Difference between revisions of "1950 AHSME Problems"

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== Problem 4 ==
 
== Problem 4 ==
  
 +
Reduced to lowest terms, <math> \frac{a^{2}-b^{2}}{ab} </math> - <math> \frac{ab-b^{2}}{ab-a^{2}} </math> is equal to:
  
 +
<math> \textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 4|Solution]]
 
[[1950 AHSME Problems/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
 +
 +
If five geometric means are inserted between <math>8</math> and <math>5832</math>, the fifth term in the gemetric series:
 +
 +
<math> \textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 5|Solution]]
 
[[1950 AHSME Problems/Problem 5|Solution]]
  
 
== Problem 6 ==
 
== Problem 6 ==
 +
 +
The values of <math>y</math> which will satisfy the equations <math> 2x^{2}+6x+5y+1=0, 2x+y+3=0 </math> may be found by solving:
 +
 +
<math> \textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations} </math>
  
 
[[1950 AHSME Problems/Problem 6|Solution]]
 
[[1950 AHSME Problems/Problem 6|Solution]]
  
 
== Problem 7 ==
 
== Problem 7 ==
 +
 +
If the digit <math>1</math> is placed after a two digit number whose tens' digit is <math>t</math>, and units' digit is <math>u</math>, the new number is:
 +
 +
<math> \textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 100t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers} </math>
  
 
[[1950 AHSME Problems/Problem 7|Solution]]
 
[[1950 AHSME Problems/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
 +
 +
If the radius of a circle is increased <math>100\%</math>, the area is increased:
 +
 +
<math> \textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these} </math>
  
 
[[1950 AHSME Problems/Problem 8|Solution]]
 
[[1950 AHSME Problems/Problem 8|Solution]]
  
 
== Problem 9 ==
 
== Problem 9 ==
 +
 +
The area of the largest triangle that can be inscribed in a semi-circle whose radius is <math>r</math> is:
 +
 +
<math> \textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2} </math>
  
 
[[1950 AHSME Problems/Problem 9|Solution]]
 
[[1950 AHSME Problems/Problem 9|Solution]]
  
 
== Problem 10 ==
 
== Problem 10 ==
 +
 +
After rationalizing the numerator of <math> \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} </math>, the denominator in simplest form is:
 +
 +
<math> \textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers} </math>
  
 
[[1950 AHSME Problems/Problem 10|Solution]]
 
[[1950 AHSME Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
 +
 +
If in the formula <math> C =\frac{en}{R+nr} </math>, <math>n</math> is increased while <math>e</math>, <math>R</math> and <math>r</math> are kept constant, then <math>C</math>:
 +
 +
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\\ \textbf{(E)}\ \text{Decreases and then increases} </math>
  
 
[[1950 AHSME Problems/Problem 11|Solution]]
 
[[1950 AHSME Problems/Problem 11|Solution]]
  
 
== Problem 12 ==
 
== Problem 12 ==
 +
 +
As the number of sides of a polygon increases from <math>3</math> to <math>n</math>, the sum of the exterior formed by extending each side in succession:
 +
 +
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ \text{Becomes }(n-3)\text{ straight angles} </math>
  
 
[[1950 AHSME Problems/Problem 12|Solution]]
 
[[1950 AHSME Problems/Problem 12|Solution]]
  
 
== Problem 13 ==
 
== Problem 13 ==
 +
 +
The roots of <math> (x^{2}-3x+2)(x)(x-4)=0 </math> are:
 +
 +
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 0\text{ and }4\qquad\textbf{(C)}\ 1\text{ and }2\qquad\textbf{(D)}\ 0,1,2\text{ and }4\qquad\textbf{(E)}\ 1,2\text{ and }4 </math>
  
 
[[1950 AHSME Problems/Problem 13|Solution]]
 
[[1950 AHSME Problems/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
 +
 +
For the simultaneous equations <cmath>2x-3y=8\\6y-4x=9</cmath>
 +
 +
<math> \textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions} </math>
  
 
[[1950 AHSME Problems/Problem 14|Solution]]
 
[[1950 AHSME Problems/Problem 14|Solution]]
  
 
== Problem 15 ==
 
== Problem 15 ==
 +
 +
The real factors of <math>x^2+4</math> are:
 +
 +
<math> \textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent} </math>
  
 
[[1950 AHSME Problems/Problem 15|Solution]]
 
[[1950 AHSME Problems/Problem 15|Solution]]
  
 
== Problem 16 ==
 
== Problem 16 ==
 +
 +
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:
 +
 +
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
  
 
[[1950 AHSME Problems/Problem 16|Solution]]
 
[[1950 AHSME Problems/Problem 16|Solution]]
  
 
== Problem 17 ==
 
== Problem 17 ==
 +
 +
The formula which expresses the relationship between <math>x</math> and <math>y</math> as shown in the accompanying table is:
 +
 +
<cmath> \begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular} </cmath>
 +
 +
<math> \textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 17|Solution]]
 
[[1950 AHSME Problems/Problem 17|Solution]]
  
 
== Problem 18 ==
 
== Problem 18 ==
 +
 +
Of the following
 +
(1) <math> a(x-y)=ax-ay </math>
 +
(2) <math> a(x-y)=ax-ay </math>
 +
(3) <math> \log (x-y)=\log x-\log y </math>
 +
(4) <math> \frac{\log x}{\log y}=\log{x}-\log{y} </math>
 +
(5) <math> a(xy)=ax\times ay </math>
 +
 +
<math> \textbf{(A)}\ \text{Only 1 and 4 are true}\qquad\\ \textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\ \textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\ \textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\ \textbf{(E)}\ \text{Only 1 is true} </math>
  
 
[[1950 AHSME Problems/Problem 18|Solution]]
 
[[1950 AHSME Problems/Problem 18|Solution]]
  
 
== Problem 19 ==
 
== Problem 19 ==
 +
 +
If <math>m</math> men can do a job in <math>d</math> days, then <math>m+r</math> men can do the job in:
 +
 +
<math> \textbf{(A)}\ d+r\text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 19|Solution]]
 
[[1950 AHSME Problems/Problem 19|Solution]]
  
 
== Problem 20 ==
 
== Problem 20 ==
 +
 +
When <math>x^{13}-1</math> is divided by <math>x-1</math>, the remainder is:
 +
 +
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math>
  
 
[[1950 AHSME Problems/Problem 20|Solution]]
 
[[1950 AHSME Problems/Problem 20|Solution]]
  
 
== Problem 21 ==
 
== Problem 21 ==
 +
 +
The volume of a rectangular solid each of whose side, front, and bottom faces are <math> 12\text{ in}^{2} </math>, <math> 8\text{ in}^{2} </math>, and <math> 6\text{ in}^{2} </math> respectively is:
 +
 +
<math> \textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 21|Solution]]
 
[[1950 AHSME Problems/Problem 21|Solution]]
  
 
== Problem 22 ==
 
== Problem 22 ==
 +
 +
Successive discounts of <math>10\%</math> and <math>20\%</math> are equivalent to a single discount of:
 +
 +
<math> \textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 22|Solution]]
 
[[1950 AHSME Problems/Problem 22|Solution]]
  
 
== Problem 23 ==
 
== Problem 23 ==
 +
 +
A man buys a house for <math>10000</math> dollars and rents it. He puts <math> 12\frac{1}{2}\% </math> of each month's rent aside for repairs and upkeep; pays <math>325</math> dollars a year taxes and realizes <math> 5\frac{1}{2}\% </math> on his investment. The monthly rent is:
 +
 +
<math> \textbf{(A)}\ 64.82\text{ dollars}\qquad\textbf{(B)}\ 83.33\text{ dollars}\qquad\textbf{(C)}\ 72.08\text{ dollars}\qquad\textbf{(D)}\ 45.83\text{ dollars}\qquad\textbf{(E)}\ 177.08\text{ dollars} </math>
  
 
[[1950 AHSME Problems/Problem 23|Solution]]
 
[[1950 AHSME Problems/Problem 23|Solution]]
  
 
== Problem 24 ==
 
== Problem 24 ==
 +
 +
The equation <math> x+\sqrt{x-2}=4 </math> has:
 +
 +
<math> \textbf{(A)}\ \text{2 real roots}\qquad\textbf{(B)}\ \text{1 real and 1 imaginary root}\qquad\textbf{(C)}\ \text{2 imaginary roots}\qquad\textbf{(D)}\ \text{No roots}\qquad\textbf{(E)}\ \text{1 real root} </math>
  
 
[[1950 AHSME Problems/Problem 4|Solution]]
 
[[1950 AHSME Problems/Problem 4|Solution]]
  
 
== Problem 25 ==
 
== Problem 25 ==
 +
 +
The value of <math> \log_{5}\frac{(125)(625)}{25} </math> is equal to:
 +
 +
<math> \textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 25|Solution]]
 
[[1950 AHSME Problems/Problem 25|Solution]]
  
 
== Problem 26 ==
 
== Problem 26 ==
 +
 +
if <math> \log_{10}{m}= b-\log_{10}{n} </math>, then <math>m=</math>
 +
 +
<math> \textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n} </math>
  
 
[[1950 AHSME Problems/Problem 26|Solution]]
 
[[1950 AHSME Problems/Problem 26|Solution]]
  
 
== Problem 27 ==
 
== Problem 27 ==
 +
 +
A car travels <math>120</math> miles from <math>A</math> to <math>B</math> at <math>30</math> miles per hour but returns the same distance at <math>40</math> miles per hour. The average speed for the round trip is closest to:
 +
 +
<math> \textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph} </math>
  
 
[[1950 AHSME Problems/Problem 27|Solution]]
 
[[1950 AHSME Problems/Problem 27|Solution]]
  
 
== Problem 28 ==
 
== Problem 28 ==
 +
 +
Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>.  <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was:
 +
 +
<math> \textbf{(A)}\ 4\text{ mph}\qquad\textbf{(B)}\ 8\text{ mph}\qquad\textbf{(C)}\ 12\text{ mph}\qquad\textbf{(D)}\ 16\text{ mph}\qquad\textbf{(E)}\ 20\text{ mph} </math>
  
 
[[1950 AHSME Problems/Problem 28|Solution]]
 
[[1950 AHSME Problems/Problem 28|Solution]]
  
 
== Problem 29 ==
 
== Problem 29 ==
 +
 +
A manufacturer built a machine which will address <math>500</math> envelopes in <math>8</math> minutes. He wishes to build another machine so that when both are operating together they will address <math>500</math> envelopes in <math>2</math> minutes. The equation used to find how many minutes <math>x</math> it would require the second machine to address <math>500</math> envelopes alone is:
 +
 +
<math> \textbf{(A)}\ 8-x=2\qquad\textbf{(B)}\ \frac{1}{8}+\frac{1}{x}=\frac{1}{2}\qquad\textbf{(C)}\ \frac{500}{8}+\frac{500}{x}=500\qquad\textbf{(D)}\ \frac{x}{2}+\frac{x}{8}=1\qquad\\ \textbf{(E)}\ \text{None of these answers} </math>
  
 
[[1950 AHSME Problems/Problem 29|Solution]]
 
[[1950 AHSME Problems/Problem 29|Solution]]
  
 
== Problem 30 ==
 
== Problem 30 ==
 +
 +
From a group of boys and girls, <math>15</math> girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>5</math> girls for each boy. The number of girls in the beginning was:
 +
 +
<math> \textbf{(A)}\ 40\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 30|Solution]]
 
[[1950 AHSME Problems/Problem 30|Solution]]
  
 
== Problem 31 ==
 
== Problem 31 ==
 +
 +
John ordered <math>4</math> pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by <math>50\%</math>. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:
 +
 +
<math> \textbf{(A)}\ 4:1\qquad\textbf{(B)}\ 2:1\qquad\textbf{(C)}\ 1:4\qquad\textbf{(D)}\ 1:2\qquad\textbf{(E)}\ 1:8 </math>
  
 
[[1950 AHSME Problems/Problem 31|Solution]]
 
[[1950 AHSME Problems/Problem 31|Solution]]
  
 
== Problem 32 ==
 
== Problem 32 ==
 +
 +
A <math>25</math> foot ladder is placed against a vertical wall of a building. The foot of the ladder is <math>7</math> feet from the base of the building. If the top of the ladder slips <math>4</math> feet, then the foot of the ladder will slide:
 +
 +
<math> \textbf{(A)}\ 9\text{ ft}\qquad\textbf{(B)}\ 15\text{ ft}\qquad\textbf{(C)}\ 5\text{ ft}\qquad\textbf{(D)}\ 8\text{ ft}\qquad\textbf{(E)}\ 4\text{ ft} </math>
  
 
[[1950 AHSME Problems/Problem 32|Solution]]
 
[[1950 AHSME Problems/Problem 32|Solution]]
  
 
== Problem 33 ==
 
== Problem 33 ==
 +
 +
The number of circular pipes with an inside diameter of <math>1</math> inch which will carry the same amount of water as a pipe with an inside diameter of <math>6</math> inches is:
 +
 +
<math> \textbf{(A)}\ 6\pi\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 36\pi </math>
  
 
[[1950 AHSME Problems/Problem 33|Solution]]
 
[[1950 AHSME Problems/Problem 33|Solution]]
  
 
== Problem 34 ==
 
== Problem 34 ==
 +
 +
When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
 +
 +
<math> \textbf{(A)}\ 5\text{ in}\qquad\textbf{(B)}\ 2\frac{1}{2}\text{ in}\qquad\textbf{(C)}\ \frac{5}{\pi}\text{ in}\qquad\textbf{(D)}\ \frac{5}{2\pi}\text{ in}\qquad\textbf{(E)}\ \frac{\pi}{5}\text{ in} </math>
  
 
[[1950 AHSME Problems/Problem 34|Solution]]
 
[[1950 AHSME Problems/Problem 34|Solution]]
  
 
== Problem 35 ==
 
== Problem 35 ==
 +
 +
In triangle <math>ABC</math>, <math>AC=24</math> inches, <math>BC=10</math> inches, <math>AB=26</math> inches. The radius of the inscribed circle is:
 +
 +
<math> \textbf{(A)}\ 26\text{ in}\qquad\textbf{(B)}\ 4\text{ in}\qquad\textbf{(C)}\ 13\text{ in}\qquad\textbf{(D)}\ 8\text{ in}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 35|Solution]]
 
[[1950 AHSME Problems/Problem 35|Solution]]
  
 
== Problem 36 ==
 
== Problem 36 ==
 +
 +
A merchant buys goods at <math>25\%</math> of the list price. He desires to mark the goods so that he can give a discount of <math>20\%</math> on the marked price and still clear a profit of <math>25\%</math> on the selling price. What percent of the list price must he mark the goods?
 +
 +
<math> \textbf{(A)}\ 125\%\qquad\textbf{(B)}\ 100\%\qquad\textbf{(C)}\ 120\%\qquad\textbf{(D)}\ 80\%\qquad\textbf{(E)}\ 75\% </math>
  
 
[[1950 AHSME Problems/Problem 36|Solution]]
 
[[1950 AHSME Problems/Problem 36|Solution]]
  
 
== Problem 37 ==
 
== Problem 37 ==
 +
 +
If <math> y =\log_{a}{x} </math>, <math>a>1</math>, which of the following statements is incorrect?
 +
 +
<math> \textbf{(A)}\ \text{If }x=1,y=0\qquad\\ \textbf{(B)}\ \text{If }x=a,y=1\qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}\qquad\\ \textbf{(D)}\ \text{If }0<x<z,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}\qquad\\ \textbf{(E)}\ \text{Only some of the above statements are correct} </math>
  
 
[[1950 AHSME Problems/Problem 37|Solution]]
 
[[1950 AHSME Problems/Problem 37|Solution]]
  
 
== Problem 38 ==
 
== Problem 38 ==
 +
 +
If the expression <math> \begin{pmatrix}a & c\\ d & b\end{pmatrix} </math> has the value <math>ab-cd</math> for all values of <math>a, b, c</math> and <math>d</math>, then the equation <math> \begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3 </math>:
 +
 +
<math> \textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.} </math>
  
 
[[1950 AHSME Problems/Problem 38|Solution]]
 
[[1950 AHSME Problems/Problem 38|Solution]]
  
 
== Problem 39 ==
 
== Problem 39 ==
 +
 +
Given the series <math> 2+1+\frac{1}{2}+\frac{1}{4}+... </math> and the following five statements:
 +
(1) the sum increases without limit
 +
(2) the sum decreases without limit
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(3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small
 +
(4) the difference between the sum and 4 can be made less than any positive quantity no matter how small
 +
(5) the sum approaches a limit
 +
Of these statments, the correct ones are:
 +
 +
<math> \textbf{(A)}\ \text{Only }3\text{ and }4\qquad\textbf{(B)}\ \text{Only }5\qquad\textbf{(C)}\ \text{Only }2\text{ and }4\qquad\textbf{(D)}\ \text{Only }2,3\text{ and }4\qquad\textbf{(E)}\ \text{Only }4\text{ and }5 </math>
  
 
[[1950 AHSME Problems/Problem 39|Solution]]
 
[[1950 AHSME Problems/Problem 39|Solution]]
  
 
== Problem 40 ==
 
== Problem 40 ==
 +
 +
The limit of <math> \frac{x^{2}-1}{x-1} </math> as <math>x</math> approaches <math>1</math> as a limit is:
 +
 +
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ \text{Indeterminate}\qquad\textbf{(C)}\ x-1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1 </math>
  
 
[[1950 AHSME Problems/Problem 40|Solution]]
 
[[1950 AHSME Problems/Problem 40|Solution]]
  
 
== Problem 41 ==
 
== Problem 41 ==
 +
 +
The least value of the function <math>ax^2+bx+c</math> with <math>a>0</math> is:
 +
 +
<math> \textbf{(A)}\ -\frac{b}{a}\qquad\textbf{(B)}\ -\frac{b}{2a}\qquad\textbf{(C)}\ b^{2}-4ac\qquad\textbf{(D)}\ \frac{4ac-b^{2}}{4a}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 41|Solution]]
 
[[1950 AHSME Problems/Problem 41|Solution]]
  
 
== Problem 42 ==
 
== Problem 42 ==
 +
 +
The equation <math> x^{x^{x}}...=2 </math> is satisfied when <math>x</math> is equal to:
 +
 +
<math> \textbf{(A)}\ \infty\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt[4]{2}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 42|Solution]]
 
[[1950 AHSME Problems/Problem 42|Solution]]
  
 
== Problem 43 ==
 
== Problem 43 ==
 +
 +
The sum to infinity of <math> \frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+... </math> is:
 +
 +
<math> \textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{1}{24}\qquad\textbf{(C)}\ \frac{5}{48}\qquad\textbf{(D)}\ \frac{1}{16}\qquad\textbf{(E)}\ \text{None of these} </math>
  
 
[[1950 AHSME Problems/Problem 43|Solution]]
 
[[1950 AHSME Problems/Problem 43|Solution]]
  
 
== Problem 44 ==
 
== Problem 44 ==
 +
 +
The graph of <math>y=\log x</math>
 +
 +
<math> \textbf{(A)}\text{Cuts the }y\text{-axis}\qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis}\qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis}\qquad\\ \textbf{(D)}\ \text{Cuts neither axis}\qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the origin} </math>
  
 
[[1950 AHSME Problems/Problem 44|Solution]]
 
[[1950 AHSME Problems/Problem 44|Solution]]
  
 
== Problem 45 ==
 
== Problem 45 ==
 +
 +
The number of diagonals that can be drawn in a polygon of <math>100</math> sides is:
 +
 +
<math> \textbf{(A)}\ 4850\qquad\textbf{(B)}\ 4950\qquad\textbf{(C)}\ 9900\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 8800 </math>
  
 
[[1950 AHSME Problems/Problem 45|Solution]]
 
[[1950 AHSME Problems/Problem 45|Solution]]
  
 
== Problem 46 ==
 
== Problem 46 ==
 +
 +
In triangle <math>ABC</math>, <math>AB=12</math>, <math>AC=7</math>, and <math>BC=10</math>. If sides <math>AB</math> and <math>AC</math> are doubled while <math>BC</math> remains the same, then:
 +
 +
<math> \textbf{(A)}\ \text{The area is doubled}\qquad\\ \textbf{(B)}\ \text{The altitude is doubled}\qquad\\ \textbf{(C)}\ \text{The area is four times the original area}\qquad\\ \textbf{(D)}\ \text{The median is unchanged}\qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0} </math>
  
 
[[1950 AHSME Problems/Problem 46|Solution]]
 
[[1950 AHSME Problems/Problem 46|Solution]]
  
 
== Problem 47 ==
 
== Problem 47 ==
 +
 +
A rectangle inscribed in a triangle has its base coinciding with the base <math>b</math> of the triangle. If the altitude of the triangle is <math>h</math>, and the altitude <math>x</math> of the rectangle is half the base of the rectangle, then:
 +
 +
<math> \textbf{(A)}\ x=\frac{1}{2}h\qquad\textbf{(B)}\ x=\frac{bh}{b+h}\qquad\textbf{(C)}\ x=\frac{bh}{2h+b}\qquad\textbf{(D)}\ x=\sqrt{\frac{hb}{2}}\qquad\\ \textbf{(E)}\ x=\frac{1}{2}b </math>
  
 
[[1950 AHSME Problems/Problem 47|Solution]]
 
[[1950 AHSME Problems/Problem 47|Solution]]
  
 
== Problem 48 ==
 
== Problem 48 ==
 +
 +
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
 +
 +
<math> \textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity} </math>
  
 
[[1950 AHSME Problems/Problem 48|Solution]]
 
[[1950 AHSME Problems/Problem 48|Solution]]
  
 
== Problem 49 ==
 
== Problem 49 ==
 +
 +
A triangle has a fixed base <math>AB</math> that is <math>2</math> inches long. The median from <math>A</math> to side <math>BC</math> is <math>1 \frac{1}{2}</math> inches long and can have any position emanating from <math>A</math>. The locus of the vertex <math>C</math> of the triangle is:
 +
 +
<math> \textbf{(A)}\ \text{A straight line }AB,1\frac{1}{2}\text{ inches from }A\qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches}\qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches}\qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along }BA\qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus} </math>
  
 
[[1950 AHSME Problems/Problem 49|Solution]]
 
[[1950 AHSME Problems/Problem 49|Solution]]
  
 
== Problem 50 ==
 
== Problem 50 ==
 +
 +
A privateer discovers a merchantman <math>10</math> miles to leeward at <math>11\text{:}45 \text{ a.m.}</math> and with a good breeze bears down upon her at <math>11</math> mph, while the merchantman can only make <math>8</math> mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only <math>17</math> miles while the merchantman makes <math>15</math>. The privateer will overtake the merchantman at:
 +
 +
<math> \textbf{(A)}\ 3\text{:}45\text{ p.m.}\qquad\textbf{(B)}\ 3\text{:}30\text{ p.m.}\qquad\textbf{(C)}\ 5\text{:}00\text{ p.m.}\qquad\textbf{(D)}\ 2\text{:}45\text{ p.m.}\qquad\textbf{(E)}\ 5\text{:}30\text{ p.m.} </math>
  
 
[[1950 AHSME Problems/Problem 50|Solution]]
 
[[1950 AHSME Problems/Problem 50|Solution]]

Revision as of 19:49, 31 August 2011

Problem 1

If $64$ is divided into three parts proportional to $2$, $4$, and $6$, the smallest part is:

$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

Problem 2

Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 3

The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 4

Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab}$ - $\frac{ab-b^{2}}{ab-a^{2}}$ is equal to:

$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 5

If five geometric means are inserted between $8$ and $5832$, the fifth term in the gemetric series:

$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 6

The values of $y$ which will satisfy the equations $2x^{2}+6x+5y+1=0, 2x+y+3=0$ may be found by solving:

$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$

Solution

Problem 7

If the digit $1$ is placed after a two digit number whose tens' digit is $t$, and units' digit is $u$, the new number is:

$\textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 100t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers}$

Solution

Problem 8

If the radius of a circle is increased $100\%$, the area is increased:

$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$

Solution

Problem 9

The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:

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

Solution

Problem 10

After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:

$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

Problem 11

If in the formula $C =\frac{en}{R+nr}$, $n$ is increased while $e$, $R$ and $r$ are kept constant, then $C$:

$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\\ \textbf{(E)}\ \text{Decreases and then increases}$

Solution

Problem 12

As the number of sides of a polygon increases from $3$ to $n$, the sum of the exterior formed by extending each side in succession:

$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ \text{Becomes }(n-3)\text{ straight angles}$

Solution

Problem 13

The roots of $(x^{2}-3x+2)(x)(x-4)=0$ are:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 0\text{ and }4\qquad\textbf{(C)}\ 1\text{ and }2\qquad\textbf{(D)}\ 0,1,2\text{ and }4\qquad\textbf{(E)}\ 1,2\text{ and }4$

Solution

Problem 14

For the simultaneous equations \[2x-3y=8\\6y-4x=9\]

$\textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions}$

Solution

Problem 15

The real factors of $x^2+4$ are:

$\textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent}$

Solution

Problem 16

The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Problem 17

The formula which expresses the relationship between $x$ and $y$ as shown in the accompanying table is:

\[\begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular}\]

$\textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 18

Of the following (1) $a(x-y)=ax-ay$ (2) $a(x-y)=ax-ay$ (3) $\log (x-y)=\log x-\log y$ (4) $\frac{\log x}{\log y}=\log{x}-\log{y}$ (5) $a(xy)=ax\times ay$

$\textbf{(A)}\ \text{Only 1 and 4 are true}\qquad\\ \textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\ \textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\ \textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\ \textbf{(E)}\ \text{Only 1 is true}$

Solution

Problem 19

If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:

$\textbf{(A)}\ d+r\text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 20

When $x^{13}-1$ is divided by $x-1$, the remainder is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

Problem 21

The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$, $8\text{ in}^{2}$, and $6\text{ in}^{2}$ respectively is:

$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 22

Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:

$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 23

A man buys a house for $10000$ dollars and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325$ dollars a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent is:

$\textbf{(A)}\ 64.82\text{ dollars}\qquad\textbf{(B)}\ 83.33\text{ dollars}\qquad\textbf{(C)}\ 72.08\text{ dollars}\qquad\textbf{(D)}\ 45.83\text{ dollars}\qquad\textbf{(E)}\ 177.08\text{ dollars}$

Solution

Problem 24

The equation $x+\sqrt{x-2}=4$ has:

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

Solution

Problem 25

The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:

$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 26

if $\log_{10}{m}= b-\log_{10}{n}$, then $m=$

$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution

Problem 27

A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:

$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$

Solution

Problem 28

Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:

$\textbf{(A)}\ 4\text{ mph}\qquad\textbf{(B)}\ 8\text{ mph}\qquad\textbf{(C)}\ 12\text{ mph}\qquad\textbf{(D)}\ 16\text{ mph}\qquad\textbf{(E)}\ 20\text{ mph}$

Solution

Problem 29

A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes alone is:

$\textbf{(A)}\ 8-x=2\qquad\textbf{(B)}\ \frac{1}{8}+\frac{1}{x}=\frac{1}{2}\qquad\textbf{(C)}\ \frac{500}{8}+\frac{500}{x}=500\qquad\textbf{(D)}\ \frac{x}{2}+\frac{x}{8}=1\qquad\\ \textbf{(E)}\ \text{None of these answers}$

Solution

Problem 30

From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 31

John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $50\%$. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:

$\textbf{(A)}\ 4:1\qquad\textbf{(B)}\ 2:1\qquad\textbf{(C)}\ 1:4\qquad\textbf{(D)}\ 1:2\qquad\textbf{(E)}\ 1:8$

Solution

Problem 32

A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:

$\textbf{(A)}\ 9\text{ ft}\qquad\textbf{(B)}\ 15\text{ ft}\qquad\textbf{(C)}\ 5\text{ ft}\qquad\textbf{(D)}\ 8\text{ ft}\qquad\textbf{(E)}\ 4\text{ ft}$

Solution

Problem 33

The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:

$\textbf{(A)}\ 6\pi\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 36\pi$

Solution

Problem 34

When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:

$\textbf{(A)}\ 5\text{ in}\qquad\textbf{(B)}\ 2\frac{1}{2}\text{ in}\qquad\textbf{(C)}\ \frac{5}{\pi}\text{ in}\qquad\textbf{(D)}\ \frac{5}{2\pi}\text{ in}\qquad\textbf{(E)}\ \frac{\pi}{5}\text{ in}$

Solution

Problem 35

In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$\textbf{(A)}\ 26\text{ in}\qquad\textbf{(B)}\ 4\text{ in}\qquad\textbf{(C)}\ 13\text{ in}\qquad\textbf{(D)}\ 8\text{ in}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 36

A merchant buys goods at $25\%$ of the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?

$\textbf{(A)}\ 125\%\qquad\textbf{(B)}\ 100\%\qquad\textbf{(C)}\ 120\%\qquad\textbf{(D)}\ 80\%\qquad\textbf{(E)}\ 75\%$

Solution

Problem 37

If $y =\log_{a}{x}$, $a>1$, which of the following statements is incorrect?

$\textbf{(A)}\ \text{If }x=1,y=0\qquad\\ \textbf{(B)}\ \text{If }x=a,y=1\qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}\qquad\\ \textbf{(D)}\ \text{If }0<x<z,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}\qquad\\ \textbf{(E)}\ \text{Only some of the above statements are correct}$

Solution

Problem 38

If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$, then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$:

$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.}$

Solution

Problem 39

Given the series $2+1+\frac{1}{2}+\frac{1}{4}+...$ and the following five statements: (1) the sum increases without limit (2) the sum decreases without limit (3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small (4) the difference between the sum and 4 can be made less than any positive quantity no matter how small (5) the sum approaches a limit Of these statments, the correct ones are:

$\textbf{(A)}\ \text{Only }3\text{ and }4\qquad\textbf{(B)}\ \text{Only }5\qquad\textbf{(C)}\ \text{Only }2\text{ and }4\qquad\textbf{(D)}\ \text{Only }2,3\text{ and }4\qquad\textbf{(E)}\ \text{Only }4\text{ and }5$

Solution

Problem 40

The limit of $\frac{x^{2}-1}{x-1}$ as $x$ approaches $1$ as a limit is:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \text{Indeterminate}\qquad\textbf{(C)}\ x-1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1$

Solution

Problem 41

The least value of the function $ax^2+bx+c$ with $a>0$ is:

$\textbf{(A)}\ -\frac{b}{a}\qquad\textbf{(B)}\ -\frac{b}{2a}\qquad\textbf{(C)}\ b^{2}-4ac\qquad\textbf{(D)}\ \frac{4ac-b^{2}}{4a}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 42

The equation $x^{x^{x}}...=2$ is satisfied when $x$ is equal to:

$\textbf{(A)}\ \infty\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt[4]{2}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 43

The sum to infinity of $\frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+...$ is:

$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{1}{24}\qquad\textbf{(C)}\ \frac{5}{48}\qquad\textbf{(D)}\ \frac{1}{16}\qquad\textbf{(E)}\ \text{None of these}$

Solution

Problem 44

The graph of $y=\log x$

$\textbf{(A)}\text{Cuts the }y\text{-axis}\qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis}\qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis}\qquad\\ \textbf{(D)}\ \text{Cuts neither axis}\qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the origin}$

Solution

Problem 45

The number of diagonals that can be drawn in a polygon of $100$ sides is:

$\textbf{(A)}\ 4850\qquad\textbf{(B)}\ 4950\qquad\textbf{(C)}\ 9900\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 8800$

Solution

Problem 46

In triangle $ABC$, $AB=12$, $AC=7$, and $BC=10$. If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:

$\textbf{(A)}\ \text{The area is doubled}\qquad\\ \textbf{(B)}\ \text{The altitude is doubled}\qquad\\ \textbf{(C)}\ \text{The area is four times the original area}\qquad\\ \textbf{(D)}\ \text{The median is unchanged}\qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$

Solution

Problem 47

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$\textbf{(A)}\ x=\frac{1}{2}h\qquad\textbf{(B)}\ x=\frac{bh}{b+h}\qquad\textbf{(C)}\ x=\frac{bh}{2h+b}\qquad\textbf{(D)}\ x=\sqrt{\frac{hb}{2}}\qquad\\ \textbf{(E)}\ x=\frac{1}{2}b$

Solution

Problem 48

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:

$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

Solution

Problem 49

A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1 \frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is:

$\textbf{(A)}\ \text{A straight line }AB,1\frac{1}{2}\text{ inches from }A\qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches}\qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches}\qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along }BA\qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$

Solution

Problem 50

A privateer discovers a merchantman $10$ miles to leeward at $11\text{:}45 \text{ a.m.}$ and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only $17$ miles while the merchantman makes $15$. The privateer will overtake the merchantman at:

$\textbf{(A)}\ 3\text{:}45\text{ p.m.}\qquad\textbf{(B)}\ 3\text{:}30\text{ p.m.}\qquad\textbf{(C)}\ 5\text{:}00\text{ p.m.}\qquad\textbf{(D)}\ 2\text{:}45\text{ p.m.}\qquad\textbf{(E)}\ 5\text{:}30\text{ p.m.}$

Solution

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