("m",(-2,.4),E);MP("n",(4,.4),W);MP("c",(4,.4),E);MP("a",(9.9,.4),W); MP("y",(-.2,8.8),SW);MP("w",(.1,8.8),S);MP("b",(.7,9),SE); [/asy]$

$\textbf{(A)}\ x + z = a + b\qquad \textbf{(B)}\ y + z = a + b\qquad \textbf{(C)}\ m + x = w + n\qquad \\ \textbf{(D)}\ x + z + n = w + c + m\qquad \textbf{(E)}\ x + y + n = a + b + m$

Solution

Problem 30

If $xy = b$ and $\frac{1}{x^2} + \frac{1}{y^2} = a$ , then $(x + y)^2$ equals:

$\textbf{(A)}\ (a + 2b)^2\qquad \textbf{(B)}\ a^2 + b^2\qquad \textbf{(C)}\ b(ab + 2)\qquad \textbf{(D)}\ ab(b + 2)\qquad \textbf{(E)}\ \frac{1}{a} + 2b$

Solution

Problem 31

The altitude drawn to the base of an isosceles triangle is $8$ , and the perimeter $32$ . The area of the triangle is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 32\qquad \textbf{(E)}\ 24$

Solution

Problem 32

With $ $1000$ a rancher is to buy steers at $ $25$ each and cows at $ $26$ each. If the number of steers $s$ and the number of cows $c$ are both positive integers, then:

$\textbf{(A)}\ \text{this problem has no solution}\qquad\\ \textbf{(B)}\ \text{there are two solutions with }{s}\text{ exceeding }{c}\qquad \\ \textbf{(C)}\ \text{there are two solutions with }{c}\text{ exceeding }{s}\qquad \\ \textbf{(D)}\ \text{there is one solution with }{s}\text{ exceeding }{c}\qquad \\ \textbf{(E)}\ \text{there is one solution with }{c}\text{ exceeding }{s}$

Solution

Problem 33

For one root of $ax^2 + bx + c = 0$ to be double the other, the coefficients $a,\,b,\,c$ must be related as follows:

$\textbf{(A)}\ 4b^2 = 9c\qquad \textbf{(B)}\ 2b^2 = 9ac\qquad \textbf{(C)}\ 2b^2 = 9a\qquad \\ \textbf{(D)}\ b^2 - 8ac = 0\qquad \textbf{(E)}\ 9b^2 = 2ac$

Solution

Problem 34

The numerator of a fraction is $6x + 1$ , then denominator is $7 - 4x$ , and $x$ can have any value between $-2$ and $2$ , both included. The values of $x$ for which the numerator is greater than the denominator are:

$\textbf{(A)}\ \frac{3}{5} < x \le 2\qquad \textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad \textbf{(C)}\ 0 < x \le 2\qquad \\ \textbf{(D)}\ 0 \le x \le 2\qquad \textbf{(E)}\ -2 \le x \le 2$

Solution

Problem 35

A triangle is formed by joining three points whose coordinates are integers. If the $x$ -coordinate and the $y$ -coordinate each have a value of $1$ , then the area of the triangle, in square units:

$\textbf{(A)}\ \text{must be an integer}\qquad \textbf{(B)}\ \text{may be irrational}\qquad \textbf{(C)}\ \text{must be irrational}\qquad \textbf{(D)}\ \text{must be rational}\qquad \\ \textbf{(E)}\ \text{will be an integer only if the triangle is equilateral.}$

Solution

Problem 36

The sides of a triangle are $30$ , $70$ , and $80$ units. If an altitude is dropped upon the side of length $80$ , the larger segment cut off on this side is:

$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

Solution

Problem 37

The first term of an arithmetic series of consecutive integers is $k^2 + 1$ . The sum of $2k + 1$ terms of this series may be expressed as:

$\textbf{(A)}\ k^3 + (k + 1)^3\qquad \textbf{(B)}\ (k - 1)^3 + k^3\qquad \textbf{(C)}\ (k + 1)^3\qquad \\ \textbf{(D)}\ (k + 1)^2\qquad \textbf{(E)}\ (2k + 1)(k + 1)^2$

Solution

Problem 38

Let $r$ be the distance from the origin to a point $P$ with coordinates $x$ and $y$ . Designate the ratio $\frac{y}{r}$ by $s$ and the ratio $\frac{x}{r}$ by $c$ . Then the values of $s^2 - c^2$ are limited to the numbers:

$\textbf{(A)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both excluded}\qquad\\ \textbf{(B)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both included}\qquad \\ \textbf{(C)}\ \text{between }{-1}\text{ and }{+1}\text{, both excluded}\qquad \\ \textbf{(D)}\ \text{between }{-1}\text{ and }{+1}\text{, both included}\qquad \\ \textbf{(E)}\ {-1}\text{ and }{+1}\text{ only}$

Solution

Problem 39

We may say concerning the solution of $|x|^2 + |x| - 6 =0$ that:

$\textbf{(A)}\ \text{there is only one root}\qquad \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad \textbf{(E)}\ \text{the product of the roots is }{-6}$

Solution

Problem 40

Given $a_0 = 1$ , $a_1 = 3$ , and the general relation $a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n$ for $n \ge 1$ . Then $a_3$ equals:

$\textbf{(A)}\ \frac{13}{27}\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ -17$

Solution

Problem 41

The roots of $Ax^2 + Bx + C = 0$ are $r$ and $s$ . For the roots of $x^2+px +q =0$

to be $r^2$ and $s^2$ , $p$ must equal:

$\textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\ \textbf{(D)}\ B^2 - 2C\qquad \textbf{(E)}\ 2C - B^2$

Solution

Problem 42

In a circle with center $O$ , chord $\overline{AB}$ equals chord $\overline{AC}$ . Chord $\overline{AD}$ cuts $\overline{BC}$ in $E$ . If $AC = 12$ and $AE = 8$ , then $AD$ equals:

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 18$

Solution

Problem 43

$\overline{AB}$ is the hypotenuse of a right triangle $ABC$ . Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$ . The length of $\overline{AB}$ is:

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$

Solution

Problem 44

Given the true statements: (1) If $a$ is greater than $b$ , then $c$ is greater than $d$ (2) If $c$ is less than $d$ , then $e$ is greater than $f$ . A valid conclusion is:

$\textbf{(A)}\ \text{If }{a}\text{ is less than }{b}\text{, then }{e}\text{ is greater than }{f}\qquad \\ \textbf{(B)}\ \text{If }{e}\text{ is greater than }{f}\text{, then }{a}\text{ is less than }{b}\qquad \\ \textbf{(C)}\ \text{If }{e}\text{ is less than }{f}\text{, then }{a}\text{ is greater than }{b}\qquad \\ \textbf{(D)}\ \text{If }{a}\text{ is greater than }{b}\text{, then }{e}\text{ is less than }{f}\qquad \\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 45

A check is written for $x$ dollars and $y$ cents, $x$ and $y$ both two-digit numbers. In error it is cashed for $y$ dollars and $x$ cents, the incorrect amount exceeding the correct amount by $$17.82$ . Then:

$\textbf{(A)}\ {x}\text{ cannot exceed }{70}\qquad \\ \textbf{(B)}\ {y}\text{ can equal }{2x}\qquad\\ \textbf{(C)}\ \text{the amount of the check cannot be a multiple of }{5}\qquad \\ \textbf{(D)}\ \text{the incorrect amount can equal twice the correct amount}\qquad \\ \textbf{(E)}\ \text{the sum of the digits of the correct amount is divisible by }{9}$

Solution

Problem 46

For values of $x$ less than $1$ but greater than $-4$ , the expression $\frac{x^2 - 2x + 2}{2x - 2}$ has:

$\textbf{(A)}\ \text{no maximum or minimum value}\qquad \\ \textbf{(B)}\ \text{a minimum value of }{+1}\qquad \\ \textbf{(C)}\ \text{a maximum value of }{+1}\qquad \\ \textbf{(D)}\ \text{a minimum value of }{-1}\qquad \\ \textbf{(E)}\ \text{a maximum value of }{-1}$

Solution

Problem 47

$ABCD$ is a rectangle (see the accompanying diagram) with $P$ any point on $\overline{AB}$ . $\overline{PS} \perp \overline{BD}$ and $\overline{PR} \perp \overline{AC}$ . $\overline{AF} \perp \overline{BD}$ and $\overline{PQ} \perp \overline{AF}$ . Then $PR + PS$ is equal to:

$[asy] draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--cycle,dot); draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); [/asy]$

$\textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT + AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF$

Solution

Problem 48

Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A$ , and on $\overline{AB}$ . $D$ is a point $4$ units from $B$ , and on $\overline{AB}$ . $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$ :

$\textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\\ \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \\ \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \\ \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \\ \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.$

Solution

Problem 49

In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:

$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$

Solution

Problem 50

In this diagram a scheme is indicated for associating all the points of segment $\overline{AB}$ with those of segment $\overline{A'B'}$ , and reciprocally. To described this association scheme analytically, let $x$ be the distance from a point $P$ on $\overline{AB}$ to $D$ and let $y$ be the distance from the associated point $P'$ of $\overline{A'B'}$ to $D'$ . Then for any pair of associated points, if $x = a,\, x + y$ equals:

$[asy] draw((0,-3)--(0,3),black+linewidth(.75)); draw((1,-2.5)--(5,-2.5),black+linewidth(.75)); draw((3,2.5)--(4,2.5),black+linewidth(.75)); draw((1,-2.5)--(4,2.5),black+linewidth(.75)); draw((5,-2.5)--(3,2.5),black+linewidth(.75)); draw((2.6,-2.5)--(3.6,2.5),black+linewidth(.75)); dot((0,2.5));dot((1,2.5));dot((2,2.5));dot((3,2.5));dot((4,2.5));dot((5,2.5)); dot((0,-2.5));dot((1,-2.5));dot((2,-2.5));dot((3,-2.5));dot((4,-2.5));dot((5,-2.5)); MP("D",(0,2.5),NW);MP("A",(3,2.5),N);MP("P",(3.5,2.5),N);MP("B",(4,2.5),N); MP("D'",(0,-2.5),NW);MP("B'",(1,-2.5),NW);MP("P'",(2.25,-2.5),N);MP("A'",(5,-2.5),NE); MP("0",(0,2.5),SE);MP("1",(1,2.5),SE);MP("2",(2,2.5),SE);MP("3",(3,2.5),SE);MP("4",(4,2.5),SE);MP("5",(5,2.5),SE); MP("0",(0,-2.5),SE);MP("1",(1,-2.5),SE);MP("2",(2,-2.5),SE);MP("3",(3,-2.5),SE);MP("4",(4,-2.5),SE);MP("5",(5,-2.5),SE); [/asy]$

$\textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a - 51\qquad \textbf{(C)}\ 17 - 3a\qquad \textbf{(D)}\ \frac {17 - 3a}{4}\qquad \textbf{(E)}\ 12a - 34$

Solution

1958 AHSC (Problems • Answer Key • Resources)
Preceded by 1957 AHSC	Followed by 1959 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+{{AHSC 50 Problems
+|year=1958
+}}
 == 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>
-For integers <math>a, b</math> and <math>c</math>, define <math>\boxed a,b,c</math> to mean <math>a^b-b^c+c^a</math>. Then <math>\boxed 1,-1,2</math> equals
+[[1958 AHSME Problems/Problem 1|Solution]]
-<math>\text{(A)} \ -4 \qquad \text{(B)} \ -2 \qquad \text{(C)} \ 0 \qquad \text{(D)} \ 2 \qquad \texxt{(E)} \4</math>
-[[1959 AHSME Problems/Problem 1|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 2|Solution]]
+[[1958 AHSME Problems/Problem 2|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 3|Solution]]
+[[1958 AHSME Problems/Problem 3|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 4|Solution]]
+[[1958 AHSME Problems/Problem 4|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 5|Solution]]
+[[1958 AHSME Problems/Problem 5|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 6|Solution]]
+[[1958 AHSME Problems/Problem 6|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 7|Solution]]
+[[1958 AHSME Problems/Problem 7|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 8|Solution]]
+[[1958 AHSME Problems/Problem 8|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 9|Solution]]
+[[1958 AHSME Problems/Problem 9|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 10|Solution]]
+[[1958 AHSME Problems/Problem 10|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 11|Solution]]
+[[1958 AHSME Problems/Problem 11|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 12|Solution]]
+[[1958 AHSME Problems/Problem 12|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 13|Solution]]
+[[1958 AHSME Problems/Problem 13|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 14|Solution]]
+[[1958 AHSME Problems/Problem 14|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 15|Solution]]
+[[1958 AHSME Problems/Problem 15|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 16|Solution]]
+[[1958 AHSME Problems/Problem 16|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 17|Solution]]
+[[1958 AHSME Problems/Problem 17|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 18|Solution]]
+[[1958 AHSME Problems/Problem 18|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 19|Solution]]
+[[1958 AHSME Problems/Problem 19|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 20|Solution]]
+[[1958 AHSME Problems/Problem 20|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 21|Solution]]
+<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 ==
+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:
-[[1959 AHSME Problems/Problem 22|Solution]]
+<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 ==
+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>
-[[1959 AHSME Problems/Problem 23|Solution]]
+[[1958 AHSME Problems/Problem 23|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 24|Solution]]
+[[1958 AHSME Problems/Problem 24|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 25|Solution]]
+[[1958 AHSME Problems/Problem 25|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 26|Solution]]
+[[1958 AHSME Problems/Problem 26|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 27|Solution]]
+[[1958 AHSME Problems/Problem 27|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 28|Solution]]
+[[1958 AHSME Problems/Problem 28|Solution]]
 == 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>
-[[1959 AHSME Problems/Problem 29|Solution]]
+<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 ==
+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>
-[[1959 AHSME Problems/Problem 30|Solution]]
+[[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 ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{AHSME 50p box|year=1958|before=[[1957 AHSME|1957 AHSC]]|after=[[1959 AHSME|1959 AHSC]]}}
+{{MAA Notice}}

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Difference between revisions of "1958 AHSME Problems"

Latest revision as of 13:21, 20 February 2020

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

See also