Difference between revisions of "1959 AHSME Problems"
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− | ==Problem | + | {{AHSC 50 Problems |
− | + | |year=1959 | |
+ | }} | ||
+ | == Problem 1== | ||
− | <math> \ | + | Each edge of a cube is increased by <math>50</math>%. The percent of increase of the surface area of the cube is: |
+ | <math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 1|Solution]] | ||
+ | |||
+ | == Problem 2== | ||
+ | Through a point <math>P</math> inside the <math>\triangle ABC</math> a line is drawn parallel to the base <math>AB</math>, dividing the triangle into two equal areas. | ||
+ | If the altitude to <math>AB</math> has a length of <math>1</math>, then the distance from <math>P</math> to <math>AB</math> is: | ||
+ | <math>\textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 2|Solution]] | ||
+ | |||
+ | == Problem 3== | ||
+ | If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification: | ||
+ | <math>\textbf{(A)}\ \text{rhombus} \qquad\textbf{(B)}\ \text{rectangles} \qquad\textbf{(C)}\ \text{square} \qquad\textbf{(D)}\ \text{isosceles trapezoid}\qquad\textbf{(E)}\ \text{none of these} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 3|Solution]] | ||
+ | |||
+ | == Problem 4== | ||
+ | If <math>78</math> is divided into three parts which are proportional to <math>1, \frac13, \frac16,</math> the middle part is: | ||
+ | <math>\textbf{(A)}\ 9\frac13 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 17\frac13 \qquad\textbf{(D)}\ 18\frac13\qquad\textbf{(E)}\ 26 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 4|Solution]] | ||
+ | |||
+ | == Problem 5== | ||
+ | The value of <math>\left(256\right)^{.16}\left(256\right)^{.09}</math> is: | ||
+ | |||
+ | <math>\textbf{(A)}\ 4 \qquad \\ | ||
+ | \textbf{(B)}\ 16\qquad \\ | ||
+ | \textbf{(C)}\ 64\qquad \\ | ||
+ | \textbf{(D)}\ 256.25\qquad \\ | ||
+ | \textbf{(E)}\ -16 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 5|Solution]] | ||
+ | |||
+ | == Problem 6== | ||
+ | |||
+ | |||
+ | Given the true statement: If a quadrilateral is a square, then it is a rectangle. | ||
+ | It follows that, of the converse and the inverse of this true statement is: | ||
+ | |||
+ | <math>\textbf{(A)}\ \text{only the converse is true} \qquad \\ | ||
+ | \textbf{(B)}\ \text{only the inverse is true }\qquad \\ | ||
+ | \textbf{(C)}\ \text{both are true} \qquad \\ | ||
+ | \textbf{(D)}\ \text{neither is true} \qquad \\ | ||
+ | \textbf{(E)}\ \text{the inverse is true, but the converse is sometimes true} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 6|Solution]] | ||
+ | |||
+ | == Problem 7== | ||
+ | |||
+ | The sides of a right triangle are <math>a</math>, <math>a+d</math>, and <math>a+2d</math>, with <math>a</math> and <math>d</math> both positive. The ratio of <math>a</math> to <math>d</math> is: | ||
+ | <math>\textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 1:4 \qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 3:1\qquad\textbf{(E)}\ 3:4 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 7|Solution]] | ||
+ | |||
+ | == Problem 8== | ||
+ | |||
+ | The value of <math>x^2-6x+13</math> can never be less than: | ||
+ | |||
+ | <math>\textbf{(A)}\ 4 \qquad | ||
+ | \textbf{(B)}\ 4.5 \qquad | ||
+ | \textbf{(C)}\ 5\qquad | ||
+ | \textbf{(D)}\ 7\qquad | ||
+ | \textbf{(E)}\ 13 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 8|Solution]] | ||
+ | |||
+ | == Problem 9== | ||
+ | A farmer divides his herd of <math>n </math>cows among his four sons so that one son gets one-half the herd, | ||
+ | a second son, one-fourth, a third son, one-fifth, and the fourth son, <math>7</math> cows. Then <math>n</math> is: | ||
+ | <math>\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 9|Solution]] | ||
+ | |||
+ | == Problem 10== | ||
+ | In <math>\triangle ABC</math> with <math>\overline{AB}=\overline{AC}=3.6</math>, a point <math>D</math> is taken on <math>AB</math> at a distance <math>1.2</math> from <math>A</math>. | ||
+ | Point <math>D</math> is joined to <math>E</math> in the prolongation of <math>AC</math> so that <math>\triangle AED</math> is equal in area to <math>ABC</math>. Then <math>\overline{AE}</math> is: | ||
+ | <math>\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 10|Solution]] | ||
+ | |||
+ | == Problem 11== | ||
+ | The logarithm of <math>.0625</math> to the base <math>2</math> is: | ||
+ | <math>\textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 11|Solution]] | ||
+ | |||
+ | == Problem 12== | ||
+ | By adding the same constant to <math>20,50,100</math> a geometric progression results. The common ratio is: | ||
+ | <math>\textbf{(A)}\ \frac53 \qquad\textbf{(B)}\ \frac43\qquad\textbf{(C)}\ \frac32\qquad\textbf{(D)}\ \frac12\qquad\textbf{(E)}\ \frac{1}3 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 12|Solution]] | ||
+ | |||
+ | == Problem 13== | ||
+ | The arithmetic mean (average) of a set of <math>50</math> numbers is <math>38</math>. If two numbers, namely, <math>45</math> and <math>55</math>, are discarded, the mean of the remaining set of numbers is: | ||
+ | <math>\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 13|Solution]] | ||
+ | |||
+ | == Problem 14== | ||
+ | Given the set <math>S</math> whose elements are zero and the even integers, positive and negative. | ||
+ | Of the five operations applied to any pair of elements: (1) addition (2) subtraction | ||
+ | (3) multiplication (4) division (5) finding the arithmetic mean (average), those elements that only yield elements of <math>S</math> are: | ||
+ | <math>\textbf{(A)}\ \text{all} \qquad\textbf{(B)}\ 1,2,3,4\qquad\textbf{(C)}\ 1,2,3,5\qquad\textbf{(D)}\ 1,2,3\qquad\textbf{(E)}\ 1,3,5 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 14|Solution]] | ||
+ | |||
+ | == Problem 15== | ||
+ | In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: | ||
+ | <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 15|Solution]] | ||
+ | |||
+ | == Problem 16== | ||
+ | The expression<math> \frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},</math> when simplified is: | ||
+ | <math>\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 16|Solution]] | ||
+ | |||
+ | == Problem 17== | ||
+ | If <math>y=a+\frac{b}{x}</math>, where <math>a</math> and <math>b</math> are constants, and if <math>y=1</math> when <math>x=-1</math>, and <math>y=5 </math> when <math>x=-5</math>, then <math>a+b</math> equals: | ||
+ | <math>\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 17|Solution]] | ||
+ | |||
+ | == Problem 18== | ||
+ | The arithmetic mean (average) of the first <math>n</math> positive integers is: | ||
+ | <math>\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 18|Solution]] | ||
+ | |||
+ | == Problem 19== | ||
+ | With the use of three different weights, namely <math>1</math> lb., <math>3</math> lb., and <math>9</math> lb., how many objects of different weights can be weighed, | ||
+ | if the objects is to be weighed and the given weights may be placed in either pan of the scale? | ||
+ | <math>\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 19|Solution]] | ||
+ | |||
+ | == Problem 20== | ||
+ | It is given that <math>x</math> varies directly as <math>y</math> and inversely as the square of <math>z</math>, and that <math>x=10</math> when <math>y=4</math> and <math>z=14</math>. Then, when <math>y=16</math> and <math>z=7</math>, <math>x</math> equals: | ||
+ | <math>\textbf{(A)}\ 180\qquad | ||
+ | \textbf{(B)}\ 160\qquad | ||
+ | \textbf{(C)}\ 154\qquad | ||
+ | \textbf{(D)}\ 140\qquad | ||
+ | \textbf{(E)}\ 120 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 20|Solution]] | ||
+ | |||
+ | == Problem 21== | ||
+ | If<math> p</math> is the perimeter of an equilateral <math>\triangle</math> inscribed in a circle, the area of the circle is: | ||
+ | <math>\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 21|Solution]] | ||
+ | |||
+ | == Problem 22== | ||
+ | The line joining the midpoints of the diagonals of a trapezoid has length <math>3</math>. If the longer base is <math>97,</math> then the shorter base is: | ||
+ | <math>\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89 </math> | ||
[[1959 AHSME Problems/Problem 22|Solution]] | [[1959 AHSME Problems/Problem 22|Solution]] | ||
+ | |||
+ | == Problem 23== | ||
+ | The set of solutions of the equation <math>\log_{10}\left( a^2-15a\right)=2</math> consists of | ||
+ | <math>\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 23|Solution]] | ||
+ | |||
+ | == Problem 24== | ||
+ | A chemist has m ounces of salt that is <math>m</math>% salt. How many ounces of salt must he add to make a solution that is <math>2m</math>% salt? | ||
+ | |||
+ | <math>\textbf{(A)}\ \frac{m}{100+m} \qquad\textbf{(B)}\ \frac{2m}{100-2m}\qquad\textbf{(C)}\ \frac{m^2}{100-2m}\qquad\textbf{(D)}\ \frac{m^2}{100+2m}\qquad\textbf{(E)}\ \frac{2m}{100+2m}</math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 24|Solution]] | ||
+ | |||
+ | == Problem 25== | ||
+ | The symbol <math>|a|</math> means <math>+a</math> if <math>a</math> is greater than or equal to zero, and <math>-a</math> if a is less than or equal to zero; the symbol <math><</math> means "less than"; | ||
+ | the symbol <math>></math> means "greater than." | ||
+ | The set of values <math>x</math> satisfying the inequality <math>|3-x|<4</math> consists of all <math>x</math> such that: | ||
+ | <math>\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1</math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 25|Solution]] | ||
+ | |||
+ | == Problem 26== | ||
+ | The base of an isosceles triangle is <math>\sqrt 2</math>. The medians to the leg intersect each other at right angles. The area of the triangle is: | ||
+ | <math>\textbf{(A)}\ 1.5 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.5\qquad\textbf{(E)}\ 4 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 26|Solution]] | ||
+ | |||
+ | == Problem 27== | ||
+ | Which one of the following is not true for the equation<math> ix^2-x+2i=0</math>, where <math>i=\sqrt{-1}</math> | ||
+ | <math>\textbf{(A)}\ \text{The sum of the roots is 2} \qquad \\ | ||
+ | \textbf{(B)}\ \text{The discriminant is 9}\qquad \\ | ||
+ | \textbf{(C)}\ \text{The roots are imaginary}\qquad \\ | ||
+ | \textbf{(D)}\ \text{The roots can be found using the quadratic formula}\qquad \\ | ||
+ | \textbf{(E)}\ \text{The roots can be found by factoring, using imaginary numbers} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 27|Solution]] | ||
+ | |||
+ | == Problem 28== | ||
+ | |||
+ | In triangle <math>ABC</math>, <math>AL</math> bisects angle <math>A</math>, and <math>CM</math> bisects angle <math>C</math>. Points <math>L</math> and <math>M</math> are on <math>BC</math> and <math>AB</math>, respectively. The sides of <math>\triangle ABC</math> are <math>a</math>, <math>b</math>, and <math>c</math>. Then <math>\frac{AM}{MB}=k\frac{CL}{LB} </math> where <math>k</math> is: | ||
+ | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 28|Solution]] | ||
+ | |||
+ | == Problem 29== | ||
+ | On a examination of <math>n</math> questions a student answers correctly <math>15</math> of the first <math>20</math>. Of the remaining questions he answers one third correctly. | ||
+ | All the questions have the same credit. If the student's mark is 50%, how many different values of <math>n</math> can there be? | ||
+ | <math>\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 29|Solution]] | ||
+ | |||
+ | == Problem 30== | ||
+ | <math>A</math> can run around a circular track in <math>40</math> seconds. <math>B</math>, running in the opposite direction, meets <math>A</math> every <math>15</math> seconds. | ||
+ | What is <math>B</math>'s time to run around the track, expressed in seconds? | ||
+ | <math>\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 30|Solution]] | ||
+ | |||
+ | == Problem 31== | ||
+ | A square, with an area of <math>40</math>, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is: | ||
+ | <math>\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 120\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 200 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 31|Solution]] | ||
+ | |||
+ | == Problem 32== | ||
+ | The length <math>l</math> of a tangent, drawn from a point <math>A</math> to a circle, is <math>\frac43 </math>of the radius <math>r</math>. The (shortest) distance from A to the circle is: | ||
+ | <math>\textbf{(A)}\ \frac{1}{2}r \qquad\textbf{(B)}\ r\qquad\textbf{(C)}\ \frac{1}{2}l\qquad\textbf{(D)}\ \frac23l \qquad\textbf{(E)}\ \text{a value between r and l.} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 32|Solution]] | ||
+ | |||
+ | == Problem 33== | ||
+ | A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. | ||
+ | Let <math>S_n</math> represent the sum of the first <math>n</math> terms of the harmonic progression; for example <math>S_3</math> represents the sum of | ||
+ | the first three terms. If the first three terms of a harmonic progression are <math>3,4,6</math>, then: | ||
+ | <math>\textbf{(A)}\ S_4=20 \qquad\textbf{(B)}\ S_4=25\qquad\textbf{(C)}\ S_5=49\qquad\textbf{(D)}\ S_6=49\qquad\textbf{(E)}\ S_2=\frac{1}2 S_4 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 33|Solution]] | ||
+ | |||
+ | == Problem 34== | ||
+ | Let the roots of<math>x^2-3x+1=0</math> be <math>r</math> and <math>s</math>. Then the expression <math>r^2+s^2 </math>is: | ||
+ | <math>\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 34|Solution]] | ||
+ | |||
+ | == Problem 35== | ||
+ | The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to". | ||
+ | In the equation <math>(x-m)^2-(x-n)^2=(m-n)^2; m</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is: | ||
+ | <math>\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 35|Solution]] | ||
+ | |||
+ | == Problem 36== | ||
+ | The base of a triangle is <math>80</math>, and one side of the base angle is <math>60^\circ</math>. The sum of the lengths of the other two sides is <math>90</math>. The shortest side is: | ||
+ | <math>\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 12 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 36|Solution]] | ||
+ | |||
+ | == Problem 37== | ||
+ | When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> becomes: | ||
+ | <math>\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 37|Solution]] | ||
+ | |||
+ | == Problem 38== | ||
+ | If <math>4x+\sqrt{2x}=1</math>, then <math>x</math>: | ||
+ | <math>\textbf{(A)}\ \text{is an integer} \qquad\textbf{(B)}\ \text{is fractional}\qquad\textbf{(C)}\ \text{is irrational}\qquad\textbf{(D)}\ \text{is imaginary}\qquad\textbf{(E)}\ \text{may have two different values} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 38|Solution]] | ||
+ | |||
+ | == Problem 39== | ||
+ | Let S be the sum of the first nine terms of the sequence <math>x+a, x^2+2a, x^3+3a, \cdots.</math> | ||
+ | Then S equals: | ||
+ | <math>\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 39|Solution]] | ||
+ | |||
+ | == Problem 40== | ||
+ | In <math>\triangle ABC</math>, <math>BD</math> is a median. <math>CF</math> intersects <math>BD</math> at <math>E</math> so that <math>\overline{BE}=\overline{ED}</math>. Point <math>F</math> is on <math>AB</math>. Then, if <math>\overline{BF}=5</math>, | ||
+ | <math>\overline{BA}</math> equals: | ||
+ | <math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 40|Solution]] | ||
+ | |||
+ | == Problem 41== | ||
+ | On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> inches is tangent to the line, the other | ||
+ | two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: | ||
+ | <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 41|Solution]] | ||
+ | |||
+ | == Problem 42== | ||
+ | Given three positive integers <math>a,b,</math> and <math>c</math>. Their greatest common divisor is <math>D</math>; their least common multiple is <math>m</math>. | ||
+ | Then, which two of the following statements are true? | ||
+ | <math>\text{(1)}\ \text{the product MD cannot be less than abc} \qquad \\ | ||
+ | \text{(2)}\ \text{the product MD cannot be greater than abc}\qquad \\ | ||
+ | \text{(3)}\ \text{MD equals abc if and only if a,b,c are each prime}\qquad \\ | ||
+ | \text{(4)}\ \text{MD equals abc if and only if a,b,c are each relatively prime in pairs} \text{ (This means: no two have a common factor greater than 1.)}</math> | ||
+ | <math>\textbf{(A)}\ 1,2 \qquad\textbf{(B)}\ 1,3\qquad\textbf{(C)}\ 1,4\qquad\textbf{(D)}\ 2,3\qquad\textbf{(E)}\ 2,4 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 42|Solution]] | ||
+ | |||
+ | == Problem 43== | ||
+ | The sides of a triangle are <math>25,39</math>, and <math>40</math>. The diameter of the circumscribed circle is: | ||
+ | <math>\textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40</math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 43|Solution]] | ||
+ | |||
+ | == Problem 44== | ||
+ | The roots of <math>x^2+bx+c=0</math> are both real and greater than <math>1</math>. Let <math>s=b+c+1</math>. Then <math>s</math>: | ||
+ | <math>\textbf{(A)}\ \text{may be less than zero}\qquad\textbf{(B)}\ \text{may be equal to zero}\qquad \textbf{(C)}\ \text{must be greater than zero}\qquad\textbf{(D)}\ \text{must be less than zero}\qquad | ||
+ | \textbf{(E)}\text{ must be between -1 and +1}</math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 44|Solution]] | ||
+ | |||
+ | == Problem 45== | ||
+ | If <math>\left(\log_3 x\right)\left(\log_x 2x\right)\left( \log_{2x} y\right)=\log_{x}x^2</math>, then <math> y</math> equals: | ||
+ | <math>\textbf{(A)}\ \frac92\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 81 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 45|Solution]] | ||
+ | |||
+ | == Problem 46== | ||
+ | A student on vacation for <math>d</math> days observed that (1) it rained <math>7</math> times, morning or afternoon (2) when it rained in the afternoon, | ||
+ | it was clear in the morning (3) there were five clear afternoons (4) there were six clear mornings. Then <math>d</math> equals: | ||
+ | <math>\textbf{(A)}\ 7\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 46|Solution]] | ||
+ | |||
+ | == Problem 47== | ||
+ | Assume that the following three statements are true: | ||
+ | (I). All freshmen are human. (II). All students are human. (III). Some students think. | ||
+ | Given the following four statements: | ||
+ | <math>\textbf{(1)}\ \text{All freshmen are students.}\qquad \\ | ||
+ | \textbf{(2)}\ \text{Some humans think.}\qquad \\ | ||
+ | \textbf{(3)}\ \text{No freshmen think.}\qquad \\ | ||
+ | \textbf{(4)}\ \text{Some humans who think are not students.}</math> | ||
+ | Those which are logical consequences of I,II, and III are: | ||
+ | <math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 2,3\qquad\textbf{(D)}\ 2,4\qquad\textbf{(E)}\ 1,2 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 47|Solution]] | ||
+ | |||
+ | == Problem 48== | ||
+ | Given the polynomial <math>a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n</math>, where <math>n</math> is a positive integer or zero, and <math>a_0</math> is a positive integer. | ||
+ | The remaining <math>a</math>'s are integers or zero. Set <math>h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|</math>. [See example 25 for the meaning of <math>|x|</math>.] | ||
+ | The number of polynomials with <math>h=3</math> is: | ||
+ | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9 </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 48|Solution]] | ||
+ | |||
+ | == Problem 49== | ||
+ | For the infinite series <math>1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots</math> let <math>S</math> be the (limiting) sum. Then <math>S</math> equals: | ||
+ | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac27\qquad\textbf{(C)}\ \frac67\qquad\textbf{(D)}\ \frac{9}{32}\qquad\textbf{(E)}\ \frac{27}{32} </math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 49|Solution]] | ||
+ | |||
+ | == Problem 50== | ||
+ | A club with <math>x</math> members is organized into four committees in accordance with these two rules: | ||
+ | |||
+ | <cmath>\text{(1)}\ \textup{Each member belongs to two and only two committees}\qquad \\ | ||
+ | \text{(2)}\ \textup{Each pair of committees has one and only one member in common}</cmath> | ||
+ | |||
+ | Then <math>x</math>: | ||
+ | |||
+ | <math>\textbf{(A)} \ \textup{cannot be determined} \qquad \\ | ||
+ | \textbf{(B)} \ \textup{has a single value between 8 and 16} \qquad \\ | ||
+ | \textbf{(C)} \ \textup{has two values between 8 and 16} \qquad \\ | ||
+ | \textbf{(D)} \ \textup{has a single value between 4 and 8} \qquad \\ | ||
+ | \textbf{(E)} \ \textup{has two values between 4 and 8}</math> | ||
+ | |||
+ | [[1959 AHSME Problems/Problem 50|Solution]] | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | * [[AMC 12 Problems and Solutions]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 50p box|year=1959|before=[[1958 AHSME|1958 AHSC]]|after=[[1960 AHSME|1960 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 14:20, 20 February 2020
1959 AHSC (Answer Key) Printable version: | AoPS Resources • PDF | ||
Instructions
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Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 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
Each edge of a cube is increased by %. The percent of increase of the surface area of the cube is:
Problem 2
Through a point inside the a line is drawn parallel to the base , dividing the triangle into two equal areas. If the altitude to has a length of , then the distance from to is:
Problem 3
If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification:
Problem 4
If is divided into three parts which are proportional to the middle part is:
Problem 5
The value of is:
Problem 6
Given the true statement: If a quadrilateral is a square, then it is a rectangle. It follows that, of the converse and the inverse of this true statement is:
Problem 7
The sides of a right triangle are , , and , with and both positive. The ratio of to is:
Problem 8
The value of can never be less than:
Problem 9
A farmer divides his herd of cows among his four sons so that one son gets one-half the herd, a second son, one-fourth, a third son, one-fifth, and the fourth son, cows. Then is:
Problem 10
In with , a point is taken on at a distance from . Point is joined to in the prolongation of so that is equal in area to . Then is:
Problem 11
The logarithm of to the base is:
Problem 12
By adding the same constant to a geometric progression results. The common ratio is:
Problem 13
The arithmetic mean (average) of a set of numbers is . If two numbers, namely, and , are discarded, the mean of the remaining set of numbers is:
Problem 14
Given the set whose elements are zero and the even integers, positive and negative. Of the five operations applied to any pair of elements: (1) addition (2) subtraction (3) multiplication (4) division (5) finding the arithmetic mean (average), those elements that only yield elements of are:
Problem 15
In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is:
Problem 16
The expression when simplified is:
Problem 17
If , where and are constants, and if when , and when , then equals:
Problem 18
The arithmetic mean (average) of the first positive integers is:
Problem 19
With the use of three different weights, namely lb., lb., and lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale?
Problem 20
It is given that varies directly as and inversely as the square of , and that when and . Then, when and , equals:
Problem 21
If is the perimeter of an equilateral inscribed in a circle, the area of the circle is:
Problem 22
The line joining the midpoints of the diagonals of a trapezoid has length . If the longer base is then the shorter base is:
Problem 23
The set of solutions of the equation consists of
Problem 24
A chemist has m ounces of salt that is % salt. How many ounces of salt must he add to make a solution that is % salt?
Problem 25
The symbol means if is greater than or equal to zero, and if a is less than or equal to zero; the symbol means "less than"; the symbol means "greater than." The set of values satisfying the inequality consists of all such that:
Problem 26
The base of an isosceles triangle is . The medians to the leg intersect each other at right angles. The area of the triangle is:
Problem 27
Which one of the following is not true for the equation, where
Problem 28
In triangle , bisects angle , and bisects angle . Points and are on and , respectively. The sides of are , , and . Then where is:
Problem 29
On a examination of questions a student answers correctly of the first . Of the remaining questions he answers one third correctly. All the questions have the same credit. If the student's mark is 50%, how many different values of can there be?
Problem 30
can run around a circular track in seconds. , running in the opposite direction, meets every seconds. What is 's time to run around the track, expressed in seconds?
Problem 31
A square, with an area of , is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is:
Problem 32
The length of a tangent, drawn from a point to a circle, is of the radius . The (shortest) distance from A to the circle is:
Problem 33
A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. Let represent the sum of the first terms of the harmonic progression; for example represents the sum of the first three terms. If the first three terms of a harmonic progression are , then:
Problem 34
Let the roots of be and . Then the expression is:
Problem 35
The symbol means "greater than or equal to"; the symbol means "less than or equal to". In the equation is a fixed positive number, and is a fixed negative number. The set of values x satisfying the equation is:
Problem 36
The base of a triangle is , and one side of the base angle is . The sum of the lengths of the other two sides is . The shortest side is:
Problem 37
When simplified the product becomes:
Problem 38
If , then :
Problem 39
Let S be the sum of the first nine terms of the sequence Then S equals:
Problem 40
In , is a median. intersects at so that . Point is on . Then, if , equals:
Problem 41
On the same side of a straight line three circles are drawn as follows: a circle with a radius of inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:
Problem 42
Given three positive integers and . Their greatest common divisor is ; their least common multiple is . Then, which two of the following statements are true?
Problem 43
The sides of a triangle are , and . The diameter of the circumscribed circle is:
Problem 44
The roots of are both real and greater than . Let . Then :
Problem 45
If , then equals:
Problem 46
A student on vacation for days observed that (1) it rained times, morning or afternoon (2) when it rained in the afternoon, it was clear in the morning (3) there were five clear afternoons (4) there were six clear mornings. Then equals:
Problem 47
Assume that the following three statements are true: (I). All freshmen are human. (II). All students are human. (III). Some students think. Given the following four statements: Those which are logical consequences of I,II, and III are:
Problem 48
Given the polynomial , where is a positive integer or zero, and is a positive integer. The remaining 's are integers or zero. Set . [See example 25 for the meaning of .] The number of polynomials with is:
Problem 49
For the infinite series let be the (limiting) sum. Then equals:
Problem 50
A club with members is organized into four committees in accordance with these two rules:
Then :
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1958 AHSC |
Followed by 1960 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.