Difference between revisions of "1954 AHSME Problems"

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== Problem 1 ==
+
== Problem 1==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The square of <math>5-\sqrt{y^2-25}</math> is:
  
 +
<math>\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25}  </math>
 +
 
 
[[1954 AHSME Problems/Problem 1|Solution]]
 
[[1954 AHSME Problems/Problem 1|Solution]]
  
== Problem 2 ==
+
== Problem 2==
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
 
  
 +
The equation <math>\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0</math> can be transformed by eliminating fractions to the equation <math>x^2-5x+4=0</math>.
 +
The roots of the latter equation are <math>4</math> and <math>1</math>. Then the roots of the first equation are:
 +
 +
<math>\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root}  </math>
 +
 
 
[[1954 AHSME Problems/Problem 2|Solution]]
 
[[1954 AHSME Problems/Problem 2|Solution]]
  
== Problem 3 ==
+
== Problem 3==
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
If <math>x</math> varies as the cube of <math>y</math>, and <math>y</math> varies as the fifth root of <math>z</math>, then <math>x</math> varies as the nth power of <math>z</math>, where n is:
  
 +
<math>\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 3|Solution]]
 
[[1954 AHSME Problems/Problem 3|Solution]]
  
== Problem 4 ==
+
== Problem 4==
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
If the Highest Common Divisor of <math>6432</math> and <math>132</math> is diminished by <math>8</math>, it will equal:
  
 +
<math>\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4    </math>
 +
 
 
[[1954 AHSME Problems/Problem 4|Solution]]
 
[[1954 AHSME Problems/Problem 4|Solution]]
  
== Problem 5 ==
+
== Problem 5==
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
 
 +
A regular hexagon is inscribed in a circle of radius <math>10</math> inches. Its area is:
  
 +
<math>\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.}  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 5|Solution]]
 
[[1954 AHSME Problems/Problem 5|Solution]]
  
== Problem 6 ==
+
== Problem 6==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is:
  
 +
<math>\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16}  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 6|Solution]]
 
[[1954 AHSME Problems/Problem 6|Solution]]
  
== Problem 7 ==
+
== Problem 7==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A housewife saved <math>\textdollar{2.50} in buying a dress on sale. If she spent </math>\textdollar{25} for the dress, she saved about:
  
 +
<math>\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\% </math> 
 +
 
 
[[1954 AHSME Problems/Problem 7|Solution]]
 
[[1954 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 base of a triangle is twice as long as a side of a square and their areas are the same.
 +
Then the ratio of the altitude of the triangle to the side of the square is:
  
 +
<math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4  </math>
 +
 
 
[[1954 AHSME Problems/Problem 8|Solution]]
 
[[1954 AHSME Problems/Problem 8|Solution]]
  
== Problem 9 ==
+
== Problem 9==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A point <math>P</math> is outside a circle and is <math>13</math> inches from the center. A secant from <math>P</math> cuts the circle at <math>Q</math> and <math>R</math>
 +
so that the external segment of the secant <math>PQ</math> is <math>9</math> inches and <math>QR</math> is <math>7</math> inches. The radius of the circle is:
  
 +
<math>\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7" </math>
 +
 
 
[[1954 AHSME Problems/Problem 9|Solution]]
 
[[1954 AHSME Problems/Problem 9|Solution]]
  
== Problem 10 ==
+
== Problem 10==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The sum of the numerical coefficients in the expansion of the binomial <math>(a+b)^8</math> is:
  
 +
<math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7 </math>   
 +
 
 
[[1954 AHSME Problems/Problem 10|Solution]]
 
[[1954 AHSME Problems/Problem 10|Solution]]
  
== Problem 11 ==
+
== Problem 11==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A merchant placed on display some dresses, each with a marked price. He then posted a sign “<math>\frac{1}{3}</math> off on these dresses.”
 +
The cost of the dresses was <math>\frac{3}{4}</math> of the price at which he actually sold them. Then the ratio of the cost to the marked price was:
  
 +
<math>\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}  </math>
 +
 
 
[[1954 AHSME Problems/Problem 11|Solution]]
 
[[1954 AHSME Problems/Problem 11|Solution]]
  
== Problem 12 ==
+
== Problem 12==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The solution of the equations <math>\begin{align*}2x-3y&=7 \ 4x-6y &=20</math> is:
  
 +
<math>\textbf{(A)}\ x=18, y=12 \qquad \textbf{(B)}\ x=0, y=0 \qquad \textbf{(C)}\ \text{There is no solution} \ \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad\textbf{(E)}\ x=8, y=5 </math> 
 +
 
 
[[1954 AHSME Problems/Problem 12|Solution]]
 
[[1954 AHSME Problems/Problem 12|Solution]]
  
== Problem 13 ==
+
== Problem 13==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off
 +
by the sides of the quadrilateral, their sum will be:
  
 +
<math>\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 13|Solution]]
 
[[1954 AHSME Problems/Problem 13|Solution]]
  
== Problem 14 ==
+
== Problem 14==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
When simplified <math>\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}</math> equals:
  
 +
<math>\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2}  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 14|Solution]]
 
[[1954 AHSME Problems/Problem 14|Solution]]
  
== Problem 15 ==
+
== Problem 15==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math>\log 125</math> equals:
  
 +
<math>\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25  \
 +
\textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)    </math>
 +
 
 
[[1954 AHSME Problems/Problem 15|Solution]]
 
[[1954 AHSME Problems/Problem 15|Solution]]
  
== Problem 16 ==
+
== Problem 16==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
If <math>f(x) = 5x^2 - 2x - 1</math>, then <math>f(x + h) - f(x)</math> equals:
  
 +
<math>\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h </math> 
 +
 
 
[[1954 AHSME Problems/Problem 16|Solution]]
 
[[1954 AHSME Problems/Problem 16|Solution]]
  
== Problem 17 ==
+
== Problem 17==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The graph of the function <math>f(x) = 2x^3 - 7</math> goes:
  
 +
<math>\textbf{(A)}\ \text{up to the right and down to the left} \ \textbf{(B)}\ \text{down to the right and up to the left}\ \textbf{(C)}\ \text{up to the right and up to the left}\ \textbf{(D)}\ \text{down to the right and down to the left}\ \textbf{(E)}\ \text{none of these ways.}    </math>
 +
 
 
[[1954 AHSME Problems/Problem 17|Solution]]
 
[[1954 AHSME Problems/Problem 17|Solution]]
  
== Problem 18 ==
+
== Problem 18==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
Of the following sets, the one that includes all values of <math>x</math> which will satisfy <math>2x - 3 > 7 - x</math> is:
  
 +
<math>\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0 </math> 
 +
 
 
[[1954 AHSME Problems/Problem 18|Solution]]
 
[[1954 AHSME Problems/Problem 18|Solution]]
  
== Problem 19 ==
+
== Problem 19==
 +
 
 +
If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other} </math>
  
 
[[1954 AHSME Problems/Problem 19|Solution]]
 
[[1954 AHSME Problems/Problem 19|Solution]]
  
== Problem 20 ==
+
== Problem 20==
 +
 
 +
The equation <math>x^3+6x^2+11x+6=0</math> has:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math>
  
 
[[1954 AHSME Problems/Problem 20|Solution]]
 
[[1954 AHSME Problems/Problem 20|Solution]]
  
== Problem 21 ==
+
== Problem 21==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The roots of the equation <math>2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5</math> can be found by solving:
  
 +
<math> \textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0 </math>
 +
 
[[1954 AHSME Problems/Problem 21|Solution]]
 
[[1954 AHSME Problems/Problem 21|Solution]]
  
== Problem 22 ==
+
== Problem 22==
 +
 
 +
The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>,
 +
since division by zero is not allowed. For other values of <math>x</math>:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.}\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.} </math>
  
 
[[1954 AHSME Problems/Problem 22|Solution]]
 
[[1954 AHSME Problems/Problem 22|Solution]]
  
== Problem 23 ==
+
== Problem 23==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
If the margin made on an article costing <math>C</math> dollars and selling for <math>S</math> dollars is <math>M=\frac{1}{n}C</math>, then the margin is given by:
 +
 
 +
<math> \textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S </math>
  
 
[[1954 AHSME Problems/Problem 23|Solution]]
 
[[1954 AHSME Problems/Problem 23|Solution]]
  
== Problem 24 ==
+
== Problem 24==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The values of <math>k</math> for which the equation <math>2x^2-kx+x+8=0</math> will have real and equal roots are:
 +
 
 +
<math> \textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9} </math>
  
 
[[1954 AHSME Problems/Problem 24|Solution]]
 
[[1954 AHSME Problems/Problem 24|Solution]]
  
== Problem 25 ==
+
== Problem 25==
 +
 
 +
The two roots of the equation <math>a(b-c)x^2+b(c-a)x+c(a-b)=0</math> are <math>1</math> and:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)} </math>
  
 
[[1954 AHSME Problems/Problem 25|Solution]]
 
[[1954 AHSME Problems/Problem 25|Solution]]
  
== Problem 26 ==
+
== Problem 26==
 +
 
 +
The straight line <math>\overline{AB}</math> is divided at <math>C</math> so that <math>AC=3CB</math>. Circles are described on <math>\overline{AC}</math>
 +
and <math>\overline{CB}</math> as diameters and a common tangent meets <math>AB</math> produced at <math>D</math>. Then <math>BD</math> equals:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii} </math>
  
 
[[1954 AHSME Problems/Problem 26|Solution]]
 
[[1954 AHSME Problems/Problem 26|Solution]]
  
== Problem 27 ==
+
== Problem 27==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A right circular cone has for its base a circle having the same radius as a given sphere.
 +
The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
 +
 +
<math> \textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{2}{1}\qquad\textbf{(E)}\ \sqrt{\frac{5}{4}} </math>
  
 
[[1954 AHSME Problems/Problem 27|Solution]]
 
[[1954 AHSME Problems/Problem 27|Solution]]
  
== Problem 28 ==
+
== Problem 28==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
If <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, the value of <math>\frac{3mr-nt}{4nt-7mr}</math> is:
 +
 
 +
<math> \textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3} </math>
  
 
[[1954 AHSME Problems/Problem 28|Solution]]
 
[[1954 AHSME Problems/Problem 28|Solution]]
  
== Problem 29 ==
+
== Problem 29==
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
If the ratio of the legs of a right triangle is <math>1: 2</math>, then the ratio of the corresponding segments of
 +
the hypotenuse made by a perpendicular upon it from the vertex is:
 +
 
 +
<math> \textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5 </math>
  
 
[[1954 AHSME Problems/Problem 29|Solution]]
 
[[1954 AHSME Problems/Problem 29|Solution]]
  
== Problem 30 ==
+
== Problem 30==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math>A</math> and <math>B</math> together can do a job in <math>2</math> days; <math>B</math> and <math>C</math> can do it in four days; and <math>A</math> and <math>C</math> in <math>2\frac{2}{5}</math> days.
 +
The number of days required for A to do the job alone is:
  
 +
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 30|Solution]]
 
[[1954 AHSME Problems/Problem 30|Solution]]
  
== Problem 31 ==
+
== Problem 31==
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
In <math>\triangle ABC</math>, <math>AB=AC</math>, <math>\angle A=40^\circ</math>. Point <math>O</math> is within the triangle with <math>\angle OBC \cong \angle OCA</math>.
 +
The number of degrees in <math>\angle BOC</math> is:
  
 +
<math>\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}</math>   
 +
 
 
[[1954 AHSME Problems/Problem 31|Solution]]
 
[[1954 AHSME Problems/Problem 31|Solution]]
  
== Problem 32 ==
+
== Problem 32==
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The factors of <math>x^4+64</math> are:
 +
 
 +
<math> \textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8) </math>
  
 
[[1954 AHSME Problems/Problem 32|Solution]]
 
[[1954 AHSME Problems/Problem 32|Solution]]
  
== Problem 33 ==
+
== Problem 33==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A bank charges <math>\textdollar{6}</math> for a loan of <math>\textdollar{120}</math>. The borrower receives <math>\textdollar{114}</math> and
 +
repays the loan in <math>12</math> easy installments of <math>\textdollar{10}</math> a month. The interest rate is approximately:
  
 +
<math>\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%    </math>
 +
 
 
[[1954 AHSME Problems/Problem 33|Solution]]
 
[[1954 AHSME Problems/Problem 33|Solution]]
  
== Problem 34 ==
+
== Problem 34==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The fraction <math>\frac{1}{3}</math>:
 +
 
 +
<math> \textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9} </math>
  
 
[[1954 AHSME Problems/Problem 34|Solution]]
 
[[1954 AHSME Problems/Problem 34|Solution]]
  
== Problem 35 ==
+
== Problem 35==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
In the right triangle shown the sum of the distances <math>BM</math> and <math>MA</math> is equal to the sum of the distances <math>BC</math> and <math>CA</math>.
 +
If <math>MB = x, CB = h</math>, and <math>CA = d</math>, then <math>x</math> equals:
 +
 
 +
<asy>
 +
defaultpen(linewidth(.8pt)+fontsize(10pt));
 +
dotfactor=4;
 +
draw((0,0)--(8,0)--(0,5)--cycle);
 +
label("C",(0,0),SW);
 +
label("A",(8,0),SE);
 +
label("M",(0,5),N);
 +
dot((0,3.5));
 +
label("B",(0,3.5),W);
 +
label("$x$",(0,4.25),W);
 +
label("$h$",(0,1),W);
 +
label("$d$",(4,0),S);</asy>
 +
 
 +
<math> \textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h </math>
  
 
[[1954 AHSME Problems/Problem 35|Solution]]
 
[[1954 AHSME Problems/Problem 35|Solution]]
  
== Problem 36 ==
+
== Problem 36==
 +
 
 +
A boat has a speed of <math>15</math> mph in still water. In a stream that has a current of <math>5</math> mph it travels a certain
 +
distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8} </math>
  
 
[[1954 AHSME Problems/Problem 36|Solution]]
 
[[1954 AHSME Problems/Problem 36|Solution]]
  
== Problem 37 ==
+
== Problem 37==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
Given <math>\triangle PQR</math> with <math>\overline{RS}</math> bisecting <math>\angle R</math>, <math>PQ</math> extended to <math>D</math> and <math>\angle n</math> a right angle, then:
  
 +
<asy>
 +
path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)
 +
{
 +
pair M,N;
 +
path mark;
 +
M=t*0.03*unit(A-B)+B;
 +
N=t*0.03*unit(C-B)+B;
 +
if(flip)
 +
mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));
 +
else
 +
mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));
 +
return mark;
 +
}
 +
unitsize(1.5cm);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
 +
pair P=(0,0), R=(3,2), Q=(4,0);
 +
pair S0=bisectorpoint(P,R,Q);
 +
pair Sp=extension(P,Q,S0,R);
 +
pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);
 +
pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);
 +
draw(P--R--Q);
 +
draw(R--Sp);
 +
draw(P--D--M);
 +
draw(anglemark2(Sp,P,R,17));
 +
label("$p$",P+(0.35,0.1));
 +
draw(anglemark2(R,Q,P,11));
 +
label("$q$",Q+(-0.17,0.1));
 +
draw(anglemark2(R,Np,D,8,true));
 +
label("$n$",Np+(+0.12,0.07));
 +
draw(anglemark2(R,M,D,13,true));
 +
label("$m$",M+(+0.25,0.03));
 +
draw(anglemark2(M,D,P,29));
 +
label("$d$",D+(-0.75,0.095));
 +
pen f=fontsize(10pt);
 +
label("$R$",R,N,f);
 +
label("$P$",P,S,f);
 +
label("$S$",Sp,S,f);
 +
label("$Q$",Q,S,f);
 +
label("$D$",D,S,f);</asy>
 +
 +
<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) </math>
 +
<math> \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad\textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad\textbf{(E)}\ \text{none of these is correct} </math>
 
[[1954 AHSME Problems/Problem 37|Solution]]
 
[[1954 AHSME Problems/Problem 37|Solution]]
  
== Problem 38 ==
+
== Problem 38==
  
If <math>\log 2 = .3010</math> and <math>\log 3 = .4771</math>, the value of <math>x</math> when <math>3^{x+3} = 135</math> is approximately
+
If <math>\log 2=.3010</math> and <math>\log 3=.4771</math>, the value of <math>x</math> when <math>3^{x+3}=135</math> is approximately:
  
<math> \textbf{(A) \ } 5 \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63</math>
+
<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63   </math>  
 +
 
 +
[[1954 AHSME Problems/Problem 38|Solution]]
  
[[1954 AHSME Problems/Problem 38|Solution]]
+
== Problem 39==
  
== Problem 39 ==
+
The locus of the midpoint of a line segment that is drawn from a given external point <math>P</math> to a given circle with center <math>O</math> and radius <math>r</math>, is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r </math>
  
 
[[1954 AHSME Problems/Problem 39|Solution]]
 
[[1954 AHSME Problems/Problem 39|Solution]]
  
== Problem 40 ==
+
== Problem 40==
 +
 
 +
If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} </math>
  
 
[[1954 AHSME Problems/Problem 40|Solution]]
 
[[1954 AHSME Problems/Problem 40|Solution]]
  
== Problem 41 ==
+
== Problem 41==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The sum of all the roots of <math>4x^3-8x^2-63x-9=0</math> is:
  
 +
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0  </math> 
 +
 
 
[[1954 AHSME Problems/Problem 41|Solution]]
 
[[1954 AHSME Problems/Problem 41|Solution]]
  
== Problem 42 ==
+
== Problem 42==
 +
 
 +
Consider the graphs of <cmath>(1): y=x^2-\frac{1}{2}x+2</cmath> and <cmath>(2) y=x^2+\frac{1}{2}x+2</cmath> on the same set of axis.
 +
These parabolas are exactly the same shape. Then:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math>
  
 
[[1954 AHSME Problems/Problem 42|Solution]]
 
[[1954 AHSME Problems/Problem 42|Solution]]
  
== Problem 43 ==
+
== Problem 43==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The hypotenuse of a right triangle is <math>10</math> inches and the radius of the inscribed circle is <math>1</math> inch. The perimeter of the triangle in inches is:
  
 +
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30 </math>   
 +
 
 
[[1954 AHSME Problems/Problem 43|Solution]]
 
[[1954 AHSME Problems/Problem 43|Solution]]
  
== Problem 44 ==
+
== Problem 44==
  
<math> \textbf{(A) \ }  \qquad \textbf{(B) \ }  \qquad \textbf{(C) \ }  \qquad \textbf{(D) \ }  \qquad \textbf{(E) \ } </math>
+
A man born in the first half of the nineteenth century was <math>x</math> years old in the year <math>x^2</math>. He was born in:
  
 +
<math>\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806 </math>   
 +
 
 
[[1954 AHSME Problems/Problem 44|Solution]]
 
[[1954 AHSME Problems/Problem 44|Solution]]
  
== Problem 45 ==
+
== Problem 45==
 +
 
 +
In a rhombus, <math>ABCD</math>, line segments are drawn within the rhombus, parallel to diagonal <math>BD</math>,
 +
and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment
 +
as a function of its distance from vertex <math>A</math>. The graph is:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<math> \textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.} </math>
  
 
[[1954 AHSME Problems/Problem 45|Solution]]
 
[[1954 AHSME Problems/Problem 45|Solution]]
  
== Problem 46 ==
+
== Problem 46==
 +
 
 +
In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> equals:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
<asy>
 +
unitsize(5cm);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
 +
dotfactor=3;
 +
pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);
 +
pair O=(0,3/8);
 +
draw((-2/3,9/16)--(2/3,9/16));
 +
draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));
 +
draw(Circle(O,3/16));
 +
draw((-2/3,0)--(2/3,0));
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,N);
 +
label("$\frac{3}{8}$",O);
 +
draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));
 +
draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));
 +
label("$\frac{1}{2}$",(.5,.25));
 +
draw((.5,.33)--(.5,.5),EndArrow(3));
 +
draw((.5,.17)--(.5,0),EndArrow(3));
 +
label("$x$",midpoint((.5,.5)--(.5,9/16)));
 +
draw((.5,5/8)--(.5,9/16),EndArrow(3));
 +
label("$60^{\circ}$",(0.01,0.12));
 +
dot(A);
 +
dot(B);
 +
dot(C);</asy>
 +
 
 +
<math> \textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}" </math>
  
 
[[1954 AHSME Problems/Problem 46|Solution]]
 
[[1954 AHSME Problems/Problem 46|Solution]]
  
== Problem 47 ==
+
== Problem 47==
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
At the midpoint of line segment <math>AB</math> which is <math>p</math> units long, a perpendicular <math>MR</math> is erected with length <math>q</math> units.
 +
An arc is described from <math>R</math> with a radius equal to <math>\frac{1}{2}AB</math>, meeting <math>AB</math> at <math>T</math>. Then <math>AT</math> and <math>TB</math> are the roots of:
 +
 
 +
<math> \textbf{(A)}\ x^2+px+q^2=0\ \textbf{(B)}\ x^2-px+q^2=0\ \textbf{(C)}\ x^2+px-q^2=0\ \textbf{(D)}\ x^2-px-q^2=0\ \textbf{(E)}\ x^2-px+q=0 </math>
  
 
[[1954 AHSME Problems/Problem 47|Solution]]
 
[[1954 AHSME Problems/Problem 47|Solution]]
  
== Problem 48 ==
+
== Problem 48==
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
A train, an hour after starting, meets with an accident which detains it a half hour, after which it
 +
proceeds at <math>\frac{3}{4}</math> of its former rate and arrives <math>3\tfrac{1}{2}</math> hours late.
 +
Had the accident happened <math>90</math> miles farther along the line, it would have arrived only <math>3</math> hours late. The length of the trip in miles was:
  
 +
<math>\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550    </math>
 +
 
 
[[1954 AHSME Problems/Problem 48|Solution]]
 
[[1954 AHSME Problems/Problem 48|Solution]]
  
== Problem 49 ==
+
== Problem 49==
  
<math> \textbf{(A) \ \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
The difference of the squares of two odd numbers is always divisible by <math>8</math>. If <math>a>b</math>, and <math>2a+1</math> and <math>2b+1</math> are the odd numbers,
 +
to prove the given statement we put the difference of the squares in the form:
 +
 
 +
<math> \textbf{(A)}\ (2a+1)^2-(2b+1)^2\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\ \textbf{(D)}\ 4(a-b)(a+b+1)\ \textbf{(E)}\ 4(a^2+a-b^2-b) </math>
  
 
[[1954 AHSME Problems/Problem 49|Solution]]
 
[[1954 AHSME Problems/Problem 49|Solution]]
  
== Problem 50 ==
+
== Problem 50==
 +
 
 +
The times between <math>7</math> and <math>8</math> o'clock, correct to the nearest minute, when the hands of a clock will form an angle of <math>84^{\circ} are:
  
<math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+
</math> \textbf{(A)}\ \text{7: 23 and 7: 53}\qquad\textbf{(B)}\ \text{7: 20 and 7: 50}\qquad\textbf{(C)}\ \text{7: 22 and 7: 53}\\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad\textbf{(E)}\ \text{7: 21 and 7: 49} $
  
 
[[1954 AHSME Problems/Problem 50|Solution]]
 
[[1954 AHSME Problems/Problem 50|Solution]]

Revision as of 14:16, 14 October 2014

Problem 1

The square of $5-\sqrt{y^2-25}$ is:

$\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25}$

Solution

Problem 2

The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$. The roots of the latter equation are $4$ and $1$. Then the roots of the first equation are:

$\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root}$

Solution

Problem 3

If $x$ varies as the cube of $y$, and $y$ varies as the fifth root of $z$, then $x$ varies as the nth power of $z$, where n is:

$\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8$

Solution

Problem 4

If the Highest Common Divisor of $6432$ and $132$ is diminished by $8$, it will equal:

$\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

Problem 5

A regular hexagon is inscribed in a circle of radius $10$ inches. Its area is:

$\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.}$

Solution

Problem 6

The value of $\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}$ is:

$\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16}$

Solution

Problem 7

A housewife saved $\textdollar{2.50} in buying a dress on sale. If she spent$\textdollar{25} for the dress, she saved about:

$\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\%$

Solution

Problem 8

The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:

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

Solution

Problem 9

A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:

$\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7"$

Solution

Problem 10

The sum of the numerical coefficients in the expansion of the binomial $(a+b)^8$ is:

$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7$

Solution

Problem 11

A merchant placed on display some dresses, each with a marked price. He then posted a sign “$\frac{1}{3}$ off on these dresses.” The cost of the dresses was $\frac{3}{4}$ of the price at which he actually sold them. Then the ratio of the cost to the marked price was:

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

Solution

Problem 12

The solution of the equations $\begin{align*}2x-3y&=7 \ 4x-6y &=20$ (Error compiling LaTeX. Unknown error_msg) is:

$\textbf{(A)}\ x=18, y=12 \qquad \textbf{(B)}\ x=0, y=0 \qquad \textbf{(C)}\ \text{There is no solution} \\ \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad\textbf{(E)}\ x=8, y=5$

Solution

Problem 13

A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, their sum will be:

$\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ$

Solution

Problem 14

When simplified $\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}$ equals:

$\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2}$

Solution

Problem 15

$\log 125$ equals:

$\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25  \\ \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$

Solution

Problem 16

If $f(x) = 5x^2 - 2x - 1$, then $f(x + h) - f(x)$ equals:

$\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h$

Solution

Problem 17

The graph of the function $f(x) = 2x^3 - 7$ goes:

$\textbf{(A)}\ \text{up to the right and down to the left} \\ \textbf{(B)}\ \text{down to the right and up to the left}\\ \textbf{(C)}\ \text{up to the right and up to the left}\\ \textbf{(D)}\ \text{down to the right and down to the left}\\ \textbf{(E)}\ \text{none of these ways.}$

Solution

Problem 18

Of the following sets, the one that includes all values of $x$ which will satisfy $2x - 3 > 7 - x$ is:

$\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0$

Solution

Problem 19

If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

$\textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other}$

Solution

Problem 20

The equation $x^3+6x^2+11x+6=0$ has:

$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$

Solution

Problem 21

The roots of the equation $2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5$ can be found by solving:

$\textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0$

Solution

Problem 22

The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$, since division by zero is not allowed. For other values of $x$:

$\textbf{(A)}\ \text{The expression takes on many different values.}\ \textbf{(B)}\ \text{The expression has only the value 2.}\ \textbf{(C)}\ \text{The expression has only the value 1.}\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.}\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 23

If the margin made on an article costing $C$ dollars and selling for $S$ dollars is $M=\frac{1}{n}C$, then the margin is given by:

$\textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S$

Solution

Problem 24

The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are:

$\textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9}$

Solution

Problem 25

The two roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are $1$ and:

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

Solution

Problem 26

The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$. Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$. Then $BD$ equals:

$\textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Solution

Problem 27

A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:

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

Solution

Problem 28

If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, the value of $\frac{3mr-nt}{4nt-7mr}$ is:

$\textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3}$

Solution

Problem 29

If the ratio of the legs of a right triangle is $1: 2$, then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:

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

Solution

Problem 30

$A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days. The number of days required for A to do the job alone is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$

Solution

Problem 31

In $\triangle ABC$, $AB=AC$, $\angle A=40^\circ$. Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$. The number of degrees in $\angle BOC$ is:

$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$

Solution

Problem 32

The factors of $x^4+64$ are:

$\textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8)$

Solution

Problem 33

A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$. The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:

$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$

Solution

Problem 34

The fraction $\frac{1}{3}$:

$\textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9}$

Solution

Problem 35

In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$. If $MB = x, CB = h$, and $CA = d$, then $x$ equals:

[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; draw((0,0)--(8,0)--(0,5)--cycle); label("C",(0,0),SW); label("A",(8,0),SE); label("M",(0,5),N); dot((0,3.5)); label("B",(0,3.5),W); label("$x$",(0,4.25),W); label("$h$",(0,1),W); label("$d$",(4,0),S);[/asy]

$\textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h$

Solution

Problem 36

A boat has a speed of $15$ mph in still water. In a stream that has a current of $5$ mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:

$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8}$

Solution

Problem 37

Given $\triangle PQR$ with $\overline{RS}$ bisecting $\angle R$, $PQ$ extended to $D$ and $\angle n$ a right angle, then:

[asy] path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) {  pair M,N;  path mark;  M=t*0.03*unit(A-B)+B;  N=t*0.03*unit(C-B)+B;  if(flip)  mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));  else  mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));  return mark; } unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair P=(0,0), R=(3,2), Q=(4,0); pair S0=bisectorpoint(P,R,Q); pair Sp=extension(P,Q,S0,R); pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp); pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R); draw(P--R--Q); draw(R--Sp); draw(P--D--M); draw(anglemark2(Sp,P,R,17)); label("$p$",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label("$q$",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label("$n$",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label("$m$",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label("$d$",D+(-0.75,0.095)); pen f=fontsize(10pt); label("$R$",R,N,f); label("$P$",P,S,f); label("$S$",Sp,S,f); label("$Q$",Q,S,f); label("$D$",D,S,f);[/asy]

$\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q)$ $\textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad\textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad\textbf{(E)}\ \text{none of these is correct}$ Solution

Problem 38

If $\log 2=.3010$ and $\log 3=.4771$, the value of $x$ when $3^{x+3}=135$ is approximately:

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63$

Solution

Problem 39

The locus of the midpoint of a line segment that is drawn from a given external point $P$ to a given circle with center $O$ and radius $r$, is:

$\textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r$

Solution

Problem 40

If $\left (a+\frac{1}{a} \right )^2=3$, then $a^3+\frac{1}{a^3}$ equals:

$\textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3}$

Solution

Problem 41

The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0$

Solution

Problem 42

Consider the graphs of \[(1): y=x^2-\frac{1}{2}x+2\] and \[(2) y=x^2+\frac{1}{2}x+2\] on the same set of axis. These parabolas are exactly the same shape. Then:

$\textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).}$

Solution

Problem 43

The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$

Solution

Problem 44

A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$. He was born in:

$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$

Solution

Problem 45

In a rhombus, $ABCD$, line segments are drawn within the rhombus, parallel to diagonal $BD$, and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $A$. The graph is:

$\textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.}$

Solution

Problem 46

In the diagram, if points $A, B$ and $C$ are points of tangency, then $x$ equals:

[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); pair O=(0,3/8); draw((-2/3,9/16)--(2/3,9/16)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); draw(Circle(O,3/16)); draw((-2/3,0)--(2/3,0)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$\frac{3}{8}$",O); draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3)); draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3)); label("$\frac{1}{2}$",(.5,.25)); draw((.5,.33)--(.5,.5),EndArrow(3)); draw((.5,.17)--(.5,0),EndArrow(3)); label("$x$",midpoint((.5,.5)--(.5,9/16))); draw((.5,5/8)--(.5,9/16),EndArrow(3)); label("$60^{\circ}$",(0.01,0.12)); dot(A); dot(B); dot(C);[/asy]

$\textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}"$

Solution

Problem 47

At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $\frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:

$\textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x^2+px-q^2=0\\ \textbf{(D)}\ x^2-px-q^2=0\\ \textbf{(E)}\ x^2-px+q=0$

Solution

Problem 48

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:

$\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550$

Solution

Problem 49

The difference of the squares of two odd numbers is always divisible by $8$. If $a>b$, and $2a+1$ and $2b+1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:

$\textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b)$

Solution

Problem 50

The times between $7$ and $8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $84^{\circ} are:$ \textbf{(A)}\ \text{7: 23 and 7: 53}\qquad\textbf{(B)}\ \text{7: 20 and 7: 50}\qquad\textbf{(C)}\ \text{7: 22 and 7: 53}\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad\textbf{(E)}\ \text{7: 21 and 7: 49} $

Solution

See also

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