Difference between revisions of "1959 AHSME Problems"

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== Problem 1==
 
== Problem 1==
  
Each edge of a cube is increased by 50 \%. The percent of increase of the surface area of the cube is:
+
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>  
+
 
 +
<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 </math>2<math> is:
+
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>
+
<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 </math>20,50,100<math> a geometric progression results. The common ratio is:
+
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>   
+
<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 </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:
+
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>     
+
<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 </math>S<math> whose elements are zero and the even integers, positive and negative.  
+
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 </math>S<math> are:
+
(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>  
+
<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>
+
<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</math> \frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},<math> when simplified is:
+
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>   
+
<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 </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:
+
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>   
+
</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</math> positive integers is:
+
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>     
+
</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</math> lb., <math>3</math> lb., and <math>9</math> lb., how many objects of different weights can be weighed,  
+
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>   
+
</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</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> x equals:
+
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 &#036;triangle</math> inscribed in a circle, the area of the circle is:
+
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>   
+
</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</math>. If the longer base is <math>97,</math> then the shorter base is:
+
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>   
+
</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</math> consists of
+
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>  
+
</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</math>% salt. How many ounces of salt must he add to make a solution that is <math>2m</math>% salt?
+
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>
+
</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|</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>|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 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:
+
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>   
+
</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</math>. The medians to the leg intersect each other at right angles. The area of the triangle is:
+
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>   
+
</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>
 
&#036;textbf{(A)}\ \text{The sum of the roots is 2} \qquad
 
&#036;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==</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:
+
== 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>   
+
</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</math> questions a student answers correctly <math>15</math> of the first <math>20</math>. Of the remaining questions he answers one third correctly.  
+
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?
+
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>
+
</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> 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.  
+
</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?
+
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>  
+
</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</math>, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is:
+
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>   
+
</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</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:
+
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>
+
</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</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  
+
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:
+
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>
+
</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</math> be <math>r</math> and <math>s</math>. Then the expression <math>r^2+s^2 </math>is:
+
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>
+
</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</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to".
+
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</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is:
+
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>
+
</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</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:
+
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>   
+
</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)</math> becomes:
+
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>   
+
</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</math>, then <math>x</math>:
+
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>     
+
</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.</math>
+
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>   
+
</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</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>,  
+
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> equals:
+
</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>   
+
</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</math> inches is tangent to the line, the other  
+
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>     
+
</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,</math> and <math>c</math>. Their greatest common divisor is <math>D</math>; their least common multiple is <math>m</math>.  
+
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.)}</math>
+
\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>
+
</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</math>, and <math>40</math>. The diameter of the circumscribed circle is:
+
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>     
+
</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</math> are both real and greater than <math>1</math>. Let <math>s=b+c+1</math>. Then <math>s</math>:
+
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}</math>     
+
\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</math>, then <math> y</math> equals:
+
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>   
+
</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</math> days observed that (1) it rained <math>7</math> times, morning or afternoon (2) when it rained in the afternoon,  
+
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:
+
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>   
+
</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.}</math>
+
\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>   
+
</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</math>, where <math>n</math> is a positive integer or zero, and <math>a_0</math> is a positive integer.  
+
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 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:
+
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>     
+
</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</math> let <math>S</math> be the (limiting) sum. Then <math>S</math> equals:
+
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>   
+
</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</math> members is organized into four committees in accordance with these two rules:
+
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

Problem 1

Each edge of a cube is increased by $50$%. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$

Solution

Problem 2

Through a point $P$ inside the $\triangle ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal areas. If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is: $\textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8}$

Solution

Problem 3

If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification: $\textbf{(A)}\ \text{rhombus} \qquad\textbf{(B)}\ \text{rectangles} \qquad\textbf{(C)}\ \text{square} \qquad\textbf{(D)}\ \text{isosceles trapezoid}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 4

If $78$ is divided into three parts which are proportional to $1, \frac13, \frac16,$ the middle part is: $\textbf{(A)}\ 9\frac13 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 17\frac13 \qquad\textbf{(D)}\ 18\frac13\qquad\textbf{(E)}\ 26$

Solution

Problem 5

The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:

$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$

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:

$\textbf{(A)}\ \text{only the converse is true} \qquad \\ \textbf{(B)}\ \text{only the inverse is true }\qquad  \\ \textbf{(C)}\ \text{both are true} \qquad \\ \textbf{(D)}\ \text{neither is true} \qquad \\ \textbf{(E)}\ \text{the inverse is true, but the converse is sometimes true}$

Solution

Problem 7

The sides of a right triangle are $a$, $a+d$, and $a+2d$, with $a$ and $d$ both positive. The ratio of $a$ to $d$ is: $\textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 1:4 \qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 3:1\qquad\textbf{(E)}\ 3:4$

Solution

Problem 8

The value of $x^2-6x+13$ can never be less than:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 13$

Solution

Problem 9

A farmer divides his herd of $n$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, $7$ cows. Then $n$ is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$

Solution

Problem 10

In $\triangle ABC$ with $\overbar{AB}=\overbar{AC}=3.6$ (Error compiling LaTeX. ! Undefined control sequence.), a point $D$ is taken on $AB$ at a distance $1.2$from $A$. Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$. Then $\overbar{AE}$ (Error compiling LaTeX. ! Undefined control sequence.) is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$

Solution

Problem 11

The logarithm of $.0625$ to the base $2$ is: $\textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2$

Solution

Problem 12

By adding the same constant to $20,50,100$ a geometric progression results. The common ratio is: $\textbf{(A)}\ \frac53 \qquad\textbf{(B)}\ \frac43\qquad\textbf{(C)}\ \frac32\qquad\textbf{(D)}\ \frac12\qquad\textbf{(E)}\ \frac{1}3$

Solution

Problem 13

The arithmetic mean (average) of a set of $50$ numbers is $38$. If two numbers, namely, $45$ and $55$, are discarded, the mean of the remaining set of numbers is: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$

Solution

Problem 14

Given the set $S$ 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 $S$ are: $\textbf{(A)}\ \text{all} \qquad\textbf{(B)}\ 1,2,3,4\qquad\textbf{(C)}\ 1,2,3,5\qquad\textbf{(D)}\ 1,2,3\qquad\textbf{(E)}\ 1,3,5$

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: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$

Solution

Problem 16

The expression$\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified is: $\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution

Problem 17

If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5 when$x=-5$, then$a+b$equals:$\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$positive integers is:$\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.)1$lb.,$3$lb., and$9$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?$\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.)x$varies directly as$y$and inversely as the square of$z$, and that$x=10$when$y=4$and$z=14$. Then, when$y=16$and$z=7$,$x$equals:$\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$is the perimeter of an equilateral$\triangle$inscribed in a circle, the area of the circle is:$\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.)3$. If the longer base is$97,$then the shorter base is:$\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$consists of$\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|$means$+a$if$a$is greater than or equal to zero, and$-a$if a is less than or equal to zero; the symbol$<$means "less than";  the symbol$>$means "greater than." The set of values$x$satisfying the inequality$|3-x|<4$consists of all$x$such that:$\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$. The medians to the leg intersect each other at right angles. The area of the triangle is:$\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=0$, where$i=\sqrt{-1}$&#036;textbf{(A)}\ \text{The sum of the roots is 2} \qquad \textbf{(B)}\ \text{The discriminant is 9}\qquad \textbf{(C)}\ \text{The roots are imaginary}\qquad \textbf{(D)}\ \text{The roots can be found using the quadratic formula}\qquad \textbf{(E)}\ \text{The roots can be found by factoring, using imaginary numbers}$

Solution

== Problem 28==$M are on BC and AB, respectively. The sides of$\triangle ABC$are$a,b,$and$c$. Then$\frac{\overbar{AM}}{\overbar{MB}}=k\frac{\overbar{CL}}{\overbar{LB}} $where$k$is:$\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.)n$questions a student answers correctly$15$of the first$20$. 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$can there be?$\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.)A$can run around a circular track in$40$seconds.$B$, running in the opposite direction, meets$A$every$15$seconds.  What is$B$'s time to run around the track, expressed in seconds?$\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$, is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius, is:$\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.)l$of a tangent, drawn from a point$A$to a circle, is$\frac43 $of the radius$r$. The (shortest) distance from A to the circle is:$\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_n$represent the sum of the first$n$terms of the harmonic progression; for example$S_3$represents the sum of  the first three terms. If the first three terms of a harmonic progression are$3,4,6$, then:$\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=0$be$r$and$s$. Then the expression$r^2+s^2 $is:$\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$means "greater than or equal to"; the symbol$\le$means "less than or equal to". In the eqeuation$(x-m)^2-(x-n)^2=(m-n)^2; m$is a fixed positive number, and$n$is a fixed negative number. The set of values x satisfying the equation is:$\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.)80$, and one side of the base angle is$60^\circ$. The sum of the lengths of the other two sides is$90$. The shortest side is:$\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)$becomes:$\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}=1$, then$x$:$\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.$Then S equals:$\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 ABC$,$BD$is a median.$CF$intersects$BD$at$E$so that$\overbar{BE}=\overbar{ED}$. Point$F$is on$AB$. Then, if$\overbar{BF}=5$,$\overbar{BA}$equals:$\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$inches is tangent to the line, the other  two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:$\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,$and$c$. Their greatest common divisor is$D$; their least common multiple is$m$.  Then, which two of the following statements are true?$\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.)25,39$, and$40$. The diameter of the circumscribed circle is:$\textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40$[[1959 AHSME Problems/Problem 43|Solution]]

== Problem 44== The roots of$ (Error compiling LaTeX. ! Missing $ inserted.)x^2+bx+c=0$are both real and greater than$1$. Let$s=b+c+1$. Then$s$:$\textbf{(A)}\ \text{may be less than zero}\qquad\textbf{(B)}\ \text{may be equal to zero}\qquad \textbf{(C)}\ \text{must be greater than zero}\qquad\textbf{(D)}\ \text{must be less than zero}\qquad \textbf{(E)}\text{ must be between -1 and +1}$[[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$, then$ y$equals:$\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.)d$days observed that (1) it rained$7$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$d$equals:$\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.}$Those which are logical consequences of I,II, and III are:$\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_n$, where$n$is a positive integer or zero, and$a_0$is a positive integer.  The remaining$a$'s are integers or zero. Set$h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$. [See example 25 for the meaning of$|x|$.]  The number of polynomials with$h=3$is:$\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}-\cdots$let$S$be the (limiting) sum. Then$S$equals:$\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 \\

Solution

See also

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