Difference between revisions of "1959 AHSME Problems"
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== Problem 1== | == Problem 1== | ||
− | Each edge of a cube is increased by 50 | + | Each edge of a cube is increased by <math>50</math>%. The percent of increase of the surface area of the cube is: |
− | \textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750 | + | <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]] | [[1959 AHSME Problems/Problem 1|Solution]] | ||
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== Problem 5== | == Problem 5== | ||
The value of <math>\left(256\right)^{.16}\left(256\right)^{.09}</math> is: | 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>\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]] | [[1959 AHSME Problems/Problem 5|Solution]] | ||
== Problem 6== | == Problem 6== | ||
+ | |||
+ | |||
Given the true statement: If a quadrilateral is a square, then it is a rectangle. | 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: | 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> | + | |
+ | <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]] | [[1959 AHSME Problems/Problem 6|Solution]] | ||
== Problem 7== | == 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: | 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> | <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> | ||
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== Problem 8== | == Problem 8== | ||
+ | |||
The value of <math>x^2-6x+13</math> can never be less than: | 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> | + | |
+ | <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]] | [[1959 AHSME Problems/Problem 8|Solution]] | ||
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== Problem 11== | == Problem 11== | ||
− | The logarithm of <math>.0625 to the base < | + | 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]] | [[1959 AHSME Problems/Problem 11|Solution]] | ||
== Problem 12== | == Problem 12== | ||
− | By adding the same constant to < | + | 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]] | [[1959 AHSME Problems/Problem 12|Solution]] | ||
== Problem 13== | == Problem 13== | ||
− | The arithmetic mean (average) of a set of < | + | 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]] | [[1959 AHSME Problems/Problem 13|Solution]] | ||
== Problem 14== | == Problem 14== | ||
− | Given the set < | + | 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 | 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 < | + | (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]] | [[1959 AHSME Problems/Problem 14|Solution]] | ||
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== Problem 15== | == 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: | 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]] | [[1959 AHSME Problems/Problem 15|Solution]] | ||
== Problem 16== | == Problem 16== | ||
− | The expression< | + | 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]] | [[1959 AHSME Problems/Problem 16|Solution]] | ||
== Problem 17== | == Problem 17== | ||
− | If < | + | 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 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>\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]] | [[1959 AHSME Problems/Problem 17|Solution]] | ||
== Problem 18== | == Problem 18== | ||
− | The arithmetic mean (average) of the first <math>n< | + | 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>\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]] | [[1959 AHSME Problems/Problem 18|Solution]] | ||
== Problem 19== | == Problem 19== | ||
− | With the use of three different weights, namely <math>1< | + | 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? | 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>\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]] | [[1959 AHSME Problems/Problem 19|Solution]] | ||
== Problem 20== | == Problem 20== | ||
− | It is given that <math>x< | + | 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: |
− | \textbf{(A)}\ 180\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 154\qquad\textbf{(D)}\ 140\qquad\textbf{(E)}\ 120 <math> | + | </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]] | [[1959 AHSME Problems/Problem 20|Solution]] | ||
== Problem 21== | == Problem 21== | ||
− | If</math> p<math> is the perimeter of an equilateral | + | 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>\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]] | [[1959 AHSME Problems/Problem 21|Solution]] | ||
== Problem 22== | == Problem 22== | ||
− | The line joining the midpoints of the diagonals of a trapezoid has length <math>3< | + | 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>\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== | == Problem 23== | ||
− | The set of solutions of the equation <math>\log_{10}\left( a^2-15a\right)=2< | + | 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>\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]] | [[1959 AHSME Problems/Problem 23|Solution]] | ||
== Problem 24== | == Problem 24== | ||
− | A chemist has m ounces of salt that is <math>m< | + | 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>\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]] | [[1959 AHSME Problems/Problem 24|Solution]] | ||
== Problem 25== | == Problem 25== | ||
− | The symbol <math>|a|< | + | 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>>< | + | the symbol </math>><math> means "greater than." |
− | The set of values <math>x< | + | 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>\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]] | [[1959 AHSME Problems/Problem 25|Solution]] | ||
== Problem 26== | == Problem 26== | ||
− | The base of an isosceles triangle is <math>\sqrt 2< | + | 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>\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]] | [[1959 AHSME Problems/Problem 26|Solution]] | ||
== Problem 27== | == Problem 27== | ||
− | Which one of the following is not true for the equation ix^2-x+2i=0, where i=\sqrt{-1} | + | Which one of the following is not true for the equation</math> ix^2-x+2i=0<math>, where </math>i=\sqrt{-1}<math> |
$textbf{(A)}\ \text{The sum of the roots is 2} \qquad | $textbf{(A)}\ \text{The sum of the roots is 2} \qquad | ||
\textbf{(B)}\ \text{The discriminant is 9}\qquad | \textbf{(B)}\ \text{The discriminant is 9}\qquad | ||
\textbf{(C)}\ \text{The roots are imaginary}\qquad | \textbf{(C)}\ \text{The roots are imaginary}\qquad | ||
\textbf{(D)}\ \text{The roots can be found using the quadratic formula}\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> | + | \textbf{(E)}\ \text{The roots can be found by factoring, using imaginary numbers} </math> |
[[1959 AHSME Problems/Problem 27|Solution]] | [[1959 AHSME Problems/Problem 27|Solution]] | ||
− | == Problem 28==< | + | == Problem 28==<math>M are on BC and AB, respectively. The sides of </math>\triangle ABC<math> are </math>a,b,<math> and </math>c<math>. Then </math>\frac{\overbar{AM}}{\overbar{MB}}=k\frac{\overbar{CL}}{\overbar{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>\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]] | [[1959 AHSME Problems/Problem 28|Solution]] | ||
== Problem 29== | == Problem 29== | ||
− | On a examination of <math>n< | + | 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< | + | 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>\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]] | [[1959 AHSME Problems/Problem 29|Solution]] | ||
== Problem 30== | == Problem 30== | ||
− | <math>A< | + | </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< | + | 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>\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]] | [[1959 AHSME Problems/Problem 30|Solution]] | ||
== Problem 31== | == Problem 31== | ||
− | A square, with an area of <math>40< | + | 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>\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]] | [[1959 AHSME Problems/Problem 31|Solution]] | ||
== Problem 32== | == Problem 32== | ||
− | The length <math>l< | + | 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>\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]] | [[1959 AHSME Problems/Problem 32|Solution]] | ||
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== Problem 33== | == Problem 33== | ||
A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. | A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. | ||
− | Let <math>S_n< | + | 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< | + | 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>\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]] | [[1959 AHSME Problems/Problem 33|Solution]] | ||
== Problem 34== | == Problem 34== | ||
− | Let the roots of<math>x^2-3x+1=0< | + | 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>\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]] | [[1959 AHSME Problems/Problem 34|Solution]] | ||
== Problem 35== | == Problem 35== | ||
− | The symbol <math>\ge< | + | The symbol </math>\ge<math> means "greater than or equal to"; the symbol </math>\le<math> means "less than or equal to". |
− | In the eqeuation <math>(x-m)^2-(x-n)^2=(m-n)^2; m< | + | In the eqeuation </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>\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]] | [[1959 AHSME Problems/Problem 35|Solution]] | ||
== Problem 36== | == Problem 36== | ||
− | The base of a triangle is <math>80< | + | 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>\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]] | [[1959 AHSME Problems/Problem 36|Solution]] | ||
== Problem 37== | == Problem 37== | ||
− | When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)< | + | 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>\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]] | [[1959 AHSME Problems/Problem 37|Solution]] | ||
== Problem 38== | == Problem 38== | ||
− | If <math>4x+\sqrt{2x}=1< | + | 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>\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]] | [[1959 AHSME Problems/Problem 38|Solution]] | ||
== Problem 39== | == 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.< | + | 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: | 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\qquaud\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a < | + | </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\qquaud\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]] | [[1959 AHSME Problems/Problem 39|Solution]] | ||
== Problem 40== | == Problem 40== | ||
− | In <math>\triangle ABC< | + | 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>\overbar{BE}=\overbar{ED}<math>. Point </math>F<math> is on </math>AB<math>. Then, if </math>\overbar{BF}=5<math>, |
− | <math>\overbar{BA}< | + | </math>\overbar{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>\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]] | [[1959 AHSME Problems/Problem 40|Solution]] | ||
== Problem 41== | == Problem 41== | ||
− | On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4< | + | 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: | 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>\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]] | [[1959 AHSME Problems/Problem 41|Solution]] | ||
== Problem 42== | == Problem 42== | ||
− | Given three positive integers <math>a,b,< | + | 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? | Then, which two of the following statements are true? | ||
− | <math>\text{(1)}\ \text{the product MD cannot be less than abc} \qquad \\ | + | </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{(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{(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.)}< | + | \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>\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]] | [[1959 AHSME Problems/Problem 42|Solution]] | ||
== Problem 43== | == Problem 43== | ||
− | The sides of a triangle are <math>25,39< | + | 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>\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]] | [[1959 AHSME Problems/Problem 43|Solution]] | ||
== Problem 44== | == Problem 44== | ||
− | The roots of <math>x^2+bx+c=0< | + | 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 | + | </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}< | + | \textbf{(E)}\text{ must be between -1 and +1}<math> |
[[1959 AHSME Problems/Problem 44|Solution]] | [[1959 AHSME Problems/Problem 44|Solution]] | ||
== Problem 45== | == Problem 45== | ||
− | If <math>\left(\log_3 x\right)\left(\log_x 2x\right)\left( \log_{2x} y\right)=\log_{x}x^2< | + | 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>\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]] | [[1959 AHSME Problems/Problem 45|Solution]] | ||
== Problem 46== | == Problem 46== | ||
− | A student on vacation for <math>d< | + | 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< | + | 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>\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]] | [[1959 AHSME Problems/Problem 46|Solution]] | ||
Line 308: | Line 327: | ||
(I). All freshmen are human. (II). All students are human. (III). Some students think. | (I). All freshmen are human. (II). All students are human. (III). Some students think. | ||
Given the following four statements: | Given the following four statements: | ||
− | <math>\textbf{(1)}\ \text{All freshmen are students.}\qquad \\ | + | </math>\textbf{(1)}\ \text{All freshmen are students.}\qquad \\ |
\textbf{(2)}\ \text{Some humans think.}\qquad \\ | \textbf{(2)}\ \text{Some humans think.}\qquad \\ | ||
\textbf{(3)}\ \text{No freshmen think.}\qquad \\ | \textbf{(3)}\ \text{No freshmen think.}\qquad \\ | ||
− | \textbf{(4)}\ \text{Some humans who think are not students.}< | + | \textbf{(4)}\ \text{Some humans who think are not students.}<math> |
Those which are logical consequences of I,II, and III are: | 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>\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]] | [[1959 AHSME Problems/Problem 47|Solution]] | ||
== Problem 48== | == Problem 48== | ||
− | Given the polynomial <math>a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n< | + | 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< | + | 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< | + | 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>\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]] | [[1959 AHSME Problems/Problem 48|Solution]] | ||
== Problem 49== | == Problem 49== | ||
− | For the infinite series <math>1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots< | + | 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>\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]] | [[1959 AHSME Problems/Problem 49|Solution]] | ||
== Problem 50== | == Problem 50== | ||
− | A club with <math>x | + | A club with </math>x$ members is organized into four committees in accordance with these two rules: |
− | \text{(1)}\ \text{Each member belongs to two and only two committees}\qquad | + | \text{(1)}\ \text{Each member belongs to two and only two committees}\qquad \\ |
\text{(2)}\ \text{Each pair of committees has one and only one member in common} | \text{(2)}\ \text{Each pair of committees has one and only one member in common} | ||
Then x: | Then x: | ||
− | \textbf{(A)} \ \text{cannont be determined} \qquad | + | \textbf{(A)} \ \text{cannont be determined} \qquad \\ |
− | \textbf{(B)} \ \text{has a single value between 8 and 16} \qquad | + | \textbf{(B)} \ \text{has a single value between 8 and 16} \qquad \\ |
− | \textbf{(C)} \ \text{has two values between 8 and 16} \qquad | + | \textbf{(C)} \ \text{has two values between 8 and 16} \qquad \\ |
− | \textbf{(D)} \ \text{has a single value between 4 and 8} \qquad | + | \textbf{(D)} \ \text{has a single value between 4 and 8} \qquad \\ |
− | \textbf{(E)} \ \text{has two values between 4 and 8} \qquad | + | \textbf{(E)} \ \text{has two values between 4 and 8} \qquad \\ |
[[1959 AHSME Problems/Problem 50|Solution]] | [[1959 AHSME Problems/Problem 50|Solution]] |
Revision as of 07:15, 11 October 2014
Contents
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 $\overbar{AB}=\overbar{AC}=3.6$ (Error compiling LaTeX. ! Undefined control sequence.), 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 $\overbar{AE}$ (Error compiling LaTeX. ! Undefined control sequence.) 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
x=-5
a+b
\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 $[[1959 AHSME Problems/Problem 17|Solution]]
== Problem 18==
The arithmetic mean (average) of the first$ (Error compiling LaTeX. ! Missing $ inserted.)n\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} $[[1959 AHSME Problems/Problem 18|Solution]]
== Problem 19==
With the use of three different weights, namely$ (Error compiling LaTeX. ! Missing $ inserted.)13
9
\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7 $[[1959 AHSME Problems/Problem 19|Solution]]
== Problem 20==
It is given that$ (Error compiling LaTeX. ! Missing $ inserted.)xy
z
x=10
y=4
z=14
y=16
z=7
x
\textbf{(A)}\ 180\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 154\qquad\textbf{(D)}\ 140\qquad\textbf{(E)}\ 120 $[[1959 AHSME Problems/Problem 20|Solution]]
== Problem 21==
If$ (Error compiling LaTeX. ! Missing $ inserted.) p\triangle
\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} $[[1959 AHSME Problems/Problem 21|Solution]]
== Problem 22==
The line joining the midpoints of the diagonals of a trapezoid has length$ (Error compiling LaTeX. ! Missing $ inserted.)397,
\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89 $[[1959 AHSME Problems/Problem 22|Solution]]
== Problem 23==
The set of solutions of the equation$ (Error compiling LaTeX. ! Missing $ inserted.)\log_{10}\left( a^2-15a\right)=2\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} $[[1959 AHSME Problems/Problem 23|Solution]]
== Problem 24== A chemist has m ounces of salt that is$ (Error compiling LaTeX. ! Missing $ inserted.)m$% salt. How many ounces of salt must he add to make a solution that is$ (Error compiling LaTeX. ! Missing $ inserted.)2m$% salt?$ (Error compiling LaTeX. ! Missing $ inserted.)\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}$[[1959 AHSME Problems/Problem 24|Solution]]
== Problem 25==
The symbol$ (Error compiling LaTeX. ! Missing $ inserted.)|a|+a
a
-a
<
>
x
|3-x|<4
x
\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$[[1959 AHSME Problems/Problem 25|Solution]]
== Problem 26==
The base of an isosceles triangle is$ (Error compiling LaTeX. ! Missing $ inserted.)\sqrt 2\textbf{(A)}\ 1.5 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.5\qquad\textbf{(E)}\ 4 $[[1959 AHSME Problems/Problem 26|Solution]]
== Problem 27==
Which one of the following is not true for the equation$ (Error compiling LaTeX. ! Missing $ inserted.) ix^2-x+2i=0i=\sqrt{-1}
== Problem 28==\triangle ABC
a,b,
c
\frac{\overbar{AM}}{\overbar{MB}}=k\frac{\overbar{CL}}{\overbar{LB}}
k
\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} $[[1959 AHSME Problems/Problem 28|Solution]]
== Problem 29==
On a examination of$ (Error compiling LaTeX. ! Missing $ inserted.)n15
20$. Of the remaining questions he answers one third correctly.
All the questions have the same credit. If the student's mark is 50%, how many different values of$ (Error compiling LaTeX. ! Missing $ inserted.)n
\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved} $[[1959 AHSME Problems/Problem 29|Solution]]
== Problem 30==$ (Error compiling LaTeX. ! Missing $ inserted.)A40
B
A
15
B
\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55 $[[1959 AHSME Problems/Problem 30|Solution]]
== Problem 31==
A square, with an area of$ (Error compiling LaTeX. ! Missing $ inserted.)40\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 120\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 200 $[[1959 AHSME Problems/Problem 31|Solution]]
== Problem 32==
The length$ (Error compiling LaTeX. ! Missing $ inserted.)lA
\frac43
r
\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.} $[[1959 AHSME Problems/Problem 32|Solution]]
== Problem 33==
A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression.
Let$ (Error compiling LaTeX. ! Missing $ inserted.)S_nn
S_3
3,4,6
\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 $[[1959 AHSME Problems/Problem 33|Solution]]
== Problem 34==
Let the roots of$ (Error compiling LaTeX. ! Missing $ inserted.)x^2-3x+1=0r
s
r^2+s^2
\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} $[[1959 AHSME Problems/Problem 34|Solution]]
== Problem 35==
The symbol$ (Error compiling LaTeX. ! Missing $ inserted.)\ge\le
(x-m)^2-(x-n)^2=(m-n)^2; m
n
\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} $[[1959 AHSME Problems/Problem 35|Solution]]
== Problem 36==
The base of a triangle is$ (Error compiling LaTeX. ! Missing $ inserted.)8060^\circ
90
\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 12 $[[1959 AHSME Problems/Problem 36|Solution]]
== Problem 37==
When simplified the product$ (Error compiling LaTeX. ! Missing $ inserted.)\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)\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)} $[[1959 AHSME Problems/Problem 37|Solution]]
== Problem 38==
If$ (Error compiling LaTeX. ! Missing $ inserted.)4x+\sqrt{2x}=1x
\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} $[[1959 AHSME Problems/Problem 38|Solution]]
== Problem 39==
Let S be the sum of the first nine terms of the sequence$ (Error compiling LaTeX. ! Missing $ inserted.)x+a, x^2+2a, x^3+3a, \cdots.\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\qquaud\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a $[[1959 AHSME Problems/Problem 39|Solution]]
== Problem 40==
In$ (Error compiling LaTeX. ! Missing $ inserted.)\triangle ABCBD
CF
BD
E
\overbar{BE}=\overbar{ED}
F
AB
\overbar{BF}=5
\overbar{BA}
\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these} $[[1959 AHSME Problems/Problem 40|Solution]]
== Problem 41==
On the same side of a straight line three circles are drawn as follows: a circle with a radius of$ (Error compiling LaTeX. ! Missing $ inserted.)4\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12 $[[1959 AHSME Problems/Problem 41|Solution]]
== Problem 42==
Given three positive integers$ (Error compiling LaTeX. ! Missing $ inserted.)a,b,c
D
m
\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.)}$$ (Error compiling LaTeX. ! Missing $ inserted.)\textbf{(A)}\ 1,2 \qquad\textbf{(B)}\ 1,3\qquad\textbf{(C)}\ 1,4\qquad\textbf{(D)}\ 2,3\qquad\textbf{(E)}\ 2,4 $[[1959 AHSME Problems/Problem 42|Solution]]
== Problem 43==
The sides of a triangle are$ (Error compiling LaTeX. ! Missing $ inserted.)25,3940
\textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40$[[1959 AHSME Problems/Problem 43|Solution]]
== Problem 44==
The roots of$ (Error compiling LaTeX. ! Missing $ inserted.)x^2+bx+c=01
s=b+c+1
s
\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}$[[1959 AHSME Problems/Problem 44|Solution]]
== Problem 45==
If$ (Error compiling LaTeX. ! Missing $ inserted.)\left(\log_3 x\right)\left(\log_x 2x\right)\left( \log_{2x} y\right)=\log_{x}x^2 y
\textbf{(A)}\ \frac92\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 81 $[[1959 AHSME Problems/Problem 45|Solution]]
== Problem 46==
A student on vacation for$ (Error compiling LaTeX. ! Missing $ inserted.)d7
d
\textbf{(A)}\ 7\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 $[[1959 AHSME Problems/Problem 46|Solution]]
== Problem 47==
Assume that the following three statements are true:
(I). All freshmen are human. (II). All students are human. (III). Some students think.
Given the following four statements:$ (Error compiling LaTeX. ! Missing $ inserted.)\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.}\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 2,3\qquad\textbf{(D)}\ 2,4\qquad\textbf{(E)}\ 1,2 $[[1959 AHSME Problems/Problem 47|Solution]]
== Problem 48==
Given the polynomial$ (Error compiling LaTeX. ! Missing $ inserted.)a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_nn
a_0
a
h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|
|x|
h=3
\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9 $[[1959 AHSME Problems/Problem 48|Solution]]
== Problem 49==
For the infinite series$ (Error compiling LaTeX. ! Missing $ inserted.)1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdotsS
S
\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac27\qquad\textbf{(C)}\ \frac67\qquad\textbf{(D)}\ \frac{9}{32}\qquad\textbf{(E)}\ \frac{27}{32} $[[1959 AHSME Problems/Problem 49|Solution]]
== Problem 50== A club with$ (Error compiling LaTeX. ! Missing $ inserted.)x$ members is organized into four committees in accordance with these two rules: \text{(1)}\ \text{Each member belongs to two and only two committees}\qquad \\ \text{(2)}\ \text{Each pair of committees has one and only one member in common} Then x: \textbf{(A)} \ \text{cannont be determined} \qquad \\ \textbf{(B)} \ \text{has a single value between 8 and 16} \qquad \\ \textbf{(C)} \ \text{has two values between 8 and 16} \qquad \\ \textbf{(D)} \ \text{has a single value between 4 and 8} \qquad \\ \textbf{(E)} \ \text{has two values between 4 and 8} \qquad \\
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.