Difference between revisions of "1953 AHSME Problems"

m
Line 6: Line 6:
 
\textbf{(B)}\ 150 \text{ oranges} \qquad
 
\textbf{(B)}\ 150 \text{ oranges} \qquad
 
\textbf{(C)}\ 200\text{ oranges}\  
 
\textbf{(C)}\ 200\text{ oranges}\  
textbf{(D)}\ \text{an infinite number of oranges}\qquad
+
\textbf{(D)}\ \text{an infinite number of oranges}\qquad
 
\textbf{(E)}\ \text{none of these} </math>   
 
\textbf{(E)}\ \text{none of these} </math>   
 
   
 
   
Line 13: Line 13:
 
==Problem 2==
 
==Problem 2==
 
   
 
   
A refrigerator is offered at sale at &#036; 250.00 less successive discounts of <math>20</math> \% and <math>15</math> \%. The sale price of the refrigerator is:  
+
A refrigerator is offered at sale at &#036; 250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:
  
<math></math>\textbf{(A)}\ 35\% \text{ less than } &#036;250.00 \qquad
+
<math>\text{(A) 35\% less than 250.00} \qquad
\textbf{(B)}\ 65\% \text{ of } &#036;250.00 \qquad
+
\text{(B) 65\% of 250.00} \qquad
\textbf{(C)}\ 77\%\text{ of }&#036;250.00\qquad
+
\text{(C) 77\% of 250.00} \qquad \
\textbf{(D)}\ 68\%\text{ of }&#036;250.00\qquad
+
\text{(D) 68\% of 250.00} \qquad
\textbf{(E)}\ \text{none of these <math>
+
\text{(E) none of these} </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 2|Solution]]
 
[[1953 AHSME Problems/Problem 2|Solution]]
Line 25: Line 25:
 
==Problem 3==
 
==Problem 3==
  
The factors of the expression </math>x^2+y^2<math> are:  
+
The factors of the expression <math>x^2+y^2</math> are:  
  
</math>\textbf{(A)}\ (x+y)(x-y) \qquad
+
<math>\textbf{(A)}\ (x+y)(x-y) \qquad
 
\textbf{(B)}\ (x+y)^2 \qquad
 
\textbf{(B)}\ (x+y)^2 \qquad
\textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\ \textbf{(D)}\ (x+iy)(x-iy)\qquad
+
\textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\  
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(D)}\ (x+iy)(x-iy)\qquad
 +
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
[[1953 AHSME Problems/Problem 3|Solution]]
+
[[1953 AHSME Problems/Problem 3|Solution]]
 
   
 
   
 
==Problem 4==
 
==Problem 4==
  
The roots of </math>x(x^2+8x+16)(4-x)=0<math> are:  
+
The roots of <math>x(x^2+8x+16)(4-x)=0</math> are:  
  
</math>\textbf{(A)}\ 0 \qquad
+
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B)}\ 0,4 \qquad
 
\textbf{(B)}\ 0,4 \qquad
 
\textbf{(C)}\ 0,4,-4 \qquad
 
\textbf{(C)}\ 0,4,-4 \qquad
 
\textbf{(D)}\ 0,4,-4,-4 \qquad
 
\textbf{(D)}\ 0,4,-4,-4 \qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 4|Solution]]
 
[[1953 AHSME Problems/Problem 4|Solution]]
 +
 
==Problem 5==
 
==Problem 5==
  
If </math>\log_6 x=2.5<math>, the value of </math>x<math> is:  
+
If <math>\log_6 x=2.5</math>, the value of <math>x</math> is:  
  
</math>\textbf{(A)}\ 90 \qquad
+
<math>\textbf{(A)}\ 90 \qquad
 
\textbf{(B)}\ 36 \qquad
 
\textbf{(B)}\ 36 \qquad
 
\textbf{(C)}\ 36\sqrt{6} \qquad
 
\textbf{(C)}\ 36\sqrt{6} \qquad
 
\textbf{(D)}\ 0.5 \qquad
 
\textbf{(D)}\ 0.5 \qquad
\textbf{(E)}\ \text{none of these} <math>
+
\textbf{(E)}\ \text{none of these} </math>
  
 
[[1953 AHSME Problems/Problem 5|Solution]]
 
[[1953 AHSME Problems/Problem 5|Solution]]
Line 59: Line 61:
 
==Problem 6==
 
==Problem 6==
  
Charles has </math>5q + 1<math> quarters and Richard has </math>q + 5<math> quarters. The difference in their money in dimes is:  
+
Charles has <math>5q + 1</math> quarters and Richard has <math>q + 5</math> quarters. The difference in their money in dimes is:  
  
</math>\textbf{(A)}\ 10(q - 1) \qquad
+
<math>\textbf{(A)}\ 10(q - 1) \qquad
 
\textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad
 
\textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad
 
\textbf{(C)}\ \frac {2}{5}(q - 1) \  
 
\textbf{(C)}\ \frac {2}{5}(q - 1) \  
 
\textbf{(D)}\ \frac{5}{2}(q-1)\qquad
 
\textbf{(D)}\ \frac{5}{2}(q-1)\qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
  
 
[[1953 AHSME Problems/Problem 6|Solution]]
 
[[1953 AHSME Problems/Problem 6|Solution]]
Line 71: Line 73:
 
==Problem 7==
 
==Problem 7==
  
The fraction </math>\frac{\sqrt{a^2+x^2}-(x^2-a^2)/\sqrt{a^2+x^2}}{a^2+x^2}<math> reduces to:  
+
The fraction <math>\frac{\sqrt{a^2+x^2}-(x^2-a^2)/\sqrt{a^2+x^2}}{a^2+x^2}</math> reduces to:  
  
</math>\textbf{(A)}\ 0 \qquad
+
<math>\textbf{(A)}\ 0 \qquad
 
\textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad
 
\textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad
 
\textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
 
\textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
 
\textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
 
\textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
\textbf{(E)}\ \frac{2x^2}{a^2+x^2}  <math>   
+
\textbf{(E)}\ \frac{2x^2}{a^2+x^2}  </math>   
  
 
[[1953 AHSME Problems/Problem 7|Solution]]
 
[[1953 AHSME Problems/Problem 7|Solution]]
Line 83: Line 85:
 
==Problem 8==
 
==Problem 8==
  
The value of </math>x<math> at the intersection of </math>y=\frac{8}{x^2+4}<math> and </math>x+y=2<math> is:  
+
The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is:  
  
</math>\textbf{(A)}\ -2+\sqrt{5} \qquad
+
<math>\textbf{(A)}\ -2+\sqrt{5} \qquad
 
\textbf{(B)}\ -2-\sqrt{5} \qquad
 
\textbf{(B)}\ -2-\sqrt{5} \qquad
 
\textbf{(C)}\ 0 \qquad
 
\textbf{(C)}\ 0 \qquad
 
\textbf{(D)}\ 2 \qquad
 
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
  
 
[[1953 AHSME Problems/Problem 8|Solution]]
 
[[1953 AHSME Problems/Problem 8|Solution]]
Line 95: Line 97:
 
==Problem 9==
 
==Problem 9==
  
The number of ounces of water needed to reduce </math>9<math> ounces of shaving lotion containing </math>50<math> \% alcohol to a lotion containing </math>30<math> \% alcohol is:  
+
The number of ounces of water needed to reduce <math>9</math> ounces of shaving lotion containing <math>50</math> \% alcohol to a lotion containing <math>30</math> \% alcohol is:  
  
</math>\textbf{(A)}\ 3 \qquad
+
<math>\textbf{(A)}\ 3 \qquad
 
\textbf{(B)}\ 4 \qquad
 
\textbf{(B)}\ 4 \qquad
 
\textbf{(C)}\ 5 \qquad
 
\textbf{(C)}\ 5 \qquad
 
\textbf{(D)}\ 6 \qquad
 
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 7    <math>
+
\textbf{(E)}\ 7    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 9|Solution]]
 
[[1953 AHSME Problems/Problem 9|Solution]]
Line 107: Line 109:
 
==Problem 10==
 
==Problem 10==
  
The number of revolutions of a wheel, with fixed center and with an outside diameter of </math>6<math> feet, required to cause a point on the rim to go one mile is:  
+
The number of revolutions of a wheel, with fixed center and with an outside diameter of <math>6</math> feet, required to cause a point on the rim to go one mile is:  
  
</math>\textbf{(A)}\ 880 \qquad
+
<math>\textbf{(A)}\ 880 \qquad
 
\textbf{(B)}\ \frac{440}{\pi} \qquad
 
\textbf{(B)}\ \frac{440}{\pi} \qquad
 
\textbf{(C)}\ \frac{880}{\pi} \qquad
 
\textbf{(C)}\ \frac{880}{\pi} \qquad
 
\textbf{(D)}\ 440\pi\qquad
 
\textbf{(D)}\ 440\pi\qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 10|Solution]]
 
[[1953 AHSME Problems/Problem 10|Solution]]
Line 119: Line 121:
 
==Problem 11==
 
==Problem 11==
  
A running track is the ring formed by two concentric circles. It is </math>10<math> feet wide. The circumference of the two circles differ by about:  
+
A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about:  
  
</math>\textbf{(A)}\ 10\text{ feet} \qquad
+
<math>\textbf{(A)}\ 10\text{ feet} \qquad
 
\textbf{(B)}\ 30\text{ feet} \qquad
 
\textbf{(B)}\ 30\text{ feet} \qquad
 
\textbf{(C)}\ 60\text{ feet} \qquad
 
\textbf{(C)}\ 60\text{ feet} \qquad
\textbf{(D)}\ 100\text{ feet}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(D)}\ 100\text{ feet}\ \textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 11|Solution]]
 
[[1953 AHSME Problems/Problem 11|Solution]]
Line 130: Line 132:
 
==Problem 12==
 
==Problem 12==
  
The diameters of two circles are </math>8<math> inches and </math>12<math> inches respectively. The ratio of the area of the smaller to the area of the larger circle is:  
+
The diameters of two circles are <math>8</math> inches and <math>12</math> inches respectively. The ratio of the area of the smaller to the area of the larger circle is:  
  
</math>\textbf{(A)}\ \frac{2}{3} \qquad
+
<math>\textbf{(A)}\ \frac{2}{3} \qquad
 
\textbf{(B)}\ \frac{4}{9} \qquad
 
\textbf{(B)}\ \frac{4}{9} \qquad
 
\textbf{(C)}\ \frac{9}{4} \qquad
 
\textbf{(C)}\ \frac{9}{4} \qquad
 
\textbf{(D)}\ \frac{1}{2}\qquad
 
\textbf{(D)}\ \frac{1}{2}\qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 12|Solution]]
 
[[1953 AHSME Problems/Problem 12|Solution]]
Line 144: Line 146:
 
A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:  
 
A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:  
  
</math>\textbf{(A)}\ 36\text{ inches} \qquad
+
<math>\textbf{(A)}\ 36\text{ inches} \qquad
 
\textbf{(B)}\ 9\text{ inches} \qquad
 
\textbf{(B)}\ 9\text{ inches} \qquad
 
\textbf{(C)}\ 18\text{ inches}\  
 
\textbf{(C)}\ 18\text{ inches}\  
 
\textbf{(D)}\ \text{not obtainable from these data}\qquad
 
\textbf{(D)}\ \text{not obtainable from these data}\qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 13|Solution]]
 
[[1953 AHSME Problems/Problem 13|Solution]]
Line 154: Line 156:
 
==Problem 14==
 
==Problem 14==
  
Given the larger of two circles with center </math>P<math> and radius </math>p<math> and the smaller with center </math>Q<math> and radius </math>q<math>. Draw </math>PQ<math>. Which of the following statements is false?  
+
Given the larger of two circles with center <math>P</math> and radius <math>p</math> and the smaller with center <math>Q</math> and radius <math>q</math>. Draw <math>PQ</math>. Which of the following statements is false?  
  
</math>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\  
+
<math>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\  
 
\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\  
 
\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\  
 
\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\  
 
\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\  
\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\ \textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 14|Solution]]
 
[[1953 AHSME Problems/Problem 14|Solution]]
Line 167: Line 169:
 
A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:  
 
A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:  
  
</math>\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\  
+
<math>\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\  
\textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\ \textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 15|Solution]]
 
[[1953 AHSME Problems/Problem 15|Solution]]
Line 174: Line 176:
 
==Problem 16==
 
==Problem 16==
  
Adams plans a profit of </math>10<math> \% on the selling price of an article and his expenses are </math>15<math> \% of sales. The rate of markup on an article that sells for &#036; </math>5.00<math> is:  
+
Adams plans a profit of <math>10</math> \% on the selling price of an article and his expenses are <math>15</math> \% of sales. The rate of markup on an article that sells for &#036; <math>5.00</math> is:  
  
</math>\textbf{(A)}\ 20\% \qquad
+
<math>\textbf{(A)}\ 20\% \qquad
 
\textbf{(B)}\ 25\% \qquad
 
\textbf{(B)}\ 25\% \qquad
 
\textbf{(C)}\ 30\% \qquad
 
\textbf{(C)}\ 30\% \qquad
 
\textbf{(D)}\ 33\frac {1}{3}\% \qquad
 
\textbf{(D)}\ 33\frac {1}{3}\% \qquad
\textbf{(E)}\ 35\%  <math>
+
\textbf{(E)}\ 35\%  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 16|Solution]]
 
[[1953 AHSME Problems/Problem 16|Solution]]
Line 186: Line 188:
 
==Problem 17==
 
==Problem 17==
  
A man has part of &#036; </math>4500<math> invested at </math>4<math> \% and the rest at </math>6<math> \%. If his annual return on each investment is the same, the average rate of interest which he realizes of the &#036;4500 is:  
+
A man has part of &#036; <math>4500</math> invested at <math>4</math> \% and the rest at <math>6</math> \%. If his annual return on each investment is the same, the average rate of interest which he realizes of the &#036;4500 is:  
  
</math>\textbf{(A)}\ 5\% \qquad
+
<math>\textbf{(A)}\ 5\% \qquad
 
\textbf{(B)}\ 4.8\% \qquad
 
\textbf{(B)}\ 4.8\% \qquad
 
\textbf{(C)}\ 5.2\% \qquad
 
\textbf{(C)}\ 5.2\% \qquad
 
\textbf{(D)}\ 4.6\% \qquad
 
\textbf{(D)}\ 4.6\% \qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 17|Solution]]
 
[[1953 AHSME Problems/Problem 17|Solution]]
Line 198: Line 200:
 
==Problem 18==
 
==Problem 18==
  
One of the factors of </math>x^4+4<math> is:  
+
One of the factors of <math>x^4+4</math> is:  
  
</math>\textbf{(A)}\ x^2+2 \qquad
+
<math>\textbf{(A)}\ x^2+2 \qquad
 
\textbf{(B)}\ x+1 \qquad
 
\textbf{(B)}\ x+1 \qquad
 
\textbf{(C)}\ x^2-2x+2 \qquad
 
\textbf{(C)}\ x^2-2x+2 \qquad
 
\textbf{(D)}\ x^2-4\  
 
\textbf{(D)}\ x^2-4\  
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 18|Solution]]
 
[[1953 AHSME Problems/Problem 18|Solution]]
Line 210: Line 212:
 
==Problem 19==
 
==Problem 19==
  
In the expression </math>xy^2<math>, the values of </math>x<math> and </math>y<math> are each decreased </math>25<math> \%; the value of the expression is:  
+
In the expression <math>xy^2</math>, the values of <math>x</math> and <math>y</math> are each decreased <math>25</math> \%; the value of the expression is:  
  
</math>\textbf{(A)}\ \text{decreased } 50\% \qquad
+
<math>\textbf{(A)}\ \text{decreased } 50\% \qquad
 
\textbf{(B)}\ \text{decreased }75\%\  
 
\textbf{(B)}\ \text{decreased }75\%\  
 
\textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad
 
\textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad
\textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\ \textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\ \textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 19|Solution]]
 
[[1953 AHSME Problems/Problem 19|Solution]]
Line 221: Line 223:
 
==Problem 20==
 
==Problem 20==
  
If </math>y=x+\frac{1}{x}<math>, then </math>x^4+x^3-4x^2+x+1=0<math> becomes:  
+
If <math>y=x+\frac{1}{x}</math>, then <math>x^4+x^3-4x^2+x+1=0</math> becomes:  
  
</math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad
+
<math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad
 
\textbf{(B)}\ x^2(y^2+y-3)=0\  
 
\textbf{(B)}\ x^2(y^2+y-3)=0\  
 
\textbf{(C)}\ x^2(y^2+y-4)=0 \qquad
 
\textbf{(C)}\ x^2(y^2+y-4)=0 \qquad
\textbf{(D)}\ x^2(y^2+y-6)=0\ \textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(D)}\ x^2(y^2+y-6)=0\ \textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 20|Solution]]
 
[[1953 AHSME Problems/Problem 20|Solution]]
Line 232: Line 234:
 
==Problem 21==
 
==Problem 21==
  
If </math>\log_{10} (x^2-3x+6)=1<math>, the value of </math>x<math> is:  
+
If <math>\log_{10} (x^2-3x+6)=1</math>, the value of <math>x</math> is:  
  
</math>\textbf{(A)}\ 10\text{ or }2 \qquad
+
<math>\textbf{(A)}\ 10\text{ or }2 \qquad
 
\textbf{(B)}\ 4\text{ or }-2 \qquad
 
\textbf{(B)}\ 4\text{ or }-2 \qquad
 
\textbf{(C)}\ 3\text{ or }-1 \qquad
 
\textbf{(C)}\ 3\text{ or }-1 \qquad
\textbf{(D)}\ 4\text{ or }-1\ \textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(D)}\ 4\text{ or }-1\ \textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 21|Solution]]
 
[[1953 AHSME Problems/Problem 21|Solution]]
Line 243: Line 245:
 
==Problem 22==
 
==Problem 22==
  
The logarithm of </math>27\sqrt[4]{9}\sqrt[3]{9}<math> to the base </math>3<math> is:  
+
The logarithm of <math>27\sqrt[4]{9}\sqrt[3]{9}</math> to the base <math>3</math> is:  
  
</math>\textbf{(A)}\ 8\frac{1}{2} \qquad
+
<math>\textbf{(A)}\ 8\frac{1}{2} \qquad
 
\textbf{(B)}\ 4\frac{1}{6} \qquad
 
\textbf{(B)}\ 4\frac{1}{6} \qquad
 
\textbf{(C)}\ 5 \qquad
 
\textbf{(C)}\ 5 \qquad
 
\textbf{(D)}\ 3 \qquad
 
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 22|Solution]]
 
[[1953 AHSME Problems/Problem 22|Solution]]
Line 255: Line 257:
 
==Problem 23==
 
==Problem 23==
  
The equation </math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5<math> has:  
+
The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:  
  
</math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \  
+
<math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \  
 
\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad
 
\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad
\textbf{(D)}\ \text{two true roots}\ \textbf{(E)}\ \text{two extraneous roots}  <math>
+
\textbf{(D)}\ \text{two true roots}\ \textbf{(E)}\ \text{two extraneous roots}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 23|Solution]]
 
[[1953 AHSME Problems/Problem 23|Solution]]
Line 265: Line 267:
 
==Problem 24==
 
==Problem 24==
  
If </math>a,b,c<math> are positive integers less than </math>10<math>, then </math>(10a + b)(10a + c) = 100a(a + 1) + bc<math> if:  
+
If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if:  
  
</math>\textbf{(A)}\ b + c = 10 \qquad
+
<math>\textbf{(A)}\ b + c = 10 \qquad
 
\textbf{(B)}\ b = c \qquad
 
\textbf{(B)}\ b = c \qquad
 
\textbf{(C)}\ a + b = 10 \qquad
 
\textbf{(C)}\ a + b = 10 \qquad
 
\textbf{(D)}\ a = b \  
 
\textbf{(D)}\ a = b \  
\textbf{(E)}\ a+b+c = 10  <math>
+
\textbf{(E)}\ a+b+c = 10  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 24|Solution]]
 
[[1953 AHSME Problems/Problem 24|Solution]]
Line 279: Line 281:
 
In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:  
 
In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:  
  
</math>\textbf{(A)}\ 1 \qquad
+
<math>\textbf{(A)}\ 1 \qquad
 
\textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad
 
\textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad
 
\textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad
 
\textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad
 
\textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad
 
\textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad
\textbf{(E)}\ \frac{2}{\sqrt{5}}    <math>
+
\textbf{(E)}\ \frac{2}{\sqrt{5}}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 25|Solution]]
 
[[1953 AHSME Problems/Problem 25|Solution]]
Line 289: Line 291:
 
==Problem 26==
 
==Problem 26==
  
The base of a triangle is </math>15<math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:  
+
The base of a triangle is <math>15</math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:  
  
</math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad
+
<math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad
 
\textbf{(B)}\ 10\text{ inches} \qquad
 
\textbf{(B)}\ 10\text{ inches} \qquad
 
\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad
 
\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad
\textbf{(D)}\ 7.5\text{ inches}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(D)}\ 7.5\text{ inches}\  
 +
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 26|Solution]]
 
[[1953 AHSME Problems/Problem 26|Solution]]
Line 300: Line 303:
 
==Problem 27==
 
==Problem 27==
  
The radius of the first circle is </math>1<math> inch, that of the second </math>\frac{1}{2}<math> inch, that of the third </math>\frac{1}{4}<math> inch and so on indefinitely.  
+
The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely.  
 
The sum of the areas of the circles is:  
 
The sum of the areas of the circles is:  
  
</math>\textbf{(A)}\ \frac{3\pi}{4} \qquad
+
<math>\textbf{(A)}\ \frac{3\pi}{4} \qquad
 
\textbf{(B)}\ 1.3\pi \qquad
 
\textbf{(B)}\ 1.3\pi \qquad
 
\textbf{(C)}\ 2\pi \qquad
 
\textbf{(C)}\ 2\pi \qquad
 
\textbf{(D)}\ \frac{4\pi}{3}\qquad
 
\textbf{(D)}\ \frac{4\pi}{3}\qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 27|Solution]]
 
[[1953 AHSME Problems/Problem 27|Solution]]
Line 313: Line 316:
 
==Problem 28==
 
==Problem 28==
  
In </math>\triangle ABC<math>, sides </math>a,b<math> and </math>c<math> are opposite </math>\angle{A},\angle{B}<math> and </math>\angle{C}<math> respectively. </math>AD<math> bisects </math>\angle{A}<math> and meets </math>BC<math> at </math>D<math>.  
+
In <math>\triangle ABC</math>, sides <math>a,b</math> and <math>c</math> are opposite <math>\angle{A},\angle{B}</math> and <math>\angle{C}</math> respectively. <math>AD</math> bisects <math>\angle{A}</math> and meets <math>BC</math> at <math>D</math>.  
Then if </math>x = \overline{CD}<math> and </math>y = \overline{BD}<math> the correct proportion is:  
+
Then if <math>x = \overline{CD}</math> and <math>y = \overline{BD}</math> the correct proportion is:  
  
</math>\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad
+
<math>\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad
 
\textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad
 
\textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad
 
\textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad
 
\textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad
\textbf{(E)}\ \frac{x}{y}=\frac{c}{b  <math>
+
\textbf{(E)}\ \frac{x}{y}=\frac{c}{b} </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 28|Solution]]
 
[[1953 AHSME Problems/Problem 28|Solution]]
Line 325: Line 328:
 
==Problem 29==
 
==Problem 29==
  
The number of significant digits in the measurement of the side of a square whose computed area is </math>1.1025<math> square inches to  
+
The number of significant digits in the measurement of the side of a square whose computed area is <math>1.1025</math> square inches to  
 
the nearest ten-thousandth of a square inch is:  
 
the nearest ten-thousandth of a square inch is:  
  
</math>\textbf{(A)}\ 2 \qquad
+
<math>\textbf{(A)}\ 2 \qquad
 
\textbf{(B)}\ 3 \qquad
 
\textbf{(B)}\ 3 \qquad
 
\textbf{(C)}\ 4 \qquad
 
\textbf{(C)}\ 4 \qquad
 
\textbf{(D)}\ 5 \qquad
 
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ 1    <math>
+
\textbf{(E)}\ 1    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 29|Solution]]
 
[[1953 AHSME Problems/Problem 29|Solution]]
Line 338: Line 341:
 
==Problem 30==
 
==Problem 30==
  
A house worth &#036; </math>9000<math> is sold by Mr. A to Mr. B at a </math>10<math> \% loss. Mr. B sells the house back to Mr. A at a </math>10<math> \% gain.  
+
A house worth &#036; <math>9000</math> is sold by Mr. A to Mr. B at a <math>10</math> \% loss. Mr. B sells the house back to Mr. A at a <math>10</math> \% gain.  
 
The result of the two transactions is:  
 
The result of the two transactions is:  
  
</math>\textbf{(A)}\ \text{Mr. A breaks even} \qquad
+
<math>\textbf{(A)}\ \text{Mr. A breaks even} \qquad
 
\textbf{(B)}\ \text{Mr. B gains }&#036;900 \qquad
 
\textbf{(B)}\ \text{Mr. B gains }&#036;900 \qquad
 
\textbf{(C)}\ \text{Mr. A loses }&#036;900\ \textbf{(D)}\ \text{Mr. A loses }&#036;810\qquad
 
\textbf{(C)}\ \text{Mr. A loses }&#036;900\ \textbf{(D)}\ \text{Mr. A loses }&#036;810\qquad
\textbf{(E)}\ \text{Mr. B gains }&#036;1710  <math>
+
\textbf{(E)}\ \text{Mr. B gains }&#036;1710  </math>
  
  
Line 351: Line 354:
 
==Problem 31==
 
==Problem 31==
  
The rails on a railroad are </math>30<math> feet long. As the train passes over the point where the rails are joined, there is an audible click.  
+
The rails on a railroad are <math>30</math> feet long. As the train passes over the point where the rails are joined, there is an audible click.  
 
The speed of the train in miles per hour is approximately the number of clicks heard in:  
 
The speed of the train in miles per hour is approximately the number of clicks heard in:  
  
</math>\textbf{(A)}\ 20\text{ seconds} \qquad
+
<math>\textbf{(A)}\ 20\text{ seconds} \qquad
 
\textbf{(B)}\ 2\text{ minutes} \qquad
 
\textbf{(B)}\ 2\text{ minutes} \qquad
 
\textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad
 
\textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad
\textbf{(D)}\ 5\text{ minutes}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(D)}\ 5\text{ minutes}\ \textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 31|Solution]]
 
[[1953 AHSME Problems/Problem 31|Solution]]
Line 365: Line 368:
 
Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:  
 
Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:  
  
</math>\textbf{(A)}\ \text{a square} \qquad
+
<math>\textbf{(A)}\ \text{a square} \qquad
 
\textbf{(B)}\ \text{a rectangle} \qquad
 
\textbf{(B)}\ \text{a rectangle} \qquad
 
\textbf{(C)}\ \text{a parallelogram with unequal sides}\ \textbf{(D)}\ \text{a rhombus}\qquad
 
\textbf{(C)}\ \text{a parallelogram with unequal sides}\ \textbf{(D)}\ \text{a rhombus}\qquad
\textbf{(E)}\ \text{a quadrilateral with no special properties}    <math>
+
\textbf{(E)}\ \text{a quadrilateral with no special properties}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 32|Solution]]
 
[[1953 AHSME Problems/Problem 32|Solution]]
Line 374: Line 377:
 
==Problem 33==
 
==Problem 33==
  
The perimeter of an isosceles right triangle is </math>2p<math>. Its area is:  
+
The perimeter of an isosceles right triangle is <math>2p</math>. Its area is:  
  
</math>\textbf{(A)}\ (2+\sqrt{2})p \qquad
+
<math>\textbf{(A)}\ (2+\sqrt{2})p \qquad
 
\textbf{(B)}\ (2-\sqrt{2})p \qquad
 
\textbf{(B)}\ (2-\sqrt{2})p \qquad
 
\textbf{(C)}\ (3-2\sqrt{2})p^2\  
 
\textbf{(C)}\ (3-2\sqrt{2})p^2\  
 
\textbf{(D)}\ (1-2\sqrt{2})p^2\qquad
 
\textbf{(D)}\ (1-2\sqrt{2})p^2\qquad
\textbf{(E)}\ (3+2\sqrt{2})p^2    <math>
+
\textbf{(E)}\ (3+2\sqrt{2})p^2    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 33|Solution]]
 
[[1953 AHSME Problems/Problem 33|Solution]]
Line 386: Line 389:
 
==Problem 34==
 
==Problem 34==
  
If one side of a triangle is </math>12<math> inches and the opposite angle is </math>30^{\circ}<math>, then the diameter of the circumscribed circle is:  
+
If one side of a triangle is <math>12</math> inches and the opposite angle is <math>30^{\circ}</math>, then the diameter of the circumscribed circle is:  
  
</math>\textbf{(A)}\ 18\text{ inches} \qquad
+
<math>\textbf{(A)}\ 18\text{ inches} \qquad
 
\textbf{(B)}\ 30\text{ inches} \qquad
 
\textbf{(B)}\ 30\text{ inches} \qquad
 
\textbf{(C)}\ 24\text{ inches} \qquad
 
\textbf{(C)}\ 24\text{ inches} \qquad
\textbf{(D)}\ 20\text{ inches}\ \textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(D)}\ 20\text{ inches}\ \textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 34|Solution]]
 
[[1953 AHSME Problems/Problem 34|Solution]]
Line 397: Line 400:
 
==Problem 35==
 
==Problem 35==
  
If </math>f(x)=\frac{x(x-1)}{2}<math>, then </math>f(x+2)<math> equals:  
+
If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals:  
  
</math>\textbf{(A)}\ f(x)+f(2) \qquad
+
<math>\textbf{(A)}\ f(x)+f(2) \qquad
 
\textbf{(B)}\ (x+2)f(x) \qquad
 
\textbf{(B)}\ (x+2)f(x) \qquad
 
\textbf{(C)}\ x(x+2)f(x) \qquad
 
\textbf{(C)}\ x(x+2)f(x) \qquad
\textbf{(D)}\ \frac{xf(x)}{x+2}\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}    <math>
+
\textbf{(D)}\ \frac{xf(x)}{x+2}\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 35|Solution]]
 
[[1953 AHSME Problems/Problem 35|Solution]]
Line 408: Line 411:
 
==Problem 36==
 
==Problem 36==
  
Determine </math>m<math> so that </math>4x^2-6x+m<math> is divisible by </math>x-3<math>. The obtained value, </math>m<math>, is an exact divisor of:  
+
Determine <math>m</math> so that <math>4x^2-6x+m</math> is divisible by <math>x-3</math>. The obtained value, <math>m</math>, is an exact divisor of:  
  
</math>\textbf{(A)}\ 12 \qquad
+
<math>\textbf{(A)}\ 12 \qquad
 
\textbf{(B)}\ 20 \qquad
 
\textbf{(B)}\ 20 \qquad
 
\textbf{(C)}\ 36 \qquad
 
\textbf{(C)}\ 36 \qquad
 
\textbf{(D)}\ 48 \qquad
 
\textbf{(D)}\ 48 \qquad
\textbf{(E)}\ 64    <math>
+
\textbf{(E)}\ 64    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 36|Solution]]
 
[[1953 AHSME Problems/Problem 36|Solution]]
Line 420: Line 423:
 
==Problem 37==
 
==Problem 37==
  
The base of an isosceles triangle is </math>6<math> inches and one of the equal sides is </math>12<math> inches.  
+
The base of an isosceles triangle is <math>6</math> inches and one of the equal sides is <math>12</math> inches.  
 
The radius of the circle through the vertices of the triangle is:  
 
The radius of the circle through the vertices of the triangle is:  
  
</math>\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad
+
<math>\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad
 
\textbf{(B)}\ 4\sqrt{3} \qquad
 
\textbf{(B)}\ 4\sqrt{3} \qquad
 
\textbf{(C)}\ 3\sqrt{5} \qquad
 
\textbf{(C)}\ 3\sqrt{5} \qquad
 
\textbf{(D)}\ 6\sqrt{3}\qquad
 
\textbf{(D)}\ 6\sqrt{3}\qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 37|Solution]]
 
[[1953 AHSME Problems/Problem 37|Solution]]
Line 433: Line 436:
 
==Problem 38==
 
==Problem 38==
  
If </math>f(a)=a-2<math> and </math>F(a,b)=b^2+a<math>, then </math>F(3,f(4))<math> is:  
+
If <math>f(a)=a-2</math> and <math>F(a,b)=b^2+a</math>, then <math>F(3,f(4))</math> is:  
  
</math>\textbf{(A)}\ a^2-4a+7 \qquad
+
<math>\textbf{(A)}\ a^2-4a+7 \qquad
 
\textbf{(B)}\ 28 \qquad
 
\textbf{(B)}\ 28 \qquad
 
\textbf{(C)}\ 7 \qquad
 
\textbf{(C)}\ 7 \qquad
 
\textbf{(D)}\ 8 \qquad
 
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 11    <math>
+
\textbf{(E)}\ 11    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 38|Solution]]
 
[[1953 AHSME Problems/Problem 38|Solution]]
Line 445: Line 448:
 
==Problem 39==
 
==Problem 39==
  
The product, </math>\log_a b \cdot \log_b a<math> is equal to:  
+
The product, <math>\log_a b \cdot \log_b a</math> is equal to:  
  
</math>\textbf{(A)}\ 1 \qquad
+
<math>\textbf{(A)}\ 1 \qquad
 
\textbf{(B)}\ a \qquad
 
\textbf{(B)}\ a \qquad
 
\textbf{(C)}\ b \qquad
 
\textbf{(C)}\ b \qquad
 
\textbf{(D)}\ ab \qquad
 
\textbf{(D)}\ ab \qquad
\textbf{(E)}\ \text{none of these}    <math>
+
\textbf{(E)}\ \text{none of these}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 39|Solution]]
 
[[1953 AHSME Problems/Problem 39|Solution]]
Line 459: Line 462:
 
The negation of the statement "all men are honest," is:  
 
The negation of the statement "all men are honest," is:  
  
</math>\textbf{(A)}\ \text{no men are honest} \qquad
+
<math>\textbf{(A)}\ \text{no men are honest} \qquad
 
\textbf{(B)}\ \text{all men are dishonest} \  
 
\textbf{(B)}\ \text{all men are dishonest} \  
 
\textbf{(C)}\ \text{some men are dishonest}\qquad
 
\textbf{(C)}\ \text{some men are dishonest}\qquad
\textbf{(D)}\ \text{no men are dishonest}\ \textbf{(E)}\ \text{some men are honest}  <math>
+
\textbf{(D)}\ \text{no men are dishonest}\ \textbf{(E)}\ \text{some men are honest}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 40|Solution]]
 
[[1953 AHSME Problems/Problem 40|Solution]]
Line 468: Line 471:
 
==Problem 41==
 
==Problem 41==
  
A girls' camp is located </math>300<math> rods from a straight road. On this road, a boys' camp is located </math>500<math> rods from the girls' camp.  
+
A girls' camp is located <math>300</math> rods from a straight road. On this road, a boys' camp is located <math>500</math> rods from the girls' camp.  
 
It is desired to build a canteen on the road which shall be exactly the same distance from each camp.  
 
It is desired to build a canteen on the road which shall be exactly the same distance from each camp.  
 
The distance of the canteen from each of the camps is:  
 
The distance of the canteen from each of the camps is:  
  
</math>\textbf{(A)}\ 400\text{ rods} \qquad
+
<math>\textbf{(A)}\ 400\text{ rods} \qquad
 
\textbf{(B)}\ 250\text{ rods} \qquad
 
\textbf{(B)}\ 250\text{ rods} \qquad
 
\textbf{(C)}\ 87.5\text{ rods} \qquad
 
\textbf{(C)}\ 87.5\text{ rods} \qquad
\textbf{(D)}\ 200\text{ rods}\ \textbf{(E)}\ \text{none of these} <math>
+
\textbf{(D)}\ 200\text{ rods}\ \textbf{(E)}\ \text{none of these} </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 41|Solution]]
 
[[1953 AHSME Problems/Problem 41|Solution]]
Line 481: Line 484:
 
==Problem 42==
 
==Problem 42==
  
The centers of two circles are </math>41<math> inches apart. The smaller circle has a radius of </math>4<math> inches and the larger one has a radius of </math>5<math> inches.  
+
The centers of two circles are <math>41</math> inches apart. The smaller circle has a radius of <math>4</math> inches and the larger one has a radius of <math>5</math> inches.  
 
The length of the common internal tangent is:  
 
The length of the common internal tangent is:  
  
</math>\textbf{(A)}\ 41\text{ inches} \qquad
+
<math>\textbf{(A)}\ 41\text{ inches} \qquad
 
\textbf{(B)}\ 39\text{ inches} \qquad
 
\textbf{(B)}\ 39\text{ inches} \qquad
 
\textbf{(C)}\ 39.8\text{ inches} \qquad
 
\textbf{(C)}\ 39.8\text{ inches} \qquad
\textbf{(D)}\ 40.1\text{ inches}\ \textbf{(E)}\ 40\text{ inches} <math>
+
\textbf{(D)}\ 40.1\text{ inches}\ \textbf{(E)}\ 40\text{ inches} </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 42|Solution]]
 
[[1953 AHSME Problems/Problem 42|Solution]]
Line 493: Line 496:
 
==Problem 43==
 
==Problem 43==
  
If the price of an article is increased by percent </math>p<math>, then the decrease in percent of sales must not exceed </math>d<math> in order to yield the same income.  
+
If the price of an article is increased by percent <math>p</math>, then the decrease in percent of sales must not exceed <math>d</math> in order to yield the same income.  
The value of </math>d<math> is:  
+
The value of <math>d</math> is:  
  
</math>\textbf{(A)}\ \frac{1}{1+p} \qquad
+
<math>\textbf{(A)}\ \frac{1}{1+p} \qquad
 
\textbf{(B)}\ \frac{1}{1-p} \qquad
 
\textbf{(B)}\ \frac{1}{1-p} \qquad
 
\textbf{(C)}\ \frac{p}{1+p} \qquad
 
\textbf{(C)}\ \frac{p}{1+p} \qquad
 
\textbf{(D)}\ \frac{p}{p-1}\qquad
 
\textbf{(D)}\ \frac{p}{p-1}\qquad
\textbf{(E)}\ \frac{1-p}{1+p}    <math>
+
\textbf{(E)}\ \frac{1-p}{1+p}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 43|Solution]]
 
[[1953 AHSME Problems/Problem 43|Solution]]
Line 507: Line 510:
  
 
In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and  
 
In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and  
obtains </math>8<math> and </math>2<math> for the roots. Another student makes a mistake only in the coefficient of the first degree term and  
+
obtains <math>8</math> and <math>2</math> for the roots. Another student makes a mistake only in the coefficient of the first degree term and  
find </math>-9<math> and </math>-1<math> for the roots. The correct equation was:  
+
find <math>-9</math> and <math>-1</math> for the roots. The correct equation was:  
  
</math>\textbf{(A)}\ x^2-10x+9=0 \qquad
+
<math>\textbf{(A)}\ x^2-10x+9=0 \qquad
 
\textbf{(B)}\ x^2+10x+9=0 \qquad
 
\textbf{(B)}\ x^2+10x+9=0 \qquad
 
\textbf{(C)}\ x^2-10x+16=0\  
 
\textbf{(C)}\ x^2-10x+16=0\  
 
\textbf{(D)}\ x^2-8x-9=0\qquad
 
\textbf{(D)}\ x^2-8x-9=0\qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 44|Solution]]
 
[[1953 AHSME Problems/Problem 44|Solution]]
Line 520: Line 523:
 
==Problem 45==
 
==Problem 45==
  
The lengths of two line segments are </math>a<math> units and </math>b<math> units respectively. Then the correct relation between them is:  
+
The lengths of two line segments are <math>a</math> units and <math>b</math> units respectively. Then the correct relation between them is:  
  
</math>\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad
+
<math>\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad
 
\textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad
 
\textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad
 
\textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad
 
\textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad
\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}  <math>
+
\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 45|Solution]]
 
[[1953 AHSME Problems/Problem 45|Solution]]
Line 532: Line 535:
  
 
Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and  
 
Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and  
saved a distance equal to </math>\frac{1}{2}<math> the longer side. The ratio of the shorter side of the rectangle to the longer side was:  
+
saved a distance equal to <math>\frac{1}{2}</math> the longer side. The ratio of the shorter side of the rectangle to the longer side was:  
  
</math>\textbf{(A)}\ \frac{1}{2} \qquad
+
<math>\textbf{(A)}\ \frac{1}{2} \qquad
 
\textbf{(B)}\ \frac{2}{3} \qquad
 
\textbf{(B)}\ \frac{2}{3} \qquad
 
\textbf{(C)}\ \frac{1}{4} \qquad
 
\textbf{(C)}\ \frac{1}{4} \qquad
 
\textbf{(D)}\ \frac{3}{4}\qquad
 
\textbf{(D)}\ \frac{3}{4}\qquad
\textbf{(E)}\ \frac{2}{5}    <math>
+
\textbf{(E)}\ \frac{2}{5}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 46|Solution]]
 
[[1953 AHSME Problems/Problem 46|Solution]]
Line 544: Line 547:
 
==Problem 47==
 
==Problem 47==
  
If </math>x>0<math>, then the correct relationship is:  
+
If <math>x>0</math>, then the correct relationship is:  
  
</math>\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad
+
<math>\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad
 
\textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \  
 
\textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \  
 
\textbf{(C)}\ \log(1+x) > x\qquad
 
\textbf{(C)}\ \log(1+x) > x\qquad
 
\textbf{(D)}\ \log (1+x) < x\qquad
 
\textbf{(D)}\ \log (1+x) < x\qquad
\textbf{(E)}\ \text{none of these}  <math>
+
\textbf{(E)}\ \text{none of these}  </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 47|Solution]]
 
[[1953 AHSME Problems/Problem 47|Solution]]
Line 559: Line 562:
 
then the ratio of the smaller base to the larger base is:  
 
then the ratio of the smaller base to the larger base is:  
  
</math>\textbf{(A)}\ \frac{1}{2} \qquad
+
<math>\textbf{(A)}\ \frac{1}{2} \qquad
 
\textbf{(B)}\ \frac{2}{3} \qquad
 
\textbf{(B)}\ \frac{2}{3} \qquad
 
\textbf{(C)}\ \frac{3}{4} \qquad
 
\textbf{(C)}\ \frac{3}{4} \qquad
 
\textbf{(D)}\ \frac{3}{5}\qquad
 
\textbf{(D)}\ \frac{3}{5}\qquad
\textbf{(E)}\ \frac{2}{5}    <math>
+
\textbf{(E)}\ \frac{2}{5}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 48|Solution]]
 
[[1953 AHSME Problems/Problem 48|Solution]]
Line 569: Line 572:
 
==Problem 49==
 
==Problem 49==
  
The coordinates of </math>A,B<math> and </math>C<math> are </math>(5,5),(2,1)<math> and </math>(0,k)<math> respectively.  
+
The coordinates of <math>A,B</math> and <math>C</math> are <math>(5,5),(2,1)</math> and <math>(0,k)</math> respectively.  
The value of </math>k<math> that makes </math>\overline{AC}+\overline{BC}<math> as small as possible is:  
+
The value of <math>k</math> that makes <math>\overline{AC}+\overline{BC}</math> as small as possible is:  
  
</math>\textbf{(A)}\ 3 \qquad
+
<math>\textbf{(A)}\ 3 \qquad
 
\textbf{(B)}\ 4\frac{1}{2} \qquad
 
\textbf{(B)}\ 4\frac{1}{2} \qquad
 
\textbf{(C)}\ 3\frac{6}{7} \qquad
 
\textbf{(C)}\ 3\frac{6}{7} \qquad
 
\textbf{(D)}\ 4\frac{5}{6}\qquad
 
\textbf{(D)}\ 4\frac{5}{6}\qquad
\textbf{(E)}\ 2\frac{1}{7}    <math>
+
\textbf{(E)}\ 2\frac{1}{7}    </math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 49|Solution]]
 
[[1953 AHSME Problems/Problem 49|Solution]]
Line 582: Line 585:
 
==Problem 50==
 
==Problem 50==
  
One of the sides of a triangle is divided into segments of </math>6<math> and </math>8<math> units by the point of tangency of the inscribed circle.  
+
One of the sides of a triangle is divided into segments of <math>6</math> and <math>8</math> units by the point of tangency of the inscribed circle.  
If the radius of the circle is </math>4<math>, then the length of the shortest side of the triangle is:  
+
If the radius of the circle is <math>4</math>, then the length of the shortest side of the triangle is:  
  
</math>\textbf{(A)}\ 12\text{ units} \qquad
+
<math>\textbf{(A)}\ 12\text{ units} \qquad
 
\textbf{(B)}\ 13\text{ units} \qquad
 
\textbf{(B)}\ 13\text{ units} \qquad
 
\textbf{(C)}\ 14\text{ units} \qquad
 
\textbf{(C)}\ 14\text{ units} \qquad
 
\textbf{(D)}\ 15\text{ units}\qquad
 
\textbf{(D)}\ 15\text{ units}\qquad
\textbf{(E)}\ 16\text{ units}$
+
\textbf{(E)}\ 16\text{ units}</math>
 
    
 
    
 
[[1953 AHSME Problems/Problem 50|Solution]]
 
[[1953 AHSME Problems/Problem 50|Solution]]

Revision as of 21:34, 8 October 2014

Problem 1

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $ $1.00$, he must sell:

$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\  \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 2

A refrigerator is offered at sale at $ 250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:

$\text{(A) 35\% less than 250.00} \qquad \text{(B) 65\% of 250.00} \qquad \text{(C) 77\% of 250.00} \qquad \\ \text{(D) 68\% of 250.00} \qquad \text{(E) none of these}$

Solution

Problem 3

The factors of the expression $x^2+y^2$ are:

$\textbf{(A)}\ (x+y)(x-y) \qquad \textbf{(B)}\ (x+y)^2 \qquad \textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\\  \textbf{(D)}\ (x+iy)(x-iy)\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 4

The roots of $x(x^2+8x+16)(4-x)=0$ are:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 5

If $\log_6 x=2.5$, the value of $x$ is:

$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ 36\sqrt{6} \qquad \textbf{(D)}\ 0.5 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 6

Charles has $5q + 1$ quarters and Richard has $q + 5$ quarters. The difference in their money in dimes is:

$\textbf{(A)}\ 10(q - 1) \qquad \textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad \textbf{(C)}\ \frac {2}{5}(q - 1) \\  \textbf{(D)}\ \frac{5}{2}(q-1)\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 7

The fraction $\frac{\sqrt{a^2+x^2}-(x^2-a^2)/\sqrt{a^2+x^2}}{a^2+x^2}$ reduces to:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad \textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(E)}\ \frac{2x^2}{a^2+x^2}$

Solution

Problem 8

The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:

$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 9

The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ \% alcohol to a lotion containing $30$ \% alcohol is:

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

Solution

Problem 10

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:

$\textbf{(A)}\ 880 \qquad \textbf{(B)}\ \frac{440}{\pi} \qquad \textbf{(C)}\ \frac{880}{\pi} \qquad \textbf{(D)}\ 440\pi\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 11

A running track is the ring formed by two concentric circles. It is $10$ feet wide. The circumference of the two circles differ by about:

$\textbf{(A)}\ 10\text{ feet} \qquad \textbf{(B)}\ 30\text{ feet} \qquad \textbf{(C)}\ 60\text{ feet} \qquad \textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 12

The diameters of two circles are $8$ inches and $12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:

$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{4}{9} \qquad \textbf{(C)}\ \frac{9}{4} \qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 13

A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:

$\textbf{(A)}\ 36\text{ inches} \qquad \textbf{(B)}\ 9\text{ inches} \qquad \textbf{(C)}\ 18\text{ inches}\\  \textbf{(D)}\ \text{not obtainable from these data}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 14

Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q$. Draw $PQ$. Which of the following statements is false?

$\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\  \textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\  \textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\\  \textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 15

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:

$\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\  \textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 16

Adams plans a profit of $10$ \% on the selling price of an article and his expenses are $15$ \% of sales. The rate of markup on an article that sells for $ $5.00$ is:

$\textbf{(A)}\ 20\% \qquad \textbf{(B)}\ 25\% \qquad \textbf{(C)}\ 30\% \qquad \textbf{(D)}\ 33\frac {1}{3}\% \qquad \textbf{(E)}\ 35\%$

Solution

Problem 17

A man has part of $ $4500$ invested at $4$ \% and the rest at $6$ \%. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:

$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 18

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

$\textbf{(A)}\ x^2+2 \qquad \textbf{(B)}\ x+1 \qquad \textbf{(C)}\ x^2-2x+2 \qquad \textbf{(D)}\ x^2-4\\  \textbf{(E)}\ \text{none of these}$

Solution

Problem 19

In the expression $xy^2$, the values of $x$ and $y$ are each decreased $25$ \%; the value of the expression is:

$\textbf{(A)}\ \text{decreased } 50\% \qquad \textbf{(B)}\ \text{decreased }75\%\\  \textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad \textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 20

If $y=x+\frac{1}{x}$, then $x^4+x^3-4x^2+x+1=0$ becomes:

$\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\\  \textbf{(C)}\ x^2(y^2+y-4)=0 \qquad \textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 21

If $\log_{10} (x^2-3x+6)=1$, the value of $x$ is:

$\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}\ 3\text{ or }-1 \qquad \textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 22

The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:

$\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 23

The equation $\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5$ has:

$\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\  \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad \textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}$

Solution

Problem 24

If $a,b,c$ are positive integers less than $10$, then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:

$\textbf{(A)}\ b + c = 10 \qquad \textbf{(B)}\ b = c \qquad \textbf{(C)}\ a + b = 10 \qquad \textbf{(D)}\ a = b \\  \textbf{(E)}\ a+b+c = 10$

Solution

Problem 25

In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{2}{\sqrt{5}}$

Solution

Problem 26

The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:

$\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(B)}\ 10\text{ inches} \qquad \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad \textbf{(D)}\ 7.5\text{ inches}\\  \textbf{(E)}\ \text{none of these}$

Solution

Problem 27

The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:

$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 28

In $\triangle ABC$, sides $a,b$ and $c$ are opposite $\angle{A},\angle{B}$ and $\angle{C}$ respectively. $AD$ bisects $\angle{A}$ and meets $BC$ at $D$. Then if $x = \overline{CD}$ and $y = \overline{BD}$ the correct proportion is:

$\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad \textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad \textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad \textbf{(E)}\ \frac{x}{y}=\frac{c}{b}$

Solution

Problem 29

The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to the nearest ten-thousandth of a square inch is:

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

Solution

Problem 30

A house worth $ $9000$ is sold by Mr. A to Mr. B at a $10$ \% loss. Mr. B sells the house back to Mr. A at a $10$ \% gain. The result of the two transactions is:

$\textbf{(A)}\ \text{Mr. A breaks even} \qquad \textbf{(B)}\ \text{Mr. B gains }&#036;900 \qquad \textbf{(C)}\ \text{Mr. A loses }&#036;900\\ \textbf{(D)}\ \text{Mr. A loses }&#036;810\qquad \textbf{(E)}\ \text{Mr. B gains }&#036;1710$


Solution

Problem 31

The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. The speed of the train in miles per hour is approximately the number of clicks heard in:

$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 32

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:

$\textbf{(A)}\ \text{a square} \qquad \textbf{(B)}\ \text{a rectangle} \qquad \textbf{(C)}\ \text{a parallelogram with unequal sides}\\ \textbf{(D)}\ \text{a rhombus}\qquad \textbf{(E)}\ \text{a quadrilateral with no special properties}$

Solution

Problem 33

The perimeter of an isosceles right triangle is $2p$. Its area is:

$\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\\  \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2$

Solution

Problem 34

If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$, then the diameter of the circumscribed circle is:

$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 35

If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals:

$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$

Solution

Problem 36

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$. The obtained value, $m$, is an exact divisor of:

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution

Problem 37

The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is:

$\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 38

If $f(a)=a-2$ and $F(a,b)=b^2+a$, then $F(3,f(4))$ is:

$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$

Solution

Problem 39

The product, $\log_a b \cdot \log_b a$ is equal to:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 40

The negation of the statement "all men are honest," is:

$\textbf{(A)}\ \text{no men are honest} \qquad \textbf{(B)}\ \text{all men are dishonest} \\  \textbf{(C)}\ \text{some men are dishonest}\qquad \textbf{(D)}\ \text{no men are dishonest}\\ \textbf{(E)}\ \text{some men are honest}$

Solution

Problem 41

A girls' camp is located $300$ rods from a straight road. On this road, a boys' camp is located $500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:

$\textbf{(A)}\ 400\text{ rods} \qquad \textbf{(B)}\ 250\text{ rods} \qquad \textbf{(C)}\ 87.5\text{ rods} \qquad \textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 42

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. The length of the common internal tangent is:

$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Solution

Problem 43

If the price of an article is increased by percent $p$, then the decrease in percent of sales must not exceed $d$ in order to yield the same income. The value of $d$ is:

$\textbf{(A)}\ \frac{1}{1+p} \qquad \textbf{(B)}\ \frac{1}{1-p} \qquad \textbf{(C)}\ \frac{p}{1+p} \qquad \textbf{(D)}\ \frac{p}{p-1}\qquad \textbf{(E)}\ \frac{1-p}{1+p}$

Solution

Problem 44

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and find $-9$ and $-1$ for the roots. The correct equation was:

$\textbf{(A)}\ x^2-10x+9=0 \qquad \textbf{(B)}\ x^2+10x+9=0 \qquad \textbf{(C)}\ x^2-10x+16=0\\  \textbf{(D)}\ x^2-8x-9=0\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 45

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:

$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$

Solution

Problem 46

Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was:

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

Solution

Problem 47

If $x>0$, then the correct relationship is:

$\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad \textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\  \textbf{(C)}\ \log(1+x) > x\qquad \textbf{(D)}\ \log (1+x) < x\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 48

If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:

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

Solution

Problem 49

The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$

Solution

Problem 50

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$, then the length of the shortest side of the triangle is:

$\textbf{(A)}\ 12\text{ units} \qquad \textbf{(B)}\ 13\text{ units} \qquad \textbf{(C)}\ 14\text{ units} \qquad \textbf{(D)}\ 15\text{ units}\qquad \textbf{(E)}\ 16\text{ units}$

Solution


See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png