Difference between revisions of "1955 AHSME Problems"

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== Problem 1 ==
+
== Problem 1==
 +
Which one of the following is not equivalent to <math>0.000000375</math>?
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000} </math>  
  
 
[[1955 AHSME Problems/Problem 1|Solution]]
 
[[1955 AHSME Problems/Problem 1|Solution]]
 +
== Problem 2==
  
== Problem 2 ==
+
The smaller angle between the hands of a clock at <math>12:25</math> p.m. is:
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
<math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ </math>  
  
 
[[1955 AHSME Problems/Problem 2|Solution]]
 
[[1955 AHSME Problems/Problem 2|Solution]]
 +
== Problem 3==
  
== Problem 3 ==
+
If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the original ten numbers:
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math>
  
 
[[1955 AHSME Problems/Problem 3|Solution]]
 
[[1955 AHSME Problems/Problem 3|Solution]]
 +
== Problem 4==
 +
The equality <math>\frac{1}{x-1}=\frac{2}{x-2}</math> is satisfied by:
  
== Problem 4 ==
+
<math> \textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0 </math>
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
 
  
 
[[1955 AHSME Problems/Problem 4|Solution]]
 
[[1955 AHSME Problems/Problem 4|Solution]]
 +
== Problem 5==
  
== Problem 5 ==
+
<math>5y</math> varies inversely as the square of <math>x</math>. When <math>y=16, x=1</math>. When <math>x=8, y</math> equals:
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024   
  
 
[[1955 AHSME Problems/Problem 5|Solution]]
 
[[1955 AHSME Problems/Problem 5|Solution]]
 +
== Problem 6==
  
== Problem 6 ==
+
A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents} </math>
  
 
[[1955 AHSME Problems/Problem 6|Solution]]
 
[[1955 AHSME Problems/Problem 6|Solution]]
 +
== Problem 7==
  
== Problem 7 ==
+
If a worker receives a <math>20</math>% cut in wages, he may regain his original pay exactly by obtaining a raise of:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{20%}\qquad\textbf{(B)}\ \text{25%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{%}\qquad\textbf{(D)}\ &#036;20\qquad\textbf{(E)}\ &#036;25 </math>
  
 
[[1955 AHSME Problems/Problem 7|Solution]]
 
[[1955 AHSME Problems/Problem 7|Solution]]
  
== Problem 8 ==
+
== Problem 8==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The graph of <math>x^2-4y^2=0</math>:
 +
 
 +
<math> \textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist} </math>
  
 
[[1955 AHSME Problems/Problem 8|Solution]]
 
[[1955 AHSME Problems/Problem 8|Solution]]
 +
== Problem 9==
  
== Problem 9 ==
+
A circle is inscribed in a triangle with sides <math>8, 15</math>, and <math>17</math>. The radius of the circle is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ </math>
  
 
[[1955 AHSME Problems/Problem 9|Solution]]
 
[[1955 AHSME Problems/Problem 9|Solution]]
 +
== Problem 10==
  
== Problem 10 ==
+
How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40} </math>
  
 
[[1955 AHSME Problems/Problem 10|Solution]]
 
[[1955 AHSME Problems/Problem 10|Solution]]
 +
== Problem 11==
  
== Problem 11 ==
+
The negation of the statement "No slow learners attend this school" is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend this school}\\ \textbf{(E)}\ \text{No slow learners do not attend this school} </math>
  
 
[[1955 AHSME Problems/Problem 11|Solution]]
 
[[1955 AHSME Problems/Problem 11|Solution]]
 +
== Problem 12==
  
== Problem 12 ==
+
The solution of <math>\sqrt{5x-1}+\sqrt{x-1}=2</math> is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0 </math>
  
 
[[1955 AHSME Problems/Problem 12|Solution]]
 
[[1955 AHSME Problems/Problem 12|Solution]]
 +
== Problem 13==
  
== Problem 13 ==
+
The fraction <math>\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}</math> is equal to:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2 </math>
  
 
[[1955 AHSME Problems/Problem 13|Solution]]
 
[[1955 AHSME Problems/Problem 13|Solution]]
 +
== Problem 14==
  
== Problem 14 ==
+
The length of rectangle <math>R</math> is <math>10</math>% more than the side of square <math>S</math>. The width of the rectangle is <math>10</math>% less than the side of the square.
 +
The ratio of the areas, <math>R:S</math>, is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200 </math>  
  
 
[[1955 AHSME Problems/Problem 14|Solution]]
 
[[1955 AHSME Problems/Problem 14|Solution]]
 +
== Problem 15==
  
== Problem 15 ==
+
The ratio of the areas of two concentric circles is <math>1: 3</math>. If the radius of the smaller is <math>r</math>, then the difference between the
 +
radii is best approximated by:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math>\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r    </math>  
  
 
[[1955 AHSME Problems/Problem 15|Solution]]
 
[[1955 AHSME Problems/Problem 15|Solution]]
 +
== Problem 16==
  
== Problem 16 ==
+
The value of <math>\frac{3}{a+b}</math> when <math>a=4</math> and <math>b=-4</math> is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless} </math>
  
 
[[1955 AHSME Problems/Problem 16|Solution]]
 
[[1955 AHSME Problems/Problem 16|Solution]]
 +
== Problem 17==
  
== Problem 17 ==
+
If <math>\log x-5 \log 3=-2</math>, then <math>x</math> equals:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25} </math>
  
 
[[1955 AHSME Problems/Problem 17|Solution]]
 
[[1955 AHSME Problems/Problem 17|Solution]]
 +
== Problem 18==
  
== Problem 18 ==
+
The discriminant of the equation <math>x^2+2x\sqrt{3}+3=0</math> is zero. Hence, its roots are:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary} </math>
  
 
[[1955 AHSME Problems/Problem 18|Solution]]
 
[[1955 AHSME Problems/Problem 18|Solution]]
 +
== Problem 19==
  
== Problem 19 ==
+
Two numbers whose sum is <math>6</math> and the absolute value of whose difference is <math>8</math> are roots of the equation:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0 </math>
  
 
[[1955 AHSME Problems/Problem 19|Solution]]
 
[[1955 AHSME Problems/Problem 19|Solution]]
 +
== Problem 20==
  
== Problem 20 ==
+
The expression <math>\sqrt{25-t^2}+5</math> equals zero for:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5 </math>
  
 
[[1955 AHSME Problems/Problem 20|Solution]]
 
[[1955 AHSME Problems/Problem 20|Solution]]
 +
== Problem 21==
  
== Problem 21 ==
+
Represent the hypotenuse of a right triangle by <math>c</math> and the area by <math>A</math>. The altitude on the hypotenuse is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2} </math>
  
 
[[1955 AHSME Problems/Problem 21|Solution]]
 
[[1955 AHSME Problems/Problem 21|Solution]]
 +
== Problem 22==
  
== Problem 22 ==
+
On a \texdollar{10000} order a merchant has a choice between three successive discounts of <math>20</math>%, <math>20</math>%, and <math>10</math>% and
 +
three successive discounts of <math>40</math>%, <math>5</math>%, and <math>5</math>%. By choosing the better offer, he can save:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ &#036;440\qquad\textbf{(C)}\ &#036;330\qquad\textbf{(D)}\ &#036;345\qquad\textbf{(E)}\ &#036;360 </math>
  
 
[[1955 AHSME Problems/Problem 22|Solution]]
 
[[1955 AHSME Problems/Problem 22|Solution]]
 +
== Problem 23==
  
== Problem 23 ==
+
In checking the petty cash a clerk counts <math>q</math> quarters, <math>d</math> dimes, <math>n</math> nickels, and <math>c</math> cents.
 +
Later he discovers that <math>x</math> of the nickels were counted as quarters and <math>x</math> of the dimes were counted as cents.
 +
To correct the total obtained the clerk must:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents} </math>
  
 
[[1955 AHSME Problems/Problem 23|Solution]]
 
[[1955 AHSME Problems/Problem 23|Solution]]
 +
== Problem 24==
  
== Problem 24 ==
+
The function <math>4x^2-12x-1</math>:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10} </math>
  
 
[[1955 AHSME Problems/Problem 24|Solution]]
 
[[1955 AHSME Problems/Problem 24|Solution]]
 +
== Problem 25==
  
== Problem 25 ==
+
One of the factors of <math>x^4+2x^2+9</math> is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these} </math>
  
 
[[1955 AHSME Problems/Problem 25|Solution]]
 
[[1955 AHSME Problems/Problem 25|Solution]]
 +
== Problem 26==
  
== Problem 26 ==
+
Mr. A owns a house worth \textdollar{10000}. He sells it to Mr. <math>B</math> at <math>10</math>% profit. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a <math>10</math>% loss. Then:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes }&#036;100\qquad\textbf{(C)}\ \text{Mr. A makes }&#036;1000\\ \textbf{(D)}\ \text{Mr. B loses }&#036;100\qquad\textbf{(E)}\ \text{none of the above is correct} </math>
  
 
[[1955 AHSME Problems/Problem 26|Solution]]
 
[[1955 AHSME Problems/Problem 26|Solution]]
 +
== Problem 27==
  
== Problem 27 ==
+
If <math>r</math> and <math>s</math> are the roots of <math>x^2-px+q=0</math>, then <math>r^2+s^2</math> equals:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2 </math>
  
 
[[1955 AHSME Problems/Problem 27|Solution]]
 
[[1955 AHSME Problems/Problem 27|Solution]]
 +
== Problem 28==
  
== Problem 28 ==
+
On the same set of axes are drawn the graph of <math>y=ax^2+bx+c</math> and the graph of the equation obtained by replacing <math>x</math> by <math>-x</math> in the given equation.
 +
If <math>b \neq 0</math> and <math>c \neq 0</math> these two graphs intersect:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis} </math>
  
 
[[1955 AHSME Problems/Problem 28|Solution]]
 
[[1955 AHSME Problems/Problem 28|Solution]]
 +
== Problem 29==
  
== Problem 29 ==
+
In the figure, <math>PA</math> is tangent to semicircle <math>SAR</math>; <math>PB</math> is tangent to semicircle <math>RBT</math>; <math>SRT</math> is a straight line;
 +
the arcs are indicated in the figure. <math>\angle APB</math> is measured by:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<asy>
 +
unitsize(1.2cm);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
 +
dotfactor=3;
 +
pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0);
 +
pair A=O1+2*dir(60), B=O2+dir(85);
 +
pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2;
 +
pair P=extension(A,Pa,B,Pb);
 +
pair[] dots={Sp,R,T,A,B,P};
 +
draw(P--P+5*(A-P));
 +
draw(P--P+5*(B-P));
 +
clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle);
 +
draw(Arc(O1,2,0,180)--cycle);
 +
draw(Arc(O2,1,0,180)--cycle);
 +
dot(dots);
 +
label("$S$",Sp,S);
 +
label("$R$",R,S);
 +
label("$T$",T,S);
 +
label("$A$",A,NE);
 +
label("$B$",B,N);
 +
label("$P$",P,NNE);
 +
label("$a$",midpoint(Arc(O1,2,0,60)),SW);
 +
label("$b$",midpoint(Arc(O2,1,85,180)),SE);
 +
label("$c$",midpoint(Arc(O1,2,60,180)),SE);
 +
label("$d$",midpoint(Arc(O2,1,0,85)),SW);</asy>
 +
 
 +
<math> \textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b </math>
  
 
[[1955 AHSME Problems/Problem 29|Solution]]
 
[[1955 AHSME Problems/Problem 29|Solution]]
 +
== Problem 30==
  
== Problem 30 ==
+
Each of the equations <math>3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}</math> has:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root} </math>
  
 
[[1955 AHSME Problems/Problem 30|Solution]]
 
[[1955 AHSME Problems/Problem 30|Solution]]
 +
== Problem 31==
  
== Problem 31 ==
+
An equilateral triangle whose side is <math>2</math> is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides.
 +
If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2} </math>
  
 
[[1955 AHSME Problems/Problem 31|Solution]]
 
[[1955 AHSME Problems/Problem 31|Solution]]
 +
== Problem 32==
  
== Problem 32 ==
+
If the discriminant of <math>ax^2+2bx+c=0</math> is zero, then another true statement about <math>a, b</math>, and <math>c</math> is that:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \textbf{(E)}\ \text{only b is negative and a and c are positive} </math>
  
 
[[1955 AHSME Problems/Problem 32|Solution]]
 
[[1955 AHSME Problems/Problem 32|Solution]]
 +
== Problem 33==
  
== Problem 33 ==
+
Henry starts a trip when the hands of the clock are together between <math>8</math> a.m. and <math>9</math> a.m.
 +
He arrives at his destination between <math>2</math> p.m. and <math>3</math> p.m. when the hands of the clock are exactly <math>180^\circ</math> apart. The trip takes:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43-7/11 min.}\qquad\textbf{(C)}\ \text{5 hr. 16-4/11 min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these} </math>
  
 
[[1955 AHSME Problems/Problem 33|Solution]]
 
[[1955 AHSME Problems/Problem 33|Solution]]
 +
== Problem 34==
  
== Problem 34 ==
+
A <math>6</math>-inch and <math>18</math>-inch diameter pole are placed together and bound together with wire.
 +
The length of the shortest wire that will go around them is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi </math>
  
 
[[1955 AHSME Problems/Problem 34|Solution]]
 
[[1955 AHSME Problems/Problem 34|Solution]]
 +
== Problem 35==
  
== Problem 35 ==
+
Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles.
 +
The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy.
 +
The original number of marbles:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38} </math>
  
 
[[1955 AHSME Problems/Problem 35|Solution]]
 
[[1955 AHSME Problems/Problem 35|Solution]]
 +
== Problem 36==
  
== Problem 36 ==
+
A cylindrical oil tank, lying horizontally, has an interior length of <math>10</math> feet and an interior diameter of <math>6</math> feet.
 +
If the rectangular surface of the oil has an area of <math>40</math> square feet, the depth of the oil is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5} </math>
  
 
[[1955 AHSME Problems/Problem 36|Solution]]
 
[[1955 AHSME Problems/Problem 36|Solution]]
 +
== Problem 37==
  
== Problem 37 ==
+
A three-digit number has, from left to right, the digits <math>h, t</math>, and <math>u</math>, with <math>h>u</math>.
 +
When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of r.
 +
The next two digits, from right to left, are:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5} </math>
  
 
[[1955 AHSME Problems/Problem 37|Solution]]
 
[[1955 AHSME Problems/Problem 37|Solution]]
 +
== Problem 38==
  
== Problem 38 ==
+
Four positive integers are given. Select any three of these integers, find their arithmetic average,
 +
and add this result to the fourth integer. Thus the numbers <math>29, 23, 21</math>, and <math>17</math> are obtained. One of the original integers is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17   
  
 
[[1955 AHSME Problems/Problem 38|Solution]]
 
[[1955 AHSME Problems/Problem 38|Solution]]
 +
== Problem 39==
  
== Problem 39 ==
+
If <math>y=x^2+px+q</math>, then if the least possible value of <math>y</math> is zero <math>q</math> is equal to:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q </math>
  
 
[[1955 AHSME Problems/Problem 39|Solution]]
 
[[1955 AHSME Problems/Problem 39|Solution]]
 +
== Problem 40==
  
== Problem 40 ==
+
The fractions <math>\frac{ax+b}{cx+d}</math> and <math>\frac{b}{d}</math> are unequal if:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc </math>
  
 
[[1955 AHSME Problems/Problem 40|Solution]]
 
[[1955 AHSME Problems/Problem 40|Solution]]
 +
== Problem 41==
  
== Problem 41 ==
+
A train traveling from Aytown to Beetown meets with an accident after <math>1</math> hr. It is stopped for <math>\frac{1}{2}</math> hr.,
 +
after which it proceeds at four-fifths of its usual rate, arriving at Beetown <math>2</math> hr. late.
 +
If the train had covered <math>80</math> miles more before the accident, it would have been just <math>1</math> hr. late.
 +
The usual rate of the train is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{20 mph}\qquad\textbf{(B)}\ \text{30 mph}\qquad\textbf{(C)}\ \text{40 mph}\qquad\textbf{(D)}\ \text{50 mph}\qquad\textbf{(E)}\ \text{60 mph} </math>
  
 
[[1955 AHSME Problems/Problem 41|Solution]]
 
[[1955 AHSME Problems/Problem 41|Solution]]
 +
== Problem 42==
  
== Problem 42 ==
+
If <math>a, b</math>, and <math>c</math> are positive integers, the radicals <math>\sqrt{a+\frac{b}{c}}</math> and <math>a\sqrt{\frac{b}{c}}</math> are equal when and only when:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{2}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1 </math>
  
 
[[1955 AHSME Problems/Problem 42|Solution]]
 
[[1955 AHSME Problems/Problem 42|Solution]]
 +
== Problem 43==
  
== Problem 43 ==
+
The pairs of values of <math>x</math> and <math>y</math> that are the common solutions of the equations <math>y=(x+1)^2</math> and <math>xy+y=1</math> are:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and 2 imaginary pairs} </math>
  
 
[[1955 AHSME Problems/Problem 43|Solution]]
 
[[1955 AHSME Problems/Problem 43|Solution]]
 +
== Problem 44==
  
== Problem 44 ==
+
In circle <math>O</math> chord <math>AB</math> is produced so that <math>BC</math> equals a radius of the circle. <math>CO</math> is drawn and extended to <math>D</math>.
 +
<math>AO</math> is drawn. Which of the following expresses the relationship between <math>x</math> and <math>y</math>?
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ \qquad \textbf{(C) \ \qquad \textbf{(D) \ } \qquad \textbf{(E) \ } </math>
+
<asy>
 +
size(200);defaultpen(linewidth(0.7)+fontsize(10));
 +
pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);
 +
draw(O--A--C--D);
 +
dot(A^^B^^C^^D^^O);
 +
pair point=O;
 +
label("$A$", A, dir(point--A));
 +
label("$B$", B, dir(point--B));
 +
label("$C$", C, dir(point--C));
 +
label("$D$", D, dir(point--D));
 +
label("$O$", O, dir(285));
 +
label("$x$", O+0.1*dir(172.5), dir(172.5));
 +
label("$y$", C+0.4*dir(187.5), dir(187.5));
 +
draw(Circle(O,1));</asy>
 +
 
 +
 
 +
<math> \textbf{(A)}\ x=3y\\ \textbf{(B)}\ x=2y\\ \textbf{(C)}\ x=60^\circ\\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y\\ \textbf{(E)}\ x=2y\text{ or }x=3y\text{, depending upon the length of }AB </math>
  
 
[[1955 AHSME Problems/Problem 44|Solution]]
 
[[1955 AHSME Problems/Problem 44|Solution]]
 +
== Problem 45==
  
== Problem 45 ==
+
Given a geometric sequence with the first term <math>\neq 0</math> and <math>r \neq 0</math> and an arithmetic sequence with the first term <math>=0</math>.
 +
A third sequence <math>1,1,2\ldots</math> is formed by adding corresponding terms of the two given sequences.
 +
The sum of the first ten terms of the third sequence is:
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given} </math>
  
 
[[1955 AHSME Problems/Problem 45|Solution]]
 
[[1955 AHSME Problems/Problem 45|Solution]]
 +
== Problem 46==
  
== Problem 46 ==
+
The graphs of <math>2x+3y-6=0, 4x-3y-6=0, x=2</math>, and <math>y=\frac{2}{3}</math> intersect in:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points} </math>
  
 
[[1955 AHSME Problems/Problem 46|Solution]]
 
[[1955 AHSME Problems/Problem 46|Solution]]
 +
== Problem 47==
  
== Problem 47 ==
+
The expressions <math>a+bc</math> and <math>(a+b)(a+c)</math> are:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0 </math>
  
 
[[1955 AHSME Problems/Problem 47|Solution]]
 
[[1955 AHSME Problems/Problem 47|Solution]]
 +
== Problem 48==
  
== Problem 48 ==
+
Given <math>\triangle ABC</math> with medians <math>AE, BF, CD</math>; <math>FH</math> parallel and equal to <math>AE</math>; <math>BH and HE</math> are drawn; <math>FE</math> extended meets <math>BH</math> in <math>G</math>.
 +
Which one of the following statements is not necessarily correct?
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH </math>
  
 
[[1955 AHSME Problems/Problem 48|Solution]]
 
[[1955 AHSME Problems/Problem 48|Solution]]
 +
== Problem 49==
  
== Problem 49 ==
+
The graphs of <math>y=\frac{x^2-4}{x-2}</math> and <math>y=2x</math> intersect in:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points} </math>
  
 
[[1955 AHSME Problems/Problem 49|Solution]]
 
[[1955 AHSME Problems/Problem 49|Solution]]
 +
== Problem 50==
  
== Problem 50 ==
+
In order to pass <math>B</math> going <math>40</math> mph on a two-lane highway, <math>A</math>, going <math>50</math> mph, must gain <math>30</math> feet.
 +
Meantime, <math>C, 210</math> feet from <math>A</math>, is headed toward him at <math>50</math> mph. If <math>B</math> and <math>C</math> maintain their speeds,
 +
then, in order to pass safely, <math>A</math> must increase his speed by:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph} </math>
 +
   
 +
[[1955 AHSME Problems/Problem 50|Solution]]
  
[[1955 AHSME Problems/Problem 50|Solution]]
 
  
 
== See also ==
 
== See also ==
 
* [[AHSME]]
 
* [[AHSME]]
 
* [[AHSME Problems and Solutions]]
 
* [[AHSME Problems and Solutions]]
 +
* [[1950 AHSME]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:43, 13 October 2014

== Problem 1==

Which one of the following is not equivalent to $0.000000375$?

$\textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000}$

Solution

Problem 2

The smaller angle between the hands of a clock at $12:25$ p.m. is:

$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$

Solution

Problem 3

If each number in a set of ten numbers is increased by $20$, the arithmetic mean (average) of the original ten numbers:

$\textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2}$

Solution

Problem 4

The equality $\frac{1}{x-1}=\frac{2}{x-2}$ is satisfied by:

$\textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0$

Solution

Problem 5

$5y$ varies inversely as the square of $x$. When $y=16, x=1$. When $x=8, y$ equals:

\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024

Solution

Problem 6

A merchant buys a number of oranges at $3$ for $10$ cents and an equal number at $5$ for $20$ cents. To "break even" he must sell all at:

$\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}$

Solution

Problem 7

If a worker receives a $20$% cut in wages, he may regain his original pay exactly by obtaining a raise of:

$\textbf{(A)}\ \text{20%}\qquad\textbf{(B)}\ \text{25%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{%}\qquad\textbf{(D)}\ &#036;20\qquad\textbf{(E)}\ &#036;25$ (Error compiling LaTeX. ! File ended while scanning use of \text@.)

Solution

Problem 8

The graph of $x^2-4y^2=0$:

$\textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist}$

Solution

Problem 9

A circle is inscribed in a triangle with sides $8, 15$, and $17$. The radius of the circle is:

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$

Solution

Problem 10

How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?

$\textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40}$

Solution

Problem 11

The negation of the statement "No slow learners attend this school" is:

$\textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend this school}\\ \textbf{(E)}\ \text{No slow learners do not attend this school}$

Solution

Problem 12

The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:

$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$

Solution

Problem 13

The fraction $\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}$ is equal to:

$\textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2$

Solution

Problem 14

The length of rectangle $R$ is $10$% more than the side of square $S$. The width of the rectangle is $10$% less than the side of the square. The ratio of the areas, $R:S$, is:

$\textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200$

Solution

Problem 15

The ratio of the areas of two concentric circles is $1: 3$. If the radius of the smaller is $r$, then the difference between the radii is best approximated by:

$\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$

Solution

Problem 16

The value of $\frac{3}{a+b}$ when $a=4$ and $b=-4$ is:

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless}$

Solution

Problem 17

If $\log x-5 \log 3=-2$, then $x$ equals:

$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution

Problem 18

The discriminant of the equation $x^2+2x\sqrt{3}+3=0$ is zero. Hence, its roots are:

$\textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary}$

Solution

Problem 19

Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

$\textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0$

Solution

Problem 20

The expression $\sqrt{25-t^2}+5$ equals zero for:

$\textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5$

Solution

Problem 21

Represent the hypotenuse of a right triangle by $c$ and the area by $A$. The altitude on the hypotenuse is:

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

Solution

Problem 22

On a \texdollar{10000} order a merchant has a choice between three successive discounts of $20$%, $20$%, and $10$% and three successive discounts of $40$%, $5$%, and $5$%. By choosing the better offer, he can save:

$\textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ &#036;440\qquad\textbf{(C)}\ &#036;330\qquad\textbf{(D)}\ &#036;345\qquad\textbf{(E)}\ &#036;360$

Solution

Problem 23

In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:

$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents}$

Solution

Problem 24

The function $4x^2-12x-1$:

$\textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10}$

Solution

Problem 25

One of the factors of $x^4+2x^2+9$ is:

$\textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 26

Mr. A owns a house worth \textdollar{10000}. He sells it to Mr. $B$ at $10$% profit. Mr. $B$ sells the house back to Mr. $A$ at a $10$% loss. Then:

$\textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes }&#036;100\qquad\textbf{(C)}\ \text{Mr. A makes }&#036;1000\\ \textbf{(D)}\ \text{Mr. B loses }&#036;100\qquad\textbf{(E)}\ \text{none of the above is correct}$ (Error compiling LaTeX. ! Missing $ inserted.)

Solution

Problem 27

If $r$ and $s$ are the roots of $x^2-px+q=0$, then $r^2+s^2$ equals:

$\textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2$

Solution

Problem 28

On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation. If $b \neq 0$ and $c \neq 0$ these two graphs intersect:

$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis}$

Solution

Problem 29

In the figure, $PA$ is tangent to semicircle $SAR$; $PB$ is tangent to semicircle $RBT$; $SRT$ is a straight line; the arcs are indicated in the figure. $\angle APB$ is measured by:

[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); pair A=O1+2*dir(60), B=O2+dir(85); pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2; pair P=extension(A,Pa,B,Pb); pair[] dots={Sp,R,T,A,B,P}; draw(P--P+5*(A-P)); draw(P--P+5*(B-P)); clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle); draw(Arc(O1,2,0,180)--cycle); draw(Arc(O2,1,0,180)--cycle); dot(dots); label("$S$",Sp,S); label("$R$",R,S); label("$T$",T,S); label("$A$",A,NE); label("$B$",B,N); label("$P$",P,NNE); label("$a$",midpoint(Arc(O1,2,0,60)),SW); label("$b$",midpoint(Arc(O2,1,85,180)),SE); label("$c$",midpoint(Arc(O1,2,60,180)),SE); label("$d$",midpoint(Arc(O2,1,0,85)),SW);[/asy]

$\textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b$

Solution

Problem 30

Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:

$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$

Solution

Problem 31

An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$\textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2}$

Solution

Problem 32

If the discriminant of $ax^2+2bx+c=0$ is zero, then another true statement about $a, b$, and $c$ is that:

$\textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \textbf{(E)}\ \text{only b is negative and a and c are positive}$

Solution

Problem 33

Henry starts a trip when the hands of the clock are together between $8$ a.m. and $9$ a.m. He arrives at his destination between $2$ p.m. and $3$ p.m. when the hands of the clock are exactly $180^\circ$ apart. The trip takes:

$\textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43-7/11 min.}\qquad\textbf{(C)}\ \text{5 hr. 16-4/11 min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 34

A $6$-inch and $18$-inch diameter pole are placed together and bound together with wire. The length of the shortest wire that will go around them is:

$\textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi$

Solution

Problem 35

Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

$\textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38}$

Solution

Problem 36

A cylindrical oil tank, lying horizontally, has an interior length of $10$ feet and an interior diameter of $6$ feet. If the rectangular surface of the oil has an area of $40$ square feet, the depth of the oil is:

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5}$

Solution

Problem 37

A three-digit number has, from left to right, the digits $h, t$, and $u$, with $h>u$. When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of r. The next two digits, from right to left, are:

$\textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5}$

Solution

Problem 38

Four positive integers are given. Select any three of these integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers $29, 23, 21$, and $17$ are obtained. One of the original integers is:

\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17

Solution

Problem 39

If $y=x^2+px+q$, then if the least possible value of $y$ is zero $q$ is equal to:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q$

Solution

Problem 40

The fractions $\frac{ax+b}{cx+d}$ and $\frac{b}{d}$ are unequal if:

$\textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc$

Solution

Problem 41

A train traveling from Aytown to Beetown meets with an accident after $1$ hr. It is stopped for $\frac{1}{2}$ hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown $2$ hr. late. If the train had covered $80$ miles more before the accident, it would have been just $1$ hr. late. The usual rate of the train is:

$\textbf{(A)}\ \text{20 mph}\qquad\textbf{(B)}\ \text{30 mph}\qquad\textbf{(C)}\ \text{40 mph}\qquad\textbf{(D)}\ \text{50 mph}\qquad\textbf{(E)}\ \text{60 mph}$

Solution

Problem 42

If $a, b$, and $c$ are positive integers, the radicals $\sqrt{a+\frac{b}{c}}$ and $a\sqrt{\frac{b}{c}}$ are equal when and only when:

$\textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{2}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1$

Solution

Problem 43

The pairs of values of $x$ and $y$ that are the common solutions of the equations $y=(x+1)^2$ and $xy+y=1$ are:

$\textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and 2 imaginary pairs}$

Solution

Problem 44

In circle $O$ chord $AB$ is produced so that $BC$ equals a radius of the circle. $CO$ is drawn and extended to $D$. $AO$ is drawn. Which of the following expresses the relationship between $x$ and $y$?

[asy] size(200);defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0); draw(O--A--C--D); dot(A^^B^^C^^D^^O); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(285)); label("$x$", O+0.1*dir(172.5), dir(172.5)); label("$y$", C+0.4*dir(187.5), dir(187.5)); draw(Circle(O,1));[/asy]


$\textbf{(A)}\ x=3y\\ \textbf{(B)}\ x=2y\\ \textbf{(C)}\ x=60^\circ\\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y\\ \textbf{(E)}\ x=2y\text{ or }x=3y\text{, depending upon the length of }AB$

Solution

Problem 45

Given a geometric sequence with the first term $\neq 0$ and $r \neq 0$ and an arithmetic sequence with the first term $=0$. A third sequence $1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

$\textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given}$

Solution

Problem 46

The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$, and $y=\frac{2}{3}$ intersect in:

$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$

Solution

Problem 47

The expressions $a+bc$ and $(a+b)(a+c)$ are:

$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$

Solution

Problem 48

Given $\triangle ABC$ with medians $AE, BF, CD$; $FH$ parallel and equal to $AE$; $BH and HE$ are drawn; $FE$ extended meets $BH$ in $G$. Which one of the following statements is not necessarily correct?

$\textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH$

Solution

Problem 49

The graphs of $y=\frac{x^2-4}{x-2}$ and $y=2x$ intersect in:

$\textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points}$

Solution

Problem 50

In order to pass $B$ going $40$ mph on a two-lane highway, $A$, going $50$ mph, must gain $30$ feet. Meantime, $C, 210$ feet from $A$, is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:

$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$

Solution


See also

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